text:
User:Egm4313.s12.team18.crete

=Andrew Crete=

Useful Links
Team 18 Course Wiki Course Website Report Table

Report 1
R1 #1.1 pg1.4

Given: Spring-dashpot system in parallel with a mass and applied force f(t). Derive the equation of motion.

Solution:

Using methods of kinematics to note that the displacement of the dashpot is equivalent to that of the spring:
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$$  \displaystyle y=y_k=y_c\! $$     (1)
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Using methods of kinetics, specifically Newton's 2nd Law:
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$$  \displaystyle my''+f_I=f(t)\! $$     (2)
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$$  \displaystyle f_I=f_k+f_c\! $$     (3)
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Constitutive Relations of forces for spring and dashpot:
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$$  \displaystyle f_k=ky_k=ky\! $$     (4)
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$$  \displaystyle f_c=cv=cy_c'\!=cy' $$     (5)
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From (3), (4), and (5):
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$$  \displaystyle f_I=ky+cy' $$     (6)
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Inserting (2) into (6):
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$$  \displaystyle my''+cy'+ky=f(t) $$     (7)
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Report 2
R2.5 K 2011 p.59 pbs. 16-17 Find an ODA of form $$y''+ay'+by=0$$ for the given basis:  #16
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Given: $$ \displaystyle e^{2.6x}, e^{-4.3x} $$
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From the given, we can see that this is a scenario of two distinct roots of form:
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$$  \displaystyle \lambda_1=\frac{1}{2}(-a+\sqrt{a^2-4b})=2.6 $$     (1)
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$$  \displaystyle \lambda_2=\frac{1}{2}(-a-\sqrt{a^2-4b})=-4.3 $$     (2)
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Introducing a variable $$c$$ to simplify calculations:

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$$  \displaystyle c=\sqrt{a^2-4b} $$     (3)
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Substituting (3) into (1) and (2):
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$$  \displaystyle \lambda_1=\frac{1}{2}(-a+c)=2.6 $$     (4)
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$$  \displaystyle \lambda_2=\frac{1}{2}(-a-c)=-4.3 $$     (5)
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From (4) and (5) we have two equations and two unknowns. Begin by solving for $$a$$ via adding (4) and (5):

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$$  \displaystyle -a=-1.7 $$
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$$  \displaystyle \rightarrow a=1.7 $$     (6)
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Substituting the value from (6) into (4) and solving for $$c$$ :

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$$  \displaystyle \frac{1}{2}(-a+c)=2.6 $$
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$$  \displaystyle \rightarrow c=6.9 $$     (7)
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Substituting the value from (7) into (3) and solving for $$b$$ :

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$$  \displaystyle 6.9=\sqrt{1.7^2-4b} $$
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$$  \displaystyle b=\frac{1.7^2-6.9^2}{4} $$
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$$  \displaystyle \rightarrow b=-11.18 $$     (8)
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Substituting the values from (6) and (8) into the general form yields:

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$$  \displaystyle y''+1.6y'-11.18y=0 $$
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 #17
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Given: $$ \displaystyle e^{-\sqrt{5}x}, xe^{-\sqrt{5}x} $$
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From the given we can see that this is a scenario of a real double root:

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$$  \displaystyle \lambda=-\frac{1}{2}a $$     (1)
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$$  \displaystyle a^2-4b=0 $$     (2)
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From the given basis:

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$$  \displaystyle e^{\frac{-ax}{2}} $$     (3)
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Setting (3) equal to (1) and solving for $$a$$ :

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$$  \displaystyle e^{\frac{-ax}{2}}=e^{-\sqrt{5}x} $$
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$$  \displaystyle \frac{a}{2}=\sqrt{5} $$
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$$  \displaystyle \rightarrow a=2\sqrt{5} $$     (4)
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Solving for $$b$$ by substituting (4) into (2):

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$$  \displaystyle {2\sqrt{5}}^2-4b=0 $$
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$$  \displaystyle b=\frac{{2\sqrt{5}}^2}{4} $$
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$$  \displaystyle \rightarrow b=\sqrt{5} $$     (5)
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Substituting the results from (4) and (5) into the general form yields:

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$$  \displaystyle y''+2\sqrt{5}y'+5y=0 $$
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Report 3
R3.6 Given:
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$$  \displaystyle y_{p,1}''-3y_{p,y}'+2y_{p,1}=r_1(x) := 4x^2 $$     (1)
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And

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$$  \displaystyle y_{p,1}''-3y_{p,y}'+2y_{p,1}=r_2(x) := -6x^5 $$     (2)
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Determining the characteristic equation:

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$$  \displaystyle r^2-3r+2=0 $$
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$$  \displaystyle \rightarrow r_1=1 $$     (3.a)
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$$  \displaystyle \rightarrow r_2=2 $$     (3.b)
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Creating a complete set of expressions for $$y_p$$:

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$$  \displaystyle y_p(x)=\sum_{j=0}^5c_jx^j $$     (4.a)
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$$  \displaystyle y_p'(x)=\sum_{j=1}^5 c_j \cdot j \cdot x^{j-1} $$     (4.b)
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$$  \displaystyle y_p'(x)=\sum_{j=2}^5 c_j \cdot j \cdot (j-1) \cdot x^{j-2} $$     (4.c)
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6 unknown coefficients, must modify (4.b) and (4.c) to get $$x^j$$ as a common variable. Using the dummy variable k:

From (4.c):

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$$  \displaystyle k=j-2 \rightarrow j=k+2 $$
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$$  \displaystyle \sum_{k=0}^3 c_{k+2} \cdot (k+2) \cdot (k+1) \cdot x^k = \sum_{j=0}^3 c_{j+2} \cdot (j+2) \cdot (j+1) \cdot x^j $$     (5.a)
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From (4.b):

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$$  \displaystyle k=j-1 \rightarrow j=k+1 $$
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$$  \displaystyle \sum_{k=0}^4 c_{k+1} \cdot (k+1) \cdot x^k = \sum_{j=0}^4 c_{j+1} \cdot (j+1) \cdot x^j $$     (5.b)
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The next step involves adding (4.a), (5.a) and (5.b). Since all three equations are summed through 3, we can combine terms to a constant summation of 3 and add the 4-5 terms individually:

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$$  \displaystyle \sum_{j=0}^3 [c_{j+2}(j+2)(j+1) - 3c_{j+1}(j+1)+2c_j]x^j - 3c_5(5)x^4 + 2[c_4x^4 + 5c_5x^5] = -6x^5 $$     (6)
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We can now equate the coefficients of the basis function for each respective order of x:

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$$  \displaystyle x^0(j=0): (2c_2-3c_1+2c_0)x^0 $$     (7.a)
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$$  \displaystyle x^1(j=1): (6c_3 - 6c_2 + 2c_1)x^1=0 $$     (7.b)
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$$  \displaystyle x^2(j=2): (12c_4 - 9c_3 + 2c_2)x^2=0 $$     (7.c)
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$$  \displaystyle x^3(j=3): (20c_5 - 12c_4 + 2c_3)x^3=0 $$     (7.d)
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$$  \displaystyle x^4: -15c_5x^4+2c_4x^4=0 $$     (7.e)
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$$  \displaystyle x^4: 2c_5x^5=-6x^5 $$     (7.f)
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From (7.f):

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$$  \displaystyle 2c_5=-6 \rightarrow c_5=-3 $$     (8)
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Substituting (8) into (7.e):
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$$  \displaystyle -15(-3) + 2c_4=0 \rightarrow c_4=-22.5 $$     (9)
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Substituting (8) and (9) into (7.d):
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$$  \displaystyle 20(-3)-12(-22.5)+2c_3=0 \rightarrow c_3=-105 $$     (10)
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Substituting (8), (9), and (10) into (7.c):
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$$  \displaystyle 12(-22.5)-9(-105)+2c_2=0 \rightarrow c_2=-337.5 $$     (11)
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Substituting (8), (9), (10), and (11) into (7.b):
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$$  \displaystyle 6(-105)-6(-337.5)+2c_2=0 \rightarrow c_1=-697.5 $$     (12)
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Substituting (8), (9), (10), (11), and (12) into (7.a):
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$$  \displaystyle 2(-337.5)-3(-697.5)+2c_2=0 \rightarrow c_0=-708.75 $$     (13)
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Thus, the particular solution of (2) is:

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$$  \displaystyle y_p(x)=-3x^5-22.5x^4-105x^3-337.5x^2-697.5x-708.75 $$
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 Part 2  For the initial conditions given in (1), (7.a)-(7.e) remain unchanged. (7.f) becomes:
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$$  \displaystyle 2c_5=4x^5 \rightarrow c_5=2 $$     (14)
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Solving for the remaining constants using the same method as before yields:
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$$  \displaystyle c_0=472.5 ; c_1=465 ; c_2=225 ; c_3=70 ; c_4=15 ; c_5=2 $$     (15)
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And thus using the coefficients calculated in (15), the particular solution for (1) is: Solving for the remaining constants using the same method as before yields:
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$$  \displaystyle y_p(x)=2x^5+15x^4+70x^3+225x^2+465x+472.5 $$
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R3.9 K 2011 p.84 pbs. 13-14  Given:
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$$  \displaystyle 8y''-6y'+y=6\cosh{x} $$     (1)
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$$  \displaystyle y(0)=0.2 $$
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$$  \displaystyle y'(0)=0.05 $$
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We begin by determining the general solution of the homogeneous ODE of (1):

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$$  \displaystyle 8\lambda^2-6\lambda+1=0 $$     (2)
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By factoring we obtain:

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$$  \displaystyle (4\lambda-1)(2\lambda-1) \rightarrow \lambda=\frac{1}{4},\frac{1}{2} $$     (3)
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From (3), we can construct a general solution to (1):

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$$  \displaystyle y_h=c_1 e^{\frac{x}{4}} + e^{\frac{x}{2}} $$     (4)
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According to K. 2011 pg. 633, we can express the right hand side of (1) as:

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$$  \displaystyle r(x)=6\cosh{x}=(6)[\frac{1}{2}(e^x+e^{-x})]=r_1(x)+r_2(x) $$
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$$  \displaystyle r_1(x)=3e^x $$     (5.a)
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$$  \displaystyle r_2(x)=3e^{-x} $$     (5.b)
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From this, it is evident that we may use the sum rule. From Table 2.1, in the first line, our particular solution then becomes:

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$$  \displaystyle y_p(x)=C_1e^x+C_2e^{-x} $$     (6)
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Taking the first and second derivatives of $$y_p(x)$$:

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$$  \displaystyle y_p'=C_1e^x-C_2e^{-x} $$     (7.a)
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$$  \displaystyle y_p''=C_1e^x+C_2e^{-x} $$     (7.b)
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In order to use the Method of Undetermined Coefficients, we must put (1) into the form of $$y''+ay'+b=r(x)$$. Dividing (1) by 8 and substituting (5.a) and (5.b):

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$$  \displaystyle y''-\frac{6}{8}y'+\frac{1}{8}y=\frac{3e^x+3e^{-x}}{8} $$     (8)
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We can now substitute (6), (7.a), and (7.b) into (8):

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$$  \displaystyle (C_1e^x+C_2e^{-x})-\frac{6}{8}(C_1e^x-C_2e^{-x})+\frac{1}{8}(C_1e^x+C_2e^{-x})=\frac{3}{8}(e^x+e^{-x}) $$     (9)
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This form allows us to solve for $$C_1$$ and $$C_2$$:

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$$  \displaystyle e^x: C_1-\frac{6}{8}C_1+\frac{1}{8}C_1=\frac{3}{8} $$
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$$  \displaystyle \rightarrow C_1=1 $$     (10.a)
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$$  \displaystyle e^{-x}: C_2+\frac{6}{8}C_2+\frac{1}{8}C_2=\frac{3}{8} $$
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$$  \displaystyle \rightarrow C_2=\frac{1}{5} $$     (10.b)
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Substituting (10.a) and (10.b) into (6) yields the particular solution to (1):

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$$  \displaystyle y_p(x)=e^{-x}+\frac{1}{5}e^{-x} $$     (11)
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The complete solution to (1) is the sum of the particular solution to (1) given in (11) plus the general solution to the homogeneous ODE of (1) given in (4):

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$$  \displaystyle y(x)=c_1e^{\frac{x}{4}}+c_2e^{\frac{x}{2}}+e^x+\frac{1}{5}e^{-x} $$     (12)
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Taking the first derivative of (12):

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$$  \displaystyle y'=\frac{c_1}{4}e^{\frac{x}{4}}+\frac{c_2}{2}e^{\frac{x}{2}}+e^x-\frac{1}{5}e^{-x} $$     (13)
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Using the initial conditions given in (1) with (12) and (13):

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$$  \displaystyle y(0)=c_1+c_2+1+\frac{1}{5}=0.2 $$     (14.a)
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$$  \displaystyle y'(0)=\frac{c_1}{4}+\frac{c_2}{2}+1-\frac{1}{5}=0.05 $$     (14.b)
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From (14.a):

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$$  \displaystyle c_2=-(c_1+1) $$     (15)
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Substituting (15) into (14.b):

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$$  \displaystyle \frac{c_1}{4}-\frac{c_1}{2}-\frac{1}{2}+1-\frac{1}{5}=0.05 $$
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$$  \displaystyle \rightarrow c_1=1 $$     (16)
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Substituting (16) into (14.a):

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$$  \displaystyle 1+c_2+1+\frac{1}{5}=0.02 $$
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$$  \displaystyle \rightarrow c_2=-2 $$     (17)
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Substituting the values from (16) and (17) into (12) gives us the complete solution to the nonhomogeneous ODE given in (1):

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$$  \displaystyle y(x)=e^{\frac{x}{4}}-2e^{\frac{x}{2}}+e^x+\frac{1}{5}e^{-x} $$
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Given
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$$  \displaystyle y''+4y'+4y=e^{-2x}\sin{2x} $$     (1)
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$$  \displaystyle y(0)=1 $$
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$$  \displaystyle y'(0)=-1.5 $$
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Begin by determining the general solution of the homogeneous ODE of (1):

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$$  \displaystyle \lambda^2 + 4 \lambda + y = 0 $$     (2)
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By factoring we obtain:

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$$  \displaystyle (\lambda+2)(\lambda+2)=0 \rightarrow \lambda=-2,-2 $$     (3)
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From (3) we can construct a general solution to the homogeneous ODE of (1):

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$$  \displaystyle y_h=c_1e^{-2x}+c_2xe^{-2x} $$     (4)
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$$r(x)$$ is given in (1) as a workable solution as:

 * {| style="width:100%" border="0"

$$  \displaystyle r(x)=e^{-2x}\sin{2x} $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From this, it is evident that we may use the basic rule. From Table 2.1, in the sixth line our particular solution then becomes:

 * {| style="width:100%" border="0"

$$  \displaystyle y_p=e^{-2x}(K\cos{2x}+M\sin{2x}) $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Taking the first derivative of (6):

 * {| style="width:100%" border="0"

$$  \displaystyle y'_p=e^{-2x}(-2K\sin{2x}+2M \cos{2x})-2e^{-2x}(K\cos{2x}+M\sin{2x}) $$     (7.a)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

(7.a) can be arranged as:

 * {| style="width:100%" border="0"

$$  \displaystyle y'_p=-2e^{-2x}(K\sin{2x}-M\cos{2x}+K\cos{2x}+M\sin{2x}) $$     (7.b)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \rightarrow y'_p=e^{-2x}[2(M-K)\cos{2x}-2(M+K)\sin{2x}] $$     (7.c)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Taking the second derivative of (6):

 * {| style="width:100%" border="0"

$$  \displaystyle y_{p}''=-2e^{-2x}[2(K+M) \cos{2x} -2(K-M) \sin{2x} ]+4e^{-2x}[(K+M) \sin{2x} +(K-M) \cos{2x}] $$     (8.a)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

(8.a) can be arranged as:

 * {| style="width:100%" border="0"

$$  \displaystyle y''_p=-4e^{-2x}[(K+M)\cos{2x}+(M-K)\sin{2x}-(K+M)\sin{2x}+(M-K)\cos{2x}] $$     (8.b)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \rightarrow y''_p=4e^{-2x}(2K\sin{2x}-2M\cos{2x}) $$     (8.c)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since (1) is in a plausible form for the Method of Undetermined Coefficients, we can substitute (6), (7.c), and (8.c) into (1):

 * {| style="width:100%" border="0"

$$  \displaystyle 4e^{-2x}(2K\sin{2x}-2M\cos{2x})+(4)e^{-2x}[2(M-K)\cos{2x}-2(K+M)\sin{2x}]+(4)e^{-2x}(K\cos{2x}+M\sin{2x})=e^{-2x}\sin{2x} $$     (9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

To solve for K and M easier, we can distribute the terms of 4 on the left hand side of (9):

 * {| style="width:100%" border="0"

$$  \displaystyle e^{-2x}(8K\sin{2x}-8M\cos{2x})+e^{-2x}(8(M-K)\cos{2x}-8(K+M)\sin{2x})+e^{-2x}(4K\cos{2x}+4M\sin{2x})=e^{-2x}\sin{2x} $$     (10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Consolidating terms further:

 * {| style="width:100%" border="0"

$$  \displaystyle e^{-2x}[(8K-8K-8M+4M)\sin{2x}+(-8M+8M-8K+4K)\cos{2x}]=e^{-2x}sin{2x} $$     (11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which finally becomes:

 * {| style="width:100%" border="0"

$$  \displaystyle e^{-2x}(-4M\sin{2x}-4K\cos{2x})=e^{-2x}\sin{2x} $$     (12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From (12), we can math $$e^{-2x}sin{2}$$ and $$e^{-2x}cos{2}$$ terms to solve for K and M:

 * {| style="width:100%" border="0"

$$  \displaystyle e^{-2x}\sin{2x}: -4M=1 $$     (13.a)
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 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle e^{-2x}\cos{2x}: -4K=0 $$     (13.b)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From (13.a):

 * {| style="width:100%" border="0"

$$  \displaystyle M=-\frac{1}{4} $$     (14.a)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From (13.b):

 * {| style="width:100%" border="0"

$$  \displaystyle K=0 $$     (14.b)
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 * style="width:95%" |
 *

 * }

Substituting (14.a) and (14.b) into (6) yields the particular solution to (1):

 * {| style="width:100%" border="0"

$$  \displaystyle y_p(x)=e^{-2x}(-\frac{1}{4}\sin{2x}) $$     (15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The complete solution to (1) is the sum of the particular solution to (1) given in (15) plus the general solution to the homogeneous ODE of (1) given in (4):

 * {| style="width:100%" border="0"

$$  \displaystyle y(x)=c_1e^{-2x}+c_2xe^{-2x}+e^{-2x}(-\frac{1}{4}\sin{2x}) $$     (16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Taking the first derivative of (16):

 * {| style="width:100%" border="0"

$$  \displaystyle y'=-2c_1e^{-2x}-2c_2xe^{-2x}+c_2e^{-2x}+2e^{-2x}(-\frac{1}{4}\cos{2x})-2e^{-2x}(-\frac{1}{4}\sin{2x}) $$     (17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Consolidating terms of (17):

 * {| style="width:100%" border="0"

$$  \displaystyle y'=-2c_1e^{-2x}+(1-2x)c_2e^{-2x}+\frac{1}{2}e^{-2x}(\sin{2x}-\cos{2x}) $$     (18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using the initial conditions given in (1) with (16) and (18) to solve for $$c_1$$ and $$c_2$$:

 * {| style="width:100%" border="0"

$$  \displaystyle y(0)=c_1=1 $$     (19.a)
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 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle -2c_1+c_2-\frac{1}{2}=-1.5 $$     (19.b)
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 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$  \displaystyle \rightarrow c_2=1 $$     (19.c)
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 * style="width:95%" |
 *

 * }

Thus we have the complete solution to the nonhomogeneous ODE given in (1). Substituting (19.a) and (19.c) into (16) yields:

 * {| style="width:100%" border="0"

$$  \displaystyle y(x)=e^{-2x}+xe^{-2x}+e^{-2x}+e^{-2x}(-\frac{1}{4}\sin{2x}) $$     (20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which can be arranged for the final solution as:

 * {| style="width:100%" border="2"

$$  \displaystyle y(x)=e^{-2x}(2+x-\frac{1}{4}\sin{2x}) $$
 * style="width:95%" |
 * style="width:95%" |
 * }

title: User:Egm4313.s12.team18.crete


text:
User:Egm4313.s12.team3.pliskow/R1work

=Problem 6: Classifying ODE's and the Superposition Principle=

Problem Statement
For each of the following ODE's, determine the order and linearity. Then, determine if the superposition principle can be applied.

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$$\displaystyle {y}''=g $$     (6.1)
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 *
 * }

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$$\displaystyle m{v}'=mg-bv^{2} $$     (6.2)
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 *
 * }

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$$\displaystyle {h}'=-k\sqrt{h} $$     (6.3)
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 *
 * }

 * {| style="width:100%" border="0"

$$\displaystyle m{y}''+ky=0 $$     (6.4)
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 *
 * }

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$$\displaystyle {y}''+\omega_0^{2}y=cos(\omega t)\;\;,\;\;\omega_0 \approx \omega $$     (6.5)
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 * style="width:95%" |
 *
 * }

 * {| style="width:100%" border="0"

$$\displaystyle L{I}''+R{I}'+\frac{1}{C}I={E}' $$     (6.6)
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 *
 * }

 * {| style="width:100%" border="0"

$$\displaystyle EI{y}=f(x) $$     (6.7)
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 *
 * }

 * {| style="width:100%" border="0"

$$\displaystyle L{\theta}''+gsin\theta =0 $$     (6.8)
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 * style="width:95%" |
 *
 * }

Problem Solution
For information regarding the order and linearity of ordinary differential equations, see class notes, sec1. For information regarding the application of the superposition principle, see class notes, sec2.

Problem 6.1

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$$\displaystyle {y}''=g $$     (6.1)
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 *
 * }

 * {| style="width:100%" border="0"

This is a second order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *
 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

Homogeneous problem:
 * {| style="width:100%" border="0"

$$\displaystyle {y_h}''=0 $$     (1)
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 * style="width:95%" |
 *
 * }

Particular problem:
 * {| style="width:100%" border="0"

$$\displaystyle {y_p}''=g $$     (2)
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 * style="width:95%" |
 *
 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{y}=y_h+y_p $$     (3)
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 * style="width:95%" |
 *
 * }

Summing (1) and (2), we obtain the following:

 * {| style="width:100%" border="0"

$$\displaystyle {y_h}+{y_p}=g $$     (4)
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 *
 * }

 * {| style="width:100%" border="0"

$$\displaystyle {({y_h}'+{y_p}')}'=g $$     (5)
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 *
 * }

 * {| style="width:100%" border="0"

$$\displaystyle {(y_h+y_p)}''=g $$     (6)
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 *
 * }

 * {| style="width:100%" border="0"

$$\displaystyle {\bar{y}}''=g $$     (7)
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 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

From equation (7), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Problem 6.2

 * {| style="width:100%" border="0"

$$\displaystyle m{v}'=mg-bv^{2} $$     (6.2)
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 *

 * }

 * {| style="width:100%" border="0"

This is a second order, non-linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

Homogeneous problem:
 * {| style="width:100%" border="0"

$$\displaystyle m{v_h}'+bv_h^{2}=0 $$     (1)
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 * style="width:95%" |
 *

 * }

Particular problem:
 * {| style="width:100%" border="0"

$$\displaystyle m{v_p}'+bv_p^{2}=mg $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{v}=v_h+v_p $$     (3)
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 * style="width:95%" |
 *

 * }

Summing (1) and (2), we obtain the following:

 * {| style="width:100%" border="0"

$$\displaystyle m({v_h}'+{v_p}')+b(v_h^{2}+v_p^{2})=mg $$     (4)
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 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle m{(v_h+v_p)}'+b(v_h^{2}+v_p^{2})=mg $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle m{\bar{v}}'+b(v_h^{2}+v_p^{2})=mg $$     (6)
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 * style="width:95%" |
 *

 * }

We cannot substitute any further since $$ b(v_h^{2}+v_p^{2}) \neq b(v_h+v_p)^{2} $$.

 * {| style="width:100%" border="0"

$$ \bar{v} $$ is NOT also a solution to the original ODE, and thus the superposition principle can't be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Problem 6.3

 * {| style="width:100%" border="0"

$$\displaystyle {h}'=-k\sqrt{h} $$     (6.3)
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 *

 * }

 * {| style="width:100%" border="0"

This is a first order, non-linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For a homogeneous equation, the superposition principle applies if a linear combination of any two solutions is also a solution to the original ODE. This is tested below.

Suppose $$h_1$$ is one solution:
 * {| style="width:100%" border="0"

$$\displaystyle {h_1}'+k\sqrt{h_1}=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And $$h_2$$ is another solution:
 * {| style="width:100%" border="0"

$$\displaystyle {h_2}'+k\sqrt{h_2}=0 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We define any linear combination of these solutions:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{h}=c_1h_1+c_2h_2 $$     (3)
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 * style="width:95%" |
 *

 * }

Summing (1) and (2), we obtain the following:

 * {| style="width:100%" border="0"

$$\displaystyle {h_1}'+{h_2}'+k\sqrt{h_1}+k\sqrt{h_2}=0 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle {(h_1+h_2)}'+k(\sqrt{h_1}+\sqrt{h_2})=0 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle {\bar{h}}'+k(\sqrt{h_1}+\sqrt{h_2})=0 $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We cannot substitute any further since $$ k(\sqrt{h_1}+\sqrt{h_2})) \neq k\sqrt{h_1+h_2} $$.

 * {| style="width:100%" border="0"

$$ \bar{h} $$ is NOT also a solution to the original ODE, and thus the superposition principle can't be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Problem 6.4

 * {| style="width:100%" border="0"

$$\displaystyle m{y}''+ky=0 $$     (6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

This is a second order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For a homogeneous equation, the superposition principle applies if a linear combination of any two solutions is also a solution to the original ODE. This is tested below.

Suppose $$y_1$$ is one solution:
 * {| style="width:100%" border="0"

$$\displaystyle m{y_1}''+ky_1=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And $$y_2$$ is another solution:
 * {| style="width:100%" border="0"

$$\displaystyle m{y_2}''+ky_2=0 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We define any linear combination of these solutions:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{y}=c_1y_1+c_2y_2 $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Summing (1) and (2), we obtain the following:

 * {| style="width:100%" border="0"

$$\displaystyle m({y_1}+{y_2})+k(y_1+y_2)=0 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle m{(y_1+y_2)}''+k(y_1+y_2)=0 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle m{\bar{y}}''+k\bar{y}=0 $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

From equation (6), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Problem 6.5

 * {| style="width:100%" border="0"

$$\displaystyle {y}''+\omega_0^{2}y=cos(\omega t)\;\;,\;\;\omega_0 \approx \omega $$     (6.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

This is a second order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

Homogeneous problem:
 * {| style="width:100%" border="0"

$$\displaystyle {y_h}''+\omega_0^{2}y_h=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Particular problem:
 * {| style="width:100%" border="0"

$$\displaystyle {y_p}''+\omega_0^{2}y_p=cos(\omega t) $$ (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{y}=y_h+y_p $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Summing (1) and (2), we obtain the following:

 * {| style="width:100%" border="0"

$$\displaystyle {y_h}+{y_p}+\omega_0^{2}y_h+\omega_0^{2}y_p=cos(\omega t) $$ (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle {(y_h+y_p)}''+\omega_0^{2}(y_h+y_p)=cos(\omega t) $$ (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle {\bar{y}}''+\omega_0^{2}(\bar{y})=cos(\omega t) $$ (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

From equation (6), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Problem 6.6

 * {| style="width:100%" border="0"

$$\displaystyle L{I}''+R{I}'+\frac{1}{C}I={E}' $$     (6.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

This is a second order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

Homogeneous problem:
 * {| style="width:100%" border="0"

$$\displaystyle L{I_h}''+R{I_h}'+\frac{1}{C}I_h=0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Particular problem:
 * {| style="width:100%" border="0"

$$\displaystyle L{I_p}''+R{I_p}'+\frac{1}{C}I_p={E}' $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{I}=I_h+I_p $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Summing (1) and (2), we obtain the following:

 * {| style="width:100%" border="0"

$$\displaystyle L({I_h}+{I_p})+R({I_h}'+{I_p}')+\frac{1}{C}(I_h+I_p)={E}' $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle L{(I_h+I_p)}''+R{(I_h+I_p)}'+\frac{1}{C}(I_h+I_p)={E}' $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle L{\bar{I}}''+R{\bar{I}}'+\frac{1}{C}\bar{I}={E}' $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

From equation (6), we can see that $$ \bar{I} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Problem 6.7

 * {| style="width:100%" border="0"

$$\displaystyle EI{y}=f(x) $$     (6.7)
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 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

This is a fourth order, linear ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
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 *

 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

Homogeneous problem:
 * {| style="width:100%" border="0"

$$\displaystyle EI{y_h}=0 $$     (1)
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 *

 * }

Particular problem:
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$$\displaystyle EI{y_p}=f(x) $$     (2)
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 *

 * }

We define the sum of the homogeneous solution and particular solution:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{y}=y_h+y_p $$     (3)
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 *

 * }

Summing (1) and (2), we obtain the following:

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$$\displaystyle EI({y_h}'+{y_p}')=f(x) $$     (4)
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 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle EI{(y_h+y_p)}=f(x) $$     (5)
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 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle EI{\bar{y}}=f(x) $$     (6)
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 *

 * }

 * {| style="width:100%" border="0"

From equation (6), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
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 *

 * }

Problem 6.8

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$$\displaystyle L{\theta}''+gsin\theta =0 $$     (6.8)
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 *

 * }

 * {| style="width:100%" border="0"

This is a second order, non-linear ODE.
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For a homogeneous equation, the superposition principle applies if a linear combination of any two solutions is also a solution to the original ODE. This is tested below.

Suppose $$\theta_1$$ is one solution:
 * {| style="width:100%" border="0"

$$\displaystyle L{\theta_1}''+gsin\theta_1 =0 $$     (1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And $$\theta_2$$ is another solution:
 * {| style="width:100%" border="0"

$$\displaystyle L{\theta_2}''+gsin\theta_2 =0 $$     (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We define any linear combination of these solutions:
 * {| style="width:100%" border="0"

$$\displaystyle \bar{\theta}=c_1\theta_1+c_2\theta_2 $$     (3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Summing (1) and (2), we obtain the following:

 * {| style="width:100%" border="0"

$$\displaystyle L({\theta_1}+{\theta_2})+g(sin\theta_1+sin\theta_2)=0 $$     (4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle L{(\theta_1+\theta_2)}''+g(sin\theta_1+sin\theta_2)=0 $$     (5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle L{\bar{\theta}}''+g(sin\theta_1+sin\theta_2)=0 $$     (6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We cannot substitute any further since $$ sin\theta_1+sin\theta_2 \neq sin(\theta_1+\theta_2) $$.

 * {| style="width:100%" border="0"

$$ \bar{\theta} $$ is NOT also a solution to the original ODE, and thus the superposition principle can't be applied.
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 *

 * }

=Contributing Members for This Project=

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Egm4313.s12.team3.pliskow/R1work


title: File:R1.1.png


text:
User:Jpking10

My name is James Phillips King. I have interests in mathematics and computer science.

http://www.jamespking.com

title: User:Jpking10


text:
Talk:Java Programming/Arrays

The section "1.1.1.2 Things you can do with arrays" is somewhat confusing because you do not need to subtract 1 from the length in order for the example program to work.

title: Talk:Java Programming/Arrays


text:
Dominant group/Intellectual Merit

For the term dominant group, the intellectual merit “is chiefly concerned with scientific merit as judged by scientists”.

With respect to grant applications to the U.S. National Endowment for the Humanities (NEH), "[k]nowledgeable persons outside NEH will read each application and advise the agency about its merits. NEH staff comments on matters of fact or on significant issues that otherwise would be missing from these reviews, then makes recommendations to the National Council on the Humanities. The National Council meets at various times during the year to advise the NEH chairman on grants. The Chairman takes into account the advice provided by the review process and, by law, makes all funding decisions." Bold added.

Dominant group
Here, the evidence needed is that which demonstrates the concepts embodied in the term dominant group, and its synonyms, explored through research, specifically, original research about these, have intellectual merit or quality.

Intellectual as a word may be an adjective or a noun.

Def.

 * 1) "Belonging to, or performed by, the intellect; mental or cognitive; as, intellectual powers, activities, etc."
 * 2) "Endowed with intellect; having the power of understanding; having capacity for the higher forms of knowledge or thought; characterized by intelligence or mental capacity; as, an intellectual person."
 * 3) "Suitable for exercising the intellect; formed by, and existing for, the intellect alone; perceived by the intellect; as, intellectual employments."
 * 4) "Relating to the understanding; treating of the mind; as, intellectual philosophy, sometimes called "mental" philosophy."
 * 5) "(archaic, poetic) Spiritual.", as an adjective,

is called an intellectual applicant or application.

Def.

 * 1) "An intelligent, learned person, especially one who discourses about learned matters."
 * 2) "(archaic) The intellect or understanding; mental powers or faculties.", as a noun,

is called an intellectual.

Merit
Def.

 * 1) "Something deserving positive recognition." "His reward for his merit was a check for $50."
 * 2) "Something worthy of a high rating."
 * 3) "A claim to commendation or reward."
 * 4) "The quality of deserving reward.", as a noun,

is called a merit.

Def.

 * 1) "To earn or to deserve."
 * 2) "To be worthy or deserving."
 * 3) "(obsolete, rare) To reward.", as a verb,

is called to merit. The results of the review have been encouraging enough to merit further investigation.

Quality
Def.

 * 1) "Level of excellence"
 * "This school is well-known for having teachers of high quality."
 * "Quality of life is usually determined by health, education, and income."
 * 1) "A property or an attribute that differentiates a thing or person."
 * "One of the qualities of pure iron is that it does not rust easily."
 * "While being impulsive can be great for artists, it is not a desirable quality for engineers."
 * "Security, stability, and efficiency are good qualities of an operating system."
 * 1) "(archaic) High social position."
 * "A peasant is not allowed to fall in love with a lady of quality."
 * "Membership of this golf club is limited to those of quality and wealth."
 * 1) "The degree to which a man-made object or system is free from bugs and flaws, as opposed to scope of functions or quantity of items."
 * "To identify quality try asking, "what does it feel like?"." as a noun,

is called a quality.

 * "Adjectives often applied to "quality": high, good, excellent, exceptional, great, outstanding, satisfactory, acceptable, sufficient, adequate, poor, low, bad, inferior, dubious, environmental, visual, optical, industrial, total, artistic, educational, physical, musical, chemical, spiritual, intellectual, architectural, mechanical." as a usage note.

Def. "Being of good worth, well made, fit for purpose." as an adjective,
 * "We only sell quality products."
 * "That was a quality game by Jim Smith."
 * "A quality system ensures products meet customer requirements."

is called a quality applicant or application.

Sciences
Def. "[o]ne whose activities make use of the scientific method to answer questions regarding the measurable universe" is called a scientist.

"A scientist may be involved in original research, or make use of the results of the research of others."

Def. "[a] method of discovering knowledge about the natural world based in making falsifiable predictions (hypotheses), testing them empirically, and developing peer-reviewed theories that best explain the known data" is called a scientific method.

Scientific method is a two-word scientific term.

Humanities
Even in a science, there are questions that endure after centuries, millennia. Intellectual merit or quality is not limited to scientists. Humanity may have created the concepts behind dominant group for only self-interest or composed them in response to something thrust upon us in ancient times.

"Evaluators [for NEH] are asked to apply the following criteria when judging the quality of applications."

1. Intellectual quality
 * Does the proposed course explore an enduring question that lends itself to sustained and open inquiry?
 * Does the proposal make a persuasive case for the historical persistence of the question?
 * Is there a well-articulated relationship between the enduring question to be considered and the works to be studied by the faculty member(s) in preparing the envisioned course?
 * Are intellectual pluralism and balance evident in the question to be considered and in the works proposed for the course?
 * Does the approach to the question suggest more than one plausible answer?
 * Does the course draw on works from a range of historical periods?
 * Does the course emphasize extensive reading?
 * Is the proposal clear, free of jargon, and accessible to nonspecialists?

2. Feasibility
 * Does the course development plan require the faculty member(s) to read extensively from a body of primary and scholarly literature that reaches significantly beyond their current expertise?
 * Is the plan of work well designed?
 * Does the course development plan enable the faculty member(s) to teach the course without relying on guest presenters (including any project team members)?
 * Is the envisioned course well suited to its intended audience?
 * Are the envisioned core readings ambitious but realistic for undergraduates at the applicant institution?
 * Are ancillary activities, if any, well conceived?

3. Impact
 * Is the proposed course new?
 * Does the course have the potential to foster intellectual community—anchored in an enduring question—among students in the course?
 * Has the institution committed to having the course offered at least twice during the grant period by each faculty member involved in developing it?
 * Are the dissemination plans appropriate?

Advancing knowledge and understanding
“How important is the proposed activity to advancing knowledge and understanding within its own field or across different fields?”

The current activity is exploratory in nature and is resulting in an advance of knowledge and understanding about the term dominant group and its uses within various fields. The identification of its own field may result; thereby, in allowing differentiation of other fields so as to determine how knowledge is advanced across those different fields. It's like differential equations. Although differential equations is a subject within mathematics, it's spread to other subjects such as physics with ensuing use advances knowledge within physics. Differential equations is a two-word scientific term.

The proposed activity consists of finding the origin of dominant group or one of its relative synonyms, defining the two-word scientific/technical term perhaps from context (lexical pragmatics) (or perhaps rigorously), determining why scientists outside biology (especially evolution or entomology) use the term, and verifying its divergence and radiance by examples.

The field of dominant group appears to be regions, or the science of regions. Within a region in any science, including the social sciences, there may be a dominant group. Each region is defined by its limits. These limits in turn may allow for a dominant group. Change the limits (characteristics) for any region and the dominant group may go extinct. If a regional genome in biology lacks the potency to take advantage of any change in regional characteristics, there will be no dominant biological group.

Advancing knowledge in each field of use
Part of the proposed activity is to determine how dominant group advances knowledge and understanding within each field which may be considered its own field or across all these different fields.

Dominant group is already being used as a scientific/technical term to advance knowledge and understanding across a great many fields.

"How important is the proposed activity to advancing knowledge and understanding within each of the fields or across these different fields?"

In theory, dominant group in any field may have at least two uses: (1) a group of field-based entities, sources, or objects, or (2) a dominant group in some way associated with that field.

In practice, depending upon the scientist's intent, dominant group may be the bad group that engages in monopolistic practices, or socially negative behavior such as discrimination, abuse, punishment, and additional possible criminal activity against other demographic groups.

It can also serve as an empirical identifier in observations. At one extreme, it is 100 % of the effect or phenomenon under study. At the other, it is a minority group effect that perhaps has some natural or artificial unfair advantage. In a society a dominant group often has an unfair advantage such as numbers, military power, or assets and money.

Evolution
Hypothesis: As an evolutionary process, a dominant group may be a force for extinction by driving a number of groups to extinction and a force for speciation when spatial or temporal dispersion creates diversity and isolation which may eventually produce new groups. As a force for extinction a particular dominant group may also through its actions on other groups ultimately produce its own extinction event. This departure from a zone then allows other groups to fill the niche.

Changes in the characteristics or properties of a region may force the dominant group into extinction; thereby, making room for the next dominant group. However, some changes may not allow a dominant group to emerge. Additionally, properties of dominance may be appearance only. No competition for resources may have occurred.

Lexical pragmatics
From a metadefinitional point of view each use of dominant group has a relationship between members of the dominant group, a population from which the dominant group is a subset, a criterion for dominance, and a region, range, distribution or "in their own country". But, each of these four structures may have their meaning in the context within which the author or speaker places the two-word humanities or scientific/technical term.

Oldest two-word terms
It may be one of the oldest two-word terms. Some of its synonyms predate 1826: for example, the phrase “die Dominanten Religionen von ganz Europa” occurs in 1726.

Origins
The proposed activity consists of finding the origin of dominant group or one of its synonyms. The fundamental concepts encompassed by the totality of synonyms may have very early origins in primitive languages.

1826 (Kirby): "Groups, according to their range, may be denominated either predominant, dominant, sub-dominant, or quiescent."

1840 (Shuckard): "The Ants and the Staphylini have been supposed to represent each other in the tropical and temperate zones. In the temperate zone, and especially in our own country, the Staphylini are a dominant group, and the ants a secondary one."

1857 (Tocqueville): “The fact that a group is egoistic and dominant proves that it is well formed and that it approaches the make-up of a man.”

1859 (Darwin): The earliest use is on page 343, “The dominant species of the larger dominant groups tend to leave many modified descendants, and thus new sub-groups and groups are formed.”

1860 (Darwin): “Under the many conditions of life which this world affords, any group which is numerous in individuals and species and is widely distributed, may properly be called dominant" [a dominant group]. [Letter 110. To W.H. Harvey, August, 1860]

1944 (Simpson): "In the history of life it is a striking fact that major changes in the taxonomic groups occupying various ecological positions do not, as a rule, result from direct competition of the groups concerned in each case and the survival of the fittest, as most students would assume a priori. On the contrary, the usual sequence is for one dominant group to die out, leaving the zone empty, before the other group becomes abundant." . "GG Simpson's so-called evolutionary species concept contains undefinable criteria and is useless in praxis.

1960 (Huxley): "Improved organization gives biological advantage. Accordingly, the new type becomes a successful or dominant group. It spreads and multiplies and differentiates into a multiplicity of branches. This new biological success is usually achieved at the biological expense of the older dominant group from which it sprang or whose place it had usurped."

1971 (Travis): The earlier theory of evolutionary progressivism (Social Darwinism) puts forth several concepts including "dominant group".

2001 (Mayr): "Other previously dominant groups of organisms that also became extinct at the end of the Cretaceous are many marine taxa, such as most nautiloids and the ammonities, both of whom had been previously highly successful organisms." Page 320: "Ernst Mayr is the biologist largely responsible for shaping the modern synthesis of genetics and evolutionary theory."

Own field
The first question to answer is "What is dominant group's own field?"

It is a scientific or technical term, specifically a two-word scientific or technical term. Its overall field is linguistics, or technical language, terminology. It is and refers to an entity. An overall field allowing application to other fields is regions or the science of regions, usually spatial, but not necessarily always spatial. Spatial science or the science of space may be the appropriate field for dominant group.

Proposer qualifications
“How well qualified is the proposer (individual or team) to conduct the project? (If appropriate, the reviewer will comment on the quality of prior work.)”

Much of the explorational data for reviewer evaluation sits within the resource structure and substructure of the dominant group research project displayed in the template entitled "Dominant group" near the bottom of this page. Possible organizational conformity may increase the probability of positive reviewer comment.

Initial efforts using the scientific method may be found within the effort to demonstrate proof of concept.

Transformative concepts
“To what extent does the proposed activity suggest and explore creative, original, or potentially transformative concepts?”

Def. "research that has the capacity to revolutionize existing fields, create new subfields, cause paradigm shifts, support discovery, and lead to radically new technologies” is called transformative research.

"[T]wo-word glossary items are the most common technical terms". Dominant group is a likely two-word glossary item captured by data mining algorithms. A first-principle's demonstration that dominant group is a two-word scientific/technical term yields one test standard for data mining algorithms to find.

Dominant group serves as an indicator that original research has been conducted, especially when it appears in a primary source.

As a two-word scientific/technical term, is dominant group a member of the dominant group of two-word scientific/technical terms?

This exploratory investigation into dominant group and its usage has the potential to demonstrate that dominant group
 * 1) when identified is a causative force for change that by its nature requires further investigation,
 * 2) should be no longer used as a scientific/technical term because its meanings are unclear and vague, or
 * 3) identifies an inhibiting or moderating force that works against 'high-risk, high-reward' "research with an inherent high degree of uncertainty and the capability to produce a major impact on important problems in biomedical/behavioral research", or in other scientific/technical fields.

Conception and organization
“How well conceived and organized is the proposed activity?”

This exploratory effort uses the template system (the dominant group template is near the bottom of this page) to organize and keep readily available the efforts already begun and those still needed.

Access to resources
“Is there sufficient access to resources?"

Part of the challenge of this type of experimental inquiry is that authors of today or even in the 1800s are likely to use dominant group to describe the past when the term itself was actually not used by an earlier author.

University level access, including its medial research center, (both of which are available local to the principal investigator, PI) to much earlier original documents may be required to confirm the term earlier than 1826, its current date of specific use in English by Kirby. Local availability for interlibrary loan of needed resources also exists.

As Google scholar, for example, displays older and older manuscripts or books, especially before 1826, the probability increases of finding the term or eliminating its use by specific authors.

Web-based resources (including e-mail) are available to the PI.

Hypotheses

 * 1) Dominant group may have as much merit as original research itself.

title: Dominant group/Intellectual Merit


text:
User:Egm4313.s12.team13.Neisha

http://en.wikiversity.org/wiki/User:Egm4313.s12.team13.Neisha.R1

http://en.wikiversity.org/wiki/User:Egm4313.s12.team13.Neisha.R2

http://en.wikiversity.org/wiki/User:Egm4313.s12.team13.Neisha.R2.1

title: User:Egm4313.s12.team13.Neisha


text:
User:Egm6322.s12.TEAM1.yang.HW1

Problem 2: Fourier Coefficients
 Solved without assistance

Find
A. Verify Eq 2.1 is valid. B. Recover from Eq2.2 the Fourier coefficients found with the trigonometric identities given.

Solution
A. Verify Eq 2.1 is valid.

if $$ n=m$$, then:

if $$ n\ne m$$, then:

B. Recover from Eq2.2 the Fourier coefficients found with the trigonometric identities given.

From the definition of the coefficients in the lecture, we can get: Then, the Fourier coefficients can be found from above:

Problem 3: Alternative Coefficients
 Solved without assistance

Given & Find
show that, equivalent to (2)-(3)P.42-10, the Fourier series can be written as

$$f(\theta )=\operatorname{Re}(\sum\limits_{n=0}^{+\infty }{{e}^{in\omega \;\theta \;}})$$

An advantage of (2)-(3)P.42-10 is a simple and elegant computation of the coefficients as shown in (1) p.42-12.

Solution
Starting from eqation (1) in P42-10, the following can be derived.

in which:

$${{\overline{a}}_{n}}={{a}_{n}}-i{{b}_{n}}$$, for n>0.

In comparison, the coefficients in equation (1) p.42-12 is obviously more simple and elegant than equation(3.1) above.

Problem 7: Computing Coefficients
 Solved without assistance

Given
$$q(x)=-4-7x+3{{x}^{2}}+2{{x}^{3}}-5{{x}^{4}}+{{x}^{5}}$$

Find
Find $${{d}_{i}}$$ such that: $$q(x)=\sum\limits_{i=0}^{n}{{{d}_{i}}{{P}_{i}}(x)}\in {{\Rho }_{n}}$$ plot q(x) in two figures

Solution
According the Lecture 42-15, components can be computed by:

And the Legendre polynomials are known as: And, The components can be computed by substituting Eq 7.2 and Eq 7.3 into Eq 7.1.

=References=

title: User:Egm6322.s12.TEAM1.yang.HW1


text:
User:Egm4313.s12.team13

=Team Members= Frank Portillo Amy Mclarty Kendal Ryser Neisha Ramnarine Steven Rossmeissl Gabrielle Steinberg Ryan Hagberg

=Reports= Report 1 Report 2

title: User:Egm4313.s12.team13


text:
Dominant group/Broader impacts

What are the broader impacts of the proposed activity regarding dominant group?

The broader impacts are "concerned with issues of education, infrastructure, diversity, and societal benefit." This along with intellectual merit “are criteria for scientists to be judged by scientists.”

"Among the major recommendations of the [National Academy of Public Administration] NAPA panel is that ‘there is a need to improve the conceptual clarity of the objectives of the new criteria as well as the language used in stating them ... This is true of the language of Criterion 2 [broader impacts], in particular’"

"[I]t may ... be simply that ... different scientific communities (i.e. disciplines) interpret Criterion 2, differently. ... Some disciplines, for example, may see themselves as purely scientific and, therefore, necessarily unconcerned with the broader impacts of their research."

"While it may not be possible to predict specific potential applications for one's research, one should be able to discuss the value or applicability of the line of inquiry or research area."

Universals
The evidence needed demonstrates the concept dominant group, specifically research or original research, focused on it, has broader merits.

Advancing discovery and training
“How well does the activity advance discovery and understanding while promoting teaching, training, and learning?”

Ideally, at least one advanced graduate student should be provided full support for several years to work on the project completing a doctoral dissertation using project data. Involve undergraduates and volunteers in field effort and subsequent analyses.

Exploring modern languages, languages on the verge of extinction, and extinct languages for terms that are either relative or exact synonyms (translations) for dominant group advances discovery of meaning and terminology.

For example: "In January 2008, a coalition of over 40 civil society groups endorsed a statement of principles calling for precautionary action related to nanotechnology." "The group has urged action based on eight principles. They are 1) A Precautionary Foundation 2) Mandatory Nano-specific Regulations 3) Health and Safety of the Public and Workers 4) Environmental Protection 5) Transparency 6) Public Participation 7) Inclusion of Broader Impacts and 8) Manufacturer Liability."

"Nanomedicines are just beginning to enter drug regulatory processes, but within a few decades could comprise a dominant group within the class of innovative pharmaceuticals, the current thinking of government safety and cost-effectiveness regulators appearing to be that these products give rise to few if any nano-specific issues. " Bold added.

"Ethnocide is when a dominant political group attempts to purposely put an end to a people’s traditional way of life. Linguicide (linguistic genocide) is when such a dominant group tries to extinguish the language of a minority group, say by punishing anyone caught speaking it." Exploring the meaning and use of dominant group in such context increases understanding of the forces at work characterized by the terms inclusion.

Here's an example of how searching and using the power of the internet can bring about remarkable discoveries.

In an article that appeared in the American Scientist (May-June 2012) issue entitled "Herschel and the Puzzle of Infra-red", "Jack White mentions that it is not known who coined the term "infrared."" "A Google Books search for "infra-red" finds two articles published in April 1874, both of which use the term in the context of Edmond Bacquerel's treatise on light." There is an 1867 work using the French infra-rouge and one in English near the same time using "infra-red", "having translated it from the French."

The author responds that "ultra red" and "infra-red" appear in a paper from 1873, researched in 1960 "in the dark ages before the Internet. Rosenberg's find is a reminder of the Internet's amazing, growing power to search original works in different languages."

Underrepresented group participation
“How well does the proposed activity broaden the participation of underrepresented groups (e.g., gender, ethnicity, disability, geographic, etc.)?”

The proposed activity is ongoing at Wikiversity.

"Wikiversity is a Wikimedia Foundation project devoted to learning resources, learning projects, and research for use in all levels, types, and styles of education from pre-school to university, including professional training and informal learning. We invite teachers, students, and researchers to join us in creating open educational resources and collaborative learning communities. To learn more about Wikiversity, try a guided tour or start editing now."

Infrastructure enhancement
“To what extent will it enhance the infrastructure for research and education, such as facilities, instrumentation, networks, and partnerships?”

Information may be made available in various venues:
 * 1) databases or contributions to other databases of all institutions, agencies, and individuals who manage similar databases,
 * 2) contributions to ongoing network analyses, and
 * 3) information sharing and access.

Wikiversity is the premier online research and education database, facility, network and partnership of the WMF. That the proposed activity is ongoing at Wikiversity makes the information generated available through access and sharing.

"Just as an FYI, Wikiversity has been placed among the 'top picks' for this Google+ education list. It has been shared over 2000 times since its original posting."

The proposed activity would partner NSF with Wikiversity to enhance the infrastructure for research and education, including where possible, facilities, instrumentation, and networks within Wikiversity and between Wikiversity and NSF.

Broad dissemination
“Will the results be disseminated broadly to enhance scientific and technological understanding?”

Have the project produce scientific publications such as
 * 1) a doctoral dissertation,
 * 2) an article in a regional journal,
 * 3) another article in a national or international journal, and
 * 4) possibly an article in a magazine widely used in university classes and read by general audiences internationally.

Disseminate results to the professional community through presentations at
 * 1) national scientific association meetings,
 * 2) local avocational groups, and
 * 3) government and cultural resource meetings.

Present the results to the general public through various outlets such as websites and local community lectures.

An additional application for funding may be made to the NSF program for Sharing Science with Public audiences.

Dominant group is an ongoing original research project that is being open-sourced here at Wikiversity both as a learning resource and to yield the broadest possible dissemination.

Benefits to society
“What may be the benefits of the proposed activity to society?"

The research effort should be designed
 * 1) to benefit multiple audiences,
 * 2) have a substantial educational impact,
 * 3) reach out to local communities, where appropriate, and
 * 4) include a significant collaboration with these same communities.

"Dominant group(s)", designated vaguely by the term, and the associated ideologies of exclusion, serve as apparent focused power structures that increase existing disparities in wealth and status, while marginalizing or disenfranchising "Others".

This proposed activity explores the two-word term "dominant group" to increase awareness of what dominant groups are linguistically as well as socially and scientifically. As the original research effort is ongoing at Wikiversity it benefits all audiences that explore learning through Wikiversity. A successful collaboration between NSF and Wikiversity helps to disseminate the societal benefits of NSF to any local community where some access to the internet and computer terminals in various forms exists such as at pre-schools, elementary and secondary schools, high schools and universities.

Dominant group is a two-word term that occurs at least once in some 270 English Wikipedia entries. These usages may be original research, plagiarism, copyright violations, properly cited uses, or simply mistakes in usage by editors and contributors. Or, dominant group because of its long history may have become a commonly used two-word term that crosses the barrier between common language words that are in a dictionary and the vocabulary of specialists, scholars, and experts. On the general web as sampled by Google, dominant group yields about 513,000 results. Many of these relate right back to the ongoing research at Wikiversity or to the earlier dominant group entries on Wikipedia that were deleted as being original research.

As a potential indicator of copyright violation, dominant group may serve to help improve major web-based resources such as Wikipedia. On Wikipedia dominant group occurs in some 270 entries and as a split term such as dominant ethnic group in some 15,700 entries. Overall the English Wikipedia contains some 5,074,967 entries.

The term "dominant group" occurs in all major fields of active discovery. Examples of the use of the term and discussion about and the implications of a dominant group are presented in various lectures and articles here at Wikiversity:
 * 1) Anthropology
 * 2) Archaeology
 * 3) Astronomy
 * 4) Astrophysics
 * 5) Attribution and copyright
 * 6) Cameroon
 * 7) Control group
 * 8) Feinburg, Walter (1998) Common schools uncummon identities. national unity and cultural difference.
 * 9) Economics
 * 10) Geology
 * 11) Gift economy
 * 12) Lamarckism
 * 13) Literature
 * 14) Materials science
 * 15) Metagenomes
 * 16) Metadefinition
 * 17) Motivation and emotion/Book/Toxic workplace
 * 18) NAU-POS254-Radical challenges to Enlightenment Ideologies
 * 19) Original research inquiry
 * 20) Planetary science
 * 21) Pragmatics
 * 22) Proof of concept
 * 23) Radiation astronomy
 * 24) Radiation chemistry
 * 25) Radiation entities
 * 26) Radiative dynamo
 * 27) Reading log McRuer, Tatum, Hjörne & Säljö
 * 28) Reflections on Tatum
 * 29) Regional astronomy
 * 30) Rigorous definition
 * 31) Semantics
 * 32) Semiotics
 * 33) Terminology
 * 34) Theoretical astronomy
 * 35) Theoretical radiation astronomy
 * 36) Transcription start site
 * 37) Visual astronomy
 * 38) X-ray astronomy

Articles or presentations of the research outside Wikiversity will be prepared openly here at Wikiversity as a teaching and training resource. Publication of such resources in open-access journals or scientific society journals benefits multiple audiences.

Before the benefits of the term and its relative synonyms is assessed, there is a need for some form of control group to serve as comparison. It should be noted that the two-word scientific term control group is a relative synonym of dominant group.

Hypotheses

 * 1) Dominant group has a broader impact due to its use in every major field.

title: Dominant group/Broader impacts


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User:Egm4313.s12.team15.moncrief

James Moncrief

title: User:Egm4313.s12.team15.moncrief


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User:Egm4313.s12.team4

Andy Jones

Brittany Jacobs

Daniel Friedman

Jonathan Di Roma

Jordan CaLoste

Neel Patel

Suzy Lorenzo

Report 1

title: User:Egm4313.s12.team4


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User:Egm4313.s12.team13.rossmeissl/R1work

Problem Statement
Using the circuit given, the governing circuit equation (4.0), and the capacitance equation (4.1), derive alternate equations (4.2) and (4.3).

Circuit Equation (4.0),

Capacitance Equation (4.1),

Alternate Equation (4.2),

$$ LCI''+RCI'+\frac{1}{C}I = V' $$

Alternate Equation (4.3),

$$ LCQ''+RCQ'+\frac{1}{C}Q = V $$

Solution
First take the derivative of the governing circuit equation (4.0) to get (4.4)

$$ V' = LC\frac{{d}^{3}v_{c}}{d{t}^{3}}+RC\frac{{d}^{2}v_{c}}{d{t}^{2}}+\frac{{d}^{}v_{c}}{d{t}^{}} $$

Since L, R, and C are constants with respect to time

Then, since

$$ I = C\frac{dv_{c}}{dt} $$

that means

$$ I' = C\frac{{d}^{2}v_{c}}{d{t}^{2}} $$

And

$$ I'' = C\frac{{d}^{3}v_{c}}{d{t}^{3}} $$

Substituting these into the derivative of the circuit equation (4.4) gives the first alternate equation (4.2)

$$ V' = LCI''+RCI'+\frac{1}{C}I $$

Then, since

$$ Q = C{v}_{c} $$

$$ Q' = C\frac{d{v}_{c}}{dt} $$

$$ Q'' = C\frac{{d}^{2}{v}_{c}}{d{t}^{2}} $$

Now substituting these into the appropriate places in the circuit equation (4.0) gives the second alternate equation (4.3)

$$ LCQ''+RCQ'+\frac{1}{C}Q = V $$


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User:Ab,the girl

Hello! i'm ab,the girl! Barrington levy i love,kromestar,i can't get this song outta my head http://www.youtube.com/watch?v=hDd-jTGri2E this is good!:) Thinking of doing a StarCraf II all nighter on friday. I think i could hack it seeing how little ive played in the last month :)

title: User:Ab,the girl


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User:Egm6322.s12.team2.steele.m2

EGM 6322 Team 2 Report 1

=R8.2 Fourier Series, Euler's Identity, and Fourier Coefficients= Question from 42-11 to 42-12: Verify (4). $$\int_0^T e^{i(n-m) \omega \theta} d\theta = $$ $$ \begin{cases} 0, & \text{if} n \ne m \\ T, & \text{if} n=m \end{cases}$$ Recover from {1} the Fourier coefficients found with the trigonometric identities above. From lecture notes [[media:pea1.f11.mtg42.djvu|Mtg 42]]

Solution
Apply Euler's identity. $$e^{ikx} = \cos kx + i\sin kx$$

$$\int_0^T f(\theta)e^{-im\omega \theta} d\theta = \sum_{n=-\infty}^{+\infty} \tilde a_n \, \int_0^T e^{i(n-m) \omega \theta} d\theta$$ The first part of the solution requires the derivation of (4) from p. 42-11. First, let n=m and substitute into the derivation with the Euler identity. The solution requires expansion of the terms in the integral into terms of sine and cosine. The key is to recognize that the expansion of the Euler identity for the positive and negative exponents as shown below. $$e^{-in\omega \theta} = \cos{n\omega \theta} - i\sin{n\omega \theta}$$ $$e^{in\omega \theta} = \cos{n\omega \theta} + i\sin{n\omega \theta}$$ $$e^{in\omega \theta} + e^{-in\omega \theta} = \cos{n\omega \theta} + \cos{n\omega \theta} - i\sin{n\omega \theta} + i\sin{n\omega \theta}$$ Note that the imaginary terms above go to zero. $$e^{in\omega \theta} + e^{-in\omega \theta} = 2\cos{n\omega \theta}$$

Divide the exponential summation by 2, and we have. $$(e^{in\omega \theta} + e^{-in\omega \theta})/2 = \cos{n\omega \theta}$$ Similarly, the pattern can be repeated for the sine term but leaves the imaginary component.

$$e^{in\omega \theta} - e^{-in\omega \theta} = \cos{n\omega \theta} - \cos{n\omega \theta} + i\sin{n\omega \theta} + i\sin{n\omega \theta}$$ Note that the real terms above go to zero. $$e^{in\omega \theta} + e^{-in\omega \theta} = 2i\sin{n\omega \theta}$$

This final term can be divided by 2i to get the equivalent value below. $$(e^{in\omega \theta} - e^{-in\omega \theta})/2 = \sin{n\omega \theta}$$ The Euler identity applies to the integral below. $$\int_0^T e^{i(n-m) \omega \theta} d\theta = \int_0^T{\cos{(n-m) \omega \theta} + i\sin{(n-m) \omega \theta}}$$ Note that if n=m then the following values occur. $$\cos{(n-m) \omega \theta} = \cos{0} = 1$$ $$\sin{(n-m) \omega \theta} = \sin{0} = 0$$ By substituting the values back into the integral, a simple integration is achieved for m=n. $$ \begin{align} \int_0^T e^{i(n-m) \omega \theta} d\theta & = \int_0^T{\cos{(n-m) \omega \theta} + i\sin{(n-m) \omega \theta}} \\ & = \int_0^T{(1 + 0)} \\ & = T \end{align}$$ Next, we concentrate on the condition where n does not equal m below. In this case, apply the substitution $$\text{T} = 2\pi / \omega$$. $$\begin{align} \int_0^T e^{i(n-m) \omega \theta} d\theta & = \int_0^T{\cos{(n-m) \omega \theta} + i\sin{(n-m) \omega \theta}} \\ & = \frac{1}{(n-m)\omega}(\sin{(n-m)\omega \theta} - i\cos{(n-m)\omega \theta} |_0^T \\ & = \frac{1}{(n-m)\omega}(\sin{(n-m)\omega \theta} - i\cos{(n-m)\omega \theta} |_0^2{\pi / \omega} \\ & = \frac{1}{(n-m)\omega}[(\sin{(n-m)\omega 2{\pi / \omega}} - i\cos{(n-m)\omega 2{\pi / \omega}} - (\sin{0} - i\cos{0})] \\ & = \frac{1}{(n-m)\omega}[(\sin{2\pi (n-m)} - i\cos{2\pi (n-m)}) + i] \\ & = \frac{1}{(n-m)\omega}[0 - i + i] \\ & = 0 \end{align}$$ The 2nd part of problem R8.2 requires the recovery of Fourier coefficients. First, the defintion below must be restated as given for the problem.  $$ \bar a_n := \begin{cases} 2 a_0, & \text{for } n=0 \\ a_n - ib_n, & \text{for } n>0 \\ a_n + ib_n, & \text{for } n<0 \end{cases}$$ Next, we revisit the equation of interest below. $$\int_0^T f(\theta)e^{-im\omega \theta} d\theta = \sum_{n=-\infty}^{+\infty} \tilde a_n \, \int_0^T e^{i(n-m) \omega \theta} d\theta$$ In the equation above, if n=m, then we can re-examine the integral above.  First, in the summation, all terms for n go to 0 except where n=m, so the summation term no longer appears. Subsequently, the integral derivation of (4) on p. 42-11 is applied. So this gives the following equation. $$\int_0^T f(\theta) e^{-in \omega \theta}d\theta = \tilde{a_n} T = \tilde{a_m} T$$ Note that the constant can be expanded into the equation from which we recover the Fourier coefficients. $$\begin{align} \tilde{a_n} & = \frac{1}{2}\bar{a_n} \\ & = \frac{1}{T} \int_0^T f(\theta) e^{-in\omega\theta} d\theta \end{align}$$

The equation can be solved in 3 parts: n=0, n>0, and n<0. Let's start with n=0 for which the Fourier coefficient is defined below. $$\begin{align} \tilde{a_0} & = \frac{1}{2}\bar{a_0} \\ & = frac{1}{2}(2a_0)\int_0^T f(\theta) e^{-in\omega \theta}d\theta \end{align}$$ Next, the Fourier coefficient for n>0 can be derived with the given definitions and the Euler identity. $$\begin{align} \tilde{a_0} & = \frac{1}{2}\bar{a_0} & = \frac{1}{2}\int_0^T f(\theta) e^{-in\omega \theta}d\theta \\ & = \frac{1}{2}\int_0^T f(\theta)(\cos{-in\omega\theta} + i\sin{-n\omega\theta} d\theta \end{align}$$ In the deriviation above for n>0, the Fourier coefficient is the real part of the complex number in the integration. $$a_n = \frac{1}{2} \mathbf{Re}\int_0^T f(\theta)(\cos{-n\omega\theta} + i\sin{-n\omega\theta}) = \frac{1}{2}\int_0^T f(\theta) \cos{-n\omega\theta}$$ Next, we derive the Fourier coefficient for n<0 by simply taking the imaginary component of the previous integral. $$a_n = \frac{1}{2}\mathbf{Im} \int_0^T f(\theta)(\cos{-n\omega\theta} + i\sin{-n\omega\theta}) = \frac{1}{2}\int_0^T f(\theta) \sin{-n\omega\theta}$$

=R8.3 Fourier Series and Real Components= Question from 42-12 Show that, equivalent to (2)-(3) p. 42-10, the Fourier series can be written as $$f(\theta) = \text{Re} \left(\sum_{n={\color{red} 0}}^{+\infty} \tilde a_n e^{in \omega \theta} \right)$$

From lecture notes [[media:pea1.f11.mtg42.djvu|Mtg 42]]

Solution
The first step is to write the original equation. $$ f(\theta) = \frac{1}{2}\sum_{n=-\infty}^{+\infty} \bar{a_n} e^{in\omega\theta} = \frac{1}{2}\sum_{n=-1}^{-\infty}(a_n + ib_n)e^{-in\omega\theta} + 2a_0 e^{in\omega\theta} + \sum_{n=1}^{\infty}(a_n - ib_n)e^{-in\omega\theta}$$ The equation can be simplified by recognizing the conjugate values. $$\bar{a_n}^* = \bar{a_{-n}}$$ The conjugate method can be applied to the summation below. $$\sum_{n=-1}^{-\infty} (a_n + ib_n) e^{in\omega\theta} = \sum_{n=1}^{+\infty} (a_n - ib_n)e^{-in\omega\theta}$$ Substitute the right hand term above back into the main equation for $$f(\theta)$$. $$ \begin{align} f(\theta) & = \frac{1}{2}(\sum_{n=1}^{\infty}(a_n - ib_n)e^{-in\omega\theta} + 2{a_0}e^{in\omega\theta} + \sum_{n=1}^{\infty}(a_n - ib_n)e^{in\omega\theta}) \\ & = \frac{1}{2}(2{a_0}e^{in\omega\theta}(e^{-in\omega\theta} + \sum_{n=1}^{\infty}(a_n - ib_n)e^{in\omega\theta}) \\ & = \frac{1}{2} (2a_0 e^{in\omega\theta} + \sum_{n=0}^{\infty}(a_n - ib_n)\cos{n\omega\theta} \\ & = \frac{1}{2} \sum_{n=0}^{\infty} \bar{a_n} \cos{n\omega\theta} \\ & = \sum_{n=0}^{\infty} \tilde{a_n} \cos{n\omega\theta} \\ & = \text{Re} \sum_{n=0}^{\infty} \tilde{a_n}(cos{n\omega\theta} + i\sin{n\omega\theta}) \\ & = \text{Re} \sum_{n=0}^{\infty} \tilde{a_n} e^{in\omega\theta} \end{align} $$ = References for Report 7 = The key to solve the problem is sum all terms in the "all" column to get the coefficient. $$A_0$$ = sum of numbers in "all" column * $$T_0$$ = 0.745999*$$T_0$$ This value above matches the value calculated in solution 9.1 by 0.7459/0.745999*100, which is almost 100%. This meets the criteria for an approximation with error < 5%. In addition, for $$A_2$$ we have the following table transcribed from the spreadsheet. Again, the key to solve the problem is sum all terms in the "all" column to get the coefficient. $$A_2$$ = sum of numbers in "all" column * $$T_0$$ = -0.7997*$$T_0$$ This value above matches the value calculated in solution 9.1 by 100%, so the error is negligible with a 100% match.

Finally, for $$A_4$$ we have the following table transcribed from the spreadsheet.

Once again we sum all terms in the "all" column to get the coefficient. $$A_4$$ = sum of numbers in "all" column * $$T_0$$ = 0.066715*$$T_0$$ This value above matches the value calculated in solution 9.1 by 0.066715/0.06671*100, which is almost 100%. The close match makes errors negligible and fulfills the criteria of an error < 5%. In summary, we have successfully calculated the coefficients by using the Gauss-Quadrature formulations.

practice

Template

=EGM6321 - Typed Notes from Principles of Engineering Analysis 1, Fall 2011 (Extra Credit)=

Text typed by Manuel Steele over Summer of 2012 with links to images in wiki categories for this class. Extra Credit PEA1

title: User:Egm6322.s12.team2.steele.m2


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User:Egm6322.s12.TEAM1.HW/HW2

Problem 1: Pose Question to be solved
 Solved without assistance

Find:
1. Without calculation, find the properties of $$ \left \{ A_j \right \} \ $$ 2. Compute the first 5 non-zero coefficients 3. Plot 3 curves on the same figure (i) the exact function (ii) Fourier-Legendre series with 3 terms (iii) Fourier-Legendre Series with 5 terms.

Solution:
1. Properties are $$ \left \{ A_j \right \} = \frac{\left \langle P_j,f \right \rangle}{\left \langle P_j,P_j \right \rangle} = \frac{2j+1}{2} \int_{ \mu = -1}^{ \mu = +1} P_j( \mu ) f ( \mu )\, d \mu \ $$

$$ f(x) \ $$ is an even function, and $$ {{P}_{j}}(x) \ $$ is odd when $$ j \ $$ is odd. so, $$ {{A}_{j}} \ $$ is zero when $$ j \ $$ is odd. 2. Start with function: $$ f( \mu )=T_0 \sinh(1- \mu^2) \ $$ For simplicity let $$ \mu = x \ $$, then we have $$ f( x )= T_0 \sinh(1- x^2) \ $$, which we apply Eq 1.2 to. Values for $$ P_0, P_2, P_4, P_6, P_8 \ $$ from F10 class lecture 36, page2 For $$ j=0 \ $$

For $$ j=2 \ $$

For $$ j=4 \ $$

For $$ j=6 \ $$

For $$ j=8 \ $$ 3.

Problem 2: show equality
 Solved without assistance

Find:
Show (1)P43-8, i.e.,show that

Solution:
Begin with $$\sum\limits_{j=0}^{\infty }{{P}_{j}}(\mu )=\sum\limits_{j=0}^{\infty }{{P}_{j}}(\mu )$$ then, we have: Further, because $$ {{P}_{j}}(\mu ) \ $$ is not always be zero, we can obtain: $$\begin{align} & {{{\bar{A}}}_{j}}-{{A}_{j}}=0,\forall j \\ & {{{\bar{A}}}_{j}}={{A}_{j}},\forall j \end{align}$$

Problem 3: Legendre homogeneous solution
 Solved without assistance

Given
$$\displaystyle {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)={{\tanh }^{-1}}x $$ (3.1) $$\displaystyle {{Q}_{1}}(x)=\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1=x{{\tanh }^{-1}}x-1 $$ (3.2)

Find
Show Eq (3.1) and Eq (3.2)

Solution
Order 0 Legendre equation is that $$\displaystyle(1-{{x}^{2}}){y}''-2x{y}'=0$$ Let z = y' $$\displaystyle {z}'+\frac{2x}{{{x}^{2}}-1}z=0$$ To know the solution, I use integrating factor method. $$\displaystyle h(x) =\exp \left[ \int_ – ^{x}{\frac{2x}{{{x}^{2}}-1}dx} \right] =\exp \left[ \int_ – ^{x}{\frac{1}{1-{{x}^{2}}}d(1-{{x}^{2}})} \right] =\exp [\log (1-{{x}^{2}})] =1-{{x}^{2}}$$  From the previous lecture (PEA1), $$\displaystyle z ={{h}^{-1}}(x)\int_ – ^{x}{h(x)*0dx} =\frac{1-{{x}^{2}}}$$  $$ y =\int{zdx}+{{k}_{1}} ={{k}_{1}}+\int_ – ^{}{\frac{1-{{x}^{2}}}dx} ={{k}_{1}}+\frac{2}\int_ – ^{}{\left( \frac{1}{1+x}+\frac{1}{1-x} \right)dx} ={{k}_{1}}+{{k}_{2}}\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) $$ $$\displaystyle Q_0$$ is one of homogeneous solutions. $$\displaystyle {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)$$  Now, We have to find what tanhx is. $$\displaystyle \tanh x = \frac{\sinh x}{\cosh x} = \frac {e^x - e^{-x}} {e^x + e^{-x}} = \frac{e^{2x} - 1} {e^{2x} + 1}$$ $$ x=\frac{1}{2}\log \left( \frac{1+\tanh x}{1-\tanh x} \right)$$ $$ {{\tanh }^{-1}}x=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) $$ These show that $$\displaystyle {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)={{\tanh }^{-1}}x$$

Order 1 Legendre equation is that $$\displaystyle (1-{{x}^{2}}){y}''-2x{y}'+2y=0$$ $$\displaystyle {y}''+\frac{2x}{({{x}^{2}}-1)}{y}'+\frac{2}{(1-{{x}^{2}})}y=0$$  using Integrating Factor Method, $$\displaystyle h(x)={{u}_{1}}^{2}(x)\exp \left[ \int_ – ^ – {{{a}_{1}}(x)dx} \right] ={{[{{P}_{1}}(x)]}^{2}}\exp \left[ \int_ – ^ – {\frac{2x}{{{x}^{2}}-1}dx} \right] ={{x}^{2}}\exp \left[ \int_ – ^ – {\frac{1}{1-{{x}^{2}}}d(1-{{x}^{2}})} \right] ={{x}^{2}}(1-{{x}^{2}})$$ $$\displaystyle {{Q}_{1}}(x) ={{u}_{2}}(x) ={{u}_{1}}(x)\int_ – ^ – {\frac{1}{h(x)}dx} =x\int{\frac{1}{{{x}^{2}}(1-{{x}^{2}})}dx} =x\int{\frac{1}+\frac{1}{(1-{{x}^{2}})}dx} =x\int{\frac{1}{(1-{{x}^{2}})}dx+}x\int{\frac{1}dx} =\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1 $$ These show that $$\displaystyle {{Q}_{1}}(x)=\frac{x}{2}\log \left( \frac{1+x}{1-x} \right)-1=x{{\tanh }^{-1}}x-1$$

Given
$$Q_n(x) = P_n(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {2n-2j+1} {(2n-j+1)j} P_{n-j}(x)$$ $$\displaystyle J = 1+2[\frac {n-1}{2}]$$ $$\displaystyle P_0(x) = 1$$ $$\displaystyle P_1(x) = x$$ $$\displaystyle P_2(x) = \frac {1} {2} (3x^2-1)$$ $$\displaystyle P_3(x) = \frac {1} {2} (5x^3-3x)$$ $$\displaystyle P_4(x) = \frac {1} {8} (35x^4 - 30x^2 + 3)$$

Find
Verify $$\displaystyle Q_0, \,\, Q_1, \,\, Q_2 $$ using above equation.

Solution
When n = 0, $$\displaystyle J = 1+2[\frac {0-1}{2}] = 0$$ $$\displaystyle Q_0(x) = P_0(x) tanh^{-1}(x) = tanh^{-1}(x) $$ When n = 1, $$\displaystyle J = 1+2[\frac {1-1}{2}] = 1$$ $$\displaystyle Q_1(x) = P_1(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^1 \frac {2-2+1} {(2-1+1)1} P_{1-1}(x)$$ $$\displaystyle Q_1(x) = P_1(x) tanh^{-1}(x) - 1 = xtanh^{-1}(x) - 1$$ When n = 2, $$\displaystyle J = 1+2[\frac {2-1}{2}] = 2$$ $$\displaystyle Q_2(x) = P_2(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^2 \frac {2n-2j+1} {(2n-j+1)j} P_{n-j}(x)$$ $$\displaystyle Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^2 \frac {2(2)-2(j)+1} {(2(2)-(j)+1)(j)} P_{2-j}(x)$$ $$\displaystyle Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - (\frac 3 2) (x) $$

Given
$$Q_n(x) = P_n(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {2n-2j+1} {(2n-j+1)j} P_{n-j}(x)$$ $$\displaystyle J = 1+2[\frac {n-1}{2}]$$

Find
Show that Qn is odd or even depending on the index n.

Solution
$$\displaystyle P_2k(x) $$ is even $$\displaystyle P_{2k+1}(x)$$ is odd

Hence, Let 2k = n = even number k = 1,2,3......  $$Q_2k(x) = P_{2k}(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {4k-2j+1} {(4k-j+1)j} P_{2k-j}(x)$$ $$\displaystyle tanh^{-1}(x)$$ is always odd. $$\displaystyle P_{2k-j}(x) $$ is odd.

$$\displaystyle Q_2k(x) = (even)(odd) - (odd) $$ When n = even, Q(x) is odd.

Let 2k+1 = n = odd number k = 1,2,3......  $$Q_{2k+1}(x) = P_{2k+1}(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {2(2k+1)-2j+1} {(2(2k+1)-j+1)j} P_{2(2k+1)-j}(x)$$ $$\displaystyle tanh^{-1}(x)$$ is always odd. $$\displaystyle P_{2(2k+1)-j}(x) $$ is even. $$\displaystyle Q_{2k+1}(x) = (odd)(odd) - (even) $$ When n = odd, Q(x) is even.

Problem 6: Orthogonality of legendre fuctions
 Solved without assistance

Given:
$$\displaystyle {{L}_{n}}:={{P}_{n}}$$, or, $$\displaystyle {{Q}_{n}}$$ For $$\displaystyle n\ne m$$ $$\begin{align} & <{{L}_{n}},{{L}_{m}}>=0 \\ & <{{L}_{n}},{{L}_{n}}>=<{{P}_{n}},{{P}_{n}}>=\frac{2}{2n+1} \\ \end{align}$$

Find:
1): $$\displaystyle <{{L}_{n}},{{L}_{n}}>=<{{P}_{n}},{{Q}_{n}}>=?$$  2): Show the following equalities:

$$\begin{align} & {{\left( {{r}_{PQ}} \right)}^{2}}={{\left( {{r}_{Q}} \right)}^{2}}[{{\left( \frac \right)}^{2}}+1-2\frac\cos \gamma ] \\ & ={{\left( {{r}_{Q}} \right)}^{2}}[{{\rho }^{2}}+1-2\rho \cos \gamma ] \end{align}$$ $$\displaystyle \cos \gamma =\cos {{\theta }_{Q}}\cos {{\theta }_{P}}\cos \left( {{\varphi }_{Q}}-{{\varphi }_{P}} \right)+\sin {{\theta }_{Q}}sin{{\theta }_{P}}$$ Recall: $$\displaystyle {{\left( {{r}_{PQ}} \right)}^{2}}={{\left( {{r}_{P}} \right)}^{2}}+{{\left( {{r}_{Q}} \right)}^{2}}-2{{r}_{P}}{{r}_{Q}}\cos \gamma $$

Solution:
1): $$\displaystyle <{{P}_{n}},{{Q}_{n}}>=\int\limits_{\mu =-1}^{\mu =+1}{{{P}_{n}}(\mu ){{Q}_{n}}(\mu )}d\mu $$  Then, we will analyze the eveness of these functions: when n is odd, $$\displaystyle {{P}_{n}}$$ is odd, $$\displaystyle {{Q}_{n}}$$ is even, so $$\displaystyle {{P}_{n}}(\mu ){{Q}_{n}}(\mu )$$ is odd. when n is even, $$\displaystyle {{P}_{n}}$$ is even, $$\displaystyle {{Q}_{n}}$$ is odd, so $$\displaystyle {{P}_{n}}(\mu ){{Q}_{n}}(\mu )$$ is odd. So, $$\displaystyle {{P}_{n}}(\mu ){{Q}_{n}}(\mu )$$ is odd whatever n is.  $$\displaystyle <{{P}_{n}},{{Q}_{n}}>=\int\limits_{\mu =-1}^{\mu =+1}{{{P}_{n}}(\mu ){{Q}_{n}}(\mu )}d\mu =0$$  2): According to the definition of spherical coordinates, we have:

$$\begin{align} & {{\left( {{r}_{PQ}} \right)}^{2}}={{\left( {{x}_{P}}-{{x}_{Q}} \right)}^{2}}+{{\left( {{y}_{P}}-{{y}_{Q}} \right)}^{2}}+{{\left( {{z}_{P}}-{{z}_{Q}} \right)}^{2}} \\ & ={{\left( {{r}_{P}}\cos {{\theta }_{P}}\cos {{\varphi }_{P}}-{{r}_{Q}}\cos {{\theta }_{Q}}\cos {{\varphi }_{Q}} \right)}^{2}}+{{\left( {{r}_{P}}\cos {{\theta }_{P}}\sin {{\varphi }_{P}}-{{r}_{Q}}\cos {{\theta }_{Q}}\sin {{\varphi }_{Q}} \right)}^{2}}+{{\left( {{r}_{P}}\sin {{\theta }_{P}}-{{r}_{Q}}\sin {{\theta }_{Q}} \right)}^{2}} \\ & ={{\left( {{r}_{P}} \right)}^{2}}{{\left( {{\cos }^{2}}{{\theta }_{P}}{{\cos }^{2}}{{\varphi }_{P}}+{{\cos }^{2}}{{\theta }_{P}}{{\sin }^{2}}{{\varphi }_{P}}+{{\sin }^{2}}{{\theta }_{P}} \right)}^{2}}+{{\left( {{r}_{Q}} \right)}^{2}}{{\left( {{\cos }^{2}}{{\theta }_{Q}}{{\cos }^{2}}{{\varphi }_{Q}}+{{\cos }^{2}}{{\theta }_{Q}}{{\sin }^{2}}{{\varphi }_{Q}}+{{\sin }^{2}}{{\theta }_{Q}} \right)}^{2}} \\ & -2{{r}_{P}}{{r}_{Q}}(\cos {{\theta }_{P}}\cos {{\varphi }_{P}}\cos {{\theta }_{Q}}\cos {{\varphi }_{Q}}+\cos {{\theta }_{P}}\sin {{\varphi }_{P}}\cos {{\theta }_{Q}}\sin {{\varphi }_{Q}}+\sin {{\theta }_{P}}\sin {{\theta }_{Q}}) \\ & ={{\left( {{r}_{P}} \right)}^{2}}+{{\left( {{r}_{Q}} \right)}^{2}}-2{{r}_{P}}{{r}_{Q}}[\cos {{\theta }_{P}}\cos {{\theta }_{Q}}(\cos {{\varphi }_{P}}\cos {{\varphi }_{Q}}+\sin {{\varphi }_{P}}\sin {{\varphi }_{Q}})+\sin {{\theta }_{P}}\sin {{\theta }_{Q}}] \\ & ={{\left( {{r}_{P}} \right)}^{2}}+{{\left( {{r}_{Q}} \right)}^{2}}-2{{r}_{P}}{{r}_{Q}}[\cos {{\theta }_{P}}\cos {{\theta }_{Q}}\cos ({{\varphi }_{P}}-{{\varphi }_{Q}})+\sin {{\theta }_{P}}\sin {{\theta }_{Q}}] \\ & ={{\left( {{r}_{Q}} \right)}^{2}}[{{\rho }^{2}}+1-2\rho \cos \gamma ] \end{align}$$ Where,

$$\displaystyle\rho =\frac$$ $$\displaystyle \cos \gamma =\cos {{\theta }_{Q}}\cos {{\theta }_{P}}\cos \left( {{\varphi }_{Q}}-{{\varphi }_{P}} \right)+\sin {{\theta }_{Q}}sin{{\theta }_{P}}$$

Problem 7: Verification
 Solved without assistance

Given:
for $$ \left | x \right \vert <1 \ $$

Find:
Use Equation 7.1 to verify equations 7.2 and 7.3

Solution
Given that: Equation 7.1 can be re-written as: and substituting $$ \alpha_i \ $$ from Eq7.3 into 7.2 you get:

We can substitute x=-x,r=-1/2 into Equation 7.5:

Here, the end of Equation 7.7 is the same to Equation 7.6.

So, we can use Equation 7.1 to verify Equations 7.2 and 7.3

Problem 8: Generate{P2,...,P6}using RR2 and from generating function expansion
 Solved without assistance

Given
1.Recurrence relation RR2:
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$$ \displaystyle (n+1)P_{n+1}-(2n+1) x P_n+nP_{n-1}=0 $$
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with:
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$$ \displaystyle P_0(x)=1 $$
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$$ \displaystyle P_1(x)=x $$
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2.Generating function for Legendre polynomials
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$$ \displaystyle \mathcal G_L(\mu,\rho)=A^{-1/2}(\mu,\rho)=(1-x)^{-1/2} $$
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$$ \displaystyle A(\mu ,\rho):=1-2\mu \rho+\rho^2=:1-x $$
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with:
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$$ \displaystyle -x:=-2\mu\rho+\rho^2 $$
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$$ \displaystyle x:=2\mu\rho-\rho^2 $$
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Find

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$$ \displaystyle {P_0,...,P_6} $$ Using both two methods
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Solution
Part I

Using RR2:
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$$ \displaystyle (n+1)P_{n+1}-(2n+1)xP_n+nP_{n-1}=0 $$
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$$ \displaystyle P_0=1, \ P_1=x $$
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n=1
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$$ \displaystyle 2P_2-3xP_1+P_0=0 $$
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$$ \displaystyle 2P_2-3x^2+1=0 $$
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$$ \displaystyle P_2=\frac{1}{2}(3x^2-1) $$
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n=2
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$$ \displaystyle 3P_3-5xP_2+2P_1=0 $$
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$$ \displaystyle 3P_3-\frac{5x}{2}(3x^2-1)+2x=0 $$
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$$ \displaystyle P_3=\frac{1}{2}(5x^3-3x) $$
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n=3
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$$ \displaystyle 4P_4-7xP_3+3P_2=0 $$
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$$ \displaystyle 4P_4-\frac{7x}{2}(5x^3-3x)+\frac{3}{2}3x^2-1=0 $$
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$$ \displaystyle P_4=\frac{1}{8}(35x^4-30x^2+3) $$
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n=4
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$$ \displaystyle 5P_5-9xP_4+4P_3=0 $$
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$$ \displaystyle 5P_5-\frac{9}{8}x(35x^4-30x^2+3)+\frac{4}{2}(5x^3-3x)=0 $$
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$$ \displaystyle P_5=\frac{1}{8}(63x^5-70x^3+15x) $$
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n=5
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$$ \displaystyle 6P_6-11xP_5+5P_4=0 $$
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$$ \displaystyle 6P_6-\frac{11}{8}x(63x^5-70x^3+15x)+\frac{5}{8}(35x^4-30x^2+3)=0 $$
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$$ \displaystyle P_6=\frac{1}{16}(231x^6-315x^4+105x^2-5) $$
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Part II

Using generating function expansion:
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$$ \displaystyle \begin{align} \mathcal G_L(\mu,\rho) &=A^{-1/2}(\mu,\rho)=(1-x)^{-1/2} \\ &=1+(-\frac{1}{2})(-x)+\frac{3}{8}(-x)^2+(-\frac{5}{16})(-x)^3+\frac{35}{128}(-x)^4... \\ &=1+\frac{1}{2}x+\frac{3}{8}x^2+\frac{5}{16}x^3+\frac{35}{128}x^4... \\ &=1+\frac{1}{2}(2\mu \rho-\rho^2)+\frac{3}{8}(4\mu^2\rho^2-4\mu \rho^3+\rho^4)+\frac{5}{16}(8\mu^3\rho^3-12\mu^2\rho^4+6\mu \rho^5-\rho^6) \\ &\ \ \ \ +\frac{35}{128}(16\mu^4\rho^4-32\mu^3\rho^5+24\mu^2\rho^6-6\mu \rho) \\ &=1+\mu \rho \\ &\ \ \ \ +\frac{1}{2}(3\mu^2-1)\rho^2 \\ &\ \ \ \ +\frac{1}{2}(5\mu^3-3\mu)\rho^3 \\ &\ \ \ \ +\frac{1}{8}(35\mu^4-30\mu^2+3)\rho^4 \\ &\ \ \ \ +\frac{1}{8}(63\mu^5-70\mu^3+15\mu)\rho^5 \\ &\ \ \ \ +\frac{1}{16}(231\mu^6-315\mu^4+105\mu^2-5)\rho^6 \end{align} $$
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So:
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$$ \displaystyle \begin{align} P_0 &= 1\\ P_1 &= \mu \\ P_2 &= \frac{1}{2}(3\mu^2-1) \\ P_3 &= \frac{1}{2}(5\mu^3-3\mu) \\ P_4 &= \frac{1}{8}(35\mu^4-30\mu^2+3) \\ P_5 &= \frac{1}{8}(63\mu^5-70\mu^3+15\mu) \\ P_6 &= \frac{1}{16}(231\mu^6-315\mu^4+105\mu^2-5) \end{align} $$
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Comparing the two series of results, we can find whatever which method we use, the Legendre polynomials we get are exactly same.(Replace variable $$\mu$$ with x )

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title: User:Egm6322.s12.TEAM1.HW/HW2