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User:Egm6321.f10.team2.reserve/HW5

= Problem 5.4 Particular Solution by using Variation of Parameters Method =

From the lecture slide Mtg 27-2

Given
Linear, nonhomogeneous, first order differential equation

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$$ \displaystyle \frac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$$ (Eq 4.1)
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Find
Particular solution using $$y\displaystyle \left( x \right)=A\left( x \right){{y}_{h}}(x)$$

To find homogeneous solution:
Homogeneous differential equation

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$$ \displaystyle \frac{d{{y}_{h}}}{dx}+P\left( x \right){{y}_{h}}=0$$ (Eq 4.2)
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and the solution of homogeneous differential equation is

 * $${{y}_{h}}=A{{e}^{-\int\limits_ – ^{x}{P\left( t \right)dt}}}=A{{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}$$

where A is a constant of integration (King, page 512).

To find particular solution:
Assuming

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$$ \displaystyle \left( x \right)=A\left( x \right){{y}_{h}}(x)=A(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}$$ (Eq 4.3)
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is solution of nonhomogeneous ODE (Eq4.1).

We determine A(x) by substituting Eq4.3 and its first derivative into Eq4.1. Differentiating Eq4.3, we obtain,

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$$ \displaystyle \frac{d}{dx}=={A}'(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}-A(x)P(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}$$ (Eq 4.4)
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Now $$\displaystyle $$ must satisfy (Eq4.1). We need to substitute $$\displaystyle $$ and $$\displaystyle $$ according to Eq4.3 and Eq4.4 into Eq4.1.

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$$ \displaystyle \frac{d}{dx}+P\left( x \right){{y}_{p}}=Q\left( x \right)$$ (Eq 4.5)
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$$ \displaystyle \left[ {A}'(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}-A(x)P(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}} \right]+P(x)\left( A(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}} \right)=Q(x)$$ (Eq 4.6)
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Second term and third term in the left side of Eq4.7 eliminate each other and we obtain

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$$ \displaystyle {A}'(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}=Q(x)$$ (Eq 4.7)
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We rearrange Eq4.7

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$$ \displaystyle {A}'(x)=Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}$$ (Eq 4.8)
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If we integrate both sides Eq4.8, A(x) will be

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$$ \displaystyle A(x)=\int{Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}dx}$$ (Eq 4.9)
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If we put Eq4.9 into Eq4.3 in order to get the particular solution, we obtain

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$$ \displaystyle \left( x \right)=A(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}={{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}\left( \int{Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}dx+C} \right)$$ (Eq 4.10)
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Eliminate the constant $$\displaystyle C$$ we can derive particular solution,

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$$ \displaystyle {{y}_{P}}={{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}\int Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}dx$$ (Eq 4.11)
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Now we can solve homework equation in Mtg10-4 as an example,

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$$ \displaystyle \frac{dy}{dx}+xy=2x+3$$ (Eq 4.12)
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Thus,

 * $$\displaystyle P\left( x \right)=x$$ and $$Q\left( x \right)=2x+3$$

If we solve the homogenous equation for $$\displaystyle P\left( x \right)=x$$,

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$$ \displaystyle {{y}_{h}}=A{{e}^{-\frac{2}}}$$
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then,

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$$ \displaystyle ={{e}^{-\int\limits_ – ^{x}{tdt}}}\int\limits_ – ^{x}{(2s+3){{e}^{\int\limits_ – ^{s}{tdt}}}ds}={{e}^{-\frac{2}}}\left[ \int\limits_ – ^{x}{2s{{e}^{^{\frac{2}}}}ds}+\int\limits_ – ^{x}{3{{e}^{^{\frac{2}}}}ds} \right]$$ (Eq 4.13)
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Integrating inside of bracket by using Wolfram Alpha, we obtain $$\displaystyle $$ shown below

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$$ \displaystyle =2+3\sqrt{\frac{\pi }{2}}.{{e}^{-\frac{2}}}.erfi\left( \frac{x}{\sqrt{2}} \right)+{{C}_{1}}{{e}^{-\frac{2}}}$$ (Eq 4.14)
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where erfi(x) is the imaginary error function.

Then we extract particular solution from above,

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$$ \displaystyle {{y}_{p}}=2+3\sqrt{\frac{\pi }{2}}.{{e}^{-\frac{2}}}.erfi\left( \frac{x}{\sqrt{2}} \right)$$ (Eq 4.15)
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= Problem 5.5 Solving Non-homogeneous Linear Second Order ODE with Constant Coefficients=

From the lecture slide Mtg 28 and Mtg 29-1

Given
The spring-dashpot system has the Equation of Motion like follows:

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$$ \displaystyle {{a}_{2}}{y}''+{{a}_{1}}{y}'+{{a}_{0}}y=f(t)$$ (Eq 5.1)
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where $$\displaystyle y$$ is a function of $$\displaystyle t$$, and $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{0}}$$ are all constants making Eq 5.1 a linear second-order ODE with constant coeffiients.

Find
(1)Find PDEs for integrating factor $$\displaystyle H(t,y)$$ using condition 2nd of exactness stated in Mtg15

(2)Solve Eq 5.1 using trial solution for integrating factor such that $$\displaystyle h(t)={{e}^{\alpha t}}$$ where $$\displaystyle \alpha $$ is some constant. .

(1) Finding PDEs for integrating factor $$\displaystyle H(t,y)$$
Multiply $$\displaystyle H(t,y)$$ to both sides of the Eq 5.1,

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$$ \displaystyle H(t,y){{a}_{2}}{y}''+H(t,y){{a}_{1}}{y}'+H(t,y){{a}_{0}}y=H(t,y)f(t)$$ (Eq 5.2)
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If we assume firstly $$\displaystyle H(t,y)$$ to be a function of only $$\displaystyle t$$(see Mtg 10), then we can just find some $$\displaystyle H(t)$$ to make the L.H.S of Eq 5.2 exact in order to change its form as,

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$$ \displaystyle \frac{d\phi }{dt}=H(t)f(t)$$ (Eq 5.3)
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then we can integrate both sides to reduce the order of Eq 5.1. However, if the assumption "$$\displaystyle H(t,y)$$ is a function of only $$\displaystyle t$$" is not satisfid, we can't integrate the R.H.S of Eq 5.3. In this case, we should move $$\displaystyle f(t)$$ from R.H.S to L.H.S and try to find any $$\displaystyle H(t,y)$$ to make the whole thing exact.

Thus let

 * $$\displaystyle F=H(t){{a}_{2}}{y}+H(t){{a}_{1}}{y}'+H(t){{a}_{0}}y=f(t,y,p){y}+g(t,y,p)$$

in which,

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& f=H(t){{a}_{2}} \\ & g=H(t){{a}_{1}}{y}'+H(t){{a}_{0}}y \\ & p={y}' \\ \end{align}$$

Then we have,

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$$ \displaystyle \begin{align} & {{f}_{tt}}={{a}_{2}}{{H}_{tt}} \\ & {{f}_{ty}}={{a}_{2}}{{H}_{ty}} \\ & {{f}_{tp}}={{f}_{yp}}=0 \\ & {{f}_{y}}={{a}_{2}}{{H}_{y}} \\ & {{f}_{yy}}={{a}_{2}}{{H}_{yy}} \\ & {{g}_{tp}}={{a}_{1}}{{H}_{t}} \\ & {{g}_{yp}}={{a}_{1}}{{H}_{y}} \\ & {{g}_{pp}}=0 \\ & {{g}_{y}}={{a}_{1}}p{{H}_{y}}+{{a}_{0}}H+{{a}_{0}}y{{H}_{y}} \\ \end{align}$$ (Eqs 5.4)
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Substituting Eqs 5.4 into (1)&(2) in Mtg15, we have,

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$$ \displaystyle {{a}_{2}}{{H}_{tt}}+2p{{a}_{2}}{{H}_{ty}}+{{p}^{2}}{{a}_{2}}{{H}_{yy}}={{a}_{1}}{{H}_{t}}-{{a}_{0}}H-{{a}_{0}}y{{H}_{y}}$$ (Eq 5.5)
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$$ \displaystyle 2{{a}_{2}}{{H}_{y}}=0$$ (Eq 5.6)
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From Eq 5.6 we can see that the basic assumption is satisfied and if we substitute Eq 5.6 in to Eq 5.5, we have,

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$$ \displaystyle {{a}_{2}}{{H}_{tt}}-{{a}_{1}}{{H}_{t}}+{{a}_{0}}H=0$$ (Eq 5.7)
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or

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$$ \displaystyle {{a}_{2}}{H}''-{{a}_{1}}{H}'+{{a}_{0}}H=0$$ (Eq 5.8)
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Since the integrating factor $$\displaystyle H(t,y)$$ turns out to be $$\displaystyle H(t)$$, the PDE governing $$\displaystyle H(t,y)$$ become an ODE given by Eq 5.8.

(2) Solve Eq 5.1 using trial solution for integrating factor such that $$\displaystyle h(t)={{e}^{\alpha t}}$$
Let's try an integrating factor like $$\displaystyle h(t)={{e}^{\alpha t}}$$ where $$\displaystyle \alpha$$ is a constant.

Multiply Eq 5.1 by $$\displaystyle h(t)={{e}^{\alpha t}}$$, we have,

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$$ \displaystyle {{e}^{\alpha t}}\left( {{a}_{2}}{y}''+{{a}_{1}}{y}'+{{a}_{0}}y \right)={{e}^{\alpha t}}f(t)$$ (Eq 5.9)
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In order to reduce the order, we integrating Eq 5.9 according to $$\displaystyle t$$ and assume that the result can be written in such form:

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$$ \displaystyle {{e}^{\alpha t}}\left( {{{\bar{a}}}_{1}}{y}'+{{{\bar{a}}}_{2}}y \right)=\int{{{e}^{\alpha t}}f(t)dt}$$ (Eq 5.10)
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where $$\displaystyle {{{\bar{a}}}_{1}}$$ and $$\displaystyle {{{\bar{a}}}_{2}}$$ are assumed to be constants.

(2.1) Determine $$\displaystyle {{{\bar{a}}}_{1}}$$ and $$\displaystyle {{{\bar{a}}}_{1}}$$ according to $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$ and $$\displaystyle {{a}_{0}}$$
Differentiate Eq 5.10 we have,

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$$ \displaystyle {{e}^{\alpha t}}\left[ {{{\bar{a}}}_{1}}{y}''+(\alpha {{{\bar{a}}}_{1}}+{{{\bar{a}}}_{0}}){y}'+\alpha {{{\bar{a}}}_{0}}y \right]={{e}^{\alpha t}}f(t)$$ (Eq 5.11)
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Compare 5.11 and 5.9 we have,

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$$ \displaystyle \begin{align} & {{{\bar{a}}}_{1}}={{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{1}}-\alpha {{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{0}}/\alpha \\ \end{align}$$ (Eqs 5.12)
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from which we can see that the validity of Eq 5.10 relies on the value of $$\displaystyle \alpha $$.

(2.2) Find quadratic equation for $$\displaystyle \alpha $$
From Eqs 5.12 we have,

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$$ \displaystyle \begin{align} & {{a}_{1}}-\alpha {{a}_{2}}={{a}_{0}}/\alpha \\ \Rightarrow & {{a}_{2}}{{\alpha }^{2}}-{{a}_{1}}\alpha +{{a}_{0}}=0 \\ \end{align}$$ (Eq 5.13)
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(2.3) Derive a reduced - order equation from Eq 5.10
From Eq 5.10 we have,

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$$ \displaystyle {y}'+\beta y=\frac{1}\int{{{e}^{\alpha t}}f(t)dt}$$ (Eq 5.14)
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where $$\displaystyle \beta ={{{\bar{a}}}_{0}}/{{{\bar{a}}}_{1}}$$

(2.4) Solve Eq 5.14 using Integrating Factor Method
From Mtg10 we know the integrating factor(for Eq 5.14) is,

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$$ \displaystyle i(t)=\exp \int{\beta dt}={{e}^{\beta t}}$$ (Eq 5.15)
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and the solution is:

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$$ \displaystyle \begin{align} y & =\frac{1}{i(t)}\int_ – ^{t}{\frac{i(\tau )}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \end{align}$$ (Eq 5.16)
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(2.5) Show the relation between $$\displaystyle \alpha $$ and $$\displaystyle \beta $$
Since,

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$$ \displaystyle \begin{align} & \beta ={{{\bar{a}}}_{0}}/{{{\bar{a}}}_{1}} \\ & {{{\bar{a}}}_{0}}={{a}_{1}}-\alpha {{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{0}}/\alpha \\ & {{{\bar{a}}}_{1}}={{a}_{2}} \\ \end{align}$$ (Eqs 5.17)
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So we have,

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$$ \displaystyle \begin{align} & \alpha \beta =\frac \\ & \alpha +\beta =\frac \\ \end{align}$$ (Eq 5.18)
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which means $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are the two roots of,

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$$ \displaystyle {{\lambda }^{2}}-(\alpha +\beta )\lambda +\alpha \beta =0$$
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or

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$$ \displaystyle {{\lambda }^{2}}-\frac\lambda +\frac=0$$ (Eq 5.19)
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which is identical to Eq 5.13. And from Eq 5.17~Eq 5.19 we can conclude that $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are inseparable pairs so that we can't distiguish them from each other through the their governing equation and the sequense of two integrating factor $$\displaystyle h(t)$$ and $$\displaystyle i(t)$$ is arbitrary. We can also conclude that the final structure of solution must be symetric with respect to $$\displaystyle \alpha $$ and $$\displaystyle \beta $$, meaning exchanging the position of $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ won't change the result.

When $$\displaystyle \alpha \ne \beta $$
Eq 5.16 can be expanded as,

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$$ \displaystyle \begin{align} y & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau }+{{k}_{1}} \right)d\tau } \\ & =\frac\left[ \int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau } \right)d\tau }+\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}{{k}_{1}}d\tau } \right] \\ & =\frac\left[ \int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}+{{k}_{2}}+\frac{\beta -\alpha }{{e}^{(\beta -\alpha )t}} \right] \\ & =\frac{(\beta -\alpha ){_{1}}}{{e}^{-\alpha t}}+\frac{{e}^{-\beta t}}+\frac{{{{\bar{a}}}_{1}}}\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \end{align}$$ (Eq 5.20)
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Note that Eq 5.20 is under the assumption that $$\alpha \ne \beta $$.

After conbination of constants, we have,

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$$ \displaystyle y={{C}_{1}}{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\beta t}}+\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}$$ (Eq 5.21)
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In which we can derive,

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$$ \displaystyle \begin{align} & y_{H}^{1}={{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \\ \end{align}$$
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However, $$\displaystyle {{y}_{P}}$$ can be furthuer deduced as,

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$$ \displaystyle \begin{align} {{y}_{P}} & =\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\int{\left( \int{{{e}^{\alpha t}}f(t)dt} \right)d\left( {{e}^{(\beta -\alpha )t}} \right)} \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\left[ {{e}^{(\beta -\alpha )t}}\int{{{e}^{\alpha t}}f(t)dt}-\int{{{e}^{(\beta -\alpha )t}}{{e}^{\alpha t}}f(t)dt} \right] \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\alpha t}}f(t)dt}-\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\beta t}}f(t)dt} \end{align}$$
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This proved my statement "the final structure of solution must be symetric with respect to $$\displaystyle \alpha $$ and $$\displaystyle \beta $$".

In sum we have (when $$\displaystyle \alpha \ne \beta $$),

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$$ \displaystyle \begin{align} & y={{C}_{1}}y_{H}^{1}+{{C}_{2}}y_{H}^{2}+{{y}_{P}} \\ & y_{H}^{1}={{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\alpha t}}f(t)dt}-\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\beta t}}f(t)dt} \\ \end{align}$$ (Eq 5.22)
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When $$\displaystyle \alpha =\beta $$
Substitute the relation $$\displaystyle \alpha =\beta $$ into Eq 5.16,

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$$ \displaystyle \begin{align} y= & \frac\int_ – ^{t}{{{e}^{(\alpha -\alpha )\tau }}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac\int_ – ^{t}{\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau }+{{k}_{1}} \right)d\tau } \\ & =\frac\left[ \int_ – ^{t}{\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau } \right)d\tau }+\int_ – ^{t}{{{k}_{1}}d\tau } \right] \\ & =\frac\left[ \int{\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}+{{k}_{2}}+{{k}_{1}}t \right] \\ & =\fract{{e}^{-\alpha t}}+\frac{{e}^{-\alpha t}}+\frac\int{\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \\ & =\fract{{e}^{-\alpha t}}+\frac{{e}^{-\alpha t}}+\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right) \end{align}$$ (Eq 5.20a)
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After conbination of constans we have

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$$ \displaystyle y={{C}_{1}}t{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\alpha t}}+\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right)$$ (Eq 5.21a)
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In which we can derive,

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$$ \displaystyle \begin{align} & y_{H}^{1}=t{{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\alpha t}} \\ & {{y}_{P}}=\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right) \\ \end{align}$$
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In sum we have (when $$\displaystyle \alpha =\beta $$),

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$$ \displaystyle \begin{align} & y={{C}_{1}}y_{H}^{1}+{{C}_{2}}y_{H}^{2}+{{y}_{P}} \\ & y_{H}^{1}=t{{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\alpha t}} \\ & {{y}_{P}}=\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right) \\ \end{align}$$ (Eq 5.22a)
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(2.7) Verify particular solution according to table in Mtg20, Fall2009
If $$\displaystyle \alpha \ne \beta$$, then from Eq 5.22 we know,

 * $$\displaystyle {{y}_{P}}=\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\alpha t}}f(t)dt}-\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\beta t}}f(t)dt}$$

If we substitute $$\displaystyle f(t)={{t}^{2}}$$ into the above equation, we have,

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$$ \displaystyle \begin{align} {{y}_{P}} & =\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\alpha t}}f(t)dt-\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\beta t}}f(t)dt \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\alpha t}}{{t}^{2}}dt-\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\beta t}}{{t}^{2}}dt \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\left( \frac{{{\alpha }^{2}}{{t}^{2}}-2\alpha t+2} \right)-\frac{(\beta -\alpha ){{a}_{2}}}\left( \frac{{{\beta }^{2}}{{t}^{2}}-2\beta t+2} \right) \\ & =\frac{{{\alpha }^{2}}{{\beta }^{2}}{{t}^{2}}-(2{{\alpha }^{2}}\beta +2\alpha {{\beta }^{2}})t+2{{\alpha }^{2}}+2{{\beta }^{2}}+2\alpha \beta } \\ & =\frac{{{(\alpha \beta )}^{2}}{{t}^{2}}-2(\alpha +\beta )t+2{{(\alpha +\beta )}^{2}}-2\alpha \beta } \\ & =\frac{\frac{{{a}_{0}}^{2}}{{{a}_{2}}^{2}}{{t}^{2}}-2\frac{{{a}_{2}}^{2}}t+2\frac{{{a}_{1}}^{2}}{{{a}_{2}}^{2}}-2\frac{{{a}_{2}}^{2}}}{\frac{{{a}_{0}}^{3}}{{{a}_{2}}^{3}}{{a}_{2}}} \\ & =\frac{1}{{t}^{2}}-\frac{2{{a}_{1}}}{{{a}_{0}}^{2}}t+\frac{2{{a}_{1}}^{2}-2{{a}_{0}}{{a}_{2}}}{{{a}_{0}}^{3}} \\ \end{align}$$ (Eq 5.23)
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 *

 * }

in which we utilize,

 * $$\displaystyle \begin{align}

\int{{{e}^{\alpha t}}{{t}^{2}}dt} & =\frac{1}{\alpha }\int{{{t}^{2}}d{{e}^{\alpha t}}} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{1}{\alpha }\int{2t{{e}^{\alpha t}}dt} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{2}\int{td}{{e}^{\alpha t}} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{2}t{{e}^{\alpha t}}+\frac{2}\int{{{e}^{\alpha t}}dt} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{2}t{{e}^{\alpha t}}+\frac{2}{{e}^{\alpha t}} \\ & =\frac{{{\alpha }^{2}}{{t}^{2}}-2\alpha t+2}{{e}^{\alpha t}} \\ \end{align}$$

Compare Eq 5.23 with Mtg20 slide of Fall 2009, we can see that Eq 5.23 has the same form with that in the slide which is a polynominal with at least order 2 according to $$\displaystyle t$$.

If $$\displaystyle \alpha=\beta$$, then from Eq 5.22a we know,

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$$ \displaystyle \begin{align} & {{y}_{P}}=\frac\left( t\int {{e}^{\alpha t}}{{t}^{2}}dt-\int {{e}^{\alpha t}}{{t}^{3}}dt \right) \\ & =\frac\left( \frac{{{e}^{\alpha t}}({{\alpha }^{2}}{{t}^{3}}-2\alpha {{t}^{2}}+2t)}-\frac{{{e}^{\alpha t}}({{\alpha }^{3}}{{t}^{3}}-3{{\alpha }^{2}}{{t}^{2}}+6\alpha t-6)} \right) \\ & =\frac{{{\alpha }^{2}}{{t}^{2}}-4\alpha t+6} \\ & =\frac{\alpha \beta }{{t}^{2}}-\frac{2(\alpha +\beta )}t+\frac{2{{(\alpha +\beta )}^{2}}-2\alpha \beta } \\ & =\frac{1}{{t}^{2}}-\frac{2{{a}_{1}}}{{{a}_{0}}^{2}}t+\frac{2{{a}_{1}}^{2}-2{{a}_{0}}{{a}_{2}}}{{{a}_{0}}^{3}} \\ \end{align}$$ (Eq 5.23a)
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 * }

We can see that Eq 5.23a still has the same form with Eq 5.23 and that in the slide Mtg20. Thus the particular solution is verified.

(2.8) Solve L2-ODE-CC with $$\displaystyle f(t)=\exp (-{{t}^{2}})$$
If $$\displaystyle f(t)=\exp (-{{t}^{2}})$$, then Eqs 5.22 and Eqs 5.22a we derive

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$$ \displaystyle \begin{align} & y_{H}^{1}=\left\{ \begin{align} & {{e}^{-\alpha t}}....................................................................................(\alpha \ne \beta ) \\ & t{{e}^{-\alpha t}}...................................................................................(\alpha =\beta ) \\ \end{align} \right. \\ & y_{H}^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\left\{ \begin{align} & \frac{\sqrt{\pi }\left[ {{e}^{\frac{4}-\alpha t}}erf(\frac{\alpha }{2}-t)-{{e}^{\frac{4}-\beta t}}erf(\frac{\beta }{2}-t) \right]}{2{{a}_{2}}(\alpha -\beta )}...........................(\alpha \ne \beta ) \\ & -\frac{1}{2{{a}_{2}}}\sqrt{\pi }{{e}^{\frac{4}-\alpha t}}\left[ (t-\frac{\alpha }{2})erf(\frac{\alpha }{2}-t)-\frac{1}{\sqrt{\pi }}{{e}^{-\frac{4}}} \right].....................(\alpha =\beta ) \\ \end{align} \right. \\ \end{align}$$ (Eq 5.23)
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 * }

It has been mentioned that governing equation of $$\alpha $$ is indentical to Eq 5.19, and it has also been mentioned that the sequence of $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ is arbitrary.

Case 1: $$\displaystyle (\gamma +1)(\gamma -2)=0$$
From the characteristic equation we can get information of $$\displaystyle \alpha $$ and $$\displaystyle \beta $$, which are the ratio of $$\displaystyle {{a}_{0}}$$ and $$\displaystyle {{a}_{1}}$$ with respect to $$\displaystyle {{a}_{2}}$$. However we can't derive information of $$\displaystyle {{a}_{2}}$$. In fact $$\displaystyle {{a}_{2}}$$ has nothing to do with characteristic equation since in the homogeneous case $$\displaystyle {{a}_{2}}$$ can always be the divisor of both sides(the R.H.S is zero so $$\displaystyle {{a}_{2}}$$ disappeared only left some ratios) and has no effect on the results if $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are fixed. But in the non-homogeneous case $$\displaystyle {{a}_{2}}$$ can be seen as part of the R.H.S(non-zero) and do have influence on the final result. As we can't get enough information to determine the value of $$\displaystyle {{a}_{2}}$$, we take it as 1:

 * $$\displaystyle {{a}_{2}}=1$$

From the characteristic equation we have the following L2-ODE-CC by refering Eq 5.13 and Eq 5.1:

 * $$\displaystyle {y}''-{y}'-2y={{e}^{-{{t}^{2}}}}$$

then

 * $$\displaystyle \begin{align}

& \left\{ \begin{align} & \alpha +\beta =\frac=-1 \\ & \alpha \beta =\frac=-2 \\ \end{align} \right. \\ & \Rightarrow \left\{ \begin{align} & \alpha =1 \\ & \beta =-2 \\ \end{align} \right. \\ \end{align}$$

Substitute these into Eq 5.23 we have the solution as, 'Bold text'
 * {| style="width:100%" border="0"

$$ \displaystyle y={{C}_{1}}{{e}^{-t}}+{{C}_{2}}{{e}^{2t}}+\frac{\sqrt{\pi }}{6}\left[ {{e}^{0.25-t}}erf(0.5-t)-{{e}^{1+2t}}erf(-1-t) \right]$$ (Eq 5.24)
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 *

 * }

where $$\displaystyle {{C}_{1}}$$ and $$\displaystyle {{C}_{2}}$$ are some constants.

Case 2: $$\displaystyle {{(\gamma -4)}^{2}}=0$$
From the characteristic equation we have the following L2-ODE-CC by refering Eq 5.13 and Eq 5.1:

 * $$\displaystyle {y}''-8{y}'+16y={{e}^{-{{t}^{2}}}}$$

then,

 * $$:\displaystyle \begin{align}

& \left\{ \begin{align} & \alpha +\beta =\frac=-8 \\ & \alpha \beta =\frac=16 \\ \end{align} \right. \\ & \Rightarrow \alpha =\beta =-4 \\ \end{align}$$

Substitute this into Eq 5.23 we have,

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

$$ \displaystyle y={{C}_{1}}t{{e}^{4t}}+{{C}_{2}}{{e}^{4t}}-\frac{\sqrt{\pi }}{2}{{e}^{4+4t}}\left[ (t+2)erf(2+t)-\frac{1}{\sqrt{\pi }}{{e}^{-{{(2+t)}^{2}}}} \right]$$ (Eq 5.25)
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 *

 * }

where $$\displaystyle {{C}_{1}}$$ and $$\displaystyle {{C}_{2}}$$ are some constants.

= Problem 5.8 L2_ODE_VC (Homogeneous) by using Reverse Engineering =

From the lecture slide Mtg 31-4

Given
Trial Solution: $$\displaystyle y=\frac$$ and Characteristic Equation: $$\displaystyle {{r}^{2}}+3=0$$

Find
Find L2_ODE_VC according to a trial solution and the characteristic equation given above

Solution
Wolfram Alpha used to find the derivatives of the trial solution $$\displaystyle y$$

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

$$ \displaystyle \left[ y=\frac \right].{{a}_{0}}(x)$$ (Eq 8.1)
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 *

 * }

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$$ \displaystyle \left[ \frac{d}{dx}\left( \frac \right)=\left( r-\frac{2}{x} \right)\left( \frac \right) \right].{{a}_{1}}(x)$$ (Eq 8.2)
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 * }

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$$ \displaystyle \left[ \frac{d{{x}^{2}}}\left( \frac \right)=\left( {{r}^{2}}-\frac{4r}{x}+\frac{6} \right)\left( \frac \right) \right].{{a}_{2}}(x)$$ (Eq 8.3)
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 *

 * }

Adding Eq8.1, Eq8.2 and Eq8.3 each other with taking common parenthesis of $$\displaystyle y$$ will be equal to $$\displaystyle y$$ multiply by characteristic equation.

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

$$ \displaystyle \frac\left[ {{a}_{2}}\left( {{r}^{2}}-\frac{4r}{x}+\frac{6} \right)+{{a}_{1}}\left( r-\frac{2}{x} \right)+{{a}_{0}} \right]=\frac\left( {{r}^{2}}+3 \right)=0$$ (Eq 8.4)
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 *

 * }

First expanding the inside of brackets and then rearranging and taking common parentheses of $$\displaystyle {{r}^{2}}$$, $$\displaystyle r$$ and 1, it will be

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

$$ \displaystyle {{r}^{2}}({{a}_{2}})+r\left( -\frac{4}{x}{{a}_{2}}+{{a}_{1}} \right)+1\left( \frac{6}{{a}_{2}}-\frac{2}{x}{{a}_{1}}+{{a}_{0}} \right)={{r}^{2}}+0.r+3$$ (Eq 8.5)
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 * style="width:95%" |
 *

 * }

From the equality of both sides, we find the values of coefficients $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{0}}$$.

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

$$ \displaystyle \begin{align} & {{a}_{2}}=1 \\ & {{a}_{1}}=\frac{4}{x} \\ & {{a}_{0}}=3+\frac{2} \\ \end{align}$$
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 *

 * }

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$$ \displaystyle {{r}^{2}}+3=0\Rightarrow {{r}_{1,2}}=\mp i\sqrt{3}$$ (Eq 8.6)
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 *

 * }

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

$$ \displaystyle {{a}_{2}}(x){y}''+{{a}_{1}}(x){y}'+{{a}_{0}}(x)y=0$$ (Eq 8.7)
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 *

 * }

is the general form of a second-order ordinary differential equation with coefficients.

If we substitute $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{0}}$$ into Eq8.7, we obtain

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

$$ \displaystyle {y}''+\frac{4}{x}{y}'+(3+\frac{2})y=0$$ (Eq 8.8)
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 *

 * }

Multiply all equation with $$\displaystyle {{x}^{2}}$$ to eliminate denominators

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$$ \displaystyle {{x}^{2}}{y}''+4x{y}'+\left( 3{{x}^{2}}+2 \right)y=0$$ (Eq 8.9)
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 *

 * }

= References =

title: User:Egm6321.f10.team2.reserve/HW5


text:
User:Jtneill/Research/Mood and mental health impacts of online social networking

See: Psychology of online social networking

 * Fields of Research (FOR): 1701 Psychology
 * Socio-Economic Objectives (SEO): code: 9299 Other Health

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


text:
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User talk:Carsrac

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title: User talk:Wmcleod


text:
User:Arided/Learner Profile

=Introduction=

I'm interested in what goes on at the nexus of "learning" and "peer production". In other words, in the interface between participatory social media and interactive, adaptive, "knowledge media". I'm currently looking at the relationship between problem solving and knowledge. My optimistic hope is that my work can be evaluated favorably for depth of insight and originality -- as part of a rich collaborative landscape.

=General background and interests=

 * mathematics (BA, 2002)
 * anthropology of mathematics
 * the economics of common resources
 * practices of peer production
 * artificial intelligence
 * computer-mediated or computer-assisted learning
 * philosophy of knowledge

=Current affiliations and activities=

 * 2010-ongoing: Ph. D. Student @ The Knowledge Media Institute, UK. (Expected completion date: Dec. 2012.) See my research outline.  Thesis title: Peer supported problem solving and mathematical knowledge.
 * 2006-ongoing: Board member at PlanetMath.org
 * 2010-ongoing: Contributor to the Planetary project, which is an effort to rebuild PlanetMath's software using contemporary web technologies
 * 2011-ongoing: working on paragogy.net, a site about peer learning, in collaboration with User:Charles Jeffrey Danoff.

=Zoomed-out future timeline=

 * Ph. D. thesis (2012)
 * "The Penmaster's Apprentice" novel (2013)
 * unknown future

=Zoomed-in next steps=

 * Work on my writing projects (listed above)
 * Work on my programming projects, involving others where this seems reasonable/possible/feasible. The projects are:
 * Planetary
 * Etherpad
 * Concept Forest
 * Arxana

=The nature of my commitments=

Right now I have to get a new "probation report" done by Dec. 15th. Current draft is at http://metameso.org/~joe/docs/probation-report-new.pdf

title: User:Arided/Learner Profile


text:
Numerical Analysis/Gaussian Quadrature

For the first example, I just want an example to show that the solution is exact for polynomials of degree $$2n-1$$, using the $$n$$th degree Legendre polynomial. I'm going to approximate $$\int_{-1}^1 x^3 dx$$.

$$\begin{align} \int_{-1}^1 x^3 dx & = \left(\frac{1}{\sqrt{3}}\right)^3 + \left(\frac{-1}{\sqrt{3}}\right)^3 \\ \frac{1}{4} 1^4 - \frac{1}{4} (-1)^4 & = \left(\frac{1}{3\sqrt{3}}\right) - \left(\frac{1}{3\sqrt{3}}\right) \\ \frac{1}{4} - \frac{1}{4} &= 0 \end{align} $$

This problem accurately illustrates the method for solving problems using the Gaussian Quadrature algorithm. Note that the zeros of the Legendre polynomials of degree $$n$$ are $$\frac{1}{\sqrt{3}}$$ and $$-\frac{1}{\sqrt{3}}$$.

We can see from the previous example that this method works quite well if we are integrating from $$-1$$ to $$1$$, but in application, we rarely want to integrate over such a simple region. So in our next example, we will show that this technique is also effective if we change the limits of integration, as seen on the Wikipedia page, then we will solve the example $$\int_4^{14} x^3 dx$$

$$ \begin{align} \int_4^{14} x^3 dx &= \frac{14-4}{2} \sum_{i=1}^2 f\left(\frac{14-4}{2} x_i + \frac{14+4}{2}\right) \\ \frac{1}{4} 14^4 - \frac{1}{4} 4^4 &= 5 \left(\left(5\frac{1}{\sqrt{3}} + 9\right)^3 +(\left(5\frac{-1}{\sqrt{3}} + 9\right)^3\right) \\ 9540 &= 5 \left(\frac{125}{3\sqrt{3}} + \frac{875}{3} + \frac{1215}{\sqrt{3}} + 729 - \frac{125}{3\sqrt{3}} + \frac{875}{3} - \frac{1215}{\sqrt{3}} + 729\right) \\ 9540 &= 5\left(1908\right)\\ 9540 &= 9540 \end{align} $$

We will next show how to solve a problem that isn't a simple polynomial. We will approximate $$\int_{-1}^1 e^x dx$$ using a two point Gaussian approximation, and discuss the error analysis.

$$ \begin{align} \int_{-1}^1 e^x dx &\approx e^{\frac{1}{\sqrt{3}}} + e^{\frac{-1}{\sqrt{3}}} \\ e - \frac{1}{e} &\approx e^{\frac{1}{\sqrt{3}}} + \frac{1}{e^{\frac{1}{\sqrt{3}}}} \end{align} $$

We will next analyze this error, looking at the actual error followed by finding the error bound. We denote the approximation by $$f_{approx}$$ and the exact solution by $$f_{exact}$$.

$$ \begin{align} \left|f_{exact} - f_{approx}\right| &= \left|e - \frac{1}{e} - e^{\frac{1}{\sqrt{3}}} - \frac{1}{e^{\frac{1}{\sqrt{3}}}}\right| \\ &= 0.00770629937787233598077 \end{align} $$

The theoretical error bound when using the Legendre polynomial method is

$$ \begin{align} \left|f_{exact}-f_{approx}\right| &\le \frac{(b-a)^{2n+1}(n!)^4}{(2n+1)\left[(2n)!\right]^3} f^{(2n)}\left(\xi\right), a<\xi<b \\ &\le \frac{2^5 2^4}{5*4!^3} e \\ &= \frac{2^9}{5*24^3} e \\ &= 0.02013542095154848322489 \end{align} $$

In our example, the actual error was well within the error bound. We also see that with only two calculations, this is a very good algorithm for approximating integrals quickly with relatively good accuracy.

Exercises
1. Show that Gaussian Quadrature can solve exactly general cubic polynomials.

a) Set up the integral: Solution: $$ \int_{-1}^1 f(x) dx = \int_{-1}^1 \left(ax^3 +bx^2 + cx + d\right) dx $$

b) Evaluate the integral: Solution: $$ \begin{align} \int_{-1}^1 f(x) dx &= \frac{a}{4} 1^4 - \frac{a}{4} (-1)^4 + \frac{b}{3} 1^3 - \frac{b}{3} (-1)^3 + \frac{c}{2} 1^2 - \frac{c}{2} (-1)^2 + d - (-d)\\ \int_{-1}^1 f(x) dx &= \frac{2}{3} b + 2d \end{align} $$

c) Evaluate the approximation: Solution: $$ \begin{align} \sum_{i=1}^2 f\left(x_i\right) &= a\left(\frac{1}{\sqrt{3}}\right)^3 + a\left(\frac{-1}{\sqrt{3}}\right)^3 + b\left(\frac{1}{\sqrt{3}}\right)^2 + b\left(\frac{-1}{\sqrt{3}}\right)^2 + c \left(\frac{1}{\sqrt{3}}\right) + c\left(\frac{-1}{\sqrt{3}}\right) + 2d\\ \sum_{i=1}^2 f\left(x_i\right) &= \frac{2}{3}b + 2d \end{align} $$

d) Do parts b and c correspond? Solution: Yes

2. Approximate $$\int_{-1}^1 sin^2(x) cos(x) dx$$ using Gaussian Quadrature

a) Evaluate the problem symbolically: Solution: $$ \begin{align} \int_{-1}^1 sin^2(x)cos(x) dx &= \int_{-1}^1 cos x dx + \int_{-1}^1 cos^3 x dx \\ &= \frac{2 sin^3(1)}{3} \\ &\approx 0.397215491060637 \end{align} $$

b) Evaluate the approximation: Solution: $$ \begin{align} \sum_{i=1}^2 f\left(x_i\right) &= sin^2\left(\frac{1}{\sqrt{3}}\right) cos\left(\frac{1}{\sqrt{3}}\right) + sin^2\left(\frac{-1}{\sqrt{3}}\right) cos\left(\frac{-1}{\sqrt{3}}\right) \\ &\approx 0.49923418313484188 \end{align} $$

c) Find the actual error Solution: $$ \left|f_{exact} - f_{approx}\right| \approx 0.102018692074204 $$

d) Find the error bound Solution: $$ \begin{align} \left|f_{exact}-f_{approx}\right| &\le \frac{(b-a)^{2n+1}(n!)^4}{(2n+1)\left[(2n)!\right]^3} f^{(2n)}\left(\xi\right), a<\xi<b \\ &\le \frac{2^5 2^4}{5*4!^3} f^4(\xi) \\ &\le \frac{2^9}{5*24^3} * 20.1824 \\ &\le 0.14949925 \end{align} $$

d) Is the actual error less than the error bound? Solution: Yes

3. If $$ x_1, x_2,...., x_n$$ are the roots of the $$n$$th Legendre Polynomial $$P_n(x)$$ and that for each $$i=1,2,....,n$$, the numbers $$c_i$$ are defined by

$$ c_i = \int_{-1}^1 \prod_{j=1,j\neq i}^n \frac{x-x_j}{x_i-x_j} dx $$

Prove that if $$P(x)$$ is any polynomial of degree less than $$2n$$, then

$$ \int_{-1}^1 P(x) dx = \sum_{i=1}^n c_i P(x_i). $$

Proof:

The set that is relevant to this proof is the set of Legendre Polynomials, a collection orthogonal polynomials $${P_0(x), P_1(x),...,P_n(x),...,}$$ with properties:
 * 1) For each  $$ n, P_n(x) $$ is a monic polynomial of degree $$ n $$.
 * 2) $$\int_{-1}^1 {P(x)}\times{P_n(x)}dx = 0$$ whenever $$P(x)$$ is a polynomial of degree less that $$n$$.

First we use polynomial division, to get $$ Q(x) = \frac{P(x)}{P_n(x)}$$ with a remainder term $$R(x)$$, so that $$P(x) = Q(x)\times P_n(x) + R(x)$$ and the two polynomials $$Q(x)$$ and $$R(x)$$ are of degree less than $$n$$ and $$P_n(x)$$ is of degree $$n$$. Then using this fact we can rewrite the original integral into the form

 * $$\int_{-1}^1 P(x)dx = \int_{-1}^1 {P_n(x)}\times{Q(x)}dx + \int_{-1}^1 R(x)dx$$.

Since the degree of $$P(x)$$ is less than twice the degree of the Legendre polynomial, then the degree of $$Q(x)$$ is less than the degree of $$P_n(x)$$. So (by Legendre property 2 started above), $$\int_{-1}^1 Q(x)\times P_n(x) dx = 0$$. Thus we see that $$\int_{-1}^1 P(x) dx = \int_{-1}^1 R(x) dx$$. We do not know what $$R(x)$$ is, but we know it has degree less than $$n$$ so it equals the polynomial that interpolates it at the values $$ x_1, x_2,...., x_n$$. We will use the Lagrange polynomial form. Since $$x_i$$ is a roots of $$P_n$$ for each $$i= 1, 2, ..., n$$, we have

 * $$P(x_i) = Q(x_i)\times P_n(x_i) + R(x_i)= Q(x_i)\times 0 + R(x_i)= R(x_i)$$

Since $$R(x)$$ is a polynomial of degree less than $$n$$, then Lagrange polynomial for $$R(x)$$ is

 * $$R(x) = \sum_{i=1}^n \left(P(x_i) \prod_{j=1,j\neq i}^n \frac{x-x_j}{x_i-x_j}\right)$$.

We next integrate this polynomial by

 * $$\int_{-1}^1 P(x) dx = \int_{-1}^1 R(x) dx = \int_{-1}^1 \left(\sum_{i=1}^n \left(P(x_i) \prod_{j=1,j\neq i}^n \frac{x-x_j}{x_i-x_j}\right)\right) dx$$.

Since $$P(x_i)$$ is a constant, we will pull that outside the integral, giving us

 * $$\int_{-1}^1 P(x) dx = \left(\sum_{i=1}^n P(x_i)\left(\int_{-1}^1 \prod_{j=1,j\neq i}^n \frac{x-x_j}{x_i-x_j} dx\right)\right)$$.

We then note that

 * $$c_i = \int_{-1}^1 \prod_{j=1,j\neq i}^n \frac{x-x_j}{x_i-x_j} dx$$,

so we will substitute $$c_i$$ into our integration, giving us

 * $$ \int_{-1}^1 P(x) dx = \sum_{i=1}^n c_i P(x_i).$$

This completes the proof.

Gaussian Quadrature Example
I realized that there was insufficient information after the derived and solved sample on Gaussian Quadrature thus i took the pain to edit this wikiversity page by adding a solved example to the information already on there and below is what i factored in.

Find the constants C_0, C_1, and x_1 so that the quadrature formula

\int_{0}^1 f (x)\,dx = C_0 f(0)+ C_1 f(x_1). $$

has the  highest  possible  degree  of  precision.

Solution

Since there  are  three  unknowns,  C_0,  C_1  and  x_1,  we  will  expect  the  formula  to  be  exact  for

f (x) = 1,  x,  and \,     \ x^2

$$ Thus

f (x)= 1,\ \int_{0}^1 f (x)\,dx= 1 = C_0 + C_1\  \qquad                  \ Equation 1,   \qquad

f (x)= x,\ \int_{0}^1 f (x)\,dx= \frac{1}{2} = C_1x_1\  \qquad             \ Equation 2. \qquad

f (x)= x^2,\ \int_{0}^1 f (x)\,dx= \frac{1}{3} = C_1x_1^2. $$

Equation 2 and 3 will yield.

\frac{c_1x_1}{c_1x_1^2} = x_1 = \frac{2}{3}. \qquad  C_1=\frac{3}{4}. \qquad C_0= \frac{1}{4}. $$ Hence

\int_{0}^1 f (x)\,dx = \frac{1}{4}f(0)+ \frac{3}{4}f(\frac{2}{3}) $$ Now,

f(x)=x^3.\qquad \int_{0}^1 x^3\,dx = \frac{1}{4} $$ And

\frac{1}{4}(0)^3 + \frac{3}{4}(\frac{2}{3})^3=\frac{2}{9}. $$ Thus the degree of the precision is 2

Example Quiz
{Why do we use the zeros of the Legendre polynomials in this method? - By convention for consistency + Because Legendre polynomials of degree $$n$$ are orthogonal to any other polynomial with degree less than $$n$$ - Because they make $$f(x) = 0$$ when approximating $$f(x)$$ - None of these
 * type=""}

{The integral of a 6th degree polynomial can be approximated exactly using the 3rd degree Legendre polynomial, True or False - True + False
 * type=""}

{Calculate the error bound for a general 3rd degree polynomial using a 2nd order Legendre polynomial - 10.56 + 0 - 4.13 - 50.25
 * type=""}

{Calculate the error bound for $$\int_4^8 x^6 + 5x^5 + 27x^4 + 108x + 276 dx$$ using a 3rd degree Legendre polynomial + 5.85 - 0.13 - 28.83 - 12.44
 * type=""}

{What are the failure conditions for this algorithm? - The $$n$$th derivative of $$f(x)$$ must be nonzero + $$f(x)$$ is discontinuous over the range of integration - $$f(x)$$ is constant over the range - The method never fails, it just can suffer from poor accuracy
 * type=""}

title: Numerical Analysis/Gaussian Quadrature


text:
User talk:Krunchlolee

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Talk:Motivation and emotion/Book/2010/Sleep and emotion

Hey there!
Nice to see you are making progress on your TB chapter! I see you have quoted my boy Ekman in your theory. Might I suggest you make his name into a link to my TB page?

Cheers!--MichelleK 00:43, 4 November 2010 (UTC)

Love your work! MichelleK 22:12, 7 November 2010 (UTC)

-- Jtneill - Talk - c 13:34, 13 December 2010 (UTC)


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