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User talk:Wiki13

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title: User talk:Wiki13


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An algorithm for global optimization inspired by collective animal behavior

By Erik Cuevas, Mauricio González, Daniel Zaldivar and Marco Pérez-Cisneros

Introduction
Global Optimization (GO) has yielded remarkable applications to many areas of science, engineering, economics and others through mathematical modelling1. In general, the goal is to find a global optimum of an objective function which has been defined over a given search space. Global optimization algorithms are usually broadly divided into deterministic and metaheuristic2. Since deterministic methods only provide a theoretical guarantee of locating a local minimum of the objective function, they often face great difficulties in solving global optimization problems3. On the other hand, metaheuristic methods are usually faster in locating a global optimum4. They virtuously adapt to black-box formulations and extremely ill-behaved functions, whereas deterministic methods usually require some theoretical assumptions about the problem formulation and its analytical properties (such as Lipschitz continuity)5.

Several metaheuristic algorithms have been developed by a combination of rules and randomness mimicking several phenomena.

Many studies have been inspired by animal behavior phenomena for developing optimization techniques. For instance, the Particle swarm optimization (PSO) algorithm which models the social behavior of bird flocking or fish schooling6. PSO consists of a swarm of particles which move towards best positions, seen so far, within a searchable space of possible solutions. Another behavior-inspired approach is the Ant Colony Optimization (ACO) algorithm proposed by Dorigo et al.7, which simulates the behavior of real ant colonies. Main features of the ACO algorithm are the distributed computation, the positive feedback and the constructive greedy search. Recently, a new metaheuristic approach which is based on the animal behavior while hunting has been proposed in8. Such algorithm considers hunters as search positions and preys as potential solutions.

Just recently, the concept of individual-organization9,10 has been widely referenced to understand collective behavior of animals. The central principle of individual-organization is that simple repeating interactions between individuals can produce complex behavioral patterns at group level9,11,12. Such inspiration comes from behavioral patterns previously seen in several animal groups. Examples include ant pheromone trail networks, aggregation of cockroaches and the migration of fish schools, all of which can be accurately described in terms of individuals following simple sets of rules13. Some examples of these rules12,14 are keeping the current position (or location) for best individuals, local attraction or repulsion, random movements and competition for the space within of a determined distance.

On the other hand, new studies15-17 have also shown the existence of collective memory in animal groups. The presence of such memory establishes that the previous history of the group structure influences the collective behavior exhibited in future stages. According to such principle, it is possible to model complex collective behaviors by using simple individual rules and configuring a general memory.

In this paper, a metaheuristic algorithm for global optimization called the Collective Animal Behavior (CAB) is introduced. In this algorithm, the searcher agents emulate a group of animals that interact to each other based on simple behavioral rules which are modeled as mathematical operators. Such operations are applied to each agent considering that the complete group has a memory storing their own best positions seen so far, by using a competition principle. The proposed approach has been compared to other well-known optimization methods. The results confirm a high performance of the proposed method for solving various benchmark functions.

Biologic fundamentals
The remarkable collective behavior of organisms such as swarming ants, schooling fish and flocking birds has long captivated the attention of naturalists and scientists. Despite a long history of scientific research, the relationship between individuals and group-level properties has just recently begun to be deciphered18.

Grouping individuals often have to make rapid decisions about where to move or what behavior to perform in uncertain and dangerous environments. However, each individual typically has only a relatively local sensing ability19. Groups are, therefore, often composed of individuals that differ with respect to their informational status and individuals are usually not aware of the informational state of others20, such as whether they are knowledgeable about a pertinent resource or about a threat.

Animal groups are based on a hierarchic structure21 which considers different individuals according to a fitness principle called Dominance22 which is the domain of some individuals within a group that occurs when competition for resources leads to confrontation. Several studies23,24 have found that such animal behavior lead to more stable groups with better cohesion properties among individuals.

Recent studies have begun to elucidate how repeated interactions among grouping animals scale to collective behavior. They have remarkably revealed that collective decision-making mechanisms across a wide range of animal group types, from insects to birds (and even among humans in certain circumstances) seem to share similar functional characteristics9,13,25. Furthermore, at a certain level of description, collective decision-making by organisms shares essential common features such as a general memory. Although some differences may arise, there are good reasons to increase communication between researchers working in collective animal behavior and those involved in cognitive science12.

Despite the variety of behaviors and motions of animal groups, it is possible that many of the different collective behavioral patterns are generated by simple rules followed by individual group members. Some authors have developed different models, one of them, known as the self-propelled particle (SPP) model, attempts to capture the collective behavior of animal groups in terms of interactions between group members which follow a diffusion process26-29.

The dynamical spatial structure of an animal group can be explained in terms of its history24. Despite such a fact, the majority of studies have failed in considering the existence of memory in behavioral models. However, recent research15,30 have also shown the existence of collective memory in animal groups. The presence of such memory establishes that the previous history of the group structure influences the collective behavior which is exhibited in future stages. Such memory can contain the location of special group members (the dominant individuals) or the averaged movements produced by the group.

According to these new developments, it is possible to model complex collective behaviors by using simple individual rules and setting a general memory. In this work, the behavioral model of animal groups inspires the definition of novel evolutionary operators which outline the CAB algorithm. A memory is incorporated to store best animal positions (best solutions) considering a competition-dominance mechanism.

Collective Animal Behavior Algorithm (CAB)
The CAB algorithm assumes the existence of a set of operations that resembles the interaction rules that model the collective animal behavior. In the approach, each solution within the search space represents an animal position. The “fitness value” refers to the animal dominance with respect to the group. The complete process mimics the collective animal behavior.

The approach in this paper implements a memory for storing best solutions (animal positions) mimicking the aforementioned biologic process. Such memory is divided into two different elements, one for maintaining the best locations at each generation (Mh) and the other for storing the best historical positions during the complete evolutionary process (Mg).

Description of the CAB algorithm
Following other metaheuristic approaches, the CAB algorithm is an iterative process that starts by initializing the population randomly (generated random solutions or animal positions). Then, the following four operations are applied until a termination criterion is met (i.e. the iteration number NI):

 * 1) Keep the position of the best individuals.
 * 2) Move from or to nearby neighbors (local attraction and repulsion).
 * 3) Move randomly.
 * 4) Compete for the space within a determined distance (update the memory).

Initializing the population
The algorithm begins by initializing a set A of Np animal positions (A={a1,a2,...aN p }). Each animal position ai is a D-dimensional vector containing parameter values to be optimized. Such values are randomly and uniformly distributed between the pre-specified lower initial parameter bound ajlow and the upper initial parameter bound ajhigh.

ai,j=ajlow+rand(0,1)•(ajhigh-ajlow);                  (1)

j=1,2,...,D;   i=1,2,...,Np.

with j and i being the parameter and individual indexes respectively. Hence,aj,i is the jth parameter of the ith individual.

All the initial positions A are sorted according to the fitness function (dominance) to form a new individual set X={x1,x2,...,xN pp }, so that we can choose the best B positions and store them in the memory Mg and Mh. The fact that both memories share the same information is only allowed at this initial stage.

Keep the position of the best individuals
Analogous to the biological metaphor, this behavioral rule, typical from animal groups, is implemented as an evolutionary operation in our approach. In this operation, the first B elements ({a1,a2,...,aB}), of the new animal position set A, are generated. Such positions are computed by the values contained inside the historical memory Mh, considering a slight random perturbation around them. This operation can be modeled as follows:

al=mhl+v;          (2)

where lε{1,2,...,B} while mhl represents the l-element of the historical memory Mh. v is a random vector with a small enough length.

Move from or to nearby neighbors
From the biological inspiration, animals experiment a random local attraction or repulsion according to an internal motivation. Therefore, we have implemented new evolutionary operators that mimic such biological pattern. For this operation, a uniform random number rm is generated within the range [0,1]. If rm is less than a threshold H, a determined individual position is moved (attracted or repelled) considering the nearest best historical position within the group (i.e. the nearest position in Mh); otherwise, it goes to the nearest best location within the group for the current generation (i.e. the nearest position in Mg). Therefore such operation can be modeled as follows:

ai=xi±r•(mhnearest-xi) with probability H                 (3) ai=xi±r•(mgnearest-xi) with probability 1-H

where iε{B+1,B+2,...,Np}, mhnearest and mgnearest represent the nearest elements of Mh and Mg to xi, while r is a random number.

Move randomly
Following the biological model, under some probability P, one animal randomly changes its position. Such behavioral rule is implemented considering the next expression:

ai=r with probability P             (4) ai=xi with probability (1-P)

being iε{B+1,B+2,...,Np} and r a random vector defined in the search space. This operator is similar to re-initialize the particle in a random position, as it is done by Eq. (1).

Compete for the space within of a determined distance (update the memory)
Once the operations to keep the position of the best individuals, such as moving from or to nearby neighbors and moving randomly, have all been applied to the all Np animal positions, generating Np new positions, it is necessary to update the memory Mh.

In order to update de memory Mh, the concept of dominance is used. Animals that interact within a group maintain a minimum distance among them. Such distance ρ depends on how aggressive the animal behaves 22,30. Hence, when two animals confront each other inside such distance, the most dominant individual prevails meanwhile other withdraw.

In the proposed algorithm, the historical memory Mh is updated considering the following procedure:

 * 1) The elements of Mh and Mg are merged into MU(MU=MhυMh).
 * 2) Each element mUi of the memory MU is compared pair-wise to remaining memory elements ({mU1,mU2,...,mU2B-1}). If the distance between both elements is less than ρ, the element getting a better performance in the fitness function prevails meanwhile the other is removed.
 * 3) From the resulting elements of MU (from step 2), it is selected the B best value to build the new Mh.

Unsuitable values of ρ yield a lower convergence rate, a longer computational time, a larger function evaluation number, the convergence to a local maximum or to an unreliable solution. The ρ value is computed considering the following equation: ρ=  Π   undefinedD(ajhigh-ajlow)/(10•D)             (5)

where ajhigh and ajlow represent the pre-specified lower and upper bound of the j-parameter respectively, in an D-dimensional space.

Computational procedure
The computational procedure for the proposed algorithm can be summarized as follows:

Step 1:	Set the parameters Np, B, H,P and NI.

Step 2:	Generate randomly the position set A={a1,a2,...,aN p } using Eq.1

Step 3:	Sort A according to the objective function (dominance) to build X={x1,x2,...,xN p }.

Step 4:	Choose the first B positions of X and store them into the memory Mg.

Step 5:	Update Mh according to Section 3.1.5. (during the first iteration:Mh=Mg).

Step 6:	Generate the first B positions of the new solution set A({a1,a2,...,aB}. Such positions correspond to the elements of Mh making a slight random perturbation around them.

ahl+v; being v a random vector of a small enough length.

Step 7:	Generate the rest of the A elements using the attraction, repulsion and random movements. for i=B+1:Np if (r1<P) then attraction and repulsion movement {if(r2<H) then ai=xi-r•(mhnearest-xi)                         else if                         ai=xi-r•(mgnearest-xi)                         } else if                 random movement {                      ai=r }            end for where r1,r2,rε rand(0,1)

Step 8:	If NI is completed, the process is finished; otherwise go back to step 3. The best value in Mh represents the global solution for the optimization problem.


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text:
University of Florida/Egm4313/s12.team11.imponenti/R4.4

Part 3
Solved by Luca Imponenti

Find $$y_n(x)\!$$, for $$n=4,7,11\!$$ such that:
 * $$y_n''-3y_n'+2y_n=r_n(x)\!$$

for $$x\!$$ in $$[0.9,3]\!$$ with the initial conditions found.

Plot $$y_n(x)\!$$ for $$n=4,7,11\!$$ for $$x\!$$ in $$[0.9,3]\!$$.

Homogeneous Solution
The homogeneous case is shown below:
 * $$y''_h-3y'_h+2y_h=0\!$$

This equation has the following roots:

$$\lambda_1=1, \lambda_2=2\!$$

Which gives yields the homogeneous solution
 * $$y_h=c_1e^x+c_2e^{2x}\!$$

General Solution, n=4
Using the taylor series approximation from earlier with $$n=4\!$$ we have

$$r_4(x)=log(2)+\frac{(x-1)}{2ln(10)}-\frac{(x-1)^2}{8ln(10)}+\frac{(x-1)^3}{24ln(10)}-\frac{(x-1)^4}{64ln(10)}\!$$

We know the particular solution, $$y_{p4}(x)\!$$, ve will have this form:

$$y_{p4}(x)=a_4(x-1)^4+a_3(x-1)^3+a_2(x-1)^2+a_1(x-1)+a_0\!$$

taking the derivatives of this solution

$$y'_{p4}(x)=\frac{d}{dx}y_{p4}(x)=4a_4(x-1)^3+3a_3(x-1)^2+2a_2(x-1)+a_1\!$$

and

$$y''_{p4}(x)=\frac{d}{dx}y'_{p4}(x)=12a_4(x-1)^3+6a_3(x-1)^2+2a_2\!$$

Plugging the above equations into the original ODE yields the following matrix equation:

 * $$ \begin{bmatrix}

2 & 0 & 0 & 0 & 0\\ -12 & 2 & 0 & 0 & 0\\ 12 & -9 & 2 & 0 & 0\\ 0 & 6 & -6 & 2 & 0\\ 0 & 0 & 2 & -3 & 2\end{bmatrix}* \begin{bmatrix} a_4 \\ a_3\\ a_2\\ a_1\\ a_0\end{bmatrix}= \begin{bmatrix} -\frac{1}{64ln(10)}\\ \frac{1}{24ln(10)}\\ -\frac{1}{8ln(10)}\\ \frac{1}{2ln(10)}\\ log(2)\end{bmatrix}\!$$

The unknown vector $$a\!$$ can be easily solved by forward substitution,the following values were calculated in matlab:

$$a_4=-.0034, a_3=-.0113, a_2=-.0577, a_1=-.1624, a_0=.1624\!$$

So the particular solution $$y_{p4}\!$$ is

$$y_{p4}=0.1624-0.1624*(x-1)-.0577*(x-1)^2-.0113*(x-1)^3-.0034*(x-1)^4\!$$

We can now find the general solution for n=4, $$y_4(x)\!$$.

$$y_4(x)=y_h(x)+y_{p4}(x)\!$$

$$y_4(x)=c_1e^x+c_2e^{2x}+0.1624-0.1624*(x-1)\!$$ $$-.0577*(x-1)^2-.0113*(x-1)^3-.0034*(x-1)^4\!$$

Solving using the initial conditions yields;

$$y_4(x)=.0595e^x-.0076e^{2x}+0.1624-0.1624*(x-1)\!$$ $$-.0577*(x-1)^2-.0113*(x-1)^3-.0034*(x-1)^4\!$$

General Solution, n=7
Using the taylor series approximation from earlier with $$n=7\!$$ we have

$$r_7(x)=log(2)+\frac{(x-1)}{2ln(10)}-\frac{(x-1)^2}{8ln(10)}+\frac{(x-1)^3}{24ln(10)}-\frac{(x-1)^4}{64ln(10)}+\frac{(x-1)^5}{160ln(10)}-\frac{(x-1)^6}{384ln(10)} +\frac{(x-1)^7}{896ln(10)}\!$$

In a similar fashion we construct a matrix equation for n=7:

 * $$ \begin{bmatrix}

2 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ -21 & 2 & 0 & 0 & 0 & 0 & 0 & 0\\ 42 & -18 & 2 & 0 & 0 & 0 & 0 & 0\\ 0 & 30 & -15 & 2 & 0 & 0 & 0 & 0\\ 0& 0 & 20 & -12 & 2 & 0 & 0 & 0\\ 0& 0 & 0 & 12 & -9 & 2 & 0 & 0\\ 0& 0 & 0 & 0 & 6 & -6 & 2 & 0\\ 0& 0 & 0 & 0 & 0 & 2 & -3 & 2\end{bmatrix}* \begin{bmatrix} a_7 \\ a_6 \\ a_5 \\ a_4 \\ a_3\\ a_2\\ a_1\\ a_0\end{bmatrix}= \begin{bmatrix} \frac{1}{896ln(10)}\\ -\frac{1}{384ln(10)}\\ \frac{1}{160ln(10)}\\ -\frac{1}{64ln(10)}\\ \frac{1}{24ln(10)}\\ -\frac{1}{8ln(10)}\\ \frac{1}{2ln(10)}\\ log(2)\end{bmatrix}\!$$

Solving:

$$a_7=.0002, a_6=.0020, a_5=.0141, a_4=.0725, a_3=.3034, a_2=.9029, a_1=1.9072, a_0=2.1084\!$$

So the particular solution $$y_{p7}\!$$ is

$$y_{p7}=2.1084+1.9072*(x-1)+.9029*(x-1)^2+.3034*(x-1)^3+.0725*(x-1)^4+.0141*(x-1)^5+.0020*(x-1)^6+.0002*(x-1)^7\!$$

We can now find the general solution for n=7, $$y_7(x)\!$$.

$$y_7(x)=y_h(x)+y_{p7}(x)\!$$

$$y_7(x)=c_1e^x+c_2e^{2x}+2.1084+1.9072*(x-1)+.9029*(x-1)^2+\!$$ $$.3034*(x-1)^3+.0725*(x-1)^4+.0141*(x-1)^5+.0020*(x-1)^6+.0002*(x-1)^7\!$$

Solving using our initial conditions yields

$$y_7(x)=-.7271e^x+.0233e^{2x}+2.1084+1.9072*(x-1)+\!$$ $$.9029*(x-1)^2+.3034*(x-1)^3+.0725*(x-1)^4+.0141*(x-1)^5+\!$$ $$.0020*(x-1)^6+.0002*(x-1)^7\!$$

General Solution, n=11
Using the taylor series approximation from earlier with $$n=11\!$$ we have

$$r_{11}(x)=log(2)+\frac{(x-1)}{2ln(10)}-\frac{(x-1)^2}{8ln(10)}+\frac{(x-1)^3}{24ln(10)}-\frac{(x-1)^4}{64ln(10)}+\frac{(x-1)^5}{160ln(10)}-\frac{(x-1)^6}{384ln(10)}+\!$$

$$\frac{(x-1)^7}{896ln(10)}-\frac{(x-1)^8}{2048ln(10)}+\frac{(x-1)^9}{4608ln(10)}-\frac{(x-1)^{10}}{10240ln(10)}+\frac{(x-1)^{11}}{22528ln(10)}\!$$

Finally, we write out the matrix equation for n=11:

$$A* \begin{bmatrix} a_{11}\\ a_{10}\\ a_9\\ a_8\\ a_7 \\ a_6 \\ a_5 \\ a_4 \\ a_3\\ a_2\\ a_1\\ a_0\end{bmatrix}= \begin{bmatrix} \frac{1}{22528ln(10)}\\ -\frac{1}{10240ln(10)}\\ \frac{1}{4608ln(10)}\\ -\frac{1}{2048ln(10)}\\ \frac{1}{896ln(10)}\\ -\frac{1}{384ln(10)}\\ \frac{1}{160ln(10)}\\ -\frac{1}{64ln(10)}\\ \frac{1}{24ln(10)}\\ -\frac{1}{8ln(10)}\\ \frac{1}{2ln(10)}\\ log(2)\end{bmatrix}\!$$

Solving the system in matlab:

$$a_{11}=0, a_{10}=.0002, a_9=.0019, a_8=.0181, a_7=.15, a_6=1.0675, a_5=6.4597,\!$$

$$a_4=32.4318, a_3=130.0033, a_2=390.3968, a_1=781.289, a_0=781.6873\!$$

So the particular solution $$y_{p11}\!$$ is

$$y_{p11}=781.6873+781.289*(x-1)+390.3968*(x-1)^2+130.0033*(x-1)^3+32.4318*(x-1)^4+\!$$

$$6.4597*(x-1)^5+1.0675*(x-1)^6+.15*(x-1)^7+.0181*(x-1)^8+.0019*(x-1)^9+.0002*(x-1)^{10}\!$$

We can now find the general solution for n=11, $$y_11(x)\!$$.

$$y_{11}(x)=y_h(x)+y_{p11}(x)\!$$

$$y_{11}(x)=c_1e^x+c_2e^{2x}+781.6873+781.289*(x-1)\!$$ $$+390.3968*(x-1)^2+130.0033*(x-1)^3+32.4318*(x-1)^4\!$$ $$6.4597*(x-1)^5+1.0675*(x-1)^6+.15*(x-1)^7+.0181*(x-1)^8+.0019*(x-1)^9+.0002*(x-1)^{10}\!$$

Solving using our initial conditions yields

$$y_{11}(x)=-287.5907e^x+.05e^{2x}+781.6873+781.289*(x-1)\!$$ $$+390.3968*(x-1)^2+130.0033*(x-1)^3+32.4318*(x-1)^4\!$$ $$6.4597*(x-1)^5+1.0675*(x-1)^6+\!$$ $$.15*(x-1)^7+.0181*(x-1)^8+.0019*(x-1)^9+.0002*(x-1)^{10}\!$$

Plot
$$y_4\!$$ shown in red

$$y_7\!$$ shown in blue

$$y_{11}\!$$ shown in green

Part 4
Solved by Luca Imponenti

Use the matlab command ode45 to integrate numerically $$y''-3y'+2y=r(x)\!$$ with $$r(x)=log(1+x)\!$$

and the initial conditions from Part 3 to obtain the numerical solution for y(x).

Plot y(x) in the same figure as above.

Matlab Solution
The numerical solution calculated using the matlab ode45 command is shown below:

ans = 0.2788   0.2854    0.2923    0.2997    0.3074    0.3229    0.3401    0.3592    0.3804    0.4040    0.4302    0.4595    0.4921    0.5285    0.5691    0.6145    0.6651    0.7218    0.7850    0.8557    0.9346    1.0228    1.1213    1.2313    1.3542    1.4914    1.6445    1.8155    2.0063    2.2193    2.4569    2.7219    3.0175    3.3471    3.7146    4.1243    4.5809    5.0898    5.6568    6.2885    6.9921    7.3442    7.7142    8.1032    8.5119

Plot
Plotting the aboved vector of y-values,along with the results from earlier yields the following graph:

where the answer calculated in matlab is shown in yellow.

Egm4313.s12.team11.imponenti (talk) 08:04, 14 March 2012 (UTC)


text:
User:Egm4313.s12.team8.lucas/R4.2

=R4.2=

Problem Statement
Consider the L2-ODE-CC with sinx as its excitation.

$$\ y''-3y'+2y=r(x)=sinx$$

Initial Conditions:

$$\ y(0)=0$$, $$\ y'(0)=1$$

Part 1) Use the Taylor series for sinx to reproduce the figure on p. 7-24 in lecture notes.

Part 2) Find the overall solution to the L2-ODE-CC above by substituting the Taylor series approximations for sinx at n=3,5, and 9 for the excitation equation. Plot the solutions over the interval $$\ [0,4 \pi]$$.

Part 3) Find the exact overall solution to the L2-ODE-CC above by using the function sinx as the excitation equation. Plot the solution and compare it to the solutions from Part 2.

Part 1
The Taylor series expansion of sinx takes the form:

$$\ sinx=\sum_{j=0}^{k}\frac{{}(-1)^{j}x^{2j+1}}{(2j+1)!}$$

Expanding this summation for j=6 gives:

$$\ sinx=\frac{x}{1!}-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\frac{x^{9}}{9!}-\frac{x^{11}}{11!}+\frac{x^{13}}{13!}$$

$$\ sinx=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\frac{x^{7}}{5040}+\frac{x^{9}}{362880}-\frac{x^{11}}{39916800}+\frac{x^{13}}{6227020800}$$

By seperating and graphing each of the terms, we can see how the sum of the polynomials equal the sin function.

 * The value for n in the plot is equal to 2j+1, not the number of terms in the series.

Part 2
To find the general solution to the ODE, we first must find the homogeneous solution:

$$\ y''-3y'+2y=0$$

$$\ \lambda^2-3\lambda+2=0$$

$$\ (\lambda-1)(\lambda-2)=0$$

$$\ \lambda_{1}-1=0  =>  \lambda_{1}=1$$

$$\ \lambda_{2}-2=0  =>  \lambda_{2}=2$$

$$\ y_{h}(x)=c_{1}e^x+c_{2}e^{2x}$$

n=3
Because n=3, the excitation equation approximating sinx becomes:

$$\ r(x)=x-\frac{x^{3}}{3!}$$

The general form of $$\ y_p$$ for a polynomial excitation equation is a polynomial with the same highest degree.

$$\ y_{p}(x)=Ax^3+Bx^2+Cx+D$$

Differentiating $$\ y_p$$ and its derivative gives:

$$\ y'_{p}(x)=3Ax^2+2Bx+C$$

$$\ y''_{p}(x)=6Ax+2B$$

By substituting these equations into the original ODE we get:

$$\ y''-3y'+2y=r(x)=x-\frac{x^{3}}{3!}$$

$$\ [6Ax+2B]-3[3Ax^2+2Bx+C]+2[Ax^3+Bx^2+Cx+D]=x-\frac{x^{3}}{3!}$$

Combining like terms gives:

$$\ 2Ax^3+(2B-9A)x^2+(6A-6B+2C)x+(2B-3C+2D)=x-\frac{x^{3}}{3!}$$

Equating like terms gives:

$$\ 2Ax^3=-\frac{x^{3}}{3!}   =>   A=-\frac{1}{12}$$

$$\ (2B-9A)x^2=0      =>     B=-\frac{3}{8}$$

$$\ (6A-6B+2C)x=x         =>      C=-\frac{3}{8}$$

$$\ (2B-3C+2D)=0         =>    D=-\frac{1}{16}$$

So the particular solution becomes:

$$\ y_{p}(x)=-\frac{1}{12}x^3-\frac{3}{8}x^2-\frac{3}{8}x-\frac{1}{16}$$

The general solution is the sum of the homogeneous and particular soluions:

$$\ y(x)=c_{1}e^x+c_{2}e^{2x}-\frac{1}{12}x^3-\frac{3}{8}x^2-\frac{3}{8}x-\frac{1}{16}$$

Differentiating gives:

$$\ y'(x)=c_{1}e^x+2c_{2}e^{2x}-\frac{1}{4}x^3-\frac{3}{4}x^2-\frac{3}{8}$$

Applying initial conditions gives:

$$\ y(0)=1=c_{1}+c_{2}-\frac{1}{16}$$

$$\ y'(0)=0=c_{1}+2c_{2}-\frac{3}{8}$$

Solving the system of equations gives:

$$\ c_{1}=2$$

$$\ c_{2}=-\frac{13}{16}$$

The final approximated solution for n=3 is:

$$\ y(x)=2e^x-\frac{13}{16}e^{2x}-\frac{1}{12}x^3-\frac{3}{8}x^2-\frac{3}{8}x-\frac{1}{16}$$

n=5
Because n=5, the excitation equation approximating sinx becomes:

$$\ r(x)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}$$

The general form of $$\ y_p$$ for a polynomial excitation equation is a polynomial with the same highest degree.

$$\ y_{p}(x)=Ax^5+Bx^4+Cx^3+Dx^2+Ex+F$$

Differentiating $$\ y_p$$ and its derivative gives:

$$\ y'_{p}(x)=5Ax^4+4Bx^3+3Cx^2+2Dx+E$$

$$\ y''_{p}(x)=20Ax^3+12Bx^2+6Cx+2D$$

By substituting these equations into the original ODE we get:

$$\ y''-3y'+2y=r(x)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}$$

$$\ [20Ax^3+12Bx^2+6Cx+2D]-3[5Ax^4+4Bx^3+3Cx^2+2Dx+E]+2[Ax^5+Bx^4+Cx^3+Dx^2+Ex+F]=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}$$

Combining like terms gives:

$$\ 2Ax^5+(2B-15A)x^4+(2C-12B+20A)x^3+(2D-9C+12B)x^2+(2E-6D+6C)x+(2F-3E+2D)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}$$

Equating like terms gives:

$$\ 2Ax^5=\frac{x^{5}}{5!}   =>   A=\frac{1}{240}$$

$$\ (2B-15A)x^4=0      =>     B=\frac{1}{32}$$

$$\ (2C-12B+20A)x^3=-\frac{x^{3}}{3!}    =>      C=\frac{1}{16}$$

$$\ (2D-9C+12B)x^2=0         =>    D=\frac{3}{32}$$

$$\ (2E-6D+6C)x=x       =>      E=\frac{19}{32}$$

$$\ (2F-3E+2D)=0       =>       F=\frac{51}{64}$$

So the particular solution becomes:

$$\ y_{p}(x)=\frac{1}{240}x^5+\frac{1}{32}x^4+\frac{1}{16}x^3+\frac{3}{32}x^2+\frac{19}{32}x+\frac{51}{64}$$

The general solution is the sum of the homogeneous and particular soluions:

$$\ y(x)=c_{1}e^x+c_{2}e^{2x}+\frac{1}{240}x^5+\frac{1}{32}x^4+\frac{1}{16}x^3+\frac{3}{32}x^2+\frac{19}{32}x+\frac{51}{64}$$

Differentiating gives:

$$\ y'(x)=c_{1}e^x+2c_{2}e^{2x}+\frac{1}{48}x^4+\frac{1}{8}x^3+\frac{3}{16}x^2+\frac{3}{16}x+\frac{19}{32}$$

Applying initial conditions gives:

$$\ y(0)=1=c_{1}+c_{2}+\frac{51}{64}$$

$$\ y'(0)=0=c_{1}+2c_{2}+\frac{19}{32}$$

Solving the system of equations gives:

$$\ c_{1}=1$$

$$\ c_{2}=-\frac{51}{64}$$

The final approximated solution for n=5 is:

$$\ y(x)=2e^x-\frac{51}{64}e^{2x}+\frac{1}{240}x^5+\frac{1}{32}x^4+\frac{1}{16}x^3+\frac{3}{32}x^2+\frac{19}{32}x+\frac{51}{64}$$

n=9
Because n=9, the excitation equation approximating sinx becomes:

$$\ r(x)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\frac{x^{9}}{9!}$$

The general form of $$\ y_p$$ for a polynomial excitation equation is a polynomial with the same highest degree.

$$\ y_{p}(x)=Ax^9+Bx^8+Cx^7+Dx^6+Ex^5+Fx^4+Gx^3+Hx^2+Jx+K$$

Differentiating $$\ y_p$$ and its derivative gives:

$$\ y'_{p}(x)=9Ax^8+8Bx^7+7Cx^6+6Dx^5+5Ex^4+4Fx^3+3Gx^2+2Hx+J$$

$$\ y''_{p}(x)=72Ax^7+63Bx^6+42Cx^5+30Dx^4+20Ex^3+Fx^2+6Gx+2H$$

By substituting these equations into the original ODE we get:

$$\ y''-3y'+2y=r(x)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\frac{x^{9}}{9!}$$

$$\ [72Ax^7+63Bx^6+42Cx^5+30Dx^4+20Ex^3+Fx^2+6Gx+2H]-3[9Ax^8+8Bx^7+7Cx^6+6Dx^5+5Ex^4+4Fx^3+3Gx^2+2Hx+J]+2[Ax^9+Bx^8+Cx^7+Dx^6+Ex^5+Fx^4+Gx^3+Hx^2+Jx+K]=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\frac{x^{9}}{9!}$$

Combining like terms gives:

$$\ 2Ax^9+(2B-27A)x^8+(2C-24B+72A)x^7+(2D-21C+63B)x^6+(2E-18D+42C)x^5+(2F-15E+30D)x^4+(2G-12F+20E)x^3+(2H-9G+12F)x^2+(2J-6H+6G)x+(2K-3J+2H)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\frac{x^{9}}{9!}$$

Equating like terms gives:

$$\ 2Ax^9=\frac{x^{9}}{9!}   =>   A=\frac{1}{725760}$$

$$\ (2B-27A)x^8=0      =>     B=\frac{1}{53760}$$

$$\ (2C-24B+72A)x^7=-\frac{x^{7}}{7!}    =>      C=\frac{1}{13440}$$

$$\ (2D-21C+63B)x^6=0         =>    D=\frac{1}{3840}$$

$$\ (2E-18D+42C)x^5=\frac{x^{5}}{5!}     =>    E=\frac{19}{3840}$$

$$\ (2F-15E+30D)x^4=0        =>       F=\frac{17}{512}$$

$$\ (2G-12F+20E)x^3=-\frac{x^{3}}{3!}     =>      G=\frac{17}{256}$$

$$\ (2H-9G+12F)x^2=0           =>         H=\frac{51}{512}$$

$$\ (2J-6H+6G)x=x          =>       J=\frac{307}{512}$$

$$\ (2K-3J+2H)=0             =>     K=\frac{819}{1024}$$

So the particular solution becomes:

$$\ y_{p}(x)=\frac{1}{725760}x^9+\frac{1}{53760}x^8+\frac{1}{13440}x^7+\frac{1}{3840}x^6+\frac{19}{3840}x^5+\frac{17}{512}x^4+\frac{17}{256}x^3+\frac{51}{512}x^2+\frac{307}{512}x+\frac{819}{1024}$$

The general solution is the sum of the homogeneous and particular soluions:

$$\ y(x)=c_{1}e^x+c_{2}e^{2x}+\frac{1}{725760}x^9+\frac{1}{53760}x^8+\frac{1}{13440}x^7+\frac{1}{3840}x^6+\frac{19}{3840}x^5+\frac{17}{512}x^4+\frac{17}{256}x^3+\frac{51}{512}x^2+\frac{307}{512}x+\frac{819}{1024}$$

Differentiating gives:

$$\ y'(x)=c_{1}e^x+2c_{2}e^{2x}+\frac{1}{80640}x^8+\frac{1}{6720}x^7+\frac{1}{1920}x^6+\frac{1}{640}x^5+\frac{19}{640}x^4+\frac{17}{128}x^3+\frac{51}{256}x^2+\frac{51}{256}x^2+\frac{307}{512}$$

Applying initial conditions gives:

$$\ y(0)=1=c_{1}+c_{2}+\frac{819}{1024}$$

$$\ y'(0)=0=c_{1}+2c_{2}+\frac{307}{512}$$

Solving the system of equations gives:

$$\ c_{1}=1$$

$$\ c_{2}=-\frac{819}{1024}$$

The final approximated solution for n=9 is:

$$\ y(x)=e^x-\frac{819}{1024}e^{2x}+\frac{1}{725760}x^9+\frac{1}{53760}x^8+\frac{1}{13440}x^7+\frac{1}{3840}x^6+\frac{19}{3840}x^5+\frac{17}{512}x^4+\frac{17}{256}x^3+\frac{51}{512}x^2+\frac{307}{512}x+\frac{819}{1024}$$

This image is an unzoomed graph of the three general solutions to the L2-ODE-CC above on the interval $$\ [0,4 \pi]$$.

This image is a zoomed picture of the left side of the graph above. Without zooming, it is impossible to tell the differences, whereas the differences are obvious here.

This image is a zoomed picture of the right side of the graph above. In this graph, the three solutions are labeled by their power. (n=3,5,9)

Part 3
In finding the exact overall solution, we no longer use Taylor series approximations. Instead we use sinx as the excitation and solve accordingly.

$$\ y''-3y'+2y=sinx$$

Our choice for $$\ y_{p}$$ is based on r(x) and can be found in K 2011 p.82 Table 2.1.

$$\ y_{p}(x)=Asinx+Bcosx$$

Differentiating $$\ y_{p}(x)$$ and its derviate yields:

$$\ y'_{p}(x)=-Bsinx+Acosx$$

$$\ y''_{p}(x)=-Asinx-Bcosx$$

Then we substitute the derivatives into the ODE:

$$\ [-Asinx-Bcosx]-3[-Bsinx+Acosx]+2[Asinx+Bcosx]=sinx$$

Combining like terms:

$$\ (3B+A)sinx+(B-3A)cosx=sinx$$

Equating like terms:

$$\ (3B+A)sinx=sinx  => 3B+A=1$$

$$\ (B-3A)cosx=0     => B-3A=0$$

Solving the system of equations gives:

A=0.1 B=0.3

$$\ y_{p}(x)=\frac{1}{10}sinx+\frac{3}{10}cosx$$

The general solution becomes:

$$\ y(x)=y_{h}(x)+y_{p}(x)=c_{1}e^x+c_{2}e^{2x}+\frac{1}{10}sinx+\frac{3}{10}cosx$$

Differentiating the general solution:

$$\ y'(x)=c_{1}e^x+2c_{2}e^{2x}+\frac{1}{10}cosx-\frac{3}{10}sinx$$

Apply initial conditions:

$$\ y(0)=c_{1}e^{0}+c_{2}e^{0}+\frac{1}{10}sin0+\frac{3}{10}cos0=c_{1}+c_{2}+\frac{3}{10}=1 =>  c_{1}+c_{2}=\frac{7}{10}$$

$$\ y'(0)=c_{1}e^{0}+2c_{2}e^{0}+\frac{1}{10}cos0-\frac{3}{10}sin0=0 => c_{1}+2c_{2}=-\frac{1}{10}$$

Solving the system of equatiions:

$$\ c_{1}=1.5$$

$$\ c_{2}=-\frac{8}{10}$$

The final, exact, general solution is:

$$\ y(x)=1.5e^x-\frac{8}{10}e^{2x}+\frac{1}{10}sinx+\frac{3}{10}cosx$$

This image shows the unzoomed graph of all three approximations and the exact solution to the L2-ODE-CC on the interval $$\ [0,4 \pi]$$. However, it is hard to distinguish the graphs without zooming in.

This image is a zoomed part of the graph above. The n=9 solution (green) is much more accurate than the n=5 solution (blue) and the n=3 solution (red) as it is closer to the exact solution (magenta).

title: User:Egm4313.s12.team8.lucas/R4.2


text:
University of Florida/Egm4313/s12.team11.R4

Intermediate Engineering Analysis Section 7566 Team 11 Due date: March 14, 2012.

Problem Statement
Obtain equations (2), (3), (n-2), (n-1), (n), and set up the matrix A as in (1) p.7-21 for the general case, with the matrix coefficients for rows 1, 2, 3, (n-2), (n-1), n, filled in, as obtained from equations (1), (2), (3), (n-2), (n-1), (n).

Given
As shown in p.7-21, the first equation is:

$$ 2C_2+ac_1+bc_0=d_0 \! $$ (1) p.7-21

According to p.7-20, the general form of the series is:

$$ \sum_{j=0}^{n-2}[c_{j+2}(j+2)(j+1)+ac_{j+1}+bc_j]x^j+ac_nnx^{n-1}+b[c_{n-1}x^{n-1}+c_nx^n]=\sum_{j=0}^{n}d_jx^j \! $$ (2) p. 7-20

From (2) p.7-20, we can obtain n+1 equations for n+1 unknown coefficients $$ {c_0,..., c_n} \! $$.

After referring to p.7-22, it can be determined that the matrix to be set up is of the following form:

$$ A =\begin{bmatrix} X && X && X && 0 && 0 & 0\\ 0 && X && X && 0 && 0 & 0\\ 0 && 0 && X && 0 && 0 & 0\\ 0 && 0 && 0 && X && X & X \\ 0 && 0 && 0 && 0 && X & X\\ 0 && 0 && 0 && 0 && 0 & X \end{bmatrix} \! $$

where the rows signify the coefficients $$ c_0, c_1, c_2, c_{n-2}, c_{n-1}, c_n \! $$, and the columns signify $$ d_0, d_1, d_2, d_{n-2}, d_{n-1}, d_n \! $$.

Solution
Building the coefficient matrix as shown in p.7-22 of the class notes, we can begin to solve for the coefficients as follows:

Equation associated with $$ d_0 \! $$:

j=0: $$ d_0=2C_2+ac_1+bc_0 \! $$ (1)

Equation associated with $$ d_1 \! $$:

j=1: $$ d_1=6c_3+2ac_2+bc_1 \! $$ (2)

Equation associated with $$ d_2 \! $$:

j=2: $$ d_2=12c_4+3ac_3+bc_2 \! $$ (3)

Equation associated with $$ d_{n-2} \! $$:

j=n-2: $$ d_{n-2}=[c_n(n)(n-1)+ac_{n-1}(n-1)+bc_{n-2}] \! $$ (n-2)

Equation associated with $$ d_{n-1} \! $$:

j=n-1: $$ d_{n-1}=ac_nn+bc_{n-1} \! $$ (n-1)

Equation associated with $$ d_n \! $$:

j=n: $$ d_n=bc_n \! $$ (n)

Using all of the above equations, (1), (2), (3), (n-2), (n-1), (n), we can then determine the A matrix to be:

$$                              A =\begin{bmatrix} b && a && 2 && 0 && 0 & 0\\ 0 && b && 2a && 0 && 0 & 0\\ 0 && 0 && b && 0 && 0 & 0\\ 0 && 0 && 0 && b && a(n-1) & n(n-1) \\ 0 && 0 && 0 && 0 && b & an\\ 0 && 0 && 0 && 0 && 0 & b \end{bmatrix} \! $$

Solved by Gonzalo Perez

Part 1
Solved by Jonathan Sheider

Problem Statement
Given: $$ y^{''} - 3y^{'} +2y = sin(x)\!$$ With initial conditions: $$ y(0) = 1, y'(0) = 0 \!$$ Using the Taylor series for sin(x), reproduce the graph in the lecture notes p.7-24.

Solution
The Taylor series expansion for sin(x) that is plotted with n = 13 is as follows: $$ y = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!} + \frac{x^{13}}{13!} - \frac{x^{15}}{15!} + \frac{x^{17}}{17!} - \frac{x^{19}}{19!} + \frac{x^{21}}{21!} - \frac{x^{23}}{23!} + \frac{x^{25}}{25!} \!$$ Using MatLab, the following Taylor series expansion was plotted and coded: The plot created with n = 13 (i.e. 13 terms) is as follows:

Part 2
Solved by Jonathan Sheider

Problem Statement
Given: $$ y^{''} - 3y^{'} +2y = r(x)\!$$ With initial conditions: $$ y(0) = 1, y'(0) = 0 \!$$ Letting $$ r(x) \!$$ equal the truncated Taylor series of $$ sin(x) \!$$, i.e. $$ r(x) = \sum_{k=0}^{n} \frac{(-1)^kt^{2k+1}}{(2k+1)!} \!$$ Find the overall solution $$ y_{n}(x) \!$$ for $$ n = 3,5,9 \!$$ and plot these solutions on the interval from $$ [0,4\pi ] \!$$

Solution
First finding the homogenous solution to the ODE: The characteristic equation: $$ \lambda^2 -3\lambda +2 = 0\!$$ $$ (\lambda -2)(\lambda -1) = 0\!$$ Therefore, $$ \lambda = 1,2\!$$ And the homogenous solution is: $$ y_{h} = c_{1}e^{x} + c_{2}e^{2x} \!$$ Next the particular solution will be evaluated. For n = 3: The excitation is therefore: $$ r_{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}\!$$ Therefore the particular solution has a form: $$ y_{p} = K_{7}x^{7} + K_{6}x^{6} + K_{5}x^{5} + K_{4}x^{4} + K_{3}x^{3} + K_{2}x^{2} + K_{1}x + K_{0} \!$$ Differentiating: $$ y_{p}^{'} = 7K_{7}x^{6} + 6K_{6}x^{5} + 5K_{5}x^{4} + 4K_{4}x^{3} + 3K_{3}x^{2} + 2K_{2}x + 1K_{1} \!$$ $$ y_{p}^{''} = 42K_{7}x^{5} + 30K_{6}x^{4} + 20K_{5}x^{3} + 12K_{4}x^{2} + 6K_{3}x + 2K_{2} \!$$ Plugging these values into the ODE and equating them to the excitation: $$ 42K_{7}x^{5} + 30K_{6}x^{4} + 20K_{5}x^{3} + 12K_{4}x^{2} + 6K_{3}x + 2K_{2} \!$$ $$ - 3(7K_{7}x^{6} + 6K_{6}x^{5} + 5K_{5}x^{4} + 4K_{4}x^{3} + 3K_{3}x^{2} + 2K_{2}x + 1K_{1})\!$$ $$ + 2(K_{7}x^{7} + K_{6}x^{6} + K_{5}x^{5} + K_{4}x^{4} + K_{3}x^{3} + K_{2}x^{2} + K_{1}x + K_{0}) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} \!$$ $$ 42K_{7}x^{5} + 30K_{6}x^{4} + 20K_{5}x^{3} + 12K_{4}x^{2} + 6K_{3}x + 2K_{2} \!$$ $$ - 21K_{7}x^{6} -18K_{6}x^{5} -15K_{5}x^{4} -12K_{4}x^{3} -9K_{3}x^{2} -6K_{2}x -3K_{1}\!$$ $$ + 2K_{7}x^{7} + 2K_{6}x^{6} + 2K_{5}x^{5} + 2K_{4}x^{4} + 2K_{3}x^{3} + 2K_{2}x^{2} + 2K_{1}x + 2K_{0} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} \!$$ To solve this, we will use an upper triangular matrix, and solve using Matlab. The matrix is as follows: $$ \begin{bmatrix} 2 &-3  &2  &0  & 0 & 0 & 0 &0\\  0&  2  &-6  &6  & 0 & 0 & 0 &0\\  0&  0&  2  &-9  &12  & 0 &0 &0\\  0&  0&  0&  2  &-12  &20  &0 &0\\  0&  0&  0& 0 &  2  &-15  &30 &0\\  0&  0&  0& 0 & 0 &  2  &-18  &42  \\  0& 0 & 0 &0  &0 & 0&  2 &-21\\  0&0  &0  & 0 & 0 &0  &0  & 2 \end{bmatrix} \begin{bmatrix} K_{0} \\ K_{1} \\ K_{2} \\ K_{3} \\ K_{4} \\ K_{5} \\ K_{6} \\ K_{7} \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ -\frac{1}{3!} \\ 0 \\ \frac{1}{5!} \\ 0 \\ -\frac{1}{7!} \end{bmatrix}\!$$ The answer is calculated in Matlab: Therefore: $$ y_{p} = \frac{-1}{10080}x^{7} + \frac{-1}{960}x^{6} + \frac{-1}{320}x^{5} + \frac{-1}{128}x^{4} + \frac{-19}{192}x^{3} + \frac{-51}{128}x^{2} + \frac{-51}{128}x + \frac{-51}{256} \!$$ The final solution is then found to be: $$ y_{n} = c_{1}e^{x} + c_{2}e^{2x} + \frac{-1}{10080}x^{7} + \frac{-1}{960}x^{6} + \frac{-1}{320}x^{5} + \frac{-1}{128}x^{4} + \frac{-19}{192}x^{3} + \frac{-51}{128}x^{2} + \frac{-51}{128}x + \frac{-51}{256} \!$$ Evaluating at the initial conditions, we have: $$ y_{n}(0) = c_{1} + c_{2} + \frac{-51}{256} = 1 \!$$ $$ y_{n}^{'}(0) = c_{1} + 2c_{2} + \frac{-51}{128} = 0\!$$ Solving this system of equations, we find: $$ c_{1} = 2, c_{2} = \frac{-205}{256} \!$$ Therefore the final solution is: $$ y_{n} = 2e^{x} + \frac{-205}{256}e^{2x} + \frac{-1}{10080}x^{7} + \frac{-1}{960}x^{6} + \frac{-1}{320}x^{5} + \frac{-1}{128}x^{4} + \frac{-19}{192}x^{3} + \frac{-51}{128}x^{2} + \frac{-51}{128}x + \frac{-51}{256} \!$$ Plotting this equation in Matlab yields:

For n = 5: The excitation is therefore: $$ r_{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!}\!$$ Therefore the particular solution has a form: $$ y_{p} = K_{11}x^{11} + K_{10}x^{10}+ K_{9}x^{9} + K_{8}x^{8} + K_{7}x^{7}+ K_{6}x^{6} + K_{5}x^{5} + K_{4}x^{4} + K_{3}x^{3} + K_{2}x^{2} + K_{1}x + K_{0} \!$$ Differentiating: $$ y_{p}^{'} = 11K_{11}x^{10} + 10K_{10}x^{9}+ 9K_{9}x^{8} + 8K_{8}x^{7}+ 7K_{7}x^{6} + 6K_{6}x^{5} + 5K_{5}x^{4} + 4K_{4}x^{3} + 3K_{3}x^{2} + 2K_{2}x + 1K_{1} \!$$ $$ y_{p}^{''} = 110K_{11}x^{9} + 90K_{10}x^{8}+ 72K_{9}x^{7} + 56K_{8}x^{6}+ 42K_{7}x^{5} + 30K_{6}x^{4} + 20K_{5}x^{3} + 12K_{4}x^{2} + 6K_{3}x + 2K_{2} \!$$ Plugging these values into the ODE and equating them to the excitation, and then solving them by using a upper triangular matrix as before (just as in the n=3 example) in Matlab: Therefore: $$ y_{p} = \frac{-1}{79833600}x^{11} + \frac{-1}{4838400}x^{10} + \frac{-1}{967680}x^{9} + \frac{-1}{215040}x^{8} + \frac{-7}{59419}x^{7} + \frac{-17}{15360}x^{6} \!$$ $$ + \frac{-17}{5120}x^5 + \frac{-17}{2048}x^4 + \frac{-307}{3072}x^3 + \frac{-819}{2048}x^2 + \frac{-819}{2048}x + \frac{-819}{4096} \!$$ The final solution is then found to be: $$ y_{n} = c_{1}e^{x} + c_{2}e^{2x} + \frac{-1}{79833600}x^{11} + \frac{-1}{4838400}x^{10} + \frac{-1}{967680}x^{9} + \frac{-1}{215040}x^{8} + \frac{-7}{59419}x^{7} + \frac{-17}{15360}x^{6} \!$$ $$ + \frac{-17}{5120}x^5 + \frac{-17}{2048}x^4 + \frac{-307}{3072}x^3 + \frac{-819}{2048}x^2 + \frac{-819}{2048}x + \frac{-819}{4096} \!$$ Evaluating at the initial conditions, we have: $$ y_{n}(0) = c_{1} + c_{2} + \frac{-819}{4096} = 1 \!$$ $$ y_{n}^{'}(0) = c_{1} + 2c_{2} + \frac{-819}{2048} = 0\!$$ Solving this system of equations, we find: $$ c_{1} = 2, c_{2} = \frac{-3277}{4096} \!$$ Therefore the final solution is: $$ y_{n} = 2e^{x} + \frac{-3277}{4096}e^{2x} + \frac{-1}{79833600}x^{11} + \frac{-1}{4838400}x^{10} + \frac{-1}{967680}x^{9} + \frac{-1}{215040}x^{8} + \frac{-7}{59419}x^{7} + \frac{-17}{15360}x^{6} \!$$ $$ + \frac{-17}{5120}x^5 + \frac{-17}{2048}x^4 + \frac{-307}{3072}x^3 + \frac{-819}{2048}x^2 + \frac{-819}{2048}x + \frac{-819}{4096} \!$$

Plotting this equation in Matlab yields:

For n = 9: The excitation is therefore: $$ r_{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!} + \frac{x^{13}}{13!} - \frac{x^{15}}{15!} + \frac{x^{17}}{17!} - \frac{x^{19}}{19!}\!$$ Therefore the particular solution has a form: $$ y_{p} = K_{19}x^{19}+ K_{18}x^{18} + K_{17}x^{17} + K_{16}x^{16} + K_{15}x^{15} + K_{14}x^{14} + K_{13}x^{13} + K_{12}x^{12} + K_{11}x^{11} + K_{10}x^{10} \!$$ $$ + K_{9}x^{9} + K_{8}x^{8} + K_{7}x^{7}+ K_{6}x^{6} + K_{5}x^{5} + K_{4}x^{4} + K_{3}x^{3} + K_{2}x^{2} + K_{1}x + K_{0} \!$$ Differentiating: $$ y_{p}^{'} = 19K_{19}x^{18}+ 18K_{18}x^{17} + 17K_{17}x^{16} + 16K_{16}x^{15} + 15K_{15}x^{14} + 14K_{14}x^{13} + 13K_{13}x^{12} + 12K_{12}x^{11} + 11K_{11}x^{10} + 10K_{10}x^{9} \!$$ $$ + 9K_{9}x^{8} + 8K_{8}x^{7}+ 7K_{7}x^{6} + 6K_{6}x^{5} + 5K_{5}x^{4} + 4K_{4}x^{3} + 3K_{3}x^{2} + 2K_{2}x + 1K_{1} \!$$ $$ y_{p}^{''} = 342K_{19}x^{17}+ 306K_{18}x^{16} + 272K_{17}x^{15} + 240K_{16}x^{14} + 210K_{15}x^{13} + 182K_{14}x^{12} + 156K_{13}x^{11} + 132K_{12}x^{10} + 110K_{11}x^{9} + 90K_{10}x^{8} \!$$ $$ +72K_{9}x^{7} + 56K_{8}x^{6}+ 42K_{7}x^{5} + 30K_{6}x^{4} + 20K_{5}x^{3} + 12K_{4}x^{2} + 6K_{3}x + 2K_{2} \!$$ Plugging these values into the ODE and equating them to the excitation, and then solving them by using an upper triangular matrix as before (just as in the n=3 example) in Matlab: Therefore: $$ y_{p} = \frac{-1}{2.433E17}x^{19} + \frac{-1}{8.536E15}x^{18} + \frac{-1}{9.485E14}x^{17} + \frac{-1}{1.112E14}x^{16} + \frac{-1}{2.202E12}x^{15} + \frac{-1}{1.094E11}x^{14} \!$$ $$ + \frac{-1}{1.563E10}x^{13} + \frac{-1}{2.404E9}x^{12} + \frac{-1}{6.657E7}x^{11} + \frac{-1}{4.537E6}x^{10} + \frac{-1}{9.074E5}x^9 + \frac{-4}{8.066E5}x^8 + \frac{-38}{3.192E5}x^7 \!$$ $$ + \frac{-73}{657010}x^{6} + \frac{-218}{65401}x^{5} + \frac{-218}{65401}x^{4} + \frac{-3441}{49157}x^{3} + \frac{-1198}{4061}x^2 + \frac{-1431}{8177}x + \frac{145}{4462} \!$$ The final solution is then found to be: $$ y_{n} = c_{1}e^{x} + c_{2}e^{2x} + \frac{-1}{2.433E17}x^{19} + \frac{-1}{8.536E15}x^{18} + \frac{-1}{9.485E14}x^{17} + \frac{-1}{1.112E14}x^{16} + \frac{-1}{2.202E12}x^{15} + \frac{-1}{1.094E11}x^{14} \!$$ $$ + \frac{-1}{1.563E10}x^{13} + \frac{-1}{2.404E9}x^{12} + \frac{-1}{6.657E7}x^{11} + \frac{-1}{4.537E6}x^{10} + \frac{-1}{9.074E5}x^9 + \frac{-4}{8.066E5}x^8 + \frac{-38}{3.192E5}x^7 \!$$ $$ + \frac{-73}{657010}x^{6} + \frac{-218}{65401}x^{5} + \frac{-218}{65401}x^{4} + \frac{-3441}{49157}x^{3} + \frac{-1198}{4061}x^2 + \frac{-1431}{8177}x + \frac{145}{4462} \!$$ Evaluating at the initial conditions, we have: $$ y_{n}(0) = c_{1} + c_{2} + \frac{145}{4462} = 1 \!$$ $$ y_{n}^{'}(0) = c_{1} + 2c_{2} + \frac{-1431}{8177} = 0\!$$ Solving this system of equations, we find: $$ c_{1} = 1.76, c_{2} = -0.7925 \!$$ Therefore the final solution is: $$ y_{n} = 1.76e^{x} + -0.7925e^{2x} + \frac{-1}{2.433E17}x^{19} + \frac{-1}{8.536E15}x^{18} + \frac{-1}{9.485E14}x^{17} + \frac{-1}{1.112E14}x^{16} + \frac{-1}{2.202E12}x^{15} + \frac{-1}{1.094E11}x^{14} \!$$ $$ + \frac{-1}{1.563E10}x^{13} + \frac{-1}{2.404E9}x^{12} + \frac{-1}{6.657E7}x^{11} + \frac{-1}{4.537E6}x^{10} + \frac{-1}{9.074E5}x^9 + \frac{-4}{8.066E5}x^8 + \frac{-38}{3.192E5}x^7 \!$$ $$ + \frac{-73}{65701}x^{6} + \frac{-218}{65401}x^{5} + \frac{-218}{65401}x^{4} + \frac{-3441}{49157}x^{3} + \frac{-1198}{4061}x^2 + \frac{-1431}{8177}x + \frac{145}{4462} \!$$ Plotting this equation in Matlab yields:

Note: On the interval from $$ 0 \!$$ to $$ 4\pi\!$$, the resulting plots of the solutions, on such a large scale (note that the Y-axis is forced to be on the order of $$ 10^{10}\!$$!) each plot looks almost exactly identical. It is not until you are able to zoom in a lot do you see the very slight change in the curve of each plot on this interval.

--Egm4313.s12.team11.sheider (talk) 01:09, 14 March 2012 (UTC)

Part 3
Solved by Daniel Suh

Problem Statement
Using the particular solution from Table 2.1 on Kreyszig, find the overall solution $$y(x)\!$$, and plot it compared with $$y_{n}(x)\!$$ for n = 3,5,9

$$y''-3y'+2y = r(x)\!$$ $$r(x)=sin(x)\!$$ Initial Conditions: $$y(0)=1\!$$, $$y'(0)=0\!$$

Homogenous Solution
To find $$y_{h}(x)\!$$, $$y''-3y'+2y = r(x)\!$$ $$\lambda^{2}-3\lambda+2=0\!$$ $$(\lambda-2)(\lambda-1)=0\!$$ $$\lambda_{1}=2, \lambda_{2}=1\!$$ $$y_{h}(x)=c_{1}e^{2x}+c_{2}e^{x}\!$$

Particular Solution
To find $$y_{p}(x)\!$$, Using Table 2.1, we find that for $$r(x)=sin(x)\!$$, $$y_{p}=Kcos (\omega x)+Msin (\omega x)\!$$ $$y'_{p}=-Ksin (\omega x)+Mcos (\omega x)\!$$ $$y''_{p}=-Kcos (\omega x)-Msin (\omega x)\!$$

Plugging it back into the equation, $$(-Kcos (x)-Msin (x))-3(-Ksin (x)+Mcos (x))+2(Kcos (x)+Msin (x)) = sin(x)\!$$ $$(3K+M)sin(x)+(K-3M)cos(x) = sin(x)\!$$

Solving for coefficients, $$(3K+M) = 1\!$$ $$(K-3M) = 0\!$$ $$K=\frac{3}{10}\!$$ $$M=\frac{1}{10}\!$$ $$y_{p}(x)=\frac{3}{10}cos(x)+\frac{1}{10}sin(x)\!$$

Overall Solution
$$y(x)=y_{h}(x)+y_{p}(x)\!$$ $$y(x)=c_{1}e^{2x}+c_{2}e^{x}+\frac{3}{10}cos(x)+\frac{1}{10}sin(x)\!$$

Solving with initial conditions, $$y(x)=c_{1}e^{2x}+c_{2}e^{x}+\frac{3}{10}cos(x)+\frac{1}{10}sin(x)\!$$ $$y'(x)=2c_{1}e^{2x}+c_{2}e^{x}-\frac{3}{10}sin(x)+\frac{1}{10}cos(x)\!$$

$$1=c_{1}+c_{2}+\frac{3}{10}\!$$ $$0=2c_{1}+c_{2}+\frac{1}{10}\!$$ $$c_{1}=\frac{-4}{5}\!$$ $$c_{2}=\frac{3}{2}\!$$

$$y(x)=\frac{-4}{5}e^{2x}+\frac{3}{2}e^{x}+\frac{3}{10}cos(x)+\frac{1}{10}sin(x)\!$$ $$\!$$

Plot
Plotting in matlab yields: As shown on the graph, the two plots are exactly the same.

Created by [| Daniel Suh] 20:43, 13 March 2012 (UTC)

Part 1
Solved by Francisco Arrieta

Problem Statement

 * Develop $$ \log(x+1) \!$$ in Taylor Series about $$ x=0 \!$$ to reproduce the figure on page 7-25

Given

 * $$ f(x)= \sum_{n=0}^\infty \frac{f^{(n)}(\hat x)}{n!}(x- \hat x)^n \!$$

Solution
For n=4:
 * $$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)} \!$$

For n=7:
 * $$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)} \!$$

For n=11:
 * $$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})} \!$$

For n=16:
 * $$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})}\!$$
 * $$ -\frac{x^{12}}{\log(10^{12})}+\frac{x^{13}}{\log(10^{13})}-\frac{x^{14}}{\log(10^{14})}+\frac{x^{15}}{\log(10^{15})}-\frac{x^{16}}{\log(10^{16})} \!$$

Using MATLAB

--Egm4313.s12.team11.arrieta (talk) 00:44, 14 March 2012 (UTC)

Part 2
Solved by: --Egm4313.s12.team11.vargas.aa (talk) 01:28, 14 March 2012 (UTC)

Problem Statement
Given: $$y''-3y'+2y=r(x)\!$$

where $$r(x)=\log (1+x)\!$$

With initial conditions: $$ y(-\frac{3}{4}) = 1, y'(-\frac{3}{4}) = 0 \!$$ Find the overall solution $$ y_{n}(x) \!$$ for $$ n = 4,7,11 \!$$ and plot these solutions on the interval from $$ [-\frac{3}{4},3] \!$$

Solution
First we find the homogeneous solution to the ODE: The characteristic equation is: $$\lambda^2-3\lambda+2=0\!$$ $$(\lambda-2)(\lambda-1)=0\!$$ Then, $$\lambda=1,2\!$$ Therefore the homogeneous solution is: $$y_h=C_1e^(2x)+c_2e^x\!$$

Now to find the particulate solution For n=4

$$ r(x)=\sum_{0}^{n}-\frac{(-1)^nx^n}{n\ln (10)} \!$$

$$ r(x)=\frac{x}{\ln(10)}-\frac{x^2}{2\ln(10)}+\frac{x^3}{3\ln(10)}-\frac{x^4}{4\ln(10)} \!$$

We can then use a matrix to organize the known coefficients:

$$ \begin{bmatrix} 2 &-3  &2  &0  & 0\\  0&  2  &-6  &6  & 0\\  0&  0&  2  &-9  &12  \\  0&  0&  0&  2  &-12\\  0&  0&  0& 0 &  2  \\   \end{bmatrix} \begin{bmatrix} K_{0} \\ K_{1} \\ K_{2} \\ K_{3} \\ K_{4} \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{1}{ln(10)} \\ \frac{1}{2ln(10)} \\ \frac{1}{3ln(10)} \\ \frac{1}{4ln(10)} \end{bmatrix}\!$$ Then, using MATLAB and the backlash operator we can solve for these unknowns: Therefore $$ y_{p4}=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4 \!$$ Superposing the homogeneous and particulate solution we get $$ y_{n}=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4+C_1e^{2x}+C_2e^{x} \!$$ Differentiating: $$y'_{n}=3.7458+3.1486x+1.1943x^2+0.2172x^3+2C_1e^{2x}+C_2e^{x}\!$$ Evaluating at the initial conditions: $$y(-0.75)=0.9698261719+0.231301601C_!+0.4723665527C_2=1\!$$ $$y'(-0.75)=1.9645125+0.4462603203C_1+0.4723665527C_2\!$$ We obtain: $$ C_1=-4.46\!$$ $$ C_2=0.055 \!$$ Finally we have: $$ y_{n}=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4-4.46e^{2x}+0.055e^{x} \!$$

For n=7

$$ r(x)=\sum_{0}^{n}-\frac{(-1)^nx^n}{n\ln (10)} \!$$

$$ r(x)=\frac{x}{\ln(10)}-\frac{x^2}{2\ln(10)}+\frac{x^3}{3\ln(10)}-\frac{x^4}{4\ln(10)}+\frac{x^5}{5\ln(10)}-\frac{x^6}{6\ln(10)}+\frac{x^7}{7\ln(10)} \!$$

We can then use a matrix to organize the known coefficients:

$$ \begin{bmatrix} 2 &-3  &2  &0  & 0 &0 &0 &0\\  0&  2  &-6  &6  & 0 &0 &0 &0\\  0&  0&  2  &-9  &12  &0 &0 &0\\  0&  0&  0&  2  &-12 &20 &0 &0 \\  0&  0&  0& 0 &  2 &-15 &30 &0\\ 0 &0 &0 &0 &0 &2 &-18 &42\\ 0 &0 &0 &0 &0 &0  &2 &-21\\ 0 &0 &0 &0 &0 &0 &0 &2\\    \end{bmatrix} \begin{bmatrix} K_{0} \\ K_{1} \\ K_{2} \\ K_{3} \\ K_{4} \\ K_{5} \\ K_{6} \\ K_{7} \end{bmatrix} \begin{bmatrix} 0 \\ \frac{1}{ln(10)} \\ \frac{1}{2ln(10)} \\ \frac{1}{3ln(10)} \\ \frac{1}{4ln(10)} \\ \frac{1}{5ln(10)} \\ \frac{1}{6ln(10)} \\ \frac{1}{7ln(10)} \end{bmatrix}\!$$ Then, using MATLAB and the backlash operator we can solve for these unknowns: Therefore $$ y_{p7}=377.4833+375.3933x+185.6066x^2+60.5479x^3+14.4946x^4+2.6492x^5+0.3619x^6+0.0310x^7 \!$$ Superposing the homogeneous and particulate solution we get $$ y_{n}=377.4833+375.3933x+185.6066x^2+60.5479x^3+14.4946x^4+2.6492x^5+0.3619x^6+0.0310x^7+C_1e^{2x}+c_2e^x \!$$ Differentiating: $$y'_{n}=375.3933+371.213x+181.644x^2+57.9784x^3+13.46x^4+2.1714x^5+0.214x^6+2C_1e^{2x}+C_2e^x\!$$ Evaluating at the initial conditions: $$y(-0.75)=178.816+0.2231301601C_!+0.4723665527C_2=1\!$$ $$y'(-0.75)=178.413+0.4462603203C_1+0.4723665527C_2\!$$ We obtain: $$ C_1=-2.6757\!$$ $$ C_2=-375.173 \!$$ Finally $$ y_{n}=377.4833+375.3933x+185.6066x^2+60.5479x^3+14.4946x^4+2.6492x^5+0.3619x^6+0.0310x^7+-2.6757e^{2x}-375.173e^x \!$$ For n=11 $$ r(x)=\sum_{0}^{n}-\frac{(-1)^nx^n}{n\ln (10)} \!$$

$$ r(x)=\frac{x}{\ln(10)}-\frac{x^2}{2\ln(10)}+\frac{x^3}{3\ln(10)}-\frac{x^4}{4\ln(10)}+\frac{x^5}{5\ln(10)}-\frac{x^6}{6\ln(10)}+\frac{x^7}{7\ln(10)} \!$$

We can then use a matrix to organize the known coefficients: $$ \begin{bmatrix} K_{0} \\ K_{1} \\ K_{2} \\ K_{3} \\ K_{4} \\ K_{5} \\ K_{6} \\ K_{7} \\ K_{8} \\ K_{9} \\ K_{10} \\ K_{11} \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{1}{ln(10)} \\ \frac{1}{2ln(10)} \\ \frac{1}{3ln(10)} \\ \frac{1}{4ln(10)} \\ \frac{1}{5ln(10)} \\ \frac{1}{6ln(10)} \\ \frac{1}{7ln(10)} \\ \frac{1}{8ln(10)} \\ \frac{1}{9ln(10)} \\ \frac{1}{10ln(10)} \\ \frac{1}{11ln(10)} \end{bmatrix}\!$$ Then, using MATLAB and the backlash operator we can solve for these unknowns: Therefore $$ y_{p11}=1753158.594+1752673.419x+875851.535x^2+291627.134x^3+72745.1129x^4+14484.362x^5+2392.510x^6+335.632x^7+40.417x^8\!$$ $$+4.1499x^9+0.3474x^(10)+0.0197x^(11) \!$$ Superposing the homogeneous and particulate solution we get $$ y_{n}=1753158.594+1752673.419x+875851.535x^2+291627.134x^3+72745.1129x^4+14484.362x^5+2392.510x^6+335.632x^7+40.417x^8\!$$ $$+4.1499x^9+0.3474x^(10)+0.0197x^(11)+C_1e^{2x}+C_2e^(x) \!$$ Differentiating: $$y'_{n}=0.2167 x^10+3.474 x^9+37.3491 x^8+323.336 x^7+2349.42 x^6+14355.1 x^5+72421.8 x^4+290980. x^3+874881. x^2\!$$ $$+1.7517x10^6 x+1.75267x10^6+2C_1e^{2x}+C_2e^x\!$$ Evaluating at the initial conditions: $$y(-0.75)=828254+0.2231301601C_!+0.4723665527C_2=1\!$$ $$y'(-0.75)=828145+0.4462603203C_1+0.4723665527C_2=0\!$$ We obtain: $$ C_1=-484.022\!$$ $$ C_2=-1753750 \!$$ Finally $$ y_{n}=1753158.594+1752673.419x+875851.535x^2+291627.134x^3+72745.1129x^4+14484.362x^5+2392.510x^6+335.632x^7+40.417x^8\!$$ $$+4.1499x^9+0.3474x^(10)+0.0197x^(11)-484.022e^{2x}-1753750e^x \!$$

Part 3
Plot for part 2 and part 3. Please note that because of the broad scale, it is almost impossible to distinguish between the graphs for n=4,7,11 nad the ODE45 operator. Only when zoomed in is there any noticeable difference

Problem Statement
Find n sufficiently high so that $$y_n(x_1), y'_n(x_1)$$ do not differ from the numerical solution by more than $$10^{-5}$$ at $$x_1=0.9$$

Solution
Using a program in MATLAB that iteratively added terms onto the taylor series of $$log(1+x)$$, terms were added until the error between the exact answer and the series was less than $$ 10^{-5}$$.

It was found after trial and error that $$ n=39 $$ for the error to be of a magnitude of $$ 10^{-5}$$. This error found was 9.7422e-005

Similarly, for $$y'_n(x_1)$$.

It was found after trial and error that $$ n=74 $$ for the error to be of a magnitude of $$ 10^{-5}$$. This error found was 9.3967e-005

Problem Statement
Develop $$log(1+x)$$ in Taylor series about $$\hat{x}=1$$ for $$ n=4,7,11$$ and plot these truncated series vs. the exact function. What is now the domain of convergence by observation?

Solution
A MATLAB program was created, which calculated the Taylor series of each n value, along with the exact function, then plotted these together to show the comparison of all the series. Below is the Taylor series for $$n=7$$ expanded at $$\hat{x}=1$$. $$\frac{x - 1}{2\, \ln\!\left(10\right)} - \frac{{\left(x - 1\right)}^2}{8\, \ln\!\left(10\right)} + \frac{{\left(x - 1\right)}^3}{24\, \ln\!\left(10\right)} - \frac{{\left(x - 1\right)}^4}{64\, \ln\!\left(10\right)} + \frac{{\left(x - 1\right)}^5}{160\, \ln\!\left(10\right)} - \frac{{\left(x - 1\right)}^6}{384\, \ln\!\left(10\right)} + \frac{\ln\!\left(2\right)}{\ln\!\left(10\right)} $$

It can be seen by observation that the domain of convergence has shifted to the right one unit. --egm4313.s12.team11.gooding (talk) 03:48, 14 March 2012 (UTC)

Part 3
Solved by Luca Imponenti

Find $$y_n(x)\!$$, for $$n=4,7,11\!$$ such that:
 * $$y_n''-3y_n'+2y_n=r_n(x)\!$$

for $$x\!$$ in $$[0.9,3]\!$$ with the initial conditions found.

Plot $$y_n(x)\!$$ for $$n=4,7,11\!$$ for $$x\!$$ in $$[0.9,3]\!$$.

Homogeneous Solution
The homogeneous case is shown below:
 * $$y''_h-3y'_h+2y_h=0\!$$

This equation has the following roots:

$$\lambda_1=1, \lambda_2=2\!$$

Which gives yields the homogeneous solution
 * $$y_h=c_1e^x+c_2e^{2x}\!$$

General Solution, n=4
Using the taylor series approximation from earlier with $$n=4\!$$ we have

$$r_4(x)=log(2)+\frac{(x-1)}{2ln(10)}-\frac{(x-1)^2}{8ln(10)}+\frac{(x-1)^3}{24ln(10)}-\frac{(x-1)^4}{64ln(10)}\!$$

We know the particular solution, $$y_{p4}(x)\!$$, ve will have this form:

$$y_{p4}(x)=a_4(x-1)^4+a_3(x-1)^3+a_2(x-1)^2+a_1(x-1)+a_0\!$$

taking the derivatives of this solution

$$y'_{p4}(x)=\frac{d}{dx}y_{p4}(x)=4a_4(x-1)^3+3a_3(x-1)^2+2a_2(x-1)+a_1\!$$

and

$$y''_{p4}(x)=\frac{d}{dx}y'_{p4}(x)=12a_4(x-1)^3+6a_3(x-1)^2+2a_2\!$$

Plugging the above equations into the original ODE yields the following matrix equation:

 * $$ \begin{bmatrix}

2 & 0 & 0 & 0 & 0\\ -12 & 2 & 0 & 0 & 0\\ 12 & -9 & 2 & 0 & 0\\ 0 & 6 & -6 & 2 & 0\\ 0 & 0 & 2 & -3 & 2\end{bmatrix}* \begin{bmatrix} a_4 \\ a_3\\ a_2\\ a_1\\ a_0\end{bmatrix}= \begin{bmatrix} -\frac{1}{64ln(10)}\\ \frac{1}{24ln(10)}\\ -\frac{1}{8ln(10)}\\ \frac{1}{2ln(10)}\\ log(2)\end{bmatrix}\!$$

The unknown vector $$a\!$$ can be easily solved by forward substitution,the following values were calculated in matlab:

$$a_4=-.0034, a_3=-.0113, a_2=-.0577, a_1=-.1624, a_0=.1624\!$$

So the particular solution $$y_{p4}\!$$ is

$$y_{p4}=0.1624-0.1624*(x-1)-.0577*(x-1)^2-.0113*(x-1)^3-.0034*(x-1)^4\!$$

We can now find the general solution for n=4, $$y_4(x)\!$$.

$$y_4(x)=y_h(x)+y_{p4}(x)\!$$

$$y_4(x)=c_1e^x+c_2e^{2x}+0.1624-0.1624*(x-1)\!$$ $$-.0577*(x-1)^2-.0113*(x-1)^3-.0034*(x-1)^4\!$$

Solving using the initial conditions yields;

$$y_4(x)=.0595e^x-.0076e^{2x}+0.1624-0.1624*(x-1)\!$$ $$-.0577*(x-1)^2-.0113*(x-1)^3-.0034*(x-1)^4\!$$

General Solution, n=7
Using the taylor series approximation from earlier with $$n=7\!$$ we have

$$r_7(x)=log(2)+\frac{(x-1)}{2ln(10)}-\frac{(x-1)^2}{8ln(10)}+\frac{(x-1)^3}{24ln(10)}-\frac{(x-1)^4}{64ln(10)}+\frac{(x-1)^5}{160ln(10)}-\frac{(x-1)^6}{384ln(10)} +\frac{(x-1)^7}{896ln(10)}\!$$

In a similar fashion we construct a matrix equation for n=7:

 * $$ \begin{bmatrix}

2 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ -21 & 2 & 0 & 0 & 0 & 0 & 0 & 0\\ 42 & -18 & 2 & 0 & 0 & 0 & 0 & 0\\ 0 & 30 & -15 & 2 & 0 & 0 & 0 & 0\\ 0& 0 & 20 & -12 & 2 & 0 & 0 & 0\\ 0& 0 & 0 & 12 & -9 & 2 & 0 & 0\\ 0& 0 & 0 & 0 & 6 & -6 & 2 & 0\\ 0& 0 & 0 & 0 & 0 & 2 & -3 & 2\end{bmatrix}* \begin{bmatrix} a_7 \\ a_6 \\ a_5 \\ a_4 \\ a_3\\ a_2\\ a_1\\ a_0\end{bmatrix}= \begin{bmatrix} \frac{1}{896ln(10)}\\ -\frac{1}{384ln(10)}\\ \frac{1}{160ln(10)}\\ -\frac{1}{64ln(10)}\\ \frac{1}{24ln(10)}\\ -\frac{1}{8ln(10)}\\ \frac{1}{2ln(10)}\\ log(2)\end{bmatrix}\!$$

Solving:

$$a_7=.0002, a_6=.0020, a_5=.0141, a_4=.0725, a_3=.3034, a_2=.9029, a_1=1.9072, a_0=2.1084\!$$

So the particular solution $$y_{p7}\!$$ is

$$y_{p7}=2.1084+1.9072*(x-1)+.9029*(x-1)^2+.3034*(x-1)^3+.0725*(x-1)^4+.0141*(x-1)^5+.0020*(x-1)^6+.0002*(x-1)^7\!$$

We can now find the general solution for n=7, $$y_7(x)\!$$.

$$y_7(x)=y_h(x)+y_{p7}(x)\!$$

$$y_7(x)=c_1e^x+c_2e^{2x}+2.1084+1.9072*(x-1)+.9029*(x-1)^2+\!$$ $$.3034*(x-1)^3+.0725*(x-1)^4+.0141*(x-1)^5+.0020*(x-1)^6+.0002*(x-1)^7\!$$

Solving using our initial conditions yields

$$y_7(x)=-.7271e^x+.0233e^{2x}+2.1084+1.9072*(x-1)+\!$$ $$.9029*(x-1)^2+.3034*(x-1)^3+.0725*(x-1)^4+.0141*(x-1)^5+\!$$ $$.0020*(x-1)^6+.0002*(x-1)^7\!$$

General Solution, n=11
Using the taylor series approximation from earlier with $$n=11\!$$ we have

$$r_{11}(x)=log(2)+\frac{(x-1)}{2ln(10)}-\frac{(x-1)^2}{8ln(10)}+\frac{(x-1)^3}{24ln(10)}-\frac{(x-1)^4}{64ln(10)}+\frac{(x-1)^5}{160ln(10)}-\frac{(x-1)^6}{384ln(10)}+\!$$

$$\frac{(x-1)^7}{896ln(10)}-\frac{(x-1)^8}{2048ln(10)}+\frac{(x-1)^9}{4608ln(10)}-\frac{(x-1)^{10}}{10240ln(10)}+\frac{(x-1)^{11}}{22528ln(10)}\!$$

Finally, we write out the matrix equation for n=11:

$$A* \begin{bmatrix} a_{11}\\ a_{10}\\ a_9\\ a_8\\ a_7 \\ a_6 \\ a_5 \\ a_4 \\ a_3\\ a_2\\ a_1\\ a_0\end{bmatrix}= \begin{bmatrix} \frac{1}{22528ln(10)}\\ -\frac{1}{10240ln(10)}\\ \frac{1}{4608ln(10)}\\ -\frac{1}{2048ln(10)}\\ \frac{1}{896ln(10)}\\ -\frac{1}{384ln(10)}\\ \frac{1}{160ln(10)}\\ -\frac{1}{64ln(10)}\\ \frac{1}{24ln(10)}\\ -\frac{1}{8ln(10)}\\ \frac{1}{2ln(10)}\\ log(2)\end{bmatrix}\!$$

Solving the system in matlab:

$$a_{11}=0, a_{10}=.0002, a_9=.0019, a_8=.0181, a_7=.15, a_6=1.0675, a_5=6.4597,\!$$

$$a_4=32.4318, a_3=130.0033, a_2=390.3968, a_1=781.289, a_0=781.6873\!$$

So the particular solution $$y_{p11}\!$$ is

$$y_{p11}=781.6873+781.289*(x-1)+390.3968*(x-1)^2+130.0033*(x-1)^3+32.4318*(x-1)^4+\!$$

$$6.4597*(x-1)^5+1.0675*(x-1)^6+.15*(x-1)^7+.0181*(x-1)^8+.0019*(x-1)^9+.0002*(x-1)^{10}\!$$

We can now find the general solution for n=11, $$y_11(x)\!$$.

$$y_{11}(x)=y_h(x)+y_{p11}(x)\!$$

$$y_{11}(x)=c_1e^x+c_2e^{2x}+781.6873+781.289*(x-1)\!$$ $$+390.3968*(x-1)^2+130.0033*(x-1)^3+32.4318*(x-1)^4\!$$ $$6.4597*(x-1)^5+1.0675*(x-1)^6+.15*(x-1)^7+.0181*(x-1)^8+.0019*(x-1)^9+.0002*(x-1)^{10}\!$$

Solving using our initial conditions yields

$$y_{11}(x)=-287.5907e^x+.05e^{2x}+781.6873+781.289*(x-1)\!$$ $$+390.3968*(x-1)^2+130.0033*(x-1)^3+32.4318*(x-1)^4\!$$ $$6.4597*(x-1)^5+1.0675*(x-1)^6+\!$$ $$.15*(x-1)^7+.0181*(x-1)^8+.0019*(x-1)^9+.0002*(x-1)^{10}\!$$

Plot
$$y_4\!$$ shown in red

$$y_7\!$$ shown in blue

$$y_{11}\!$$ shown in green

Part 4
Solved by Luca Imponenti

Use the matlab command ode45 to integrate numerically $$y''-3y'+2y=r(x)\!$$ with $$r(x)=log(1+x)\!$$

and the initial conditions from Part 3 to obtain the numerical solution for y(x).

Plot y(x) in the same figure as above.

Matlab Solution
The numerical solution calculated using the matlab ode45 command is shown below:

ans = 0.2788   0.2854    0.2923    0.2997    0.3074    0.3229    0.3401    0.3592    0.3804    0.4040    0.4302    0.4595    0.4921    0.5285    0.5691    0.6145    0.6651    0.7218    0.7850    0.8557    0.9346    1.0228    1.1213    1.2313    1.3542    1.4914    1.6445    1.8155    2.0063    2.2193    2.4569    2.7219    3.0175    3.3471    3.7146    4.1243    4.5809    5.0898    5.6568    6.2885    6.9921    7.3442    7.7142    8.1032    8.5119

Plot
Plotting the aboved vector of y-values,along with the results from earlier yields the following graph:

where the answer calculated in matlab is shown in yellow.

Egm4313.s12.team11.imponenti (talk) 08:04, 14 March 2012 (UTC)


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Nuclear power greener/A Critique of Nuclear Power in the Fuel and Energy Complex of Ukraine

(Review Paper) Cited in

Points Made
This article studies the condition and outlooks of Ukrainian nuclear-power energy 25 years after the accident on the Chernobyl NPS and the recent accidents at Fukushima nuclear plant.

Methods
Research is being conducted on the following areas: •Prospective deep burial place for wastes •Metallic corps—the critical element of a nuclear reactor which determines term of exploitation and expiration dates. Experimental research is performed by developed countries to estimate the state of the metallic corps and predict terms of exploitation •Main ways of nuclear power developments are analyzed •Issues related to Nuclear Fuel Exhaust (NFE) are analyzed •Problems dealing with radioactive waste

Results
•Experimental research proved that the exploitation term of the metallic corps can be prolonged from 10 to 20 years. Ukraine would benefit if its corps of reactors had the same characteristic to allow for the prolongation of exploitation.

•Some of the main ways that Ukraine should implement for its nuclear power developments were identified to be the following -In order to increase the level of nuclear safety, reconstruction and modernization of its operating units is needed -An extension of exploitation of operation power units -Action of new reactor options -Ukraine should create its own NFC -Infrastructure is needed to deal with radioactive waste -Creation of an ecologically safe system -Development and production of systems that deal with atomic-industrial complexities

•In order for Chernobyl NPS to improve its growth and success in the future implementation of the steps to improve usage time and plant development is needed.


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User:Egm4313.s12.team1.armanious/scrap

Statement
Find the scalar product , the magnitude of f and g, and the angle between f and g for (see R7.3 Lect. 11-1 pg. 8): 1) $$f(x)=\cos x, g(x)=x, \text{for} -2\leq x \leq 10\!$$ 2) $$f(x)=\frac12(3x^2-1), g(x)=\frac12(5x^3-3x), \text{for} -1\leq x \leq +1\!$$

Solution
The scalar product, the function equivalent of the vector dot product, of two functions can be found in the following way:
 * {| style="width:100%" border="0"

$$  \displaystyle :=\int^b_af(x)g(x)dx $$     (3.0) The magnitude of a function is defined as the square root of the scalar product between the function and itself:
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$$  \displaystyle ||f||:=^{1/2}=\left [\int^b_af^2(x)dx \right ]^{1/2} $$     (3.1) The cosine of the angle between two functions can be found in the following way:
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$$  \displaystyle \cos \theta = \frac{}{||f||||g||} $$     (3.2) Note that these equations are very similar to their vector counterparts. Part 1
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$$  \displaystyle f(x)=\cos x, g(x)=x, \text{for} -2\leq x \leq 10 $$     (3.3) The scalar product of f and g can be found using (3.0):
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$$  \displaystyle =\int_{-2}^{10}x\cos (x) dx=\left [ x\sin(x)+cos(x) \right ]_{-2}^{10} $$     (3.4) Solving this yields the scalar product:
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$$  \displaystyle (10\sin10+\cos10)-(-2\sin(-2)+\cos(-2))=-7.68 $$     (3.5) To find the magnitude of f, the scalar product of f with itself must first be found:
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$$  \displaystyle =\int_{-2}^{10}\cos^{2}(x)dx=\int_{-2}^{10}\left (\frac12\cos(2x)+\frac12 \right )dx=\left [ \frac14\sin(2x)+\frac12x \right ]_{-2}^{10} $$     (3.6) Solving this yields:
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$$  \displaystyle \left [ \frac14\sin(2x)+\frac12x \right ]_{-2}^{10}=\left ( \frac14\sin(20)+5 \right )-\left ( \frac14\sin(-4)-1 \right )=6.04 $$     (3.7) Therefore the magnitude of f is found to be:
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$$  \displaystyle $$     (3.8) A similar approach can be taken for the magnitude of g:
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 * f||=^{1/2}=\sqrt{6.04}=2.46
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$$  \displaystyle <g,g>=\int_{-2}^{10}x^2dx=\left [ \frac{x^3}{3} \right ]_{-2}^{10} $$     (3.9)
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$$  \displaystyle \left [ \frac{x^3}{3} \right ]_{-2}^{10}=\frac{1000}3-\frac{-8}{3}=\frac{1008}{3}=336 $$     (3.10)
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$$  \displaystyle $$     (3.11) Using the above information with (3.2), the cosine of the angle between f and g is found to be
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 * g||=<g,g>^{1/2}=\sqrt{336}=18.33
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$$  \displaystyle \cos \theta = \frac{<f,g>}{||f||||g||}=\frac{-7.68}{(2.46)(18.33)}=-0.1705 $$     (3.12) Therefore, the angle between the functions f and g is
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$$  \displaystyle \theta =\cos^{-1}(-0.1705)=1.74rad=99.8^{\circ} $$     (3.13) Part 2
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$$  \displaystyle f(x)=\frac12(3x^2-1), g(x)=\frac12(5x^3-3x), \text{for} -1\leq x \leq +1 $$     (3.14) The scalar product of f and g can be found using (3.0):
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$$  \displaystyle <f,g>=\int_{-1}^{1}\frac14\left ( 3x^2-1 \right )\left ( 5x^3-3x \right ) dx=\frac14\int_{-1}^{1}\left ( 15x^5-14x^3+3x \right ) dx=\frac14\left [ \frac{15}{6}x^6-\frac{14}{4}x^4+\frac32x^2 \right ]_{-1}^{1} $$     (3.15) Solving this yields the scalar product:
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$$  \displaystyle \frac14\left [ \frac52x^6-\frac72x^4+\frac32x^2 \right ]_{-1}^{1}=\frac14\left ( \frac12-\frac12 \right )=0 $$     (3.16) To find the magnitude of f, the scalar product of f with itself must first be found:
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$$  \displaystyle <f,f>=\int_{-1}^{1}\left (\frac12\left (3x^2-1 \right )\right)^2 dx=\frac14\int_{-1}^{1}\left (9x^4-6x^2+1  \right )dx=\frac14\left [ \frac95x^5-2x^3+x \right ]_{-1}^{1} $$     (3.17) Solving this yields:
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$$  \displaystyle \frac14\left [ \left( \frac95-2+1\right )-\left (-\frac95+2-1 \right ) \right ] =\frac14\left ( \frac45+\frac45 \right )=\frac25 $$     (3.18) Therefore the magnitude of f is found to be:
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$$  \displaystyle $$     (3.19) A similar approach can be taken for the magnitude of g:
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 * f||=<f,f>^{1/2}=\sqrt{0.4}=0.632
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$$  \displaystyle <g,g>=\int_{-1}^{1}\left (\frac12\left (5x^3-3x \right )\right)^2 dx=\frac14\int_{-1}^{1}\left (25x^6-15x^4+9x^2  \right )dx=\frac14\left [ \frac{25}{7}x^7-3x^5+3x^3 \right ]_{-1}^{1} $$     (3.20)
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$$  \displaystyle \frac12 \left (\frac{25}{7}-3+3 \right ) =\frac{25}{14} $$     (3.21)
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$$  \displaystyle $$     (3.22) Using the above information with (3.2), the cosine of the angle between f and g is found to be
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 * g||=<g,g>^{1/2}=\sqrt{\frac{25}{14}}=1.336
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$$  \displaystyle \cos \theta = \frac{<f,g>}{||f||||g||}=\frac{0}{(0.632)(1.336)}=0 $$     (3.23) '''Because the scalar product of the two functions is zero, the two functions are orthogonal. That is:'''
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$$  \displaystyle \theta =\frac{\pi}{2}rad=90^{\circ} $$     (3.24)
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Author
Solved and Typed By - Egm4313.s12.team1.armanious (talk) 23:48, 21 April 2012 (UTC) Reviewed By -

title: User:Egm4313.s12.team1.armanious/scrap


text:
User:MechanusSunrise

I have a B.A. in Anthropology focusing in sociocultural anthropology.

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Licensing
for school project

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University of Florida/Egm4313/s12 Report 4, Problem 4.3

Problem Statement
Given: $$y''-3y'+2y=r(x)\!$$

where $$r(x)=\log (1+x)\!$$

With initial conditions: $$ y(-\frac{3}{4}) = 1, y'(-\frac{3}{4}) = 0 \!$$ Find the overall solution $$ y_{n}(x) \!$$ for $$ n = 4,7,11 \!$$ and plot these solutions on the interval from $$ [-\frac{3}{4},3] \!$$

Solution
First we find the homogeneous solution to the ODE: The characteristic equation is: $$\lambda^2-3\lambda+2=0\!$$ $$(\lambda-2)(\lambda-1)=0\!$$ Then, $$\lambda=1,2\!$$ Therefore the homogeneous solution is: $$y_h=C_1e^(2x)+c_2e^x\!$$

Now to find the particulate solution For n=4

$$ r(x)=\sum_{0}^{n}-\frac{(-1)^nx^n}{n\ln (10)} \!$$

$$ r(x)=\frac{x}{\ln(10)}-\frac{x^2}{2\ln(10)}+\frac{x^3}{3\ln(10)}-\frac{x^4}{4\ln(10)} \!$$

We can then use a matrix to organize the known coefficients:

$$ \begin{bmatrix} 2 &-3  &2  &0  & 0\\  0&  2  &-6  &6  & 0\\  0&  0&  2  &-9  &12  \\  0&  0&  0&  2  &-12\\  0&  0&  0& 0 &  2  \\   \end{bmatrix} \begin{bmatrix} K_{0} \\ K_{1} \\ K_{2} \\ K_{3} \\ K_{4} \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{1}{ln(10)} \\ \frac{1}{2ln(10)} \\ \frac{1}{3ln(10)} \\ \frac{1}{4ln(10)} \end{bmatrix}\!$$ Then, using MATLAB and the backlash operator we can solve for these unknowns: Therefore $$ y_{p4}=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4 \!$$ Superposing the homogeneous and particulate solution we get $$ y_{n}=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4+C_1e^{2x}+C_2e^{x} \!$$ Differentiating: $$y'_{n}=3.7458+3.1486x+1.1943x^2+0.2172x^3+2C_1e^{2x}+C_2e^{x}\!$$ Evaluating at the initial conditions: $$y(-0.75)=0.9698261719+0.231301601C_!+0.4723665527C_2=1\!$$ $$y'(-0.75)=1.9645125+0.4462603203C_1+0.4723665527C_2\!$$ We obtain: $$ C_1=-4.46\!$$ $$ C_2=0.055 \!$$ Finally we have: $$ y_{n}=4.0444+3.7458x+1.5743x^2+0.3981x^3+0.0543x^4-4.46e^{2x}+0.055e^{x} \!$$

For n=7

$$ r(x)=\sum_{0}^{n}-\frac{(-1)^nx^n}{n\ln (10)} \!$$

$$ r(x)=\frac{x}{\ln(10)}-\frac{x^2}{2\ln(10)}+\frac{x^3}{3\ln(10)}-\frac{x^4}{4\ln(10)}+\frac{x^5}{5\ln(10)}-\frac{x^6}{6\ln(10)}+\frac{x^7}{7\ln(10)} \!$$

We can then use a matrix to organize the known coefficients:

$$ \begin{bmatrix} 2 &-3  &2  &0  & 0 &0 &0 &0\\  0&  2  &-6  &6  & 0 &0 &0 &0\\  0&  0&  2  &-9  &12  &0 &0 &0\\  0&  0&  0&  2  &-12 &20 &0 &0 \\  0&  0&  0& 0 &  2 &-15 &30 &0\\ 0 &0 &0 &0 &0 &2 &-18 &42\\ 0 &0 &0 &0 &0 &0  &2 &-21\\ 0 &0 &0 &0 &0 &0 &0 &2\\    \end{bmatrix} \begin{bmatrix} K_{0} \\ K_{1} \\ K_{2} \\ K_{3} \\ K_{4} \\ K_{5} \\ K_{6} \\ K_{7} \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{1}{ln(10)} \\ \frac{1}{2ln(10)} \\ \frac{1}{3ln(10)} \\ \frac{1}{4ln(10)} \\ \frac{1}{5ln(10)} \\ \frac{1}{6ln(10)} \\ \frac{1}{7ln(10)} \end{bmatrix}\!$$ Then, using MATLAB and the backlash operator we can solve for these unknowns: Therefore $$ y_{p7}=377.4833+375.3933x+185.6066x^2+60.5479x^3+14.4946x^4+2.6492x^5+0.3619x^6+0.0310x^7 \!$$ Superposing the homogeneous and particulate solution we get $$ y_{n}=377.4833+375.3933x+185.6066x^2+60.5479x^3+14.4946x^4+2.6492x^5+0.3619x^6+0.0310x^7+C_1e^(2x)+c_2e^x \!$$ Differentiating: $$y'_{n}=375.3933+371.213x+181.644x^2+57.9784x^3+13.46x^4+2.1714x^5+0.214x^6+2C_1e^(2x)+C_2e^x\!$$ Evaluating at the initial conditions: $$y(-0.75)=178.816+0.2231301601C_!+0.4723665527C_2=1\!$$ $$y'(-0.75)=178.413+0.4462603203C_1+0.4723665527C_2\!$$ We obtain: $$ C_1=-2.6757\!$$ $$ C_2=-375.173 \!$$


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User:Egm4313.s12.team11.arrieta/Report 4

Part 1
Solved by Francisco Arrieta

Problem Statement

 * Develop $$ \log(x+1) \!$$ in Taylor Series about $$ x=0 \!$$ to reproduce the figure on page 7-25

Given

 * $$ f(x)= \sum_{n=0}^\infty \frac{f^{(n)}(\hat x)}{n!}(x- \hat x)^n \!$$

Solution
For n=4:
 * $$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)} \!$$

For n=7:
 * $$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)} \!$$

For n=11:
 * $$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})} \!$$

For n=16:
 * $$ \log(x+1) = \frac{x}{\log(10)}-\frac{x^2}{\log(10^2)}+\frac{x^3}{\log(10^3)}-\frac{x^4}{\log(10^4)}+\frac{x^5}{\log(10^5)}-\frac{x^6}{\log(10^6)}+\frac{x^7}{\log(10^7)}-\frac{x^8}{\log(10^8)}+\frac{x^9}{\log(10^9)}-\frac{x^{10}}{\log(10^{10})}+\frac{x^{11}}{\log(10^{11})}\!$$
 * $$ -\frac{x^{12}}{\log(10^{12})}+\frac{x^{13}}{\log(10^{13})}-\frac{x^{14}}{\log(10^{14})}+\frac{x^{15}}{\log(10^{15})}-\frac{x^{16}}{\log(10^{16})} \!$$

Using MATLAB

--Egm4313.s12.team11.arrieta (talk) 00:44, 14 March 2012 (UTC)


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Summary
n=11 matrixes for R4.3 part 2. Egm4313.s12.team5.north (talk) 00:29, 14 March 2012 (UTC)John NorthEgm4313.s12.team5.north (talk) 00:29, 14 March 2012 (UTC)

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Talk:Dominant group/Metagenome/Term test

Wikipedia "dominant group" examples
20. "The colt more than lived up to the lofty expectations on the Rowley Mile by delivering one of the most dominant Group One performances in racing history." from the entry Frankel (horse). The sentence using dominant group was deleted from the entry. --Marshallsumter (talk) 20:35, 29 July 2012 (UTC)
 * 1) "Dominant group" is a technical term with specific meanings that include "majority" or "plurality" among several others. None of these meanings is clearly noted. Without a citation to the primary literature this example is either original research, a guess, or plagiarism. Plagiarism is unlikely as no source is found on Google scholar.
 * 2) As a source in the primary literature exists, this is likely plagiarism. The entry Euenantiornithes has been redirected to Enantiornithes and the sentence containing dominant group was not carried over to second entry. --Marshallsumter (talk) 20:20, 29 July 2012 (UTC)
 * 3) This example is straight from Simpson's contribution to the modern synthesis, without citation, therefore, plagiarism. Marshallsumter (talk) 00:34, 14 March 2012 (UTC)
 * 4) This example is okay because the sentence is followed by citations. If "dominant group" is not used in the sources, this might be misuse but not plagiarism or copyright violation.
 * 5) Okay because the sentence is followed by citations.
 * 6) No citations, probably original research.
 * 7) Okay, citations provided.
 * 8) Probably original research, no citations. The term was put into the article at its creation on January 10, 2005. Only one article on Google scholar predates and includes similar usage. Looks more like plagiarism or possible copyright violation.  --Marshallsumter (talk) 23:16, 11 July 2012 (UTC)


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User:Egm4313.s12.team8.koester/R4.4.2

Problem Statement
Develop $$ \displaystyle log(1+x)$$ in a Taylor Series about â=1.

Compute for n= 4, 7, 11.

Plot these three functions along with the real solution.

Find the domain of convergence.

Solution
Taylor Series expansions follow the form:
 * $$ \ \sum_{n = 0}^{\infty} \frac { f^n(a) (x-a)^n }{n!}\ $$

For our problem this translates to:
 * $$ \ \sum_{n = 1}^{\infty} (-1)^{n+1}\frac { (x-1)^n }{n}\ $$

We start the summation at 1 because log(0) is not defined.

For n=4 terms this expands to:
 * $$ (x-1)- \frac {(x-1)^2}{2} + \frac {(x-1)^3}{3} - \frac {(x-1)^4}{4} $$

For n=7 terms this expands to:
 * $$ (x-1)- \frac {(x-1)^2}{2} + \frac {(x-1)^3}{3} - \frac {(x-1)^4}{4} + \frac {(x-1)^5}{5} - \frac {(x-1)^6}{6} + \frac {(x-1)^7}{7}  $$

For n=11 terms this expands to:
 * $$ (x-1)- \frac {(x-1)^2}{2} + \frac {(x-1)^3}{3} - \frac {(x-1)^4}{4} + \frac {(x-1)^5}{5} - \frac {(x-1)^6}{6} + \frac {(x-1)^7}{7} - \frac {(x-1)^8}{8} + \frac {(x-1)^9}{9} - \frac {(x-1)^{10}}{10} + \frac {(x-1)^{11}}{11} $$

Graph
Now these 3 functions will be plotted against the actual solution:

It can then be found from observation of the graph that the domain of convergence is (0,2).

title: User:Egm4313.s12.team8.koester/R4.4.2


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UCSportStudies

Welcome
Welcome to the Wikiversity Page for the Discipline of Sport Studies at the University of Canberra.

About Sport Studies at the University of Canberra
Sport Studies offers educational programs and research projects in a wide range of sport related fields.

For comprehensive information about Sport Studies at the University of Canberra see this Wikiversity page.

Ben Rattray

Ben's Wikiversity page is here.

Information about Ben's research output can be found here.

Keith Lyons

Keith Lyons was appointed Director of the National Institute of Sport Studies at the University of Canberra (UCNISS) in March 2009. He has a particular interest in the use of educational technology in sport and physical education.

Information about Keith's work can be found here.

Information about Keith's publications and presentations can be found here.

In Semester 1, 2012, Keith coordinated and taught the Sport Coaching Pedagogy unit. SlideShare resources for the unit can be found here. In Semester 2, 2011, Keith taught the Business, Politics and Sport unit. SlideShare resources for the unit can be found here.

title: UCSportStudies


text:
Nuclear power greener/A critique of Unsafe at Any Dose

(Review Paper) Cited in

NO Radiation is okay radiation
Doctors opinions are not as well known in the nuclear power industry because generally they do not get as much or even no attention from congress on the issue. Helen Caldicott focuses on dangers that we are not aware of nor do we see yet from the impacts of nuclear radiation. Many physicists advocate for nuclear power because they focus on the external less harmful sources of radiation. Physicists also do not acknowledge that radiation is cumulative and the harmful effects may not surface until later in life or generations later. There are many long term effects that will take generations to appear from incidents such as Chernobyl or Fukushima.

Rational
Helen Caldicott is a doctor advocating that no exposure to radiation is the only safe method for avoiding the short and long term effects of radiation exposure. After Chernobyl many advocates for nuclear power argued that there were few deaths afterwards and few genetic abnormalities in offspring from those who survived the incident. The International Atomic Energy Agency said there were 4000 deaths from cancer where as a report published in 2009 by the New York Academy of Sciences said there were almost one million deaths due to cancer or other diseases in the aftermath of Chernobyl. Quite a large difference??? The high doses of radiation from Chernobyl caused many miscarriages which also means that the accounted offspring of the survivors that appeared to have no genetic abnormalities is not a reliable source of data because we are not aware of what genetic disorders or abnormalities of the miscarried fetuses. Also most mutations that are caused by radiation are recessive genes therefore it could take many generations before two recessive genes come together to form a child that has a certain disease. We are not seeing the effects of Chernobyl or Fukushima yet and probably won't see the effects until many generations down the road. Right now there are 2,600 genetic diseases on record that could be caused by radiation induced mutation. This number will rise, especially because we continue to increase the artificial exposure we are exposed to. What physicists fail to recognize is that radiation is cumulative and effects do not appear immediately. It takes years for certain types of cancers to surface, for example it takes 5 to 10 years for Leukemia to emerge and other types of cancer can take anywhere from 15 to 60 years to emerge.

Results
Helen Caldicott states that "nuclear power is neither clean, nor suitable, nor an alternative to fossil fuels". She continues this with stating that it contributes greatly to global warming. In conclusion the arguments presented by physicists for nuclear power do not account for the damages of radiation that we missed seeing due to the high number of miscarriages after Chernobyl as well as does not account for the damages that we have not witnessed yet due to the recessive nature of the radiation - induced genes, making it a process that takes generations before all the damaging and severely life threatening effects will showcase. Nuclear power is unnecessary being that we can meet all of our energy needs through solar, wind and geothermal energy as well as conservation. Nuclear power is only putting lives now and lives generations from now at risk of many genetic diseases. There is no good amount of radiation exposure, especially considering any exposure is cumulative throughout a lifespan.


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Controversies in Science/Are humans causing global warming/A critique of A mathematical analysis of the divergence problem in dendroclimatology

(Review Paper) Cited in

Points Made
Divergences appear across much of the upper northern hemisphere but are not limited to there. The growth of trees reduced in response to warmer temperatures in Alaska after 1950, Canada and Siberia after 1970. It was also found that in tundra areas the photosynthesis rate increased and in the forest areas the photosynthesis rate decreased in a study that was done between 1982 and 2003. The decrease was most likely due to moisture limitations at higher temperatures. Another point that was made is that trees may experience a decrease sensitivity to temperature or a negative response to warming because of reasons such as draught, frost damage, seasonal temperature patterns change, or a change in snow pack. The decreased sensitivity changes the growth rate, this cannot be automatically associated with the climate changes.

Methods
Using a mathematical equation of r=mT+b (r=ring property, T=temperature, m&b = parameters) would provide a completely linear line on a graph, which is what has been assumed from the short calibration period they have to work with. However Loehle says there is no way this is possible! The graph would not be linear if all data was used. There are other contributing factors that play a role in the radius growth between tree rings from decade to decade. The mathematical equation they used to prove the linear line is neither reliable nor accurate.

Results
This proves that there are other environmental factors outside of climate changes that effect tree growth. The other factors are things such as draught, frost damage, change in seasonal temperature patterns, and change in snow packs. We cannot assume that an equation built off a short period of time equates to a linear line for decades to come. The equation is misleading. Environmental factors play a huge role in the diversions that occur within tree growth.

 * This is a very good point. Archaeologists use dendrochronology to look at past environments, however it is a vague measurement. To conclude something like humans causing global warming, there would need to be much more evidence than just analyzing trees. Cthom4751 (talk) 17:28, 13 April 2012 (UTC)


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User:Intercog

My name is Jon Mason & I am the director & principal consultant for InterCog Pty Ltd, a consulting company based in Australia. I am exploring the potential of Wikiuniversity as a space for program development within the Centre for School Leadership & Learning Development at Charles Darwin University.

Contact Details
jon [at] intercog [dot] net

Projects

 * Mobile Education

title: User:Intercog


text:
Mobile Education

The aim of this project is to gather relevant resources & identify useful activities associated with mobile education that will help guide school leaders in utilising mobile technologies for educational programs.

A distinction is made here between learning and education. While both terms share much in common, education typically implies the delivery of (organizational perspective) or access to (individual perspective) formally developed and accredited courses, ultimately for formal recognition and qualification. Learning, on the other hand, truly takes place anywhere and at any time. The methods and issues that affect learning do of course inform this project and therefore the considerations of people working in formal education spaces, such as school/college or university students, teachers or administrators.

Topics

 * 1) /Terminology/
 * 2) /Case Studies/
 * 3) /Bring Your Own Device (BYOD)/
 * 4) /Issues/
 * 5) /Policies/
 * 6) /Quotes/

Further Inquiry

 * Stanford Mobile Inquiry-based Learning Environment (SMILE)
 * Shih, J. L., Chuang, C. W., & Hwang, G. J. (2010). An Inquiry-based Mobile Learning Approach to Enhancing Social Science Learning Effectiveness. Educational Technology & Society, 13 (4), 50–62.International Forum of Educational Technology & Society (IFETS).

title: Mobile Education


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Mobile Education/Quotes

A society which is mobile, which is full of channels for the distribution of a change occurring anywhere, must see to it that its members are educated to personal initiative and adaptability. Otherwise, they will be overwhelmed by the changes in which they are caught and whose significance or connections they do not perceive. John Dewey, 1916. Democarcy and Education

title: Mobile Education/Quotes