University of Florida/Egm4313/s12.team5.R5

Question
Find $$ R_c $$ for the following series: 1. $$ r(x)=\sum_{k=0}^\infty(k+1)kx^k $$ 2. $$ r(x)=\sum_{k=0}^\infty \frac{(-1)^k}{\gamma^k}x^{2k} $$ $$ \gamma= \text{constant} $$ Use (2)-(3)p.7-31 to find $$ R_c $$ for the Taylor series of 3. $$ sinx   $$ at $$  \hat x=0 $$ 4. $$ log(1+x)  $$ at  $$ \hat x=0 $$ 5. $$ log(1+x)  $$ at  $$ \hat x=1 $$

Solution
1. $$ r(x)=\sum_{k=0}^\infty(k+1)kx^k = 0 + 2x + 6x^2... $$ Using L'Hospital's rule and (2)p.7-31, and setting $$ d_k=(k+1)k $$: $$ R_c = \left [\lim_{k \to \infty } \left|\frac{(k+2)(k+1)}{(k+1)k}\right|\right]^{-1} = \left [\lim_{k \to \infty } \left|\frac{k+2}{k}\right|\right]^{-1} = \left [\lim_{k \to \infty } \left|\frac{1}{1}\right|\right]^{-1} = 1 $$

2.$$ r(x)=\sum_{k=0}^\infty \frac{(-1)^k}{\gamma^k}x^{2k} = 1 - \frac{1}{\gamma}x^2 + \frac{1}{\gamma^2}x^4... $$ Using (3)p.7-31 and setting $$ d_k=\frac{(-1)^k}{\gamma^k} $$: $$ R_c= \lim_{k \to \infty} \left[ \sqrt[k] {\left|\frac{(-1)^k}{\gamma^k} \right|} \right]^{-1}=\lim_{k \to \infty} \frac{\sqrt[k]{|\gamma^k}|}{\sqrt[k]{|(-1)^k|}} = \lim_{k \to \infty} \frac{\sqrt{|\gamma|}}{\sqrt{|-1|}} = \sqrt{|\gamma|}$$

3. Taylor series about $$ \hat x = 0 $$: $$ sinx = x + \frac{1}{6}x^3 + \frac{1}{120}x^5... \Rightarrow \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}x^{2k+1} $$ Using (2)p.7-31 and setting $$ d_k=\frac{(-1)^k}{(2k+1)!} $$: $$ R_c=\left [\lim_{k \to \infty } \left|\frac{\frac{(-1)^{k+1}}{(2(k+1)+1)!}}{\frac{(-1)^k}{(2k+1)!}}\right|\right]^{-1} = \left[\lim_{k \to \infty}\left |\frac{-1}{(2k+3)(2k+2)} \right|\right]^{-1} = \infty $$

4. Taylor series about $$ \hat x = 0 $$: $$log(1+x)= 0 + x - \frac{x^2}{2} + \frac{x^3}{3}... \Rightarrow \sum_{n = 0}^{\infty }\frac{(-1)^k (x)^{k+1} }{(k+1)} $$ Using L'Hospital's rule and (2)p.7-31, and setting $$ d_k=\frac{(-1)^k}{(k+1)} $$: $$ R_c=\left [\lim_{k \to \infty } \left| \frac{\frac{(-1)^{k+1}}{(k+2)}}{\frac{(-1)^k}{(k+1)}} \right|\right]^{-1} = \left [\lim_{k \to \infty } \left| \frac{(-1)(k+1)}{(k+2)} \right|\right]^{-1} = \left [\lim_{k \to \infty } \left| \frac{-1}{1} \right|\right]^{-1} = 1 $$

5. Taylor series about $$ \hat x=1 $$: $$ log (1+x)=log(2) + \frac{(x-1)}{2} - \frac{(x-1)^2}{8} + \frac{(x-1)^3}{24}... \Rightarrow log(2) + \sum_{n = 0}^{\infty } \frac{(-1)^{k+1}(x-1)^k }{k2^k} $$ Using L'Hospital's rule and (2)p.7-31, and setting $$ d_k=\frac{(-1)^{k+1}}{k2^k} $$: $$ R_c=\left [\lim_{k \to \infty } \left| \frac{\frac{(-1)^{k+2}}{(k+1)2^{k+1}}}{\frac{(-1)^{k+1}}{k2^k}} \right|\right]^{-1} = \left [\lim_{k \to \infty } \left| \frac{-1k}{2(k+1)} \right|\right]^{-1} = \left [\lim_{k \to \infty } \left| \frac{-1}{2} \right| \right]^{-1} = 2 $$

Author
This problem is solved and uploaded by Radina Dikova

Question
Determine whether the following pairs of functions are linearly independent:

$$ \displaystyle \ 1. f(x) = x^2, g(x) = x^4\ $$

$$ \displaystyle \ 2. f(x) = \cos(x), g(x) = \sin(3x)\ $$

First use the Wronskian method, then use the Gramian method.

Solution
The Wronskian is defined as: $$ \displaystyle \ W(f,g) := \det \begin{bmatrix} f &g \\ f' & g' \end{bmatrix} = fg' -gf'\ $$

If $$ \displaystyle \ W(f,g) \neq 0\ $$, then the functions f and g are linearly independent.

The Gramian is defined as: $$ \displaystyle \ \Gamma (f,g) : = \det \begin{bmatrix}  & \\ &  \end{bmatrix}\ $$

Where $$ \displaystyle \  := \int_{a}^{b} f(x)g(x)dx\ $$

If $$ \displaystyle \ \Gamma(f,g) \neq 0\ $$, then the functions f and g are linearly independent.

1. Using Wronskian.

$$ \displaystyle \ W(f,g) = fg' - gf' = x^2 (4x^3) - x^4 (2x) = 4x^5 - 2x^5 = 2x^5 \neq 0 $$

Therefore, f and g are linearly independent.

2. Using Wronskian.

$$ \displaystyle \ W(f,g) = fg' - gf' = \cos(x) (3\cos(3x)) - \sin(3x) (-\sin(x)) \neq 0 $$

Therefore, f and g are linearly independent.

1. Using Gramian with an interval of [-1,1]

$$ \displaystyle \  = \int_{-1}^{1} x^2(x^2)dx = \frac{1}{5} x^5 \mid_{-1}^{1} = \frac{2}{5}\ $$

$$ \displaystyle \  =  = \int_{-1}^{1} x^2(x^4)dx = \frac{1}{7} x^7 \mid_{-1}^{1} = \frac{2}{7}\ $$

$$ \displaystyle \  = \int_{-1}^{1} x^4(x^4)dx = \frac{1}{9} x^9 \mid_{-1}^{1} = \frac{2}{9}\ $$

$$ \displaystyle \ \Gamma (f,g) =   -   = \frac{2}{5} (\frac{2}{9}) - \frac{2}{7} (\frac{2}{7}) = \frac{4}{45} - \frac{4}{49} \neq 0\ $$

Therefore, f and g are linearly independent.

2. Using Gramian with an interval of [-1,1]

$$ \displaystyle \  = \int_{-1}^{1} \cos(x) (\cos(x)) dx = \int_{-1}^{1} \cos^2(x) dx = \frac{1}{2} \int_{-1}^{1} 1 + \cos(2x) dx = \frac {1}{2} (x + \frac{1}{2} \sin(2x)) \mid_{-1}^{1} = 1.4546\ $$

$$ \displaystyle  =  = \int_{-1}^{1} \sin(3x) \cos(x) dx = 0\ $$

$$ \displaystyle <g,g> = <g,g> = \int_{-1}^{1} \sin^2(3x) dx = \frac{1}{2} \int_{-1}^{1} 1-cos(6x) dx = \frac{1}{2} (x - \frac{1}{6} \sin(6x)) \mid_{-1}{1} = 1.04657\ $$

$$ \displaystyle \ \Gamma(f,g) = <f,f> <g,g> - <f,g> <g,f> = 1.4546 (1.04657 ) - 0(0) \neq 0\ $$

Therefore, f and g are linearly independent.

Author
This problem was solved and uploaded by David Herrick.

Question
Verify that $$b_1$$ and $$b_2$$ in (1)-(2) p.7-34 are linearly independent using the Gramian

Solution
The given Grammian:


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

$$  \displaystyle \Gamma (b_1,b_2) = \begin{bmatrix} \left \langle b_1,b_1 \right \rangle \left \langle b_1,b_2 \right \rangle \\ \left \langle b_2,b_1 \right \rangle \left \langle b_2,b_2 \right \rangle \end{bmatrix} $$
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 * }

In reference to to (3) on p.8-9


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$$  \displaystyle \left \langle b_1,b_1 \right \rangle = (b_1 \cdot b_2) $$
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 * }

Calculating the dot products yields:


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$$  \displaystyle \left \langle b_1,b_1 \right \rangle = 4 + 49 = 53 $$
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 * }


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$$  \displaystyle \left \langle b_1,b_2 \right \rangle = 3 + 21 = 24 $$
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 * }


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$$  \displaystyle \left \langle b_2,b_1 \right \rangle = 3 + 21 = 24 $$
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 * <p style="text-align:right">
 * }


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$$  \displaystyle \left \langle b_2,b_2 \right \rangle = 2.25 + 9 = 11.25 $$
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 * }

Plugging the dot products into the Grammian yields:


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$$  \displaystyle
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\Gamma = 596.25 - 576 = 20.25 \ne 0

$$


 * }

Therefore, b1 and b2 are linearly independent.

Author
Solved and uploaded by Derik Bell

Question
Show that: $$ y_p (x) = \sum_{i=0}^{n} (y_{p,i} (x)) $$ is the particular solution to: $$ y' ' + p(x) y' + q(x) y = r(x) $$ Discuss the choice of particular solutions in the table on p8-3. In other words, for r(x) = kcos(wx), why would you need to have both cos(wx) and sin(wx) in the particular solution?

Solution
For a single excitation that satisfies $$ r_i (x) $$, $$ y_{p,i} ' ' + p(x) y_{p,i} ' + q(x) y_{p,i} = r_i (x) $$ for example: $$ y_{p0} ' ' + p(x) y_{p0} ' + q(x) y_{p0} = r_0 (x) $$ $$ y_{p1} ' ' + p(x) y_{p1} ' + q(x) y_{p1} = r_1 (x) $$ $$ y_{p2} ' ' + p(x) y_{p2} ' + q(x) y_{p2} = r_2 (x) $$ and so on until... $$ y_{p,n} ' ' + p(x) y_{p,n} ' + q(x) y_{p,n} = r_n (x) $$ where by linearity: $$ r(x) = \sum_{i=0}^n r_i (x) $$ Since $$ y_{p,i} $$ is the solution to a single iteration of r(x) and $$ r(x) = \sum_{i=1}^n r_i (x) $$, then by linearity, the solution to r(x) is: $$ y_p (x) = \sum_{i=1}^n y_{p,i} $$ Part 2: kcos(wx) is a periodic function. As shown by (3) on p8-2, any periodic function can be broken down into a fourier trigonometric series: $$ r(x) = a_0 + \sum_{n=1}^\infty [a_n cos(nwx) + b_n sin(nwx)] $$ r(x) can be further broken down as the sum of: $$ r_a=a_0 $$ $$ r_b = \sum_{n=1}^\infty a_n cos(nwx) $$ $$ r_c = \sum_{n=1}^\infty b_n sin(nwx) $$ Where $$ r(x) = r_a + r_b + r_c $$ Since r(x) is expressed in terms of cos(x) and sin(x) the particular solution, which is also a sum of the individual particular solutions for each iteration of $$ r_a $$, $$ r_b $$, and $$ r_c $$, needs to be in terms of sin(x) and cos(x) as well. That applies to all periodic functions as shown on p8-2, which sin(x) is as well. Therefore that justifies why the particular solutions for kcos(wx), ksin(wx), $$ ke^{\alpha x} \cos{wx} $$, and $$ ke^{\alpha x} \sin{wx} $$ must all include both cos(x) and sin(x).

Author
This problem was solved and uploaded by John North.

Question

 * 1. Show that cos(7x) and sin(7x) are linearly independent using the Wronskian and the Gramian (integrate over 1 period)
 * 2. Find 2 equations for the two unknowns M,N and solve for M,N
 * 3. Find the overall solution y(x) that corresponds to the initial condition y(0)=1, y'(0)=0. Plot the solution over 3 periods.

Solution
(1)

First, using Wronskian:


 * For 2 functions, f and g, the Wrosnkian is defined as $$\displaystyle W(f,g):=det\begin{bmatrix}

f &g \\ f'&g' \end{bmatrix}=fg'-gf' $$


 * Where f and g are linearly independent if $$\displaystyle W(f,g)\neq 0 $$


 * For $$\displaystyle f=cos(7x), g=sin(7x),f'=-7sin(7x),g'=7cos(7x) $$


 * Then, $$\displaystyle W(f,g):=det\begin{bmatrix}

cos(7x) &sin(7x) \\ -7sin(7x)&7cos(7x) \end{bmatrix}=7cos^{2}(7x)+7sin^{2}(7x)\neq 0 $$

Therefore, f and g are linearly independent

Second, using Gramian:


 * Consider two functions, f and g, where the scalar product is defined as


 * $$\displaystyle <f,g>:=\int_{a}^{b}f(x)g(x)dx $$


 * And the Gramian defined as


 * $$\displaystyle \Gamma (f,g):=det\begin{bmatrix}

<f,f> &<f,g> \\ <g,f> &<g,g> \end{bmatrix} $$


 * Then f and g are linearly indepdent if $$\displaystyle \Gamma (f,g)\neq 0 $$


 * For f=cos(7x) and g=sin(7x) and integrating over one period ($$\displaystyle \frac{2\pi }{7} $$)


 * $$\displaystyle <f,f>=\int_{0}^{\frac{2\pi }7{}}cos^{2}(7x)dx $$


 * Letting $$\displaystyle u=7x, du=7dx $$ and changing limits of integration by plugging in old limits into "u" equation


 * $$\displaystyle <f,f>=\frac{1}{7}\int_{0}^{2\pi }cos^{2}(u)du=\frac{1}{7}[\frac{u}{2}+\frac{1}{4}sin2u\mathbf{\mid} _{0}^{2\pi } ]=\frac{\pi }{7} $$


 * $$\displaystyle <g,g>=\int_{0}^{\frac{2\pi }{7}}sin^{2}(7x)dx $$


 * Letting $$\displaystyle u=7x,du=7dx $$ and changing the limits of integration by plugging in old limits into "u" equations


 * $$\displaystyle <g,g>=\frac{1}{7}\int_{0}^{2\pi }sin^{2}(u)du=\frac{1}{7}[\frac{u}{2}-\frac{1}{4}sin2u\mid _{0}^{2\pi }]=\frac{\pi }{7} $$


 * $$\displaystyle <f,g>=<g,f>=\int_{0}^{\frac{2\pi }{7}}cos(7x)sin(7x)dx $$


 * From Kreyszig p.479, it is apparent that sin and cos are orthogonal to each other, so the above integration will equal zero


 * $$\displaystyle <f,g>=<g,f>=\int_{0}^{\frac{2\pi }{7}}cos(7x)sin(7x)dx=0 $$


 * Plugging in the results of each integral into the Gramian


 * $$\displaystyle \boldsymbol{\Gamma }(f,g)=det\begin{bmatrix}

<f,f> &<f,g> \\ <g,f> &<g,g> \end{bmatrix}=det\begin{bmatrix} \frac{\pi }{7} &0 \\ 0& \frac{\pi }{7} \end{bmatrix}=\frac{\pi ^{2}}{49}\neq 0 $$

Therefore, f and g are linearly independent

(2)


 * Given $$\displaystyle y''-3y'-10y=3cos(7x) $$


 * And particular solutions of the form $$\displaystyle y_{p}(x)=Mcos(7x)+Nsin(7x),y'_{p}(x)=-M7sin(7x)+N7cos(7x),y''_{p}(x)=-M49cos(7x)-N49sin(7x) $$


 * Plug particular solutions back into original ODE and collect like terms


 * $$\displaystyle -M49cos(7x)-N49sin(7x)+M21sin(7x)-N21cos(7x)-M10cos(7x)-N10sin(7x)=3cos(7x) $$


 * $$\displaystyle -59Mcos(7x)-59Nsin(7x)+21Msin(7x)-21Ncos(7x)=3cos(7x) $$


 * Equating coefficients


 * $$\displaystyle -59M=3\rightarrow M=-\frac{3}{59} $$
 * $$\displaystyle -21N=3\rightarrow N=-\frac{1}{7} $$

$$\displaystyle M=-\frac{3}{59},N=-\frac{1}{7} $$

(3)

The overall solution $$\displaystyle y(x)=y_{p}(x)+y_{h}(x) $$ consists of the particular solution and homogeneous soloution


 * Homogeneous solotuion


 * $$\displaystyle y''-3y'-10y=0 $$


 * $$\displaystyle a^{2}-4b=3^{2}-4(10)=49>0 $$ so we have distinct real roots


 * $$\displaystyle \lambda _{1,2}=\frac{1}{2}[-a\pm \sqrt{a^{2}-4b}]=\frac{1}{2}[3\pm7 ]=5,-2 $$


 * $$\displaystyle y_{h}(x)=c_{1}e^{5x}+c_{2}e^{-2x} $$


 * Using initial conditions $$\displaystyle y(0)=1,y'(0)=1 $$ and $$\displaystyle y'_{h}(x)=5c_{1}e^{5x}-2c_{2}e^{-2x} $$


 * $$\displaystyle 5c_{1}-2c_{2}=0 $$
 * $$\displaystyle c_{1}+c_{2}=1 $$


 * Solving the two equations for the two unknowns yields $$\displaystyle c_{1}=\frac{2}{7},c_{2}=\frac{5}{7} $$


 * $$\displaystyle y_{h}(x)=\frac{2}{7}e^{5x}+\frac{5}{7}e^{-2x} $$


 * Particular solution


 * $$\displaystyle y_{p}(x)=Kcos(\omega x)+Msin(\omega x) $$ where $$\displaystyle \omega =7, K=M=-\frac{3}{59},M=N=-\frac{1}{7} $$


 * $$\displaystyle y_{p}(x)=-\frac{3}{59}cos(7x)-\frac{1}{7}sin(7x) $$


 * Giving us an overall solution of

$$\displaystyle y(x)=y_{h}(x)+y_{p}(x)=\frac{2}{7}e^{5x}+\frac{5}{7}e^{-2x}-\frac{3}{59}cos(7x)-\frac{1}{7}sin(7x) $$


 * Plotting the solution over three periods $$\displaystyle P=\frac{2\pi }{7} \Rightarrow 3P=\frac{6\pi }{7} $$

Matlabcode EDU>> x=0:0.001:(6*pi)/7; EDU>> y=(2/7).*exp(5.*x)+(5/7).*exp(-2.*x)-(3/59).*cos(7.*x)-(1/7).*sin(7.*x); EDU>> plot(x,y)

Author
Solved and uploaded by Joshua House

Question
Find the solution to the following initial condition problem, and plot it over 3 periods. $$ \displaystyle y'' + 4y' + 13y = 2e^{-2x}cos(3x) $$ $$ \displaystyle y_h(x) = e^{-2x}(Acos(3x)+Bsin(3x)) $$ $$ \displaystyle y_p(x) = xe^{-2x}(Mcos(3x)+Nsin(3x)) $$ $$ \displaystyle y(0) = 1, y'(0) = 0 $$

Solution
First, we take the first and second derivative of the particular solution: $$ \displaystyle y'_p(x) = e^{-2x}(-3Msin(3x)-2Mcos(3x)+3Ncos(3x)-2Nsin(3x)) = e^{-2x}[(-3M-2N)sin(3x)+(-2M+3N)cos(3x)] $$ $$ \displaystyle y''_p(x) = e^{-2x}(-9Mcos(3x)+6Msin(3x)-9Nsin(3x)-6Ncos(3x))-2e^{-2x}(-3Msin(3x)-2Mcos(3x)+3Ncos(3x)-2Nsin(3x)) $$ $$ \displaystyle = 2e^{-2x}[(12M-5N)sin(3x)+(13M-12N)cos(3x)]$$ Now, we plug the particular solution derivatives into the initial equation: $$ \displaystyle y'' + 4y' + 13y = 2e^{-2x}cos(3x)$$ $$ \displaystyle 2e^{-2x}[(12M-5N)sin(3x)+(13M-12N)cos(3x)] + 4[e^{-2x}[(-3M-2Nsin(3x)+(-2M+3N)cos(3x)]]$$ $$ \displaystyle + 13[e^{-2x}[Mcos(3x)+Nsin(3x)]] = 2e^{-2x}cos(3x) $$ $$ \displaystyle e^{-2x}[(24M-10N)sin(3x)+(26M-24N)cos(3x)+(-12M-8N)sin(3x)+(-8M+12N)cos(3x)+13Mcos(3x)+13Nsin(3x)] $$ $$ \displaystyle = 2e^{-2x}cos(3x) $$  $$ \displaystyle (12M-5N)sin(3x) + (31M-12N)cos(3x) = 2cos(3x) $$  Now, we equate the coefficients of sin(3x) and cos(3x) to determine the unknown coefficients M and N:  $$ \displaystyle 12M-5N = 0 $$ $$ \displaystyle 31M-12N = 2 $$ $$ \displaystyle N = \frac{24}{11}, M = \frac{10}{11} $$  Therefore, the particular solution is:  $$ \displaystyle y_p(x) = e^{-2x}[ \frac{10}{11}cos(3x) + \frac{24}{11}sin(3x)] $$  Now, we focus on the homogeneous part of the solution. It is given to us as:  $$ \displaystyle y_h(x) = e^{-2x}[Acos(3x)+Bsin(3x)] $$  As you can see, this is identical to the particular solution, except that M is now A and N is now B. Therefore, the first derivative of the homogenous solution will be in the same form as the first derivative of the particular solution: $$ \displaystyle y'_h(x) = e^{-2x}(-3Asin(3x)-2Acos(3x)+3Bcos(3x)-2Bsin(3x)) = e^{-2x}[(-3A-2B)sin(3x)+(-2A+3B)cos(3x)] $$ Remembering the two initial conditions, y(0) = 1 and y'(0) = 0, we apply these to the homogenous equations: $$ \displaystyle y_h(0) = e^{-2x}[Acos(3x)+Bsin(3x)] = 1 $$ $$ \displaystyle y'_h(x) = e^{-2x}[(-3A-2B)sin(3x)+(-2A+3B)cos(3x)] = 0 $$ This yields two equations that we can use to solve for the coefficients A and B: $$ \displaystyle A = 1 $$ $$ \displaystyle -2A+3B = 0 \Rightarrow 3B=2 \Rightarrow B = \frac{3}{2} $$ Therefore, the homoegenous solution is: $$ \displaystyle y_h(x) = e^{-2x}[cos(3x) + \frac{3}{2}sin(3x)] $$ We find the final solution, y(x), by adding the homogenous and particular solutions as seen below: $$ \displaystyle y(x) = y_h(x) + y_p(x) = e^{-2x}[cos(3x) + \frac{3}{2}sin(3x)] + e^{-2x}[ \frac{10}{11}cos(3x) + \frac{24}{11}sin(3x)] $$ $$ \displaystyle y(x) = e^{-2x}[ \frac{21}{11}cos(3x) + \frac{81}{22}sin(3x)] $$

Matlab Code: x= 0:0.001:(2*pi/3);

y = exp(-2.*x)*(21/11).*cos(3.*x) + exp(-2.*x)*(81/22).*sin(3.*x);

plot(x,y)



Author
This problem was solved and uploaded by Will Knapper

Question
$$ v=4e_1+2e_2=c_1b_1+c_2b_2 $$ The oblique basis vectors are: $$ b_1=2e_1+7e_2 $$ $$ b_2=1.5e_1+3e_2 $$ 1. Find the components $$ c_1, c_2 $$ using the Gram matrix as in (1)p.8-11. 2. Verify the results by using (1)-(2)p.7c-34 in (2)p8-11, and rely on the non-zero determinant of the matrix of components of $$ b_1, b_2 $$ relative to the basis $$ e_1 , e_2 $$, as discussed on p.7c-34.

Solution
1. Using Gram matrix as in (1) p.8-10: $$ \begin{bmatrix} <b_1,b_1> & <b_1,b_2> \\ <b_2,b_1> & <b_2,b_2> \end{bmatrix} $$ $$ \begin{Bmatrix} c_1 \\ c_2 \end{Bmatrix} $$ = $$ \begin{Bmatrix} <b_1, v> \\ <c_2, v> \end{Bmatrix} $$ From (3)p.8-9 we know that $$ <b_i, b_j>=b_i \cdot b_j $$. Solving the various components we get: $$ \begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix} $$ $$ \begin{Bmatrix} c_1 \\ c_2 \end{Bmatrix} $$ = $$ \begin{Bmatrix} 22 \\ 12 \end{Bmatrix} $$ In order to solve for $$ c_1, c_2 $$ we need to calculate the inverse of $$ \begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix} $$. This gives is the Gram matrix as used in (1)p.8-11: $$ c= \Gamma^{-1}d $$ $$ \begin{bmatrix} 0.55556 & -1.18519 \\ -1.18519 & 2.61728 \end{bmatrix} $$ $$ \begin{Bmatrix} 22 \\ 12 \end{Bmatrix} $$ = $$ \begin{Bmatrix} c_1 \\ c_2 \end{Bmatrix} $$ Thus: $$ c_1= -2 ; c_2= 5.333= \frac{16}{3} $$

2. Plugging in $$ b_1, b_2 $$ into $$ v $$ we get: $$ c_1(2e_1+7e_2)+c_2(1.5e_1+3e_2)=4e_1+2e_2 $$ $$ 2c_1e_1+7c_1e_2+1.5c_2e_1+3c_2e_2=4e_1+2e_2 $$ $$ e_1(2c_1+1.5c_2)+e_2(7c_1+3c_2)=4e_1+2e_2 $$ Separating into components: $$ 2c_1+1.5c_2=4 $$ and $$ 7c_1+3c_2=2 $$ Solving the two linearly independent equations we get: $$ \ $$ $$ c_1= -2 ; c_2= \frac{16}{3} $$

Author
This problem was solved and uploaded by Radina Dikova

Question
Find the integral (see R5.9)

$$ \displaystyle \ \int x^nlog(1+x)dx\ $$

using integration by parts and then with the help of General binomial theorem.

$$ \displaystyle \ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k}   x^{n-k}y^k \ $$

$$ \ \binom{n}{k} = \frac{n!}{k!(n-k)!} = \frac{n(n-1)...(n-k+1)}{k!} \ $$

Solution
The indefinite integral of $$ \ x^n \ $$ is

$$ \ \frac{x^{n+1}}{n+1} + C \ $$

For n = 0, we get $$ \ \int x^0 = x \ $$

For n = 1, we get $$ \ \int x^1 = \frac{x^2}{2} \ $$

And the indefinite integral of $$ \ log(1+x) \ $$ is

$$ \ (x+1)log(1+x) - x + C \ $$

Integration by parts

- For n = 0 $$ \ \int x^0log(1+x)dx = log(1+x)[x+1] - x \ $$

- For n = 1 $$ \ \int x^1log(1+x)dx = \frac{1}{2}*[(x - log(1+x))+ x^2log(1+x)- \frac{x^2}{2}] \ $$

Author
Solved and uploaded by Mike Wallace

Question
Consider the L2-ODE-CC(5)p.7b-7 with log(1+x) as excitation:

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

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

and the initial conditions

$$ \ y(-\frac{3}{4}) = 1, y'(-\frac{3}{4}) = 0 \ $$

Part One
Project the excitation r(x) on the polynomial basis $$ \ [b_j(x) = x^j, j = 0, 1,...,n] \ $$

i.e., find $$ \ d_j \ $$ such that

r(x) ≈ $$ \ r_n(x) = \sum_{j = 0}^{n} d_jx^j \ $$

for x in $$ \ [- \frac{3}{4}, 3 ], \ $$ and for n = 3, 6, 9

Plot $$ \ r(x), r_n(x) \ $$ to show uniform approximation and convergence

Part Two
In a separate series of plots, compare the approximation of the function log(1+x) by 2 methods:

A. Projection on polynomial basis (1) p.8-17

B. Taylor series expansion about x = 0

Observe and discuss the pros and cons of each method

Find $$ \ y_n(x) \ $$ such that:

$$ \ y_n'' + ay_n' + by_n = r_n(x) \ $$

with the same initial conditions (2) p.7c-28

Plot $$ \ y_n(x) \ $$ for n = 3, 6, 9, for x in $$ \ [- \frac{3}{4}, 3]. \ $$

In a series of separate plots, compare the results obtained with the projected excitation on polynomial basis to those with truncated Taylor series of the excitation. Plot also the numerical solution as a baseline for comparison.

Solution
Using $$ \ y'' - 3y'+ 2y = log(1+x) \ $$

For n = 0

$$ \ {b_0} = {x^0} \ $$

$$ \ r = log(1+x) = C_0x^0 \ $$

$$ \ d_0 = <x^0, log(1+x)>  = \int_{-\frac{3}{4}}^3 log(1+x)dx = 2.14175 \ $$

$$ \ <x^0, x^0> = \int_{-\frac{3}{4}}^3 x^0x^0dx = x|_{-\frac{3}{4}}^3 = \frac{15}{4} \ $$

$$ \ d_0 = C_0<x^0, x^0> = C_0(\frac{15}{4}) = 2.14175 \ $$

$$ \ C_0 = 0.57113 \ $$

--

$$ \ Y = Y_h + Y_p = C_1e^{2x} + C_2e^x + 0.57113 \ $$

$$ \ Y' = 2C_1e^{2x} + C_2e^x \ $$

$$ \ y(\frac{3}{4}) = C_1e^{\frac{6}{4}} + C_2e^{\frac{3}{4}} + 0.57113 = 1 \ $$

$$ \ y'(-\frac{3}{4}) = -\frac{6}{4}C_1e^{-\frac{6}{4}} + C_2e^{-\frac{3}{4}} = 0 \ $$

Subtracting Y from Y' in order to find the coefficients

$$ \ C_1e^{-\frac{6}{4}} - 0.57113 = -1 \ $$

$$ \ C_1 = - -1.92205 \ $$

$$ \ C_2 = 1.81582 \ $$

For n = 0 the final solution will be

$$ \ Y = -1.92205e^{2x} + 1.81582e^x + 0.57113 \ $$

--

For n = 1

$$ \ C_0 = -.09399, C_1 = 0.591217 \ $$

We need to find the homogeneous Y

$$ \ \lambda^2 - 3\lambda + 2 = 0 => (\lambda-2)(\lambda-1) = 0 \ $$

$$ \ \lambda_{1,2} = 2, 1 \ $$

$$ \ Y_h = C_1e^{2x} + C_2e^x \ $$

$$ \ Y = Y_h + Y_p = C_1e^{2x} + C_2e^x - 0.09399 + 0.591217x \ $$

Solving for the initial conditions

$$ \ y(\frac{3}{4}) = C_1e^{\frac{6}{4}} + C_2e^{\frac{3}{4}} - 0.09399 + 0.591217(\frac{3}{4}) = 1 \ $$

$$ \ Y' = = 2C_1e^{2x} + C_2e^x + 0.591217 \ $$

$$ \ y'(-\frac{3}{4}) = -\frac{6}{4}C_1e^{-\frac{6}{4}} + C_2e^{-\frac{3}{4}} + 0.591217 = 0 \ $$

This gives coefficients of:

$$ \ C_1 = -9.5398, C_2 = 7.761 \ $$

-- For n = 1 the final solution will be;

$$ \ Y = -9.5398e^{2x} + 7.761e^x + 0.591217x - 0.09399 \ $$

Author
This problem was solved and uploaded by Mike Wallace

Contribution Summary
Problem 2 was solved and Problem 5 was proofread by David Herrick

Problem 5 was solved and uploaded by Joshua House

Problems 1 and 7 were solved and uploaded by Radina Dikova

Problem 3 was solved and uploaded by Derik Bell

Problem 4 was solved and uploaded by John North

Problem 8 and 9 were solved and uploaded by Mike Wallace

Problem 6 was solved and uploaded by William Knapper