Question #223176

Question1 Given A= [■(2&2&1@1&3&1@1&2&2)] 1.1 find the eigenvalues of A 1.2. Determine an eigenvector for the eigenvalue = 5 Question2 A periodic function f(x) with period 2 π is defined by : f(x)= [ (x+π)/2 -π˂ X˂ 0, (x-π)/2 0˂ X˂ π] Determine the Fourier series for the periodic function


1
Expert's answer
2021-08-06T11:35:48-0400

1.


A=(221131122)A=\begin{pmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{pmatrix}

det(AλI)=2λ2113λ1122λ=0\det(A-\lambda I)=\begin{vmatrix} 2-\lambda & 2 & 1 \\ 1 & 3-\lambda & 1 \\ 1 & 2 & 2-\lambda \\ \end{vmatrix}=0

(2λ)2(3λ)+2+23+λ4+2λ4+2λ=0(2-\lambda)^2(3-\lambda)+2+2-3+\lambda-4+2\lambda-4+2\lambda=0

(44λ+λ2)(3λ)+5λ7=0(4-4\lambda+\lambda^2)(3-\lambda)+5\lambda-7=0

124λ12λ+4λ2+3λ2λ3+5λ7=012-4\lambda-12\lambda+4\lambda^2+3\lambda^2-\lambda^3+5\lambda-7=0

λ3+7λ211λ+5=0-\lambda^3+7\lambda^2-11\lambda+5=0

λ2(λ1)+6λ(λ1)5(λ1)=0-\lambda^2(\lambda-1)+6\lambda(\lambda-1)-5(\lambda-1)=0


(λ1)(λ26λ+5)=0-(\lambda-1)(\lambda^2-6\lambda+5)=0

(λ1)2(λ5)=0-(\lambda-1)^2(\lambda-5)=0

The roots are λ1=1,λ2=1,λ3=5.\lambda_1=1, \lambda_2=1, \lambda_3=5.

These are the eigenvalues.

λ=5\lambda=5


(2λ2113λ1122λ)=(321121123)\begin{pmatrix} 2-\lambda & 2 & 1 \\ 1 & 3-\lambda & 1 \\ 1 & 2 & 2-\lambda \\ \end{pmatrix}=\begin{pmatrix} -3 & 2 & 1 \\ 1 & -2 & 1 \\ 1 & 2 & -3 \end{pmatrix}

R1=R1/(3)R_1=R_1/(-3)


(12/31/3121123)\begin{pmatrix} 1 & -2/3 & -1/3 \\ 1 & -2 & 1 \\ 1 & 2 & -3 \end{pmatrix}

R2=R2R1R_2=R_2-R_1


(12/31/304/34/3123)\begin{pmatrix} 1 & -2/3 & -1/3 \\ 0 & -4/3 & 4/3 \\ 1 & 2 & -3 \end{pmatrix}

R3=R3R1R_3=R_3-R_1


(12/31/304/34/308/38/3)\begin{pmatrix} 1 & -2/3 & -1/3 \\ 0 & -4/3 & 4/3 \\ 0 & 8/3 & -8/3 \end{pmatrix}

R2=3R2/4R_2=-3R_2/4


(12/31/301108/38/3)\begin{pmatrix} 1 & -2/3 & -1/3 \\ 0 & 1 & -1 \\ 0 & 8/3 & -8/3 \end{pmatrix}

R1=R1+2R2/3R_1=R_1+2R_2/3


(10101108/38/3)\begin{pmatrix} 1 & 0 & -1\\ 0 & 1 & -1 \\ 0 & 8/3 & -8/3 \end{pmatrix}

R3=R38R2/3R_3=R_3-8R_2/3


(101011000)\begin{pmatrix} 1 & 0 & -1\\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}

(101011000)(x1x2x3)=(000)\begin{pmatrix} 1 & 0 & -1\\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 0\\ 0 \\ 0 \end{pmatrix}

If we take x3=t,x_3=t, then x1=t,x2=t.x_1=t, x_2=t.


x=(ttt)=(111)t\vec x=\begin{pmatrix} t\\ t\\ t \end{pmatrix}=\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}tv=(111)\vec v=\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}

This is the eigenvector.



λ=1\lambda=1

(2λ2113λ1122λ)=(121121121)\begin{pmatrix} 2-\lambda & 2 & 1 \\ 1 & 3-\lambda & 1 \\ 1 & 2 & 2-\lambda \\ \end{pmatrix}=\begin{pmatrix} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 1 & 2 & 1 \end{pmatrix}

R2=R2R1R_2=R_2-R_1

(121000121)\begin{pmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 1 & 2 & 1 \end{pmatrix}

R3=R3R1R_3=R_3-R_1

(121000000)\begin{pmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}


(121000000)(x1x2x3)=(000)\begin{pmatrix} 1 & 2 & 1\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 0\\ 0 \\ 0 \end{pmatrix}

If we take x2=t,x3=s,x_2=t, x_3=s, then x1=2ts.x_1=-2t-s.

x=(2tsts)=(210)t+(101)s\vec x=\begin{pmatrix} -2t-s\\ t\\ s \end{pmatrix}=\begin{pmatrix} -2\\ 1\\ 0 \end{pmatrix}t+\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}s

u=(210),w=(101)\vec u=\begin{pmatrix} -2\\ 1\\ 0 \end{pmatrix}, \vec w=\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}

These are the eigenvectors.


2.


a0=2πππf(x)dxa_0=\dfrac{2}{\pi}\displaystyle\int_{-\pi}^{\pi}f(x)dx

=2ππ0x+π2dx+2π0πxπ2dx=\dfrac{2}{\pi}\displaystyle\int_{-\pi}^{0}\dfrac{x+\pi}{2}dx+\dfrac{2}{\pi}\displaystyle\int_{0}^{\pi}\dfrac{x-\pi}{2}dx

=2π[x24+πx2]0πd+2π[x24πx2]π0=\dfrac{2}{\pi}[\dfrac{x^2}{4}+\dfrac{\pi x}{2}]\begin{matrix} 0 \\ -\pi & d \end{matrix}+\dfrac{2}{\pi}[\dfrac{x^2}{4}-\dfrac{\pi x}{2}]\begin{matrix} \pi \\ 0 \end{matrix}

=π2π2=0=\dfrac{\pi}{2}-\dfrac{\pi}{2}=0



an=1πππf(x)cosnxdxa_n=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi}f(x)\cos{nx}dx

=1ππ0x+π2cosnxdx+1π0πxπ2cosnxdx=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{0}\dfrac{x+\pi}{2}\cos{nx}dx+\dfrac{1}{\pi}\displaystyle\int_{0}^{\pi}\dfrac{x-\pi}{2}\cos{nx}dx

=cosπn12πn2+cosπn12πn2=0=-\dfrac{\cos{\pi n}-1}{2\pi n^2}+\dfrac{\cos{\pi n}-1}{2\pi n^2}=0



bn=1πππf(x)sinnxdxb_n=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi}f(x)\sin{nx}dx

=1ππ0x+π2sinnxdx+1π0πxπ2sinnxdx=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{0}\dfrac{x+\pi}{2}\sin{nx}dx+\dfrac{1}{\pi}\displaystyle\int_{0}^{\pi}\dfrac{x-\pi}{2}\sin{nx}dx

=12n12n=1n=-\dfrac{1}{2n}-\dfrac{1}{2n}=-\dfrac{1}{n}




f(x)=n=11nsinnxf(x)=-\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{n}\sin{nx}



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