Diagonalising Matrices / Recurrence Relations

[Ted123]

Homework Statement

[PLAIN]http://img530.imageshack.us/img530/6672/linn.jpg

The Attempt at a Solution

For parts (a) and (b) I've found the eigenvalues to be -\frac{1}{3} and -1 with corresponding eigenvectors \begin{bmatrix} -1 \\ 3 \end{bmatrix} and \begin{bmatrix} -1 \\ 1 \end{bmatrix} respectively.

Now for part (c) I know there is a way of solving this by diagonalising matrices but I can't remember the method.

The recurrence relation can be written as \begin{bmatrix} a_n \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} -\frac{4}{3} & -\frac{1}{3} \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n-1} \\ a_{n-2} \end{bmatrix}

We can diagonalise A = \begin{bmatrix} -\frac{4}{3} & -\frac{1}{3} \\ 1 & 0 \end{bmatrix} by:

letting D = \begin{bmatrix} -\frac{1}{3} & 0 \\ 0 & -1 \end{bmatrix} and P = \begin{bmatrix} -1 & -1 \\ 3 & 1 \end{bmatrix} so that we have A= PDP^{-1}

Now how do I find a_n from here?

 

[micromass]

I'll give a hint:

\left( \begin{array}{c} a_n\\ a_{n-1} \end{array}\right)=\left( \begin{array}{cc} -\frac{4}{3} & -\frac{1}{3}\\ 1 & 0 \end{array}\right)\left( \begin{array}{c} a_{n-1}\\ a_{n-2} \end{array}\right)=\left( \begin{array}{cc} -\frac{4}{3} & -\frac{1}{3}\\ 1 & 0 \end{array}\right)^2\left( \begin{array}{c} a_{n-2}\\ a_{n-3} \end{array}\right)=??\left( \begin{array}{c} a_1\\ a_0 \end{array}\right)

 

[Ted123]

I'll give a hint:

\left( \begin{array}{c} a_n\\ a_{n-1} \end{array}\right)=\left( \begin{array}{cc} -\frac{4}{3} & -\frac{1}{3}\\ 1 & 0 \end{array}\right)\left( \begin{array}{c} a_{n-1}\\ a_{n-2} \end{array}\right)=\left( \begin{array}{cc} -\frac{4}{3} & -\frac{1}{3}\\ 1 & 0 \end{array}\right)^2\left( \begin{array}{c} a_{n-2}\\ a_{n-3} \end{array}\right)=??\left( \begin{array}{c} a_1\\ a_0 \end{array}\right)

\begin{bmatrix} a_n \\ a_{n-1} \end{bmatrix} = A^{n-1} \begin{bmatrix} a_1 \\ a_0 \end{bmatrix}

so how do I use the diagonalisation? EDIT: I see...

 

[micromass]

You use the diagonalization to calculate A^{n-1}. For example

A^3=(PDP^{-1})(PDP^{-1})(PDP^{-1})=PD^3P^{-1}

 

[Ted123]

You use the diagonalization to calculate A^{n-1}. For example

A^3=(PDP^{-1})(PDP^{-1})(PDP^{-1})=PD^3P^{-1}

I get \begin{bmatrix} a_n \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} -\frac{1}{3}\left( -\frac{1}{3} \right) ^{n-1} + (-1)^{n-1} \\ \left( -\frac{1}{3} \right) ^{n-1} - (-1)^{n-1} \end{bmatrix}

Reading off the top component, is this the same as the printed formula?

 

[micromass]

Well, it requires some reworking. But it's the same thing. So it looks like you've solved it!

 

[Ted123]

Well, it requires some reworking. But it's the same thing. So it looks like you've solved it!

There's actually a further part to the question:

[PLAIN]http://img703.imageshack.us/img703/273/lin0.jpg

After multiplying through how do I 'compare coefficients'?

 

[micromass]

What happens if you multiply

(x^2+4x+3)(a_0+a_1x+a_2x^2+a_3x^3+...)

Multiply these things out. What you get, should be equal to x.

 

[Ted123]

What happens if you multiply

(x^2+4x+3)(a_0+a_1x+a_2x^2+a_3x^3+...)

Multiply these things out. What you get, should be equal to x.

This gives:

2x = 3a_0 + (4a_0 + 3a_1)x + (a_0 + 4a_1 + 3a_2)x^2 + (a_1 +4a_2 +3a_3)x^3 + (a_2 +4a_3)x^4 + a_3x^5 + \cdots

If we equate coefficients we see that

3a_0 = 0 \Rightarrow a_0 = 0

4a_0 + 3a_1 = 2 \Rightarrow 3a_1 = 2 \Rightarrow a_1 = \frac{2}{3}

I could keep going forever but how do I show that the coefficients satisfy the recurrence relation?

 

[micromass]

In general, what will the coefficient of xⁿ be? You'll see your recurrence relation popping up...

 

[Ted123]

In general, what will the coefficient of xⁿ be? You'll see your recurrence relation popping up...

In general the coefficient of x^n is 3a_n + 4a_{n-1} + a_{n-2}

How does the link with the recurrence relation?

 

[micromass]

Yes. So what does a_n equal?

 

[Ted123]

Yes. So what does a_n equal?

Well for n\in \mathbb{N} \backslash \{1\} ,

3a_n + 4a_{n-1} + a_{n-2} = 0

so a_n = -\frac{4}{3} a_{n-1} - \frac{1}{3} a_{n-2}

which is precisely the recurrence relation!

 

[Ted123]

Well for n\in \mathbb{N} \backslash \{1\} ,

3a_n + 4a_{n-1} + a_{n-2} = 0

so a_n = -\frac{4}{3} a_{n-1} - \frac{1}{3} a_{n-2}

which is precisely the recurrence relation!

The characteristic polynomial is given by

c_A (\lambda ) = \begin{vmatrix} \displaystyle -\frac{4}{3} - \lambda & \displaystyle -\frac{1}{3} \\ 1 & -\lambda \end{vmatrix} =0

\Rightarrow (-\frac{4}{3} - \lambda ) \left(-\lambda \right) - ( -\frac{1}{3} ) ( 1) = 0

\displaystyle \Rightarrow \frac{4}{3} \lambda + \lambda ^2 + \frac{1}{3} = 0

\displaystyle \Rightarrow 3\lambda ^2 + 4\lambda + 1=0

\displaystyle \Rightarrow (3\lambda +1)(\lambda +1)=0

\displaystyle \Rightarrow \lambda = -\frac{1}{3} or \lambda = -1

\displaystyle \therefore -\frac{1}{3} and -1 are eigenvalues of A

The eigenvectors

\begin{bmatrix} x \\ y \end{bmatrix} of A with eigenvalue -1/3 satisfy the vector equation

\begin{bmatrix} -1 & -\frac{1}{3} \\ 1 & \frac{1}{3} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

which is equivalent to the single linear equation x = -\frac{y}{3}

Hence the eigenvectors of A with eigenvalue -1/3 are \begin{bmatrix} -\frac{y}{3} \\ y \end{bmatrix} where y \neq 0

 

[Ted123]

The characteristic polynomial is given by

c_A (\lambda ) = \begin{vmatrix} \displaystyle -\frac{4}{3} - \lambda & \displaystyle -\frac{1}{3} \\ 1 & -\lambda \end{vmatrix} =0

\Rightarrow (-\frac{4}{3} - \lambda ) \left(-\lambda \right) - ( -\frac{1}{3} ) ( 1) = 0

\displaystyle \Rightarrow \frac{4}{3} \lambda + \lambda ^2 + \frac{1}{3} = 0

\displaystyle \Rightarrow 3\lambda ^2 + 4\lambda + 1=0

\displaystyle \Rightarrow (3\lambda +1)(\lambda +1)=0

\displaystyle \Rightarrow \lambda = -\frac{1}{3} or \lambda = -1

\displaystyle \therefore -\frac{1}{3} and -1 are eigenvalues of A

The eigenvectors

\begin{bmatrix} x \\ y \end{bmatrix} of A with eigenvalue -1/3 satisfy the vector equation

\begin{bmatrix} -1 & -\frac{1}{3} \\ 1 & \frac{1}{3} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

which is equivalent to the single linear equation x = -\frac{y}{3}

Hence the eigenvectors of A with eigenvalue -1/3 are \begin{bmatrix} -\frac{y}{3} \\ y \end{bmatrix} where y \neq 0

Then

AP = \begin{bmatrix} -\frac{4}{3} & -\frac{1}{3} \\ 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 3 & 1 \end{bmatrix}= \begin{bmatrix} \frac{1}{3} & 1 \\ -1 & -1 \end{bmatrix}

and

PD = \begin{bmatrix} -1 & -1 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -\frac{1}{3} & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} & 1 \\ -1 & -1 \end{bmatrix}

and P is invertible since \text{det}(P) = (-1)(1) - (-1)(3) - -1+3=2 \neq 0

Now \begin{bmatrix} a_n \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} -\frac{1}{3} & 0 \\ 0 & -1 \end{bmatrix} ^{n-1} \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{3}{2} & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} a_1 \\ a_0 \end{bmatrix}

= \begin{bmatrix} a_n \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} (-\frac{1}{3})^{n-1} & 0 \\ 0 & (-1)^{n-1} \end{bmatrix} \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{3}{2} & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} \frac{2}{3} \\ 0 \end{bmatrix}

= \begin{bmatrix} a_n \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} -(-\frac{1}{3})^{n-1} & -(-1)^{n-1} \\ 3(-\frac{1}{3})^{n-1} & (-1)^{n-1} \end{bmatrix} \begin{bmatrix} \frac{1}{3} \\ -1 \end{bmatrix}

\begin{bmatrix} a_n \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} -\frac{1}{3} ( -\frac{1}{3} ) ^{n-1} + (-1)^{n-1} \\ ( -\frac{1}{3} ) ^{n-1} - (-1)^{n-1} \end{bmatrix}

Reading off the top component we see that a_n = -\frac{1}{3} ( -\frac{1}{3} ) ^{n-1} + (-1)^{n-1}

a_n = -\frac{1}{3} ( -\frac{1}{3} ) ^n ( -\frac{1}{3} ) ^{-1} + (-1)^n (-1)^{-1}

a_n = ( -\frac{1}{3} ) ^n - (-1)^n

a_n = \frac{(-1)^n}{3^n} - (-1)^n

a_n = (-1)^n \left( \frac{1}{3^n} - 1 \right)

 

[Ted123]

{\bf p}(t) = (f(t)\cos\,t,f(t)\sin\,t)

{\bf p}'(t) = (f'(t)\cos\,t - f(t)\sin\,t , f'(t)\sin\,t + f(t) \cos\,t)

\displaystyle \oint_C -\frac{y}{x^2+y^2} \; dx + \frac{x}{x^2+y^2} \; dy

\displaystyle = \int^{\pi}_{-\pi} \left( -\frac{y}{x^2+y^2} \frac{dx}{dt} + \frac{x}{x^2+y^2} \frac{dy}{dt} \right) \; dt

\displaystyle = \int^{\pi}_{-\pi} \bigg \{ \left( -\frac{f(t)\sin\,t}{f^2(t)\cos^2 \, t} + f^2(t) \sin^2 \, t \right) \left( f'(t) \cos \, t - f(t)\sin \, t \right)

+ \left( \frac{f(t)\cos \, t}{f^2(t)\cos^2 \, t} + f^2(t) \sin^2 \, t \right) \left( f'(t)\sin \, t + f(t)\cos \, t \right) \bigg \} \; dt

\displaystyle = \int^{\pi}_{-\pi} \left( -\frac{f(t)f'(t) \sin\,t\cos\,t}{f^2(t)} + \frac{f^2(t)\sin^2 t}{f^2(t)} + \frac{f(t)f'(t) \sin\,t\cos\,t}{f^2(t)} + \frac{f^2(t)\cos^2 t}{f^2(t)} \right) \;dt

\displaystyle = \int^{\pi}_{-\pi} \left( \sin^2 t + \cos^2 t \right) \;dt

\displaystyle = \int^{\pi}_{-\pi} \;dt = \left[ t \right]^{\pi}_{-\pi} = 2\pi

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2(a) Call the 3 sections of the curve C in the order described in the question C_1, C_2 ,C_3 respectively.

Then \displaystyle \int_C = \int_{C_1+C_2+C_3} = \int_{C_1} + \int_{C_2} + \int_{C_3}

So parametrising C_1 , C_2 , C_3 as follows,

C_1 : For 0 \leq t \leq 1\;,\;p(t) = (1-t)(0,0) + t(R,0) = (tR,0)

p'(t) = (R,0)

C_2 : For 0 \leq t \leq \frac{\pi}{2}\;,\;p(t) = (R\cos\,t , R\sin\,t)

p'(t) = (-R\sin\,t , R\cos\,t)

C_3 : For 0 \leq t \leq 1\;,\;p(t) = (1-t)(0,R) + t(0,0) = (0,R-Rt)

p'(t) = (0,-R)

Hence \displaystyle \int_{C_1} = \int_{C_1} \left( -xy^2 \frac{dx}{dt} + x^2y \frac{dy}{dt} \right)\;dt

\displaystyle = \int^1_0 \bigg \{ (-tR)(0)(R) + (R^2)(0)(0) \bigg \}\;dt = 0

\displaystyle \int_{C_2} = \int^{\frac{\pi}{2}}_0 \bigg \{ (-R\cos\,t)(R^2\sin\,t)(-R\sin\,t) + (R^2\cos^2t)(R\sin\,t)(R\cos\,t) \bigg \} \;dt

\displaystyle = R^4 \int^{\frac{\pi}{2}}_0 \left( \cos\,t\,\sin^3t + \sin\,t\,\cos^3\,t \right) \;dt

\displaystyle = R^4 \left( \frac{1}{4} + \frac{1}{4} \right) = \frac{R^4}{2}

\displaystyle \int_{C_3} = \int^1_0 \bigg \{ (0)\left( (R-Rt)^2 \right)(0) + (0)(R-Rt)(-R) \bigg \} \;dt =0

Hence \displaystyle \int_C = 0 + \frac{R^4}{2} + 0 = \frac{R^4}{2}

\sin^3\,t and \cos^3\,t

2(b) Let P = -xy^2 and Q = x^2y

Then Green's theorem says that \displaystyle \oint_C P\;dx + Q\;dy = \iint_S \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \;dxdy

So (calling the integral I) \displaystyle I = \iint_S (2xy + 2xy) \;dxdy = \iint_S 4xy\;dxdy