Assignment 16

course Mth 174

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Physics II

06-24-2009

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18:45:53

Query problem 10.5.12 (3d edition 10.5.12) (was 9.5.12) period 1 fn defined by f(x) = x if 0 < x < 1

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RESPONSE -->

I used x = bt/2pi for b = 2 so x = t/pi

f(x) = 0 for -pi < x < 0

f(x) = 1 for 0 < x < pi

x = t/pi so t = pix

I used the Fourier polynomial fr the square wave function:

F(x) =1/2 + 2/pi sin(t) + 2/3pi sin(3t) + 2/5pi sin(5t)

I then substituted t = pi x

F(x) =1/2 + 2/pi sin(pi x) + 2/3pi sin(3pi x) + 2/5pi sin(5pi x)

We first shift the interval (- pi , pi) to the interval (0, 1).

 

To shift the interval (-pi, pi) to (-1/2, 1/2) you would replace x by 2 pi x. To then shift the interval to (0, 1) you would in addition substitute x-1/2 for x. The function sin(k x) would become sin(2 pi k ( x - 1/2) ) = sin(2 pi kx - k pi). For even k this would be just sin(2 pi k x); for odd k this would be -sin(2 pi k x).

 

The integral of the function itself over the interval is 1, so you would have a0 = 1/2.

 

The integral of x * sin(2 pi k x) from 0 to 1 is easily found by integration by parts to be 1 / (2 pi k) (details of the integration: let u = x, dv = sin(2 pi k x) dx; v = -1 / (2 pi k) cos(2 pi k x), so the integral of v du is a multiple of sin(2 pi k x) and hence yields 0 at both limits; the u v term is x cos(2 pi k x) / (2 pi k), which yields 0 at the left limit and -1 / (2 pi k) at the right limit).

 

It follows that the first five terms of the series would be 1/2, -1/(2 pi), -1/(4 pi), -1 / (6 pi) and -1 / (8 pi).

 

This yields the Fourier polynomial

 

1/2 - (1/pi)sin(2pix) - 1/(2pi)sin(4pix) - 1/(3pi)sin(6pix) - 1/(4pi)sin(8pix).

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18:46:06

what is the fourth degree Fourier polynomial?

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RESPONSE -->

F(x) =1/2 + 2/pi sin(pi x) + 2/3pi sin(3pi x) + 2/5pi sin(5pi x)

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18:47:05

Describe the graph of this polynomial on [0,1).

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RESPONSE -->

The graph is a square wave that reaches f(x) = 1 and f(x) = 0

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18:47:41

What substitution did you use to compensate for the fact that the period of the function is not 2 `pi?

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RESPONSE -->

x = bt/ 2pi= t/pi

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19:06:30

Query problem 10.5.24 (3d edition 10.5.24) (was 9.5.24) integral of cos^2(mx) from -`pi to `pi is `pi

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RESPONSE -->

When I integrated I got 0 instead of pi for an answer and am not sure where I went wrong. This is what I did:

integral from pi to -pi cos^2(mx)dx

u = mx

du=m

at x = pi, u =m* pi

at x = -pi, u = -m* pi

1/m integral from m*pi to -m*pi cos^2x dx

= 1/m * 1/2cosxsinx + 3/2sinx from m*pi to -m * pi

=1/m * 1/2cos(m*pi) sin(m*pi) + 3/2sin(m*pi) - 1/m * 1/2cos(-m*pi)sin(-m*pi) + 3/2sin(-m*pi)

I wasn't sure what to do from here

The antiderivative of cos^2(mx) is 1 / (2 m) cos(mx) sin(mx) + x / 2.

Evaluated at x = pi this gives us

1 / (2 m) cos(m pi) sin(m pi) + pi/2.

Since sin(m pi) = 0, the result is just pi/2.

Evaluated at x = -pi we get - pi/2.

So the integral from - pi to pi is just

int(cos^2(mx), x from -pi to pi) = pi / 2 - (-pi / 2) = pi.

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19:06:54

which formula from the table did you used to establish your result and what substitution did you use?

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RESPONSE -->

I used formula Iv-18

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19:09:08

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

I had a hard time understanding example 3 in the book (page 507 4th edition) where it talks about the energy theorem. I don't understand exactly what Asub0 means and why A^2 sub 0 = 2a^2sub0 = 1/2, etc. I also did not understand the graph made from these expressions and what it means.

if you email me with this question then I can send you an explanation which includes pictures

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&#This looks very good. Let me know if you have any questions. &#