Assignment 16

course Mth 174

j奉CDُSٰ¸assignment #016

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Physics II

08-06-2009

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17:54:24

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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20:21:25

what is the fourth degree Fourier polynomial?

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0+(1/pi)-(1/(2pi))+(1/(3pi))-(1/(4pi)) = 7/(12pi)

You appear to have found the coefficients. However you don't add the coefficients, you use then to construct a Fourier polynomial in 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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20:27:31

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

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The graph starts at about .32 for the first point. It then drops for the second point to about .16. Rises to about .27 and then drops back down to about .19. It is basically a sine wave with decreasing amplitude.

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20:58:30

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

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Instead of simply using sin(2pi), I substituted ((2pikx)/b) to compensate for the different period.

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21:32:32

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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21:33:13

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

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I used formula number 18 with n=2 and some positive integer for m.

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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21:33:19

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

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&#Your work looks good. See my notes. Let me know if you have any questions. &#