Assignment 16 Query

course Mth 174

Əѐassignment #016

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VJ]

Physics II

11-28-2007

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14:07:28

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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I have no idea if this is right, but here's what I did:

To find a_0, I took the period = 1 = b and used the formulas on p. 509.

a_0 = int( f(x) dx,0,1)

= int ( x dx, 0,1)

= (1/2)x^2 from 1 to 0

= 1/2

I set the range of integration from 0 to 1 because the given value was f(x) = x for 0 <= x < 1.

a_1 = 2 * int(x cos(2x 'pi), 0, 1)

= 2(x/(2'pi) sin(2x 'pi) - int(1/(2'pi) sin(2x 'pi) ) )

= 2(x/(2'pi) sin(2x 'pi) - ( 1/(2'pi) )^2 cos(2x 'pi) )

= 2( 1/(2'pi) )^2

= 2 / (4'pi^2) = 1/(2 'pi^2)

I used integration by substitution. To find the next three a Fourier coefficients, I noticed that the n value from x cos(2*x*n*'pi) would turn up in the coefficient of the final cos function ( ( 1/(2*n*'pi) )^2 ) and be squared along with the rest, giving the following functions:

a_2 = 1 / (8 'pi^2)

a_3 = 1 / (18 'pi^2)

a_4 = 1 / (32 'pi^2)

For the b Fourier coefficients, I used the same basic process to get:

b_1 = 2 * int( x sin(2x 'pi), 0, 1) = -1/(2'pi)

b_2 = -1/(4'pi)

b_3 = -1/(6'pi)

b_4 = -1/(8'pi)

This gives a Fourier polynomial of the following:

f(x) ~ 1/2 + 1/(2 pi^2) * cos(x) + 1/(8 pi^2) * cos(2x) + 1/(18 pi^2) * cos(3x) + 1/(32 pi^2) * cos(4x) - 1/(2pi) * sin(x) - 1/(4pi) * sin(2x) - 1/(6pi) * sin(3x) - 1/(8pi) * sin(4x)

As for the graph, between 0 and 1 it looks like a line with a negative slope over the interval, but a mild slope.

Good attempt but you didn't shift quite correctly. Your function would be f(x) ~ 1/2 + 1/(2 pi^2) * cos(2 pi x) + 1/(8 pi^2) * cos(2* 2 pi x) + 1/(18 pi^2) * cos(3 * 2 pi x) etc.; however because of the shift there is a factor of pi involved in your constants, and you end up with pi rather than pi^2. See the details below and let me know if you have questions:

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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14:07:43

what is the fourth degree Fourier polynomial?

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See above. I don't think I got it right, but I don't know what I missed.

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14:07:47

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

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14:08:20

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

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I couldn't get the substition method to work, which is probably why I didn't get the right answer.

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14:14:33

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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I used formula IV.18 to find my result:

int( cos^x (mx), -'pi, 'pi)

= 1/2 cos(mx) sin(mx)

Which is probably wrong considering that this integral equals 0.

I believe the table gives

sin(mx)cos(m x)/(2 m) + x/2.

For integer m we have sin(m * pi) = sin( -m * pi) = 0, so the sin * cos term gives you zero.

The x/2 evaluated at the limits gives you pi / 2 - (-pi/2) = pi.

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14:15:09

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

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Formula IV.18. I substituted mx for x.

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14:15:50

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

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I have had a very difficult time with these things, and I just don't know what I am missing.

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