Assignment 16

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course MTH 174

4:10 p.m. 7/20/10

nGyޑSassignment #016

016. `query 16

Physics II

07-20-2010

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15:57:10

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 -->

Here we have to shift the interval, first by using 2pix and then to move from (-1/2,1/2) to (0,1) interval we use (x - 1/2). So we have 2pi(x-1/2). When

we plug this into our formula we get: cos (2pi k (x-1/2)) and sin (2pi k (x-1/2). An equivalent, and simpler, expression is cos (2pi k x) and sin (2 pi k

x). It is easy to see that a_o= 1/2. Then it is beneficial to find a general form of the cos and sin integrals using k over our interval. The integral is

evaluated using integration by parts and then using the values of our interval we arrive at the general form of: -1/(pi*k) for the b_k and 0 for all a_k.

Using this we get: b_1= -1/pi, b_2= -1/2pi b_3= -1/3pi and b_4= -1/4pi.

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15:58:55

what is the fourth degree Fourier polynomial?

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RESPONSE -->

After all of the above we put our values into the Fourier equation and we have: f(x) = 1/2 - 1/pi sin(2pi x) - 1/2pi sin (4pi x) - 1/3pi sin(6pi x) - 1/4pi

sin (8pi x).

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16:03:03

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

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RESPONSE -->

The graph is a periodic function and on our interval it is increasing linearly. It's y-intercept is 1/2 and then has a local min about x=0.0968 of 0.0143

and a local max at about x=0.904 of 0.985. The function is also periodic along the line of f(x)=x like a sin wave with its closest values near the middle

of the interval and farthest away at the ends.

good description

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16:03:51

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 -->

First 2 pi x and then 2pi(x-1/2) to accommodate the [0,1) aspect of the interval.

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16:05:48

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 -->

Here I used IV- 18 from the table. The integral is thus: 1/2 cos (mx) sin (mx) + x/2. Evaluating this over the interval gives: (0 + pi/2) - ( 0 - pi/2) =

pi.

Good. Just for comparison:

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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16:06:07

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

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RESPONSE -->

The formula was IV-18 with n= 2.

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16:09:04

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

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RESPONSE -->

The first question was pretty difficult for me and took a little while to figure it all out. I also learned Fourier series are important in thermodynamics.

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Very good work.

There is a symmetry-based explanation of the last question, which expands into the general question of the integral of sin(mx) cos(nx), at the link below. The explanation includes some graphs. Also a graph of the Fourier approximation of the first question.

It's worth a look at

http://vhcc2.vhcc.edu/dsmith/GenInfo/qa_query_etc/cal2/query_16.htm.

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