course Mth 174
sbxc{xEassignment #016
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Physics II
04-07-2007
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assignment #016
g|xmۃ˳
Physics II
04-07-2007
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13:09:56
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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13:31:31
what is the fourth degree Fourier polynomial?
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Substituting: g(t) = f(bt/(2pi)) = f(x) b=1
g(t) = f(t/(2pi) = f(x), -pi<=t<=pi
x = t/(2pi) t = 2pi*x
a0 = 1/b*integral f(x)dx (-b/2 to b/2)
a0 = 1/1*integral x dx (-1/2 to 1/2)
a0 = x^2/2 (-1/2 to 1/2)
a0 = 0
a1 = 2/b*integral f(x)cos(2pix/b) dx (-b/2 to b/2)
a1 = 2/1*integral x*cos(2pix/1) dx (-1/2 to 1/2)
a1 = 0
b1 = 2/b*integral f(x)sin(2pix/b) dx (-b/2 to b/2)
b1 = 2/1*integral x sin(2pix) dx (-1/2 to 1/2)
b1 = 2[-1/(2pi)*xcost(2pix) + 1/(4pi^2)*1sin(2pix)] (-1/2 to 1/2) (using rule #15, p(x)=x, a=2pi)
b1 = 1/pi
Using similar methods:
a2 = 0
b2 = -1/(2pi)
a3 = 0
b3 = 1/(3pi)
a4 = 0
b4 = -1/(4pi)
g4(t) = (1/pi)sin(t) - 1/(2pi)sin(2t) + 1/(3pi)sin(3t) - 1/(4pi)sin(4t)
Substituting:
f4(x) = (1/pi)sin(2pix) - 1/(2pi)sin(4pix) + 1/(3pi)sin(6pix) - 1/(4pi)sin(8pix)
Since any of these sine functions is odd, with period 1, the above gives you an odd function of period 1. However since the period is 1, the function will endlessly repeat its values on the interval (-1/2, 1/2), running from y = -1/2 to y = 1/2 on each such interval. So it won't be equal to x on the entire interval from 0 to 1, though this will be the case from 0 to 1/2.
If you shift this graph 1/2 unit to the right and add 1/2 to the y values you get the desired function, which would be
1/2 + (1/pi)sin(2pi(x-1/2)) - 1/(2pi)sin(4pi(x-1/2)) + 1/(3pi)sin(6pi(x-1/2)) - 1/(4pi)sin(8pi(x-1/2)), or
1/2 - (1/pi)sin(2pix) - 1/(2pi)sin(4pix) - 1/(3pi)sin(6pix) - 1/(4pi)sin(8pix).
You changed the period but did not shift the function.
The correct solution:
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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13:46:51
Describe the graph of this polynomial on [0,1).
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On [0,1):
Graph starts at (0,0), increases to about (0.2,0.2), decreases very slightly to about (0.25,0.22), increases to a maximum at about (0.4,0.5), decreases to (0,0), then decreases to about (0.6,-0.5), increases to about (0.75,-0.2) decreases very slightly to about (-0.8,-0.22), then increases to (1,0). The parts of the graph from (0,0) to (0.5,0) and (0.5,0) to (1,0) are mirrored about x=0.5 and across the x-axis.
This is an accurate description of the graph of your solution. However as noted above, it needs to be shifted both horizontally and vertically to approximate the given function.
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13:47:05
What substitution did you use to compensate for the fact that the period of the function is not 2 `pi?
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See first response.
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13:47:47
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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13:56:21
which formula from the table did you used to establish your result and what substitution did you use?
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I substituted u = mx, du = m dx, dx = du/m
Integral cos^2(u) du/m (-pi to pi)
= (1/m)*Integral cos^2(*u) du (-pi to pi)
Using formula #18, n=2:
= (1/m)[(1/2)cos(u)sin(u) + (1/2)*Integral cos^0(u) du] (-pi to pi)
= (1/m)[(1/2)cos(u)sin(u) + (1/2)u] ((-pi to pi)
Substituting back and factoring out 1/2:
=1(2m)[cos(mx)sin(mx) + mx] (-pi to pi)
=1/(2m)[(0+mpi) - (0-mpi)] (since sin(pi * any integer) = 0)
=1/(2m)*2mpi
=pi
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14:04:46
Query Add comments on any surprises or insights you experienced as a result of this assignment.
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Problems 9 and 10 definitely illustrated the improvement in approximation of the function with increasing terms in the Fourier series. Problem #18: ""Justify the formula b(sub k) = (1/pi)integral f(x)sin(kx) dx (-pi to pi)...""
I was able to complete this problem using the methodology on p.511-512. But I don't totally understand their justification for reducing the right side
to one term (section above first boxed formula on p. 512.
sin(kx) cos(mx) is the product of an even and an odd function, so is itself an odd function. Its integral over any region symmetric with respect to the origin must be 0.
Note that cos(kx) goes through k complete cycles between x = -pi and x = pi, while cos(mx) goes through m complete cycles.
Assuming k > m (which can be assumed without loss of generality since the values of k and m can be interchanged without changing the integrand), cos(kx) and cos(mx) will go 'in and out of phase' k - m times between x = -pi and x = pi, each time on an interval of length 2 pi / (k - m). This is easy to see by graphing two cosine functions for different integer values of k and m.
Partition the interval [-2 pi, 2 pi] into subintervals each having length 2 pi / (k - m). On each such subinterval the product cos(kx) * cos(m x) is antisymmetric with respect to the midpoint of the interval and hence contributes 0 to the integral. Since the interval [-pi, pi] is partitioned into such subintervals, the integral from -pi to pi is 0.
This can be easily but somewhat messily 'nailed down' with algebraic-trigonometric manipulations, which however don't add much to this geometric explanation.
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This looks good. Let me know if you have questions.