#$&* course mth 174 5/31 5:00 pm Calculus IIAsst # 16
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Given 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). (recall that replacing x with x - b shifts the graph b units in the x direction) 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). The graph below depicts this function on the approximate interval (-1, 2) in red, as well as the function y = x (in blue). On the interval (0, 1) the two functions agree very well. You can see that the Fourier series repeats with period 1. . **** What substitution did you use to compensate for the fact that the period of the function is not 2 `pi?
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&&*** The period is 1 and not 2 pi. We are told that for the period of 1, f(x) = x for values between 0 and 1. So instead of f(x) = x we use f(X)= 2 ‘pi kx/the period So 2 ‘ pi kx/1 or just 2 ‘pi k x instead of x for values of f(x) between 0 and 1********** 17:19:41 student solution: f has period b = 1, so I let x = bt/2'pi = t/2'pi &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ###I had some difficulty with this problem but I think I ended up with the correct answers and work. I had one question noted above regarding the a terms always being 0 and thus not included in the series. The other problem I have is getting from the coefficients of ½, -1/(2 pi) , -1/(4 pi), -1/(6 pi) , -1/(8 ‘pi) To the first 5 terms of ½ - 1/pi*(sin (2 ‘pi x)) - 1/2pi*(sin (4 ‘pi x)) - 1/3pi*(sin (6 ‘pi x)) - 1/4pi*(sin(8 pi x) I understand we are taking the values at pi, 2 pi, 3 pi, 4 pi, etc but I’m still a little unclear on where the term - 1/pi*(sin (2 ‘pi x)) I thought the answer should be ½ - 1/pi*(sin (2 ‘pi x)) - 1/2pi*(sin (4 ‘pi x)) - 1/3pi*(sin (6 ‘pi x)) - 1/4pi*(sin(8 pi x) Based on the coefficients we found above??????????
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21:32:53 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Integral from -pi to pi of cos^2 x we want to see if the value is pi I begin by letting u=mx and du=mdx so 1/m u = dx So 1/m * integral from -pi to pi cos^2u So this could be writing at 1/m * integral from -pi to pi of ½ + (1/2) cos 2u So I have 1/(2m) integral of 1 + 1/2m integral cos 2u Evaluating I get x/(2m) + 1/(4m)*sin 2u x/(2m) + 1/(4m) sin 2mx so evaluating at pi I get pi/(2m) and evaluating at -pi I get (-pi/m) so I have pi/(2m) - (-pi / 2m) which would just give us pi so the statement above is true. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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21:32:53 ......!!!!!!!! STUDENT QUESTION on 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. The function depicted below is a superposition of sine and cosine functions. The function is defined on the interval -pi < x < pi, which corresponds to the interval between the vertical lines near the left and right sides of the picture. By the methods of this section we can find the Fourier series for this function, which has coefficients a_1 = 1 and a_5 = .3. All other coefficients are 1. Thus the Fourier series for this function is f(x) = 1 cos(1x) + .3 cos(5x), or just f(x) = cos(x) + .3 cos(5x). Here's a graph of y = cos(x): Here's a graph of y = .3 cos(5x) Here's a graph of both functions together. If the y coordinates of each point of this graph are added they will give us the y coordinates of the original graph, so we say that the sum of these two graphs is the original graph. The amplitude of the cos(x) term is greater than that of the cos(5x) term, so we say that the cos(x) term has more energy. The energy is in fact determined by amplitude. The first term has amplitude 1, so its energy relative energy is 1. The first term has amplitude .3, so its relative energy is .3. If we had a general Fourier series with a_k and b_k terms, then they would be the coefficients of a cosine and a sine function. • Think of the cosine and the sine contributions as being represented by the legs of a right triangle. The hypotenuse represents what we get when we combine them. The hypotenuse would be sqrt(a_k^2 + b_k^2). • The actual argument is a little more specific than this, but this at least makes it plausible that the energy term is sqrt(a_k^2 + b_k^2). The energy for the k = 0 term is simply sqrt(A_0^2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!