course Mth 174 Əѐassignment #016
......!!!!!!!!...................................
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
......!!!!!!!!...................................
RESPONSE --> 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.
.................................................
......!!!!!!!!...................................
14:07:43 what is the fourth degree Fourier polynomial?
......!!!!!!!!...................................
RESPONSE --> See above. I don't think I got it right, but I don't know what I missed.
.................................................
......!!!!!!!!...................................
14:07:47 Describe the graph of this polynomial on [0,1).
......!!!!!!!!...................................
RESPONSE --> .
.................................................
......!!!!!!!!...................................
14:08:20 What substitution did you use to compensate for the fact that the period of the function is not 2 `pi?
......!!!!!!!!...................................
RESPONSE --> I couldn't get the substition method to work, which is probably why I didn't get the right answer.
.................................................
......!!!!!!!!...................................
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
......!!!!!!!!...................................
RESPONSE --> 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.
.................................................
......!!!!!!!!...................................
14:15:09 which formula from the table did you used to establish your result and what substitution did you use?
......!!!!!!!!...................................
RESPONSE --> Formula IV.18. I substituted mx for x.
.................................................
......!!!!!!!!...................................
14:15:50 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> I have had a very difficult time with these things, and I just don't know what I am missing.
.................................................
"