Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Compute its cosine Fourier series ?
see image for question
- az_lenderLv 72 months agoFavorite Answer
omega = pi/3.
a0 is simply the average value of the function on the interval -3 to 3; which is the same as the average value of e^x on [0,3]. The integral from x = 0 to 3 of e^x dx is
e^3 - e^0 = 19.086; divide by 3 and you get 6.362.
a0 = 6.362.
an = (1/3) * integral from x = -3 to +3 of
e^x * cos(n*pi*x/3) dx. [call this LINE ONE]
This should be integrated by parts.
Let u = cos(n*pi*x/3) and dv = e^x dx, so
du = (n*pi/3) [-sin(n*pi*x/3)] dx and v = e^x.
Now we have
uv - integral of v du
= e^x*cos(n*pi*x/3) +
+ (n*pi/3) integral of e^x*sin(n*pi*x/3) dx.
A second application of integration by parts is necessary:
u2 = sin(n*pi*x/3) dx and dv2 = e^x dx, so
du2 = (n*pi/3)*cos(n*pi*x/3) dx and v2 = e^x.
Now we have
e^x*cos(n*pi*x/3) + n*(pi/3)*e^x*sin(n*pi*x/3) -
n^2*(pi^2/9)*integral of e^x*cos(n*pi*x/3) dx.
Note that the whole expression on the last two lines is equal to the expression on LINE ONE above, so now you can combine the two like terms and obtain
[1 + n^2*pi^2/9] * integral of e^x*cos(n*pi*x/3) dx
= e^x*cos(n*pi*x/3) + n*(pi/3)*e^x*sin(n*pi*x/3).
Which means that the integral is equal to
[e^x*cos(n*pi*x/3) + n*(pi/3)*e^x*sin(n*pi*x/3)] /
[1 + n^2*pi^2/9].
Now we can plug in the limits (3 and -3) to evaluate "an" --
2[e^3*cos(n*pi) - 1*cos(0)]/[1 + n^2*pi^2/9].
The sine went away because sin(n*pi) is always 0.
The use of 0 as the lower limit, and the "2" outside the first square bracket, come about because f(x) was created by reflecting the function e^x through the y axis.
We need to multiply both the numerator and the denominator by 9, obtaining
an = 18[e^3*cos(n*pi) - 1]/(9 + n^2*pi^2).
Obviously cos(n*pi) is just 1 or -1, depending on whether n is even or odd, so we could write:
an = 18[e^3*(-1)^n - 1] / (9 + n^2*pi^2).
That's my final answer to your third question.
I am encouraged by the resemblance of my "an" to the expression that appears in your fourth and final question, but my head is spinning now. I'm 75, that's my excuse...