Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

Compute its cosine Fourier series ?

see image for question

Relevance
• 2 months ago

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" --

you get

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).