Find the first 6 terms of the Fourier serie

Question

Find the first 6 terms of the Fourier series of the periodic function 𝑓(𝑥) of period 𝑇 = 2𝜋

defined by: 𝑓(𝑥) = 𝑥 − 𝜋 < 𝑥 < 𝜋

Answer 1

I do not know which format you are to create, but let me do

the sine-cosine form, as others can be derived from that.

In general, the n-th term of the Fourier series is of the form

a_ncos(2πnx/P) + b_nsin(2πnx/P)), where P is your period 2π.

Therefore the n-th term simplifies to

a_ncos(nx) + b_nsin(nx)

For P = 2π and f(x) = x the coefficients a_n and b_n are (by the Fourier Theorem)

a_n = 1/π ∫x cos(nx) dx

b_n = 1/π ∫x sin(nx) dx

where the integrals are over the given interval from -π to π.

Let's first evaluate the indefinite integrals

A_n(x) = ∫x cos(nx) dx

B_n(x) = ∫x sin(nx) dx

The case n=0 is special, as the integrals reduce to

A0(x) = ∫x dx = x²/2 + C

B0(x) = ∫0 dx = C

In case n > 0, for both A_n(x) and B_n(x) use integration by parts of the form

∫uv' = uv - ∫u'v

For A_n(x) we have u = x, v' = cos(nx). Hence v = sin(nx)/n.

A_n(x) = x sin(nx)/n - ∫sin(nx)/n dx

= x sin(nx)/n + cos(nx)/n² + C

For B_n(x) we have u = x, v' = sin(nx). Hence v = -cos(nx)/n.

B_n(x) = -x cos(nx)/n - ∫-cos(nx)/n dx

= -x cos(nx)/n + sin(nx)/n² + C

Now we can calculate the coefficients a_n, b_n:

a_n = (A(π) - A(-π))/π

b_n = (B(π) - B(-π))/π

a0 = (π²/2 - π²/2)/π = 0

b0 = 0

For n > 0

a_n = ( π sin(πn)/n + cos(πn)/n² - (-πsin(-πn)/n + cos(-πn)/n²))/π = 0

b_n = (-πcos(πn)/n + sin(πn)/n² - (πcos(-πn)/n + sin(-πn)/n²))/π = 2×(-1)ⁿ/n

The first six terms are of the Fourier series for f(x)=x are

0 - 2 sin(x)/1 + 2 sin(2x)/2 - 2 sin(3x)/3 + 2 sin(4x)/4 - 2 sin(5x)/5