find the Fourier series expansion function f(x)=x(2π-x)
Answers
Answer:
Step-by-step explanation:
Derivation of Fourier series expansion of a
function defined in [−π, π]:
In Fourier series expansion, we would like to write the function as a series in
sine and cosine terms in the form:
f(x) = a0
2
+
X∞
n=1
ancos nx + bn sin nx
For finding the above unknown co-efficients a0, an and bn in the Fourier series
expansion of a function, one need to recall the value of certain integrals:
1. Z π
−π
sin mx dx = 0 for any integer m.
2. Z π
−π
cos mx dx = 0 for any integer m.
3. Z π
−π
sin mxcos nx dx = 0 for any integers m and n.
4. Z π
−π
sin mxsin nx dx = 0 for integers m 6= n.
5. Z π
−π
cos mxcos nx dx = 0 for integers m 6= n.
6. Z π
−π
sin mxsin nx dx = π when the integers m = n.
7. Z π
−π
cos mxcos nx dx = π when the integers m = n.
[All the above integrals easily follow by evaluating using integration by parts]
Now suppose f(x) = a0
2 +
X∞
j=1
aj cos jx + bj sin jx.
To find a0:
Observe that
Z π
−π
f(x) dx =
a0
2
Z π
−π
dx +
X∞
j=1
aj
Z π
−π
cos jx dx + bj
Z π
−π
sin jx dx
=
a0
2
2 π +
X∞
j=1
(0 + 0)
This implies that
a0 =
1
π
Z π
−π
f(x) dx