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determination of Fourier coefficients∗
pahio†
2013-03-22 1:41:03
Suppose that the real function f may be presented as sum of the Fourier
series:
f (x) =
∞
X
a0
+
(am cos mx + bm sin mx)
2
m=0
(1)
Therefore, f is periodic with period 2π. For expressing the Fourier coefficients
am and bm with the function itself, we first multiply the series (1) by cos nx
(n ∈ Z) and integrate from −π to π. Supposing that we can integrate termwise,
we may write
Z π
Z
Z π
Z π
∞ X
a0 π
f (x) cos nx dx =
cos nx dx +
am
cos mx cos nx dx + bm
sin mx cos nx dx .
2 −π
−π
−π
−π
m=0
(2)
When n = 0, the equation (2) reads
Z π
a0
· 2π = πa0 ,
f (x) dx =
2
−π
(3)
since in the sum of the right hand side, only the first addend is distinct from
zero.
When n is a positive integer, we use the product formulas of the trigonometric identities, getting
Z π
Z
1 π
[cos(m − n)x + cos(m + n)x] dx,
cos mx cos nx dx =
2 −π
−π
Z
Z π
1 π
sin mx cos nx dx =
[sin(m − n)x + sin(m + n)x] dx.
2 −π
−π
The latter expression vanishes always, since the sine is an odd function. If
m 6= n, the former equals zero because the antiderivative consists of sine terms
∗ hDeterminationOfFourierCoefficientsi
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which vanish at multiples of π; only in the case m = n we obtain from it a
non-zero result π. Then (2) reads
Z π
f (x) cos nx dx = πan
(4)
−π
to which we can include as a special case the equation (3).
By multiplying (1) by sin nx and integrating termwise, one obtains similarly
Z π
f (x) sin nx dx = πbn .
(5)
−π
The equations (4) and (5) imply the formulas
Z
1 π
f (x) cos nx dx (n = 0, 1, 2, . . .)
an =
π −π
and
bn
1
=
π
Z
π
f (x) sin nx dx (n = 1, 2, 3, . . .)
−π
for finding the values of the Fourier coefficients of f .
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