finding Fourier series of |x sin(x)|
Asked 7 years, 11 months ago
Modified 6 years, 7 months ago
I have this function: f (x)
1
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= |x sin(x)|
Now, since this is an even function I know the bk
≡ 0
.
I tried calculating the ak coefficients and got
2(−1)
ak =
2
k
k+1
− 1
so what I'm not sure about is this:
1. is this correct? I tried finding Fourier series calculator but failed...
2. if so, what am I to do about k
= 1
?
thanks in advance
fourier-series
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edited Nov 12, 2015 at 23:22
asked Nov 12, 2015 at 20:24
user169852
CIsForCookies
241
1
The function x
20:49
↦ x sin x
1
2
12
is even, and your function f is odd. – Christian Blatter Nov 12, 2015 at
There are two functions, your first definition is odd and it is not the same as |x sin(x) . So which
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function are you interested
in? – Fabian Nov 12, 2015 at 20:50
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I realize my mistake.
Pls assume
the |xsinx|with
is the
function
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Policy.and not the previous def...
– CIsForCookies Nov 12, 2015 at 21:21
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1 Answer
Your answer is correct for n
the calculation.
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≠ 1
. In order to see what happens when n
= 1
, let's review
4
Assuming you are defining the function on the interval [−π, π] and extending
periodically from there, and assuming the form
∞
a0
f (x) =
+ ∑[an cos(nx)
2
n=1
+ bn sin(nx)]
then, as you said, bn
= 0
for all n because f is even. The formula for an is
π
1
an =
∫
π
f (x) cos(nx) dx
−π
Since your f is even, so is f (x) cos(nx) , so we can integrate over [0, π] and double the
result:
π
2
an =
∫
π
f (x) cos(nx) dx
0
On the interval [0, π] , we have |x sin(x)|
when computing the integral:
so we can shed the absolute values
π
2
an =
= x sin(x)
∫
π
x sin(x) cos(nx) dx
0
Using the trig identity sin(a) cos(b)
1
=
2
[sin(a + b)
, this becomes
+ sin(a − b)]
an =
π
1
(∫
π
x sin((n + 1)x) dx
0
π
− ∫
x sin((n − 1)x) dx)
0
Note that if n = 1, the integrand in the second integral becomes x sin(0x) = 0 , so only
the first integral contributes to the answer in that case. The first integral doesn't have any
special case to worry about, because we are only interested in n ≥ 0.
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π
∫
x sin((n + 1)x) dx
0
n
π cos(πn)
=
π(−1)
=
(n + 1)
and for the second integral (when n
≠ 1
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n + 1
):
π
∫
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x sin((n − 1)x) dx
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∫
((
)
)
0
n
π cos(πn)
π(−1)
=
=
n − 1
(n − 1)
So, for n
= 1
we have
n
(−1)
a1 =
and for n
≠ 1
=
n + 1
−1
2
we have
n
n
(−1)
an =
(−1)
−
n + 1
n − 1
2(−1)
=
2
n
n+1
− 1
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answered Nov 12, 2015 at 23:21
user169852
So when n=1 and the first integral can't be resolved (denominator = 0) I just ignore the first integral
and only use the second? And if so, how the series should look like? a0/2 + a1 + sum {from 2 to inf}
? – CIsForCookies Nov 12, 2015 at 23:38
2
The formula for the second integral, π(−1)n /(n − 1) , is only true if n ≠ 1 . But the integral can
still be resolved when n = 1 . In that case, we have sin((n − 1)x) = sin(0x) = 0 , so the
integrand is zero, so the integral is zero, which is why we only use the formula for the first integral
when n
= 1
∞
. Yes, you can write the result as a0 /2 + a1 cos(x) + ∑n=2 an cos . – user169852 Nov
(nx)
12, 2015 at 23:41
I had a doubt.Why can't we do sin(b) cos(a)
=
1
2
[sin(a + b)
here? – Aladdin Apr 7, 2020 at 4:40
− sin(a − b)]
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