Mixing
Find mistake in Fourer series calculation.
$begingroup$
The function $f$ is $3$-periodic and $$f(t)=left{
begin{array}{ll}
t,quad &0leq tleq1\
1,quad &1<t<2\
3-t,quad &2leq t leq 3
end{array}
right. $$
Expand $f$ as a (real) Fourier series.
Attempt: I simply use the formulae for $a_0,a_n$ and $b_n$. The value of $L$ is $2L=3Leftrightarrow L=3/2.$ this gives
$$a_0=frac{1}{2L}intlimits_{-L}^{L}f(t)dt=frac{1}{3}cdot2=frac{2}{3},$$
which also is easy to see by drawing $f(t)$. Now
begin{align}
a_n &=frac{2}{3}left(intlimits_{-3/2}^{-1}cosleft(frac{2pi nt}{3}right)dt-intlimits_{-1}^{0}tcosleft(frac{2pi nt}{3}right)dt+intlimits_{0}^{1}tcosleft(frac{2pi nt}{3}right)dt+intlimits_{1}^{3/2}cosleft(frac{2pi nt}{3}right)right)\
&=frac{sin(pi n)-sinleft(frac{2pi n}{3}right)}{pi n},
end{align}
since the two middle integrals cancel and the two outer are the same. For $b_n$ I have, with identical integrals as above but replacing $cos$ by $sin$, that $b_n=0$. This means that
begin{align}
f(t)&=frac{2}{3}-sum_{n=1}^{infty}frac{sin(pi n)-sinleft(frac{2pi n}{3}right)}{pi n}cosleft(frac{2pi nt}{3}right)\
&=frac{2}{3}-frac{2}{pi}sum_{n=1}^{infty}frac{sin{left(frac{2pi n}{3}right)}}{n}cosleft(frac{2pi nt}{3}right),
end{align}
since $sin(pi n)=0 forall ninmathbb{Z}.$ However the book wants the answer to be
$$f(t)=frac{2}{3}-frac{3}{pi^2}sum_{n=1}^{infty}frac{1-cosleft(frac{2pi n}{3}right)}{n^2}cosleft(frac{2pi nt}{3}right).$$
Can someone help me find the mistake?
proof-verification fourier-analysis fourier-series
asked Jan 24 at 17:36
ParsevalParseval
3,0241719
$endgroup$
add a comment |
$begingroup$
The function $f$ is $3$-periodic and $$f(t)=left{
begin{array}{ll}
t,quad &0leq tleq1\
1,quad &1<t<2\
3-t,quad &2leq t leq 3
end{array}
right. $$
Expand $f$ as a (real) Fourier series.
Attempt: I simply use the formulae for $a_0,a_n$ and $b_n$. The value of $L$ is $2L=3Leftrightarrow L=3/2.$ this gives
$$a_0=frac{1}{2L}intlimits_{-L}^{L}f(t)dt=frac{1}{3}cdot2=frac{2}{3},$$
which also is easy to see by drawing $f(t)$. Now
begin{align}
a_n &=frac{2}{3}left(intlimits_{-3/2}^{-1}cosleft(frac{2pi nt}{3}right)dt-intlimits_{-1}^{0}tcosleft(frac{2pi nt}{3}right)dt+intlimits_{0}^{1}tcosleft(frac{2pi nt}{3}right)dt+intlimits_{1}^{3/2}cosleft(frac{2pi nt}{3}right)right)\
&=frac{sin(pi n)-sinleft(frac{2pi n}{3}right)}{pi n},
end{align}
since the two middle integrals cancel and the two outer are the same. For $b_n$ I have, with identical integrals as above but replacing $cos$ by $sin$, that $b_n=0$. This means that
begin{align}
f(t)&=frac{2}{3}-sum_{n=1}^{infty}frac{sin(pi n)-sinleft(frac{2pi n}{3}right)}{pi n}cosleft(frac{2pi nt}{3}right)\
&=frac{2}{3}-frac{2}{pi}sum_{n=1}^{infty}frac{sin{left(frac{2pi n}{3}right)}}{n}cosleft(frac{2pi nt}{3}right),
end{align}
since $sin(pi n)=0 forall ninmathbb{Z}.$ However the book wants the answer to be
$$f(t)=frac{2}{3}-frac{3}{pi^2}sum_{n=1}^{infty}frac{1-cosleft(frac{2pi n}{3}right)}{n^2}cosleft(frac{2pi nt}{3}right).$$
Can someone help me find the mistake?
proof-verification fourier-analysis fourier-series
asked Jan 24 at 17:36
ParsevalParseval
3,0241719
$endgroup$
add a comment |
$begingroup$
The function $f$ is $3$-periodic and $$f(t)=left{
begin{array}{ll}
t,quad &0leq tleq1\
1,quad &1<t<2\
3-t,quad &2leq t leq 3
end{array}
right. $$
Expand $f$ as a (real) Fourier series.
Attempt: I simply use the formulae for $a_0,a_n$ and $b_n$. The value of $L$ is $2L=3Leftrightarrow L=3/2.$ this gives
$$a_0=frac{1}{2L}intlimits_{-L}^{L}f(t)dt=frac{1}{3}cdot2=frac{2}{3},$$
which also is easy to see by drawing $f(t)$. Now
begin{align}
a_n &=frac{2}{3}left(intlimits_{-3/2}^{-1}cosleft(frac{2pi nt}{3}right)dt-intlimits_{-1}^{0}tcosleft(frac{2pi nt}{3}right)dt+intlimits_{0}^{1}tcosleft(frac{2pi nt}{3}right)dt+intlimits_{1}^{3/2}cosleft(frac{2pi nt}{3}right)right)\
&=frac{sin(pi n)-sinleft(frac{2pi n}{3}right)}{pi n},
end{align}
since the two middle integrals cancel and the two outer are the same. For $b_n$ I have, with identical integrals as above but replacing $cos$ by $sin$, that $b_n=0$. This means that
begin{align}
f(t)&=frac{2}{3}-sum_{n=1}^{infty}frac{sin(pi n)-sinleft(frac{2pi n}{3}right)}{pi n}cosleft(frac{2pi nt}{3}right)\
&=frac{2}{3}-frac{2}{pi}sum_{n=1}^{infty}frac{sin{left(frac{2pi n}{3}right)}}{n}cosleft(frac{2pi nt}{3}right),
end{align}
since $sin(pi n)=0 forall ninmathbb{Z}.$ However the book wants the answer to be
$$f(t)=frac{2}{3}-frac{3}{pi^2}sum_{n=1}^{infty}frac{1-cosleft(frac{2pi n}{3}right)}{n^2}cosleft(frac{2pi nt}{3}right).$$
Can someone help me find the mistake?
proof-verification fourier-analysis fourier-series
asked Jan 24 at 17:36
ParsevalParseval
3,0241719
$endgroup$
The function $f$ is $3$-periodic and $$f(t)=left{
begin{array}{ll}
t,quad &0leq tleq1\
1,quad &1<t<2\
3-t,quad &2leq t leq 3
end{array}
right. $$
Expand $f$ as a (real) Fourier series.
Attempt: I simply use the formulae for $a_0,a_n$ and $b_n$. The value of $L$ is $2L=3Leftrightarrow L=3/2.$ this gives
$$a_0=frac{1}{2L}intlimits_{-L}^{L}f(t)dt=frac{1}{3}cdot2=frac{2}{3},$$
which also is easy to see by drawing $f(t)$. Now
begin{align}
a_n &=frac{2}{3}left(intlimits_{-3/2}^{-1}cosleft(frac{2pi nt}{3}right)dt-intlimits_{-1}^{0}tcosleft(frac{2pi nt}{3}right)dt+intlimits_{0}^{1}tcosleft(frac{2pi nt}{3}right)dt+intlimits_{1}^{3/2}cosleft(frac{2pi nt}{3}right)right)\
&=frac{sin(pi n)-sinleft(frac{2pi n}{3}right)}{pi n},
end{align}
since the two middle integrals cancel and the two outer are the same. For $b_n$ I have, with identical integrals as above but replacing $cos$ by $sin$, that $b_n=0$. This means that
begin{align}
f(t)&=frac{2}{3}-sum_{n=1}^{infty}frac{sin(pi n)-sinleft(frac{2pi n}{3}right)}{pi n}cosleft(frac{2pi nt}{3}right)\
&=frac{2}{3}-frac{2}{pi}sum_{n=1}^{infty}frac{sin{left(frac{2pi n}{3}right)}}{n}cosleft(frac{2pi nt}{3}right),
end{align}
since $sin(pi n)=0 forall ninmathbb{Z}.$ However the book wants the answer to be
$$f(t)=frac{2}{3}-frac{3}{pi^2}sum_{n=1}^{infty}frac{1-cosleft(frac{2pi n}{3}right)}{n^2}cosleft(frac{2pi nt}{3}right).$$
Can someone help me find the mistake?
proof-verification fourier-analysis fourier-series
proof-verification fourier-analysis fourier-series
asked Jan 24 at 17:36
ParsevalParseval
3,0241719
asked Jan 24 at 17:36
ParsevalParseval
3,0241719
edited Jan 24 at 19:54
Parseval
asked Jan 24 at 17:36
ParsevalParseval
3,0241719
asked Jan 24 at 17:36
ParsevalParseval
3,0241719
asked Jan 24 at 17:36
ParsevalParseval
3,0241719
3,0241719
add a comment |
add a comment |
2 Answers
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$begingroup$
Hint: What is $f(t)$ for $-1 le t le 0$? Is it $f(t)=+t$ or $f(t)=-t$?
Btw, you could evaluate $a_n$ as
$$a_n = frac{2}{T} int_{0}^{T} f(t) cosleft(frac{2pi}{T}tright) dt,$$
which could save some of your time in figuring out what $f(t)$ should be for $t<0$.
In response to the comment,
$$a_n = frac{2}{T}int_{-T/2}^{T/2} f(t)cosleft(frac{2pi}{T}tright) dt = frac{2}{T}int_{-T/2}^{0} f(t)cosleft(frac{2pi}{T}tright) dt + frac{2}{T}int_{0}^{T/2} f(t)cosleft(frac{2pi}{T}tright) dt.$$
For the first integral, substitute $t = y+T$ to get
$$int_{-T/2}^{0} f(t)cosleft(frac{2pi}{T}tright) dt = int_{T/2}^{T} f(y)cosleft(frac{2pi}{T}yright) dy.$$
Therefore,
$$a_n = frac{2}{T}int_{0}^{T} f(t)cosleft(frac{2pi}{T}tright) dt.$$
answered Jan 24 at 17:50
Math LoverMath Lover
14.1k31437
$endgroup$
$begingroup$
Well, ok I see one error there, should be $f(t)=-t$. Could you explain why that formula for $a_n$ works? What is $T$?
$endgroup$
– Parseval
Jan 24 at 18:06
$begingroup$
The two middle integrals are opposite in sign, so those terms don't cancel out, they add up. If you evaluate those integrals (integrate by parts) I think you will get the desired $1/n^2$ term.
$endgroup$
– Aditya Dua
Jan 24 at 20:11
$begingroup$
@Parseval Edited the answer to address your question.
$endgroup$
– Math Lover
Jan 24 at 22:49
add a comment |
$begingroup$
since the two middle integrals cancel...
They don't. Our $f(t)=begin{cases}1&-frac32 le tle -1\-t&-1le tle 0\t&0le tle 1\1&1le tle frac32end{cases}$ is even. Integrate it against an odd function like $sinleft(frac{2pi nt}{3}right)$ and we'll get zero by odd symmetry. Integrate it against something even like $cosleft(frac{2pi nt}{3}right)$, and the parts for negative $t$ will mirror the parts for positive $t$ and reinforce each other, doubling the total value. By even symmetry,
begin{align*}a_n &= frac23int_{-frac32}^{frac32}f(t)cosleft(frac{2pi nt}{3}right),dt = frac43int_{0}^{frac32}f(t)cosleft(frac{2pi nt}{3}right),dt\
a_n &= frac43int_0^1 tcosleft(frac{2pi nt}{3}right),dt+frac43int_1^{frac32}cosleft(frac{2pi nt}{3}right),dt\
a_n &= frac43left[frac{3t}{2pi n}sinleft(frac{2pi nt}{3}right)+frac{9}{4pi^2 n^2}cosleft(frac{2pi nt}{3}right)right]_0^1 +frac43left[frac{3}{2pi n}sinleft(frac{2pi nt}{3}right)right]_1^{frac32}\
a_n &= frac2{pi n}sinleft(frac{2pi n}{3}right) +frac{3}{pi^2 n^2}left(1-cosleft(frac{2pi n}{3}right)right) + frac2{pi n}left(sinpi n - sinleft(frac{2pi n}{3}right)right)\
a_n &= frac{3}{pi^2 n^2}+frac2{pi n}(sin(pi n))-frac{3}{pi n^2}cosleft(frac{2pi n}{3}right)\
a_n &= begin{cases}frac23& n=0\ 0& 3mid n,n>0\ frac{9}{2pi^2 n^2}& 3nmid nend{cases}end{align*}
In that last step, we use that $sin(pi n) = 0$ and $cosleft(frac{2pi n}{3}right)$ has a cycle of values $1,-frac12,-frac12$; subtract that from $1$ and it becomes $0,frac32,frac32$. Also, we incorporate the previously calculated (by a different method) $a_0$. The book's version of the answer applies the $sin(pi n)=0$ fact, but doesn't do anything with the $cosleft(frac{2pi n}{3}right)$ term.
So there it is. Incorporate the middle part of the integral as it should be for this even function, and we get the right answer.
answered Jan 24 at 21:58
jmerryjmerry
14.6k1632
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2 Answers
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2 Answers
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votes
$begingroup$
Hint: What is $f(t)$ for $-1 le t le 0$? Is it $f(t)=+t$ or $f(t)=-t$?
Btw, you could evaluate $a_n$ as
$$a_n = frac{2}{T} int_{0}^{T} f(t) cosleft(frac{2pi}{T}tright) dt,$$
which could save some of your time in figuring out what $f(t)$ should be for $t<0$.
In response to the comment,
$$a_n = frac{2}{T}int_{-T/2}^{T/2} f(t)cosleft(frac{2pi}{T}tright) dt = frac{2}{T}int_{-T/2}^{0} f(t)cosleft(frac{2pi}{T}tright) dt + frac{2}{T}int_{0}^{T/2} f(t)cosleft(frac{2pi}{T}tright) dt.$$
For the first integral, substitute $t = y+T$ to get
$$int_{-T/2}^{0} f(t)cosleft(frac{2pi}{T}tright) dt = int_{T/2}^{T} f(y)cosleft(frac{2pi}{T}yright) dy.$$
Therefore,
$$a_n = frac{2}{T}int_{0}^{T} f(t)cosleft(frac{2pi}{T}tright) dt.$$
answered Jan 24 at 17:50
Math LoverMath Lover
14.1k31437
$endgroup$
$begingroup$
Well, ok I see one error there, should be $f(t)=-t$. Could you explain why that formula for $a_n$ works? What is $T$?
$endgroup$
– Parseval
Jan 24 at 18:06
$begingroup$
The two middle integrals are opposite in sign, so those terms don't cancel out, they add up. If you evaluate those integrals (integrate by parts) I think you will get the desired $1/n^2$ term.
$endgroup$
– Aditya Dua
Jan 24 at 20:11
$begingroup$
@Parseval Edited the answer to address your question.
$endgroup$
– Math Lover
Jan 24 at 22:49
add a comment |
$begingroup$
Hint: What is $f(t)$ for $-1 le t le 0$? Is it $f(t)=+t$ or $f(t)=-t$?
Btw, you could evaluate $a_n$ as
$$a_n = frac{2}{T} int_{0}^{T} f(t) cosleft(frac{2pi}{T}tright) dt,$$
which could save some of your time in figuring out what $f(t)$ should be for $t<0$.
In response to the comment,
$$a_n = frac{2}{T}int_{-T/2}^{T/2} f(t)cosleft(frac{2pi}{T}tright) dt = frac{2}{T}int_{-T/2}^{0} f(t)cosleft(frac{2pi}{T}tright) dt + frac{2}{T}int_{0}^{T/2} f(t)cosleft(frac{2pi}{T}tright) dt.$$
For the first integral, substitute $t = y+T$ to get
$$int_{-T/2}^{0} f(t)cosleft(frac{2pi}{T}tright) dt = int_{T/2}^{T} f(y)cosleft(frac{2pi}{T}yright) dy.$$
Therefore,
$$a_n = frac{2}{T}int_{0}^{T} f(t)cosleft(frac{2pi}{T}tright) dt.$$
answered Jan 24 at 17:50
Math LoverMath Lover
14.1k31437
$endgroup$
$begingroup$
Well, ok I see one error there, should be $f(t)=-t$. Could you explain why that formula for $a_n$ works? What is $T$?
$endgroup$
– Parseval
Jan 24 at 18:06
$begingroup$
The two middle integrals are opposite in sign, so those terms don't cancel out, they add up. If you evaluate those integrals (integrate by parts) I think you will get the desired $1/n^2$ term.
$endgroup$
– Aditya Dua
Jan 24 at 20:11
$begingroup$
@Parseval Edited the answer to address your question.
$endgroup$
– Math Lover
Jan 24 at 22:49
add a comment |
$begingroup$
Hint: What is $f(t)$ for $-1 le t le 0$? Is it $f(t)=+t$ or $f(t)=-t$?
Btw, you could evaluate $a_n$ as
$$a_n = frac{2}{T} int_{0}^{T} f(t) cosleft(frac{2pi}{T}tright) dt,$$
which could save some of your time in figuring out what $f(t)$ should be for $t<0$.
In response to the comment,
$$a_n = frac{2}{T}int_{-T/2}^{T/2} f(t)cosleft(frac{2pi}{T}tright) dt = frac{2}{T}int_{-T/2}^{0} f(t)cosleft(frac{2pi}{T}tright) dt + frac{2}{T}int_{0}^{T/2} f(t)cosleft(frac{2pi}{T}tright) dt.$$
For the first integral, substitute $t = y+T$ to get
$$int_{-T/2}^{0} f(t)cosleft(frac{2pi}{T}tright) dt = int_{T/2}^{T} f(y)cosleft(frac{2pi}{T}yright) dy.$$
Therefore,
$$a_n = frac{2}{T}int_{0}^{T} f(t)cosleft(frac{2pi}{T}tright) dt.$$
answered Jan 24 at 17:50
Math LoverMath Lover
14.1k31437
$endgroup$
Hint: What is $f(t)$ for $-1 le t le 0$? Is it $f(t)=+t$ or $f(t)=-t$?
Btw, you could evaluate $a_n$ as
$$a_n = frac{2}{T} int_{0}^{T} f(t) cosleft(frac{2pi}{T}tright) dt,$$
which could save some of your time in figuring out what $f(t)$ should be for $t<0$.
In response to the comment,
$$a_n = frac{2}{T}int_{-T/2}^{T/2} f(t)cosleft(frac{2pi}{T}tright) dt = frac{2}{T}int_{-T/2}^{0} f(t)cosleft(frac{2pi}{T}tright) dt + frac{2}{T}int_{0}^{T/2} f(t)cosleft(frac{2pi}{T}tright) dt.$$
For the first integral, substitute $t = y+T$ to get
$$int_{-T/2}^{0} f(t)cosleft(frac{2pi}{T}tright) dt = int_{T/2}^{T} f(y)cosleft(frac{2pi}{T}yright) dy.$$
Therefore,
$$a_n = frac{2}{T}int_{0}^{T} f(t)cosleft(frac{2pi}{T}tright) dt.$$
answered Jan 24 at 17:50
Math LoverMath Lover
14.1k31437
edited Jan 24 at 22:49
answered Jan 24 at 17:50
Math LoverMath Lover
14.1k31437
answered Jan 24 at 17:50
Math LoverMath Lover
14.1k31437
answered Jan 24 at 17:50
Math LoverMath Lover
14.1k31437
14.1k31437
$begingroup$
Well, ok I see one error there, should be $f(t)=-t$. Could you explain why that formula for $a_n$ works? What is $T$?
$endgroup$
– Parseval
Jan 24 at 18:06
$begingroup$
The two middle integrals are opposite in sign, so those terms don't cancel out, they add up. If you evaluate those integrals (integrate by parts) I think you will get the desired $1/n^2$ term.
$endgroup$
– Aditya Dua
Jan 24 at 20:11
$begingroup$
@Parseval Edited the answer to address your question.
$endgroup$
– Math Lover
Jan 24 at 22:49
add a comment |
$begingroup$
Well, ok I see one error there, should be $f(t)=-t$. Could you explain why that formula for $a_n$ works? What is $T$?
$endgroup$
– Parseval
Jan 24 at 18:06
$begingroup$
The two middle integrals are opposite in sign, so those terms don't cancel out, they add up. If you evaluate those integrals (integrate by parts) I think you will get the desired $1/n^2$ term.
$endgroup$
– Aditya Dua
Jan 24 at 20:11
$begingroup$
@Parseval Edited the answer to address your question.
$endgroup$
– Math Lover
Jan 24 at 22:49
$begingroup$
Well, ok I see one error there, should be $f(t)=-t$. Could you explain why that formula for $a_n$ works? What is $T$?
$endgroup$
– Parseval
Jan 24 at 18:06
$begingroup$
Well, ok I see one error there, should be $f(t)=-t$. Could you explain why that formula for $a_n$ works? What is $T$?
$endgroup$
– Parseval
Jan 24 at 18:06
$begingroup$
The two middle integrals are opposite in sign, so those terms don't cancel out, they add up. If you evaluate those integrals (integrate by parts) I think you will get the desired $1/n^2$ term.
$endgroup$
– Aditya Dua
Jan 24 at 20:11
$begingroup$
The two middle integrals are opposite in sign, so those terms don't cancel out, they add up. If you evaluate those integrals (integrate by parts) I think you will get the desired $1/n^2$ term.
$endgroup$
– Aditya Dua
Jan 24 at 20:11
$begingroup$
@Parseval Edited the answer to address your question.
$endgroup$
– Math Lover
Jan 24 at 22:49
$begingroup$
@Parseval Edited the answer to address your question.
$endgroup$
– Math Lover
Jan 24 at 22:49
add a comment |
$begingroup$
since the two middle integrals cancel...
They don't. Our $f(t)=begin{cases}1&-frac32 le tle -1\-t&-1le tle 0\t&0le tle 1\1&1le tle frac32end{cases}$ is even. Integrate it against an odd function like $sinleft(frac{2pi nt}{3}right)$ and we'll get zero by odd symmetry. Integrate it against something even like $cosleft(frac{2pi nt}{3}right)$, and the parts for negative $t$ will mirror the parts for positive $t$ and reinforce each other, doubling the total value. By even symmetry,
begin{align*}a_n &= frac23int_{-frac32}^{frac32}f(t)cosleft(frac{2pi nt}{3}right),dt = frac43int_{0}^{frac32}f(t)cosleft(frac{2pi nt}{3}right),dt\
a_n &= frac43int_0^1 tcosleft(frac{2pi nt}{3}right),dt+frac43int_1^{frac32}cosleft(frac{2pi nt}{3}right),dt\
a_n &= frac43left[frac{3t}{2pi n}sinleft(frac{2pi nt}{3}right)+frac{9}{4pi^2 n^2}cosleft(frac{2pi nt}{3}right)right]_0^1 +frac43left[frac{3}{2pi n}sinleft(frac{2pi nt}{3}right)right]_1^{frac32}\
a_n &= frac2{pi n}sinleft(frac{2pi n}{3}right) +frac{3}{pi^2 n^2}left(1-cosleft(frac{2pi n}{3}right)right) + frac2{pi n}left(sinpi n - sinleft(frac{2pi n}{3}right)right)\
a_n &= frac{3}{pi^2 n^2}+frac2{pi n}(sin(pi n))-frac{3}{pi n^2}cosleft(frac{2pi n}{3}right)\
a_n &= begin{cases}frac23& n=0\ 0& 3mid n,n>0\ frac{9}{2pi^2 n^2}& 3nmid nend{cases}end{align*}
In that last step, we use that $sin(pi n) = 0$ and $cosleft(frac{2pi n}{3}right)$ has a cycle of values $1,-frac12,-frac12$; subtract that from $1$ and it becomes $0,frac32,frac32$. Also, we incorporate the previously calculated (by a different method) $a_0$. The book's version of the answer applies the $sin(pi n)=0$ fact, but doesn't do anything with the $cosleft(frac{2pi n}{3}right)$ term.
So there it is. Incorporate the middle part of the integral as it should be for this even function, and we get the right answer.
answered Jan 24 at 21:58
jmerryjmerry
14.6k1632
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$begingroup$
since the two middle integrals cancel...
They don't. Our $f(t)=begin{cases}1&-frac32 le tle -1\-t&-1le tle 0\t&0le tle 1\1&1le tle frac32end{cases}$ is even. Integrate it against an odd function like $sinleft(frac{2pi nt}{3}right)$ and we'll get zero by odd symmetry. Integrate it against something even like $cosleft(frac{2pi nt}{3}right)$, and the parts for negative $t$ will mirror the parts for positive $t$ and reinforce each other, doubling the total value. By even symmetry,
begin{align*}a_n &= frac23int_{-frac32}^{frac32}f(t)cosleft(frac{2pi nt}{3}right),dt = frac43int_{0}^{frac32}f(t)cosleft(frac{2pi nt}{3}right),dt\
a_n &= frac43int_0^1 tcosleft(frac{2pi nt}{3}right),dt+frac43int_1^{frac32}cosleft(frac{2pi nt}{3}right),dt\
a_n &= frac43left[frac{3t}{2pi n}sinleft(frac{2pi nt}{3}right)+frac{9}{4pi^2 n^2}cosleft(frac{2pi nt}{3}right)right]_0^1 +frac43left[frac{3}{2pi n}sinleft(frac{2pi nt}{3}right)right]_1^{frac32}\
a_n &= frac2{pi n}sinleft(frac{2pi n}{3}right) +frac{3}{pi^2 n^2}left(1-cosleft(frac{2pi n}{3}right)right) + frac2{pi n}left(sinpi n - sinleft(frac{2pi n}{3}right)right)\
a_n &= frac{3}{pi^2 n^2}+frac2{pi n}(sin(pi n))-frac{3}{pi n^2}cosleft(frac{2pi n}{3}right)\
a_n &= begin{cases}frac23& n=0\ 0& 3mid n,n>0\ frac{9}{2pi^2 n^2}& 3nmid nend{cases}end{align*}
In that last step, we use that $sin(pi n) = 0$ and $cosleft(frac{2pi n}{3}right)$ has a cycle of values $1,-frac12,-frac12$; subtract that from $1$ and it becomes $0,frac32,frac32$. Also, we incorporate the previously calculated (by a different method) $a_0$. The book's version of the answer applies the $sin(pi n)=0$ fact, but doesn't do anything with the $cosleft(frac{2pi n}{3}right)$ term.
So there it is. Incorporate the middle part of the integral as it should be for this even function, and we get the right answer.
answered Jan 24 at 21:58
jmerryjmerry
14.6k1632
$endgroup$
add a comment |
$begingroup$
since the two middle integrals cancel...
They don't. Our $f(t)=begin{cases}1&-frac32 le tle -1\-t&-1le tle 0\t&0le tle 1\1&1le tle frac32end{cases}$ is even. Integrate it against an odd function like $sinleft(frac{2pi nt}{3}right)$ and we'll get zero by odd symmetry. Integrate it against something even like $cosleft(frac{2pi nt}{3}right)$, and the parts for negative $t$ will mirror the parts for positive $t$ and reinforce each other, doubling the total value. By even symmetry,
begin{align*}a_n &= frac23int_{-frac32}^{frac32}f(t)cosleft(frac{2pi nt}{3}right),dt = frac43int_{0}^{frac32}f(t)cosleft(frac{2pi nt}{3}right),dt\
a_n &= frac43int_0^1 tcosleft(frac{2pi nt}{3}right),dt+frac43int_1^{frac32}cosleft(frac{2pi nt}{3}right),dt\
a_n &= frac43left[frac{3t}{2pi n}sinleft(frac{2pi nt}{3}right)+frac{9}{4pi^2 n^2}cosleft(frac{2pi nt}{3}right)right]_0^1 +frac43left[frac{3}{2pi n}sinleft(frac{2pi nt}{3}right)right]_1^{frac32}\
a_n &= frac2{pi n}sinleft(frac{2pi n}{3}right) +frac{3}{pi^2 n^2}left(1-cosleft(frac{2pi n}{3}right)right) + frac2{pi n}left(sinpi n - sinleft(frac{2pi n}{3}right)right)\
a_n &= frac{3}{pi^2 n^2}+frac2{pi n}(sin(pi n))-frac{3}{pi n^2}cosleft(frac{2pi n}{3}right)\
a_n &= begin{cases}frac23& n=0\ 0& 3mid n,n>0\ frac{9}{2pi^2 n^2}& 3nmid nend{cases}end{align*}
In that last step, we use that $sin(pi n) = 0$ and $cosleft(frac{2pi n}{3}right)$ has a cycle of values $1,-frac12,-frac12$; subtract that from $1$ and it becomes $0,frac32,frac32$. Also, we incorporate the previously calculated (by a different method) $a_0$. The book's version of the answer applies the $sin(pi n)=0$ fact, but doesn't do anything with the $cosleft(frac{2pi n}{3}right)$ term.
So there it is. Incorporate the middle part of the integral as it should be for this even function, and we get the right answer.
answered Jan 24 at 21:58
jmerryjmerry
14.6k1632
$endgroup$
since the two middle integrals cancel...
They don't. Our $f(t)=begin{cases}1&-frac32 le tle -1\-t&-1le tle 0\t&0le tle 1\1&1le tle frac32end{cases}$ is even. Integrate it against an odd function like $sinleft(frac{2pi nt}{3}right)$ and we'll get zero by odd symmetry. Integrate it against something even like $cosleft(frac{2pi nt}{3}right)$, and the parts for negative $t$ will mirror the parts for positive $t$ and reinforce each other, doubling the total value. By even symmetry,
begin{align*}a_n &= frac23int_{-frac32}^{frac32}f(t)cosleft(frac{2pi nt}{3}right),dt = frac43int_{0}^{frac32}f(t)cosleft(frac{2pi nt}{3}right),dt\
a_n &= frac43int_0^1 tcosleft(frac{2pi nt}{3}right),dt+frac43int_1^{frac32}cosleft(frac{2pi nt}{3}right),dt\
a_n &= frac43left[frac{3t}{2pi n}sinleft(frac{2pi nt}{3}right)+frac{9}{4pi^2 n^2}cosleft(frac{2pi nt}{3}right)right]_0^1 +frac43left[frac{3}{2pi n}sinleft(frac{2pi nt}{3}right)right]_1^{frac32}\
a_n &= frac2{pi n}sinleft(frac{2pi n}{3}right) +frac{3}{pi^2 n^2}left(1-cosleft(frac{2pi n}{3}right)right) + frac2{pi n}left(sinpi n - sinleft(frac{2pi n}{3}right)right)\
a_n &= frac{3}{pi^2 n^2}+frac2{pi n}(sin(pi n))-frac{3}{pi n^2}cosleft(frac{2pi n}{3}right)\
a_n &= begin{cases}frac23& n=0\ 0& 3mid n,n>0\ frac{9}{2pi^2 n^2}& 3nmid nend{cases}end{align*}
In that last step, we use that $sin(pi n) = 0$ and $cosleft(frac{2pi n}{3}right)$ has a cycle of values $1,-frac12,-frac12$; subtract that from $1$ and it becomes $0,frac32,frac32$. Also, we incorporate the previously calculated (by a different method) $a_0$. The book's version of the answer applies the $sin(pi n)=0$ fact, but doesn't do anything with the $cosleft(frac{2pi n}{3}right)$ term.
So there it is. Incorporate the middle part of the integral as it should be for this even function, and we get the right answer.
answered Jan 24 at 21:58
jmerryjmerry
14.6k1632
edited Jan 25 at 12:13
answered Jan 24 at 21:58
jmerryjmerry
14.6k1632
answered Jan 24 at 21:58
jmerryjmerry
14.6k1632
answered Jan 24 at 21:58
jmerryjmerry
14.6k1632
14.6k1632
add a comment |
add a comment |
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