Something about Fourier expansion
Suppose f is continuity on $[-pi,+pi],period ~T=2pi$
$$f(x)sim frac{a_0}{2}+sum_{n=1}^{+infty}a_n cos nx +b_nsin nx$$
$$F(x)=int_0^x[f(t)-frac{a_0}{2}]dt$$
1)Calculate the F(x) Fourier expansion
2)proof
$$int_a^bf(t)dt=int_a^b frac{a_0}{2}dt +sum_{n=1}^{infty}int_a^b a_n cos nx +b_nsin nx dt$$
My attempt
1)$a_n=int_{-pi}^{pi}f(t)cos ntdt$
$b_n=int_{-pi}^{pi}f(t)sin nt dt$
Then I don’t know what to do next
2)
I think it’s a little like uniformly convergence’character
But I don’t have idea to proof this
real-analysis calculus integration functional-analysis analysis
asked 2 days ago
jacksonjackson
758
add a comment |
Suppose f is continuity on $[-pi,+pi],period ~T=2pi$
$$f(x)sim frac{a_0}{2}+sum_{n=1}^{+infty}a_n cos nx +b_nsin nx$$
$$F(x)=int_0^x[f(t)-frac{a_0}{2}]dt$$
1)Calculate the F(x) Fourier expansion
2)proof
$$int_a^bf(t)dt=int_a^b frac{a_0}{2}dt +sum_{n=1}^{infty}int_a^b a_n cos nx +b_nsin nx dt$$
My attempt
1)$a_n=int_{-pi}^{pi}f(t)cos ntdt$
$b_n=int_{-pi}^{pi}f(t)sin nt dt$
Then I don’t know what to do next
2)
I think it’s a little like uniformly convergence’character
But I don’t have idea to proof this
real-analysis calculus integration functional-analysis analysis
asked 2 days ago
jacksonjackson
758
add a comment |
Suppose f is continuity on $[-pi,+pi],period ~T=2pi$
$$f(x)sim frac{a_0}{2}+sum_{n=1}^{+infty}a_n cos nx +b_nsin nx$$
$$F(x)=int_0^x[f(t)-frac{a_0}{2}]dt$$
1)Calculate the F(x) Fourier expansion
2)proof
$$int_a^bf(t)dt=int_a^b frac{a_0}{2}dt +sum_{n=1}^{infty}int_a^b a_n cos nx +b_nsin nx dt$$
My attempt
1)$a_n=int_{-pi}^{pi}f(t)cos ntdt$
$b_n=int_{-pi}^{pi}f(t)sin nt dt$
Then I don’t know what to do next
2)
I think it’s a little like uniformly convergence’character
But I don’t have idea to proof this
real-analysis calculus integration functional-analysis analysis
asked 2 days ago
jacksonjackson
758
Suppose f is continuity on $[-pi,+pi],period ~T=2pi$
$$f(x)sim frac{a_0}{2}+sum_{n=1}^{+infty}a_n cos nx +b_nsin nx$$
$$F(x)=int_0^x[f(t)-frac{a_0}{2}]dt$$
1)Calculate the F(x) Fourier expansion
2)proof
$$int_a^bf(t)dt=int_a^b frac{a_0}{2}dt +sum_{n=1}^{infty}int_a^b a_n cos nx +b_nsin nx dt$$
My attempt
1)$a_n=int_{-pi}^{pi}f(t)cos ntdt$
$b_n=int_{-pi}^{pi}f(t)sin nt dt$
Then I don’t know what to do next
2)
I think it’s a little like uniformly convergence’character
But I don’t have idea to proof this
real-analysis calculus integration functional-analysis analysis
real-analysis calculus integration functional-analysis analysis
asked 2 days ago
jacksonjackson
758
asked 2 days ago
jacksonjackson
758
asked 2 days ago
jacksonjackson
758
asked 2 days ago
jacksonjackson
758
asked 2 days ago
jacksonjackson
758
758
add a comment |
add a comment |
1 Answer
1
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oldest
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Let $$F(x) sim frac{c_0}{2}+sum_{n=1}^{infty}c_n cos nx +d_nsin nx.$$ Since $F$ is $2pi$-periodic $C^1$, note that the Fourier series of $F$ converges uniformly to $F$ and therefore equality holds ($alpha$-Holder continuity is one of the sufficient conditions for uniform convergence, see e.g. this earlier post). Now, we can find $c_n$ and $d_n$ as follows.
$$
pi c_n = int_{-pi}^pi F(t)cos nt dt = frac{sin nt}{n}F(t)|^{pi}_{-pi}-int_{-pi}^pi (f(t)-frac{a_0}{2})frac{sin nt}{n}dt = -pi frac{b_n}{n}, quad nneq 0,
$$
$$
pi d_n = int_{-pi}^pi F(t)sin nt dt = frac{-cos nt}{n}F(t)|^{pi}_{-pi}+int_{-pi}^pi (f(t)-frac{a_0}{2})frac{cos nt}{n}dt = pi frac{a_n}{n}.
$$ This gives
$$
F(x) = frac{c_0}{2}+sum_{n=1}^{infty}left(-frac{b_n}{n}right)cos nx +frac{a_n}{n}sin nxtag{*}.
$$ where $frac{c_0}{2} = sum_{n=1}^infty frac{b_n}{n}$ from $F(0) = 0$. It can be also calculated explicitly:
$$begin{eqnarray}
pi c_0 = int_{0}^{2pi} F(t) dt& =& (t-pi)F(t)|^{2pi}_{0}-int_{0}^{2pi}(t-pi) (f(t)-frac{a_0}{2})dt \&=&int_{0}^{2pi}left(sum_{n=1}^infty frac{2}{n}sin ntright) (f(t)-frac{a_0}{2})dt \
&=&2pisum_{n=1}^infty frac{b_n}{n}.
end{eqnarray}$$
For (b), note that the stament is equivalent to
$$
F(b) - F(a) = sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx.
$$ But this is obvious from $(*)$:
$$begin{eqnarray}
sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx &=&sum_{n=1}^{infty}frac{a_n}{n}(sin nb-sin na)+left(-frac{b_n}{n}right)(cos nb-cos na)\& =& F(b) - F(a).
end{eqnarray}$$
answered 2 days ago
SongSong
6,050318
i don’t understand this why F is $in C^1$then F is uniformly convergence
– jackson
2 days ago
And the five line integration by part is wrong
– jackson
2 days ago
It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
– jackson
2 days ago
The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
– jackson
yesterday
You are absolutely right ... Sorry for the confusion.
– Song
yesterday
add a comment |
Your Answer
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Let $$F(x) sim frac{c_0}{2}+sum_{n=1}^{infty}c_n cos nx +d_nsin nx.$$ Since $F$ is $2pi$-periodic $C^1$, note that the Fourier series of $F$ converges uniformly to $F$ and therefore equality holds ($alpha$-Holder continuity is one of the sufficient conditions for uniform convergence, see e.g. this earlier post). Now, we can find $c_n$ and $d_n$ as follows.
$$
pi c_n = int_{-pi}^pi F(t)cos nt dt = frac{sin nt}{n}F(t)|^{pi}_{-pi}-int_{-pi}^pi (f(t)-frac{a_0}{2})frac{sin nt}{n}dt = -pi frac{b_n}{n}, quad nneq 0,
$$
$$
pi d_n = int_{-pi}^pi F(t)sin nt dt = frac{-cos nt}{n}F(t)|^{pi}_{-pi}+int_{-pi}^pi (f(t)-frac{a_0}{2})frac{cos nt}{n}dt = pi frac{a_n}{n}.
$$ This gives
$$
F(x) = frac{c_0}{2}+sum_{n=1}^{infty}left(-frac{b_n}{n}right)cos nx +frac{a_n}{n}sin nxtag{*}.
$$ where $frac{c_0}{2} = sum_{n=1}^infty frac{b_n}{n}$ from $F(0) = 0$. It can be also calculated explicitly:
$$begin{eqnarray}
pi c_0 = int_{0}^{2pi} F(t) dt& =& (t-pi)F(t)|^{2pi}_{0}-int_{0}^{2pi}(t-pi) (f(t)-frac{a_0}{2})dt \&=&int_{0}^{2pi}left(sum_{n=1}^infty frac{2}{n}sin ntright) (f(t)-frac{a_0}{2})dt \
&=&2pisum_{n=1}^infty frac{b_n}{n}.
end{eqnarray}$$
For (b), note that the stament is equivalent to
$$
F(b) - F(a) = sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx.
$$ But this is obvious from $(*)$:
$$begin{eqnarray}
sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx &=&sum_{n=1}^{infty}frac{a_n}{n}(sin nb-sin na)+left(-frac{b_n}{n}right)(cos nb-cos na)\& =& F(b) - F(a).
end{eqnarray}$$
answered 2 days ago
SongSong
6,050318
i don’t understand this why F is $in C^1$then F is uniformly convergence
– jackson
2 days ago
And the five line integration by part is wrong
– jackson
2 days ago
It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
– jackson
2 days ago
The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
– jackson
yesterday
You are absolutely right ... Sorry for the confusion.
– Song
yesterday
add a comment |
Let $$F(x) sim frac{c_0}{2}+sum_{n=1}^{infty}c_n cos nx +d_nsin nx.$$ Since $F$ is $2pi$-periodic $C^1$, note that the Fourier series of $F$ converges uniformly to $F$ and therefore equality holds ($alpha$-Holder continuity is one of the sufficient conditions for uniform convergence, see e.g. this earlier post). Now, we can find $c_n$ and $d_n$ as follows.
$$
pi c_n = int_{-pi}^pi F(t)cos nt dt = frac{sin nt}{n}F(t)|^{pi}_{-pi}-int_{-pi}^pi (f(t)-frac{a_0}{2})frac{sin nt}{n}dt = -pi frac{b_n}{n}, quad nneq 0,
$$
$$
pi d_n = int_{-pi}^pi F(t)sin nt dt = frac{-cos nt}{n}F(t)|^{pi}_{-pi}+int_{-pi}^pi (f(t)-frac{a_0}{2})frac{cos nt}{n}dt = pi frac{a_n}{n}.
$$ This gives
$$
F(x) = frac{c_0}{2}+sum_{n=1}^{infty}left(-frac{b_n}{n}right)cos nx +frac{a_n}{n}sin nxtag{*}.
$$ where $frac{c_0}{2} = sum_{n=1}^infty frac{b_n}{n}$ from $F(0) = 0$. It can be also calculated explicitly:
$$begin{eqnarray}
pi c_0 = int_{0}^{2pi} F(t) dt& =& (t-pi)F(t)|^{2pi}_{0}-int_{0}^{2pi}(t-pi) (f(t)-frac{a_0}{2})dt \&=&int_{0}^{2pi}left(sum_{n=1}^infty frac{2}{n}sin ntright) (f(t)-frac{a_0}{2})dt \
&=&2pisum_{n=1}^infty frac{b_n}{n}.
end{eqnarray}$$
For (b), note that the stament is equivalent to
$$
F(b) - F(a) = sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx.
$$ But this is obvious from $(*)$:
$$begin{eqnarray}
sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx &=&sum_{n=1}^{infty}frac{a_n}{n}(sin nb-sin na)+left(-frac{b_n}{n}right)(cos nb-cos na)\& =& F(b) - F(a).
end{eqnarray}$$
answered 2 days ago
SongSong
6,050318
i don’t understand this why F is $in C^1$then F is uniformly convergence
– jackson
2 days ago
And the five line integration by part is wrong
– jackson
2 days ago
It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
– jackson
2 days ago
The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
– jackson
yesterday
You are absolutely right ... Sorry for the confusion.
– Song
yesterday
add a comment |
Let $$F(x) sim frac{c_0}{2}+sum_{n=1}^{infty}c_n cos nx +d_nsin nx.$$ Since $F$ is $2pi$-periodic $C^1$, note that the Fourier series of $F$ converges uniformly to $F$ and therefore equality holds ($alpha$-Holder continuity is one of the sufficient conditions for uniform convergence, see e.g. this earlier post). Now, we can find $c_n$ and $d_n$ as follows.
$$
pi c_n = int_{-pi}^pi F(t)cos nt dt = frac{sin nt}{n}F(t)|^{pi}_{-pi}-int_{-pi}^pi (f(t)-frac{a_0}{2})frac{sin nt}{n}dt = -pi frac{b_n}{n}, quad nneq 0,
$$
$$
pi d_n = int_{-pi}^pi F(t)sin nt dt = frac{-cos nt}{n}F(t)|^{pi}_{-pi}+int_{-pi}^pi (f(t)-frac{a_0}{2})frac{cos nt}{n}dt = pi frac{a_n}{n}.
$$ This gives
$$
F(x) = frac{c_0}{2}+sum_{n=1}^{infty}left(-frac{b_n}{n}right)cos nx +frac{a_n}{n}sin nxtag{*}.
$$ where $frac{c_0}{2} = sum_{n=1}^infty frac{b_n}{n}$ from $F(0) = 0$. It can be also calculated explicitly:
$$begin{eqnarray}
pi c_0 = int_{0}^{2pi} F(t) dt& =& (t-pi)F(t)|^{2pi}_{0}-int_{0}^{2pi}(t-pi) (f(t)-frac{a_0}{2})dt \&=&int_{0}^{2pi}left(sum_{n=1}^infty frac{2}{n}sin ntright) (f(t)-frac{a_0}{2})dt \
&=&2pisum_{n=1}^infty frac{b_n}{n}.
end{eqnarray}$$
For (b), note that the stament is equivalent to
$$
F(b) - F(a) = sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx.
$$ But this is obvious from $(*)$:
$$begin{eqnarray}
sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx &=&sum_{n=1}^{infty}frac{a_n}{n}(sin nb-sin na)+left(-frac{b_n}{n}right)(cos nb-cos na)\& =& F(b) - F(a).
end{eqnarray}$$
answered 2 days ago
SongSong
6,050318
Let $$F(x) sim frac{c_0}{2}+sum_{n=1}^{infty}c_n cos nx +d_nsin nx.$$ Since $F$ is $2pi$-periodic $C^1$, note that the Fourier series of $F$ converges uniformly to $F$ and therefore equality holds ($alpha$-Holder continuity is one of the sufficient conditions for uniform convergence, see e.g. this earlier post). Now, we can find $c_n$ and $d_n$ as follows.
$$
pi c_n = int_{-pi}^pi F(t)cos nt dt = frac{sin nt}{n}F(t)|^{pi}_{-pi}-int_{-pi}^pi (f(t)-frac{a_0}{2})frac{sin nt}{n}dt = -pi frac{b_n}{n}, quad nneq 0,
$$
$$
pi d_n = int_{-pi}^pi F(t)sin nt dt = frac{-cos nt}{n}F(t)|^{pi}_{-pi}+int_{-pi}^pi (f(t)-frac{a_0}{2})frac{cos nt}{n}dt = pi frac{a_n}{n}.
$$ This gives
$$
F(x) = frac{c_0}{2}+sum_{n=1}^{infty}left(-frac{b_n}{n}right)cos nx +frac{a_n}{n}sin nxtag{*}.
$$ where $frac{c_0}{2} = sum_{n=1}^infty frac{b_n}{n}$ from $F(0) = 0$. It can be also calculated explicitly:
$$begin{eqnarray}
pi c_0 = int_{0}^{2pi} F(t) dt& =& (t-pi)F(t)|^{2pi}_{0}-int_{0}^{2pi}(t-pi) (f(t)-frac{a_0}{2})dt \&=&int_{0}^{2pi}left(sum_{n=1}^infty frac{2}{n}sin ntright) (f(t)-frac{a_0}{2})dt \
&=&2pisum_{n=1}^infty frac{b_n}{n}.
end{eqnarray}$$
For (b), note that the stament is equivalent to
$$
F(b) - F(a) = sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx.
$$ But this is obvious from $(*)$:
$$begin{eqnarray}
sum_{n=1}^infty int_a^b left(a_n cos nx +b_nsin nxright)dx &=&sum_{n=1}^{infty}frac{a_n}{n}(sin nb-sin na)+left(-frac{b_n}{n}right)(cos nb-cos na)\& =& F(b) - F(a).
end{eqnarray}$$
answered 2 days ago
SongSong
6,050318
edited yesterday
answered 2 days ago
SongSong
6,050318
answered 2 days ago
SongSong
6,050318
answered 2 days ago
SongSong
6,050318
6,050318
i don’t understand this why F is $in C^1$then F is uniformly convergence
– jackson
2 days ago
And the five line integration by part is wrong
– jackson
2 days ago
It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
– jackson
2 days ago
The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
– jackson
yesterday
You are absolutely right ... Sorry for the confusion.
– Song
yesterday
add a comment |
i don’t understand this why F is $in C^1$then F is uniformly convergence
– jackson
2 days ago
And the five line integration by part is wrong
– jackson
2 days ago
It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
– jackson
2 days ago
The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
– jackson
yesterday
You are absolutely right ... Sorry for the confusion.
– Song
yesterday
i don’t understand this why F is $in C^1$then F is uniformly convergence
– jackson
2 days ago
i don’t understand this why F is $in C^1$then F is uniformly convergence
– jackson
2 days ago
And the five line integration by part is wrong
– jackson
2 days ago
And the five line integration by part is wrong
– jackson
2 days ago
It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
– jackson
2 days ago
It should be$pi c_0=tF(t)|_{-pi}^{pi}-int_{-pi}^{+pi}(f(t)-frac{a_0}{2}) dt$ and I don’t know how to proof it equal to 0
– jackson
2 days ago
The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
– jackson
yesterday
The five line the left of $c_n$ should be $pi$?because $c_n=frac{1}{pi}int_0^{2pi}F(t)cos ntdt$ so do the next line ,am I right?
– jackson
yesterday
You are absolutely right ... Sorry for the confusion.
– Song
yesterday
You are absolutely right ... Sorry for the confusion.
– Song
yesterday
add a comment |
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