Coefficients in Generalised Fourier series Given By $c_k = langle f, psi_k rangle$? $c_k$ Implicitly Chosen So That Sum Converges?

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My notes state the following:
The coefficients $c_k$ in the generalised Fourier series
$$f(x) = sumlimits_k = 1^infty c_k phi_k(x)$$
, with respect to the orthonormal set, $ psi_k$,
$$f(x) = sumlimits_k = 0^infty c_k psi_k(x)$$
, are given by $c_k = langle f, psi_k rangle$.
Note: We have implicitly assumed that it is possible to choose the coefficients $c_k$ so that the sum
$$sumlimits_k = 0^N c_k psi_k(x)$$
converges to $f(x)$ as $N to infty$.
I have two questions about this:
Why are the coefficients $c_k$ in the generalised Fourier series, with respect to the orthonormal set, given by $c_k = langle f, psi_k rangle$? Can someone please demonstrate this?
How is it that we have implicitly assumed that it is possible to choose the coefficient $c_k$ so that the sum converges to $f(x)$ as $N to infty$? I don't see why this is the case.
I would greatly appreciate it if people could please take the time to clarify this.
real-analysis functional-analysis fourier-analysis fourier-series inner-product-space
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up vote
5
down vote
favorite
My notes state the following:
The coefficients $c_k$ in the generalised Fourier series
$$f(x) = sumlimits_k = 1^infty c_k phi_k(x)$$
, with respect to the orthonormal set, $ psi_k$,
$$f(x) = sumlimits_k = 0^infty c_k psi_k(x)$$
, are given by $c_k = langle f, psi_k rangle$.
Note: We have implicitly assumed that it is possible to choose the coefficients $c_k$ so that the sum
$$sumlimits_k = 0^N c_k psi_k(x)$$
converges to $f(x)$ as $N to infty$.
I have two questions about this:
Why are the coefficients $c_k$ in the generalised Fourier series, with respect to the orthonormal set, given by $c_k = langle f, psi_k rangle$? Can someone please demonstrate this?
How is it that we have implicitly assumed that it is possible to choose the coefficient $c_k$ so that the sum converges to $f(x)$ as $N to infty$? I don't see why this is the case.
I would greatly appreciate it if people could please take the time to clarify this.
real-analysis functional-analysis fourier-analysis fourier-series inner-product-space
You have $phi_k$ and then $psi_k?$
â zhw.
Aug 21 at 21:56
@zhw. Yes. I copied it straight from the notes.
â The Pointer
Aug 21 at 23:21
Do you know the basics of Lebesgue integration and $L^2$ spaces?
â zhw.
Aug 24 at 18:10
@zhw. No, thatâÂÂs beyond my current level.
â The Pointer
Aug 24 at 21:27
You understand uniform convergence?
â zhw.
Aug 24 at 21:47
 |Â
show 1 more comment
up vote
5
down vote
favorite
up vote
5
down vote
favorite
My notes state the following:
The coefficients $c_k$ in the generalised Fourier series
$$f(x) = sumlimits_k = 1^infty c_k phi_k(x)$$
, with respect to the orthonormal set, $ psi_k$,
$$f(x) = sumlimits_k = 0^infty c_k psi_k(x)$$
, are given by $c_k = langle f, psi_k rangle$.
Note: We have implicitly assumed that it is possible to choose the coefficients $c_k$ so that the sum
$$sumlimits_k = 0^N c_k psi_k(x)$$
converges to $f(x)$ as $N to infty$.
I have two questions about this:
Why are the coefficients $c_k$ in the generalised Fourier series, with respect to the orthonormal set, given by $c_k = langle f, psi_k rangle$? Can someone please demonstrate this?
How is it that we have implicitly assumed that it is possible to choose the coefficient $c_k$ so that the sum converges to $f(x)$ as $N to infty$? I don't see why this is the case.
I would greatly appreciate it if people could please take the time to clarify this.
real-analysis functional-analysis fourier-analysis fourier-series inner-product-space
My notes state the following:
The coefficients $c_k$ in the generalised Fourier series
$$f(x) = sumlimits_k = 1^infty c_k phi_k(x)$$
, with respect to the orthonormal set, $ psi_k$,
$$f(x) = sumlimits_k = 0^infty c_k psi_k(x)$$
, are given by $c_k = langle f, psi_k rangle$.
Note: We have implicitly assumed that it is possible to choose the coefficients $c_k$ so that the sum
$$sumlimits_k = 0^N c_k psi_k(x)$$
converges to $f(x)$ as $N to infty$.
I have two questions about this:
Why are the coefficients $c_k$ in the generalised Fourier series, with respect to the orthonormal set, given by $c_k = langle f, psi_k rangle$? Can someone please demonstrate this?
How is it that we have implicitly assumed that it is possible to choose the coefficient $c_k$ so that the sum converges to $f(x)$ as $N to infty$? I don't see why this is the case.
I would greatly appreciate it if people could please take the time to clarify this.
real-analysis functional-analysis fourier-analysis fourier-series inner-product-space
asked Aug 17 at 23:10
The Pointer
2,7372830
2,7372830
You have $phi_k$ and then $psi_k?$
â zhw.
Aug 21 at 21:56
@zhw. Yes. I copied it straight from the notes.
â The Pointer
Aug 21 at 23:21
Do you know the basics of Lebesgue integration and $L^2$ spaces?
â zhw.
Aug 24 at 18:10
@zhw. No, thatâÂÂs beyond my current level.
â The Pointer
Aug 24 at 21:27
You understand uniform convergence?
â zhw.
Aug 24 at 21:47
 |Â
show 1 more comment
You have $phi_k$ and then $psi_k?$
â zhw.
Aug 21 at 21:56
@zhw. Yes. I copied it straight from the notes.
â The Pointer
Aug 21 at 23:21
Do you know the basics of Lebesgue integration and $L^2$ spaces?
â zhw.
Aug 24 at 18:10
@zhw. No, thatâÂÂs beyond my current level.
â The Pointer
Aug 24 at 21:27
You understand uniform convergence?
â zhw.
Aug 24 at 21:47
You have $phi_k$ and then $psi_k?$
â zhw.
Aug 21 at 21:56
You have $phi_k$ and then $psi_k?$
â zhw.
Aug 21 at 21:56
@zhw. Yes. I copied it straight from the notes.
â The Pointer
Aug 21 at 23:21
@zhw. Yes. I copied it straight from the notes.
â The Pointer
Aug 21 at 23:21
Do you know the basics of Lebesgue integration and $L^2$ spaces?
â zhw.
Aug 24 at 18:10
Do you know the basics of Lebesgue integration and $L^2$ spaces?
â zhw.
Aug 24 at 18:10
@zhw. No, thatâÂÂs beyond my current level.
â The Pointer
Aug 24 at 21:27
@zhw. No, thatâÂÂs beyond my current level.
â The Pointer
Aug 24 at 21:27
You understand uniform convergence?
â zhw.
Aug 24 at 21:47
You understand uniform convergence?
â zhw.
Aug 24 at 21:47
 |Â
show 1 more comment
4 Answers
4
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oldest
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up vote
3
down vote
accepted
The statement about convergence is vague. But here is the basic idea: Suppose $psi_k$ is orthonormal on $[a,b],$ and there are constants $c_k$ such that
$$f(x) =sum_k=1^inftyc_kpsi_k(x),,,xin [a,b].$$
Then the $c_k$ are given by the formula
$$c_k = int_a^b f(x)psi_k(x),dx, ,,k=1,2,dots.$$
Proof: Fix $k=k_0.$ Then
$$int_a^b f(x)psi_k_0(x),dx = int_a^b left (sum_k=1^inftyc_kpsi_k(x)right )psi_k_0(x),dx$$ $$ =int_a^b left (sum_k=1^inftyc_kpsi_k(x)psi_k_0(x)right),dx = sum_k=1^inftyc_kint_a^b psi_k(x)psi_k_0(x),dx$$ $$ = c_k_0int_a^b psi_k_0(x)psi_k_0(x),dx = c_k_0cdot 1 = c_k_0.$$
If you've never seen this kind of argument before, the first expression may look like it's pulled out of a hat. But it's the thing that works. We get to the second expression by simple substitution for $f.$ The third expression should explain itself. In the fourth expression we have switched summation and integration. This needs justification, but with your permission we just sail on through. The rest of the expressions come about by orthonormality: Those integrals equal $0$ except when $k=k_0,$ where it equals $1.$
I'm hoping this answers question 1. For question 2. the answer is simple: If we don't have coefficients $c_k$ such that $sum c_kpsi_k =f$ in some manner, then how could we substitute $sum c_kpsi_k$ in for $f$ in the second integral?
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up vote
2
down vote
If the orthonormal set is complete then the $c_k$'s exist for any square integrable function $f$. Otherwise there exist square integrable functions for which the coefficients do not exist. If the orthonormal set is complete then the expansion $f=sum c_k psi_k$ is valid in $L^2$ norm. This means that if $S_n=sum_k=1^n c_k psi_k$ then $int |S_n-f|^2 to 0$. This implies $int f(x) psi _j (x), dx =lim int f(x) S_n(x), dx$ Using orhtonormality we see that $int f(x) s_n(x), dx=c_j$ for all $n geq j$ which gives the formula for $c_j$.
1
Thanks for the answer. Can you please rephrase this in a way that's easier for a novice to understand?
â The Pointer
Aug 17 at 23:43
1
The arguments in my answer are general ones valid for orthonormal bases in an inner product spaces; they are not specific to functions. My suggestion for you is to take a look at inner product spaces and orthonormal bases. There are plenty of places where you can find these including Wikipedia pages.
â Kavi Rama Murthy
Aug 17 at 23:47
add a comment |Â
up vote
1
down vote
For your second question, you just need to recall the definition of convergence for series:
Given the series $sum_k=0^infty v_k$, define the sequence $s_N$ of partial sums as $s_N=sum_k=0^N v_k$. Then, it is said that the series converges to $S$, or that $sum_k=0^infty v_k=S$ if $s_Nto S$.
So, if you say that $f(x)=sum_k=0^infty c_kpsi_k$, by definition you're saying that the sequence $s_N=sum_k=0^N c_kpsi_k$ converges to $f(x)$. Then, you're assuming the existence of such $c_k$'s.
To answer your first question, you just need to use the linearity of the inner product:
$$(f,psi_k)=(sum_n=0^infty c_npsi_n,psi_k)=sum_n=0^infty c_n(psi_n,psi_k)=c_k,$$
where the last equation is a consequence of the set $psi_k$ being orthonormal, since $(psi_n,psi_k)=0$ if $nneq k$ and $(psi_k,psi_k)=1$.
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up vote
1
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When defining an orthonormal set $psi_i(x)$ we have that$$int_Dpsi_i(x)psi_j^*(x)dx=begincases1&,i=j\0&,ine jendcases$$where $D$ is the common domain of all $psi_i(x)$ (take it $Bbb R$) and $*$ operator denotes the complex conjugate. Using this definition $$f(x)=sum_n=1^inftyc_npsi_n(x)$$therefore $$int_Df(x)psi_k^*(x)dx=sum_n=1^inftyc_nint_Dpsi_n(x)psi_k^*(x)dx=sum_n=1^inftyc_ndelta_n-k=c_k$$which completes the proof of question 1.
The answer to the question 2 is a little hard and tricky. As far as I know there are no known necessary and sufficient conditions to ensure existence of such coefficients neither in classical Fourier Series nor in the general case (for example in the case of Fourier Series we may choose some non-Reimann integrable periodic function such as Dirichlet function), but we have some sufficient conditions called Dirichlet conditions after a German mathematician Peter Gustav Lejeune Dirichlet. You may refer to it in https://en.wikipedia.org/wiki/Dirichlet_conditions
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The statement about convergence is vague. But here is the basic idea: Suppose $psi_k$ is orthonormal on $[a,b],$ and there are constants $c_k$ such that
$$f(x) =sum_k=1^inftyc_kpsi_k(x),,,xin [a,b].$$
Then the $c_k$ are given by the formula
$$c_k = int_a^b f(x)psi_k(x),dx, ,,k=1,2,dots.$$
Proof: Fix $k=k_0.$ Then
$$int_a^b f(x)psi_k_0(x),dx = int_a^b left (sum_k=1^inftyc_kpsi_k(x)right )psi_k_0(x),dx$$ $$ =int_a^b left (sum_k=1^inftyc_kpsi_k(x)psi_k_0(x)right),dx = sum_k=1^inftyc_kint_a^b psi_k(x)psi_k_0(x),dx$$ $$ = c_k_0int_a^b psi_k_0(x)psi_k_0(x),dx = c_k_0cdot 1 = c_k_0.$$
If you've never seen this kind of argument before, the first expression may look like it's pulled out of a hat. But it's the thing that works. We get to the second expression by simple substitution for $f.$ The third expression should explain itself. In the fourth expression we have switched summation and integration. This needs justification, but with your permission we just sail on through. The rest of the expressions come about by orthonormality: Those integrals equal $0$ except when $k=k_0,$ where it equals $1.$
I'm hoping this answers question 1. For question 2. the answer is simple: If we don't have coefficients $c_k$ such that $sum c_kpsi_k =f$ in some manner, then how could we substitute $sum c_kpsi_k$ in for $f$ in the second integral?
add a comment |Â
up vote
3
down vote
accepted
The statement about convergence is vague. But here is the basic idea: Suppose $psi_k$ is orthonormal on $[a,b],$ and there are constants $c_k$ such that
$$f(x) =sum_k=1^inftyc_kpsi_k(x),,,xin [a,b].$$
Then the $c_k$ are given by the formula
$$c_k = int_a^b f(x)psi_k(x),dx, ,,k=1,2,dots.$$
Proof: Fix $k=k_0.$ Then
$$int_a^b f(x)psi_k_0(x),dx = int_a^b left (sum_k=1^inftyc_kpsi_k(x)right )psi_k_0(x),dx$$ $$ =int_a^b left (sum_k=1^inftyc_kpsi_k(x)psi_k_0(x)right),dx = sum_k=1^inftyc_kint_a^b psi_k(x)psi_k_0(x),dx$$ $$ = c_k_0int_a^b psi_k_0(x)psi_k_0(x),dx = c_k_0cdot 1 = c_k_0.$$
If you've never seen this kind of argument before, the first expression may look like it's pulled out of a hat. But it's the thing that works. We get to the second expression by simple substitution for $f.$ The third expression should explain itself. In the fourth expression we have switched summation and integration. This needs justification, but with your permission we just sail on through. The rest of the expressions come about by orthonormality: Those integrals equal $0$ except when $k=k_0,$ where it equals $1.$
I'm hoping this answers question 1. For question 2. the answer is simple: If we don't have coefficients $c_k$ such that $sum c_kpsi_k =f$ in some manner, then how could we substitute $sum c_kpsi_k$ in for $f$ in the second integral?
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The statement about convergence is vague. But here is the basic idea: Suppose $psi_k$ is orthonormal on $[a,b],$ and there are constants $c_k$ such that
$$f(x) =sum_k=1^inftyc_kpsi_k(x),,,xin [a,b].$$
Then the $c_k$ are given by the formula
$$c_k = int_a^b f(x)psi_k(x),dx, ,,k=1,2,dots.$$
Proof: Fix $k=k_0.$ Then
$$int_a^b f(x)psi_k_0(x),dx = int_a^b left (sum_k=1^inftyc_kpsi_k(x)right )psi_k_0(x),dx$$ $$ =int_a^b left (sum_k=1^inftyc_kpsi_k(x)psi_k_0(x)right),dx = sum_k=1^inftyc_kint_a^b psi_k(x)psi_k_0(x),dx$$ $$ = c_k_0int_a^b psi_k_0(x)psi_k_0(x),dx = c_k_0cdot 1 = c_k_0.$$
If you've never seen this kind of argument before, the first expression may look like it's pulled out of a hat. But it's the thing that works. We get to the second expression by simple substitution for $f.$ The third expression should explain itself. In the fourth expression we have switched summation and integration. This needs justification, but with your permission we just sail on through. The rest of the expressions come about by orthonormality: Those integrals equal $0$ except when $k=k_0,$ where it equals $1.$
I'm hoping this answers question 1. For question 2. the answer is simple: If we don't have coefficients $c_k$ such that $sum c_kpsi_k =f$ in some manner, then how could we substitute $sum c_kpsi_k$ in for $f$ in the second integral?
The statement about convergence is vague. But here is the basic idea: Suppose $psi_k$ is orthonormal on $[a,b],$ and there are constants $c_k$ such that
$$f(x) =sum_k=1^inftyc_kpsi_k(x),,,xin [a,b].$$
Then the $c_k$ are given by the formula
$$c_k = int_a^b f(x)psi_k(x),dx, ,,k=1,2,dots.$$
Proof: Fix $k=k_0.$ Then
$$int_a^b f(x)psi_k_0(x),dx = int_a^b left (sum_k=1^inftyc_kpsi_k(x)right )psi_k_0(x),dx$$ $$ =int_a^b left (sum_k=1^inftyc_kpsi_k(x)psi_k_0(x)right),dx = sum_k=1^inftyc_kint_a^b psi_k(x)psi_k_0(x),dx$$ $$ = c_k_0int_a^b psi_k_0(x)psi_k_0(x),dx = c_k_0cdot 1 = c_k_0.$$
If you've never seen this kind of argument before, the first expression may look like it's pulled out of a hat. But it's the thing that works. We get to the second expression by simple substitution for $f.$ The third expression should explain itself. In the fourth expression we have switched summation and integration. This needs justification, but with your permission we just sail on through. The rest of the expressions come about by orthonormality: Those integrals equal $0$ except when $k=k_0,$ where it equals $1.$
I'm hoping this answers question 1. For question 2. the answer is simple: If we don't have coefficients $c_k$ such that $sum c_kpsi_k =f$ in some manner, then how could we substitute $sum c_kpsi_k$ in for $f$ in the second integral?
edited Aug 24 at 23:05
answered Aug 24 at 22:50
zhw.
66.7k42872
66.7k42872
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up vote
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If the orthonormal set is complete then the $c_k$'s exist for any square integrable function $f$. Otherwise there exist square integrable functions for which the coefficients do not exist. If the orthonormal set is complete then the expansion $f=sum c_k psi_k$ is valid in $L^2$ norm. This means that if $S_n=sum_k=1^n c_k psi_k$ then $int |S_n-f|^2 to 0$. This implies $int f(x) psi _j (x), dx =lim int f(x) S_n(x), dx$ Using orhtonormality we see that $int f(x) s_n(x), dx=c_j$ for all $n geq j$ which gives the formula for $c_j$.
1
Thanks for the answer. Can you please rephrase this in a way that's easier for a novice to understand?
â The Pointer
Aug 17 at 23:43
1
The arguments in my answer are general ones valid for orthonormal bases in an inner product spaces; they are not specific to functions. My suggestion for you is to take a look at inner product spaces and orthonormal bases. There are plenty of places where you can find these including Wikipedia pages.
â Kavi Rama Murthy
Aug 17 at 23:47
add a comment |Â
up vote
2
down vote
If the orthonormal set is complete then the $c_k$'s exist for any square integrable function $f$. Otherwise there exist square integrable functions for which the coefficients do not exist. If the orthonormal set is complete then the expansion $f=sum c_k psi_k$ is valid in $L^2$ norm. This means that if $S_n=sum_k=1^n c_k psi_k$ then $int |S_n-f|^2 to 0$. This implies $int f(x) psi _j (x), dx =lim int f(x) S_n(x), dx$ Using orhtonormality we see that $int f(x) s_n(x), dx=c_j$ for all $n geq j$ which gives the formula for $c_j$.
1
Thanks for the answer. Can you please rephrase this in a way that's easier for a novice to understand?
â The Pointer
Aug 17 at 23:43
1
The arguments in my answer are general ones valid for orthonormal bases in an inner product spaces; they are not specific to functions. My suggestion for you is to take a look at inner product spaces and orthonormal bases. There are plenty of places where you can find these including Wikipedia pages.
â Kavi Rama Murthy
Aug 17 at 23:47
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If the orthonormal set is complete then the $c_k$'s exist for any square integrable function $f$. Otherwise there exist square integrable functions for which the coefficients do not exist. If the orthonormal set is complete then the expansion $f=sum c_k psi_k$ is valid in $L^2$ norm. This means that if $S_n=sum_k=1^n c_k psi_k$ then $int |S_n-f|^2 to 0$. This implies $int f(x) psi _j (x), dx =lim int f(x) S_n(x), dx$ Using orhtonormality we see that $int f(x) s_n(x), dx=c_j$ for all $n geq j$ which gives the formula for $c_j$.
If the orthonormal set is complete then the $c_k$'s exist for any square integrable function $f$. Otherwise there exist square integrable functions for which the coefficients do not exist. If the orthonormal set is complete then the expansion $f=sum c_k psi_k$ is valid in $L^2$ norm. This means that if $S_n=sum_k=1^n c_k psi_k$ then $int |S_n-f|^2 to 0$. This implies $int f(x) psi _j (x), dx =lim int f(x) S_n(x), dx$ Using orhtonormality we see that $int f(x) s_n(x), dx=c_j$ for all $n geq j$ which gives the formula for $c_j$.
answered Aug 17 at 23:27
Kavi Rama Murthy
22.9k2933
22.9k2933
1
Thanks for the answer. Can you please rephrase this in a way that's easier for a novice to understand?
â The Pointer
Aug 17 at 23:43
1
The arguments in my answer are general ones valid for orthonormal bases in an inner product spaces; they are not specific to functions. My suggestion for you is to take a look at inner product spaces and orthonormal bases. There are plenty of places where you can find these including Wikipedia pages.
â Kavi Rama Murthy
Aug 17 at 23:47
add a comment |Â
1
Thanks for the answer. Can you please rephrase this in a way that's easier for a novice to understand?
â The Pointer
Aug 17 at 23:43
1
The arguments in my answer are general ones valid for orthonormal bases in an inner product spaces; they are not specific to functions. My suggestion for you is to take a look at inner product spaces and orthonormal bases. There are plenty of places where you can find these including Wikipedia pages.
â Kavi Rama Murthy
Aug 17 at 23:47
1
1
Thanks for the answer. Can you please rephrase this in a way that's easier for a novice to understand?
â The Pointer
Aug 17 at 23:43
Thanks for the answer. Can you please rephrase this in a way that's easier for a novice to understand?
â The Pointer
Aug 17 at 23:43
1
1
The arguments in my answer are general ones valid for orthonormal bases in an inner product spaces; they are not specific to functions. My suggestion for you is to take a look at inner product spaces and orthonormal bases. There are plenty of places where you can find these including Wikipedia pages.
â Kavi Rama Murthy
Aug 17 at 23:47
The arguments in my answer are general ones valid for orthonormal bases in an inner product spaces; they are not specific to functions. My suggestion for you is to take a look at inner product spaces and orthonormal bases. There are plenty of places where you can find these including Wikipedia pages.
â Kavi Rama Murthy
Aug 17 at 23:47
add a comment |Â
up vote
1
down vote
For your second question, you just need to recall the definition of convergence for series:
Given the series $sum_k=0^infty v_k$, define the sequence $s_N$ of partial sums as $s_N=sum_k=0^N v_k$. Then, it is said that the series converges to $S$, or that $sum_k=0^infty v_k=S$ if $s_Nto S$.
So, if you say that $f(x)=sum_k=0^infty c_kpsi_k$, by definition you're saying that the sequence $s_N=sum_k=0^N c_kpsi_k$ converges to $f(x)$. Then, you're assuming the existence of such $c_k$'s.
To answer your first question, you just need to use the linearity of the inner product:
$$(f,psi_k)=(sum_n=0^infty c_npsi_n,psi_k)=sum_n=0^infty c_n(psi_n,psi_k)=c_k,$$
where the last equation is a consequence of the set $psi_k$ being orthonormal, since $(psi_n,psi_k)=0$ if $nneq k$ and $(psi_k,psi_k)=1$.
add a comment |Â
up vote
1
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For your second question, you just need to recall the definition of convergence for series:
Given the series $sum_k=0^infty v_k$, define the sequence $s_N$ of partial sums as $s_N=sum_k=0^N v_k$. Then, it is said that the series converges to $S$, or that $sum_k=0^infty v_k=S$ if $s_Nto S$.
So, if you say that $f(x)=sum_k=0^infty c_kpsi_k$, by definition you're saying that the sequence $s_N=sum_k=0^N c_kpsi_k$ converges to $f(x)$. Then, you're assuming the existence of such $c_k$'s.
To answer your first question, you just need to use the linearity of the inner product:
$$(f,psi_k)=(sum_n=0^infty c_npsi_n,psi_k)=sum_n=0^infty c_n(psi_n,psi_k)=c_k,$$
where the last equation is a consequence of the set $psi_k$ being orthonormal, since $(psi_n,psi_k)=0$ if $nneq k$ and $(psi_k,psi_k)=1$.
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1
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up vote
1
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For your second question, you just need to recall the definition of convergence for series:
Given the series $sum_k=0^infty v_k$, define the sequence $s_N$ of partial sums as $s_N=sum_k=0^N v_k$. Then, it is said that the series converges to $S$, or that $sum_k=0^infty v_k=S$ if $s_Nto S$.
So, if you say that $f(x)=sum_k=0^infty c_kpsi_k$, by definition you're saying that the sequence $s_N=sum_k=0^N c_kpsi_k$ converges to $f(x)$. Then, you're assuming the existence of such $c_k$'s.
To answer your first question, you just need to use the linearity of the inner product:
$$(f,psi_k)=(sum_n=0^infty c_npsi_n,psi_k)=sum_n=0^infty c_n(psi_n,psi_k)=c_k,$$
where the last equation is a consequence of the set $psi_k$ being orthonormal, since $(psi_n,psi_k)=0$ if $nneq k$ and $(psi_k,psi_k)=1$.
For your second question, you just need to recall the definition of convergence for series:
Given the series $sum_k=0^infty v_k$, define the sequence $s_N$ of partial sums as $s_N=sum_k=0^N v_k$. Then, it is said that the series converges to $S$, or that $sum_k=0^infty v_k=S$ if $s_Nto S$.
So, if you say that $f(x)=sum_k=0^infty c_kpsi_k$, by definition you're saying that the sequence $s_N=sum_k=0^N c_kpsi_k$ converges to $f(x)$. Then, you're assuming the existence of such $c_k$'s.
To answer your first question, you just need to use the linearity of the inner product:
$$(f,psi_k)=(sum_n=0^infty c_npsi_n,psi_k)=sum_n=0^infty c_n(psi_n,psi_k)=c_k,$$
where the last equation is a consequence of the set $psi_k$ being orthonormal, since $(psi_n,psi_k)=0$ if $nneq k$ and $(psi_k,psi_k)=1$.
answered Aug 20 at 3:54
Sergio Enrique Yarza Acuña
600314
600314
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When defining an orthonormal set $psi_i(x)$ we have that$$int_Dpsi_i(x)psi_j^*(x)dx=begincases1&,i=j\0&,ine jendcases$$where $D$ is the common domain of all $psi_i(x)$ (take it $Bbb R$) and $*$ operator denotes the complex conjugate. Using this definition $$f(x)=sum_n=1^inftyc_npsi_n(x)$$therefore $$int_Df(x)psi_k^*(x)dx=sum_n=1^inftyc_nint_Dpsi_n(x)psi_k^*(x)dx=sum_n=1^inftyc_ndelta_n-k=c_k$$which completes the proof of question 1.
The answer to the question 2 is a little hard and tricky. As far as I know there are no known necessary and sufficient conditions to ensure existence of such coefficients neither in classical Fourier Series nor in the general case (for example in the case of Fourier Series we may choose some non-Reimann integrable periodic function such as Dirichlet function), but we have some sufficient conditions called Dirichlet conditions after a German mathematician Peter Gustav Lejeune Dirichlet. You may refer to it in https://en.wikipedia.org/wiki/Dirichlet_conditions
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When defining an orthonormal set $psi_i(x)$ we have that$$int_Dpsi_i(x)psi_j^*(x)dx=begincases1&,i=j\0&,ine jendcases$$where $D$ is the common domain of all $psi_i(x)$ (take it $Bbb R$) and $*$ operator denotes the complex conjugate. Using this definition $$f(x)=sum_n=1^inftyc_npsi_n(x)$$therefore $$int_Df(x)psi_k^*(x)dx=sum_n=1^inftyc_nint_Dpsi_n(x)psi_k^*(x)dx=sum_n=1^inftyc_ndelta_n-k=c_k$$which completes the proof of question 1.
The answer to the question 2 is a little hard and tricky. As far as I know there are no known necessary and sufficient conditions to ensure existence of such coefficients neither in classical Fourier Series nor in the general case (for example in the case of Fourier Series we may choose some non-Reimann integrable periodic function such as Dirichlet function), but we have some sufficient conditions called Dirichlet conditions after a German mathematician Peter Gustav Lejeune Dirichlet. You may refer to it in https://en.wikipedia.org/wiki/Dirichlet_conditions
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up vote
1
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up vote
1
down vote
When defining an orthonormal set $psi_i(x)$ we have that$$int_Dpsi_i(x)psi_j^*(x)dx=begincases1&,i=j\0&,ine jendcases$$where $D$ is the common domain of all $psi_i(x)$ (take it $Bbb R$) and $*$ operator denotes the complex conjugate. Using this definition $$f(x)=sum_n=1^inftyc_npsi_n(x)$$therefore $$int_Df(x)psi_k^*(x)dx=sum_n=1^inftyc_nint_Dpsi_n(x)psi_k^*(x)dx=sum_n=1^inftyc_ndelta_n-k=c_k$$which completes the proof of question 1.
The answer to the question 2 is a little hard and tricky. As far as I know there are no known necessary and sufficient conditions to ensure existence of such coefficients neither in classical Fourier Series nor in the general case (for example in the case of Fourier Series we may choose some non-Reimann integrable periodic function such as Dirichlet function), but we have some sufficient conditions called Dirichlet conditions after a German mathematician Peter Gustav Lejeune Dirichlet. You may refer to it in https://en.wikipedia.org/wiki/Dirichlet_conditions
When defining an orthonormal set $psi_i(x)$ we have that$$int_Dpsi_i(x)psi_j^*(x)dx=begincases1&,i=j\0&,ine jendcases$$where $D$ is the common domain of all $psi_i(x)$ (take it $Bbb R$) and $*$ operator denotes the complex conjugate. Using this definition $$f(x)=sum_n=1^inftyc_npsi_n(x)$$therefore $$int_Df(x)psi_k^*(x)dx=sum_n=1^inftyc_nint_Dpsi_n(x)psi_k^*(x)dx=sum_n=1^inftyc_ndelta_n-k=c_k$$which completes the proof of question 1.
The answer to the question 2 is a little hard and tricky. As far as I know there are no known necessary and sufficient conditions to ensure existence of such coefficients neither in classical Fourier Series nor in the general case (for example in the case of Fourier Series we may choose some non-Reimann integrable periodic function such as Dirichlet function), but we have some sufficient conditions called Dirichlet conditions after a German mathematician Peter Gustav Lejeune Dirichlet. You may refer to it in https://en.wikipedia.org/wiki/Dirichlet_conditions
answered Aug 26 at 22:03
Mostafa Ayaz
9,6633730
9,6633730
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You have $phi_k$ and then $psi_k?$
â zhw.
Aug 21 at 21:56
@zhw. Yes. I copied it straight from the notes.
â The Pointer
Aug 21 at 23:21
Do you know the basics of Lebesgue integration and $L^2$ spaces?
â zhw.
Aug 24 at 18:10
@zhw. No, thatâÂÂs beyond my current level.
â The Pointer
Aug 24 at 21:27
You understand uniform convergence?
â zhw.
Aug 24 at 21:47