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:



  1. 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?


  2. 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.







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  • You have $phi_k$ and then $psi_k?$
    – zhw.
    Aug 21 at 21:56










  • @zhw. Yes. I copied it straight from the notes.
    – The Pointer
    Aug 21 at 23:21










  • Do you know the basics of Lebesgue integration and $L^2$ spaces?
    – zhw.
    Aug 24 at 18:10










  • @zhw. No, that’s beyond my current level.
    – The Pointer
    Aug 24 at 21:27










  • You understand uniform convergence?
    – zhw.
    Aug 24 at 21:47














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:



  1. 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?


  2. 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.







share|cite|improve this question




















  • You have $phi_k$ and then $psi_k?$
    – zhw.
    Aug 21 at 21:56










  • @zhw. Yes. I copied it straight from the notes.
    – The Pointer
    Aug 21 at 23:21










  • Do you know the basics of Lebesgue integration and $L^2$ spaces?
    – zhw.
    Aug 24 at 18:10










  • @zhw. No, that’s beyond my current level.
    – The Pointer
    Aug 24 at 21:27










  • You understand uniform convergence?
    – zhw.
    Aug 24 at 21:47












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:



  1. 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?


  2. 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.







share|cite|improve this question












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:



  1. 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?


  2. 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.









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 17 at 23:10









The Pointer

2,7372830




2,7372830











  • You have $phi_k$ and then $psi_k?$
    – zhw.
    Aug 21 at 21:56










  • @zhw. Yes. I copied it straight from the notes.
    – The Pointer
    Aug 21 at 23:21










  • Do you know the basics of Lebesgue integration and $L^2$ spaces?
    – zhw.
    Aug 24 at 18:10










  • @zhw. No, that’s beyond my current level.
    – The Pointer
    Aug 24 at 21:27










  • You understand uniform convergence?
    – zhw.
    Aug 24 at 21:47
















  • You have $phi_k$ and then $psi_k?$
    – zhw.
    Aug 21 at 21:56










  • @zhw. Yes. I copied it straight from the notes.
    – The Pointer
    Aug 21 at 23:21










  • Do you know the basics of Lebesgue integration and $L^2$ spaces?
    – zhw.
    Aug 24 at 18:10










  • @zhw. No, that’s beyond my current level.
    – The Pointer
    Aug 24 at 21:27










  • You understand uniform convergence?
    – zhw.
    Aug 24 at 21:47















You have $phi_k$ and then $psi_k?$
– zhw.
Aug 21 at 21:56




You have $phi_k$ and then $psi_k?$
– zhw.
Aug 21 at 21:56












@zhw. Yes. I copied it straight from the notes.
– The Pointer
Aug 21 at 23:21




@zhw. Yes. I copied it straight from the notes.
– The Pointer
Aug 21 at 23:21












Do you know the basics of Lebesgue integration and $L^2$ spaces?
– zhw.
Aug 24 at 18:10




Do you know the basics of Lebesgue integration and $L^2$ spaces?
– zhw.
Aug 24 at 18:10












@zhw. No, that’s beyond my current level.
– The Pointer
Aug 24 at 21:27




@zhw. No, that’s beyond my current level.
– The Pointer
Aug 24 at 21:27












You understand uniform convergence?
– zhw.
Aug 24 at 21:47




You understand uniform convergence?
– zhw.
Aug 24 at 21:47










4 Answers
4






active

oldest

votes

















up vote
3
down vote



accepted
+100










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?






share|cite|improve this answer





























    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 $c_j$.






    share|cite|improve this answer
















    • 1




      Thanks for the answer. Can you please rephrase this in a way that's easier for a novice to understand?
      – 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.
      – Kavi Rama Murthy
      Aug 17 at 23:47

















    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 $(psi_k,psi_k)=1$.






    share|cite|improve this answer



























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      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






      share|cite|improve this answer




















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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
        +100










        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?






        share|cite|improve this answer


























          up vote
          3
          down vote



          accepted
          +100










          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?






          share|cite|improve this answer
























            up vote
            3
            down vote



            accepted
            +100







            up vote
            3
            down vote



            accepted
            +100




            +100




            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?






            share|cite|improve this answer














            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?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 24 at 23:05

























            answered Aug 24 at 22:50









            zhw.

            66.7k42872




            66.7k42872




















                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 $c_j$.






                share|cite|improve this answer
















                • 1




                  Thanks for the answer. Can you please rephrase this in a way that's easier for a novice to understand?
                  – 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.
                  – Kavi Rama Murthy
                  Aug 17 at 23:47














                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 $c_j$.






                share|cite|improve this answer
















                • 1




                  Thanks for the answer. Can you please rephrase this in a way that's easier for a novice to understand?
                  – 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.
                  – Kavi Rama Murthy
                  Aug 17 at 23:47












                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 $c_j$.






                share|cite|improve this answer












                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 $c_j$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                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?
                  – 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.
                  – Kavi Rama Murthy
                  Aug 17 at 23:47












                • 1




                  Thanks for the answer. Can you please rephrase this in a way that's easier for a novice to understand?
                  – 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.
                  – 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?
                – 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?
                – 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.
                – 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.
                – Kavi Rama Murthy
                Aug 17 at 23:47










                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 $(psi_k,psi_k)=1$.






                share|cite|improve this answer
























                  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 $(psi_k,psi_k)=1$.






                  share|cite|improve this answer






















                    up vote
                    1
                    down vote










                    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 $(psi_k,psi_k)=1$.






                    share|cite|improve this answer












                    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 $(psi_k,psi_k)=1$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 20 at 3:54









                    Sergio Enrique Yarza Acuña

                    600314




                    600314




















                        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






                        share|cite|improve this answer
























                          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






                          share|cite|improve this answer






















                            up vote
                            1
                            down vote










                            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






                            share|cite|improve this answer












                            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







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 26 at 22:03









                            Mostafa Ayaz

                            9,6633730




                            9,6633730






















                                 

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