How Does a Fourier $sin$/$cos$ Series Arise From a “Normal” Fourier Series? How Does This Relate to the...












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I am told that Fourier showed that we can represent an arbitrary continuous function, $f(x)$, as a convergent series in the elementary trigonometric functions



$$f(x) = sum_{k = 0}^infty a_k cos(kx) + b_k sin(kx)$$



Also, suppose that ${phi_n(x)}^infty_{n = 0}$ is a set of orthogonal functions with respect to a weight function $w(x)$ on the interval $(a, b)$. And let $f(x)$ be an arbitrary function defined on $(a, b)$. Then the generalised Fourier series is



$$f(x) = sum_{k = 0}^infty c_k phi_k (x)$$



I have the following questions relating to this:




  1. How does a Fourier $sin$/$cos$ series arise from a "normal" Fourier series $f(x) = sum_{k = 0}^infty a_k cos(kx) + b_k sin(kx)$?


  2. How does this relate to the generalised Fourier series $f(x) = sum_{k = 0}^infty c_k phi_k (x)$?



I would greatly appreciate clarification on this.



EDIT: When I say Fourier $sin$/$cos$ Series, I'm referring to what is known as "Fourier sine series" and "Fourier cosine series".










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$endgroup$












  • $begingroup$
    What do you mean by a Fourier $text{sin/cos}$ series? Do you mean a series that contains only the sine or cosine terms?
    $endgroup$
    – aghostinthefigures
    Aug 19 '18 at 19:35










  • $begingroup$
    @aghostinthefigures Sorry, I'm referring to what is known as "Fourier sine series" and "Fourier cosine series".
    $endgroup$
    – The Pointer
    Aug 19 '18 at 19:37












  • $begingroup$
    Got it; thanks!
    $endgroup$
    – aghostinthefigures
    Aug 19 '18 at 19:39
















4












$begingroup$


I am told that Fourier showed that we can represent an arbitrary continuous function, $f(x)$, as a convergent series in the elementary trigonometric functions



$$f(x) = sum_{k = 0}^infty a_k cos(kx) + b_k sin(kx)$$



Also, suppose that ${phi_n(x)}^infty_{n = 0}$ is a set of orthogonal functions with respect to a weight function $w(x)$ on the interval $(a, b)$. And let $f(x)$ be an arbitrary function defined on $(a, b)$. Then the generalised Fourier series is



$$f(x) = sum_{k = 0}^infty c_k phi_k (x)$$



I have the following questions relating to this:




  1. How does a Fourier $sin$/$cos$ series arise from a "normal" Fourier series $f(x) = sum_{k = 0}^infty a_k cos(kx) + b_k sin(kx)$?


  2. How does this relate to the generalised Fourier series $f(x) = sum_{k = 0}^infty c_k phi_k (x)$?



I would greatly appreciate clarification on this.



EDIT: When I say Fourier $sin$/$cos$ Series, I'm referring to what is known as "Fourier sine series" and "Fourier cosine series".










share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you mean by a Fourier $text{sin/cos}$ series? Do you mean a series that contains only the sine or cosine terms?
    $endgroup$
    – aghostinthefigures
    Aug 19 '18 at 19:35










  • $begingroup$
    @aghostinthefigures Sorry, I'm referring to what is known as "Fourier sine series" and "Fourier cosine series".
    $endgroup$
    – The Pointer
    Aug 19 '18 at 19:37












  • $begingroup$
    Got it; thanks!
    $endgroup$
    – aghostinthefigures
    Aug 19 '18 at 19:39














4












4








4





$begingroup$


I am told that Fourier showed that we can represent an arbitrary continuous function, $f(x)$, as a convergent series in the elementary trigonometric functions



$$f(x) = sum_{k = 0}^infty a_k cos(kx) + b_k sin(kx)$$



Also, suppose that ${phi_n(x)}^infty_{n = 0}$ is a set of orthogonal functions with respect to a weight function $w(x)$ on the interval $(a, b)$. And let $f(x)$ be an arbitrary function defined on $(a, b)$. Then the generalised Fourier series is



$$f(x) = sum_{k = 0}^infty c_k phi_k (x)$$



I have the following questions relating to this:




  1. How does a Fourier $sin$/$cos$ series arise from a "normal" Fourier series $f(x) = sum_{k = 0}^infty a_k cos(kx) + b_k sin(kx)$?


  2. How does this relate to the generalised Fourier series $f(x) = sum_{k = 0}^infty c_k phi_k (x)$?



I would greatly appreciate clarification on this.



EDIT: When I say Fourier $sin$/$cos$ Series, I'm referring to what is known as "Fourier sine series" and "Fourier cosine series".










share|cite|improve this question











$endgroup$




I am told that Fourier showed that we can represent an arbitrary continuous function, $f(x)$, as a convergent series in the elementary trigonometric functions



$$f(x) = sum_{k = 0}^infty a_k cos(kx) + b_k sin(kx)$$



Also, suppose that ${phi_n(x)}^infty_{n = 0}$ is a set of orthogonal functions with respect to a weight function $w(x)$ on the interval $(a, b)$. And let $f(x)$ be an arbitrary function defined on $(a, b)$. Then the generalised Fourier series is



$$f(x) = sum_{k = 0}^infty c_k phi_k (x)$$



I have the following questions relating to this:




  1. How does a Fourier $sin$/$cos$ series arise from a "normal" Fourier series $f(x) = sum_{k = 0}^infty a_k cos(kx) + b_k sin(kx)$?


  2. How does this relate to the generalised Fourier series $f(x) = sum_{k = 0}^infty c_k phi_k (x)$?



I would greatly appreciate clarification on this.



EDIT: When I say Fourier $sin$/$cos$ Series, I'm referring to what is known as "Fourier sine series" and "Fourier cosine series".







real-analysis functional-analysis fourier-analysis fourier-series orthogonality






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edited Aug 19 '18 at 19:39







The Pointer

















asked Aug 19 '18 at 18:03









The PointerThe Pointer

2,62321538




2,62321538












  • $begingroup$
    What do you mean by a Fourier $text{sin/cos}$ series? Do you mean a series that contains only the sine or cosine terms?
    $endgroup$
    – aghostinthefigures
    Aug 19 '18 at 19:35










  • $begingroup$
    @aghostinthefigures Sorry, I'm referring to what is known as "Fourier sine series" and "Fourier cosine series".
    $endgroup$
    – The Pointer
    Aug 19 '18 at 19:37












  • $begingroup$
    Got it; thanks!
    $endgroup$
    – aghostinthefigures
    Aug 19 '18 at 19:39


















  • $begingroup$
    What do you mean by a Fourier $text{sin/cos}$ series? Do you mean a series that contains only the sine or cosine terms?
    $endgroup$
    – aghostinthefigures
    Aug 19 '18 at 19:35










  • $begingroup$
    @aghostinthefigures Sorry, I'm referring to what is known as "Fourier sine series" and "Fourier cosine series".
    $endgroup$
    – The Pointer
    Aug 19 '18 at 19:37












  • $begingroup$
    Got it; thanks!
    $endgroup$
    – aghostinthefigures
    Aug 19 '18 at 19:39
















$begingroup$
What do you mean by a Fourier $text{sin/cos}$ series? Do you mean a series that contains only the sine or cosine terms?
$endgroup$
– aghostinthefigures
Aug 19 '18 at 19:35




$begingroup$
What do you mean by a Fourier $text{sin/cos}$ series? Do you mean a series that contains only the sine or cosine terms?
$endgroup$
– aghostinthefigures
Aug 19 '18 at 19:35












$begingroup$
@aghostinthefigures Sorry, I'm referring to what is known as "Fourier sine series" and "Fourier cosine series".
$endgroup$
– The Pointer
Aug 19 '18 at 19:37






$begingroup$
@aghostinthefigures Sorry, I'm referring to what is known as "Fourier sine series" and "Fourier cosine series".
$endgroup$
– The Pointer
Aug 19 '18 at 19:37














$begingroup$
Got it; thanks!
$endgroup$
– aghostinthefigures
Aug 19 '18 at 19:39




$begingroup$
Got it; thanks!
$endgroup$
– aghostinthefigures
Aug 19 '18 at 19:39










3 Answers
3






active

oldest

votes


















1












$begingroup$

The "normal" Fourier series is simply a specific case of the generalized Fourier series for which $$phi_k = {sin(kx),cos(kx)}, w(x) = 1$$ where $k$ is appropriately defined based on the domain of $f$ precisely to make each function in the family orthogonal to each other.



Consequently, the coefficients ${a_k, b_k}$ of the "normal" Fourier series are calculated in the same way as the generalized ones $c_k$, through an inner product with the "target" function $f$.



The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series since



$$f(x) = sum_{k = 0}^infty a_k cos(kx) + b_k sin(kx) = sum_{k = 0}^infty a_k cos(kx) + sum_{k = 0}^infty b_k sin(kx)$$



and only one of them may be all that is required to approximate $f$ when either $a_k$ or $b_k$ are collectively $0$. This is equivalent to saying that $f(x)$ is orthogonal to either all $sin(kx)$ or all $cos(kx)$ on the function domain.



Such a situation can usually be qualitatively deduced before inner products are calculated; for example, a symmetric function $f(x) = f(-x)$ defined on a symmetric domain $(-a, a)$ will have $a_k = 0$ and thus can be fully approximated with a Fourier cosine series. This is described here and here for sine and cosine series respectively.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the clear answer. Can you please elaborate on this: "The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series ..." I'm not totally clear on what you mean by this (How does Fourier sine/cosine series arise from the "normal" Fourier series?)
    $endgroup$
    – The Pointer
    Aug 20 '18 at 18:15






  • 1




    $begingroup$
    Will update the answer shortly to clarify this.
    $endgroup$
    – aghostinthefigures
    Aug 20 '18 at 18:29










  • $begingroup$
    Did you forget? :P
    $endgroup$
    – The Pointer
    Aug 23 '18 at 2:22










  • $begingroup$
    The added content is there; I don’t see how I can say more in the answer without being repetitive.
    $endgroup$
    – aghostinthefigures
    Aug 23 '18 at 3:06










  • $begingroup$
    Oh, sorry, I didn’t notice.
    $endgroup$
    – The Pointer
    Aug 23 '18 at 3:07



















0












$begingroup$

I don't know whether you are asking about compact fouier series or exponential fouier series,I will tell about compact fouier series $c_n=sqrt {a_n^2+b_n^2}$ & $theta_n=-tan^{-1} {frac {a_n}{b_n}}$ so compact fouier series becomes $x(t)=c_0+sum_{i=1}^{infty}c_n cos ({ n*omega_o*t+theta_n})$.



Hint:-
$cos(a+b)=cos a*cos b-sin a*sin b$



Note:-
Fourier series is used for representing only periodic signals only ,for arbitrary signal we use fouier transform.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    You can expand an integrable function $f$ on $[-pi,pi]$ in a Fourier series as you have written in your first equation. You can also start with a function on $[0,pi]$ and extend it to an even function on $[-pi,pi]$, and the resulting Fourier series will have only $cos$ terms. Or you can extend $f$ to be an odd function on $[-pi,pi]$, and the resulting Fourier series will have only $sin$ terms. There are pointwise issues that arise at $0$ when you expand in this way, but this is not so difficult to analyze.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      The "normal" Fourier series is simply a specific case of the generalized Fourier series for which $$phi_k = {sin(kx),cos(kx)}, w(x) = 1$$ where $k$ is appropriately defined based on the domain of $f$ precisely to make each function in the family orthogonal to each other.



      Consequently, the coefficients ${a_k, b_k}$ of the "normal" Fourier series are calculated in the same way as the generalized ones $c_k$, through an inner product with the "target" function $f$.



      The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series since



      $$f(x) = sum_{k = 0}^infty a_k cos(kx) + b_k sin(kx) = sum_{k = 0}^infty a_k cos(kx) + sum_{k = 0}^infty b_k sin(kx)$$



      and only one of them may be all that is required to approximate $f$ when either $a_k$ or $b_k$ are collectively $0$. This is equivalent to saying that $f(x)$ is orthogonal to either all $sin(kx)$ or all $cos(kx)$ on the function domain.



      Such a situation can usually be qualitatively deduced before inner products are calculated; for example, a symmetric function $f(x) = f(-x)$ defined on a symmetric domain $(-a, a)$ will have $a_k = 0$ and thus can be fully approximated with a Fourier cosine series. This is described here and here for sine and cosine series respectively.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thanks for the clear answer. Can you please elaborate on this: "The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series ..." I'm not totally clear on what you mean by this (How does Fourier sine/cosine series arise from the "normal" Fourier series?)
        $endgroup$
        – The Pointer
        Aug 20 '18 at 18:15






      • 1




        $begingroup$
        Will update the answer shortly to clarify this.
        $endgroup$
        – aghostinthefigures
        Aug 20 '18 at 18:29










      • $begingroup$
        Did you forget? :P
        $endgroup$
        – The Pointer
        Aug 23 '18 at 2:22










      • $begingroup$
        The added content is there; I don’t see how I can say more in the answer without being repetitive.
        $endgroup$
        – aghostinthefigures
        Aug 23 '18 at 3:06










      • $begingroup$
        Oh, sorry, I didn’t notice.
        $endgroup$
        – The Pointer
        Aug 23 '18 at 3:07
















      1












      $begingroup$

      The "normal" Fourier series is simply a specific case of the generalized Fourier series for which $$phi_k = {sin(kx),cos(kx)}, w(x) = 1$$ where $k$ is appropriately defined based on the domain of $f$ precisely to make each function in the family orthogonal to each other.



      Consequently, the coefficients ${a_k, b_k}$ of the "normal" Fourier series are calculated in the same way as the generalized ones $c_k$, through an inner product with the "target" function $f$.



      The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series since



      $$f(x) = sum_{k = 0}^infty a_k cos(kx) + b_k sin(kx) = sum_{k = 0}^infty a_k cos(kx) + sum_{k = 0}^infty b_k sin(kx)$$



      and only one of them may be all that is required to approximate $f$ when either $a_k$ or $b_k$ are collectively $0$. This is equivalent to saying that $f(x)$ is orthogonal to either all $sin(kx)$ or all $cos(kx)$ on the function domain.



      Such a situation can usually be qualitatively deduced before inner products are calculated; for example, a symmetric function $f(x) = f(-x)$ defined on a symmetric domain $(-a, a)$ will have $a_k = 0$ and thus can be fully approximated with a Fourier cosine series. This is described here and here for sine and cosine series respectively.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thanks for the clear answer. Can you please elaborate on this: "The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series ..." I'm not totally clear on what you mean by this (How does Fourier sine/cosine series arise from the "normal" Fourier series?)
        $endgroup$
        – The Pointer
        Aug 20 '18 at 18:15






      • 1




        $begingroup$
        Will update the answer shortly to clarify this.
        $endgroup$
        – aghostinthefigures
        Aug 20 '18 at 18:29










      • $begingroup$
        Did you forget? :P
        $endgroup$
        – The Pointer
        Aug 23 '18 at 2:22










      • $begingroup$
        The added content is there; I don’t see how I can say more in the answer without being repetitive.
        $endgroup$
        – aghostinthefigures
        Aug 23 '18 at 3:06










      • $begingroup$
        Oh, sorry, I didn’t notice.
        $endgroup$
        – The Pointer
        Aug 23 '18 at 3:07














      1












      1








      1





      $begingroup$

      The "normal" Fourier series is simply a specific case of the generalized Fourier series for which $$phi_k = {sin(kx),cos(kx)}, w(x) = 1$$ where $k$ is appropriately defined based on the domain of $f$ precisely to make each function in the family orthogonal to each other.



      Consequently, the coefficients ${a_k, b_k}$ of the "normal" Fourier series are calculated in the same way as the generalized ones $c_k$, through an inner product with the "target" function $f$.



      The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series since



      $$f(x) = sum_{k = 0}^infty a_k cos(kx) + b_k sin(kx) = sum_{k = 0}^infty a_k cos(kx) + sum_{k = 0}^infty b_k sin(kx)$$



      and only one of them may be all that is required to approximate $f$ when either $a_k$ or $b_k$ are collectively $0$. This is equivalent to saying that $f(x)$ is orthogonal to either all $sin(kx)$ or all $cos(kx)$ on the function domain.



      Such a situation can usually be qualitatively deduced before inner products are calculated; for example, a symmetric function $f(x) = f(-x)$ defined on a symmetric domain $(-a, a)$ will have $a_k = 0$ and thus can be fully approximated with a Fourier cosine series. This is described here and here for sine and cosine series respectively.






      share|cite|improve this answer











      $endgroup$



      The "normal" Fourier series is simply a specific case of the generalized Fourier series for which $$phi_k = {sin(kx),cos(kx)}, w(x) = 1$$ where $k$ is appropriately defined based on the domain of $f$ precisely to make each function in the family orthogonal to each other.



      Consequently, the coefficients ${a_k, b_k}$ of the "normal" Fourier series are calculated in the same way as the generalized ones $c_k$, through an inner product with the "target" function $f$.



      The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series since



      $$f(x) = sum_{k = 0}^infty a_k cos(kx) + b_k sin(kx) = sum_{k = 0}^infty a_k cos(kx) + sum_{k = 0}^infty b_k sin(kx)$$



      and only one of them may be all that is required to approximate $f$ when either $a_k$ or $b_k$ are collectively $0$. This is equivalent to saying that $f(x)$ is orthogonal to either all $sin(kx)$ or all $cos(kx)$ on the function domain.



      Such a situation can usually be qualitatively deduced before inner products are calculated; for example, a symmetric function $f(x) = f(-x)$ defined on a symmetric domain $(-a, a)$ will have $a_k = 0$ and thus can be fully approximated with a Fourier cosine series. This is described here and here for sine and cosine series respectively.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Aug 20 '18 at 19:05

























      answered Aug 19 '18 at 19:53









      aghostinthefiguresaghostinthefigures

      1,2641217




      1,2641217












      • $begingroup$
        Thanks for the clear answer. Can you please elaborate on this: "The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series ..." I'm not totally clear on what you mean by this (How does Fourier sine/cosine series arise from the "normal" Fourier series?)
        $endgroup$
        – The Pointer
        Aug 20 '18 at 18:15






      • 1




        $begingroup$
        Will update the answer shortly to clarify this.
        $endgroup$
        – aghostinthefigures
        Aug 20 '18 at 18:29










      • $begingroup$
        Did you forget? :P
        $endgroup$
        – The Pointer
        Aug 23 '18 at 2:22










      • $begingroup$
        The added content is there; I don’t see how I can say more in the answer without being repetitive.
        $endgroup$
        – aghostinthefigures
        Aug 23 '18 at 3:06










      • $begingroup$
        Oh, sorry, I didn’t notice.
        $endgroup$
        – The Pointer
        Aug 23 '18 at 3:07


















      • $begingroup$
        Thanks for the clear answer. Can you please elaborate on this: "The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series ..." I'm not totally clear on what you mean by this (How does Fourier sine/cosine series arise from the "normal" Fourier series?)
        $endgroup$
        – The Pointer
        Aug 20 '18 at 18:15






      • 1




        $begingroup$
        Will update the answer shortly to clarify this.
        $endgroup$
        – aghostinthefigures
        Aug 20 '18 at 18:29










      • $begingroup$
        Did you forget? :P
        $endgroup$
        – The Pointer
        Aug 23 '18 at 2:22










      • $begingroup$
        The added content is there; I don’t see how I can say more in the answer without being repetitive.
        $endgroup$
        – aghostinthefigures
        Aug 23 '18 at 3:06










      • $begingroup$
        Oh, sorry, I didn’t notice.
        $endgroup$
        – The Pointer
        Aug 23 '18 at 3:07
















      $begingroup$
      Thanks for the clear answer. Can you please elaborate on this: "The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series ..." I'm not totally clear on what you mean by this (How does Fourier sine/cosine series arise from the "normal" Fourier series?)
      $endgroup$
      – The Pointer
      Aug 20 '18 at 18:15




      $begingroup$
      Thanks for the clear answer. Can you please elaborate on this: "The Fourier sine and cosine series thus appear as individual "parts" of the "normal" Fourier series ..." I'm not totally clear on what you mean by this (How does Fourier sine/cosine series arise from the "normal" Fourier series?)
      $endgroup$
      – The Pointer
      Aug 20 '18 at 18:15




      1




      1




      $begingroup$
      Will update the answer shortly to clarify this.
      $endgroup$
      – aghostinthefigures
      Aug 20 '18 at 18:29




      $begingroup$
      Will update the answer shortly to clarify this.
      $endgroup$
      – aghostinthefigures
      Aug 20 '18 at 18:29












      $begingroup$
      Did you forget? :P
      $endgroup$
      – The Pointer
      Aug 23 '18 at 2:22




      $begingroup$
      Did you forget? :P
      $endgroup$
      – The Pointer
      Aug 23 '18 at 2:22












      $begingroup$
      The added content is there; I don’t see how I can say more in the answer without being repetitive.
      $endgroup$
      – aghostinthefigures
      Aug 23 '18 at 3:06




      $begingroup$
      The added content is there; I don’t see how I can say more in the answer without being repetitive.
      $endgroup$
      – aghostinthefigures
      Aug 23 '18 at 3:06












      $begingroup$
      Oh, sorry, I didn’t notice.
      $endgroup$
      – The Pointer
      Aug 23 '18 at 3:07




      $begingroup$
      Oh, sorry, I didn’t notice.
      $endgroup$
      – The Pointer
      Aug 23 '18 at 3:07











      0












      $begingroup$

      I don't know whether you are asking about compact fouier series or exponential fouier series,I will tell about compact fouier series $c_n=sqrt {a_n^2+b_n^2}$ & $theta_n=-tan^{-1} {frac {a_n}{b_n}}$ so compact fouier series becomes $x(t)=c_0+sum_{i=1}^{infty}c_n cos ({ n*omega_o*t+theta_n})$.



      Hint:-
      $cos(a+b)=cos a*cos b-sin a*sin b$



      Note:-
      Fourier series is used for representing only periodic signals only ,for arbitrary signal we use fouier transform.






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        I don't know whether you are asking about compact fouier series or exponential fouier series,I will tell about compact fouier series $c_n=sqrt {a_n^2+b_n^2}$ & $theta_n=-tan^{-1} {frac {a_n}{b_n}}$ so compact fouier series becomes $x(t)=c_0+sum_{i=1}^{infty}c_n cos ({ n*omega_o*t+theta_n})$.



        Hint:-
        $cos(a+b)=cos a*cos b-sin a*sin b$



        Note:-
        Fourier series is used for representing only periodic signals only ,for arbitrary signal we use fouier transform.






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          I don't know whether you are asking about compact fouier series or exponential fouier series,I will tell about compact fouier series $c_n=sqrt {a_n^2+b_n^2}$ & $theta_n=-tan^{-1} {frac {a_n}{b_n}}$ so compact fouier series becomes $x(t)=c_0+sum_{i=1}^{infty}c_n cos ({ n*omega_o*t+theta_n})$.



          Hint:-
          $cos(a+b)=cos a*cos b-sin a*sin b$



          Note:-
          Fourier series is used for representing only periodic signals only ,for arbitrary signal we use fouier transform.






          share|cite|improve this answer











          $endgroup$



          I don't know whether you are asking about compact fouier series or exponential fouier series,I will tell about compact fouier series $c_n=sqrt {a_n^2+b_n^2}$ & $theta_n=-tan^{-1} {frac {a_n}{b_n}}$ so compact fouier series becomes $x(t)=c_0+sum_{i=1}^{infty}c_n cos ({ n*omega_o*t+theta_n})$.



          Hint:-
          $cos(a+b)=cos a*cos b-sin a*sin b$



          Note:-
          Fourier series is used for representing only periodic signals only ,for arbitrary signal we use fouier transform.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 19 '18 at 19:31

























          answered Aug 19 '18 at 19:23









          Ch.Siva Ram KishoreCh.Siva Ram Kishore

          457




          457























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              $begingroup$

              You can expand an integrable function $f$ on $[-pi,pi]$ in a Fourier series as you have written in your first equation. You can also start with a function on $[0,pi]$ and extend it to an even function on $[-pi,pi]$, and the resulting Fourier series will have only $cos$ terms. Or you can extend $f$ to be an odd function on $[-pi,pi]$, and the resulting Fourier series will have only $sin$ terms. There are pointwise issues that arise at $0$ when you expand in this way, but this is not so difficult to analyze.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You can expand an integrable function $f$ on $[-pi,pi]$ in a Fourier series as you have written in your first equation. You can also start with a function on $[0,pi]$ and extend it to an even function on $[-pi,pi]$, and the resulting Fourier series will have only $cos$ terms. Or you can extend $f$ to be an odd function on $[-pi,pi]$, and the resulting Fourier series will have only $sin$ terms. There are pointwise issues that arise at $0$ when you expand in this way, but this is not so difficult to analyze.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You can expand an integrable function $f$ on $[-pi,pi]$ in a Fourier series as you have written in your first equation. You can also start with a function on $[0,pi]$ and extend it to an even function on $[-pi,pi]$, and the resulting Fourier series will have only $cos$ terms. Or you can extend $f$ to be an odd function on $[-pi,pi]$, and the resulting Fourier series will have only $sin$ terms. There are pointwise issues that arise at $0$ when you expand in this way, but this is not so difficult to analyze.






                  share|cite|improve this answer









                  $endgroup$



                  You can expand an integrable function $f$ on $[-pi,pi]$ in a Fourier series as you have written in your first equation. You can also start with a function on $[0,pi]$ and extend it to an even function on $[-pi,pi]$, and the resulting Fourier series will have only $cos$ terms. Or you can extend $f$ to be an odd function on $[-pi,pi]$, and the resulting Fourier series will have only $sin$ terms. There are pointwise issues that arise at $0$ when you expand in this way, but this is not so difficult to analyze.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 22 '18 at 19:20









                  DisintegratingByPartsDisintegratingByParts

                  59.5k42580




                  59.5k42580






























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