how to compute the coefficients for half-range expansion of Fourier Series?












0














If a function $f$ is defined on $[-L, L]$, then its Fourier series is given by
$$
begin{aligned} a _ { 0 } & = frac { 1 } { L } int _ { - L } ^ { L } f left( x right) d x \
a _ { n } & = frac { 1 } { L } int _ { - L } ^ { L } f left( x right) cos left( frac { n pi x } { L } right) d x \
b _ { n } & = frac { 1 } { L } int _ { - L } ^ { L } f left( x right) sin left( frac { n pi x } { L } right) d x
end{aligned}
$$



This means that the even though the function is defined on $[L, L]$, the fourier series representation is over $(-infty, infty)$ and with period $2L$.
However, if a function is defined on $[0, 2L]$, then the formulas changes slightly,
$$
begin{aligned} a _ { 0 } & = frac { 1 } { L } int _ { 0 } ^ { 2 L } f left( x right) d x \
a _ { n } & = frac { 1 } { L } int _ { 0 } ^ { 2 L } f left( x right) cos left( frac { n pi x } { L } right) d \ b _ { n } & = frac { 1 } { L } int _ { 0 } ^ { 2 L } f left( x right) sin left( frac { n pi x} { L } right) d end{aligned}
$$



See the difference in the limits of integration. I see how the equations are derived but I am missing the fundamental connection of how the "period" comes into play here. In particular, I am not understanding the statement




For f(x) periodic with period 2L, any interval $(x_0,x_0+2L)$ can be
used, with the choice being one of convenience or personal preference







fourier-series







share|cite|improve this question



























    0














    If a function $f$ is defined on $[-L, L]$, then its Fourier series is given by
    $$
    begin{aligned} a _ { 0 } & = frac { 1 } { L } int _ { - L } ^ { L } f left( x right) d x \
    a _ { n } & = frac { 1 } { L } int _ { - L } ^ { L } f left( x right) cos left( frac { n pi x } { L } right) d x \
    b _ { n } & = frac { 1 } { L } int _ { - L } ^ { L } f left( x right) sin left( frac { n pi x } { L } right) d x
    end{aligned}
    $$



    This means that the even though the function is defined on $[L, L]$, the fourier series representation is over $(-infty, infty)$ and with period $2L$.
    However, if a function is defined on $[0, 2L]$, then the formulas changes slightly,
    $$
    begin{aligned} a _ { 0 } & = frac { 1 } { L } int _ { 0 } ^ { 2 L } f left( x right) d x \
    a _ { n } & = frac { 1 } { L } int _ { 0 } ^ { 2 L } f left( x right) cos left( frac { n pi x } { L } right) d \ b _ { n } & = frac { 1 } { L } int _ { 0 } ^ { 2 L } f left( x right) sin left( frac { n pi x} { L } right) d end{aligned}
    $$



    See the difference in the limits of integration. I see how the equations are derived but I am missing the fundamental connection of how the "period" comes into play here. In particular, I am not understanding the statement




    For f(x) periodic with period 2L, any interval $(x_0,x_0+2L)$ can be
    used, with the choice being one of convenience or personal preference







    fourier-series







    share|cite|improve this question

























      0












      0








      0







      If a function $f$ is defined on $[-L, L]$, then its Fourier series is given by
      $$
      begin{aligned} a _ { 0 } & = frac { 1 } { L } int _ { - L } ^ { L } f left( x right) d x \
      a _ { n } & = frac { 1 } { L } int _ { - L } ^ { L } f left( x right) cos left( frac { n pi x } { L } right) d x \
      b _ { n } & = frac { 1 } { L } int _ { - L } ^ { L } f left( x right) sin left( frac { n pi x } { L } right) d x
      end{aligned}
      $$



      This means that the even though the function is defined on $[L, L]$, the fourier series representation is over $(-infty, infty)$ and with period $2L$.
      However, if a function is defined on $[0, 2L]$, then the formulas changes slightly,
      $$
      begin{aligned} a _ { 0 } & = frac { 1 } { L } int _ { 0 } ^ { 2 L } f left( x right) d x \
      a _ { n } & = frac { 1 } { L } int _ { 0 } ^ { 2 L } f left( x right) cos left( frac { n pi x } { L } right) d \ b _ { n } & = frac { 1 } { L } int _ { 0 } ^ { 2 L } f left( x right) sin left( frac { n pi x} { L } right) d end{aligned}
      $$



      See the difference in the limits of integration. I see how the equations are derived but I am missing the fundamental connection of how the "period" comes into play here. In particular, I am not understanding the statement




      For f(x) periodic with period 2L, any interval $(x_0,x_0+2L)$ can be
      used, with the choice being one of convenience or personal preference







      fourier-series







      share|cite|improve this question













      If a function $f$ is defined on $[-L, L]$, then its Fourier series is given by
      $$
      begin{aligned} a _ { 0 } & = frac { 1 } { L } int _ { - L } ^ { L } f left( x right) d x \
      a _ { n } & = frac { 1 } { L } int _ { - L } ^ { L } f left( x right) cos left( frac { n pi x } { L } right) d x \
      b _ { n } & = frac { 1 } { L } int _ { - L } ^ { L } f left( x right) sin left( frac { n pi x } { L } right) d x
      end{aligned}
      $$



      This means that the even though the function is defined on $[L, L]$, the fourier series representation is over $(-infty, infty)$ and with period $2L$.
      However, if a function is defined on $[0, 2L]$, then the formulas changes slightly,
      $$
      begin{aligned} a _ { 0 } & = frac { 1 } { L } int _ { 0 } ^ { 2 L } f left( x right) d x \
      a _ { n } & = frac { 1 } { L } int _ { 0 } ^ { 2 L } f left( x right) cos left( frac { n pi x } { L } right) d \ b _ { n } & = frac { 1 } { L } int _ { 0 } ^ { 2 L } f left( x right) sin left( frac { n pi x} { L } right) d end{aligned}
      $$



      See the difference in the limits of integration. I see how the equations are derived but I am missing the fundamental connection of how the "period" comes into play here. In particular, I am not understanding the statement




      For f(x) periodic with period 2L, any interval $(x_0,x_0+2L)$ can be
      used, with the choice being one of convenience or personal preference







      fourier-series




      fourier-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 10 '18 at 2:16









      Tyler Hilton

      1,40121949




      1,40121949






















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          0














          Suppose $T$ is a period of $f$, then $underline {text{for each }A in mathbb R }$ there is some $n in mathbb Z$ s.t. $nT leqslant A < (n+1)T$, then $A = nT +s$ for some $s in [0, T)$, so
          $$
          boxed {int_A^{T+A} f (t)mathrm dt}= int_{nT+s}^{(n+1)T+s} f(t )mathrm dt =int_{s-T}^s f(u+(n+1)T)mathrm du [t = u + (n+1)T] = int_{s-T}^s f(u)mathrm du ;[text{periodicity}] = left(int_{s-T}^0 + int_0^sright) f = int_{s-T}^0 f(t+T)mathrm dt +int_0^s f ;[text{periodicity}] = left(int_s^T + int_0^sright) f = boxed {int_0^T f};.
          $$






          share|cite|improve this answer





















          • Thanks, I get this but I don't get how the limits of integration has changed to $2L$ while the coefficient is still $1/L$ and even inside the trig function, the division is by $L$.
            – Tyler Hilton
            Dec 10 '18 at 4:44










          • Then you could just extend $f$ on to $overline fcolon mathbb R to mathbb R$ by periodicity $2L$ , then apply the formula for $overline f$ on $[-L, L]$, then change the limits of integration.
            – xbh
            Dec 10 '18 at 4:50













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          0














          Suppose $T$ is a period of $f$, then $underline {text{for each }A in mathbb R }$ there is some $n in mathbb Z$ s.t. $nT leqslant A < (n+1)T$, then $A = nT +s$ for some $s in [0, T)$, so
          $$
          boxed {int_A^{T+A} f (t)mathrm dt}= int_{nT+s}^{(n+1)T+s} f(t )mathrm dt =int_{s-T}^s f(u+(n+1)T)mathrm du [t = u + (n+1)T] = int_{s-T}^s f(u)mathrm du ;[text{periodicity}] = left(int_{s-T}^0 + int_0^sright) f = int_{s-T}^0 f(t+T)mathrm dt +int_0^s f ;[text{periodicity}] = left(int_s^T + int_0^sright) f = boxed {int_0^T f};.
          $$






          share|cite|improve this answer





















          • Thanks, I get this but I don't get how the limits of integration has changed to $2L$ while the coefficient is still $1/L$ and even inside the trig function, the division is by $L$.
            – Tyler Hilton
            Dec 10 '18 at 4:44










          • Then you could just extend $f$ on to $overline fcolon mathbb R to mathbb R$ by periodicity $2L$ , then apply the formula for $overline f$ on $[-L, L]$, then change the limits of integration.
            – xbh
            Dec 10 '18 at 4:50


















          0














          Suppose $T$ is a period of $f$, then $underline {text{for each }A in mathbb R }$ there is some $n in mathbb Z$ s.t. $nT leqslant A < (n+1)T$, then $A = nT +s$ for some $s in [0, T)$, so
          $$
          boxed {int_A^{T+A} f (t)mathrm dt}= int_{nT+s}^{(n+1)T+s} f(t )mathrm dt =int_{s-T}^s f(u+(n+1)T)mathrm du [t = u + (n+1)T] = int_{s-T}^s f(u)mathrm du ;[text{periodicity}] = left(int_{s-T}^0 + int_0^sright) f = int_{s-T}^0 f(t+T)mathrm dt +int_0^s f ;[text{periodicity}] = left(int_s^T + int_0^sright) f = boxed {int_0^T f};.
          $$






          share|cite|improve this answer





















          • Thanks, I get this but I don't get how the limits of integration has changed to $2L$ while the coefficient is still $1/L$ and even inside the trig function, the division is by $L$.
            – Tyler Hilton
            Dec 10 '18 at 4:44










          • Then you could just extend $f$ on to $overline fcolon mathbb R to mathbb R$ by periodicity $2L$ , then apply the formula for $overline f$ on $[-L, L]$, then change the limits of integration.
            – xbh
            Dec 10 '18 at 4:50
















          0












          0








          0






          Suppose $T$ is a period of $f$, then $underline {text{for each }A in mathbb R }$ there is some $n in mathbb Z$ s.t. $nT leqslant A < (n+1)T$, then $A = nT +s$ for some $s in [0, T)$, so
          $$
          boxed {int_A^{T+A} f (t)mathrm dt}= int_{nT+s}^{(n+1)T+s} f(t )mathrm dt =int_{s-T}^s f(u+(n+1)T)mathrm du [t = u + (n+1)T] = int_{s-T}^s f(u)mathrm du ;[text{periodicity}] = left(int_{s-T}^0 + int_0^sright) f = int_{s-T}^0 f(t+T)mathrm dt +int_0^s f ;[text{periodicity}] = left(int_s^T + int_0^sright) f = boxed {int_0^T f};.
          $$






          share|cite|improve this answer












          Suppose $T$ is a period of $f$, then $underline {text{for each }A in mathbb R }$ there is some $n in mathbb Z$ s.t. $nT leqslant A < (n+1)T$, then $A = nT +s$ for some $s in [0, T)$, so
          $$
          boxed {int_A^{T+A} f (t)mathrm dt}= int_{nT+s}^{(n+1)T+s} f(t )mathrm dt =int_{s-T}^s f(u+(n+1)T)mathrm du [t = u + (n+1)T] = int_{s-T}^s f(u)mathrm du ;[text{periodicity}] = left(int_{s-T}^0 + int_0^sright) f = int_{s-T}^0 f(t+T)mathrm dt +int_0^s f ;[text{periodicity}] = left(int_s^T + int_0^sright) f = boxed {int_0^T f};.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 2:36









          xbh

          5,6551522




          5,6551522












          • Thanks, I get this but I don't get how the limits of integration has changed to $2L$ while the coefficient is still $1/L$ and even inside the trig function, the division is by $L$.
            – Tyler Hilton
            Dec 10 '18 at 4:44










          • Then you could just extend $f$ on to $overline fcolon mathbb R to mathbb R$ by periodicity $2L$ , then apply the formula for $overline f$ on $[-L, L]$, then change the limits of integration.
            – xbh
            Dec 10 '18 at 4:50




















          • Thanks, I get this but I don't get how the limits of integration has changed to $2L$ while the coefficient is still $1/L$ and even inside the trig function, the division is by $L$.
            – Tyler Hilton
            Dec 10 '18 at 4:44










          • Then you could just extend $f$ on to $overline fcolon mathbb R to mathbb R$ by periodicity $2L$ , then apply the formula for $overline f$ on $[-L, L]$, then change the limits of integration.
            – xbh
            Dec 10 '18 at 4:50


















          Thanks, I get this but I don't get how the limits of integration has changed to $2L$ while the coefficient is still $1/L$ and even inside the trig function, the division is by $L$.
          – Tyler Hilton
          Dec 10 '18 at 4:44




          Thanks, I get this but I don't get how the limits of integration has changed to $2L$ while the coefficient is still $1/L$ and even inside the trig function, the division is by $L$.
          – Tyler Hilton
          Dec 10 '18 at 4:44












          Then you could just extend $f$ on to $overline fcolon mathbb R to mathbb R$ by periodicity $2L$ , then apply the formula for $overline f$ on $[-L, L]$, then change the limits of integration.
          – xbh
          Dec 10 '18 at 4:50






          Then you could just extend $f$ on to $overline fcolon mathbb R to mathbb R$ by periodicity $2L$ , then apply the formula for $overline f$ on $[-L, L]$, then change the limits of integration.
          – xbh
          Dec 10 '18 at 4:50




















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