how to compute the coefficients for half-range expansion of Fourier Series?
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
asked Dec 10 '18 at 2:16
Tyler Hilton
1,40121949
add a comment |
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
asked Dec 10 '18 at 2:16
Tyler Hilton
1,40121949
add a comment |
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
asked Dec 10 '18 at 2:16
Tyler Hilton
1,40121949
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
asked Dec 10 '18 at 2:16
Tyler Hilton
1,40121949
asked Dec 10 '18 at 2:16
Tyler Hilton
1,40121949
asked Dec 10 '18 at 2:16
Tyler Hilton
1,40121949
asked Dec 10 '18 at 2:16
Tyler Hilton
1,40121949
asked Dec 10 '18 at 2:16
Tyler Hilton
1,40121949
1,40121949
add a comment |
add a comment |
1 Answer
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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
$$
boxed {int_A^{T+A} f (t)mathrm dt}= int_{nT+s}^{(n+1)T+s} f(t )mathrm dt =int_{s-T}^s f(u+(n+1)T)mathrm du [t = u + (n+1)T] = int_{s-T}^s f(u)mathrm du ;[text{periodicity}] = left(int_{s-T}^0 + int_0^sright) f = int_{s-T}^0 f(t+T)mathrm dt +int_0^s f ;[text{periodicity}] = left(int_s^T + int_0^sright) f = boxed {int_0^T f};.
$$
answered Dec 10 '18 at 2:36
xbh
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
add a comment |
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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
$$
boxed {int_A^{T+A} f (t)mathrm dt}= int_{nT+s}^{(n+1)T+s} f(t )mathrm dt =int_{s-T}^s f(u+(n+1)T)mathrm du [t = u + (n+1)T] = int_{s-T}^s f(u)mathrm du ;[text{periodicity}] = left(int_{s-T}^0 + int_0^sright) f = int_{s-T}^0 f(t+T)mathrm dt +int_0^s f ;[text{periodicity}] = left(int_s^T + int_0^sright) f = boxed {int_0^T f};.
$$
answered Dec 10 '18 at 2:36
xbh
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
add a comment |
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
$$
boxed {int_A^{T+A} f (t)mathrm dt}= int_{nT+s}^{(n+1)T+s} f(t )mathrm dt =int_{s-T}^s f(u+(n+1)T)mathrm du [t = u + (n+1)T] = int_{s-T}^s f(u)mathrm du ;[text{periodicity}] = left(int_{s-T}^0 + int_0^sright) f = int_{s-T}^0 f(t+T)mathrm dt +int_0^s f ;[text{periodicity}] = left(int_s^T + int_0^sright) f = boxed {int_0^T f};.
$$
answered Dec 10 '18 at 2:36
xbh
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
add a comment |
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
$$
boxed {int_A^{T+A} f (t)mathrm dt}= int_{nT+s}^{(n+1)T+s} f(t )mathrm dt =int_{s-T}^s f(u+(n+1)T)mathrm du [t = u + (n+1)T] = int_{s-T}^s f(u)mathrm du ;[text{periodicity}] = left(int_{s-T}^0 + int_0^sright) f = int_{s-T}^0 f(t+T)mathrm dt +int_0^s f ;[text{periodicity}] = left(int_s^T + int_0^sright) f = boxed {int_0^T f};.
$$
answered Dec 10 '18 at 2:36
xbh
5,6551522
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
$$
boxed {int_A^{T+A} f (t)mathrm dt}= int_{nT+s}^{(n+1)T+s} f(t )mathrm dt =int_{s-T}^s f(u+(n+1)T)mathrm du [t = u + (n+1)T] = int_{s-T}^s f(u)mathrm du ;[text{periodicity}] = left(int_{s-T}^0 + int_0^sright) f = int_{s-T}^0 f(t+T)mathrm dt +int_0^s f ;[text{periodicity}] = left(int_s^T + int_0^sright) f = boxed {int_0^T f};.
$$
answered Dec 10 '18 at 2:36
xbh
5,6551522
answered Dec 10 '18 at 2:36
xbh
5,6551522
answered Dec 10 '18 at 2:36
xbh
5,6551522
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
add a comment |
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
add a comment |
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