Mixing
Fourier Transform of derivative of $xf(x)$?
$begingroup$
What is the Fourier transform of $$frac{partial}{partial x}(xf(x))$$
Shall I separate the product first? Then what is the Fourier transform of $$xfrac{partial}{partial x}f(x)$$
calculus fourier-analysis fourier-transform
asked Jan 25 at 0:07
kinder chenkinder chen
1434
$endgroup$
add a comment |
$begingroup$
What is the Fourier transform of $$frac{partial}{partial x}(xf(x))$$
Shall I separate the product first? Then what is the Fourier transform of $$xfrac{partial}{partial x}f(x)$$
calculus fourier-analysis fourier-transform
asked Jan 25 at 0:07
kinder chenkinder chen
1434
$endgroup$
$begingroup$
For your original question - assuming things are nice, you can integrate by parts to get $-it$ times Fourier transform of $xf(x)$.
$endgroup$
– herb steinberg
Jan 25 at 0:52
add a comment |
$begingroup$
What is the Fourier transform of $$frac{partial}{partial x}(xf(x))$$
Shall I separate the product first? Then what is the Fourier transform of $$xfrac{partial}{partial x}f(x)$$
calculus fourier-analysis fourier-transform
asked Jan 25 at 0:07
kinder chenkinder chen
1434
$endgroup$
What is the Fourier transform of $$frac{partial}{partial x}(xf(x))$$
Shall I separate the product first? Then what is the Fourier transform of $$xfrac{partial}{partial x}f(x)$$
calculus fourier-analysis fourier-transform
calculus fourier-analysis fourier-transform
asked Jan 25 at 0:07
kinder chenkinder chen
1434
asked Jan 25 at 0:07
kinder chenkinder chen
1434
asked Jan 25 at 0:07
kinder chenkinder chen
1434
asked Jan 25 at 0:07
kinder chenkinder chen
1434
asked Jan 25 at 0:07
kinder chenkinder chen
1434
1434
$begingroup$
For your original question - assuming things are nice, you can integrate by parts to get $-it$ times Fourier transform of $xf(x)$.
$endgroup$
– herb steinberg
Jan 25 at 0:52
add a comment |
$begingroup$
For your original question - assuming things are nice, you can integrate by parts to get $-it$ times Fourier transform of $xf(x)$.
$endgroup$
– herb steinberg
Jan 25 at 0:52
$begingroup$
For your original question - assuming things are nice, you can integrate by parts to get $-it$ times Fourier transform of $xf(x)$.
$endgroup$
– herb steinberg
Jan 25 at 0:52
$begingroup$
For your original question - assuming things are nice, you can integrate by parts to get $-it$ times Fourier transform of $xf(x)$.
$endgroup$
– herb steinberg
Jan 25 at 0:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Using this definition of the Fourier Transform
$$F(s) = mathscr{F}left{f(x)right} = int_{-infty}^infty f(x) e^{-2pi i sx} space dx$$
and these two theorems
$$mathscr{F}left{dfrac{d}{dx}f(x)right} = 2pi i sF(s)$$
$$mathscr{F}left{-2pi i xf(x)right} = dfrac{d}{ds}F(s)= F'(s)$$
one can derive the answer
$$begin{align*}mathscr{F}left{dfrac{d}{dx}left(xf(x)right)right} &= 2pi i smathscr{F}left{xf(x)right} \
\
&= - smathscr{F}left{-2pi i xf(x)right} \
\
&= -sdfrac{d}{ds}F(s)\
\
&= -s F'(s)
end{align*}$$
Update to add a proof of the 2nd theorem
$$begin{align*}mathscr{F}^{-1}left{dfrac{d}{ds}F(s)right} &= int_{-infty}^infty {left[dfrac{d}{ds}F(s)right]e^{2pi i xs}}ds \
\
&= int_{-infty}^infty {lim_{Delta s to 0}{dfrac{F(s+Delta s) - F(s)}{Delta s}}e^{2pi i xs}}ds \
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[int_{-infty}^infty {F(s+Delta s)e^{2pi i xs}}ds -int_{-infty}^infty {F(s)e^{2pi i xs}}dsright]}\
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[int_{-infty}^infty {F(s')e^{2pi i x(s'-Delta s)}}ds' -f(x)right]}\
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[e^{-2pi i xDelta s}int_{-infty}^infty {F(s')e^{2pi i xs'}}ds'-f(x)right]}\
\
&= lim_{Delta s to 0}{dfrac{e^{-2pi i xDelta s}f(x)-f(x)}{Delta s}}\
\
&= lim_{Delta s to 0}{dfrac{e^{-2pi i x(0+Delta s)}-e^{-2pi i x0}}{Delta s}f(x)}\
\
&= left(dfrac{d}{ds}e^{-2pi i xs}right)biggr|_{s=0} space f(x)\
\
mathscr{F}^{-1}left{dfrac{d}{ds}F(s)right} &= -2pi i x f(x)\
\
dfrac{d}{ds}F(s) &= mathscr{F}left{-2pi i x f(x)right}\
\
end{align*}$$
answered Jan 26 at 20:33
Andy WallsAndy Walls
1,764139
$endgroup$
$begingroup$
Thx, can you elaborate your 2nd theorem about $F'(s)$ pls?
$endgroup$
– kinder chen
Jan 27 at 1:33
$begingroup$
Well, it's the dual of the first theorem. There is a symmetry between the Fourier Transform and the Inverse Fourier Transform with the normalization convention I normally use. Although the symmetry isn't perfect as evinced by the negative sign, which pops up in duals of Fourier Transform theorems.
$endgroup$
– Andy Walls
Jan 27 at 2:56
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Using this definition of the Fourier Transform
$$F(s) = mathscr{F}left{f(x)right} = int_{-infty}^infty f(x) e^{-2pi i sx} space dx$$
and these two theorems
$$mathscr{F}left{dfrac{d}{dx}f(x)right} = 2pi i sF(s)$$
$$mathscr{F}left{-2pi i xf(x)right} = dfrac{d}{ds}F(s)= F'(s)$$
one can derive the answer
$$begin{align*}mathscr{F}left{dfrac{d}{dx}left(xf(x)right)right} &= 2pi i smathscr{F}left{xf(x)right} \
\
&= - smathscr{F}left{-2pi i xf(x)right} \
\
&= -sdfrac{d}{ds}F(s)\
\
&= -s F'(s)
end{align*}$$
Update to add a proof of the 2nd theorem
$$begin{align*}mathscr{F}^{-1}left{dfrac{d}{ds}F(s)right} &= int_{-infty}^infty {left[dfrac{d}{ds}F(s)right]e^{2pi i xs}}ds \
\
&= int_{-infty}^infty {lim_{Delta s to 0}{dfrac{F(s+Delta s) - F(s)}{Delta s}}e^{2pi i xs}}ds \
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[int_{-infty}^infty {F(s+Delta s)e^{2pi i xs}}ds -int_{-infty}^infty {F(s)e^{2pi i xs}}dsright]}\
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[int_{-infty}^infty {F(s')e^{2pi i x(s'-Delta s)}}ds' -f(x)right]}\
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[e^{-2pi i xDelta s}int_{-infty}^infty {F(s')e^{2pi i xs'}}ds'-f(x)right]}\
\
&= lim_{Delta s to 0}{dfrac{e^{-2pi i xDelta s}f(x)-f(x)}{Delta s}}\
\
&= lim_{Delta s to 0}{dfrac{e^{-2pi i x(0+Delta s)}-e^{-2pi i x0}}{Delta s}f(x)}\
\
&= left(dfrac{d}{ds}e^{-2pi i xs}right)biggr|_{s=0} space f(x)\
\
mathscr{F}^{-1}left{dfrac{d}{ds}F(s)right} &= -2pi i x f(x)\
\
dfrac{d}{ds}F(s) &= mathscr{F}left{-2pi i x f(x)right}\
\
end{align*}$$
answered Jan 26 at 20:33
Andy WallsAndy Walls
1,764139
$endgroup$
$begingroup$
Thx, can you elaborate your 2nd theorem about $F'(s)$ pls?
$endgroup$
– kinder chen
Jan 27 at 1:33
$begingroup$
Well, it's the dual of the first theorem. There is a symmetry between the Fourier Transform and the Inverse Fourier Transform with the normalization convention I normally use. Although the symmetry isn't perfect as evinced by the negative sign, which pops up in duals of Fourier Transform theorems.
$endgroup$
– Andy Walls
Jan 27 at 2:56
add a comment |
$begingroup$
Using this definition of the Fourier Transform
$$F(s) = mathscr{F}left{f(x)right} = int_{-infty}^infty f(x) e^{-2pi i sx} space dx$$
and these two theorems
$$mathscr{F}left{dfrac{d}{dx}f(x)right} = 2pi i sF(s)$$
$$mathscr{F}left{-2pi i xf(x)right} = dfrac{d}{ds}F(s)= F'(s)$$
one can derive the answer
$$begin{align*}mathscr{F}left{dfrac{d}{dx}left(xf(x)right)right} &= 2pi i smathscr{F}left{xf(x)right} \
\
&= - smathscr{F}left{-2pi i xf(x)right} \
\
&= -sdfrac{d}{ds}F(s)\
\
&= -s F'(s)
end{align*}$$
Update to add a proof of the 2nd theorem
$$begin{align*}mathscr{F}^{-1}left{dfrac{d}{ds}F(s)right} &= int_{-infty}^infty {left[dfrac{d}{ds}F(s)right]e^{2pi i xs}}ds \
\
&= int_{-infty}^infty {lim_{Delta s to 0}{dfrac{F(s+Delta s) - F(s)}{Delta s}}e^{2pi i xs}}ds \
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[int_{-infty}^infty {F(s+Delta s)e^{2pi i xs}}ds -int_{-infty}^infty {F(s)e^{2pi i xs}}dsright]}\
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[int_{-infty}^infty {F(s')e^{2pi i x(s'-Delta s)}}ds' -f(x)right]}\
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[e^{-2pi i xDelta s}int_{-infty}^infty {F(s')e^{2pi i xs'}}ds'-f(x)right]}\
\
&= lim_{Delta s to 0}{dfrac{e^{-2pi i xDelta s}f(x)-f(x)}{Delta s}}\
\
&= lim_{Delta s to 0}{dfrac{e^{-2pi i x(0+Delta s)}-e^{-2pi i x0}}{Delta s}f(x)}\
\
&= left(dfrac{d}{ds}e^{-2pi i xs}right)biggr|_{s=0} space f(x)\
\
mathscr{F}^{-1}left{dfrac{d}{ds}F(s)right} &= -2pi i x f(x)\
\
dfrac{d}{ds}F(s) &= mathscr{F}left{-2pi i x f(x)right}\
\
end{align*}$$
answered Jan 26 at 20:33
Andy WallsAndy Walls
1,764139
$endgroup$
$begingroup$
Thx, can you elaborate your 2nd theorem about $F'(s)$ pls?
$endgroup$
– kinder chen
Jan 27 at 1:33
$begingroup$
Well, it's the dual of the first theorem. There is a symmetry between the Fourier Transform and the Inverse Fourier Transform with the normalization convention I normally use. Although the symmetry isn't perfect as evinced by the negative sign, which pops up in duals of Fourier Transform theorems.
$endgroup$
– Andy Walls
Jan 27 at 2:56
add a comment |
$begingroup$
Using this definition of the Fourier Transform
$$F(s) = mathscr{F}left{f(x)right} = int_{-infty}^infty f(x) e^{-2pi i sx} space dx$$
and these two theorems
$$mathscr{F}left{dfrac{d}{dx}f(x)right} = 2pi i sF(s)$$
$$mathscr{F}left{-2pi i xf(x)right} = dfrac{d}{ds}F(s)= F'(s)$$
one can derive the answer
$$begin{align*}mathscr{F}left{dfrac{d}{dx}left(xf(x)right)right} &= 2pi i smathscr{F}left{xf(x)right} \
\
&= - smathscr{F}left{-2pi i xf(x)right} \
\
&= -sdfrac{d}{ds}F(s)\
\
&= -s F'(s)
end{align*}$$
Update to add a proof of the 2nd theorem
$$begin{align*}mathscr{F}^{-1}left{dfrac{d}{ds}F(s)right} &= int_{-infty}^infty {left[dfrac{d}{ds}F(s)right]e^{2pi i xs}}ds \
\
&= int_{-infty}^infty {lim_{Delta s to 0}{dfrac{F(s+Delta s) - F(s)}{Delta s}}e^{2pi i xs}}ds \
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[int_{-infty}^infty {F(s+Delta s)e^{2pi i xs}}ds -int_{-infty}^infty {F(s)e^{2pi i xs}}dsright]}\
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[int_{-infty}^infty {F(s')e^{2pi i x(s'-Delta s)}}ds' -f(x)right]}\
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[e^{-2pi i xDelta s}int_{-infty}^infty {F(s')e^{2pi i xs'}}ds'-f(x)right]}\
\
&= lim_{Delta s to 0}{dfrac{e^{-2pi i xDelta s}f(x)-f(x)}{Delta s}}\
\
&= lim_{Delta s to 0}{dfrac{e^{-2pi i x(0+Delta s)}-e^{-2pi i x0}}{Delta s}f(x)}\
\
&= left(dfrac{d}{ds}e^{-2pi i xs}right)biggr|_{s=0} space f(x)\
\
mathscr{F}^{-1}left{dfrac{d}{ds}F(s)right} &= -2pi i x f(x)\
\
dfrac{d}{ds}F(s) &= mathscr{F}left{-2pi i x f(x)right}\
\
end{align*}$$
answered Jan 26 at 20:33
Andy WallsAndy Walls
1,764139
$endgroup$
Using this definition of the Fourier Transform
$$F(s) = mathscr{F}left{f(x)right} = int_{-infty}^infty f(x) e^{-2pi i sx} space dx$$
and these two theorems
$$mathscr{F}left{dfrac{d}{dx}f(x)right} = 2pi i sF(s)$$
$$mathscr{F}left{-2pi i xf(x)right} = dfrac{d}{ds}F(s)= F'(s)$$
one can derive the answer
$$begin{align*}mathscr{F}left{dfrac{d}{dx}left(xf(x)right)right} &= 2pi i smathscr{F}left{xf(x)right} \
\
&= - smathscr{F}left{-2pi i xf(x)right} \
\
&= -sdfrac{d}{ds}F(s)\
\
&= -s F'(s)
end{align*}$$
Update to add a proof of the 2nd theorem
$$begin{align*}mathscr{F}^{-1}left{dfrac{d}{ds}F(s)right} &= int_{-infty}^infty {left[dfrac{d}{ds}F(s)right]e^{2pi i xs}}ds \
\
&= int_{-infty}^infty {lim_{Delta s to 0}{dfrac{F(s+Delta s) - F(s)}{Delta s}}e^{2pi i xs}}ds \
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[int_{-infty}^infty {F(s+Delta s)e^{2pi i xs}}ds -int_{-infty}^infty {F(s)e^{2pi i xs}}dsright]}\
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[int_{-infty}^infty {F(s')e^{2pi i x(s'-Delta s)}}ds' -f(x)right]}\
\
&= lim_{Delta s to 0}{dfrac{1}{Delta s}left[e^{-2pi i xDelta s}int_{-infty}^infty {F(s')e^{2pi i xs'}}ds'-f(x)right]}\
\
&= lim_{Delta s to 0}{dfrac{e^{-2pi i xDelta s}f(x)-f(x)}{Delta s}}\
\
&= lim_{Delta s to 0}{dfrac{e^{-2pi i x(0+Delta s)}-e^{-2pi i x0}}{Delta s}f(x)}\
\
&= left(dfrac{d}{ds}e^{-2pi i xs}right)biggr|_{s=0} space f(x)\
\
mathscr{F}^{-1}left{dfrac{d}{ds}F(s)right} &= -2pi i x f(x)\
\
dfrac{d}{ds}F(s) &= mathscr{F}left{-2pi i x f(x)right}\
\
end{align*}$$
answered Jan 26 at 20:33
Andy WallsAndy Walls
1,764139
edited Jan 27 at 17:29
answered Jan 26 at 20:33
Andy WallsAndy Walls
1,764139
answered Jan 26 at 20:33
Andy WallsAndy Walls
1,764139
answered Jan 26 at 20:33
Andy WallsAndy Walls
1,764139
1,764139
$begingroup$
Thx, can you elaborate your 2nd theorem about $F'(s)$ pls?
$endgroup$
– kinder chen
Jan 27 at 1:33
$begingroup$
Well, it's the dual of the first theorem. There is a symmetry between the Fourier Transform and the Inverse Fourier Transform with the normalization convention I normally use. Although the symmetry isn't perfect as evinced by the negative sign, which pops up in duals of Fourier Transform theorems.
$endgroup$
– Andy Walls
Jan 27 at 2:56
add a comment |
$begingroup$
Thx, can you elaborate your 2nd theorem about $F'(s)$ pls?
$endgroup$
– kinder chen
Jan 27 at 1:33
$begingroup$
Well, it's the dual of the first theorem. There is a symmetry between the Fourier Transform and the Inverse Fourier Transform with the normalization convention I normally use. Although the symmetry isn't perfect as evinced by the negative sign, which pops up in duals of Fourier Transform theorems.
$endgroup$
– Andy Walls
Jan 27 at 2:56
$begingroup$
Thx, can you elaborate your 2nd theorem about $F'(s)$ pls?
$endgroup$
– kinder chen
Jan 27 at 1:33
$begingroup$
Thx, can you elaborate your 2nd theorem about $F'(s)$ pls?
$endgroup$
– kinder chen
Jan 27 at 1:33
$begingroup$
Well, it's the dual of the first theorem. There is a symmetry between the Fourier Transform and the Inverse Fourier Transform with the normalization convention I normally use. Although the symmetry isn't perfect as evinced by the negative sign, which pops up in duals of Fourier Transform theorems.
$endgroup$
– Andy Walls
Jan 27 at 2:56
$begingroup$
Well, it's the dual of the first theorem. There is a symmetry between the Fourier Transform and the Inverse Fourier Transform with the normalization convention I normally use. Although the symmetry isn't perfect as evinced by the negative sign, which pops up in duals of Fourier Transform theorems.
$endgroup$
– Andy Walls
Jan 27 at 2:56
add a comment |
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$begingroup$
For your original question - assuming things are nice, you can integrate by parts to get $-it$ times Fourier transform of $xf(x)$.
$endgroup$
– herb steinberg
Jan 25 at 0:52