Looking for Fourier transform of x^-(n/m), where n,m are integers but n/m is not
Jay R. Yablon
the need to evaluate the Fourier transform of the general form:
F(x^-(n/m)) = $ x^-(n/m) exp^-x.w dx (1)
For example, what might be F(x^-(10/3)), which contains a cubed root,
and is for the underlying function (x^10)^-(1/3)?
At the link
http://74.125.93.132/search?q=cache:fmiTYnSjDbgJ:en.wikipedia.org/wiki/Fourier_transform+fourier+transform&cd=1&hl=en&ct=clnk&gl=us,
there is a very complete table including (#308):
F(x^n) = i^n sqrt(2pi) delta^(n) (w) (2)
where delta^(n) is the n-th distribution derivative of the Dirac delta
function. And (#310):
F(1/x^n) = F(x^-n) = -i sqrt(pi/2) [(-iw)^n-1/(n-1)!] sgn(w) (3)
And even the square root (#311):
F(1/sqrt(x)) = F(x^-(1/2)) = 1/sqrt(w) (4)
But I see nothing with the form (1), which involves a root other than
the square root, nor is it immediately clear to me how to easily
calculate (1) using the various ingredients in the table shown in this
link. And, I'd prefer to avoid having to convolve anything, because
that just replaces one integral with another.
Any help is appreciated -- preferably a link or reference to a table
which does have this function, or a good boost on how to combine the
ingredients at the above link to get to (1).
Thanks,
Jay
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.roadrunner.com/~jry/FermionMass.htm
Jay R. Yablon
line 310 of:
http://74.125.95.132/search?q=cache:fmiTYnSjDbgJ:en.wikipedia.org/wiki/Fourier_transform+fourier+transform&cd=1&hl=en&ct=clnk&gl=us,
for the Fourier transform F(1/x^n), where n-->m/n where each of m and n
are integers.
The "fractional factorial" threw me at first, but it seems as if one can
sensibly define this Fourier transform for non-integer factorials using
the gamma function, as described in both
http://74.125.95.132/search?q=cache:UFun5xqdxlgJ:en.wikipedia.org/wiki/Factorial+factorial&cd=1&hl=en&ct=clnk&gl=us
and http://en.wikipedia.org/wiki/Gamma_function. (There are also a
number of gamma function calculators for this infinitely-recursive
function when you get to other than half-integer factorials, see, e.g.,
http://functions.wolfram.com/webMathematica/FunctionEvaluation.jsp?name=Gamma)
Am I correct that one can use line 310 of that first reference, with m/n
type factorials evaluated via the gamma function, to evaluate the
Fourier transform F(1/x^n), ?
Thanks,
Jay
"Jay R. Yablon" <jya...@nycap.rr.com> wrote in message
news:7mdij5F...@mid.individual.net...
Jay R. Yablon
> Per the query below, it seems to me that the answer may in fact be on
> line 310 of:
> http://74.125.95.132/search?q=cache:fmiTYnSjDbgJ:en.wikipedia.org/wiki/Fourier_transform+fourier+transform&cd=1&hl=en&ct=clnk&gl=us,
> for the Fourier transform F(1/x^n), where n-->m/n where each of m and
> n
> are integers.
>
> The "fractional factorial" threw me at first, but it seems as if one
> can
> sensibly define this Fourier transform for non-integer factorials
> using
> the gamma function, as described in both
> http://74.125.95.132/search?q=cache:UFun5xqdxlgJ:en.wikipedia.org/wiki/Factorial+factorial&cd=1&hl=en&ct=clnk&gl=us
> and http://en.wikipedia.org/wiki/Gamma_function. (There are also a
> number of gamma function calculators for this infinitely-recursive
> function when you get to other than half-integer factorials, see,
> e.g.,
> http://functions.wolfram.com/webMathematica/FunctionEvaluation.jsp?name=Gamma)
>
> Am I correct that one can use line 310 of that first reference, with
> m/n
> type factorials evaluated via the gamma function, to evaluate the
> Fourier transform F(1/x^n), ?
>
> Thanks,
>
> Jay
I have calculated this out, and believe that the above is correct.
That is, referring to
http://74.125.95.132/search?q=cache:fmiTYnSjDbgJ:en.wikipedia.org/wiki/Fourier_transform+fourier+transform&cd=1&hl=en&ct=clnk&gl=us,
#311 is indeed a special case of #310 for n=1/2.
To see this, substitute n=1/2 into #310. Use -.5!=sqrt(pi) (You can
deduce that from what is on
http://74.125.95.132/search?q=cache:UFun5xqdxlgJ:en.wikipedia.org/wiki/Factorial+factorial&cd=1&hl=en&ct=clnk&gl=us).
Use sqrt(i)=(1/sqrt(2))(1+/-i).
The result you get from #310 is (with no |abs| sign)"
Fourier(1/x^.5) = .5sgn(w)[+/-1+i] (1/sqrt(w)) (1)
Now, this is for the integral from x=-oo to +oo. The first term is for
the +x half of the integral, the second for the -x half (which is why
the i shows up, because that is what happens for square roots of
negative numbers).
If you now take |x| and |w|, then sgn(w)=1, the real part of (1) gets
duplicated twice over and so the factor of .5 is doubled, and up to the
+/- sign, one has, precisely, #311.
So, #310 DOES work for non-integer n, using non-integer factorials, and
#311 is indeed a special case of this. Do the full calculation using
n=1/2 in #310 and you will
see that this is correct.
So, based on this known special case, it appears that #310 applies for
fractional n as well as integer n, using the fractional factorials and
the associated Gamma functions.
Jay.
Oh No
>In the course of working through a physics exercise, I have come across
>the need to evaluate the Fourier transform of the general form:
>
>F(x^-(n/m)) = $ x^-(n/m) exp^-x.w dx (1)
This isn't actually the FT, which has an i in the exponent. Its a bit
different also from a gamma function, which you look at elsewhere. Note
the difference in the exponent for the gamma function, and also that the
integration in the gamma function runs from 0 to oo.
>
>For example, what might be F(x^-(10/3)), which contains a cubed root,
>and is for the underlying function (x^10)^-(1/3)?
>
>At the link
>http://74.125.93.132/search?q=cache:fmiTYnSjDbgJ:en.wikipedia.org/wiki/
>Fourier_transform+fourier+transform&cd=1&hl=en&ct=clnk&gl=us,
>there is a very complete table including (#308):
>
>F(x^n) = i^n sqrt(2pi) delta^(n) (w) (2)
>
>where delta^(n) is the n-th distribution derivative of the Dirac delta
>function. And (#310):
>
>F(1/x^n) = F(x^-n) = -i sqrt(pi/2) [(-iw)^n-1/(n-1)!] sgn(w) (3)
>
>And even the square root (#311):
>
>F(1/sqrt(x)) = F(x^-(1/2)) = 1/sqrt(w) (4)
>
>But I see nothing with the form (1), which involves a root other than
>the square root, nor is it immediately clear to me how to easily
>calculate (1) using the various ingredients in the table shown in this
>link. And, I'd prefer to avoid having to convolve anything, because
>that just replaces one integral with another.
Its seems highly improbable to me that either this FT, or the function
you wrote down, could be evaluated except by numerical methods.
T
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Jay R. Yablon
"Oh No" <No...@charlesfrancis.wanadoo.co.uk> wrote in message
news:u61PlyB5...@charlesfrancis.wanadoo.co.uk...
> Thus spake Jay R. Yablon <jya...@nycap.rr.com>
>>In the course of working through a physics exercise, I have come
>>across
>>the need to evaluate the Fourier transform of the general form:
>>
>>F(x^-(n/m)) = $ x^-(n/m) exp^-x.w dx (1)
>
> This isn't actually the FT, which has an i in the exponent.
That is a typo.
> Its a bit
> different also from a gamma function, which you look at elsewhere.
> Note
> the difference in the exponent for the gamma function, and also that
> the
> integration in the gamma function runs from 0 to oo.
But that has nothing to do with the Fourier transform. Look at my later
posts. The gamma fucntion is pretenent only to the extent of needing to
evaluate a fractional factorial.
>>
>>For example, what might be F(x^-(10/3)), which contains a cubed root,
>>and is for the underlying function (x^10)^-(1/3)?
>>
>>At the link
>>http://74.125.93.132/search?q=cache:fmiTYnSjDbgJ:en.wikipedia.org/wiki/
>>Fourier_transform+fourier+transform&cd=1&hl=en&ct=clnk&gl=us,
>>there is a very complete table including (#308):
>>
>>F(x^n) = i^n sqrt(2pi) delta^(n) (w) (2)
>>
>>where delta^(n) is the n-th distribution derivative of the Dirac delta
>>function. And (#310):
>>
>>F(1/x^n) = F(x^-n) = -i sqrt(pi/2) [(-iw)^n-1/(n-1)!] sgn(w) (3)
>>
>>And even the square root (#311):
>>
>>F(1/sqrt(x)) = F(x^-(1/2)) = 1/sqrt(w) (4)
>>
>>But I see nothing with the form (1), which involves a root other than
>>the square root, nor is it immediately clear to me how to easily
>>calculate (1) using the various ingredients in the table shown in this
>>link. And, I'd prefer to avoid having to convolve anything, because
>>that just replaces one integral with another.
>
> Its seems highly improbable to me that either this FT, or the
> function
> you wrote down, could be evaluated except by numerical methods.
Look at my later posts. Referring to
http://74.125.95.132/search?q=cache:fmiTYnSjDbgJ:en.wikipedia.org/wiki/Fourier_transform+fourier+transform&cd=1&hl=en&ct=clnk&gl=us,
#311 is indeed a special case of #310 for n=1/2. While it is not a
general proof, this makes it highly plausible that that #311 can be
applied to fractional exponents.
Jay
Ken S. Tucker
> "Oh No" <N...@charlesfrancis.wanadoo.co.uk> wrote in message
>
> news:u61PlyB5...@charlesfrancis.wanadoo.co.uk...
>
> > Thus spake Jay R. Yablon <jyab...@nycap.rr.com>
> Look at my later posts. Referring tohttp://74.125.95.132/search?q=cache:fmiTYnSjDbgJ:en.wikipedia.org/wik...,
> #311 is indeed a special case of #310 for n=1/2. While it is not a
> general proof, this makes it highly plausible that that #311 can be
> applied to fractional exponents.
> Jay
Personally I think you're well grounded. Check this out,
http://en.wikipedia.org/wiki/Fractional_calculus
I agree with Charles to write a program to do that integral,
just adding bits, over a finite range provides a test for
infinities.
Regards
Ken
Jay R. Yablon
somebody's attention, because somebody changed entry #310 at
http://en.wikipedia.org/wiki/Fourier_transform to expressly deal with
F(1/|x|^a), *with a being a non-integer,* by writing this expressly in
terms of the Gamma function which is designed for non-integer
factorials.
This entry, which makes sense to me, if correct, tells us that aside
from a constant coefficient C which is a function of the Gamma function,
the "answer" I was looking for is:
Fourier(1/|x|^a) = C |w|^(a-1) (1)
So, whatever |x|^a is, it gets inverted, turned into w (omega), and
drops down one power.
Substantively, this is the same as the result I have been discussing,
but for the exact magnitude of the constant coefficient.
Importantly, it is finite, convergent.
Jay
Jay R. Yablon
Thanks so much for showing me the link below to fractional calculus. It
could not have come at a better time!
Jay
"Ken S. Tucker" <dyna...@vianet.on.ca> wrote in message
news:875ed802-6997-4491...@g23g2000yqh.googlegroups.com...
Ken S. Tucker
> Ken,
> Thanks so much for showing me the link below to fractional calculus. It
> could not have come at a better time!
> Jay
You're welcome Jay.
Could be there's some nifty math awaiting a physics application.
I find the "fractional derivative exponent", such as "a" at this link,
http://en.wikipedia.org/wiki/Fractional_calculus#Half_derivative_of_a_simple_function
can be variable. Then another expression of physics has "a" as a
function of space or time. That seems to happen in GR when time
and length (clocks and rods vary in length) are affected by G_uv=T_uv.
Example:
The 43" of arc/century of the residual semimajor axis rotation of
Mercury's orbit is given in a Newtonian formula such as,
acceleration = Mass/r^a , a=2.000 000 18.
with "r" being the distance from the Sun. As "r"=> oo, a=>2.
Following de Broglie's Wave Mechanics, it looks reasonable to
convert the Sun's mass into an extended Wave, in accord with
T_uv, then the field effects G_uv are equateable to said Wave.
Regards
Ken S. Tucker
Oh No
>> Thus spake Jay R. Yablon <jya...@nycap.rr.com>
>>>In the course of working through a physics exercise, I have come
>>>across
>>>the need to evaluate the Fourier transform of the general form:
>>>
>>>F(x^-(n/m)) = $ x^-(n/m) exp^-x.w dx (1)
>>
>> This isn't actually the FT, which has an i in the exponent.
>
>That is a typo.
ok, it just confused me as to what your meaning was.
>
>> Its a bit
>> different also from a gamma function, which you look at elsewhere.
>>Note
>> the difference in the exponent for the gamma function, and also that
>>the
>> integration in the gamma function runs from 0 to oo.
>
>But that has nothing to do with the Fourier transform. Look at my
>later posts. The gamma fucntion is pretenent only to the extent of
>needing to evaluate a fractional factorial.
>
>
>Look at my later posts. Referring to
>http://74.125.95.132/search?q=cache:fmiTYnSjDbgJ:en.wikipedia.org/wiki/
>Fourier_transform+fourier+transform&cd=1&hl=en&ct=clnk&gl=us,
>#311 is indeed a special case of #310 for n=1/2. While it is not a
>general proof, this makes it highly plausible that that #311 can be
>applied to fractional exponents.
>
I'm sure it can.