Talk:Abel transform

Fixed apparent error in proof of inverse transform David s graff 17:51, 2 October 2006 (UTC)

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As someone trying to use this article as an intro to the Abel transform, I found parts of the geometrical interpretation a little unclear. Since

$x=r\,\cos {\theta }=r{\frac {x}{r}}={\frac {r\,{\sqrt {r^{2}-y^{2}}}}{r}}$

Shouldn't

$dx={\frac {{\sqrt {r^{2}-y^{2}}}\,dr}{r}}$ rather than :$dx={\frac {r\,dr}{\sqrt {r^{2}-y^{2}}}}$? Sunderfil (talk) 00:03, 24 September 2009 (UTC)

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It's been awhile since you posted your question, but the answer is no, your expression is incorrect. Since

$x={\sqrt {r^{2}-y^{2}}}$

then

$dx={\frac {r\,dr}{\sqrt {r^{2}-y^{2}}}}$

Mac John Concord (talk) 21:55, 1 December 2010 (UTC)Mac John Concord

Apparently the inversion formula is wrong

In the section "Verification of the inverse Abel transform" the step where the derivative is taken formally, it is not taken into acount, that the lower bound of the integral is also changing. Hence the proof is wrong. I tried to verify the formula by an example. I used $f(r)=1$ for $r\in [0,1]$ and $f(r)=0$ otherwise and got the result ${\frac {1}{2}}(1-r^{2})$ after applying the abel transform and the given inversion formula. Hence the inversion formula is incorrect. — Preceding unsigned comment added by 82.119.166.92 (talk) 12:21, 20 February 2013 (UTC)

you are wrong. The extra term is zero. https://en.m.wikipedia.org/wiki/Differentiation_under_the_integral_sign AManWithNoPlan (talk) 04:05, 26 April 2015 (UTC)