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Oct
31
comment How to express this operator through derivetives?
Works! Except in non-pole points it does not give the original function, but this is even better. Also often need to use lomit on the quotinent.
Oct
31
comment Ascribing values to Gamma of negative integers
@Claude Leibovici or, without using complex numbers: Limit[(Gamma[-n + x] + Gamma[-n - x])/2, x -> 0]
Oct
31
comment Ascribing values to Gamma of negative integers
@Claude Leibovici It gives Gamma[-n] for all other real values except negative integers.
Oct
31
comment Ascribing values to Gamma of negative integers
@Claude Leibovici Mathematica code: Limit[Re[Gamma[-n + I x]], x -> 0]
Oct
31
comment Ascribing values to Gamma of negative integers
@Henry For instance it gives $(1/t)^{-1}=\log|x|+\gamma$ (which is by the way, consistent with discrete integral of $1/t$ which is $\psi(t)$ and asymptotically approaches $\log|x|+\gamma$ at $x\to+\infty$
Oct
31
comment Ascribing values to Gamma of negative integers
@Henry good point indeed... But on the other hand they are consistent with Fourier differintegral: $$(t^n)^{(s)}|_{t=1}=\frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{e^{- i \omega }}{(-i\omega)^s} \int_{-\infty}^{+\infty}t^n e^{i\omega t}dt \, d\omega=\Gamma(s+1)(-1)^s t^{n-s}$$
Oct
17
comment Dirac Delta definition in non-standard analysis?
Delta is even function so its derivative should be odd. That is zero at x=0.
Oct
17
comment Dirac Delta definition in non-standard analysis?
@Hurkyl if derivative of delta is negative at x<0 and positive at x>0, this means that delta is non-positive around and in zero, which would not give positive integral. It is evident you made a mistake.
Oct
17
comment Dirac Delta definition in non-standard analysis?
@Hurkyl it seems you confused the right and left parts of the function in this definition of derivative. Its derivative should be positive at negative part and negative at positive area, otherwise its integral would be negative. Also note that this definition has a major disadvantage: delta is even function so its derivative shoud be odd. And also it should be 0 in x=0 because delta has maximum there.
Oct
17
comment Dirac Delta definition in non-standard analysis?
it is differentiable in a sense. At least, the notion of derivative of dirac delta (and further derivatives) is widely used. See my answer for the for that I found is employed in non-standard analysis. Its derivative thus would be $-\frac{2 w^3 z e^{-(w z)^2}}{\sqrt{\pi }}$
Oct
17
comment Dirac Delta definition in non-standard analysis?
this has only a few properties of Dirac Delta and does not have others (i.e. differentiability).
Oct
17
comment Dirac Delta definition in non-standard analysis?
How would u differentiate such delta function?
Oct
17
comment Dirac Delta definition in non-standard analysis?
@Semiclassical this is what is non-standard analysis. And also $\delta(0)=\frac{\omega}{\sqrt{\pi}}$
Oct
15
comment Obtaining generating function via Fourier transform
@Semiclassical yes but I wonder whether it is possible using Fourier transform.
Oct
14
comment Fourier transform of exponent?
@Inquisitive this exactly follows from the first formula in the question so what?
Oct
14
comment Fourier transform of exponent?
@Inquisitive because integral of that function along the real line is 0.
Oct
6
comment Are there any functions which were proposed as elementary by mathematicians but not considered elementary now?
@Lucian that's ok.
Sep
11
comment How to solve the following delay differential equation?
what is $g_1$?.
Sep
2
comment What does this expression give?
@Mhenni Benghorbal no
Aug
27
comment Is there an easier way to find the “natural” integration constant?
"When h>0, it does not converge, but the most logical value would be extending it using it the geometric series formula. When h<0, the series converges, and the h cancels the negative." - so, you have just "discovered" that $-\int_{-\infty}^x e^x$ converges, while $-\int_{x}^{\infty} e^x$ does not. Very novel!