Binomial approximation using Mellin transform

[bdforbes]

I know how to derive the binomial approximation for $(1+\alpha x)^{\gamma}$ using a Mellin transform, but for $(1-\alpha x)^{\gamma}$ the method appears to fail because I can't take x to infinity.

Here is the basics of the method. Take the Mellin transform of $(1+\alpha x)^{\gamma}$:

$$M(p) = \int^\infty_0 (1+\alpha x)^{-\gamma}x^{p-1}dx$$

Do some change of variables trickery:

$$\begin{align*}<br /> M(p) &= \alpha^{-p}\int^1_0(1-z)^{\gamma-p-1}z^{p-1} dz\\<br /> &=\alpha^{-p}\frac{\Gamma(\gamma-p)\Gamma(p)}{\Gamma(\gamma)}<br /> \end{align*}$$

Use the inverse Mellin transform and close the integral to the left:

$$\begin{align*}<br /> (1+\alpha x)^{-\gamma}&=\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}(x\alpha)^{-p}\frac{\Gamma(\gamma-p)\Gamma(p)}{\Gamma(\gamma)}dp\\<br /> &=\sum_{n=0}^{\infty}(\alpha x)^n \frac{(-1)^n}{n!}\frac{\Gamma(\gamma+n)}{\Gamma(\gamma)}<br /> \end{align*}$$

But taking the Mellin transform of $(1-\alpha x)^{\gamma}$ fails immediately:

$$M(p) = \int^\infty_0 (1-\alpha x)^{-\gamma}x^{p-1}dx$$

The integrand will become complex quite quickly. I've tried playing around with changes of variables, but I can't figure it out.

Can this method be adapted for this purpose?

 

[rrogers]

Second case: In the forward transform you will have a pole when
$$x=\frac{1}{\alpha}$$.
You have to rearrange the integral limits to avoid this; the typical way is to use the Heavyside function
$$H(1-x \alpha )$$

First case: You need to condition your inversion transform.
Closing to the left when
$$x\alpha <1$$ and to the right when
$$x \alpha >1$$ .
You changed the sign of $$\gamma$$ midway; using the Mellin xform expression you need $$\gamma>0$$

Ray

 

[bdforbes]

using the Mellin xform expression you need $\gamma>0$

Does this mean this method is only applicable to expressions of the form $(1+\alpha x)^{-\gamma}$ with $\gamma>0$?

Second case: In the forward transform you will have a pole when $x=\frac{1}{\alpha}$.
You have to rearrange the integral limits to avoid this; the typical way is to use the Heavyside function $H(1-x\alpha)$

So I would effectively end up with this?

$$M\{H(1-x\alpha)(1-\alpha x)^{-\gamma}\}(p)=\int\limits_0^{1/\alpha}(1-\alpha x)^{-\gamma}x^{p-1}dx$$

How would I proceed from here?

 

[rrogers]

Does this mean this method is only applicable to expressions of the form $(1+\alpha x)^{-\gamma}$ with $\gamma>0$?



So I would effectively end up with this?

$$M\{H(1-x\alpha)(1-\alpha x)^{-\gamma}\}(p)=\int\limits_0^{1/\alpha}(1-\alpha x)^{-\gamma}x^{p-1}dx$$

How would I proceed from here?

Good question. Up till now I was just following along some presentations in some books. Extending the results requires some thought. Life is nothing but problems ; actually they point to some areas that I feel I have to tip-toe through, but that I should have clearly in mind.
I am not an expert. What are your goals? At this point I would examine your xform result; it is a form of the Beta function. Now looking at the xform, it is meromorphic. It would seem that this form could be proved to have an inverse transform for all values . If that is so, then can we identify the associated series with the binomial power? For some values of the vertical imaginary integration axis some extra terms might come and go; but these are usually just dropping or adding terms to the underlying function.
I can try to be explicit later. I would appreciate your thoughts.

Ray

 

[bdforbes]

At this point I would examine your xform result; it is a form of the Beta function. Now looking at the xform, it is meromorphic. It would seem that this form could be proved to have an inverse transform for all values . If that is so, then can we identify the associated series with the binomial power? For some values of the vertical imaginary integration axis some extra terms might come and go; but these are usually just dropping or adding terms to the underlying function.

I had considered that earlier. We arrive at the result which we know is correct for all values of x, yet the derivation only applies to certain values of x. It would be great if we could make an argument at this stage to extend to all values of x.

 

[rrogers]

Examining the ascending $$|\beta x|<1$$ and descending series for $$|\beta x|>1$$ leads to the conclusion that the inverse Mellin transform exists and is reasonable for both $$\lambda <0$$ and $$\beta <0$$ .

The remaining problem is to prove that the inversion series corresponds to your original expression in these domains. One way is to use the convolution property; but that looks to be complicated. Another way is to use a Taylor series expansion comparison at zero;between the inversion and the original formula. Or using the "Direct Mapping" theorem; that looks to be the best way.
That's actually a hidden form of Taylor series. I have to check something though.
BTW: If you have the Taylor series the original problem would be solved:)

Ray

 

[rrogers]

Sorry about the last post. I didn't see yours. I will think about what you said.

Ray

 

[rrogers]

I have examined the underlying process and have an elementary solution; i.e. without Mellin xforms.
The Mellin process should follow the same outline; two cases for the power being >,< 0. In the Mellin case I think it is educational as I think it talks to the Inverse Mellin xform of the Analytic continuation of the Beta function.
If your interested in partial notes I can email a file (it's a little long and windy). Otherwise wait until I finish writing things up.
There are minor restrictions on x, -1<x ; but scaling can deal with that if I don't find a more elegant solution.

Ray

 

[bdforbes]

Thanks Ray, I'll be interested to see the results. I'm quite busy at the moment, so I'm happy to wait until you've had the chance to write it up.