Interpolation problem

**Vinod** · Jul 13th 2013, 04:51 AM

Show that B (m + 1, n) = $(-1)^m \Delta^m \left(\frac 1n \right)$ where m is a positive integer, n > 0

Some problem solving hint is required from any member of this forum.

**chiro** · Jul 13th 2013, 06:45 PM

Hey Vinod.

What is the definition of delta?

**Vinod** · Jul 14th 2013, 08:21 AM

Hi Chiro,
Thanks for your prompt response. Yesterday, I found difficult to understand the answer provided in the book.But today I revised interpolation topic. As a result, I understood the answer to the problem
Ans: We have

$\int_0^{\infty}e^{-nx}dx=\frac1n$

$\triangle^m \int_0^{\infty}e^{-nx}dx=\triangle^m\left(\frac1n\right)$

$\rightarrow \int_0^{\infty}\triangle^m e^{-nx}dx=\triangle^m\left(\frac1n\right)$
But
$\triangle^m e^{-nx}=\triangle^{m-1} (e^{-(n+1)x}- e^{-nx})$

$=\triangle^{m-1} e^{-nx}(e^{-x}-1)$
$=e^{-nx}(e^{-x}-1)^m$
Hence
$\int_0^{\infty} e^{-nx}(e^{-x}-1)^m dx=\triangle^m\left(\frac1n\right)$

Putting z = $e^{-x}$ ,we obtain

$(-1)^m \int_0^{\infty}z^{n-1}(1-z)^m dz=\triangle^m\left(\frac1n\right)$

$\rightarrow. B(m+1,n)=(-1)^m\triangle^m\left(\frac1n\right)$

,

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