Thread: Interpolation problem

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

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

Hey Vinod.

What is the definition of delta?

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

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

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

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

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

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

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

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



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