Here's the question you clicked on:
badreferences
Let \(n\in\mathbb N\). For\[e^xf_n(x)=\sum_{k=1}^\infty\frac{k^nx^k}{\left(k-1\right)!}\]show that \(f_n(x)\) is a polynomial of degree \(n+1\) with integer coefficients. Tricky question.
take derivative\[e^x=\sum_{k=1}^\infty\frac{x^{k-1}}{(k-1)!}\]multiply by x\[xe^x=\sum_{k=1}^\infty\frac{x^{k}}{(k-1)!}\]and take derivative like this n times and multiply by x again every time finally we will have something like this\[e^xP(x)=\sum_{k=1}^\infty\frac{k^nx^{k}}{(k-1)!}\]
actually better to write\[e^xP_n(x)=\sum_{k=1}^\infty\frac{k^nx^{k}}{(k-1)!}\]
lets prove by induction that degree of \(P_n(x)\) is n+1 \[P_1(x)=x+x^2\]lets say degree of \(P_n(x)\) is n+1 and prove that degree of \(P_{n+1}(x)\) is n+2\[P_{n+1}(x)=xe^{-x}[e^xP_n(x)]'=x(P_n^'(x)+P_n(x))\] which is from degree 1+n+1=n+2
@mukushla can u show me the derivative of |dw:1348828878035:dw|
\[e^x(1+x)=\sum_{k=1}^\infty\frac{kx^{k-1}}{(k-1)!}\]
OH....... NOW, I got it.... THANX @mukushla
http://openstudy.com/study#/updates/50650f91e4b08d185211d261 Solution in last post.