## anonymous 4 years ago 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.

1. anonymous

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$e^xP(x)=\sum_{k=1}^\infty\frac{k^nx^{k}}{(k-1)!}$

2. anonymous

actually better to write$e^xP_n(x)=\sum_{k=1}^\infty\frac{k^nx^{k}}{(k-1)!}$

3. anonymous

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

4. anonymous

@mukushla can u show me the derivative of |dw:1348828878035:dw|

5. anonymous

$e^x(1+x)=\sum_{k=1}^\infty\frac{kx^{k-1}}{(k-1)!}$

6. anonymous

OH....... NOW, I got it.... THANX @mukushla

7. anonymous