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.

Question

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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