If p(x) = $\sum_{k=0}^{10} k^{2} x^{ \underline{k} }$, evaluate $\bigtriangleup ^{6}$ p(x) $| _{x=0}$.

Question

If p(x) = $\sum_{k=0}^{10} k^{2} x^{ \underline{k} }$, evaluate $\bigtriangleup ^{6}$ p(x) $| _{x=0}$.

Not sure how to start with this. I know that p(x) = $\sum_{k=0}^{n} \bigtriangleup ^{k}$p(0) $\frac{{x}^{\underline{k}}}{k!}$

If you can help clear this question up, I'd be grateful!

Answer 1

HINT: You should know that $\Delta x^{\underline k}=kx^{\underline{k-1}}$ if $k\ge 0$ and that $\Delta 0=0$; if not, you should prove them. In particular this means that $\Delta^6x^{\underline k}=0$ if $k<6$. You also know that $\Delta cf(x)=c\Delta f(x)$ if $c$ is a constant. If you put those pieces together, you should be able to write down $\Delta^6 p(x)$ fairly easily:

• What is $\Delta^6 k^2x^{\underline k}$ when $k<6$?
• What is $\Delta^6 k^2x^{\underline k}$ when $k\ge 6$?

Then substitute $x=0$. Note that substituting $x=0$ wipes out every non-constant term of a polynomial in $x$.