Any help with the proof I have posted for the following question is greatly appreciated; I have listed my particular issue at the end. Thank you!
Let $p \geq 0$. For $n=1,2, \dots,$ find $P_n(x)$ about $x=0$ of $f(x)=(1+x)^p$ and find an upper bound depending on $n,p$ for the error on $x \in [0,1]$.
Since the solution depends heavily on Taylor's theorem, I will state it here:
Suppose $f \in C[a,b]$, $f^{n+1} \in C[a,b]$ and $x_0 \in [a,b]$. For every $x \in [a,b]$, there exists a number $\xi(x)$ between $x_0$ and $x$ such that $$f(x)=P_n(x)+R_n(x)$$ where $$P_n(x)=\sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$$ $$R_n(x)=\frac{f^{(n+1)}(\xi(x))}{(n+1)!}(x-x_0)^{n+1}$$
As $f(x)$ describes the binomial distribution, we have $$P_n(x)=\sum_{k=0}^n\frac{p(p-1)\cdots (p-k+1)}{n!} x^k$$
We will now find an upper bound on $n,p$ for the error on $x \in [0,1]$. Using $f^{(n+1)}$ obtained earlier, we have: $$R_n(x)=\frac{p(p-1)\cdots (p-n)(1+\xi(x))^{p-n-1}}{(n+1)!}x^{n+1}$$
\begin{eqnarray*}
\max_{x \in [0,1]}|R_n(x)|& = & \max_{x \in [0,1]} \left|\frac{p(p-1)\cdots (p-n)(1+\xi(x))^{p-n-1}}{(n+1)!}x^{n+1}\right| \\
& = & \max_{x \in [0,1], \xi \in [0,x]} \left|\frac{p(p-1)\cdots (p-n)(1+\xi)^{p-n-1}}{(n+1)!}x^{n+1}\right| \\
& = & \left|\frac{p(p-1)\cdots (p-n)}{(n+1)!}\right| \max_{x \in [0,1], \xi \in [0,x]} \left|(1+\xi)^{p-n-1} x^{n+1}\right| \\
& \leq & \left|\frac{p(p-1)\cdots (p-n)}{(n+1)!}\right| \max_{x \in [0,1], \xi \in [0,1]} \left|(1+\xi)^{p-n-1} x^{n+1}\right| \\
& \leq & \left|\frac{p(p-1)\cdots (p-n)}{(n+1)!}\right| \max_{\xi \in [0,1]} \left|(1+\xi)^{p-n-1} \right| \\
& \leq & \left|\frac{p(p-1)\cdots (p-n)}{(n+1)!}\right| \left|2^{p-n-1}\right|
\end{eqnarray*}
Now, I have a major problem with the transition from the second last step to the last step. Specifically, is the maximum of $\left|2^{p-n-1}\right|$ always achieved when $\xi=1$? It seems to me that for example, if $p=1, n=1$, then $p-n-1<0$, and the previous condition fails. Any advice would be helpful here. Should I take it on a case-by-case basis? Is there anything else wrong with the proof? Thanks for reading my large post, I hope that I have provided enough context.
Best Answer
I think you will have to do a case based analysis. If p is an integer, then for $$n \geq p\ , R_n =0 .$$ and $$n \leq p\ , \xi =1 .$$ gives maximum.
If not, then $ p-n\geq 1$ gives maximum for $\xi =1 $.
Or else,if $ p-n\leq 1$ gives maximum for $\xi =0$.