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Let F: [a,b]-> R be infinitely differentiable on [a,b]. Let $|F^{(n)}(x)| \leq C$, where $F^{(n)}(x)$ is the $n^\text{th}$ derivative of $F$, and $C$ is a nonnegative constant. Let $P_n(x)$ be the Taylor polynomial of order $n$ of $F$, centered around $y$. Prove the sequence of functions $P_n$(x) uniformly converges to $F(x)$ on $[a,b]$.

To start off, for all $\varepsilon>0$ there's a natural number $N$ such that $|P_n(x)-F(x)|< \varepsilon$ when $n \geq N$.

I know F can be represented as a Taylor series, but there can be error $E$ in this representation, so let $F = P_n(x)-E$. So $|E|< \varepsilon$. $E = F^{(n+1)}(x)\frac{(x-y)}{(n+1)!}$ since it's part of the Taylor polynomial. So $|F^{(n+1)}(x)\frac{(x-y)^{n+1}}{(n+1)!}| < \varepsilon$. This is where I hit a brick wall. How can I get an expression for N out of this? All I can think of is that $(x-y) \leq (b-a)$.

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You can use the integral representation of the remainder:

$$R_n(x)=\int_y^x\frac{f^{(n+1)}(t)}{n!}(x-t)^ndt$$

Then $$|R_n(x)|\leqslant C\int_y^x\frac{|x-t|^n}{n!}dt$$

That is, you need to show $R_n\to 0$ uniformly. Alternatively, you can use Lagrange's form of the remainder $$R_n(x)={f^{(n+1)}(\xi)}\frac{(x-y)^{n+1}}{(n+1)!}$$ for some intermediate point depending on $y$ and $x$.

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  • $\begingroup$ Thank you for your response Pedro, but unfortunately I cannot use the integral since we have not yet covered the notion of an integral in class. $\endgroup$ Nov 7, 2013 at 22:27
  • $\begingroup$ But I think I am onto something; just to make sure, $F^{(n+1)}(x)$ in my original post is a constant, right? By the way, the x should actually be y in this case. $\endgroup$ Nov 7, 2013 at 22:32

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