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Let$$P_n(x)=\sum_{k=0}^{n}(-1)^k\frac{x^{2k+1}}{(2k+1)!}$$

let $c_n$ be the number of real zeros of $P_n$.

determine$$\lim_{n \rightarrow \infty}\frac{c_n}{2n+1}$$

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Hint: Proof that $c_n$ is nodecreinsing and $n<c_n\leq 2n+1$. Use the sandwiche theorem in $\frac{n}{2n+1}<\frac{c_n}{2n+1}\leq 1$. –  Elias Dec 17 '12 at 18:12

2 Answers 2

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This is a partial answer that gives a lower bound for the number of real roots. According to Taylor's theorem for any $x \in \mathbb{R}$ there exists $\xi \in [-x, x]$ such that

$$ \sin(x) = P_n(x) + \frac{\frac{\partial^{2n+3} \sin}{\partial x^{2n+3}}(\xi)\,x^{2n+3}}{(2n + 3)!} $$

and in particular $$\left|\sin(x) - P_n(x)\right| \leq \frac{|x|^{2n+3}}{(2n+3)!}.$$ So if $\sin(x)$ goes from $1$ to $-1$ on some interval $[a, b]$ (or from $-1$ to $1$) and

$$\frac{|x|^{2n+3}}{(2n+3)!} < 1$$

on $[a,b]$ then $P_n(x)$ must have a root on $[a,b]$. Now use that $$e \left(\frac{n}{e}\right)^n \leq n!$$ to see that this inequality certainly holds for $|x| \leq \frac{2n+3}{e}$. This shows that $P_n(x)$ has at least $$ \frac{4n+6}{\pi e} - 2$$ real roots.

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By Fundamental Teorem of Algebra we have the number of real zeros of $P_n(x)$ is no more that $2n+1$. Use the fellowing facts:

1) All polinom $P(x)$ of grau odd have almost $1$ real zero.

2) If in $P_n(x)=\sum_{k=1}^{2n+1}a_k\cdot x^{k}$ we have $a_k\in\mathbb{R}$ for all $k\in\mathbb{N}$ them the number of no real complex zeros of $P_n(x)$ are even and the number of real zeros of $P_n(x)$ are odd.

3) Proof that for all $P_n$ that $|P(x)|>0$ for all $x > 2n+1$.

4) Proof that for all $P_n$ and $\epsilon>0 $ that exist $a=a(\epsilon,n)>0$ sholt that

$$ |\sin(x)-P_n(x)|<\epsilon $$

for all $x\in[-a,+a]$ and $P(x)$ have the number of zeros of $\sin$ in $[-a,+a]$.

5) Use the fact 4) to proof that $c_n\leq c_{n+1}$.

6) Use the facts 2) and 3) to proof that $c_n<c_{n+1}$

Now I think with this last fact you can find a way to demonstrate this result. Good Look.

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