Question: Find a number N s.t $\forall x>0 $ and $ n\ge N: |\sum_{k=0}^n\frac {(-1)^kx^k}{k!}|<2$
Thoughts We used the remainder of the taylor series of $e^{-x}$, after some bounding we got stuck. Would love someone's help finishing up.
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1I'm afraid we can't help you finishing up, such an $N$ does not exist. $$\sum_{k=0}^n \frac{(-1)^kx^k}{k!}$$ is, for $n \geqslant 1$, a non-constant polynomial, and thus unbounded on $(0,\infty)$. – Daniel Fischer Jan 21 '14 at 20:33
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Your problem currently asks for single value of $N$ which would work for all $x$ at the same time (i.e. $(\exists N)(\forall x)(\forall n\geq N)$). As Daniel pointed out, this is impossible. However, if you asked for $N$ after being given $x$, the problem would become more interesting (i.e. $(\forall x)(\exists N)(\forall n\geq N)$)... – Peter Košinár Jan 21 '14 at 21:05
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I think maybe that's what the writer of this question meant – jreing Jan 22 '14 at 07:32
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As Daniel Fischer said, it is impossible to choose $N$ independently of $x$. But after $x>0$ is fixed, we can find $N$ as follows: since $|e^{-x}|<1$, it suffices to bound the remainder $r_n(x)=\sum_{k=n+1}^\infty\frac {(-1)^kx^k}{k!}$ by $1$. The Lagrange form of remainder yields $$ |r_n(x)|\leq \frac{x^{n+1}}{(n+1)!}$$ According to Stirling, $$(n+1)!\ge \sqrt{2\pi}(n+1)^{n+3/2}e^{-n-1}$$ Thus, $$ |r_n(x)|\leq \left(\frac{ ex }{ n+1}\right)^{n+1} $$ and choosing $N=\lfloor ex\rfloor$ does the job.
you can call me Al
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