Numerical Analysis - Upper bound for $|R|$

Question

I am asked to find an upper bound for $|R|$ valid for all $x\in[-1,1]$ that is independent of $x$ and $\xi$.

Given that,

$$R(x)=\frac{|x|^6}{6!}e^\xi$$

for $x\in[-1,1]$ where $\xi$ is between $x$ and $0$. I began with stating that

$$|R_n(x)|=\frac{|x^{n+1}e^{c_x}|}{(n+1)!},$$ $$=\frac{|x|^{n+1}e^{c_x}}{(n+1)!},$$ because $e^z > 0$ for all $z$ and $$|R_n(x)|\leq\frac{e^{c_x}}{(n+1)!},$$ because $|x|\leq 1$ for all $x\in[-1,1]$

Does this satisfy the question or will I need more?

Doesn't the trivial upper bound $\dfrac e{6!}$ work for you? — lhf, Apr 12 '16 at 01:51
I think so, I arrived at $|R|=\frac{e}{6!}$, however. And this is the correct approach, is it not? — cryptomath, Apr 12 '16 at 01:55

score 1 · Accepted Answer · answered Apr 12 '16 at 01:58

1

Since both $x,\xi \in [-1,1]$, we have $|x|\le 1$ and $|\xi|\le 1$ and so $$ |R(x)|= \left|\frac{|x|^6}{6!}e^\xi\right| \le \frac{1}{6!} e^1=\frac{e}{6!} $$

answered Apr 12 '16 at 01:58

lhf

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Numerical Analysis - Upper bound for $|R|$

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