Find an interval $[0, b]$ over which $e^x$ can be approximated by $1+x+\frac{x^2}{2!}$ to three decimal-place accuracy How to find an interval $[0, b]$ over which $e^x$
can be approximated by $1+x+\frac{x^2}{2!}$ to three decimal-place
accuracy.
I feel we have to find $b$ from the inequality $|R_2(x)|\leq |\frac{Mx^3}{3!}|\leq |\frac{Mb^3}{6}|<10^{-4}$ for $0 \leq x \leq b$ and $M=\sup_{[0,b]}|e^x|$. But I stuck here at $|e^b b^3|<6 \cdot 10^{-4}$.
 A: So your analysis so far is correct. You want to find $b>0$ such that
$$e^b b^3 < 6\cdot 10^{-4}.$$
Notice that $e<3$ so that $e^b b^3<3^b b^3$. So it's enough to find $b>0$ such that
$$3^b b^3 < 6\cdot 10^{-4}.$$
This can't be solved analytically, but we can just substitute some values for $b$. Trying $b=0.1$ we get $0.0012\ldots$ which is not small enough, and for $b=0.01$ we get $0.000001\ldots$ which is less than $6\cdot 10^{-4}$.
A: For your curiosity
Making the problem more general, for $n$ terms, you need to solve for $b$
$$\frac 1{(n+1)!}\, b^{n+1} \,e^b \leq \epsilon$$ The only explicit solution is given in terms of Lambert function
$$b \leq (n+1)\,\, W\left(\frac{1 }{n+1}\Big[\epsilon  (n+1)!\Big]^{\frac{1}{n+1}}\right)$$ Expanded as a series around $\epsilon=0$, an approximation  will be
$$b\sim \Big[\epsilon  (n+1)!\Big]^{\frac{1}{n+1}}-\frac 1{n+1}\Big[\epsilon  (n+1)!\Big]^{\frac{2}{n+1}}$$ Applied to your case ($n=2$, $\epsilon=10^{-4}$, this would give
$$b \sim \frac{30 \sqrt[3]{5}-\sqrt[3]{3}}{100\times 15^{2/3}}=0.081972$$ while the exact value is $0.0820673$
