How can the value of $e$ be found to five-place accuracy using Higher Order Mean Value Theorem. Hint: Use the fact that $e\approx 3$.

Question

Higher Order Mean Value Theorem equates to as follows:

$$f(x)=f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + ... + \frac{f^n(a)(x-a)^n}{n!} + \frac{f^{(n+1)}(a)(x-a)^{(n+1)}}{(n+1)!}$$

I know that,

$$\\e = \sum_{n=0}^{\infty }\left( \frac{1}{n!} \right)\\$$

But how can I calculate the value of e using $e\approx3$ ? I couldn't find a starting point.

Following method can be used but it doesn't use $e\approx3$:

Let

$f(x)=e^{x}$

Considering x=1, it becomes:

$f(x)=e^{1}=e$

Now summing the Higher Order MVT Equation gives by taking $x=1$ & $a=0$,

$$\sum_{n=0}^{10}\frac{\left(\frac{d^{n}}{da^{n}}\left(e^{a} \right)\left(x-a \right)^{n} \right)}{n!}$$

Resulting in,

$$\frac{9864101}{3628800}= 2.71828$$

Answer 1

Instead, we shall use $e<3.$

$$e=\sum_{n=0}^{\infty}\frac{1}{n!}=\sum_{n=0}^m \frac{1}{n!} + \sum_{n=m+1}^\infty \frac{1}{n!}$$

$$ \sum_{n=m+1}^\infty \frac{1}{n!} < \frac{1}{(m+1)!}(\sum_{n=0}^\infty\frac{1}{n!})=\frac{1}{(m+1)!}e<\frac{3}{(m+1)!}$$

Therefore, we have $0<e-\sum_{n=1}^m\frac{1}{n!}<\frac{3}{(m+1)!}$ for all $m$.

When $m=8$, $\frac{3}{9!}<10^{-5}$, hence we know that $e$ agrees with $\sum_{n=1}^8\frac{1}{n!}=2.71827...$ at least up to $2.7182$.

Answer 2

Just for your curiosity.

Starting from @Just a user's answer, suppose that you want to know $m$ such that $$\frac{3}{(m+1)!} \leq \epsilon \implies (m+1)!~ \geq ~\frac 3 \epsilon$$ ANswering this question @Gary provided a superb approximation of the inverse of the factorial function in terms of Lambert function.

Applied to your case, $$m \sim \frac{{\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)}}{{W\left( {\frac{1}{e}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right)}} - \frac{3}{2} \qquad \text{where}\qquad y=\frac 3 \epsilon$$ Suppose that we want $\epsilon=10^{-20}$, this would give as a real $m=20.5739$ that is to say $\lceil m\rceil=21$.