Calculate the number e using Taylor polynomial and Lagrange Remainder

Question

I am supposed to determine the value of number e with exactness (mistake) less than $10^{-8}$, using Taylor polynomial and Lagrange Remainder.

My hint is that $T_{n}(e^{x},0)(x)$.

I know the formula for Lagrange Remainder, as well as for Taylor polynomial of e, but I am not able to combine it.

Can anyone help me?

check the various errors for different values of n, once you get error less than 10^-8, you can calculate the value of e for that n, using the taylor polynomial — rhombicosicodecahedron, May 01 '19 at 17:48
I do not understand, can you please explain in which way I should get various errors? @rhombicosicodecahedron — Shelley, May 01 '19 at 17:53
You have to find $n$ such that the absolute value of the reminder is less than $10^{-8}$, right ? — Dlem, May 01 '19 at 17:58
I am supposed to find the value of e with exactness (mistake) less than 10^{-8} @DLeMeur — Shelley, May 01 '19 at 18:00
More precisely : $e = T_n(1) + R_n(1)$ for all $n$. So if $|R_n(1)|\leq 10^{-8}$, $T_n(1)$ gives the approximation you want. — Dlem, May 01 '19 at 18:02

score 1 · Accepted Answer · answered May 01 '19 at 18:02

Let the $n$th Taylor polynomial of $e^x$ be $T_n$ with corresponding remainder $R_n$. To approximate the value of $e$, you want to focus on evaluating the Taylor polynomial at $x = 1$, i.e., $T_x(1)$ because $e^1 = e$.

Your goal is to find some positive integer $n$ such that $|T_n(1) - e^1| = |R_n(1)| < 10^{-8}$. You said you have a way to calculate the remainders, so the easiest way to solve your problem would be to evaluate $|R_n(1)|$ for each $n = 1, 2, 3...$ until you get a value less than $10^{-8}$.

Once you find a suitable $n$, evaluate $T_n(1)$ for that $n$ to get your approximation.

score 1 · Answer 2 · answered May 01 '19 at 18:49

$$e=\sum_{k=0}^\infty \ \frac{1}{k!}=\sum_{k=0}^N\ \frac{1}{k!}+\sum_{k=N+1}^\infty\ \frac{1}{k!}=\sum_{k=0}^N\ \frac{1}{k!}+\frac{1}{(N+1)!}+\frac{1}{(N+2)!}+...$$

Note the remainder term is :

$$\frac{1}{(N)!}[\frac{1}{N+1}+\frac{1}{(N+1)(N+2)}+\frac{1}{(N+1)(N+2)(N+3)}+...]<\frac{1}{N!}[\frac{1}{(N+1)}+\frac{1}{(N+1)^2}+\frac{1}{(N+1)^3}+...]=\frac{1}{N\cdot N!}$$

So you can probably get away with $\frac{1}{(N+1)!}<10^{-8}$.

$N\approx 11. $

score 1 · Answer 3 · 2019-05-01T19:06:12.207

1

The Lagrange remainder can be expressed as

$$\frac{t^{n+1}e^t}{(n+1)!}$$ where $0<t<1$ and is bounded by

$$\frac1{(n+1)!}.$$

Then use $$11!<10^8<12!$$

$$\sum_{n=0}^{11}\dfrac1{n!}=\frac{13563139}{4989600}.$$

edited May 01 '19 at 19:06

answered May 01 '19 at 18:57

Calculate the number e using Taylor polynomial and Lagrange Remainder

3 Answers3