Calculate ch(0.2) to the nearest 0.01

Question

Help me calculate ch(0.2) to the nearest 0.01. I tried to rewrite ch as a series but I still don't know how to evaluate it and what to do with factorial

Help me please. it's very important

Answer 1

You have received a naswer from impartialmale who explains a lot of things. So, I shall just focus on the details.

If you expand your formula, you have $$ch(x)=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}=x+\frac{x^3 }{6}+\frac{x^5}{120}+\frac{x^7}{5040}+\frac{x^9}{362880}+\frac{x^{11}}{39916800}+...$$ Now, make $x=0.2$ in the above formula and compute the terms one after each other and add them up to have the partial sums. You will see when you need to stop.

Answer 2

Using the following theorem related to Dalambe' s criteria (see the reference: http://ecademy.agnesscott.edu/~lriddle/apcalculus/approxSeries.pdf):

Theorem Let $\{a_n\}$ be a positive decreasing sequence such that $$ \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=L<1. $$
Then

(1) If $\displaystyle\frac{a_{n+1}}{a_n}$ decreases to the limit $L$ then $$ S_n+a_n\left(\frac{L}{1-L}\right)<S<S_n+\frac{a_{n+1}}{1-\frac{a_{n+1}}{a_n}}. $$ (2) If $\displaystyle\frac{a_{n+1}}{a_n}$ increases to the limit $L$ then $$ S_n+\frac{a_{n+1}}{1-\frac{a_{n+1}}{a_n}}<S<S_n+a_n\left(\frac{L}{1-L}\right). $$ Here $$ S=\sum_{n=0}^{\infty}a_n, \quad S_n=\sum_{k=0}^{n}a_k, $$ to calculate $$ ch(0.2)=\sum_{n=0}^{\infty}\frac{(0.2)^{2n+1}}{(2n+1)!} $$