In lecture, we went through solving a Taylor error bound for arcsine. I followed most of it except for where it talks about the odds divided by the evens divided by $2n+1$ gaining in accuracy by a factor of 1/10 for each successive term (see bolded sentence below), which comes from this part of the series:
$$ \sum_{x=1}^n \frac{1}{2n+1} \prod_{k=1}^n \frac{2n-1}{2n}$$
Where does the last factor of 1/10 come from? This is not obvious to me from the above expression and I am wondering whether someone could help me see what's going on.
Lecture notes:
$$\arcsin(x) = x + \sum \limits_{n=1}^{\infty} \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1) } { 2 \cdot 4 \cdot 6 \cdots (2n) } \cdot \frac{ x^{2n+1} }{ 2n+1 } $$
Asuming we have the Taylor expansion for arcsin(x),
$$\arcsin(\frac{1}{10}) = \frac{1}{10} + \frac{1}{2} \cdot \frac{1}{3} \cdot (\frac{1}{10})^3 + \frac{3}{8}\cdot \frac{1}{5} \cdot (\frac{1}{10})^5 + \frac{5}{16}\cdot\frac{1}{7}\cdot(\frac{1}{10})^7 + \cdots + E_N$$
The Taylor error bound is
$$E_N < \frac{C}{(N+1)!} (\frac{1}{10})^{N+1} $$
$$\left( \frac{d}{dx} \right)^{N+1} \arcsin(x) < C$$
$$0 \le x \le \frac{1}{10}$$
$C$ is an upper bound for the $N+1$st derivative of arcsin(x) for all $x$ between 0 and 1/10.
Even though we don't have a good bound for the $N+1$st derivative, if we look at the terms in this series, we can make a good guess.
Each step we gain 1/100 fold increase in accuracy because of $\left( \frac{1}{10} \right)^n$
If we look at the coefficients, the $2n+1$ and the product of odds over the product of evens, then we're picking up another factor of 10 in the denominator. We claim that $a_{n+2}$, the next term in the series, is less than the previous term, $a_n$, divided by 1000.
We're picking up three decimal places of accuracy with each subsequent term. That means that if we want to get within $10^{-10}$, it's going to suffice choose $N$ bigger than or equal to 7. The first four terms suffice to approximate $\arcsin(\frac{1}{10})$ within $10^{-10}$
$$N \ge 7$$
The assertion in the notes is, roughly speaking, false.
Let $\arcsin x=\sum a_{2n+1} x^{2n+1}$. Then $$\lim_{n\to\infty}\frac{a_{2n+3}}{a_{2n+1}}=1.$$ More informally, in the long run we are picking up nothing "in the denominator." In the case $x=\frac{1}{10}$, for large $n$, almost all the gain in accuracy comes from the extra $\frac{1}{10^2}$.