I don't need to find the exact values, rather an approximation for error in Taylor series expansion.
I have seen the following lines.
|En(x)| < (1/700!)*e^100*100^701
|En(x)| < 100^-200
To predict the error, its not necessary to find the very exact value. It is only require to make a good approximation. Can anyone please explain how these prediction can be made in a simple way?
On
Let me assume that you want the approximate value of $$y=\frac{e^a a^{b+1}}{b!}$$ where $b$ is large and $a<b$. Take logarithms $$\log(y)=a+(b+1)\log(a)-\log(b!)$$ Now, use Stirling approximation for large values of $b$; this gives $$\log(b!)=b (\log (b)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left(b\right)\right)+O\left(\frac{1}{b}\right)$$ Then $$\log(y)\approx a+b+(1+b)\log(a)-(b+\frac 12)\log(b)-\frac 12\log(2\pi)$$
Using $a=100$ and $b=700$, this would give $$\log(y)\approx -561.726\implies \log_{10}(y)\approx -243.955\implies y\approx 1.10999 \times 10^{-244}$$ while the exact value would be $y\approx 1.10986 \times 10^{-244}$.
As you can see, it is not bad at all.
If we reduce the problem to $$y=\frac{e^{100}\, 100^{100 k+1}}{(100 k)!}$$ using the same method, we should arrive to $$\log(y)\approx 100 k-100 k \log (k)+\frac{1}{2} \log \left(\frac{50}{\pi k}\right)+100 \approx -100 (k \log (k)-k-1)$$
By the inequality, Factorial Inequality problem $\left(\frac n2\right)^n > n! > \left(\frac n3\right)^n$ , we have that $$\frac{e^{100}\cdot 100^{701}}{700!}<\frac{3^{700}e^{100}\cdot 100^{701}}{700^{700}}<(3^8/7^7)^{100}\cdot 10<\frac{10}{125^{100}}<10^{-200}. $$ where we used the inequality $e<3$.
Note that the exact value is approximately equal to $10^{-244}$, so the second expression $|E_n(x)| < 100^{-200}$ seems to be wrong.