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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?

Mikasa
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Md Didarul Islam
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  • Could you add parentheses in the first expression ? Is it $\frac{e^{100} 100^{701}}{700!}$ or something else ? – Claude Leibovici Sep 24 '16 at 06:11
  • yes. that is what it is. – Md Didarul Islam Sep 24 '16 at 06:24
  • There is a formula for the error in a Taylor series expansion. That formula has pieces in it that depend on the function being expanded and its derivatives. To make good approximations, you have to know something about those derivatives. So you can't make general rules about how to make these approximations – they depend opn the function. If you have a particular expansion in mind, maybe we could show you where the estimates for that particular function come from. – Gerry Myerson Sep 24 '16 at 06:39
  • Actually i am not worried about the taylor expansion or error function, I know the expansion formula or error formula. I am only bothered to make the calculation without calculator. The two lines I have written in the question. How can I go from the first line to the second line. I know those two values are close. The professor wrote directly from 1st to 2nd. I am just curious, how he did it. It looked somewhat simple. But i dont know the trick! – Md Didarul Islam Sep 24 '16 at 06:43
  • @Md Didarul Islam The second estimate is not implied by the first one. Take a look to my answer. – Robert Z Sep 24 '16 at 07:02

2 Answers2

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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 $$\log(y)\approx 100 k-100 k \log (k)+\frac{1}{2} \log \left(\frac{50}{\pi k}\right)+100 \approx -100 (k \log (k)-k-1)$$

Claude Leibovici
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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.

Community
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Robert Z
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