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How to compute $e^x$ ($2.71218...$) to its any power with any shortcut or a method. I want to know a method to calculate in big powers like $e^{50}$ not small powers, For eg-$0.02$ (using Taylor series or Feymenn method.) If you want to give any alternative method prescribed above for finding small powers, You could give.

Key Flex
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Piyush Choudhury
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    Maybe you could use $e^{x+y}=e^xe^y$ and a method for small powers. For example $e^{50} = \prod_{i=1}^{500} e^{0.1}$. – humanStampedist Dec 04 '18 at 15:12
  • $e^x =\sum_{n\geq 0}\frac{x^n}{n!}$, so the partial sums will give good approximations as long as you sum enough terms (you can figure how far you need to go using the Lagrange remainder formula) – Ivo Terek Dec 04 '18 at 15:13
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    What is the Feymenn method in this context? Can you provide a link? – callculus42 Dec 04 '18 at 15:16
  • $e^3 \simeq 20$... – the_candyman Dec 04 '18 at 15:18
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    Maybe calculate $x=e^{50/64}$ by the usual methods; then to calculate $x^2, x^4, x^8, x^{16}, x^{32}, x^{64}=e^{50}$ requires only one additional multiplication each. – MJD Dec 04 '18 at 15:20
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    @IvoTerek: That is a dreadful method in real life. You need so many terms that (1) it takes forever, and (2) rounding errors accumulate unacceptably. (And it's totally useless if $x$ is large and negative.) – TonyK Dec 04 '18 at 15:24
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    $50 = 2 + 2^4 + 2^5$ so $e^{50} = e^2 \cdot e^{2^4} \cdot e^{2^5}$ where $e^{2^k}$ are computed from $e$ by repeated squaring. – Robert Israel Dec 04 '18 at 15:40
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    Many comments without any reaction of the OP. – callculus42 Dec 04 '18 at 17:23

1 Answers1

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Since $$ \cdots \;{4 \over {10}} < {{21} \over {50}} < \log _{10} e = 0.43429 \cdots < {{22} \over {50}} < {4 \over 9} < {5 \over {10}}\; \cdots $$ and you can find many other better bounds, depending on the precision that you need.

Then for instance you can get $$ 10^{\,21} < e^{\,50} < 10^{\,22} $$ or better $$ e^{\,50} \approx 10^{\,21} \cdot 10^{\,{{14} \over {1000}}50} = 10^{\,21} \cdot 10^{\,{7 \over {10}}} = 10^{\,22} \cdot 10^{\, - {3 \over {10}}} \approx 10^{\,22} {1 \over {\root 3 \of {10} }} \approx {1 \over 2}10^{\,22} $$

G Cab
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