Approximating $e^{-x}$

Question

So we all know that the Taylor series for $e^x$ is $1 + x + \frac{x^2}{2} + \frac{x^3}{6}+\ldots$ Similarly for $e^{-x}$ it comes comes out to be $1 - x + \frac{x^2}{2} - \frac{x^3}{6}+\ldots$ Now for the problem I have at hand $x \gg1$. I am confused as how does the series converge for a very large $x$ because according to the Taylor series you should get a very large negative value for large $x$.

Also, we know that when $x$ tends to infinity, $e^{-x}$ should tend to zero. Is there any way to approximate $e^{-x}$ for large but finite values?

Thanks for your help

Answer 1

The Taylor series converges for all $x$. If $x=50$, say, the terms will grow until the one $\frac {50^{50}}{50!}\approx 2.9 \cdot 10^{20}$ The next term is $-\frac {50}{51}$ of this and they continue decreasing. They will cancel very delicately so the final result is $e^{-50}\approx 1.9\cdot 10^{-22}$

You are correct that the Taylor series is not a useful way to get the answer. One way is to use repeated squaring to get the value for integral exponents. Then if you wanted $\exp(-50.12345)$ you could use the Taylor series to get $\exp(0.12345)$ I would guess that computer libraries start with $2^x$ because they can handle the integer part of $x$ easily. Once you have one exponential, you can use the properties of logarithms to get the rest.

Answer 2

As Ross explains, this Taylor series converges for all $x$, but it's probably the case that your Taylor polynomial did not have enough terms to illustrate that yet.

As for approximating, $e^{-50}=(e^{-1})^{50}$. If you want to keep using Taylor series somehow, you can use it to approximate $e^{-1}$. Only using up to the tenth degree approximant, we have $e^{-1}\approx0.367879464\ldots$ and Taylor's Theorem proves that this approximation is within $\frac{1}{11!}$ of correct. Also it's an overestimate. In fact since the series is alternating, we can squeeze $e^{-1}$ between two adjacent approximants, like bere between the degree $10$th and degree $11$th.

$$0.367879464\ldots-\frac{1}{11!}<e^{-1}<0.367879464\ldots$$

Raising to the $50$th power

$$\left(0.367879464\ldots-\frac{1}{11!}\right)^{50}<e^{-50}<(0.367879464\ldots)^{50}$$

These outer expressions are relatively quickly computed by repeated squaring, writing $50=32+16+2$. You'd have to square five times because of the $32$.

$$1.92875\times10^{-22}<e^{-50}<1.92876\times10^{-22}$$