How to calculate $e^x$ with a standard calculator

Is there a simple method for calculating the $e^x$ ($x\in\mathbb{R}$) with a basic add/subtract/multiply/divide calculator that converges in reasonable time, preferably without having to memorize coefficients as in the case of the power series?

I've found one for n^th roots with Newton iteration that's pretty much dead simple and I'd really like to know what else can be done.

EDIT: I'm not really that excited about approximation unless the end result is completely accurate. There are some good ideas here, but this answer is the simplest option for a full-accuracy result using only a +/-/×/÷ calculator.

I also have a strong preference for algorithms which can be directly evaluated step-by-step on such a calculator. Otherwise pencil and paper or an excellent memory must be exercised.

Solution 1:

This is what I would do to calculate $e^x$ with perfect 8-digit accuracy.

1. Take $\lfloor x\rfloor$ and $b = x - \lfloor x\rfloor$, the whole and fractional parts respectively. ($e^x = e^{\lfloor x \rfloor + (x - \lfloor x\rfloor)} = e^{\lfloor x \rfloor} e^{x - \lfloor x \rfloor}$)

2. Use the $(((b/4 + 1)b/3 + 1)b/2 + 1)b + 1$ pattern to calculate $e^{x - \lfloor x\rfloor}$. (If the fractional part of the exponent is .4 or less, only terms up to $b/7$ are needed for 8-digit accuracy. Worst-case scenario ($b \rightarrow 1$) terms up to $b/10$ are needed.) This method is easy to implement on a simple calculator, especially ones with a memory slot to quickly reinsert $b$.

3. When that is finished, multiply by $e$ ($\approx 2.71828183$ by memorization) and press the equals button $\lfloor x\rfloor$ times to repeat multiply. The result is $e^x$.

I've analyzed the number of terms required for full accuracy. Here is the chart (click to view full image):

Basically, if terms up to $b/t$ are needed to fully calculate $e^x$ up to 8-digit precision (7 digits after the decimal point), accounting for rounding (the $\frac{1}{2}$ term), the relation between $x$ and $t$ is given by $\sqrt[t]{\frac{1}{2}10^{-7}t!} = x$.

Solution 2:

To echo Robert's and lhf's comments, Padé works nicely for approximation purposes. The $[2,2]$ approximant is easily remembered, in particular:

$$\exp(z)\approx\frac{(z+3)^2+3}{(z-3)^2+3}$$

but of course it is trivial to generate higher-order approximants (yet the coefficients aren't as easy to remember).

Here then is how to make this approximation slightly more practical: this hinges on the usual identity

$$\exp(2z)=\exp(z)^2$$

The "repeated squaring" idea is as follows, for the case of $z > 0$:

1. Keep halving the argument $z$ until $|z| < 0.1$. Note how many times you halved the argument. Call it $n$.

2. Evaluate the $[2,2]$ approximant at this reduced argument.

3. Square the result of the previous step $n$ times.

The result of this should be good to eight or so digits. For negative argument, negate the argument first, apply the repeated squaring method, and reciprocate the result afterwards. ($\exp(-z)=\frac1{\exp(z)}$)

Solution 3:

If $x$ is not too big, you could try approximating $e^x$ as $(1 + x/n)^n$ for some large $n$ (not too huge, or you run into roundoff error). Somewhat better is $(1 + x/n + x^2/(2 n^2))^n$. For example, with $x = 1$ and $n = 1000$, $(1 + 1/1000 + 1/(2\cdot 1000^2))^{1000} \approx 2.718281376$; the correct value is $e = 2.717281828$.

[EDIT: ] If you don't want powers, you might try the continued fraction $$ e^x = 1 + \cfrac{x}{1 - \cfrac{x}{2 + \cfrac{x}{3 - \cfrac{x}{2 + \cfrac{x}{5 - \cfrac{x}{2 + \cdots}}}}}}$$

Solution 4:

You may want to try a Padé approximant. See also https://mathoverflow.net/questions/41226/pade-approximant-to-exponential-function/41228#41228.