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Is there a method to mentally evaluate $e^{-x}$ for $x>0$? Just to have an idea when computing probabilities or anything that is an exponential function of some parameters.

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You could compute its Taylor series. The Taylor series for $e^{-x}$ will always converge to $e^{-x}$, although for $x$ not close to zero, it might not be practical to mentally compute a higher order Taylor approximation. –  ShawnD Oct 10 '11 at 5:50
    
What I do is pretend $e=2$. Maybe there is a more precise way which is almost as easy? –  Dan Brumleve Oct 10 '11 at 5:51
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Another trick is $e^{-x} = 10^{-x/\ln(10)} \simeq 10^{-0.4x}.$ –  Gerben Oct 10 '11 at 6:21
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In the book "Surely You Are Joking, Mr Feynman!" Richard Feynman described how he computed $e^x$ mentally to several significant digits by using some simple approximations. I don't have the book at hand to give more details about his method, but it might be worth looking at. –  Dilip Sarwate Oct 10 '11 at 11:42

5 Answers 5

up vote 16 down vote accepted
+50

If $x$ is "small enough", I like using the $(2,2)$ Padé approximant, which is easily cast into a memorable form:

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

For $|x| < \frac12$, the absolute difference between the approximant and the true function is $< 8\times 10^{-5}$. Pretty good approximation for a mere rational function... of course, as with all such approximants, the further you go from $0$, the less accurate it becomes.

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+1: I just became a fortune-teller. I foresee a bunch of freshman calculus students doing battle with this next Spring. The Taylor series agree up to $x^4$. With quintic terms it is $1$ vs. $5/6$. –  Jyrki Lahtonen Oct 10 '11 at 9:48
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@Jyrki: Just in case you've forgotten, Padé approximants are rational functions precisely designed to have power series expansions whose first few terms match the first few terms of the power series of the function they're approximating. ;) Since this is a $(2,2)$ approximant (the two numbers denote respectively the degrees of the numerator and denominator), it is expected to agree up to the $2+2=4$-th order term. Neat, eh? –  J. M. Oct 10 '11 at 9:59
    
Thanks for the reminder. I've seen that described in the context of Berlekamp-Massey algorithm (see e.g. the 3rd reference here. –  Jyrki Lahtonen Oct 10 '11 at 15:04
    
@Jyrki: Tightly connected indeed. Toeplitz matrices are very tricky that way... (I'm still figuring out the intricate web myself.) –  J. M. Oct 10 '11 at 15:06
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Yes, I used it as a question testing their ability to manipulate power series. Thanks! –  Jyrki Lahtonen Aug 15 '12 at 6:45

Dilip was asking in the comments about "Feynman's method"; since this is too long for a comment, I shall be quoting the relevant paragraphs here:

Lucky Numbers

One day at Princeton I was sitting in the lounge and overheard some mathematicians talking about the series for $e^x$, which is $1+x+x^2/2!+x^3/3!$ Each term you get by multiplying the preceding term by $x$ and dividing by the next number. For example, to get the next term after $x^4/4!$ you multiply that term by $x$ and divide by $5$. It's very simple.

When I was a kid I was excited by series, and had played with this thing. I had computed $e$ using that series, and had seen how quickly the new terms became very small.

I mumbled something about how it was easy to calculate $e$ to any power using that series (you just substitute that power for $x$).

"Oh yeah?" they said. "Well, then what's $e$ to the $3.3$?" said some joker—I think it was Tukey.

I say, "That's easy. It's $27.11$."

Tukey knows it isn't so easy to compute all that in your head. "Hey! How'd you do that?"

Another guy says, "You know Feynman, he's just faking it. It's not really right."

They go to get a table, and while they're doing that, I put on a few more figures. "$27.1126$," I say.

They find it in the table. "It's right! But how'd you do it!"

"I just summed the series."

"Nobody can sum the series that fast. You must just happen to know that one. How about $e$ to the $3$?"

"Look," I say. "It's hard work! Only one a day!"

"Hah! It's a fake!" they say, happily.

"All right," I say, "It's $20.085$."

They look in the book as I put a few more figures on. They're all excited now, because I got another one right.

Here are these great mathematicians of the day, puzzled at how I can compute $e$ to any power! One of them says, "He just can't be substituting and summing—it's too hard. There's some trick. You couldn't do just any old number like $e$ to the $1.4$."

I say, "It's hard work, but for you, OK. It's $4.05$."

As they're looking it up, I put on a few more digits and say, "And that's the last one for the day!" and walk out.

What happened was this: I happened to know three numbers—the logarithm of $10$ to the base $e$ (needed to convert numbers from base $10$ to base $e$), which is $2.3026$ (so I knew that $e$ to the $2.3$ is very close to $10$), and because of radioactivity (mean-life and half-life), I knew the $\log$ of $2$ to the base $e$, which is $.69315$ (so I also knew that $e$ to the $.7$ is nearly equal to $2$). I also knew $e$ (to the $1$), which is $2.71828$.

The first number they gave me was $e$ to the $3.3$, which is $e$ to the $2.3$—ten—times $e$, or $27.18$. While they were sweating about how I was doing it, I was correcting for the extra $.0026$—$2.3026$ is a little high.

I knew I couldn't do another one; that was sheer luck. But then the guy said $e$ to the $3$: that's $e$ to the $2.3$ times $e$ to the $.7$, or ten times two. So I knew that it was $20.$something, and while they were worrying about how I did it, I adjusted for the $.693$.

Now I was sure I couldn't do another one, because the last one was again by sheer luck. But the guy said $e$ to the $1.4$, which is $e$ to the $.7$ times itself. So all I had to do is fix up $4$ a little bit!

They never did figure out how I did it.

...

There's less to it than meets the eye. ;)

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I just typed this up since I don't have a digital copy of any of Feynman's books. Please correct anything that I may have transcribed wrongly... –  J. M. Oct 12 '11 at 18:05
    
how exactly do you think he corrected for the "small errors"? I can see how he got $e^{3.3} = e^{2.3} \times e \approx 27.18$, but I don't see how he was able to say $27.1126$. Similarly for other approximations. –  picakhu Aug 15 '12 at 20:24

$e^{-x} \approx 2^{-1.44x} \approx 10^{-0.43x}$

where $\log_2(e) \approx 1.44$ and $\log_{10}(e) \approx 0.43$.

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doesn't this nail down to how to compute $10^{-x}$? –  ACAC Oct 13 '11 at 20:28
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@Bob: in general that's hard, but if you just need an order-of-magnitude estimate, it's trivial. For example: $10^{13.2315} = 10^{13} \times 10^{0.2315}.$ The latter factor is hard to calculate, but for a numerical estimate it's often not important. You can of course refine such estimates by going back to base $e$ or 2: since $10 \simeq 2^{3.3},$ $10^{0.2315} \simeq 2^{0.7} \simeq 1.7$. –  Gerben Oct 14 '11 at 10:32

As a supplement to Henry's answer, consider these rational approximations to $\log_{10}(e)$ and $\log_{2}(e)$.

Two of the approximants for the continued fraction for $\log_{10}(e)$ are $\frac{3}{7}$ (low and not as good as $.43$) and $\frac{10}{23}$ (high but better than $.43$). So if it makes computation easier, you can try $$ 10^{-10x/23}<e^{-x}<10^{-3x/7}\tag{1} $$ when $x>0$. The order in $(1)$ is reversed for $x<0$.

Two of the approximants for the continued fraction for $\log_{2}(e)$ are $\frac{10}{7}$ (low and not as good as $1.44$) and $\frac{13}{9}$ (high but better than $1.44$). So if it makes computation easier, you can try $$ 2^{-13x/9}<e^{-x}<2^{-10x/7}\tag{2} $$ when $x>0$. The order in $(2)$ is reversed for $x<0$.

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A diagram may also help...enter image description here

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I'll make sure to print this out and carry it with me. –  The Chaz 2.0 Oct 12 '11 at 18:21
    
@pic: actually, it's an iPhone. Please keep me out of such comments. –  The Chaz 2.0 Aug 16 '12 at 0:42

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