Finding the exponential of a very small number

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Hi
I wish to find the exact exponential of numbers in the range of 10^-23. Matlab gives me the answer as 1 for exp(3.528393650226818e-23). This is not right. How do I accomplish this?

Accepted Answer

John D'Errico
John D'Errico on 11 Oct 2017
Edited: John D'Errico on 11 Oct 2017
Multiple ways to solve this problem. Sorry, but quad precision is not an option. There is no way to achieve quad precision, at least without recourse to some other tool, or exporting the results to a language where quad precision is available.
Brute force, without resource to the symbolic toolbox ... Use my HPF toolbox, which is free to download. I'll use 100 decimal digits of precision, even though
DefaultNumberOfDigits 100
s = hpf('-6.471606349773182e-23');
d = hpf('-6.604939683106516e-23');
See that HPF has no problem in evaluating expressions like those that you have.
exp(s)
ans =
0.9999999999999999999999352839365022681800000020940844373212284440930971466223949024568510358194049670
log(1+exp(s))
ans =
0.6931471805599453094171997634264277021655001348837763634509871205166792020337447156058633269050605702
So A is just
A = hpf('1.449056603773585e+19')*(log(1+exp(s))-log(1+exp(d)))
A =
0.000009660377358490614968553143308576941771731705532455911053800738735222612516743226005743174365407550000
As a check with the symbolic toolbox, it yields:
s = sym('-6.471606349773182e-23');
d = sym('-6.604939683106516e-23');
A = sym('1.449056603773585e+19')*(log(1+exp(s))-log(1+exp(d)))
A =
0.0000096603773584906138365242807537049
In fact, the symbolic toolbox loses some precision here, because it used limited precision to do some of those computations. And, since you only provided those numbers to about 20 significant digits, what can you expect? But it does reasonably well. The exact result, reporting it as 32 significant digits is:
0.0000096603773584906149685531433085769
Can we get that result by using only numerical methods in double precision? Brute force will of course fail, as you found.
exp(-6.471606349773182e-23)
ans =
1
A = 1.449056603773585e+19*(log(1+exp(s))-log(1+exp(d)))
A =
0
So in order to get a viable result for A using double precision, one needs to resort to numerical analysis. We need to understand that even if we could compute this for very tiny numbers s and d, the results log(1+exp(s)) and log(1+exp(d)) are very close to each other. Using HPF, the exact results out to 40 decimal digits are:
log(1+exp(s))
ans =
0.6931471805599453094171997634264277021655
log(1+exp(d))
ans =
0.6931471805599453094171990967597610354955
So, when we subtract the two results, we will see massive subtractive cancellation. HPF is able to give the exact results here because I did the computations in sufficiently high precision, but then only reported the first 40 digits.
Going back to double precision, even tools in MATLAB like log1p must fail, because exp(s) returns exactly 1. As far as MATLAB is concerned, with doubles for s and d, these two results are identical.
log(1+exp(s)) == log(1+exp(d))
ans =
logical
1
Ok, so can we do something? Yes. With some mental effort. Note that is s is TINY, then exp(s) is essentially 1. And then we lose accuracy because of the massive subtractive cancellation, because both logs are returning essentially log(2).
What if we computed the result of
log(1+exp(x)) - log(2)
Of course, we need to do this computation accurately, for very tiny x. First see that
log(1+exp(x)) - log(2) = log((1+exp(x))/2)
When we do the subtraction, those -log(2) terms will exactly cancel each other out. But they enable us to do the computation in double precision. We will need to use a Taylor series. I'll use the symbolic toolbox here, because I'm too lazy to do the algebra on paper.
syms x
taylor(log((1+exp(x))/2))
ans =
-x^4/192 + x^2/8 + x/2
For tiny values of x of magnitude smaller than eps, we can truncate that series to one term. But we can see what happens if we use 2 terms anyway.
logapprox1 = @(x) x/2;
logapprox2 = @(x) x/2 + x.^2/8;
s = -6.471606349773182e-23;
d = -6.604939683106516e-23;
format long g
A = 1.449056603773585e+19*(logapprox1(s)-logapprox1(d))
A =
9.66037735849058e-06
A = 1.449056603773585e+19*(logapprox2(s)-logapprox2(d))
A =
9.66037735849058e-06
As you can see, the two results are both reasonably accurate. Remember, the exact result (using scientific notation) was:
9.6603773584906149685531433085769e-6
So our double precision computation was correct out to almost the full precision available using doubles.
The trick was to use finesse. Brute force only works if you are willing and able to do the computation using the proper tools that will support brute force. As I showed, even the symbolic toolbox missed a few digits out there at the end.
  1 Comment
Walter Roberson
Walter Roberson on 11 Oct 2017
sym('-6.471606349773182e-23') turns out to represent the values as some kind of internal binary fraction that is not as simple as moving the decimal place and using an appropriate power of 10 as denominator. I did some analysis of the representation error in a post a few months ago, but I would have to dig it out.

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More Answers (3)

Christoph F.
Christoph F. on 10 Oct 2017
Edited: Christoph F. on 10 Oct 2017
> This is not right.
Is is as close to the correct answer as can be represented with the 64-bit floating point format that Matlab uses.
exp(3.528393650226818e-23) is approximately 1 + 3.528393650226818e-23 (Taylor series, terminated after the first term), which will also give the result "1".
If 64-bit floating point is not enough precision for your purpose, you should look for something that supports 80-bit or quadruple precision (128-bit) floating point data types. Or redefine your task (i.e. how many digits are required?), since a computer will never give you an exact value for a number that has no finite representation.
  3 Comments
Himanshi Rani
Himanshi Rani on 10 Oct 2017
I have to actually find the value of A to enough precision:
A= (1.449056603773585e+19)*(log(1+exp(s))-log(1+exp(d)))
where s= -6.471606349773182e-23 and d= -6.604939683106516e-23
By default i.e without using the symbolic toolbox or the method described by Jan Simon, I get A as 0

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Walter Roberson
Walter Roberson on 10 Oct 2017
Edited: Walter Roberson on 11 Oct 2017
taylor log(1+exp(x)) to order 5 (so you get the ^4 terms). evaluate at symbol x1 and at symbol x2 and subtract the two -- which will cancel out a log(2) . Multiply by 1.449056603773585e+19 . factor to get
-75471698113207552 * ( x1 - x2) * ( x1^3 + x1^2*x2 + x1*x2^2 - 24*x1 + x2^3 - 24*x2 - 96)
You could work with that, or you could note that in your range 24*x1 or 24*x2 are going to be noise compared to 96, so taking into account floating point accuracy you could reduce this to
-75471698113207552 * ( x1 - x2) * ( - 96)
Multiply out the two constants...
7245283018867924992 * (x1 - x2)
For your s and d, this formula is off in at most the 16th decimal place.
The 5th order taylor series differs from a high precision answer starting at 2^(-70)
  3 Comments
Walter Roberson
Walter Roberson on 11 Oct 2017
No, I was right. You have
A= (1.449056603773585e+19)*(log(1+exp(s))-log(1+exp(d)))
which is a constant times log(1+exp(s)) - log(1+exp(d))
As s approaches 0, exp(s) approaches 1, so you approach
log(1+1) - log(1+exp(d))
which gets you the log(2) . The same argument for exp(d) with small d, so you headed towards something that very closely approximates log(2) minus something else that very closely approximates log 2. When you taylor the general form log(1+exp(x)) near 0 to get a bit more accuracy, one of the elements in the sum is going to be that expression evaluated at exactly 0, giving log(2); that part of the sum is independent of s or d, so when you subtract the two taylor forms, the log(2) cancel out, leaving only the terms that follow.
taylor(log(1+exp(x)),x,'Order',3)
is x^2/8 + x/2 + log(2) . Do not use direct precision arguments with respect to the log(2) because we are going to cancel those out as described above, so we are left considering x^2/8 + x/2 as the effective value. When x is in the order of 1e-23, then we know that x^2/8 will be on the order of 1e-47 which is negligible compared to 1e-23/2 . We can therefore see that an order 2 approximation is enough: near 1e-23, log(1+exp(x)) is approximately x/2 + log(2) . Substitute in s and d, subtract, and we see that log(1+exp(s)) - log(1+exp(d)) is going to be approximately s/2 - d/2 . All of which is to be multiplied by (1.449056603773585e+19)

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Jan
Jan on 10 Oct 2017
The more exact numerical output is - as expected:
1.000000000000000000000035283936502268180000000622478087548...
If you do not want to use VPA and symbolical solutions, you can apply some mathematics also. Look at the Taylor expansion:
exp(x) = 1 + x + x^2/2 + x^3/6 + ...
For x very near to 0.0 this is 1 + x. As Cristoph has explained already 1 + 5.52e-23 cannot be represented "accurately" with the 64 bit IEEE 754 doubles. But perhaps you can proceed with your calculations using the 2 variables:
c1 = 1
c2 = x
and consider, that you need the sum of both.
  2 Comments
Jan
Jan on 10 Oct 2017
Do you have the Symbolic Toolbox? Unfortunately I do not have it. For a suggestion it would be useful if you explain exactly, what you want to achieve.

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