Question

Does anyone know how to make the following Matlab code approximate the exponential function more accurately when dealing with large and negative real numbers?

For example when x = 1, the code works well, when x = -100, it returns an answer of 8.7364e+31 when it should be closer to 3.7201e-44.

The code is as follows:

s=1
a=1;
y=1;
for k=1:40
    a=a/k;
    y=y*x;
    s=s+a*y;
end
s

Any assistance is appreciated, cheers.

EDIT:

Ok so the question is as follows:

Which mathematical function does this code approximate? (I say the exponential function.) Does it work when x = 1? (Yes.) Unfortunately, using this when x = -100 produces the answer s = 8.7364e+31. Your colleague believes that there is a silly bug in the program, and asks for your assistance. Explain the behaviour carefully and give a simple fix which produces a better result. [You must suggest a modification to the above code, or it’s use. You must also check your simple fix works.]

So I somewhat understand that the problem surrounds large numbers when there is 16 (or more) orders of magnitude between terms, precision is lost, but the solution eludes me.

Thanks

EDIT:

So in the end I went with this:

s = 1;
x = -100;
a = 1;
y = 1;
x1 = 1;

for k=1:40
    x1 = x/10;
    a = a/k;
    y = y*x1;
    s = s + a*y;
end

s = s^10;
s

Not sure if it’s completely correct but it returns some good approximations.

exp(-100) = 3.720075976020836e-044
s = 3.722053303838800e-044

After further analysis (and unfortunately submitting the assignment), I realised increasing the number of iterations, and thus increasing terms, further improves efficiency. In fact the following was even more efficient:

s = 1;
x = -100;
a = 1;
y = 1;
x1 = 1;

for k=1:200
    x1 = x/200;
    a = a/k;
    y = y*x1;
    s = s + a*y;
end

s = s^200;
s

Which gives:

exp(-100) = 3.720075976020836e-044
s = 3.720075976020701e-044

Answer 1

As John points out in a comment, you have an error inside the loop. The y = y*k line does not do what you need. Look more carefully at the terms in the series for exp(x).

Anyway, I assume this is why you have been given this homework assignment, to learn that series like this don’t converge very well for large values. Instead, you should consider how to do range reduction.

For example, can you use the identity

exp(x+y) = exp(x)*exp(y)

to your advantage? Suppose you store the value of exp(1) = 2.7182818284590452353…

Now, if I were to ask you to compute the value of exp(1.3), how would you use the above information?

exp(1.3) = exp(1)*exp(0.3)

But we KNOW the value of exp(1) already. In fact, with a little thought, this will let you reduce the range for an exponential down to needing the series to converge rapidly only for abs(x) <= 0.5.

Edit: There is a second way one can do range reduction using a variation of the same identity.

exp(x) = exp(x/2)*exp(x/2) = exp(x/2)^2

Thus, suppose you wish to compute the exponential of large number, perhaps 12.8. Getting this to converge acceptably fast will take many terms in the simple series, and there will be a great deal of subtractive cancellation happening, so you won’t get good accuracy anyway. However, if we recognize that

12.8 = 2*6.4 = 2*2*3.2 = ... = 16*0.8

then IF you could efficiently compute the exponential of 0.8, then the desired value is easy to recover, perhaps by repeated squaring.

exp(12.8)
ans =
          362217.449611248

a = exp(0.8)
a =
          2.22554092849247
a = a*a;
a = a*a;
a = a*a;
a = a*a
          362217.449611249

exp(0.8)^16
ans =
          362217.449611249

Note that WHENEVER you do range reduction using methods like this, while you may incur numerical problems due to the additional computations necessary, you will usually come out way ahead due to the greatly enhanced convergence of your series.