[erlang-questions] How to calculate series expansion e^x
Richard A. O'Keefe
Wed Feb 25 04:27:58 CET 2015
On 25/02/2015, at 11:00 am, Harit Himanshu <harit.subscriptions@REDACTED> wrote:
> I am trying to learn Functional Programming and Erlang by practicing problems available online.
> One question where I don't know about solving it is
> The series expansion of ex is given by:
> 1 + x + x2/2! + x3/3! + x4/4! + .......
> Evaluate e^x for given values of x, by using the above expansion for the first 10 terms.
> The problem statement could be found here
> Can someone guide me how this could be achieved in Erlang?
The first 10 terms of exp(X) are
X X^2 X^3 X^9
1 + --- + --- + --- + ... + ---
1 2 6 9!
This is a polynomial. We can evaluate it cheaply using
Horner's Rule. Any time you have a polynomial to
evaluate, you should think about using Horner's Rule.
X / X / X / X \ \ \
1 + --- * | 1 + --- * | 1 + --- * ... | 1 + --- | ...| |
1 \ 2 \ 3 \ 9 / / /
We can see a building block here: 1 + (X/n)*Rest.
So let's make a function for the building block:
step(X, N, Rest)
when is_float(X), is_float(N), is_float(Rest) ->
1.0 + (X/N)*Rest.
There are 9 steps, which is small enough to do by hand:
(NINE steps to get TEN terms. I hope that's clear.)
when is_float(X) ->
step(X, 1.0, step(X, 2.0, step(X, 3.0,
step(X, 4.0, step(X, 5.0, step(X, 6.0,
step(X, 7.0, step(X, 8.0, step(X, 9.0, 1.0))))))))).
I should point out that this is a lousy way to compute
exp(X) for abs(X) "large". Waving hands vaguely, you
want X^10/10! "small". The relative error for X = 1.0
is -1.1e-7, for X = 2.0 is -4.6e-5, and for X = 4.0 is
a scary -0.0081.
One technique that's used in cases like this is
RANGE REDUCTION. e^x = 2^(1.442695... * x).
To begin, scale x by log of e to the base 2,
and separate that into an integer part N and a fraction
part F. e^x is then 2^(N+F) = 2^N*2^F. Since F is
"small", we can use something like this polynomial.
And then we can use the equivalent of C's ldexp() to
handle the 2^N* part. Or we could if Erlang _had_ an
equivalent of ldexp().
You don't really need the is_float/1 tests; I put them in
so that the compiler could generate better native code.
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