[erlang-questions] How to calculate series expansion e^x

[Harit Himanshu]

Richard and Alex

Thanks a lot for your help. I got help up with my day job and could not try
it out. Thanks a lot for your help on this problem.
I tried a different version and it looks like following. Let me know if you
see any issues with that

-module(solution).
-export([main/0]).

main() ->
  {ok, [N]} = io:fread("", "~d"),
  Data = read_input([], N),
  [io:format("~p~n", [e_x(X)]) || X <- Data].

read_input(Input, N) ->
  case N of
    0 -> lists:reverse(Input);
    _ -> {ok, [Num]} = io:fread("", "~s"),
      {NumFloat, _} = string:to_float(Num),
      read_input([NumFloat | Input], N - 1)
  end.

e_x(X) ->
  1 + lists:sum([math:pow(X, N) / fact(N) || N <- lists:seq(1, 9)]).

fact(N) ->
  case N of
    0 -> 1;
    1 -> 1;
    _ -> N * fact(N - 1)
  end.

On Wed, Feb 25, 2015 at 4:38 PM, Alex Alvarez <eajam@REDACTED> wrote:

>  Or just simply...
>
>
> -module(e_fun).
>
> -export ([start/2]).
>
> start (N, X) ->
>   1 + step (N - 1, 1, X, 1.0, 0).
>
> step (0, _, _, _, S) -> S;
> step (N, NN, X, Z, S) ->
>     ZZ = Z * X/NN,
>   step (N - 1, NN + 1, X, ZZ, S + ZZ).
>
>
> Here I'm just reusing the each previous term (ZZ) of the Taylor series to
> compute the next term and then add it up to the sum (S).
>
> Cheers,
> Alex
>
>
>
> On 02/24/2015 10:27 PM, Richard A. O'Keefe wrote:
>
> On 25/02/2015, at 11:00 am, Harit Himanshu <harit.subscriptions@REDACTED> <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.)
>
> exp(X)
>   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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