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

[Harit Himanshu]

Thanks Guido, That is more useful.

On Tue, Mar 3, 2015 at 3:32 PM, Guido Rumi <guido.rumi@REDACTED> wrote:

> I would change
>
> fact(N) ->
>   case N of
>     0 -> 1;
>     1 -> 1;
>     _ -> N * fact(N - 1)
>   end.
>
> to
>
> fact(N) ->
>   fact_tr(N, 1).
>
> fact_tr(N, Acc) ->
>   case N of
>     0 -> Acc;
>     1 -> Acc;
>     _ -> fact_tr   (N - 1, N * Acc)
>   end.
>
> To make it more efficient
> http://stackoverflow.com/questions/33923/what-is-tail-recursion
>
> On Tue, Mar 3, 2015 at 6:32 PM, Harit Himanshu <
> harit.subscriptions@REDACTED> wrote:
>
>> 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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>>
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