@jacob05

Impressive script, i just learned something.

The loop in your script started from 2 till 5001... can you explain why?

jacob05:
cc: Yemoll

I think your solution or your friend's isn't quite it...

Firstly, using a factorial finding function to solve this problem is wrong and naive. e.g I believe $yyy increases linearly with loop and its factorial gets calculated as the $ii increase to 5000.. try finding the factorial of 5000.. without //bcscale(25);

Lastly, bcscale(25) ; , makes the solution very likely to be inaccurate as the result of all bc functions are trimmed or scaled to 25.

I'm not saying mine is correct or is not ... ... You judge... I'm just making my point..

Hi @jacob05,

I designed the php script @slyrox uploaded for the problem... as @Lyphiard rightly mentioned that the problem @OP posted came from a competition in USA which we solved in our team before sharing it nairaland to develop other programmers on this platform.

The accepted solution/result on the competition portal was "5.08845978368001251890" which is the same as the result of our script/algorithm and different from the solution/result of your script/algorithm "5.088459783680012467179075225".

However, this is how we solved it or how we got the algorithm;

Looking at the equation posted;
x^2 + (5/3)x^3 + (23/12)x^4 + (119/60)x^5 ... + (2*[5000!-1]/[5000!])x^5000
it can easily be translated into a summation equation
which is to sum the result of ((2 * ([a+1]! – 1) / [a+1]! ) * x^(a+1) from a = 1 till 5000 where x is a constant
so for a = 1 we have x^2
a = 2 we have (5/3)x^3
sum the results till a = 5000

All we did was to translate the deduced summation equation into a script/algorithm. there was no reason behind using php, I could have used several other lang java, C#, VB, C, ruby, python

The result of the problem was also requested in the nearest 20 decimal places. why i added bcscale(25) is to control the number of trailing digits after the decimal and not affect the integrity of the 20th digit after the decimal. then i took off the trailing 5 digits after the 20th decimal

I will be willing to learn if you comeup with a much quicker or less-lame solution with regards to the posted problem. Thanks

BTW, My understanding of your script is that 5000 = 5000!; correct me if i am wrong.