Can Euler's formula accurately calculate arcsine?

[Marie Cury]

1. I use Euler formula of arctan to calculate arcsine
2. This equation\arctan x = \frac{x}{1+x^2} \sum_{n=0}^\infty \prod_{k=1}^n \frac{2k x^2}{(2k+1)(1+x^2)}.
If I input 0.9999999999, I will not be able to get the expected result. If input = 0.9 then it is pretty correct, but 0.99 is definitely wrong and more wrong when the decimal digits get higher till uncomputable.
Could someone help me ?

 

[Marie Cury]

2 weeks already and everyone stopped breathing into my thread!
you ignore my post I guess!

 

[gabbagabbahey]

It sounds like you just need a better calculator, try using Mathematica.

 

[Avodyne]

It wasn't clear what you were asking. It's an infinite series, so you must be truncating it at some point, and so of course the result is not exact.

 

[kamerling]

1. I use Euler formula of arctan to calculate arcsine



2. This equation\arctan x = \frac{x}{1+x^2} \sum_{n=0}^\infty \prod_{k=1}^n \frac{2k x^2}{(2k+1)(1+x^2)}.

If I input 0.9999999999, I will not be able to get the expected result. If input = 0.9 then it is pretty correct, but 0.99 is definitely wrong and more wrong when the decimal digits get higher till uncomputable.
Could someone help me ?

I don't see what the problem is, I get at least 4 digits correct If I evaluate the sum for n<=13 and 10 digits correct for n<=32. (I don't think n=0 should count).
Because \frac{2k x^2}{(2k+1)(1+x^2)} is smaller than 1/2 for all x, the error must halve for each increase of n.

I used the following python program

rom math import *

x = 0.9999
limit = 20

sum = 0
for n in range (1, limit):
    prod = 1
    for k in range (1, n):
        prod *= 2*k*x*x/((2*k+1)*(1+x*x))
    sum += prod
    print n, atan(x), sum * x / (1+x*x)

1 0.785348160897 0.4999999975
2 0.785348160897 0.666649995833
3 0.785348160897 0.733303328833
4 0.785348160897 0.761866186262
5 0.785348160897 0.77455952004
6 0.785348160897 0.780328640213
7 0.785348160897 0.782991044782
8 0.785348160897 0.784233375995
9 0.785348160897 0.784817943983
10 0.785348160897 0.785094816918
11 0.785348160897 0.785226647987
12 0.785348160897 0.785289691324
13 0.785348160897 0.785319949099
14 0.785348160897 0.7853345162
15 0.785348160897 0.785341547891
16 0.785348160897 0.785344949982
17 0.785348160897 0.785346599315
18 0.785348160897 0.78534740034
19 0.785348160897 0.785347789989
20 0.785348160897 0.785347979799

Converges nicely as you can see.