MAPLE output for calculating Pi using the Taylor Series for 6*arcsin(x) at x=1/2.
We know that the arcsin(1/2) = Pi/6.
> 6*arcsin(1/2); evalf(%);
So, we can get Pi by finding 6 times the Taylor Series for arcsin(x), evaluating the series at x=1/2.
How can we find the Taylor Series for arcsin(x)?
Well, the derivative of the arcsin(x) is:
> f:=diff(arcsin(x),x);
So, if we find the Taylor Series for f(x), we can integrate it to get the Taylor Series for arcsin(x).
Let's find out a few derivatives for f(x) and plug in a=0.
> fp:=diff(f,x); subs(x=0, fp);
> fpp:=diff(fp,x); subs(x=0, fpp);
> fppp:=diff(fpp,x); subs(x=0, fppp);
> fpppp:=diff(fppp,x); subs(x=0, fpppp);
> fppppp:=diff(fpppp,x); subs(x=0, fppppp);
> fpppppp:=diff(fppppp,x); subs(x=0, fpppppp);
So, the beginning of the Taylor Series for 1/sqrt(1-x^2) is:
> taylor(1/sqrt(1-x^2), x=0, 7);
We can integrate this to get the Taylor Series for arcsin(x):
> taylor(arcsin(x), x=0, 8);
Now that we have the Taylor Series for the arcsin(x), we can find 6 times it, and evaluate it at x=1/2 in order to get Pi.
The 7th order Taylor Series for 6*arcsin(x), and its value at x=1/2 are:
> mtaylor(6*arcsin(x), x=0, 8); evalf(subs(x=1/2, mtaylor(6*arcsin(x), x=0, 8)));
The 9th order Taylor Series for 6*arcsin(x), and its value at x=1/2 are:
> mtaylor(6*arcsin(x), x=0, 10); evalf(subs(x=1/2, mtaylor(6*arcsin(x), x=0, 10)));
The 11th order Taylor Series for 6*arcsin(x), and its value at x=1/2 are:
> mtaylor(6*arcsin(x), x=0, 12); evalf(subs(x=1/2, mtaylor(6*arcsin(x), x=0, 12)));
Here are some higher order evaluations:
> evalf(subs(x=1/2, mtaylor(6*arcsin(x), x=0, 20)));
> evalf(subs(x=1/2, mtaylor(6*arcsin(x), x=0, 24)));
Finally, using (only) a 25th order Taylor Series, we get Pi correct to eight places.
> evalf(subs(x=1/2, mtaylor(6*arcsin(x), x=0, 26))); evalf(Pi-%);