# Maths - Infinite series

If we take an infinite series, for instance each term may have a higher polynomial term. This series may possibly converge or diverge, if the series converges then it can be useful for calculating numerical values of functions.

Here are some series that may be useful:

 sin(x) x - x3/3! + x5/5! ... +(-1)rx2r+1/(2r+1)! all values of x cos(x) 1 - x2/2! + x4/4! ... +(-1)rx2r/(2r)! all values of x ln(1+x) x - x2/2! + x3/3! ... +(-1)r+1xr/(r)! -1 < x <= 1 exp(x) 1 + x1/1! + x2/2! + x3/3! ... + xr/(r)! all values of x exp(-x) 1 - x1/1! + x2/2! - x3/3! ... all values of x e 1 + 1/1! + 2/2! + 3/3! =2.718281828 sinh(x) x + x3/3! + x5/5! ... +x2r+1/(2r+1)! all values of x cosh(x) 1 + x2/2! + x4/4! ... +x2r/(2r)! all values of x f(x) f(0) + x f'(0) + x2/2! f''(0)+ x3/3! f'''(0) ... Maclaurins series f(x + h) f(x) + h f'(x) + h2/2! f''(x)+ h3/3! f'''(x) ... Taylors series

## Example of use of Taylors series

We want to approximate x-(1/2) close to 1 as required for reorthogonalising a matrix as required here.

 f = x-(1/2) f(1) = 1 f' = -(1/2) * x-(3/2) f'(1) = -1/2 f'' =(3/4) * x-(5/2) f''(1) = 3/4

so using taylors series gives:

f(x + h) = f(x) + h f'(x) + h2/2! f''(x)+ h3/3! f'''(x) ...

since we want to find the value at 1 let h = x - 1 so that we can evaluate near h=0

f(1 + h) = f(1) + h f'(1) + h2/2! f''(1)+ h3/3! f'''(1) ...

f(1 + h) = 1 - h /2 + h2 3/8 ...

substituting h = x - 1 gives:

f(1 + h) = f(1 + x - 1) = f(x) = 1 - (x - 1) /2 + (x - 1)2 3/8 ...

= (3 - x)/2 ...

See message from Magnus