Taylor Series of arctang (x) = tang^(-1) (x)
Calculus
Level
3
1 - (1/3) + (1/5) - (1/7) + (1/9) - (1/11) + ... = G ;Aproximate G up to 3 decimal places
The answer is 0.785.
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1 solution
1/(1 + x^2) = 1 - x^2 + x^4 - x^6 + x^8 -... if absolute value of x is less than 1. Let's integrate.. Then arctang(x) = tang^(-1)(x) = C + x - x^3/3 + x^5/5 - x^7/7 +...(C constant) if absolute value of x is less than 1; arctang(0) = 0 implies that C=0. Now, applying Abel Theorem and Dirichlet Criterium we can do lim x->1 arctg x = arctang (1) = Pi/4 = 3,14159../4 = 0,785... = 1 - 1/3 + 1/5 - 1/7 + ....