How do you find the taylor polynomial of degree 4 of the function #f(x)= arctan(x^1)# at a=0?

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Answer 1 · 2017-03-09T01:35:22

#arctanx ~= x- x^3/3+x^5/5-x^7/7+x^9/9#

Start from the integral:

#arctanx = int_0^x (dt)/(1+t^2)#

Now note that the integrand function is in the form of the sum of a geometric series of ratio #-t^2#:

#1/(1+t^2) = sum_(n=0)^oo (-t^2)^n = sum_(n=0)^oo (-1)^nt^(2n)#

which has radius of convergence #R=1#. Thus in the interval #(-1,1)# we can integrate term by term and have:

#arctanx = sum_(n=0)^oo int_0^x(-1)^nt^(2n)dt = sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1)#

For #n=4# we have:

#arctanx = x- x^3/3+x^5/5-x^7/7+x^9/9+R_4(x)#