How do you use the Maclaurin series to write the first four non-zero terms of #(x^3)arctan(x^7)#?

Question

5051 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-26T00:08:26

#x^3arctan(x^7)=x^10-x^24/3+x^38/5-x^52/7+...+(-1)^nx^(14n+10)/(2n+1)+...#

Start with the basic Maclaurin series:

#1/(1-x)=sum_(n=0)^oox^n#

Replace #x# with #-x^2#:

#1/(1+x^2)=sum_(n=0)^oo(-x^2)^n=sum_(n=0)^oo(-1)^nx^(2n)#

Integrate this:

#intdx/(1+x^2)=sum_(n=0)^oo(-1)^nintx^(2n)dx#

#arctan(x)=sum_(n=0)^oo(-1)^nx^(2n+1)/(2n+1)#

Replace #x# with #x^7#:

#arctan(x^7)=sum_(n=0)^oo(-1)^n(x^7)^(2n+1)/(2n+1)=sum_(n=0)^oo(-1)^nx^(14n+7)/(2n+1)#

Multiply by #x^3#:

#x^3arctan(x^7)=x^3sum_(n=0)^oo(-1)^nx^(14n+7)/(2n+1)=sum_(n=0)^oo(-1)^nx^(14n+10)/(2n+1)#

Writing out the terms from #n=0# to #n=3#:

#x^3arctan(x^7)=x^10-x^24/3+x^38/5-x^52/7+...#