Question #7a6ca

2 Answers
Jan 13, 2017

#x^2ln (1+x^3) = sum_(n=1)^oo (-1)^(n-1) (x^(3n+2))/n#

Explanation:

Take the MacLaurin series for #ln(1+t)#, which is well known:

#ln (1+t) = sum_(n=1)^oo (-1)^(n-1) t^n/n#

Substitute: #t=x^3#

#ln (1+x^3) = sum_(n=1)^oo (-1)^(n-1) (x^3)^n/n = sum_(n=1)^oo (-1)^(n-1) (x^(3n))/n#

Now multiply by #x^2# term by term:

#x^2ln (1+x^3) = sum_(n=1)^oo (-1)^(n-1) (x^2 x^(3n))/n = sum_(n=1)^oo (-1)^(n-1) (x^(3n+2))/n#

Jan 13, 2017

#x^2ln(1+x^3) = -sum_(n=1)^(oo) (-1)^(n)x^(3n + 2)/(n)#

#= x^5 - x^8/2 + x^11/3 - x^14/4 + . . . #


Recall that

#sum_(n=0)^(oo) t^n = 1/(1-t)#,

which is the derivative of #-ln(1-t)#. So, substitute #t = -x^3# to get

#sum_(n=0)^(oo) (-x^3)^n = 1/(1 + x^3)#,

(whose derivative has no sign change), so that

#intsum_(n=0)^(oo) (-1)^(n)(x^3)^(n)dx = sum_(n=0)^(oo) int(-1)^(n)(x^3)^ndx#

#= sum_(n=0)^(oo) [(-1)^(n)1/(n+1)(x^3)^(n+1)] + C#

#= sum_(n=1)^(oo) [(-1)^(n-1)1/(n)x^(3n)] + C#

#= -sum_(n=1)^(oo) [(-1)^(n)1/(n)x^(3n)] + C#

#= ln(1 + x^3)#, #x >= 0#,

where #C = ln(1 + 0^3) = 0#.

Since the sum depends only on #n#, we can multiply through by #x^2# to get the result:

#color(blue)(x^2ln(1+x^3) = -sum_(n=1)^(oo) (-1)^(n)x^(3n + 2)/(n))#

#= color(blue)(x^5 - x^8/2 + x^11/3 - x^14/4 + . . . )#

whereas that for #ln(1+x^3)#, if you wish to compare, was:

#ln(1+x^3) = x^3 - x^6/2 + x^9/3 - x^12/4 + . . . #

which indeed was just multiplied by #x^2# to get to #x^2ln(1+x^3)#.