How do you find the power series for #f(x)=xln(1+2x)# and determine its radius of convergence?

1 Answer
Aug 20, 2017

The power series is #sum_(n=0)^oo((-1)^n2^(n+1))/(n+1)x^(n+2)#, its radius of convergence is #R=1/2#.

Explanation:

Start with a basic power series:

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

Replace #x# with #-2x#:

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

#1/(1+2x)=sum_(n=0)^oo(-1)^n2^nx^n#

Integrate both sides. In the series, we integrate just the part with #x#.

#intdx/(1+2x)=sum_(n=0)^oo(-1)^n2^nintx^ndx#

#1/2ln(1+2x)=C+sum_(n=0)^oo(-1)^n2^nx^(n+1)/(n+1)#

Don't forget the #1//2# in the integral. Absolute value bars aren't needed for #ln(1+2x)# since we're only working in #x>=0#.

Add a constant of integration. Fortunately, letting #x=0# shows that #C=0#, so this isn't really an issue.

The next step to reaching the "goal function" of #xln(1+2x)#. To do this, multiply both sides by #2x#.

#xln(1+2x)=2xsum_(n=0)^oo((-1)^n2^n)/(n+1)x^(n+1)#

Bring #2x# into the series:

#xln(1+2x)=sum_(n=0)^oo((-1)^n2^(n+1))/(n+1)x^(n+2)#

#xln(1+2x)=2x^2-2x^3+8/3x^4-4x^5+...#


For the radius of convergence of the series #suma_n#, find the ratio #abs(a_(n+1)/a_n)#. Here, #a_n=((-1)^n2^(n+1))/(n+1)x^(n+2)#.

#abs(a_(n+1)/a_n)=abs((((-1)^(n+1)2^(n+2))/(n+2)x^(n+3))/(((-1)^n2^(n+1))/(n+1)x^(n+2)))=abs((-1)^(n+1)/(-1)^n2^(n+2)/2^(n+1)x^(n+3)/x^(n+2)(n+1)/(n+2))#

#abs(a_(n+1)/a_n)=abs(-2x(n+1)/(n+2))#

Find its limit as #nrarroo#:

#lim_(nrarroo)abs(a_(n+1)/a_n)=lim_(nrarroo)abs(-2x(n+1)/(n+2))#

The limit only concerns how #n# changes, so #abs(2x)# can be extracted from the limit (we can also disregard the #-1# as we're working within absolute values):

#lim_(nrarroo)abs(a_(n+1)/a_n)=2absxlim_(nrarroo)abs((n+1)/(n+2))#

The limit is #1#:

#lim_(nrarroo)abs(a_(n+1)/a_n)=2absx#

The ratio test states that #suma_n# converges when #lim_(nrarroo)abs(a_(n+1)/a_n)<1#. So, our interval of convergence will occur when:

#2absx<1#

Or:

#absx<1/2#

Thus, the radius of convergence is #R=1/2#.