How do you find the power series for #f(x)=xln(1+2x)# and determine its radius of convergence?
1 Answer
The power series is
Explanation:
Start with a basic power series:
#1/(1-x)=sum_(n=0)^oox^n#
Replace
#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
#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
Add a constant of integration. Fortunately, letting
The next step to reaching the "goal function" of
#xln(1+2x)=2xsum_(n=0)^oo((-1)^n2^n)/(n+1)x^(n+1)#
Bring
#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
#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
#lim_(nrarroo)abs(a_(n+1)/a_n)=lim_(nrarroo)abs(-2x(n+1)/(n+2))#
The limit only concerns how
#lim_(nrarroo)abs(a_(n+1)/a_n)=2absxlim_(nrarroo)abs((n+1)/(n+2))#
The limit is
#lim_(nrarroo)abs(a_(n+1)/a_n)=2absx#
The ratio test states that
#2absx<1#
Or:
#absx<1/2#
Thus, the radius of convergence is