Question

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

Answer 1

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+...`

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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`.