How do you find the taylor series for ln(1+x^2)?

1 Answer
Jan 14, 2018

ln(1+x^2)=sum_(n=1)^oo(-1)^(n+1)/nx^(2n)=x^2-x^4/2+x^6/3-x^8/4...

Convergent when |x|<1

Explanation:

We start by working out a taylor series for ln(1+x). I will be expanding around 0, so it will be a Maclaurin series.

The general formula for a Maclaurin series is:
f(x)=sum_(n=0)^oof^n(0)/(n!)x^n

This means we need to work out the nth derivative of ln(1+x). Let's start by taking some derivatives and see what values they produce at x=0:

f(x)=ln(1+x)-> 0

f'(x)=1/(1+x)-> 1

f''(x)=-1/(1+x)^2-> -1

f^3(x)=2/(1+x)^3-> 2

f^4(x)=-6/(x+1)^4-> -3

f^5(x)=24/(x+1)^5-> 24
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f^n(x)=(-1)^(n+1)((n-1)!)/(x+1)^(n)-> (-1)^(n+1)(n-1)!

We can now plug this into the Maclaurin expansion (note that we ignore the first term, since it is 0):
ln(1+x)=sum_(n=1)^oo((-1)^(n+1)(n-1)!)/(n!)x^n

This simplifies to:
ln(1+x)=sum_(n=1)^oo(-1)^(n+1)/nx^n

Now that we have a series for ln(1+x), we can replace all the x's with x^2 to get a series for ln(1+x^2):
ln(1+x^2)=sum_(n=1)^oo(-1)^(n+1)/n(x^2)^n=

=sum_(n=1)^oo(-1)^(n+1)/nx^(2n)=x^2-x^4/2+x^6/3-x^8/4...

which is the series we were looking for.

To test the convergence, we can use the ratio test:
lim_(n->oo)-(x^(2(n+1))/(n+1))/(x^(2n)/n)=lim_(n->oo)-(x^2cancel(x^(2n)))/(n+1)*n/cancel(x^(2n))=

=lim_(n->oo)-(nx^2)/(n+1)=-x^2

So, we need to look at when |-x^2|<1:
|-x^2|<1

x^2<1

sqrt(x^2)< sqrt1

|x|<1

which means the series is convergent when |x|<1