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

Answer 1

#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#

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# #color(white)(.....................)⋮# #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#
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Answer 2

To find the Taylor series for ( \ln(1+x^2) ), you start by finding the derivatives of ( \ln(1+x^2) ) with respect to ( x ) and evaluating them at ( x = 0 ) to find the coefficients of the series. The formula for the Taylor series expansion of a function ( f(x) ) about ( x = a ) is:

[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n ]

For ( \ln(1+x^2) ), the first few derivatives are:

[ f(x) = \ln(1+x^2) ] [ f'(x) = \frac{2x}{1+x^2} ] [ f''(x) = \frac{2(1-x^2)}{(1+x^2)^2} ] [ f'''(x) = \frac{-8x}{(1+x^2)^3} ]

Evaluating these derivatives at ( x = 0 ), we get:

[ f(0) = \ln(1) = 0 ] [ f'(0) = \frac{0}{1} = 0 ] [ f''(0) = \frac{2}{1} = 2 ] [ f'''(0) = \frac{0}{1} = 0 ]

So, the Taylor series expansion for ( \ln(1+x^2) ) about ( x = 0 ) is:

[ \ln(1+x^2) = 0 + 0(x-0) + \frac{2}{2!}(x-0)^2 + 0(x-0)^3 + \cdots ]

[ = x^2 + \frac{x^4}{2} + \frac{x^6}{3} + \cdots ]

Therefore, the Taylor series for ( \ln(1+x^2) ) is:

[ \ln(1+x^2) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{2n}}{n} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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