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

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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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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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