# How do you find the Taylor series of #f(x)=ln(x)# ?

Keep on working in this vein until you reach the degree that has been asked for. Usually during high school and early university, you won't need it beyond about the sixth degree as it gets quite time-consuming to calculate by hand.

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To find the Taylor series of ( f(x) = \ln(x) ), we first find the derivatives of ( f(x) ) up to the desired order at a chosen center ( c ). Then, we evaluate these derivatives at ( c ) and use them to form the terms of the Taylor series. The general formula for the Taylor series expansion of ( f(x) ) centered at ( c ) is:

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

For ( f(x) = \ln(x) ), the derivatives are:

[ f'(x) = \frac{1}{x}, \quad f''(x) = -\frac{1}{x^2}, \quad f'''(x) = \frac{2}{x^3}, \quad f^{(4)}(x) = -\frac{6}{x^4}, ]

and so on.

Choose a center ( c ) and substitute these derivatives into the Taylor series formula. The most common choice for ( c ) would be ( c = 1 ) since ( \ln(1) = 0 ) and it simplifies the calculations. Thus, the Taylor series expansion of ( f(x) = \ln(x) ) centered at ( x = 1 ) would be:

[ \ln(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n} (x - 1)^n ]

This series converges to ( \ln(x) ) for ( x ) in the interval ( (0, 2] ).

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To find the Taylor series of ( f(x) = \ln(x) ), we can use the Taylor series expansion formula:

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

where ( f^{(n)}(a) ) denotes the ( n )-th derivative of ( f(x) ) evaluated at ( x = a ).

For ( f(x) = \ln(x) ), we choose a point ( a ) around which to expand the series. Let's choose ( a = 1 ) for simplicity. Then, we need to compute the derivatives of ( \ln(x) ) at ( x = 1 ).

- ( f(x) = \ln(x) )
- ( f'(x) = \frac{1}{x} )
- ( f''(x) = -\frac{1}{x^2} )
- ( f'''(x) = \frac{2}{x^3} )
- ( f^{(4)}(x) = -\frac{6}{x^4} )
- Continuing this pattern, we find ( f^{(n)}(x) = (-1)^{n+1} \frac{(n-1)!}{x^n} ) for ( n \geq 2 ).

Now, we substitute these derivatives into the Taylor series expansion formula centered at ( x = 1 ):

[ \ln(x) = \ln(1) + \frac{1}{1!}(x-1)^1 + \frac{-1}{2!}(x-1)^2 + \frac{2}{3!}(x-1)^3 - \frac{6}{4!}(x-1)^4 + \cdots ]

Simplifying, we get:

[ \ln(x) = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \cdots ]

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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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