# What is the power series of #f(x)= ln(5-x)^2#? What is its radius of convergence?

with radius of convergence

Start from the sum of a geometric series:

Integrating term by term, we obtain a series with the same radius of convergence:

Now using the properties of logarithms we have:

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The power series expansion of ( f(x) = \ln((5 - x)^2) ) is:

[ \ln((5 - x)^2) = -2 \sum_{n=1}^{\infty} \frac{(x - 5)^n}{n} ]

The radius of convergence ( R ) of a power series can be found using the ratio test:

[ R = \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right| ]

For the given power series, the general term ( a_n ) is ( \frac{(x - 5)^n}{n} ). Applying the ratio test:

[ \lim_{n \to \infty} \left| \frac{\frac{(x - 5)^n}{n}}{\frac{(x - 5)^{n+1}}{n+1}} \right| ]

[ = \lim_{n \to \infty} \left| \frac{n+1}{n} \cdot \frac{1}{|x - 5|} \right| ]

[ = |x - 5| ]

So, for the series to converge, ( |x - 5| < 1 ). Therefore, the radius of convergence is ( R = 1 ).

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The power series of ( f(x) = \ln(5-x)^2 ) is: [ \sum_{n=0}^{\infty} \frac{2(-1)^n}{n+1}(x-5)^{n+1} ]

Its radius of convergence is the distance from the center of the series to the nearest point at which the series converges. In this case, the radius of convergence is ( R = 5 ).

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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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