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

Question

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

Evelyn Agard · Accepted Answer

#ln(5-x)^2 = 2ln5 -2/5sum_(n=0)^oo x^(n+1)/(5^n(n+1))#

with radius of convergence #R=5#.

Start from the sum of a geometric series: #sum_(n=0)^oo x^n = 1/(1-x)# with radius of convergence #R=1#. Integrating term by term, we obtain a series with the same radius of convergence: #sum_(n=0)^oo int_0^x t^n = int_0^x dt/(1-t)# #sum_(n=0)^oo x^(n+1)/(n+1) = -ln(1-x)# Now using the properties of logarithms we have: #ln(5-x)^2 = 2ln(5(1-x/5)) =2ln5 +2ln(1-x/5)# and substituting #x/5# to #x# in the series above we get: #ln(5-x)^2 = 2ln5 -2/5sum_(n=0)^oo x^(n+1)/(5^n(n+1))# converging for #abs (x/5) < 1#, that is with radius of convergence #R=5#

Isabella Ackerman · Answer

undefined 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 $.

Elijah Abram · Answer

undefined 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 $.