Question

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that Rn(x) → 0.] f(x) = ln(x), a = 3 f(x) = ln(3) + ∞ n = 1 Find the associated radius of convergence R. R =

Answer 1

Answer:

The Taylor series is $\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}$.

The radius of convergence is $R=3$.

Step-by-step explanation:

The Taylor expansion.

Recall that as we want the Taylor series centered at $a=3$ its expression is given in powers of $(x-3)$. With this in mind we need to do some transformations with the goal to obtain the asked Taylor series from the Taylor expansion of $\ln(1+x)$.

Then,

$\ln(x) = \ln(x-3+3) = \ln(3(\frac{x-3}{3} + 1 )) = \ln 3 + \ln(1 + \frac{x-3}{3}).$

Now, in order to make a more compact notation write $\frac{x-3}{3}=y$. Thus, the above expression becomes

$\ln(x) = \ln 3 + \ln(1+y).$

Notice that, if x is very close from 3, then y is very close from 0. Then, we can use the Taylor expansion of the logarithm. Hence,  

$\ln(x) = \ln 3 + \ln(1+y) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{y^n}{n}.$

Now, substitute $\frac{x-3}{3}=y$ in the previous equality. Thus,

$\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.$

Radius of convergence.

We find the radius of convergence with the Cauchy-Hadamard formula:

$R^{-1} = \lim_{n\rightarrow\infty} \sqrt[n]{|a_n|},$

Where $a_n$ stands for the coefficients of the Taylor series and $R$ for the radius of convergence.

In this case the coefficients of the Taylor series are

$a_n = \frac{(-1)^{n+1}}{ n3^n}$

and in consequence $|a_n| = \frac{1}{3^nn}$. Then,

$\sqrt[n]{|a_n|} = \sqrt[n]{\frac{1}{3^nn}}$

Applying the properties of roots

$\sqrt[n]{|a_n|} = \frac{1}{3\sqrt[n]{n}}.$

Hence,

$R^{-1} = \lim_{n\rightarrow\infty} \frac{1}{3\sqrt[n]{n}} =\frac{1}{3}$

Recall that

$\lim_{n\rightarrow\infty} \sqrt[n]{n}=1.$

So, as $R^{-1}=\frac{1}{3}$ we get that $R=3$.