Represent $f(x)=\ln x$ as power series, in powers of $(x-4)$ I was requested to represent $f(x)=\ln x$ as a power series, in powers of $(x-4)$. I only started studying power series today, and though I was able to represent different functions as power series I can't seem to work this one out, specially with the condition that the series must be in powers of $(x-4)$. 
Any ideas on how to solve this problem?
 A: Note that we have the formula $$\ln(1+x)=\sum_1^\infty \frac {(-1)^n x^n}{n}$$
Thus $$\ln x = \ln (4+x-4) = \ln (4(1+\frac {x-4}{4}))=$$
$$\ln 4+\ln(1+\frac {x-4}{4})=$$
$$ \ln 4 +\sum _1^\infty \frac {(-1)^n(x-4)^n}{n4^n}$$
A: Here's something to get you started. You are asked to provide a power series representation of $ln(x)$ centered at $x=4$ and provide the radius and interval of convergence of this power series. Here I would directly use Taylor's formula, which states that a power series representation is given by $$\sum_{n=0}^{\infty} \frac{f^{(k)}(4) (x-4)^k}{k!}$$
So you need to start computing the derivatives of $ln(x)$: $$f'(x)= \frac{1}{x}, f''(x)=-\frac{1}{x^2}, f^{(3)}(x)= \frac{2}{x^3}, f^{(4)}(x)=-\frac{2 \times 3}{x^4},... $$ 
Once you have the general formula for $$f^{(k)}(x)= \frac{(-1)^{k-1}(k-1)!}{x^k}$$ then you plug it in and use the ratio test to determine the interval and radius of convergence of this power series representation.
A: There are a few ways to go about this.
The most direct way.
$f(x) = f(x-a) + f'(x-a) (x-a) + \frac {f''(x-a)}{2!} (x-a)^2 + \cdots + \frac {f^{(n)}(x-a)}{n!} (x-a)^n + \cdots$
I hope that this looks familiar to you.
We set $a = 4$
$f(4) = \ln 4\\
f'(x) = \frac {1}{x}, f'(4) = \frac {1}{4}\\
f''(x) = -\frac {1}{x^2}, f''(4) = -\frac {1}{4^2}\\
f^{(n)}(x) = (-1)^{n-1} \frac {1}{x^n}, f^{(n)}(4) = (-1)^{n-1} \frac {(n-1)!}{4^n}$
And you can verify this last line via induction.
$f(x) = \ln 4 + \sum_\limits{n=1}^{\infty} (-1)^{n-1}\frac {(x-a)^n}{n4^n}$
And there is the tricky way to do this.
$f'(x) = \frac {1}{x}\\
f'(x) = \frac {1}{4 + (x-4)}\\
f'(x) = \frac 14 \frac {1}{1 + \frac {x-4}{4}}$
This can be thought of as the sum of a geometric series.
i.e $\frac {1}{1 - a} = \sum_\limits{n=0}^\infty a^n$ if $|a|< 1$
$f'(x) = \frac 14  \sum_\limits{n=0}^\infty (-1)^n \left(\frac {x-4}{4}\right)^n$
$f(x) - f(4) =  \int \sum_\limits{n=0}^\infty (-1)^n \left(\frac {x-4}{4}\right)^n\ \frac {dx}{4}\\
 = \sum_\limits{n=0}^\infty (-1)^n \frac {1}{(n+1)}\left(\frac {x-4}{4}\right)^{n+1}
\\ f(x) - \ln 4  = \sum_\limits{n=1}^\infty (-1)^{n-1} \frac {(x-4)^n}{n4^n}$
And still, the series will only converge if $|\frac {x-4}{4}| < 1$ 
And, if you have been working with power series for a while you might say.
I know $\ln (1-x) = -\sum_\limits{n=1}^\infty \frac {x^n}{n}\\
\ln (1+x) = \sum_\limits{n=1}^\infty (-1)^n \frac{x^n}{n}$
From doing exercises like the one above.
$\ln x = \sum_\limits{n=1}^\infty (-1)^n\frac {(x-1)^n}{n}$ substituting $(x-1)$ in for $x$ in the second equation above.
To center at 4.
$\ln x = \ln (4\frac {x}{4}) = \ln 4 + \ln \frac {x}{4}$
And substitute into what we had for $\ln x$
$\ln x = \ln 4 + \sum_\limits{n=1}^\infty (-1)^n\frac {(\frac{x}{4}-1)^n}{n}\\
\ln 4 + \sum_\limits{n=1}^\infty (-1)^n\frac {(x-1)^n}{n4^n}\\
$
A: $$f(x)=\ln (x)=\ln (4)-\sum_{n=1}^\infty \dfrac{(-1)^n}{n\cdot 4^n}\cdot (x-4)^n,$$
power series that converges if $|x-4|<4$.
