I am working on this problem.
Find Taylor series of function $f(x)=\ln(x)$ at $a = 6$.
$$f(x) =\sum_{n=0}^\infty c_n (x- 6)^n$$
I seem to be having trouble with the interval of convergence can someone show me how to do it.
[Math] Taylor series convergence with natural logs
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Best Answer
$$ \ln(x) = \int\frac{1}{x}dx\\ \frac{1}{x} = \frac{1}{6 - (6 - x)} = \frac{1}{6}\frac{1}{1 - \frac{-(x - 6)}{6}} = \frac{1}{6}\sum_0^\infty (-1)^n\frac{(x - 6)^n}{6^n} $$
This is a geometric series and thus converges when the common ratio: $\frac{x - 6}{6}$ is smaller than one:
$$ \left|\frac{x - 6}{6}\right| < 1 \\ |x - 6| < 6 \\ 0 < x < 12 $$
Oh, and if you want the taylor series then you need to integrate the above:
$$ \ln(x) = \ln(6) + \frac{1}{6}\sum_0^\infty (-1)^n\frac{(x - 6)^{n + 1}}{(n+1)6^n} = \ln(6) -\sum_1^\infty \frac{(-1)^n}{n}\frac{(x - 6)^n}{6^{n}} $$
Notice that the addition of $\ln(6)$ is necessary to make this correct (since if you plug in $x = 6$ to the sum, you get zero). Note that the sum became negative because you get $(-1)^{n - 1} = -(-1)^n$ and the $\frac{1}{6}$ went away because you have $\frac{1}{6^{n - 1}} = \frac{1}{6^n\frac{1}{6}} = \frac{6}{6^n}$ which cancels the divide by $6$ in front of the original sum.