Question

Could someone please show how to find a Taylor polynomial for ln(x) centered around x=2? (Along with a brief methodology/explanation, if possible...)

Answer 1

\[f(x)=\ln(x)\\
f'(x)=\frac{1}{x}=x^{-1}\\
f''(x)=-\frac{1}{x^2}=x^{-2}\\
f'''(x)=2x^{-3}\] first probably good to look for a pattern

Answer 2

then we have to evaluate each of these at \(x=2\) and finally divide each by \(n!\) where \[c_n=\frac{f^{(n)}(2)}{n!}\]

Answer 3

\[f(2)=\ln(2)\] for \(c_0\) no pattern there it is just the first term

Answer 4

\[f'(2)=\frac{1}{2}\\
f''(2)=-\left(\frac{1}{2}\right)^2\]
\[f'''(2)=2\times \left(\frac{1}{2}\right)^3\]

Answer 5

pretty clear that it alternates
also maybe it is clear that
\[f^{(n)}{2}=(n-1)!\left(\frac{1}{2}\right)^n\]

Answer 6

when you divide by \(n!\) you will be left with
\[c_n=\frac{(-1)^n\left(\frac{1}{2}\right)^n}{n}\]

Answer 7

guess i lost @james238867

Answer 8

I'm still here. I'm actually understanding your explanation. Thanks!

Answer 9

oh ok then i guess we are more or less done
really it amounts to looking for a pattern, the less arithmetic you do the better, helps to see what the pattern is

Answer 10

final answer will look something like
\[\sum\frac{(-1)^n\left(\frac{1}{2}\right)^n(x-2)^n}{n}\]

Answer 11

there are other ways to write it, you could put the \(2^n\) in the denominator

Answer 12

also you should check that it is \((-1)^n\) or \((-1)^{n+1}\) i.e. check that the terms you want to be negative are in fact negative

Answer 13

notice that you really don't want to do arithmetic as i said, and even though it is generally stupid to use the power rule for derivatives, you want to in this case
\[f'(x)=x^{-1}\\
f''(x)=-x^{-2}\\
f'''(x)-2x^{-3}\\
f^{(4)}(x)=-3\times -2x^{-4}\] etc
helps to get the pattern down

Answer 14

I see. You've answered the question really completely. Thanks!

Answer 15

yw