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I know the pattern of the Taylor series but I do not know how to formulate the formula.

The given is $f(x) = \log_3(2x-1)$ at $x = 1$.

$f^{(0)}(x) = \log_3(2x-1)$

$f^{(1)}(x) = \frac{1}{\log3} \frac{2}{(2x-1)}$

$f^{(2)}(x) = \frac{1}{\log3} \frac{-4}{(2x-1)^2}$

$f^{(3)}(x) = \frac{1}{\log3} \frac{16}{(2x-1)^3}$

$f^{(4)}(x) = \frac{1}{\log3} \frac{-96}{(2x-1)^4}$

$f^{(0)}(1) = \log_3(1)$

$f^{(1)}(1) = \frac{1}{\log3} \cdot 2$

$f^{(2)}(1) = \frac{1}{\log3} \cdot (-4)$

$f^{(3)}(1) = \frac{1}{\log3} \cdot 16$

$f^{(4)}(1) = \frac{1}{\log3} \cdot (-96)$

I have a problem on how to formulate the numerator part of the Taylor series. I am uncertain on what to do with pattern $2, -4, 16, -96,\ldots$.

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    $\begingroup$ Around $x=$ what ? $\endgroup$ Jun 9 '21 at 9:23
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    $\begingroup$ If the Taylor expansion should be done around $1$, just write $x=1+h$, express $\log_3$ in terms of $\log=\ln$, and use the Taylor expansion formula for $-\ln(1-y)$, obtained e.g. by using the one for its derivative. $\endgroup$
    – dan_fulea
    Jun 9 '21 at 10:25
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    $\begingroup$ Hint: $$ \log _3 (2x - 1) = \frac{1}{{\log 3}}\log (2x - 1) = \frac{1}{{\log 3}}\log (1 + 2(x - 1)). $$ $\endgroup$
    – Gary
    Jun 9 '21 at 11:07
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I think the doubt is how to generalize the Taylor expansion of the given function around a point $x$. We note that every time we differentiate a function, the term will be multiplied by $2$ (coming from $2x-1$), and alternatively, $-1$ gets multiplied. We will also get a factorial term coming from the power of the term $(2x-1)$. Thus, except for the first term, we can write $$f^{(n)}(x)=\frac{(-1)^{n+1}}{log(3)}\frac{2^n}{(2x-1)^n}(n-1)!$$ Hope this helps.

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