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 $2, -4, 16, -96,\ldots$.

  • 2
    $\begingroup$ Around $x=$ what ? $\endgroup$ Jun 9 '21 at 9:23
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.