Re: Taylor Series Problem

$\displaystyle f(x)=ln(x)$

$\displaystyle f'(x)=\frac{1}{x}$

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

$\displaystyle f^3(x)=\frac{2}{x^3}$

$\displaystyle f^4(x)=\frac{-6}{x^4}$

$\displaystyle f^5(x)=\frac{24}{x^5}$

Substitute $\displaystyle x=1$ into all of these to get (respectively) $\displaystyle 0, 1, -1, 2, -6, -24$

Then, just substitute into the formula:

$\displaystyle f(x)\approx f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^2}{2!}+\frac{f^3(a)(x-a)^3}{3!}+\frac{f^4(a)(x-a)^4}{4!}+\frac{f^5(a)(x-a)^5}{5!}$

So, for example, for this term:

$\displaystyle \frac{f^4(a)(x-a)^4}{4!}$

We'd substitute in as:

$\displaystyle \frac{-6(x-1)^4}{4!}$

$\displaystyle =\frac{-1}{4}(x-1)^4$

Hope this helps.

Re: Taylor Series Problem

Thank you! That clears up a lot of my confusion.

Re: Taylor Series Problem

Quote:

Originally Posted by

**bobsanchez** Find the first five terms of the Taylor series for f(x) = lnx about x = 1.

Okay, so I know the general formula for a Taylor series, and roughly how to do this, but I've seen conflicting stuff in my notes and whenever I've looked at examples (it's not really "conflicting", I just don't understand, I'm sure).

Could someone walk me through this one so I can make sure I understand the proper steps and everything? I really appreciate it.

A fast solution starts from the well known series expansion...

$\displaystyle \ln (1+\xi)= \xi - \frac{\xi^{2}}{2} + \frac{\xi^{3}}{3} - \frac{\xi^{4}}{4} + ... $ (1)

... and now You set in (1) $\displaystyle x=1 + \xi \implies \xi= x-1$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$