# Thread: Taylor series of a logarithm

1. ## Taylor series of a logarithm

why we do tailor around 1 and not 0
?

2. Originally Posted by transgalactic

why we do tailor around 1 and not 0
?
Hint:
Spoiler:
Assuming that $\displaystyle f$ is analytic at $\displaystyle 0$ then $\displaystyle f(x)=\sum_{j=0}^{\infty}\frac{f^{(j)}(0)x^j}{j!}$ where (and this is meant as emphasis) $\displaystyle f^{(0)}(0)=f(0)$

P.S. I don't see any fabric.

3. Originally Posted by transgalactic

why we do tailor around 1 and not 0
?
Because log z is singular (not defined) at zero.

4. the question states how to get over loss of significance problem in here
f=ln x -ln y?

f=ln x -ln y=ln z
the tailor series for ln z is:
$\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k+1}(Z-1)}{k}$
but how this expression equals z-1 ?

5. Originally Posted by transgalactic
the question states how to get over loss of significance problem in here
f=ln x -ln y?

f=ln x -ln y=ln z
the tailor series for ln z is:
$\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k+1}(Z-1)}{k}$
but how this expression equals z-1 ?
It doesn't. But, $\displaystyle \ln(z){\color{red}\overset{z\to1}{\sim}}z-1$

6. why??

how you got to this conclution that it close to z-1
?

7. Originally Posted by transgalactic
why??

how you got to this conclution that it close to z-1
?
Your expression is equal to $\displaystyle \ln(z)$. So, $\displaystyle \lim_{z\to1}\frac{\ln(z)}{z-1}=\lim_{x\to0}\frac{\ln(1-x)}{x}=1$. Another easier way is to write out the terms of your series and notice that $\displaystyle \left|(z-1)-\sum_{n=1}\frac{(-1)^n(z-1)^n}{n}\right|<\varepsilon$ by taking $\displaystyle |z-1|<\delta$ for a suitable $\displaystyle \delta$