<s>[INST] $\ln $ and Taylor Series Expansion (what went wrong) Edited
Problem
I'm trying to express $\ln{(1-(\frac{N}{K})^{\frac{1}{4}})}$ in terms of $\ln N$, where $K$ is a constant and $1 \leq N \leq K$. This also implies $\frac{N}{K} \leq 1$. 

Anyone knows if it is possible to represent $\ln{(1-(\frac{N}{K})^{\frac{1}{4}})}$ in terms of $\ln N$ ?
  Cuz as seen below, using the Taylor Expansion just returns the original term back to us and it didn't seem to help a lot.  

Taylor Series Expansion Used
The Taylor series I'll be using is: (refer to this Website for Taylor Series Expansions)
$$\ln (1+x) = x- \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots $$
where $-1 \leq x \leq 1$. 
Solving
Then, observe that $\ln{(1-(\frac{N}{K})^{\frac{1}{4}})} = \ln{(1-(\frac{N}{K})^{\frac{1}{8}}) (1+(\frac{N}{K})^{\frac{1}{8}})} = \ln(1-\frac{N}{K})^{\frac{1}{8}} + \ln (1+\frac{N}{K})^{\frac{1}{8}}$. 
By the series expansion: we get: 
$$\ln(1-\frac{N}{K})^{\frac{1}{8}} = -(\frac{N}{K})^{\frac{1}{8}}-\frac{(\frac{N}{K})^\frac{1}{4}}{2} - \frac{(\frac{N}{K})^\frac{3}{8}}{3}- \frac{(\frac{N}{K})^\frac{1}{2}}{4} - \ldots$$
and
$$\ln(1+\frac{N}{K})^{\frac{1}{8}} = (\frac{N}{K})^{\frac{1}{8}}-\frac{(\frac{N}{K})^\frac{1}{4}}{2} + \frac{(\frac{N}{K})^\frac{3}{8}}{3}- \frac{(\frac{N}{K})^\frac{1}{2}}{4} - \ldots$$
Thus taking the sum of the 2 logarithms above, we get:
$$\ln{(1-(\frac{N}{K})^{\frac{1}{4}})} =- (\frac{N}{K})^\frac{1}{4} - \frac{(\frac{N}{K})^\frac{1}{2}}{2} - \frac{(\frac{N}{K})^\frac{3}{4}}{3} - \ldots $$

Notice that if we were to differentiate the series, we get: 
$\frac{d}{dx} x + \frac{1}{2}(x^2) + \frac{1}{3}(x^3) + \ldots = 1+x+x^2+ \ldots = \frac{1}{1-x}$, for $|x| \leq 1$. 
Hence, $x + \frac{1}{2}(x^2) + \frac{1}{3}(x^3) + \ldots = \int \frac{1}{1-x} dx = -\ln{(1-x)} + C$  for some arbitrary constant $C$. 

Hence, $\ln{(1-(\frac{N}{K})^{\frac{1}{4}})} = -[(\frac{N}{K})^\frac{1}{4} + \frac{(\frac{N}{K})^\frac{1}{2}}{2} + \frac{(\frac{N}{K})^\frac{3}{4}}{3} - \ldots]  = \int -\frac{1}{1-(\frac{N}{K})^\frac{1}{4}} d\frac{N}{K} = \ln {(1-(\frac{N}{K})^\frac{1}{4})} + C$.
We get back the same term!
 [/INST] $$
\left(1-\left(\frac{N}{K}\right)^{1/2}\right)\left(1+\left(\frac{N}{K}\right)^{1/2}\right) = \left(1-\frac{N}{K}\right)\neq \left(1-\left(\frac{N}{K}\right)^{1/4}\right)
$$
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