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 $-1 \leq x \leq 1$.


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!

  • 1
    $\begingroup$ $\bigl(1-x^{1/2}\bigr)\bigl(1+x^{1/2}\bigr) = 1 - \bigl(x^{1/2}\bigr)^2 = 1-x \neq 1-x^{1/4}$ $\endgroup$ – Daniel Fischer Jul 1 '15 at 11:12
  • $\begingroup$ @DanielFischer Literally word for word. :/ +1 $\endgroup$ – Chinny84 Jul 1 '15 at 11:13
  • $\begingroup$ Corrected the mistake. $\endgroup$ – Happytreat Jul 1 '15 at 12:47
  • $\begingroup$ You could write $\ln\left(1-\left(\frac NK\right)^{1/4}\right)=\ln(1 - e^{(\ln N - \ln K)/4}).$ I doubt that will make you happy, however. $\endgroup$ – David K Jul 1 '15 at 13:48

Note that instead of going through the steps of adding the Taylor series for $\ln\left(1+\left(\frac NK\right)^{1/8}\right)$ and $\ln\left(1-\left(\frac NK\right)^{1/8}\right)$, you could have derived the series

$$\ln\left(1-\left(\frac NK\right)^{\frac14}\right) = -\left(\frac NK\right)^\frac14 - \frac{\left(\frac NK\right)^\frac12}{2} - \frac{\left(\frac NK\right)^\frac34}{3} - \ldots $$

simply by applying the known Taylor series expansion of $\ln(1+x)$ with $x = -\left(\frac NK\right)^{1/4}.$

Your method of differentiating and then integrating the series is a partial confirmation that the series is the correct Taylor series of $\ln\left(1-\left(\frac NK\right)^{1/4}\right).$ (You would completely confirm the correctness of the series if you showed that $C=0$.) So it's hardly surprising that this manipulation of the series gives you back what you started with (plus the pesky unknown constant $C$).

Of course you conceivably could have started with a much more complicated expression that happened to give the same Taylor series and therefore was equal to $\ln\left(1-\left(\frac NK\right)^{1/4}\right).$ Getting back the same expression is a clue that perhaps the expression $\ln\left(1-\left(\frac NK\right)^{1/4}\right)$ is already about as simple as it should be.

There's no reason to expect $\ln N$ to show up anywhere in any of these calculations unless you force it to occur by substituting $N = e^{\ln N}$. In fact,

$$\ln\left(1-\left(\frac NK\right)^{\frac14}\right) = \ln\left(1-e^{(\ln N - \ln K)/4}\right),$$

so unless you have a nicer way of writing $\ln\left(1-e^{(x - \ln K)/4}\right)$ in terms of $x$ then I think this may be as good as it gets.

  • $\begingroup$ True... Actually I am also suspecting there isn't a simpler way to represent the $\ln$ term too... Thanks! $\endgroup$ – Happytreat Jul 1 '15 at 14:18

$$ \left(1-\left(\frac{N}{K}\right)^{1/2}\right)\left(1+\left(\frac{N}{K}\right)^{1/2}\right) = \left(1-\frac{N}{K}\right)\neq \left(1-\left(\frac{N}{K}\right)^{1/4}\right) $$

  • $\begingroup$ oops, thanks for pointing that out guys. Any idea whether there is a way to represent $\ln (1 -( \frac{N}{K}) ^\frac{1}{4})$ in terms of $\ln N$??? $\endgroup$ – Happytreat Jul 1 '15 at 12:38

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