# A complex log inequality

Standard

Let $z$ be a complex number. The power series expansion of $\text{log}(1+z)$ about $z_0=0$ is given by

$\displaystyle{\text{log}(1+z) = \sum_{n=1}^\infty (-1)^{n-1}\frac{z^n}{n} = z-\frac{z^2}{2} + \ldots}$

which has radius of convergence 1. If $|z|<1$ then

$\displaystyle{\left|1-\frac{\log(1+z)}{z}\right|\leq \frac{|z|}{2(1-|z|)}}$

If we further assume $|z|<1/2$ then

$\displaystyle{\left|1-\frac{\log(1+z)}{z}\right|\leq \frac{1}{2}}$

This gives,

$\displaystyle{\frac{|z|}{2}\leq |\log(1+z)|\leq \frac{3|z|}{2}, \quad |z|<\frac{1}{2}}$

I just wanted to see how this inequality will appear graphically, so here are the plots made using SageMath 7.5.1 (by fixing the real part of z to zero and varying the imaginary part till 1/2)

i = CDF.0; p1 = plot(lambda t: abs(log(1+t*i)), 0, 0.5, rgbcolor=(0.8,0,0),legend_label=’ $|log(1+z)|$’, thickness=2); p2 = plot(lambda t: abs((t*i)/2), 0, 0.5, rgbcolor=(0,0.8,0), legend_label=’$|z|/2$’, thickness=2); p3 = plot(lambda t: abs(3*(t*i)/2), 0, 0.5, rgbcolor=(0,0,0.8), legend_label=’ $3|z|/2$’, thickness=2); p1+p2+p3

I tried to get a graph where this inequlaity fails (i.e. the plots intersect), but failed to do so.

# Solving Logarithmic Equations

Standard

While reading John Derbyshire’s Prime Obsession I came across the following statement (clearly explained on pp. 74):

Any positive power of $\log(x)$ eventually increases more slowly than any positive power of $x$.

It is easy to prove this (existence) analytically, by taking derivative to compare slopes. But algebraically it implies that (for example):

There are either no real solution or two real solutions of the equation
$\log(x) = x^\varepsilon$
for any given $\varepsilon>0$.

Now the question that arises is “How to find this $x$?” I had no idea about how to solve such logarithmic equations, so I took help of Google and discovered this Mathematic.SE post. So, we put $\log(x)=y$ and re-write the equation as:

$y=e^{y\varepsilon}$

Now to be able to use Lambert W function (also called the product logarithm function) we need to re-write the above equation, but I have failed to do so.

But using WolframAlpha I was able to solve $\log(x)=x^2$ to get $x=e^{\frac{-W(-2)}{2}}$ (which is an imaginary number, i.e. no real solution of this equation) but I was not able to figure out the steps involved. So if you have any idea about the general method or the special case of higher exponents, please let me know.