At which parameter value $c>0$ do the number of solutions of $\log(1+x^2)=x^c$ change?

72 Views Asked by At

I'm looking at the functions $x\mapsto \log(1+x^2)$ and $x\mapsto x^c,\ c>0$ on the interval $\mathbb R^+_0$.

I'm interested in the properties of $$\log(1+x^2)=x^c.$$ Graphically, for small $c$, the function $x^c$ is concave and intersects $\log(1+x^2)$ once or twice. Then at some value $c=c_1$, they don't seem to intersect anymore and then at a value $c=c_2$ they start touching again.

What are $c_1, c_2$, and, generally, what is the enclosed area between 0 and the first intersection point of the two functions, as a function of $c$?

enter image description here

1

There are 1 best solutions below

3
On BEST ANSWER

Consider the function $f_c(x)=\log(1+x^2)-x^c$.

Its derivative with respect to $x$ is $f_c'(x)=\frac{2x}{1+x^2}-cx^{c-1}$ We want to find $x$ and $c$ such that $f_c(x)=f_c'(x)=0$.

We can eliminate $c$ from the equations. $$c=\frac{2x^2}{(1+x^2)(-\log(1+x^2))}.$$

Replacing in $f_c(x)=0$ we get $$\log(1+x^2)-e^{\frac{2x^2\log(x)}{(1+x^2)(-\log(1+x^2))}}=0$$

which can be solved numerically. From the values of $x$ we can then get those of $c$.

Since the solution(s) will be obtained by numerical methods, it would be convenient to think better what equation(s) to use.