Friday 20 December 2013

real analysis - Functions whose derivative is the inverse of that function

Everyone knows that there are at least three functions whose derivative is the function itself, namely $e^x, \ 0$ and $-e^{x}$. ( are there more?)



I was drawing some polynomials and their derivatives and noted that sometimes it was almost like the inverse. This lead me to ask this question: is there a function whose derivative is the inverse of that function?




Well, I figured that at least some kind of answer can be found to be of the form $a x^b$.



Lets solve this:



$f(x) = ax^b, f'(x) = abx^{b-1}$. Then $$f \circ f'(x) = a^{b+1}b^bx^{(b-1)b}=x=a^b b x^{(b-1)b}=f' \circ f (x).$$



Thus $b(b-1) = 1 \iff b^2-b-1=0 \iff b = \phi \vee 1-\phi,\ \phi = \frac{1+\sqrt 5}{2}$



We also see that $ab^{b-1}=1$, because both the multipliers must be one. Thus we get $a= \frac{1}{b^{b-1}}$. If $b=\phi$, we get $a=\phi^{\phi-1}$. If $b=1-\phi, \ a=(1-\phi)^{\phi}$




Thus two functions that satisfy the condition are $\phi^{\phi-1} x^\phi$ and $ (1-\phi)^{\phi}x^{1-\phi}$.



I would like to know if there are more functions like these, and do these functions have any 'interesting' properties, like exponential function, apart from this one condition about inverse being the derivative?

No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...