Derivative equals inverse

In this video, I find a function whose derivative equals to its inverse function. The answer is quite surprising!

[drpeyam]

Typo: At the end, the 1/f should be 1/f’

[adrianmiranda]

I submitted this problem for a high school maths challenge and it made it in haha. More recently I saw a post on mathoverflow on which it was actually proven to be unique.

[johnkampmeyer]

Great video as always, Dr. Peyam!

This inspired me to come up with a very interesting generalization to this problem, namely f^(n)(x) = f^(-1)(x) (the n-th derivative of f is equal to the inverse of f). If we use the solution f(x) = a*x^b as you have done in your video, then it is not hard to show that f^(n)(x) = (a*Gamma(b+1)/Gamma(b-n+1))*x^(b-n), and f^(-1)(x) = a^(-1/b)*x^(1/b) as before. I use the Gamma function for a possible fractional derivative generalization :). Equating the exponents of x gives us the equation b - n = 1/b, or b^2 - n*b - 1 = 0, which is precisely the family of quadratic equations which gives us the metallic ratios! For example, n = 1 leads us to the golden ratio and 1 minus the golden ratio as possible exponents, as you have derived, and n = 2 (the second derivative of f is equal to the inverse of f) leads us to the silver ratio and -1/(silver ratio) as possible exponents of the solution to the differential equation!

I now solve for the constant a. We have, by equating the exponents of either side of our equation, that a*Gamma(b+1)/Gamma(b-n+1) = a^(-1/b), or = 1. b is a solution to b - n = 1/b, so 1 + 1/b = b - n + 1. This implies that = 1, or a =

I thought this was too interesting not to share :)

[pierreabbat]

You're going to have trouble raising -φ to the -φ power, because that's a negative number raised to an irrational power. You can't do it without complex numbers, but the negative reals are the branch cut of the logarithm function. x^(1-φ) has to have a branch cut somewhere, and it may not be possible to place the branch cut so that everywhere off the branch cut, the derivative equals the inverse.

Wow! The Phi! ... Nobody expects the Spanish Inquisition ... ... golden ratio ;-P

[leonardromano]

You found two special solutions. Solving the equation in the end: f"f'=1 gives rise to infinitely many solutions for f, easily found by integrating the ODE twice, resulting in two integration constants. I guess it is just nice to see how this equation is related to phi as the only monomial solutions are the ones presented by you :)

[PackSciences]

I don't understand what you wrote at 12:00
Because (f^-1)' = 1/(f' circ f^-1) = 1/(f^-1 circ f^-1) = 1/(f' circ f')
You said you erased the f because "evaluated at f inverse or something", I don't understand which steps you did at that particular moment.

[khajiit]

for the first case i like this form better:

(1/phi)^1/phi * x^phi

because

1/phi = phi-1

don't know if it makes much difference in the working out but it just looks nicer to me.

[torresfan]

Hey Dr. Peyam, love your work ! I would just like to add that we could actually a whole class of functions for the ODE given above by using some something known as Laisant's Formula, which states( under the assumption that both the original and inverse functions are differentiable) that int ( f^-1(x)) dx = xf^-1(x) - Fof(f^-1(x)) + C, where F is the integral of f(x)
Applying this to the ODE we end up with the class of functions satisfying this ODE as,
f(x)= xf^-1(x)- (int f(x) dx ) of f^-1(x) + C

[nicholaspatel]

Could we integrate with respect to x to get f(x) = int (f^-1(x)) dx then to get the integral use the substitution u=f^-1(x), followed by integration by parts in the ‘u world’ to obtain a solveable differential equation?

[Arup]

Dr. Is there any function f(x) for which f(f(x))=e^x

[yoavcarmel]

it's me! thank you for this video :)

[PeterBarnes]

You can also try to solve it using a Taylor polynomial, which, after some struggling to understand the correct application of multinomial theorem, only gives you the following equality:
Given
f(x)= {sum over 'n' from 0 to infinity}( a_n * x^n )
where f'(x) = f^-1(x) ;

x = {sum over 'n' from 0 to infinity}[ (a_n * n!)* {sum over m number of terms 'k_t' whose sum is n}( {product over 't' from 0 to m} [ (a_m ^ k_t)*(m ^ k_t)*(x ^ (k_t * (m-1))) / (k_t)! ] ) ]

You could try to divide by x, and you could try to turn the product into a wierd combination of summations and other awkward functions, but the real trouble is that you can't 'solve' for a_n, not in any intuitive way, at least using this level of calculus.

(And don't even start on convergence. Think about that to yourselves, so that I can be left out of it.)

[maxwellmogadam]

i love watching different people solve the same problem

[valeriuok]

Both solutions are complex value functions. In fact, there is no solution for real value functions. An invertible, derivable function is strictly monotonic and so is its inverse. On the other hand, the derivative must be strictly positive or negative (and so will the inverse), which means the original function will not be defined on the entire set of real numbers.

[digxx]

Interestingly there doesn't seem to be a place where to put an integration constant... Shouldn't a first order differential equation have at least one integration constant?

[aswinibanerjee]

If we take the general case and go further we can show
Here f"(x) is double derivative of f(x)

[osirisapex]

Very interesting problem! Nice job as usual! 👍

[husklyman]

is it possible that:

f^-1(x)=1/f(x)
?

[factsheet]

Very nice video with a cute solution as always :)