A fascinating differential equation: when the 2nd derivative equals the inverse function

[@lucasmichaudel1294]

This is very much incomplete as you assumed one form of solution but never showed it was the only possible one

[@wychan7574]

A general solution derived from solving the equation is not possible, so only a particular solution is conjectured and verified.

[@gnhernandez]

And there is a theorem that says that a second order differential equation has 2 independent unique solutions. So if you find them you are good

[@antarctic214]

​@@gnhernandez This does not apply here because we are not working with a linear ODE (the equation contains the inverse of the target function)

[@oscarlam5381]

Then he has to that it is not possible ​@wychan7574

[@maklovitz]

1:58 you have to also check separate cases: B=0 and B=1, because after double derivative the form wont be x^(B-2)

[@jonathanterry1983]

This is trivial: for b=0, f-inv(x) is undefined because it is a vertical line at x=a. For b=1, f-inv(x) = x/a = f''(x) = 0 x/a = 0 1/a = 0 (for all non-zero values of x) --> a is undefined

[@insane_mind]

thanks for correcting that part

[@maths_505]

Thank you for pointing it out brother. SUIIIIIIIIIIIIIIIIII

[@dexthefish96]

that was fun! your reasoning is clear and easy to follow. the whole process felt very natural.

[@Serghey_83]

Problem: y · y" = 1, where y = f(x) Solution: erfi(±√(ln(y) + α)) = √(2/π) · x · exp(α) + β, where α, β are constants & y = y(x). erfi(x) = 2 / √π · integral [0, x] exp(t²)dx

[@poorgrammar3136]

f^(-1) is an inverse function, not 1/f

[@tonypalmeri722]

I usually don't "feel satisfied" with finding the function(s) satisfying a differential equation until I plug the function(s) back into the equation and confirm. In this case, of course, evaluating the 2nd derivative, and also the inverse of the function and confirming that they are indeed, equal. The *other* thing I am always curious to see is the *graphical* interpretation of "f" and in this case it's 2nd derivative and inverse.

[@vitorbordini5246]

It would be interesting to generalize it,i.e,find a function f such that its n-th derivative is equal to its inverse

[@maths_505]

Someone has drafted a solution as part of a reply to another comment here.

[@Loyis]

Can't believe we got differential equations loss meme before GTA6

[@MathsScienceandHinduism]

This is a famous one on YT

[@MrWael1970]

Very cool. Thank you.

[@dr.aaqibiqbal2340]

Which software are you using for writing ?

[@stefanalecu9532]

Did you reupload this or what?

[@maths_505]

There was an error that needed fixing

[@worldnotworld]

Beautiful. Can we be sure these are the only solutions? I'm gonna try it with y=C1e^xC2... And it crashed in a hurry; you end up with a exponential function of x on one side and a logarithm on the other; no good at all. Still, might there be other classes of solutions? Is there any way of proving there are or there are not?

[@maths_505]

@@worldnotworld I'm not sure how to prove that the set of solutions I found is complete. But tbh I can't thing of any other functions that could work so the most rational thing to do is leave it to the viewer as an exercise 😂

[@daliasprints9798]

I would look for whether any general uniqueness theorems apply.

[@worldnotworld]

@@maths_505 I wonder whether linear combinations of the solutions you found, or more general polynomials with "weird" non-integer exponents, might work. But you're leaving this for me as an exercise, so wish me luck!

[@worldnotworld]

@@daliasprints9798 I don't know about those. Any pointers?

[@maths_505]

Perhaps weird looking functions could work. Not sure about linear combinations though since the equation is non linear.

[@Scientificus]

what about f'' = -f^(-1) ?

[@bozzisimone]

D_n(f(x))=f^(-1)(x)

[@deweiter]

Assume f(x) is equal a*x^b. That means, that D_n(f(x)) = a * (b!/(b-n)!) * x^(b-n) and f^(-1)(x) = a^(-1/b) * x^(1/b). Therefore, b - n= 1/b and a * (b!/(b-n)!)=a^(-1/b). First is quadratic equation, which gives us b = n/2 ± sqqt(n^2+4)/2. Second one gives us a^(1+1/b)=b!/(b-n)! and, therefore, a = (b!/(b-n)!)^(1-1/(b+1))) Putting all that back in the function gives us f(x) = ( (n/2 ± sqrt(n^2+4)/2)! / (-n/2 ± sqrt(n^2+4)/2)! )^(1 - 1/(1 + n/2 ± sqrt(n^2+4))) * x^(n^2 ± sqrt(n^2+4))

[@NichaelCramer]

Pretty nice, but it would be nice if there were a way to approach the solution in a more general way, for example if you didn’t know the starting facts (I.e. the facts about the characteristics of functions of power functions). But to be fair 1] I realize it’s probably reasonable to assume that anyone who would realistically be trying to solve a problem like this could be expected to know those facts, and 2] just about the only thing I remember from my differential equations class (all those many years ago), is that, once you get beyond completely trivia examples, it was almost always likely that a fair amount of clever/educated guessing would be involved…

[@aaronkaw4857]

Just rotate the negative sign.

[@aayushiajith.]

Where do u get these questions from

[@aayushiajith.]

Where do u find a lot of questions

[@r4_in_space]

Ok, math wizard.

[@rob876]

Shouldn't the solution have 2 constants in it?

[@lol1991]

Still got the dad joke, cool

[@tunistick8044]

does that mean that we have grown up? 😔😔

[@enthdegree]

Is that the only solution??

[@GeraldPreston1]

bruh seriously that was my first first comment and you had to reupload

[@maths_505]

Sorry bro there was a pretty bad error there

[@CM63_France]

Hi, I aggree, linear combination wouldn't work here. "ok, cool" : 0:53 , 1:55 , 5:09 , 6:05 , 8:49 , "terribly sorry about that" : 6:31 .

[@apnakaamkrelala]

Ohk cool feels like heaven now 😌🤌✨

[@mcalkis5771]

Hmm. For this I gotta try to find some more solutions. They gotta exist.

[@topquark22]

This is cool. But doesn't a 2nd order differential equation always have 2 free parameters (constants)?

[@thomaspeck4537]

That is generally the case, but I would not expect that to necessarily apply when the inverse function is involved.

[@CM63_France]

Hi, OK, cool! Still at hotel because of tree fall onto the rail track. Hope will be back to home tomorrow.

[@maths_505]

Have a safe trip bro

[@andreisoceanu4320]

I did not quite understand why do we assume it is a polynomial function only

[@maths_505]

The derivatives of power functions are power functions. Likewise, the inverse of a power function is a power function.

[@juniorcyans2988]

Awesome🎉🎉🎉🎉

[@NonTwinBrothers]

alpha hare

[@robertsandy3794]

Now all you need is a root beer!

[@insouciantFox]

I thought the soln to x³-x-1=0 was the silver ratio

[@maths_505]

Nah bro it's the solution to x²-2x-1=0

[@mikecaetano]

Break out the root beer!

[@Nottherealbegula4]

Rootful video

[@kingzenoiii]

oh my~

[@MatthisDayer]

' = i

[@maths_505]

Not today😂

[@Hexer1985]

So in conclusion, you just made a guess and proved that it fits the initial condition. But the more interesting thing is: How did you know that your guess will is valid solution? I mean, guessing the solution and prove that in a KZbin video makes only sense if you know that the guessed result will work... So, you should have at least a basic idea what to look for based on the initial condition. Please explain how do you get there.

[@maths_505]

Rewatch the first couple minutes of the video, I explained exactly why a power function will work.

[@GranHerrmanno]

Ok cool 😂

[@ATYPICALJEEASPIRANT]

lmao it was enjoyable

[@worldnotworld]

OK Kewl

[@Mario_Altare]

Third to like, first to see

[@maths_505]

That is quite interesting 😂

[@comdo777]

asnwer=1 isit

[@comdo777]

f=(2/-2)

[@Player_is_I]

Math on meth ✅

[@jackkalver4644]

I learned that a differential equation doesn’t have to have an infinite number of solutions.

[@enthdegree]

He didn't prove that, he made up this arbitrary family of functions and then chugged through some totally mechanical algebra to find a member of the family that solves it. Says nothing about the solutions in general

[@manansharma9164]

First

[@victorbilyk5778]

Fascinating? Are you joking? Why don't you look for the first integral? Multiply both sides of the equation by f' and reduce it to the first order one, which is easily solvable in terms of quadratures.