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

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[@maths_505]

15 percent off everything using the code MATHS505 on Advanced MathWear: my-store-ef6c0f.creator-spring.com/ Complex analysis lectures: kzbin.info/aero/PLVkOfIPb514EP3CjWQQ-JmKpIiNoEUS0k&si=rdmwOkQ6Vxg64yrE If you like the videos and would like to support the channel: www.patreon.com/Maths505 You can follow me on Instagram for write ups that come in handy for my videos: instagram.com/maths.505?igshid=MzRlODBiNWFlZA== My LinkedIn: www.linkedin.com/in/kamaal-mirza-86b380252

[@tiagobeaulieu1745]

Know what's even crazier? The function (with the non-complex parameters) and its inverse / derivative intersect at phi! But wait, there's more... when you take the area between the function and its inverse from 0 to phi, you get... 1/phi !!! Phi doesn't often show up, but when it appears in a problem, it's everywhere!

[@Cheesy_33]

well, when you're looking for something that much, of course you're going to find it

[@isaacpianos5208]

When it rains it pours

[@kikivoorburg]

@@Cheesy_33 I’m usually on board with the “phi isn’t everywhere” talk, but given that it plays a role in the solution, why is it strange that it would also show up in intersections and integrals? Edit: I’d left a comment about beta being phi, only to realise the video already brought that up. Whoops! Removed my comment and edited this one

[@hqTheToaster]

Ah, yes, Phi, the Tomas the Tank Engine of math.

[@penguin9257]

cancel the f(x) on both sides we get ' = ^-1

[@drslyone]

Unrelated to your problem, but I'll give a fun fact anyway. There are about 1.61 km in a mile, and so 1/1.61 ~ 0.62 miles for 1 km. Since this number is close to the golden ratio, it almost has the 1/phi = phi - 1 property.

[@BridgeBum]

I have used the Fibonacci sequence (and therefore Phi) as a mental shortcut way of translating km to MI and vice versa for years, it's actually a nice practical application based on this coincidence.😊

[@kikones34]

@@BridgeBum But then what do you do if you need to convert a number that's not in the Fibonacci sequence? :P

[@BridgeBum]

@@kikones34 I estimate with something near or multiply by a constant. For example 10 isn't in the sequence but 5 is, so 5 to 8 becomes 10 to 16.

[@lacryman5541]

A plot of the graph would be nice and a plot of the derivative in order to check that they are symmetrical relatively to y=x

[@michaelzumpano7318]

That was really neat. I would have tried to make headway with a gamma, matrix or series solution to the problem, but your choice of a power function really made it simpler. I really look forward to your videos.

[@IndranilBiswas_]

Great video. Really liked the use of power function. The fact that Golden ratio appears here points to all the intricate connected-ness of math ♥️

[@maths_505]

Indeed

[@MathPhysicsEngineering]

If you liked this video sure you will like my channel as well I have lots of similar content

[@edmundwoolliams1240]

Are there any other solutions? If so, perhaps you could define a function that satifies that differential equation and do a video deriving its properties and do a graph sketch?

[@Avgur_Smile]

There is a very stange explanation in video. Decision can be found in a very simple way. There is no the initial condition in the task. So decision can be found up to an indefinite constant df/dx = f^{-1} => f * df = dx => 1/2 * d(f^2) = dx => 1/2 * f^2 = x + C, where C is indefinite constant => f = \pm \sqrt{x + C} Here I have used annotations and symbols from LaTeX: \pm is plus-minus, \sqrt is square root. That's all.

[@hansgerhard7364]

f^{-1} is the inverse function, not just 1/f(x). Your solution doesn't fulfill the equation. The derivative of sqrt(x) is 1/(2 sqrt(x)), while the inverse function of sqrt(x) is x^2.@@Avgur_Smile

[@Yusuf-cg2zv]

Let g equal to inverse of f(x). So derivative of inverse g equal to g. With Using the inverse derivative property you can show these are the only solutions

[@illumexhisoka6181]

A few days ago I solved a similar equation when I was bored during a lecture what is a function that the drivative of the inverse is the inverse of the drivative And I don't need to tell you that it got too complex

[@maths_505]

Here so that no one can say third

[@vybs9235]

Imagine this being a chain

[@illumexhisoka6181]

@@vybs9235 sorry I Broke it

[@suey1690]

117th

[@vybs9235]

@@illumexhisoka6181 yeah dude

[@peterpan1886]

Third. Haha, I defy your rules.

[@jmdawlat]

Great video! It would have been even better if you had shown a plot of the function, and explored whether there were any interesting geometric consequences.

[@padraiggluck2980]

From having encountered this equation so many times as soon as I see x^2-x-1=0 I know we have the golden ration.

[@ingenuity23-yg4ev]

i give you 5/phi stars for this video😂! Beautiful solution development, loved how phi was just everywhere in this question

[@blacklistnr1]

While I like this solution and how phi popups up! Based on the title I wanted an exploration of the derivative=inverse space not just guess a form and tactically solve it. This is like posting a "Trip to the Alps" video with pictures of just your feet on the ground!

[@richardsayfer3979]

Great video! And the solution is beautiful as always❤

[@Avgur_Smile]

I think author should study roots of calculus. It's a very stange explanation. Decision can be found in very simple way. There is no the initial condition in the task. So decision can be found up to an indefinite constant df/dx = f^{-1} => f * df = dx => 1/2 * d(f^2) = dx => 1/2 * f^2 = x + C, where C is indefinite constant => f = \pm \sqrt{x + C} Here I have used symbols from LaTeX: \pm is plus-minus, \sqrt is square root. That's all.

[@smaari]

Thank you for such beautiful question and solution. I placed your solution on desmos and the only thing I would add to it is that for the inverse function x^(1/phi)*(phi-1)^((1-phi)/phi), x>=0. Thank you again, it is a fantastic video.

[@Avgur_Smile]

It's a very stange explanation. Decision can be found in very simple way. There is no the initial condition in the task. So decision can be found up to an indefinite constant df/dx = f^{-1} => f * df = dx => 1/2 * d(f^2) = dx => 1/2 * f^2 = x + C, where C is indefinite constant => f = \pm \sqrt{x + C} Here I have used symbols from LaTeX: \pm is plus-minus, \sqrt is square root. That's all.

[@MrWael1970]

It is very interesting. Thank you

[@ilyacherel5852]

functions f and g defined on ]-k/2 ; +∞] by f(x) = sqrt(2x + k) and g(x) = -sqrt(2x+k) with k in R also work

[@Toto-cm5ux]

It's not all the set of solutions but it's very cool!

[@abdulllllahhh]

this exact problem popped up in my head while shitting earlier, what a coincidence but also thank u for doing this for me

[@saraandsammyb.9599]

I loved this!!! Is there anyway that you can make a lecture series explaining the content like those crazy integrals with beta functions and crazy series'?

[@maths_505]

That's pretty much the whole channel 😂

[@saraandsammyb.9599]

@@maths_505 no I know you explain them but for people who don't know all of that math. Like for instance I know up to calculus 2 but I cant understand most of the stuff even though it looks like I should. Because we never learned stuff like the beta function, euler macheroni constant etc.., But I mean like a course explaining stuff that isnt so trivial like above the calc 2 level? or do you have any ideas for content I could watch to learn that stuff to be able to understand most of your stuff?

[@Cybrtronlazr]

​@saraandsammyb.9599 the course after calc 2, multivariable calculus (calc 3) doesn’t cover any of that stuff either, so I am interested in that stuff, too. Reply if you find anything good!

[@connorkurrack265]

look for uni level advanced calc@@Cybrtronlazr

[@kowalguitar]

What is the program?

[@meraldlag4336]

Great video

[@hqTheToaster]

What about where the derivative of the function f of (x squared) equals one over (f of (square root of x) + conjugate of f of (square root of x))?

[@DavidMFChapman]

Fascinating! I collect problems where GR crops up.

[@antonk.653]

This is so strange. It almost satisfies a harmonic oscillation, because if you just changed f(x) so that its 2nd derivative f ' ' (x) yields -f(x) by adding a complex exponent somewhere, you could solve ordinary differential equations like this.

[@drdca8263]

Thinking about how to try to find all possible solutions: from f’ = f^{-1} f’’ = (f^{-1})’ and (f^{-1})’(x) = 1/(f’(f^{-1}(x))) applying the original condition, this is 1/(f’(f’(x))) And, I suppose we could continue in this way, to find f’’’ in terms of f’, And generally find f^{(n)} in terms of f’ . Hm. Could this let us obtain constraints on a power series for f? If f has a fixed point, then f inverse also has a fixed point, and so f’ has a fixed point.

[@YouTube_username_not_found]

Did you find anything new?

[@rounaksarkar6493]

How do we know that there are no other solution? And if there are other solutions then what are some examples?

[@marvin.marciano]

√2x has 1/√2x as inverse and 1/√2x as derivative

[@SuperSilver316]

I haven’t looked at this too deeply, but is a linear combination of solutions also a solution? No right cause this problem is nonlinear, and that breaks the superposition principle. Also the inverse of that linear combination might not equal its derivative, but I would have to look more deeply.

[@maths_505]

Yeah the linear combination is definitely not gonna work and because of the same reason, the complex conjugate of the 2nd solution.

[@mohamednejighnimi4860]

the superposition principle applies only to linear systems

[@jordanraddick505]

I was going to ask how we knew that this was the only solution, but you answered at the end that we don't. But can we even know the full set of solutions? Isn't it possible that between one and countably infinite weirdboi special functions or noncomputable functions also have this property? Or am I missing something?

[@OmnipotentEntity]

Because the second solution is complex exponential, which is defined in terms of the complex logarithm, you have countably infinite solutions from the second solution.

[@OmnipotentEntity]

Excuse me, not complex exponential, but *has* a complex exponential.

[@mathunt1130]

You started off with an ansatz and played around with it. What you could have done is differentiated it again, to obtain a second-order differential equation without the inverse function, and investigated that.

[@TheEternalVortex42]

How does differentiation remove the inverse function?

[@mathunt1130]

@@TheEternalVortex42 write y=f^{-1}(x). Take f() of both sides and use implicit differentiation to obtain f'(y)dy/dx=1

[@s.p.a.3583]

Can u try to solve, the derivate of the inverse = the inverse of the derivate?

[@ajamessssss]

Nice!

[@math_qz_2]

Good video

[@Halleluyah83]

Найти такую функцию, в каждой точке точке области её определения производная в данной точке равна обратной функции в этой точке.

[@Jet-Pack]

Cool but how does the plot of the function look like?

[@estebanrodriguez5409]

Graphs at the end would have been a good idea

[@josepherhardt164]

Two beads of mustard per hot dog. That's the Gulden Ratio.

[@terdragontra8900]

theres almost definitely a one parameter family of solutions, but my guess is the others dont have a nice form

[@plebeian_]

mb solve f'' * f'=1

[@gagadaddy8713]

f(x)=1/3sqrt(2x+a)^(3/2)+b

[@banrtv9446]

why doesnt e^x work for this?

[@chrislynch6545]

He mentions it but the inverse of e^x is log x. You might be thinking of the antiderivative

[@newwaveinfantry8362]

How can you know that only a polynomial can solve this?

[@maths_505]

Other solutions are welcome if you can find em. But I think that's unlikely; differentiating the equation again gives a 2nd order DE so I don't think the given DE will have more solutions.

[@sonobox-lu6mr]

It's not a polynomial, it's a power function.

[@burakmuezzinoglu3730]

Reis yargı dağıtmışsın, şiir gibi

[@abhrakantasamui4344]

Michael Penn has done it few years back 😅 but okay... what is your favorite function u have every seen or wonder about ...

[@maths_505]

Just did some KZbin hunting. Turns out, Dr Peyam was the one who first made a video on this DE so he beat us both to it.

[@TheEternalVortex42]

kzbin.info/www/bejne/ZnrPiKx_lrKLp6s

[@jonathanv.hoffmann3089]

🙏🙏🙏

[@vybs9235]

Here so that no one can say first

[@pacotaco1246]

That wont stop people with fingers and a disregard for sequential order

[@vybs9235]

@@pacotaco1246 well the time helps I guess plus it's a joke so who cares LoL

[@user-ky4qs2ib2q]

​@@pacotaco1246 first 🥇

[@pacotaco1246]

@@user-ky4qs2ib2q the professy is fufilled. Huzzah!

[@ben_adel3437]

I tried to write a function that is ax+b=f(x) And then took the inverse which is (x-b)/a and then said this should equal the derivative which is just a And then wrote it like (x-b)/a=a And then found out that a²=x-b and then a=±√(x-b) therefore any ±√(x-b)x+b would satisfy this equation idk really know how derivatives work and where i made a wrong thing

[@simeonsurfer5868]

It won't work well because a dépend of x, wich means that the derivative must include the derivative of a, and you don't do that here.

[@pooydragon5398]

Your derivative is a constant where as your inverse is a linear function. Those two cannot be equal.

[@angelommv]

' = -1

[@notnot4396]

I’ve never heard anyone say “a by b” to mean “a divided by b”, I would always hear/say “a over b” and would interpret “a by b” to mean “a times b”. Really threw me off

[@ManhNguyen-oy2zg]

Cool

[@MrGyulaBacsi]

you showed one solution. could you prove that there are no other solutions?

[@taterpun6211]

Here so that no one can say 335th

[@maths_505]

😂😂😂

[@TanmaY_Integrates]

By profession are you a teacher or professor or PhD student!?

[@maths_505]

Full time KZbinr and masters student.

[@TanmaY_Integrates]

@@maths_505 yeah I knew it! You shall make a video on your PhD research. Is masters and PhD same?

[@maths_505]

@@TanmaY_Integrates I'll enroll in a PhD program once I'm done with my masters. And yes I'll make videos on what I'll be studying there.

[@maths_505]

@@TanmaY_Integrates they are different programs

[@ArminVollmer]

That topic had Michael Penn first.

[@mart4640]

Indeed though, it was incredibly cool when I first saw it there!

[@maths_505]

Did some searching on KZbin. Turns out, Dr Peyam was the first to make a video on this particular DE. So he beat us both to it 😂 honestly not surprised; Peyam is definitely the best math KZbinr alongside 3b1b.

[@davidseed2939]

draw β more clearly… difficult to follow also your φ looks like ρ

[@worldOFfans]

this problem is overPHIlled

[@felipegiglio2047]

what kind of solution is this? you didnt prove those are the only solutions to the problem, which is supposed to be the hard part

[@chonpincher]

It was good enough by 5:40.

[@justrandomthings8158]

Here so that no one can say second

[@vybs9235]

Lmao 😂😂

[@paulor.r.correia1789]

👏👏👏👏👏🇧🇷🇧🇷🇧🇷

[@justinbrentwood1299]

There's also f(x)=0

[@deweiter]

No, there isn't, because if f(x)=0, then inverse of this function doesn't exist

[@justinbrentwood1299]

@@deweiter I'm pretty sure it does. The idea behind an inverse is that the ordered pairs of the function are swapped. For example, for f(x)=x+5, (0,5), (1,6), and (2,7) are ordered pairs. For it's inverse, the ordered pairs would be (5,0), (6,1), and (7,2). And of you find the inverse (f^-1(x)=x-5), you can even verify it. So for f(x)=0, its ordered pairs are (C, 0), for all real numbers C. Its inverse, therefore, has ordered pairs of (0, C). It's inverse isn't a function, and it's only defined on the domain of {0}, but it exists. On 2nd thought though, f^-1(x) can equal something other than f'(x), so it doesn't satisfy the problem.

[@bjornfeuerbacher5514]

@@justinbrentwood1299 "It's inverse isn't a function" Indeed. And hence it's irrelevant for the problem here, since when solving differential equations, one always looks for functions as solutions.

[@justinbrentwood1299]

@@bjornfeuerbacher5514 The answer to differential equations can be non functions. For example, dy/dx = 3x^2 / (2y) ... y^2 = x^3 - 8

[@bjornfeuerbacher5514]

@@justinbrentwood1299 "The answer to differential equations can be non functions." Since when? In all books I've ever seen, it was always stated that the solutions of differential equations are functions. y^2 = x^3 - 8 yields _two_ functions as solutions of your differential equation: y = +sqrt(x³ - 8) and y = -sqrt(x³ - 8).

[@trmbrnoir]

Third

[@rennoc6478]

So beautiful

[@gdzephyriac2766]

Here so that no one can say i:th

[@maths_505]

Legend🔥

[@dylanbyrne8478]

phi count was atleasst 500

[@illumexhisoka6181]

Forth

[@fionnohora3724]

This is copied straight from Michael penns video

[@borhen-di6ik]

we can take a counter example, like f(x)= ln(x) and we know f '(x) = 1/x and f ^(-1) (x) = exp(x) than 1/x = exp(x) isn't true because if x=1 than 1 = exp(1) it's impossible than f '(x) = f ^(-1) (x) not true . My name is Borhane Eddine Bouchniba and I Study applied mathematics in Tunisia in Faculty of Sciences of Monastir, Thank you so Much

[@JoseSanchezLopez-yf3lo]

my dude your reasoning only shows that the function YOU chose is not a solution. that doesn't mean there can't be any (there are at least two as the video shows)

[@borhen-di6ik]

I mean it's wrong in general case or it is true with a condition, but what is the condition ?

[@hamzaiqbal7178]

Here so no one can say third

[@illumexhisoka6181]

Someone already said that

[@maths_505]

Already taken

[@maths_505]

And 4th is occupied too so better be quick😂

[@Avgur_Smile]

It's a very stange explanation. Decision can be found in very simple way. There is no the initial condition in the task. So decision can be found up to an indefinite constant df/dx = f^{-1} => f * df = dx => 1/2 * d(f^2) = dx => 1/2 * f^2 = x + C, where C is indefinite constant => f = \pm \sqrt{x + C} Here I have used symbols from LaTeX: \pm is plus-minus, \sqrt is square root. That's all.

[@markjeeninga6024]

You consider the multiplicative inverse. The problem considers the functional inverse. You forgot the factor 2, but your solution for the multiplicative inverse-problem is otherwise correct.

[@misterenter-iz7rz]

g(x)=f^(-1)(x), f'(x)=g(x), x-y-z, g(f(x))=x, g'(f(x))f'(x)=1, g'(f(x))g(x)=1, g'(f(x))=1/g(x), g'(g^(-1)(x))=1/g(x), .....😅

[@user-kv9ru1wf5l]

A very big nothing

[@mmmmmmok5292]

i lose 1 braincell every time he pronounces φ as Faee

[@Vijay1989]

Lousy presentation with barely legible handwriting

[@user-sr1he9fp1s]

the fact that you used beta and then changed it to phi :(

[@plebeian_]

mb solve f'' * f'=1

[@probiem]

We cannot write that equation by differentiating both sides wrt x. Because derivative of f inverse wrt y is equal to 1 over derivative of f wrt x. So independent and dependent variables are different.