the greatest functional equation of all time.

🌟Support the channel🌟

🌟my other channels🌟

🌟My Links🌟

🌟How I make Thumbnails🌟

🌟Suggest a problem🌟

[HagenvonEitzen]

Another idea: From the case x=y, we see that xf(x) is a fixpoint. In particular, f has *at least* one fixpoint. What if we plug in two fixpoints x and y? Then x² (f(x)+f(y))=(x+y) f(f(x)y) becomes f(xy)=x² and by symmetry also =y², in other words x=y, i.e., f has *exactly* one fixpoint. Then xf(x) must be constant, i.e., f(x)=c/x. Plug this in and solve x²(c/x+c/y)=(x+y)c/(yc/x) for c

[goodplacetostop]

7:35 Homework
7:46 To je res dober kraj za ustavitev… Btw, one of my friends spent one year at University of Ljubljana. She liked the country but she had some health issues which impacted her grades.

[-P]

Watching Janja Garmbret is just so motivating, all in all the bouldering competitions are my favourite sport events to follow. I hope she comes back as well as the YT livestreams..

[DenisBencic]

Greetings from Slovenia :) So glad to hear you include Janja Garnbret!

[mrminer]

7:30 When fixing one variable but not the other, it is necessary to say, " . . . but Y remains wild and free."

[danfoster]

Here's what's amusing. If you add 0 to the domain and range, the the function f(x)=0 is the only solution (fairly easily proved, by letting x=1 and y=0 and following your nose). Very strange, that the addition of one point to the domain and range can cause the functional equation to be satisfied by something completely different. [Also: If you only add 0 to the range, then the function f=0 is a second solution; if you only add 0 to the domain, then there is no solution. Very weird.]

[aweebthatlovesmath]

I love functional equation. Especially when professor penn uploads them!

[jacemandt]

2:02 Helpful tip just in case anyone thought that maybe 2=0 😄

[faisalal-faisal]

My approach: from x=y, we find that f(xf(x))=xf(x), i.e. xf(x) is a fixed point. Special case: f(f(1))=f(1). Fixed points are sometimes useful in solving functional equations. So let's suppose we consider an arbitrary fixed point z. Plug x=z(=f(z)) and y=1 into the equation to get z^2(z+f(1))=(z+1)z => z^2+(f(1)-1)z-1=0.

This last equation is true for ALL fixed points of f, so in particular for z=f(1). Plug that in to find f(1)=1 (using the fact that f(1)>0). But then the equation becomes z^2-1=0, hence z=1 (because z>0). Using now the fixed point z=xf(x), we get f(x)=1/x. And indeed this is a solution to the functional equation.

[skerJG]

Here's another fun way to do it! Suppose there exists an a such that f(a) = 1. Setting x = a gives a^2(1+f(y)) = (y + a)f(y) and we can solve for f(y) = a^2/(y - a^2 + a).

Then setting x = y (like you did) shows xf(x) is a fixed point, which means x = 1 is a fixed point, so f(1) = 1 = a^2/(1 - a^2 + a), so 1 - a^2 + a = a^2, which gives the quadratic you got (but with a the input instead of the output!) 2a^2 - a - 1 = 0, which gives a = 1.

[myrus]

When is the collab with Janna Garmbret coming?

[potawatomi]

You’re amazing, Michael. Always love watching your videos.

[varegulDac]

i absolutely love when you do functional equations and integrals 💞💞

[MDNQ-udty]

by symmetry/commutativity: x^2*f(f(y)x) = y^2*f(f(x)y)

This suggests f(x) = 1/x because

(x/y)^2 * f(f(y)x) = f(f(x)y)

if we invert x and y to get

(y/x)^2 * f(f(1/y)/x) = f(f(1/x)/y)

Hence if f inverts then it will simply invert inversions(reciprocals) and all that matters is that it all works out which it does. One can almost read off this solution after reducing by symmetry. It isn't very rigorous and requires a bit of insight but it's quick. If you want you can multiple the last two together to get

f(f(y)x) * f(f(1/y)/x) = f(f(1/x)/y) * f(f(x)y)

And this is more obvious that inversion is a symmetry in the functional equation(if f is a homomorphism then it will have to carry inverses to inverses). This doesn't give us all functions but it's another quick and dirty approach.

[vizart]

I was surprised to see the result. This ones different yet still simple.

[abrahammekonnen]

I'm finally starting to get how to approach functional equations!

Makes me wonder if it's even possible to have a sort of unified theory of functional equations. (If that even makes any sense.)

[deslomator]

I' m used to seeing Janja in thumbnails, but it took me some time to realize this wasn't a World Cup video

[mrfork]

So (0, oo) means that 0 is not included. Sorry i didn't know the american/english notation for interval ( in french [ to include, ] to exclude left bound).
Thanks for your quick answer.

[sprk]

Didn't expect to hear Janja get mentioned on this channel lol, though I thought i heard she took the rest of the season off to project La Dura Dura.

[CharnelMouse]

Some notes from solving before playing the video, so I didn't know f was only on (0, infinity):
Given the presence of f(af(b)), I went straight to looking for a monomial solution. The presence of an x+y term means x and y have the same dimension. Let the dimension of x and y be [x], and [f(x)] = [x]^a. Plugging this into the equation gives a+2 = 1+a(a+1), or a = 1 or -1. We therefore know we're looking at the form f(x) = Ax + B/x. The cases x=y=1 and x=y=i give B+A and B-A as fixed points, and A=0, B=1 follow as the only monomial solution. Not a thorough proof, but it works on C\{0}.