I Solved A Functional Equation

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[@ferlywahyu342]

What's math editor you used bro ?

[@user-kp2rd5qv8g]

Using plane polar coordinates, we get f(r) = f(x) f( y). So, obviously f is an exponential function as the product of two fs returns an f. Since the RHS should be independent of theta, f(x) should be e^kx^2, which makes the RHS e^k(x^2+y^2) = e^kr^2 = f(r). So, f(x) = e^kx^2.

[@user-sq4cd9od5m]

very nice 👌

[@phill3986]

👍👍👍

[@go_gazelle]

What exactly is k?
What does f(y) equal?
Why not check to see if:
f( sqrt(x² + y²) ) is in fact equal to whatever f(y) is multiplied by your solution for f(x)?

[@YouTube_username_not_found]

One can prove that f should be even if one includes negative values.

[@Khashayarissi-ob4yj]

With luck and With regards

[@Blaqjaqshellaq]

No matter the value of k, f(0) will be 1!

[@user-gq4vk8iy3c]

No we can't define f(0) as f:R+→R+

[@kikilolo6771]

6:32 no it is false. The function must have another condition like continuous or monotone. And if that is the case, you MUST sepcify it at the begining of the video because it is the same as defining a system of 2 equations and randomly add a third one out of nowhere. Moreover the ensemble of the function isnt somethong anecdotical as you treat it in most of your videos. I'm sorry but I'm really upset.

[@MrGeorge1896]

Starting with looking for trivial solutions we get 3 possibilities:
f(x) = 0
f(x) = 1
f(x) =
1 if x equal to zero
0 if x not equal to zero
But what about non-trival solutions?
with y = 0: f(|x|) = f(x) f(0)
Two important observations:
f(0) = 1 and f(x) = f(-x)
Now we set x = y and get
f(√2 x) = f²(x)
f(2 x) = f²(√2 x) = f⁴(x)
f(4 x) = f ^ 16 (x)
f(8 x) = f ^ 64 (x) and so on...
if we set f(1) = c we can see that f(x) = c ^ (x²) satisfies this sequence.