A Quintic Functional Equation Maybe

[@MrGeorge1896]

Second method but a little bit easier: We will use both y = x + 1/x and x³ + 1/x³ = y³ - 3y as shown in the video but also the binomial formula for n=5: y⁵ = (x + 1/x)⁵ = x⁵ + 1/x⁵ + 5(x³ + 1/x³) + 10(x + 1/x) so x⁵ + 1/x⁵ = y⁵ -5 (y³ - 3y) - 10y = y⁵ - 5y³ +15y - 10y = y⁵ - 5y³ + 5y.

[@barberickarc3460]

x + 1/x has so many useful properties huh, maybe do a video on polynomials with symmetric coefficients? Could be a lecture could just be a regular video. I remember solving a problem on aplusbi with this method, the z = z ^11 one i think

[@phill3986]

AKA palindrome polynomials

[@mcwulf25]

Yeah he gave us an 11th power!

[@mystychief]

In the solution f(0)=0, but in the first equation f(0) is not attainable. x+1/x cannot even be between -2 and +2 in the real world. I suppose it's not a problem.

[@HarmonicEpsilonDelta]

The solution should have included “for any x in the range of the function of g(y)=y+1/y we have f(x)=…” it does not seem that bad for me

[@samueldeandrade8535]

Your observation doesn't make much sense. Given a functional equation, F(a(x)) = b(x) the domain of a(x) may have NOTHING to do with the domain of F(x). There is no reason to talk about it based on the domain of a(x). One expected condition is just domain of F(x) must contain the range of a(x) I don't know why there is always someone that care about details like this.

[@samueldeandrade8535]

​​@@HarmonicEpsilonDelta indeed, the solution has to include your observation. For example, F(x²)=x² All this equation is saying is that F(x)=x, for x≥0 Both functions F(x)=x, F(x)=|x| satisfy the given equation.

[@sevenpheonix3293]

I believe this problem can be generally solved as f(x + 1/x) = x^n + 1/x^n for any positive integer n with the Chebyeshev polynomials. More specifically, the solution is f(x) = 2Tn(x/2), where Tn is the nth Chebyeshev polynomial of the first kind. this could probably be proved with a polar form complex substitution & de moivre’s formula, and in fact i reckon a similar relation could be obtained for f(x - 1/x) = x^n - 1/x^n!

[@derwolf7810]

Alternatively using the binomial formula for n-th grade, completing the hypercube (of grades 5 and 3; similar to completing the square), and fiirst reparameterize f to replace x with y: ==> f(y + 1/y) := y^5 + 1/y^5 x := y + 1/y ==> f(x) = f(y + 1/y) = y^5 + 1/y^5 = (y^5 + 5 y^3 + 10 y + 10 1/y + 5 1/y^3 + 1/y^5) - (5 y^3 + 10 y + 10 1/y + 5 1/y^3) = (y + 1/y)^5 - 5(y^3 + 2 y + 2 1/y + 1/y^3) = (y + 1/y)^5 - 5(y^3 + 3 y + 3 1/y + 1/y^3) + 5 (y + 1/y) = (y + 1/y)^5 - 5 (y + 1/y)^3 + 5 (y + 1/y) = x^5 - 5 x^3 + 5 x

[@mcwulf25]

I prefer the first two methods. There's something nice about working with a variable plus its reciprocal.

[@mystychief]

The function with argument plus or minus the golden ratio and plus or minus 1/(the golden ratio) give maxima and minima (4 total).

[@Foamea45]

10:44 it is( 2/(z+(z^2-4)^1/2)^5

[@Algorithmic26]

use binomial expand

[@phill3986]

😊🎉😊👍👍👍😊🎉😊

[@Jerry_Ntk]

Someone please explain to me why f(y) is equal to f(x) even though we have y=x+(1/x)

[@lagomoof]

One letter has been substituted for another. If we take the discovered function f(y) and then reset y to be x instead of x+1/x, we'll get the same discovered function but with all y's changed to x's. If we pull some other variable out of the ether, let's say w, and set the y in f(y) to be equal to w, we get f(w) which would be the discovered function with all y's changed to w's etc. so if we do that with x, we'll get f(x).

[@mcwulf25]

It's not the same x. You can put anything in the parentheses. f(😊) = 😊^2 - 5😊 + 5

[@samueldeandrade8535]

By abuse of notation. Which actually causes mistakes. For example, take the functional equation F(x²)=x² If you just change variables, y=x², you get F(y)=y So, can we conclude the solution is F(x)=x? The answer is NO. the given equation just implies F(x)=x, for x≥0 How F is defined for x

[@samueldeandrade8535]

Just to give another explanation, if we have F(y(x)) = g(y) it is WRONG to write F(x) = g(x) This only makes sense if the funtion/variable y = y(x): U→R is surjective Then you can use any letter that you want instead of y. But, if y=y(x) is not surjective, if its range misses any point, you can't.

[@lagomoof]

@@samueldeandrade8535 Substituting one letter for another, provided the incoming letter isn't already in use anywhere in a formula, is nowhere near as bad as putting a formula in its place. It's only a letter swap here and that can't run into the same problem(s) you're describing.

[@cosmolbfu67]

x^5 + 1/(x^5) = (x+1/x)^5 -5((x+1/x)^3 -3(x+1/x)) -10(x+1/x) So f(x) = x^5 -5(x^3 -3x) -10x = x^5 -5(x^3) +5x

[@rob876]

x + 1/x = y x^2 + 1/x^2 = y^2 - 2 (x + 1/x)(x^2 + 1/x^2)^2 = x^3 + 1/x^3 + x + 1/x => y(y^2 - 2) = x^3 + 1/x^3 + y => x^3 + 1/x^3 = y^3 - 3y x^4 + 1/x^4 = y^4 - 4y^2 + 2 (x + 1/x)(x^4 + 1/x^4) = x^5 + 1/x^5 + x^3 + 1/x^3 => y(y^4 - 4y^2 + 2) = x^5 + 1/x^5 + y^3 - 3y => x^5 + 1/x^5 = y^5 - 5y^3 + 5y so f(y) = y^5 - 5y^3 + 5y and hence f(x) = x^5 - 5x^3 + 5x