Solving A Nice Functional Equation

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[ShortsOfSyber]

I really like this solution from a colleague on twitter:

[xiangge]

A simpler solution: since x^4+1/x^4 = (x^2+1/x^2)^2 - 2 = ((x+1/x)^2 - 2)^2 - 2; we can get f(x) = (x^2 - 2)^2 - 2 right away. Simplify the right-hand side, we can get f(x) = x^4 -4x^2 +2

[popitripodi]

I solved it using the first method!!! Nice and easy❤❤❤❤thank you!!!

[Cristian_AQ_EV]

It's easy. I see 2 ways of solving this:
1). We express (x^4 +1/x^4) in terms of (x + 1/x)

2). Let (x + 1/x) = t, then we find x in terms of t.

[kingoreo]

I don’t think i can ever agree with the whole “2nd method first” thing. You thought of it second, yes, but that order is arbitrary. You could have thought of anything first. Indexing by appearance rather than ideation seems so much more logical to me.

(Love the videos by the way)

[alain]

I used the first method, a little clever but so easy. The second is heavy artillery.

[PunmasterSTP]

Oh wow I had no idea about the last method you shared. That was very cool, and I’ll need to share that with my friends!

[demenion]

You would know immediately that the two solutions to x²-zx+1=0 are inverses of each other since the constant term is equal to the product of the roots. You could also see it from x+1/x=z since the equation is symmetric under replacing x <-> 1/x

[md.mustakim]

It can also be done by using square formula.

[GulibleKarma]

to figure out the function, the question to be asked is: "what did i have to do to take the input and get to the output"

for this case, how do i get x+(1/x) to x^4 + (1/x^4)?

(x+(1/x))^2 =
(x^2 + 1/(x^2) + 2) (square input)
x^2 + (1/x^2) (subtract 2 from result)
(x^2 + 1/x^2) =
x^4 + 1/(x^4) + 2 (again square result)
x^4 + 1/(x^4) (subtract 2 from result again)

putting our steps into an expression, we get:

((x^2 - 2)^2 - 2)=
(x^4 - 2x^2 + 4 - 2)
= x^4 - 2x^2 +2

f(x) = x^4 - 2x^2 + 2

[Laggron]

Things went a bit differently for me. I developed x + 1/x to the power 4:
(x + 1/x)^4 = x^4 + 4x^2 + 6 + 4x^-2 + x^-4
Rearranging: x^4 + x^-4 = (x+1/x)^4 -4x^2 -4x^-2 - 6
Developing this on the side: (x + 1/x)^2 = x^2 + 2 + x^-2
Then I factorize my previous expression: x^4 + x^-4 = (x + 1/x)^4 - 4(x + 1/x)^2 + 2

[alextang]

1st method is the most suitable way to solve this kind of question because x^n+1/x^n (say f(x, n)) can be expressed in terms of x+1/x (say f(x, 1)) easily. In some books, you can also find the general formula for f(x, n) in terms of f(x, 1). 😋😋😋😋😋😋

[Blaqjaqshellaq]

The solution can also be presented as f(x)=[(x^2) - 2]^2 - 2.

[someperson]

Assuming that x belongs to Reals\{0}, the range of x + 1/x is {x| |x| >= 2, x real}. In that
case the answer should be that f(x) = x^4 - 4x^2 + 2, for |x| >= 2, where x is real, and the
value of f(x), for |x| < 2, where x is real, is undetermined from the given. On the other
hand, if x belongs to Complex\{0}, then f(z) = z^4 - 4z^2 + 2, for all complex z.

[arrrhoo]

another way is put x=tan(y), Then f(x)=f(2/sin(2y).

In right hand side equation write tan(y) in terms of 2/sin(2y).

Now substitute 2/sin(2y) as z in both LHS and RHS.

We will get the formula for f(z)

[abirmahbub]

I can solve it within 30 second 😂😂
But thanks for giving me new things 🧡

[scottleung]

I got traces of the correct answer, but with a strange remainder.

[Songvbm]

at 7:57, how z is replaced by x ? Otherwise the assumption was x+1/x=z ..

[canaDavid]

This is a bad equation. For x in range <-2, 2>, f(x) can be *literally* anything. One would have to say that x can be in C to have anything differently.

[gatosimple]

Para que me sirven y que aplicación tiene en el mundo real.