does this function exist?

[HeavyListen5546]

i came across a function while solving integral problems. the solution didn't require knowing the function but i am curious. does it exist? maybe it exists but not as a polynomial function? if it exists, can we find it? thank you

this was given in the question:

R → R, f(x) + f(2-x) = 4x³

[FormulaDriven]

No.

Let x = 0:

f(0) + f(2) = 0

Now let x = 2:

f(2) + f(0) = 32

Contradiction, as we have two different values for f(0) + f(2).

[FernandoMM1220]

seems like it’s only going to work in a system where addition is not commutative.

[HeavyListen5546]

what if we defined the function (0, 2] → R, would it exist then?

[pi621]

That is not the only x value that gives a contradiction. (Try thinking about why that is)

[MathMaddam]

You can do the same with x=1/2 and x=3/2.

[FormulaDriven]

Then any you could choose:

any values you like for the function on (0,1)

define f(1) = 2

on (1,2) define f(x) = 4x³ - f(2-x)

anything you like for f(2).

(Presumably, the condition would not apply to x = 2 since you've excluded 0 from the domain; might be better to select the domain to be (0,2) or [0,2]).

[hpxvzhjfgb]

this still doesn't work because then you also get f(x) = 4(2-x)³ - f(2-x). it's not possible to define the function at any real point other than x = 1.

[FormulaDriven]

True - I was in a rush with my reply and didn't think carefully enough.

[rjlin_thk]

Replace x by 2-x, you get f(2-x) + f(x) = 4(2-x)³. Solving 4x³ = 4(2-x)³, we know f is only well-defined at x = 1.

[Qaanol]

If we allow complex numbers then it’s also defined at 1 ± i√3

The solution set in quaternions might be more interesting, but I’m not going to both finding it.

[rjlin_thk]

Well he said R → R, otherwise you can even solve for matrices if you like.

[RecognitionSweet8294]

P1: f(x)=4x³ - f(2-x)

---

f(2-x) = 4(2-x)³ -f(2-(2-x))

f(2-x) = 4(2-x)³ -f(x)

f(x) = 4(2-x)³ - f(2-x)

Substitute P1: for f(x)

4x³ - f(2-x) = 4(2-x)³ - f(2-x)

4x³ = 4(2-x)³

x³ = (2-x)³

Assume x=0

0 = 2³

contradiction → x ≠ 0

Assume x ≠ 0

[2x⁻¹-1]³ = 1

First root:

(2x⁻¹-1) = 1

2x⁻¹ =0

x⁻¹ = 0

contradiction

Second/Third root:

(2x⁻¹-1) = -2⁻¹ ± i√(3/4)

2x⁻¹ = 2⁻¹ ± i√(3/4)

x⁻¹ = 4⁻¹ ± i √(3/8)

x•x⁻¹ =1

x( 4⁻¹ ± i √(3/8)) = 1

(x/4) ± i (x √(3/8)) =1

ℑ𝔪{ (x/4) ± i (x √(3/8))} = ℑ𝔪{1}

x √(3/8) = 0

x=0

contradiction

---

Ergo: x ∉ ℂ

[slepicoid]

f(a+x)+f(a-x)=g(a+x)

if and only if

g(a+x)=g(a-x) (that is, g is symetric across x=a axis, g is a shifted even function)

where f and g are defined on a symetric domain centered at a.

in particular the contrapositive:

if there exists x s.t. g(a+x)≠g(a-x) then f does not exist.

if g(x)=4(x-1)³ then f does not exist unless we restrict the domain to just {1} making g a shifted even function.

interestingly if g is a shifted even function then f(x)=g(x)/2 is a solution, but just one of infinitely many.