Solving a Homemade Functional Equation in Two Ways

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[aayanansari]

At first, it seemed one of those complicated problems which you show, but after factoring the terms in the RHS and simplifying the ones in the LHS, the problem became quite easy and I was able to solve it.
Btw your uploads are much appreciated 🙂

[satrajitghosh]

Writing y for ( 2x^2 + x)/4 one gets
4y = x* (2x+1)
Hereby
4*x^4 + 4*x^3 + x^2 = x^2*(2x+1)^2
= (4y)^2
Therefore f(y).= 16*y^2
That is the answer

[christianthomas]

easy: just square (x**2/2 + x/4), you got: 1/16*(4x**4 + 4x**3 +x**2) which gives: 4x**4 + 4x**3 + x**2 = 16 (x**2/2 +x4)**2 which is egual to (4(x**2/2 + x/4))**2 and by identification: f(u) = (4u)**2. therefoe: f(x) = 16x**2

[boguslawszostak]

This is easier to understand when we substitute t for x, because that's how we are used to describing parametric functions.
f ((t ^ 2 + t) / 4) = 4t ^ 4 + 4t ^ 3 + t ^ 2
we usually write functions using the notations f (x) = y
with these markings we have
x = (t ^ 2 + t) / 4; y = 4t ^ 4 + 4t ^ 3 + t ^ 2
4 x = t (2 t + 1);
y = t ^ 2 (2 t + 1) ^ 2 = (t (2 t + 1)) ^ 2

after substituting t (2 t + 1) = 4 x we get
f (x) = y = (4x) ^ 2 = 16 x ^ 2

[damiennortier]

My method is similar to your second method :
Take out 1/4 : f(1/4(2x^2 + x)) = 4x^2 + 4x^3 + x^2.
Now, notice that the left side is (2x^2 + x)^2 so we can substitute 4y = 2x^2 + x (why 4y? Because of the 1/4...)
F(y) = (4y)^2 = 16y^2

[Polpaccio]

Quite nice.
But I found the 1st method resolution too much convoluted. There's no need to calculate X^3 and X^4, because you can take out X^2 in the RHS so you just need X and X^2. Less steps means less possibilities of a mistake.

[vitowidjojo]

Love how you say that is homemade equation.

[mcwulf]

Solved in one line. y = x ^2/3 + x/4.
Multiply by 4 and square it to get RHS. (4y)^2 = 16y^2.

[moeberry]

2nd method is way faster but I appreciate the Brute Force Method too.

[scottleung]

The first method is painful, but the second one is pure genius!

[SuperYoonHo]

Gee It is awesome!!! Thanks!!! This is the best birthday ever(except the rain)

[zunaidparker]

You need to specify the domain of your function in these types of problems. By giving only the problem as described, assuming a real x, the function is not defined for most of the negative real line.

[hosugabi]

Everything is fine with this problem, but you could define the statement of the problem more precisely, I refer to the domain of f. If f:R->R, then the equation may have solutions only for x values which are in the image of x->x^2/2+x/4, the argument of f in the initial equation. For the any other value of x, f(x) is an arbitrary function, because there aren't any constraints.
It might seem trivial, but you should mention this case.

[krishangkrishna]

I kinda did what you did in the first method.
Calling (xx/2)+ (x/4)=p
I divided f(p) by p and got the 2 deg quotient and 0 remainder.
Then i divided the quotient by p and got a constant quotient.
So, f(p) became 16pp.
Ig i kinda wrote f(p) in base p

[ckdlinked]

Amazing, you are making me back to high school, what a nice really !!!

[bosorot]

3rd method is long division x^2/2 + x/4 with 4x^4+4x^3+x^2 . will get 8x^2+4x . repeat the process and get 16 .

[stevenlitvintchouk]

I did it something like the 2nd method here, except I simplified the equation immediately with substitution. From the left side, I guessed that 2*x^2 + x would turn out to be a factor of the right side. So I immediately let u = 2*x^2 + x, so the whole thing becomes f(u/4) = u^2

[RAG]

All in favour of the second method, since it involved some mathematical thinking instead of hacking it out. All one needed was to factorise the expression and get (2x^2+x)^2.

[antonyqueen]

The right side of the equation was shouting very load from the beginning “FACTORISE ME”!

[professorpoke]

I put f(x) = ax² + bx + c
And using given condition, I got
f(x) = 16x²