help me with a tough maths question

[h.potter]

Solve this question in functions which seems to be diifficult?

the question is
if g(x) is a polynomial satisfying

g(x)g(y)=g(x)+g(y)+g(xy) - 2

for all real x and y, and g(2)=5
then find g(3).

the answer is 10
please tell me how to do this ques.

My sir told me that this ques. could be solved in a quarter of page.
But any advice is appreciated
thanks

[jemidiah]

I'll give you two hints as to how to answer this (your teacher could tell you, but he doesn't seem to want to, so I'll respect that and only get you halfway or so).

Hint #1: g(x) is a quadratic. In my opinion this should be given to you, but... it's not, from the problem statement you gave us. There are two reasons that I think g(x) is a quadratic. First, it looks like the problem would otherwise have an infinite number of solutions. Second, the problem seems to be designed around a quadratic. I'd be interested in knowing why your teacher didn't say what order polynomial you're dealing with.

Hint #2: g(0) = 1, g(1) = 2.

From these two pieces of information you could probabably figure out g(3), but to truly solve the problem you need to figure out how to figure out g(0) and g(1).


Good luck!

[h.potter]

its not that easy

if it was given that g(x) is quadratic, then simply putting three values of x such as 2, 1, 0, 1/2, 4(all of these can be found) would have solved
I can prove that its quadratic by diffrenciation, but it gets too long.
I was looking for a shorter solution

[VBAhack]

I solved it by assumming a certain degree for g. By comparing g(x)g(y)=g(x)+g(y)+g(xy) - 2 term by term you can figure it out.

[jemidiah]

This one is bugging me... I'd like to see your proofs, since I know I've been overthinking this.


I tried to make a generalized polynomial and do term-by-term comparisons, but then I got into infinite series that just looked too nasty to be useful.

Then I tried calculus, but kept ending back up at the beginning of the whole mess.

I also tried visualizing the operation of plugging values in for x which helped me to understand why some of my previous efforts had just let me in circles (no new information that can get you to g(3) is added by just plugging in values).

Now I think that since every piece of information about g has been used except for the fact that it is a polynomial, there must be some property of a quadratic that is unique to quadratics, and is demonstrated by the relationship given. However, I don't know what that property is. Maybe symmetry?

[jemidiah]

So I've worked on it some more and have gotten closer with Calculus and some odd reasoning. Maybe it could be cleaned up or squeezed in to fit in a quarter of a page... I'm not sure. If you take out the parentheses and write small on the algebra then I think you could.

Let g be of (finite) degree n.
Then...

g(x)g(2) = g(x) + g(2) + g(2x) - 2
4g(x) = 3 + g(2x)
4g'(x) = 2g'(2x)
4g''(x) = 4g''(2x)

Note that, for all x, g''(x) = g''(2x), so that g''(x) must be constant or zero (otherwise there would be x's at the extremes that, when doubled, would have 2nd derivatives that would have to differ from each other, so that for some x, g''(x) <> g''(2x)--a contradiction).

Since g''(x) is constant or zero, g'(x) is zero, or linear. Since 2g'(x) = g'(2x), g'(x) must be linear or always zero, except that g'(x) is not always zero, since g(0) = 1,* and g(2) = 5, and g is smooth. Thus, g'(x) is linear, meaning g(x) is quadratic.

*g(0)g(2) = g(0) + g(2) + g(0) - 2 --> 3g(0) = 3 --> g(0) = 1

Since 4g'(x) = 2g'(2x), 4g'(0) = 2g'(0), or g'(0) = 0. Since g is quadratic and is centered at x=0, g = a(x-0)^2 + b. Since g(0) = 1, b=1; since g(2) = 5, a=1. So g(x) = x^2 + 1, and g(3) = 10. Q.E.D.


Edit: I've tried to shorten it as much as possible here. It's not as readable or understandable, but I think it's pretty rigorous.


Suppose g is of finite degree n, g(x)g(2) = g(x) + g(2) + g(2x) - 2, and g(2) = 5.
Then 4g(x) = 3 + g(2x) --> 4g'(x) = 2g'(2x) --> 4g'(0) = 2g'(0) --> g'(0) = 0, and also 4g''(x) = 4g''(2x).
Since g''(x) = g''(2x), g''(x) is constant or zero.
Then g'(x) is linear, constant, or zero. However, g(0) = 1*, g(2) = 5, g is smooth, and g'(0) = 0, so g'(x) is not a constant, but is linear.
Thus g(x) is quadratic. Since g'(0) = 0, g is centered at zero, so g=ax^2 + b.
Since g(0) = 1, b=1; g(2) = 5, a=2. Thus g(x) = x^2 + 1 and g(3) = 10. Q.E.D.

*g(0)g(2) = g(0) + g(2) + g(0) - 2 -> 3g(0) = 3 -> g(0) = 1


Edit 2: Fixed a small mistake