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 this could be solved in a quarter of a page

- Jun 23rd 2007, 04:27 AMh.potterSolve 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 this could be solved in a quarter of a page - Jun 23rd 2007, 05:40 AMCaptainBlack
- Jun 23rd 2007, 06:52 AMh.potter
what you did is only correct if g(1) is not 2.

g(1) is 2 which can be calculated very easily

and besides, if this was true,

then,

g(n)g(1)=g(3)+g(1)+g(3)-2

g(n)(g(1)-2)=g(1)-2

g(n)=1(according to you)

which means for any value of n, g(n) will be 1, which is not possible.

for finding g(1)

g(1)g(2)=g(1)+g(2)+g(2)-2

5g(1)=g(1)+8

g(1)=2 - Jun 23rd 2007, 07:48 AMtopsquark
$\displaystyle g(1) = 2 $ not so easily.

$\displaystyle g(1)g(1) = g(1) + g(1) + g(1 \cdot 1) - 2$

$\displaystyle g^2(1) = 3g(1) - 2$

$\displaystyle g^2(1) - 3g(1) + 2 = 0$

$\displaystyle (g(1) - 1)(g(1) - 2) = 0$

So $\displaystyle g(1) = 1$ or $\displaystyle g(1) = 2$. If $\displaystyle g(1) = 1$ then

$\displaystyle g(n)g(1) = g(n) + g(1) + g(n \cdot 1) - 2$

$\displaystyle g(n) = 2g(n) - 2$

Thus $\displaystyle g(n) = 1$ for all n. But the problem states $\displaystyle g(2) = 5$, so this cannot be.

Thus $\displaystyle g(1) = 2$.

-Dan - Jun 23rd 2007, 08:08 AMh.potter
for finding g(1)

g(1)g(2)=g(1)+g(2)+g(2)-2

5g(1)=g(1)+8

g(1)=2

this is also correct

what wrong do you find with this?? - Jun 23rd 2007, 08:28 AMtopsquark
g(x) is a polynomial, so

$\displaystyle g(x) = \sum_{n = 0}^N a_nx^n$ where N may be infinite. So

$\displaystyle g(x)g(y) = g(x) + g(y) + g(xy) - 2$

implies:

$\displaystyle \sum_{m, n = 0}^Na_ma_nx^my^n = \sum_{n = 0}^N a_n(x^n + y^n + x^ny^n) - 2$

$\displaystyle a_0^2 + a_0a_1x + a_1a_0y + a_1^2xy + ... = 3a_0 + a_1(x + y + xy) + ... - 2$

We can find the coefficients by matching powers of x and y.

Let's look at the $\displaystyle x^0$ coefficient.

$\displaystyle a_0a_0 = 3a_0 - 2$

Thus

$\displaystyle a_0^2 - 3a_0 + 2 = 0$

$\displaystyle (a_0 - 1)(a_0 - 2) = 0$

Thus

$\displaystyle a_0 = 1$ or $\displaystyle a_0 = 2$.

Now

$\displaystyle a_0a_1x + a_0a_1y + a_1^2xy = a_1(x + y + xy)$

Thus

$\displaystyle a_0a_1 = a_1$

$\displaystyle a_1^2 = a_1$

The first tells us that $\displaystyle a_0 = 1$, and the second tells us $\displaystyle a_1 = 1$ or $\displaystyle a_1 = 0$.

Now

$\displaystyle [a_0a_2 x^2 + a_0a_2 y^2 + a_2^2x^2y^2] + [a_1a_2 xy^2 + a_1a_2 x^2y] = [a_2(x^2 + y^2 + x^2y^2)] $

This gives:

$\displaystyle a_0a_2 = a_2$

$\displaystyle a_2^2 = a_2$

$\displaystyle a_1a_2 = 0$

The first tells us nothing new. The second says $\displaystyle a_2 = 1$ or $\displaystyle a_2 = 0$. The third implies that at least one of $\displaystyle a_1 = 0$ or $\displaystyle a_2 = 0$.

There are no contradictions so far, but it's getting*really*messy. The good news is that, in general, I'm getting that $\displaystyle a_0 = 1$ and some set of $\displaystyle a_n = 1$ or $\displaystyle a_n = 0$ for n > 0. Since we need $\displaystyle g(2) = 5$ we must have $\displaystyle g(x) = 1 + 1 \cdot x^2$.

This fits with both our $\displaystyle g(1) = 2$ and $\displaystyle g(2) = 5$. So I would say that

$\displaystyle g(3) = 1 + 3^2 = 10$ as we expected.

-Dan - Jun 23rd 2007, 08:32 AMtopsquark
- Jun 23rd 2007, 09:41 PMh.potter
thanks for the answer:)

you can get a more decisive answer by considering only n as 0, 1 and a general n and putting x or y=2

but basically using your solution

i never thought of doing questions tthis way, thanks

but i'm looking for a shorter solution if someone can help:D