Solve this question in functions which seems to be diifficult?

**h.potter** · June 23rd 2007, 04:27 AM

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

**CaptainBlack** · June 23rd 2007, 05:40 AM

Originally Posted by h.potter

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

First, we have:

$ g(6)g(1) = g(6) + g(1) + g(6) -2 $

so:

$ g(6)[g(1)-2]=g(1)-2 $

so $g(6)=1$

Now:

$ g(2)g(3)=g(2)+g(3)+g(6)-2 $

in this quation only g(3) is an unknown, so it can be solved for.

RonL

**h.potter** · June 23rd 2007, 06:52 AM

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

**topsquark** · June 23rd 2007, 07:48 AM

Originally Posted by h.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

$g(1) = 2$ not so easily.

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

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

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

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

So $g(1) = 1$ or $g(1) = 2$ . If $g(1) = 1$ then
$g(n)g(1) = g(n) + g(1) + g(n \cdot 1) - 2$

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

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

Thus $g(1) = 2$ .

-Dan

**h.potter** · June 23rd 2007, 08:08 AM

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??

**topsquark** · June 23rd 2007, 08:28 AM

g(x) is a polynomial, so
$g(x) = \sum_{n = 0}^N a_nx^n$ where N may be infinite. So
$g(x)g(y) = g(x) + g(y) + g(xy) - 2$
implies:

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

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

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

Let's look at the $x^0$ coefficient.
$a_0a_0 = 3a_0 - 2$

Thus
$a_0^2 - 3a_0 + 2 = 0$

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

Thus
$a_0 = 1$ or $a_0 = 2$ .

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

Thus
$a_0a_1 = a_1$
$a_1^2 = a_1$

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

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

This gives:
$a_0a_2 = a_2$
$a_2^2 = a_2$
$a_1a_2 = 0$

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

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

This fits with both our $g(1) = 2$ and $g(2) = 5$ . So I would say that
$g(3) = 1 + 3^2 = 10$ as we expected.

-Dan

**topsquark** · June 23rd 2007, 08:32 AM

Originally Posted by h.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??

Two things. At first I had decided your derivation of the value of g(1) wasn't general enough. However I now think you are correct and that is was good enough.

-Dan

**h.potter** · June 23rd 2007, 09:41 PM

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

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