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
- 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

where N may be infinite. So

implies:

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

Let's look at the coefficient.

Thus

Thus

or .

Now

Thus

The first tells us that , and the second tells us or .

Now

This gives:

The first tells us nothing new. The second says or . The third implies that at least one of or .

There are no contradictions so far, but it's getting*really*messy. The good news is that, in general, I'm getting that and some set of or for n > 0. Since we need we must have .

This fits with both our and . So I would say that

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