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Thread: Solve this question in functions which seems to be diifficult?

  1. #1
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    Red face 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 this could be solved in a quarter of a page
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  2. #2
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    Quote Originally Posted by h.potter View Post
    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:

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

    so:

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

    so $\displaystyle g(6)=1$

    Now:

    $\displaystyle
    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
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  3. #3
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    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
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by h.potter View Post
    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
    $\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
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  5. #5
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    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??
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  6. #6
    Forum Admin topsquark's Avatar
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    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
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by h.potter View Post
    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
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  8. #8
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    Thumbs up

    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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