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Math Help - 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:

    <br />
g(6)g(1) = g(6) + g(1) + g(6) -2<br />

    so:

    <br />
g(6)[g(1)-2]=g(1)-2<br />

    so g(6)=1

    Now:

    <br />
g(2)g(3)=g(2)+g(3)+g(6)-2<br />

    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
    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
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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
    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
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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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