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A dumb question, but i'm just curious

  1. Nov 2, 2004 #1
    Well, this problem is kinda dumb; it's in the GIF image file that I attached...
    a,b,c,d,...,z are all real constants, and 'n' is natural; I was wondering if
    such a function would exist...for the real constants a,b,c,d,...,z
    (filename is polyprobgif.gif and sumformul.gif)
     

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    Last edited: Nov 2, 2004
  2. jcsd
  3. Nov 2, 2004 #2

    JasonRox

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    How does it get to z?

    Is it n-25 = 0 so zx^0 = z?
     
  4. Nov 2, 2004 #3

    AKG

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    Of course. Let [itex]P_n(\mathbb{R})[/itex] be the space of polynomials of degree less than or equal to n with real coefficients. Since your notation is confusing, instead of a, b, c, ..., z being real constants, let's call them [itex]a_0, a_1, \dots , a_n[/itex]. Then f is an operator on [itex]P_n(\mathbb{R})[/itex] which maps a polynomial [itex]b_nx^n + b_{n - 1}x^{n - 1} + \dots + b_0[/itex] to [itex]a_nb_nx^n + \dots + a_0b_0[/itex].
     
  5. Nov 2, 2004 #4
    What would be the equation/formula for this function? By the way, can there exist some value 'phi' whose product with the sum will yield this series? <<the attached file phigif.gif has the problem>>--what would be the formula to get the value of phi (possibly some math operations with a,b,c,d,e,f...?)?
     

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    Last edited: Nov 2, 2004
  6. Nov 2, 2004 #5

    shmoe

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    Your notation is still confusing, is n=25? If that's not what you mean, have a look at AKG's post and the use of subscripts to deal with an arbitrary number of coefficients.

    You have [tex]\Phi(x)f(x)=g(x)[/tex] (can I assume your dot means multiply?) and you want to find [tex]\Phi(x)[/tex]. Just take [tex]{\mbox\Phi(x)=g(x)/f(x)}[/tex] for x not a zero of f, whatever you like at the zeros of f, and hope that all the zeros of f are also zeros of g.

    edit-You said "value of [tex]\Phi[/tex]" multiple times, if you were hoping that [tex]\Phi[/tex] was a constant (and not dependant on x) then I should say you'd need all the coefficients on the right to be equal (otherwise it's impossible). Then take [tex]\Phi[/tex] to be this common value.
     
    Last edited: Nov 2, 2004
  7. Nov 3, 2004 #6

    AKG

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    Could you please clarify what you're asking? You're asking for a function f such that:

    [tex]f(b_nx^n + b_{n - 1}x^{n - 1} + \dots + b_0) = a_nb_nx^n + a_{n - 1}b_{n - 1}x^{n - 1} + \dots + a_0b_0,[/tex]

    where, in your example,

    [tex]b_n = b_{n - 1} = \dots = b_0 = 1[/tex]

    and

    [tex]a_n = a,\ a_{n - 1} = b,\ a_{n - 2} = c,\ a_0 = z?[/tex]

    Is this not what you're asking? If so, then I've already given the answer:

    [tex]f(b_nx^n + \dots + b_0) = a_nb_nx^n + \dots + a_0b_0[/tex]

    This function takes its argument from the space of n degree (or less) polynomials with real coefficients, and returns another polynomial from the same space but with different coefficients. It is a linear operator. If we choose

    [tex]\{x^n,\ x^{n - 1},\ \dots ,\ x,\ 1\}[/tex]

    to be a basis of our space, then we can express polynomials as vectors by their coefficients. So, we'd have:

    [tex]f(b_n,\ b_{n - 1},\ \dots ,\ b_0) = (a_nb_n,\ a_{n - 1}b_{n - 1},\ \dots ,\ a_0b_0)[/tex]

    Specifically, if we have the vector (polynomial) like the one in your example, namely, (1, 1, 1, ..., 1), then we get:

    [tex]f(1,\ 1,\ 1,\ \dots ,\ 1) = (a_n,\ a_{n - 1},\ \dots ,\ a_0)[/tex]

    Just like you wanted. Just recall that I'm replacing

    [tex]a,\ b,\ c,\ d,\ \dots ,\ z[/tex]

    with the less confusing

    [tex]a_n,\ a_{n - 1},\ \dots ,\ a_0,[/tex]

    since I assume you didn't mean for there to only be 26 numbers, just n of them.
     
    Last edited: Nov 3, 2004
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