1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Solving a certain polynomial

  1. Oct 28, 2009 #1
    So I have an equation:
    h(x) = [tex]\sum a_{i}x^{i}[/tex] from i=0 to d.

    I know [tex]a_{i}[/tex] and x.
    I am trying to prove that there is a y where g(x) = [tex]\sum a_{i}y^{i}[/tex] from i=0 to d, g(x) = h(x), and y does not = x.

    How do I do this? Sorry for the bad use of Latex.
  2. jcsd
  3. Oct 28, 2009 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In general? You can't, for example, the polynomial x3 is 1-1. It's unclear what you mean here, g(x) is not actually a function of x, and in fact seems to be h(y). So what you really want is to find x and y so that h(x)=h(y) right?
  4. Oct 28, 2009 #3
    Yes, sorry. That was a typo. That is g(y), not g(x).
  5. Oct 28, 2009 #4


    User Avatar
    Science Advisor

    It's still not clear what you want to prove. If you have [itex]h(x)= \sum_{i=0}^d a_i x^i[/itex] and you replace the variable x by any y, you get [itex]h(y)= \sum_{i=0}^d a_iy^i[/itex]. It is the same function, just written differently.

    If you mean x and y to be specific numbers and want to prove that there exist [itex]y\ne x[/itex] such that h(y)= h(x), you can't- it is not, in general, true. As Office Shredder says, polynomials can be one-to-one. His example of h(x)= x3/sup] shows that.
  6. Oct 28, 2009 #5
    I am looking for a y in terms of [tex]a_{i}[/tex] and [tex]x^{i}[/tex] in a solution that contains no polynomial equation. The most important thing I think is to figure out the properties of the summation of [tex]a_{i}[/tex] (which I don't know at all). I know I can't just divide that out, but there has to be a way to separate it from the summation of y.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook