Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parameters of the polynom

  1. Oct 25, 2006 #1
    I hope this is a correct section.

    I have a function that looks like this:
    y = a1*x1*a2*x2*a3*x3....an*xn

    All the parameters of the polynom (a1, a2, a3, ... an) are unknown.
    Is there a way to find them using a small-medium (less than n) set of examples where each example is a set of xs (x1, x2, x3, ... xn) and y.
    I know that there is no way to find these values perfectly but accurate enough answer (each new example raises the accuracy) will be ok.
    Is it possible to do the same thing with n or more examples perfectly? and what about infinite number of examples.

    BTW, why doesn't this latex work:
    [tex] y = x_1a_1x_2a_2x_3a_3...x_na_n [/tex]
    as well as this:
    Last edited: Oct 25, 2006
  2. jcsd
  3. Oct 25, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    If you write that correctly, then all you need is one collection of values where the x_i's are nonzero. This will give you the value of the product of the a_i's, which is the best you can do.

    As for the LaTeX, try:
    [tex]y = x_1a_1x_2a_2x_3a_3\ldots x_na_n[/tex]
  4. Oct 31, 2006 #3
    you latex is not working because you have to set your subscripts in brackets as follows

    [tex] y = x_{1}a_{1}x_{2}a_{2}x_{3}a_{3}...x_{n}a_{n} [/tex]

    On that note simply use the commutative proerty of multiplication to rewrite this as

    [tex] y = \prod_{i=1}^nx_{i}\prod_{i=1}^ma_{i}= XA [/tex]

    [tex] y/X = A [/tex]

    Thus what you are asking is if the a' can be uniquely determined by knowledge of the divisors of y. No not really. The fundamental theorem of arithmetic says that there exists a set of primes such that

    [tex] y = \prod_{i=1}^np_{i}^{k_{n}} [/tex]

    Therefore there are as many possible sets of a's in your above formula as there are ways of grouping the prime factors of y up to multiplicity. That is you have something like [itex]\prod_{i=1}^nk_{n}[/itex], sets of a's that satisfy your formula. now if y is prime then you have either a = 1 or a = p, simple enough. If you are thinking of that construction over the real numbers there are an uncountable infinity of sets of a's. But let's suppose we are over the natural numbers (which seems clear), well then if you pick all of the x's and the y and try to pair them up with a's, you might not be able to match the pairs [itex](a_{i},x_{i})[/itex] evenly. The question is not really a simple one. Supposee I have y = 2*3*5*7 let a[1] = 2 and a[2] = 3 then I can write (2*5)(3*7), or (2*7)(3*5), 5 and sven are uniquely determined but the pairings are not. Suppose we have y = 2*3*5 and we want a[1] = 2 and a[2] = 3, then we have to invoke infinite multiplicity of 1 to write (2*5)(3*1) or (2*1)(3*5), one and five are unique here but what if we thought about a[1] = 2, a[2] = 3, a[3] = 5, then we would have only one way to pair these as (2*1)(3*1)(5*1).
    Last edited: Oct 31, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook