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Lindemann–Weierstrass theorem?

  1. Jul 28, 2008 #1
    Lindemann–Weierstrass theorem??

    Let X1, X2, X3,...Xn be an increasing sequence of real numbers. Prove that the n exponential functions e^x1t,e^x2t.....e^xnt are linearly independent.

    My question is how? I look at examples but how do I go about explaining this?


    Show that the set {1, sqrt(2), sqrt(3), sqrt(6)} is linearly independent over the rational numbers
  2. jcsd
  3. Jul 28, 2008 #2
    Re: Lindemann–Weierstrass theorem??

    (sorry, ignore this post)
  4. Jul 28, 2008 #3
    Re: Lindemann–Weierstrass theorem??

    I think if you just keep picking values of t, you will get a system of linear equations that is overdetermined (and unsolvable since the columns of rows of constants in the matrix are linearly independent).
    Last edited: Jul 28, 2008
  5. Jul 29, 2008 #4


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    Re: Lindemann–Weierstrass theorem??

    Both of those look pretty straight forward from the definition of "independent".

    Given [itex]a_1e^{x_1t}+ a_2e^{x_2t}+ \cdot\cdot\cdot+ a_ne^{x_nt}= 0[/itex]
    for all t, prove that [itex]a_1= a_2= \cdot\cdot\cdot= a_n= 0[/itex]
    I suspect it would be simplest to prove this by induction on n.

    For the second one, you want to prove that if [itex]a+ b\sqrt{2}+ c\sqrt{3}+ d\sqrt{6}= 0[/itex], and a, b, c are rational numbers, then a= b= c= d= 0. You will, of course, use the fact that [itex]\sqrt{2}[/itex] and [itex]\sqrt{3}[/itex] are not rational.
  6. Jul 29, 2008 #5
    Re: Lindemann–Weierstrass theorem??

    How would one go from the kth case to the k+1th?
  7. Jul 29, 2008 #6


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    Re: Lindemann–Weierstrass theorem??

    If we multiply both sides of

    [tex]a_1e^{x_1t}+ a_2e^{x_2t}+ \cdots + a_ne^{x_nt}= 0[/tex]

    by [itex]e^{-x_n t}[/itex], we get

    [tex]a_1e^{(x_1 - x_n)t}+ a_2e^{(x_2 - x_n)t}+ \cdots + a_ne^{(x_{n-1} - x_n)t} + a_n = 0.[/tex]

    Now let [itex]t \to \infty[/itex].
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