Lindemann–Weierstrass theorem?

1. Jul 28, 2008

phantomprime

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?

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Show that the set {1, sqrt(2), sqrt(3), sqrt(6)} is linearly independent over the rational numbers

2. Jul 28, 2008

uman

Re: Lindemann–Weierstrass theorem??

(sorry, ignore this post)

3. Jul 28, 2008

maze

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
4. Jul 29, 2008

HallsofIvy

Re: Lindemann–Weierstrass theorem??

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

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

For the second one, you want to prove that if $a+ b\sqrt{2}+ c\sqrt{3}+ d\sqrt{6}= 0$, and a, b, c are rational numbers, then a= b= c= d= 0. You will, of course, use the fact that $\sqrt{2}$ and $\sqrt{3}$ are not rational.

5. Jul 29, 2008

maze

Re: Lindemann–Weierstrass theorem??

How would one go from the kth case to the k+1th?

6. Jul 29, 2008

morphism

Re: Lindemann–Weierstrass theorem??

If we multiply both sides of

$$a_1e^{x_1t}+ a_2e^{x_2t}+ \cdots + a_ne^{x_nt}= 0$$

by $e^{-x_n t}$, we get

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

Now let $t \to \infty$.