How to Prove the Linear Independence of Exponential Functions?

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Discussion Overview

The discussion centers around proving the linear independence of a set of exponential functions, specifically the functions e^x1t, e^x2t, ..., e^xnt, where X1, X2, ..., Xn is an increasing sequence of real numbers. The scope includes theoretical reasoning and mathematical proofs related to linear independence.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests using the Lindemann–Weierstrass theorem as a basis for proving the linear independence of the exponential functions.
  • Another participant proposes that by selecting various values of t, an overdetermined system of linear equations can be formed, indicating the linear independence of the functions.
  • A different participant mentions that to prove linear independence, one can start with the equation a_1e^{x_1t} + a_2e^{x_2t} + ... + a_ne^{x_nt} = 0 and show that all coefficients must be zero, hinting at a potential inductive proof.
  • One participant inquires about the transition from the kth case to the k+1th case in the inductive proof.
  • Another participant suggests manipulating the original equation by multiplying both sides by e^{-x_n t} and taking the limit as t approaches infinity to derive a new form of the equation.

Areas of Agreement / Disagreement

Participants express various approaches to proving linear independence, but there is no consensus on a single method or resolution of the proposed ideas. Multiple competing views remain regarding the best approach to the proof.

Contextual Notes

Some participants reference definitions and properties of linear independence, but the discussion does not resolve the mathematical steps or assumptions necessary for the proofs presented.

phantomprime
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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
 
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(sorry, ignore this post)
 


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).
 
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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.
 


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


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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