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First of all let me apologize for my bad English. I have the set $\{e^x, e^{2x},e^{3x},e^{4x}\}$ and I have to demonstrate that it is linearly independent. I know that there is not a set of scalars that give the zero vector, except for the trivial solution. Thank you and sorry again.

levap
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juan camilo
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  • So, write out what it means to be linearly independent and see what you can do for different values of $x$. – The Count Apr 19 '17 at 14:30

2 Answers2

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Let $f (x)=\alpha e^x+\beta e^{2x}+\gamma e^{3x}+\delta e^{4x} $. You want to show that if $f (x)=0$ for all $x $, then $\alpha=\beta=\gamma=\delta=0$.

If $f=0$, then in particular $0=f (0)=f'(0)=f''(0)=f^{(3)}(0) $. This will give you four equations on your coefficients.

Martin Argerami
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    An approach that produces slightly nicer equations is as follows: note that $f(x) = g(e^x)$, where $g(x) = \alpha x + \beta x^2 + \gamma x^3 + \delta x^4$. If $f = 0$, then $g = 0$. So, $0 = g'(0) = g''(0) = g'''(0)$. – Ben Grossmann Apr 19 '17 at 14:34
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Assume that $ae^x + be^{2x} + ce^{3x} + de^{4x} = 0$. Let us differentiate the equation three times to get:

$$ ae^x + 2be^{2x} + 3ce^{3x} + 4de^{4x} = 0, \\ ae^x + 4be^{2x} + 9ce^{3x} + 16de^{4x} = 0, \\ ae^x + 8be^{2x} + 27ce^{3x} + 64de^{4x} = 0. $$

Plugging in $x = 0$ in each of the four equations we get

$$ a + b + c + d = 0, \\ a + 2b + 3c + 4d = 0, \\ a + 4b + 9c + 16d = 0, \\ a + 8b + 27c + 64d = 0. $$

Can you show that the only solution of this system of equations is $a = b = c = d = 0$?

levap
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    You can do it by brute force but this generalizes via https://en.wikipedia.org/wiki/Vandermonde_matrix. – levap Apr 19 '17 at 14:34