Question 608556
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Presuming you mean linearly independent, let *[tex \LARGE V] be a vector space of all functions of a real variable *[tex \LARGE t].


Suppose *[tex \LARGE a] and *[tex \LARGE b] are two real numbers such that


*[tex \LARGE \ \ \ \ \ \ \ \ \ ae^t\ +\ be^{2t}\ =\ 0\ \forall\ t]


If *[tex \LARGE e^t] and *[tex \LARGE e^{2t}] are linearly independent then *[tex \LARGE a\ =\ 0] and *[tex \LARGE b\ =\ 0]


Divide by *[tex \LARGE e^t] since *[tex \LARGE e^t] is never zero


*[tex \LARGE \ \ \ \ \ \ \ \ \ be^{t}\ =\ -a]


hence *[tex \LARGE be^t] must be independent of *[tex \LARGE t] which only occurs when *[tex \LARGE b\ =\ 0].  Therefore *[tex \LARGE a\ =\ 0] as well.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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