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The problem is as follows:

Consider the space of all functions continuous on $[-1,1]$.

Given three elements of this space, $f_1(t) = e^t$, $f_2(t) = e^{2t}$, $f(3) = e^{-t}$.

Determine if function $f_4(t) = e^{-2t}$ belongs to $span[f_1,f_2,f_3]$

I am not really sure how to approach this problem. I understand how to determine if a vector is in a span of other vectors, but not these exponential functions.

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    $\begingroup$ You could try to evaluate the linear combination $a_1 f_1(t) + a_2 f_2(t) + a_3 f_3(t)$ at different $t$, then choose the $a_i$ such that this gives exactly the same as $f_4(t)$ as what you would do in the vector sense. See if that works out. $\endgroup$ – Stan Tendijck Feb 26 at 23:57
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Hint: suppose $e^{-2t}=ae^{t}+be^{2t}+ce^{-t}$. Put $t=0$ to get $a+b+c=1$. Differentiate the equation and put $t=0$ and differentiate the equation two more times and put $t=0$. Can you now get a contradiction?

Alternatively, put $y=e^{t}$ and get a polynomial equation in $y$. This equation has infinitely many solutions so all the coefficients are $0$. [ The equation is $by^{4}+ay^{3}+cy-10$. Every number $y$ of the form $e^{t}$ with $-1 \leq t \leq 1$ satisfies this equation. This implies that this fourth degree polynomial has infinitely many solutions. This is a contradiction.

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  • $\begingroup$ I tried what you suggested, and was unable to prove a contradiction. I am not sure how writing the equation in polynomial form helps me $\endgroup$ – Myydraal Feb 27 at 2:09
  • $\begingroup$ @Myydraal I have added an explanation. $\endgroup$ – Kabo Murphy Feb 27 at 5:08
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Hint: If $f_4=\alpha f_1+\beta f_2+\gamma f_3$, then, multiplying both sides by $e^{2t}$, you'll get$$1=\alpha e^{3t}+\beta e^{5t}+\gamma e^t.$$Can you take it from here?

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