# Determine if function in span of other functions

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.

• 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. – Stan Tendijck Feb 26 at 23:57

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.
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?