Show that if $a\sin x + b\cos x + ce^x$ is the zero function for constants $a, b, c$ then $a = b = c = 0$

Question

Suppose that $a\sin x + b\cos x + ce^x$ is the zero function. Prove that $a=b=c=0$.

Does the zero function simply mean that $a\sin x + b\cos x + ce^x = 0$? I am going under the assumption that it is.

Geometrically speaking, I think it is pretty clear that $a=b=c=0$ must be true if $a\sin x + b\cos x + ce^x = 0$ as there are no x-values that make the equation true otherwise.

Is there a purely algebraic way of saying this that does not utilize showing a graph?

Answer 1

Let $x=0$.

Thus, $$b+c=0$$

Let $x=\frac{\pi}{2}.$

Thus, $$a+c\cdot e^{\frac{\pi}{2}}=0.$$

Now, substitute $x=-\frac{\pi}{2}$ and solve this system.

We obtain: $$-a+c\cdot e^{-\frac{\pi}{2}}=0.$$ Thus, $$c\left(e^{\frac{\pi}{2}}+e^{\frac{\pi}{2}}\right)=0$$ or $$c=0,$$ which gives $a=0$ and $b=0$.

Answer 2

If you expand sin, cos and exp into series, and then add them and get one series for the sum, then function is 0 if and only if coefficient on every x^n will be 0, and thus a=b=c=0 will be the only solution for that.

Answer 3

Assume $c \neq 0$. Then $x \longmapsto e^x$ is a linear combination of $\sin$ and $\cos$, thus is periodic. It is impossible.

Thus $c=0$, and $a\cos +b\sin=0$. Then evaluate at $0$ and $\pi/2$.

Answer 4

The Wronskian (determinant) addresses exactly this question, and applies just as well to an arbitary number of ($(n - 1)$-times differentiable) functions: The functions $f_1, \ldots, f_n$ are linearly independent if and only if the determinant $$W[f_1, \ldots, f_n]$$ of the matrix $$\pmatrix{f_1 & \cdots & f_n \\ f_1' & \cdots & f_n' \\ \vdots & & \vdots \\ f_1^{(n - 1)} & \cdots & f_n^{(n - 1)}}$$ is identically zero.

In our case, $f_1 = \sin, f_2 = \cos, f_3 = \exp$, and computing directly gives $$W[\sin, \cos, \exp](x) = -2 e^x ,$$ which is not (identically) zero, so the three functions are linearly independent, that is, the only solution $(a, b, c)$ to $$a \sin x + b \cos x + c e^x = 0$$ the zero solution.

Answer 5

Given

$a\sin x + b\cos x + ce^x = 0, \tag 1$

we multiply through by $e^{-x}$, which yields

$ae^{-x}\sin x + be^{-x}\cos x + c = 0; \tag 2$

letting $x \to \infty$, we find that

$c = 0; \tag 3$

we are left with

$a\sin x + b\cos x = 0; \tag 4$

now set

$x = \dfrac{\pi}{2}, \tag 5$

and obtain

$a = a\sin \dfrac{\pi}{2} + b\cos \dfrac{\pi}{2} = 0; \tag 6$

thus

$a = 0; \tag 7$

in a similar manner, choosing

$x = 0 \tag 8$

we find

$b = a \sin 0 + b \cos 0 = 0 \tag 9$

as well.

We have in fact established that $\sin x$, $\cos x$, and $e^x$ are linearly independent.