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$$f(x)= 1 - e^x\sin(x)$$ $$g(x)= 1 + e^x\cos(x)$$

Prove that between any two roots of $f$ there exists at least one root of $g$.

I know Rolle's Theorem and the Intermediate Value Theorem (I think) need to be used. Can someone show a step by step proof for this?

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One can calculate that $f'(x) = -e^x(\sin(x)+\cos(x))= f(x)-g(x)$. The roots of this function are $x = \pi n - \frac{\pi}{4}$ for $n \in \mathbb{Z}$. With a little bit of work, you can show that these are never roots of $f$, but I'm guessing this would be implied in the question (if $f$ and $f'$ were to share a root, then $f$ and $g$ share this root too).

Now, let $x_1$ and $x_2$ be two consecutive roots of $f$. Then $f'(x_1) \neq 0$, assume that $f'(x_1) > 0$ (the other case is similar). Thus $f$ is positive on the interval $(x_1,x_2)$, and thus $f'(x_2) < 0$.

Now, $f'(x_1) = f(x_1) - g(x_1) = -g(x_1) > 0$ and thus $g(x_1) < 0$. Similarly, we find that $g(x_2) > 0$. Now, apply Bolzano's theorem, a corollary of the intermediate value theorem:

If $g$ is a continuous function and $x_1,x_2 \in \mathbb{R}$ such that $g(x_1)g(x_2) < 0$, then there exists $x_3 \in (x_1,x_2)$ such that $g(x_3) = 0$.

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  • $\begingroup$ Hi there. Thanks heaps for the reply. Makes more sense now. But I'm still confused as to how to really write the proof. $\endgroup$ – Jim Wilson May 15 '16 at 4:57
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You can consider that $$ \eqalign{ & f(x) = 1 - e^{\,x} \sin (x) \cr & g(x) = 1 + e^{\,x} \cos (x) = 1 - e^{\,x} \sin (x - \pi /2) \cr} $$ so that$$ \eqalign{ & f(x) = 0\quad \to \quad \sin (x) = e^{\, - x} \cr & g(x) = 0\quad \to \quad \sin (x - \pi /2) = e^{\, - x} \cr} $$ and since the roots of $f(x)$ and $g(x)$ cannot be separated by less than $\pi$, it follows ...

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