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What methods/theorems are commonly used when trying to prove that a function has exactly one root within a given interval $(a,b)$, or that it has no roots?

I have the function $f(x)=\dfrac1x-\dfrac{\cos x}{\sin x}$. It is a strictly increasing function that is not defined for $a$ nor $b$....

Would just like a hint.

gebruiker
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Mother
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  • root = crosses the x axis. – Paul Sep 19 '14 at 13:35
  • It's a bit more advanced than that, buddy. I need to use the mean value theorem and intermedia value theorem; only problem is that the function is not defined for a nor b, so I'm not sure how I can do some of the calculations. – Mother Sep 19 '14 at 13:38
  • Can you tell us anything else about this function? Can you give us the explicit form of $f(x)$? If not, do we know that the function is continuous? Do we know that it has a limit at a or b, and what those limits are? – Ben Grossmann Sep 19 '14 at 13:41

2 Answers2

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Using the expression of $f$ you gave in a comment above to explain a method:

$$\frac{1}{x}-\frac{\cos x}{\sin x} = \frac{\sin x - x\cos x}{x\sin x} = \frac{x-\frac{x^3}{6}-(x-\frac{x^3}{2}) +o(x^3)}{x\sin x} = \frac{\frac{2x^3}{6}+o(x^3)}{x^2+o(x^2)}\sim \frac{x}{3} $$ when $x\to 0$. Your function can thus be extended by continuity on $a=0$. Show you can also do that at $b=\pi$ (not even necessary in your case), and then apply the mean value theorem or your machinery of choice.

Clement C.
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We have $f(x) = 1/x - \cos(x)/\sin(x)$. Note that, with the restriction of $x \in (0,\pi), x \neq \pi/2$, $$ 1/x - \cos(x)/\sin(x) = 0 \iff\\ \frac{\sin(x) - x\cos(x)}{x \sin(x)} = 0 \iff\\ \sin(x) - x\cos(x) = 0 \iff\\ \tan(x) - x = 0 $$ So, it suffices to show that $g(x) = \tan(x) - x$ has no roots on this interval. In order to do so, it suffices to note that $g$ is continuous and strictly increasing on the intervals where it's defined, and that we have the following limits: $$ \lim_{x \to 0^+}g(x) = 0\\ \lim_{x \to \pi/2^-}g(x) = \infty\\ \lim_{x \to \pi/2^+}g(x) = -\infty\\ \lim_{x \to \pi^-}g(x) < 0 $$

Ben Grossmann
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  • Hi. I was thinking about the fact that it's strictly increasing: can I combine that with the limits so that if I have (a,b) - and it's not defined for a and b, but the limit as I approach a from the right = negative number and the limit as I approach b from the right = positive number, then as it's strictly increasing within (a,b), it must have at least one root? Can I use this argument with the limits? – Mother Sep 19 '14 at 13:56
  • Nope, the conclusion you should reach is that there are no roots on the interval. Namely, use these facts to show that $g$ (and therefore $f$) can't have any zeros on $(0,\pi/2)$ or $(\pi/2,\pi)$. Then note that $f(\pi/2)$ isn't zero. So, $f$ can't have any zeros on the interval. – Ben Grossmann Sep 19 '14 at 14:04
  • Sorry, you misunderstood me. I got the point that it has no roots on the interval. Let me rephrase: Imagine a function that is increasing on the interval (0,a), but it is not defined for x = 0. How can I "prove" that it is always positive on the interval (0,a), if I cannot show that f(0) > 0? This is why I was asking if I could show something using limits, but I am not sure. – Mother Sep 19 '14 at 14:08
  • You can. See my answer below: $f$ can be extended by continuity into a $g$ which is continuous, has $g(0)=0$, and $g(x)=f(x)$ for all $x \in (0,\pi)$. You can also prove similarly that $g$ is strictly increasing on $[0,\pi)$, which gives you what you want. – Clement C. Sep 19 '14 at 14:10
  • @Daniel the problem with your question is that the way that you "prove" however many roots a function has varies wildly depending on the convenient properties of the function. I will, however, say this: if $f$ is defined on $(0,a)$, if it is strictly increasing wherever it is defined, and if $\lim_{x \to 0}f(x) = 0$, then we can indeed say that $f$ will have no roots on $(0,a)$. – Ben Grossmann Sep 19 '14 at 14:12
  • All right thanks. I think I understand it somewhat now. – Mother Sep 19 '14 at 14:14