Prove that $f(x)=2x-L\cos(Lx)$ has only one root in range $x\in [-1, 1]$ when $L \in [0, \pi/2]$

It's simple to show that it has at least one root using Intermediate value theorem, but I find it difficult to show that it has only one root.

  • $\begingroup$ is that $\;\cos LX\;$ ? Because you wrote twice $\;coxLX\;$ ... $\endgroup$ – DonAntonio Jul 7 '19 at 10:36
  • $\begingroup$ @DonAntonio, of course, sorry $\endgroup$ – Mr.OY Jul 7 '19 at 10:38

For $x<0$ both terms in $f$ are negative, so that there can be no root in $[-1,0]$. If $x>\frac L2$, the first term dominates the second, as $\cos u\le 1$. Thus the roots can only occur inside $(0,\frac\pi4]$. On this interval $f'(x)=2+L^2\sin(Lx)$ is positive, so that $f$ is monotonous on this interval and can only have one root there and thus in the whole of $[-1,1]$.

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  • $\begingroup$ Why if both terms are negative there can't be a zero in $\;[-1,0]\;$ ? We have $\;2x-L\cos Lx\;$ , not $\;2x+L\cos Lx\;$ . In the latter case we'd have a sum of two negative, but in the former (and original) case, we have a negative minus a negative = negative plus positive... $\endgroup$ – DonAntonio Jul 7 '19 at 14:48
  • $\begingroup$ @DonAntonio : The cosine is positive on $(-\frac\pi2,\frac\pi2)$, where the argument for the given interval $[-1,1]$ and for the given values of $L\in[0,\frac\pi2]$ falls into. Remember, it is a symmetric function, while the sine is the anti-symmetric one. $\endgroup$ – Lutz Lehmann Jul 7 '19 at 15:04
  • $\begingroup$ Of course...but it was you who wrote that for $\;x<0\;$ both terms in $\;f\;$ are negative...:) $\endgroup$ – DonAntonio Jul 7 '19 at 15:40
  • $\begingroup$ @DonAntonio : Yes, both $2x$ and $-L\cos(Lx)$ are negative, so that the sum is negative and never zero. $\endgroup$ – Lutz Lehmann Jul 7 '19 at 16:28
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    $\begingroup$ @Mr.OY Yes, that too is sufficient and removes some complexity. I was originally trying to isolate the root interval more using both of the cosine estimates $1-x^2/2\le \cos x\le 1$, but that does not really help further for a manual argument. $\endgroup$ – Lutz Lehmann Jul 9 '19 at 20:44

Hint (but not a solution):

  1. try drawing the graph (Desmos is a good tool; include a slider for the value $L$)

  2. For any particular $L$, split into two cases: $x < 0$ and $x \ge 0$. On one of these intervals, $f$ is monotone, so the main question is why there are no double-roots in the other interval.


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