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

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.

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

## 2 Answers

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 $$[-1,1]$$. • 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... – DonAntonio Jul 7 '19 at 14:48
• @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. – Lutz Lehmann Jul 7 '19 at 15:04
• Of course...but it was you who wrote that for $\;x<0\;$ both terms in $\;f\;$ are negative...:) – DonAntonio Jul 7 '19 at 15:40
• @DonAntonio : Yes, both $2x$ and $-L\cos(Lx)$ are negative, so that the sum is negative and never zero. – Lutz Lehmann Jul 7 '19 at 16:28
• @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. – 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.