How many solutions has the equation $\sin x= \frac{x}{100}$ ?

Question

Usually when I was asked to solve this type of problem, I would solve it graphically but this one seems to be trickier. It doesn't seem wise to put $f(x)=\sin x$ and $g(x)=\frac{x}{100}$ in the same graph and then counting all the intersection points. What would be some algebraic methods to solve this?

Another illustration: http://www.wolframalpha.com/input/?i=plot+(sin+x)+-+x%2F100 — CiaPan, Jun 08 '16 at 18:09

Bernard · Answer 1 · 2016-07-11T08:09:39.987

5

First, we may suppose $x\ge 0$ since both sides are odd functions.

Using the Intermediate value theorem, there'll be two non-negative solutions on each interval $]2k\pi,2(k+1)\pi[$ as long as $\frac x{100}\le 1$, i.e. $x\le 100$. There results the number of non-negative solutions is equal to $2\times \biggl\lfloor \dfrac{50}{\pi}\biggr\rfloor=32$.

Hence, by symmetry, the total number of roots is $\;\color{red}{63}$.

edited Jul 11 '16 at 08:09

answered Jun 05 '16 at 16:22

Bernard

175,478

Isn't $2\times \biggl\lfloor \dfrac{50}{\pi}\biggr\rfloor=30$ ? – lmc Jun 05 '16 at 19:04
And the $0$ solution? – Bernard Jun 05 '16 at 19:16
Symmetry allows to limit one's investigations to $x>0$, but you have to count all real solutions nevertheless. – Christian Blatter Jun 08 '16 at 18:20

Steven Alexis Gregory · Accepted Answer · 2016-06-09T20:20:59.147

4

Outside of the interval $[-100, 100]$ there are no solutions.

$x = 100\;$ occurs close to the right end of the interval $[30 \pi, 32 \pi]$

$x = -100\;$ occurs close to the left end of the interval $[-32 \pi, -30 \pi]$

Each interval $[2n\pi, (2n+2)\pi]$ will have 2 solutions for $n = -32$ to $n = 30$ except $x = 0$ will get counted twice.

So there are $2*32-1 = 63\;$ solutions.

edited Jun 09 '16 at 20:20

answered Jun 05 '16 at 16:41

Steven Alexis Gregory

26,769

I agree that each interval $[2n\pi, (2n+2)\pi]$ will have 2 solutions. I don't understand how you got 59. Don't we have 32 intervals, so there should be $2*32-1=63$ solutions? – lmc Jun 05 '16 at 19:25
@Lucija97 - Yes I fixed it. Thanks.. – Steven Alexis Gregory Jun 08 '16 at 18:00
@Lucija97 - Do you know that you could have edited my answer too? – Steven Alexis Gregory Jun 08 '16 at 20:51
1

No, I'm new here. – lmc Jun 09 '16 at 19:26
2

@Lucija97 - Well I'm not and it seems I can't count. – Steven Alexis Gregory Jun 09 '16 at 20:19

How many solutions has the equation $\sin x= \frac{x}{100}$ ?

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