MHB Find the number of real numbers that satisfy the given equation

[anemone]

How many real numbers $x$ satisfy $\sin x=\dfrac{x}{100}$?

 

[kaliprasad]

How many real numbers $x$ satisfy $\sin x=\dfrac{x}{100}$?

Spoiler

first let us look at positive x and same number of solutions shall be for -ve x

as $\sin,x$ and $x$ both are odd functions.

as $\sin\,x$ is less than 1 so x shall be less than 100 and so if we draw a sin curve there shall be $\dfrac{100}{2\pi}$ or 15.91 ( around 16) units so there shall be 16 parts( 1/2 oscilaltions) in positive side and 16 in -ve side.

for each on the curve y = x shall intersect the sin curve 2 times so 16 in positive side including zero) so there is 1 value at 0, 15 positive values and 15 -ve values or 31 values

 

[Greg]

Spoiler

$-100\lt x\lt100$. The positive peaks of $\sin(x)$, between 0 and 100, are at $\frac{(4k+1)\pi}{2},\;k\in\mathbb{Z},\,0\le k\le15$. Since the zero of $\sin(x)$ immediately following $\frac{61\pi}{2}$ is at $31\pi$ and $31\pi\lt100$, we count 16 peaks and 32 intersections of $\sin(x)$ and $\frac{x}{100}$. By symmetry, there are 32 intersections on the interval $-100\lt x\lt0$, so we count 64 intersections. Adding 1 for the intersection at $x=0$ gives a total of 65 intersections, so there are 65 real numbers for which $\sin(x)=\frac{x}{100}$$$\text{ }$$

 

[anemone]

Hi kaliprasad and greg1313,

Thank you for participating in this challenge but the correct answer should be 63...(Nod)

 

[kaliprasad]

Hi kaliprasad and greg1313,

Thank you for participating in this challenge but the correct answer should be 63...(Nod)

Thanks anemone,
I realize my mistake. I am not making an excuse but it was a counting mistake.

Spoiler

in 16 intervals x cuts it 2times one for sin x going from 0 to 1 and another for going down from 1 to 0 so 2 times and 16 intervals give 32 ( including a zero) so 31 positive values and 31 -ve values and one zero so 63

 

[Greg]

Thanks.$$\text{ }$$

Spoiler

I should have checked the intersection points around 0. 65 - 2 = 63.

 

[anemone]

Thanks anemone,
I realize my mistake. I am not making an excuse but it was a counting mistake.

Spoiler

in 16 intervals x cuts it 2times one for sin x going from 0 to 1 and another for going down from 1 to 0 so 2 times and 16 intervals give 32 ( including a zero) so 31 positive values and 31 -ve values and one zero so 63

You're welcome, Kali! You know, even if you made some other blunders but not the counting error, I would still give you multiple chances to fix your solution because you are not only a great friend to me and to MHB, you are one of the very great and precious challenge problem solvers too!:)

Thanks.$$\text{ }$$

Spoiler

I should have checked the intersection points around 0. 65 - 2 = 63.

You're welcome, greg1313! I am so glad that I have now another member who likes to participate in my challenge problems, thanks!

 

[Greg]

I am so glad that I have now another member who likes to participate in my challenge problems, thanks!

Thank you for presenting an opportunity to learn. :)