1

I think this question is mostly about having a good eye on specific properties. Here is my logic

Since the term on the RHS is $99\sin (\pi(x+1))$. For any integer values of $x$, the term will end up being $99\sin (n\pi)$, which is obviously zero. That gives us one solution.

Now if we go for non integer solutions, the possibilities seem endless. I am might not know some properties required to sovle this question, so please help me with that.

Aditya
  • 6,191
  • Notice that while it is true that the RHS will be zero for all integers, the LHS will be zero when $x=-1$. So not every integer will be a solution, but rather only when $x=-1$. – PCeltide Apr 13 '20 at 17:46

4 Answers4

1

I think the best strategy for questions like these are to look at the properties of their graphs. If we bring $99$ from the RHS to the LHS, we have:

$\frac{(x + 1)}{99} = sin(\pi(x + 1)) = sin(\pi x + \pi)$

Now it is evident from the LHS of the above equation, that we're trying to draw a line $y = \frac{x}{99} + \frac{1}{99}$ with $y$-intercept of $\frac{1}{99}$. Also, our RHS is bounded between $1$ and $-1$. So for $x \geq 99$ and $x < -100$, we will definitely have no solution. That's enough with the LHS for now, let's shift our focus to RHS. We have a sine wave which is positively shifted by $\pi$, and covers a period in $[x, x + 2]$. Moreover, in a period, a sine wave covers every value between $[-1, 1]$ so it will intersect the line exactly twice in a period. This is a major clue, we just need to now look carefully at how many times those periods would be completed in the positive and the negative $x$-axis given the above constraints, and we'll have our answer.

Thus for the positive $x$-axis we'll have about $\lfloor \frac{99}{2} \rfloor * 2 = 98$ real solutions. For the negative $x$-axis, similarly, $\lfloor \frac{-100}{2} \rfloor * 2 = 100$ for a total of $198 + 1$ real solutions counting the one additional solution due to $x = 0$.

Shikhar Jaiswal
  • 410
1

By a shift in the equation, it will be altered to $$x=99\sin \pi x$$which has $x=0$ as one of its roots. Since this equation is symmetric w.r.t. the origin, the number of its roots will become:$$N=2\times \text{Number of positive roots} +1$$hence we only try to find the number of its positive roots.

Clearly $x$ and $99\sin \pi x$ coincide for $x>0$ only when $\sin \pi x>0$, or $$2k <x<2k+1$$ and $x\le 99$. This yields us a maximal value for $k$ for which we must have $$2k \le 99\\2k+1\le 99$$which gives us $k\le 49$. Since the first $k=49$ positive peaks of $99\sin \pi x$ coincide with $x$, each of which meeting it twice. For $x\ge 2\times 49+1=99$ however, there is no answer since we must have $$x>2\times 49+2 >99$$hence the total number of positive roots becomes $99$ and the total number of roots will then be $199$.

Mostafa Ayaz
  • 31,924
  • Could you please clarify how you shifted the equation exactly? – Aditya Apr 13 '20 at 18:17
  • You could say let $u = x+1$ then replace $x+1$ with $u$ everywhere you see it. – Doug M Apr 13 '20 at 18:22
  • It is just saying that an equation like $x+a=f(x+a)$ is essentially the same $x=f(x)$, since horizontal shifting does not change the characteristics of an equation (including its number of roots). Here $f(x)=99\sin \pi x$ and $a=1$. – Mostafa Ayaz Apr 13 '20 at 18:22
  • @DougM, yes. That was possibly a better idea :) – Mostafa Ayaz Apr 13 '20 at 18:25
  • I am have a few more questions, but I think you made a mistake in the 12th or 13th line, where it should have been $x<100$ – Aditya Apr 13 '20 at 18:32
  • There was an extra $\pi$ in line $12$ and I removed it, thanks. Except it, which mistake are you pointing at? We had so far investigated all the roots $x$ for $0<x<99$. So, I targeted $x\ge 99$. – Mostafa Ayaz Apr 13 '20 at 18:34
  • I meant the line where you wrote $x<1$. I don’t know whether it’s wrong, I just think it is – Aditya Apr 13 '20 at 18:38
  • Oops! Sorry for that. I also fixed another typo..... – Mostafa Ayaz Apr 13 '20 at 18:40
  • Okay, so with that cleared up, let me get to the actual problems. If $2k<x<2k+1$, why is 99 greater than both $2k$ and $2k+1$ – Aditya Apr 13 '20 at 18:49
  • This assumption is made to find the positive peaks of $\sin$ that coincide with $x$ in exactly two distinct points. We have implicitly considered that this equation has no integer solution except $x=0$. – Mostafa Ayaz Apr 13 '20 at 18:51
  • Wait I thought $x=-1$ is the single integer solution – Aditya Apr 14 '20 at 14:54
  • 1
    We are analyzing the solutions of $x=99\sin \pi x$ which are the same of $x+1=99\sin \pi(x+1)$ with an extra shift. – Mostafa Ayaz Apr 14 '20 at 15:09
1

For generality, assume the equation is

$$t=a\sin t,\>\>\>a>1$$

The number of roots is given analytically by

$$N(a)= 3+4\cdot \text{Int}\left[\frac1{2\pi\csc^{-1}a}-\frac14\right]\tag1$$

where Int[] denotes integer values. So, for the equation under consideration $x+1=99\sin (\pi (x+1))$, we have $a=99\pi$. Plug into (1) to obtain the number of roots $N(99\pi)=199$.

PS: For the derivation of (1), refer to What is the number of solutions of $x = a\sin(x)$

Quanto
  • 97,352
0

A simplier approach

The number of real solution to the equation $x+1 = 99\sin(\pi(x+1))$

Say $x+1 = n$

The equation becomes $n = 99\sin(\pi*n)$

$n/99 = \sin(\pi*n)$

$\arcsin(n/99) = \pi*n$

$\arcsin(m) = 99\pi*m$

Where $n= 99m$

Remember that $\arcsin(m)$ is real if only $m ≤ 1$ and $m ≥ -1$

Therefore $-1 ≤ m ≤ 1$

$-1 ≤ n/99 ≤ 1$

$-1 ≤ (x+1)/99 ≤ 1$

$-100 ≤ x ≤ 98$

This is the range of values $x$ must take for a real solution to occur

The graph of our function $x+1 = 99\sin(\pi(x+1))$ is offcourse periodic because of the $\sin$, the straight line cuts it at every linear displacement

So the number of real roots here is the number of elements in the set of $x$, where $-100 ≤ x ≤ 98$

Counting all the numbers here including 0 gives us $199$

Aderinsola Joshua
  • 992