Number of points of non-differentiabilty of $f(x)=(\arccos x)|\sin (kx)|$

Question

Consider a function $$f(x)=(\arccos x)|\sin(kx)|$$ where $k\in \mathbb{N}$ and $k\leq 100$.

If for four consecutive values of $k=\{k_1,k_2,k_3,k_4\}$ ,

$f(x)$ has equal number of points of non-differentiabilty,

then which of the following statement(s) is/are true

(1) One of the possible values of $k_1+k_2+k_3+k_4$ is $94$.

(2) One of the possible values of $k_1+k_2+k_3+k_4$ is $182$.

(3) One of the possible values of $k_1+k_2+k_3+k_4$ is $358$.

(4) $\sum_{k=1}^{100}$(number of points of non-differentiability of $f(x)$) is $3216$.

My Attempt

Clearly $f(x)$ is defined for $x\in [-1,1]$ and is non differentiable when $\sin (kx)=0$ i.e $x=\frac{k\pi}{n}$ .

Since $-1\leq \frac{k\pi}{n}\leq 1$ we have $-\frac{n}{\pi}\leq k\leq\frac{n}{\pi}$

From here onwards I am not able to think further

Answer 1

The number of points of non-differentiability on $(0,1]$ is given by $[k/\pi]$, where [] is the floor function. By symmetry and the fact that $\{0\}$ is a point of non-differentiability, the number of relevant points for $n$ and $n+1$ are different if and only if $[n/\pi]$ and $[(n+1)/\pi]$ are different. Now, for the first three options, we can back-calculate the value of $k_1$ and see if $[k_i/pi]$ has the same values. This only works for option 1, as $_1+_2+_3+_4=94$ implies $k_1=22$ and corresponding $[k_i/\pi]$'s have the same value. But does not work for the other two options. (Actually, you can find that $k_1$ has to be a multiple of $22$ for four consecutive integers to have the same value for $[k_i/pi]$ as $\pi$ is close to $\frac{22}{7}$.)

For the 4th option, first, observe that the total number of points of non-differentiability for $k$ is $2*[k/\pi]+1$. Next, to calculate $\sum_{1\le k \le 100} [k/\pi]$, we exploit the pattern at multiples of 22. Until $21$, it's $(1+2+..+6)*3 = 63$. From $22$ until $25$, it's $7*4 = 28$. Then on, every $18$ numbers have a similar pattern as $\{4,5,...,21\}$ and the next four have that of $\{22,23,24,25\}$. Accounting for the jumps, You can count the sum as $$\sum_k [k/\pi] = 63+28+... (7*18+63)+(7*4+28)+... (14*18+63)+(14*4+28)+... (21*18+63)+(21*4+28)+... 29*3+30*3+31*3 = 1558.$$ Thus, $\sum_{1\le k \le 100} (2*[k/\pi]+1) = 3216$, as given.

Answer 2

This answer uses the fact that $$3.14\lt\pi\lt\frac{22}{7}$$ (see here and here)

$f(x)$ is defined only for $x\in [-1,1]$ and is non differentiable when $\sin (kx)=0$ i.e $kx=m\pi$ where $m$ is an integer.

Since $-1\leq \frac{m\pi}{k}\leq 1$, we have $\frac{-k}{\pi}\le m\le \frac{k}{\pi}$. So, the number of such integers $m$ is given by $g(k):=2\lfloor\frac{k}{\pi}\rfloor+1$.

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(1)

Since $k_2=k_1+1,k_3=k_1+2$ and $k_4=k_1+3$, we have $94=k_1+(k_1+1)+(k_1+2)+(k_2+3)$ which implies $k_1=22$. Now, since $7\lt \frac{22}{\pi}\lt\frac{25}{\pi}\lt 8$, we have $g(22)=g(23)=g(24)=g(25)=15$, so (1) is correct.

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(2)

Similarly, we get $k_1=44$. Now, since $14\lt \frac{44}{\pi}\lt\frac{47}{\pi}\lt 15$, we have $g(44)=g(45)=g(46)=g(47)=29$, so (2) is correct.

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(3)

Similarly, we get $k_1=88$. Now, since $28\lt \frac{88}{\pi}\lt\frac{91}{\pi}\lt 29$, we have $g(88)=g(89)=g(90)=g(91)=57$, so (3) is correct.

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(4)

We can get the value of $g(k)$ using the inequality $3.14\lt\pi\lt\frac{22}{7}$.

• For $m=0,1,\cdots, 6$, we have $g(3m+1)=g(3m+2)=g(3m+3)=2m+1$ since $$m\lt\frac{3m+1}{\pi}\lt\frac{3m+3}{\pi}\lt m+1$$ (For $m=0$, this holds. For $1\le m\le 6$, this holds since $3\lt\pi\lt 3+\frac 16\le 3+\frac 1m$.)

• $g(22)=g(23)=g(24)=g(25)=15$ since $7\lt \frac{22}{\pi}\lt\frac{25}{\pi}\lt 8$

• For $m=8,9,\cdots, 13$, we have $g(3m+2)=g(3m+3)=g(3m+4)=2m+1$ since $$m\lt\frac{3m+2}{\pi}\lt\frac{3m+4}{\pi}\lt m+1$$ (This holds since $3+\frac{1}{m+1}\le 3+\frac{1}{8+1}\lt\pi\lt 3+\frac{2}{13}\le 3+\frac 2m$)

• $g(44)=g(45)=g(46)=g(47)=29$ since $14\lt \frac{44}{\pi}\lt\frac{47}{\pi}\lt 15$

• For $m=15,16,\cdots, 20$, we have $g(3m+3)=g(3m+4)=g(3m+5)=2m+1$ since $$m\lt\frac{3m+3}{\pi}\lt\frac{3m+5}{\pi}\lt m+1$$ (This holds since $3+\frac{2}{m+1}\le 3+\frac{2}{15+1}\lt\pi\lt 3+\frac{3}{20}\le 3+\frac 3m$)

• $g(66)=g(67)=g(68)=g(69)=43$ since $21\lt \frac{66}{\pi}\lt\frac{69}{\pi}\lt 22$

• For $m=22,23,\cdots, 27$, we have $g(3m+4)=g(3m+5)=g(3m+6)=2m+1$ since $$m\lt\frac{3m+4}{\pi}\lt\frac{3m+6}{\pi}\lt m+1$$ (This holds since $3+\frac{3}{m+1}\le 3+\frac{3}{22+1}\lt\pi\lt 3+\frac{4}{27}\le 3+\frac 4m$)

• $g(88)=g(89)=g(90)=g(91)=57$ since $28\lt \frac{88}{\pi}\lt\frac{91}{\pi}\lt 29$

• For $m=29,30,31$, we have $g(3m+5)=g(3m+6)=g(3m+7)=2m+1$ since $$m\lt\frac{3m+5}{\pi}\lt\frac{3m+7}{\pi}\lt m+1$$ (This holds since $3+\frac{4}{m+1}\le 3+\frac{4}{29+1}\lt \pi\lt 3+\frac{5}{31}\le 3+\frac 5m$)

Therefore, we finally get $$\begin{align}&\sum_{k=1}^{100}\text{(number of points of non-differentiability of $f(x)$)} \\\\&=\sum_{k=1}^{100}g(k) \\\\&=3\sum_{m=0}^{31}(2m+1)+(15+29+43+57) \\\\&=3\times 2\times\frac{31\times 32}{2}+3\times 32+144 \\\\&=3216\end{align}$$

So, (4) is correct.