Tricky Points of Non-Differentiability

Calculus Level 3

f ( x ) = ∏ i = 1 ∞ 1 + 2 cos ⁡ ( 2 x 3 i ) 3 f(x) = \prod_{i=1}^{\infty} \frac{1+ 2\cos\left( \frac{2x}{3^i} \right)}{3}

Let f ( x ) f(x) denote an function as described above. The number of points where ∣ x f ( x ) ∣ + ∣ ∣ x − 2 ∣ − 1 ∣ |xf(x)| + | |x-2|-1| is non-differentiable in x ∈ ( 0 , 3 π ) x \in (0,3\pi) is k k , find k 2 k^2 .


The answer is 25.

Kunal Gupta by Kunal Gupta , 23, India

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2 solutions

Chew-Seong Cheong
May 18, 2015

Let g ( x ) = ∣ x f ( x ) ∣ + ∣ ∣ x − 2 ∣ − 1 ∣ g(x) = |xf(x)|+||x-2|-1| . g ( x ) g(x) is non-differentiable or discontinuous , when the value within an absolute function ∣ ⋅ ∣ |\cdot \space | passes through 0 0 , for x ∈ ( 0 , 3 π ) x \in (0, 3\pi) .

That is when: { x f ( x ) = 0 ⟹ f ( x ) = 0 , since x > 0 x − 2 = 0 ⟹ x = 2 ∣ x − 2 ∣ − 1 = 0 ⟹ x = 1 , 3 \begin{cases} xf(x) = 0 & \implies f(x) = 0 \text{, since } x > 0 \\ x - 2 = 0 & \implies x = 2 \\ |x-2|-1 = 0 & \implies x = 1, 3 \end{cases}

When f ( x ) = 0 f(x) = 0 , we have:

1 + 2 cos ⁡ 2 x 3 i = 0 cos ⁡ 2 x 3 i = − 1 2 2 x 3 i = 2 n π ± 2 3 π where n = 0 , 1 , 2 , … 2 x 3 = 2 n ± 2 3 π For x ∈ ( 0 , 3 π ) , i can only be 1 x = 3 n π ± π = π , 2 π for x ∈ ( 0 , 3 π ) \begin{aligned} 1 + 2\cos{\frac{2x}{3^i}} & = 0 \\ \cos{\frac{2x}{3^i}} & = -\frac 12 \\ \frac {2x}{3^i} & = 2n\pi \pm \frac 23 \pi & \small \color{#3D99F6} \text{where }n = 0, 1, 2, \dots \\ \frac {2x}3 & = 2n \pm \frac 23 \pi & \small \color{#3D99F6} \text{For }x \in (0, 3\pi) \text{, }i \text{ can only be 1} \\ x & = 3n\pi \pm \pi \\ & = \pi, 2\pi & \small \color{#3D99F6} \text{for }x \in (0, 3\pi) \end{aligned}

Therefore, g ( x ) g(x) is non-differentiable when x = { 1 , 2 , 3 , π , 2 π } ⟹ k = 5 x = \{1,2,3,\pi, 2\pi\} \implies k = 5 and k 2 = 25 k^2 = \boxed{25} .

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I am not sure about some details of the solution:

  • Checking the factors of f f for zeros, I get

1 + 2 cos ⁡ ( 2 x 3 k ) 3 = 0 ⇒ cos ⁡ ( 2 x 3 k ) = − 1 2 ⇒ 2 x 3 k = ± 2 π 3 + 2 π n , n ∈ Z ⇒ x = ± 3 k − 1 π + 3 k π n \begin{aligned} \frac{ 1+2\cos\left(\frac{2x}{3^k}\right) }{ 3 }&=0&\Rightarrow&&\cos\left(\frac{2x}{3^k}\right)&=-\frac{1}{2}&\Rightarrow&&\frac{2x}{3^k}&=\pm\frac{2\pi}{3}+2\pi n,&&n\in\mathbb{Z}\\\\ &&&&&&\Rightarrow&& x&=\pm 3^{k-1}\pi+ 3^k\pi n \end{aligned}

This still leads to a total of two zeros in ( 0 ; 3 π ) (0;\:3\pi) :

k = 1 : x ∈ { π ; 2 π } ⊂ ( 0 ; 3 π ) \begin{aligned} k=1:&&x\in\{\pi;\:2\pi\}\subset(0;\:3\pi) \end{aligned}

For k > 1 k>1 , no factor has another zero in ( 0 ; 3 π ) (0;\:3\pi) . Together with { 1 ; 2 ; 3 } ⊂ ( 0 ; 3 π ) \{1;\:2;\;3\}\subset(0;\:3\pi) , we have 5 points where f might not be differentiable!

  • How do we know that f ( x ) f(x) is differentiable in the first place? It is defined has a limit of smooth functions, but such limits must not be differentiable at all (counter-example: Weierstrass function ).

As Indraneel Mukhopadhyaya already pointed out, one can use

1 + 2 cos ⁡ ( 2 x ) = { x = n π : 3 else: sin ⁡ ( 3 x ) sin ⁡ ( x ) \begin{aligned} 1+2\cos(2x)&=\begin{cases} x=n\pi:&3\\\\ \text{else:}&\frac{\sin(3x)}{\sin(x)} \end{cases} \end{aligned}

to show that f ( x ) f(x) does indeed converge to

f ( x ) = lim ⁡ n → ∞ ∏ k = 1 n 1 + 2 cos ⁡ ( 2 x 3 k ) 3 = { x = 0 : 1 else: sin ⁡ x x \begin{aligned} f(x)=\lim_{n\rightarrow\infty}\prod_{k=1}^n\frac{1+2\cos\left(\frac{2x}{3^k}\right)}{3}=\begin{cases} x=0:&1\\\\ \text{else:}&\frac{\sin{x}}{x} \end{cases} \end{aligned}

That's very important, because only now we know that f ( x ) f(x) is differentiable!

  • Now it is easy to prove that the given function is not differentiable at x ∈ { 1 ; 2 ; 3 ; π ; 2 π } x\in\{1;\:2;\:3;\:\pi;\:2\pi\} : Just calculate all leftside and rightside limits and find that they are not equal!

Carsten Meyer - 2 years, 3 months ago

Using the fact that 1+2cos2x=(sin3x)/(sinx),we can find that f(x)=(sinx)/x.Then we can construct a piecewise table for g(x) (g(x) is the function whose points of non differentiability we have to check).By checking the continuity of derivative of g(x) we find the points of discontinuity of derivative of g(x) at x=1,2,3,pi and 2pi.

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