Triggy Sum

Calculus Level 5

sin ⁡ x 1 3 + sin ⁡ 2 x 2 3 + sin ⁡ 3 x 3 3 + ⋯ \large \dfrac{\sin{x}}{1^3}+\dfrac{\sin{2x}}{2^3}+\dfrac{\sin{3x}}{3^3}+\cdots

What is the maximum value of the above expression?

π 3 36 \dfrac{\pi^3}{36} π 3 18 \dfrac{\pi^3}{18} π 3 18 2 \dfrac{\pi^3}{18\sqrt{2}} π 3 18 3 \dfrac{\pi^3}{18\sqrt{3}}

Digvijay Singh by Digvijay Singh , 21, India

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1 solution

Hassan Abdulla
May 9, 2018

let f ( x ) = ∑ k = 1 ∞ sin ⁡ ( k x ) k 3 f ′ ( x ) = ∑ k = 1 ∞ cos ⁡ ( k x ) k 2 f ′ ′ ( x ) = ∑ k = 1 ∞ − sin ⁡ ( k x ) k ∑ k = 1 ∞ sin ⁡ ( k x ) k is the Fourier series expansion of the function f ( x ) = π − x 2 for x ∈ ( 0 , 2 π ) ) so f ′ ′ ( x ) = 1 2 x − π 2 f ′ ( x ) = 1 4 x 2 − π 2 x + C // integrate both side f ′ ( 0 ) = ∑ k = 1 ∞ 1 k 2 = ζ ( 2 ) = π 2 6 ⇒ C = π 2 6 ⇒ f ′ ( x ) = 1 4 x 2 − π 2 x + π 2 6 f ( x ) = 1 12 x 3 − π 4 x 2 + π 2 6 x + C // integrate both side f ( 0 ) = 0 ⇒ C = 0 ⇒ f ( x ) = 1 12 x 3 − π 4 x 2 + π 2 6 x 1 4 x 2 − π 2 x + π 2 6 = 0 ⇒ 3 x 2 − 6 π + 2 π 2 = 0 ⇒ x 6 π ± 36 π 2 − 24 π 2 6 = π ( 3 ± 1 3 ) //critical points f ( π ( 3 + 1 3 ) ) = 1 12 π 3 ( 3 + 1 3 ) 3 − π 3 4 ( 3 + 1 3 ) 2 + π 3 6 ( 3 + 1 3 ) = − π 3 18 3 f ( π ( 3 − 1 3 ) ) = 1 12 π 3 ( 3 − 1 3 ) 3 − π 3 4 ( 3 − 1 3 ) 2 + π 3 6 ( 3 − 1 3 ) = π 3 18 3 so the maximum value of the expression is π 3 18 3 \text{let } f\left( x \right) =\sum _{ k=1 }^{ \infty }{ \frac { \sin { \left( kx \right) } }{ { k }^{ 3 } } } \\ f^{ ' }\left( x \right) =\sum _{ k=1 }^{ \infty }{ \frac { \cos { \left( kx \right) } }{ { k }^{ 2 } } } \\ f^{ '' }\left( x \right) =\sum _{ k=1 }^{ \infty }{ \frac { -\sin { \left( kx \right) } }{ { k } } } \\ \sum _{ k=1 }^{ \infty }{ \frac { \sin { \left( kx \right) } }{ { k } } } \text{ is the Fourier series expansion of the function } f(x)=\frac { \pi -x }{ 2 } \text{ for } x\in (0,2\pi )) \\ \text{so } f^{ '' }\left( x \right) =\frac { 1 }{ 2 } x-\frac { \pi }{ 2 } \\ f^{ ' }\left( x \right) =\frac { 1 }{ 4 } x^{ 2 }-\frac { \pi }{ 2 } x+C \color{#D61F06} \text{ // integrate both side} \\ f^{ ' }\left( 0 \right) =\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 2 } } } =\zeta \left( 2 \right) =\frac { \pi ^{ 2 } }{ 6 } \Rightarrow C=\frac { \pi ^{ 2 } }{ 6 } \Rightarrow f^{ ' }\left( x \right) =\frac { 1 }{ 4 } x^{ 2 }-\frac { \pi }{ 2 } x+\frac { \pi ^{ 2 } }{ 6 } \\ f\left( x \right) =\frac { 1 }{ 12 } x^{ 3 }-\frac { \pi }{ 4 } x^{ 2 }+\frac { \pi ^{ 2 } }{ 6 } x+C\qquad \color{#D61F06} \text{ // integrate both side} \\ f\left( 0 \right) =0\Rightarrow C=0\Rightarrow f\left( x \right) =\frac { 1 }{ 12 } x^{ 3 }-\frac { \pi }{ 4 } x^{ 2 }+\frac { \pi ^{ 2 } }{ 6 } x\\ \frac { 1 }{ 4 } x^{ 2 }-\frac { \pi }{ 2 } x+\frac { \pi ^{ 2 } }{ 6 } =0\Rightarrow 3x^{ 2 }-6\pi +2\pi ^{ 2 }=0\Rightarrow x\frac { 6\pi \pm \sqrt { 36\pi ^{ 2 }-24\pi ^{ 2 } } }{ 6 } =\pi \left( \frac { \sqrt { 3 } \pm 1 }{ \sqrt { 3 } } \right) \color{#D61F06} \text{ //critical points} \\ f\left( \pi \left( \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } } \right) \right) =\frac { 1 }{ 12 } { \pi }^{ 3 }\left( \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } } \right) ^{ 3 }-\frac { { \pi }^{ 3 } }{ 4 } \left( \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } } \right) ^{ 2 }+\frac { { \pi }^{ 3 } }{ 6 } \left( \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } } \right) =\frac { -{ \pi }^{ 3 } }{ 18\sqrt { 3 } } \\ f\left( \pi \left( \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } } \right) \right) =\frac { 1 }{ 12 } { \pi }^{ 3 }\left( \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } } \right) ^{ 3 }-\frac { { \pi }^{ 3 } }{ 4 } \left( \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } } \right) ^{ 2 }+\frac { { \pi }^{ 3 } }{ 6 } \left( \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } } \right) =\frac { { \pi }^{ 3 } }{ 18\sqrt { 3 } } \\ \large \text{so the maximum value of the expression is }\frac { { \pi }^{ 3 } }{ 18\sqrt { 3 } }

20 Helpful 16 Interesting 18 Brilliant 5 Confused

I want to I me where the negative in step 3 went, it's not there in step 5.

jonathan nicholson - 2 years, 1 month ago

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f ′ ′ ( x ) = ∑ k = 1 ∞ − sin ⁡ ( k x ) k = − ∑ k = 1 ∞ sin ⁡ ( k x ) k = − ( π − x 2 ) = x − π 2 f^{ '' }\left( x \right) =\sum _{ k=1 }^{ \infty }{ \frac { -\sin { \left( kx \right) } }{ { k } } }=-\sum _{ k=1 }^{ \infty }{ \frac { \sin { \left( kx \right) } }{ { k } } }=-\left ( \frac { \pi -x }{ 2 } \right )= \frac { x -\pi }{ 2 }

Hassan Abdulla - 2 years ago

max. value of f(x) tends to infinity as x tends to infinity since f(x) is a cubic equation with positive co-efficient of highest degree term?? why do we calculate local maxima and minima here??

Hrithik Thakur - 1 year, 8 months ago

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it's a periodic function so it's repeat it self

see step 4

Hassan Abdulla - 1 year, 8 months ago

It's fine now... Thanks

Hrithik Thakur - 1 year, 8 months ago

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