Mocking the J

Algebra Level 3

j = 1 ( cos j π 2 j + i sin j π 2 j ) = ? \large \prod_{j=1}^\infty \left(\cos \frac{j\pi}{2^{j}} + i \sin \frac{j\pi}{2^{j}} \right) = \, ?

Clarification : i = 1 i = \sqrt{-1} .


Inspiration .


The answer is 1.0.

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1 solution

Nihar Mahajan
Mar 31, 2016

j = 1 ( cos j π 2 j + i sin j π 2 j ) = j = 1 e i j π 2 j ( ) \large \prod_{j=1}^\infty (\cos \frac{j\pi}{2^{j}} + i \sin \frac{j\pi}{2^{j}} )\\ \large =\prod_{j=1}^\infty e^{i\frac{j\pi}{2^{j}}}~~~~(\color{#302B94}{* }) = e i π ( j = 1 j 2 j ) \Large =e^{i \pi \large{(\displaystyle \sum_{j=1}^{\infty} \frac{j}{2^j})}} = e 2 i π = 1 ( ) \large{ =e^{2i\pi}=\huge\boxed {1}}~~~~( \color{#302B94}{** })

Use cos x + i sin x = e i x Note: j = 1 j 2 j denotes the sum of an infinite Arithmetic-Geometric Progression Geometric Progression: first term as well as common ratio is 1 2 Arithmetic Progression: first term as well as common difference is 1 \color{#3D99F6}{\boxed{\color{forestgreen}{\boxed{\small {\color{#302B94}{* \text{Use }\cos x+ i\sin x=e^{ix}}} \\ \small {\color{#302B94}{** \text{Note: }\displaystyle \sum_{j=1}^{\infty} \frac{j}{2^j} \text{ denotes the sum of an} \\ \text{ infinite Arithmetic-Geometric Progression}\\ \text{Geometric Progression: first term as well as common ratio is }\frac{1}{2} \\ \text{Arithmetic Progression: first term as well as common difference is } 1 }}}}}}

Partial Credits: To Rishabh Cool, for the LaTeX.

Moderator note:

Nice approach of simplifying the expression.

Beautiful solution ±1

Harsh Shrivastava - 5 years, 2 months ago

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Haha, I just copy pasted Rishabh's presentation from inspiration problem :P (Of course, I did make appropriate changes xD)

Nihar Mahajan - 5 years, 2 months ago

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In the last line of the boxed text, "Common difference" should be at the place of "Common ratio".

Aditya Sky - 5 years, 2 months ago

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@Aditya Sky Thanks, Edited.

Nihar Mahajan - 5 years, 2 months ago

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