Deriving the Area of a Circle using Sine of Infinite-Frequency

#Geometry

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Use \includegraphics[width = (your choice), scale = 2.0 (your choice) ]

Nikola Alfredi - 1 year, 4 months ago

brilliant

Nahom Assefa - 1 year, 4 months ago

From the perspective of the unit circle, a signal oscillating at the Nyquist frequency would simply produce DC or 0Hz, since pi radians of rotation is equal to 0 on the Y axis. But that's not exactly what's happening.Write My Essay

Dennis Kemp - 1 year, 3 months ago

Can you please tell how area of circle will be equal to distance between endpoints of the integration?

Ayush Bihani - 1 year ago

After you increase the frequency of the Sine function to infinity (with the amplitude bound by the contours of the circle as described above), every point on the wave also becomes a point on the surface of the circle itself (at frequency = infinity this is true).

So if you take the straight-line distance of the wave (which is every point on the surface) it is inevitably going to be equal to the area of the surface itself since every point is accounted for in this logic.

Jake Maason - 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$