Trigonometric Sums

Graph for reference

Like most of my notes, this one starts with curiosity. I saw this problem and went to Desmos to solve it. Then, I got a bit curious about trigonometric functions and what that pattern would result in (y=sin⁡(x)+sin⁡(2x)+sin⁡(3x)+sin⁡(4x)y=\sin(x)+\sin(2x)+\sin(3x)+\sin(4x)); the resulting graph looked a bit like the graph of a heart rate monitor in hospitals. I used ∑\sum to make it easier, giving me the function f(x)=∑n=1asin⁡(n×x)f(x)= \displaystyle\sum_{n=1}^a \sin(n\times x) Since Desmos can't handle infinity, I used the next best thing: a=99a=99. This gave me a graph that was a bit strange: it was so squished together that it looked like a solid shape instead of a continuous line; the shape looked suspiciously like the graph of the cotangent function. After messing around a bit, I found the the two graphs that contained this strange function/shape: y=cot⁡(x4)2y=\frac{\cot\left(\frac{x}{4}\right)}{2}

y=cot⁡(x4+π2)2y=\frac{\cot\left(\frac{x}{4}+\frac{\pi}{2}\right)}{2} I was baffled by the connection between the two functions, sine and cotangent. However, this is only half of the story.

I went through the same process with the cosine function, giving me g(x)=∑n=1bcos⁡(n×x)g(x)= \displaystyle\sum_{n=1}^b \cos(n\times x) b=99b=99 This time, the graph/shape looked suspiciously like the secant/cosecant function. Once again, after some messing and mapping, I managed to get the two defining functions of this new cosine sum: y=csc⁡(x2)−12y=\frac{\csc\left(\frac{x}{2}\right)-1}{2}

y=csc⁡(x2)+1−2y=\frac{\csc\left(\frac{x}{2}\right)+1}{-2}

My question to the brilliant community of Brilliant: How on Earth are these functions related? They seemed to pop up out of nowhere.

#Calculus

Note by Blan Morrison
3 years ago

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Comments

Graph for reference

First,

sin⁡x+sin⁡2x+sin⁡3x+⋯+sin⁡nx=sin⁡x2(sin⁡x+sin⁡2x+sin⁡3x+⋯+sin⁡nx)sin⁡x2=sin⁡x2sin⁡x+sin⁡x2sin⁡2x+sin⁡x2sin⁡3x+⋯+sin⁡x2sin⁡nxsin⁡x2\sin {x}+\sin {2x}+\sin {3x}+\cdots +\sin {nx}\\ =\dfrac { \sin { \frac { x }{ 2 } } (\sin {x}+\sin {2x}+\sin {3x}+\cdots +\sin {nx}) }{ \sin { \frac { x }{ 2 } } } \\ =\frac { \sin { \frac { x }{ 2 } } \sin { x } +\sin { \frac { x }{ 2 } } \sin { 2x }+\sin { \frac { x }{ 2 } } \sin { 3x }+\cdots +\sin { \frac { x }{ 2 } } \sin { nx } } { \sin { \frac { x }{ 2 } } }

Using the Product-to-Sum Formula:

=cos⁡(x2−x)−cos⁡(x2+x)+cos⁡(x2−2x)−cos⁡(x2+2x)⋯+cos⁡(x2−nx)−cos⁡(x2+nx)2sin⁡x2=cos⁡(−x2)−cos⁡(3x2)+cos⁡(−3x2)−cos⁡(5x2)⋯+cos⁡(x2−nx)−cos⁡(x2+nx)2sin⁡x2=\frac { \cos { (\frac { x }{ 2 } -x) } -\cos { (\frac { x }{ 2 } +x) } +\cos { (\frac { x }{ 2 } -2x) } -\cos { (\frac { x }{ 2 } +2x) } \cdots +\cos { (\frac { x }{ 2 } -nx) } -\cos { (\frac { x }{ 2 } +nx) } }{ 2\sin { \frac { x }{ 2 } } } \\ =\frac { \cos { (-\frac { x }{ 2 } ) } -\cos { (\frac { 3x }{ 2 } ) } +\cos { (-\frac { 3x }{ 2 } ) } -\cos { (\frac { 5x }{ 2 } ) } \cdots +\cos { (\frac { x }{ 2 } -nx) } -\cos { (\frac { x }{ 2 } +nx) } }{ 2\sin { \frac { x }{ 2 } } }

Since cos(a)=cos(−a)cos{(a)}=cos{(-a)}, we can cancel the terms:

=cos⁡(x2)−cos⁡(x2+nx)2sin⁡x2=\frac { \cos { (\frac { x }{ 2 } ) } -\cos { (\frac { x }{ 2 } +nx) } }{ 2\sin { \frac { x }{ 2 } } }

Now, as nn approaches infinity, cos⁡(x2+nx)\cos { (\frac { x }{ 2 } +nx) } fluctuates continuously and infinitely frequently between −1-1 and 11, and the whole expression fluctuates continuously and infinitely frequently between

cos⁡(x2)+12sin⁡x2\frac { \cos { (\frac { x }{ 2 } ) } +1 }{ 2\sin { \frac { x }{ 2 } } } and cos⁡(x2)−12sin⁡x2\frac { \cos { (\frac { x }{ 2 } ) } -1 }{ 2\sin { \frac { x }{ 2 } } }

near a point.

Using the Double-Angle Formula:

cos⁡(x2)+12sin⁡x2=2cos⁡2x4−1+12⋅2sin⁡x4cos⁡x4=cos⁡2x42sin⁡x4cos⁡x4=cot⁡x42\frac { \cos { (\frac { x }{ 2 } ) } +1 }{ 2\sin { \frac { x }{ 2 } } } \\ =\frac { 2\cos ^{ 2 }{ \frac { x }{ 4 } } -1+1 }{ 2\cdot 2\sin { \frac { x }{ 4 } } \cos { \frac { x }{ 4 } } } \\ =\frac { \cos ^{ 2 }{ \frac { x }{ 4 } } }{ 2\sin { \frac { x }{ 4 } } \cos { \frac { x }{ 4 } } } \\ =\frac { \cot { \frac { x }{ 4 } } }{ 2 }

Similarly:

cos⁡(x2)−12sin⁡x2=1−2sin⁡2x4−12⋅2sin⁡x4cos⁡x4=−sin⁡2x42sin⁡x4cos⁡x4=−tan⁡x42=cot⁡(x4+π2)2\frac { \cos { (\frac { x }{ 2 } ) } -1 }{ 2\sin { \frac { x }{ 2 } } } \\ =\frac { 1-2\sin ^{ 2 }{ \frac { x }{ 4 } } -1 }{ 2\cdot 2\sin { \frac { x }{ 4 } } \cos { \frac { x }{ 4 } } } \\ =\frac { -\sin ^{ 2 }{ \frac { x }{ 4 } } }{ 2\sin { \frac { x }{ 4 } } \cos { \frac { x }{ 4 } } } \\ =\frac { -\tan { \frac { x }{ 4 } } }{ 2 } =\frac { \cot { (\frac { x }{ 4 } +\frac { \pi } { 2 } ) } }{ 2 }

So the expression fluctuates frequently between cot⁡x42\frac { \cot { \frac { x }{ 4 } } }{ 2 } and cot⁡(x4+π2)2\frac { \cot { (\frac { x }{ 4 } +\frac { \pi } { 2 } ) } }{ 2 } .

In other words, ∑k=1nsin⁡(k×x)\displaystyle\sum_{k=1}^n \sin(k\times x) looks "sandwiched" between cot⁡x42\frac { \cot { \frac { x }{ 4 } } }{ 2 } and cot⁡(x4+π2)2\frac { \cot { (\frac { x }{ 4 } +\frac { \pi } { 2 } ) } }{ 2 } .

In the same way, we can also find the functions that "sandwich" ∑k=1ncos⁡(k×x)\displaystyle\sum_{k=1}^n \cos(k\times x).

icreiuhe Zhu - 3 years ago

Looks tough- glad I’m in Geometry

Ella Collard - 3 years ago
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