proof of sine series

Feb 2010
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hey what is the proof of sine and cos series
 

Prove It

MHF Helper
Aug 2008
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hey what is the proof of sine and cos series
Assume you can write \(\displaystyle \sin{x}\) as a polynomial.

Then

\(\displaystyle \sin{x} = c_0 + c_1x + c_2x^2 + c_3x^3 + c_4x^4 + \dots\).

Let \(\displaystyle x = 0\) and you can see \(\displaystyle c_0 = 0\).


Differentiate both sides:

\(\displaystyle \cos{x} = c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + 5c_5x^4 + \dots\).

Let \(\displaystyle x = 0\) and you can see \(\displaystyle c_1 = 1\).


Differentiate both sides:

\(\displaystyle -\sin{x} = 2c_2 + 3\cdot 2c_3x + 4\cdot 3c_4x^2 + 5\cdot 4c_5x^3 + 6\cdot 5 c_6x^4 + \dots\).

Let \(\displaystyle x = 0\) and you can see \(\displaystyle c_2 = 0\).


Differentiate both sides:

\(\displaystyle -\cos{x} = 3\cdot 2c_3 + 4\cdot 3\cdot 2c_4x + 5\cdot 4\cdot 3c_5x^2 + 6\cdot 5\cdot 4c_6x^3 + 7\cdot 6\cdot 5c_x^4 + \dots\).

Let \(\displaystyle x = 0\) and you can see \(\displaystyle c_3 = -\frac{1}{3\cdot 2} = -\frac{1}{3!}\).


Differentiate both sides:

\(\displaystyle \sin{x} = 4!c_4 + 5!c_5x + 6\cdot 5\cdot 4\cdot 3c_6x^2 + 7\cdot 6\cdot 5\cdot 4c_7x^3 + 8\cdot 7\cdot 6\cdot 5c_8x^4 + \dots\).

Let \(\displaystyle x = 0\) and you can see \(\displaystyle c_4 = 0\).


Differentiate both sides:

\(\displaystyle \cos{x} = 5!c_5 + 6!c_6x + 7\cdot 6\cdot 5\cdot 4 \cdot 3c_7x^2 + 8\cdot 7\cdot 6\cdot 5 \cdot 4c_8x^3 + 9\cdot 8\cdot 7\cdot 6\cdot 5c_9 x^4 + \dots\).

Let \(\displaystyle x = 0\) and you can see \(\displaystyle c_5 = \frac{1}{5!}\).


I think you can see that if we continue this way, we will find

\(\displaystyle \sin{x} = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + \dots - \dots\)

\(\displaystyle = \sum_{n = 1}^{\infty}\frac{(-1)^{n - 1}x}{(2n - 1)!}\).


A similar process can be used to find the series for \(\displaystyle \cos{x}\). Alternatively, you can differentiate both sides of the series for \(\displaystyle \sin{x}\).
 

Laurent

MHF Hall of Honor
Aug 2008
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Paris, France
hey what is the proof of sine and cos series
It depends on your definition of sine and cos. Usually, we define the complex exponential as \(\displaystyle \exp(z)=\sum_{n=0}^\infty \frac{z^n}{n!}\) and then \(\displaystyle \cos(x)\) and \(\displaystyle \sin(x)\) are the real and imaginary parts of \(\displaystyle \exp(ix)\), which gives their series expansion.

This is the most convenient definition of cos and sine (and it gives the series immediately) but there are other ways to procede without complex numbers. For instance using a differential equation: \(\displaystyle \cos\) can be defined as the unique solution of \(\displaystyle y''+y=0\) with \(\displaystyle y(0)=1\) and \(\displaystyle y'(0)=0\) (if we prove that it indeed exists and is unique). Then we prove that the power series yields another solution, so it must be equal to \(\displaystyle \cos\). There are probably other ways, depending what you know about sine and cosine.
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
It depends on your definition of sine and cos. Usually, we define the complex exponential as \(\displaystyle \exp(z)=\sum_{n=0}^\infty \frac{z^n}{n!}\) and then \(\displaystyle \cos(x)\) and \(\displaystyle \sin(x)\) are the real and imaginary parts of \(\displaystyle \exp(ix)\), which gives their series expansion.

This is the most convenient definition of cos and sine (and it gives the series immediately) but there are other ways to procede without complex numbers. For instance using a differential equation: \(\displaystyle \cos\) can be defined as the unique solution of \(\displaystyle y''+y=0\) with \(\displaystyle y(0)=1\) and \(\displaystyle y'(0)=0\) (if we prove that it indeed exists and is unique). Then we prove that the power series yields another solution, so it must be equal to \(\displaystyle \cos\). There are probably other ways, depending what you know about sine and cosine.
Better yet, define \(\displaystyle \sin{x}\) and \(\displaystyle \cos{x}\) as the vertical and horizontal lengths on the unit circle made by angle \(\displaystyle x\). Then it demonstrates the NEED to have a series representation, so that the function can be evaluated at any value of \(\displaystyle x\).
 

Laurent

MHF Hall of Honor
Aug 2008
1,174
769
Paris, France
Better yet, define \(\displaystyle \sin{x}\) and \(\displaystyle \cos{x}\) as the vertical and horizontal lengths on the unit circle made by angle \(\displaystyle x\).
Out of sheer curiosity, or just to tickle you: how would you define the angle \(\displaystyle x\)? and prove the series expansion from there?

I guess the historical definition would be that the angle \(\displaystyle x\) defines an arc of length \(\displaystyle x\) on the unit circle. This leads to the "definition" (needs some care) \(\displaystyle \psi(\sin x)=x\) for \(\displaystyle \sin x\), where \(\displaystyle \psi(u)=\int_0^{u}\frac{1}{\sqrt{1-t^2}}dt\). The function \(\displaystyle \psi\) is obviously differentiable, and \(\displaystyle \psi'(u)=\frac{1}{\sqrt{1-u^2}}\). Since \(\displaystyle \sin\) is the inverse function of \(\displaystyle \psi\) (on some intervals), it is differentiable as well and \(\displaystyle \sin' x=\frac{1}{\psi'(\sin x)}=\sqrt{1-\sin^2 x}=\cos x\) if \(\displaystyle x\) is on the good interval. Similarly, we would find \(\displaystyle \cos'x=-\sin x\) and conclude that \(\displaystyle \cos\) satisfies the equation \(\displaystyle y''+y=0\), \(\displaystyle y(0)=1,y'(0)=0\). We could also compute the derivatives at 0 and use the Taylor formula with integral remainder (to be rigorous) for a proof with the ideas of your first post.
 

Prove It

MHF Helper
Aug 2008
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Out of sheer curiosity, or just to tickle you: how would you define the angle \(\displaystyle x\)? and prove the series expansion from there?

I guess the historical definition would be that the angle \(\displaystyle x\) defines an arc of length \(\displaystyle x\) on the unit circle. This leads to the "definition" (needs some care) \(\displaystyle \psi(\sin x)=x\) for \(\displaystyle \sin x\), where \(\displaystyle \psi(u)=\int_0^{u}\frac{1}{\sqrt{1-t^2}}dt\). The function \(\displaystyle \psi\) is obviously differentiable, and \(\displaystyle \psi'(u)=\frac{1}{\sqrt{1-u^2}}\). Since \(\displaystyle \sin\) is the inverse function of \(\displaystyle \psi\) (on some intervals), it is differentiable as well and \(\displaystyle \sin' x=\frac{1}{\psi'(\sin x)}=\sqrt{1-\sin^2 x}=\cos x\) if \(\displaystyle x\) is on the good interval. Similarly, we would find \(\displaystyle \cos'x=-\sin x\) and conclude that \(\displaystyle \cos\) satisfies the equation \(\displaystyle y''+y=0\), \(\displaystyle y(0)=1,y'(0)=0\). We could also compute the derivatives at 0 and use the Taylor formula with integral remainder (to be rigorous) for a proof with the ideas of your first post.
The angle \(\displaystyle x\) is the number of lengths of the radius on the circumference of the circle, with the positive angle being the anticlockwise direction. In the case of the unit circle, since the radius is \(\displaystyle 1\), the angle is just the arc length. Since you could go around the unit circle as many times as you like, you will be able to have an arc length greater than (or less than if you go in the other direction) \(\displaystyle 2\pi\).