functional DE definition of sin, cos

[baxtrom]

Hello fellow hobby-scientists,

the normal differential equation-based definition of the trigonometric functions sin and cos are as solutions to the well-known second order DE y''(x) = -y(x). It occured to me that they can also be defined by a simple first order functional differential equation on the form

 

y'(x) = y(-x),

 

which for y(0) = 1 has the solution y(x) = sin(x) + cos(x). This obviously means it is possible to construct sin(x) as 0.5*(y(x) - y(-x)) and cos(x) as 0.5*(y(x) + y(-x)). As an attempted short section discussing this fact was swiftly and mercilessly rejected from the wikipedia article on trigonometric functions I'd like to ask you guys what you think. Does this definition add to the understanding of the nature of the sine and cosine functions and their apparent relationship to the exponential function (which is defined by y'(x) = y(+x), y(0) = 1..) ?

 

As this is my first entry, try to be nice

[DrRocket]

Hello fellow hobby-scientists,

the normal differential equation-based definition of the trigonometric functions sin and cos are as solutions to the well-known second order DE y''(x) = -y(x). It occured to me that they can also be defined by a simple first order functional differential equation on the form

 

y'(x) = y(-x),

 

which for y(0) = 1 has the solution y(x) = sin(x) + cos(x). This obviously means it is possible to construct sin(x) as 0.5*(y(x) - y(-x)) and cos(x) as 0.5*(y(x) + y(-x)). As an attempted short section discussing this fact was swiftly and mercilessly rejected from the wikipedia article on trigonometric functions I'd like to ask you guys what you think. Does this definition add to the understanding of the nature of the sine and cosine functions and their apparent relationship to the exponential function (which is defined by y'(x) = y(+x), y(0) = 1..) ?

 

As this is my first entry, try to be nice

 

 

The fundamental relationship of the sine and cosine to the exponential function is Euler's formula :

 

 

[math] e^{ix} = \displaystyle \sum_0^ \infty \dfrac {(ix)^n}{n!} = cos(x) + i \ sin(x)[/math]

 

So [math]cos(x) = Re(e^{ix})[/math] and [math]sin(x) = Im(e^{ix})[/math]

 

This is the way that sine and cosine are defined in a rigorous analytical treatment -- see for instance Rudin's Real and Complex Analysis.

 

I don't see how defining sine and cosine in terms of your differential equation adds clarity. For instance, without going back to a power series I don't see a clear path via the DE to showing that the solutions are periodic.

[baxtrom]

The fundamental relationship of the sine and cosine to the exponential function is Euler's formula :

 

 

[math] e^{ix} = \displaystyle \sum_0^ \infty \dfrac {(ix)^n}{n!} = cos(x) + i \ sin(x)[/math]

 

So [math]cos(x) = Re(e^{ix})[/math] and [math]sin(x) = Im(e^{ix})[/math]

 

This is the way that sine and cosine are defined in a rigorous analytical treatment -- see for instance Rudin's Real and Complex Analysis.

 

I don't see how defining sine and cosine in terms of your differential equation adds clarity. For instance, without going back to a power series I don't see a clear path via the DE to showing that the solutions are periodic.

 

Oui, that's the classical relationship using complex math. I just thought it was interesting to note that there is also a very simple first order DE with obvious similarities to the exponential DE which results in a trig solution. Just look at them, they belong together

 

[math] \frac{d y}{d x} = y(x) [/math],

which results in the exponential solution, and

 

[math] \frac{d y}{d x} = y(-x) [/math],

which gives a trigonometric solution. (I don't remember how to get the d:s straight in LaTeX ).

 

Perhaps it's just the romantic in me - a cold, rational mathematician would hardly give this pair of DEs a second thought. Still, a function class [math] y_{\nu}(x) [/math] which satisfies [math] y'(x) = y(\nu x ) [/math] is, if I'm not mistaken, only convergent for [math] \nu \in [-1 \ldots 1] [/math] and includes the trigonometric function above for [math] \nu = -1 [/math] and the exponential for [math] \nu = 1 [/math]. Obviously it's possible to construct a function class which contains any arbitrary pair of otherwise unrelated functions, yet still I find this apparant simple relationship between the exp and the trig functions in the real domain interesting. But wikipedia editors thought otherwise!

[timo]

Regardless of the quality of your idea or how interesting it is: It is not appropriate for Wikipedia. Wikipedia is not supposed to be a collection of thoughts that the authors find interesting. It is supposed to reflect mainstream knowledge, which an original idea of yours is not (by definition). Quality of an idea is not necessarily a criterion (->fascism).

[baxtrom]

Regardless of the quality of your idea or how interesting it is: It is not appropriate for Wikipedia. Wikipedia is not supposed to be a collection of thoughts that the authors find interesting. It is supposed to reflect mainstream knowledge, which an original idea of yours is not (by definition). Quality of an idea is not necessarily a criterion (->fascism).

 

Ah, that was what the editor meant by "unencyclopedic". It's an "original idea"! Well, perhaps "fascism" is a better topic for wikipedia than "original" thoughts on the relationship between mathematical functions. Good point! Still I can't imagine such a simple idea to be original.