About sin & cos

  • #1

Main Question or Discussion Point

with the series representation of sin or cos as a starting point (you don't know nothing else about those functions), how to prove:

a. they are periodic.

b. the value of the period.
 

Answers and Replies

  • #2
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Are you talking about the Taylor series expansion of sin and cos?

If you are talking about them, then the period must be 2pi since the Taylor expansion is based on the use of radians as a measure of angles.

If something is periodic you can always check if f(x) = f(x+kP) where P is the period and k=0,1,2,3...
 
  • #3
jbunniii
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Show that the series expansions satisfy the differential equation that uniquely defines sine and cosine:

[tex]y'' = -y[/tex]

sine is the unique solution satisfying [itex]y(0) = 0[/itex] and [itex]y'(0) = 1[/itex].
cosine is the unique solution satisfying [itex]y(0) = 1[/itex] and [itex]y'(0) = 0[/itex].

You can differentiate the series term by term because they are power series that converge everywhere.
 
  • #4
Mute
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I don't think those replies do not show what the OP asks to show. The OP wants to know how to use the power series to demonstrate periodicity of the functions. Showing that the power series satisfy some other definition of cos and sin presumably defeats the purpose of the question.

Are you talking about the Taylor series expansion of sin and cos?

If you are talking about them, then the period must be 2pi since the Taylor expansion is based on the use of radians as a measure of angles.

If something is periodic you can always check if f(x) = f(x+kP) where P is the period and k=0,1,2,3...
All you know are that you have some functions s(x) and c(x), defined by their power series. x is just a dimensionless argument. It need not be interpreted as a radian. Even if you did interpret it as a radian, you cannot automatically deduce the period is [itex]2\pi[/itex]. [itex]\tan\theta[/itex]'s argument is in radians, but its period isn't [itex]2\pi[/itex].


Show that the series expansions satisfy the differential equation that uniquely defines sine and cosine:

[tex]y'' = -y[/tex]

sine is the unique solution satisfying [itex]y(0) = 0[/itex] and [itex]y'(0) = 1[/itex].
cosine is the unique solution satisfying [itex]y(0) = 1[/itex] and [itex]y'(0) = 0[/itex].

You can differentiate the series term by term because they are power series that converge everywhere.
Again, in terms of the OP's problem, that shows the functions s(x) and c(x) satisfy that differential equation, but under the conditions of the problem we don't know about the functions sin or cos, so this is in some sense the first discovery of these functions, and we haven't proven periodicity yet. Of course, perhaps this is the way it actually gets done - the method which I think the OP is asking for (see below) looks intractable to me.


For the OP: what I might try, in place of showing the power series satisfy some other definition of sine or cosine, is to use the definition of periodicity, f(x + P) = f(x) and the binomial expansion. e.g.,

[tex]s(x+P) = \sum_{k=0}^\infty \frac{(-1)^k(x+P)^{2k+1}}{(2k+1)!} = \sum_{k=0}^\infty \sum_{m=0}^{2k+1} \frac{(-1)^k}{(2k+1)!}\frac{(2k+1)!}{m!(2k+1-m)!} x^m P^{2k+1-m}[/tex]

I would then switch the order of summation (you'll have to figure out how to change the limits appropriately), and write the series as

[tex]\sum_{m=?}^? \frac{x^m}{m!}\left[\sum_{k=?}^{?} \frac{(-1)^k}{(2k+1-m)!} P^{2k+1-m}\right][/tex]

Then, show that for P = 2pi this reduces to the original series. In principle I think this should work, but in practice it may be too hard to do. A quick attempt at it give me a hypergeometric function for the inner function, and good luck dealing with that... especially since there's a cosine in the argument! I may have screwed something up, though.
 
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  • #5
jbunniii
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I suggested proving that the functions to which the series converge satisfy the indicated differential equation because there's no obvious way that I can see to prove periodicity directly from the series expansion.

Yes, you then have to show that any solution to the differential equation has period [itex]2\pi[/itex]. To do that is not entirely straightforward, but at least it can be done. See, for example, this thread:

https://www.physicsforums.com/showthread.php?t=274664

There's also a nice treatment in Knopp, "Theory and Application of Infinite Series." You can see the relevant section at Google Books, section 24, page 198. Hopefully this link works:

http://preview.tinyurl.com/2ddxs5f
 
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  • #6
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Mute++
 
  • #7
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Rudin works all this out in the Prologue of "Real and Complex Analysis", so maybe you should just get a copy of the book and read it. Here is an outline of his approach:

1. Define [tex]\exp(z)[/tex] by its series representation (for complex z).

2. Deduce [tex]\exp(a) \; \exp(b) = \exp(a+b)[/tex]

3. Define sin and cos by [tex]\cos t = Re(\exp(i t))[/tex] and [tex]\sin t = Im(\exp(i t))[/tex]-- in other words, by their series representations.

4. Show there is a smallest positive [tex]t_0[/tex] such that [tex]\cos t_0 = 0[/tex].
Define [tex]\pi[/tex] by [tex]\pi = 2 t_0[/tex].

5. Show that [tex]\exp(z + 2 \pi i) = \exp(z)[/tex]

6. [tex]\exp(\pi i / 2) = i[/tex] and [tex]\exp(z) = 1[/tex] if and only if [tex]z / (2 \pi i)[/tex] is an integer.

7. [tex]\exp[/tex] is a periodic function with period [tex]2 \pi i[/tex]

Periodicity of sin and cos then follow.
 

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