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i want to know if exist or if you can develop or give me ideas of a proof to show that the least number for which sine is periodic is $2\pi$ $$\neg \{\exists n\in \mathbb{R} \wedge n < 2\pi: \sin(x)=\sin(x+n)\}$$ $$\neg \{\exists n\in \mathbb{R} \wedge n < 2\pi: \cos(x)=\sin(x+n)\}$$

Is not enough to proof it by definition of fundamental period, i am wondering for a proof that avoids definitions of certain properties (i am not avoiding properties, just definitions, in order to show that from the inner core of geometry and logic). Can we build trigonometry without the definition of fundamental period?

We are allowed to use ALL THE THEORY we know about, analysis, logic, model theory, geometry, combinatorics, even topology, if you want, we just may try to avoid if it`s possible the definition of the periodicity. If that is a just a definition then....

I am not taking out the rest of the axioms of geometry, i am just trying to figure out wether or not the periodicity of trigonometrical functions is an axiom; or you can use any axiom, property, theorem that do not depends on that fact to proof it. Off course, we could not use Fourier series because that theory depends mainly on the periodicity of those functions, if we go further, i think that the sine function depends integrally on the existence of a least bound of periodicity.

Thank you very much

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    $\begingroup$ If we are going to pick nits, what about $n=-2\pi$? But more seriously, what definition of the sine function is the starting point? $\endgroup$ – Harald Hanche-Olsen Apr 3 '13 at 11:15
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    $\begingroup$ If you want a Principia Mathematica style proof, you are going to have to be extremely explicit about the axioms you are allowed to use. What are your "axioms of trigonometry"? Which is being taken out? (I'm also sure @Asaf would appreciate being notified about your response.) $\endgroup$ – user642796 Apr 3 '13 at 11:26
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    $\begingroup$ Every axiom that I know? So the axioms "The periodicity of cosine is $2\pi$" and "$\sin \theta = \cos ( \frac{\pi}{2} - \theta )$" are allowed? $\endgroup$ – user642796 Apr 3 '13 at 11:36
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    $\begingroup$ Periodicity of $\tan$?;) $\endgroup$ – Eckhard Apr 3 '13 at 11:40
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    $\begingroup$ 't Seems you are misunderstanding what a "Princ. Math.-style proof" is all about. The idea is to give a definite set of rules, and then use them to derive and formally found e.g. analysis, trigonometry etc. etc. by defining the corresponding notions in the limited language and showing that they agree with what we expect. $\endgroup$ – Lord_Farin Apr 3 '13 at 11:42
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Ok at first I will say what definitions I use. \begin{align*} \exp(x)&=\sum_{k=0}^\infty \frac{x^k}{k!}\\ \sin(x)&=\frac{1}{2i} \left(\exp(ix)-\exp(-ix)\right)\\ \cos(x)&=\frac{1}{2}\left(\exp(ix)+\exp(-ix)\right) \end{align*} In special with those definitions I get the so called euler forlmula: $$\exp(ix)=\cos(x) +i \cdot \sin(x)$$ Taking a unit circle in the complex plane, one can see that $\cos(x)$ is the projection on the real axis of $\exp(ix)$, when $x$ is real.

For the next step we will need the functional equation of the exponential function

$$\exp(x+y)=\exp(x)\cdot \exp(y)$$

We prove this equation, for the proof we will need the binomial theorem and the cauchy product. At first we recall the Binomial Theorem $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k} $$ The cauchy product is a special way to multiply two series, when both series are absolut convergent. As our series is absolute convergent for all $x$ we will just write the cauchy product $$\left(\sum_{k=0}^\infty a_k\right) \cdot \left( \sum_{n=0}^\infty b_n\right)= \sum_{k=0}^\infty \sum_{n=0}^k a_{k-n} b_n $$

With those informations \begin{align*} \exp(A+B)&=\sum_{j=0}^\infty \frac{1}{j!} (A+B)^j \\ &=\sum_{j=0}^\infty \frac{1}{j!} \sum_{k=0}^j \binom{j}{k} A^{j-k} B^k\\ &=\sum_{j=0}^\infty \sum_{k=0}^j \frac{1}{(j-k)!} A^{j-k} \frac{1}{k!} B^k\\ &=\left(\sum_{j=0}^\infty \frac{1}{j!} A^j\right) \cdot \left(\sum_{k=0}^\infty \frac{1}{k!} B^k \right)\\ &=\exp(A) \cdot \exp(B) \end{align*}

(I copied myself here, I made this proof for the matrixexponential function)

Now we look at \begin{align*} \cos(x+y)+i\cdot \sin(x+y)&=\exp(i(x+y))\\ &= \exp(ix) \cdot \exp(iy)\\ &= (\cos(x)+i\sin(x)) \cdot (\cos(y)+i\sin(y))\\ &= \cos(x)\cos(y) - \sin(x)\sin(y) + i (\sin(x)\cos(y)+\sin(y)\cos(x)) \end{align*}

Looking only at the imaginary part we see that $$\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$$

Another definition I use

$\pi$ is the smallest positive zero of the $\sin$ function.

Now let's assume that $$\sin(n+x)=\sin(x)$$ for all $x$, then we have $$\sin(x+n)=\sin(x)\cos(n)+\sin(n)\cos(x)=\sin(x)$$ At first we look at the special case $x=0$, with the series above we know that $$0=0\cdot \cos(n)+ \sin(n)\cdot 1$$ Ok so we know that $\sin(n)=0$ must be true hence $n$ must be an integer multiple of $\pi$, because else $0<n-\pi<\pi$ and $$\sin(n-\pi)=\sin(n)\cos(-\pi) + \sin(-\pi)\cos(n)=0$$ would be a smaller positive zero than $\pi$ in contradiction to that $\pi$ is the smallest positive zero.

Now we try if $$\sin(\pi+\frac{\pi}{2})=\sin(\frac{\pi}{2})$$ \begin{align*} \sin(\pi+\frac{\pi}{2})&= \sin(\pi)\cos(\frac{\pi}{2})+\sin(\frac{\pi}{2})\cdot \cos(\pi)\\ &=-\sin(\frac{\pi}{2})\\ &\neq \sin(\frac{\pi}{2}) \end{align*} Hence $\pi$ can't be the period of $\sin$ and the next possible choice $2\pi$ must be the period.

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  • $\begingroup$ Wow, that is a big step, i would beg you to complement your answer with Cauchy product and binomial product $\endgroup$ – Sebastian Valencia Apr 3 '13 at 11:53
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    $\begingroup$ For me is not clear why if sin(n)=0, n must be a multiple of $\pi$, is n an integer multiple? $\endgroup$ – Sebastian Valencia Apr 3 '13 at 11:56
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    $\begingroup$ We definied $\pi$ to be the smallest positive number such that $sin(x)=0$. If $n$ is not an integer multiple you have $\sin(n-\pi)=\sin(n) \cos(-\pi) + \sin(n) \cos(-\pi)=0$ but $\pi$ and $n$ are both zeroes of the sin function, and hence $\sin(n-\pi)$ would be a zero of the sin function and be smaller than $\pi$ $\endgroup$ – Dominic Michaelis Apr 3 '13 at 12:02
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    $\begingroup$ @DominicMichaelis, that proves that there can be no zero of $\sin$ between $\pi$ and $2\pi$, which is considerably weaker than the statement that $n$ must be a multiple of $\pi$. (It's also sufficient, though, once you show that $\pi$ isn't a period and $2\pi$ is). $\endgroup$ – Peter Taylor Apr 3 '13 at 12:32
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    $\begingroup$ @SebastianGriotberg updated my answer sry it took a bit longer $\endgroup$ – Dominic Michaelis Apr 3 '13 at 12:53
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Maybe you want something like this: If $f\colon \mathbb R\to\mathbb R$ is a function then the set of periods $$P:=\{p\in\mathbb R\mid \forall x\in\mathbb R\colon f(x+p)=f(x)\} $$ is an ideal in the $\mathbb Z$-module $\mathbb R$ (i.e. sums and differences of periods are also periods). Then $P\cap(a,b)=\emptyset$ with $a<b$ implies that $P=0$ or $P=p\mathbb Z$ for some $p\ge b-a$ (which is a more explicit way of saying that any non-principal ideal in $\mathbb R$ is dense).

Apply these facts to the sine function, about which we assume

  1. $\sin\colon\mathbb R\to\mathbb R$ has $2\pi$ as a period
  2. $\sin x>0$ for $0<x<\pi$
  3. $\sin 0 =0$
  4. $\exists x\in\mathbb R\colon \sin x<0$

Then we have $P\cap(0,\pi)=\emptyset$ because $\sin(x)>0$ for $0<x<\pi$ and hence $\sin(0+x)\ne\sin(0)=0$ for such $x$. Therefore $P=p\mathbb Z$ with $p\ge\pi$ and we have $2\pi=kp$ for some $k\in\mathbb Z$. This leaves only the cases $p=\pi$ and $p=2\pi$. Since $\sin x\ge0$ for $0\le x\le\pi$, the former would imply $\sin x\ge0$ for all $x$, which is absurd.

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  • $\begingroup$ Clarify me a thing, those assumptions, were made for the proof, but why it is needed that for the ideal, p must be greater than pi $\endgroup$ – Sebastian Valencia Apr 3 '13 at 12:09
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    $\begingroup$ @SebastianGriotberg One $>$ had to be replaced with $\ge$, thanks. Thus $p=\pi$ is not immediately excluded by checkingonly $\sin 0$ vs. $\sin p$ as of course it happens that $\sin\pi=\sin 0$. But it is excluded for many other easons - in fact conditions 2 and 3 above suffice, but we can be very specific using $\sin(\frac\pi2+\pi)=-1\ne1=\sin\frac\pi2$ if we like. $\endgroup$ – Hagen von Eitzen Apr 3 '13 at 15:12

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