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Pi is the ratio of circumference of a circle to its diameter.

Okay. Got that, easy enough. Now, why does the following equality hold true?

$$ \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots$$

Why do we get closer to Pi from summing up many rational fractions? Granted, we can do a little bit of calculus here, find out that it all comes down to either fundamental theorem of calculus or Taylor series. Either route you pick the key here is the fact that derivatives of inverse trigonometric functions are rational or at least non-trigonometric.

If we dig deeper, it becomes clear that it all depends on our choice of units for trigonometry. Why are radians the right way to measure angles? It all boils down to the well-known limit:

$$\lim\limits_{x \to 0} \frac{\sin x}{x} = 1 $$

The above equality is true if and only if $ x $ is measured in radians. Why does sine have to be $ 2\pi $-periodic for that limit to equal 1?

My guess is that period of sine is $ 2\pi $ because it's the only value "makes" that limit equal 1. That's the only sane explanation I could devise. Isn't there a more obscure mathematical connection here? Resolving the problem via falling in a virtuous circle is the only way to go here?

Still, why is it $ 2\pi $ and not, say, $ \sqrt{42} $? I apologize in advance for insufficient clarity of my writing, English is not my mother tongue.

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    $\begingroup$ Maybe this is helpful Why is radian so common in maths?. $\endgroup$ – kingW3 Jul 15 '17 at 15:11
  • $\begingroup$ ...radians are a measure of length around the unit circle, so it seems natural to use them as a measure of an angle around the unit circle as well. $\endgroup$ – Franklin Pezzuti Dyer Jul 15 '17 at 15:11
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    $\begingroup$ IMO, a better question would be "Why do we use degrees?" It appears nowhere in mathematics AFAIK, and radians aren't bad at all... $\endgroup$ – Simply Beautiful Art Jul 15 '17 at 15:14
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    $\begingroup$ I think your issue is in the statement "Got that, easy enough". The definition of $\pi$ as ratio of circumference to the diameter of a circle is not easy enough but rather it is made to seem like it is easy enough. The idea that a circumference can be measured is intuitive but difficult to establish and define in a rigorous manner. If you take all the trouble to handle these things rigorously, the series of $\pi/4$ comes into the category of "Got that, easy enough". $\endgroup$ – Paramanand Singh Jul 15 '17 at 16:26
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    $\begingroup$ Also the limit at the end of your question is independent of the unit of measurement of angles. The symbol $\sin x$ just represents a very popular and interesting function of a real variable $x$ and contrary to its interesting and popular nature, it's definition is significantly difficult compared to simpler and less interesting function like $x^{2}$. When one studies circular functions in calculus it is best to ditch the idea of "sine of an angle" and instead focus on "sine of a real number $x$" similar to "square root of a real number $x$". $\endgroup$ – Paramanand Singh Jul 15 '17 at 16:39
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There are two common ways of defining $π$, and they are interestingly and non-trivially equivalent. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $


If you wish to define $π$ as the ratio of a circle's circumference to diameter, you would first have to define arc length, which is highly non-trivial. You also need to prove that every circle has the same ratio, otherwise your choice of definition is ill-founded. After you do that properly, you can then take a point that traces the unit circle from $(1,0)$ anti-clockwise at speed $1$, and then define $(\cos(t),\sin(t))$ (for real $t$) as the coordinates of that point at time $t$ (equivalently after it has traced an arc of length $t$). Then you can prove that $\lfrac{\sin(t)}{t} \to 1$ as $t \to 0$. Many textbooks give a 'proof' that squeezes between areas, but this is invalid unless you also have proven (or assumed without proof) that the area of the unit circle is $π$. No, it is not obvious and not easy without some calculus. But it should be now clear why the limit is completely determined by your choice of definition of $π$. If you go this route, you also will have trouble defining the complex-valued trigonometric functions later in a natural and intuitive way, which is why I prefer the other common approach below.


The main alternative is to define $\exp,\cos,\sin$ expressly to solve useful differential equations. $\exp$ is the unique complex-valued function such that $\exp' = \exp$ and $\exp(0) = 1$. The motivation is that using such a function $\exp$ we can solve $\lfrac{dy}{dx}+f(x)y = g(x)$ for variables $x,y$ given functions $f,g$, by considering $\lfrac{d}{dx}\Big(y·\exp(\int f(x)\ dx)\Big)$. Also, simple harmonic motion requires solving $f'' = f$ for a real function $f$. By considering the desired functions' behaviour around $0$, we can discover them intuitively by using a polynomial approximation as described here. It is non-trivial to prove that taking the limit (the infinite series) produces the functions we want, and the linked post shows an elementary proof.

The same intuition suggests the series expansions for $\cos,\sin$ as well, and we can observe that we can define $\cos,\sin$ via $\cos(z) = \lfrac12(\exp(iz)+\exp(-iz))$ and $\sin(z) = \lfrac1{2i}(\exp(iz)-\exp(-iz))$ to get the desired expansions, and then can easily prove all their properties based on the properties of $\exp$. For example $\sin' = \cos$ and $\cos' = -\sin$ (by the chain rule on the above definition). It is a matter of preference whether one proves general facts of power series to get termwise differentiation, but here I am describing a bare-hands elementary approach.

I shall sketch how to get all the basic properties of $\exp$. From $\lfrac{d}{dz}(\exp(z)·\exp(-z)) = 0$ we get $\exp(z) \exp(-z) = \exp(0)·\exp(-0) = 1$ (by a suitable application of Rolle's theorem). Similarly $\lfrac{d}{dz}(\exp(z+w)·\exp(-z)) = 0$ for any constant $w$, and hence $\exp(z+w) = \exp(z)·\exp(w)$. Thus $1 = \exp(0) = \exp(it)·\exp(-it) = (\cos(t)+i\sin(t))·(\cos(t)-i\sin(t)) = \cos(t)^2+\sin(t)^2$. Now for every real $t$ we can see from the infinite series that $\cos(t),\sin(t)$ are also real and hence are the $x,y$-coordinates of $\exp(it)$. Thus $|\exp(it)| = 1$. Next we have $\left|\lfrac{d}{dt}(\exp(it))\right| = |i\exp(it)| = 1$ and so $\exp(it)$ travels along the unit circle at constant unit speed.

At this point, if you wish you can use arc length to define $π$ as in the first method (with its caveats) and get $\exp(i2π) = 1$, but I shall show a purely analytic alternative approach that literally follows the intuitive idea that $p = \exp(it)$ traces the unit circle as the real parameter $t$ increases. Note that $p$ cannot stay within the top-right quadrant (namely $Re(p) > 0$ and $Im(p) > 0$), otherwise by the extreme value theorem it would reach some left-most position but at that point we would have $Re(\lfrac{dp}{dt}) = Re(ip) = -Im(p) < 0$ and so it must continue moving left-ward (by Rolle's theorem again). Since $Im(\lfrac{dp}{dt}) = Im(ip) = Re(p) > 0$ in that quadrant, $p$ must first exit the quadrant by crossing the imaginary axis rather than the real axis. That crossing point must be $i$, and we can define $u$ to be the smallest positive root of $\cos$ so that $i = \exp(iu)$. Then $\exp(i4u) = i^4 = 1$, and therefore $\exp$ has period $4ui$. One is free to define $π = 2u$.


Whichever property you use to define $π$, it takes non-trivial work to prove the properties that you did not use to define it. So there is no circularity really, and the elegance of the relation between the exponential function and trigonometric functions actually has quite complex underpinnings, which makes it marvelous!

Also, there could have been a misconception in your question of why an infinite sum of rationals can reach $π$. The fact is that every real number can be reached by some infinite sum of rationals. The decimal representation is one common instance, where each real number is represented by some infinite sum $n+\sum_{k=0}^\infty a_k 10^{-k}$ where $n$ is an integer and $a_0,a_1,\cdots$ are integers in the range $[0..9]$. It is only because $π$ is special that we are interested in all sorts of infinite series that sum to it. The one you mention is an especially curious instance. If you want only infinite sums of rationals following some algorithmic pattern, then what you get is exactly the computable reals. Nearly all mathematical constants that mathematicians have defined are in fact computable, but some like Chaitin's constant are not.

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  • $\begingroup$ +1 for using decimals to demonstrate that every real number is the sum of an infinite series of rational terms. I wonder why many people find this hard to believe when almost everyone is familiar with the idea of decimals. $\endgroup$ – Paramanand Singh Jul 16 '17 at 8:12
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    $\begingroup$ @ParamanandSingh: It's because decimals are intrinsically non-trivial. For instance, how exactly does one multiply two infinite decimals? No computable algorithm would be able to determine the digits in finite time, given two decimals as digit-sequence programs. This is already the case for addition, so already one must sort of have the notion of equivalence classes even when using decimal reals. People don't think enough about what decimals mean, so they stumble on infinite series too. And your comment made me notice a silly error that I've fixed. =) $\endgroup$ – user21820 Jul 16 '17 at 9:27
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Concerning the limit$$\displaystyle\lim_{x\to0}\frac{\sin x}x=1,$$you stated that it “is true if and only if $x$ is measured in radians”. This is false. The sine function is a function from the reals into the reals. So, when, in the context of this limit, someone speaks of $\sin x$, then $x$ is a number, not an angle, and thefore asserting that “$x$ is measured in radians” makes no sense.

Of course, originally the sine was the sine of an angle, not the sine of a number. But we are dealing with the limit $\lim_{x\to0}\frac{\sin x}x$ here and, in this context, the sine has always meant a function from the real numbers into the real numbers. That's how Euler dealt with this limit in his Introductio in analysin infinitorum, published in 1748. I suggest that you read the short article Why the sine has a simple derivative, by V. Frederick Rickey.

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  • $\begingroup$ I agree with the intended meaning behind your answer. But there are multiple ways in history to define the trigonometric functions, and that makes things messier here. Specifically, you are right about $\sin$ as defined by the Taylor series, for examples, where there's no notion of angles. However, historically $\sin(x)$ was defined geometrically, and hence required the notion of angles. It then becomes important to set down how angles are measured. Of course, in that case the limit has no meaning because the "$x$" can't be both a length and an angle. $\endgroup$ – user21820 Jul 21 '17 at 8:53
  • $\begingroup$ And I'd be glad to upvote your answer if you clarified that issue, since it's indeed an important issue to know exactly what type of objects we are dealing with. $\endgroup$ – user21820 Jul 21 '17 at 8:56
  • $\begingroup$ I am not here chasing for upvotes. Besides, since I wrote that we are dealing with numbers and only with numbers here, I have no idea about how could I possibly explain better “exactly what type of objects we are dealing with” even if my life depended on it. $\endgroup$ – José Carlos Santos Jul 21 '17 at 16:21
  • $\begingroup$ I meant that you have already explained well "exactly what type of objects we are dealing with". The only thing I'm saying is that your answer assumes without stating that $\sin$ is defined as a function from reals to reals. That was surely not how it was originally defined geometrically, which is the true source of the confusion in the asker's inquiry. If you could clarify a bit in that direction, your answer would be complementary to mine. I don't mind editing your answer or mine, if you wish, since I just wanted to give you the credit of pointing out the issue in the question. $\endgroup$ – user21820 Jul 22 '17 at 5:38
  • $\begingroup$ @user21820 Happy now? $\endgroup$ – José Carlos Santos Jul 22 '17 at 8:03
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Radian as a unit for an angle was chosen because it's natural: it's the length of the corresponding arc in a unit circle, so the crucial inequality $\sin x<x<\tan x$ (proving that limit) for $x\in(0,\pi/2)$ can be seen immediately from geometry. If you choose another unit, $\pi$ won't go away, and if it's degrees, you'll just get an annoying factor of $\pi/180$ in a lot of equations. Because choosing another unit, you'd have a function $\sin_a(x)=\sin ax$ (if it's degrees, $a=\pi/180)$, and $\sin_a(x)/x\rightarrow a$ as $x\rightarrow0.$

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Well, $2\pi$ is the circumference of a circle of radius $1$. Hence lengths of arcs of the unit circle are fractions of $2\pi$. For example, if I walk one fifth of the way around the circumference of the unit circle, I have walked a distance of $\frac{1}{5}\cdot2\pi$. When radian measure is used, angle measure coincides with arc-length measure.

Now the statement $$ \lim_{x\to0}\frac{\sin x}{x}=1 $$ (with $x$ in radians) is closely related to the phenomenon that the length of a chord and the length of the arc with the same endpoints as the chord approach each other as the angle subtended by the arc approaches zero. If radian measure is used, the length of the arc subtended by the angle $2x$ is also $2x$, while the length of the corresponding chord is $2\sin x$. Therefore the ratio $\frac{2\sin x}{2x}$ approaches $1$ as the angle approaches $0$.

If some other measure of angle were used, the length of the arc of the unit circle subtended by the angle $2x$ would no longer be $2x$, but some other proportionality constant times $x$. For example, if degree measure were used, the arc length would be $\frac{2\pi}{360}\cdot2x$. The chord length would still be $2\sin x$, however. In this situation the ratio $$\frac{2\sin x}{\frac{2\pi}{360}\cdot2x} $$ would approach $1$ as the angle approached $0$.

Another way to answer your question: the period of $\sin$ is $2\pi$ when radian measure is used because when you have walked a distance $2\pi$ around the circumference of the unit circle, you have returned to the point where you started.

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