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$\exp:x\mapsto \sum\limits_{n=0}^{+\infty}\cfrac{1}{n!}x^n$

$\cos:x\mapsto \Re\left(\exp \left(i x\right)\right)=\sum\limits_{n=0}^{+\infty}\cfrac{\left(-1\right)^n}{\left(2n\right)!}x^{2n}$

$\sin:x\mapsto \Im\left(\exp \left(i x\right)\right)=\sum\limits_{n=0}^{+\infty}\cfrac{\left(-1\right)^n}{\left(2n+1\right)!}x^{2n+1}$

I've never really learned how these functions were constructed. I learned a lot of things with their differential equation (for $\exp$) or unit circle (for $\cos$ and $\sin$) definition and then when I learned about series, the teacher just said "we could use those series as definitions and find all the properties you know about those functions but we don't have enough time" but we still did some as exercises on series.

I do get how we get most of the properties: mostly Cauchy products and a few other things. But one thing I do not get is how we find the specific values.

For example, $\sin(\pi)=0$ or maybe this is used as a definition for $\pi$. But then how do you find $\sin\left(\cfrac{\pi}{2}\right)=1$? And how do we even prove there is such a number $\pi$?

J. M. ain't a mathematician
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xavierm02
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3 Answers3

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We give a brief sketch on how to introduce trigonometric functions through series. We will assume that basic properties of series have been proved. Yes, we need to define $\pi$. We will define it in a way close in spirit to what you did. Our aim is to set up some useful machinery, enough to fully answer your question about $\sin(\pi/2)$.

The Pythagorean Identity: By properties of series, the derivative of $\sin x$ is $\cos x$, and the derivative of $\cos x$ is $-\sin x$. Now differentiate $\sin^2 x+\cos^2 x$. We get $2\sin x\cos x-2\cos x\sin x$. This is $0$, so $\sin^2 x+\cos^2 x$ is constant. It is easy to show that the constant is $1$.

The Addition Laws: We prove the Addition Laws for sine and cosine. Fix $y$, and let $$\begin{align}f(x)&=\sin(x+y)-\left(\sin x\cos y+\cos x\sin y\right) \\ g(x)&=\cos(x+y)-\left(\cos x\cos y-\sin x\sin y\right)\end{align}$$ It is easy to verify that $f'(x)=g(x)$ and $g'(x)=-f(x)$. Now let $S(x)=f^2(x)+g^2(x)$. We find, as in an earlier calculation, that $S'(x)=0$. So $S(x)$ is constant. But easily by substitution $S(0)=0$. It follows that $S(x)$ is identically $0$, and the Addition Laws for sine and cosine follow.

On Defining $\pi$: We can use the series to estimate $\cos 1$, and show that it is strictly between $1/2$ and $13/24$. From this it follows by $\cos 2x=2\cos^2 x-1$ that $\cos 2$ is negative. By continuity, there is a smallest positive number $q$ such that $\cos q=0$. Define $\pi$ to be $2q$.

The Value of $\sin(\pi/2)$: Since $\cos q=0$, we have by the definition of $\pi$ that $\cos(\pi/2)=0$. By the Pythagorean Identity, we have $\sin(\pi/2)=\pm 1$. But it cannot be $-1$, for since $\pi/2$ is the smallest positive zero of $\cos x$, the sine function is increasing from $0$ to $\pi/2$.

Remark: It looks like a lot of work, but once the basics have been established, the rest goes smoothly. We proved the full Addition Laws instead of the fragment needed for $\sin(\pi/2)$ to make it clear that other standard facts are not difficult to derive.

André Nicolas
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  • So we never actually get a value from the series. We deduce formulas from the series and then try to combine the good formulas to get an equation we can solve by using previously known values. Right? – xavierm02 Apr 02 '13 at 18:20
  • In the calculation I used $\sin(0)=0$, $\cos(0)=1$. These are obtained directly from the series. Then $\pi$ is defined as the smallest positive real $b$ such that $\sin b=0$. That there is such a $b$ can be proved by a series manipulation. But it is easier to first prove the Addition Laws. I could add proof of existence of smallest positive $b$ with $\sin b=0$. Not hard, but takes a few steps. – André Nicolas Apr 02 '13 at 18:33
  • Ok thank you very much, I got the idea :) I somehow thought that you had to compute values with some ninja technique using the series but if it's done by proving formulas first and then using them, I can (probably) do it :) – xavierm02 Apr 02 '13 at 18:49
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    @xavierm02: I made a couple of substantial changes. Changed the definition of $\pi$. Sorry, but it saves a few lines, makes the exposition smoother. And I added a full proof that the object $\pi$ exists. Interesting methodological point is that cosine is easier to work with than sine, it should be the main function, with sine secondary. For example $\cos(nx)$ is given by nicer formulas than $\sin(nx)$. – André Nicolas Apr 02 '13 at 19:08
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A related problem. The idea of Taylor series is to approximate a function in terms of simpler functions (polynomials), that we can multiply, add, and subtract easily. For instance,

$$ \sin(\pi)\approx \pi-\frac{\pi^3}{3!}+\frac{\pi^5}{5!}-\frac{\pi^7}{7!}+\frac{\pi^9}{9!} = 0.00692526980. $$

Note that, the more number of terms you take the closer you get to the exact value $0$.

Community
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Mhenni Benghorbal
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  • But how do you prove it tends to $0$? I'm not asking about how to numerically get the value. I'm asking about how you prove it. – xavierm02 Apr 02 '13 at 15:43
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    I’d say that you need a definition of $\pi$ first. If you’re using the classical one about circles, then you need to connect $\sin$ and $\cos$ to circles. As @André has said, a convenient definition of $\pi$ is the smallest positive number whose sine is zero; but then your question is already answered. – Lubin Apr 02 '13 at 18:39
  • @Lubin: I do not think the OP is interested in that and we don't want to make things complicated more than it should be. Thanks for the comment. – Mhenni Benghorbal Apr 02 '13 at 19:15
  • @xavierm02: The radius of convergence of the sine series is infinity and you can prove that using the ratio test or the root test. – Mhenni Benghorbal Apr 02 '13 at 19:18
  • I know that. It's just that I wondered how you calculated specific values exactly and I thought you had to use some magic trick on the series or somehow identify it with another series and deduce the value from it. Whereas in fact you just get some equations and deduce the values from it. – xavierm02 Apr 02 '13 at 19:37
  • @xavierm02: I just explained to you in my answer the basic idea behind Taylor series. – Mhenni Benghorbal Apr 02 '13 at 19:38
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There are various ways to define $\cos$ and $\sin$, be it as solutions of differential equations, via the exponential function, or as infinite series. In all cases you have to prove something of the following sort: There is a real number $\tau$ between ${3\over2}$ and ${8\over5}$, say, such that $\cos t>0$ for $0\leq t<\tau$ and $\cos \tau=0$. This number $\tau$ is called ${\pi\over2}$. From Euler's formula (or similar) it then follows that $\exp(2\pi i)=1$, which in turn implies the periodicity of $\cos$ and $\sin$.

For the proof you have to use properties of $\cos$ and $\sin$ stemming from the differential equation, or estimates obtained from looking at the exponential series. There is no appeal to elementary euclidean geometry involved.

A proper treatment can be found here: Walter Rudin, Principles of mathematical analysis, chapter 8, section 3: The trigonometric functions.

Christian Blatter
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