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This is something I was shaky about when I took calculus, real analysis, and then complex analysis. Specifically, is the following chain of definitions circular in any way?

  1. Define the set $\mathbb{N}$ of natural numbers, define basic arithmetic operations
  2. Extend $\mathbb{N}$ to $\mathbb{Z}$, then to $\mathbb{Q}$, and extend the arithmetic operations to these sets as well
  3. Use the arithmetic operations to define distance and use that to define convergence, limit, and related concepts
  4. Use 3 to extend $\mathbb{Q}$ to $\mathbb{R}$
  5. Define cosine and sine on $\mathbb{R}$ using the series definitions and then define other trig functions using cosine and sine
  6. Define $e=\lim(1+1/n)^n$ and define $\pi$ as $\ldots$
  7. Use 5 and 6 to derive the usual properties of trig functions, of $e$, and of $\pi$
  8. Define $e^x$, $x$ real, as $\lim(1+x/n)^n$
  9. Finally, define, for $z=x+iy$, $e^z=e^x(\cos y+i\sin y)$, and derive properties of $e^z$ from 7 and 8 and complex arithmetic operations.

I also have 3 auxiliary questions: (i) what is a standard / convenient definition for $\pi$ above? (ii) how does one fork off from 5 and 6 above to get the geometric interpretations of $\pi$ and the trig functions? (iii) is there a book (or a few books) that give students a sort of big picture ("big" for a guy like me) view above?

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  • $\begingroup$ $\pi$ can be defined as the unique number $x \in (3,4)$ such that $sin(x)=0$, as inferred from the series definition. $\endgroup$ – Andrea May 19 '16 at 0:00
  • $\begingroup$ I take $2\pi$ to be the ratio the circumference of a circle and its radius. $\endgroup$ – ncmathsadist May 19 '16 at 0:50
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    $\begingroup$ Another definition is $$\pi := 2 \cdot \int_{-1}^1 \sqrt{1-x^2} \ dx$$ $\endgroup$ – MathematicsStudent1122 May 19 '16 at 2:45
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    $\begingroup$ The definition in $(8)$ works for complex $x$ also and it is easily shown to be equivalent to the definition $(9)$. See math.stackexchange.com/a/1668179/72031 for the proof of going from $(8)$ to $(9)$ for complex $x$. $\endgroup$ – Paramanand Singh May 19 '16 at 10:37
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    $\begingroup$ Rudin's Principles of Mathematical Analysis (aka "Baby Rudin") takes pretty much precisely this approach, with all the basic theory, through power series, developed before the exponential and trigonometric functions are formally introduced. (They are occasionally used earlier in examples and exercises, but segregated from the formal theory.) $\endgroup$ – Barry Cipra May 19 '16 at 12:23
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The definition of $\pi$ is based on the route you take to define the circular functions $\sin x, \cos x$. The traditional approach based on the circle (that's why the name circular functions) is rigorous/intuitive/fruitful/easy. Many believe that the geometric definition based on circle is not rigorous, but I have shown in this answer that it is a fully rigorous approach. However, rigor always comes at a price and you need to pay it in any one of the following manner:

1) Establish that any arc of a circle has a length. This is based on the fact that the function $f(x) = \sqrt{1 - x^{2}}$ is monotone on $[0, 1]$ and the definition of arc-length. If we use this approach then $\pi$ is defined as $$\pi = 2\int_{0}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$$

2) Establish that any sector of a circle has an area. This is based on the fact that the function $f(x) = \sqrt{1 - x^{2}}$ is continuous on $[0, 1]$ and the definition of area of a plane region. If we use this approach then $\pi$ is defined as $$\pi = 4 \int_{0}^{1}\sqrt{1 - x^{2}}\,dx$$ If you use the series definition of $\sin x, \cos x$ then $\pi$ can be defined as $2\alpha$ where $\alpha$ is the least positive root of $\cos x = 0$.

Your second question deals with the geometric interpretation of $\pi$ and trigonometric functions. This is easy when you start off with series representation of circular functions as definition (well anything is easy if you use a powerful tool like power series and hence it is preferable not to use them as definitions). You need to establish $\sin^{2}x + \cos^{2}x = 1$ and then we can easily prove $$\pi = 2\int_{0}^{1}\frac{dx}{\sqrt{1 - x^{2}}}$$ Further with some effort we can show the monotone nature of $\sin x, \cos x$ on specific intervals and thus define inverse trigonometric functions. Then we can show that if $P = (\cos \theta, \sin\theta)$ is a point on the circle $x^{2} + y^{2} = 1$ then area of corresponding sector is $\theta/2$ and length of corresponding arc is $\theta$. This gives the geometric interpretation.

I have presented various approaches to the theory of exponential, logarithmic functions and circular functions on my blog. Most of the material there is a development of the ideas I found in my most favorite book "A Course of Pure Mathematics". This book not only gives the big picture (which may be available in other books too), but also the fine details (which are normally avoided or presented in terse fashion in many books).

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    $\begingroup$ Why the downvote?? $\endgroup$ – Paramanand Singh May 19 '16 at 11:52
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Your list seems to be ok. Although I would rather take the series definition $e^x = \sum\limits_{n=0}^\infty \frac {x^n}{n!} $. This gives you a definition for exponentials of complex numbers right from the beginning. Your limit definition for Euler's number follows easily. But of course your way works fine too.

As for $\pi$, a "standard" definition is that $\pi/2$ the smallest positive root of cosine. It's not hard to show the series for cosine converges everywhere. It's obviously 1 at $x=0$ and negative eventually, so it must have a root.

With this in hand all the geometric stuff can be developed with a little effort (but more than I care to type on my phone).

One great place to see a careful development of all of this stuff is Tom Apostol's Calculus. He has detailed proofs for almost all of this. About the only thing he shies away from is a full fledged construction of the reals (he sketches Dedekind cuts but omits some tedious details).

Edit: Cosine has a root.

Let $\cos(x) = \sum\limits_{n=0}^\infty \dfrac{(-1)^nx^{2n}}{(2n)!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\cdots$.

This means $\cos(0)=1$.

Consider $\cos(3) = \sum\limits_{n=0}^\infty \dfrac{(-1)^n3^{2n}}{(2n)!} = 1 - \dfrac{3^2}{2!}+\dfrac{3^4}{4!}-\dfrac{3^6}{6!}+\cdots$.

Let $n=2k-1$. Then $\dfrac{(-1)^n3^{2n}}{(2n)!}+\dfrac{(-1)^{n+1}3^{2(n+1)}}{(2(n+1))!} = -\dfrac{3^{4k-2}}{(4k-2)!}+\dfrac{3^{4k}}{(4k)!}=\dfrac{3^{4k-2}}{(4k-2)!}\left(-1+\dfrac{9}{(4k-1)(4k)}\right)$.

When $k \geq 3$, the above quantity is negative.

Therfore, $\cos(3)=1-\dfrac{9}{2}+\dfrac{81}{24}+\mbox{negative stuff}+\mbox{more negative stuff}+\cdots$ so $\cos(3)=-0.125+\mbox{negative stuff}<0$.

Therefore, since we "already know" cosine is continuous and we "know" the intermediate value theorem (as is the case in Apostol), cosine must have a root between $x=0$ and $x=3$.

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    $\begingroup$ Rudin's Real and Complex Analysis starts exactly with this construction in the Prologue. $\endgroup$ – lhf May 19 '16 at 2:53
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    $\begingroup$ Is it obvious from its series definition that $\cos x$ is eventually negative? $\endgroup$ – Barry Cipra May 19 '16 at 11:08
  • $\begingroup$ Maybe not "obvious" but it's also not terribly difficult. If I remember correctly, this is done in Apostol. I think it takes less than a page to establish this. $\endgroup$ – Bill Cook May 19 '16 at 13:38
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    $\begingroup$ @BillCook, if you first derive the identity $\sin^2x+\cos^2x=1$ (from $\exp(x+y)=\exp(x)\exp(y)$ and $\exp(ix)=\cos x+i\sin x$), then you can argue from the derivative formulas $(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$ that $\cos x$ can't be forever positive. (It starts out concave down, so to stay above the $x$-axis, it would need an inflection point....) $\endgroup$ – Barry Cipra May 24 '16 at 14:43

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