68
$\begingroup$

This question is inspired by another one, asking to prove that something approximately equal to $1.2$ is bigger than something approximately equal to $0.9$. The numerical answer to this question was (expectedly) downvoted, though in my opinion it is the most reasonable approach to this kind of problems (${\tiny \text{which I personally find completely useless}}$).

My question will consist of 2 parts:

  1. Prove (without calculator) that $e<\pi$;

  2. Explain what do we learn from the proof/what makes this problem interesting.

Edit: Existing answers only confirm my point of view about various weird inequalitites. Fortunately there is $3$ between $e$ and $\pi$, otherwise the things would be very boring.

$\endgroup$
  • 28
    $\begingroup$ Please do say how do you define both these numbers. $\endgroup$ – DonAntonio Sep 10 '13 at 16:18
  • 4
    $\begingroup$ @DonAntonio Use any definition you like. $\endgroup$ – Start wearing purple Sep 10 '13 at 16:21
  • 2
    $\begingroup$ I always thoght that comparing $e^\pi$ with $\pi^e$ (without calculator) was the only funny specimen of "this kind of problems". $\endgroup$ – Hagen von Eitzen Sep 10 '13 at 16:30
  • 2
    $\begingroup$ @HagenvonEitzen somehow i never remember which is bigger and have to redo the argument everytime... $\endgroup$ – Jean-Sébastien Sep 10 '13 at 17:23
  • 1
    $\begingroup$ @Jean-Sébastien Which grows faster, $x^e$ or $e^x$? Now, just remember that the inequality is the same for all $x$ (that is, $e^x$ is always greater than $x^e$). $\endgroup$ – Akiva Weinberger Dec 4 '15 at 20:01
117
+50
$\begingroup$

Inscribe a regular hexagon in a circle of radius $1$. Since a straight line is the shortest distance between two points the circumference of the circle is longer than the circumference of the hexagon. We take the definition of $\pi$ as half the circumference of the unit circle.

Putting all this together we obtain $2\pi \gt 6$ or $\pi \gt 3$

We take $e$ as the sum $1+1+\frac 12+\frac 1{3!}+\cdots$ which converges absolutely and which, after the first three terms, is term by term less than the sum $1+1+\frac 12+\frac 1{2^2}+\cdots$ since the later terms in the second sum are obtained by dividing the previous term by $2$, and in the first sum by $n\gt 2$ (crudely for $n\ge 3$ we have $n!\gt 2^{n-1}$).

Summing the geometric series we have $e\lt 3 \lt\pi$.

What do we learn - well how easy it is to make an estimate depends on the definition. The geometric definition of $\pi$ lends itself to a good enough estimate. There are different ways of defining $e$ too, but the sum offers a range of possibilities for estimating, particularly as the terms decrease very quickly. But the geometric definition for $\pi$ requires assumed knowledge about a straight line as the shortest distance between two points, which seems obvious - yet conceals the trickiness of defining the length of a curve - so this looks simpler than it is.

$\endgroup$
  • 2
    $\begingroup$ Fast, simple, elegant... Love it! $\endgroup$ – Barranka Sep 17 '13 at 18:29
  • 1
    $\begingroup$ Not to mention honest about the difficulties of using arc length in definitions of basic things. $\endgroup$ – Ryan Reich Sep 19 '13 at 23:26
  • $\begingroup$ Note that it would be possible also to define $\frac {\pi}2$ as the least zero of the infinite series for $\cos x$ - which is, of course, an exponential series. And with some work it is possible to prove in this way that $\pi \gt 3$. However there is still some work to do to establish that the various different things called $\pi$ are actually the same. $\endgroup$ – Mark Bennet Sep 22 '13 at 20:00
  • $\begingroup$ Very creative and interesting way to look at the problem. $\endgroup$ – Kirthi Raman Dec 24 '13 at 15:23
69
$\begingroup$

Use $e<3$ and $\pi>3$.

The first follows from $e:=\lim \left(1+\frac1n\right)^n$ quickly, the second from comparing a circle with its inscribed hexagon.

$\endgroup$
  • 8
    $\begingroup$ can you give a hint to a quick proof of e<3 using the above definition? $\endgroup$ – Phani Raj Sep 10 '13 at 17:39
  • 4
    $\begingroup$ Since $(1+1/n)^n<(1+1/n)^{n+1}$, it's enough to prove that the latter is decreasing. It starts from 4, then 3.375, then 3.16..., then 3.05..., then 2.98..., already smaller than 3. $\endgroup$ – pts Sep 10 '13 at 19:34
  • 20
    $\begingroup$ addendum to @pts comment: to see that $$\left({{n+1}\over{n}}\right)^{{{1}\over{n+1}}}$$ is decreasing we show that $${{\left({{n+1}\over{n}}\right)^{{{1}\over{n+1}}}}\over{\left({{n+2 }\over{n+1}}\right)^{{{1}\over{n+2}}}}} \gt 1$$ or equivalently $$\left({{\left({{n+1}\over{n}}\right)^{{{1}\over{n+1}}}}\over{\left( {{n+2}\over{n+1}}\right)^{{{1}\over{n+2}}}}}\right)^{\left(n+1 \right)\,\left(n+2\right)} \gt 1$$ The latter is the product of two terms $ \gt 1$ $$\left({n+1}\over{n} \right)$$ and $$\left({(n+1)^2}\over{n(n+2)} \right)^{(n+1)}$$ $\endgroup$ – miracle173 Sep 10 '13 at 22:51
  • 52
    $\begingroup$ @miracle173: That is an impressive use of 600 characters $\endgroup$ – Eric Stucky Sep 22 '13 at 21:31
35
$\begingroup$

$$e =\sum_{n=0}^\infty \frac{1}{n!}= 2+\sum_{n=2}^\infty \frac{1}{n!}< 2+\sum_{n=2}^\infty \frac{1}{2^{n-1}}=3$$

By inscribing a regular hexagon in a circle, and noting its perimeter is less that that of the circle, we have $6r < 2 \pi r$ or $\pi > 3$.

$\endgroup$
18
$\begingroup$

Use absolutely convergent sum definitions for $e$ and $\pi$:

$$e^x=1+\sum_{n=1}^\infty{x^n \over n!}$$

and

$${\pi^2 \over 6} = \sum_{n=1}^\infty{1 \over n^2}$$

Then assume $e \ge \pi$, so $e^2 \ge \pi^2$, which means

$$1+\sum_{n=1}^\infty{2^n \over n!} \ge 6\sum_{n=1}^\infty{1 \over n^2}$$

$$\iff 1+\sum_{n=1}^\infty{{2^n\over n!}-{1\over n^2}} \ge 5\sum_{n=1}^\infty{1 \over n^2}$$

For all $n \ge 8$, ${2^n \over n!} \lt {1\over n^2}$, so $\sum_{n=8}^\infty{{2^n\over n!}-{1\over n^2}} \lt 0$. This means that

$$1+\sum_{n=1}^7{{2^n \over n!}-{1\over n^2}} \gt 5\sum_{n=1}^\infty{1\over n^2}$$

or more simply,

$$1+\sum_{n=1}^7{2^n \over n!} \gt 5\sum_{n=1}^\infty{1\over n^2}$$

$$\iff 1+2/1+4/2+8/6+16/24+32/120+64/720+128/5040 \gt 5+5/4+5/9+5/16+5/25+5/36+5/49+...$$

$$\iff 1/4+1/9+1/15+4/45+8/315 \gt 5/16+5/36+5/64+5/49+5/81+...$$

Which is clearly impossible given that paired LHS and RHS terms are all such that (LHS term) $\lt$ (RHS term), and that the LHS has only the terms shown. Therefore, our initial assumption is false, meaning that $e\lt\pi$.

It is difficult to say what can be learned from this. I think the clearest comparison would be if there were a definite limit statement for the value of $\pi$, rather than a geometric one, for which there is no "natural" geometric equivalent for $e$. Nevertheless, this sum comparison at least shows the difference between $\dfrac {2^n}{n!}$ and $\dfrac {1}{n^2}$.

$\endgroup$
9
$\begingroup$

Here is one direct way to demonstrate this inequality using only series:

$e=3 - \displaystyle\sum _{k=0}^{\infty}\dfrac{k+1}{(k+3)!}$

which is asymptotic from above, and

$\pi=3+2 \displaystyle\sum _{k=1}^{\infty } \frac{k (5 k+3) (2 k-1)! k!}{2^{k-1} (3 k+2)!}$

which is asymptotic from below.

In a glance, we see that $e$ equals 3 minus a positive quantity while $\pi$ equals 3 plus a positive quantity. Thus $e<3<\pi$ and $e<\pi$.

A benefit here is that the proof lends itself to being thought of in a dynamic sense; one can intuitively appreciate that as the series converge to their respective targets, they move away from the number 3 in opposite directions.

What is interesting about the question is that in mathematics, proving "obvious" things often requires an unexpected amount of thought. I am always struck by the wonderful range of approaches people employ in answering even simple questions. This underscores the fundamentally creative aspect of mathematics.

References for $e$ series:

http://functions.wolfram.com/Constants/E/06/01/01/ (Last entry) http://www.brotherstechnology.com/docs/Improving_Convergence_(CMJ-2004-01).pdf http://www.brotherstechnology.com/math/e-formulas.html

References for $\pi$ series:

http://mathworld.wolfram.com/PiFormulas.html Eq (29) http://www.pi314.net/eng/schrogosper.php (States equivalence of nested expression and series) http://dspace.mit.edu/bitstream/handle/1721.1/6088/AIM-304.pdf?sequence=2

$\endgroup$
9
$\begingroup$

Another way of showing that $e<3$, using $ \int_1^e \frac{1}{x} dx = 1$ (similar to the answer by dfeuer)

For any $x>0$ we have :

$$0\le (x-2)^2 = x^2-4x + 4 \implies \frac{1}{x} \ge 1-\frac{x}{4} \tag{1}$$

Then we can bound the integral by the area below the line: $$\int_1^3 \frac{1}{x} \, dx \ge \int_1^3 \left(1 - \frac{x}{4}\right) \, dx = 2 \frac{3/4 + 1/4}{2} =1 \tag{2}$$

enter image description here

This implies that $e \le 3$ (further, the inequality can be made strict by noting that $(1)$ is strict except for the single point $x=2$)

That $\pi >3$ can be seen easily by bounding the circle perimeter with an inscribed hexagon.

$\endgroup$
7
$\begingroup$

$\log e = 1$, where $$\log x = \int_1^x\!\frac 1 t\,dt.$$

By the hexagon approach (which I would frame in terms of estimating the integral $\int_{-1}^1\sqrt{1-x^2}\,dx$), $\pi > 3$, so $\log \pi > \log 3$.

Calculating the lower sum with an appropriate partition should do it.

$\endgroup$
2
$\begingroup$

Use the Leibniz formula for $\pi$: $$ \frac{\pi}{4} = \sum_{j=0}^\infty \frac{(-1)^j}{2j+1} \text{.} $$ This is an alternating sum where each partial sum is alternately an upper and lower bound. The first few terms constrain $\pi$ to these ranges:

[0,4]
[2.666..., 4]
[2.666..., 3.4666...]
[2.895..., 3.4666...]
[2.895..., 3.33968...]
[2.9760..., 3.33968...]
[2.9760..., 3.28374...]
[3.01707..., 3.28374...]

at which point we know $\pi$ is greater than $3$.

A similar method is this descending series for $\mathrm{e}$: $$ \mathrm{e} = 3 + \sum_{k=2}^\infty \frac{-1}{k!(k-1)k} \text{.} $$ Now after one term, we know $\mathrm{e} < 3-\frac{1}{4} < 3$ and the desired inequality follows.

For determining which of two unknown numbers given only by series is larger, it is helpful to have a series whose partial sums are bounds of some sort. Since we know we want $\mathrm{e} < \pi$ it is helpful to have a decreasing series for $\mathrm{e}$, an increasing series for $\pi$, and/or alternating series for either or both.

Further lesson: It is very nice to have intervals (or at least bounds) on unfamiliar numbers. Alternating series are quite handy for this.

Other series for this method: \begin{align} \mathrm{e}^{-1} &= \sum_{k=0}^\infty \frac{(-1)^k}{k!} &&\text{gives $\mathrm{e}<3$ after $5$ terms} \\ \frac{\pi^2}{8} &= \sum_{k=1}^\infty \frac{1}{(2k-1)^2} &&\text{gives $3 < \pi$ after 4 terms} \end{align}

Further further lesson: the more representations of a thing you know, the more you are able to show about it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.