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$\pi$ and $e$ are constants which we stumble on everytime we are doing math. I wonder if there is some book which exploits this fact and contains exercises whose solution involved these constants.

J. W. Tanner
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mau
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  • Any theoretical statistics book is bound to make use of those constants. Actually any book on integration techniques will as well. – qwr Jul 29 '24 at 20:25
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    Try googling "books problems e pi". – copper.hat Jul 29 '24 at 20:25
  • "e: The Story of a Number" by Eli Maor – Jean Marie Jul 29 '24 at 21:23
  • A history of $\pi$ by Peter Beckmann – Jean Marie Jul 29 '24 at 21:27
  • So you want a book where all the exercise solutions contain pi or $e$? Why do you want such a book? Those constants arise naturally in certain contexts. What you're talking about is contrived. Which is not a bad thing, but it's a bit random and I don't see what the motivation could be. – Adam Rubinson Jul 29 '24 at 22:30
  • @copper.hat I tried a similar search, but with no result. – mau Jul 30 '24 at 17:37
  • @JeanMarie: I know of both books, but they focus on the history behind the constants (and in the case of Beckmann to bash whoever he did not like) – mau Jul 30 '24 at 17:38
  • @AdamRubinson: I am writing a booklet about these constants (and &varphi, added for good or bad), and I would like to add some questions which lead naturally to them. For example I can use the problem of the secretary who puts letters in a random envelope, or the method allegedly chosen by Kepler to find a wife, to get (an approximation of) e; the idea is actually to show that such numbers are quite common even in problems not contrived. – mau Jul 30 '24 at 17:45
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    About an "uprising" of $e$ in a place where it wasn't expected, see this answer to a question of mine. – Jean Marie Jul 30 '24 at 19:31
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    Not a recommendation, but Google returns https://www.amazon.com/%CF%80-Three-fundamental-numbers-mathematics/dp/B08XL7YX9M. I don't know if this has problems: https://www.worldscientific.com/worldscibooks/10.1142/6278#t=aboutBook – copper.hat Jul 30 '24 at 20:18
  • @Adam Rubinson Please give it as an answer. It is appropriate because OP has said that he/she writes a booklet of exercises about these numbers. I would be happy (as others hopefully) to learn about this new 'uprising" of $e$ and probably upvote this answer. – Jean Marie Jul 31 '24 at 14:38
  • @JeanMarie Ok - done. I kind of already wrote the answer in the comments, but now I deleted this comment and so hopefully everyone has forgotten what I wrote, so that they can discover the answer for themselves. – Adam Rubinson Jul 31 '24 at 14:50

2 Answers2

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As requested in the comments, here is a question I came up with, and after investigating it, realised the answer has to do with one of the mathematical constants $\pi$ or $e$ (although I will not reveal which one).

Is there $a_1\in \left( \frac{1}{2}, 1 \right),$ and a sequence $a_{n+1}=(n+1)a_n-1\ \forall\ n\in\mathbb{N},\ $ such that $\frac{1}{n+1} < a_n < \frac{1}{n}\ \forall\ n\in\mathbb{N}?$

When I first came up with this problem, I thought the answer was "no", and that the problem was related to the irregularity of distributions problem. However, this is not the case, and I will leave it as an exercise to the reader to discover the truth.

Adam Rubinson
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  • I have programmed your sequence with the following Matlab code : a=exp(1)-2 % your initialization) ;A=[a]; for n=1:20 a=(n+1)*a-1 A=[A,a]; end 1./A and what I obtain is indeed a sequence whose integer parts are 1,2,3,...15, and suddenly, this sequence behaves in a foolish way. How do you explain it ? – Jean Marie Jul 31 '24 at 16:52
  • @JeanMarie "I obtain a sequence with integer part $1,2,3,\ldots, 15.$" I'm not sure what you mean by this. If $a_1 = e-2,$ then $a_2=0.436\ldots,\ a_3=0.309\ldots, \ a_4=0.238\ldots.$ So they all satisfy the inequality $\frac{1}{n+1} < a_n < \frac{1}{n}\ $. To prove this will always happen, consider: $a_1=e-2 = \displaystyle\sum_{n=2}^{\infty}\frac{1}{n!}.$ – Adam Rubinson Jul 31 '24 at 19:36
  • Explanation : instead of $\frac{1}{n+1} < a_n < \frac{1}{n}$, I am working on the inverses $n < \frac{1}{a_n} < n+1$; it is why I consider the integer part of $ \frac{1}{a_n} $ – Jean Marie Jul 31 '24 at 20:10
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    @JeanMarie you don't need to do that though. You're doing it as an aside? – Adam Rubinson Aug 01 '24 at 10:26
  • In fact, the justification is based on the following "Horner scheme" : $e-2=\frac12(1+\frac13(1+\frac14(1+(\cdots))))$ – Jean Marie Aug 02 '24 at 15:33
  • [...Ctd] See in particular Engel expansion – Jean Marie Aug 03 '24 at 15:56
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Precalculus would be a good place to start

https://openstax.org/details/books/precalculus-2e/

Chapters 4 and 5.

vallev
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