$\pi$ 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.
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$\begingroup$ Any theoretical statistics book is bound to make use of those constants. Actually any book on integration techniques will as well. $\endgroup$– qwrCommented Jul 29 at 20:25
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1$\begingroup$ Try googling "books problems e pi". $\endgroup$– copper.hatCommented Jul 29 at 20:25
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$\begingroup$ "e: The Story of a Number" by Eli Maor $\endgroup$– Jean MarieCommented Jul 29 at 21:23
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1$\begingroup$ About an "uprising" of $e$ in a place where it wasn't expected, see this answer to a question of mine. $\endgroup$– Jean MarieCommented Jul 30 at 19:31
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1$\begingroup$ Not a recommendation, but Google returns amazon.com/%CF%80-Three-fundamental-numbers-mathematics/dp/…. I don't know if this has problems: worldscientific.com/worldscibooks/10.1142/6278#t=aboutBook $\endgroup$– copper.hatCommented Jul 30 at 20:18
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2 Answers
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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.
answered Jul 31 at 14:50
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$\begingroup$ 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 ? $\endgroup$ Commented Jul 31 at 16:52
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$\begingroup$ @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: $a_1=e-2 = \displaystyle\sum_{n=2}^{\infty}\frac{1}{n!}.$ $\endgroup$ Commented Jul 31 at 19:36
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$\begingroup$ 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} $ $\endgroup$ Commented Jul 31 at 20:10
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1$\begingroup$ @JeanMarie you don't need to do that though. You're doing it as an aside? $\endgroup$ Commented Aug 1 at 10:26
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$\begingroup$ In fact, the justification is based on the following "Horner scheme" : $e-2=\frac12(1+\frac13(1+\frac14(1+(\cdots))))$ $\endgroup$ Commented Aug 2 at 15:33
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Precalculus would be a good place to start
https://openstax.org/details/books/precalculus-2e/
Chapters 4 and 5.
