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1. • CommentRowNumber1.
   • CommentAuthorKeithEPeterson
   • CommentTimeDec 8th 2016
   • (edited Dec 8th 2016)
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   Author: KeithEPeterson
   Format: MarkdownItexYesterday I defined a function that replaces decimal digits equal to $y$ of the number $x$ with the arbitrary number $z$. Assuming I got it right: $sgn(x)\cdot\sum_{n=-\left\lfloor log_{10} |x| \right\rfloor}\frac{\sum_{y=\left \lfloor 10^{n-1}|x|\right\rfloor-10\left\lfloor10^{n-2}|x|\right\rfloor}z+\sum_{y\ne\left \lfloor 10^{n-1}|x|\right\rfloor-10\left\lfloor10^{n-2}|x|\right\rfloor}\left \lfloor 10^{n-1}|x|\right\rfloor-10\left\lfloor10^{n-2}|x|\right\rfloor}{10^{n}}$ Maybe I could implement the lambda calculus over the Reals for giggles... Have any of you partaken in recreational math? If yes, what kind of stuff were you working on?

   Yesterday I defined a function that replaces decimal digits equal to $y$ of the number $x$ with the arbitrary number $z$. Assuming I got it right: $sgn(x)\cdot\sum_{n=-\left\lfloor log_{10} |x| \right\rfloor}\frac{\sum_{y=\left \lfloor 10^{n-1}|x|\right\rfloor-10\left\lfloor10^{n-2}|x|\right\rfloor}z+\sum_{y\ne\left \lfloor 10^{n-1}|x|\right\rfloor-10\left\lfloor10^{n-2}|x|\right\rfloor}\left \lfloor 10^{n-1}|x|\right\rfloor-10\left\lfloor10^{n-2}|x|\right\rfloor}{10^{n}}$

   Maybe I could implement the lambda calculus over the Reals for giggles…

   Have any of you partaken in recreational math? If yes, what kind of stuff were you working on?

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeDec 8th 2016
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   Author: Todd_Trimble
   Format: MarkdownItexWell, yes! For example there are problem sections in journals like the American Mathematical Monthly, and I often take a crack at those, or at Putnam type problems, etc. etc. Sometimes I'll work on silly things just for kicks, like characterizing the triangular squares or evaluating tricky integrals. Generally speaking I have a taste for generatingfunctionology, binomial identities, finite difference calculus... Answering questions on MO or Mathematics StackExchange can often have a recreational feel. To be honest, I don't make a huge distinction between "recreational math" and "serious math". For me both involve a certain amount of playing and jiggering around, to see what works. Concrete questions and abstract considerations are frequently mixed together; it's a big potpourri. This is very true in number theory for example, and I do a certain amount of diddling around, writing things up for my own amusement (recently it was cooking up a home-made proof of Lagrange's four square theorem, avoiding the usual pigeonhole arguments). One of my first loves in mathematics was continued fractions, and the old stuff you might find in HAKMEM has a certain appeal to me. I would like to understand better some of Gosper's algorithms for doing arithmetic of continued fractions from an algebraic/coalgebraic perspective -- that's the type of thing I mean when I say that abstract and the concrete are all mixed together. But I'm hardly alone in this. It's pretty clear to me that a lot of category theorists love this type of fooling around. The late Steve Schanuel is an excellent example, as is Peter Freyd, and probably Michael Barr, and probably Andreas Blass, to name a few examples just off the top of my head.

   Well, yes! For example there are problem sections in journals like the American Mathematical Monthly, and I often take a crack at those, or at Putnam type problems, etc. etc. Sometimes I’ll work on silly things just for kicks, like characterizing the triangular squares or evaluating tricky integrals. Generally speaking I have a taste for generatingfunctionology, binomial identities, finite difference calculus…

   Answering questions on MO or Mathematics StackExchange can often have a recreational feel.

   To be honest, I don’t make a huge distinction between “recreational math” and “serious math”. For me both involve a certain amount of playing and jiggering around, to see what works. Concrete questions and abstract considerations are frequently mixed together; it’s a big potpourri. This is very true in number theory for example, and I do a certain amount of diddling around, writing things up for my own amusement (recently it was cooking up a home-made proof of Lagrange’s four square theorem, avoiding the usual pigeonhole arguments).

   One of my first loves in mathematics was continued fractions, and the old stuff you might find in HAKMEM has a certain appeal to me. I would like to understand better some of Gosper’s algorithms for doing arithmetic of continued fractions from an algebraic/coalgebraic perspective – that’s the type of thing I mean when I say that abstract and the concrete are all mixed together. But I’m hardly alone in this. It’s pretty clear to me that a lot of category theorists love this type of fooling around. The late Steve Schanuel is an excellent example, as is Peter Freyd, and probably Michael Barr, and probably Andreas Blass, to name a few examples just off the top of my head.

3. • CommentRowNumber3.
   • CommentAuthorKeithEPeterson
   • CommentTimeDec 8th 2016
   • (edited Dec 8th 2016)
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   Author: KeithEPeterson
   Format: MarkdownItexWouldn't a continued fraction be the cardinality class of $\infinity$-groupoids? Consider the groupoids with the cardinality of the golden ratio for instance. One would think they should have a beautiful self-recursive interpretation if given a topology. Edit: Yeah. The cardinality of such a groupoid is given by the hairy expression, $|\phi| = \sum_{iso classes of objects[\phi]}1+\frac{1}{\sum_{iso classes of objects[\phi]}1+\frac{1}{\sum_{iso classes of objects [\phi]}1+\frac{1}{\sum_{iso classes of objects[\phi]}1+\frac{1}{\ddots}}}}$, showing that there is one such groupoid, and what this thing really is: A groupoid that has a disjoint object next to a groupoid, that has a disjoint object next to a groupoid,... Who says math can't be fun?

   Wouldn’t a continued fraction be the cardinality class of $\infinity$-groupoids? Consider the groupoids with the cardinality of the golden ratio for instance. One would think they should have a beautiful self-recursive interpretation if given a topology.

   Edit: Yeah. The cardinality of such a groupoid is given by the hairy expression, $|\phi| = \sum_{iso classes of objects[\phi]}1+\frac{1}{\sum_{iso classes of objects[\phi]}1+\frac{1}{\sum_{iso classes of objects [\phi]}1+\frac{1}{\sum_{iso classes of objects[\phi]}1+\frac{1}{\ddots}}}}$,

   showing that there is one such groupoid, and what this thing really is: A groupoid that has a disjoint object next to a groupoid, that has a disjoint object next to a groupoid,…

   Who says math can’t be fun?

4. • CommentRowNumber4.
   • CommentAuthorKeithEPeterson
   • CommentTimeDec 8th 2016
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   Author: KeithEPeterson
   Format: MarkdownItexActually, the page [continued fraction](https://ncatlab.org/nlab/show/continued+fraction) on nlab doesn't mention groupoid cardinality. I feel it should.

   Actually, the page continued fraction on nlab doesn’t mention groupoid cardinality. I feel it should.

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeDec 8th 2016
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   Author: Mike Shulman
   Format: MarkdownItexI certainly never do any math that I don't find fun!

   I certainly never do any math that I don’t find fun!

6. • CommentRowNumber6.
   • CommentAuthorKeithEPeterson
   • CommentTimeJan 1st 2017
   • (edited Jan 2nd 2017)
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   Author: KeithEPeterson
   Format: MarkdownItexOh, now here's a nice continued fraction expression for $\pi$, $$ \pi =2+\frac{2+\frac{2+\frac{2+\frac{\vdots }{1+\frac{2}{1+\frac{2}{\ddots }}}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{\ddots }}}}}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{\ddots }}}}}}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{1+\frac{2}{ \ddots }}}}}}}} ,$$ it follows from Viète's formula and Engel expansion. Edit: math typo in the expression, thank you Todd for pointing it out.

   Oh, now here’s a nice continued fraction expression for $\pi$,

   $$\pi =2+\frac{2+\frac{2+\frac{2+\frac{\vdots }{1+\frac{2}{1+\frac{2}{\ddots }}}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{\ddots }}}}}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{\ddots }}}}}}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{1+\frac{2}{ \ddots }}}}}}}} ,$$

   it follows from Viète’s formula and Engel expansion.

   Edit: math typo in the expression, thank you Todd for pointing it out.

7. • CommentRowNumber7.
   • CommentAuthorTodd_Trimble
   • CommentTimeJan 2nd 2017
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   Author: Todd_Trimble
   Format: MarkdownItexThat doesn't look right to me. Isn't this claiming $\pi = 2 + \frac{\pi}{\sqrt{2}}$?

   That doesn’t look right to me. Isn’t this claiming $\pi = 2 + \frac{\pi}{\sqrt{2}}$?

8. • CommentRowNumber8.
   • CommentAuthorKeithEPeterson
   • CommentTimeJan 2nd 2017
   • (edited Jan 2nd 2017)
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   Author: KeithEPeterson
   Format: MarkdownItexIf,$\frac{a b c d}{e f g h} = \frac{d+\frac{c+\frac{b+\frac{a}{e}}{f}}{g}}{h}$, and if $\frac \pi2 = \frac 2{\sqrt 2} \cdot \frac 2{\sqrt{2+\sqrt 2}} \cdot \frac2{\sqrt{2+\sqrt{2+\sqrt 2}}} \cdots$, then $\pi = ( \frac 2{\sqrt 2} \cdot \frac 2{\sqrt{2+\sqrt 2}} \cdot \frac2{\sqrt{2+\sqrt{2+\sqrt 2}}} \cdots)\cdot\frac2 1$, with, $\sqrt{a+ \sqrt{a+ \sqrt{a+ \sqrt{a+ \cdots }}}} = 1+\frac{a}{1+\frac{a}{1+\frac{a}{}\ddots }}$, then, $\pi =2+\frac{2+\frac{2+\frac{2+\frac{\vdots }{1+\frac{2}{1+\frac{2}{\ddots }}}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{\ddots }}}}}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{\ddots }}}}}}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{1+\frac{2}{ \ddots }}}}}}}} ,$ Edit: Sorting out fraction.

   If,$\frac{a b c d}{e f g h} = \frac{d+\frac{c+\frac{b+\frac{a}{e}}{f}}{g}}{h}$,

   and if $\frac \pi2 = \frac 2{\sqrt 2} \cdot \frac 2{\sqrt{2+\sqrt 2}} \cdot \frac2{\sqrt{2+\sqrt{2+\sqrt 2}}} \cdots$,

   then $\pi = ( \frac 2{\sqrt 2} \cdot \frac 2{\sqrt{2+\sqrt 2}} \cdot \frac2{\sqrt{2+\sqrt{2+\sqrt 2}}} \cdots)\cdot\frac2 1$,

   with, $\sqrt{a+ \sqrt{a+ \sqrt{a+ \sqrt{a+ \cdots }}}} = 1+\frac{a}{1+\frac{a}{1+\frac{a}{}\ddots }}$,

   then, $\pi =2+\frac{2+\frac{2+\frac{2+\frac{\vdots }{1+\frac{2}{1+\frac{2}{\ddots }}}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{\ddots }}}}}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{\ddots }}}}}}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{1+\frac{2}{ \ddots }}}}}}}} ,$

   Edit: Sorting out fraction.

9. • CommentRowNumber9.
   • CommentAuthorTodd_Trimble
   • CommentTimeJan 2nd 2017
   • (edited Jan 2nd 2017)
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   Author: Todd_Trimble
   Format: MarkdownItex[never mind]

   [never mind]

10. • CommentRowNumber10.
    • CommentAuthorKeithEPeterson
    • CommentTimeJan 2nd 2017
    • (edited Jan 2nd 2017)
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    Author: KeithEPeterson
    Format: MarkdownItexYou are right by the way, by switching the values 1 and 2 in the infinite expansion, it became $\sqrt2$. Which is baaaaad.

    You are right by the way, by switching the values 1 and 2 in the infinite expansion, it became $\sqrt2$. Which is baaaaad.

11. • CommentRowNumber11.
    • CommentAuthorTodd_Trimble
    • CommentTimeJan 2nd 2017
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    Author: Todd_Trimble
    Format: MarkdownItex> $\pi =2+\frac{2+\frac{2+\frac{2+\frac{\vdots }{1+\frac{2}{1+\frac{2}{\ddots }}}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{\ddots }}}}}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{\ddots }}}}}}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{1+\frac{2}{ \ddots }}}}}}}} ,$ I believe > $x = 1+\frac{2}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{1+\frac{2}{ \ddots }}}}}}}$ reduces to $x = 2$, since it satisfies $x = 1 + \frac{2}{x}$. As a result, it sure looks to me that the cited expression for $\pi$ (let's denote it $y$) satisfies the equation $y = 2 + \frac{y}{2}$.

    $\pi =2+\frac{2+\frac{2+\frac{2+\frac{\vdots }{1+\frac{2}{1+\frac{2}{\ddots }}}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{\ddots }}}}}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{\ddots }}}}}}}{1+\frac{2}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{1+\frac{2}{ \ddots }}}}}}}} ,$

    I believe

    $x = 1+\frac{2}{1+\frac{2}{1+\frac{2}{{1+\frac{2}{1+\frac{2}{1+\frac{2}{ \ddots }}}}}}}$

    reduces to $x = 2$, since it satisfies $x = 1 + \frac{2}{x}$. As a result, it sure looks to me that the cited expression for $\pi$ (let’s denote it $y$) satisfies the equation $y = 2 + \frac{y}{2}$.

12. • CommentRowNumber12.
    • CommentAuthorTodd_Trimble
    • CommentTimeJan 2nd 2017
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    Author: Todd_Trimble
    Format: MarkdownItexWhile we're on the topic, let me recall that continued fractions which exhibit the type of regularity seen in some of your expressions are fixed points of fractional linear transformations over $\mathbb{Z}$, i.e. satisfy $x = \frac{a x + b}{c x + d}$ where $a, b, c, d \in \mathbb{Z}, a d - b c = \pm 1$. Such $x$ thus belong to quadratic number fields. Therefore it might be tricky getting nice expressions for e.g. $\sqrt{2 +\sqrt{2}}$ which aren't quadratic surds. (So in general, check your work carefully.)

    While we’re on the topic, let me recall that continued fractions which exhibit the type of regularity seen in some of your expressions are fixed points of fractional linear transformations over $\mathbb{Z}$, i.e. satisfy $x = \frac{a x + b}{c x + d}$ where $a, b, c, d \in \mathbb{Z}, a d - b c = \pm 1$. Such $x$ thus belong to quadratic number fields. Therefore it might be tricky getting nice expressions for e.g. $\sqrt{2 +\sqrt{2}}$ which aren’t quadratic surds. (So in general, check your work carefully.)

13. • CommentRowNumber13.
    • CommentAuthorKeithEPeterson
    • CommentTimeJan 2nd 2017
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    Author: KeithEPeterson
    Format: MarkdownItexWow, I defined the value 2.5, not $\pi$. Bravo me. Not so fun with math. But mistakes are part of math.

    Wow, I defined the value 2.5, not $\pi$. Bravo me. Not so fun with math.

    But mistakes are part of math.

14. • CommentRowNumber14.
    • CommentAuthorTodd_Trimble
    • CommentTimeJan 2nd 2017
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    Author: Todd_Trimble
    Format: MarkdownItexWell, I got $y = 4$, not $y = 2.5$. ;-)

    Well, I got $y = 4$, not $y = 2.5$. ;-)

15. • CommentRowNumber15.
    • CommentAuthorKeithEPeterson
    • CommentTimeJan 2nd 2017
    • (edited Jan 2nd 2017)
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    Author: KeithEPeterson
    Format: MarkdownItexUgh, I hate ascending fractions!!!

    Ugh, I hate ascending fractions!!!

16. • CommentRowNumber16.
    • CommentAuthorKeithEPeterson
    • CommentTimeJan 2nd 2017
    • (edited Jan 2nd 2017)
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    Author: KeithEPeterson
    Format: MarkdownItexThis is just proof that counting down from infinities is a bad idea. And let me be clear, this is a really bad attempt at a hack. Edit: No one makes mistakes in math, right? I mean, especially not in front of a classroom. And especially especially not when sending in work to be refereed.

    This is just proof that counting down from infinities is a bad idea.

    And let me be clear, this is a really bad attempt at a hack.

    Edit: No one makes mistakes in math, right? I mean, especially not in front of a classroom. And especially especially not when sending in work to be refereed.

17. • CommentRowNumber17.
    • CommentAuthorTodd_Trimble
    • CommentTimeJan 2nd 2017
    • (edited Jan 2nd 2017)
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    Author: Todd_Trimble
    Format: MarkdownItexRe #15: I think you probably want $\frac{b + \sqrt{b^2 + 4a}}{2}$, right? And here, we should get $\frac{1 + \sqrt{1 + 8}}{2} = 2$, as I had said before. (Since my comments, such as this, seem to be made irrelevant by repeated edits/erasures by Keith, I think what I shall do is not comment for a while on his posts.)

    Re #15: I think you probably want $\frac{b + \sqrt{b^2 + 4a}}{2}$, right? And here, we should get $\frac{1 + \sqrt{1 + 8}}{2} = 2$, as I had said before.

    (Since my comments, such as this, seem to be made irrelevant by repeated edits/erasures by Keith, I think what I shall do is not comment for a while on his posts.)