Wikipedia:Reference desk/Archives/Mathematics/2023 May 12

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May 12

Number sets and pros and cons

Do successively larger number sets have pros and cons that are improved or worsened by the next number set?

For example, the natural numbers, integers, rational numbers, real numbers, complex numbers, quaternions, octonions, and sedenions have pros and cons as follows:

Pros and cons of various number sets

Number set	Pros	Cons
Natural numbers ($\mathbb {N}$)	Natural numbers can always be added or multiplied. Mathematical induction can easily be done with natural numbers.	Natural numbers cannot always be subtracted. Differences m - n of natural numbers m and n with m ≥ n cannot be expressed as sums of natural numbers.
Integers ($\mathbb {Z}$)	Integers can always be subtracted. Differences of integers can always be expressed as sums.	Integers cannot always be divided. Quotients m / n of integers m and n with n | m cannot be expressed as products of integers. Induction with integers would require a reversed arrow of time, by not only going forward to the future (from P(n) to P(n + 1)), but also backward to the past (from P(n) to P(n - 1)).
Rational numbers ($\mathbb {Q}$)	Rational numbers can always be divided (unless one is dividing by zero, of course). Quotients of rational numbers can always be expressed as products.	The rational numbers are not complete as either a poset (Dedekind completeness) or a metric space (metric completeness). Nonnegative rational numbers do not always have rational square roots. The rational numbers do not include important constants like π and e.
Real numbers ($\mathbb {R}$)	The real numbers are complete both as a poset (Dedekind completeness) and as a metric space (metric completeness). Nonnegative real numbers always have real square roots. The real numbers include important constants like π and e. The exponential function on the real numbers is injective.	The real numbers are not an algebraically closed field (in particular, square roots of negative numbers do not exist in the real numbers).
Complex numbers ($\mathbb {C}$)	Multiplication of complex numbers is still commutative. The complex numbers are an algebraically closed field. Every complex number has exactly two complex square roots.	The complex numbers cannot be turned into an ordered field. The exponential function on the complex numbers is not injective.
Quaternions ($\mathbb {H}$)	Multiplication of quaternions is still associative.	Multiplication of quaternions is no longer commutative. Negative real numbers have infinitely many square roots in the quaternions.
Octonions ($\mathbb {O}$)	Multiplication of octonions is still alternative. The octonions still do not contain any zero divisors.	Multiplication of octonions is no longer associative.
Sedenions ($\mathbb {S}$)	Multiplication of sedenions is still power-associative.	Multiplication of sedenions is no longer alternative. The sedenions contain zero divisors.

GeoffreyT2000 (talk) 06:03, 12 May 2023 (UTC)

• To me it's a weird question. Pros and cons??? These things are what they are. They are necessarily what they are. They could not, in any possible world, be different (though first-order logic is too weak to capture this).
  So what sense does it make to speak of pros and cons? Relative to what? --Trovatore (talk) 06:12, 12 May 2023 (UTC)

  Just as a point of order, the notion that the "could not...be different", isn't quite right, especially for the more esoteric number systems. For example, is there any world where quaternions don't exist? Sure; quaternions are useful, but they are a construct that solves a problem, not an inevitable conclusion one arrives at by following a line of thinking that is inevitable from first principles. It's been said that the only parts of math that exist outside of human invention are counting numbers, and that everything else is an invention for convenience, or as Leopold Kronecker put it, "God made the integers, all else is the work of man". Every other number system was invented to make mathematics more efficient, or to solve a problem; back to the quaternion example, complex numbers allow for vastly simplified two-dimensional mechanics. Working with the square root of negative 1, and Euler's formula takes a very messy trigonometry problem and makes it much friendlier to work with; strictly speaking neither cartesian coordinates nor complex numbers are necessary, but they sure are freaking useful. However, it only works in two dimensions. Hamilton saw this problem, and worked out a solution for three dimensions, and that's the quaternions, but that's also not strictly necessary either. It's all "made up", none of it is "inevitable" or "necessary". It's just "what is useful". There certainly exists universes which are perfectly consistent in which no one invented quaternions. Similarly, there are any arbitrary number of potential number systems which haven't yet been invented, but could. --Jayron32 14:49, 12 May 2023 (UTC)

  No, there is no world where quaternions don't exist. Hope this helps.

  That's a bit glib, of course. I'm a Platonist. For what it's worth, I didn't start out that way. It took a long time into my Ph.D. studies, being dragged along little by little, seeing how empty the alternatives are and how things fit together in ways that are not easily explained by the "invention" narrative.

  But I can't expect you to have followed the same path, and we aren't going to settle it here. Interested readers can start with our philosophy of mathematics article. --Trovatore (talk) 15:54, 12 May 2023 (UTC)

• One could say that several extensions were "invented" to overcome a limitation (listed as a "con" above): the integers allow solving $x+7=3,$ the rationals allow solving $7x=3,$ the reals allow solving $x^{7}=3,$ the complex numbers allow solving $x^{2}+7=3.$ It is a bit strange (IMO) to label these limitations as "cons". Is it a con of screwdrivers that they cannot saw wood? Or of saws that they don't turn screws? I think of these sets and their attendant theories as tools in the mathematician's toolbox; each tool has its uses and some tools are better suited for certain jobs than others. Extending a system making it suitable for one type of job may entail losing some capability: the extension to the uncountable real numbers means use of mathematical induction becomes impossible. (Also, in general, the theory of a more powerful system is more complicated, so being more powerful is at the same time a pro and a con.) There are many other extensions of the naive number concept: p-adic numbers, transfinite numbers, ordinal numbers, cardinal numbers, infinitesimals, hyperreal numbers, surreal numbers. These cannot be ordered in a chain in which each is viewed as an extension of the previous one. Each number system has its uses and each has its limitations.  --Lambiam 06:55, 12 May 2023 (UTC)  I forgot to mention the Gaussian integers. Denoting these as $\mathbb {Z[i]} ,$ we have the canonical injections $\mathbb {Z} \hookrightarrow \mathbb {Z[i]} \hookrightarrow \mathbb {C} .$ Yet they are incomparable to $\mathbb {Q}$ and $\mathbb {R} ,$ which are also situated between $\mathbb {Z}$ and $\mathbb {C} .$  --Lambiam 13:36, 12 May 2023 (UTC)
• The pros and cons thing is nonsense; each number set was invented (either explicitly by a specific mathematician at a specific time, or implicitly as humanity sort of settled on it as a "thing") to serve a specific purpose, which is to say there was some deficiency in mathematics (some operation the old set couldn't do, or something like that) that the prior sets didn't handle, so the concept of what counted as a number was expanded to include more things. It represents an overall growth in the understanding of how to think about what a number is, but there's no "pros and cons". --Jayron32 11:48, 12 May 2023 (UTC)

"Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk." -- Leopold Kronecker. Does that count as a Pro? --RDBury (talk) 17:12, 12 May 2023 (UTC)

It counts as a piece of nonsense. --Trovatore (talk) 17:39, 12 May 2023 (UTC)

I thought Giuseppe Peano created the natural numbers. Perhaps the SF created the negative integers.  --Lambiam 18:35, 12 May 2023 (UTC)

Sum of the reciprocals of numbers in a set

For given positive integer n, let the sequence A_n be the sequence n*(generalized pentagonal numbers)+1, i.e. n+1, 2*n+1, 5*n+1, 7*n+1, 12*n+1, 15*n+1, 22*n+1, 26*n+1, ...

For all positive integer n, find the sum of the reciprocals of all numbers in the sequence A_n 2001:B042:4005:4133:18C0:328F:4CBA:E181 (talk) 20:05, 12 May 2023 (UTC)

Umm ... why? --Trovatore (talk) 00:00, 13 May 2023 (UTC)

If you just want a formula, it's $-{\frac {8\pi \sin({\frac {\pi }{3}}{\sqrt {\frac {n-24}{n}}})}{{\sqrt {n(n-24)}}(1-2\cos({\frac {\pi }{3}}{\sqrt {\frac {n-24}{n}}}))}}$, or any equivalent formulation thereof. I can try to draft up how I got to this, but it make take a little while. GalacticShoe (talk) 01:59, 13 May 2023 (UTC)

For $n<24$ this formula sends us on a little trip through the complex domain. For $n=24$, the formula gives us $0/0.$ For $n=25$ the result appears to be negative, which is curious for the sum of an everywhere positive series.  --Lambiam 06:13, 13 May 2023 (UTC)

At $n=25$ the expression yields $-{\frac {8\pi \sin({\frac {\pi }{15}})}{5-10\cos({\frac {\pi }{15}})}}\approx 1.0928$. For $n<24$, though the complex values yielded in the numerator and denominator are unfortunate, the expression is still real valued. For example, $n=23$ yields $-{\frac {8\pi \sinh({\frac {\pi }{3{\sqrt {23}}}})}{{\sqrt {23}}(1-2\cosh({\frac {\pi }{3{\sqrt {23}}}}))}}\approx 1.1007$. I think the only really problematic value here is indeed the indeterminate value given at $n=24$, though the limit does exist there and is equal to ${\frac {\pi ^{2}}{9}}\approx 1.0966$. GalacticShoe (talk) 11:47, 13 May 2023 (UTC)

Sorry, I overlooked the unary minus sign.  --Lambiam 15:25, 13 May 2023 (UTC)

OK, this is pretty impressive if it's real. I half thought it might be a joke. You want to give us a 30000-foot view of the ideas involved? I'm not super-interested in the plug-and-chug part, just the technique. --Trovatore (talk) 19:40, 13 May 2023 (UTC)

Yeah, for sure! I confess that the most involved part, which is converting the sum into a closed-form expression, was done through WolframAlpha. However, after spending some time looking at WolframAlpha's answer (and learning more about digamma functions than I think I will ever use in my entire life), I will supply what I think to be how it was done in-engine. Also, I know you don't want to go super into the nitty-gritty, but I will include some specific expressions just to highlight what's happening.

So first, note that the generalized pentagonal numbers are given by:

$p_{k}={\frac {3k^{2}-k}{2}}$

This means that the expression sought is:

$\sum _{k=-\infty }^{\infty }{\frac {1}{{\frac {3k^{2}-k}{2}}n+1}}$

or:

$\sum _{k=-\infty }^{\infty }{\frac {2}{(3k^{2}-k)n+2}}$.

Using WolframAlpha, specifically, splitting this sum into:

$(\sum _{k=-\infty }^{0}{\frac {2}{(3k^{2}-k)n+2}})+(\sum _{k=0}^{\infty }{\frac {2}{(3k^{2}-k)n+2}})-1$

it's possible to convert this sum into an expression using the digamma function $\psi (n)$. Specifically:

${\frac {2(\psi ({\frac {1}{6}}{\sqrt {\frac {n-24}{n}}}-{\frac {1}{6}})-\psi (-{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}}+{\frac {1}{6}})+\psi ({\frac {1}{6}}{\sqrt {\frac {n-24}{n}}}+{\frac {1}{6}}))-\psi (-{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}}-{\frac {1}{6}}))}{\sqrt {n(n-24)}}}-1$

Using the digamma reflection formula and recurrence formulas to find:

$\psi (f(n))-\psi (-f(n))=-{\frac {1}{f(n)}}-\pi \cot(\pi f(n))$

this can be converted into a completely elementary expression. The last set of steps which resulted in the formula supplied above was just a bunch of expression manipulation and some trig identities, nothing particularly complicated or worth mentioning.

Now, as to how I think WolframAlpha got that expression (and I must note here, I would have never thought of doing this had I not used WolframAlpha):

${\frac {2}{(3k^{2}-k)n+2}}={\frac {2}{3n}}{\frac {1}{k^{2}-{\frac {1}{3}}k+{\frac {2}{3n}}}}={\frac {2}{3n}}{\frac {1}{(k-({\frac {1}{6}}-{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}}))(k-({\frac {1}{6}}+{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}}))}}$

Splitting this up with partial fraction decomposition yields:

${\frac {2}{\sqrt {n(n-24)}}}({\frac {1}{k-({\frac {1}{6}}+{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}})}}-{\frac {1}{k-({\frac {1}{6}}-{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}})}})$

or:

${\frac {2}{\sqrt {n(n-24)}}}({\frac {1}{k+(-{\frac {1}{6}}-{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}})}}-{\frac {1}{k+(-{\frac {1}{6}}+{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}})}})$

Now, when taking the sum of this over $k$ from $0$ to $\infty$, we can use the digamma series formula:

$\psi (z)=-\gamma +\sum _{k=0}^{\infty }({\frac {1}{k+1}}-{\frac {1}{k+z}})$

to obtain:

$\psi (z_{1})-\psi (z_{2})=\sum _{k=0}^{\infty }{\frac {1}{k+z_{2}}}-\sum _{k=0}^{\infty }{\frac {1}{k+z_{1}}}$

yielding the closed-form expression for that half of the infinite series. Meanwhile, for $k$ from $-\infty$ to $0$, we can flip it to a sum from $0$ to $\infty$ again by just taking terms with $-k$ instead to yield:

${\frac {2}{\sqrt {n(n-24)}}}({\frac {1}{k+({\frac {1}{6}}-{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}})}}-{\frac {1}{k+({\frac {1}{6}}+{\frac {1}{6}}{\sqrt {\frac {n-24}{n}}})}})$

The full closed-form expression for the result follows.

GalacticShoe (talk) 22:36, 13 May 2023 (UTC)

Generalizing the earlier comment about $n=23$, using the fact that $\cosh(ix)=\cos(x)$, $\sinh(ix)=i\sin(x)$, $\cosh(-x)=\cosh(x)$, and $\sinh(-x)=-\sinh(x)$, we can make this formula work for $n<24$ without having to work with complex values. In particular, it becomes:

$-{\frac {8\pi \sinh({\frac {\pi }{3}}{\sqrt {\frac {24-n}{n}}})}{{\sqrt {n(24-n)}}(1-2\cosh({\frac {\pi }{3}}{\sqrt {\frac {24-n}{n}}}))}}$

Although this still only exists at $n=24$ in terms of limits. Not sure if there's an expression that combines everything together without having to deal with complex values, and that evaluates to a non-$0/0$ expression at $24$. GalacticShoe (talk) 14:56, 15 May 2023 (UTC)

The $\operatorname {sinc}$ function solves the $0/0$ issue, but not explicitly the complex excursion. One could use the function

$\operatorname {singamajig} (x)=\sum _{i=0}^{\infty }{\tfrac {(-x)^{i}}{(2i+1)!}},$

and a similar replacement for the $\cos$ bit, but these functions have no standard closed-form notation.  --Lambiam 19:54, 15 May 2023 (UTC)

On the off chance I had a seach for singamajig on Google - and it exists!, albeit as a species of various cute fluffy toys that sing. There's also cosymajig which are sustainable fabrics.😀 NadVolum (talk) 21:01, 15 May 2023 (UTC)