Wikipedia:Reference desk/Archives/Mathematics/2023 July 8

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July 8

Identity as an involution

Sure, excluding the identity element of a group as an involution has pros and cons. One of the pros is that the order of an involution would always be exactly equal to 2. Some of the cons are that involutions would not necessarily be preserved by homomorphisms and that excluding the identity would needlessly complicate the formula for the number of involutions in a direct product (in G x H, it would be mn+m+n rather than mn, where m and n are the number of involutions in G and H respectively).

So, do most sources on group theory exclude the identity element of a group as an involution? Also, are there any sources that include the identity as an involution? GeoffreyT2000 (talk) 01:59, 8 July 2023 (UTC)

Let me start by saying that very few English sources use the term involution as a term for certain kinds of elements of a group. If they do, it means "element of order 2", which occurs so often that it would be convenient to have a shorter term. The rare exception defining the term in this sense is found in this textbook. The convenience would be spoiled by including the identity element. That said, this use is potentially confusing. An endofunction $f$ is called an involution if it is its own inverse, which is equivalent to $f\circ f=id.$ The set of bijective endofunctions on a given domain constitute a group under the operation of function composition. Clearly, the identity element of the group is its own inverse and so is an involution in the more common sense, but not in the group-theoretic sense. In particular, permutations are conventionally considered both elements of a group and simultaneously bijective endofunctions on a finite domain, so confusion may readily arise. In fact, it can be seen "in action" in the textbook referred to above here, in the proof of Theorem 7.5. There, the term cannot have the "order 2" sense, since the identity permutation is in $T_{0}.$  --Lambiam 09:58, 8 July 2023 (UTC)

Involutions play an important role in the classification of finite simple groups. So that area the word "involution" is used quite frequently. They don't come up that much in general group theory though. They do come up in certain combinatorial problems and if you search "involution" in the OEIS there are over 500 results. In particular, OEIS: A000085, the number of involutions in S_n. In the context of simple groups an involution is understood to not include the identity since it's a trivial case and doesn't produce much information about the group in question. But the identity does seem to be included in combinatorial contexts such as the OEIS entry linked to above. So apparently it comes down to context and convenience. --RDBury (talk) 15:49, 8 July 2023 (UTC)

The sense of involution given at A000085 is "self-inverse permutation", the same sense as used for bijective endomorphisms. $\mathrm {S} _{1}$ contains precisely one member, the identity permutation, which has order 1, and no elements of order 2, yet the count in A00085 equals 1. This only confirms that it is confusing to define involution as "element of order 2". The sequence giving the number of elements of order 2 in $\mathrm {S} _{n}$ is OEIS A001189.  --Lambiam 18:19, 8 July 2023 (UTC)

Fitting an power curve to data

I have deduced a process fits the curve y = ((ax + 1)^b) - 1, in which y = 0 at x = 0. I have two other measured data points, y at x1 and x2. SoI need y to equal k1 at x = x1 and also equal to k2 (larger than k1) at x = x2. I'm getting pretty rusty on math as I'm 75. How do I derive formulas for a and b? Dionne Court (talk) 02:53, 8 July 2023 (UTC)

One doesn't get rusty on maths by old age, but by lack of practice! The system of equations

$(ax_{1}+1)^{b}-1=y_{1},$

$(ax_{2}+1)^{b}-1=y_{2},$

in which $a$ and $b$ are the unknowns, cannot be solved algebraically. This requires the use of a numerical method. The simplest approaches are unstable for such equations; it helps if one has a priori limits on the values of the unknowns. Presumably the $x_{i}$ and $y_{i}$ of the measured data points are positive, and $y$ increases with $x$, in which case we may assume that $a$ and $b$ are either both positive or both negative.

If the value of $b$ is known, we can compute quantities $z_{1}$ and $z_{2}$ by using:

$z_{i}=(y_{i}+1)^{1/b}-1.$

It is easy to see that then $z_{i}=ax_{i}.$ To verify that the value of $b$ used to compute the $z_{i}$ is correct, we can check whether the equality $x_{1}z_{2}-x_{2}z_{1}=0$ holds. Note now that if we do not know $b$ but define

$f(b)=x_{1}z_{2}-x_{2}z_{1},$

in which the $z_{i}$ themselves depend on $b$ as described above, we have transformed the problem into finding a zero of a univariate function. This can be approached in many ways, but if you don't have a maths package doing that automatically, I recommend the regula falsi for its robustness and simplicity. Once $b$ is found this way, we have $a=x_{i}/z_{i},$ where a sanity check is that both choices give the same result.  --Lambiam 10:43, 8 July 2023 (UTC)

[edit 07.11]. I assume you want $a>0$. Let’s start from some necessary conditions of solvability for

$(ax_{1}+1)^{b}=y_{1}+1$

$(ax_{2}+1)^{b}=y_{2}+1.$

Of course, we may and do assume without loss of generality $0<x_{1}<x_{2}$. Assume therefore

$a>0,\;b\in \mathbb {R} ,\,0<x_{1}<x_{2},\;y_{1}+1>0,\;y_{2}+1>0$

satisfy the given equations. It is clear that if $y_{1}<y_{2}$, resp. if $y_{1}>y_{2}$, we must have $b>0$, resp. $b<0$ (the trivial case $y_{1}=y_{2}$ implies $y_{1}=y_{2}=b=0$, with any $a$; so we can assume $y_{1}\neq y_{2}$). Since the function $(t+1)^{1/t}$ is decreasing for $t>0$, we have, taking for $t$ the values $ax_{1}<ax_{2}$,

$(y_{1}+1)^{1/abx_{1}}=(ax_{1}+1)^{1/ax_{1}}>(ax_{2}+1)^{1/ax_{2}}=(y_{2}+1)^{1/abx_{2}}.$

We then raise both sides to the power $ab$; in the case $y_{1}<y_{2}$ this exponent $ab$ is positive and we get $(y_{1}+1)^{1/x_{1}}>(y_{2}+1)^{1/x_{2}};$ in the case $y_{1}>y_{2}$ the inequality is reversed and in both cases ${\frac {(y_{2}+1)^{1/x_{2}}-(y_{1}+1)^{1/x_{1}}}{y_{2}-y_{1}}}<0,$

that is therefore a necessary condition of existence of a solution $(a,b)\in \mathbb {R} _{+}\times \mathbb {R}$ for given data $0<x_{1}<x_{2}$ and $y_{1}>-1,y_{2}>-1$. Let’s show these conditions are also sufficient.

Note that we can eliminate $b$ and write a single equation for $a$. Indeed we have

$b={\frac {\log(y_{1}+1)}{\log(ax_{1}+1)}}={\frac {\log(y_{2}+1)}{\log(ax_{2}+1)}}.$

So if we put $q:={\frac {\log(y_{1}+1)}{\log(y_{2}+1)}}$ we have

$(ax_{2}+1)^{q}=ax_{1}+1$

Now that in case $y_{1}<y_{2}$ (resp. $y_{1}>y_{2}$), on the left there is a concave increasing (resp. a convex increasing) function (because $0<q<1$, resp. $q>1$) and it equals the affine function on the right exactly at one value $a>0$ provided the derivative at $0,$ $x_{2}q$ is larger (resp. smaller) than $x_{1}$ (the derivative of the affine function). But this is the exactly the above solvability condition, so it is also sufficient.

Example: consider $x_{1}=1,x_{2}=2,y_{1}=e,y_{2}=\pi$. These data are OK because the condition is satisfied, and will give a $b>0$ because $y_{1}<y_{2}$. [Wolfram Alpha][1] gives:

$a=4634.0814..$

$b=0.15557..$.

Note that one the necessary condition is verified, if you want to solve the equation by yourself, there are plenty of possible approaches to solve the above scalar equation for $a$. pma 12:14, 11 July 2023 (UTC)