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April 2

Besides converting a Maclaurin series into a Taylor series, are there other mathematical topics involving a well known formula - turning into another well known formula - by converting [every appearance of] some variable x into (x-a)?

Btw, although the variable $a$ in a Taylor series, is universally quantified, my question in the header allows $a$ to be some constant (and also allows $a$ to be existentially quantified).

Additionally, does this kind of conversion, of $x$ into $(x-a),$ as made in the case of Maclaurin/Taylor series (or in any analogous case the header is asking about), have an accepted name, implying the conversion of $x$ into $(x-a)?$ 147.235.210.76 (talk) 06:56, 2 April 2025 (UTC)

How about $\cos(x-\pi /2)=\sin(x)$. 101.119.184.57 (talk) 12:53, 2 April 2025 (UTC)

What's the (first) well known formula trurning into another well known formula, by converting [every appearance of] x in the first formula into $(x-\pi /2)$ in the second formula? 147.235.210.76 (talk) 13:10, 2 April 2025 (UTC)

Would you consider the formula for Jaccard similarity index: $J_{A,B}={{|A\cap B|} \over {|A\cup B|}}$. What if we J in a formula? We can make up a new formula $S_{A,B}={{2J_{A,B}} \over {1+J_{A,B}}}$. But, if we do substitution on that, we get $S_{A,B}={{2|A\cap B|} \over {|A|+|B|}}$, which is an entirely different formula, the Dice-Sorensen similarity coefficient. Of course, you can go the other way and replace the S formula to get back to the J formula. It isn't just replacing units like changing radians to degrees. It is completely changing the formula. One produces an index. One produces a coefficient. 68.187.174.155 (talk) 13:24, 2 April 2025 (UTC)

Taylor published his formula in 1715, in his book Methodus Incrementorum Directa et Inversa. Maclaurin, who knew of Taylor's work, published his version in 1742, crediting Taylor, in his book Treatise of Fluxions. So, historically, the Maclaurin series is a special case of the Taylor series, obtained by instantiating $a=0.$  ​‑‑Lambiam 13:34, 2 April 2025 (UTC)

History is well known, but my question in the header is asking in principle, about "converting [every appearance of] some variable x into (x-a)", as we do when we pass from the Maclaurin formula to the Taylor formula (although the latter preceded the former). 147.235.210.76 (talk) 14:58, 2 April 2025 (UTC)

There is the general notion of a change of variables, of which this is a very simple instance.  ​‑‑Lambiam 21:20, 2 April 2025 (UTC)

Yep, but my question was: "does this kind of conversion, of $x$ into $(x-a)$... have an accepted name, implying the conversion of $x$ into $(x-a)?$" 147.235.210.76 (talk) 21:35, 2 April 2025 (UTC)

No, not every simple routine step, such as substituting $2y$ for $y,$ or replacing $2(x+a)-x$ by $x+2a,$ and so on and so forth, has been given a special name. This one is not an exception.  ​‑‑Lambiam 06:14, 3 April 2025 (UTC)

All right. 147.235.210.76 (talk) 06:33, 3 April 2025 (UTC)

April 5

Why was maths created?

Why was math's created? Many kids may think that math's gives you useless time that you have to spend on homework. But if you see in our everyday lives, it is very useful from taking care of your money to teaching your kids your tactics for real life math's. It is essential that you learn math's to the best of your abilities! Remember never give up try your best! 203.30.15.85 (talk) 08:58, 5 April 2025 (UTC)

The origins of mathematics as a human endeavour are shrouded in prehistory, but from the earliest written records on their is evidence of mathematical activity, which has three major sources. One is practical, as a tool for bookkeeping (inventories, debts, taxes) and for measuring the areas of plots of land. One is scientific, for studying the motion of the stars and planets. And then there is the quest for understanding: Why do some patterns appear to hold? How can we be certain? What are the consequences? For more, see History of mathematics.  ​‑‑Lambiam 10:19, 5 April 2025 (UTC)

Does a binary operation, that becomes zero when its two operands are identical, have an accepted name?

e.g. $f(x,y)=g(x)-g(y),$ for any function $g.$

79.177.147.187 (talk) 20:25, 5 April 2025 (UTC)

I don't think so. The additive operation of a Boolean ring has this property, but I did not spot a text defining Boolean rings and mentioning this property that gave it a name. It is also not an element of Category:Properties of binary operations.  ​‑‑Lambiam 22:56, 5 April 2025 (UTC)

In some contexts this is called the "alternating property", see for example Lie algebra#Definition of a Lie algebra. Usually though, an algebraic system would have to have more structure to be considered "interesting". --RDBury (talk) 04:13, 6 April 2025 (UTC)

Saying that an operation "is alternating", or "has the alternating property", would generally not be understood as having this meaning outside the context of Lie algebras. I suspect its use in that context comes from the use for alternating multilinear maps.  ​‑‑Lambiam 06:46, 6 April 2025 (UTC)

It’s a standard term in linear algebra: Bilinear form#Symmetric, skew-symmetric, and alternating forms. 100.36.106.199 (talk) 02:20, 7 April 2025 (UTC)

Which is a special case (2-linear) of alternating k-linear maps, with the Lie bracket – additionally satisfying the Jacobi identity – as an instance. As far as I know, in the sense of f (..., v,..., v,...) = 0, the term is used exclusively for linear maps. Another special case is that in which the arguments are the column vectors of a matrix and the operation is the determinant. In that case is it is a consequence of the sign flipping when two vectors are swapped:

det(..., u, v,...) = −det(..., v, u,...).

Perhaps this property is the origin of the somewhat mysterious term.  ​‑‑Lambiam 06:53, 7 April 2025 (UTC)

April 8

Lindemann–Weierstrass theorem (Baker's reformulation)

Let $z_{1},\ldots ,z_{n}$ be distinct algebraic numbers. Then:
Version A: The numbers ${\text{e}}^{z_{1}},\ldots ,{\text{e}}^{z_{n}}$ are linearly independent over ${\overline {\mathbb {Q} }}$.
Version B: Over $\mathbb {Q}$.

My questions are so:
1. A implies B of course. But does B imply A? (I myself think not.)
2. Can one prove the special case $n=2$ (Lindemann's theorem) independently? יהודה שמחה ולדמן (talk) 04:48, 8 April 2025 (UTC)

The question is not quite clear. "Over $\mathbb {Q}$" is not a statement. What is the implied quantification in the question "does B imply A?"?  ​‑‑Lambiam 09:35, 8 April 2025 (UTC)

It means "the same statement but with the algebraic numbers replaced by the rational numbers". 100.36.106.199 (talk) 11:38, 8 April 2025 (UTC)

I think the answer to question 1 is affirmative, with some work, because given one linear equation with algebraic coefficients you actually have a system of such equations and you should be able to eliminate variables. 100.36.106.199 (talk) 11:50, 8 April 2025 (UTC)

What about question 2? יהודה שמחה ולדמן (talk) 12:03, 8 April 2025 (UTC)

Linear independence over Q clearly does not imply linear independence over the algebraic closure. Of course, using the hypothesis that the exponents are distinct algebraic numbers, one can prove the stronger statement, and therefore also the weaker one. Both being true, the question of whether one implies the other is somewhat moot. Tito Omburo (talk) 12:06, 8 April 2025 (UTC)

Then why are there proofs about the weak statement (version B)?

It makes Lindemann's theorem (the special case) weaker too:

If $z\neq 0$ is algebraic, then ${\text{e}}^{z}$ is irrational. יהודה שמחה ולדמן (talk) 14:58, 8 April 2025 (UTC)

These are easier to prove, but since linear independence over Q doesn't imply linear independence over the algebraic closure, the statements are strictly weaker. Tito Omburo (talk) 16:03, 8 April 2025 (UTC)

What about the case $n=2$ (Lindemann's theorem)? Can we prove it independently? יהודה שמחה ולדמן (talk) 18:53, 8 April 2025 (UTC)

Tito, I think you're wrong about this. Linear independence over Q for a given set of numbers doesn't imply linear independence over Q-bar for that same set of numbers, but this does not mean that linear independence over Q for all collections of numbers of a given type is weaker than the same for Q-bar. Consider just the baby case of {0, z} for algebraic z. Suppose there's a linear relation over Q-bar, a*1 + b*e^z = 0. Well then either b = 0 (trivial) or e^z is algebraic; in the latter case, e^z satisfies a polynomial relation with coefficients in Q. This polynomial relation is a linear relation between e^0, e^z, e^(2z), etc. with rational coefficients. By the weaker version of the theorem, this relation is trivial, so the original linear relation over Q-bar must also be trivial, QED. In other words, applying the weak version to the larger set recovers the strong version for a smaller set; I'm pretty sure the same thing happens in general. 100.36.106.199 (talk) 23:30, 8 April 2025 (UTC)

I stand corrected. This seems quite plausible. Tito Omburo (talk) 23:35, 8 April 2025 (UTC)

Can these be true?

1. $0^{0}=1$ (exponentiation)
2. $0!=1$ (factorial)
3. $\varphi (0)=0$ (Euler's totient function)
4. $\sigma _{1}(0)=-{\frac {1}{12}}$ (divisor function)
5. $\Phi _{0}(x)=1$ (cyclotomic polynomial)

220.132.216.52 (talk) 08:09, 8 April 2025 (UTC)

1. See Zero to the power of zero.
2. This is the standard definition.
3. The totient function is defined on the positive natural numbers, so asking for the value of $\varphi (0)$ is as meaningless as asking for the value of $\varphi ({\sqrt {-2}}).$ An author could – theoretically – define an "extended" or "generalized" totient function defined on a larger domain, but should have a good reason for doing so.
4. Basically the same as for 3. Also, $\textstyle {1{+}2{+}3{+}...=-{\frac {1}{12}}}$ is not true in the standard definition of infinite series.
5. Same as for 3. Is there any interesting property of the cyclotomic polynomials that remains true under this definition when a variable conventionally ranging over the positive natural numbers is allowed to assume the value $0,$ but fails to hold when defining, e.g., $\Phi _{0}(x)=4\,$?

 ​‑‑Lambiam 09:23, 8 April 2025 (UTC)

I would argue, following some points raised in this Math Stack Exchange discussion, that $\varphi (0)=0$ makes some kind of sense. The sum is empty, and it would preserve the property that $a\mid b\implies \varphi (a)\mid \varphi (b)$ when $b=0$. But I also wonder when you would ever need to define it. Double sharp (talk) 07:37, 9 April 2025 (UTC)

On the other hand, if $\phi (n)$ is defined as the order of the group of units in the ring $Z/nZ$, then $\phi (0)=2$, which of course violates this property. Tito Omburo (talk) 13:17, 9 April 2025 (UTC)

That is also true. I suppose the situation with $\varphi (0)$ may then be that there are reasonable arguments for multiple values and it doesn't come up often enough to argue for any one in particular. Double sharp (talk) 08:44, 17 April 2025 (UTC)

Well, what is the Dedekind psi function of 0? 220.132.216.52 (talk) 02:21, 10 April 2025 (UTC)

For the empty set, can these be true?

1. sum = 0
2. product = 1
3. GCD = 0
4. LCM = 1
5. union = empty set
6. intersection = the set containing all things
7. sup = -∞
8. inf = +∞
9. shortest common supersequence = empty string
10. longest common subsequence = an infinite string containing all strings as subsequences

220.132.216.52 (talk) 02:20, 10 April 2025 (UTC)

While one is usually taught that the empty set is the unique set with no elements, in the context of variables running over a set one is (tacitly) assumed to know the type of the elements of the set, also for an empty set. Most of the above are true (with the usual definitions for the operations involved) under some assumption on the types and false under some other assumption. All are meaningless under most type assumptions.

As an illustration, let $V\subseteq \mathbb {R} ^{3}$ be a finite set of vectors in Euclidean three-dimensional space, and consider the resultant vector $\textstyle {\mathbf {s} =\sum _{\mathbf {v} \in V}\mathbf {v} }.$ Then we should always have that $\mathbf {s} \in \mathbb {R} ^{3},$ shouldn't we? But if $V$ is empty, and the sum of the empty set is $0,$ which as we know is an element of $\mathbb {R} ,$ we are led to the conclusion that $\mathbb {R}$ and $\mathbb {R} ^{3}$ are not totally separate spaces but are tethered to each other at a common origin, together with all other vector spaces. To avoid such nonsense, we should agree that the sum of an empty set of vectors is not just "zero" but a null vector, specifically, the null vector of the vector space in which the elements of the set are assumed to live, even if there aren't any. More generally, the set should be a subset of some additive group or semigroup, and then "zero" means the identity element of the addition operation of the (semi)group.

Some of the operations listed assume that the sets are ordered or partially ordered sets. Then we should not just know the type of elements but also the ordering on these elements, since one can define different orderings on a given set of elements.

For $\mathbb {N} _{0}={0,1,2,...},$ the usual ordering is the total order in which $0<1<2<\cdots ,$ but in some contexts the partial order of the divisibility relation. This relation makes the set a complete lattice, specifically a division lattice, whose bottom $\bot$ (least element) is $1,$ while its top $\top$ (greatest element) is $0.$

Now:

1. Handled above; in many contexts (sets of integers / rational numbers / real numbers / complex numbers) this is the standard since $0$ is the identity element of addition.
2. Similar to case 1 if you replace "addition" by "multiplication".
3. Using divisibility, the GCD operation returns the infimum of the elements, so the GCD of the empty set would be the infimum of the empty set, which is the top of the division lattice, $0.$
4. Dually, the LCM of the empty set is the bottom of the division lattice, $1.$
5. This follows from the standard definition.
6. This is problematic. There is no such thing as a universal set, "the set containing all things". If it existed, every other set would be a subset, so no set could have a greater cardinality. But, by Cantor's theorem, its cardinality would be strictly less than that of its power set. So there needs to be a restriction on the universe ${\mathcal {U}}$ of which the element sets are assumed to be subsets, and then the intersection is ${\mathcal {U}}.$
7. The supremum of the empty set is the bottom of the (semi-)complete (semi-)lattice from which the elements are taken. If it has no bottom, the empty set has no supremum. If we embed the real numbers in the extended real numbers, that bottom is $-\infty .$ But if we are considering the non-negative reals, the bottom is $0,$ and likewise for $\mathbb {N} _{0}.$
8. Dual to the supremum.
9. This follows from the standard definition.
10. No. Even if we restrict the alphabet to $\{{\mathsf {0}},{\mathsf {1}}\},$ there are infinitely many infinite strings containing all finite strings as subset, so this is undefined. In some context you might just pick any of the candidates, e.g.
        ${\mathsf {0100011011000001010011100101110111}}\cdots$
    and declare it to be the lucky winner, but there is no good reason to prefer this over its complement ${\mathsf {1011100100111}}\cdots$.

 ​‑‑Lambiam 11:32, 10 April 2025 (UTC)

April 10

Kakeya Set

Hi. I have some questions about Kakeya Sets, just for fun.

1. Can a Kakeya set exist, given the additional restriction that it contain exactly one line segment for each angle? (Beyond the obvious answer of a circle)
2. Can such a set be made arbitrarily large? (my intuition was yes picturing a spiral shape, but I have a gut feeling that my intuition is wrong. )

Duomillia (talk) 00:32, 10 April 2025 (UTC)

I have an intuitive gut feeling that intuitions and gut feelings are the same thing. -- Jack of Oz ^{[pleasantries]} 00:57, 10 April 2025 (UTC)

My intuition tells me that we must not indulge in gut feelings.  ​‑‑Lambiam 11:39, 10 April 2025 (UTC)

My gut tells me not to eat ice cream after pizza. —Tamfang (talk) 18:29, 10 April 2025 (UTC)

Isn't the closure of the deltoid shown as the first illustration in our article Kakeya set an example meeting your additional restriction? Also, isn't every superset of a Kakeya set, up to $\mathbb {R} ^{2},$ also a Kakeya set?  ​‑‑Lambiam 11:47, 10 April 2025 (UTC)

To clarify, for 1. when you say "exactly one line segment for each angle", do you mean line segments centered at the origin? The example of the circle you gave contains multiple line segments for any particular orientation centered at any given point within the circle. GalacticShoe (talk) 17:02, 10 April 2025 (UTC)

If you think of the line segment as being oriented, if it makes an angle $\alpha$ with the horizontal (oriented from $-\infty$ to $+\infty$), turning it around by half a turn changes the angle to $\alpha +\pi$ (modulo $2\pi$). So when making a full turn in a disk of diameter $1$ it attains each angle $0\leq \alpha <2\pi$ precisely once.  ​‑‑Lambiam 17:26, 10 April 2025 (UTC)

Ah I see where I went off the rails, I failed to notice the "unit" part of "unit line segment" in the definition of a Kakeya set, in which case yeah the unit circle would clearly work. I imagine the Reuleaux triangle, or any Reuleaux polygon, would be another example. GalacticShoe (talk) 17:35, 10 April 2025 (UTC)

Reduced Dedekind psi function?

The Euler's totient function is:

$\varphi (n)={\begin{cases}1&{\text{if }}n=1\\(p-1)p^{r-1}&{\text{if }}n=p^{r}{\text{ with }}p{\text{ prime}}\\\varphi (a)\varphi (b)&{\text{if }}n=ab{\text{ with }}gcd(a,b)=1\\\end{cases}}$

and we have the “reduced” Euler's totient function: (sequence A011773 in the OEIS)

$\lambda (n)={\begin{cases}1&{\text{if }}n=1\\(p-1)p^{r-1}&{\text{if }}n=p^{r}{\text{ with }}p{\text{ prime}}\\lcm(\lambda (a),\lambda (b))&{\text{if }}n=ab{\text{ with }}gcd(a,b)=1\\\end{cases}}$

Also, the Dedekind psi function is:

$\psi (n)={\begin{cases}1&{\text{if }}n=1\\(p+1)p^{r-1}&{\text{if }}n=p^{r}{\text{ with }}p{\text{ prime}}\\\psi (a)\psi (b)&{\text{if }}n=ab{\text{ with }}gcd(a,b)=1\\\end{cases}}$

and we have the “reduced” Dedekind psi function:

$\gamma (n)={\begin{cases}1&{\text{if }}n=1\\(p+1)p^{r-1}&{\text{if }}n=p^{r}{\text{ with }}p{\text{ prime}}\\lcm(\gamma (a),\gamma (b))&{\text{if }}n=ab{\text{ with }}gcd(a,b)=1\\\end{cases}}$

the values of the “reduced” Dedekind psi function $\gamma (1)$ to $\gamma (24)$ are 1, 3, 4, 6, 6, 12, 8, 12, 12, 6, 12, 12, 14, 24, 12, 24, 18, 12, 20, 6, 8, 12, 24, 12, but this sequence does not in OEIS, thus does this function exist in number theory? 220.132.216.52 (talk) 02:54, 10 April 2025 (UTC)

If the sequence is not in the OEIS that's usually a good indication that it's not well known. If it were it would be much more likely to be found there than here. --RDBury (talk) 13:53, 15 April 2025 (UTC)

April 14

Unit usage

1. Do Canadians ever measure room space in square metres?
2. Do Canadians use metric units to measure size of things? Are licence plates measured in millimetres there?
3. Is it so that even in French-speaking Canada, most people give their height in feet/inches and their weight in pounds?
4. Does United Kingdom use kilometre and km/h in any official purposes?--40bus (talk) 20:17, 14 April 2025 (UTC)

   These questions are not about mathematics. Try the Miscellaneous section.  ​‑‑Lambiam 22:03, 14 April 2025 (UTC)

We have an article, Metrication in Canada, which I assume would answer most of these questions. Otherwise it's more a culture question than a math question. --RDBury (talk) 08:32, 15 April 2025 (UTC)

April 16

Bayes’ theorem with multiple evidences of different weights

How is Bayes’ theorem calculated when multiple pieces of evidence that support or rebut a hypothesis with different strength amounts have to be considered? Here is an example scenario.

There are two jars, each filled with 20 balls. The left jar has 18 green balls and two orange balls. The right jar has five green balls and 15 orange balls. Each ball weighs about 70 grams regardless of color. Each jar sits on top of its own electronic scale. You close your eyes and are given one of the balls at random. The ball is green, but the right jar’s scale reading is 70 grams lighter than what it previously was. How would this be computed? Primal Groudon (talk) 21:38, 16 April 2025 (UTC)

We use the following notations for various events:

$L$: the ball was drawn from the left jar.

$G$: the ball is green.

$W$: the right jar is lighter than before.

(These each have complementary events, which if they need to be denoted can be done by using an overbar; for example, ${\overline {G}}$ would mean: the ball is orange. Then $P(G\cap {\overline {G}})=0$ and $P(G\cup {\overline {G}})=1.$)

In the set up, we know by prior real-world knowledge that jars do not get spontaneously lighter, so $P(L\cap W)=0.$

If we try to apply Bayes' theorem we do not directly get anywhere. Ii is easier to go back to basics, the definition of conditional probability:

$P(A|B)={\frac {P(A\cap B)}{P(B)}},{\text{ provided that }}P(B)>0.$

The role of $A$ – the event whose likelihood is to be determined – is taken by $L,$ while $B$ – the observed event – is $G\cap W.$ Then we get:

$P(L|G\cap W)={\frac {P(L\cap G\cap W)}{P(G\cap W)}}.$

We know that $P(L\cap W)=0.$ Thus, a fortiory, $P(L\cap G\cap W)=0,$ and so $P(L|G\cap W)=0.$

The problem in applying Bayes' theorem is that it uses the conditional probability with the events swapped, $P(G\cap W|L),$ which is not directly available. It can likewise be determined by applying the basic definition of conditional probability; however, this route is an unnecessary detour.  ​‑‑Lambiam 22:51, 16 April 2025 (UTC)

What’s the minimal size of a nonce leakage so that the private can be recovered from a single signature in practice ?

There’re a lot of papers on how to recover a private key from a nonce leakage in a ᴇᴄᴅꜱᴀ signature. But the less bits are known the more signatures are required.

Now if I don’t know anything about private key, how much higher order or lower order bits leakage are required at minimum in order to recover a private key from a single signature ? I’m interested in secp256k1. 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 22:37, 16 April 2025 (UTC)

According to the abstract of this paper, the answer is: 12 bits.  ​‑‑Lambiam 23:09, 16 April 2025 (UTC)

The question is how to do it in pratice ? The paper seems only theoritical and do not seems to speak on how to implement it. This might means for actually doing it that the number of bits is larger. 37.171.242.50 (talk) 10:17, 17 April 2025 (UTC)

If you have the capability of mounting side-channel attacks to give you 12 bits, surely you are also able to use their method of attack, which they qualify as being "very practical", to obtain the coveted secret key.  ​‑‑Lambiam 13:56, 17 April 2025 (UTC)

I just temporarirly uploaded it to https://jumpshare.com/s/236VsVoUccTfoWhSv9go. It seems to require several signatures and not just 1. 78.246.6.147 (talk) 15:08, 17 April 2025 (UTC)