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September 19

Property of a matrix

If I raise a permutation matrix to a high power it is still a permutation matrix (all values 0 or 1). If I raise a stochastic matrix to a high power it is still a stochastic matrix (all values <=1). But there are a lot of matrices raised to a high power have elements with very large values. Is there a name for the property of a matrix that when raised to a high power remain reasonably valued? RJFJR (talk) 22:12, 19 September 2024 (UTC)[reply]

The spectrum of such a matrix is inside the closed unit disk (a necessary but not sufficient condition). A matrix whose spectrum is in the open unit disk always satisfies this property. I'm not sure there is a general name for this property though. Tito Omburo (talk) 23:07, 19 September 2024 (UTC)[reply]

That's it. Thank you! RJFJR (talk) 02:08, 20 September 2024 (UTC)[reply]

Note that when the spectral radius is strictly less than

1,

the powers of the matrix tend to the null matrix, which is not very exciting. I also vaguely remember seeing a theorem that states or implies that when the spectral radius is equal to

1,

the growth of the entries is polynomial in the value of the exponent. --Lambiam 21:55, 20 September 2024 (UTC)[reply]

September 20

$A def = B + C$

1. Since the expression in the title is a definition, we can conclude, that A exists if and only if both B and C exist as well.

2. However, we can't conclude that if B or C exist then A exists. Check: A denotes the total price, B denotes the price of the first product, and C denotes the price of the second product: if only the price $B$ of the first product exists, we still can't conclude - that the price $C$ of the second product exists - nor that the total price $A$ exists. The defintion $A{\overset {\underset {\mathrm {def} }{}}{=}}B+C$ only tells us (besides the relation $A{\overset {\underset {\mathrm {def} }{}}{=}}B+C$ ), that the total price $A$ exists if and only if both - the price $B$ of the first product exists - and the price $C$ of the second product exists.

3. For concluding, that if any value of the above three values A,B,C exists then all of them exist, it's sufficient to write down three definitions:

A{\overset {\underset {\mathrm {def} }{}}{=}}B+C.

B{\overset {\underset {\mathrm {def} }{}}{=}}A-C.

C{\overset {\underset {\mathrm {def} }{}}{=}}A-B.

4. Question: as I've indicated, it's sufficient (to write down all three defintions), but is there any shorter notation, to make sure that if any value of the above three values A,B,C exists then - all of them exist - and satisfy $A{\overset {\underset {\mathrm {def} }{}}{=}}B+C?$

HOTmag (talk) 13:26, 20 September 2024 (UTC)[reply]

Are you asking about a shorter notation? I may be looking mighty foolish, but isn't it circular and therefore meaningless to write the system of three definitions like you have above? Or rather, wouldn't writing down the first be exactly as meaningful as writing down all three, since both situations only relate A, B, C to one another? I get they're definitions and not merely statements, but I'm not really seeing the difference here. You'd have to define B, C in terms other than A to reach a more meaningful domain. Again, I could look totally foolish right now, since I've never answered a question here before. Remsense ‥ 论 13:34, 20 September 2024 (UTC)[reply]

It seems you haven't read 2#. HOTmag (talk) 13:40, 20 September 2024 (UTC)[reply]

I don't really understand it, no. You're defining A, B, C as the prices of products, but then you're writing abstract definitions of them in terms of each other. Am I missing something? Oh, did you mean to say A is the total? It makes more sense to me that way. Remsense ‥ 论 13:43, 20 September 2024 (UTC)[reply]

In any case, I still don't understand what the extra two definitions achieve: either B and C are defined outside of A or they're not, right? If we wanted to know them in terms of A, we already got that in the first definition. Remsense ‥ 论 13:45, 20 September 2024 (UTC)[reply]

I've just added an addition to 2#, to make it clear. HOTmag (talk) 13:49, 20 September 2024 (UTC)[reply]

I return to my initial question then, are you just looking for a shorter notation for this? Remsense ‥ 论 13:52, 20 September 2024 (UTC)[reply]

Yep. I've just made it clear in 4# (thanks to your question). HOTmag (talk) 13:55, 20 September 2024 (UTC)[reply]

Sorry for being slow on the uptake. I'm not sure if this needs to fit into any particular system or paradigm: anything wrong with

\forall (B,C)\in \mathbb {R} ^{+},\exists A=B+C

. Sorry if that hurts anyone to see, haven't flexed these muscles in a while Remsense ‥ 论 14:31, 20 September 2024 (UTC)[reply]

Not to bug you, but only since I'm relatively unsure of myself in this area—was this answer something like what you were looking for? Remsense ‥ 论 20:10, 20 September 2024 (UTC)[reply]

You limit the set of the Bs and the Cs to be the positive integers, but my question is general, without limiting anything. HOTmag (talk) 02:05, 22 September 2024 (UTC)[reply]

You gave the example of prices, so that's what I picked the positive reals* based on. Clearly, you can replace the set with whatever you want. Remsense ‥ 论 02:08, 22 September 2024 (UTC)[reply]

If A denotes the total price, B denotes the price of the first product, and C denotes the price of the second product, does your definition let us deduce the other definition:

C{\overset {\underset {\mathrm {def} }{}}{=}}A-B

which I would like to deduce, bearing in mind that not all products have a price? Note: Since the latter is a defintion of C, then the existence of A and of B must be derived from the very existence of C. HOTmag (talk) 02:21, 22 September 2024 (UTC)[reply]

It seems the bit about "deducing a definition" articulates a fundamental confusion you have about what you're trying to accomplish. Definitions are stated, not deduced. Remsense ‥ 论 05:34, 22 September 2024 (UTC)[reply]

By asking whether a new definition can be deduced from an old definition, I mean whether a new claim, that was presented before as an additional definition, can be derived as consequence, from a given assumption that was presened before as an old definition.

In our case, the given assumption, was presented before as an old definition:

A def = B + C

. The new claim, was presened before as an additional definition:

C{\overset {\underset {\mathrm {def} }{}}{=}}A-B.

The question is, whether we can deduce the latter from the former, i.e. whether we can deduce the new claim from the given assumption. HOTmag (talk) 13:08, 22 September 2024 (UTC)[reply]

If the values we assign to B and C are logically independent from one another, then no. That's what logical independence means.Remsense ‥ 论 13:17, 22 September 2024 (UTC)[reply]

This is totally confused. The addition operation

+

is special in that it is operand-wise strictly monotonic and therefore has, at least in the integers and reals (but not in the natural numbers) an operand-wise inverse, which we can denote using the subtraction operation

-

. In general, this is not possible.

Defining some quantity

A

by an equation of the form

A=f(B,C)

only makes sense if all terms in the right-hand side are defined. It is not just that this fails to define

A

if

B

or

C

is not defined. It just does not make sense. And if

B

is not defined, writing something like

B~{\overset {\underset {\mathrm {def} }{}}{=}}~\cdots

only increases the confusion.

There is another issue in which the definedness of a defined term depends on the definedness of another term, namely when defining a function. Suppose we define a new function

f

using existing known functions with a limited domain. For example, we may define real-valued function

f

on the real numbers by the equation

f(x)={\sqrt {1-x}}+{\sqrt {1+x}}.

For

f(x)

to be defined by this equation for some given value of

x,

it is necessary that both

{\sqrt {1-x}}

and

{\sqrt {1+x}}

are defined. This is more a matter of common sense than anything else, and I see no need for some notational device to express this dependency. --Lambiam 22:29, 20 September 2024 (UTC)[reply]

If A denotes the total price, B denotes the price of the first product, and C denotes the price of the second product, does the definition

A def = B + C

make senee? If it does, can you deduce the definition:

C{\overset {\underset {\mathrm {def} }{}}{=}}A-B?

HOTmag (talk) 02:13, 22 September 2024 (UTC)[reply]

You can't define something that already has a meaning. So if A, B and C all have independent meanings, as shown by the use of "denote", then none of them can be defined in terms of the other two. It may be true that A = B + C based on these meanings, but that's not a definition. In any case, you can't deduce a definition. If A has no independent meaning then you can define it to be anything, B + C, B - C, or B * C. --RDBury (talk) 05:29, 22 September 2024 (UTC)[reply]

Well, I'm presenting an analogous question (taken from a discipline very close to arithmetic), in my following thread. I hope my new question explains also my old question, but if my old question is still not clear, you can ignore it, and focus on my new question. HOTmag (talk) 13:38, 22 September 2024 (UTC)[reply]

September 22

How can we briefly characterize a given set of vectors, as "linearly dependent - every proper sub set being linearly independent", while we only refer to the vectors rather than to their set?

For example: S is a set of the following vectors:

A=(1,1,0),

B=(1,0,0),

C=(0,1,0).

Note: A=B+C, and B=A-C, and C=A-B, so the set S is linearly dependent.

Using A,B,C only, i.e without using S, what's the shortest description, claiming that the set S is linearly dependent but every proper sub set of S is linearly independent? HOTmag (talk) 13:30, 22 September 2024 (UTC)[reply]

@HOTmag: I'd just say 'S has k linearly independent elements' (in the give example k=2). --CiaPan (talk) 14:46, 22 September 2024 (UTC)[reply]

Is your response a suggestion of rephrasing my question?

If it's intended to be an answer, then please note: My condition requires to be "using A,B,C only, i.e without using S". Additionally, where does your description claim, that S is linearly dependent? HOTmag (talk) 14:56, 22 September 2024 (UTC)[reply]

One way to characterise the set is "a set of vectors, any one of which can be written in terms of the others in a unique way". The set is just A, B and C, i.e. any property of them is a property of the set of them. --2A04:4A43:900F:F4C3:49F4:4EFB:C442:608F (talk) 15:15, 22 September 2024 (UTC)[reply]

Since my condition requires to be "using A,B,C only, i.e without using S", so I guess you mean the following: "Each vector, can be written as a unique linear combination of the other vectors". Thanks. HOTmag (talk) 15:52, 22 September 2024 (UTC)[reply]

Assuming the vectors are

v_{1},v_{2},\dots ,v_{k}\in R^{n}

, form the matrix

V

whose columns are

v_{1},\dots ,v_{k}

. The stated condition is then:

the matrix of $k\times k$ minors of $V$ is zero, and
each of the columns of the matrix of $(k-1)\times (k-1)$ minors of V has a non-zero entry.

- Tito Omburo (talk) 17:36, 22 September 2024 (UTC)[reply]

Your description, both uses sets, and also becomes longer than the original one indicated in the title. HOTmag (talk) 08:41, 23 September 2024 (UTC)[reply]

You can say, "each of the sets {A,B}, {A,C} and {B,C} is linearly independent". --Lambiam 21:05, 22 September 2024 (UTC)[reply]

You also have to add that the set {A,B,C} is linearly dependent, but then the description - both uses sets, and also becomes longer than the original one indicated in the title. HOTmag (talk) 08:40, 23 September 2024 (UTC)[reply]

There is a unique linear combination generating 0, and in this linear combination all coefficients are nonzero.2404:2000:2000:8:FDE8:8311:95E3:654D (talk) 00:00, 23 September 2024 (UTC)[reply]

Yes, and by symbolic notation the description even becomes shorter. Thanks. HOTmag (talk) 08:42, 23 September 2024 (UTC)[reply]

September 23

Converting map scale into height and width

How can I calculate the height and width in meters of a future geographic map of the world based on its desired scale in cm and km? 212.180.235.46 (talk) 15:35, 23 September 2024 (UTC)[reply]

For a map of the world, you will have to specify which map projection it will use, and then to which part(s) of the projection the scale is to apply (and for some projections, how much of the globe to include). Mapping a spherical surface onto a plane surface cannot be done with a constant scale factor. Eg: you might specify a Mercator projection, between latitudes +/-85deg, with a scale of 1:10,000,000 at the equator, which would give you a map about 4m wide along the equator, and (judging by the illustration for Mercator) slightly more than that high. Other projections will vary considerably. -- Verbarson ^talk_edits 18:07, 23 September 2024 (UTC)[reply]

My own map with van der Grinten projection and equatorial scale of 1:20 000 000 for example is 1.18 m high and 1.94 m wide. Out of curiosity, I started to think about dimensions of imaginative humongous world maps. With that as benchmark, if my calculations are correct, a 1:160 000 scale world map, for example, would be 147 m long and 242 m wide. 212.180.235.46 (talk) 20:35, 23 September 2024 (UTC)[reply]

The scale of a map is the ratio between the distance between two points on the map and that between the corresponding points on the ground. Any map projection necessarily distorts the distances, so the scale you get if you take New York and San Francisco as the two points is generally substantially different from what you get if you take Berlin and Moscow. For maps of the whole Earth it is non-trivial to decide on a "nominal scale". In spite of the name, the van der Grinten projection is not a geometric projection in which points on the map are found by following rays projected along a straight line out of a shrunk globe. The equatorial and central-meridional scales are well defined, though. The length ℓ_eq of the equator is approximately 40,000 km. If the equatorial scale is chosen to be 1 :

S

_eq, the width of the map is ℓ_eq /

S

_eq. So for

S

_eq = 160,000 this comes out as about 250 m. The length ℓ_mer of a meridian, measured from one pole to the other, is about 40,000 km. Letting

S

_cm stand for the central-meridional scale factor, the height of the map will be ℓ_mer /

S

_cm. If these scale factors are chosen to be equal, the map will have a height of half its width. For a circular map as shown in the article,

S

_cm = 2

S

_eq. --Lambiam 10:14, 24 September 2024 (UTC)[reply]

September 26

Given the optimal ate pairing e(A,B)=y is to possible to determine I and J such as e(I,J)=2y or even e(I,J)=3y?

Simple question : let’s say I have a pairing friendly curve having a very large trace, and that I have a pairing with points A∈ $G_{1}$ and B∈ $G_{2}$ such as the optimal ate pairing e(A,B)=y, then is it possible to fully determine 2 point I and J such as e(I,J) equals a target multiple of the finite’s field element y ?

Or does it requires full pairing or full generalized Miller’s inversion and thus would be impossible in practice on a curve like bn254 ? 2A01:E0A:401:A7C0:9CB:33F3:E8EB:8A5D (talk) 15:51, 26 September 2024 (UTC)[reply]

September 27

Uses of Pascal's tetrahedron

The following is a use of Pascal's triangle:

To find how many ways there are to make a total of $n$ circles all black or red, the formula is just $2^{n}$ . For example, there are $2^{6}=64$ ways to make a group of 6 circles, all black or red, classified by whether each circle is black or red. An example is black-red-black-red-black-red.

But how about finding the number of ways to make a group of 6 circles, all black or red, classified by how many are black or red. To find out how many ways there are to make a total of $n$ circles all black or red classified by how many are black and how many are red, you use the n+1th row of Pascal's triangle. For $n=6$ , this means we use the seventh row, which is $1,6,15,20,15,6,1$ . This means that there is one way to color 6 circles where all of them are black, 6 where 5 are black and one is red, 15 where 4 are black and 2 are red, 20 where 3 are black and 3 are red, 15 where 2 are black and 4 are red, 6 where one is black and 5 are red, and one where all 6 are red.

How about a similar application for Pascal's tetrahedron?? Here is the seventh layer of the tetrahedron:

1 6 6 15 30 15 20 60 60 20 15 60 90 60 15 6 30 60 60 30 6 1 6 15 20 15 6 1

Just as the seventh row of Pascal's triangle can be used for the classification of ways to make 6 circles all of which are black or red classified by how many are black and how many are red, it is likewise true that the seventh layer of Pascal's tetrahedron can be used for... Georgia guy (talk) 14:06, 27 September 2024 (UTC)[reply]

It counts how many ways there are to make 6 circles all of which are black, red, or green, classified by how many are black, how many are red, and how many are green. For example, if you want there to be 2 of each colour, you get

{6 \choose 2,2,2}={6 \choose 2}{4 \choose 2}{2 \choose 2}=90

ways, which is exactly the middle entry of this layer. Double sharp (talk) 14:55, 27 September 2024 (UTC)[reply]

I would call 1, 6, 15, ... the sixth row, and the top row, with a single 1, the zeroth row. That's what Wikipedia's does in the articles I've seen at least. Anyway, this should be in the article Pascal's pyramid, and can be further extended to Pascal's simplex. The article on the Multinomial theorem is relevant here as well. --RDBury (talk) 17:25, 27 September 2024 (UTC)[reply]

RDBury, is the top row "1" not really a row?? Georgia guy (talk) 21:02, 27 September 2024 (UTC)[reply]

Well, you can call it whatever you want and index it as you like; I'm not going to get into the philosophy of rows. But the formulas are simpler if you start with the top "1" as row 0. --RDBury (talk) 00:55, 28 September 2024 (UTC)[reply]

RDBury, does that statement parallel the statement that trigonometry is simpler if you use radians as opposed to degrees?? Georgia guy (talk) 01:04, 28 September 2024 (UTC)[reply]

The OP was calling what most people call row 6 "the seventh row" and I thought that was worth pointing out for future reference. Getting caught up in what counts as a "row" and whether statements are parallel is a matter for philosophy and linguistics. --RDBury (talk) 02:20, 28 September 2024 (UTC)[reply]

September 28

Bitcoin price rigging

I have been day trading bitcoin for almost a year now. I read articles from bitcoin news to try and guess when the dips and dives will happen and so far it has helped. I just read and article about bitcoin possibly losing freedom to governments and large institutions being able to rig the price. This has happened recently with the German government of Saxony selling off a hoard of seized coin, The Mt. Gox dispersal, and the US dept of Justice mysteriously moving 30/230k of its seized hoard two days after Trumps bitcoin speech in Nashville raised the price. https://mpost.io/u-s-unloads-2-billion-in-bitcoin-from-silk-road-seizure/

Is it possible for a whale (large bitcoin hodler) to make smaller transactions in a short time to move the price for a larger transaction at a later time? On the upswing this is called 'pump and dump' and 'poop and scoop' on the down swing. Both are illegal in most market trades.

This post may fit better in Humanities where finance is listed or IT where crypto may belong. 2604:3D08:5E7A:6A00:D94:3638:168B:18A0 (talk) 08:38, 28 September 2024 (UTC)[reply]

I'm pretty sure that this isn't a math question. But the article on Pump and dump does explicitly mention cryptocurrencies. --RDBury (talk) 13:22, 28 September 2024 (UTC)[reply]

The reason I posted in math is because I am wondering if a small buy or sell in a quick time frame will actually move the price enough for traders like me who watch the minute candle scale. I use MetaTrader 4 where the price moves in microseconds every time a buy/sell happens. Is there a formula for volume needed to move the price in a small time frame or article about time/volume/price/ ratio calculations?

Pump and dump, Bear raid, Short and distort, and Uptick rule, all help to explain how it is possible for whales to control the price. Fear_of_missing_out#Investing is the main cause of up-spikes since I have been investing and most are followed by dives. https://www.weforum.org/agenda/2024/08/explainer-carry-trades-and-how-they-impact-global-markets/ The carry trade crash the 1st week of August caused another huge dive. Foreign_exchange_market#Carry_trade 2604:3D08:5E7A:6A00:D94:3638:168B:18A0 (talk) 17:48, 28 September 2024 (UTC)[reply]

The time needed to move the needle with high-frequency trading is a combination of the latency of the connection between the computers of the flash traders and those of the exchanges, which depends on the current state of the networking technology and the physical distance between the traders and the exchange, as well as any regulations enforcing a time lag. It then will take time before small-time traders can see the needle having moved. A mathematical model will require too many parameters to be of practical use. --Lambiam 08:51, 29 September 2024 (UTC)[reply]

September 29

Is this a possible Groth16/ZkSnark verifier‑side simplification ?

Hello,

The verification algorithm is already simple but I was thinking about some costly environments like blockchains having low block limits. This might be naïve thinking but I was wondering at possibility : normally the prover gives 3 elliptic curves points to the verifier A ; B ; C When public inputs are used C/the inputs vector is split.

But as a simplification part, why not completely ditch the C parts of the proof when public inputs are used ? That way, the verifier would have to compute 1 pairing in less for verifying the proof. I’m meaning e(C,verifying_key_part). It seems to me the requirement to pair with public inputs would still ensure the security of the system… Is it because skipping that pairing would allow to forge public inputs ? As far I understand, a malicious prover would still have to satisfy all constraints of the quadratic arithmetic program and thus would have to use public inputs satisfying constraints. Or is it because it would be impossible to rework the protocol to have the prover being able to produce proofs that verify ?

Or even maybe both of the assumptions above ? 2A01:E0A:401:A7C0:9CB:33F3:E8EB:8A5D (talk) 11:51, 29 September 2024 (UTC)[reply]

September 30

How to find max(x^y) for each r where x + y = r and x, y are positive integers

Drawing fractal canopy diagrams had me wondering for a given SVG file size, I could add more branches or more depth. Below are two examples:

To make the tree as dense as possible, I wish to maximise the number of leaf nodes. In other words, how can I find max(x^y) for each r where x + y = r and x, y are positive integers?

I realise I can use calculus but I'm unsure what equation I should differentiate.

Thanks, cmɢʟee⎆τaʟκ 00:50, 30 September 2024 (UTC)[reply]

Unfortunately, this is actually quite complicated. We can rewrite the problem as

\max(x^{r-x})

, and consider all real

x

instead of just integers. Notice that

\ln(x^{r-x})=(r-x)\ln(x)

. Because

\ln

is monotonic over

x

,

x^{r-x}

is maximized when

(r-x)\ln(x)

is. Its derivative is

-\ln(x)+(r-x)/x

, which is

0

when

r=x+x\ln(x)

. This is a nice closed-form expression, but it's for

r

in terms of

x

. Inverting it is complicated, but Wolfram Alpha gives us

x=r/W(er)

where

W

is the Lambert W function. While this is sort of a closed-form expression, it's still unfortunately annoying to work with since

W

is implicitly defined as

W(z)=w\Leftrightarrow z=we^{w}

. In any case,

x=r/W(er)

is irrational for all positive rational

r\neq 1

. The proof of this is annoying so I've put it below, but ultimately what it implies is that when

r>1

, as far as I can tell, the best you can do is either round

x

for convenience, or take floor/ceiling of

x

and compare values to get the max over integers.

Proof that

r/W(er)

is irrational for positive rational

r\neq 1

Suppose $r\in \mathbb {Q} ^{+}\backslash \{1\}$ , and let $w:=W(er)$ .
$W(er)=0\Rightarrow r=0$ and $W(er)=1\Rightarrow r=1$ , so $w\neq 0,1$ as well.
By definition, $er=we^{w}$ . Since $w\neq 0$ , we can rearrange to get $r/w=e^{w-1}$ .
Lindemann's 1882 theorem implies that if $k$ is nonzero rational, then $e^{k}$ is not only irrational, but transcendental as well.
If $w$ is rational, then since $w\neq 1$ , $e^{w-1}$ is transcendental, while $r/w$ is rational, which is contradictory.
$w$ must be irrational and is nonzero, so $r/w$ is irrational.
We conclude that $r\in \mathbb {Q} ^{+}\backslash \{1\}$ implies $r/W(er)$ is irrational.

GalacticShoe (talk) 02:41, 30 September 2024 (UTC)[reply]

Here is a numerical recipe (Newton's method) for solving

r=x+x\ln x

for real-valued

x

:

Set $x=9$
Iterate the replacement $x=x^{*}$ until convergence, where
$x^{*}={\frac {x+r}{2+\ln x}}.$

In practice (

r<90

), two iterations will bring you close enough; then test

\lfloor x\rfloor

and

\lceil x\rceil

to get the optimal integer value. --Lambiam 10:17, 30 September 2024 (UTC)[reply]

Thanks so much, @GalacticShoe and @Lambiam. I thought analytically I was heading into a dead end. Guess solving it iteratively is still the best for small values. cmɢʟee⎆τaʟκ 11:45, 30 September 2024 (UTC)[reply]

P.S. It seems there's an interesting trend:

x=1 for r=2	(1 term):	1¹.
x=2 for r=3 to 4	(2 terms):	2¹ and 2².
x=3 for r=5 to 7	(3 terms):	3², 3³ and 3⁴.
x=4 for r=8 to 11	(4 terms):	4⁴, 4⁵, 4⁶ and 4⁷.

x changes when r is the x–1th triangular number + 2. Serendipity? cmɢʟee⎆τaʟκ 17:06, 30 September 2024 (UTC)[reply]

Unfortunately, serendipity. The number of

r

for each

x

is the sequence OEIS:A108414. Although it starts with

1,2,3,4

, it quickly levels out. The inverse triangular number function you're looking for is

\lceil ({\sqrt {8r-7}}-1)/2\rceil \approx {\sqrt {2r}}

, while

x=r/W(er)

grows faster than

r/\ln(r)

, which in turn grows faster than

{\sqrt {2r}}

, hence the number of terms slowing down in growth. GalacticShoe (talk) 03:46, 1 October 2024 (UTC)[reply]

Note that, unlike for T_n, these first differences in the r-values for which x changes do not only always increase but may even decrease. --Lambiam 03:58, 1 October 2024 (UTC)[reply]

After some testing, it seems that rounding suffices at least for r < 100, possibly more. To simplify it, applying Newton's method twice, we can get this approximation which maximizes

x^{y}

for these values of

x+y=r

:

x=round\left({\frac {9+5.197225r}{2.373852+4.197225\ln(9+r)}}\right)

GalacticShoe (talk) 04:14, 1 October 2024 (UTC)[reply]

October 1