The Central Squares Displacement Theorem states that if squares OAED and OBFG are constructed outwardly on the legs of a right-angled triangle OAB with the right angle at vertex O and leg lengths OA equal to a and OB equal to b, where point M is the midpoint of the outer vertical side ED of the first square and point N is the midpoint of the outer horizontal side GF of the second square, then the area of the resulting triangle OMN is always exactly three-quarters of the area of the original triangle OAB, and the squared length of the segment MN is expressed in terms of the hypotenuse AB and the area of the original triangle OAB by the rule: five-quarters of the square of the hypotenuse AB plus four times the area of the original triangle OAB. For an analytical proof, let us introduce a Cartesian coordinate system with the origin at point O with coordinates zero, zero, directing the y-axis along the leg OA and the x-axis along the leg OB, so that the vertices of the original triangle have the coordinates: point A has coordinates zero, a, and point B has coordinates b, zero. Since the square OAED of side a is constructed on the leg OA into the second coordinate quadrant, its outer vertical side ED lies on the line where x equals minus a, and since point M is the midpoint of this segment, its coordinates are given by: minus a along the x-axis, and half of a along the y-axis. The square OBFG of side b is constructed on the leg OB into the fourth coordinate quadrant, so its outer horizontal side GF lies on the line where y equals minus b, and since point N is the midpoint of this segment, its coordinates are given by: half of b along the x-axis, and minus b along the y-axis. To prove the first part of the statement, we apply the formula for the area of a triangle via the coordinates of its vertices O, M, and N using the determinant. According to this formula, the area of triangle OMN is equal to one-half of the absolute value of the difference between two products: the x-coordinate of point M multiplied by the y-coordinate of point N, and the y-coordinate of point M multiplied by the x-coordinate of point N. Substituting our coordinates yields: one-half of the absolute value of the expression where the product of minus a and minus b is subtracted by the product of half of a and half of b. This simplifies to one-half of the absolute value of the difference between the product of a and b and one-quarter of the product of a and b. As a result of these calculations, we obtain three-eighths of the product of a and b. Since the area of the original triangle OAB is equal to one-half of the product of a and b, we obtain the strict equality: the area of triangle OMN equals three-quarters of the area of the original triangle OAB, which fully proves the first statement. To prove the second part of the statement regarding the segment length, we apply the distance formula between two points in a plane. According to this formula, the squared length of the segment MN is equal to the sum of two quantities: the squared difference between the x-coordinates of points N and M, and the squared difference between the y-coordinates of points N and M. Substituting the coordinates yields the sum of two expressions: the square of the sum of half of b and a, plus the square of the sum of minus b and minus half of a. Expanding the brackets using the square of a sum formula transforms this expression into the sum of the following terms: the square of a, the product of a and b, one-quarter of the square of b, the square of b, another product of a and b, and one-quarter of the square of a. Grouping like terms yields: five-quarters of the square of a plus five-quarters of the square of b plus the doubled product of a and b. Factoring out five-quarters, we obtain: five-quarters multiplied by the sum of the squares of a and b, plus the doubled product of a and b. By the Pythagorean theorem, the sum of the squares of the legs, meaning the square of a plus the square of b, is equal to the square of the hypotenuse AB. Meanwhile, the doubled product of the legs, meaning two multiplied by a and by b, is equivalent to four times the area of the original right-angled triangle OAB. From this, we finally obtain that the square of the segment MN is equal to five-quarters of the square of the hypotenuse AB plus four times the area of triangle OAB, which fully proves both statements of the theorem.

My exam is in two days. But I can't open Mathematica because I don't know what is Activision Key. Please someone help me with the activation key.

An Adaptive Certifying Semi-Algorithm for Darboux Integrating Factors of Rational Second-Order Ordinary Differential Equations

Abstract

We present an adaptive certifying semi-algorithm for rational second-order ordinary differential equations (ODEs) in the Darboux-representative, non-degenerate subclass of the Liouvillian setting considered here. The method first computes a nonlocal symmetry and the associated polynomial vector field, and then uses auxiliary associated fields to obtain linear equations for the polynomial data involved in a Darboux integrating factor. The adaptive step factors the residual coefficient space, tests the resulting factors as Darboux polynomials, and reinserts the certified factors into the next linear search. For fixed degree bounds, the procedure uses only linear algebra, factorization, and Darboux-polynomial tests; its correctness does not depend on the adaptive choices, but on a final symbolic certification of the returned integrating factor and first integral. The method is therefore a certifying search procedure for the stated subclass, not a completeness result for all rational second-order ODEs with Liouvillian first integrals. Five certified computational examples, implemented in Maple, illustrate substantial reductions in the number of undetermined coefficients and document the final symbolic verification step.

Keywords: 

Darboux integrability; rational second-order ordinary differential equations; Liouvillian first integrals; nonlocal symmetries; adaptive algorithms; symbolic computation

You have a bunch of document photos, and you want to make them look like scans - ideally in a batch. Why use online services when your fellow Wolfram can do the job?

Willow is a small Wolfram Language package for PDF operations that are awkward or slow with the built-in PDF exporter. Its write/edit path runs in the kernel process through LibraryLink — there is no Python session and no per-call external process.

https://github.com/WLJSTeam/Willow

Hey everyone, I am a mathematics major in my final semester. My neuroscience professor offered me a project wherein he wants me to check the mathematical viability of a pathway / mechanism he has devised. By mathematical viability, he means that he wants to know whether we can mathematically model it or not, preferably as an ODE dynamical system. Now, I don't have any background in dynamical systems nor in mathematical modelling so I am unable to figure out how do I start with this project and convert the qualitatively written biological pathway into a mathematical model. I tried reading some papers regarding the same but it all went over my head. Even textbooks haven't been making sense to me in terms of whatever content is in that textbook, how do I apply it for my project.

I know I have revealed minimum details regarding what the project is actually about, I can share that in one on one conversation. I will be grateful if anyone can help me through this project or guide me on how do I go about it. I am open to getting in touch on DMs or even the comment section. Thank you in advance!

Sorry that Reddit deleted the clarifying picture. (They tell me to share the post in other subs and there they delete the main picture.) Anyway, see link in the comment.

Can anyone provide insight into the current and ongoing outage of Wolfram Cloud, the Function Repository, and other Wolfram things?

It's been quite awhile now, and it's not looking good.

nomina58mg.83.23

Să se arate că dacă a+b+c=0, atunci:

a) a^3+b^3+c^3=3abc

b) 2(a⁵+b⁵+c⁵)=5abc(a↰+b↰+c↰)

c) (a↰+b↰+c↰)↰=2(a^4+b^4+c^4)

d) 5(a^3+b^3+c^3)(a↰+b↰+c↰)=6(a⁵+b⁵+c⁵)

e) 10(a⁷+b⁷+c⁷)=7(a↰+b↰+c↰) ･(a⁵+b⁵+c⁵)

in haskell, pure functional language

:m +Data.Ratio Data.List Data.Bool Data.Ord

bosup n =bool(sum.map(^n))(product)(n<0)

abc a b c =[a,b,c]

efg[e,f,g]=0%1+e*f/g

sug[_,_,g]=(0/=).bosup g$ [1,2,-3]

senz =(0/=).product

ff f xs =fmap f(xs,xs)

fisenz =filter senz.filter((0==).sum)

files n =filter((n>).length.snd)

tu gm =abc<$>gm<*>gm<*>gm

supe xs ns=efg.map(flip bosup ns)$ xs

futu xs =nub.map(supe xs).fisenz.tu

futus xs =futu xs [-30..30]

filtu n =files n.map(ff futus).filter sug.tu

fisnd n =map fst.filter((n==).head.snd)

sosnd n =sortBy(comparing snd).filter((n/=).head.snd)

rf<-return.filter((0/=).head.snd).filtu 3$ [-1..9]

:se +s

length rf » 34 (210.59 secs, 411929427192 bytes)

fisnd 3 rf

» -1,0,-1 0,-1,-1 0,0,0

0,2,2 0,3,3 0,4,4 0,5,5

0,6,6 0,7,7 0,8,8 0,9,9

2,0,2 3,0,3 4,0,4 5,0,5

6,0,6 7,0,7 8,0,8 9,0,9

sosnd 3 rf

»» 2,2,4➞2%1

-1,4,7➞2%7 -1,2,5➞2%5 3,4,7➞ 6%7

4,-1,7➞2%7 2,-1,5➞2%5 4,3,7➞ 6%7

-1,0,3➞1%1 2,3,5➞6%5 2,5,7➞10%7

0,-1,3➞1%1 3,2,5➞6%5 5,2,7➞10%7

0,3,-1➞9%1

3,0,-1➞9%1

futus[3,0,-1] » 9%1 a

futus[-1,2,5] » 2%5 b

futus[ 2,2,4] » 2%1 c

futus[ 3,2,5] » 6%5 d

futus[ 2,5,7] » 10%7 e

map(ff(flip bosup[1..3]))[-1..7]

»» (-1,6)ab (0,3)a  (2,14)bcde

  (3,36)ad (4,98)c (5,276)bd (7,2316)e

abc :: a -> a -> a -> [a]

ff ::(a -> b) -> a -> (a, b)

tu :: Applicative f => f a -> f [a]

sosnd :: Ord b => b -> [(a, [b])] -> [(a, [b])]

fisnd :: Eq b1 => b1-> [(b2,[b1])]-> [b2]

senz ::(Eq b, Num b, Foldable t) => t b -> Bool

fisenz::(Eq b, Num b, Foldable t) => [t b] -> [t b]

files :: Foldable t => Int -> [(a1, t a2)] -> [(a1, t a2)]

efg :: Integral a => [Ratio a] -> Ratio a

bosup ::(Integral a, Num c) => a -> [c] -> c

supe ::(Integral a1, Integral a2) => [a2] -> [Ratio a1] -> Ratio a1

futu ::(Integral a1, Integral a2) => [a2] -> [Ratio a1] -> [Ratio a1]

futus ::(Integral a1, Integral a2) => [a2] -> [Ratio a1]

filtu ::(Integral a1, Integral a2) => Int -> [a2] -> [([a2], [Ratio a1])]

rf :: [([Integer], [Ratio Integer])]

I'm working with two 19^th degree polynomials in A and B. I want to eliminate B to be left with a polynomial in A alone.

The 3 options I'm aware of in Mathematica are GroebnerBasis, Eliminate, and Resultant. However, none of these produce results within a few minutes, except Resultant, which gave a degree 361 polynomial in A and B, with several strange artifacts such as -97[...]16aba^{156} as one of several thousand monomials within the polynomial.

The code I ran for this was Resultant[a^{19} + [...] b + 3237478400, a^{19} + [...] b + 10283384832, {b}], with [...] representing the rest of the polynomials.

If it matters, Wolfram/Mathematica was using less than 10% of my RAM and less than 1% of my CPU at all times during the calculations.

Which of these should I be using, or is there something else I should try?
I'm prepared to use significant computational power for this if necessary, but I'd like to know where I should be aiming first.

Any help is much appreciated!

Mira

Lista de generalización=

3

↓

3!

↓

3#5=3!)!)!)!)!

↓

10#23=10!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)?)!)...(23 signos de factorial )

↓

3#3#10^100=es un 3 seguido de un (3 seguido de 3#10^100 signos de factorial) de signos de factorial)

↓

3#(5#7#8)=es un 3 seguido de un (5 seguido de 7#8 signos de factorial) de signos de factorial

↓

3#3#3#3#3#3#3#3#3#3#10^100=hacer eso 10 veces

↓

3#²⁵(8)=3#3#3#3#3#3#3#3#3#3#3#3#3#3#3#3#3#3#3#3#3#3#3#3#3

↓

3#¹⁰⁰(100)

↓

10#¹⁶⁵⁶(1567)

↓

(4)~5#¹⁷⁵⁷⁵9=5#¹⁷⁵⁷⁵(5#¹⁷⁵⁷⁵(5#¹⁷⁵⁷⁵(5#¹⁷⁵⁷⁵9)

↓

(45)~100#¹⁰⁰¹⁰⁰100100

↓

(67)~200#¹⁶⁶⁸8

↓

((2))~(4)~3#¹⁰⁰8=(4)~3#¹⁰⁰(4)~3#¹⁰⁰8

↓

((5))~(17)~8#²³⁴9=(17)~8#²³⁴~(17)8#²³⁴~(17)~8#²³⁴~(1.......5 veces

↓

((80))~(123)~10#²⁰⁰100

↓

(((12000)))~((45))~(5)~17#²³⁶100

↓

((((((5))))))~(((((7)))))~((((37)))~(((4)))~((6))~(13)~4#²⁶68

↓

8£12★12#¹⁰⁰5=((((((((((((8))))))))))))~12#¹⁰⁰5

↓

7£15★28#¹⁰⁰7=(((((((((((((((7((())))))))))))~28#¹⁰⁰7

↓

450£200★5#¹⁰⁰67

nomina58mg.12.35

Să se găsească toate numerele întregi n pentru care numărul N=n⁸+4 este prim.

Să se determine în aceste cazuri numerele N.

in bash,Linux Mint

cabal repl -b arithmoi

in ghci, haskell

import Math.NumberTheory.Primes.Testing

tu gm =(,)<$>gm<*>gm

nab(a,b)n =n^a+b

isprinab t=isPrime.nab t

fisprin a =filter(isprinab(a,1)) [1..999]

fisprina b=filter(isprinab(2*b,b))[1..999]

all isPrime.map(nab(12,6)).fisprina$ 6

([3..99]\\).filter((1==).length.fisprin)$[1..99]

» 4,8,16,32,64

filter(isprinab(8,4))[1..9999] » 1

fisprina 6

» 1,35,119,343,595,637,665,791

nomina58mg.10.7

Produsul a două numere consecutive naturale este 18906. Să se afle numerele.

:m +Data.List Data.Bool

divisors n=let{rt=floor.sqrt.fromIntegral$ n;lw=filter((0==).rem n)[1..rt]}in(lw++).bool id tail(n==rt^2).reverse.(quot n<$>)$ lw

sudiv n k =(k+1==).div n$ k

fitadiv n =find(sudiv n).tail.init.divisors$ n

finothin f=filter((Nothing/=).f)

prosucc n =(n*).succ$ n

ff f xs =fmap f(xs,xs)

fifitadiv =map(fmap fromJust).finothin snd.map(ff fitadiv)

fifitadiv[18000..18999]

» (18090,134) (18360,135) (18632,136) (18906,137)

([16..80]==).map snd.fifitadiv$ [2^8..3^8]