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Posts tagged #Polynomials

Original post on mathstodon.xyz

I've been making plots of the sets of roots of some polynomials. They look more interesting than I'd expected! And I've been making sounds from them, too. I have a bunch on this page of my website (I'll probably be adding more). Click on the images to embiggen them, and if you're in a hurry, the […]

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This LEGO cuboid represents f(x)=(x+0)(x+3)(x-1)
=x³+2x²-3x ( blue cube x³+ red cuboid 3x²+ white cuboid -1x² + white cuboid 3*-1*x. The + sign represents the directions forward, right and up, the - sign represents backward, left and down.
Half of its surface equals 3x² (blue squares) +2*3x (red rectangles) + 2*-1x (white rectangles)+ 3*-1*x (white rectangle) =3x²+4x-3;
Epsilon, eff off ; )
A cube has six boundaries, squares. So the derivative f'(x) of f(x)=x³ equals 3x². A square has four boundaries, edges. Therefore the derivative of f(x)=x² is f'(x)=2x. An edge has two boundaries, points. So g'(x) equals 1, if g(x)=x. Points have no boundaries: no boundary, no derivative. (A tesseract has eight boundaries, cubes. Half of them are 4x³ etc.

This LEGO cuboid represents f(x)=(x+0)(x+3)(x-1) =x³+2x²-3x ( blue cube x³+ red cuboid 3x²+ white cuboid -1x² + white cuboid 3*-1*x. The + sign represents the directions forward, right and up, the - sign represents backward, left and down. Half of its surface equals 3x² (blue squares) +2*3x (red rectangles) + 2*-1x (white rectangles)+ 3*-1*x (white rectangle) =3x²+4x-3; Epsilon, eff off ; ) A cube has six boundaries, squares. So the derivative f'(x) of f(x)=x³ equals 3x². A square has four boundaries, edges. Therefore the derivative of f(x)=x² is f'(x)=2x. An edge has two boundaries, points. So g'(x) equals 1, if g(x)=x. Points have no boundaries: no boundary, no derivative. (A tesseract has eight boundaries, cubes. Half of them are 4x³ etc.

drawn with this ingenious tool/toy:
https://www.matheretter.de/rechner/gfplot
Calculus of f(x)=x³+2x²-3x (blue graph), f'(x)=3x²+4x-3, pink, f''(x)=6x+4, turquoise, f'''(x)=6, green, F(x)=x^4/4+2x³/3-3x²/2, purple

drawn with this ingenious tool/toy: https://www.matheretter.de/rechner/gfplot Calculus of f(x)=x³+2x²-3x (blue graph), f'(x)=3x²+4x-3, pink, f''(x)=6x+4, turquoise, f'''(x)=6, green, F(x)=x^4/4+2x³/3-3x²/2, purple

Calculus of polynomials without epsilon
Half of the boundaries of an n-dimensional cuboid represent the slope of the graph of the corresponding function!
One x² represents one third of half the boundaries of a cuboid AND the area between function graph and x-axis, the antiderivative. It is shockingly simple, isn't it?

Calculus of polynomials without epsilon Half of the boundaries of an n-dimensional cuboid represent the slope of the graph of the corresponding function! One x² represents one third of half the boundaries of a cuboid AND the area between function graph and x-axis, the antiderivative. It is shockingly simple, isn't it?

Pears aren't more important than apples;)
#Calculus matters!
Speaking of calculus of #polynomials: It could be soo easy. This approach is suitable for people with #dyscalculia.
Polynomials:
n-dimensional cuboids.
Epsilon, eff off!
With #ALText: we are #strongerTogether, and #inclusion is #fairness.

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String Theory Inspires a Brilliant, Baffling New Math Proof | Quanta Magazine Years ago, an audacious Fields medalist outlined a sweeping program that, he claimed, could be used to resolve a major problem in algebraic geometry. Other mathematicians had their doubts. Now he says...

www.quantamagazine.org/string-theor... #stringtheory #algebraicgeometry #equation #manifolds #polynomials #proof #math #maths #iteachmath #MathSky

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The maximal length of the Erdős–Herzog–Piranian lemniscate in high degree I’ve just uploaded to the arXiv my preprint The maximal length of the Erdős–Herzog–Piranian lemniscate in high degree. This paper resolves (in the asymptotic regime) an old question about the polynomial lemniscates attached to monic polynomials of a given degree , and specifically the question of bounding the arclength of such lemniscates. For instance, when […]
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Follow the scientific journey from Newton’s gravitational law to the development of orthogonal polynomials by Legendre, Laplace, and Gauss in SIAM's The Birth and Early Developments of Orthogonal Polynomials.

📚Find out more: tinyurl.com/4594f28t

#SIAM #Earth #Polynomials

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A plot of all roots of all 15th degree polynomials with terms ±1/(n+1)x^n.  The roots are tiny blue dots on a white background.  They form a ring around the origin at the center of the plot, with something like fractal filigree around the outside of the ring.  The roots are symmetric: each quadrant is identical.

A plot of all roots of all 15th degree polynomials with terms ±1/(n+1)x^n. The roots are tiny blue dots on a white background. They form a ring around the origin at the center of the plot, with something like fractal filigree around the outside of the ring. The roots are symmetric: each quadrant is identical.

#Noisevember! I revisited the Littlewood polynomial sound from day 2 of Noisevember. I thought to investigate a different sort of polynomial. Here, instead of polynomials with coefficients all ±1, the polynomials have coefficient ±1/(n+1) on the x^n term. As […]

[Original post on mathstodon.xyz]

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When Your Simple Math Fact Gets Brutally Upgraded

When Your Simple Math Fact Gets Brutally Upgraded

When Your Simple Math Fact Gets Brutally Upgraded

#Mathematics #Pascaltriangle #Mathhumor #Overexplaining #Polynomials

sciencehumor.io/math-memes/when-your-sim...

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Collinear Fractals & Bandt’s Conjecture: Hidden Geometry of Polynomial Roots | ICERM 2025
Collinear Fractals & Bandt’s Conjecture: Hidden Geometry of Polynomial Roots | ICERM 2025 YouTube video by Bernat Espigulé

Our group member Bernat Espigule's talk at @icerm.bsky.social Illustrating Mathematics: Reunion/Expansion.
Collinear Fractals and Bandt’s Conjecture. Fractal Fract. 2024, 8, 725. doi.org/10.3390/frac...
Interactive demo: complextrees.com/collinear/
@univgirona.bsky.social
#fractals #polynomials

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Hi fam,
have a fulfilling weekend!
Some joker marked my account, e.g. this post about #polynomials & #calculus, which is a textbook example of #inclusion, as spam. This is incredibly unfair; So I'm asking you to share this: bsky.app/profile/paul... (w #ALText) as a protest.
#education #mathematics

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An nice explanation of a discovery which got some coverage in the popular press last spring.
#mathematics #math #Polynomials #Algebra
Source:https://buff.ly/jQOPnY6

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Understanding a function that finds the degree of a multi-variate polynomial The following function finds the degree of a multi-variate polynomial. PolyDeg[expr_]:=expr//ToList//Exponent[#,Variables[#]]&/@#&//Plus@@@#&//Max; Here the function ToList is the same...

Understanding a function that finds the degree of a multi-variate polynomial – mathematica.stackexchange.com The following function finds the degree of a multi-variate polynomial. PolyDeg[expr_]:...

#programming #polynomials #pure-function

Origin | Interest | Match

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A significant theme was the deep relationship between Gaussian integration, orthogonal polynomials, and weight functions, crucial concepts in approximation theory. #Polynomials 5/6

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@jenxofficial.bsky.social asked: TX for your interest!
I use #ALText to be #fair ( #inclusion for the visually impaired) and because there are up to 8.3k characters, #authorsofBluesky can use!
#actuallyautistic #autism #autismawareness
#calculus #polynomials w/o #epsilon:
bsky.app/profile/paul...!

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Visualization of the famous Yang Hui or Pascal Triangle, dimensions 0, 1, 2 (relevant for the Pythagoras theorem 2.0) and 3
In the not yet published novel "Fluten", a combination of thriller and textbook, the characters Sheldon and Hasan, students at the fictional University of Aachen, develop this elegant formula and call it the first Shalan theorem. In German, “Fluten” has a double meaning. It is both the plural of flood and a verb.

Pythagoras: a²+b²=c²
=a²+2ad+d²
=(a+d)²
Cheap trick: Outsourcing of d² leads to d²(2a/d+1), 2a/d+1 becomes m²
Solve for a
2a/d+1=m²
a/d=(m²-1)/2
a=d(m²-1)/2
First Shalan theorm
(d(m²-1)/2)²+d²m²=(d(m²+1)/2)²
Take any natural number >2 and call it dm. "d", as the distance between a and c, must be natural. But not "m" which must be >1.
HERE I FORGOT SOMETHING IMPORTANT; SOME PRIME NUMBERS CAN BE THE HYPOTENUSE!
17=2(4²+1)/2 So m=4, d=2 In this case d must be even.
The great thing about this quadratic formula for lazy people is that all you have to do is square m!
And this is how it works: Let dm=24
Find two factors that give the product 24
d=12, m=2
(12(2²-1)/2)²+24²=(12(2²+1)/2)²
=>18²+24²=(18+12)²
=30²
d=18,m=4/3
(18(16/9-1)/2)²+24²=(a+d)²
18(7/9)/2=7
=>7²+24²=25² 7,24,25 is a primitive Pythagorean triple. With d=2,m=12, you get the other primitive triple 24,143,145: 143²+24²=145²
d=16,m=3/2
8(9-4)/4=10, 10²+24²=26²
d=8,m=3
32²+24²=40²
d=6,m=4
45²+24²=51²
d=4,m=6
70²+24²=74²
If dm is a prime number, d must be 1:
dm=17
=> 288/2=a, 144²+17²=(144+1)²
This is shockingly simple, isn't it?

Visualization of the famous Yang Hui or Pascal Triangle, dimensions 0, 1, 2 (relevant for the Pythagoras theorem 2.0) and 3 In the not yet published novel "Fluten", a combination of thriller and textbook, the characters Sheldon and Hasan, students at the fictional University of Aachen, develop this elegant formula and call it the first Shalan theorem. In German, “Fluten” has a double meaning. It is both the plural of flood and a verb. Pythagoras: a²+b²=c² =a²+2ad+d² =(a+d)² Cheap trick: Outsourcing of d² leads to d²(2a/d+1), 2a/d+1 becomes m² Solve for a 2a/d+1=m² a/d=(m²-1)/2 a=d(m²-1)/2 First Shalan theorm (d(m²-1)/2)²+d²m²=(d(m²+1)/2)² Take any natural number >2 and call it dm. "d", as the distance between a and c, must be natural. But not "m" which must be >1. HERE I FORGOT SOMETHING IMPORTANT; SOME PRIME NUMBERS CAN BE THE HYPOTENUSE! 17=2(4²+1)/2 So m=4, d=2 In this case d must be even. The great thing about this quadratic formula for lazy people is that all you have to do is square m! And this is how it works: Let dm=24 Find two factors that give the product 24 d=12, m=2 (12(2²-1)/2)²+24²=(12(2²+1)/2)² =>18²+24²=(18+12)² =30² d=18,m=4/3 (18(16/9-1)/2)²+24²=(a+d)² 18(7/9)/2=7 =>7²+24²=25² 7,24,25 is a primitive Pythagorean triple. With d=2,m=12, you get the other primitive triple 24,143,145: 143²+24²=145² d=16,m=3/2 8(9-4)/4=10, 10²+24²=26² d=8,m=3 32²+24²=40² d=6,m=4 45²+24²=51² d=4,m=6 70²+24²=74² If dm is a prime number, d must be 1: dm=17 => 288/2=a, 144²+17²=(144+1)² This is shockingly simple, isn't it?

This is just a screenshot of the alternative text of the photo that shows a visualization of the Yang Hui or Pascal Tringle, dimensions 0, 1, 2, 3.. 

I FORGOT SOMETHING IMPORTANT; SOME PRIME NUMBERS CAN BE THE HYPOTENUSE!
17=2(4²+1)/2 So m=4, d=2 In this case d must be even.

This is just a screenshot of the alternative text of the photo that shows a visualization of the Yang Hui or Pascal Tringle, dimensions 0, 1, 2, 3.. I FORGOT SOMETHING IMPORTANT; SOME PRIME NUMBERS CAN BE THE HYPOTENUSE! 17=2(4²+1)/2 So m=4, d=2 In this case d must be even.

Visualization of enlarged n-dimensional cubes, dimensions 0,1,2,3
(a+b)⁰=1
(a+b)¹=1a+1b
(a+b)²=1a²+2ab+1b²
(a+b)³=1a³+3a²b+3ab²+b³

Visualization of enlarged n-dimensional cubes, dimensions 0,1,2,3 (a+b)⁰=1 (a+b)¹=1a+1b (a+b)²=1a²+2ab+1b² (a+b)³=1a³+3a²b+3ab²+b³

My screenshot reads:
Today I have one of my Cassandra days. But my truth, which no one wants to hear, is such a beautiful one. And I wouldn't have pushed Apollon off the edge of the bed.

For Pythagoras 2.0 you only need a single number greater than two and you can immediately tell which Pythagorean triples it belongs to!
This is so much less complicated than the traditional methods.
And you only have to square one number. Which doesn't always have to be natural!

The Shalan theorems deserve more attention.

Pythagoras theorem: a²+b²=c²
1st Shalan theorem
d(m²-1)/2 as "a"
dm as "b", d(m²+1)/2 as "c" with c=a+d

In my pinned tweet I was guilty of an omission: prime numbers can not only appear as legs of right-angled triangles, but some can also become hypotenuses. I forgot to mention that. Those prime numbers, successors of even square numbers (16+1=17)* or half of the successors of odd square numbers ((25+1)/2=13)**, can be a hypotenuse; If so, d must be even.
*d=2,m=4: 2(4²-1)/2=15
=> 15²+8²=17²
**d=8,m=3/2: 8(9/4-4/4)/2=5
=> 5²+12²=13²

My screenshot reads: Today I have one of my Cassandra days. But my truth, which no one wants to hear, is such a beautiful one. And I wouldn't have pushed Apollon off the edge of the bed. For Pythagoras 2.0 you only need a single number greater than two and you can immediately tell which Pythagorean triples it belongs to! This is so much less complicated than the traditional methods. And you only have to square one number. Which doesn't always have to be natural! The Shalan theorems deserve more attention. Pythagoras theorem: a²+b²=c² 1st Shalan theorem d(m²-1)/2 as "a" dm as "b", d(m²+1)/2 as "c" with c=a+d In my pinned tweet I was guilty of an omission: prime numbers can not only appear as legs of right-angled triangles, but some can also become hypotenuses. I forgot to mention that. Those prime numbers, successors of even square numbers (16+1=17)* or half of the successors of odd square numbers ((25+1)/2=13)**, can be a hypotenuse; If so, d must be even. *d=2,m=4: 2(4²-1)/2=15 => 15²+8²=17² **d=8,m=3/2: 8(9/4-4/4)/2=5 => 5²+12²=13²

Hi fam,
have a fulfillig day!
I hope that this reaches many users; The three Shalan theorems and the consequences - that #polynomials are just building instructions for n-dimensional cuboids - can revolutionize #mathematics #teaching.
#math #maths #education #Pythagoras #sharingisthenewlearning

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#プログラミング勉強中
#programming
#エルミート多項式
#herimite #polynomials
#julia #julia言語 #julialang
#jupyternotebook
#線形代数の半歩先
#講談社

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#プログラミング勉強中
#programming
#エルミート多項式
#herimite #polynomials
#julia #julia言語 #julialang
#jupyternotebook
#線形代数の半歩先
#講談社

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Mathematician solves algebra’s oldest problem using intriguing new number sequences A UNSW mathematician has built an algebraic solution to an equation that was once believed impossible to solve.

Mathematician Norman Wildberger solves Algebra's oldest problem using novel number sequences [via UNSW] 🧪✖️📐📈✔️

"His method uses novel sequences of numbers that represent complex geometric relationships."

www.unsw.edu.au/newsroom/new...

#Combinatorics #higher #order #polynomials #algebra #solution

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Mathematician solves algebra's oldest problem using intriguing new number sequences A UNSW Sydney mathematician has discovered a new method to tackle algebra's oldest challenge—solving higher polynomial equations.

phys.org/news/2025-05... #MTBoS #iteachmath #math #maths #algebra #polynomials

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"RFK Jr. is just an ignorant asshole", says an autistic kid that became an adult who holds down a job, pays taxes, contributes 2 cute formulas and a revolution in #mathematics #education to society.

#actuallyautistic #autism #autismawareness #autismacceptance #neurodivergent #polynomials #calculus

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Polynôme annulateur scindé à racines simples et diagonalisabilité.

#matrice #polynomials #diagonalisation

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How to find roots of polynomials in numerical analysis using the bisection method with Maple
How to find roots of polynomials in numerical analysis using the bisection method with Maple YouTube video by tondekush

youtu.be/twfHYd0e9vM
#roots #polynomials #maplesoft #maple #numericalAnalysis #mathematics #math #equations

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My QT is a reply to Catriona Agg ‪@catrionaagg.bsky.social‬ who wrote:
"If you teach maths (at any level), why not join in with
#MathsToday this term? The idea is to share a little thing from your classroom..."
#sharingisthenewlearning #SharingIsCaring #Calculus #polynomials #tesseract

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Exploring the sum and product of polynomial roots: Teaching students how coefficients reveal the relationships between roots and equations. A key concept bridging algebra and deeper mathematical insight! #MathsEducation #Polynomials

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His work makes interesting progress on the long-standing Aaronson-Ambainis conjecture at the intersection of #quantum computing and #Boolean function analysis. Thanks to Sreejata, we now know this conjecture is true for random restrictions of bounded #polynomials.

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A problem about the irreducibility of a polynomial (Cyclotomic polynomial). This is traditionally proved by using Eisenstein's criterion. Here we use the technique of "reduction modulo p".
#mathematics, #algebra, #polynomials, #factorization, #irreducibility

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The cuboid made of LEGOs represents
f(x)=(x+0)(x+3)(x-1)
=x³+2x²-3x. It is made of a blue cube x³, a red cuboid 3x², a white cuboid -1x² and a white cuboid  3*-1*x. The + sign represents the directions forward and right (and up), the - sign represents (backward, left and) down. So f(x) becomes x³+3x²-1x²-3x=x³+2x²-3x
Half of its surface equals 3x² (blue square)+2*3x (red rectangles) + 2*-1x (white rectangles)x+3*-1 (white rectangle)=3x²+4x-3. The slope of the function, the first derivative, can be described by half of the surface of the cuboid; Epsilon go home ; )
Half the boundaries of a square are 2x, so the slope of g(x)=x² is simply g'(x)=2x. Half the boundary of an edge x is one point 1. If there are n/m edges, the slope is represented by n/m
One cuboid represents only one fourth of half the boundaries of a four-dimensional cube. Therefore the anti-derivative of h(x)=x³ is H(x)=x⁴/4. One x² represents only one third of half the boundaries of a cube. The anti-derivative of i(x)=x² is I(x)=x³/3. And so on.
Bonus feature: You can find the x-value of middle inflection point or the vertex easily; This is how it works:
f(x)=a1x^n+a2x⁽n-1)+...+a(n+1)x⁰ The x-value of the vertex/middle inflection point is just -a2/n*a1. The x-value of the onlyinflection point of f(x)=1x³+2x²-3x is -2/1*3=-2/3 It's shockingly simple, isn't it?

The cuboid made of LEGOs represents f(x)=(x+0)(x+3)(x-1) =x³+2x²-3x. It is made of a blue cube x³, a red cuboid 3x², a white cuboid -1x² and a white cuboid 3*-1*x. The + sign represents the directions forward and right (and up), the - sign represents (backward, left and) down. So f(x) becomes x³+3x²-1x²-3x=x³+2x²-3x Half of its surface equals 3x² (blue square)+2*3x (red rectangles) + 2*-1x (white rectangles)x+3*-1 (white rectangle)=3x²+4x-3. The slope of the function, the first derivative, can be described by half of the surface of the cuboid; Epsilon go home ; ) Half the boundaries of a square are 2x, so the slope of g(x)=x² is simply g'(x)=2x. Half the boundary of an edge x is one point 1. If there are n/m edges, the slope is represented by n/m One cuboid represents only one fourth of half the boundaries of a four-dimensional cube. Therefore the anti-derivative of h(x)=x³ is H(x)=x⁴/4. One x² represents only one third of half the boundaries of a cube. The anti-derivative of i(x)=x² is I(x)=x³/3. And so on. Bonus feature: You can find the x-value of middle inflection point or the vertex easily; This is how it works: f(x)=a1x^n+a2x⁽n-1)+...+a(n+1)x⁰ The x-value of the vertex/middle inflection point is just -a2/n*a1. The x-value of the onlyinflection point of f(x)=1x³+2x²-3x is -2/1*3=-2/3 It's shockingly simple, isn't it?

Calculus of f(x)=x³+2x²-3x (blue graph)
f'(x)=3x²+4x-3, pink, equals the first derivative and describes the slope of f(x) At the point (-2/3;f(-2/3)) the slope of f(x) is minimal.  f''(x)=6x+4, turquoise, 2nd derivative, describes the slope of f'(x), f'''(x)=6, green, the 3rd derivative, describes the slope of f''(x)
F(x)=x⁴/4+2x³/3-3x²/2+c, purple, is the anti-derivative of f(x) and describes the area between the graph of f(x) and the x-axis.

Calculus of f(x)=x³+2x²-3x (blue graph) f'(x)=3x²+4x-3, pink, equals the first derivative and describes the slope of f(x) At the point (-2/3;f(-2/3)) the slope of f(x) is minimal. f''(x)=6x+4, turquoise, 2nd derivative, describes the slope of f'(x), f'''(x)=6, green, the 3rd derivative, describes the slope of f''(x) F(x)=x⁴/4+2x³/3-3x²/2+c, purple, is the anti-derivative of f(x) and describes the area between the graph of f(x) and the x-axis.

A Christmas Gift for Graduation Preparation or
#Calculus for Lazy Ones
#haiku #poem

Polynomials:
n-dimensional cuboids.
Epsilon go home!

Graphs drawn with this ingenious toy/tool: www.matheretter.de/rechner/gfplot
#mathematics #math #polynomials #Polynome #sharingisthenewlearning #SharingIsCaring

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many of the recent successes in #AI #ML are due to #structured low-rank representations!

but...What's the connection between #lowrank adapters, #tensor networks, #polynomials and #circuits?

join our #AAAI25 workshop to know the answer!

and 2 more days to submit!
👇👇👇
april-tools.github.io/colorai/

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Why should We use Orthogonal Polynomials? Orthogonal Polynomials for Data Science

Why should We use Orthogonal Polynomials?
rahulbhadani.medium.com/why-should-w...
#DataScience #Regression #Polynomials #orthogonalpolynomial #statistics #mathematics #models #numericalanalysis
#academicsky

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Oh, people this is one of the best approaches to #polynomials and such out there. Love that Dan put it on Desmos!

(x+2)(2x-3) is multiple two lines, so... why not do that geometrically? why do we get that quadratic? fun! #mtbos

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Oy. Another weekend of math I don’t remember. #modernparenting #polynomials #remoteschooling #ixl

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