euler s pentagonal number theorem
play

Eulers Pentagonal Number Theorem Dan Cranston September 28, 2011 - PowerPoint PPT Presentation

Dec 05, 2022 •133 likes •1.12k views

Eulers Pentagonal Number Theorem Dan Cranston September 28, 2011 Introduction Introduction Triangular Numbers: 1 , 3 , 6 , 10 , 15 , 21 , 28 , 36 , 45 , 55 , ... Introduction Triangular Numbers: 1 , 3 , 6 , 10 , 15 , 21 , 28 , 36 , 45 , 55 ,

  1. Euler’s Pentagonal Number Theorem Dan Cranston September 28, 2011

  2. Introduction

  3. Introduction Triangular Numbers: 1 , 3 , 6 , 10 , 15 , 21 , 28 , 36 , 45 , 55 , ...

  4. Introduction Triangular Numbers: 1 , 3 , 6 , 10 , 15 , 21 , 28 , 36 , 45 , 55 , ... Square Numbers: 1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100 , ...

  5. Introduction Triangular Numbers: 1 , 3 , 6 , 10 , 15 , 21 , 28 , 36 , 45 , 55 , ... Square Numbers: 1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100 , ... Pentagonal Numbers: 1 , 5 , 12 , 22 , 35 , 51 , 70 , 92 , 117 , 145 , ...

  6. Generalized Pentagonal Numbers

  7. Generalized Pentagonal Numbers The k th pentagonal number, P ( k ), is the k th partial sum of the arithmetic sequence a n = 1 + 3( n − 1) = 3 n − 2.

  8. Generalized Pentagonal Numbers The k th pentagonal number, P ( k ), is the k th partial sum of the arithmetic sequence a n = 1 + 3( n − 1) = 3 n − 2. k (3 n − 2) = 3 k 2 − k � P ( k ) = 2 n =1

  9. Generalized Pentagonal Numbers The k th pentagonal number, P ( k ), is the k th partial sum of the arithmetic sequence a n = 1 + 3( n − 1) = 3 n − 2. k (3 n − 2) = 3 k 2 − k � P ( k ) = 2 n =1 ◮ P (8) = 92, P (500) = 374 , 750, etc.

  10. Generalized Pentagonal Numbers The k th pentagonal number, P ( k ), is the k th partial sum of the arithmetic sequence a n = 1 + 3( n − 1) = 3 n − 2. k (3 n − 2) = 3 k 2 − k � P ( k ) = 2 n =1 ◮ P (8) = 92, P (500) = 374 , 750, etc. and P (0) = 0.

  11. Generalized Pentagonal Numbers The k th pentagonal number, P ( k ), is the k th partial sum of the arithmetic sequence a n = 1 + 3( n − 1) = 3 n − 2. k (3 n − 2) = 3 k 2 − k � P ( k ) = 2 n =1 ◮ P (8) = 92, P (500) = 374 , 750, etc. and P (0) = 0. ◮ Extend domain, so P ( − 8) = 100, P ( − 500) = 375 , 250, etc.

  12. Generalized Pentagonal Numbers The k th pentagonal number, P ( k ), is the k th partial sum of the arithmetic sequence a n = 1 + 3( n − 1) = 3 n − 2. k (3 n − 2) = 3 k 2 − k � P ( k ) = 2 n =1 ◮ P (8) = 92, P (500) = 374 , 750, etc. and P (0) = 0. ◮ Extend domain, so P ( − 8) = 100, P ( − 500) = 375 , 250, etc. ◮ { P (0) , P (1) , P ( − 1) , P (2) , P ( − 2) , ... } = { 0 , 1 , 2 , 5 , 7 , ... } is an increasing sequence.

  13. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers.

  14. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p ( n ) denote the number of partitions of n .

  15. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p ( n ) denote the number of partitions of n . ◮ 3 = 2+1 = 1+1+1,

  16. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p ( n ) denote the number of partitions of n . ◮ 3 = 2+1 = 1+1+1, so p (3) = 3.

  17. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p ( n ) denote the number of partitions of n . ◮ 3 = 2+1 = 1+1+1, so p (3) = 3. ◮ 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1,

  18. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p ( n ) denote the number of partitions of n . ◮ 3 = 2+1 = 1+1+1, so p (3) = 3. ◮ 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1, so p (4) = 5.

  19. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p ( n ) denote the number of partitions of n . ◮ 3 = 2+1 = 1+1+1, so p (3) = 3. ◮ 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1, so p (4) = 5. ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1,

  20. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p ( n ) denote the number of partitions of n . ◮ 3 = 2+1 = 1+1+1, so p (3) = 3. ◮ 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1, so p (4) = 5. ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p (5) = 7.

  21. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p ( n ) denote the number of partitions of n . ◮ 3 = 2+1 = 1+1+1, so p (3) = 3. ◮ 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1, so p (4) = 5. ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p (5) = 7. ◮ 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1,

  22. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p ( n ) denote the number of partitions of n . ◮ 3 = 2+1 = 1+1+1, so p (3) = 3. ◮ 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1, so p (4) = 5. ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p (5) = 7. ◮ 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p (6) = 11.

  23. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p ( n ) denote the number of partitions of n . ◮ 3 = 2+1 = 1+1+1, so p (3) = 3. ◮ 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1, so p (4) = 5. ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p (5) = 7. ◮ 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p (6) = 11. Each summand in a certain partition is called a part.

  24. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p ( n ) denote the number of partitions of n . ◮ 3 = 2+1 = 1+1+1, so p (3) = 3. ◮ 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1, so p (4) = 5. ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p (5) = 7. ◮ 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p (6) = 11. Each summand in a certain partition is called a part. So 3 has 1 part, 2 + 1 has 2 parts, and 1 + 1 + 1 has 3 parts.

  25. Partition Numbers We identify a partition of n by its Ferrers diagram.

  26. Partition Numbers We identify a partition of n by its Ferrers diagram. A partition with its parts in decreasing size from top to bottom is in standard form.

  27. Partition Numbers We identify a partition of n by its Ferrers diagram. A partition with its parts in decreasing size from top to bottom is in standard form. Three different partitions of 9: 5 + 3 + 1 4 + 3 + 2 4 + 3 + 1 + 1

  28. Special Partition Numbers p d ( n ) = number of partitions of n into distinct parts

  29. Special Partition Numbers p d ( n ) = number of partitions of n into distinct parts ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p d (5) = 3.

  30. Special Partition Numbers p d ( n ) = number of partitions of n into distinct parts ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p d (5) = 3. ◮ 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p d (6) = 4.

  31. Special Partition Numbers p d ( n ) = number of partitions of n into distinct parts ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p d (5) = 3. ◮ 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p d (6) = 4. p e ( n ) = number of partitions of n into an even number of distinct parts

  32. Special Partition Numbers p d ( n ) = number of partitions of n into distinct parts ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p d (5) = 3. ◮ 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p d (6) = 4. p e ( n ) = number of partitions of n into an even number of distinct parts; similar for p o ( n ), so p e ( n ) + p o ( n ) = p d ( n )

  33. Special Partition Numbers p d ( n ) = number of partitions of n into distinct parts ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p d (5) = 3. ◮ 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p d (6) = 4. p e ( n ) = number of partitions of n into an even number of distinct parts; similar for p o ( n ), so p e ( n ) + p o ( n ) = p d ( n ) ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p e (5) = 2.

  34. Special Partition Numbers p d ( n ) = number of partitions of n into distinct parts ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p d (5) = 3. ◮ 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p d (6) = 4. p e ( n ) = number of partitions of n into an even number of distinct parts; similar for p o ( n ), so p e ( n ) + p o ( n ) = p d ( n ) ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p e (5) = 2. ( p o (5) = 1)

  35. Special Partition Numbers p d ( n ) = number of partitions of n into distinct parts ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p d (5) = 3. ◮ 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p d (6) = 4. p e ( n ) = number of partitions of n into an even number of distinct parts; similar for p o ( n ), so p e ( n ) + p o ( n ) = p d ( n ) ◮ 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p e (5) = 2. ( p o (5) = 1) ◮ 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p e (6) = 2. ( p o (6) = 2)

  36. Pentagonal Number Theorem Main Theorem ∞ 1 − x − x 2 + x 5 + x 7 − x 12 − x 15 + x 22 + x 26 + ... � (1 − x m ) = m =1

Recommend

Diophantine Equations Involving the Euler Totient Function Number Theory Seminar, Dalhousie

Diophantine Equations Involving the Euler Totient Function Number Theory Seminar, Dalhousie

Diophantine Equations Involving the Euler Totient Function Number Theory Seminar, Dalhousie University J.C. Saunders Ben-Gurion University of the Negev December 20, 2019 The Euler Totient Function For a natural number n , the Euler totient

362 views • 25 slides
Euler characteristic. ( K ) = f 0 ( K ) f 1 ( K ) + f 2 ( K ) . . . 1

Euler characteristic. ( K ) = f 0 ( K ) f 1 ( K ) + f 2 ( K ) . . . 1

Euler characteristic. ( K ) = f 0 ( K ) f 1 ( K ) + f 2 ( K ) . . . 1 d v ( K 2 ) = , 6 v K 2 where d v is the number of edges entering v . Stiefel-Whitney classes. n W n ( K ) = (mod 2) . n K

360 views • 13 slides
Orthogonal Polynomials for Bernoulli and Euler Polynomials Lin JIU Dalhousie University Number

Orthogonal Polynomials for Bernoulli and Euler Polynomials Lin JIU Dalhousie University Number

Orthogonal Polynomials for Bernoulli and Euler Polynomials Lin JIU Dalhousie University Number Theory Seminar Jan. 14th, 2019 Acknowledgment Diane Shi Tianjin University Objects Bernoulli numbers B n : Euler numbers E n :

598 views • 32 slides
Random Conical tilt 3D reconstruction Central section theorem Euler angles Principle

Random Conical tilt 3D reconstruction Central section theorem Euler angles Principle

Random Conical tilt 3D reconstruction Central section theorem Euler angles Principle of conical tilt series Missing cone artefact Multivariate statistical analysis Early 3D studies and negative staining problems

938 views • 47 slides
Formalizing an Analytic Proof of the Prime Number Theorem Dedicated to Mike Gordon John Harrison

Formalizing an Analytic Proof of the Prime Number Theorem Dedicated to Mike Gordon John Harrison

Formalizing an Analytic Proof of the Prime Number Theorem Dedicated to Mike Gordon John Harrison Intel Corporation TTVSI (Mikefest) Royal Society, London Wed 26th March 2008 (10:00 10:45) 0 What is the prime number theorem? Let ( x )

495 views • 28 slides
Generalized Intermediate Value Theorem Intermediate Value Theorem Theorem Intermediate Value

Generalized Intermediate Value Theorem Intermediate Value Theorem Theorem Intermediate Value

Generalized Intermediate Value Theorem Intermediate Value Theorem Theorem Intermediate Value Theorem Suppose f is continuous on [ a , b ] and let N be any number between f ( a ) and f ( b ) , where f ( a ) = f ( b ) . Then there exists a

59 views • 4 slides
1 The Euler system of equations The Euler system of equations describing flows of incompressible

1 The Euler system of equations The Euler system of equations describing flows of incompressible

On integration of the equations of incompressible fluid flow Valery Dryuma, e-mail: valdryum@gmail.com Institute of Mathematics and Computer Science, AS RM, Kishinev 1 The Euler system of equations The Euler system of equations describing

284 views • 3 slides
Objectives Congruences: Modular Arithmetic Euler Totient Function Fermats Little

Objectives Congruences: Modular Arithmetic Euler Totient Function Fermats Little

Introduction to Number Theory Debdeep Mukhopadhyay Assistant Professor Department of Computer Science and Engineering Indian Institute of Technology Kharagpur INDIA -721302 Objectives Congruences: Modular Arithmetic Euler Totient

666 views • 15 slides
Understanding the Prime Number Theorem Misunderstood Monster or Beautiful Theorem? Liam Fowl

Understanding the Prime Number Theorem Misunderstood Monster or Beautiful Theorem? Liam Fowl

Introduction Complex Plane Complex functions and Analytic Continuation Gamma Function Laplace Transform Zeta Function Understanding the Prime Number Theorem Misunderstood Monster or Beautiful Theorem? Liam Fowl September 5, 2014 Liam Fowl

618 views • 20 slides
Todays Agenda Wrap up of Number Theory (Sec. 3.7) Fermats Little Theorem Public

Todays Agenda Wrap up of Number Theory (Sec. 3.7) Fermats Little Theorem Public

Todays Agenda Wrap up of Number Theory (Sec. 3.7) Fermats Little Theorem Public Key Cryptography (RSA) Strings and Languages (Chap. 12) Based on Rosen and slides by K. Busch 1 Fermats little theorem: For any prime and

357 views • 14 slides
Stability of Transonic Shock Solutions for Euler-Poisson and Euler Equations Chunjing XIE

Stability of Transonic Shock Solutions for Euler-Poisson and Euler Equations Chunjing XIE

Stability of Transonic Shock Solutions for Euler-Poisson and Euler Equations Chunjing XIE University of Michigan June 8, 2011 Nonlinear Hyperbolic PDEs, Dispersive and Transport Equations: Analysis and Control, SISSA, Italy Joint work with

632 views • 28 slides
Euler Characteristics of Categories and Homotopy Colimits Thomas M. Fiore joint work with

Euler Characteristics of Categories and Homotopy Colimits Thomas M. Fiore joint work with

Introduction Finiteness Obstructions and Euler Characteristics for Categories Classifying I -Spaces Homotopy Colimit Formula and the Inclusion-Exclusion Principle Comparison with Leinsters Notions Applications and Summary Euler

357 views • 34 slides
Euler, Plato & balloons! Euler : The master of us all! Born in Basel, Switzerland in 1707

Euler, Plato & balloons! Euler : The master of us all! Born in Basel, Switzerland in 1707

Euler, Plato & balloons! Euler : The master of us all! Born in Basel, Switzerland in 1707 to a pastor. Got his Master of Philosophy degree at the ripe young age of 16. Moved to St Petersburg, Russia to become Professor of Physics

969 views • 55 slides
Kelvin-Helmholtz instability above Richardson number 1 / 4 J P Parker, C P Caulfield, R R Kerswell

Kelvin-Helmholtz instability above Richardson number 1 / 4 J P Parker, C P Caulfield, R R Kerswell

Kelvin-Helmholtz instability above Richardson number 1 / 4 J P Parker, C P Caulfield, R R Kerswell September 4, 2019 Miles-Howard theorem The local/gradient Richardson number is defined as / z Ri g = g ( u / z ) 2 . Theorem

189 views • 15 slides
6.2: Runge Kutta Methods (RKM) (A) 2nd Order RKM (or Improved Euler Method) Failure of Euler

6.2: Runge Kutta Methods (RKM) (A) 2nd Order RKM (or Improved Euler Method) Failure of Euler

6.2: Runge Kutta Methods (RKM) (A) 2nd Order RKM (or Improved Euler Method) Failure of Euler Method: Only slope on left end of interval [ t, t + h ] is used. Iteration Scheme Improvement: Given t, y ( t ), Start: y 0 , t 0 compute slope

646 views • 7 slides
Rigidity of boundary actions Kathryn Mann Brown University The PSL 2 ( R ) character variety

Rigidity of boundary actions Kathryn Mann Brown University The PSL 2 ( R ) character variety

Rigidity of boundary actions Kathryn Mann Brown University The PSL 2 ( R ) character variety Theorem (Goldman 88) Hom( 1 g , PSL 2 ( R )) has 4 g 3 connected components, classified by the Euler number. -2g+2 ... -1 0 1 ...

431 views • 24 slides
System Modeling: Complex Number and Harmonic Motion Prof. Seungchul Lee Industrial AI Lab.

System Modeling: Complex Number and Harmonic Motion Prof. Seungchul Lee Industrial AI Lab.

System Modeling: Complex Number and Harmonic Motion Prof. Seungchul Lee Industrial AI Lab. Complex Number 2 Complex Number Add 3 Euler's Formula Complex number in complex exponential 4 Complex Number Multiply 5 Geometrical

358 views • 34 slides
Theorem: If a range space S = ( P, R ) over a set P of size n has VC-dimension k , then the number

Theorem: If a range space S = ( P, R ) over a set P of size n has VC-dimension k , then the number

Theorem: If a range space S = ( P, R ) over a set P of size n has VC-dimension k , then the number of distinct ranges satisfies k n | R | . i i =0 Proof: Here is a sketch of the proof, which is induction on n and k . (Details

342 views • 5 slides
Interplay between the Beale-Kato-Majda theorem and the analyticity-strip method to investigate

Interplay between the Beale-Kato-Majda theorem and the analyticity-strip method to investigate

Interplay between the Beale-Kato-Majda theorem and the analyticity-strip method to investigate numerically the incompressible Euler singularity problem Miguel Bustamante and Marc Brachet " s w o fl d n a s e l c i t r a p

1.02k views • 45 slides
On the minimal coloring number of even-parallels of links Eri Matsudo Nihon University Graduate

On the minimal coloring number of even-parallels of links Eri Matsudo Nihon University Graduate

Introduction Theorem Proof of Theorem 2 On the minimal coloring number of even-parallels of links Eri Matsudo Nihon University Graduate School of Integrated Basic Sciences Nihon University, December 20, 2016 1 / 12 Introduction Theorem

534 views • 15 slides
Lecture 6: Graphs. Graphs! Euler Definitions: model. Euler Again!! Konigsberg bridges problem.

Lecture 6: Graphs. Graphs! Euler Definitions: model. Euler Again!! Konigsberg bridges problem.

Lecture 6: Graphs. Graphs! Euler Definitions: model. Euler Again!! Konigsberg bridges problem. Can you make a tour visiting each bridge exactly once? c Konigsberg bridges by Bogdan Gius a - License. B A D C Can you draw a

713 views • 14 slides
From Euler, Schwarz, Ritz and Galerkin Brachystochrone Euler Lagrange to Modern Computing

From Euler, Schwarz, Ritz and Galerkin Brachystochrone Euler Lagrange to Modern Computing

Euler, Schwarz, Ritz, Galerkin Martin J. Gander Before Ritz From Euler, Schwarz, Ritz and Galerkin Brachystochrone Euler Lagrange to Modern Computing Laplace Riemann Schwarz Runge/Ritz Ritz Martin J. Gander Vaillant Prize Chladni

506 views • 34 slides
Lecture 25: Project Euler Day 2 Project Euler: 104 The Fibonacci sequence is defined by the

Lecture 25: Project Euler Day 2 Project Euler: 104 The Fibonacci sequence is defined by the

Lecture 25: Project Euler Day 2 Project Euler: 104 The Fibonacci sequence is defined by the recurrence relation: F n = F n 1 + F n 2 , where F 1 = 1 and F 2 = 1. It turns out that F 541 , which contains 113 digits, is the first Fibonacci

502 views • 5 slides
LOSS OF SMOOTHNESS AND ENERGY CONSERVING ROUGH WEAK SOLUTIONS FOR THE 3 d EULER EQUATIONS Claude

LOSS OF SMOOTHNESS AND ENERGY CONSERVING ROUGH WEAK SOLUTIONS FOR THE 3 d EULER EQUATIONS Claude

DISCRETE AND CONTINUOUS doi:10.3934/dcdss.2010.3.185 DYNAMICAL SYSTEMS SERIES S Volume 3 , Number 2 , June 2010 pp. 185197 LOSS OF SMOOTHNESS AND ENERGY CONSERVING ROUGH WEAK SOLUTIONS FOR THE 3 d EULER EQUATIONS Claude Bardos Laboratory J.

152 views • 13 slides

More recommend