Bachelor of Ecole Polytechnique

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Bachelor of Ecole Polytechnique Computational Mathematics, year 2, semester 1

Lecturer: Lucas Gerin (send mail) (mailto:[email protected]

Symbolic computation 3: Linear recurrences

Exercise 1. A periodic sequence

# execute this part to modify the css style from IPython.core.display import HTML def css_styling():

styles = open("./style/custom2.css").read() return HTML(styles)

css_styling()

## loading python libraries

# necessary to display plots inline:

%matplotlib inline

# load the libraries

import matplotlib.pyplot as plt # 2D plotting library

import numpy as np # package for scientific computing from pylab import *

from math import * # package for mathematics (pi, arctan, sqrt, factorial ...) import sympy as sympy # package for symbolic computation

from sympy import *

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Do it yourself. Let be a real number, we de�ne the sequence as follows: and for

Using SymPy , prove that for every the sequence is periodic.

a ∉ {1, −1} ( u

n

)

n≥0

= a

u

₀

n ≥ 1

= .

u

n

1 + u

_n−1

1 − u

_n−1

a ( ) u

n

Answers. According to the above script we have

for every . Since is de�ned by a simple recurrence of order one this su�ces to prove that

for every , i.e. is periodic with period at most . Indeed we can prove eq. ( ) by a simple induction:

For we trust SymPy.

Assume eq. ( ) is true for some . We write

where . By the induction hypothesis

and the proof is done.

= u

₄

u

₀

a u

_n

=

u

_n+4

u

n

(⋆)

n ( ) u

_n

4 ⋆

n = 0

⋆ n

= = f ( ) u

_n+1+4

u

_n+5

u

_n+4

f ( x ) = (1 + x )/(1 − x )

f ( u

_n+4

) = f ( ) = u

n

u

_n+1

Exercise 2. A number theory theorem with Sympy

u_0 = a

u_1 = -(a + 1)/(a - 1) u_2 = -1/a

u_3 = (a - 1)/(a + 1) u_4 = a

u_5 = -(a + 1)/(a - 1) u_6 = -1/a

u_7 = (a - 1)/(a + 1) u_8 = a

u_9 = -(a + 1)/(a - 1) u_10 = -1/a

u_11 = (a - 1)/(a + 1)

# # We compute the first few terms:

def InductionFormula(x):

return (1+x)/(1-x) var('a')

Sequence=[a]

print('u_0 = a') for i in range(1,12):

Sequence.append(simplify(InductionFormula(Sequence[-1]))) print('u_'+str(i)+' = '+str(Sequence[-1]))

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Do it yourself.

With Sympy simplify the following expression:

1.

Prove the following:

Theorem. Every integer can be written as the sum of three cubes of rational numbers.

To save you time in the proof you can use the following notations:

2.

+ +

(x⁶+ 45 − 81 + 27x⁴ x² ) 3x( + 3x² )²

3

(− + 30 − 9x⁴ x² ) 3x( + 3)x²

3

(−12 + 36xx³ ) ( + 3x² )²

3

n

p q r s t u

= x

⁶

+ 45 − 81 + 27, x

⁴

x

²

= 3 x ( + 3 , x

²

)

²

= − + 30 − 9, x

⁴

x

²

= 3 x ( + 3), x

²

= −12 + 36 x

³

x ,

= ( + 3 x

²

)

²

Answers. Let be an integer. The above script shows that

where are integers. Thus

Therefore can be written as a sum of cubes of rationals.

x 8 x = ( p / q )

³

+ ( r / s )

³

+ ( t / u )

³

, p , q , r , s , t , u

x = ( p /2 q )

³

+ ( r /2 s )

³

+ ( t /2 u )

³

. x

Exercise 3. Solving a recurrence (almost) by hand

8*x var('x')

A=(x**6+45*x**4-81*x**2+27)/(3*x*(x**2+3)**2) B=(-x**4+30*x**2-9)/(3*x*(x**2+3))

C=(-12*x**3+36*x)/((x**2+3)**2) simplify(A**3+B**3+C**3)

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Do it yourself. Let be the sequence de�ned by

The goal of the exercise is to �nd a formula for .

For this purpose we introduce the sequence de�ned by

for some parameters . Use SymPy to �nd .

To solve a system of equations with SymPy with unknowns , the syntax is solve([x-y-2,3*y+x],[x,y])

1.

Prove with SymPy that this sequence satis�es the recurrence ( ): for every we have

2.

( ) u

n

{ u

₀

u

n

= 1, = 2 u

_n−1

+ 3 n

²

(∀ n ≥ 1) (⋆) u

_n

( ) v

n

{ v

₀

= , = , = , = u

₀

v

₁

u

₁

v

₂

u

₂

v

₃

u

₃

= α + a + bn + c (∀ n ≥ 0) v

n

2

ⁿ

n

²

α , a , b , c α , a , b , c

x , y

( ) v

n

⋆ n

= 2 + 3 . v

n

v

_n−1

n

²

Answers. According to the above script we get

With this choice of parameters,

α = 19, a = −3,

and

b = −12,

coincide for the �rst values.

c = −18.

( ) u

_n

( ) v

_n

{c: -18, alpha: 19, b: -12, a: -3}

# --- Question 1 --- var('n',integer=True)

var('alpha a b c') def v(n,alpha,a,b,c):

return alpha*2**n +a*n**2 +b*n +c def u(n):

if n==0:

return 1 else:

return u(n-1)*2+3*n*n Solutions=solve(

# Quantities which are to be zero:

[v(0,alpha,a,b,c)-u(0), # initial condition v(1,alpha,a,b,c)-u(1),

v(2,alpha,a,b,c)-u(2), v(3,alpha,a,b,c)-u(3), ],

# Unknowns:

[alpha,a,b,c]

)

print(Solutions)

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Answers. Question 2. According to the above script we proved that the formula is true: for every we have that

Finally we have proved that the solution of ( ) is given by

n v

_n

− 2 v

_n−1

= 3 . n

²

= 19 × − 3 − 12 ⋆ n − 18.

v

n

2

ⁿ

n

²

Solving recurrences with SymPy: rsolve

With these coefficients the simplification of v(n)-2v(n-1) gives : 3*n*

*2

#--- Question 2 ---

# We fix parameters according to Question 1:

c_s=-18 alpha_s=19 b_s=-12 a_s=-3 var('n')

print('With these coefficients the simplification of v(n)-2v(n-1) gives : ' +str(simplify(v(n,alpha_s,a_s,b_s,c_s)-2*v(n-1,alpha_s,a_s,b_s,c_s))))

# We can check that u and v coincide for the first values:

#print([u(n) for n in range(10)])

#print([v(n,alpha_s,a_s,b_s,c_s) for n in range(10)])

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The function rsolve

We will now see how to use Sympy to obtain explicit formulas for some sequences de�ned by linear recurrences. More precisely, we will see how to obtain an explicit formula for in two cases:

Linear recurrence of order one: this is a sequence is de�ned by

where are some given constants and is an arbitrary function.

1.

Linear recurrence of order two: this is a sequence is de�ned by

where are some given constants and is an arbitrary function.

2.

Some known examples �t in this settings:

Geometric sequences: .

1.

Arithmetic sequences: .

2.

The Fibonacci sequence: .

3.

The following script shows how to solve the linear recurrence of a geometric sequence, using function rsolve :

u

_n

( u

n

)

n≥0

{ u

₀

u

n

= a ,

= α u

_n−1

+ f ( n ), ( n ≥ 1),

a , α f

( u

_n

)

n≥0



  u

₀

u

₁

u

_n

= a ,

= b ,

= α u

_n−1

+ β u

_n−2

+ f ( n ), ( n ≥ 2),

a , , b , α , β f

= a , = r u

₀

u

n

u

_n−1

= a , = + r u

₀

u

n

u

_n−1

= 1, = 1, = +

F

₁

F

₂

F

n

F

_n−1

F

_n−2

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Exercise 4. Application: Fibonacci with rsolve

Do it yourself.

Use SymPy to solve the recurrence of the Fibonacci sequence and �nd an explicit formula for .

1.

(Theory) Use the formula obtained at Question 1 to prove that

(Hint: Set .)

2.

F

n

= φ = .

n→+∞

lim F

n

F

_n−1

1 + 5 √ ⎯⎯

θ = (1 − √ 5 ⎯⎯ )/2 2

---

Example 1 (order one): u(n)=3*u(n-1), u(0)=5 The formula for u(n) is 5*3**n

---

Example 2 (order one): 2v(n)=v(n-1)+2**n, v(0)=1 The formula for v(n) is 2*2**n/3 + 2**(-n)/3 ---

Example 3 (order two): w(n)=2*w(n-1)+w(n-2)-1, w(0)=1,w(1)=1

The formula for w(n) is (1 + sqrt(2))**n/4 + (-sqrt(2) + 1)**n/4 + 1/2

# First example: a geometric sequence print('---')

print('Example 1 (order one): u(n)=3*u(n-1), u(0)=5')

u = Function('u') # declares the name of the sequence n = symbols('n',integer=True) # declares the variable

f = u(n)-3*u(n-1) # the expression which is to be zero

ExplicitFormula=rsolve(f,u(n),{u(0):5}) # solves f=0 with a given initial condition print('The formula for u(n) is '+str(ExplicitFormula)+'')

# Second example with a non-constant term print('---')

print('Example 2 (order one): 2v(n)=v(n-1)+2**n, v(0)=1')

v = Function('v') # declares the name of the sequence n = symbols('n',integer=True) # declares the variable

f = 2*v(n)-v(n-1)-2**n # the expression which is to be zero ExplicitFormula2=rsolve(f,v(n),{v(0):1})

print('The formula for v(n) is '+str(ExplicitFormula2)+'')

# Third example: order two

print('---')

print('Example 3 (order two): w(n)=2*w(n-1)+w(n-2)-1, w(0)=1,w(1)=1') w = Function('w') # declares the name of the sequence n = symbols('n',integer=True) # declares the variable

f =w(n)-2*w(n-1)-w(n-2)+1 # the expression which is to be zero ExplicitFormula3=rsolve(f,w(n),{w(0):1,w(1):1})

print('The formula for w(n) is '+str(ExplicitFormula3)+'')

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Answers.

We put the above answer in : 1.

The factor cancels out so we obtain:

Since the numerator tends to , the denominator to one.

2.

𝙻𝚊𝚝𝚎𝚇

= ( ⁻ ) ⁼ ^{( − ) .}

F

n

2

⁻ⁿ

5 √ 5 ⎯⎯ ( 1 + √ 5 ⎯⎯ )

ⁿ

( − √ 5 ⎯⎯ + 1 )

ⁿ

√ 5 ⎯⎯

5 φ

ⁿ

θ

ⁿ

√5

F

_n 5

F

_n−1

= φ

ⁿ

− θ

ⁿ

,

− φ

ⁿ⁻¹

θ

ⁿ⁻¹

= φ − θ ( θ / φ )

ⁿ⁻¹

, (we divide everything by .)

1 − θ ( θ / φ )

ⁿ⁻¹

φ

ⁿ⁻¹

| θ / φ | < 1 φ

Remark. The output of rsolve is an expression which depends on the symbolic variable n . If we want to evaluate this expression (for instance for ) we must write:

n = 10

Exercise 5. Application: Another example of order two

Do it yourself. Let be de�ned by

Use rsolve to �nd an explicit formula for .

( ) J

n

 .

  J

₀

J

₁

J

n

= 1 = 2

= 2 J

_n−2

+ 5 for every n ≥ 2 J

_n

The formula for u(n) is \frac{2^{- n}}{5} \sqrt{5} \left(\left(1 + \sqr t{5}\right)^{n} - \left(- \sqrt{5} + 1\right)^{n}\right)

10th Fibonacci number = sqrt(5)*(-(-sqrt(5) + 1)**10 + (1 + sqrt(5))**1 0)/5120

After simplification : 55

u = Function('u') # declares the name of the sequence n = symbols('n',integer=True) # declares the variable

f = u(n)-u(n-1)-u(n-2) # the expression which is to be zero ExplicitFormula=simplify(rsolve(f,u(n),{u(1):1,u(2):1}))

Formula2=latex(simplify(ExplicitFormula))

print('The formula for u(n) is '+str(Formula2)+'')

Value=ExplicitFormula.subs(n,10)

print('10th Fibonacci number = '+str(Value))

print('After simplification : '+str(simplify(Value)))

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Answers. The explicit formula for

J

n is

( 7 + 12 ) + ( −7 + 12 ) − 5 2

ⁿ²⁻²

√ 2 ⎯⎯ ( ⁻ ^√ ² ^⎯⎯ )

ⁿ

4 √ 2 ⎯⎯

Do it yourself.(Theory) According to the formula found by SymPy , what happens when for the sequence ?

n → +∞

_√^J₂ⁿⁿ

Answers. We rewrite the above formula as

Therefore oscillates between and .

= ( 2 + 5 + (−1 (−2 + 5) ) − 1.

J

n

√ 2 ⎯⎯

ⁿ¹₄

√ 2 ⎯⎯ )

ⁿ

√ 2 ⎯⎯

Jn

√2ⁿ

10/4 = 5/2 4 √ 2 ⎯⎯ /4 = √ 2 ⎯⎯

Sympy says the formula for J(n) is 2**(n/2 - 2)*(7*sqrt(2) + 12) + (-sq rt(2))**n*(-7*sqrt(2) + 12)/4 - 5

2^{\frac{n}{2} - 2} \left(7 \sqrt{2} + 12\right) + \frac{\left(- \sqrt{

2}\right)^{n}}{4} \left(- 7 \sqrt{2} + 12\right) - 5

J = Function('J') # declares the name of the sequence n = symbols('n',integer=True) # declares the variable

f = J(n)-2*J(n-2)-5 # the expression which is to be zero ExplicitFormula=rsolve(f,J(n),{J(0):1,J(1):2})

Formula=simplify(ExplicitFormula)

print('Sympy says the formula for J(n) is '+str(Formula)+'') print(latex(simplify(Formula)))

Bachelor of Ecole Polytechnique

Symbolic computation 3: Linear recurrences

Table of contents

Exercise 1. A periodic sequence

a ∉ {1, −1} ( u

)

= a

u

n ≥ 1

= .

u

1 + u

1 − u

a ( ) u

= u

u

a u

=

u

u

(⋆)

n ( ) u

4 ⋆

n = 0

⋆ n

= = f ( ) u

u

u

f ( x ) = (1 + x )/(1 − x )

f ( u

) = f ( ) = u

u

Exercise 2. A number theory theorem with Sympy

n

p q r s t u

= x

+ 45 − 81 + 27, x

x

= 3 x ( + 3 , x

)

= − + 30 − 9, x

x

= 3 x ( + 3), x

= −12 + 36 x

x ,

= ( + 3 x

)

x 8 x = ( p / q )

+ ( r / s )

+ ( t / u )

, p , q , r , s , t , u

x = ( p /2 q )

+ ( r /2 s )

+ ( t /2 u )

. x

Exercise 3. Solving a recurrence (almost) by hand

( ) u

{ u

u

= 1, = 2 u

+ 3 n

(∀ n ≥ 1) (⋆) u

( ) v

{ v

= , = , = , = u

v

u

v

u

v

u

= α + a + bn + c (∀ n ≥ 0) v

2

n

α , a , b , c α , a , b , c

x , y

( ) v

⋆ n

= 2 + 3 . v

v

= ( ⁻ ) ⁼ ^{( − ) .}