Symbolic computing 1: Proofs with SymPy

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1 Bachelor of Ecole Polytechnique Computational Mathematics, year 2, semester Lecturer: Lucas Gerin (send mail) Symbolic computing : Proofs with SymPy π Table of contents Introduction to SymPy Let SymPy do the proof Archimedes vs SymPy Solving equations (Bonus) Decomposition of fractions # 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 *

2 Introduction to SymPy Remark. From now on every Lab is graded and is individual. You must upload your.pynb le each Thursday on the moodle, before :59. Using python library SymPy we can perform exact computations. For instance, compare the following scripts: print('square root of two is '+str(np.sqrt(2))+'') print('square root of two is '+str(sqrt(2))+'') square root of two is square root of two is sqrt(2) One can expand or simplify expressions: print('the square root of 40 is '+str(sqrt(40))+'') print(expand((sqrt(3)+sqrt(2))**20)) var('x') # We declare a 'symbolic' variable print(simplify((x**2-2*x + )/(x-))) the square root of 40 is 2*sqrt(0) *sqrt(6) x - Do it yourself. (A) ϕ = + 5 F = ϕ4 ϕ 2 Set. Use SymPy to simplify. +ϕ 7 phi=(+sqrt(5))/2 formula=(phi**4-phi)/(phi**7+) print("f = "+str(formula)) print("simplified F = "+str(simplify(formula))) F = (-sqrt(5)/2 - /2 + (/2 + sqrt(5)/2)**4)/( + (/2 + sqrt(5)/2)**7 ) simplified F = -4*sqrt(5)/29 + 4/29 With Sympy one can also obtain Taylor expansions of functions with series :

3 # Real variable var('x') Expression=cos(x) print('expansion of cos(x) at x=0: '+str(expression.series(x,0))) # integer variable var('n',integer=true) Expression=cos(/n) print('expansion of cos(/n) when n->+oo: '+str(expression.series(n,oo))) # oo mean Expansion of cos(x) at x=0: - x**2/2 + x**4/24 + O(x**6) Expansion of cos(/n) when n->+oo: /(24*n**4) - /(2*n**2) + + O(n** (-6), (n, oo)) SymPy can also compute with "big O's". (By default (x) is considered for x 0.) var('x') simplify((x+o(x**3))*(x+x**2+o(x**3))) x**2 + x**3 + O(x**4) Remark. A nice feature of Sympy is that you can export formulas in instance: LateX. For phi=(+sympy.sqrt(5))/2 formula=expand(phi**) print(formula) print(latex(formula)) 99/2 + 89*sqrt(5)/2 \frac{99}{2} + \frac{89 \sqrt{5}}{2} Warning: Fractions such as /4 Sympy from evaluating the expression. For example: must be introduced with Rational(,4) to keep print('(/4)^3 = '+str((/4)**3)) print('(/4)^3 = '+str(rational(,4)**3)) (/4)^3 = (/4)^3 = /64 Let SymPy do the proofs Exercise. A simple (?) recurrence

4 Do it yourself. (B) Let a, b be two real numbers, we de ne the sequence (u n ) n 0 as follows: and for u 0 = a, u = b n 2 u n + u n =. u n 2 5 u n a, b. Write a short program which returns the rst values of in terms of. The output should be something like: u_0 = a u_ = b u_2 = (b + )/a Use SymPy to make and prove a conjecture for the asymptotic behaviour of the sequence ( u n ). def InductionFormula(x,y): return (+x)/y var('a b') Sequence=[a,b] print('u_0 = a') print('u_ = b') for i in range(2,3): Sequence.append(simplify(InductionFormula(Sequence[-],Sequence[-2]))) #Sequence.append(InductionFormula(Sequence[-],Sequence[-2])) print('u_'+str(i)+' = '+str(sequence[-])) u_0 = a u_ = b u_2 = (b + )/a u_3 = (a + b + )/(a*b) u_4 = (a + )/b u_5 = a u_6 = b u_7 = (b + )/a u_8 = (a + b + )/(a*b) u_9 = (a + )/b u_0 = a u_ = b u_2 = (b + )/a Answers.. See the cell above. 2. If a, b 0, the sequence is well de ned and we observe that u 5 = u 0 and u 6 = u. Since the sequence is de ned by induction, this implies that the sequence is periodic: u n+5 = u n for every n. So if we trust Sympy the proof is done.

5 Exercise 2. A number theory theorem with Sympy Do it yourself. (C).. With Sympy simplify the following expression: 3 x x 4 8 x x x x x + + ( 3x( x 2 + 3) 2 ) ( 3x( x 2 + 3) ) ( ( x 2 + 3) 2 ) 2. Prove the following: Theorem. Every integer n can be written as the sum of three cubes of rational numbers. 3 3 var('x') A=((x**6+45*x**4-8*x**2+27)/(3*x*(x**2+3)**2))**3 B=((-x**4+30*x**2-9)/(3*x*(x**2+3)))**3 C=((-2*x**3+36*x)/((x**2+3)**2))**3 simplify(a+b+c) 8*x Answers. If we put (which are integers) then the above script shows that Therefore p = n n 4 8 n , q = 3n( n ) 2, r = n n 2 9, s = n3( n 2 + 3), t = 2 n n, u = ( n 2 + 3) 2 8n = (p/q) 3 + (r/s) 3 + (t/u) 3. n = (p/2q) 3 + (r/2s) 3 + (t/2u) 3.

6 Exercise 3. What if Archimedes had known Sympy? n n 3 2 n For, let be a regular -gon with radius. Here is : Archimedes (IIIrd century BC) used the fact that n to obtain good approximations of. 2π gets closer and closer to the unit circle Do it yourself. (D) Let L n be the length of any side of n. Compute L and use the following picture to write L n+ as a function of L n : O OC = (OB) is the bisector of ˆDOC. ˆOAC is a right angle. is the center of the circle,.

7 OB CB = BD n+ OAC A 2 = O A 2 + A C 2 = O A 2 + ( L n /2). BAC is also rectangle at A, therefore Ln+ 2 = B C 2 = A B 2 + B C 2 = ( OA ) 2 + ( L n /2). Answers. As is the bisector we have that, which both are sides of. Besides, is rectangle at. By Pythagora's theorem Putting everything together we obtain L n+ = 2 2 ( L n /2) 2. Do it yourself. (E). Write a script which computes exact expressions for the rst values L, L 2,, L n. (First try for small n 's.) 2. Find a sequence a n such that a n L n converges to π (we don't ask for a proof). Deduce some good algebraic approximations of π. Export your results in Latex in order to get nice formulas. (In order to check your formulas, you may compute numerical evaluations. With SymPy, a numerical evaluation is obtained with N(expression).) SuccessiveApproximations=[] p=8 for n in range(,p): OldValue=SuccessiveApproximations[-] NewValue=expand(sqrt(2-2*sqrt(-(OldValue**2)*Rational(,4)))) SuccessiveApproximations.append(NewValue) print(latex(simplify(3*(2**n)*newvalue))) print(n(newvalue*3*2**(n))) 6 \sqrt{- \sqrt{3} + 2} \sqrt{- \sqrt{\sqrt{3} + 2} + 2} \sqrt{- \sqrt{\sqrt{\sqrt{3} + 2} + 2} + 2} \sqrt{- \sqrt{\sqrt{\sqrt{\sqrt{3} + 2} + 2} + 2} + 2} \sqrt{- \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3} + 2} + 2} + 2} + 2} + 2} \sqrt{- \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3} + 2} + 2} + 2} + 2} + 2} + 2} \sqrt{- \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3} + 2} + 2} + 2} + 2} + 2} + 2} + 2}

8 Answers. When n goes large, n gets closer and closer to the unit circle. As the perimeter of is, we expect that 3 2π, 2 n L n n 3 2 n L n a n = 3 2 n n = 8 therefore we choose. For we obtain: π = Solving equations Exercise 4. Solving equations with Sympy: the easy case One can solve equations with Sympy. The following script shows how to solve x 2 = x + : var('x') # we declare variable SetOfSolutions=solve(x**2-x-,x) print(setofsolutions) [/2 + sqrt(5)/2, -sqrt(5)/2 + /2] We will use solve to handle a more complicated equation. Let n be an integer, we are interested in solving the equation x 3 + nx =. ( ) Do it yourself. (Theory) (F) n ( ) [0, ] S n x x 3 x nx 0 x. Prove that for every, Equation has a unique real solution in the interval. This solution will be denoted by. 2. On the same gure, plot, and for and for several (large) values of n. What can we conjecture for the limit S n? 3. (Bonus) Prove that S n /n, and deduce that S n = n + o(/n).

9 RangeOf_x=np.arange(0,,0.0) plt.plot(rangeof_x,rangeof_x**3,label='$x\mapsto x$**3') for n in [0, 20, 30]: f=[-n*x for x in RangeOf_x] plt.plot(rangeof_x,f,label='n ='+str(n)+' ') plt.xlim(0, ),plt.ylim(-, ) plt.xlabel('value of x') plt.legend() plt.title('location of S_n') plt.show() Answers.. The map is continuous and increasing on, since Besides, x f (x) = x 3 + nx [0, ] f (x) = 3 x 2 + n > n > 0. f (0) = 0 3 n 0 =, f () = 3 + n = n > 0. By the intermediate value theorem, this implies that there is a unique S n (0, ) such that f ( S n ) = 0, i.e. ( S n ) 3 + n S nn =. +, the solution of Equation 2. On the gure above we observe that when ( ) seems to get closer and closer to zero. We therefore conjecture lim S n n + 3. In fact, in Question ) one could be more precise: By the intermediate value theorem, this implies that S n (0, /n). If we use the fact that ( S n ) 3 + n S n =, we obtain since S n 0. Thus S n = n + o(/n). = 0. f (0) = 0 3 n 0 =, f (/n) = (/n) 3 + n /n = (/n) 3 > 0. n S n = ( S n ) 3 = + o(),

10 Do it yourself. (G). Write a script which computes the exact expression of S n. 2. In the previous questions it is proved that S n = n + o(/n). Use series to obtain the next orders in the asymptotic behaviour of S n (up to (/ n 5 ) ). var('x') var('n',integer=true) # Question. Solutions=solve(x**3+n*x-,x) Sn=Solutions[0] # The two other solutions are complex numbers print("sn = "+str(sn)) # Question 2. Taylor=series(Sn,n,oo,5) print("taylor expansion when epsilon -> 0 : "+str(taylor)) Sn = -n/(3*(sqrt(n**3/27 + /4) + /2)**(/3)) + (sqrt(n**3/27 + /4) + /2)**(/3) Taylor expansion when epsilon -> 0 : -/n**4 + /n + O(n**(-5), (n, oo) ) Answers. According to the above script, S n = + (/ n 5 ). n n 4 Exercise 5: Solving equations and "big O's" We consider the following equation: X 5 3εX 4 = 0, ( ) ε ε > 0 where is a positive parameter. A quick analysis shows that if is small enough then ( ) has a unique real solution, that we denote by S ε. The degree of this equation is too high to be solved by SymPy : var('x') var('e') solve(x**5-3*e*x**4-,x) [] Indeed, SymPy needs help to handle this equation. f (x) = x 5 3εx 4 ε In the above script we plotted the function for some small. This suggests that. lim ε 0 S ε =

11 RangeOf_x=np.arange(0,2,0.0) def ToBeZero(x,eps): return x**5+x**4*(-3*eps) - eps=0.05 plt.plot(rangeof_x,[tobezero(x,eps) for x in RangeOf_x],label='x**5+x**4*(-3*eps)-') plt.xlim(0, 2) plt.ylim(-, ) plt.plot([-2,2],[0,0]) plt.plot([,],[-2,2]) plt.xlabel('value of x') plt.title('location of S_n, with eps ='+str(eps)) plt.legend() plt.show() Do it yourself. (H). We admit that can be written as = + rε + s + ), S ε ε 2 ( ε 3 S ε r, s r, s for some real. Use SymPy to nd. (You can use any SymPy function already seen in this notebook.) var('r') var('s') var('eps') Expression=ToBeZero(+r*eps+s*eps**2+O(eps**3),eps) Simple=simplify(Expression) print(simple) solve([-3+5*r,5*s-2*r+0*r**2],[r,s]) -3*eps + 5*eps*r + 5*eps**2*s - 2*eps**2*r + 0*eps**2*r**2 + O(eps**3 ) [(3/5, 8/25)]

12 + rε + s ε 2 + ( ε 3 ) Answers. If we plug into equation ( ) we obtain (with the script): 0 = 3ε + 5rε + 5sε 2 2r ε r 2 ε 2 + O( ε 3 ). ( ) ε 0 = 3 + 5r + 5sε 2rε + 0r 2 ε + O( ε 2 ), 3 + 5r = 0 ε 0 r = 3/5 If we divide equation ( ) by we obtain which yields by letting, i.e.. ε 2 If we plug this into ( ) and divide by we obtain 0 = 5s 2r + 0 r 2 + O(ε), which gives, i.e.. 5s 2r + 0 r 2 = 0 s = 8/25 Exercise 6: Decomposition of fractions (Bonus). Do it yourself. (J) Set With SymPy nd reals (You can use solve, simplify, expand. Remember to use Rational(,) for fractions.) f (x) =. 2x 3 0x 2 30x + 54 a, b, c, α, β, γ such that a b c f (x) = + +. x α x β x γ a b c Answers. First, we observe that and must 2x 3 0x 2 30x+54 x α + + x β x γ have the same domain of de nition. Therefore {Roots of 2x 3 0x 2 30x + 54} = {α, β, γ}. In the script below we obtain: α = 3, β = 7 + 4, γ = var('x a b c') Num=2*x**3-0*x**2-30*x + 54 Roots=solve(Num,x) print(roots) [alpha, beta, gamma] = Roots [-3, -sqrt(7) + 4, sqrt(7) + 4]

13 Answers. We must have a b = x 3 0x 2 30x + 54 x + 3 x c x 4 7 x = 0,, 2 for. We obtain with the following script that x x (x + 3) # Left-hand side: def LHS(x): return Rational(,2*x**3-0*x**2-30*x + 54) # Right-hand side: def RHS(x,a,b,c): return a/(x-alpha)+b/(x-beta)+c/(x-gamma) # we identify a,b,c by solving the following system: Solutions=solve([LHS(0)-RHS(0,a,b,c),LHS()-RHS(,a,b,c),LHS(2)-RHS(2,a,b,c)],[a,b,c] print(solutions) astar=solutions[a] bstar=solutions[b] cstar=solutions[c] anum=n(astar) bnum=n(bstar) cnum=n(cstar) # to get a nice formula: var('x') print(latex(rhs(x,astar,bstar,cstar))) {c: -/68 + sqrt(7)/68, b: -sqrt(7)/68 - /68, a: /84} \frac{- \frac{\sqrt{7}}{68} - \frac{}{68}}{x \sqrt{7}} + \frac {- \frac{}{68} + \frac{\sqrt{7}}{68}}{x \sqrt{7}} + \frac{}{8 4 \left(x + 3\right)}