Numerical Algorithms.

• Solving linear equations is well known.
• And we can do it to a fairly high degree of accuracy.
• But non-linear equations are another story.
• Read the second paragraph on page 259.
• We have a number of problems before we even get started.
  • Assume we want to approximate π with

    	         z3   z5   z7         (-1)n
    arctan(z) = z - -- + -- - -- + ... + ------
                     3    5    7          2n+1
    
    		= π/4 when z = 1
            

  • This will eventually have a truncation error
  • This was discovered by Madhava of Sangamagrama
  • He gives a better approximation:

                   1      1      1
    π = 121/2(1 - ---- + ---- - ---- ... 
                  3×3    5×32   7×33
    	

    i                          Ti                      π                   Error
    3
    		0.66666666666666666663	2.66666666666666666652	0.47492598692312643083
    		0.88888888888888888888	3.07920143567800389884	0.06239121791178921933
    5
    		0.86666666666666666662	3.46666666666666666648	0.32507401307687355807
    		0.91111111111111111108	3.15618147156995399643	0.01458881798016087999
    7
    		0.72380952380952380949	2.89523809523809523797	0.24635455835169786654
    		0.90582010582010582010	3.13785289159568016380	0.00373976199411295198
    ...                     ...                    ...                     ...
    1995
    		0.78514766245832072787	3.14059064983328291147	0.00100200375651020453
    		0.90689968211710892586	3.14159265358979305832	0.00000000000000005768
    1997
    		0.78564841358501076292	3.14259365434004305168	0.00100100075024993568
    		0.90689968211710892586	3.14159265358979305832	0.00000000000000005768
    1999
    		0.78514816345994823164	3.14059265383979292654	0.00099999975000018945
    		0.90689968211710892586	3.14159265358979305832	0.00000000000000005768
    
    	

  • The second approximation begins to exhibit round-off errors as well.
  • We also have problems with underflow and overflow.
  • And then there are the algorithms themselves.
  • One of the major concerns of numerical analysis is understanding and controlling such errors.
• Solving for x in f(x) = 0
  • There is, and can be, no formula such as the quadratic equation, for polynomials of degree greater than four.
  • Bisection Method:
    • Given two points a, and b.
    • WLOG assume f(a) < 0 and f(b) > 0

    • BISECTION(a,b)
      
       mid ← (a+b)/2.0
       n ← 0
       while  n < MAX_ITERATIONS and fabs(f(mid)) > MIN_ERROR do
           n ← n + 1
           if f(mid) < 0 
                 a ← mid
           else
                 b ← mid
           mid ← (a+b)/2.0
       return mid, fabs(f(mid))
      
      	     

    • Algorithm questions?
    • This is a slow algorithm
    • If there are multiple roots between the original a and b, there is no guarantee which it will find.
    • But it will converge to one root.
• Regula Falsi
  • False position.
  • Like bisection, start with a point above and a point below.
  • Rely on the intermediate value thereom to give us the root.
  • Derive the equation of the secant line.

  • REGULA_FALSI(a,b)
    
     newPos ← b - f(b)(b-a)/(f(b)-f(a)) 
     n ← 0
     while  n < MAX_ITERATIONS and fabs(f(newPos)) > MIN_ERROR do
         n ← n + 1
         if f(newPos) < 0 
               a ← newPos
         else
               b ← newPos
         newPos ← b - f(b)(b-a)/(f(b)-f(a)) 
     return newPos, fabs(f(newPos))
    
    
    	

• Newton's Method
  • Start with an initial guess that is "close enough"
  • Approximate the tangent to the curve.
  • We assume a continuous function.
  • (y-y₁) = m(x-x₁₎
  • x = x_{1 - (y - y1)/m}
  • let f'(x₁ = m (slope at current point)

  • NEWTON(a)
    
     n ← 0
     x ← a
     f_of_x ← f(x)
     f_prime_of_x ← fPrime(x)
     while  n < MAX_ITERATIONS and  fabs(f_of_x) > MIN_ERROR do
         n ← n + 1
         x ← x - f_of_x/f_prime_of_x
         f_of_x ← f(x)
         f_prime_of_x ← fPrime(x)
     return x, f_of_x
    
    	 

  • Newton's converges very quickly
  • And works as long as the initial guess is good.
  • But if the initial guess is bad, it will have problems.
• An example:
  • f(x) = x³ -5x -1
  • 
  • My Code