Recursive Methods for Finding Roots of Functions

1
Recursive Methods for Finding Roots of Functions

• Bisection
• Newtons Method

2
Intermediate Value Theorem

• Finding the root of a function will rely on this
  theorem.
• It states that for any function that is
  continuous over the interval a,b then that
  function will take on every value between f(a) to
  f(b).

3
BISECTION

• Bisection requires narrowing one root onto a
  certain interval a,b
• Testing this interval in the function f should
  give values which are opposite in sign for f(a)
  and f(b).
• Bisecting this interval and testing f((ab)/2)
  will yield another smaller interval in which to
  continue bisecting until the desired accuracy is
  reached.

4
BisectionExample
Given ,
find the root between 2 and 3 to an accuracy of
10-3.

• 1st Compare f(2) -1 and f(3) 16. By the
  Intermediate Value Theorem we are insured that at
  least one root exists between 2 and 3.

• 2nd Test f((23)/2) f(2.5). All we need to
  know is its sign. It is positive.

• 3rd, we narrow our search interval from 2,3 to
  2,2.5 since f(2) is negative and f(2.5) is
  positive.

5
BisectionExample

• Continuing this pattern to finish the problem
• iteration 2 f(2.25)gt0, new interval 2,2.25
• iteration 3 f(2.125)gt0, new interval 2,2.125
• iteration 4 f(2.0625)lt0, new interval
  2.0625,2.125
• iteration 5 f(2.09375)lt0, new interval
  2.09375,2.125
• iteration 6 f(2.109375)gt0, new interval
  2.09375,2.109375
• iteration 7 f(2.1015625)gt0, new int
  2.09375,2.1015625
• iteration 8 f(2.09765625)gt0, new int
  2.09375,2.09765625
• iteration 9 f(2.095703125)gt0, int
  2.09375,2.095703125
• Root approximation is the midpoint of this
  interval x2.094390625

6
BisectionExample

• We stop at this point because the desired
  accuracy has been reached.
• The root is always as accurate as half of the
  interval arrived at, or in this case
  (2.09503125-2.09375)/2 .0009765625

7
BisectionAccuracy

• Accuracy is always arrived at by taking half the
  current interval because the actual root could be
  no farther than that away from the midpoint.
• The initial accuracy of the midpoint is
  b-a/2
• After each iteration, this accuracy is halved.
• After n iterations accuracy is b-a/2(n1)

8
Psuedocode

• Input function, interval where root exists, and
  desired accuracy.
• f, a, b, E
• n 1
• While E gt b - a / 2n
• g (b a) / 2 //make a guess for root at
  midpoint
• if f(g) is the same sign as b
• then b g
• else a g
• endif
• n n 1
• End loop
• Display guess as g to accuracy of b - a / 2n

9
NEWTONS METHOD

• Requires you to make a relatively good guess for
  the root of f(x), x0.
• Construct a tangent line to f(x) at this point
  y - f(x0) f(x0)(x - x0)
• Find the x-intercept of this line (y 0)
  0 - f(x0) f(x0)(x - x0) OR
  x x0 - f(x0) / f(x0)
• Repeat guess for this new x.

10
Newtons MethodExample
Given ,
find the root between 2 and 3 to an accuracy of
10-3.

• 1st Guess By view the graph make an initial
  guess of x0 2.
• 2nd find new x x1 2 - f(2) / f(2) 2.1
• 3rd repeat for x 2.1

11
Newtons MethodExample

• Continuing this pattern takes much less time to
  narrow down than bisection
• iteration 2 x2 2.1 - f(2.1) / f(2.1)
  2.0945681211
• iteration 3 x3 x2 - f(x2) / f(x2)
  2.0945514817
• You can already see that by the third iteration
  the accuracy of the root is to 0.0000166394 or
  less than 10-3.
• The number of iterations will depend on how good
  your guess is.

12
Psuedocode
Input function, guess, and desired accuracy. f,
g, E n 0 While gn - gn-1 gt E n n
1 gn gn-1 - f(gn-1) / f(gn-1) End
loop Display root as gn to accuracy of gn - gn-1
13
Quirks and Exceptions toNewtons Method

• if along the way f(xn) 0 then you will get a
  horizontal line with no x-intercept.

14
Quirks and Exceptions toNewtons Method

• You may get the wrong root depending on your
  initial guess.

x0
x1
15
Quirks and Exceptions toNewtons Method

• You may get the wrong root.

x0
x1