![xn=xn−1−f ( xn−1)
xn−1−xn−2
f (xn−1)−f ( xn−2)
,
This method is also called regula falsi or method of false positions.
Again, if the points x2, x1 and x0 are not very close to the root, the
problems of slow convergence, oscillations reappear, but
convergence is guaranteed if there are real roots. The order of
convergence is not quadratic but more than linear.
Bisection method
If we take x2=
x0+x1
2
in the above ,
we get,
f (x2)−f (x1)
x2−x1
=
f (x0)−f (x1)
x0−x1
.
Or,
x0+x1
2
=x
2
=x1−f (x1)
x1−x0
f (x1)−f (x0),
The method is useful when f (x0),f (x1)are negative of each other, so
that by intermediate value theorem f (x2)has a valuef (x2)≈0 .
The iteration equation becomes.
xn+xn+1
2
=x
n+2
=xn+1−f (xn+1)
xn+1−xn
f (xn+1)−f (xn)
,
Example
Takef (x)=x3
−x−2 .
We see that if x1=2,x0=1
then f (1)=−2,f (2)=+4 of opposite signs. Since it is a continuous
function, by intermediate value theorem there is a root in the
interval [1,2].
Now x2=
x0+x1
2
=
1+2
2
=1.5,](https://image.slidesharecdn.com/iterationmethodsfornumericalsolutionofpolynomials-250302043754-58e3ac26/85/iteration-methods-for-numerical-solution-of-polynomials-docx-22-320.jpg)
![so f (1.5)=(1.5)3
−(1.5)−2=−0.125, and f (2)=+4 .
Hence we would further search for x3=
1.5+2
2
=1.75 and find f (1.75).
Continuing this iteration 15 steps we getx16=1.5214 sufficient near 0
and it could be taken as a root practically
f (1.75)=0.00008.
Picard’s Fixed point iteration process
The roots of y=f (x) are the points wheref (x)=0 or where the graph of
y=f (x)meets the X-axis, i.e., where the graphs of
y=f (x) and y=x. We may pretend to ‘solve’ this function by extracting
an fromf (x)=0 , by making it x=g(x).
Simply if we set F(x)=g(x)−x , in an interval [a, b] , then we have
F(a)−a≥0, F(b)−b≤0, as the function is continuous , it takes every value
in [a, b] so there is a point c ∈[a,b]: F(c)=0 which is a fixed point in the
interval. This is by virtue of intermediate value theorem described
elsewhere in this book.
Sufficient condition for this would beF'(x)
<0 then F(x)would be
decreasing. i.e.F'(x)
=g'(x )
−1<0, or, g' (x)
<1. In other words, slope of y=g(x)
would be less than slope of y=x. Obviously it is tacitly assumed that
has continuous derivative in the interval. The following examples
shall make this point clearer. ( it may be noted that y=x is the line on
which any function and its inverse function (if exists) meet. Since this
line does not change interchanging of y∧x.
For example let f (x)=x
2
−2x−8=0and we can write it asx=g(x)=
x
2
−8
2
from the same function. The roots off (x)=0 is the same points where
graphs y=g(x)](https://image.slidesharecdn.com/iterationmethodsfornumericalsolutionofpolynomials-250302043754-58e3ac26/85/iteration-methods-for-numerical-solution-of-polynomials-docx-23-320.jpg)
iteration methods for numerical solution of polynomials.docx
- 1. Iteration methods for approximate roots of polynomials In the process, for finding roots of polynomials, we hazard a reasonable guess for a real number root and employ an iteration formula to obtain next approximation or correction of error . A little preparation for adopting the process is required such as rational root theorem, intermediate value theorem, synthetic division etc. dealt hereafter in rudiments. Rational root theorem Suppose the polynomialan x n +an−1 x n−1 +….a0 has all integer coefficients. If not we can multiply throughout by LCM of the denominators of all coefficients. Suppose it has a rational root p q where they have no common factors, i.e., they are coprime. So an(p q ) n +an−1(p q ) n−1 +….a0=0 . Or, an p n +an−1 p n−1 q+….a0q n =0 Or, p(an p n−1 +an−1 p n−2 q+..a1 q n−1 )=−a0q n So p is a factor of left side, it must divide the right sidea0q n also. Butp does not divide q n as it does not divideq as assumed. Thereforep dividesa0 .
- 2. Similarlyan p n +an−1 p n−1 q+….a0q n =0 gives an−1 p n−1 q+……..a0 q n =−an p n . Or, q(an−1 p n−1 +……..a0q n−1 )=−an p n . Now q must divide anas it and therefore does not divide p does not divide p n . So the rational root theorem says, the denominator of the rational root must divide the leading coefficient and the numerator of the root must divide the last term. This fact can be used to determine rational roots of a polynomial if any, by analyzing factors of the leading coefficient and last term. The integer root theorem is a special case of rational root theorem when the leading coefficient is 1. If the monic polynomial has any integer roots it must appear as a factor of the last term. As an application, if a polynomial is has an integer root or rational root r it can be found out by factoring the leading coefficient and the last term. Then the factor x−r can be taken out of the polynomial reducing its degree by one just by dividing the polynomial byx−r . Descartes’s rule of signs There can be no more positive roots of the polynomial f (x) than the changes of sign in the polynomial written in descending order of power. Similarly the maximum number of negative roots of a polynomialf (x) shall be maximum number of changes of sign inf (−x) .
- 3. Since the polynomial must have pairs of imaginary roots, their number must be even, got after deducting positive and negative roots in the polynomial. Formal proof may also be given later. Explanation; Suppose a polynomial with one minus sign coefficient as in the following an xn +an−1 xn−1 −an−2 xn−2 ….a0is multiplied by a factor x−a where a is positive. In sketch + + - + + ……… + - + - - ……… - + ± ∓ ± ± ………. ± It may be observed that there is at least one more sign change from 2nd place to 3rd place. Thus for every additional positive root there is at least one more sign change in the polynomial. In other words, there can be no more positive roots of the polynomial than the changes of sign in the polynomial written in descending order of power.
- 4. Synthetic Division This is nothing but writing the process of long division in brief. An example below: Divide x 6 −2 x 4 +3 x 3 −x 2 −2x+6 by x−2. Just write the coefficients of the polynomial, including missing powers , if any. Write 2 in x−2. to its below left and make a dashed line. The first coefficient 1 is carried down, as first member of quotient. This carried down 1 multiplied by 2 in the divisor, raised up, added to 0 carried down. See above. Next, this 2 is again multiplied by 2, raised up, written below – 2 ,added to it and carried below. See below .
- 5. So the dividend is x 5 +2x 4 +2 x 3 +7 x 2 +13 x+24and remainder is last term54 . This evaluates the polynomial at x=2, as per remainder theorem (Division algorithm) f (x)=d(x)(x−a)+f (a). When the polynomial f (x) is divided byx−a , the dividend is (x) , and remainder is R=f (a). This is value f (a)of the polynomial at x=a. The polynomial is divisible byx−a if the remainder R=f (a) is 0. (Division algorithm)In fact with a little practice, this can be done all in one step. So synthetic division is used primarily for evaluation of a polynomial at a point. If R=f (a)=0,thenf (x)=d(x)(x−a) andf (x) is divisible byx−a .(Factor theorem). Synthetic division can be exhibited as a formula. Note this key example below. Let us divide a x 2 +bx+c by x−a through synthetic division as under.
- 6. In the first step a is just brought down. In the 2nd step,a is multiplied byr and placed under b. Its totalar+b is brought down and multiplied again by and the resultar 2 +br is placed under c and total ar 2 +br+c is brought down. This is the final remainder. This can be generalized to get remainder theoremf (x)=d(x)(x−a)+R(x) wheref (x) is a polynomial of degree n, andd (x), is a polynomial of degree < n. Further R(x)=f (a). In f (x)=d(x)(x−a)+f (a) ,d (x) is unique. If f (a)=0 the same result is called factor theorem. This x−a is called a factor of f (x) and a is a root . Again dividingf (x) by x−bif is banother root, we get another polynomial as quotient of degree further one degree less . If the polynomial f (x) has n roots, the process stops after n steps and further division result in fractions. Thus a polynomial of degree n has maximum n roots or n factors and no more; albeit some of them may be repeated. Since the division process gives unique quotient at each step, the polynomial has n roots and no more and can be factorized in unique way. This is unique factorization theorem in parlance with unique factorization theorem in arithmetic. In other words, if the equationf (x)=0 has more than n distinct roots, it must be an identity, not an equation and
- 7. shall be satisfied by every value of x. As an exercise you may prove the following quadratic equation is an identity and is satisfied by every value ofx . Same here below. Just find the values of the expressions for x=a ,x=b ,x=c Thus if a polynomial has n roots and no more, else every number satisfies it. In another way, called Horner’s method putx5 +2x4 +2 x3 +7 x2 +13 x+24 in nested form in successive steps as follows. x5 +2x4 +2 x3 +7 x2 +13 x+24 ¿ x 4 (x+2)+2x 3 +7 x 2 +13 x+24 ¿ x3 (x (x+2)+2)+7 x2 +13 x+24 ¿ x2 (x (x (x+2)+2)+7)+13 x+24 ¿ x(x(x(x (x+2)+2)+7)+13)+24
- 8. In each successive steps we are we are taking the highest available power to put the terms inside bracket, such as x5 +2x4 =x4 (x+2), x4 (x+2)+2x3 =x3 (x(x+2)+2),etc. In evaluating the polynomial, adding a coefficient and a multiplication by x greatly reduces the calculations such as raising to powers. Horner’s method is equivalent to synthetic division by Ruffini With a bit analysis the Horner’s method is seen to be equivalent to synthetic division a process by . Take another example .To divide Things would be clear if we write long division and synthetic division side by side. Long division. (Please ignore different colours of texts) Synthetic division:
- 9. Please compare the two to understand. In synthetic division, the coefficients of divisor g(x)are written in the left side column with their signs changed and coefficients of f (x)are written in a horizontal row as shown. The 1 in top left column is not a coefficient, but just to show that the leading coefficient of g(x)is not anything other than 1. Had it been 3, both numerator and denominator could have been divided by 3 and then the process of synthetic division would have followed. Both the dividend and the divisor would have been factored out by same factor before starting division. For example take The synthetic division is then as usual. As another example , to divide 4 x 4 −6 x 3 +3 x−5by
- 10. 2 x−1 lease see below: The third row, is the sum of the first and second rows all divided by 2. In Horner’s method if we have to divide f (x)=x3 +4 x2 −5 x+5 by x−3, we present here, long division, synthetic division and Horner’s algorithm for comparison and understanding. Horner’s nested brackets If this polynomial is to be evaluated at x=3, we would divide f (x) by x−3and the remainder would be R=f (3); f (x)=(x−3)g(x)+f (3). Long Division and synthetic division are as under
- 11. If you continue division process even after the remainder has been obtained, you get digits after the decimal place. This is also example of Paravartya sutra in Vedic Mathematics. To explain how the method of synthetic division is equivalent to Horner’s method, we show below that both are essentially equivalent to division algorithm in long division process . Please see below. Let the polynomial be f (x)=∑ i=0 n ai xi =¿¿ . To evaluate f (x0), define a new sequence of constants b0,b1 ,… so that
- 12. b0=f (x0) , value of the function at x=x0 . In view of the above, f (x)=∑ i=0 n ai xi =¿¿ and f (x0)=¿ by substituting the formula for b0,b1 ,….etc. This is nothing but the usual division algorithm, so both Horner’s method and synthetic division are equivalent to division algorithm at large. Butf (x0) is the remainder whenf (x) is divided by −x0 . By division algorithm,f (x)=g(x)(x−x0)+f (x0) .So the value of the polynomial f (x)at is the remainder f (x0)when the polynomial is divided byx−x0 . By substituting values of constants b0,b1 ,… etc. in the polynomial we get
- 13. f (x)=(b1+b2 x+b3 x2 …….bn xn−1 )( x−x0)+b0 So (x0)=b0 . We can start at bn=anand use the recursive formula until we reach b0. This is quickly achieved by doing synthetic division as described above. To find value of a polynomial in Horner’s method or by synthetic division. Given a polynomial f (x) we evaluate it at a pointx=a ,as per remainder theorem computed by Horner’s method or synthetic division. If the remainder is 0,then x=a is a root of the polynomial or x−a is a factor of the polynomial f (x) as per factor theorem. Now the reverse problem is to find out the roots or factors of the polynomial in the same method, using Horner’s method or synthetic division. The procedure is, hazard a guess for some x=x0 ,reasonably approximate root. Particularly it is helpful for finding largest root. Then divide polynomial f (x) by x−x0 and remainderf (x0) is obtained in the above method. Our aim is now to make this remainder approximate to 0 by recursively repeating the procedure. Straight and simple. The basic idea is, what happens when we reduce all roots of the polynomialf (x) uniformly by a number say, r. (If f (x)=A x3 +Bx2 +Cx+D then f (x+r) keeps the form of the polynomial unchanged ,i.e., f (x+r) ¿ A(x+r)3 +B(x+r)2 +C(x+r)+D
- 14. If we divide it by (x+r) by synthetic division or otherwise then the remainder is D). To compute a root x=r of the polynomialf (x)=0 we choose a sequence x=x0, x=x1 and x1,x=x2 …such that f (x)→0. Computer would do everything for us, but it must be given some algorithm, a routine procedure of what to do, how to do, when to start and when to stop. If x0leads to x1 and x1 leads tox2 , by means of some algorithm, and so on, two situations may arise. Either we are we are nearing a real root or getting away from it. The steps of difference x2−x1 , x1−x0 etc. depend upon what? They obviously depend upon rate of change of the function i.e. how the function changes from point to point, i.e., the differential coefficient of the function at the points, f ' (x0), f ' (x1), etc. Since ' (x0)≈ f (x1)−f (x0) x1−x0 , Or,x1−x0≈ f (x1)−f (x0) f ' (x0) . This is as if the function is linear in a small range nearx0 . (The tangent at that point has two points on the curve arbitrarily close to each other) Supposing f (x1)=0 i.e.,x1 is exactly a root the above expression becomes x1−x0≈ −f (x0) f ' (x0) , or x1≈ x0− f (x0) f ' (x0) . If f (x1)≠0we can take x2≈ x1− f (x1) f ' (x1) as new value of x1that may make f (x2)=0, or x2 a root of the equation. If not We go to x3≈ x2− f (x2) f ' (x2) so that x3may be a root.
- 15. So the formulaxn+1 ≈ xn− f (xn) f ' (xn) takes us to further and further approximation of a real number root on the condition that f ' (xn)≠0. We have to explore the conditions under which it goes near a root or goes away from it. This is Newton’s method or Newton-Raphson . Raphson further sharpened the tool. Newton’s criteria for finding upper limit of roots By Taylor’s theorem, f (x+h) ¿ f (h)+ x f ' (h)+ x 2 2! f '' (h)… x n n! f n (h), If f (h)and all derivativesf ' (h),f '' (h),… f n (h) are all positive at x=h for x>0 then f (x)>0 for ¿0 . This shows h is higher than the highest root, i.,e.,h is an upper bound for roots. We may find by trial some least integerh1such that f n−2 (h1)>0whenx>h1 .If thish1 does not makef n−2 (x)>0 , then again by trial a greater integerh2>h1 so that f n−2 (h1)>0. In this way we can find a least integer h so that f (h)and all derivativesf ' (h),f '' (h),… f n (h) are all positive at x=h for ¿0 . then f (x)>0 for x>0 . For example If f (x)=x3 −10x2 −11 x−100, Then f ' (x)=3 x2 −20 x−11,f ' ' (x)=2(3 x−10)
- 16. f '' ' (x)=6 The least integer values of x for which f ' (x)>0 is 4 , that for f ' ' (x)>0 is 8 and that for f '' ' (x)>0 is 12. Thus 12 exceeds all roots and thus integer part of highest root must be 11. Geometrically Obviously f '(x0 ) =tanΨ ≈ f (x0) x0−x1 as in the figure. It remains to see under what condition this approximation really converges to a root. Assumingf ' (x) to be continuous in addition to f ' (x)≠0 in an interval around x0 . (These conditions are obviously satisfied in case of polynomial functions.) Convergence is guaranteed if this following condition is satisfied. Supposing α is a root of the polynomial, i.e.f (α)=0 , Taylor’s theorem with residue is f (α)=0=f (xn)+f ' (xn)(α−xn) + 1 2 f '' (εn)(α−εn ) for some εn,α−xn ≤εn≤α +xn , α−xn ≈εnmay be taken as error in n-th step. Rearranging the terms,
- 17. f (xn) f ' (xn) +α−xn= −f ' ' (εn) 2f '(xn) (α−xn) 2 . Putting xn+1=xn− f (xn) f ' (xn) , the last equation becomes, α−xn+1= −f ' '(εn ) 2f ' (xn) (α−xn) 2 . If α−xn ≈εnis the error in n-th step, α−xn+1=εn+1is error in n+1st step. So εn+1= −f '' (εn ) 2f ' (xn ) εn 2 . If ¿ |f '' (εn ) 2f ' (xn)|=M <1, the errors in successive iterations would definitely converge to 0 . this is sufficient condition for f (α)=0 or to be a rootα; condition of quadratic convergence. At least good accuracy is achieved after a few steps of iteration. What has it to do with synthetic division or Horner’s method ? Synthetic division would quickly computef (xn) . If the polynomial f (x)has a double root atx=α , then the iterations oscillate ‘twice’ around the root, diminishing the ‘speed’ of convergence. Naturally increasing the interval to would help maintain the quadratic convergence. This is done by setting xn+1=xn−2 f (xn) f ' (xn) , and in case of multiplicity m of the root, we may set xn+1 ≈ xn−m f (xn) f ' (xn) , the concept called successive over relaxation. Multiple roots Suppose f (x)has a multiple root,
- 18. take f (x)=(x−α )m g(x). On differentiation, f ' (x)=m(x−α )m−1 g(x)+(x−α )m g' (x) So f (x)andf ' (x) have a common factor(x−α )m−1 . So it helps finding df ' (x) and analyzing for roots before analyzing df (x) have. Intermediate value theorem. Now about ‘guess of x=x0 . Just we may find some a,b: f (a)<0,f (b)>0 . Since f (x)is a continuous function, it takes every value for f (a)<f (x)<f (b) if the function is real valued. We may choose any point x0: a<x0<b. This is intermediate value theorem. The function must pass through every value f (a)<f (x)<f (b), obviously 0 is a value of f (a) between them iff (a)<0<f (b). Particularly, if the function is monotonically increasing f ' (x)>0in the intervala< x<b , we can start at the beginning x0=a ,or if the function is monotonically decreasingf ' (x)<0, we can start at the end point x0=b, and convergence to a root is guaranteed in this case. Formal proof may be given later.
- 19. Example 1 To find cube root of a number. Take f (x)=x 3 −a. Immediately we have xn+1=xn− f (xn) f ' (xn) =¿ xn− xn 3 −a 3 xn 2 =xn− 1 3 xn + a 3 xn 2 . A suitable initialization shall serve with a few steps. Caution not to fall in ditch by taking xn=0 which blows up with f '(x0 )=f ' (0)=0 . Example 2 In fact it is a non-example. Non-examples are as important as examples to understand anything . To find natural logarithm of numbers . Take the functionf (x)−lnx and see how to run into trouble. The function is defined only forx>e and it has a root atx=1 . We have xn+1=xn− f (xn) f ' (xn) =xn (1−ln xn). If we initialize at xn=e, the next iterate is 0 and the sequence does not proceed further. If we takexn>e , the next step ends in negative and the process halts. As Newton-Raphson method is slow converging , sometimes oscillating and sometimes fails on condition as seen above, more criteria and algorithms other than Newton’s method got devised as time elapsed .
- 20. Other helpful theorems 1) Suppose that for α is a root of that for f (x),f (α )=0. By Taylor’s expansion, f (α+h)=f (α ) yields → +h f ' (α )+ h 2 2! f ' ' (α)..., f ' (α +h)=f ' (α)+hf ' ' (α )+ h2 2! f ' '' (α )+…., If the first term h r r! f r (α ) in both series above is not 0, its sign is sign of both f (α+h) and f '(α+h) for very small values of h if h>0. Please note that they are of opposite signs otherwise. 2) If α is a multiple root, then as x increases through x=α, the terms f (α),f ' (α ),f '' (α)... are all of same sign immediately after x=α, as shown above, but are alternatively + and – immediately beforex=α . 3) As already shown the theorem for Des Cartes rule of signs.
- 21. Secant method Instead of considering tangent at the point x0,Let us consider the secant joining points on the curve f (x0) and f (x1) .Any point x2 where the secant cuts the X-axis is given by the equation of the secant. If we consider nearer to a root of the equation, then taking slopes of the segment between x2, x1 and x1, x0 to be equal, we get assuming f (x2)≈0 orx2 is very near the root, f (x2)−f (x1) x2−x1 = f (x0)−f (x1) x0−x1 , Or,x2=x1−f (x1) x1−x0 f (x1)−f (x0) , Or the recurrence relation becomes,
- 22. xn=xn−1−f ( xn−1) xn−1−xn−2 f (xn−1)−f ( xn−2) , This method is also called regula falsi or method of false positions. Again, if the points x2, x1 and x0 are not very close to the root, the problems of slow convergence, oscillations reappear, but convergence is guaranteed if there are real roots. The order of convergence is not quadratic but more than linear. Bisection method If we take x2= x0+x1 2 in the above , we get, f (x2)−f (x1) x2−x1 = f (x0)−f (x1) x0−x1 . Or, x0+x1 2 =x 2 =x1−f (x1) x1−x0 f (x1)−f (x0), The method is useful when f (x0),f (x1)are negative of each other, so that by intermediate value theorem f (x2)has a valuef (x2)≈0 . The iteration equation becomes. xn+xn+1 2 =x n+2 =xn+1−f (xn+1) xn+1−xn f (xn+1)−f (xn) , Example Takef (x)=x3 −x−2 . We see that if x1=2,x0=1 then f (1)=−2,f (2)=+4 of opposite signs. Since it is a continuous function, by intermediate value theorem there is a root in the interval [1,2]. Now x2= x0+x1 2 = 1+2 2 =1.5,
- 23. so f (1.5)=(1.5)3 −(1.5)−2=−0.125, and f (2)=+4 . Hence we would further search for x3= 1.5+2 2 =1.75 and find f (1.75). Continuing this iteration 15 steps we getx16=1.5214 sufficient near 0 and it could be taken as a root practically f (1.75)=0.00008. Picard’s Fixed point iteration process The roots of y=f (x) are the points wheref (x)=0 or where the graph of y=f (x)meets the X-axis, i.e., where the graphs of y=f (x) and y=x. We may pretend to ‘solve’ this function by extracting an fromf (x)=0 , by making it x=g(x). Simply if we set F(x)=g(x)−x , in an interval [a, b] , then we have F(a)−a≥0, F(b)−b≤0, as the function is continuous , it takes every value in [a, b] so there is a point c ∈[a,b]: F(c)=0 which is a fixed point in the interval. This is by virtue of intermediate value theorem described elsewhere in this book. Sufficient condition for this would beF'(x) <0 then F(x)would be decreasing. i.e.F'(x) =g'(x ) −1<0, or, g' (x) <1. In other words, slope of y=g(x) would be less than slope of y=x. Obviously it is tacitly assumed that has continuous derivative in the interval. The following examples shall make this point clearer. ( it may be noted that y=x is the line on which any function and its inverse function (if exists) meet. Since this line does not change interchanging of y∧x. For example let f (x)=x 2 −2x−8=0and we can write it asx=g(x)= x 2 −8 2 from the same function. The roots off (x)=0 is the same points where graphs y=g(x)
- 24. And graph of y=x meet. Simple. Having known the function x=g(x) we can choose a point x0 nearby a root, as may be known from intermediate value theorem and put the iteration equation as x1=g(x0),x2=g(x1),x3=g(x2) and so onxn+1=g(xn). In this particular example, taking x0=5, we get, x1=g(5)= 5 2 −8 2 =8.5, x2=g(8.5)= 8.52 −8 2 =32.125, and so on. This form of x=g(x)= x2 −8 2 does not work and divergesxn. Next we take x=h(x)= 2x+8 x . The iteration equation xn+1= 2 xn+8 x starting fromx0=4.5 also does not work. Next let us try x=g(x)=√2x+8 and the iteration equation xn+1=g(x)=√xn+8 . Starting atx0=5 , the successive iterations are :