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Bairstow method example ( Enter your problem )
  1. (Method-1). Algorithm Formula : `b_0=a_0+rb_1+sb_2`
  2. (Method-1). Example-1 `f(x)=x^4-3x^3+3x^2-3x+2` and `r=0.1,s=0.1`
  3. (Method-1). Example-2 `f(x)=x^4-2x^3+6x^2-2x+5` and `r=-1,s=-1`
  4. (Method-2). Algorithm Formula : `b_2=a_2-pb_1-qb_0`
  5. (Method-2). Example-1 `f(x)=x^3+x^2-x+2` and `r=-0.9, s=0.9`
  6. (Method-2). Example-2 `f(x)=x^4+x^3+2x^2+x+1` and `r=0.5, s=0.5`

2. (Method-1). Example-1 `f(x)=x^4-3x^3+3x^2-3x+2` and `r=0.1,s=0.1`
(Next example)

1. (Method-1). Algorithm Formula : `b_0=a_0+rb_1+sb_2`





Algorithm
Bairstow method Algorithm
Bairstow method is an iterative method used to find all the roots of a polynomial both the real and complex. It is based on the idea of synthetic division of the given polynomial by a quadratic function.
The method divides a polynomial, `f_n(x)=a_0 + a_1x + a_2x^2 + ... + + a_nx^n` by a quadratic function `x^2-rx-s`, where `r` and `s` are guessed values.
This division gives us a new polynomial `f_(n-2)(x)=b_2 + b_3x + b_4x^2 + ... + + b_nx^(n-2)`
and the remainder polynomial `R(x)=b_1(x-r)+b_0`
Since the quotient `f_(n-2)(x)` and the remainder `R(x)` are obtained by synthetic division the coefficients `b_i` (where i=0,1,...n) can be obtained by the following recurrence relationship
`b_n=a_n`
`b_(n-1)=a_(n-1)+rb_n`
`b_(i)=a_(i)+rb_(i+1)+sb_(i+2)` for (i=n-2,...,0)
If `x^2-rx-s` is an exact factor of `f_n(x)` then the remainder `R(x)` is 0 and the real/complex roots of `x^2-rx-s` are the roots of `f_n(x)`. It may be noted that `x^2-rx-s` is considered based on some guess values for `r,s`. So Bairstow's method reduces to determining the values of `r` and `s` such that `R(x)=0`. Since both `b_0` and `b_1` are functions of `r` and `s`, they can be expanded using Taylor series of `b_0`, `b_1` as
`b_0(r+Delta r,s+Delta s) = b_0+(del b_0)/(del r)Delta r+(del b_0)/(del s)Delta s`
`b_1(r+Delta r,s+Delta s) = b_1+(del b_1)/(del r)Delta r+(del b_1)/(del s)Delta s`
By setting both equations equal to 0
`(del b_0)/(del r)Delta r+(del b_0)/(del s)Delta s = -b_0`
`(del b_1)/(del r)Delta r+(del b_1)/(del s)Delta s = -b_1`
To solve the system of equations above, we need partial derivatives of `b_0` and `b_1` with respect to `r` and `s`. Bairstow has shown that these partial derivatives can be obtained by a synthetic division of the `b's`, means by replacing `a_i's` with `b_i's`, and `b_i's` with `c_i's`, such that
`c_n=b_n`
`c_(n-1)=b_(n-1)+rc_n`
`c_(i)=b_(i)+rc_(i+1)+sc_(i+2)` for (i=n-2,...,1)
where `(del b_0)/(del r)=c_1, (del b_0)/(del s)=(del b_1)/(del r)=c_2, (del b_1)/(del s)=c_3`
This system of equations may be written as.
`c_1 Delta r+c_2 Delta s = -b_0`
`c_2 Delta r+c_3 Delta s = -b_1`
These equations can be solved for `Delta r` and `Delta s`, which can be used to improved the initial guess of `r` and `s`.
At each step, an approximate error in r and s estimated as
`|epsilon_(a,r)|=|(Delta r)/(r)|xx100%`
`|epsilon_(a,s)|=|(Delta s)/(s)|xx100%`
If `|epsilon_(a,r)|>epsilon_(s)` or `|epsilon_(a,s)|>epsilon_(s)`, where `epsilon_(s)` is the iteration stopping error, then we repeat the process with the new guess `(r+Delta r,s+Delta s)`, Otherwise the roots of `f_(n)(x)` can be determined by
`x_(1,2)=(r+-sqrt(r^2+4s))/2`
If we want to find all the roots of `f_(n)(x)` then we have the following 3 possibilities
1. If quotient polynomial `f_(n-2)(x)` is a third order or higher order polynomial then apply the Bairstow's method again to the quotient polynomial. The previous values of `(r,s)` can be used as the starting guesses for this iteration.
2. If quotient polynomial `f_(n-2)(x)` is a quadratic function then use quadratic formula to obtain the remaining two roots of `f_(n)(x)`.
3. If quotient polynomial `f_(n-2)(x)` is a linear function say `ax+b=0` then the remaining single root is `x=(-b)/a`



This material is intended as a summary. Use your textbook for detail explanation.
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2. (Method-1). Example-1 `f(x)=x^4-3x^3+3x^2-3x+2` and `r=0.1,s=0.1`
(Next example)





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