The Newton Raphson method is also called as ____________a)Diameter methodb)Chord methodc)Secant methodd)Tangent method

The Iterative formula for Newton Raphson method is given by __________a)𝑥(1)=𝑥(0)+𝑓′(𝑥(0))𝑓(𝑥(0))x(1)=x(0)+ f(x(0))f ′ (x(0))​ b)𝑥(1)=𝑥(0)−𝑓(𝑥(0))𝑓′(𝑥(0))x(1)=x(0)− f ′ (x(0))f(x(0))​ c)𝑥(1)=𝑥(0)+𝑓(𝑥(0))𝑓′(𝑥(0))x(1)=x(0)+ f ′ (x(0))f(x(0))​ d)𝑥(1)=𝑥(0)−𝑓′(𝑥(0))𝑓(𝑥(0))x(1)=x(0)− f(x(0))f ′ (x(0))​

The basic Newton method is used for the solution of nonlinear problems

Algorithm 3.18: parallel chord methodInput: initial guess x0 ∈ R, function f : R → R, derivativef ′ : R → R, tolerance εtol > 0Output: approximation x ∈ R of a zero x∗ of f1 x

In Newton Raphson method, if the curve 𝑓(𝑥)f(x) is constant then ________.a)𝑓′(𝑥)=0f ′ (x)=0b)𝑓(𝑥)=0f(x)=0c)None of the mentionedd)𝑓′(𝑥)=𝑐f ′ (x)=c

Bairstow’s method is an iterative approach related loosely to both the Müller and Newton-Raphson methods. Before launching into a mathematical description of the technique,recall the factored form of the polynomial,f5(x) = (x + 1)(x − 4)(x − 5)(x + 3)(x − 2) (7.28)If we divided by a factor that is not a root (for example, x + 6), the quotient would be afourth-order polynomial. However, for this case, a remainder would result.On the basis of the above, we can elaborate on an algorithm for determining a root ofa polynomial: (1) guess a value for the root x = t, (2) divide the polynomial by the factorx − t, and (3) determine whether there is a remainder. If not, the guess was perfect and theroot is equal to t. If there is a remainder, the guess can be systematically adjusted and theprocedure repeated until the remainder disappears and a root is located. After this is ac-complished, the entire procedure can be repeated for the quotient to locate another root.Bairstow’s method is generally based on this approach. Consequently, it hinges on themathematical process of dividing a polynomial by a factor. Recall from our discussion ofpolynomial deflation (Sec. 7.2.2) that synthetic division involves dividing a polynomial bya factor x − t. For example, the general polynomial [Eq. (7.1)]fn(x) = a0 + a1x + a2x2 +···+ an xn (7.29)can be divided by the factor x − t to yield a second polynomial that is one order lower,fn−1(x) = b1 + b2x + b3x2 +···+ bn xn−1 (7.30)with a remainder R = b0, where the coefficients can be calculated by the recurrence rela-tionshipbn = anbi = ai + bi+1t for i = n − 1 to 0Note that if t were a root of the original polynomial, the remainder b0 would equal zero.To permit the evaluation of complex roots, Bairstow’s method divides the polynomialby a quadratic factor x2 − rx − s. If this is done to Eq. (7.29), the result is a new polyno-mialfn−2(x) = b2 + b3x +···+ bn−1xn−3 + bn xn−2with a remainderR = b1(x − r) + b0 (7.31)As with normal synthetic division, a simple recurrence relationship can be used to performthe division by the quadratic factor:bn = an (7.32a)cha01064_ch07.qxd 3/20/09 11:53 AM Page 185186 ROOTS OF POLYNOMIALSbn−1 = an−1 + rbn (7.32b)bi = ai + rbi+1 + sbi+2 for i = n − 2 to 0 (7.32c)The quadratic factor is introduced to allow the determination of complex roots. Thisrelates to the fact that, if the coefficients of the original polynomial are real, the complexroots occur in conjugate pairs. If x2 − rx − s is an exact divisor of the polynomial, complexroots can be determined by the quadratic formula. Thus, the method reduces to determin-ing the values of r and s that make the quadratic factor an exact divisor. In other words, weseek the values that make the remainder term equal to zero.Inspection of Eq. (7.31) leads us to conclude that for the remainder to be zero, b0 andb1 must be zero. Because it is unlikely that our initial guesses at the values of r and s willlead to this result, we must determine a systematic way to modify our guesses so that b0 andb1 approach zero. To do this, Bairstow’s method uses a strategy similar to the Newton-Raphson approach. Because both b0 and b1 are functions of both r and s, they can beexpanded using a Taylor series, as in [recall Eq. (4.26)]b1(r + r,s + s) = b1 +∂b1∂r r +∂b1∂ssb0(r + r,s + s) = b0 +∂b0∂r r +∂b0∂ss (7.33)where the values on the right-hand side are all evaluated at r and s. Notice that second- andhigher-order terms have been neglected. This represents an implicit assumption that −r and−s are small enough that the higher-order terms are negligible. Another way of expressingthis assumption is to say that the initial guesses are adequately close to the values of r ands at the roots.The changes, r and s, needed to improve our guesses can be estimated by settingEq. (7.33) equal to zero to give∂b1∂r r +∂b1∂ss = −b1 (7.34)∂b0∂r r +∂b0∂ss = −b0 (7.35)If the partial derivatives of the b’s can be determined, these are a system of two equationsthat can be solved simultaneously for the two unknowns, r and s. Bairstow showed thatthe partial derivatives can be obtained by a synthetic division of the b’s in a fashion simi-lar to the way in which the b’s themselves were derived:cn = bn (7.36a)cn−1 = bn−1 + rcn (7.36b)ci = bi + rci+1 + sci+2 for i = n − 2 to 1 (7.36c)where ∂b0/∂r = c1, ∂b0/∂s = ∂b1/∂r = c2, and ∂b1/∂s = c3. Thus, the partial derivativesare obtained by synthetic division of the b’s. Then the partial derivatives can be substitutedinto Eqs. (7.34) and (7.35) along with the b’s to givec2r + c3s = −b1c1r + c2s = −b0cha01064_ch07.qxd 3/20/09 11:53 AM Page 1867.5 BAIRSTOW’S METHOD 187These equations can be solved for r and s, which can in turn be employed to improve theinitial guesses of r and s. At each step, an approximate error in r and s can be estimated, as in|εa,r| =rr100% (7.37)and|εa,s| =ss100% (7.38)When both of these error estimates fall below a prespecified stopping criterion εs, the val-ues of the roots can be determined byx = r ± √r 2 + 4s2 (7.39)At this point, three possibilities exist:1. The quotient is a third-order polynomial or greater. For this case, Bairstow’s methodwould be applied to the quotient to evaluate new values for r and s. The previous valuesof r and s can serve as the starting guesses for this application.2. The quotient is a quadratic. For this case, the remaining two roots could be evaluateddirectly with Eq. (7.39).3. The quotient is a first-order polynomial. For this case, the remaining single root can beevaluated simply asx = −sr