CSIR NET MATHEMATICS | Numerical Analysis | Iterative Method Q.97(A) Q.101(B) Q.94(C)

Welcome friends, in this video we are going to discuss the problem of Numeric Analysis from CSI NET June 2019. Example number 94 from booklet C, Mars 4.75. Consider the equation x square plus ax plus b equal to 0, which has two real roots alpha and beta, then which of the following iteration scheme converges when x0 is chosen sufficiently close to alpha. First option xn plus 1 equal to minus axn plus b divided by xn if modulus alpha greater than modulus beta. Second option xn plus 1 minus xn square plus b divided by a if modulus alpha greater than 1. Third option xn plus 1 equal to minus b divided by xn plus a if modulus alpha less than modulus beta. fourth option xn plus 1 equal to minus xn square plus b divided by a if 2 alpha less than alpha plus beta. Here we are given quadratic equation x square plus ax plus b equal to 0. Now we know if the standard form ax square plus bx plus c equal to 0 then the roots are minus b plus or minus square root b square minus 4ac divided by 2. So therefore the roots of this given equation x square plus ax plus b equal to 0 is x is equal to minus a plus or minus square root a square minus 4b divided by 2. So if we compare with the standard equation then b will be a c will be b and a will be 1 and we are able to generate this solution. So first root alpha equal to minus a plus square root a square minus 4b divided by 2 and beta equal to minus a minus square root a square minus 4b divided by 2. Now alpha plus beta if we take the addition of 2 root then we have minus a minus a so minus 2a divided by 2 so we have minus a and plus square root cancel is the minus square root so alpha plus beta equal to minus a alpha into beta so we have after the multiplication a square by 4 minus square root a square minus 4b square divided by 4 therefore we have a square by 4 minus a square minus 4b divided by 4 so a square by 4 cancel with the minus a square by 4 and minus minus plus 4b divided by 4 4 4 cancel then we have So, we have two equations alpha plus beta equal to minus a and alpha into beta equal to b. So, we will use this result for the answer. Now, we will check all the options one by one. Before that just we recall the convergence criteria for iteration method is modulus g dash x less than 1. First option here gx is equal to as a function of x, xn replaced by x then we have minus ax plus b divided by x. If we rewrite then minus into a plus d divided by x. Now g dash x is equal to b upon x square. The differentiation of a is 0. And differentiation of 1 upon x is minus 1 upon x square. So minus minus plus so we have b upon x square. Now condition is g dash x less than 1. But we want the answer or x 0 close to alpha. therefore g dash alpha less than so g dash alpha is equal to b divided by alpha square less than but b equal to alpha into beta we already proved therefore alpha cancel from the numerator and denominator so beta divided by alpha less than 1 and hence modulus beta less than alpha in other alpha greater than modulus beta and hence the option one is correct now for option two and four have the same function therefore we check both the options simultaneously GX is equal to minus X and replaced by X so X square plus B divided by a if we rewrite then minus X square plus a plus B by a take the differentiation then we have minus 2x divided by a modulus ZS X less than 1 x0 is near to alpha therefore g-alpha less than 1 minus 2 alpha divided by a less than 1 therefore 2 alpha less than modulus a or in other words modulus alpha less than a by 2 so this inequality depend on the value of a if a is less than 2 if it is a equal to 1 then this modulus alpha less than 1 by 2 and therefore the second equality wrong if a is less than 2 and so the option number 2 is not always correct so it is a wrong statement. Now if we continue from this inequality and a equal to minus alpha plus beta so we put it here then 2 alpha less than alpha plus beta and which is a option number 4 so therefore option number 4 is correct now option number 3 for the option number 3 x and replaced by x then gx is equal to minus b divided by x plus a find g dash x so b divided by x plus a whole square modulus g dash x less than 1 close to alpha so modulus g dash alpha less than 1 if we replace x by alpha then b divided by x plus a power 2 is equal to b divided by alpha plus a power. Now substitute the value of a and b, b equal to alpha beta and a equal to minus alpha plus beta whole square. Alpha cancel with the minus alpha then alpha beta divided by beta square therefore beta cancel with the beta square 1 element then alpha by beta less than 1 and hence modulus alpha less than modulus beta and which is option number three therefore option number three is correct answer so this example has three correct answer first third and fourth thanks for watching for more videos like share subscribe press the bell icon of next gen math