Presentation on theme: "If the gradient of the curve close to the intersection is close to zero then the iteration will home in on the solution incredibly quickly. Solve x = x."— Presentation transcript:
If the gradient of the curve close to the intersection is close to zero then the iteration will home in on the solution incredibly quickly. Solve x = x 3 – x 2 – x + 2
The case when F`( ) = 0 If F`( ) = 0 then the relationship cannot be used as it results in e r+1 = 0. e r+1 e r 2 In this case This is called quadratic convergence as it contains a square term
rxrxr error E E E The table shows the values of x 1, x 2, x 3 etc and the associated errors e 1, e 2, e 3 Column 4 shows that the ratio is approximately constant showing that the sequence of iterations is quadratic and converges very quickly. This is because F` which is the condition for quadratic convergence. The spiral of convergence reaches the root very quickly because the gradient of the curve is close to zero. Column 5 shows that the ratio is not constant. x = x 3 – x 2 – x + 2
If e 1 and e 2 are known then e 3 can be estimated = This is approx the value of e 3 in the table below rxrxr error E E E
What equation does this solve Remove the subscripts
At the intersection the gradient of F(x) is approximately zero This is the condition for quadratic convergence
It is given that f(x) = x 2 – sin x. (i)The iteration x n+1 =, with x 1 = 0.875, is to be used to find a real root, α, of the equation f(x) = 0. Explain why this rearrangement is valid. (ii)Find x 2, x 3 and x 4, giving the answers correct to 6 decimal places. (iii)Give a reason at to whether the iteration shows linear or quadratic convergence (iv)The error e n is defined by e n = α – x n. Given that α = , correct to 6 decimal places, find e 3 and e 4. (v)Given that g(x) =, use e 3 and e 4 to estimate g′(α). (vi)Find the % error using this method compared to the exact value of g′(α).