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CHE/COSC/MATH 4340-01 Numerical Analysis
Solution of Writing Assignment 04 Due Date: Thursday, 02/26/
Problem 1. Using Taylor expansions for f (x + h) and f (x + k), derive the following approximation to f ′(x): f ′(x) ≈ k^2 f (x + h) − h^2 f (x + k) + (h^2 − k^2 ) f (x) (k − h)kh
Answer: By Taylor’s Theorem
f (x + h) = f (x) + h f ′(x) +
f ′′(x)h^2 +
f ′′′(ξh)h^3
f (x + k) = f (x) + k f ′(x) +
f ′′(x)k^2 +
f ′′′(ξk)k^3 ,
where ξh is between x and x + h and ξk is between x and x + k. Multiply the first equation by k^2 and the second equation by h^2 and subtract one from the other yielding
k^2 f (x + h) − h^2 f (x + k) = (k^2 − h^2 ) f (x) + kh(k − h) f ′(x) +
( f ′′′(ξh) − f ′′′(ξk))(kh)^2 (k − h). (0.2)
Solve for f ′(x):
f ′(x) = k^2 f (x + h) − h^2 f (x + k) + (h^2 − k^2 ) f (x) (k − h)kh
( f ′′′(ξk) − f ′′′(ξh))(kh), (0.3)
from which we obtain the desired result.
Problem 2. If the Secant method is applied to f (x) = x^2 − 2 with x 0 = 0 and x 1 = 1, what is x 2? Answer: Using Secant Method we have
xn+ 1 = xn − xn − xn− 1 x^2 n − x^2 n− 1
(x^2 n − 2 )
= xn −
xn − xn− 1 (xn − xn− 1 )(xn + xn− 1 ) (x^2 n − 2 )
= xn − x^2 n − 2 xn + xn− 1
Using this relation we get
x 2 = x 1 − x^21 − 2 x 1 + x 0
Problem 4. Assume that r is a zero of multiplicity 2 of the polynomial p(x).
- Prove that Newton’s method converges linearly. Hint: Appropriately use Taylor expansions p(r) = p(xn − en) and p′(r) = p′(xn − en). As usual en = xn − r.
Answer: Notice the error equation associated with Newton’s method can be written as
en+ 1 = en − p(xn) p′(xn)
en p′(xn) − p(xn) p′(xn)
Moreover, by Taylor’s expansion
0 = p(r) = p(xn − en) = p(xn) − p′(xn)en +
p′′(sn)e^2 n 0 = p′(r) = p′(xn − en) = p′(xn) − p′′(tn)en,
each of them gives p′(xn)en − p(xn) =
p′′(sn)e^2 n p′(xn) = p′′(tn)en,
Putting these two expressions into the error equation yields
en+ 1 =
p′′(sn)e^2 n p′′(tn)en
[ (^) p′′(sn) 2 p′′(tn)
]
en. (0.10)
This last equation establishes the linear convergence of Newton’s method for finding a zero of multiplicity 2.
- Now consider a variation of Newton’s method
xn+ 1 = xn − 2 p(xn) p′(xn)
Prove that it converges quadratically to r. Hint: Appropriately use Taylor expansions p(xn) = p(r + en) and p′(xn) = p′(r + en). Answer: For this variation, the error equation becomes
en+ 1 = en − 2 p(xn) p′(xn)
en p′(xn) − 2 p(xn) p′(xn)
By Taylor’s expansion
p(xn) = p(r + en) = p(r) + p′(r)en +
p′′(r)e^2 n +
p′′′(zn)e^3 n
=
p′′(r)e^2 n +
p′′′(zn)e^3 n,
p′(xn) = p′(r + en) = p′(r) + p′′(r)en +
p′′′(wn)e^2 n
= p′′(r)en +
p′′′(wn)e^2 n.
Using these two we get
en p′(xn) − 2 p(xn) = p′′(r)e^2 n +
p′′′(wn)e^3 n − p′′(r)e^2 n −
p′′′(zn)e^3 n =
[ 1
p′′′(wn) −
p′′′(zn)
]
e^3 n.
Furthermore, we also know that p′(xn) = p′(r + en) = p′(r) + p′′(an)en = p′′(an)en. Putting these last two identities to the error equation gives
en+ 1 =
[
1 2 p ′′′(wn) − 1 3 p ′′′(zn)
]
e^3 n p′′(an)en
[
1 2 p ′′′(wn) − 1 3 p ′′′(zn)
]
p′′(an) e^2 n. (0.14)
As the method converges, we know that wn → r, zn → r, and an → r, from which we deduce that
en+ 1 ≈
[ (^) p′′′(r) 6 p′′(r)
]
e^2 n, (0.15)
which is a statement of quadratic convergence.
