## Home Work TWO

1. 2.4

a      b

Start bisection from

1. one root
2. there are two roots with , and 0 is between these two roots
3. , as in the previous example
4. , as there is one root only
5. one root
6. , two roots with c

For the example the Newton formula is with the good starting point . This starting point will quickly leed to the solution

d

For the example the Secant formula is with the good starting points e Solutions are stated above, the example of bisection and Newton codes were provided during lectures. Good stopping criteria would be, for example, 2. 2.9 a-c

a

Suppose is given, and we now how to calculate , we want to sove for the equation . In other words, we want to find 0 of . Then the Newtow formula is indeed which leads to the formula in the text book.

b

We now need to evaluate , using only function. This gives Similarly, to evaluate we use c

Similarly, to calculate , we use Also, to calculate , we use 3. 2.10

a To answer this question, please think about convergence rates. If convergence rate is quadratic, as it is for the Newtwon method, then the error is squared at each iteration. Since the in the end of iterations, then in Method one the error is at first iteration, at second and at the third iteration. Since the error is squared, than with out shadow of a doubt the first method is Newton iterations

b I would say that the Bisection is the Method 3, since the error is halved at each iteration. Indeed, the error is initially, then , and then which is exactly a half at each iteration

c Method 3 can not be Secant, since the error is only halved at each iteration. Secant should converge faster, so I would guess that Secant is Method 2.

4. 2.22

a

Approximating the function by a second degree polynomial, we obtain We now solve where this crosses the axis to obtain There is a , since one needs to solve quadratic equation to obtain .

b The biggest disadvantage of this formula is that it has two roots. This is natural, since the parabola may cross the axis in two places. The second big disadvantage of this formula is that it may not cross at all, i.e. the quadratic equation may have no real roots. Finally, the third big disadvantage is that the formula contains , i.e. one need to know analytically the second derivative of .

c

We need to chose a solution that is closest to the solution one would obtain if were to be equal to zero. In other words, one should choose a solution that is closest to the solution obtained by the Newton method. By using the Taylor expansion, we see that the correct choice is
“+” if “-” if In other words, the sign in front of the radical should be the sign of . Then this formula will reduce to Newton formula (PLEASE MAKE SURE YOU CAN ACTUALLY SEE THIS!).

d The fastest way to derive the Halley method is to apply the Newton formula to the function since any root of that is not a root of is also a root of .

e Since the Newton formula is we can equivalently rewrite it as The resulting formula for Halley method is therefore  5. (2.27)

(a)

Since we get and Now assuming with , and keeping terms only up to inclusively, we divide these two expressions to obtain, in leading order, b

If then the above expressions are replaced by and so that Consequently, the formula for Newton method gives , which implies linear convergence.

6. (2.28)

(a) The Newton method for is given by simplifies to (b)

Using with and we change variables as we get Here the term is not taken into account.

c

Since we obtain This compares well with .

d For the case of , we write it as , to obtain This evaluates to which is remarkably close to .