Numerical AnalysisMath135A (001) Fall 2010

General philosophy

In this class we will derive algorithms (this is the easy part), we will prove rigorously that these algo-rithms converge to the right answer (this is harder) and we will study rigorously how fast they converge tothe right answer (this is also hard). A good background in calculus is needed: we will use again andagain tools such as Taylor’s theorem in order to derive error estimates.

There will be lots of coding too. The homework will be computationally intensive. You will be asked towrite codes during the midterms and the final exam (with pen and paper, not with a computer).

Main Topics covered (not inclusive)

1. Solving nonlinear equations of one variableWe will derive algorithms to solve f(x) = 0. The most important one is Newton’s method. Bigemphasis will be given on deriving rigorous error estimates.

2. IEEE Floating Point Arithmetic.We will learn how numbers are stored in the computer and what are the problems arising from thefinite precision of the computer.

3. Polynomial interpolationHow to approximate a function by a polynomial related and error estimates. This topic is needed inorder to do Numerical integration (the next topic).

4. Numerical integration

We will derive algorithms to compute∫ b

af(x)dx. Big emphasis will be given on deriving rigorous error

estimates.

5. Solving linear Systems of n equationsWe will write codes to solve Ax = b with Gaussian elimination.

MATLAB and SCILAB

MATLAB is one of the most popular software in science and engineering. It is very robust, fast and easy touse. You can buy a student version for about $100. Fortunately there is also a “free version of MATLAB”called SCILAB which does basically the same job than MATLAB but is a little less user friendly. You candownload it for free at www.scilab.org/ and it will be enough for all what we need to do.

You will be taught how to use MATLAB and SCILAB (they are basically the same language). Previousfamiliarity with these packages is NOT REQUIRED for enrollment. The first dicussion will be acrash course in MATLAB/SCILAB. Don’t miss it!

Your homework must be written in MATLAB or SCILAB. During exams you will be asked to writesome MATLAB/SCILAB codes.

1

Grading Scheme:Homework: 8%2 Midterms: 46% (23% each)Final: 46%

Tentative dates of exams:Midterm 1 : Wednesday, October 20Midterm 2: Wednesday, November 17Final : Friday, December 10

Homework:There will be about 8 homework (≈ 1% each).All numerical computation should be done in MATLAB/SCILAB.No late homework will be accepted.

Office hours:To be determined

2

Homework # 1

Due Monday, 10/04

NOTE: Each time you are ask to write a code which gives the solution with accuracy ε, it means that youhave to use the stopping criterium

|pn − pn−1| ≤ ε

Note that this does not guaranty that |pn − p| ≤ ε !!! So you are actually not sure to have found the solutionwith accuracy ε !!!

1. Let f(x) = x− 2−x.

(a) Plot with MATLAB the function f on [−5, 5]. Use the MATLAB command “GRID ON” so thatthere is a grid on the plot. If you are using SCILAB typea=get(”current axes”);a.grid=[1 1];Provide this plot. Looking at this plot, is there a zero in [0,1]? Zoom in many times and give anestimate for this zero.

(b) Use the bisection method to find the zero of f with accuracy 10−5 (provide the MATLAB code).Hoe many iterations were needed?

(c) Note that in the bisection method, if p0 stands for the initial iterate (b+ a)/2, then we have that|pn − pn−1| = (b − a)/2n+1. Use this to find analytically at what iteration your code will stop.Compare with the previous question.

2. (a) Sketch the graph of y = x and y = tanx.

(b) Use the Bisection method to find an approximation to within 10−5 to the first positive x satisfyingx = tanx (provide the MATLAB code)

3. Suppose f is a continuous function on [a, b] with f(a) > a and f(b) < b. Show that f has a fixed pointin [a, b].

4. Let f(x) = ln(x+ 2)

(a) Sketch the graph of f(x).

(b) Find f([0, 2])

(c) Is it true that f([0, 2]) ⊂ [0, 2]? What can you conclude about the existence of a fixed point?

(d) Find maxx∈[0,2] |f ′(x)|.(e) Find a number k ∈ (0, 1) such that

|f ′(x)| ≤ k for all x ∈ [0, 2]

(f) What can you conclude about using a fixed point iteration.

(g) Find the fixed point with accuracy 10−3. How many iterations were needed? (provide the MAT-LAB code)

(h) Use the theoretical result given in class (|p− pn| ≤ kn(b−a)) to estimate the number of iterationsrequired to achieve 10−3 accuracy. Why is this number different than the one obtained in g). ?

5. Let f(x) = e−x

1

(a) Sketch the graph of f(x).

(b) Find f([ 13 , 1])

(c) Is it true that f([ 13 , 1]) ⊂ [ 13 , 1]? What can you conclude about the existence of a fixed point?

(d) Find maxx∈[ 13 ,1] |f ′(x)|.

(e) Find a number k ∈ (0, 1) such that

|f ′(x)| ≤ k for all x ∈[

13, 1

](f) What can you conclude about using a fixed point iteration.

(g) Find the fixed point with accuracy 10−3. How many iterations were needed? (provide the MAT-LAB code)

(h) Use the theoretical result given in class (|p− pn| ≤ kn(b−a)) to estimate the number of iterationsrequired to achieve 10−3 accuracy.

2

Homework # 2

Due Monday, 10/11

1. Find the solution ofx3 = x2 + x+ 1

using:

(a) the bisection method (What initial interval did you choose? How did you choose it?)

(b) Newton’s method (What initial iterate did you choose? How did you choose it?)

(c) the secant method (What initial iterates did you choose? How did you choose them?)

In each case, use the stopping criterium |pn − pn−1| ≤ 10−6. How many iterates were needed in eachcase to reach this accuracy. Which method is the fastest to converge? The slowest?

(Hint: Before to do any coding, plot the function with matlab so that you can see roughly whereis the zero.)

2. Repeat problem 1., but this time with the equation

ex =1

0.1 + x2

3. Newton’s method is the commonly used method for calculating square roots on a computer.

(a) What equation would you solve in order to find√a?

(b) Show that in this case, Newton’s method reduces to the following fixed point iteration:

pn+1 =12

(pn +

a

pn

)(1)

(c) Explain with a picture why, for any initial iterate p0 > 0, Newton’s method will converge (I amnot looking for a precise answer: just draw a picture and write something short).

(d) Use iteration (1) to find√

57 with accuracy 10−10. Start with the initial iterate p0 = 57. Howmany iterates were needed?

4. In class we have proven that Newton method converges by viewing it as a fixed pointiteration. In this problem, you will prove it in a different way, without using a fixed pointiteration.

In all the problem, we will assume that

f(x), f ′(x), f ′′(x) are continousf(p) = 0, f ′(p) 6= 0

The sequence pn is defined by

pn+1 = pn −f(pn)f ′(pn)

We will prove the following statement: If p0 is chosen sufficiently close to p, then limn→∞ pn = pand the iterates have order of convergence 2.

1

(a) Use Taylor’s theorem to show that there exist a numbers ξn between p and pn such that

p− pn+1 = −(p− pn)2f ′′(ξn)2f ′(pn)

(2)

(b) Explain why it is possible to pick a (small) interval I = [p− δ, p+ δ] on which f ′(x) 6= 0.(c) Since f ′(x) 6= 0 on I, then minx∈I |f ′(x)| 6= 0 and therefore the number

M =maxx∈I |f ′′(x)|2 minx∈I |f ′(x)|

.

is well defined. Show that if |pn − p| ≤ δ, then

|p− pn+1| ≤M |p− pn|2 (3)

and therefore, multiplying both side by M :

M |p− pn+1| ≤ (M |p− pn|)2 (4)

(d) Pick the initial iterate p0 such that

|p− p0| ≤ δ (5)

|p− p0| <1M

(6)

i. Use inequalities (3) and (6) to prove that |p− p1| ≤ 1M .

ii. Use inequalities (3), (5) and (6) to prove that |p− p1| ≤ δ. ( There is a hint at the end of theproblem. )

(e) We have just proven that:

If |p− p0| ≤ δ and |p− p0| <1M

(7)

then |p− p1| ≤ δ and |p− p1| <1M

(8)

We can do it again to get

|p− p2| ≤ δ and |p− p2| <1M

and of course, iterating the argument:

|p− pn| ≤ δ and |p− pn| <1M

(9)

Use (4) and part of (9) to prove that

M |p− pn| ≤ (M |p− p0|)2n

(10)

Make sure to explain why (9) is needed.(f) Use all the above to prove the following statement: “If p0 is chosen sufficiently close to p,

then limn→∞ pn = p and the iterates have order of convergence 2 ”

Hint for (d)ii): Note that, because of (6), M |p− p0| < 1 and therefore, using (3), we get

|p− p1| ≤ (M |p− p0|) |p− p0| ≤ |p− p0|

Then you are almost done...

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Homework #3

Due Monday, 10/18

1. We have seen in class that, if f(x) has a root of multiplicity m ≥ 2, then Newton’s method convergesonly with order 1. Here is a modified version of Newton’s method:

pn+1 = pn −mf(pn)f ′(pn)

Show that, if it converges, then it has to converge with order 2.

2. Let f(x) = ex − x− 1

(a) Plot f(x) with matlab. Where is the zero (just guess with the picture)?

(b) Show that the zero is actually where you thought it was, and show that it has multiplicity 2.

(c) How many iterations are needed with Newton’s method to find this zero with accuracy 10−10?(start with p0 = 1)

(d) How many iterations are needed with the modified version of Newton’s method that we have seenin problem 1.?

3. Prove by induction that (11111 . . . 1)2 = 2n − 1 (the binary number contain n digits, all are equal toone).

4. Below is the IEEE floating point representation of three numbers

0 10000001010 10010011000000000000000000 00000000000000000000000000 (1)1 10000001010 10010011000000000000000000 00000000000000000000000000 (2)0 01111111111 01010011000000000000000000 00000000000000000000000000 (3)

(a) What are these numbers?

(b) Complete the following sentences:The “spacing” between number (1) and the closest next machine number is 2−m where m =The “spacing” between number (3) and the closest next machine number is 2−m where m =Show your reasoning (no hand-waving argument, be clear).

5. (a) What is the largest possible machine number?

(b) What is the smallest possible machine number?

(c) What is the smallest possible “spacing” between two machine numbers?

(d) What is the IEEE floating point representation of one billion? What is the “spacing” betweenone billion and the next machine number?

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Homework # 4

Due Friday, 10/29

1. (a) Use Lagrange formula to find the unique polynomial of degree ≤ 2 which passes through the threepoints (0,1), (-1,2) and (1,3).

(b) Do some algebra to write the polynomial as follow:

p(x) = a2x2 + a1x + a0

2. Consider the function f(x) = e−x. We are going to approximate it with a polynomial of degree 2 andwe are going to study how good is this approximation.

(a) Find a polynomial p2(x) = a2x2+a1x+a0 which approximate f(x) on the interval [−1, 1]. Choose

your polynomial so that p2(−1) = f(−1), p2(0) = f(0) and p2(1) = f(1). Use Lagrange formulaand then do some algebra to find a1, a2 and a3.

(b) Plot both f and p2 on matlab on the interval let say [−2, 2].(c) Write a matlab code to find the maximum distance between the line representing the polynomial

and the line representing the function f on the interval [−1, 1]. In other words, define the errorby E(x) = |f(x) − p2(x)| and find the maximum value of E on the interval [−1, 1] using yourmatlab code.

maxx∈[0,1]

E(x) =

(d) Use the error formula that we derived in class to get an upper bound on the error on the interval[−1, 1] of the type

E(x) ≤ C|(x− a)(x− b)(x− c)| for all x ∈ [−1, 1]

I want numerical values for C, a,b and c. I want the best C possible.(e) Make a sketch of the function (x− a)(x− b)(x− c) from the previous question. Find analytically

where is its max on [−1, 1] and what is this max. Then write

E(x) ≤M for all x ∈ [−1, 1]

I want a numerical value for M .(f) Does c) and e) agree with one another?(g) Use Taylor’s theorem to find the best polynomial of degree 2 which approximate f(x) around

x = 0. This means that the polynomial and f must have the same value at x = 0 and theirderivatives must match up to order 2 at x = 0. Estimate the error on the interval [−1, 1] usingthe error formula coming from Taylor’s theorem:

E(x) ≤ C(x− a)3 for all x ∈ [−1, 1]

I want numerical values for C and a. I want the best possible C. Then Use this to find a boundfor the error on the interval [−1, 1]

E(x) ≤M for all x ∈ [−1, 1]

I want a numerical value for M .(h) Which of the two polynomials do the the best approximation? The one with one node or the one

with three nodes?

1

Homework # 5

Due Friday, 11/05

1. (a) Write down the statement of the intermediate value Theorem.

(b) Use the intermediate value theorem to prove

Theorem 1. Suppose g(x) and w(x) is continuous on [a, b]. Suppose also that w(x) ≤ 0 for allx ∈ [a, b]. Then there exists θ ∈ [a, b] such that∫ b

a

g(x)w(x)dx = g(θ)∫ b

a

w(x)dx

Your proof has to be more detailed than the one I did in class. Be more precise on how to usethe intermediate value theorem in order to conclude the proof.

2. In order to derive Simpson’s rule, we proved that, if p(x) is the unique polynomial of degree ≤ 2 whichinterpolates the function f(x) at x0, x1 and x2, then∫ x2

x0

p(x)dx =h

3(f(x0) + 4f(x1) + f(x2)) where h = x1 − x0

In this problem we will do the same for the trapezoidal rule. Let p(x) be the unique polynomial ofdegree ≤ 1 which interpolate the function f(x) at x0 and x1. Prove that:∫ x1

x0

p(x)dx =h

2(f(x0) + f(x1)) where h = x1 − x0

Remark: In class we derived this result by drawing a picture and calculating the area of a trapezoid.Here I am asking you to derive this result in a more rigourous way: First use Lagrange’s formula andthen integrate the polynomial that you obtained.

3. While proving that the error in the trapezoidal rule is O(h2), we used the fact that:∫ x4

x3

(x− x3)(x− x4)dx = −16(x3

4 − 3x24x3 + 3x4x

23 − x3

3

)= − (x4 − x3)

3

6

Check that this identity is indeed true.

4. (a) Compute∫ 1

0exdx by hand.

(b) Let f(x) = ex. What aremax

x∈[0,1]|f ′′(x)| and max

x∈[0,1]

∣∣∣f (4)(x)∣∣∣

(c) Suppose you want to use the trapezoidal rule to compute∫ 1

0exdx. How many nodes points will

be necessary in order to be sure that the error is less than 10−8. To do this you will have to usethe following corollary that was proven in class:

Corollary 1. Suppose f(x) is twice continously differentiable and let M = maxx∈[a,b] |f ′′(x)|,then ∣∣∣∣∣

∫ b

a

f(x)dx− I [a,b]n (f)

∣∣∣∣∣ ≤ (b− a)M12

h2

1

(d) Use the trapezoidal rule to compute∫ 1

0exdx with the number of node points from the previous

question. What is the actual error? Smaller or bigger than 10−8? How can you explain this?Provide your matlab code.

(e) Repeat c) and d) but with Simpson’s rule. Compare the number of node points needed for thistwo method in order to reach a 10−8 precision

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Homework #6

Due Monday, 11/15

1. Let ITn (ex) be the approximation of

∫ 1

0exdx that you get by using the trapezoidal rule with n+ 1 node

points (x0, x1, . . . , xn). Similarly, let ISn (ex) be the approximation of

∫ 1

0exdx that you get by using

Simpsons rule with n + 1 node points. Define the errors:

ETn =

∣∣∣∣∫ 1

0

exdx− ITn (ex)

∣∣∣∣ ESn =

∣∣∣∣∫ 1

0

exdx− ISn (ex)

∣∣∣∣(a) Write a program which compute ET

n and ESn for n = 10, 110, 210, 310, 410, . . . , 2010 and then plot

log ETn versus log n and log ES

n versus log n. Use the command plot(, ,x) so that you just see thedata points. You should obtain lines.

(b) What are the equations of these lines (you can use the matlab command basic fitting and do alinear fitting. You will find this command in “Tools”, just above your plot.)

(c) Why are the slopes −2 and −4? How can you explain this with the error formula we have derivein class.

2. In the previous problem we computed∫ 1

0exdx with the trapezoid rule and Simpsons rule. It was not

very usefull because this integral could be computed by hand. But it was good because it allowedus to study numerically the error. In this problem you will compute

∫ 1

0e−x2

dx with Simpson’s rule.Your answer should be accurate up to 7 digits after the point (i.e: the 7 first digits after the pointshould be exact). Note that this integral can not be computed analitically. In this case we really needa computer.

(a) How many nodes points are necessary in order to be sure that the first seven digits after the pointare correct? (hint: using MAPLE, I found that maxx∈[0,1]

∣∣f (4)(x)∣∣ = 12, where f(x) = e−x2

).

(b) Do the computation with this number of node points.

3. Since Simpsons rule has degree of precision 3, if you compute∫ 1

0x3dx with Simpsons rule, then the

error should be exactly zero. Use Simpson rule to compute∫ 1

0x3dx with n = 100. What is∫ 1

0

x3dx− I [0,1]n (x3)?

How can you explain this difference between the theory and the actual computation?

4. (a) What system of nonlinear equations would you solve in order to find w1, w2, w3, x1, x2 and x3

such that the integration formula

I3(f) = w1f(x1) + w2f(x2) + w3f(x3)

is exactly equal to∫ 1

−1f(x)dx for every polynomial f(x) of degree ≤ 5.

(b) Check that the nodes and weights from the table I gave you in class do satisfy this system.

5. Gaussian Quadrature VS Simpson’s ruleLet

f(x) =1

x + 1.5

1

(a) Sketch or plot with matlab f(x) on [−1, 1].

(b) Compute∫ 1

−1f(x)dx by hand.

(c) Write a code to compute∫ 1

−1f(x)dx using Gaussian Quadrature with 7 nodes points x1, x2, . . . , x7

(hint: store the nodes and weights of the table in two vectors “node(i)” and “w(i)”). What is theerror?

(d) Write a code to compute∫ 1

−1f(x)dx using Simpson rule with nodes points x0, x1, . . . , x6 (so that

you have a total of 7 nodes points, as in question (b)). What is the error?

(e) Which method work the best?

2

Homework #7

Due Monday, 11/29

1. Log-log plot / Numerical errorLet f(x) = ex. Define the errors made by approximating f ′(2) with the forward difference and thesymmetric difference:

EF (h) =∣∣∣∣f ′(2)− f(2 + h)− f(2)

h

∣∣∣∣ ES(h) =∣∣∣∣f ′(2)− f(2 + h)− f(2− h)

2h

∣∣∣∣(a) Write a program which compute EF (h) for h = 10−1, 10−2, . . . , 10−5 and then plot log10 (EF (h))

versus log10(h). Use the command plot(, ,’x’) so that you just see the data points. Do the samething with ES(h).Remark: This time we use log10 (log base 10) instead of log. The reason for doing this issimply because the plot you obtain is more readable. For example, with the symmetric difference,you should obtain a point whose coordinates is something like (-3,-6). This means that, whenh = 10−3, the error is roughly 10−6 (which of course is what you expect since the symmetricdifference is O(h2))

(b) What are the equations of these lines ( you can use the matlab command “basic fitting” and doa linear fitting. You will find this command in “Tools”, just above your plot.)

(c) Why are the slopes 1 and 2? (We know from the theory that EF (h) ≤ M12 h and ES(h) ≤ M2

6 h2.In this question, you can assume that there exists constant C1 and C2 such that EF (h) = C1hand ES(h) = C2h

2.)

(d) Repeat question (a), but this time with h = 10−1, 10−2, . . . , 10−10. In class we did study theeffect of the finite precision of the computer when doing a symmetric difference. Does the log-logplot that you obtain for the symmetric difference roughly match this very rough analysis?

(e) Study the effect of the finite precision of the computer when doing a forward difference (just mimicwhat we did in class with the symmetric difference). From this analysis, what is the optimal hthat one should use? What is the best accuracy that one can get? Does this match what youobserve on your log-log plot?

2. Prove the following theorem and corollary:

Theorem 1. Suppose f is three time continuously differentiable, then there exist ξ1 ∈ [xn−h, xn] andξ2 ∈ [xn − 2h, xn] such that

f ′(xn)− 3f(xn)− 4f(xn − h) + f(xn − 2h)2h

=h2

3(2f ′′′(ξ2)− f ′′′(ξ1))

Corollary 1. Let M = maxx∈[a,b] |f ′′′(x)| and suppose xn = b is the end-point, then

error =∣∣∣∣f ′(xn)− 3f(xn)− 4f(xn − h) + f(xn − 2h)

2h

∣∣∣∣ ≤Mh2

1

Homework #8

Due Friday 12/03

Define the n× n matrix An and the n× 1 vector bn

An =

−2 1 0 0 0 0 . . . 0 0 0 01 −2 1 0 0 0 . . . 0 0 0 00 1 −2 1 0 0 . . . 0 0 0 00 0 1 −2 1 0 . . . 0 0 0 0...

......

......

...0 0 0 0 0 0 . . . 0 1 −2 10 0 0 0 0 0 . . . 0 0 1 −2

bn =

3333.........33

For example,

A6 =

−2 1 0 0 0 01 −2 1 0 0 00 1 −2 1 0 00 0 1 −2 1 00 0 0 1 −2 10 0 0 0 1 −2

A6 =

333333

The goal of this homework will be to solve Anx = bn for large n and study how

long it takes for the computer to do so.

Remark : The matrix An occurs, for example, when one want to solve the heat equation on a rode. Thesolution of Ax = b will be the heat distribution on the rode: x(i) will be the temperature at the nodepoint xi = ih. Typically when dealing with this type of problem, n is very large, that is why we need goodalgorithms. And, as I said in class, Gaussian elimination is a very bad one. For this specific problem, inpractice, people use algorithm that use the fact that “most” of the matrix An is filled with zeros.

1. Write a matlab code init.m (“initialize”) which creates An and bn.

2. Write a code fact.m which do the LU-factorization of a given matrix without pivoting. Given amatrix A, your code should return you a matrix in which the row multiplier mi,j are stored in thelower triangular part of the matrix , and the ui,j are stored in the upper triangular part of the matrix.The code I wrote is 8 lines long. Use your code to find the LU factorization of the matrix A6. Usethe LU command of matlab to check that your code is correct (Matlab has a build-in algorithm whichcompute the LU factorization of a matrix: just type “LU(A)”).

3. Write a code rhs.m (“rhigt hand side”) which use the mi,j computed by fact.m to do the modificationfrom b to g (I am using the same notations than in class). The code I wrote is 5 lines long. Write alast code backsub.m which use the ui,j computed by fact.m and the gi computed by rhs.m to findthe solution x of Ax = b. The code I wrote is 8 lines long. Use your four codes to solve A6x = b6.

1

If you did everything correctly, you just have to type in the matlab window:

initfactrhsbacksub

Check your answer using the “backslash” matlab command: if you want matlab to solve Ax = b,you need to type:

A\b′ if you have enter b as a row vector (b’ means transpose of b)A\b if you have enter b as a column vector.

4. Solve A500x = b500. What are x1 and x100. Use the “backslash” command of matlab to check youranswer.

5. We will now see how long it takes for each of the three part of your code to solve A500x = b500. Asyou know:

• number of operations performed in fact.m ≈ n3

3

• number of operations performed in rhs.m ≈ n2

2

• number of operations performed in backsub.m ≈ n2

2

(Here we count only the multiplications and divisions; the number of additions and subtractions arethe same.) So of course we expect fact.m to take much more time than rhs.m and backsub.m. Usethe command “tic” and “toc” of matlab to find the computing time for each of the three parts of yourcode (write tic at the beginning of each code, and toc at the end).

6. We will now solve A1000x = b1000. Note that n is twice larger than in the previous question. Thenumber of operations performed by fact.m should be how many times larger? Use the tic and toccommand of matlab to see if you get a computing time which match the theory.

7. Solve A1000x = b1000. Use the command tic and toc to find the computing time for each of the threeparts of your code. Use the backslash command of matlab to check that x(1) and x(100) are correct.

8. Solve A2000x = b2000. How long does it take? Then solve A2000x = c2000 where

c =

−33−33...−33

Since you have already solve A2000x = b2000, it should be very quick to solve A2000x = c2000: you justneed to use rhs.m and backsub.m. How long does it take?

2

Math 135AOctober 4, 2010

Homework 1

1. Plot

(a) (Plot in Wolframalpha.com)Zoomed inEstimate for zero: x=0.6412

(b) function [ a, b, k ]=hw1b(f, a, b, er)

f=inline(f); %f is a string of characters from the input on the command line

if (f(a)*f(b) >0)||(a >= b) error('Does not satisfy conditions for bisection method'); end

%k is iterations

%a is left point

%b is right point

%x(k) is midpoint

k=1; x(k)=(b+a)/2;

while(abs(b-a)/2.>er && f(a)*f(b)<0) if(f(x(k))*f(a)<0) b=x(k);

else a=x(k);

end k=k+1;

x(k)=(b+a)/2;

endAfter running the code: [a, b, k]=hw1b('x-2.^-x', -5, 5, 10.^-5), we get 0.6412 after 20 iterations

(c) Solving analytically we have,

10−5 =b− a2n+1

10−5 =10

2n+1

solving for n yields 18.9316, which is close to 20.

2. Plot

(a)

1

(b) 4.4934Matlab codefunction [ a, b, k ]=hw1b(f, a, b, er) f=inline(f); if (f(a)*f(b) >0)||(a >= b) error('Does not satisfyconditions for bisection method'); end k=1; x(k)=(b+a)/2;

(c) a−f(a) < 0 and b−f(b) > 0 we can call this new function g(x) = x−f(x), by applying the IntermediateValue theorem, there exists a number p between a and b such that g(P ) = 0, thus f(P ) = p which is the�xed point. Q.E.D.

3. plot

(a)

(b) f(0) = ln(2) > 0, f(2) = ln(4) < 2,

(c) Yes, it is true, which implies that there exists a �xed point on [0, 2]

(d) Maximum is 12

(e) 12

(f) We can use �xed point iteration to �nd the convergence

(g) 1.1445 to 1.1465, 11 iterations (used same code as above)

(h) 10.9658,

4. plot

(a)

(b) f([ 13 , 1]) ≈ [0.71653, 0.36787]

(c) It is true, there exists a �xed point on [ 13 , 1].

2

(d) The max is 1/3

(e) k=1/3

(f) Fixed point iteration allows us to see whether e−x diverges or converges

(g) between 0.5664 and 0.5684, 11 iterations[a, b, k]=hw1b('exp(-x)-x',0,2,10^-3)

(h) number of iterations 5.918

3

Math 135AOctober 11, 2010

Homework 2

1. Plot

(a) x3 = x2 + x+ 1

Bisection Method1.8393, 21 iterations for 10−5 accuracy, interval was chosen via chart between 1 and 2

(b) Initial iterate: 2, chose it via looking at chart, 1.83928676, 4 iterations

(c) Initial iterate: between 1 and2 , chose it via looking at chart, 1.8393, 7 iterations

Fastest convergence: Newtons, slowest, bisection, slowest

2. Plot

(a) Bisection: 0.6498, 18 iterations

4

(b) Newtons: 0.6498 5 iterations

(c) Secant: 0.6498, 7 iterations

Bisection is the slowest, newtons is the fastest

3. Roots

(a) f(x) = x2 − a = 0 to �nd the root

(b) Newtons Method ReductionSo we have

g(x) = x− f(x)

f ′(x)

then

g(x) = x− x2 − a2x

which

g(x) = x− x2

2x− a

2x

thereforeg(x) = x− x

2− a

2x

which

g(x) =1

2

(x− a

x

)thus

pn+1 =1

2

(pn −

a

pn

)(c) Since it's linear, as long as p0 6= 0 (since it would diverge because it's part of the denominator) it will

continue to get smaller and smaller and converge eventually to the answer.

4. Proof of the convergence of Newton's Method

(a) Since

pn+1 = pn −f(pn)

f ′(pn)

We can expand the Taylor Series First two terms with ξ

f(p) = f(pn) + (p− pn)f ′(pn) +1

2(p− pn)2f ′′(ξ)

If we set the above equation to 0 and dividing by f ′(p0)

f(pn)

f ′(pn)+ (p− pn) +

1

2(p− pn)2

f ′′(ξ)

f ′(pn)= 0

rearrange to de�ne pn

pn =f(pn)

f ′(pn)+

1

2(p− pn)2

f ′′(ξ)

f ′(pn)+ p

plugging this back into the Newtons Method Sequence de�nition of pn+1,

pn+1 =

[f(pn)

f ′(pn)+

1

2(p− pn)2

f ′′(ξ)

f ′(pn)+ p

]− f(pn)

f ′(pn)

which simpli�es to

pn+1 =1

2(p− pn)2

f ′′(ξ)

f ′(pn)+ p

thus

p− pn+1 = −(p− pn)2f ′′(ξ)

2f ′(pn)

5

(b) Since the function is continuous, we can pick a δ neighborhood in which the function isn't changingdirection, thus f ′(x) 6= 0 holds true for that neighborhood.

(c) We know that |pn−p| > |pn+1−p|, since after every iteration, the sequence is converging, and the distancefrom p gets smaller and smaller. Since |pn − p| < δ this implies (using p− pn+1 = −(p− pn) f ′′(ξ)

2f ′(pn)

|p− pn+1| ≤M |(p− pn)|2

thereforeM |p− pn+1| ≤ (M |p− pn|)2

(d) two parts

i. Substituting n = 0 into equation 3 on the homework we have

|p− p1| ≤M |p− p0|2

and using equation 6

|p− p1| ≤M |p− p0|2 <M

M2

thus|p− p1|M

≤ |p− p0|2 <1

M2

thus|p− p1|M

≤ 1

M2

and we have

|p− p1| ≤1

M

ii. Since M |p− p0| ≤ 1 and|p− p1| ≤ (M |p− p0|)|p− p0| ≤ |p− p0|

we have |p− p0| ≤ δ, then|p− p1| ≤ δ

(e) We use induction to proveM |p− pn| ≤ (M |p− p0|)2

n

soM |p− p1| ≤ (M |p− p0|)2

furthemoreM |p− p2| ≤ (M |p− p1|)2 ≤

((M |p− p0|)2

)2since |p− pn|M < 1 this implies that it converges and we can deduce

M |p− pn| ≤ (M |p− p0|)2n

(f) Using the above equationlimn→∞

(M |p− p0|)2n

= 0

since M |p− p0n| ≤ 0 < 1Because of

|p− p1| ≤M |p− p0|2

it has order of convergence 2

6

Math 135AOctober 18, 2010

Homework 3

1. Let

g(x) = x−m f(x)

f ′(x)

By proving that g′(p) = 0,we prove that pn+1 converges with order 2. So we plug in

f(x) = (x− p)mh(x) where h(p) 6= 0

and it's derivativef ′(x) = m(x− p)m−1h(x) + (x− p)mh′(x)

into the �rst equation, thus we have

g(x) = x−m (x− p)mh(x)m(x− p)m−1h(x) + (x− p)mh′(x)

which simpli�es to

g(x) = x− mh(x)(x− p)mh(x) + h′(x)(x− p)

then the derivative is

g′(x) = 1−[

mh(x)

mh(x) + h′(x)(x− p)+ (x− p) d

dx

(mh(x)

mh(x) + h′(x)(x− p)

)]we then can �nd g′(p)

g′(p) = 1−[

mh(p)

mh(p) + h′(p)(p− p)+ (p− p) d

dx

(mh(p)

mh(p) + h′(x)(p− p)

)]which simpli�es to

g′(p) = 1− mh(p)

mh(p)= 0

Thus it converges with order 2. Q.E.D.

2. Plot

(a) Here's the graph

7

The zero seems to be at x = 0

(b) f(x = 0) = e0 − 0− 1 = 1− 0− 1 = 0 thus f(0) = 0

f

f ′=ex − x− 1

ex − 1= 1− x

ex − 1

f ′(0) = f(0) = 0f ′′(0) = 1 6= 0, thus it has convergence of order 2

(c) 28 iterations

(d) 5 iterations

3. So(1)2 = 1× 20 = 21 − 1

(11)2 = 1× 21 + 1× 20 = 22 − 1

(111)2 = 1× 22 + 1× 21 + 1× 20 = 23 − 1

Therefore(111...1)2 = 2n − 1

From a more generalized way of induction, assume the above is true

1(111...1)2 = 2n−1 + 2n = 2n+1 − 1

4. IEEE Floating point

(a) 3224, -3224, 339/256≈ 1.32421875

(b) We have211 − 212

252=

1

241

8

thus m = 41Spacing

20 − 21

252

thus m = 52

5. -

(a) (−1)02211−1−1023(1− 1252 ) = 3.59× 10308

(b) −3.59× 10308

(c) −21075

(d) 10000011100

9

Math 135A10/29/2010

Homework 4

1. Lagrange formula

(a) for (0,1) (-1, 2) and (1,3) P (x) = y0l0(x) + y1l1(x) + y2l2(x) + y3l3(x), plugging in

l0(x) =(x−−1)(x− 1)

(0−−1)(−1− 1)= − (x+ 1)(x− 1)

2

l1(x) =(x− 0)(x− 1)

(−1− 0)(−1− 1)=x(x− 1)

2

l2(x) =(x− 0)(x−−1)(1− 0)(1−−1)

=x(x+ 1)

2

thus

P (x) = − (x+ 1)(x− 1)

2+ x(x− 1) +

3

2x(x+ 1)

(b) Rewriting

P (x) =1

2

(4x2 + x+ 1

)2. f(x) = e−x

(a) (−1, e), (0, 1), (1, e−1)Using Lagrange's formula

l0(x) =(x− 0)(x− 1)

(−1− 0)(−1− 1)=x(x− 1)

2

l1(x) =(x−−1)(x− 1)

(0−−1)(0− 1)= −(x− 1)(x+ 1)

l2(x) =(x−−1)(x− 0)

(1−−1)(1− 0)=x(x+ 1)

2

So

P (x) =e

2x(x− 1)− (x− 1)(x+ 1) +

x(x+ 1)

2e

which simpli�es to(e− 1)2

2ex2 − (e2 − 1)

2ex+ 1

(b) Plotting it we have,

10

(c) 0.55169254Matlab code used:f = @(x)-1*(exp(-x)-((exp(1)-1)^2/(2*exp(1))*x.^2-(exp(2)-1)/(2*exp(1))*x+1))x=fminbnd(f,-1,1)The output was 0.5517, thus

maxx∈[0, 1]

E(x) = 0.5517

(d) error

Error =f (n+1)(ξ(x))

(n+ 1)!(x− x0)...(x− xn)

=f ′′′(ξ(x))

3!(x− x0)(x− x1)(x− x2)

=f ′′′(ξ(x))

3!(x− 0)(x−−1)(x− 1)

=f ′′′(ξ(x))

3!x(x+ 1)(x− 1)

we know thatf ′′′ = −e−x

max occurs at -1 e+1=2.71828

error =2.71828

6x(x+ 1)(x− 1)

(e) Sketch

using the x value of where the maximum occurs

E(x) ≤ | − 1|√3

= 0.577350

(f) Error is close, but not quite the same, they agree with one another

(g) Taylor series approximation around f(x)

1− x+x2

2!− ξ3

6

error formula from taylors theorem, we use the Lagrange form of the remainder term

fn+1(ξ)

(n+ 1)!(x− a)n+1

since a = 0 and n = 2f3(ξ)

3!(x− a)3

11

Thus

C =1

6f3(ξ)

using ξ = −1, we haveC = e/6 = 0.45304

The error is 0.45304(x) since x=0

(h) The best approximation is the one with three nodes

12

Math 135A11/5/2010

Homework 5

1. -

(a) Intermediate Value Theorem: If f is a continuous real-valued function on the interval [a, b] and h is anumber between f(a) and f(b) then there exists a c ∈ [a, b] such that f(c) = h.

(b) w is a continuos real-valued function on the interval of integration [a, b], thus there exists a θ ∈ [a, b]such that f(θ) is between f(a) and f(b). THus

m ≤ g(x) ≤M

multiplying by w(x)mw(x) ≥ g(x)w(x) ≥Mw(x)

g(x) attains every value in m,M , and g(x)w(x) is continuous

n ≤´ bag(x)w(x)dx´ baw(x)dx

≤M

let θ ∈ [a, b] thus

g(θ)

ˆ b

a

w(x)dx =

ˆ b

a

g(x)w(x)dx

2. So we haveP (x) = ax+ bˆ x1

x0

(ax+ b) dx

1

2ax2 + bx|x1

x0=

(x1 − x0)2

(a(x1 + x0) + 2b)

f(x0) = ax0 + b

andf(x1) = ax1 + b

f(x1) + f(x0) = a(x0 + x1) + 2bˆ x1

x0

P (x)dx =

(x1 − x0

2

)(f(x0) + f(x1)) =

h

2(f(x0) + f(x1))

3. The identity is true, I checked.

ˆ x4

x3

(x− x3)(x− x4) dx = −1

6(x4 − x3)6

= x3x4x−x4x

2

2− x3x

2

2+x3

3

=1

6x(6x4x3 − 3x4x− 3x3x+ 2x2)

= −1

6(x4 − x3)6

QED

4. -

13

(a) Computing ˆ 1

0

ex dx = ex|10 = e1 − e0 = e− 1

(b) Maxmaxx∈[0, 1]

|f ′′(x)| = maxx∈[0, 1]

|f4(x)| = e1 = e

(c) Using(b− a)M

12h2 =

(1− 0)3e

12n2= 10−8

solving for n we haven ≈ 4759

Error for Simpson's error

10−8 =(b− a)180

Mh4 = 35

(d) Trapezoidal Rule: 1.718281834781444, error:6.23× 10−9

Simpsons Rule: 1.718281834141971 error:5.6× 10−9

Matlab Code: Simpson's Rule

function [F] = simpson(f,a,b,n)%f=name of function, a=initial value, b=end value, n=number of double intervals of size 2hf=inline(f);n = 2 * n;h = (b - a) / n; S = f(a);for i = 1 : 2 : n-1 x = a + h .* i; S = S + 4 * f(x); endfor i = 2 : 2 : n-2 x = a + h .* i; S = S + 2 * f(x); endS = S + f(b); F = h * S / 3;For Trapezoid rulefunction [F] = trapezoid(f,a,b,n)%f=name of function, a=initial value, b=end value, n=strips f=inline(f);h = (b - a) / n;T = 0.5*f(a); for i = 1 : 1 : n-1 x = a + h .* i; T = T + f(x); endT = T + 0.5*f(b); F = h*T;

14

Math 135A11/15/2010

Homework 6

1. Program Code

(a) (see previous homework for simpson and trapezoid function)//Simpson Rulek=10:100:2010;y=[];for i=1:21y=[y abs(exp(1)-1-simpson('exp(x)',0,1,k(i)))]; endny=log10(y); nx=log10(k); plot(nx, ny, 'o')// Trapezoid Ruleny=[];for i=1:1:21y=[y abs(exp(1)-1-trapezoid('exp(x)',0,1,k(i)))];end ny=log10(y);plot(nx, ny, 'o')

(b) Here is the chart

15

(c) The error formula for Trapezoidal rule is O(h2) where as for simpsonsrule it is O(h4). 22 = 4, simpsonsrule is better

2. -

(a) Error has to be on the order of 10−7, Using the error formula derive din class

10−7 =(b− a)12h4

180

where h = (b− a)/n, we substitute this in and solve, we have n ≈ 29

(b) Matlab code is belowsimpson('exp(-1*x*x)',0,1,29)thus we have the answer

7.333997511728708× 10−1

3. Use simpsons rule we compute with n = 100, matlab code is belowabs(0.25-simpson('x*x*x',0,1,100))the error is

5.551115123125783× 10−17

the error is not the algorithm's fault, but the computer's fault, there is a loss of precision.

4. So we have a polynomial of degree 5

I3(f) = Ix(αx5 + βx4 + γx3 + δx2 + µx+ ν)

(a) so we have

I3(x5) =

ˆ 1

−1x5dx = 0 = w1x

51 + w2x

52 + w3x

53

I3(x4) =

ˆ 1

−1x4dx =

2

5= w1x

41 + w2x

42 + w3x

43

16

I3(x3) =

ˆ 1

−1x3dx = 0 = w1x

31 + w2x

32 + w3x

33

I3(x2) =

ˆ 1

−1x2dx =

2

3= w1x

21 + w2x

22 + w3x

23

I3(x) =

ˆ 1

−1x dx = 0 = w1x1 + w2x2 + w3x3

I3(1) =

ˆ 1

−11 dx = 2 = w1 + w2 + w3

(b) (From page 276)

I[−1, 1]3 = 0.5555f(−0.7745) + .8888f(0) + .5555f(.77459) = 3.06693

the calculated integral is ˆ(x5 + x4 + x3 + x2 + x+ 1)dx = 3.06667

which agree.

5. Gaussian v. Simpson

(a) Graph

(b) Computing integral analytically we have

ˆ 1

−1

1

x+ 1.5dx = ln(2x+ 3)|1−1 = ln(5) ≈ 1.60943791243410037

(c) Gaussian Quadrature Error:The node points are ±.9491079123427584 ±.7415311855993945 ±.4058451513773971 and 0.

(d) Simpson's Rule Error is calculated thusly in scilablog(5)-simpson(-1,1,7,)the error which comes out to be .0001889, (log is the natural log here, as opposed to question 1 wherelog is base-10.)

(e) The method that works best is Gaussian quadrature, which gives us an error 11 orders of magnitudebetter.

17

Math 135A11/29/2010

Homework 7

1. Matlab code is below

(a) h=[]; for i=1:5h=[h 10^(-i)]; endef=[]; es=[]; for i=1:5ef=[ef abs(exp(2)-(exp(2+h(i))-exp(2))/h(i))];es=[es abs(exp(2)-(exp(2+h(i))-exp(2-h(i)))/(2*h(i)))];endnh=log10(h);nef=log10(ef);nes=log10(es);plot(nh, nef, 'x');plot(nh,nes,'x');Plot

18

(b) yEF= 1.0031x+ 0.57999, yEs = 1.9644x+ 0.019253

(c) The slopes are 1 and 2 because log(h) = 2 ∗ log(h)(d) So we have

h=[]; for i=1:10h=[h 10^(-i)]; endef=[]; es=[]; for i=1:10ef=[ef abs(exp(2)-(exp(2+h(i))-exp(2))/h(i))];es=[es abs(exp(2)-(exp(2+h(i))-exp(2-h(i)))/(2*h(i)))];endnh=log10(h);nef=log10(ef);nes=log10(es);plot(nh, nef, 'x');plot(nh,nes,'x');

Yes it roughly matches

(e) The optimal h is 10−6 which agrees with computational e�orts.

2. Proof

19

(a) TheoremWe do a Taylor series expansion around xn − h

f(xn − h) = f(xn)− f ′(xn)h+ f ′′(xn)h2

2− f ′′′(xn)h

3

3!

and around xn − 2h

f(xn − 2h) = f(xn)− f ′(xn)2h+ f ′′(xn)2h2

1− f ′′′(xn)8h

3

3!

let's set n = 0then we have

−3f(x0) + 4f(x1)− f(x2) = (−3 + 4− 1)f(x0) + (−4hf ′(x0) + 2hf ′(x0))

−3f(x0) + 4f(x1)− f(x2) = 2hf ′(x0) +Oh2f ′(x0) +4

3![2f ′′′(ξ2)− f ′′′(ξ1)]

Thus

f ′(x0) =3f(x0)− 4f(x1)− f(x1)

2h+ [2f ′′′(ξ2)− f ′′′(ξ1)]

23h

3

2h

Thus we have a remainder term ofh3

3(2f ′′′(ξ2)− f ′′′(ξ1))

(b) CorollaryUsing the triangle inequality it is true

error ≤ maxx∈[a, b]

h2

3|(2f ′′′(ξ2)− f ′′′(ξ1)| ≤

h2

3maxx∈[a,b]

[12f ′′′(ξ2) + |f ′′′(ξ1)|] ≤h2

33M =Mh2

20

Math 135ADecember 3, 2010

Homework 8

1. init.m coden=6;A=toeplitz([-2,1,zeros(1,n-2)]);b=3*ones(n, 1);

2. fact.m coden=max(size(A)); %n is the size of the matrix givenAo=A; % Ao is the matrix which will undergo operationm=A-A;%m is an empty matrix with same dimensions as Afor i=2:nfor j=i:nm(i,j-1)=Ao(j,i-1)/Ao(i-1,i-1);Ao(j,:)=Ao(j,:)-Ao(i-1,:)*m(i,j-1); endendU=Ao+m;% we can add these matrices because one is a lower triangular % and the other is an upper triangular matrix

3. rhs.m codebo=b; for i=2:n for j=i:n bo(j)=bo(j)-m(i,j-1)*bo(i-1); endendg=bo;backsub.m codex=zeros(n,1);x(n)=g(n)/Ao(n,n);for i=n-1:-1:1x(i)=(g(i)-Ao(i,i+1:n)*x(i+1:n))/Ao(i,i);end

4. Table from x1 to x100 (truncated at decimal)

-750 -8085 -15120 -21855 -28290 -34425 -40260 -45795 -51030 -55965

-1497 -8802 -15807 -22512 -28917 -35022 -40827 -46332 -51537 -56442

-2241 -9516 -16491 -23166 -29541 -35616 -41391 -46866 -52041 -56916

-2982 -10227 -17172 -23817 -30162 -36207 -41952 -47397 -52542 -57387

-3720 -10935 -17850 -24465 -30780 -36795 -42510 -47925 -53040 -57855

-4455 -11640 -18525 -25110 -31395 -37380 -43065 -48450 -53535 -58320

-5187 -12342 -19197 -25752 -32007 -37962 -43617 -48972 -54027 -58782

-5916 -13041 -19866 -26391 -32616 -38541 -44166 -49491 -54516 -59241

-6642 -13737 -20532 -27027 -33222 -39117 -44712 -50007 -55002 -59697

-7365 -14430 -21195 -27660 -33825 -39690 -45255 -50520 -55485 -60150

5. For A500:fact.m: Elapsed time is 1.369596 seconds.rhs.m: Elapsed time is 0.008380 seconds.backsub.m: Elapsed time is 0.011052 seconds.

21

6. The number of operations performed by fact.m should be 23 = 8 times larger. The Elapsed time is 11.404127seconds, which is 8.3266 times larger than the time it took for A500. The computing time matches the theory.

7. For A1000:fact.m: Elapsed time is 11.404127 seconds (from above)rhs.m: Elapsed time is 0.031975 seconds.backsub: Elapsed time is 0.033450 seconds.x(1) = −1.5×10−3 and x(100)=−1.3515× 105

8. For A2000

fact.m: Elapsed time is 97.856150 seconds.rhs.m: Elapsed time is 0.124420 seconds.backsub:Elapsed time is 0.089146 seconds.

Matlab code to change from b to c: for i=1:2:2000 b(i)=-b(i); endFor A2000 and c2000rhs.m: Elapsed time is 0.125618 seconds.backsub: Elapsed time is 0.084331 seconds.

22

Midterm 1

Math 135A: Numerical AnalysisProfessor: Thomas LaurentUniversity of California, Riverside

1. [2 Points] In this problem, you don't need to justify your answer. -0.5 point per mistake)

2. What is the order of convergence of the sequence you obtain if you apply Newton's method to the functionf(x) graphed below?

3. What is the order of convergence of the sequence you obtain if you apply the bisection method to the functionf(x) graphed below?

4. What is the order of convergence of the sequence you obtain if you apply Newton's method to the functionf(x) graphed below?

5. What is the order of convergence of the sequence you obtain if you apply a �xed point iteration to the functionf(x) graphed below?

6. What is the order of convergence of the sequence you obtain if you apply a �xed point iteration to the functionf(x) graphed below?

7. [4 Points] Suppose p is a �xed point of g and suppose that the sequence de�ned by pn+1 = g(pn) convergesto p. Suppose also that g is three times continuously di�erentiable and g′(p) = g′′(p) = 0. Show that thesequence {pn}n≥0 converges to p with order 3 (at least).

1

8. Let f(x) be a continuous function which maps [a, b] into itself (i.e.: f([a, b]) ⊂ [a, b]). Use the intermediatevalue theorem to prove that f has a �xed point p ∈ [a, b].

9. In this problem, we will study a variant of Newton's method. Let f(x) be a smooth function (all the derivativesexist and are continuous, so you don't need to worry about that). Let p be the zero of f(x). The sequence{pn}n≥0 is de�ned as shown on the picture below.

On the above picture the line L is tangent to the graph of f(x) at the point (p, 0) and the lines L0, L1, L2

are parallel to the line L

10. [2 points] Express pn+1 as a function of pn. (If this were Newton's method, the formula would be pn+1 =pn − f(pn)/f ′(pn). What is the formula for this variant of Newton's method?)

11. [3 points] Assuming that pn converges to p, what is the order of convergence? Circle the right answer:pnconverges linearly to ppnconverges at least quadratically to p.Justify your answer (you have to use some theorem we proved in class.)

12. [1 point] Explain in no more than three sentences why this variant of Newton's method is of no practicalinterest to �nd the zero of a function. (there is a very simple reason for that! It is just common sense, nomath needed here!)

13. [2 points]

14. What is the distance (or spacing as I called it in class) between two adjacent machine numbers around 10,000?Give an exact answer in the format 1/2n.

15. [1 point] Suppose you want to use Newton's method to �nd the zero of a function. You know that this zerois somewhere around 10,000. Does it make sense to use the stopping criteria |pn+1 − pn| ≤ 10−10?

16. Write down the simplest possible MATLAB code to solve x = 1 + 0.3 cos(x) with Newton's method. Use thestopping criteria

|pn − pn−1||pn|

≤ 10−8

choose your initial iterate to be 2. If your code compiles and gives the correct answer, then you get full credit.

Midterm 1 Solutions

1. Solutions

2. 2

3. 1

2

4. 1

5. 2

6. 1

7. Let g(p) = p Use Taylor series approximation, since g′(p) = g′′(p) = 0, then the only terms that are left is

g(pn) = g(p) +g′′′(ξ)(pn − p)3

3!

where ξ ∈ [pn, p] thus|pn+1||pn − p|3

=

∣∣∣∣g′′′(ξn)3!

∣∣∣∣because ξn ∈ [pn, p] where ξn → p

8. Let g(p) = f(p)− p use the intermediate value theorem that at one place g(p) = 0.

9. -

10. pn+1 = pn − f(pn)/f ′(p)

11. Let g(x) = x − f(x)/f ′(p) then g′(x) = 1 − f ′(x)/f ′(p) and we have g′(p) = 1 − 1 = 0 which implies pnconverges quadratically.

12. to �nd p we have to know f ′(p)

13. -

14. 213/252 = 1/239

15. Yes, it makes sense to use the stopping criteria |pn+1 − pn| ≤ 10−10

16. Matlab codep=@(x) 1+0.3*cos(x)-x;dp=@(x) -0.3*sin(x)-1;n=2;x=2;while(abs(p(x)-p(n))/abs(p(x)) <=10^(-8) || x=2)n=x;x=n-p(n)/dp(n);end

x will return the answer.

3

Midterm 2

1. [2 points] This problem is here to help you do the next oneLet f(x) be a smooth function (i.e. f has as may continuous derivatives as you want.)

2. On the picture below, draw the unique polynomial of degree zero which interpolates f(x) at x0.

3. Let p0(x) be this unique polynomial of degree zero which interpolate f(x) at x0. What is the error formula?Complete the following:For each x there exists a ξ in the interval ________ such that

f(x)− p0(x) =

Since ξdepends on x we often write ξ(x)

4. In this problem, we will study the numerical integration method described in the picture below:

The integral´ baf(x) dx is approximated by the sum of the area of the rectangles.

We will assume during all the problem that f(x) has one continuous derivative. Also we will use the notationh = b−a

n .

4

5. [3 points] As you know the formula for Simpson's rule is

In(f) =h

3(f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + . . .+ 4f(xn−1) + f(xn))

What is the formula for the integration method we are studying here?

In(f) =

Comments: No explanations needed, just give the formula.

6. [5 points] Let E0 be the error made by our integration method on the �rst interval [x0, x1]:

E0 is the area of the dark region.Prove that there exists ξ0 ∈ [x0, x1] such that E0 = f ′(ξ0)h

2/2Comments: Make sure to name the theorem you are using and to say why it is OK to use it.

7. As you know, for Simpson's rule, the error formula is

ˆ b

a

f(x) dx− In(f) = −(b− a)f (4)(ξ)

180h4

where ξ is some number between a and b. What is the error formula for our numerical integration method?Comments: There are two steps in this proof. Provide a picture to make your explanations clearer (okay onepicture: I am really looking for one speci�c picture). State which theorem you are using and say why it isOK to use it. Make sure to say in which interval lie the ξi. Your proof do not need to be more detailed thanthe one I gave in class.

8. [5 points] Find w1, w2, and x2 so that the numerical integration rule

I2(f) = w1f(0) + w2f(x2)

approximates ˆ 1

0

f(x) dx

with degree of precision 2.

In other words, you have to choose w1, w2, and x2 so that I2(p) =´ 10p(x) dx for every polynomial p(x) of

degree ≤ 2.Comment: Be careful, here we integrate from 0 to 1, and not from -1 to 1 as we have done in class withGaussian quadrature. Also the node x1 is �xed to be equal to 0.

Midterm 2 Solutions

5

1. [x0, x], Applying the mean value theorem f ′(ξ(x))(x− x0)

2. -

3. h(f(x0) + f(x2) + ...f(xn−1))

4.´ x1

x0f(x)− p0(x) dx =

´ x1

x0f ′(ξ(x))(x− x0) dx = f ′(ξ(θ))h

2

2 Using the weighted mean value theorem

f ′(θ) =

ˆ b

a

w(x)f(x)dx

where w(x) does not change sign ˆ x1

x0

f ′(ξ(x))(x− x0)dx

then θ ∈ [x0, x] such that F ′(ξ(θ)ˆ x1

x0

(x− x0) dx =

ˆ x1

x0

f ′(ξ(x))(x− x0)dx

thus

E0 =f ′(ξ(θ))h2

2

5. Error for general casen−1∑i=0

f ′(ξi)h2

2

where ξ ∈ [xi, xi+1] then

=

n−1∑i=0

f ′(ξi)

let maxx∈[a,b] f′(x) =M and minx∈[a,b] = m then

m ≤ f ′(ξi) ≤M

which implies

nm ≤n−1∑i=0

f(ξi) ≤ nM

divide the above equation by n invoke mean value theorem we have

f ′(ξ) =1

n

n−1∑i=0

f ′(ξ)

thus(a− b)2

2nf ′(ξ)

where ξ ∈ [x0, xn]

6. Let P (x) = αx2 + βx+ γ be a polynomial of degree 2, where α, β, γ = 1then

ˆ 1

0

x2 dx = w10 + w2x22 =

1

3

ˆ 1

0

x dx = w10 + w2x2 =1

2ˆ 1

0

1 dx = w1 + w2 = 1

Solving we have x2 = 23 , w1 = 1

4 and w2 = 34

6