Homework Solutions Math2130

8/13/2019 Homework Solutions Math2130

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1.1 Basic Tools for Calculus

Page 12

#1 What is the third-order Taylor polynomial for f(x) =

x+ 1, aboutx0 = 0?

Solution:

P3(x) =f(0) +xf(0) +

x2

2!f(0) +

x3

3!f(0) (1)

f(0) = 1

f(x) = 12

(x+ 1)12 so f(0) =1

2

f(x) = 14

(x+ 1)3

2 so f(0) = 14

f(x) = 3

8(x+ 1)

5

2 so f(0) =3

8

Substituting in (1) yields

P3(x) = 1 +1

2x 1

8x2 +

1

16x3.

#4 Given that

R(x) =|x|6

6! e

for x [1, 1], where is between x and 0, find and upper bound for|R|,valid for all x [1, 1], that is independent ofx and .

Solution: Note first that x [1, 1] |x| 1.|x|66!

e 1 e1

6! =

e

6! 0.00378.

#6 Given that

R(x) =|x|4

4!

11 +

1


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forx

1

2

,1

2, whereis betweenxand 0, find and upper bound for |R|,value for all x

1

2,1

2

, that is independent ofxand .

Solution: We use the following facts

(i) x

12

,1

2

|x| 1

2

(ii) for 0 < z < 1

2, y = z4 is increasing, thus for z1 < z2, we have that

z41 < z42 , e.g., z

4 < 1

24

(iii) 1

|1 +| for 1

2,1

2

is maximized for that minimizes|1 +|, i.e.

when= 12

.

|x|44!

11 +

|x|

4

4!|1 +|12

44!|1 +|

12

44!1 + 12 =

12

44!12

=

12

34

= 1

192.

#7 What is the fourth-order Taylor polynomial for 1

x+ 1about x0= 0?

Solution:

P4(x) =f(0) +xf(0) +

x2

2!f(0) +

x3

3!f(0) +

x4

4!f(iv)(0) (2)

f(0) = 1

f(x) = (x+ 1)2 so f(0) = 1f(x) = 2(x+ 1)3 so f(0) = 2

f(x) =

6(x+ 1)4 so f(0) =

6

f(iv)(x) = 24(x+ 1)5 so f(iv)(0) = 24

Substituting in (2) yields

P4(x) = 1 x+x2 x3 +x4.

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#10(c) Find the Taylor polynomial of third-order for sin(x), using: x0 =

2.

Solution:

P3(x) =f

2

+

x 2

f

2

+

x

2

22!

f

2

+

x

2

33!

f

2

(3)

f(0) = 1

f(x) = cos(x) so f(0) = 0

f(x) = sin(x) so f(0) = 1f(x) = cos(x) so f(0) = 0

Substituting into (3) yields

P3(x) = 1

x 2

22

.

#11(d) For each function below construct the third-order Taylor polynomialapproximation, using x0 = 0, and then estimate the error by computing anupper bound on the remainder, over the given interval: (d) f(x) = ln(1+ x),

x 1

2,1

2

.

Solution:

P4(x) =f(0) +xf(0) +

x2

2!f(0) +

x3

3!f(0) +

x4

4!f(iv)() (4)

for between 0 and x.

f(x) = ln(1 +x) so f(0) = ln(1) = 0

f(x) = 1

1 +x so f(0) = 1

f(x) = (x+ 1)2 so f(0) = 1f(x) = 2(x+ 1)3 so f(0) = 2

f(iv)

(x

) = 6(x

+ 1)4

= 6

(1 +x)4 sof(iv)

(

) = 6

(1 +)4


ln(1 +x) =x x2

2 +

x3

3 x

4

4(1 +)4.

3


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SoP3(x) =x

x2

2

+x3

3

. For the error bound, first note that x

1

2

,1

2 |x| 12

.

|R3(x)| |x|4

4(1 +)4

12

44(1 +)4

12

44

1 +1

2

412

4412

4 =14#12(b) Construct a Taylor polynomial approximation that is accurate towithin 103, over the indicated interval, for each of the following functions,using x0 = 0: (b) f(x) =e

x, x [0, 1].

Solution: As in example 2 of the notes, we need n such that

|Rn(x)| 103

for all x [0, 1]. xn+1(n+ 1)!e = |x|n+1(n+ 1)! e 1 e

0

(n+ 1)!=

1

(n+ 1)!

Solving 1

(n+ 1)! 103 for n, we find that n= 7, thus we must find P7(x).

Since f(n)

(x) = (1)n

ex

for all n we have that f(n)

(0) = (1)n

for all n.Thus

P7(x) = 1 x+ x2

2 x

3

3! +

x4

4! x

5

5! +

x6

6! x

7

7!.

#23 Use the Integral Mean Value Theorem to show that the pointwiseform (1.3) of the Taylor remainder (usually called the Lagrange form) fol-lows from the integral form (1.2) (usually called the Cauchy form).

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Solution: Note: (x t)n doesnt change sign for t [x0, x].

Rn(x) = 1

n!

xx0

(x t)nf(n+1)(t) dt (Integral Form)

= f(n+1)()

n!

xx0

(x t)n dt (Using Mean Value Theorem)

= f(n+1)()

n! (x t)

n+1

n+ 1

xx0

= f(n+1)()

n! 0 (x x0)

n+1

n+ 1

= (x

x0)

n+1

(n+ 1)! f(n+1)(). (pointwise form)

#24 For each function in Problem 11, use the Mean Value Theorem to finda value M such that

|f(x1) f(x2)| M|x1 x2|that is valid for all x1, x2 in the interval used in Problem 11: 11(b)f(x) =ln(1 +x) on the interval [1, 1], 11(c) f(x) = sin(x) on the interval [0, ].

Solution: From the notes,

|f(x2) f(x1)| |f()| |x2 x1|.For 11(b): [x1, x2] = [1, 1] and [1, 1].

f(x) = ln(1 +x) = f(x) = 11 +x

= |f()| = 11 +

11 1

= So Mdoes not exist.

For 11(c): [x1, x2] = [0, ] and [0, ].

f(x) = sin(x) = f(x) = cos(x)= |f()| = | cos(x)| 1 =M

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#25 A function is called monotoneon an interval if its derivative is strictly

positive or strictly negative on the interval. Suppose f is continuous andmonotone on the interval [a, b], andf(a)f(b)< 0; prove that there is exactlyone value [a, b] such that f() = 0.

Solution: Iff is monotone on [a, b] and f(a)f(b)< 0 then either

f(a)< 0 and f(b)> 0 or f(a)> 0 and f(b)< 0.

Thus, as f C([a, b]), by the Intermediate Value Theorem[a, b] suchthat f() = 0.

1.2 Error, etc.

Page 19

#1 Use Taylors Theorem to show that

1 +x = 1 +

1

2x+ O(x2),

for x sufficiently small.

Solution: x0= 0 and

f(x) = f(0) +xf(0) +x2

2f(0) +...

= f(0) +xf(0) +O

(x2) (5)

Calculating coefficients:

f(x) = (1 +x)1

2 = f(0) = 1f(x) =

1

2(1 +x)

1

2 = f(0) =12

6


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1 +x = 1 +

12

x+ O(x2).


ex = 1 +x+ O(x2),for x sufficiently small.

Solution: x0= 0 and

f(x) = f(x) =ex

= f(0) = f(0) = 1Thus, as in #1, ex = 1 +x+ O(x2).


1 cos(x)x

=1

2x+ O(x3),

for x sufficiently small.

Solution: x0= 0 and

cos(x) = 1 12!

x2 + 1

4!x4 ...

= 1 cos(x) = 12!

x2 14!

x4 +...

= 1 cos(x)x

= 1

2!x 1

4!x3 +...

= 1

2x+ O(x3).

#4 Show thatsin(x) =x + O(x3).

Solution: x0= 0 and

sin(x) =x 13!

x3 +...= x+ O(x3).

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#6 Recall the summation formula

1 +r+r2 +r3 +...+rn =n

k=0

fk =1 rn+1

1 r .

Use this to prove that

nk=0

rk = 1

1 r + O(rn+1).

Hint: What is the definition ofO notation?

Solution: nk=0

rk =1 rn+1

1 r = 1

1 r rn+1

1 r , for|r|


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for r=1

e and n= 9.

rn+1

1 r = e10

1 e1 = 7.2 105


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1.3 Computer Arithmetic

Page 27

#2 Perform the indicated computations in each of three ways: (i) Exactly; (ii)Using 3 digit decimal arithmetic, with chopping; (iii) Using 3 digit decimalarithmetic, with rounding. For both approximations, compute the absoluteerror and the relative error.

(a)1

6+

1

10 (c)

1

7+

1

10

+

1

9

Solution:

(a) (i) 16

+ 110

= 0.26

(ii)

f l

f l

1

6

+f l

1

10

= f l(f l(0.16) +f l(0.1))

= f l(0.166 + 0.100)

= f l(0.266)

= 0.266 (chopping)

absolute error = |0.26 0.266| = 0.0006relative error =

0.0006

0.26= 2.5 103 (2 significant figures)

(iii)

f l

f l

1

6

+f l

1

10

= f l(f l(0.16) +f l(0.1))

= f l(0.167 + 0.100)

= f l(0.267)

= 0.267 (rounding)

absolute error = |0.26 0.267| = 3.3 104

relative error = 0.0003

0.26= 1.25 103 (2 significant figures)

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(c) (i) 1

7

+ 1

10 +1

9

= 0.3539682

(ii)

f l

f l

1

7

+f l

1

10

= f l(f l(0.142) + 0.100)

= f l(0.242)

= 0.242

so

f l0.142 +f l1

9 = f l(0.142 + 0.111)

= f l(0.353)

= 0.353 (chopping)

absolute error = |0.3539682 0.353| = 9.683 104

relative error = 9.683... 104

0.3539682 = 2.735 103 (4 significant figures)

(iii) Rounding obtained via the same method, but as f l

1

7

= 0.143,

this leads to the answer 0.354. Thus,

absolute error = |0.3539682 0.354| = 3.175 105

relative error = 3.175... 105

0.3539682 = 8.969 105 (4 significant figures)

#4 For f(x) = ex 1

x , how many terms in the Taylor expansion are needed

to get single precision accuracy (7 decimal digits) for all x

0,1

2

? How

many terms are needed for double precision accuracy (14 decimal digits) overthis same range?

Solution: Recall

ex = 1 +x+x2

2! +...+

xn

n! +

xn+1

(n+ 1)!e,

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for between 0 and x, and x

0,1

2. Rearranging:ex 1

x = 1 +

x

2!+

x2

3! +...+

xn1

n! +

xn

(n+ 1)!e

Rn(x)

.

For single precision (i.e., rounding to t= 7 decimal places) find n such that

|Rn(x)| 5 108 ifn= 7). So we need 9 terms or n= 8.

For double precision we need n such that

|Rn(x)|


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with 3digit decimal arithmetic yields 1 1020.

#8 Assume we are using 3digit decimal arithmetic. For = 0.0001 anda1= 5, compute

a2= a0+

1

a1

for a0 equal to each of 1, 2, and 3. Comment.

Solution:

a0= 1

f l12 = 1 105

f l(5 105) = 50000f l(1 + 5 105) = f l(50001) = 50000.

a0= 2

...fl(2 + 5 105) =f l(50002) = 50000

a0= 3

...fl(3 + 5 105) =f l(50003) = 50000

a2 is the same in all three cases.

#12 What is the machine epsilon for a computer that usees binary arith-metic, 24 bits for the fraction, and rounds? What if it chops?

Solution: = 2 and t= 24: From the lecture notes

=1

21t =

1

22124 = 0.596

107 (rounding)

= 1t = 2124 = 0.119 107 (chopping)

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2.1 Horners Rule

Page 42

#1(d) Write each of the following polynomials in nested form: (d) x3+3x+2.

Solution: 2 + 3x+x3 = 2 +x(3 +x2).

#2(a) Write each of the following polynomials in nested form, but this timetake advantage of the fact that they involve only even powers ofx to mini-

mize the computations: (a) 1 12

x2 + 1

24x4.

Solution:

1 12

x2 + 1

24x4 = 1 +x2

1

2

+

1

24x2

.

#3(a) Write each of the following polynomials in nested form: (a) 1 x2 +1

2x4 1

6x6.

Solution:

1 x2

+

1

2 x4

1

6 x6

= 1 +x2(1) +12x2 16 x4

= 1 +x2

(1) +x2

1

2

1

6

x2

Note: it takes longer to work out

1 +x

x

1 +x

x

1

21

6x2

.

#7(b) Consider the polynomial

p(x) = 1(x

1) +

1

6

(x

1)(x

2) +

1

7

(x

1)(x

2)(x

4).

This can be written in nested-like form by factoring out each binomialterm as far as it will go, thus:

p(x) = 1 + (x 1)

1 + (x 2)

1

6+

1

7(x 4)

.

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Write each of the following polynomials in this kind of nested form: (b)

p(x) = 1 +67x 12 521 x 12 (x 4) +17 x 12 (x 4)(x 2).

Solution:

p(x) = 1 +67

x 1

2

5

21

x 1

2

(x 4) +1

7

x 1

2

(x 4)(x 2)

= 1 +

x 12

6

7 5

21(x 4) +1

7(x 4)(x 2)

= 1 +

x 12

6

7+ (x 4)

5

21

+

1

7(x 2)

2.2 Difference Approximations

Page 49

#2(a), (d) Compute, by hand, approximations to f(1) for each of the fol-

lowing functions, using h = 1

16and each of the derivative approximations

contained in (2.1) and (2.5): (a) f(x) = arctan(x), (d) f(x) =ex.

Solution:

For (a) f(x) = arctan(x)

Forward Difference Approximation

f(x+h) f(x)h

= arctan

1 + 116

arctan(1)116

= 0.8157

4116

= 0.485 f(1) (3 decimal places)

Central Difference Approximation

f(x+h) f(x h)2h

= arctan

1 + 1

16

arctan(1 116

)216

= 0.500 (3 decimal places)...which is better.

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Note that the answer in the book is wrong. Also, observe that f(x) =1

x2 + 1 so f(1) = 0.5.

For (d) f(x) =ex

Forward Difference Approximation

f(x+h) f(x)h

= e(1+

1

16) e1116

= 0.356 (3 decimal places)

Central Difference Approximation

f(x+h) f(x h)2h

= e(1+

1

16) +e(1 1

16)

216

= 0.368 (3 decimal places)...which is better.

Observe that f(x) = ex so f(1) = e1 0.367 to 3 decimal places.

#4 Use the approximations from this section to fill in approximations to themissing values in Table 2.4 (for x= 1 and x= 1.5).

x f(x) f(x)

1.00 1.00000000001.5 0.8862269255

Solution: Using h= 0.1Forward Difference Approximation

x= 1

f(x+ 0.1) f(x)0.1

=0.9513... 1.00

0.1

=

0.486

f(1) (3 decimal places)

x= 1.5

f(x+ 0.1) f(x)0.1

=0.8935... 0.88622...

0.1 = 0.0729 f(1.5) (3 decimal places)

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Centered Difference Approximation

x= 1.5f(x+ 0.1) f(x)

0.1 =

0.8935... 0.8872...0.2

= 0.0313 f(1.5) (3 decimal places)

#5 Use the error estimate (2.5) for the centered approximation to the firstderivative to prove that this approximation will be exactfor any quadraticpolynomial.

Solution: Let p(x) = a+ bx + cx2 so p(x) = b+ 2cx. Using the centereddifference approximation

p(x+h)

p(x

h)

2h = [a+b(x+h) +c(x+h)2]

[a+b(x

h) +c(x

h)2]

2h

= 2bh+c(x+h)2 c(x h)2

2h

= b+ c

2h[(x+h)2 (x h)2]

= b+ c

2h[(x2 + 2xh+h2) (x2 2xh+h2)]

= b+4cxh

2h= b+ 2cx

= p

(x

).

#6(a) Find coefficientsA, B, andCso that (a)f(x) =Af(x) + Bf(x + h) +Cf(x+ 2h) + O(h2). Hint: Use Taylors theorem.

Solution:

f(x+h) =f(x) +hf(x) +h2

2!f(x) +

h3

3!f(c1), (6)

for c1 between x and x+h.

f(x+ 2h) =f(x) + 2hf(x) +(2h)2

2! f(x) +

(2h)3

3! f(c2), (7)

for c2 between x and x+ 2h. Now computing 4(6)-(7) gives4f(x+h) f(x+ 2h) = 3f(x) + 2hf(x) + h

3

6[4f(c1) 8f(c2)]

= 3f(x) + 2hf(x) +2h3

3 [f(c1) 2f(c2)].

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So dividing by 2h yields:

2h

f(x+h) 12h

f(x+ 2h) = 32h

f(x) +f(x) + O(h2)or

f(x) = 32h

f(x) +2

hf(x+h) 1

2hf(x+ 2h) + O(h2).

Thus A= 32h

, B = 2

h, and C= 1

2h.

#8 Use Taylors Theorem to show that the approximation

f(x)

8f(x+h) 8f(x h) f(x+ 2h) +f(x 2h)

12h

isO(h4).

Solution: Expand to powers of h4 (with remainder termO(h5) as we thendivide by 12h:

f(x+h) =f(x) +hf(x) +h2

2!f(x) +

h3

3!f(x) +

h4

4!f(iv)(x) +

h5

5!f(v)(c1)

f(x h) =f(x) hf(x) + h2

2!f(x) h

3

3!f(x) +

h4

4!f(iv)(x) h

5

5!f(v)(c2)

f(x+ 2h) = f(x) + 2hf(x) +(2h)2

2! f(x) +

(2h)3

3! f(x) +

(2h)4

4! f(iv)(x) +

(2h)5

5! f(v)(c3)

= f(x) + 2hf(x) + 2h2f(x) +4h3

3 f(x) +

2h4

3 f(iv)(x) +

4h5

15f(v)(c3)

f(x2h) =f(x)2hf(x)+2h2f(x) 4h3

3 f(x)+

2h4

3 f(iv)(x) 4h

5

15f(v)(c4)

Thus,

8f(x+h) 8f(x h) f(x+ 2h) +f(x 2h) = 16hf(x) 4hf(x) + ch5

= 12hf(x) + ch5,

where cdepends on f(v)(ci) for i= 1, 2, 3, 4. So dividing by 12h yields

8f(x+h) 8f(x h) f(x+ 2h) +f(x 2h)12h

=f(x) +ch4,

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wherec= c

12

. Rearranging

f(x) =8f(x+h) 8f(x h) f(x+ 2h) +f(x 2h)

12h + O(h4).

#10 Use the derivative approximation from Problem 8 to fill in as much aspossible of the table in Problem 4. (just do x= 1.2)

Solution: With x= 1.2 and h= 0.1

f(1.2) = 8f(1.2 + 0.1) 8f(1.2 0.1) f(1.2 + 0.2) +f(1.2 0.2)

12(0.1)

= 8f(1.3) 8f(1.1) f(1.4) +f(1.0)1.2

= 8(0.89747...) 8(0.95135...) (0.88726...) + 1.0

1.2= 0.265 (3 decimal places)

#12 Use Taylor expansions for f(x h) to derive an O(h2) accurate approx-imation tof(x) usingf(x) andf(x h). Provide all the details of the errorestimate. Hint: Go out as far as the fourth derivative term, and then addthe two expansions.

Solution: See class notes.

2.3 Eulers Method

Page 56

#1 Use Eulers method with h = 0.25 to compute approximate solutionvalues for the initial value problem

y = sin(t+y), y(0) = 1.

You should get y4 = 1.851566895 (be sure that your calculator is set to radi-ans).

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Solution:

yn+1 = yn+h sin(tn+yn)tn = nh= 0.25n

So yn+1= yn= 0.25sin(0.25n+yn)y0 = y(0) = 1

n= 0

y1 = y0+ 0.25 sin(0 +y0)

= 1 + 0.25 sin(1)

= 1.2103677462...(

y(0.25))

n= 1

y2 = y1+ 0.25 sin(0.25 +y1)

= 1.45884498566...( y(0.5))n= 2

y3 = y2+ 0.25 sin(0.5 +y2)

= 1.6902572796...( y(0.75))n= 3

y4 = y3+ 0.25 sin(0.75 +y3)

= 1.85156689533...( y(1))#4 Use Eulers method with h = 0.25 to compute approximate solutionvalues for

y= ety, y(0) = 1.What approximate value do you get for y(1) = 0.7353256638?

Solution:

yn+1 = yn+hf(tn, yn)= yn+ 0.25e

tn+yn

= yn+ 0.25e0.25nyn,

withy0= y(0) = 1.

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n= 0

y1 = y0+ 0.25e0y0

= 1 + 0.25e1= 0.320429542885...( y(0.25))

n= 1

y2 = y1+ 0.25e0.25y1

= 0.121827147394...( y(0.5))

n= 2

y3 = y2+ 0.25e0.5y2

= 0.486730952236...( y(0.75))

n= 3

y4 = y3+ 0.25e0.75y3

= 0.812025139875...( y(1))

#6 Repeat the above with h = 0.125. What value do you now get fory8 y(1)?

Solution:

yn+1 = yn+hf(tn, yn)

= yn+ 0.125etn+yn

= yn+ 0.125e0.125nyn,

withy0= y(0) = 1.

n= 0

y1 = y0+ 0.125e0y0

= 0.660214771443...( y(0.125))

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n= 1

y2 = y1+ 0.125e0.125y1

= 0.38610503923...( y(0.25))

n= 2

y3 = y2+ 0.125e0.25y2

= 0.149966473291...( y(0.375))

n= 3

y4 = y3+ 0.125e0.375y3

= 0.061333798435...( y(0.5))

n= 4

y5 = y4+ 0.125e0.5y4

= 0.255163498508...( y(0.625))

n= 5

y6 = y5+ 0.125e0.625y5

= 0.436100739954...( y(0.75))

n= 6

y7 = y6+ 0.125e0.75y6

= 0.607194720154...( y(0.875))

n= 7

y8 = y7+ 0.125e0.875y7

= 0.770581294899...( y(1))

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2.4 Linear Interpolation

Page 62

#1(a) Use linear interpolation to find approximations to the following valuesof the error function, using the table in the text. For each case, give an upperbound on the error in the approximation: (a) erf(0.34).

Solution: Set f(x) = erf(x) and choose x0 = 0.3, and x1 = 0.4 so that[a, b] = [0.3, 0.4].

P1(x) = x x1x0 x1 f(x0) + x x0x1 x0 f(x1).

With x= 0.34,

P1(0.34) = 0.6(0.3286...)+0.4(0.4283...) = 0.3685 (2 significant figures).

For the error in the approximation we need the 2nd derivative off(x) = erf(x).From the argument on p.61 of the text

f(x) = 4x

ex2

.

Without resorting to a graphical approach this is difficult to bound. A crudeupper bound is

|f(x)| 4(0.4)

e(0.3)2

= 0.8250087277,

where we have bounded using max (numerator)

min (denominator) and the fact that y1 = x

is increasing on [0.3, 0.4] andy2= ex2 is decreasing on [0.3, 0.4]. Thus from

the error bound formula

|f(x) P1(x)| M

8(x1 x0)2

= 0.250087277

8 (0.4 0.3)2

= 0.001031 (4 significant figures),

23


24/67

whereM= max |f(x)| for all x [x0, x1].

Check: using erf(x) = 2

x0

et2

dt on my TI-89 graphing calculator I

get

|f(x) P1(x)| = 2

0.340

et2

dt 0.3685329977

= |0.369364529345... 0.3685329...|= 0.000831...,

consistent with our result above.

#2 The gamma function, denoted by (x), occurs in a number of applica-tions, most notably probability theory and the solution of certain differentialequations. It is basically the generalization of the factorial function to non-integer values, in that (n+ 1) = n!. Table 2.8 gives values of (x) for xbetween 1 and 2. Use linear interpolation to approximate values of (x) asgiven below. (a) (1.005) = 0.9971385354

x (x)

1.00 1.00000000001.10 0.9513507699

1.20 0.91816874241.30 0.89747069631.40 0.88726381751.50 0.88622692551.60 0.89351534931.70 0.90863873291.80 0.93138377101.90 0.96176583192.00 1.000000000

Solution: Choosex0= 1.00 and x1 = 1.10. Set f(x) = (x).

P1(x) = x x1

x0 x1

f(x0) +x x0

x1 x0

f(x1).

With x= 1.005

P1(1.005) = 0.95(1.00)+0.05(0.9513507699) = 0.9976 (4 significant figures).

24


25/67

#4 Construct a linear interpolating polynomial to the function f(x) = x1

using x0 = 1

2 and x1 = 1 as nodes. What is the upper bound on the error

over the interval

1

2, 1

, according to the error estimate?

Solution:

P1(x) =

x x1x0 x1

f(x0) +

x x0x1 x0

f(x1)

=

x 112 1

1

2

1+

x 1

2

1 12

11

= 3

2x.

For the error bound we need f(x):

f(x) = x2 and f(x) = 2x3 = 2x3

.

We need an upper bound for f(x) on

1

2, 1

. Noting thatx3 is increasing

on

1

2, 1

= 2

x3 is decreasing on

1

2, 1

and positive throughout, thus

|f(x)

|= 2

|x3

| 2

1

23

= 2 (21)3= 2(8) = 16 = M .So using the error bound formula

|f(x) P1(x)| M8

(x1 x0)2 =168

1 1

2

2= 2

1

4

=

1

2.

#6 Repeat the above for f(x) =x1

3 , using the interval

1

8, 1

.

Solution:

P1(x) = x x1x0 x1 f(x0) +

x x0x1 x0 f(x1)

=

x 118 1

1

8

13

+

x 1

8

1 18

11

3

= 4

7x+

3

7 or

1

7(4x+ 3).

25


26/67

Now

f(x) =1

3 x2

3 and f(x) = 2

9 x5

3 = 2

9x5

3 .

Where is |f(x)| a maximum on

1

8, 1

? f(x) is monotonic (doesnt change

sign) on

1

8, 1

, so check value at endpoints:

f

1

8

= 7.1 and f(1) = 0.2

=

|f(x)

| 7.1 =M. Thus from the error bound formula

|f(x) P1(x)| M8

(x1 x0)2 =7.18

1 1

8

2= 0.6806 (4 significant figures)

2.5 Trapezoid Rule

Page 70

#1 Apply the trapezoid rule with h=1

8, to approximate the integral

I= 10

11 +x4

dx= 0.92703733865069.

How small does h have to be to get that the error is less that 103? 106?

Solution:

f(x) = 11 +x4

, h=1

8, [a, b] = [0, 1], n=

b ah

=1 0

18

= 8.

26


27/67

i xi f(xi)

0 0 11 18

0.99987...2 14 0.99805...3 3

8 0.99025...

4 12 0.97014...5 5

8 0.93145...

6 34 0.87157...7 7

8 0.79400...

8 1 0.70710...

T8(f) = 12h(f(x0) + 2f(x1) +...+ 2f(x7) +f(x8))

=

1

2

1

8

(1 + 2(0.99987...) + 2(0.99805...) +...+ 2(0.79400...) + 0.70710...)

= 0.926115180158...

In order to estimate the error we need the 2nd derivative off(x) = (1+x4)1

2 .

f(x) = 12

(1 +x4)3

2 4x3 = 2x3(1 +x4) 32 .

Using the product rule we can show that

f(x) =6x2(x4 1)

(x4 + 1)5

2

.

This is a complicated function to get an upper bound on. We would use acombination of calculus and curve sketching techniques to deduce the follow-ing diagram:

27


28/67

Alternatively, you could get an upper bound by maximizing and minimizingthe numerator and denominator in f respectively. (Note: I wouldnt giveyou one as complicated as this in our tests!) Now recall from lecture notes

that to get an error 103 we require

|I(f) Tn(f)| b a12

h2M= h2

12(1.39275) 103, (8)

where M = max |f(x)| for all x [a, b] (as indicated in the diagram).Rearranging, we require

h2 0.08616047... or h 0.0928227...

But h must divide b a= 1, so as

n= b ah

or h= b an

= 1n 0.0928227...

= n 10.77The book would stop here.

28


29/67

so choose n= 11 =

h=

1

n= 0.09. As in (8) we seek h such that

h2

12(1.39275) 106 or h2 8.616047 106 so h 0.0029353104.

As before

h= 1

n 0.0029353104...= n 340.679...

so with n= 341, = h= 1n

= 0.00293255132...

#2 Use the trapezoid rule with h=


I= 20

sin(x) dx= 1.

How small does h have to be to get that the error is less that 103? 106?

Solution:

f(x) = sin(x), h=

4, [a, b] =

0,

2

, n=

b ah

=2 04

= 2.

i xi f(xi)

0 0 0

1 4

12

2 2 1

T2(f) = 1

2h(f(x0) + 2f(x1) +f(x2))

=

1

2

4

0 +

22

+ 1

=

2 + 1

8

0.948059...

(Note: exact answer is 1.)In order to estimate the error we need the 2nd derivative off(x) = sin(x).

f(x) = cos(x) and f(x) = sin(x)

29


30/67

Thus|f(x)| = | sin(x)| 1 =M on 0,

2 .From the error bound formula we need|I(f) Tn(f)| b a

12 h2M=

2

12h2 1 =

24h2 103 (9)

= h2 0.007639437... so h 0.08740387...

But h must divide b a= 2

, so noting that h= b a

n =

2

n 0.08740387...

= n 17.97.

So choosen

= 18 =h

=

2

n = 0.087266

....

From (9) we also seek h such that

24h2 106

= h2 7.639... 106= h 0.002763953...

As before

h=2

n 0.002763953... = n 568.315...

Choosing n= 569 = h=2

569= 0.0027606...



I=

10

x(1 x2) dx= 14

.

Feel free to use a computer program or a calculator, as you wish. How smalldoesh have to be to get that the error is less than 103? 106?

Solution:

f(x) =x(1x2), h= 18

, [a, b] = [0, 1], n= b a

h =

1 018

= 8.

30


31/67

i xi f(xi)

0 0 01 18

63512

2 141564

3 38

165512

4 1238

5 58

195512

6 342164

7 78

105512

8 1 0

T8(f) = 1

2 h(f(x0) + 2f(x1) +...+ 2f(x7) +f(x8))

=

1

2

1

8

(0 + 2

63

512

+ 2

15

64

+...+ 2

105

512

+ 0)

= 63

256= 0.2461 (4 significant figures)

Note: exact answer is 1

4

. In order to estimate the error we need the 2nd

derivative off(x) =x(1 x2).f(x) = 1 3x2 = f(x) = 6x.

So|f(x)| = 6|x| 6 =Mon [0, 1]. From the error bound formula we needh such that

|I(f) Tn(f)| b a12

h2M= h2

12(6) 103, (10)

Rearranging, we require

h2 0.002 or h 0.0447213595...

But h must divide b a= 1, so ash=

b an

= 1

n 0.0447213595...

= n 22.36...

31


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so choose n= 23 =

h=

1

23 = 0.04347826. For an error less than 106 we

needh such that

|I(f) Tn(f)| b a12

h2M= h2

12(6) 106


h2 2 106 or h 0.00141421356...

As before

h= 1

n 0.00141421356...= n 707.106...

so with n= 708, = h= 1708

= 0.00141242937...



I=

10

ln(1 +x) dx= 2 ln(2) 1.

How small does h have to be to get that the error is less than 103? 106?

Solution:

f(x) = ln(1+x), h=1

8, [a, b] = [0, 1], n=

b ah

=2 1

18

= 8.

i xi f(xi)

0 0 01 18 ln

98

2 1

4 ln

54

3 3

8 ln

118

4 12 ln

325 58 ln 138 6 34 ln 74

7 78

ln158

8 1 ln(2)

32


33/67

T8(f) = 1

2h(f(x0) + 2f(x1) +...+ 2f(x7) +f(x8))

=

1

2

1

8

0 + 2 ln

9

8

+ 2ln

5

4

+...+ 2 ln

15

8

+ ln(2)

= 0.3856 (4 significant figures)

(Note: exact answer is 0.3863 to 4 significant figures). In order to estimatethe error we need the 2nd derivative off(x) = ln(1 +x).

f(x) = 1

1 +x= f(x) = 1

(1 +x)2.

So|

f(x)|=

1

(1 +x)2 1

(1 + 0)2 = 1 = M on [0, 1]. From the error bound

formula we need h such that

|I(f) Tn(f)| b a12

h2M= h2

12(1) 103, (11)


h2 0.012 or h 0.10954451150...But h must divide b a= 1, so as

h= b a

n =

1

n0.10954451150...

= n 9.12...so choose n= 10 = h = 1

10 = 0.1. For an error less than 106 we need h

such that

|I(f) Tn(f)| b a12

h2M= h2

12(1) 106


h2 1.2 105 or h 0.003464101615...

As beforeh=

1

n 0.003464101615...= n 288.67...

so with n= 289, = h= 1289

= 0.0034602076...

33


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35/67

Back substitution yields:

x1 = 0.008495 x2= 0.1575 x3 = 0.2274 x4= 0.2009(each to 4 significant figures).

#6 Use the tridiagonal algorithm in this section to solve the following systemof equations:

1 12 0 012

13

14

00 14

15

16

0 0 16

17

x1x2x3x4

=

2253301514

Note that this is a very small change from the previous problem, since theonly difference is that f2 has changed by only

1

12. How much has the answer

changed?

Solution:

1 12 0 012

13

14

00 14

15

16

0 0 16

17

2253301514

r2 12r1 r2

1 12 0 00 1

1214

00 14

15

16

0 0 16

17

2153301514

r3 3r2 r3

1 12 0 00 112

14 0

0 0 1120

16

0 0 1617

21

3730

1514

r4+ 1033

r3 r4

1 12

0 00 112

14 0

0 0 1120

16

0 0 0 134693

21

3730

9671386

The associated linear system is:x1 +

12

x2 = 2112

x2 + 1

4x3 = 1

1120x3 +

16x4 = 3730

134693

x4 = 9671386

35


36/67

Back substitution yields:

x1= 1.004 x2 = 1.993 x3= 3.336 x4= 3.608

(each to 4 significant figures).

#8 Verify that the following system is diagonally dominant and use thealgorithm of this section to find the solution.

12

1021 0 0

14

13

113

00 15

14

121

0 0 15

16

x1x2x3x4

=

61421791565634201310

Solution: Checking for diagonal dominance:

0.5 = 1

2 >

10

21= 0.4761...

0.3 = 1

3 >

1

4+

1

13=

17

52= 0.3269...

0.25 = 1

4 >

1

5+

1

21=

26

105= 0.247...

0.16 = 1

6 h2 1

?Recall the triangle inequality, namely

|a+b

| |a

|+

|b

|,

so h2 1 =

h2+ (1) h2+ 1 < 2 < 2 + h2 = |2 +h2|

1Note: this problem is harder than I would give you in an exam!

40


41/67

Middle equations: Is|2 +h2| h2

+ 1+ h2 1? First observe thath

2 1< 0 as h


42/67

so by the IMVT a root offexists in [1, 1.5]. Set x3=1 + 1.5

2 = 1.25.

#2(a) For each of the functions listed below, do a calculation by hand (i.e.with a calculator) to find the root to an accuracy of 0 .1. This process willtake at most five iterations for all of these, and fewer for several of them: (a)f(x) =x ex2, for [a, b] = [0, 1].

Solution: Need k such that

| xk| b a2k


43/67

3.2 Newtons Method

Page 99

#1 Write down Newtons method as applied to the function f(x) = x3 2.Simplify the computation as much as possible. What has been accomplishedif we find the root of this function?

Solution:f(x) =x3 2 = f(x) = 3x2.

Thus

xn+1 = xn f(xn)f(xn)

= xn (x3n 2)3x2n

= 2(x3n+ 1)

3x2n,

wherex0 is given. If we find a root offwe have found the cube root of 2.

#4 Write down Newtons method as applied to the function f(x) =a x1.Simplify the resulting computation as much as possible. What has been ac-

complished if we find the root of this function?

Solution:

f(x) =a x1 =a 1x

= f(x) =x2 = 1x2

.

Thus


= xn

(a 1xn

)1

x2n

= xn(2 axn),

wherex0 is given. A root off corresponds to a= 1

x, or x=

1

a, a = 0.

43


44/67

#6 Do three iterations of Newtons method for f(x) = 3 ex, using x0= 1.Repeat, using x0= 2, 4, 8, 16. Comment on your results.

Solution:f(x) = 3 ex = f(x) = ex.

Thus


= xn

3 exnexn

= [(xn 1)exn

+ 3]exn

,

x0= 1

x1 = [(1 1)e1 + 3]e1= 3e1

= 1.103...

x2 = 1.098...

x3 = 1.0986122...

x0= 2

x1 = [(2 1)e2 + 3]e2= 1.406...

x2 = 1.141...

x3 = 1.0995133...

x0= 4

x1 = [(4 1)e4 + 3]e4

= 3.054...x2 = 2.196...

x3 = 1.529...

44


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46/67

3.3 How to Stop Newtons Method

Page 102

#1 Under the assumption that f() = 0 and xn , prove (3.10); be sureto provide all the details. Hint: Expand f and f in Taylor series aboutx= .

Solution: From the lecture notes we have

xnxn+1 xn = en

f(xn)f(xn)

= en f(+en)f(+en)

= en

f() +enf() +e2n

f()2!

+...

f() +enf() +e2n

f()2! +...

=

f() +enf() +e2nf()2!

+...

f() +enf()2!

+e2nf()3!

+...

Thus

limn

xnxn+1 xn

=

limn

f() +enf() +e2n f()2! +...limn

f() +en

f()2!

+e2nf()

3! +...

=f() + lim

nenf

() + limn

e2nf()

2! +...

f() + limn

enf()

2! + lim

ne2n

f()3!

+...

=f() +f() lim

nen+

f()2!

limn

e2n+...

f() + f()2!

limn en+ f

()3!

limn e

2n+...

= f() + 0 + 0 +...

f() + 0 + 0 +...= 1,

46


47/67

where we used the fact that

limn

en = limn

(xn ) = 0.

#4 Figure 3.5 shows the geometry of a planetary orbit around the sun. Theposition of the sun is given by S, the position of the planet is given by P. Letxdenote the angle defined by P0OA, measured in radians. The dotted line isa circle concentric to the ellipse and having a radius equal to the major axisof the ellipse. Let Tbe the total orbital period of the planet, and let t bethe time required for the planet to go from A to P. Then Keplers equationfrom orbital mechanics, relating xand t, is

x

sin(

x) =

2t

T .

Hereis the eccentricity of the elliptical orbit (the extent to which it deviatesfrom a circle). For an orbit of eccentricity = 0.01 (roughly equivalent to

that of the Earth), what is the value ofx corresponding to t= T

4? What is

the value ofx corresponding to t = T

8? Use Newtons method to solve the

required equation.

Figure 3.5 Orbital geometry for Problems 4 and 5

47


48/67

Solution: = 0.01 and t=T

4

leads to

x 0.01 sin(x) =2T4

T =

2.

Set f(x) =x 0.01 sin(x) 2 , then

f(0) = 2 0

so by the IMVT a root [0, ]. Set x0= 2

. Newtons method is:


= xn

xn 0.01 sin(xn) 2

(1 0.01 cos(xn))=

xncos(xn) sin(xn) 50cos(xn) 100

Iterating with x0=

2:

n = 0 x1

= 1.5807963267...

n = 1 x2= 1.5807958268...

n = 2 x3= 1.5807958268...

The method for t=T

8is done in the same way.

3.8 Secant Method

Page 123

#1 Do three steps of the secant method for f(x) = x3 2, using x0 = 0,x1 = 1.

48


49/67

Solution: The question allows us to approximate the cube root of 2 since

f(x) = 0 = x= 3

2. The secant method is:xn+1 = xn f(xn)

xn xn1

f(xn) f(xn1)

= xn (x3n 2)

xn xn1(x3n 2) (x3n1 2)

= xn (x3n 2)

xn xn1x3n x3n1

(22)

Nowb3 a3

b

a

=b2 +ab+a2,

so with b= xn and a= xn1 we havex3n x3n1xn xn1 =x

2n+xnxn1+x

2n1

soxn xn1x3n x3n1

= 1

x2n+xnxn1+x2n1

.

Substituting this into (22) yields:

xn+1= xn (x3n 2)

x2n+xnxn1+x2n1

(23)

Doing 3 steps of (23) (withx

0 = 0 andx

1= 1)n= 1

x2 = x1 (x31 2)

(x21+x0x1+x20)

= 1 (13 2)

(12 + (0)(1) + 02)= 2

n= 2

x3 = x2

(x32 2)

(x22+

x1

x2+

x21)

= 2 (23 2)

(22 + (1)(2) + 12)

= 8

7 or 1.14286 (6 significant figures)

49


50/67


51/67

n= 1

x2 = ln(1 + 0.405465108108...)

= 0.340368285804...

n= 2

x3 = ln(1 + 0.340368285804...)

= 0.292944416354...

#2 Let Y = 1

2 be fixed, and take h =

1

8. Do three steps of the following

fixed point iteration

yn+1= Y +1

2h(Y ln(Y) ynln(yn))

using y0 = Y.

Solution: With Y =1

2 and h=

1

8we have

yn+1=1

2+

1

16

1

2ln

1

2

ynln(yn)

.

Using y0= Y =1

2we have

n= 0

y1 = 1

2+

1

16

1

2ln

1

2

y0ln(y0)

= 1

2+

1

16

1

2ln

1

2

1

2ln

1

2

= 1

2+

1

16ln

1

2

= 0.543321698785...

n= 1

y2 = 1

2+

1

16

1

2ln

1

2

y1ln(y1)

= 0.542376812253...

51


52/67

n= 2

y3 = 1

2+

116

1

2ln1

2

y2ln(y2)

= 0.54239978931...

#4 Consider the fixed point iteration xn+1 = 1 +exn. Show that this it-

eration converges for any x0 [1, 2]. How many iterations does the theorypredict it will take to achieve 105 accuracy?

Solution: Observe that with g(x) = 1 +ex

g(1) = 1.367... [1, 2]g(2) = 1.135... [1, 2]

and as g(x) is monotonically decreasing on [1, 2] we know that 1 g(x) 2for all x [1, 2], i.e. g : [1, 2] [1, 2]. To show the second condition ofTheorem 3.5 consider

|g(x)| = |ex| |e1| = 0.367< 1,

for all x (1, 2), as ex is decreasing. Thus by Theorem 3.5 the iterationconverges for all x0 [1, 2]. We apply the error bound formula

|xn |

Ln

1 L

(x1 x0),

whereL=1

e = 0.367...and

x0 = 1.5

x1 = 1 +ex0

= 1 +e1.5

= 1.22313016015...,

so we seek n such that 1e

n1 1

e

(1.2231... 1.5)< 105

52


53/67

or en < 2.2830965... 105so n < ln(2.2830965... 105)

or n > 10.687..

so take n= 11.

#5(a)(c) For each function listed below, find an interval [a, b] such thatg([a, b]) [a, b]. Draw a graph of y = g(x) and y = x over this interval,and confirm that a fixed point exists there. Estimate (by eye) the valueof the fixed point, and use this as a starting value for a fixed point iter-

ation. Does the iteration converge? Explain: (a) g(x) = 12

x+2

x

, (c)

g(x) = 1 +ex.

Solution:(a) Not straightforward. Need to do some curve sketching. Notice first thatwe can work the fixed points out exactly, as for

g(x) =1

2x+

1

x=x = 1

x=

1

2x or 2 =x2 or x=

2.

Focus on the fixed point at x = 2. Noticeg(2) = 2 gives that (2, 2)is on the graph ofg. Now

g(x) =1

2 1

x2 = 0 = x=

2

are critical points. Maxima, minima, or inflection points? Look at the secondderivative.

g(x) = 2

x3 = g(

2) =

2

23

2

>0

so ((2),

2) is a local minimum. Notice also that

limx+

g(x) = + and limx

g(x) = .

This information leads to the following graph:

53


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Since g(1) = 1.5 =g(2), thus

2 g(x) 1.5 for all x [1, 2]1 g(x) 1.5 for all x [1, 2]

Thus, condition (i) in Theorem 3.5 is satisfied with [a, b] = [1, 2]. To satisfythe second condition we need to show that |g(x)|


55/67

Nowg(1) = 1 +e1 1.136...g(2) = 1 +e

2

1.135... and as g(x) is decreasing,

1 g(x) 2 for all x [1, 2]

i.e. g: [1, 2] [1, 2] and so condition (i) in Theorem 3.5 is satisfied. To satisfy(ii), consider g(x) =ex so|g(x)| = ex, which again is a decreasingfunction. And so as

|g(1)| =e1 0.367...|g(2)| =e2 0.135...

= |g(x)|


56/67


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Solution: To find P2(x) first find the Lagrange basis functions:

L0(x) = (x x1)(x x2)(x0 x1)(x0 x2)=

(x 1)(x 2)(0 1)(0 2) =

1

2(x 1)(x 2)

L1(x) = (x x0)(x x2)(x1 x0)(x1 x2)=

(x 0)(x 2)(1 0)(1 2) = x(x 2)

L2(x) = (x x0)(x x1)(x2 x0)(x2 x1)=

(x 0)(x 1)(2 0)(2 1) =

1

2x(x 1)

Thus,

P2(x) = f(x0)L0(x) +f(x1)L1(x) +f(x2)L2(x)

= x(x 2) + (8)12 x(x 1)

= x(3x 2)The sketch is difficult to draw as both graphs match well over the specifiedinterval:

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#4 Construct the quadratic polynomial that interpolates y =

x at the

nodesx0=1

4, x1 =

9

16 and x2= 1.

Solution: To find P2(x) first find the Lagrange basis functions:

L0(x) = (x x1)(x x2)(x0 x1)(x0 x2) =

x 9

16

(x 1)

14 9

16

14 1 =6515

x 9

16

(x 1)

L1(x) = (x x0)(x x2)(x1 x0)(x1 x2) =

x 1

4

(x 1)

916

14

916

1 = 25635

x 14

(x 1)

L2(x) = (x x0)(x x1)(x2

x0)(x2

x1)

= x 1

4

x 9

16

1 14 1 916 =

64

21 x 1

4x 9

16Thus,

P2(x) = f(x0)L0(x) +f(x1)L1(x) +f(x2)L2(x)

=

1

2

64

15

x 9

16

(x 1) +

3

4

256

35

x 1

4

(x 1)

+(1)

64

21

x 1

4

x 9

16

= after much tedious simplification

= 32105 x2 + 2221x+ 935#6 Find the polynomial of degree 3 that interpolates y = x3 at the nodesx0 = 0, x1 = 1, x2 = 2 and x3 = 3. (Simplify your interpolating polynomialas much as possible.) Hint: This is easy if you think about the implicationsof the uniqueness of the interpolating polynomial.

Solution: From our lecture notes we know that for distinct nodes, the inter-polating polynomial is unique. Thus the cubic polynomial that interpolatesthe cubic x3 at x0 = 0, x1 = 1, x2 = 2 and x3 = 3 must be x

3! We will do

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the calculation nonetheless.

L0(x) = (x x1)(x x2)(x x3)(x0 x1)(x0 x2)(x0 x3)=

(x 1)(x 2)(x 3)(0 1)(0 2)(0 3) =

16

(x 1)(x 2)(x 3

L1(x) = (x x0)(x x2)(x x3)(x1 x0)(x1 x2)(x1 x3)=

(x 0)(x 2)(x 3)(1 0)(1 2)(1 3) =

1

2x(x 2)(x 3)

L2(x) = (x x0)(x x1)(x x3)(x2 x0)(x2 x1)(x2 x3)=

(x 0)(x 1)(x 3)(2 0)(2 1)(2 3) =

1

2x(x 1)(x 3)

L3(x) = (x x0)(x x1)(x x2)(x3 x0)(x3 x1)(x3 x2)=

(x 0)(x 1)(x 2)(3 0)(3 1)(3 2) =

1

6x(x 1)(x 2)

Thus, after setting f(x) =x3

P3(x) = f(x0)L0(x) +f(x1)L1(x) +f(x2)L2(x) +f(x3)L3(x)

= 0 +1

2x(x 2)(x 3) + (8)

1

2

x(x 1)(x 3) + (27)

1

6

x(x 1)(x 2)

= x3 (after much simplification)

4.3 Interpolation Error

Page 180

#1 What is the error in quadratic interpolation to f(x) = x, using equallyspaced nodes on the interval

1

4, 1

?

Solution: From lecture notes we have

|f(x) P2(x)| h3

9

3M,

whereM= maxx(x0,x2)

|f(x)|. Here f(x) = xand h= 1 14

2 =

3

4 1

2=

3

8, so

x0 =1

4 x1=

1

4+

3

8=

5

8 x2 =

5

8+

3

8= 1.

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Also

f(x) =x 12 = f(x) =12

x 12

= f(x) = 14

x3

2

= f(x) =38

x5

2 ,

which is a decreasing function on

1

4, 1

. Thus

|f(x)| 38

1

4 5

2

= 3

8 45

2 = 3

8 (32) = 12 =M ,for all x

1

4, 1

. Thus

|f(x) P2(x)| 38

39

3(12) =

3

3

128 = 0.0406 (3 significant figures).

#2 Repeat the above for f(x) =x1 on

1

2, 1

.

Solution: Proceed as in #1.

f(x) =x1 and h=1 12

2 =

1

4, so

x0=1

2 x1=

3

4 x2 = 1.

Also

f(x) =x1 = f(x) = x2= f(x) = 2x3= f(x) = 6x4.

So

|f(x)

|=

6

|x|4= 96 = M , for all x

1

2, 1 .

Thus using the error estimate given at the start of #1 we have

|f(x) P2(x)| 14

39

3(96) =

3

18 = 0.0962 (3 significant figures).

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4.6 Hermite Interpolation

Page 192

#1 Construct the cubic Hermite interpolate to f(x) =

1 +x using thenodes a = 0 and b = 1. Plot the error between the interpolate and thefunction.

Solution: f(x) = (1 +x)1

2 = f(x) =12

(1 +x)1

2 . Then

L1(x) = x x2

x1

x2

= x 10

1

= 1 x

L2(x) = x x1

x2 x1 = x 01 0 =x

withL1(x) = 1 and L2(x) = 1. Now

h1(x) = [1 2(x x1)L1(x1)]L21(x)= [1 2(x 0)(1)](1 x)2= (1 + 2x)(1 x)2 (24)

h2(x) = [1 2(x x2)L2(x2)]L22(x)= [1

2(x

1)(1)]x2

= (3 2x)x2 (25)h1(x) = (x x1)L21

= (x 0)(1 x)2= x(1 x)2 (26)

h2(x) = (x x2)L22(x)= (x 1)x2 (27)

Using (24)-(27) we have

H(x) = f(x1)h1(x) +f(x1)h1(x) +f(x2)h2(x) +f

(x2)h2(x)

= 11

2 (1 + 2x)(1 x)2 +12

x(1 x)2 + 2 12 (3 2x)x2 +12

21

2 (x 1)x2

= (after simplifying)

= 0.0251x3 0.111x2 + 0.5x+ 1 (coefficients to 3 significant figures)

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#2 Construct the cubic Hermite interpolate to f(x) = sin(x) using the nodesa= 0 and b= . Plot the error between the interpolate and the function.

Solution: f(x) = sin(x) = f(x) = cos(x). Then

L1(x) = x x2

x1 x2=

x 0

= x

L2(x) = x x1

x2 x1 = x 0 0=

x

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withL1(x) =

1

and L2(x) =

1

. Now

h1(x) = [1 2(x x1)L1(x1)]L21(x)=

1 2(x 0)

1

( x)2

2

=

1 +

2

x

( x)2

2 (28)

h2(x) = [1 2(x x2)L2(x2)]L22(x)=

1 2(x )

1

x2

2

= 3 2

xx22

(29)

h1(x) = (x x1)L21= (x 0)( x)

2

2

= x( x)2

2 (30)

h2(x) = (x x2)L22(x)= (x ) x

2

2 (31)


H(x) = f(x1)h1(x) +f(x1)h1(x) +f(x2)h2(x) +f

(x2)h2(x)

= 0 + 1x( x)2

2 + 0

3 2

x

x2

2+ (1)(x ) x

2

2

= (after simplifying)

= 0x3

1

x2 +x+ 0

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#5 Construct the cubic Hermite interpolate to f(x) = x1

3 on the interval1

8, 1

. What is the maximum error as predicted by theory? What is the

actual (observed) maximum error?

Solution: f(x) =x1

3 =

f(x) =

1

3

x2

3 . Then

L1(x) = x x2

x1 x2 = x 118 1=

x 17

8

=8

7(1 x)

L2(x) = x x1

x2 x1 = x 1

8

1 18=

8

7

x 1

8

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65/67

withL1(x) =

8

7 and L2(x) =

8

7. Now

h1(x) = [1 2(x x1)L1(x1)]L21(x)

=

1 2

x 1

8

8

7

8

7

2(1 x)2

=

16

7x+

5

7

64

49

(1 x)2 (32)

h2(x) = [1 2(x x2)L2(x2)]L22(x)

=

1 2(x 1)

8

7

8

7

2x 1

8

2

= 23716

7x64

49x 1

82 (33)

h1(x) = (x x1)L21=

x 1

8

8

7

(1 x)2 (34)

h2(x) = (x x2)L22(x)

= (x 1)

64

49

x 1

8

2(35)


H(x) = f(x1)h1(x) +f(x1)h1(x) +f(x2)h2(x) +f(x2)h2(x)

=

1

8

13

16x

7 +

5

7

64

49

(1 x)2 +

1

3

1

8

23

x 18

8

7

(1 x)2

+(11

3 )

23

716

7x

64

49

x 1

8

2+

1

3

(1

2

3 )(x 1)

64

49

x 1

8

2= (after simplifying)

= 0.466x3 1.26x2 + 1.46x+ 0.336 (coefficients to 3 significant figures)From the lecture notes, the interpolation error is

f(x) H(x) = (x x1)2(x x2)2 f(4)

(x)4!

so

|f(x) H(x)|

x 18

2(x 1)2

M24 , (36)65


66/67

whereMis an upper bound for

|f(4)(x)

|on

1

8

, 1.f(x) =x

1

3 = f(x) =13

x2

3

= f(x) = 29

x5

3

= f(x) =1027

x8

3

= f(4)(x) = 8081

x11

3

so

|f(4)(x)

|= 8081

|x

| 11

3, which is a decreasing function on 1

8

, 1, thus|f(4)(x)| 80

81

1

8

113

=

80

81

(2048) =

20736

10 =M .

We also need to maximize

x 1

8

2(x1)2 on

1

8, 1

. Basic shape of

h(x) =

x 1

8

2(x 1)2 is:

so use Calculus to find the local maximum point.

h(x) = 1

32(x 1)(8x 1)(16x 9)

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Thus local extrema occur atx = 1,x =1

8andx =

9

16. Thush(x)

h

916=2401

65536 . Thus from (36)

|x13 H(x)| 240165536

2073610

124

= 3.165 (predicted).

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