# Homework Sol

## Full text

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Math 4650 Homework #4 Solutions Math 4650 Homework #4 Solutions

### ••

2.2 #7.2.2 #7. Use Theorem 2.2 to show that Use Theorem 2.2 to show that  gg((xx) ) == ππ + + 00..5sin(5sin(x/x/2)2) has a unique ﬁxed has a unique ﬁxed  point on

point on  [0[0,, 22ππ]]. . Use ﬁxedUse ﬁxed-p-point iteroint iteration to ation to ﬁnd an ﬁnd an apprapproximaoximation to tion to the ﬁxed poithe ﬁxed point nt  that is accurate to within

that is accurate to within  1010−−22

. Use Corollary 2.4 to estimate the number of iterations . Use Corollary 2.4 to estimate the number of iterations required to achieve

required to achieve 1010−−22

accuracy, and compare this theoretical estimate to the number  accuracy, and compare this theoretical estimate to the number  actually needed.

actually needed. Since

Since

11

sinsin

### ≤

1, we see that1, we see that ππ

00..55

gg((xx))

### ≤

ππ + 0+ 0..5 for every5 for every xx; in particular; in particular 00

gg((xx))

### ≤

22ππ for everyfor every xx. . SoSo gg satisﬁsatisﬁes condition (a) es condition (a) of Theorem 2.2. of Theorem 2.2. FFurthurthermorermoree we have

we have gg

((xx) ) = = 00..25cos(25cos(x/x/2), so that2), so that

gg

((xx))

### | ≤

00..25 for every25 for every xx. . ThThus condus conditiition (b)on (b) is satisﬁed as well, using

is satisﬁed as well, using kk == 1144. . So therSo there is e is a a uniunique ﬁxeque ﬁxed d poinpoint.t. For the actual approximation, we start at the midpoint of [0

For the actual approximation, we start at the midpoint of [0 ,, 22ππ], with], with pp00 == ππ. We get. We get

the sequence the sequence  p  p00 = = 33..141592654141592654  p  p11 = = 33..641592654641592654  p  p22 = = 33..626048865626048865  p  p33 = = 33..626995623626995623  p  p44 = = 33..626938795626938795  p  p55 = = 33..626942209626942209  p  p66 = = 33..626942004626942004

which rounds oﬀ to

which rounds oﬀ to pp

### ≈

33..63, and this accuracy is attained already at63, and this accuracy is attained already at pp22..

One formula in Corollary 2

One formula in Corollary 2..4 says that the error would theoretically be at most4 says that the error would theoretically be at most

p pnn

p p

11 44nn maxmax

p p00

a,a, bb

p p00

### }}

== π π 44nn..

So theoretically if we wanted the error guaranteed to at most 0

So theoretically if we wanted the error guaranteed to at most 0 ..01, we’d choose01, we’d choose nn soso that

that ππ

44nn

### ≤

00..0101,, i.e.,i.e., n n >> 44..

The other formula in Corollary 2

The other formula in Corollary 2..4 tells us to use4 tells us to use pp11

### −

p p00, and this formula gives, and this formula gives

p pnn

p p

### ≤

44 33 11 44nn(0(0..5) = 05) = 0..0101,, which gives which gives n n >> 3.3. Either way,

Either way, nn = 2 is suﬃcient so the theoretical prediction is pessimistic (as usual).= 2 is suﬃcient so the theoretical prediction is pessimistic (as usual).

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2.3 #1.2.3 #1. Let Let f f ((xx) ) == xx22

### −

66 and and  pp00 = = 11. Use Newton’s method to ﬁnd . Use Newton’s method to ﬁnd  pp22..

We have We have gg((xx) ) == xx

### −

f f ((xx)) f  f ((xx)) == xx

x x22

### −

66 22xx .. Hence

Hence pp11 == gg(1) = 3(1) = 3..5, and5, and pp22 == gg(3(3..5) = 25) = 2..607143.607143.

The actual square root of 6 is 2

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### •

2.3 #3. Let f (x) = x2

### −

6. With  p0 = 3 and  p1 = 2, ﬁnd  p3.

a. Use the Secant method. We have  pn = pn−1

f ( pn−1) pn−1

pn−2 f ( pn−1)

### −

f ( pn−2) , so that p2 = 2.4000 and p3 = 2.454545.

b. Use the method of False Position.

We have to verify the signs: f (2) =

### −

2 < 0 and f (3) = 3 > 0, so there is a root. Finding the test point is the same as the secant method: p2 = 2.4000.

Now we ﬁnd the new interval: f (2.4) =

### −

0.24, so that our new interval is [2.4, 3.0]. Using the secant method with these two points, we get p3 = 2.444444.

c. Which of (a) or (b) is closer to

### √

6?

The secant method has error 0.005055, while the false position method has error 0.005045. They are nearly indistinguishable, but false position happens to be slightly closer this time.

### •

2.3 #12. Use all three methods in this section to ﬁnd solutions to within  10−7

for the  following problems.

a. x2

4x + 4

ln x = 0 for 1

x

2 and for 2

x

4 Newton For 1

x

### ≤

2: using p0 = 1.5, we get

p0 = 1.5000000

p1 = 1.4067209

p2 = 1.4123700

p3 = 1.4123912

and all further outputs are the same. For 2

x

### ≤

4: using p0 = 3, we get

p0 = 3.0000000

p1 = 3.0591674

p2 = 3.0571061

p3 = 3.0571035

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Secant For 1

x

### ≤

2, using p0 = 1 and p1 = 2, we get  p0 = 1.0000000  p1 = 2.0000000  p2 = 1.5906161  p3 = 1.2845478  p4 = 1.4279661  p5 = 1.4136346  p6 = 1.4123782  p7 = 1.4123912

For 2

x

### ≤

4, using p0 = 2 and p1 = 4, we get

p0 = 2.0000000  p1 = 4.0000000  p2 = 2.4192186  p3 = 2.7560397  p4 = 3.3170225  p5 = 3.0097690  p6 = 3.0506707  p7 = 3.0572890  p8 = 3.0571028  p9 = 3.0571035

False p For 1

x

### ≤

2, using a = 1 and b = 2 as the initial endpoints, the resulting endpoints are [1.0000000, 2.0000000] [1.0000000, 1.5906161] [1.0000000, 1.4555373] [1.0000000, 1.4222100] [1.0000000, 1.4145936] [1.0000000, 1.4128836] [1.0000000, 1.4125012] [1.0000000, 1.4124157] [1.0000000, 1.4123967] [1.0000000, 1.4123924] [1.0000000, 1.4123914] [1.0000000, 1.4123912]

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For 2

x

### ≤

4, using a = 2 and b = 4 as the endpoints, we get [2.4192186, 4.0000000] [2.7560397, 4.0000000] [2.9360446, 4.0000000] [3.0119816, 4.0000000] [3.0407874, 4.0000000] [3.0512697, 4.0000000] [3.0550261, 4.0000000] [3.0563648, 4.0000000] [3.0568410, 4.0000000] [3.0570103, 4.0000000] [3.0570704, 4.0000000] [3.0570918, 4.0000000] [3.0570994, 4.0000000] [3.0571021, 4.0000000] [3.0571030, 4.0000000] [3.0571034, 4.0000000] [3.0571035, 4.0000000] b. x + 1

### −

2sin πx = 0 for 0

x

12 and for 12

x

### ≤

1

Here I’ll just present the answers (0 .2060351 and 0.6819748) and how long it takes. Newton’s method on [0, 12] reaches the answer in 3 steps. On [12, 1] it also reaches in 3 steps.

The secant method on [0,12] reaches the answer in 7 steps. On [ 12, 1] it reaches in 6 steps.

False position on [0, 12] reaches the answer to within 10−7

in 11 steps. On [12, 1] it reaches the answer in 13 steps.

### •

2.3 #19. The iteration equation for the Secant method can be written in the simpler   form

pn =

f ( pn−1) pn−2

f ( pn−2) pn−1

f ( pn−1)

### −

f ( pn−2)

.

Explain why, in general, this iteration equation is likely to be less accurate than the one given in Algorithm 2.4.

Both formulas involve subtracting two nearly equal numbers in both the numerator and denominator, which results in a large loss of precision. We then divide these two low-precision numbers and get a result with very low precision. However, in the actual algorithm we use this low-precision number only to correct an estimate we already have (so it’s OK if we only know the correction term to one digit, for example, since

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approximation is the low-precision value, so we have lost a bunch of decimal places of  precision by doing this.

### •

2.4 #6. Show that the following sequences converge linearly to p = 0. How large must  n be before we have

pn

p

5

### ×

10 −2 ?  a. pn = 1 n, n

### ≥

1. We have lim n→∞  pn+1  pn = lim n→∞ n n + 1 = 1,

so we have α = 1 and λ = 1. We must have n

### ≥

20 to be within the desired error. b. pn = 1 n2, n

### ≥

1. We have lim n→∞  pn+1  pn = lim n→∞ n2 (n + 1)2 = 1,

so α = 1 and λ = 1 again. We must have n

### ≥

5 to be within the desired error.

### •

2.4 #9.

a. Construct a sequence that converges to 0 of order  3.

b. Suppose α > 1. Construct a sequence that converges to 0 of order  α. The answer for both is about the same: just choose

pn = 1 2αn and we get lim n→∞  pn+1  pα n = lim n→∞ (2αn )α 2αn+1 = limn→∞1 = 1.

So λ = 1, and hence the order is actually α.

### •

2.5 #4. Let g(x) = 1 + (sin x)2 and  p(0)0 = 1. Use Steﬀensen’s method to ﬁnd  p(1)0 and   p(2)0 .

The new function is

h(x) = x

[g(x)

x] 2 g(g(x))

### −

2g(x) + x, and we have  p(n+1) 0 = h( p (n) 0 ). Therefore p(1)0 = 2.152904628 and p(2)0 = 1.873464044.

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