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18.100A PS5

Author: Eric Emer

February 25, 2013

Collaborators: Callie McRee

1 Exercise 4.3 #2

Graph showing f(x) = x3 x (above) and f(x) = x3 x 1 (below).

Using intuitive geometric reasoning based on the graph off(x), we see that x = 1/

3 is locatedat the local minima of f(x). This means that for x > 1/

3 we will have that f(x) > 0 and for

x < 1/

3 we will have that f(x) < 0. We notice:

f(x) =d

dx(x3 x 1) = 3x2 1

If we were to use Newtons Method, we would get that:

an+1 = an f(an)f(an)

= an a3n an 13a2n 1

=2a3n + 1

3a2n 1Notably, for an > 1/

3 we get that an+1 > 0, and an+1 < 0 for an < 1/

3. Suppose that

f(K) = 0, where K is a unique positive integer.

en = an K

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f2(x) = e2n0 ln e

20

We examine which one goes to zero more quickly.

limn

(e02n ln 2

e2n0

ln e20

) = 0

the So clearly, f2 goes to zero more quickly, and the sequence from Section 4.4 converges moreslowly to M than the sequence we found in this problem.

3 Problem 4-2

3.1 a

L = 0.73908513L is a root of the function f(x) = x cos x.

3.2 b

Proof. We examine the error term.

en = an LWe observe:

|an L| = | cos an1 L| = | cos an1 cos L|The mean value theorem states that f(b) f(a) = f(t)(b a) for t [a, b]. By the mean valuetheorem, we have that:

| cos an1 cos L| = | sin t||an1 L|

Because this is sequentially takings cosines, all the terms after a0 will be on the interval [1, 1]. Land t will also be on this interval. We know that sin t is a strictly increasing function on [1, 1].Therefore, we know that | sin t| < | sin1|.

|an L| < | sin1||an1 L||an L| < | sin1|n|a0 L|

We observe that | sin1| < 1, so

limn

| sin1|n|a0 L| = 0limn

|an

L

| 0

Since it is an absolute value, it must be that as n , an L.

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4 Problem 4-3

4.1 a

Proof. We examine the two bounds limits for a > 0.

A = B

1/3

limaA

B/a2 = B1/3

limaA

a = B1/3

We also examine how they change as a increases. We define two functions to represent the bounds:

f1(a) = a, f

1(a) = 1

f2(a) = B/a2, f2(a) = 2Ba3

We notice that as a increases, a strictly increases, this is trivial. We also notice that as a increases,B/a2 strictly decreases. Given that we have already shown the limits of these expressions, we canconclude that:

a B1/3 B/a2

4.2 b

Proof.

an+1 = ran + sB/a2n, r + s = 1, r > 0, s > 0

en = an A, an = en + A

en+1 = an+1 Aen+1 = ran + sB/a

2n A

We substitute for an and apply some algebra:

en+1 = r(en + A) +sA3

(en + A)2A = r(A + en) + sA

(1 + en/A)2A

4.3 c

Suppose := en/A.

sA

(1 + en/A)2 sA 1

(1 + en/A)2 sA(1 en/A + ( en

A)2)2 sA(1 2 en

A+ 3(

enA

)2)

en+1 = r(A + en) + sA(1 2 enA

+ 3(enA

)2)A = r(A + en) + sA 2sen + 3se2n/AA

en+1 = rA + ren + sA2sen + 3se2n/AA = ren2sen + 3se2n/A+ A(1 + r + s), where r + s = 1

en+1 = (r 2s)en + 3sA

e2n

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4.4 d

Because we know that e2n/A will converge faster than en, we want the squared term to be weightedmore heavily. In fact, if possible wed like to remove the degree 1 term, and keep only the seconddegree term. Therefore, we set r = 2/3, and s = 1/3. We get en+1 = e

2n/A.

Now we use Newtons Method to show that we get the same result, using f(x) = x3 B.

an+1 = an f(an)f(an)

= an a3n B3a2n

an+1 =1

3(a5n + a2nb + 3an)

en = an Ban = B + en

en+1 = an+1 B

en+1 = 13 (a5n + a2nB + 3an)B

en+1 =1

3((B + en)5 + (B + en)2B + 3(B + en))B

en+1 =1

3((B + en)5 + (B + en)2B + 3en)

We see that our recurrence formula for an from Newtons Method is,

an+1 = an a3n B3a2n

=3a3n a3n B

3a2n=

2

3(an) 1

3(

B

a2n)

giving us r = 2/3, s = 1/3.

5 Problem 5

Improve the result of Example 5.2C by showing there is a constant K > 0 such that | ln n!n ln n|