18.100a-ps5
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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
1
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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.
3
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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|
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