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Math 106: Review for Final Exam, Part II
- The heights of women in a certain country are normally distributed with a mean of 170 cm
and a standard deviation of 10 cm. What fraction of these women have heights between
165 cm and 180 cm?
We use the normal density function f(x) =
2 πs
e
−
(x−m)^2 2 s^2.
The mean is m = 170 and the standard deviation is s = 10.
So, the fraction of women between 165 and 180 cm is
2 π · 10
165
e
−
(x−170)^2 2 · (^102) ≈ 0 .5328.
- Use a second-degree Taylor polynomial to estimate
3
3
3
We choose f(x) = 3
x and x 0 = 27 because 27 is the perfect cube closest to 28.
f(x) = x
1 / 3 f(27) = 3
f
′ (x) =
x
− 2 / 3
3 x 2 / 3
f
′ (27) =
2 / 3
f
′′ (x) = −
x
− 5 / 3 = −
9 x^5 /^3
f
′′ (27) = −
9 · 275 /^3
Now plug in to the Taylor polynomial formula with x 0 = 27.
P 2 (x) = f(x 0 ) + f
′ (x 0 )(x − x 0 ) +
f
′′ (x 0 )
(x − x 0 )
2 = 3 +
(x − 27) −
(x − 27)
2
Finally, evaluate at x = 28.
3
28 ≈ P 2 (28) = 3 +
2
- What is the largest possible error that could have occurred in your previous estimate?
We know that |f(x) − Pn(x)| ≤
Kn+
(n + 1)!
|x − x 0 |
n+ .
In this case, n = 2, x 0 = 27, and x = 28.
K 3 = max of |f
′′′ (x)| on [27, 28] = max of
27 x 8 / 3
on [27, 28] =
8 / 3
Putting this all together, we have |f(x) − P 2 (x)| ≤
10 177147
3!
3
- Use a comparison to show whether each of the following converges or diverges. If an
integral converges, give a good upper bound for its value.
(a)
1
7 + 5 sin x
x 2 dx
∞
1
7 + 5 sin x
x 2
dx
1
7 + 5 sin x
x 2 dx
For all x ≥ 1, we have 0 ≤
7 + 5 sin x
x 2
x 2
x 2
because the maximum of sin x is 1.
∞
1
dx
x 2
= 12 lim t→∞
t
1
dx
x 2
= 12 lim t→∞
x
t
1
= 12 lim t→∞
[
t
]
= 12[0 − (−1)]
Therefore, the original integral in question must converge to a value less than 12.
(b)
∞
1
1 + 3x 2
3
10 x 12
dx
∞
1
1 + 3x
2
3
3
10 x 12
dx
∞
1
1 + 3x 2
3
10 x 12
dx
For x ≥ 1, we have
1 + 3x 2
3
10 x 12
2 x 3
3
10 x 12
≥ 0. (We’ve made the numerator smaller
and the denominator larger, so the new fraction is smaller.)
But
2 x 3
3
10 x 12
2 x 3
3
27 x 12
2 x 3
3 x 4
x
and we know that
∞
1
dx
x
diverges (compute for
yourself or notice that p = 1).
Therefore the original integral must also diverge.
- Decide if each of the following sequences {ak}
∞ k=1 converges or diverges.^ If a sequence
converges, compute its limit.
(a) ak = 3 +
k
ak = 3 +
k
ak = 3 +
k
Terms are 3. 1 , 3. 01 , 3. 001 , 3. 0001 , ....
lim k→∞
k
= 3, so the sequence converges to 3.
(b) ak = (−1)
k ak = (−1)
k ak = (−1)
k Terms are − 1 , 1 , − 1 , 1 , ....
lim k→∞
k doesn’t exist, so the sequence diverges.
(c) ak =
3 + 5k
7 + 2k
ak =
3 + 5k
7 + 2k
ak =
3 + 5k
7 + 2k
Terms are 8/ 9 , 13 / 11 , 18 / 13 , 23 / 15 , ....
lim k→∞
3 + 5k
7 + 2k
(by L’Hopital’s Rule or by inspection), so the sequence converges to
- Decide if each of the following series converges or diverges. If a series converges, find its
value.
(a) 3 33 ...1 + 31 + 31 + 3...01 + 301 + 301 + 3...001 + 3001 + 3001 + 3...0001 +0001 +0001 + .........
lim k→∞
ak = 3 6 = 0, so the series diverges by the nth Term Test. (We keep adding 3’s forever.)
[Compare this with the first sequence of the previous problem.]
(b) 1 + 11 + 11 + 1///2 + 12 + 12 + 1///3 + 13 + 13 + 1///4 +4 +4 + .........
This is the famous Harmonic Series, which diverges even though the terms do approach 0. We
can use the Integral Test:
∞
1
dx
x
diverges, which means that
∞ ∑
k=
k
must diverge too.
(c) 5^55 −−− 555 ///3 + 53 + 53 + 5/// 999 −−− 555 ///27 +27 +27 + .........
This is a geometric series with r = −
, so it converges to
a
1 − r
- Decide if each of the following series converges or diverges. If a series converges, find
upper and lower bounds for its value.
(a)
∞ ∑
k=
k
3
k + 1
∞ ∑
k=
k
3
k + 1
∞ ∑
k=
k
3
k + 1
[Alternating Series Test]
The terms of this series alternate in sign.
And,
3
3
3
And, lim k→∞
3
k + 1
Therefore, by the Alternating Series Test, the series must converge.
We know that any two consecutive partial sums will provide upper and lower bounds:
The integral converges, so the series must converge too.
Further, we know that
2
x(ln x) 2
dx ≤
∞ ∑
k=
k(ln(k)) 2
≤ a 2 +
2
x(ln x) 2
dx.
Therefore, our lower bound is
2
x(ln x) 2
dx =
ln 2
And our upper bound is a 2 +
∞
2
x(ln x) 2
dx =
2(ln 2) 2
ln 2
- Does the first series from the previous problem converge absolutely or conditionally?
∞ ∑
k=
k
3
k + 1
∞ ∑
k=
3
k + 1
, which diverges by the Integral Test (check for yourself).
Therefore, the first series from the previous problem converges conditionally.
- Compute the radius and interval (including endpoints) of convergence for
∑^ ∞
k=
(x + 3)
k
k · 5 k
∞ ∑
k=
(x + 3)
k
k · 5 k
∑^ ∞
k=
(x + 3)
k
k · 5 k
lim k→∞
(x+3) k+
(k+1)· 5 k+
(x+3)k k· 5 k
= lim k→∞
(x + 3) k+
(x + 3) k
k
k + 1
k
k+
Use L’Hopital on the middle fraction.
(x + 3) · 1 ·
x + 3
So, we are guaranteed convergence when
x + 3
< 1. But this is equivalent to the following.
x + 3
− 5 < x + 3 < 5
− 8 < x < 2
To check convergence at the endpoints (where the Ratio Test is inconclusive), we plug in to the series
itself.
At x = 2, we have
∑^ ∞
k=
k
k · 5 k
∑^ ∞
k=
k
k · 5 k
∑^ ∞
k=
k
, which is the Harmonic Series and thus diverges.
At x = −8, we have
∞ ∑
k=
k
k · 5 k
∞ ∑
k=
k
k · 5 k
∞ ∑
k=
k
k
, which converges by the Alternating
Series Test.
Thus, the interval of convergence is − 8 ≤ x < 2 and the radius of convergence is 5.
- Find the complete Taylor series (in summation notation) for fff(((xxx) = ln (1) = ln (1) = ln (1 −−− xxx))) about xxx = 0= 0= 0
and determine its interval of convergence.
f(x) = ln(1 − x) f(0) = 0
f
′ (x) =
1 − x
f
′ (0) = − 1
f
′′ (x) =
(1 − x) 2
f
′ (0) = − 1
f
′′′ (x) =
(1 − x) 3
f
′ (0) = − 2
f
(4) (x) =
(1 − x) 4
f
′ (0) = − 6
Now plug in to the Taylor series formula with x 0 = 0.
f(x 0 ) + f
′ (x 0 )x +
f
′′ (x 0 )
(x − x 0 )
2
f
′′′ (x 0 )
(x − x 0 )
3
x
2
x
3
= −x −
x 2
x 3
x 4
∞ ∑
k=
−x
k
k
We now find the interval of convergence as in the previous problem.
lim k→∞
−x k+
(k+1)
−xk k
= lim k→∞
−x
k+
−x k
k
k + 1
Use L’Hopital on the second fraction.
= | − x|
= |x|
So, we are guaranteed convergence for |x| < 1, which is equivalent to − 1 < x < 1. Now check the
endpoints.
At x = 1, we have
∞ ∑
k=
k
k
∞ ∑
k=
k
∞ ∑
k=
k
, which is the negative of the Harmonic Series and
thus diverges.
At x = −1, we have
∞ ∑
k=
k
k
∞ ∑
k=
k+
k
, which converges by the Alternating Series Test.
So, the interval of convergence is − 1 ≤ x < 1.
- Write the complete series equal to
1
0
e
−x 2 dx
0
e
−x 2 dx
1
0
e
−x 2 dx and show that it converges.
We know e
w = 1 + w +
w
2
w
3
- ..., so by substitution we obtain the following.
e
−x 2 = 1 + (−x
2 ) +
(−x
2 )
2
(−x
2 )
3
2
x
4
x
6
