Exam 2 Review
L’Hopital Rules:
Suppose that
lim f (x) = lim g(x) = 0.
x→u
x→u
or
lim |f (x)| = lim |g(x)| = ∞
x→u
If
0
(x)
limx→u fg0 (x)
x→u
exists in either the finite or infinite sense, then
f (x)
f 0 (x)
= lim 0
x→u g(x)
x→u g (x)
lim
Improper Integrals
Definition 1:
Z
b
b
Z
f (x) dx = lim
f (x) dx
a→−∞
−∞
Z ∞
a
b
Z
f (x) dx = lim
b→∞
a
Definition 2:
f (x) dx
a
Let f be continuous on the interval [a, b) and suppose that
lim |f (x)| = ∞.
t→b−
Then
b
Z
a
Z
f (x) dx = lim−
t→b
t
f (x) dx
a
given that this limit exists and is finite. In this case the integral converges! Otherwise it diverges.
Definition 3: let f be continuous on [a, b] except at the number c (some number between a and b), and suppose
that f (x) → ∞ as x → c. Then
Z b
Z c
Z b
f (x) dx =
f (x) dx +
f (x) dx
a
a
c
given that both integrals converge. Otherwise we say the integral on the left diverges.
Infinite Series:
The infinite series
P∞
Sn =
ak converges and has sum S if the sequence of partial sums
n
X
ak = a1 + a2 + a3 + ... + an−1 + an
k=1
converges to S. If the sequence of partial sums {Sn } diverges, then the series diverges.
Geometric Series
A geometric series
∞
X
k=1
ark−1 = a + ar + ar2 + ... =
a
1−r
ONLY IF |r| < 1! Otherwise it diverges. Observe that a is always the first term in the series.
nth-Term Test for Divergence: If
lim 6= 0 or DNE
n→∞
then we say that the infinite series
P∞
ak diverges.
Integral Test: Let f be P
continuous, positive, nonincreasing on [1, ∞) and say ak = f (k) for all positive integers
∞
k. Then the infinite series k=1 ak converges iff
Z ∞
f (x) dx
1
converges.
p-Series:
The series
∞
X
1
1
1
1
= 1 + p + p + p + ...
kp
2
3
4
k−1
where p is a constant, converges if p > 1 and diverges if p ≤ 1.
Ordinary Comparison Test: Suppose that 0 ≤ an ≤ bn for n ≥ N. Then
P
P
• If
bn converges, so does
an .
P
P
• If
an diverges , so does
bn .
Suppose that an ≥ 0, bn > 0, and
Limit Comparison Test:
an
=L
n→∞ bn
lim
If 0 < L < ∞, then
P
Ratio Test: Let
P
an and
P
bn converge or diverge together. If L=0 and
P
an be a series of positive terms and suppose that
lim
n→∞
an+1
= ρ.
an
• If ρ < 1, the series converges.
• If ρ > 1 or if limn→∞
an+1
an
= ∞, the series diverges.
• If ρ = 1, the test is inconclusive.
Alternating Series Test:
Let
∞
X
(−1)k ak = a0 − a1 + a2 − a3 + a4 + ...
k=0
be an alternating series with ak > ak+1 > 0. If
lim ak = 0
k→∞
then the series converges.
bn converges, then
P
an converges.
Absolute Convergence:
We say that
P
absolutely, then
ak converges.
P
Conditional Converges: A series
Absolute Ratio Test:
Let
P
P
ak converges absolutely if
ak converges conditionally if
P
P
|ak |. This implies that if
ak converges but
P
P
ak converges
|ak | diverges.
an be a series of nonzero terms and suppose that
lim
n→∞
|an+1 |
= ρ.
|an |
• If ρ < 1, the series converges.
• If ρ > 1, the series diverges
• If ρ = 1, the test is inconclusive.
Convergence Set of a Power Series: We call the set on which a power series
∞
X
an xn = a0 + a1 x + a2 x2 + a3 x3 + ...
n=0
converges its convergence set. This set is always an interval of one of the following three types:
• The single point x=0.
• An interval (−R, R), plus possibly one or both end points.
• The whole real line.
To find this set, the Absolute Ration Test is very useful.
Taylor’s Formula with Remainder: Let f be a function whose (n + 1) derivative f (n+1) exists for each x in
an open interval I containing a. Then for each x in I,
f (x) = f (a) + f 0 (a)(x − a) +
f 00 (a)
f 000 (a)
f n (a)
(x − a)2 +
(x − a)3 + ... +
(x − a)n + Rn (x)
2!
3!
n!
where the remainder Rn is given by
Rn (x) =
f (n+1) (c)
(x − a)n+1
(n + 1)!
and c is somewhere between x and a.
Taylor’s Theorem: Let f be a function with derivatives of all orders in some interval I containing a. Then
f (x) = f (a) + f 0 (a)(x − a) +
f 00 (a)
f 000 (a)
f n (a)
(x − a)2 +
(x − a)3 + ... +
(x − a)n + ...
2!
3!
n!
on the interval I, if and only if
f (n+1) (c)
(x − a)n+1 = 0.
n→∞ (n + 1)!
lim Rn (x) = lim
n→∞
Where is c? In the figure below, say that (a − r, a + r) is the interval where your function f has continuos
derivatives. Let a be some number in this interval. Depending on the value of x, c is ”between” the numbers x
and a.
some specific x0
?
a
|
a−r
{z
x0
}
a+r
c is somewhere here
some specific x1
?
x1
|
a−r
{z
a
}
a+r
c is somewhere here
In practice, it is not easy to determine this number c.
Important Maclaurin Series:
1.
1
1−x
= 1 + x + x2 + x3 + ...
for |x| < 1.
2. ln(1 + x) = x −
x2
2
+
x3
3
−
x4
4
3. ex = 1 + x +
x2
2!
+
x3
3!
+
x4
4!
+ ...
4. sin(x) = x −
x3
3!
+
x5
5!
−
x7
7!
+
x9
9! ...
for all x.
5. cos(x) = 1 −
x2
2!
+
x4
4!
−
x6
6!
+
x8
8! ...
for all x.
+
x5
5
+ ...
for |x| < 1.
for all x.
6. sinh(x) = x +
x3
3!
+
x5
5!
+
x7
7!
+
x9
9! ...
for all x.
7. cosh(x) = 1 +
x2
2!
+
x4
4!
+
x6
6!
+
x8
8! ...
for all x.
8. tan−1 (x) = x −
x3
3
x5
5
7
9
− x7 + x9 ...
for all x.
9. (1 + x)p = 1 + p1 x + p2 x2 + p3 x3 + ...
for |x| < 1.
+
Taylor polynomials: The Taylor polynomial of order n based at a is
Pn (x) = f (a) + f 0 (a)(x − a) +
f 00 (a)
f 000 (a)
f n (a)
(x − a)2 +
(x − a)3 + ... +
(x − a)n
2!
3!
n!
When a = 0, the Taylor polynomial if called Maclaurin polynomial.