1
Seq Sequenc uences es
Definition of Terms sequence is a function f whose f whose domain is the set of natural numbers N (or a finite Definition. A sequence is subset of N), that is, f : N 1 2 3
→ → → →
R
a1 a2 a3
.. .
n
→
an
1. The elements elements of the function function are called the terms the terms of of the sequence. 2. We denote the n the nth th term of the sequence by a n . 3. We denote the sequence by by a1 , a2 , a3 , . . . or an
∞
} { }n=1 or simply {an}.
{
4. If the domain is a finite subset of N, then it is called a finite sequence . 5. If the domain is
N,
then it is called an infinite sequence .
Definition. A sequence is defined recursively if recursively if the nth term arises from a formula that depends
on the previous terms of the sequence. 1 1 1 1 2 3 4 5
Examples: 1(a). 1, 1, , , , ; 1(e). 1, 1, Remarks:
−2 , 1 , −4 5 ; 1(i). 1,1, 1, 2, 3, 5; 1(k). 1,1, 1 , 1, 1 , 1 3
3 27 81
2
3
1. Two sequence sequencess an and bn are ar e equal if a a n = b = b n for all n all n..
{ }
{ }
2. To get a visual representation of the terms of the sequence following:
{an}, we can do either of the
(a) Plot the point p ointss corresponding to successive successive elements elements of the sequence sequence on the number number line. (b) Plot the points (n, (n, an ) on the Cartesian plane. Example:
1(a). draw plot plot on the real number number line and on the Cartesian Cartesian plane
Convergence of Sequences Definition. The limit of limit of a sequence
{an} (den (denote oted d by lim an ) is equal to L if we can make an n →∞
as close as we want to L by taking n suffici sufficien ently tly large. large. In this case, case, the sequenc sequencee is said to be convergent . Otherwise, the sequence is said to be divergent . Theorem. If f is f is a function that is continuous on [1, [1,
∞) such that f ( f (n) = a n for all n ∈ N, then
lim an = lim f ( f (x).
n→∞
x→∞
Remark: Limit theorems for functions can also be applied to limits of sequences. Theorem. (Squeeze Theorem) Let an ,
{ { } {bn} and { {cn} be sequences such that an ≤ bn ≤ cn for all values of n beyond some index N . N . If the sequen sequencces { {an} and { {cn} have a common limit L as n → ∞, then { {bn} also has the limit L as n → ∞. 3(a) 3(a).. 0 if r if r > 0; 0; 3(b). 3(b). no limi limit; t; 3(c). 3(c). 1; 3(e). 3(e). 0; 3(h). 3(h). 0; 3(k). 3(k). 0 (use (use Sque Squeez ezee Theorem, a Theorem, a n = 0 and c and c n = 1/n); /n); 4. r < 1 < 1 Examples:
| |
Definition. A sequence an is said to be
{ }
1
1. increasing if a n+1
≥ an (or (an+1 − an) ≥ 0) for all n, 2. decreasing if a n+1 ≤ a n (or (an+1 − an ) ≤ 0) for all n, 3. monotone if it is either increasing or decreasing. Examples:
5(a). increasing; 5(b). decreasing; 5(d). neither
Definition. A real number R is said to be
1. an upper bound of an if a n
{ } ≤ R for all n, 2. a lower bound of {an } if a n ≥ R for all n, 3. the least upper bound of {an } if it is the smallest upper bound of {an }, that is, R ≤ r for any upper bound r of {an }, 4. the greatest lower bound of {an } if it is the largest lower bound of {an }, that is, R ≥ r for any lower bound r of {an }. Definition. A sequence {an } is said to be 1. bounded above if it has an upper bound, 2. bounded below if it has a lower bound, 3. bounded if it is bounded above and below. Examples:
6(a). bounded below; 6(b). bounded above; 6(c). bounded; 6(d). neither; 6(e).
bounded Remarks:
1. An increasing sequence is bounded below by a 1 , and a decreasing sequence is bounded above by a 1 . 2. Not all bounded sequences are convergent. 3. Not all monotone sequences are convergent. Theorem. (Bounded Monotone Convergence Theorem) Every bounded monotone sequence is con-
vergent. Examples: 7(a). convergent; 7(c). divergent; 7(g). divergent; 7(k). divergent; 7(l). convergent
by BMCT; 7(p). convergent by BMCT (limit is 6) Remark: To test for convergence of sequences, we use one of the following methods: 1. Look at limit of f (x), where f (n) = a n . 2. Use BMCT. 3. Use the Squeeze Theorem. Homework I, 3(j), 7(f), 7(q).
2
2
Series of Constant Terms
Definition. An infinite series is an expression of the form ∞
an = a 1 + a2 + a3 + . . . + an + . . . . n=1
1. The numbers a 1 , a2 , . . . , an , . . . are called terms of the series. ∞
2. Let S n denote the sum of the first n terms of the series
an , that is, n=1
S 1 = a1 S 2 = a1 + a2 S 3 = a1 + a2 + a3 .. . S n = a1 + a2 + . . . + an We call S n the nth partial sum of the series and S n the sequence of partial sums .
{ }
∞
Definition. Let S n be the sequence of partial sums of the series
{ }
n=1
∞
{S n} converges to a limit S , then we say that S , that is,
an = a 1 + a2 + a3 + . . .. If
an is a convergent series and its sum is equal to n=1
∞
an = lim S n = S. n→∞
n=1
∞
On the other hand, if {S n } is a divergent sequence, then the series series . A divergent series has no sum. Examples: 1(a).
n(n + 1) ; 1(c). 2
an is said to be a divergent n=1
−1 if n is odd, 0 if n is even; 1(e). 1 − n +1 1 ; 4. S 2
n
>
n + 2 2
∞
arn−1 = a +ar +ar 2 +. . ., where a and r are nonzero constants
Theorem. The geometric series n=1
is
∞
arn−1 =
1. convergent if r < 1 and
| |
n=1
a
1
− r.
2. divergent if r
| | ≥ 1. ∞
Proof. If r = 1, then the geometric series becomes
a. So, S n = na and n=1
lim S n = lim na =
n→∞
n→∞
+
∞, −∞,
a > 0 a < 0
hence, the series is divergent. For r = 1, the nth partial sum is S n = a + ar + ar2 + ar3 + . . . + ar n−1 . Multiplying the equation by the common ratio r, we get
rS n = ar + ar 2 + ar 3 + . . . + ar n .
3
Therefore, S n (1
− rS n
=
− r)S n
=
a + ar + ar 2 + ar 3 + . . . + ar n−1
− arn a (1 − rn ) a (1 − rn ) 1−r
= a
S n =
ar + ar 2 + ar 3 + . . . + arn
−
Taking the limit of the sequence of partial sums, we have lim S n =
n→∞
a
1
− r , |r| < 1.
If r
| | ≥ 1, the limit does not exist.
2(a). convergent, sum is 3; 2(b). convergent, sum is 18/5; 2(c). divergent; 2(d). 1 convergent if x < 1, sum is , divergent otherwise; 3(a). 1147/495 1 x Examples:
Remarks:
| |
−
∞
1. If
∞
∞
an and
∞
n=1
an n=1
∞
±
bn are convergent, then n=1
∞
(an n=1
± bn) is convergent and
n=1
± bn )
=
bn . n=1 ∞
2. If c is a nonzero constant and
∞
an is convergent, then n=1
∞
c
(an
∞
can is also convergent and n=1
can = n=1
an . n=1 ∞
3. If c is a nonzero constant and
∞
an is divergent, then n=1
can is also divergent. n=1
4. Convergence is unaffected by deleting a finite number of terms from a series; that is, for any ∞
positive integer k,
∞
an = a 1 + a2 + a3 + . . . and n=1
an = a k + ak+1 + ak+2 + . . . are both n=k
convergent or both divergent. ∞
an is convergent, then lim an = 0.
Theorem. If
n→∞
n=1
∞
Proof. First, note that for any given series
an , n=1
an = (a1 + a2 + . . . + an−1 + an ) = S n
− S n
1,
− (a1 + a2 + . . . + an
1)
−
−
where S n denotes the nth partial sum of the series. ∞
Suppose now that
an is convergent. Then by definition, there is a number S such that n−1
lim S n = S . Therefore,
n→∞
lim an =
n→∞
= =
lim (S n
− S n 1) lim S n − lim S n n n S − S = 0. n→∞ →∞
4
−
→∞
1
−
Remark: The previous statement is not useful to determine convergence of a series. However, its
contrapositive is useful to check for divergence of a series. ∞
Theorem. (Divergence Test) If lim an = 0, then the series n→∞
an is divergent. n=1
2(i). divergent by Divergence Test; 2(k). divergent by Divergence Test; 2(l). ∞ 1 convergent, sum is 4; 2(n). divergent since is 4 times the harmonic series 4n n=1 Examples:
3
Convergence Tests for Series with Nonnegative Terms
Sometimes, it is difficult to find an expression for the nth partial sum of the series. This makes it difficult to determine whether the series is convergent or divergent. However, there are other tests to determine the convergence of a series. The following tests can be used if the terms of the series are nonnegative. ∞
Theorem. (Integral Test) Let
positive, and decreasing in [1,
an be a series with nonnegative terms. Suppose f is continuous, n=1
∞) and f (n) = an for all n ∈ N. ∞
1. If the improper integral
f (x) dx converges, then the series converges.
1 ∞
2. If the improper integral
f (x) dx diverges, then the series diverges.
1
Remarks:
1. Because convergence is not affected by the first few terms of the series, the integral and the series need not start at n = 1. We can relax the condition on f : Suppose f is continuous, positive, and decreasing in [N, ), where N N.
∞
∈
∞
2. The sum of the series is not equal to the integral:
∞
an =
n=1
1
f (x) dx.
1(i). convergent; 2. convergent when p > 1; 1(b). convergent (sum is π 2 /6); 1(e). divergent; 1(f). divergent; 1(h). convergent; 1(j). divergent Examples:
∞
Theorem. (Comparison Test) Suppose
∞
an and n=1
bn are series with nonnegative terms. n=1
∞
≤ bn for all n and
∞
1. If an
bn is convergent, then n=1
n=1
∞
≥ bn for all n and
∞
2. If an
bn is divergent, then n=1
1
1(a). convergent, compare with ∞
an is also divergent. n=1
∞
Examples:
an is also convergent.
n=1
2n−1
∞
; 1(m). convergent, compare with
1 1(o). convergent, compare with ; 1(f). divergent, compare with 2 n n=1 Remarks:
∞
1 ; n 2 n=1
1 n n=1
1. Compare with series whose convergence (or divergence) is known. 2. Application is limited because of the conditions of the comparison test; ex. 1(q). ∞
Theorem. (Limit Comparison Test) Suppose that
an and n=1
5
∞
bn are series with positive terms. n=1
an = c > 0, then either both series converge or both series diverge. n→∞ bn
1. If lim
∞ ∞ an 2. If lim = 0 and bn converges, then an also converges. n→∞ bn n=1 n=1 ∞
an 3. If lim =+ n→∞ bn
∞ and
∞
bn diverges, then n=1
an also diverges. n=1
Remarks:
an 1. No conclusion when lim = 0 and n→∞ bn an = n→∞ bn
2. No conclusion when lim
∞
bn diverges. n=1 ∞
∞ and
bn converges. n=1 ∞
∞ 1 1 Examples: 1(q). convergent, compare with ; 1(r). convergent, compare with ; n n 2 3 n=1 n=1 ∞ 1 1(s). convergent, compare with ; 1(x). convergent, discard the first 3 terms and compare n3 n=1 ∞ ∞ 1 (n + 3)3 and n! (n + 3)! n=1 n=1
Remarks:
1. Adding or subtracting a finite number of terms will not change the convergence or divergence of a series. 2. If all of the terms of a series are positive, regrouping the terms will not affect the convergence ∞ 1 of the series; ex. . n 2 n=1 3. If all the terms are positive, rearranging the terms will not affect the convergence of the series; ∞ 1 ex. . n 2 n=1 4. If there are both positive and negative terms on the series, the previous remarks cannot be ∞
applied; ex.
( 1)n . n=1
−
∞
an is a given convergent series of positive terms, its terms can be grouped in any
Theorem. If n=1
manner and the resulting series also will be convergent and will have the same sum as the given series.
4
Convergence Tests for Series with Positive and Negative Terms
The following convergence tests can be applied to series with both positive and negative terms. Definition. An alternating series is a series whose terms are alternately positive and negative. It
is of the form
∞
∞
n
( 1)n−1 bn ,
( 1) bn or n=1
−
n=1
−
where b n > 0 for all n. Theorem. (Alternating Series Test) If a given alternating series satisfies
1. lim bn = 0, and n→∞
6
2. the sequence bn is decreasing, i.e., bn+1
≤ bn for all n,
{ }
then the series is convergent. [ insert diagram of number line here ] Examples: 1. convergent; 9. divergent; 2. convergent ∞
∞
Definition. A series n=1
∞
is convergent. If
an is said to be absolutely convergent if the series of absolute values ∞
an is convergent but n=1
convergent .
n=1
∞
n=1
|an| is divergent, then
|an|
an is said to be conditionally n=1
1. conditionally convergent; 4. absolutely convergent
Examples: ∞
an is absolutely convergent, then it is convergent.
Theorem. If n=1
∞
∞
Proof. Suppose that Note that for all n,
an is absolutely convergent. Then by definition, n=1
n=1
|an| is convergent.
−|an| ≤ an ≤ |an|, which implies 0 ∞
Since n=1
≤ an + |an| ≤ 2|an| 2|an |. By the Comparison Test, ∞
|an| is convergent, then so is
convergent. Now,
n=1
∞
∞
(an + an ) is also
| |
n=1
∞
an = n=1
(an + an
| | − |an|)
n=1 ∞
=
∞
(an + an )
| |−
n=1
n=1
|an|. ∞
Both series on the right-hand side of the equation are convergent. Therefore,
an is also convern=1
gent. 5. convergent since it is absolutely convergent
Example: Remarks:
1. The sum of a conditionally convergent series is different when the terms of the series are rearranged, ex. ( 1)n−1 n n=1 ∞
−
= 1 = = =
S =
− 12 + 13 − 14 + 15 − 16 + 17 − 18 + 19 − 101 + 111 − 121 + ···
− 12 − 14 + 13 − 16 − 18 + 15 − 101 − 121 + ··· 1 1 1 1 1 1 − + − + − + ··· 2 4 6 8 10 12 1 1 1 1 1 1 1 − + − + − + ··· 2 2 3 4 5 6 1
1 S 2
2. The sum of an absolutely convergent series is fixed even if the terms of the series are rearranged. 7
3. There are two tests to determine whether a series is absolutely convergent: the Ratio Test and the Root Test . ∞
Theorem. (Ratio Test) Suppose that
an is a series with nonzero terms. n=1
1. If lim
an+1 = L < 1, then the series is absolutely convergent, and is therefore convergent. an
2. If lim
an+1 an+1 = L > 1 or lim = n→∞ an an
n→∞
n→∞
∞, then the series is divergent.
3. No conclusion can be made regarding convergence of the series if lim
n→∞
an+1 = 1. an
15. absolutely convergent, hence convergent; 17. divergent; 19. absolutely convergent, hence convergent; 21. divergent Examples:
∞
Theorem. (Root Test) Given a series
an . n=1
1/n | | = L < 1, then the series is absolutely convergent, and is therefore convergent. n 2. If lim |an |1/n = L > 1 or lim |an |1/n = ∞, then the series is divergent. n n 3. No conclusion can be made regarding the convergence of the series if lim |an |1/n = 1. n
1. If lim an →∞
→∞
→∞
→∞
22. absolutely convergent, hence convergent; 24. absolutely convergent, hence convergent; 26. divergent; 27. absolutely convergent, hence convergent Examples:
Summary for Tests for Convergence 1. Divergence Test ∞
2. Special types: geometric, p-series,
1 , alternating n! n=1
3. Comparison Test or Limit Comparison Test (series of positive terms) 4. Integral Test (series of positive terms) 5. Test for absolute convergence (series of positive and negative terms) 6. Ratio Test: factorials, rational, k n 7. Root Test: (bn )n
8
5
Power Series
We are now going to consider series whose terms are polynomials in x. By substituting a real number for x, we arrive at a series with constant terms. Our goal is to find all the possible values of x that will make the resulting series convergent. Definition of Terms ∞
Definition. A series of the form
cn (x n=0
− a)n = c0 + c1(x − a) + c2(x − a)2 + c3(x − a)3 + . . .,
where x is a variable, a is a fixed real number and the cn ’s are constants, is called a power series centered at a or a power series about a. Note that in the definition, we adopt the convention that (x a)0 = 1 even when x = a.
−
xn x (centered at x = 0); (centered at x = 0); n! n=0 n=0 ∞
Examples:
x =
−1)
∞
n
(x + 1)n (centered at n n=1 ∞
√
Remark: We can also have a power series in φ(x), where φ(x) is a real-valued function. This power
series can be expressed as ∞
cn (φ(x))n = c 0 + c1 φ(x) + c2 (φ(x))2 + c3 (φ(x))3 + . . . . n=0
However, this is rarely used in practical applications. We will only focus on power series centered at a. ∞
− a)n is convergent for x = x1 (where x1 = a), then it is n=0 absolutely convergent for all values of x for which |x − a| < |x1 − a|. cn (x
Theorem. If the power series
∞
− a)n is divergent for x = x1 (where x1 = a), then it is n=0 divergent for all values of x for which |x − a| > |x1 − a|. cn (x
Theorem. If the power series
∞
cn (x
Theorem. For a given power series n=0
− a)n, there are only three possibilities:
1. The series converges only at x = a. 2. The series converges for all real values of x. 3. There is a positive number R such that the series converges if x x a > R.
| − a| < R and diverges if
| − |
∞
Definition. Given a power series
cn (x n=0
− a)n.
1. The number R in the previous theorem is called the radius of convergence of the power series. By convention, R = 0 if the series converges at only one point and R = if the series converges for all real values.
∞
2. The interval of convergence is the set of all values of x so that the series converges. In the case where R = 0, the interval of convergence is a . In the case where R = , the interval of convergence is R.
{ }
∞
Remarks:
1. In the case where R > 0, there are four possibilities for the interval of convergence: (a
− R, a + R)
[a
− R, a + R) 9
(a
− R, a + R]
(a
− R, a + R)
2. Most of the time, the radius and interval of convergence can be obtained using the Ratio Test or the Root Test. 3. For the endpoints of the interval, the series may be absolutely convergent, conditionally convergent, or divergent. The Ratio Test or the Root Test will fail, so other convergence tests should be use to check for convergence of the series. Examples: 1(a). R = 1, ( 1, 1); 1(c). R =
∞, R; 1(d). 1(l). R = 1, [ −2, 0); 1(m). R = 1, (1, 3); 1(p). R = 2, [ −1, 3] −
R = 0, 0 ; 1(b). R =
{ }
3 2,
− 32 , 32 ;
Power Series Representations of Functions ∞
The sum of a power series
cn (x n=0
− a)n is a function
∞
f (x) =
cn (x n=0
− a)n = c0 + c1(x − a) + c2(x − a)2 + c3(x − a)3 + . . . ,
whose domain is the interval of convergence of the power series. Our goal is to represent certain types of functions as sums by manipulating a geometric series or by differentiating or integrating such a series. We start with the power series ∞
xn = 1 + x + x2 + x3 + . . . , n=0
which is a geometric series with a = 1 and common ratio r = x. This is convergent whenever x < 1 and divergent when x 1. Using the formula for the sum of a convergent geometric series,
| |
| | ≥
∞
1 1
−x
xn ,
= n=0
|x| < 1.
2. the graphs are close to each other when x < 1, otherwise the graphs are ∞ ∞ ∞ ( 1)n xn n n 2n very different; 3(a). ( 1) x , x < 1; 3(b). x , x < 1; 3(d). , x < 3; 3(e). n+1 3 n=0 n=0 n=0 ∞ n n+2 ( 1) x , x < 3 n+1 3 n=0 Remark: Power series are used to calculate function values like sin x, e x , ln x and x, which cannot be evaluated by the usual arithmetic operations.
| |
Examples:
−
−
| |
−
| |
| |
| |
√
Differentiation and Integration of Power Series ∞
cn (x
Theorem. If the power series n=0
defined by
− a)n has radius of convergence R > 0, then the function f
∞
f (x) =
cn (x n=0
− a)n = c0 + c1(x − a) + c2(x − a)2 + c3(x − a)3 + . . .
is differentiable (and therefore continuous) on the interval (a
− R, a + R) and ∞
1. f (x) = c 1 + 2c2 (x
2.
− a) + 3c3(x − a)
f (x) dx = C + c0 (x
− a) + c1
(x
2
+ 4c4 (x
− a)
3
+ . . . =
ncn (x n=1
− a)2 + c2 (x − a)3 + ··· = C + 2
3
The radii of convergence of the following power series are both R.
10
− a)n
∞
cn n=0
1
−
(x
− a)n+1 .
n + 1
Remark: The radius of convergence is the same, but the interval of convergence may change.
∞ xn , domain [ 1, 1); 3(f). nxn−1 , domain ( 1, 1); 3(f). Examples: 4(a). f (x) = n + 1 n=0 n=1 ∞ ∞ ∞ n ( 1)n−1 xn 2n−2 1 Use the item to find (substitute x = 2 ); 3(h). nx , x < 1; 3(l). ln3 + , 2n n3n n=1 n=1 n=1 ∞ 1 x < 3; 3(l). Use the item to find (substitute x = 1); 5. Show that f (x) = f (x) and n n3 n=1 ∞ xn f (0) = 1; 3(i). ( 1)n . 6. Compute up to second derivative and plug in to differential equation; n! n=0 ∞ ∞ n 2n+1 ( 1) x ( 1)n x2n+1 3(p). ; 3(n). n!(2n + 1) 2n + 1 n=0 n=0 ∞
−
− −
| |
||
−
−
−
−
Products and Quotients of Power Series Finding the products and quotients of power series is similar to finding the product and quotient of polynomials. However, it is important that the two series are absolutely convergent so that the product or quotient can be computed. Recall that only absolutely convergent series can have the terms be regrouped or rearranged and still arrive at the same sum. 4 3 31 5 x 3 2x5 Examples: e−x tan−1 x = x x + x + ; tan x = x + + + 3 30 3 15 2
6
−
···
···
Taylor and Maclaurin Series
In the previous section, we were able to find a power series representation for certain functions. In general, if a function has a power series representation, there is a method to find it. Suppose f is a function that can be represented by a power series about x = a:
− a) + c2(x − a)2 + c3(x − a)3 + c4(x − a)4 + c5(x − a)5 + . . . , |x − a| < R. If we differentiate f repeatedly, then for |x − a| < R we get f (x) = c1 + 2c2 (x − a) + 3c3 (x − a)2 + 4c4 (x − a)3 + 5c5 (x − a)4 + . . . f (x) = 1 · 2c2 + 2 · 3c3 (x − a) + 3 · 4c4 (x − a)2 + 4 · 5c5 (x − a)3 + . . . f (x) = 1 · 2 · 3c3 + 2 · 3 · 4c4 (x − a) + 3 · 4 · 5c5 (x − a)2 + . . . f (x) = c 0 + c1 (x
.. .
If we let x = a, then f (a) = c0 f (a) = c1 f (a) = 2c2 f (a) = 2 3c3 f (4)(a) = .. .
· 2 · 3 · 4c4
If we continue this process, we get the generalization f (n) (a) = 2 3 4 . . . ncn = n!cn .
· · · ·
Solving for the nth coefficient, we obtain f (n) (a) cn = . n! 11
Theorem. If f has a power series representation about a, that is if ∞
f (x) =
cn (x n=0
− a)n, |x − a| < R
f (n) (a) then its coefficients are given by the formula cn = . n! Definition. Given a real-valued function f (x). 1. The power series f (n) (a) (x n! n=0 ∞
−
f (a) a) = f (a) + (x 1! n
−
f (a) a) + (x 2!
−
f (a) a) + (x 3! 2
− a)3 + ···
is called the Taylor series of f about a. 2. The power series f (n) (0) n f (0) f (0) 2 f (0) 3 x = f (0) + x + x + x + n! 1! 2! 3! n=0 ∞
···
is called the Maclaurin series of f . ∞ ∞ ∞ 2n+1 2n 2n+1 xn n x n x nx Examples: 1(a). ; 1(b). ( 1) ; 1(c). ( 1) ; 1(d). ( 1) ; n! (2n + 1)! (2n)! (2n)! n=0 n=0 n=0 n=0 ∞ ∞ ∞ ∞ n n 3 n 2n 2n+1 (x 2) ( 1) π ( 1) π 2 2(a). e ; 2(c). x + x ; 2(f). ( 1)n (x n! 2(2n)! 3 2(2n + 1)! 3 n=0 n=0 n=0 n=0 1)n ; 3(a). e2 1 Remark: The Taylor/Maclaurin series of f is not necessarily equal to f . Example: 4. Maclaurin series expansion is 0 (use L’Hopital’s Rule), but f (x) = 0 if x = 0 ∞
− √
−
−
−
−
−
−
−
−
−
−
Theorem. Let f be a function such that f and its derivatives exist in some interval (a
Then the function is represented by its Taylor series if and only if f (n+1) (zn ) (x n→∞ (n + 1)!
lim Rn (x) = lim
n→∞
− r, a + r).
− a)n+1 = 0,
where zn is some number between x and a.
7
Binomial Series
Recall that if k is a natural number, the binomial series expansion of (a + b)k is given by (a + b)k = a k + kak−1 b +
k(k 1) k−2 2 a b + 2!
−
··· + k(k − 1) · . . n!. · (k − n + 1) ak
n n
−
b +
Note that k 0
= 1,
k n
=
k(k
− 1)(k − 2) · . . . · (k − n + 1) , n!
and so
k
k
k n
(a + b) = n=0
ak−n bn .
In particular, we have k
k
(1 + x) = n=0
k n
xn .
We now generalize the binomial series to the case where k is any real number. 12
··· + kabk
1
−
+ bk .
Theorem. (The Binomial Theorem) Let k be a real number and x < 1.
(1 + x)k =
| | k (k − 1) 2 k (k − 1)(k − 2) 3 k (k − 1) · . . . · (k − n + 1) n 1 + kx + x + x + ··· + x + ···
2! k(k 1) . . . (k = 1+ n! n=1 ∞
− · · −
3! n + 1)
n!
xn
If k is a positive integer, the binomial series terminates after a finite number of terms. Proof. For x < 1, let
| |
∞
k(k
f (x) = 1 +
− 1) · . . . · (k − n + 1) xn. n!
n=1
Then
∞
k(k
f (x) = n=1
− 1) · . . . · (k − n + 1) xn (n − 1)!
1
−
.
Multiplying f (x) by x, ∞
xf (x) =
k(k
− 1) · . . . · (k − n + 1) xn (n − 1)! n=1 k(k − 1) · . . . · (k − n + 1) n n x . ∞
=
n!
n=1
Rewriting f (x), ∞
− 1) · . . . · (k − n + 1) xn 1 (n − 1)! n=2 k(k − 1) · . . . · (k − n + 1) n k + (k − n) x .
f (x) = k +
k(k
−
∞
=
n!
n=1
The series representations for f (x) and xf (x) are absolutely convergent for x < 1, so the sum of the two series is given by
| |
∞
k(k
(1 + x)f (x) = k 1 + n=1
− 1) · . . . · (k − n + 1) xn n!
= kf (x). Therefore, f (x) f (x)
k 1 + x d d (ln f (x)) = ln(1 + x)k dx dx ln f (x) = ln(1 + x)k + C =
But f (0) = 1, so C = 0. Therefore, f (x) = (1 + x)k ∞
Examples: 1(a). 1+ ∞
|x| < 1; 1(c). x +
n=1
( 1)
n1
· 3 · 5 · . . . · (2n − 1) xn, |x| < 1; 1(b). 1+
− 2n n! n=1 1 · 3 · . . . · (2n − 1) x2n+1 · , |x| < 1; 1(e). 2n n!
2n + 1
13
∞
1 3 . . . (2n 2n n! n=1
∞
· · ·
( 1)n (n + 1)xn , x < 1 n=0
−
| |
− 1) x2n,
