6.10. THE BINOMIAL 375

6.10 The 6.10.1 Introduction This section focuses on deriving a Maclaurin series for functions of the form f (x) = (1 + x)k for any number k. We use the results we obtained in the section on Taylor and Maclaurin series and combine them with a known and useful result known as the to derive a nice formula for a Maclaurin series for f (x) = (1 + x)k for any number k.

6.10.2 The Binomial Theorem This theorem deals with expanding expressions of the form (a + b)k where k is a positive . In the case k = 2, the result is a known identity

(a + b)2 = a2 + 2ab + b2

It is also easy to derive an identity for k = 3.

(a + b)3 = a3 + 3a2b + 3ab2 + b3

There is also a formula for k in general. That formula is known as the Binomial Theorem. Before we state it, let us explain it a little bit. (a + b)k will be a sum of terms. Each term will contain a coeffi cient as well as powers of a and b. More precisely, we will have

(a + b)k =

k k 1 k (k 1) k 2 2 k (k 1) (k 2) k 3 3 a + ka − b + − a − b + − − a − b 2! 3! k (k 1) (k 2) ... (k n + 1) k n k +... + − − − a − b n! k 1 k +... + kab − + b We see from the formula that the powers of a and b are of the form aibj where i decreases from k to 0 and j increases from 0 to k. The coeffi cients k (k 1) (k 2) ... (k n + 1) − − − which appear in this expansion are called bino- n! mial coeffi cients. We use a special notation for them.

Definition 6.10.1 (Binomial coeffi cients) The binomial coeffi cients, denoted k n , are defined by:  k k (k 1) (k 2) ... (k n + 1) = − − − if n 1 n n! ≥   k = 1 0   Remark 6.10.2 The numerator of the fraction in the definition has exactly n terms. This is helpful when figuring out the coeffi cients. 376 CHAPTER 6. INFINITE SEQUENCES AND SERIES

This notation allows us to write:

Theorem 6.10.3 (Binomial Theorem) Suppose that k is a positive integer. Then k k (a + b)k = ak nbn n − n=0 X   4 Example 6.10.4 Find n for n = 0, 1, 2, 3, 4.

4 = 1  • 0 4 = 4 • 1  4 3 4 = × = 6 • 2 2!  4 3 2 4 = × × = 4 • 3 3!  4 3 2 1 4 = × × × = 1 • 4 4!  Example 6.10.5 Expand (a + b)4. From the binomial theorem, we have

4 4 (a + b)4 = a4 nbn n − n=0 X   4 4 4 4 4 = a4 + a3b + a2b2 + ab3 + b4 0 1 2 3 4           = a4 + 4a3b + 6a2b2 + 4ab3 + b4

For what follows, we will be interested in expanding (1 + x)k. In the case k is a positive integer, the binomial theorem gives us

k k (1 + x)k = xn n n=0 X   6.10.3 The Binomial Series The binomial series extends the binomial theorem for cases when k is not an 1 integer. For example, how would we expand (1 + x) 2 ? In other words, given f (x) = (1 + x)k, for any k, what is a Maclaurin series for f? We derive it like any other Maclaurin series. That is, we have

∞ xn (1 + x)k = f (n) (0) n! n=0 X 6.10. THE BINOMIAL SERIES 377

So, we need to find f (n) (0). f (x) = (1 + x)k f (0) = 1 k 1 f 0 (x) = k (1 + x) − f 0 (0) = k k 2 f 00 (x) = k (k 1) (1 + x) − f 00 (0) = k (k 1) − k 3 − f 000 (x) = k (k 1) (k 2) (1 + x) − f 000 (0) = k (k 1) (k 2) − . − . − − . . (n) k n (n) f (x) = k (k 1) ... (k n + 1) (1 + x) − f (0) = k (k 1) ... (k n + 1) − − − − Therefore, the Maclaurin series for (1 + x)k is

∞ xn (1 + x)k = k (k 1) ... (k n + 1) n! n=0 − − X ∞ k = xn n n=0 X   This series is known as the binomial series. To study its convergence, we use the ration test.

an+1 | | = an | | k (k 1) ... (k n) x n+1 n! | − − | | | (n + 1)! k (k 1) ... (k n + 1) x n | − − | | | k n = | − | x n + 1 | | So, since k n lim | − | = 1 n n + 1 →∞ we have k n lim | − | x = x n n + 1 →∞ | | | | By the , this series converges if x < 1. Convergence at the endpoints depends on the values of k and needs to| be| checked every time.

Definition 6.10.6 (Binomial Series) If x < 1 and k is any real number, then | | ∞ k (1 + x)k = xn n n=0 X   k where the coeffi cients n are the binomial coeffi cients. This series is called the binomial series.  Remark 6.10.7 This formula is very similar to the binomial theorem. In this case, we have an infinite sum. In the case of the binomial theorem (k is a k positive integer), we have a finite sum because n = 0 whenever k > n (why?).  378 CHAPTER 6. INFINITE SEQUENCES AND SERIES

1 Example 6.10.8 Expand as a . 1 + x 1 We have already done this using substitution and the power series of . We 1 x found − 1 = 1 x + x2 x3 + ... 1 + x − − ∞ = ( 1)n xn n=0 − X We just derive the series to illustrate our new technique. From the formula above, we will have

1 1 = (1 + x)− 1 + x ∞ 1 = − xn n n=0 X   1 So, we need to compute −n . Remember that there is a separate definition depending on whether n = 0 or n 1. So, we need to handle both cases. As we will see in the examples, sometimes  ≥ the formula we get for n 1 also works for n = 0, but not always. ≥ 1 − = 1 0   and 1 ( 1) ( 1 1) ( 1 2) ... ( 1 n + 1) − = − − − − − − − n n!   Remember that the numerator has n factors. So, we get

1 ( 1)n (1) (2) (3) ... (n) − = − n n!   = ( 1)n − This formula also works in the case n = 0, so we can combine both cases and get 1 ∞ = ( 1)n xn 1 + x n=0 − X Which is the same formula as we had found using a different method. 1 Example 6.10.9 Expand as a power series. √1 + x First, we notice that 1 1 = (1 + x)− 2 √1 + x 6.10. THE BINOMIAL SERIES 379

1 So, we use the formula above in the case k = . We obtain: −2 1 1 ∞ − = 2 xn √ n 1 + x n=0 X   where 1 − 2 = 1 0   and if n 1, ≥ 1 1 1 1 1 1 2 ... n + 1 − −2 −2 − −2 − −2 − 2 =       n n!   1 3 5 2n 1 ... − −2 −2 −2 − 2 =         n! The numerator has n factors. We get 1 − (1) (3) (5) ... (2n 1) 2 = ( 1)n − n − 2nn!   This formula does not work when n = 0, so we cannot combine it with what we found when n = 0. Therefore, 1 1 (1) (3) (1) (3) (5) = 1 x + x2 x3.... √1 + x − 2 222! − 233! ∞ (1) (3) (5) ... (2n 1) = 1 + ( 1)n − xn 2nn! n=1 − X 6.10.4 Problems 1. Using the binomial theorem, expand (a + b)5.

k k k k k k 2. Write and simplify the expression for 0 , 1 , 2 , 3 , k 1 , k . −       k 3. In many books, the binomial coeficients are defined by the formula = n k!   . Prove that this is the same formula as the one given in the n!(k n)! notes.− 4. Using the binomial series, find the Maclaurin series for the functions below. In each case, find the .

(a) √1 + x 380 CHAPTER 6. INFINITE SEQUENCES AND SERIES

1 (b) (2 + x)3 x (c) √4 + x2

3 5. Using the binomial series, find the first three Taylor polynomials of (1 + 2x) 4 . Graph the and the Taylor polynomials. 1 6. Consider f (x) = √1 x2 − (a) Using the binomial series, find the Maclaurin series for f. 1 (b) Using part a, find the Maclaurin series for sin− x.

6.10.5 Answers 1. Using the binomial theorem, expand (a + b)5.

(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

k k k k k k 2. Write and simplify the expression for 0 , 1 , 2 , 3 , k 1 , k . − k       = 1 0   k = k 1   k k (k 1) = − 2 2   k k (k 1) (k 2) = − − 3 3!   k = k k 1  −  k = 1 k   k 3. In many books, the binomial coeficients are defined by the formula = n k!   . Prove that this is the same formula as the one given in the n!(k n)! notes.−

4. Using the binomial series, find the Maclaurin series for the functions below. In each case, find the radius of convergence. 6.10. THE BINOMIAL SERIES 381

(a) √1 + x

(1)(3)(5)...(2n 3) x ∞ n 1 − n √1 + x = 1 + + ( 1) − x and R = 1 2 2nn! n=2 − X 1 (b) (2 + x)3

1 ∞ (n + 1) (n + 2) = ( 1)n xn and R = 2 3 2n+4 (2 + x) n=0 − X x (c) √4 + x2

x 1 ∞ (1)(3)(5)...(2n 1) = x + ( 1)n − x2n+1 and R = 2 √ 2 2 − 23n+1n! 4 + x n=1 X

5. Using the binomial series, find the first three Taylor polynomials (T1, T2, 3 4 T3) of (1 + 2x) . Graph the function and the Taylor polynomials. 3 T (x) = 1 + x 1 2 3 3 2 T2 (x) = 1 + x x 2 − 8 3 3 2 5 3 T3 (x) = 1 + x x + x 2 − 8 16

1 6. Consider f (x) = √1 x2 − (a) Using the binomial series, find the Maclaurin series for f.

1 ∞ (1)(3)(5)...(2n 1) = 1 + − x2n √ 2 2nn! 1 x n=1 − X 1 (b) Using part a, find the Maclaurin series for sin− x.

(1)(3)(5)...(2n 1) 2n+1 1 ∞ − x sin− x = x + 2nn! 2n + 1 n=1 X