2 Jan 1999

efinition

Binomial means the sum (or difference) of two terms.

Expansion of (x + y)n, where n is any rational number, is called the binomial expansion.

ositive Integral Index

If n is a positive integer, then

 (x + y)n = xn + nxn - 1y + n(n - 1) ¾¾¾¾ 2! xn - 2y2 + n(n - 1)(n - 2) ¾¾¾¾¾¾¾ 3! xn - 3y3 + ¼ + yn.

Notes:

1. There are n + 1 terms.
2. The (r + 1)th term is nCr xn - ryr.
3. The expansion of (x - y)n may be obtain by replacing y by -y.

on-positive Integral Index

If n is not a positive integer, then we have

 (1 + x)n = 1 + nx + n(n - 1) ¾¾¾¾ 2! x2 + n(n - 1)(n - 2) ¾¾¾¾¾¾¾ 3! x3 + ¼.

Notes:

1. If n is NOT a positive integer, the series is infinite.
2. The above expansion is in ascending powers of x, and is only valid (ie meaningful) when |x| < 1 (small x).
3. To expand (a + x)n in ascending powers of x, it must first be written as an(1 + x/a)n.

pecial Series

(1 - x)-1 = 1 + x + x2 + ¼ + xr + ¼.

(1 - x)-2 = 1 + 2x + 3x2 + ¼ + (r + 1)xr + ¼.

Replace x by -x to get the corresponding expansions for (1 + x)-1 and (1 + x)-2.

xamples

Example 1: Find the first four terms of (2 - x)-2, stating the set of values of x for which the expansion is valid.

Solution:

(2 - x)-2  =  2-2(1 - x/2)-2

=
 1 é -2(-3) -2(-3)(-4) ù - ê 1 + (-2)(- x/2) + ¾¾¾ (- x/2)2 + ¾¾¾¾¾ (- x/2)3 + ¼ ú 4 ë 2! 3! û

=  1/4 + (1/4)x + (3/16)x2 + (1/8)x3 + ¼.

The expansion is valid provided |x/2| < 1, ie, |x| < 2.

Example 2:  Give the binomial expansion, for small x, of (1 + x)1/4 up to and including the term in x2, and simplify the coefficients.
By putting x = 1/16 in your expression, show that  4Ö17 » 8317/4096.

Solution:

 1 (1/4)(-3/4) (1 + x)1/4 = 1 + ¾ x + ¾¾¾¾¾ x2 + ¼ 4 2! = 1 + x/4 - 3x2/32 + ¼.

Putting x = 1/16,

 1/16 3(1/16)2 (1 + 1/16)1/4 = 1 + ¾¾ - ¾¾¾¾ + ¼ 4 32 (17/16)1/4 » 8317/8192 \  4Ö17 » 8317/4096.

 5 Example 3:  Expand ¾¾¾¾¾¾¾ as a series (1 + 3x)(1 - 2x)
of ascending powers of x giving:
(a)    the first four terms,
(b)    the range of values of x for which the expansion is valid.

Solution:

 5 Expressing ¾¾¾¾¾¾¾ as partial fractions gives (1 + 3x)(1 - 2x)

 5 3 2 ¾¾¾¾¾¾¾ = ¾¾¾ + ¾¾¾ (1 + 3x)(1 - 2x) 1 + 3x 1 - 2x = 3(1 + 3x)-1 + 2(1 - 2x)-1.

 (1 + 3x)-1 = 1 - 3x + (3x)2 - (3x)3 + ¼ = 1 - 3x + 9x2 - 27x3 + ¼.

 (1 - 2x)-1 = 1 + 2x + (2x)2 + (2x)3 + ¼ = 1 + 2x + 4x2 + 8x3 + ¼.

 5 \ ¾¾¾¾¾¾¾ = 3[1 - 3x + 9x2 - 27x3 + ¼] (1 + 3x)(1 - 2x) + 2[1 + 2x + 4x2 + 8x3 + ¼] = 5 - 5x + 35x2 - 65x3 + ¼.

Expansion is valid when

 |3x| < 1 and |2x| < 1 |x| < 1/3 and |x| < 1/2 Þ  |x| < 1/3.

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