Chapter 4: Carry out binomial expansion

Biomedical engineering frequently requires the use of algebraic expansions to model physiological phenomena, analyze biomedical signals, and solve engineering problems involving small perturbations or approximations. The binomial theorem provides a powerful method to expand expressions raised to a power, especially when dealing with terms like \((1 + x)^n\), which commonly appear in biomedical instrumentation calibrations and error estimations. Mastery of binomial expansion enables biomedical engineers to simplify complex expressions for easier computation and analysis in clinical and research settings.

4.1 Basic concepts of binomial theorem

A binomial is an algebraic expression with two terms, such as x + 3 or 2a − b. Raising a binomial to a power by repeated multiplication becomes long and error-prone as the power grows. The binomial theorem gives a short rule for the expansion, using numbers called binomial coefficients that come from counting combinations. Expansions of this kind are used to approximate functions and to simplify expressions in calculations on sensor signals and mechanical systems.

4.1.1 Binomial expressions and notation

A binomial expression is the sum or difference of two terms, written (a + b)ⁿ when it is raised to a power. The letters a and b stand for the two terms, which may be numbers, variables or products such as 2x, and n is the power, a positive whole number. A difference is handled as a sum by treating the second term as negative: (x − 2)ⁿ is (x + (−2))ⁿ, with a = x and b = −2.

The binomial theorem states that

(a + b)ⁿ = C(n, 0) aⁿ + C(n, 1) a^(n−1) b + C(n, 2) a^(n−2) b² + ... + C(n, n−1) a bⁿ⁻¹ + C(n, n) bⁿ

Here C(n, r), also written nCr, is the binomial coefficient of the term containing bʳ. In every term the power of a falls by one while the power of b rises by one, and the two powers add up to n. The expansion has n + 1 terms.

Example 1: Identify a, b and n in (2x − 3)⁴.

(2x − 3)⁴ = (2x + (−3))⁴

Answer: a = 2x, b = −3 and n = 4. The expansion has 4 + 1 = 5 terms.

Example 2: Expand (x + 1)³.

Here a = x, b = 1 and n = 3. The coefficients C(3, 0), C(3, 1), C(3, 2) and C(3, 3) are 1, 3, 3 and 1.

(x + 1)³ = 1 × x³ × 1⁰ + 3 × x² × 1¹ + 3 × x × 1² + 1 × x⁰ × 1³

= x³ + 3x² + 3x + 1

Answer: x³ + 3x² + 3x + 1

Example 3: Expand (2 + y)².

Here a = 2, b = y and n = 2. The coefficients are 1, 2 and 1.

(2 + y)² = 1 × 2² × y⁰ + 2 × 2¹ × y¹ + 1 × 2⁰ × y²

= 4 + 4y + y²

Answer: 4 + 4y + y²

Example 4: Expand (1 + x)⁴.

Here a = 1, b = x and n = 4. The coefficients are 1, 4, 6, 4 and 1.

(1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴

Answer: 1 + 4x + 6x² + 4x³ + x⁴

Example 5: Expand (x − 2)³.

Here a = x, b = −2 and n = 3. The coefficients are 1, 3, 3 and 1.

(x − 2)³ = x³ + 3x²(−2) + 3x(−2)² + (−2)³

= x³ − 6x² + 3x(4) − 8

= x³ − 6x² + 12x − 8

Answer: x³ − 6x² + 12x − 8

4.1.2 Factorials and their use in binomial coefficients

The factorial of a positive whole number n, written n!, is the product of all the whole numbers from n down to 1: n! = n × (n − 1) × (n − 2) × ... × 2 × 1. By definition 0! = 1 and 1! = 1. Factorials grow very quickly: 5! = 120, but 10! = 3 628 800. A quotient of factorials is simplified by cancelling the common factors, for example 8! ÷ 6! = 8 × 7 because 8! = 8 × 7 × 6!.

Binomial coefficients are written with factorials:

C(n, r) = n! ÷ [r! × (n − r)!]

The numerator is the factorial of n, and the denominator is the product of the factorials of r and of n − r. The two numbers in the denominator always add up to n.

Example 1: Evaluate 6!, 0! and 1!.

6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

0! = 1 and 1! = 1

Answer: 720, 1 and 1.

Example 2: Simplify 8! ÷ 6!.

8! ÷ 6! = (8 × 7 × 6!) ÷ 6! = 8 × 7

Answer: 56

Example 3: Find C(5, 2).

C(5, 2) = 5! ÷ [2! × (5 − 2)!] = 5! ÷ (2! × 3!)

5! = 120, 2! = 2 and 3! = 6

C(5, 2) = 120 ÷ (2 × 6) = 120 ÷ 12

Answer: 10

Example 4: Find C(8, 5).

C(8, 5) = 8! ÷ (5! × 3!) = 40 320 ÷ (120 × 6) = 40 320 ÷ 720

Answer: 56

Example 5: Find C(6, 0) and C(10, 10).

C(6, 0) = 6! ÷ (0! × 6!) = 6! ÷ (1 × 6!) = 1

C(10, 10) = 10! ÷ (10! × 0!) = 1

Answer: Both equal 1. In general C(n, 0) = C(n, n) = 1.

4.1.3 Binomial Coefficients

Binomial coefficients are the numerical factors that multiply the terms in the expansion of (a + b)ⁿ. Three facts help to find them quickly.

• End values: C(n, 0) = 1 and C(n, n) = 1, so the first and last terms of every expansion have coefficient 1.
• Symmetry: C(n, r) = C(n, n − r), so the coefficients read the same from both ends. For example C(8, 3) = C(8, 5).
• Sum: the coefficients of (a + b)ⁿ add up to 2ⁿ, because putting a = 1 and b = 1 in the expansion gives 2ⁿ.

4.1.3.1 Definition and calculation using combinations (nCr)

A combination is a selection of r items from n different items in which the order does not matter. The number of such selections is C(n, r), written nCr and read "n choose r". It is the same number as the binomial coefficient, because each term of (a + b)ⁿ with bʳ is formed by choosing which r of the n brackets supply the b.

For example, the combinations of 2 letters from A, B and C are AB, AC and BC, so C(3, 2) = 3.

Two ways of calculating nCr:

• Factorial formula: C(n, r) = n! ÷ [r! × (n − r)!].
• Short form: write r factors from n downward on top and r! below: C(n, r) = [n × (n − 1) × ... (r factors)] ÷ r!. This avoids long factorials.

A scientific calculator has an nCr key, used by entering n, pressing nCr, entering r and pressing equals. The result should be checked against the symmetry rule.

Example 1: Find C(7, 3).

C(7, 3) = 7! ÷ (3! × 4!) = 5 040 ÷ (6 × 24) = 5 040 ÷ 144

Answer: 35

Example 2: Find C(10, 3) by the short form.

C(10, 3) = (10 × 9 × 8) ÷ (3 × 2 × 1) = 720 ÷ 6

Answer: 120

Example 3: Find C(10, 1).

C(10, 1) = 10! ÷ (1! × 9!) = 10 × 9! ÷ 9!

Answer: 10

Example 4: A technician places 3 electrodes on a phantom that has 8 marked positions, and the order of placing does not matter. How many different sets of 3 positions are possible?

Number of sets = C(8, 3) = (8 × 7 × 6) ÷ (3 × 2 × 1) = 336 ÷ 6

Answer: 56 different sets of positions.

Example 5: A test rig has 6 sensors and 2 are to be chosen for a check. How many different pairs can be chosen?

C(6, 2) = (6 × 5) ÷ (2 × 1) = 30 ÷ 2

Answer: 15 pairs.

4.1.3.2 Pascal's Triangle as a tool for finding binomial coefficients

Pascal's triangle is an arrangement of binomial coefficients in rows. Row n lists C(n, 0), C(n, 1), ..., C(n, n). Each number is the sum of the two numbers above it, and the edges are all 1. This follows from Pascal's rule:

C(n, r) = C(n − 1, r − 1) + C(n − 1, r)

The first rows are:

n = 0                 1
n = 1               1   1
n = 2             1   2   1
n = 3           1   3   3   1
n = 4         1   4   6   4   1
n = 5       1   5  10  10   5   1
n = 6     1   6  15  20  15   6   1
n = 7   1   7  21  35  35  21   7   1

Row n gives the coefficients of the expansion of (a + b)ⁿ. For (a + b)⁴ the coefficients are 1, 4, 6, 4, 1. The triangle is quick for small n, and for large n the factorial formula is quicker.

Steps for building the next row:

1. Write 1 at each end of the new row.
2. Add each pair of neighbouring numbers in the row above.
3. Write each sum between the two numbers that were added, in the new row.
4. Check the symmetry of the row and that its numbers add up to 2ⁿ.

Example 1: Find C(4, 2) with Pascal's rule.

C(4, 2) = C(3, 1) + C(3, 2) = 3 + 3

Answer: 6

Example 2: Find C(5, 3) with Pascal's rule.

C(5, 3) = C(4, 2) + C(4, 3) = 6 + 4

Answer: 10

Example 3: Find C(6, 4) by symmetry.

C(6, 4) = C(6, 2) = (6 × 5) ÷ (2 × 1) = 15

Answer: 15

Example 4: Use row 6 of the triangle to find C(7, 4).

C(7, 4) = C(6, 3) + C(6, 4) = 20 + 15

Answer: 35

Example 5: Build row 8 from row 7 and check that its sum is 2⁸.

Row 7 is 1, 7, 21, 35, 35, 21, 7, 1.

Row 8: 1, 1 + 7, 7 + 21, 21 + 35, 35 + 35, 35 + 21, 21 + 7, 7 + 1, 1

= 1, 8, 28, 56, 70, 56, 28, 8, 1

Sum = 1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 256 = 2⁸

Answer: Row 8 is 1, 8, 28, 56, 70, 56, 28, 8, 1, and it sums to 256.

Chapter Summary

This chapter introduced the basic concepts of the binomial theorem and binomial expansion, starting with binomial expressions and notation. It explained factorials and their essential role in calculating binomial coefficients, including the use of combinations denoted as \(nCr\). Pascal's Triangle was presented as a practical tool for determining binomial coefficients efficiently. The general form of the binomial expansion was then explored, with detailed expansions for powers \(n=2, 3, 4\). Special cases such as expansions of \((1 + x)\) and the use of the binomial series for negative and fractional powers were also covered. The chapter demonstrated how binomial expansion can simplify algebraic expressions and assist in solving polynomial equations. Finally, it highlighted the importance of binomial expansion in engineering applications and showed how it helps estimate values in engineering calculations.