By the end of this chapter, you will be able to:
These skills will help you solve practical engineering problems with confidence and precision in your electronics work.
Compiled by: Peter Kiprono
Electronics engineers often encounter binomial expressions when analyzing circuits, signal processing algorithms, and error approximations. Understanding how to expand these expressions quickly and accurately is crucial for simplifying complex algebraic terms encountered in filter design, amplifier gain calculations, and probability estimates in digital communications. This chapter introduces the binomial theorem, essential notation, and the use of factorials to calculate binomial coefficients, which form the foundation for carrying out binomial expansions in electronics engineering problems.
A binomial is an algebraic expression with two terms, such as a + b or 1 + x. The binomial theorem gives a way to write the power of a binomial, (a + b)ⁿ, as a sum of terms without multiplying the brackets out one by one. The sizes of the terms are decided by numbers called binomial coefficients, and these in turn are built from factorials. In electronics the theorem is used to approximate expressions such as (1 + x)ⁿ for small x, for example when a component value differs slightly from its nominal value. This topic covers the notation, factorials and binomial coefficients that the expansion is built from.
A binomial expression is a sum or difference of two terms. A power of a binomial, (a + b)ⁿ, with n a whole number, is expanded by the binomial theorem:
(a + b)ⁿ = C(n, 0) aⁿ + C(n, 1) aⁿ⁻¹ b + C(n, 2) aⁿ⁻² b² + … + C(n, n) bⁿ
In summation notation this is written (a + b)ⁿ = Σ C(n, k) aⁿ⁻ᵏ bᵏ, with the sum taken for k = 0, 1, 2, … , n. The symbol Σ means "the sum of", and k is the counter that runs through the terms.
C(n, k) is the binomial coefficient. It is also written nCk, nCr or "n choose k", and it counts the number of ways to choose k items from n. Its calculation is explained under factorials and binomial coefficients below.
The structure of the expansion can be read directly from the formula.
If the second term is subtracted, as in (a − b)ⁿ, the signs of the terms alternate, because b is replaced by −b.
Example 1: Expand (1 + x)⁴.
Here a = 1, b = x and n = 4.
k = 0: C(4, 0) × 1⁴ × x⁰ = 1
k = 1: C(4, 1) × 1³ × x = 4x
k = 2: C(4, 2) × 1² × x² = 6x²
k = 3: C(4, 3) × 1 × x³ = 4x³
k = 4: C(4, 4) × 1⁰ × x⁴ = x⁴
Answer: (1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴.
Example 2: Expand (2 + y)².
Here a = 2, b = y and n = 2.
k = 0: C(2, 0) × 2² = 4
k = 1: C(2, 1) × 2 × y = 4y
k = 2: C(2, 2) × y² = y²
Answer: (2 + y)² = 4 + 4y + y².
Example 3: Expand (2 + 0.5)³ and find its value.
Here a = 2, b = 0.5 and n = 3.
k = 0: 2³ = 8
k = 1: 3 × 2² × 0.5 = 6
k = 2: 3 × 2 × 0.5² = 1.5
k = 3: 0.5³ = 0.125
Sum: 8 + 6 + 1.5 + 0.125 = 15.625
Answer: (2 + 0.5)³ = 15.625, which agrees with 2.5³ = 15.625.
The factorial of a 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. A factorial can be built from the one before it, n! = n × (n − 1)!, which is how the table below is filled.
| n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| n! | 1 | 1 | 2 | 6 | 24 | 120 | 720 | 5040 | 40 320 | 362 880 | 3 628 800 |
Factorials grow very quickly. The binomial coefficient is written with factorials as
C(n, k) = n! / (k! (n − k)!), for 0 ≤ k ≤ n
Because n! contains (n − k)! as a factor, most of the product cancels, and the coefficient is quicker to find in the shortened form
C(n, k) = [n × (n − 1) × … × (n − k + 1)] / k!
with exactly k factors in the top line. For example, C(10, 3) = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120.
Example 1: Calculate 5!.
5! = 5 × 4 × 3 × 2 × 1 = 120
Answer: 5! = 120.
Example 2: Calculate C(5, 2).
C(5, 2) = 5! / (2! × 3!) = 120 / (2 × 6) = 120 / 12 = 10
Answer: C(5, 2) = 10.
Example 3: Calculate C(7, 3).
C(7, 3) = 7! / (3! × 4!) = 5040 / (6 × 24) = 5040 / 144 = 35
Answer: C(7, 3) = 35. A technician who has to choose 3 test points out of 7 can do so in 35 different ways.
Example 4: Calculate C(10, 0).
C(10, 0) = 10! / (0! × 10!) = 10! / (1 × 10!) = 1
Answer: C(10, 0) = 1.
Example 5: Calculate C(8, 2) with the shortened form.
C(8, 2) = (8 × 7) / (2 × 1) = 56 / 2 = 28
Answer: C(8, 2) = 28.
The binomial coefficients are the numbers that multiply the terms in a binomial expansion. They can be found from factorials, from combinations, or from Pascal's triangle. The coefficients have these properties, which are used to check results and to save work.
A combination is a selection of r items from n different items when the order does not matter. The number of such selections is written nCr, and it equals the binomial coefficient C(n, r).
nCr = n! / (r! (n − r)!)
Selecting r items to keep is the same as selecting n − r items to leave out, which is why nCr = nC(n − r). Calculators have an nCr key that gives the value directly.
The coefficient of bᵏ in (a + b)ⁿ is nCk, because each term of the expansion arises by choosing b from k of the n brackets and a from the others.
Example 1: A stores clerk can take 2 different resistor values from a set of 6. How many different pairs are possible?
6C2 = 6! / (2! × 4!) = (6 × 5) / (2 × 1) = 15
Answer: there are 15 pairs.
Example 2: Verify the symmetry 6C2 = 6C4.
6C4 = 6! / (4! × 2!) = 720 / (24 × 2) = 15
Answer: 6C2 = 6C4 = 15.
Example 3: Verify that the coefficients for n = 3 add up to 2³.
3C0 + 3C1 + 3C2 + 3C3 = 1 + 3 + 3 + 1 = 8
Answer: the sum is 8 = 2³.
Example 4: Calculate 12C3.
12C3 = (12 × 11 × 10) / (3 × 2 × 1) = 1320 / 6 = 220
Answer: 12C3 = 220.
Pascal's triangle is a triangular array in which each row gives the binomial coefficients for one value of n. Every row begins and ends with 1, and every other number is the sum of the two numbers diagonally above it. Rows and entries are counted from 0, so the entry in row n at position k is C(n, k), and row n gives the coefficients of (a + b)ⁿ.
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Row 5: 1 5 10 10 5 1
Row 6: 1 6 15 20 15 6 1
Row 7: 1 7 21 35 35 21 7 1
Row 8: 1 8 28 56 70 56 28 8 1
The triangle is quick for small n. For large n, or for a single coefficient, the factorial formula is faster. The numbers in row n add up to 2ⁿ.
Example 1: Use Pascal's triangle to find C(6, 3).
Row 6 is 1, 6, 15, 20, 15, 6, 1. Counting from position 0, the entry at position 3 is 20.
Answer: C(6, 3) = 20, which agrees with (6 × 5 × 4) / (3 × 2 × 1) = 120 / 6 = 20.
Example 2: Use Pascal's triangle to find C(4, 1).
Row 4 is 1, 4, 6, 4, 1, and the entry at position 1 is 4.
Answer: C(4, 1) = 4, which agrees with 4! / (1! × 3!) = 24 / 6 = 4.
Example 3: Write the coefficients of the expansion of (a + b)⁶.
Row 6 is 1, 6, 15, 20, 15, 6, 1.
Answer: (a + b)⁶ = a⁶ + 6a⁵b + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6ab⁵ + b⁶.
Example 4: Build row 7 from row 6 and check that its sum is 2⁷.
Row 7 starts with 1. The next entries are 1 + 6 = 7, 6 + 15 = 21, 15 + 20 = 35, 20 + 15 = 35, 15 + 6 = 21, 6 + 1 = 7, and the row ends with 1.
Answer: row 7 is 1, 7, 21, 35, 35, 21, 7, 1, and the sum is 1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = 128 = 2⁷.
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Create a free accountThis chapter introduced the basic concepts of the binomial theorem, including binomial expressions and their notation, the role of factorials in calculating binomial coefficients, and how to determine these coefficients using combinations and Pascal's Triangle. It then explored the general form of binomial expansion and demonstrated expansions for powers of two, three, and four. Special cases such as expansions of expressions in the form \((1 + x)^n\), including negative and fractional powers, were examined using the binomial series. Practical applications were discussed, highlighting how binomial expansion simplifies algebraic expressions and assists in solving polynomial equations. The chapter also covered engineering applications where binomial expansions are used to model and solve real-world problems. Finally, methods for estimating values in engineering calculations using binomial expansions were presented to enhance accuracy and efficiency.
Expand the expression \((a + b)^2\) and find the coefficient of the \(b^2\) term. (2 marks)
Calculate the value of \({}^5C_3\) (the number of combinations of 5 items taken 3 at a time). (2 marks)
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