Binomial Expansions

# Binomial Expansions ## Objectifs et prérequis This chapter aims to equip students with the ability to expand binomial expressions using Pascal’s Triangle and the Binomial Theorem, identify specific terms and coefficients, extend the theorem to rational exponents, and apply binomial approximations in practical contexts. Students should be comfortable with algebraic manipulation, factorial notation, combinations, and basic limits before proceeding. ## Introduction Binomial expansions provide a systematic method for expanding expressions of the form $(a + b)^n$, where $n$ is a rational number. This technique is foundational in algebra, calculus, and applied mathematics, enabling the approximation of functions, computation of probabilities, and analysis of series. The chapter progresses from concrete constructions (Pascal’s Triangle) to general theorems and finally to approximations, ensuring both conceptual depth and procedural fluency. ## I. Generating Pascal’s Triangle Pascal’s Triangle is a triangular array of numbers where each number is the sum of the two numbers directly above it. It provides the binomial coefficients for the expansion of $(a + b)^n$ when $n$ is a non-negative integer. > **Definition**: Pascal’s Triangle is constructed such that the entry in row $n$ and column $k$ (both starting from 0) is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, representing the number of ways to choose $k$ elements from a set of $n$. **Construction**: - Row 0: $1$ - Row 1: $1\quad 1$ - Row 2: $1\quad 2\quad 1$ - Row 3: $1\quad 3\quad 3\quad 1$ - Row 4: $1\quad 4\quad 6\quad 4\quad 1$ - Each entry: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ for $0 < k < n$, with $\binom{n}{0} = \binom{n}{n} = 1$. **Example 1**: Use Pascal’s Triangle to expand $(x + y)^4$. - Row 4: $1, 4, 6, 4, 1$ - $(x + y)^4 = 1x^4y^0 + 4x^3y^1 + 6x^2y^2 + 4x^1y^3 + 1x^0y^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$ **Exercise 1**: Use Pascal’s Triangle to expand $(2a - 3b)^3$. - Row 3: $1, 3, 3, 1$ - $(2a - 3b)^3 = 1(2a)^3(-3b)^0 + 3(2a)^2(-3b)^1 + 3(2a)^1(-3b)^2 + 1(2a)^0(-3b)^3$ - $= 8a^3 + 3(4a^2)(-3b) + 3(2a)(9b^2) + (-27b^3)$ - $= 8a^3 - 36a^2b + 54ab^2 - 27b^3$ **Correction**: Term 1: $1 \cdot 8a^3 \cdot 1 = 8a^3$ Term 2: $3 \cdot 4a^2 \cdot (-3b) = -36a^2b$ Term 3: $3 \cdot 2a \cdot 9b^2 = 54ab^2$ Term 4: $1 \cdot 1 \cdot (-27b^3) = -27b^3$ Result: $8a^3 - 36a^2b + 54ab^2 - 27b^3$ ## II. The General Binomial Theorem (Positive Integers) The Binomial Theorem generalizes the expansion of $(a + b)^n$ for any positive integer $n$, using combinations to determine coefficients. > **Theorem (Binomial Theorem for Positive Integers)**: For any positive integer $n$, > $$ > (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k > $$ > where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. **Properties**: - The expansion has $n+1$ terms. - Exponents of $a$ decrease from $n$ to $0$; exponents of $b$ increase from $0$ to $n$. - Coefficients are symmetric: $\binom{n}{k} = \binom{n}{n-k}$. - Sum of coefficients: $2^n$ (set $a = b = 1$). **Example 2**: Expand $(x + 2)^5$ using the Binomial Theorem. $$ (x + 2)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} 2^k $$ - $k=0$: $\binom{5}{0}x^5 \cdot 2^0 = 1 \cdot x^5 \cdot 1 = x^5$ - $k=1$: $\binom{5}{1}x^4 \cdot 2^1 = 5 \cdot x^4 \cdot 2 = 10x^4$ - $k=2$: $\binom{5}{2}x^3 \cdot 2^2 = 10 \cdot x^3 \cdot 4 = 40x^3$ - $k=3$: $\binom{5}{3}x^2 \cdot 2^3 = 10 \cdot x^2 \cdot 8 = 80x^2$ - $k=4$: $\binom{5}{4}x^1 \cdot 2^4 = 5 \cdot x \cdot 16 = 80x$ - $k=5$: $\binom{5}{5}x^0 \cdot 2^5 = 1 \cdot 1 \cdot 32 = 32$ - Result: $x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32$ **Exercise 2**: Find the coefficient of $x^3$ in the expansion of $(2x - 1)^6$. - General term: $T_{k+1} = \binom{6}{k} (2x)^{6-k} (-1)^k$ - We want exponent of $x$ to be 3: $6 - k = 3 \Rightarrow k = 3$ - $T_4 = \binom{6}{3} (2x)^{3} (-1)^3 = 20 \cdot 8x^3 \cdot (-1) = -160x^3$ - Coefficient: $-160$ **Correction**: $\binom{6}{3} = 20$, $(2x)^3 = 8x^3$, $(-1)^3 = -1$ $20 \times 8 \times (-1) = -160$ ## III. Finding Specific Terms and Coefficients This section focuses on identifying particular terms (e.g., middle term, constant term, term independent of $x$) without full expansion, using the general term formula. > **General Term**: In the expansion of $(a + b)^n$, the $(k+1)$-th term is given by: > $$ > T_{k+1} = \binom{n}{k} a^{n-k} b^k > $$ **Strategies**: - To find the term containing $x^m$, set exponent of $x$ in $T_{k+1}$ equal to $m$ and solve for $k$. - To find the constant term, set exponent of $x$ to zero. - Middle term(s): if $n$ even, one middle term at $k = n/2$; if $n$ odd, two middle terms at $k = (n-1)/2$ and $k = (n+1)/2$. **Example 3**: Find the term independent of $x$ in the expansion of $\left(x^2 + \frac{1}{x}\right)^6$. - General term: $T_{k+1} = \binom{6}{k} (x^2)^{6-k} \left(\frac{1}{x}\right)^k = \binom{6}{k} x^{12 - 2k} x^{-k} = \binom{6}{k} x^{12 - 3k}$ - Set exponent of $x$ to zero: $12 - 3k = 0 \Rightarrow k = 4$ - $T_5 = \binom{6}{4} x^{0} = 15 \cdot 1 = 15$ - Term independent of $x$: $15$ **Exercise 3**: In the expansion of $\left(2x^3 - \frac{3}{x}\right)^8$, find the coefficient of $x^6$. - General term: $T_{k+1} = \binom{8}{k} (2x^3)^{8-k} \left(-\frac{3}{x}\right)^k = \binom{8}{k} 2^{8-k} (-3)^k x^{24 - 3k - k} = \binom{8}{k} 2^{8-k} (-3)^k x^{24 - 4k}$ - Set exponent of $x$ to 6: $24 - 4k = 6 \Rightarrow 4k = 18 \Rightarrow k = 4.5$ → not integer → no such term? Wait: $24 - 4k = 6 \Rightarrow 4k = 18 \Rightarrow k = 4.5$ → invalid. Check: exponent from $(x^3)^{8-k}$ is $3(8-k) = 24 - 3k$ From $\left(-\frac{3}{x}\right)^k$: $(-3)^k x^{-k}$ Total exponent: $24 - 3k - k = 24 - 4k$ → correct. $24 - 4k = 6 \Rightarrow k = 4.5$ → not integer → **no term with $x^6$** But let’s double-check: maybe $x^6$ is possible? Try $k=4$: exponent = $24 - 16 = 8$ $k=5$: $24 - 20 = 4$ So $x^6$ does not occur. **Answer**: No such term; coefficient is $0$. **Correction**: Since $k$ must be integer between 0 and 8, and $24 - 4k = 6$ gives $k = 4.5$, no solution → coefficient of $x^6$ is $0$. **Example 4**: Find the middle term(s) in the expansion of $(x - 2y)^7$. - $n = 7$ (odd) → two middle terms: $k = 3$ and $k = 4$ - $T_4$ ($k=3$): $\binom{7}{3} x^{4} (-2y)^3 = 35 \cdot x^4 \cdot (-8y^3) = -280 x^4 y^3$ - $T_5$ ($k=4$): $\binom{7}{4} x^{3} (-2y)^4 = 35 \cdot x^3 \cdot 16y^4 = 560 x^3 y^4$ - Middle terms: $-280 x^4 y^3$ and $560 x^3 y^4$ **Exercise 4**: Find the constant term in the expansion of $\left(\frac{3}{x} + 2x^2\right)^9$. - General term: $T_{k+1} = \binom{9}{k} \left(\frac{3}{x}\right)^{9-k} (2x^2)^k = \binom{9}{k} 3^{9-k} 2^k x^{-(9-k)} x^{2k} = \binom{9}{k} 3^{9-k} 2^k x^{-9 + 3k}$ - Set exponent of $x$ to 0: $-9 + 3k = 0 \Rightarrow 3k = 9 \Rightarrow k = 3$ - $T_4 = \binom{9}{3} 3^{6} 2^{3} = 84 \cdot 729 \cdot 8$ - $84 \cdot 729 = 61236$; $61236 \cdot 8 = 489888$ - Constant term: $489888$ **Correction**: $\binom{9}{3} = 84$, $3^6 = 729$, $2^3 = 8$ $84 \times 729 = 61236$; $61236 \times 8 = 489888$ ## IV. Binomial Expansion for Rational Indices and Convergence The Binomial Theorem extends to rational exponents $n = \frac{p}{q}$ (where $p, q$ integers, $q \neq 0$), yielding an infinite series that converges under specific conditions. > **Theorem (Binomial Theorem for Rational Indices)**: For any rational number $n$ and $|x| < 1$, > $$ > (1 + x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k > $$ > where the generalized binomial coefficient is defined as: > $$ > \binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}, \quad \binom{n}{0} = 1 > $$ > The series converges if and only if $|x| < 1$. **Note**: This form assumes the binomial is written as $(1 + x)^n$. For $(a + b)^n$, factor out $a^n$: $(a + b)^n = a^n \left(1 + \frac{b}{a}\right)^n$, provided $a \neq 0$ and $\left|\frac{b}{a}\right| < 1$. **Properties**: - If $n$ is a positive integer, the series terminates after $n+1$ terms (matches earlier theorem). - If $n$ is not a positive integer, the series is infinite. - Convergence is strict: divergence occurs if $|x| \geq 1$. **Example 5**: Expand $(1 + x)^{-2}$ up to and including the $x^3$ term. - $n = -2$ - $\binom{-2}{0} = 1$ - $\binom{-2}{1} = \frac{-2}{1!} = -2$ - $\binom{-2}{2} = \frac{(-2)(-3)}{2!} = \frac{6}{2} = 3$ - $\binom{-2}{3} = \frac{(-2)(-3)(-4)}{3!} = \frac{-24}{6} = -4$ - $(1 + x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + \cdots$ **Exercise 5**: Expand $\frac{1}{(1 - 2x)^2}$ as a series in ascending powers of $x$ up to $x^3$, and state the values of $x$ for which the expansion is valid. - Rewrite: $\frac{1}{(1 - 2x)^2} = (1 - 2x)^{-2}$ - Use $(1 + y)^n$ with $y = -2x$, $n = -2$ - General term: $\binom{-2}{k} (-2x)^k$ - $k=0$: $\binom{-2}{0} (-2x)^0 = 1$ - $k=1$: $\binom{-2}{1} (-2x)^1 = (-2)(-2x) = 4x$ - $k=2$: $\binom{-2}{2} (-2x)^2 = (3)(4x^2) = 12x^2$ - $k=3$: $\binom{-2}{3} (-2x)^3 = (-4)(-8x^3) = 32x^3$ - Expansion: $1 + 4x + 12x^2 + 32x^3 + \cdots$ - Convergence: requires $|y| < 1 \Rightarrow |-2x| < 1 \Rightarrow |2x| < 1 \Rightarrow |x| < \frac{1}{2}$ **Correction**: $\binom{-2}{1} = -2$, times $(-2x)$ = $(-2)(-2x) = 4x$ $\binom{-2}{2} = \frac{(-2)(-3)}{2} = 3$, times $(-2x)^2 = 4x^2$ → $3 \cdot 4x^2 = 12x^2$ $\binom{-2}{3} = \frac{(-2)(-3)(-4)}{6} = \frac{-24}{6} = -4$, times $(-2x)^3 = -8x^3$ → $(-4)(-8x^3) = 32x^3$ Condition: $|-2x| < 1 \Rightarrow |x| < 0.5$ **Example 6**: Expand $\sqrt{1 + x}$ up to $x^2$ term. - $\sqrt{1 + x} = (1 + x)^{1/2}$ - $n = \frac{1}{2}$ - $\binom{1/2}{0} = 1$ - $\binom{1/2}{1} = \frac{1/2}{1!} = \frac{1}{2}$ - $\binom{1/2}{2} = \frac{(1/2)(1/2 - 1)}{2!} = \frac{(1/2)(-1/2)}{2} = \frac{-1/4}{2} = -\frac{1}{8}$ - Expansion: $1 + \frac{1}{2}x - \frac{1}{8}x^2 + \cdots$ **Exercise 6**: Expand $(8 + x)^{1/3}$ up to and including the $x^2$ term. - Factor out $8^{1/3} = 2$: $(8 + x)^{1/3} = 2 \left(1 + \frac{x}{8}\right)^{1/3}$ - Let $y = \frac{x}{8}$, $n = \frac{1}{3}$ - Expand $(1 + y)^{1/3}$: - $\binom{1/3}{0} = 1$ - $\binom{1/3}{1} = \frac{1/3}{1!} = \frac{1}{3}$ - $\binom{1/3}{2} = \frac{(1/3)(1/3 - 1)}{2!} = \frac{(1/3)(-2/3)}{2} = \frac{-2/9}{2} = -\frac{1}{9}$ - $(1 + y)^{1/3} = 1 + \frac{1}{3}y - \frac{1}{9}y^2 + \cdots$ - Substitute $y = x/8$: - $= 1 + \frac{1}{3} \cdot \frac{x}{8} - \frac{1}{9} \cdot \left(\frac{x}{8}\right)^2 + \cdots = 1 + \frac{x}{24} - \frac{x^2}{576} + \cdots$ - Multiply by 2: $2 \left(1 + \frac{x}{24} - \frac{x^2}{576} + \cdots\right) = 2 + \frac{x}{12} - \frac{x^2}{288} + \cdots$ **Correction**: $2 \times \frac{x}{24} = \frac{2x}{24} = \frac{x}{12}$ $2 \times \left(-\frac{x^2}{576}\right) = -\frac{2x^2}{576} = -\frac{x^2}{288}$ Valid for $\left|\frac{x}{8}\right| < 1 \Rightarrow |x| < 8$ ## V. Binomial Approximations When $x$ is small, higher powers of $x$ become negligible, allowing approximations of $(1 + x)^n$ by truncating the binomial series after a few terms. This is widely used in physics, engineering, and numerical analysis. > **Approximation Principle**: For $|x| \ll 1$, $(1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots$, where the approximation improves as more terms are included and $x$ approaches 0. **Common Approximations**: - First-order: $(1 + x)^n \approx 1 + nx$ (linear approximation) - Second-order: $(1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2$ (quadratic approximation) - Error is bounded by the next term in the series for $|x| < 1$. **Example 7**: Approximate $\sqrt{1.02}$ using the binomial expansion. - $\sqrt{1.02} = (1 + 0.02)^{1/2}$ - Use second-order approximation: $1 + \frac{1}{2}(0.02) + \frac{\frac{1}{2}(\frac{1}{2} - 1)}{2}(0.02)^2$ - $= 1 + 0.01 + \frac{(0.5)(-0.5)}{2}(0.0004) = 1 + 0.01 + \frac{-0.25}{2}(0.0004) = 1 + 0.01 - 0.125 \times 0.0004$ - $= 1 + 0.01 - 0.00005 = 1.00995$ - Actual value: $\sqrt{1.02} \approx 1.0099504938$ → error ≈ $4.9 \times 10^{-7}$ **Exercise 7**: Approximate $\frac{1}{\sqrt{1 - 0.04}}$ using the binomial expansion up to the $x^2$ term, and compare with the actual value. - $\frac{1}{\sqrt{1 - 0.04}} = (1 - 0.04)^{-1/2}$ - $n = -\frac{1}{2}$, $x = -0.04$ - Expansion: $(1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2$ - $= 1 + \left(-\frac{1}{2}\right)(-0.04) + \frac{(-\frac{1}{2})(-\frac{1}{2} - 1)}{2}(-0.04)^2$ - First term: $1$ - Second term: $(-\frac{1}{2})(-0.04) = 0.02$ - Third term: numerator: $(-\frac{1}{2})(-\frac{3}{2}) = \frac{3}{4}$; divided by 2: $\frac{3}{8}$; times $(-0.04)^2 = 0.0016$ → $\frac{3}{8} \times 0.0016 = 0.0006$ - Total: $1 + 0.02 + 0.0006 = 1.0206$ - Actual value: $\frac{1}{\sqrt{0.96}} = \frac{1}{0.979795897} \approx 1.020620726$ - Approximation: $1.0206$; error ≈ $0.0000207$ **Correction**: $\frac{n(n-1)}{2} = \frac{(-\frac{1}{2})(-\frac{3}{2})}{2} = \frac{\frac{3}{4}}{2} = \frac{3}{8}$ $\frac{3}{8} \times (0.0016) = 0.0006$ $1 + 0.02 + 0.0006 = 1.0206$ Actual: $1 / \sqrt{0.96} \approx 1.0206207$ → difference ≈ $2.07 \times 10^{-5}$ **Example 8**: Approximate $(1.005)^{10}$ using the binomial theorem. - $(1.005)^{10} = (1 + 0.005)^{10}$ - Use second-order: $1 + 10(0.005) + \frac{10 \cdot 9}{2}(0.005)^2$ - $= 1 + 0.05 + 45 \times 0.000025 = 1 + 0.05 + 0.001125 = 1.051125$ - Actual: $(1.005)^{10} \approx 1.051140132$ → error ≈ $1.5 \times 10^{-5}$ **Exercise 8**: Use the binomial approximation to estimate $\sqrt[3]{1.006}$ up to the $x^2$ term. - $\sqrt[3]{1.006} = (1 + 0.006)^{1/3}$ - $n = \frac{1}{3}$, $x = 0.006$ - Approximation: $1 + nx + \frac{n(n-1)}{2}x^2$ - $= 1 + \frac{1}{3}(0.006) + \frac{\frac{1}{3}(\frac{1}{3} - 1)}{2}(0.006)^2$ - $= 1 + 0.002 + \frac{\frac{1}{3}(-\frac{2}{3})}{2}(0.000036)$ - $= 1 + 0.002 + \frac{-\frac{2}{9}}{2}(0.000036)$ - $= 1 + 0.002 + \left(-\frac{1}{9}\right)(0.000036)$ - $= 1 + 0.002 - 0.000004 = 1.001996$ - Actual: $\sqrt[3]{1.006} \approx 1.001996008$ → excellent agreement **Correction**: $\frac{n(n-1)}{2} = \frac{(1/3)(-2/3)}{2} = \frac{-2/9}{2} = -1/9$ $-1/9 \times 0.000036 = -0.000004$ $1 + 0.002 - 0.000004 = 1.001996$ ## Synthèse du chapitre - Pascal’s Triangle provides binomial coefficients for positive integer exponents via recursive addition. - The Binomial Theorem for positive integers gives a finite expansion: $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k$. - The general term $T_{k+1} = \binom{n}{k} a^{n-k}b^k$ enables direct computation of specific terms without full expansion. - For rational exponents, the Binomial Theorem becomes an infinite series: $(1+x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k$, convergent only when $|x| < 1$. - Binomial approximations truncate the series after a few terms for small $x$, yielding accurate estimates: $(1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2$. - Key skills include identifying term patterns, solving for $k$ in exponent equations, applying convergence conditions, and judging approximation accuracy. ## Glossaire - **Binomial Coefficient**: The number $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ for integers $n \geq k \geq 0$, or the generalized form $\frac{n(n-1)\cdots(n-k+1)}{k!}$ for rational $n$. - **General Term**: The $(k+1)$-th term in a binomial expansion, expressed as $T_{k+1} = \binom{n}{k} a^{n-k}b^k$. - **Convergence**: The property of an infinite series to approach a finite limit; for $(1+x)^n$ with rational $n$, convergence holds iff $|x| < 1$. - **Pascal’s Triangle**: A triangular array where each entry is the sum of the two above it, providing binomial coefficients for $(a+b)^n$, $n \in \mathbb{N}$. - **Middle Term**: In $(a+b)^n$, the term(s) with $k = \lfloor n/2 \rfloor$ or $k = \lceil n/2 \rceil$; one if $n$ even, two if $n$ odd. - **Constant Term**: The term in the expansion where the exponent of the variable is zero. - **Approximation**: A simplified expression derived by truncating a binomial series, valid when the variable is small in magnitude. - **Divergence**: The failure of an infinite series to settle to a finite value; occurs in $(1+x)^n$ for rational $n$ when $|x| \geq 1$.