7.3 Q-1

Exercise 7.3 — Question 1

This exercise covers the Binomial Series for negative and fractional exponents, focusing on expanding expressions of the form $(1+x{)}^n$ where $n$ is not a positive integer.

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Key Formula

For any rational $n$ and $|x|<1$:

$(1+x{)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+\cdots$

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Technique: Factoring Before Expanding

When the expression is $(a+x{)}^n$ rather than $(1+x{)}^n$, factor out $a^n$ first:

$(a+x{)}^n=a^n{\left(1+\frac{x}{a}\right)}^n$

Then apply the binomial series to ${\left(1+\frac{x}{a}\right)}^n$, which is valid when $\left|\frac{x}{a}\right|<1$.

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Worked Examples

Example 1: Expand $(1+x{)}^{-2}$ up to the term in $x^3$

Using $n=-2$:

$(1+x{)}^{-2}=1+(-2)x+\frac{(-2)(-3)}{2!}{x}^2+\frac{(-2)(-3)(-4)}{3!}{x}^3+\cdots$

$=1-2x+3x^2-4x^3+\cdots |x|<1$

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Example 2: Expand $(1-x{)}^{-1}$ up to the term in $x^3$

Using $n=-1$ and replacing $x$ with $-x$:

$(1-x{)}^{-1}=1+x+x^2+x^3+\cdots |x|<1$

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Example 3: Expand $(3-2x{)}^{-2}$ and state the range of validity

Step 1: Factor out $3^{-2}$:

$(3-2x{)}^{-2}=3^{-2}{\left(1-\frac{2x}{3}\right)}^{-2}=\frac{1}{9}{\left(1-\frac{2x}{3}\right)}^{-2}$

Step 2: Apply the series with $n=-2$ and $u=-\frac{2x}{3}$:

${\left(1-\frac{2x}{3}\right)}^{-2}=1+2\cdot \frac{2x}{3}+3\cdot {\left(\frac{2x}{3}\right)}^2+4\cdot {\left(\frac{2x}{3}\right)}^3+\cdots$

$=1+\frac{4x}{3}+\frac{4x^2}{3}+\frac{32x^3}{27}+\cdots$

Step 3: Multiply by $\frac{1}{9}$:

$(3-2x{)}^{-2}=\frac{1}{9}+\frac{4x}{27}+\frac{4x^2}{27}+\frac{32x^3}{243}+\cdots$

Validity: $\left|\frac{2x}{3}\right|<1\Rightarrow |x|<\frac{3}{2}$

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