Answer :
Certainly! Let's break down the solution step-by-step:
### Part (a): Expanding [tex]\( \sqrt{4-x} \)[/tex]
To expand [tex]\( \sqrt{4-x} \)[/tex] in ascending powers of [tex]\( x \)[/tex] up to and including the term in [tex]\( x^3 \)[/tex]:
1. Function Representation:
We start with [tex]\( f(x) = \sqrt{4-x} \)[/tex].
2. Maclaurin Series Expansion:
We need to expand [tex]\( f(x) \)[/tex] around [tex]\( x = 0 \)[/tex]. The general form of a Maclaurin series for a function [tex]\( f(x) \)[/tex] is:
[tex]\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \][/tex]
3. Calculate the derivatives at [tex]\( x = 0 \)[/tex]:
- [tex]\( f(0) = \sqrt{4} = 2 \)[/tex]
- First derivative: [tex]\( f'(x) = \frac{d}{dx} \sqrt{4-x} = -\frac{1}{2} (4-x)^{-\frac{1}{2}} \)[/tex]. Hence, [tex]\( f'(0) = -\frac{1}{2} \cdot 4^{-\frac{1}{2}} = -\frac{1}{4} \)[/tex]
- Second derivative: [tex]\( f''(x) = \frac{d}{dx} \left( -\frac{1}{2} (4-x)^{-\frac{1}{2}} \right) = \frac{1}{4} (4-x)^{-\frac{3}{2}} \)[/tex]. Thus, [tex]\( f''(0) = \frac{1}{4} \cdot 4^{-\frac{3}{2}} = \frac{1}{32} \)[/tex]
- Third derivative: [tex]\( f'''(x) = \frac{d}{dx} \left( \frac{1}{4} (4-x)^{-\frac{3}{2}} \right) = -\frac{3}{8} (4-x)^{-\frac{5}{2}} \)[/tex]. Therefore, [tex]\( f'''(0) = -\frac{3}{8} \cdot 4^{-\frac{5}{2}} = -\frac{3}{512} \)[/tex]
4. Construct the Series:
Using the calculated derivatives, the series expansion up to [tex]\( x^3 \)[/tex] term is:
[tex]\[ \sqrt{4-x} \approx f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 \][/tex]
Substituting the derivatives:
[tex]\[ \sqrt{4-x} \approx 2 - \frac{1}{4}x + \frac{1}{32} \cdot \frac{x^2}{2} - \frac{3}{512} \cdot \frac{x^3}{6} \][/tex]
Simplifying this, we get:
[tex]\[ \sqrt{4-x} \approx 2 - \frac{1}{4}x - \frac{1}{64}x^2 - \frac{1}{512}x^3 \][/tex]
5. Range of Validity:
The expansion [tex]\( \sqrt{4 - x} \)[/tex] is valid for [tex]\( |x/4| < 1 \)[/tex], which simplifies to:
[tex]\[ |x| < 4 \][/tex]
### Part (b): Finding [tex]\( \sqrt{63} \)[/tex]
1. Substitution:
We're given [tex]\( x = \frac{1}{16} \)[/tex] to approximate [tex]\( \sqrt{63} \)[/tex]. To understand this, note that we use the expanded form [tex]\( 4 - x \)[/tex] where [tex]\( x \)[/tex] is a small perturbation to 4.
2. Plugging [tex]\( x = \frac{1}{16} \)[/tex] into the Expansion:
Using the expansion from part (a),
[tex]\[ \sqrt{4 - \frac{1}{16}} \approx 2 - \frac{1}{4} \cdot \frac{1}{16} - \frac{1}{64} \cdot \left( \frac{1}{16} \right)^2 - \frac{1}{512} \cdot \left( \frac{1}{16} \right)^3 \][/tex]
Simplifying each term:
[tex]\[ \sqrt{4 - \frac{1}{16}} \approx 2 - \frac{1}{64} - \frac{1}{64 \cdot 16} - \frac{1}{512 \cdot 16^2} \][/tex]
[tex]\[ \sqrt{4 - \frac{1}{16}} \approx 2 - \frac{1}{64} - \frac{1}{1024} - \frac{1}{8192} \][/tex]
Summing these values, we obtain:
[tex]\[ \sqrt{4 - \frac{1}{16}} \approx 2 - 0.015625 - 0.0009765625 - 0.0001220703125 \approx 2 - 0.0167236328125 \approx 1.9832763671875 \][/tex]
3. Correcting to Four Decimal Places:
The value approximates to:
[tex]\[ \sqrt{3.9375} \approx 1.9843 \][/tex]
Therefore, the expanded form of [tex]\( \sqrt{4-x} \)[/tex] is:
[tex]\[ \sqrt{4 - x} \approx 2 - \frac{1}{4}x - \frac{1}{64}x^2 - \frac{1}{512}x^3 \][/tex]
for [tex]\( |x| < 4 \)[/tex].
And [tex]\( \sqrt{63} \)[/tex] correct to four decimal places, using [tex]\( x = \frac{1}{16} \)[/tex], is:
[tex]\[ \sqrt{63} \approx 1.9843 \][/tex]
### Part (a): Expanding [tex]\( \sqrt{4-x} \)[/tex]
To expand [tex]\( \sqrt{4-x} \)[/tex] in ascending powers of [tex]\( x \)[/tex] up to and including the term in [tex]\( x^3 \)[/tex]:
1. Function Representation:
We start with [tex]\( f(x) = \sqrt{4-x} \)[/tex].
2. Maclaurin Series Expansion:
We need to expand [tex]\( f(x) \)[/tex] around [tex]\( x = 0 \)[/tex]. The general form of a Maclaurin series for a function [tex]\( f(x) \)[/tex] is:
[tex]\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \][/tex]
3. Calculate the derivatives at [tex]\( x = 0 \)[/tex]:
- [tex]\( f(0) = \sqrt{4} = 2 \)[/tex]
- First derivative: [tex]\( f'(x) = \frac{d}{dx} \sqrt{4-x} = -\frac{1}{2} (4-x)^{-\frac{1}{2}} \)[/tex]. Hence, [tex]\( f'(0) = -\frac{1}{2} \cdot 4^{-\frac{1}{2}} = -\frac{1}{4} \)[/tex]
- Second derivative: [tex]\( f''(x) = \frac{d}{dx} \left( -\frac{1}{2} (4-x)^{-\frac{1}{2}} \right) = \frac{1}{4} (4-x)^{-\frac{3}{2}} \)[/tex]. Thus, [tex]\( f''(0) = \frac{1}{4} \cdot 4^{-\frac{3}{2}} = \frac{1}{32} \)[/tex]
- Third derivative: [tex]\( f'''(x) = \frac{d}{dx} \left( \frac{1}{4} (4-x)^{-\frac{3}{2}} \right) = -\frac{3}{8} (4-x)^{-\frac{5}{2}} \)[/tex]. Therefore, [tex]\( f'''(0) = -\frac{3}{8} \cdot 4^{-\frac{5}{2}} = -\frac{3}{512} \)[/tex]
4. Construct the Series:
Using the calculated derivatives, the series expansion up to [tex]\( x^3 \)[/tex] term is:
[tex]\[ \sqrt{4-x} \approx f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 \][/tex]
Substituting the derivatives:
[tex]\[ \sqrt{4-x} \approx 2 - \frac{1}{4}x + \frac{1}{32} \cdot \frac{x^2}{2} - \frac{3}{512} \cdot \frac{x^3}{6} \][/tex]
Simplifying this, we get:
[tex]\[ \sqrt{4-x} \approx 2 - \frac{1}{4}x - \frac{1}{64}x^2 - \frac{1}{512}x^3 \][/tex]
5. Range of Validity:
The expansion [tex]\( \sqrt{4 - x} \)[/tex] is valid for [tex]\( |x/4| < 1 \)[/tex], which simplifies to:
[tex]\[ |x| < 4 \][/tex]
### Part (b): Finding [tex]\( \sqrt{63} \)[/tex]
1. Substitution:
We're given [tex]\( x = \frac{1}{16} \)[/tex] to approximate [tex]\( \sqrt{63} \)[/tex]. To understand this, note that we use the expanded form [tex]\( 4 - x \)[/tex] where [tex]\( x \)[/tex] is a small perturbation to 4.
2. Plugging [tex]\( x = \frac{1}{16} \)[/tex] into the Expansion:
Using the expansion from part (a),
[tex]\[ \sqrt{4 - \frac{1}{16}} \approx 2 - \frac{1}{4} \cdot \frac{1}{16} - \frac{1}{64} \cdot \left( \frac{1}{16} \right)^2 - \frac{1}{512} \cdot \left( \frac{1}{16} \right)^3 \][/tex]
Simplifying each term:
[tex]\[ \sqrt{4 - \frac{1}{16}} \approx 2 - \frac{1}{64} - \frac{1}{64 \cdot 16} - \frac{1}{512 \cdot 16^2} \][/tex]
[tex]\[ \sqrt{4 - \frac{1}{16}} \approx 2 - \frac{1}{64} - \frac{1}{1024} - \frac{1}{8192} \][/tex]
Summing these values, we obtain:
[tex]\[ \sqrt{4 - \frac{1}{16}} \approx 2 - 0.015625 - 0.0009765625 - 0.0001220703125 \approx 2 - 0.0167236328125 \approx 1.9832763671875 \][/tex]
3. Correcting to Four Decimal Places:
The value approximates to:
[tex]\[ \sqrt{3.9375} \approx 1.9843 \][/tex]
Therefore, the expanded form of [tex]\( \sqrt{4-x} \)[/tex] is:
[tex]\[ \sqrt{4 - x} \approx 2 - \frac{1}{4}x - \frac{1}{64}x^2 - \frac{1}{512}x^3 \][/tex]
for [tex]\( |x| < 4 \)[/tex].
And [tex]\( \sqrt{63} \)[/tex] correct to four decimal places, using [tex]\( x = \frac{1}{16} \)[/tex], is:
[tex]\[ \sqrt{63} \approx 1.9843 \][/tex]
