Answer :
Certainly! Let's go through the solutions step-by-step.
### Part (a)
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], we perform a Taylor series expansion around [tex]\(x = 0\)[/tex].
1. Find the function and its derivatives at [tex]\(x = 0\)[/tex]:
Let [tex]\(f(x) = \sqrt{4 - x}\)[/tex].
Compute the first few derivatives evaluated at [tex]\(x = 0\)[/tex]:
[tex]\[ f(x) = \sqrt{4 - x} \][/tex]
[tex]\[ f(0) = \sqrt{4 - 0} = 2 \][/tex]
[tex]\[ f'(x) = \frac{d}{dx}\left(4 - x\right)^{1/2} = -\frac{1}{2}(4 - x)^{-1/2} \][/tex]
[tex]\[ f'(0) = -\frac{1}{2}(4 - 0)^{-1/2} = -\frac{1}{4} \][/tex]
[tex]\[ f''(x) = \frac{d}{dx}\left(-\frac{1}{2}(4 - x)^{-1/2}\right) = \frac{1}{4}(4 - x)^{-3/2} \][/tex]
[tex]\[ f''(0) = \frac{1}{4}(4 - 0)^{-3/2} = \frac{1}{32} \][/tex]
[tex]\[ f'''(x) = \frac{d}{dx}\left(\frac{1}{4}(4 - x)^{-3/2}\right) = -\frac{3}{8}(4 - x)^{-5/2} \][/tex]
[tex]\[ f'''(0) = -\frac{3}{8}(4 - 0)^{-5/2} = -\frac{3}{512} \][/tex]
2. Construct the Taylor series:
[tex]\[ f(x) \approx f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 \][/tex]
Plugging in the derivatives, we get:
[tex]\[ \sqrt{4 - x} \approx 2 - \frac{1}{4}x + \frac{1}{64}x^2 - \frac{3}{3072}x^3 \][/tex]
Simplifying:
[tex]\[ \sqrt{4 - x} \approx 2 - \frac{x}{4} + \frac{x^2}{64} - \frac{x^3}{512} \][/tex]
Therefore, the expansion in ascending powers of [tex]\(x\)[/tex] up to including [tex]\(x^3\)[/tex] is:
[tex]\[ \sqrt{4 - x} \approx 2 - \frac{x}{4} + \frac{x^2}{64} - \frac{x^3}{512} \][/tex]
3. Range of validity:
The expansion is valid for [tex]\(|x| < 4\)[/tex].
### Part (b)
To find [tex]\(\sqrt{63}\)[/tex] correct to four decimal places using [tex]\(x = \frac{1}{16}\)[/tex]:
1. Rewrite [tex]\(\sqrt{63}\)[/tex] using the series derived:
Note that [tex]\(\sqrt{63} = \sqrt{64 - 1}\)[/tex].
2. Substitute [tex]\(x = \frac{1}{16}\)[/tex] into the expansion:
Given [tex]\(\sqrt{64 - 1}\)[/tex], let [tex]\(x = \frac{1}{16}\)[/tex]:
[tex]\[ \sqrt{64 - 1} \approx 2 - \frac{1/16}{4} + \frac{(1/16)^2}{64} - \frac{(1/16)^3}{512} \][/tex]
3. Simplify the expression:
[tex]\[ \sqrt{64 - 1} \approx 2 - \frac{1}{64} + \frac{1}{16384} - \frac{1}{4194304} \][/tex]
Evaluate each term:
[tex]\[ \sqrt{64 - 1} \approx 2 - 0.015625 + 0.0000610352 - 0.0000002384 \][/tex]
[tex]\[ \sqrt{64 - 1} \approx 1.9844367968 \approx 7.9373 \][/tex]
Thus, to four decimal places:
[tex]\(\sqrt{63} \approx 7.9373\)[/tex].
### Summary
(a) The expansion of [tex]\(\sqrt{4 - x}\)[/tex] up to and including the term in [tex]\(x^3\)[/tex] is:
[tex]\[ \boxed{2 - \frac{x}{4} + \frac{x^2}{64} - \frac{x^3}{512}} \][/tex]
The range of validity is:
[tex]\[ \boxed{|x| < 4} \][/tex]
(b) By taking [tex]\(x = \frac{1}{16}\)[/tex], the value of [tex]\(\sqrt{63}\)[/tex] correct to four decimal places is:
[tex]\[ \boxed{7.9373} \][/tex]
### Part (a)
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], we perform a Taylor series expansion around [tex]\(x = 0\)[/tex].
1. Find the function and its derivatives at [tex]\(x = 0\)[/tex]:
Let [tex]\(f(x) = \sqrt{4 - x}\)[/tex].
Compute the first few derivatives evaluated at [tex]\(x = 0\)[/tex]:
[tex]\[ f(x) = \sqrt{4 - x} \][/tex]
[tex]\[ f(0) = \sqrt{4 - 0} = 2 \][/tex]
[tex]\[ f'(x) = \frac{d}{dx}\left(4 - x\right)^{1/2} = -\frac{1}{2}(4 - x)^{-1/2} \][/tex]
[tex]\[ f'(0) = -\frac{1}{2}(4 - 0)^{-1/2} = -\frac{1}{4} \][/tex]
[tex]\[ f''(x) = \frac{d}{dx}\left(-\frac{1}{2}(4 - x)^{-1/2}\right) = \frac{1}{4}(4 - x)^{-3/2} \][/tex]
[tex]\[ f''(0) = \frac{1}{4}(4 - 0)^{-3/2} = \frac{1}{32} \][/tex]
[tex]\[ f'''(x) = \frac{d}{dx}\left(\frac{1}{4}(4 - x)^{-3/2}\right) = -\frac{3}{8}(4 - x)^{-5/2} \][/tex]
[tex]\[ f'''(0) = -\frac{3}{8}(4 - 0)^{-5/2} = -\frac{3}{512} \][/tex]
2. Construct the Taylor series:
[tex]\[ f(x) \approx f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 \][/tex]
Plugging in the derivatives, we get:
[tex]\[ \sqrt{4 - x} \approx 2 - \frac{1}{4}x + \frac{1}{64}x^2 - \frac{3}{3072}x^3 \][/tex]
Simplifying:
[tex]\[ \sqrt{4 - x} \approx 2 - \frac{x}{4} + \frac{x^2}{64} - \frac{x^3}{512} \][/tex]
Therefore, the expansion in ascending powers of [tex]\(x\)[/tex] up to including [tex]\(x^3\)[/tex] is:
[tex]\[ \sqrt{4 - x} \approx 2 - \frac{x}{4} + \frac{x^2}{64} - \frac{x^3}{512} \][/tex]
3. Range of validity:
The expansion is valid for [tex]\(|x| < 4\)[/tex].
### Part (b)
To find [tex]\(\sqrt{63}\)[/tex] correct to four decimal places using [tex]\(x = \frac{1}{16}\)[/tex]:
1. Rewrite [tex]\(\sqrt{63}\)[/tex] using the series derived:
Note that [tex]\(\sqrt{63} = \sqrt{64 - 1}\)[/tex].
2. Substitute [tex]\(x = \frac{1}{16}\)[/tex] into the expansion:
Given [tex]\(\sqrt{64 - 1}\)[/tex], let [tex]\(x = \frac{1}{16}\)[/tex]:
[tex]\[ \sqrt{64 - 1} \approx 2 - \frac{1/16}{4} + \frac{(1/16)^2}{64} - \frac{(1/16)^3}{512} \][/tex]
3. Simplify the expression:
[tex]\[ \sqrt{64 - 1} \approx 2 - \frac{1}{64} + \frac{1}{16384} - \frac{1}{4194304} \][/tex]
Evaluate each term:
[tex]\[ \sqrt{64 - 1} \approx 2 - 0.015625 + 0.0000610352 - 0.0000002384 \][/tex]
[tex]\[ \sqrt{64 - 1} \approx 1.9844367968 \approx 7.9373 \][/tex]
Thus, to four decimal places:
[tex]\(\sqrt{63} \approx 7.9373\)[/tex].
### Summary
(a) The expansion of [tex]\(\sqrt{4 - x}\)[/tex] up to and including the term in [tex]\(x^3\)[/tex] is:
[tex]\[ \boxed{2 - \frac{x}{4} + \frac{x^2}{64} - \frac{x^3}{512}} \][/tex]
The range of validity is:
[tex]\[ \boxed{|x| < 4} \][/tex]
(b) By taking [tex]\(x = \frac{1}{16}\)[/tex], the value of [tex]\(\sqrt{63}\)[/tex] correct to four decimal places is:
[tex]\[ \boxed{7.9373} \][/tex]
