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### Part (a): Series Expansion of [tex]\(\sqrt{4 - x}\)[/tex]
We need 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].
To start, we use the binomial series expansion for [tex]\( (1 + u)^n \)[/tex]:
[tex]\[ (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots \][/tex]
For [tex]\(\sqrt{4 - x}\)[/tex], we rewrite it as [tex]\((4(1 - \frac{x}{4}))^{1/2}\)[/tex]:
[tex]\[ \sqrt{4 - x} = \sqrt{4} \cdot \sqrt{1 - \frac{x}{4}} \][/tex]
[tex]\[ = 2 \cdot (1 - \frac{x}{4})^{1/2} \][/tex]
Now, we apply the binomial expansion for [tex]\((1 - \frac{x}{4})^{1/2}\)[/tex], where [tex]\(u = -\frac{x}{4}\)[/tex] and [tex]\(n = \frac{1}{2}\)[/tex]:
[tex]\[ (1 - \frac{x}{4})^{1/2} = 1 + (\frac{1}{2})(-\frac{x}{4}) + \frac{(\frac{1}{2})(\frac{1}{2} - 1)}{2!}(-\frac{x}{4})^2 + \frac{(\frac{1}{2})(\frac{1}{2} - 1)(\frac{1}{2} - 2)}{3!}(-\frac{x}{4})^3 + \cdots \][/tex]
Calculating each term:
1. The [tex]\(u\)[/tex] term:
[tex]\[ (\frac{1}{2})(-\frac{x}{4}) = -\frac{x}{8} \][/tex]
2. The [tex]\(u^2\)[/tex] term:
[tex]\[ \frac{(\frac{1}{2})(-\frac{1}{2})}{2!} (\frac{x^2}{16}) = \frac{-1/8}{2} \cdot \frac{x^2}{16} = -\frac{x^2}{256} \][/tex]
3. The [tex]\(u^3\)[/tex] term:
[tex]\[ \frac{(\frac{1}{2})(-\frac{1}{2})(-\frac{3}{2})}{3!} (-\frac{x}{4})^3 = \frac{-1/8 \cdot -3/2}{6} \cdot -\frac{x^3}{64} = -\frac{x^3}{2048} \][/tex]
Summarizing these terms in the expansion:
[tex]\[ (1 - \frac{x}{4})^{1/2} \approx 1 - \frac{x}{8} - \frac{x^2}{256} - \frac{x^3}{2048} \][/tex]
Multiplying by 2:
[tex]\[ \sqrt{4 - x} \approx 2 \left( 1 - \frac{x}{8} - \frac{x^2}{256} - \frac{x^3}{2048} \right) \][/tex]
[tex]\[ = 2 - \frac{x}{4} - \frac{x^2}{128} - \frac{x^3}{1024} \][/tex]
So, the expansion up to the term in [tex]\( x^3 \)[/tex] is:
[tex]\[ \boxed{2 - \frac{x}{4} - \frac{x^2}{128} - \frac{x^3}{1024}} \][/tex]
The expansion is valid for [tex]\(-4 < x < 4\)[/tex].
### Part (b): Approximate [tex]\(\sqrt{63}\)[/tex] Using [tex]\(x = \frac{1}{16}\)[/tex]
Given [tex]\(x = \frac{1}{16}\)[/tex], we substitute this into our expansion to estimate [tex]\(\sqrt{63}\)[/tex].
First calculate the expansion at [tex]\( x = \frac{1}{16} \)[/tex]:
[tex]\[ 2 - \frac{\frac{1}{16}}{4} - \frac{\left(\frac{1}{16}\right)^2}{128} - \frac{\left(\frac{1}{16}\right)^3}{1024} \][/tex]
1. [tex]\(\frac{1}{16} \div 4 = \frac{1}{64} = 0.015625\)[/tex]
2. [tex]\(\left(\frac{1}{16}\right)^2 = \frac{1}{256}\)[/tex], then [tex]\(\frac{1}{256} \div 128 = \frac{1}{32768} \approx 0.0000305\)[/tex]
3. [tex]\(\left(\frac{1}{16}\right)^3 = \frac{1}{4096}\)[/tex], then [tex]\(\frac{1}{4096} \div 1024 = \frac{1}{4194304} \approx 0.00000024\)[/tex]
So the expansion result is:
[tex]\[ \approx 2 - 0.015625 - 0.0000305 - 0.00000024 \approx 1.984344 \][/tex]
Finally, [tex]\(\sqrt{63}\)[/tex] can be approximated by considering that [tex]\(\sqrt{63} \approx \sqrt{64 - 1}\)[/tex] which is close to 8 (since [tex]\(\sqrt{64} = 8\)[/tex]) simplified by our given [tex]\( x \)[/tex] value.
Thus we get:
[tex]\[ \sqrt{63} \approx 1.984344 \][/tex]
When expressed to four decimal places, [tex]\(\sqrt{63} \approx 15.75\)[/tex].
So [tex]\(\sqrt{63} \approx \boxed{15.7500}\)[/tex] (correct to four decimal places).
### Part (a): Series Expansion of [tex]\(\sqrt{4 - x}\)[/tex]
We need 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].
To start, we use the binomial series expansion for [tex]\( (1 + u)^n \)[/tex]:
[tex]\[ (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots \][/tex]
For [tex]\(\sqrt{4 - x}\)[/tex], we rewrite it as [tex]\((4(1 - \frac{x}{4}))^{1/2}\)[/tex]:
[tex]\[ \sqrt{4 - x} = \sqrt{4} \cdot \sqrt{1 - \frac{x}{4}} \][/tex]
[tex]\[ = 2 \cdot (1 - \frac{x}{4})^{1/2} \][/tex]
Now, we apply the binomial expansion for [tex]\((1 - \frac{x}{4})^{1/2}\)[/tex], where [tex]\(u = -\frac{x}{4}\)[/tex] and [tex]\(n = \frac{1}{2}\)[/tex]:
[tex]\[ (1 - \frac{x}{4})^{1/2} = 1 + (\frac{1}{2})(-\frac{x}{4}) + \frac{(\frac{1}{2})(\frac{1}{2} - 1)}{2!}(-\frac{x}{4})^2 + \frac{(\frac{1}{2})(\frac{1}{2} - 1)(\frac{1}{2} - 2)}{3!}(-\frac{x}{4})^3 + \cdots \][/tex]
Calculating each term:
1. The [tex]\(u\)[/tex] term:
[tex]\[ (\frac{1}{2})(-\frac{x}{4}) = -\frac{x}{8} \][/tex]
2. The [tex]\(u^2\)[/tex] term:
[tex]\[ \frac{(\frac{1}{2})(-\frac{1}{2})}{2!} (\frac{x^2}{16}) = \frac{-1/8}{2} \cdot \frac{x^2}{16} = -\frac{x^2}{256} \][/tex]
3. The [tex]\(u^3\)[/tex] term:
[tex]\[ \frac{(\frac{1}{2})(-\frac{1}{2})(-\frac{3}{2})}{3!} (-\frac{x}{4})^3 = \frac{-1/8 \cdot -3/2}{6} \cdot -\frac{x^3}{64} = -\frac{x^3}{2048} \][/tex]
Summarizing these terms in the expansion:
[tex]\[ (1 - \frac{x}{4})^{1/2} \approx 1 - \frac{x}{8} - \frac{x^2}{256} - \frac{x^3}{2048} \][/tex]
Multiplying by 2:
[tex]\[ \sqrt{4 - x} \approx 2 \left( 1 - \frac{x}{8} - \frac{x^2}{256} - \frac{x^3}{2048} \right) \][/tex]
[tex]\[ = 2 - \frac{x}{4} - \frac{x^2}{128} - \frac{x^3}{1024} \][/tex]
So, the expansion up to the term in [tex]\( x^3 \)[/tex] is:
[tex]\[ \boxed{2 - \frac{x}{4} - \frac{x^2}{128} - \frac{x^3}{1024}} \][/tex]
The expansion is valid for [tex]\(-4 < x < 4\)[/tex].
### Part (b): Approximate [tex]\(\sqrt{63}\)[/tex] Using [tex]\(x = \frac{1}{16}\)[/tex]
Given [tex]\(x = \frac{1}{16}\)[/tex], we substitute this into our expansion to estimate [tex]\(\sqrt{63}\)[/tex].
First calculate the expansion at [tex]\( x = \frac{1}{16} \)[/tex]:
[tex]\[ 2 - \frac{\frac{1}{16}}{4} - \frac{\left(\frac{1}{16}\right)^2}{128} - \frac{\left(\frac{1}{16}\right)^3}{1024} \][/tex]
1. [tex]\(\frac{1}{16} \div 4 = \frac{1}{64} = 0.015625\)[/tex]
2. [tex]\(\left(\frac{1}{16}\right)^2 = \frac{1}{256}\)[/tex], then [tex]\(\frac{1}{256} \div 128 = \frac{1}{32768} \approx 0.0000305\)[/tex]
3. [tex]\(\left(\frac{1}{16}\right)^3 = \frac{1}{4096}\)[/tex], then [tex]\(\frac{1}{4096} \div 1024 = \frac{1}{4194304} \approx 0.00000024\)[/tex]
So the expansion result is:
[tex]\[ \approx 2 - 0.015625 - 0.0000305 - 0.00000024 \approx 1.984344 \][/tex]
Finally, [tex]\(\sqrt{63}\)[/tex] can be approximated by considering that [tex]\(\sqrt{63} \approx \sqrt{64 - 1}\)[/tex] which is close to 8 (since [tex]\(\sqrt{64} = 8\)[/tex]) simplified by our given [tex]\( x \)[/tex] value.
Thus we get:
[tex]\[ \sqrt{63} \approx 1.984344 \][/tex]
When expressed to four decimal places, [tex]\(\sqrt{63} \approx 15.75\)[/tex].
So [tex]\(\sqrt{63} \approx \boxed{15.7500}\)[/tex] (correct to four decimal places).
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