Answer :
Certainly! Let's break down the problems one by one.
### 1. Power Series for \( f(x) = \sqrt{16 - x^2} \)
#### About \( x = 0 \)
To find the power series representation of \( f(x) \) about \( x = 0 \), we use the binomial series expansion for a function of the form \( (1 - u)^{\frac{1}{2}} \), where \( u = \frac{x^2}{16} \).
[tex]\[ f(x) = \sqrt{16 - x^2} = 4 \sqrt{1 - \frac{x^2}{16}} \][/tex]
Using the binomial series for \( (1 - u)^{\frac{1}{2}} \), we have:
[tex]\[ (1 - u)^{\frac{1}{2}} = \sum_{n=0}^{\infty} (-1)^n \binom{\frac{1}{2}}{n} u^n \][/tex]
where
[tex]\[ \binom{\frac{1}{2}}{n} = \frac{\left( \frac{1}{2} \right) \left( \frac{1}{2} - 1 \right) \left( \frac{1}{2} - 2 \right) \cdots \left( \frac{1}{2} - (n-1)\right)}{n!} \][/tex]
Substituting \( u = \frac{x^2}{16} \):
[tex]\[ f(x) = 4 \sum_{n=0}^{\infty} (-1)^n \binom{\frac{1}{2}}{n} \left(\frac{x^2}{16} \right)^n \][/tex]
Simplifying the coefficients,
[tex]\[ f(x) = 4 \sum_{n=0}^{\infty} (-1)^n \binom{\frac{1}{2}}{n} \frac{x^{2n}}{16^n} \][/tex]
This is the power series for \( f(x) = \sqrt{16 - x^2} \) about \( x = 0 \).
### 2. Power Series for \( g(x) = (16 - x)^7 \)
#### About \( x = 0 \)
For \( g(x) = (16 - x)^7 \), we again use the binomial series:
[tex]\[ (16 - x)^7 = 16^7 \left( 1 - \frac{x}{16} \right)^7 \][/tex]
Expanding using the binomial theorem,
[tex]\[ (1 - \frac{x}{16})^7 = \sum_{n=0}^{\infty} \binom{7}{n} (-1)^n \left( \frac{x}{16} \right)^n \][/tex]
Then,
[tex]\[ g(x) = 16^7 \sum_{n=0}^{\infty} \binom{7}{n} (-1)^n \left( \frac{x}{16} \right)^n \][/tex]
Simplifying,
[tex]\[ g(x) = 16^7 \sum_{n=0}^{\infty} \binom{7}{n} (-1)^n \frac{x^n}{16^n} = \sum_{n=0}^{\infty} \binom{7}{n} (-1)^n 16^{7-n} x^n \][/tex]
This is the power series for \( g(x) = (16 - x)^7 \) about \( x = 0 \).
### 3. Power Series for \( f(x) = \sqrt{2 + x} \)
#### About \( x = 1 \)
To find the power series for \( f(x) = \sqrt{2 + x} \) about \( x = 1 \), we write \( f(x) \) in the form of a binomial expansion:
[tex]\[ f(x) = \sqrt{2 + x} = \sqrt{3 - 1 + (x-1)} = \sqrt{3 + (x-1)} \][/tex]
Expanding around \( x = 1 \),
[tex]\[ f(x) = \sqrt{3} \sqrt{1 + \frac{x-1}{3}} \][/tex]
Using the binomial series for \( (1 + u)^{\frac{1}{2}} \):
[tex]\[ (1 + u)^{\frac{1}{2}} = \sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n} u^n \][/tex]
[tex]\[ \sqrt{3} \left(1 + \frac{x-1}{3} \right)^{\frac{1}{2}} = \sqrt{3} \sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n} \left(\frac{x-1}{3}\right)^n \][/tex]
This yields,
[tex]\[ f(x) = \sqrt{3} \sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n} \left(\frac{x-1}{3}\right)^n \][/tex]
This is the power series for \( f(x) = \sqrt{2 + x} \) about \( x = 1 \).
### 4. Convergence of the Series
[tex]\[ \sum_{n=0}^{\infty} \frac{4^{n+1} (x-1)^n}{n^3 + 2n + 1} \][/tex]
To determine the convergence, consider the general term:
[tex]\[ a_n = \frac{4^{n+1} (x-1)^n}{n^3 + 2n + 1} \][/tex]
Apply the Ratio Test:
[tex]\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{4^{n+2} (x-1)^{n+1}}{(n+1)^3 + 2(n+1) + 1} \cdot \frac{n^3 + 2n + 1}{4^{n+1} (x-1)^n} \right| \][/tex]
Simplify the ratio inside the limit,
[tex]\[ = \lim_{n \to \infty} \left| 4 \cdot (x-1) \cdot \frac{n^3 + 2n + 1}{(n+1)^3 + 2(n+1) + 1} \right| = 4|x-1| \lim_{n \to \infty} \frac{n^3 + 2n + 1}{n^3(1 + \frac{3}{n} + \frac{3}{n^2} + \frac{1}{n^3})} = 4|x-1| \cdot 1 = 4|x-1| \][/tex]
For the series to converge,
[tex]\[ 4|x-1| < 1 \][/tex]
[tex]\[ |x-1| < \frac{1}{4} \][/tex]
Hence, the radius of convergence is \( \frac{1}{4} \). The interval of convergence is \( \left(1 - \frac{1}{4}, 1 + \frac{1}{4}\right) = (0.75, 1.25) \).
Thus, the series converges for [tex]\( x \)[/tex] in the interval [tex]\( (0.75, 1.25) \)[/tex].
### 1. Power Series for \( f(x) = \sqrt{16 - x^2} \)
#### About \( x = 0 \)
To find the power series representation of \( f(x) \) about \( x = 0 \), we use the binomial series expansion for a function of the form \( (1 - u)^{\frac{1}{2}} \), where \( u = \frac{x^2}{16} \).
[tex]\[ f(x) = \sqrt{16 - x^2} = 4 \sqrt{1 - \frac{x^2}{16}} \][/tex]
Using the binomial series for \( (1 - u)^{\frac{1}{2}} \), we have:
[tex]\[ (1 - u)^{\frac{1}{2}} = \sum_{n=0}^{\infty} (-1)^n \binom{\frac{1}{2}}{n} u^n \][/tex]
where
[tex]\[ \binom{\frac{1}{2}}{n} = \frac{\left( \frac{1}{2} \right) \left( \frac{1}{2} - 1 \right) \left( \frac{1}{2} - 2 \right) \cdots \left( \frac{1}{2} - (n-1)\right)}{n!} \][/tex]
Substituting \( u = \frac{x^2}{16} \):
[tex]\[ f(x) = 4 \sum_{n=0}^{\infty} (-1)^n \binom{\frac{1}{2}}{n} \left(\frac{x^2}{16} \right)^n \][/tex]
Simplifying the coefficients,
[tex]\[ f(x) = 4 \sum_{n=0}^{\infty} (-1)^n \binom{\frac{1}{2}}{n} \frac{x^{2n}}{16^n} \][/tex]
This is the power series for \( f(x) = \sqrt{16 - x^2} \) about \( x = 0 \).
### 2. Power Series for \( g(x) = (16 - x)^7 \)
#### About \( x = 0 \)
For \( g(x) = (16 - x)^7 \), we again use the binomial series:
[tex]\[ (16 - x)^7 = 16^7 \left( 1 - \frac{x}{16} \right)^7 \][/tex]
Expanding using the binomial theorem,
[tex]\[ (1 - \frac{x}{16})^7 = \sum_{n=0}^{\infty} \binom{7}{n} (-1)^n \left( \frac{x}{16} \right)^n \][/tex]
Then,
[tex]\[ g(x) = 16^7 \sum_{n=0}^{\infty} \binom{7}{n} (-1)^n \left( \frac{x}{16} \right)^n \][/tex]
Simplifying,
[tex]\[ g(x) = 16^7 \sum_{n=0}^{\infty} \binom{7}{n} (-1)^n \frac{x^n}{16^n} = \sum_{n=0}^{\infty} \binom{7}{n} (-1)^n 16^{7-n} x^n \][/tex]
This is the power series for \( g(x) = (16 - x)^7 \) about \( x = 0 \).
### 3. Power Series for \( f(x) = \sqrt{2 + x} \)
#### About \( x = 1 \)
To find the power series for \( f(x) = \sqrt{2 + x} \) about \( x = 1 \), we write \( f(x) \) in the form of a binomial expansion:
[tex]\[ f(x) = \sqrt{2 + x} = \sqrt{3 - 1 + (x-1)} = \sqrt{3 + (x-1)} \][/tex]
Expanding around \( x = 1 \),
[tex]\[ f(x) = \sqrt{3} \sqrt{1 + \frac{x-1}{3}} \][/tex]
Using the binomial series for \( (1 + u)^{\frac{1}{2}} \):
[tex]\[ (1 + u)^{\frac{1}{2}} = \sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n} u^n \][/tex]
[tex]\[ \sqrt{3} \left(1 + \frac{x-1}{3} \right)^{\frac{1}{2}} = \sqrt{3} \sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n} \left(\frac{x-1}{3}\right)^n \][/tex]
This yields,
[tex]\[ f(x) = \sqrt{3} \sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n} \left(\frac{x-1}{3}\right)^n \][/tex]
This is the power series for \( f(x) = \sqrt{2 + x} \) about \( x = 1 \).
### 4. Convergence of the Series
[tex]\[ \sum_{n=0}^{\infty} \frac{4^{n+1} (x-1)^n}{n^3 + 2n + 1} \][/tex]
To determine the convergence, consider the general term:
[tex]\[ a_n = \frac{4^{n+1} (x-1)^n}{n^3 + 2n + 1} \][/tex]
Apply the Ratio Test:
[tex]\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{4^{n+2} (x-1)^{n+1}}{(n+1)^3 + 2(n+1) + 1} \cdot \frac{n^3 + 2n + 1}{4^{n+1} (x-1)^n} \right| \][/tex]
Simplify the ratio inside the limit,
[tex]\[ = \lim_{n \to \infty} \left| 4 \cdot (x-1) \cdot \frac{n^3 + 2n + 1}{(n+1)^3 + 2(n+1) + 1} \right| = 4|x-1| \lim_{n \to \infty} \frac{n^3 + 2n + 1}{n^3(1 + \frac{3}{n} + \frac{3}{n^2} + \frac{1}{n^3})} = 4|x-1| \cdot 1 = 4|x-1| \][/tex]
For the series to converge,
[tex]\[ 4|x-1| < 1 \][/tex]
[tex]\[ |x-1| < \frac{1}{4} \][/tex]
Hence, the radius of convergence is \( \frac{1}{4} \). The interval of convergence is \( \left(1 - \frac{1}{4}, 1 + \frac{1}{4}\right) = (0.75, 1.25) \).
Thus, the series converges for [tex]\( x \)[/tex] in the interval [tex]\( (0.75, 1.25) \)[/tex].
