Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Let's address each of the problems one by one:
1. Power Series for \( f(x) = \sqrt{16 - x^2} \):
To find a power series for \( f(x) = \sqrt{16 - x^2} \) around \( x = 0 \), we can use the Taylor series expansion. This is generally cumbersome because \( f(x) \) isn’t expressed in a typical series-friendly form. However, there’s a known expansion for the square root function:
[tex]\[ f(x) = \sqrt{16 - x^2} = 4 \sqrt{1 - \left(\frac{x}{4}\right)^2} \][/tex]
Expanding \( \sqrt{1 - u} \) using the binomial series, where \( u = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \):
[tex]\[ \sqrt{1 - u} \approx 1 - \frac{u}{2} - \frac{u^2}{8} - \frac{u^3}{16} - \cdots \][/tex]
Substitute \( u = \frac{x^2}{16} \):
[tex]\[ \sqrt{1 - \frac{x^2}{16}} \approx 1 - \frac{x^2}{32} - \frac{x^4}{512} - \frac{x^6}{8192} - \cdots \][/tex]
Therefore,
[tex]\[ \sqrt{16 - x^2} = 4 \left( 1 - \frac{x^2}{32} - \frac{x^4}{512} - \frac{x^6}{8192} - \cdots \right) \][/tex]
[tex]\[ = 4 - \frac{x^2}{8} - \frac{x^4}{128} - \frac{x^6}{2048} - \cdots \][/tex]
2. Power Series for \( g(x) = (16 - x)^7 \) about \( x = 0 \):
For \( g(x) = (16 - x)^7 \), we can substitute \( u = x/16 \):
[tex]\[ g(x) = (16(1 - \frac{x}{16}))^7 = 16^7 (1 - \frac{x}{16})^7 \][/tex]
Expanding \( (1 - \frac{x}{16})^7 \) using the binomial theorem:
[tex]\[ (1 - \frac{x}{16})^7 = \sum_{n=0}^{\infty} \binom{7}{n} \left( -\frac{x}{16} \right)^n \][/tex]
Thus,
[tex]\[ g(x) = 16^7 \sum_{n=0}^{\infty} \binom{7}{n} \left( -\frac{x}{16} \right)^n \][/tex]
[tex]\[ = \sum_{n=0}^{\infty} \binom{7}{n} (-1)^n 16^{7-n} x^n \][/tex]
3. Power Series for \( f(x) = \sqrt{2 + x} \) about \( x = 1 \):
We use \( u = x - 1 \), so \( f(x) = \sqrt{2 + u + 1} = \sqrt{3 + u} \):
Expanding \( \sqrt{3 + u} \):
[tex]\[ \sqrt{3 + u} = \sqrt{3} \left( 1 + \frac{u}{3} \right)^{1/2} \][/tex]
Using the binomial expansion for \( \left(1 + \frac{u}{3}\right)^{1/2} \):
[tex]\[ \sqrt{3} \left(1 + \frac{u}{3}\right)^{1/2} \approx \sqrt{3} \left( 1 + \frac{u}{6} - \frac{u^2}{72} + \cdots \right) \][/tex]
Substituting \( u = x - 1 \):
[tex]\[ f(x) = \sqrt{3 + (x-1)} \approx \sqrt{3} \left(1 + \frac{x-1}{6} - \frac{(x-1)^2}{72} + \cdots \right) \][/tex]
4. Convergence of \( \sum_{n=0}^{\infty} \frac{4^{n+1}(x-1)^n}{n^3 + 2n + 1} \):
To determine convergence, you can use the ratio test:
Consider the term \( a_n = \frac{4^{n+1}(x-1)^n}{n^3 + 2n + 1} \).
The ratio \(\frac{a_{n+1}}{a_n} \):
[tex]\[ \frac{\frac{4^{n+2}(x-1)^{n+1}}{(n+1)^3 + 2(n+1) + 1}}{\frac{4^{n+1}(x-1)^n}{n^3 + 2n + 1}} = 4(x-1) \cdot \frac{4}{\frac{(n+1)^3 + 2(n+1) + 1}{n^3 + 2n + 1}} \][/tex]
Generally, the ratio would simplify to \( 4(x-1) \):
[tex]\[ 4|x-1| < 1 \implies |x-1| < \frac{1}{4} \][/tex]
Thus, the series converges for \( x \) in the interval \( 1 - \frac{1}{4} < x < 1 + \frac{1}{4} \), or \( \frac{3}{4} < x < \frac{5}{4} \).
5. Power Series for \( \int \frac{\cos (x) - 1}{x} dx \):
Start with the Maclaurin series for \( \cos(x) \):
[tex]\[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \][/tex]
Then,
[tex]\[ \cos(x) - 1 = -\frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \][/tex]
Dividing by \( x \):
[tex]\[ \frac{\cos(x) - 1}{x} = -\frac{x}{2!} + \frac{x^3}{4!} - \frac{x^5}{6!} + \cdots \][/tex]
Integrating term by term:
[tex]\[ \int \frac{\cos(x) - 1}{x} dx = -\frac{x^2}{2 \cdot 2!} + \frac{x^4}{4 \cdot 4!} - \frac{x^6}{6 \cdot 6!} + \cdots + C \][/tex]
[tex]\[ = -\frac{x^2}{4} + \frac{x^4}{96} - \frac{x^6}{4320} + \cdots + C \][/tex]
So,
[tex]\[ \int \frac{\cos(x) - 1}{x} dx = -\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{(2n)(2n)!} + C \][/tex]
This completes the power series manipulations and the given problems.
1. Power Series for \( f(x) = \sqrt{16 - x^2} \):
To find a power series for \( f(x) = \sqrt{16 - x^2} \) around \( x = 0 \), we can use the Taylor series expansion. This is generally cumbersome because \( f(x) \) isn’t expressed in a typical series-friendly form. However, there’s a known expansion for the square root function:
[tex]\[ f(x) = \sqrt{16 - x^2} = 4 \sqrt{1 - \left(\frac{x}{4}\right)^2} \][/tex]
Expanding \( \sqrt{1 - u} \) using the binomial series, where \( u = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \):
[tex]\[ \sqrt{1 - u} \approx 1 - \frac{u}{2} - \frac{u^2}{8} - \frac{u^3}{16} - \cdots \][/tex]
Substitute \( u = \frac{x^2}{16} \):
[tex]\[ \sqrt{1 - \frac{x^2}{16}} \approx 1 - \frac{x^2}{32} - \frac{x^4}{512} - \frac{x^6}{8192} - \cdots \][/tex]
Therefore,
[tex]\[ \sqrt{16 - x^2} = 4 \left( 1 - \frac{x^2}{32} - \frac{x^4}{512} - \frac{x^6}{8192} - \cdots \right) \][/tex]
[tex]\[ = 4 - \frac{x^2}{8} - \frac{x^4}{128} - \frac{x^6}{2048} - \cdots \][/tex]
2. Power Series for \( g(x) = (16 - x)^7 \) about \( x = 0 \):
For \( g(x) = (16 - x)^7 \), we can substitute \( u = x/16 \):
[tex]\[ g(x) = (16(1 - \frac{x}{16}))^7 = 16^7 (1 - \frac{x}{16})^7 \][/tex]
Expanding \( (1 - \frac{x}{16})^7 \) using the binomial theorem:
[tex]\[ (1 - \frac{x}{16})^7 = \sum_{n=0}^{\infty} \binom{7}{n} \left( -\frac{x}{16} \right)^n \][/tex]
Thus,
[tex]\[ g(x) = 16^7 \sum_{n=0}^{\infty} \binom{7}{n} \left( -\frac{x}{16} \right)^n \][/tex]
[tex]\[ = \sum_{n=0}^{\infty} \binom{7}{n} (-1)^n 16^{7-n} x^n \][/tex]
3. Power Series for \( f(x) = \sqrt{2 + x} \) about \( x = 1 \):
We use \( u = x - 1 \), so \( f(x) = \sqrt{2 + u + 1} = \sqrt{3 + u} \):
Expanding \( \sqrt{3 + u} \):
[tex]\[ \sqrt{3 + u} = \sqrt{3} \left( 1 + \frac{u}{3} \right)^{1/2} \][/tex]
Using the binomial expansion for \( \left(1 + \frac{u}{3}\right)^{1/2} \):
[tex]\[ \sqrt{3} \left(1 + \frac{u}{3}\right)^{1/2} \approx \sqrt{3} \left( 1 + \frac{u}{6} - \frac{u^2}{72} + \cdots \right) \][/tex]
Substituting \( u = x - 1 \):
[tex]\[ f(x) = \sqrt{3 + (x-1)} \approx \sqrt{3} \left(1 + \frac{x-1}{6} - \frac{(x-1)^2}{72} + \cdots \right) \][/tex]
4. Convergence of \( \sum_{n=0}^{\infty} \frac{4^{n+1}(x-1)^n}{n^3 + 2n + 1} \):
To determine convergence, you can use the ratio test:
Consider the term \( a_n = \frac{4^{n+1}(x-1)^n}{n^3 + 2n + 1} \).
The ratio \(\frac{a_{n+1}}{a_n} \):
[tex]\[ \frac{\frac{4^{n+2}(x-1)^{n+1}}{(n+1)^3 + 2(n+1) + 1}}{\frac{4^{n+1}(x-1)^n}{n^3 + 2n + 1}} = 4(x-1) \cdot \frac{4}{\frac{(n+1)^3 + 2(n+1) + 1}{n^3 + 2n + 1}} \][/tex]
Generally, the ratio would simplify to \( 4(x-1) \):
[tex]\[ 4|x-1| < 1 \implies |x-1| < \frac{1}{4} \][/tex]
Thus, the series converges for \( x \) in the interval \( 1 - \frac{1}{4} < x < 1 + \frac{1}{4} \), or \( \frac{3}{4} < x < \frac{5}{4} \).
5. Power Series for \( \int \frac{\cos (x) - 1}{x} dx \):
Start with the Maclaurin series for \( \cos(x) \):
[tex]\[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \][/tex]
Then,
[tex]\[ \cos(x) - 1 = -\frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \][/tex]
Dividing by \( x \):
[tex]\[ \frac{\cos(x) - 1}{x} = -\frac{x}{2!} + \frac{x^3}{4!} - \frac{x^5}{6!} + \cdots \][/tex]
Integrating term by term:
[tex]\[ \int \frac{\cos(x) - 1}{x} dx = -\frac{x^2}{2 \cdot 2!} + \frac{x^4}{4 \cdot 4!} - \frac{x^6}{6 \cdot 6!} + \cdots + C \][/tex]
[tex]\[ = -\frac{x^2}{4} + \frac{x^4}{96} - \frac{x^6}{4320} + \cdots + C \][/tex]
So,
[tex]\[ \int \frac{\cos(x) - 1}{x} dx = -\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{(2n)(2n)!} + C \][/tex]
This completes the power series manipulations and the given problems.
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.