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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.
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