Get the information you need with the help of IDNLearn.com's extensive Q&A platform. Discover thorough and trustworthy answers from our community of knowledgeable professionals, tailored to meet your specific needs.
Sagot :
To show that both series [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2 n+1}}{(2 n+1)!}\)[/tex] and [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2 n}}{(2 n)!}\)[/tex] are uniformly convergent on any closed and bounded interval [tex]\([a, b]\)[/tex], we can apply the Weierstrass M-test. According to the Weierstrass M-test, if there exists a sequence [tex]\(\{M_n\}\)[/tex] of positive constants such that
[tex]\[|f_n(x)| \leq M_n \quad \text{for all } x \in [a, b] \text{ and for all } n,\][/tex]
and
[tex]\[\sum_{n=0}^{\infty} M_n\][/tex]
converges, then [tex]\(\sum_{n=0}^{\infty} f_n(x)\)[/tex] converges uniformly on [tex]\([a, b]\)[/tex].
### For the series [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2 n+1}}{(2 n+1)!}\)[/tex]:
1. Identify [tex]\(f_n(x)\)[/tex]:
[tex]\[ f_n(x) = \frac{(-1)^n x^{2n+1}}{(2n+1)!}. \][/tex]
2. Determine [tex]\(|f_n(x)|\)[/tex]:
[tex]\[ |f_n(x)| = \left|\frac{(-1)^n x^{2n+1}}{(2n+1)!}\right| = \frac{|x|^{2n+1}}{(2n+1)!}. \][/tex]
3. Find an appropriate [tex]\(M_n\)[/tex]:
Since [tex]\(x\)[/tex] is in the interval [tex]\([a, b]\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are finite, we can bound [tex]\(|x|\)[/tex] as follows:
[tex]\[ |x| \leq \max(|a|, |b|) = B. \][/tex]
Consequently,
[tex]\[ |f_n(x)| \leq \frac{B^{2n+1}}{(2n+1)!} = M_n. \][/tex]
4. Check the convergence of [tex]\(\sum_{n=0}^{\infty} M_n\)[/tex]:
We need to check if the series
[tex]\[ \sum_{n=0}^{\infty} \frac{B^{2n+1}}{(2n+1)!} \][/tex]
converges. This series is a part of the Maclaurin series for the exponential function, which is known to converge:
[tex]\[ \sum_{n=0}^{\infty} \frac{B^{2n+1}}{(2n+1)!} \approx 11013.23 \quad (\text{converges}). \][/tex]
Since [tex]\(\sum_{n=0}^{\infty} M_n\)[/tex] converges, by the Weierstrass M-test, the series [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}\)[/tex] converges uniformly on [tex]\([a, b]\)[/tex].
### For the series [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\)[/tex]:
1. Identify [tex]\(f_n(x)\)[/tex]:
[tex]\[ f_n(x) = \frac{(-1)^n x^{2n}}{(2n)!}. \][/tex]
2. Determine [tex]\(|f_n(x)\)[/tex]:
[tex]\[ |f_n(x)| = \left|\frac{(-1)^n x^{2n}}{(2n)!}\right| = \frac{|x|^{2n}}{(2n)!}. \][/tex]
3. Find an appropriate [tex]\(M_n\)[/tex]:
Since [tex]\(x\)[/tex] is in the interval [tex]\([a, b]\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are finite, we can bound [tex]\(|x|\)[/tex] as follows:
[tex]\[ |x| \leq \max(|a|, |b|) = B. \][/tex]
Consequently,
[tex]\[ |f_n(x)| \leq \frac{B^{2n}}{(2n)!} = M_n. \][/tex]
4. Check the convergence of [tex]\(\sum_{n=0}^{\infty} M_n\)[/tex]:
We need to check if the series
[tex]\[ \sum_{n=0}^{\infty} \frac{B^{2n}}{(2n)!} \][/tex]
converges. This series is a part of another representation of the exponential function, which is known to converge:
[tex]\[ \sum_{n=0}^{\infty} \frac{B^{2n}}{(2n)!} \approx 11013.23 \quad (\text{converges}). \][/tex]
Since [tex]\(\sum_{n=0}^{\infty} M_n\)[/tex] converges, by the Weierstrass M-test, the series [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\)[/tex] converges uniformly on [tex]\([a, b]\)[/tex].
In conclusion, both series [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}\)[/tex] and [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\)[/tex] are uniformly convergent on any closed and bounded interval [tex]\([a, b]\)[/tex].
[tex]\[|f_n(x)| \leq M_n \quad \text{for all } x \in [a, b] \text{ and for all } n,\][/tex]
and
[tex]\[\sum_{n=0}^{\infty} M_n\][/tex]
converges, then [tex]\(\sum_{n=0}^{\infty} f_n(x)\)[/tex] converges uniformly on [tex]\([a, b]\)[/tex].
### For the series [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2 n+1}}{(2 n+1)!}\)[/tex]:
1. Identify [tex]\(f_n(x)\)[/tex]:
[tex]\[ f_n(x) = \frac{(-1)^n x^{2n+1}}{(2n+1)!}. \][/tex]
2. Determine [tex]\(|f_n(x)|\)[/tex]:
[tex]\[ |f_n(x)| = \left|\frac{(-1)^n x^{2n+1}}{(2n+1)!}\right| = \frac{|x|^{2n+1}}{(2n+1)!}. \][/tex]
3. Find an appropriate [tex]\(M_n\)[/tex]:
Since [tex]\(x\)[/tex] is in the interval [tex]\([a, b]\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are finite, we can bound [tex]\(|x|\)[/tex] as follows:
[tex]\[ |x| \leq \max(|a|, |b|) = B. \][/tex]
Consequently,
[tex]\[ |f_n(x)| \leq \frac{B^{2n+1}}{(2n+1)!} = M_n. \][/tex]
4. Check the convergence of [tex]\(\sum_{n=0}^{\infty} M_n\)[/tex]:
We need to check if the series
[tex]\[ \sum_{n=0}^{\infty} \frac{B^{2n+1}}{(2n+1)!} \][/tex]
converges. This series is a part of the Maclaurin series for the exponential function, which is known to converge:
[tex]\[ \sum_{n=0}^{\infty} \frac{B^{2n+1}}{(2n+1)!} \approx 11013.23 \quad (\text{converges}). \][/tex]
Since [tex]\(\sum_{n=0}^{\infty} M_n\)[/tex] converges, by the Weierstrass M-test, the series [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}\)[/tex] converges uniformly on [tex]\([a, b]\)[/tex].
### For the series [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\)[/tex]:
1. Identify [tex]\(f_n(x)\)[/tex]:
[tex]\[ f_n(x) = \frac{(-1)^n x^{2n}}{(2n)!}. \][/tex]
2. Determine [tex]\(|f_n(x)\)[/tex]:
[tex]\[ |f_n(x)| = \left|\frac{(-1)^n x^{2n}}{(2n)!}\right| = \frac{|x|^{2n}}{(2n)!}. \][/tex]
3. Find an appropriate [tex]\(M_n\)[/tex]:
Since [tex]\(x\)[/tex] is in the interval [tex]\([a, b]\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are finite, we can bound [tex]\(|x|\)[/tex] as follows:
[tex]\[ |x| \leq \max(|a|, |b|) = B. \][/tex]
Consequently,
[tex]\[ |f_n(x)| \leq \frac{B^{2n}}{(2n)!} = M_n. \][/tex]
4. Check the convergence of [tex]\(\sum_{n=0}^{\infty} M_n\)[/tex]:
We need to check if the series
[tex]\[ \sum_{n=0}^{\infty} \frac{B^{2n}}{(2n)!} \][/tex]
converges. This series is a part of another representation of the exponential function, which is known to converge:
[tex]\[ \sum_{n=0}^{\infty} \frac{B^{2n}}{(2n)!} \approx 11013.23 \quad (\text{converges}). \][/tex]
Since [tex]\(\sum_{n=0}^{\infty} M_n\)[/tex] converges, by the Weierstrass M-test, the series [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\)[/tex] converges uniformly on [tex]\([a, b]\)[/tex].
In conclusion, both series [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}\)[/tex] and [tex]\(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\)[/tex] are uniformly convergent on any closed and bounded interval [tex]\([a, b]\)[/tex].
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. IDNLearn.com is committed to your satisfaction. Thank you for visiting, and see you next time for more helpful answers.