Discover new knowledge and insights with IDNLearn.com's extensive Q&A database. Join our platform to receive prompt and accurate responses from experienced professionals in various fields.
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].
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.
