Answer :
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].
