Using the sequential definition of uniform continuity to show $sin(x)$ is uniformly continuous on $mathbb{R}$
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I want to show $sin(x)$ is uniformly continuous on $mathbb{R}$. Let ${a_{n}}$ and ${b_{n}}$ be sequences such that $lim_{ntoinfty}[b_{n} - a_{n}] = 0$. Then, we need to show $lim_{ntoinfty} |sin(b_{n}) - sin(a_{n})| = 0$. But, I cannot prove this equality. Can someone please help me?
real-analysis inequality uniform-convergence uniform-continuity
asked Nov 16 at 8:29
joseph
1258
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up vote
0
down vote
favorite
I want to show $sin(x)$ is uniformly continuous on $mathbb{R}$. Let ${a_{n}}$ and ${b_{n}}$ be sequences such that $lim_{ntoinfty}[b_{n} - a_{n}] = 0$. Then, we need to show $lim_{ntoinfty} |sin(b_{n}) - sin(a_{n})| = 0$. But, I cannot prove this equality. Can someone please help me?
real-analysis inequality uniform-convergence uniform-continuity
asked Nov 16 at 8:29
joseph
1258
2
Mean Value Theorem.
– Kavi Rama Murthy
Nov 16 at 8:30
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to show $sin(x)$ is uniformly continuous on $mathbb{R}$. Let ${a_{n}}$ and ${b_{n}}$ be sequences such that $lim_{ntoinfty}[b_{n} - a_{n}] = 0$. Then, we need to show $lim_{ntoinfty} |sin(b_{n}) - sin(a_{n})| = 0$. But, I cannot prove this equality. Can someone please help me?
real-analysis inequality uniform-convergence uniform-continuity
asked Nov 16 at 8:29
joseph
1258
I want to show $sin(x)$ is uniformly continuous on $mathbb{R}$. Let ${a_{n}}$ and ${b_{n}}$ be sequences such that $lim_{ntoinfty}[b_{n} - a_{n}] = 0$. Then, we need to show $lim_{ntoinfty} |sin(b_{n}) - sin(a_{n})| = 0$. But, I cannot prove this equality. Can someone please help me?
real-analysis inequality uniform-convergence uniform-continuity
real-analysis inequality uniform-convergence uniform-continuity
asked Nov 16 at 8:29
joseph
1258
asked Nov 16 at 8:29
joseph
1258
asked Nov 16 at 8:29
joseph
1258
asked Nov 16 at 8:29
joseph
1258
asked Nov 16 at 8:29
joseph
1258
1258
2
Mean Value Theorem.
– Kavi Rama Murthy
Nov 16 at 8:30
add a comment |
2
Mean Value Theorem.
– Kavi Rama Murthy
Nov 16 at 8:30
2
2
Mean Value Theorem.
– Kavi Rama Murthy
Nov 16 at 8:30
Mean Value Theorem.
– Kavi Rama Murthy
Nov 16 at 8:30
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
By the mean value theorem:
$sin(b_{n}) - sin(a_{n})= cos (t_n)(b_n-a_n)$ with $t_n$ between $a_n$ and $b_n$.
Can you proceed ?
answered Nov 16 at 8:32
Fred
42.8k1643
i understand now, thanks.
– joseph
Nov 16 at 8:37
add a comment |
up vote
1
down vote
By simple trigonometric manipulation:
$sin(b_n) - sin(a_n) = 2sin(frac{b_n-a_n}{2})cos(frac{b_n+a_n}{2})le 2sin(frac{b_n-a_n}{2})$.
answered Nov 16 at 10:24
Offlaw
2489
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
By the mean value theorem:
$sin(b_{n}) - sin(a_{n})= cos (t_n)(b_n-a_n)$ with $t_n$ between $a_n$ and $b_n$.
Can you proceed ?
answered Nov 16 at 8:32
Fred
42.8k1643
i understand now, thanks.
– joseph
Nov 16 at 8:37
add a comment |
up vote
3
down vote
accepted
By the mean value theorem:
$sin(b_{n}) - sin(a_{n})= cos (t_n)(b_n-a_n)$ with $t_n$ between $a_n$ and $b_n$.
Can you proceed ?
answered Nov 16 at 8:32
Fred
42.8k1643
i understand now, thanks.
– joseph
Nov 16 at 8:37
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
By the mean value theorem:
$sin(b_{n}) - sin(a_{n})= cos (t_n)(b_n-a_n)$ with $t_n$ between $a_n$ and $b_n$.
Can you proceed ?
answered Nov 16 at 8:32
Fred
42.8k1643
By the mean value theorem:
$sin(b_{n}) - sin(a_{n})= cos (t_n)(b_n-a_n)$ with $t_n$ between $a_n$ and $b_n$.
Can you proceed ?
answered Nov 16 at 8:32
Fred
42.8k1643
answered Nov 16 at 8:32
Fred
42.8k1643
answered Nov 16 at 8:32
Fred
42.8k1643
answered Nov 16 at 8:32
Fred
42.8k1643
42.8k1643
i understand now, thanks.
– joseph
Nov 16 at 8:37
add a comment |
i understand now, thanks.
– joseph
Nov 16 at 8:37
i understand now, thanks.
– joseph
Nov 16 at 8:37
i understand now, thanks.
– joseph
Nov 16 at 8:37
add a comment |
up vote
1
down vote
By simple trigonometric manipulation:
$sin(b_n) - sin(a_n) = 2sin(frac{b_n-a_n}{2})cos(frac{b_n+a_n}{2})le 2sin(frac{b_n-a_n}{2})$.
answered Nov 16 at 10:24
Offlaw
2489
add a comment |
up vote
1
down vote
By simple trigonometric manipulation:
$sin(b_n) - sin(a_n) = 2sin(frac{b_n-a_n}{2})cos(frac{b_n+a_n}{2})le 2sin(frac{b_n-a_n}{2})$.
answered Nov 16 at 10:24
Offlaw
2489
add a comment |
up vote
1
down vote
up vote
1
down vote
By simple trigonometric manipulation:
$sin(b_n) - sin(a_n) = 2sin(frac{b_n-a_n}{2})cos(frac{b_n+a_n}{2})le 2sin(frac{b_n-a_n}{2})$.
answered Nov 16 at 10:24
Offlaw
2489
By simple trigonometric manipulation:
$sin(b_n) - sin(a_n) = 2sin(frac{b_n-a_n}{2})cos(frac{b_n+a_n}{2})le 2sin(frac{b_n-a_n}{2})$.
answered Nov 16 at 10:24
Offlaw
2489
answered Nov 16 at 10:24
Offlaw
2489
answered Nov 16 at 10:24
Offlaw
2489
answered Nov 16 at 10:24
Offlaw
2489
2489
add a comment |
add a comment |
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2
Mean Value Theorem.
– Kavi Rama Murthy
Nov 16 at 8:30