Mixing
Trigonometric inequality for Theorem 3.4.4 in Stenger 1993
I'm looking at the proof of Theorem 3.4.4 in "Frank Stenger Numerical Methods based on Sinc and Analytic Functions". The book seems to be available on google books, but I will try to make the specific issue self-contained.
Let $t,d,h in mathbb{R}$ and $zin mathbb{C}$, write $z=x + iy$. Assume $-d<y<d$. Then I need the bound
$$
left lvert frac{ e^{-i pi (t-id)/h} - cos(pi z /h)}{ (t-id-z) sin( pi (t-id)/h)}-frac{ e^{-i pi (t+id)/h} - cos(pi z /h)}{ (t+id-z) sin( pi (t+id)/h)} right lvert leq frac{ e^{-pi d/h} + cosh(pi y/h)}{ d sinh(pi d / h)}
$$
It is clear that $lvert e^{-i pi (t-id)/h} lvert leq e^{- pi d /h}$ and that $lvert cos(pi z /h)) lvert leq cosh(pi y/h)$, so the triangle inequality gets us some way. I can't seem to get anywhere with the denominator though.
trigonometry inequality complex-numbers
asked May 9 '18 at 13:40
Henrik
87411226
add a comment |
I'm looking at the proof of Theorem 3.4.4 in "Frank Stenger Numerical Methods based on Sinc and Analytic Functions". The book seems to be available on google books, but I will try to make the specific issue self-contained.
Let $t,d,h in mathbb{R}$ and $zin mathbb{C}$, write $z=x + iy$. Assume $-d<y<d$. Then I need the bound
$$
left lvert frac{ e^{-i pi (t-id)/h} - cos(pi z /h)}{ (t-id-z) sin( pi (t-id)/h)}-frac{ e^{-i pi (t+id)/h} - cos(pi z /h)}{ (t+id-z) sin( pi (t+id)/h)} right lvert leq frac{ e^{-pi d/h} + cosh(pi y/h)}{ d sinh(pi d / h)}
$$
It is clear that $lvert e^{-i pi (t-id)/h} lvert leq e^{- pi d /h}$ and that $lvert cos(pi z /h)) lvert leq cosh(pi y/h)$, so the triangle inequality gets us some way. I can't seem to get anywhere with the denominator though.
trigonometry inequality complex-numbers
asked May 9 '18 at 13:40
Henrik
87411226
add a comment |
I'm looking at the proof of Theorem 3.4.4 in "Frank Stenger Numerical Methods based on Sinc and Analytic Functions". The book seems to be available on google books, but I will try to make the specific issue self-contained.
Let $t,d,h in mathbb{R}$ and $zin mathbb{C}$, write $z=x + iy$. Assume $-d<y<d$. Then I need the bound
$$
left lvert frac{ e^{-i pi (t-id)/h} - cos(pi z /h)}{ (t-id-z) sin( pi (t-id)/h)}-frac{ e^{-i pi (t+id)/h} - cos(pi z /h)}{ (t+id-z) sin( pi (t+id)/h)} right lvert leq frac{ e^{-pi d/h} + cosh(pi y/h)}{ d sinh(pi d / h)}
$$
It is clear that $lvert e^{-i pi (t-id)/h} lvert leq e^{- pi d /h}$ and that $lvert cos(pi z /h)) lvert leq cosh(pi y/h)$, so the triangle inequality gets us some way. I can't seem to get anywhere with the denominator though.
trigonometry inequality complex-numbers
asked May 9 '18 at 13:40
Henrik
87411226
I'm looking at the proof of Theorem 3.4.4 in "Frank Stenger Numerical Methods based on Sinc and Analytic Functions". The book seems to be available on google books, but I will try to make the specific issue self-contained.
Let $t,d,h in mathbb{R}$ and $zin mathbb{C}$, write $z=x + iy$. Assume $-d<y<d$. Then I need the bound
$$
left lvert frac{ e^{-i pi (t-id)/h} - cos(pi z /h)}{ (t-id-z) sin( pi (t-id)/h)}-frac{ e^{-i pi (t+id)/h} - cos(pi z /h)}{ (t+id-z) sin( pi (t+id)/h)} right lvert leq frac{ e^{-pi d/h} + cosh(pi y/h)}{ d sinh(pi d / h)}
$$
It is clear that $lvert e^{-i pi (t-id)/h} lvert leq e^{- pi d /h}$ and that $lvert cos(pi z /h)) lvert leq cosh(pi y/h)$, so the triangle inequality gets us some way. I can't seem to get anywhere with the denominator though.
trigonometry inequality complex-numbers
trigonometry inequality complex-numbers
asked May 9 '18 at 13:40
Henrik
87411226
asked May 9 '18 at 13:40
Henrik
87411226
edited Nov 20 '18 at 13:16
asked May 9 '18 at 13:40
Henrik
87411226
asked May 9 '18 at 13:40
Henrik
87411226
asked May 9 '18 at 13:40
Henrik
87411226
87411226
add a comment |
add a comment |
1 Answer
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First $|tpm id-z|ge |d|$. Also for real $a$ and $b$
$$
|sin(a+ib)|=|sin a cosh b+i cos a sinh b|=|sin a, text{coth}, b+i cos a| |sinh b |ge |sinh b|
$$
since $|text{coth}, b|ge 1$. This leads to the rhs
$$
2frac{ e^{-pi d/h} + cosh(pi y/h)}{ |d sinh(pi d / h)|}.
$$
answered Jul 24 '18 at 8:04
Andrew
9,20711945
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
First $|tpm id-z|ge |d|$. Also for real $a$ and $b$
$$
|sin(a+ib)|=|sin a cosh b+i cos a sinh b|=|sin a, text{coth}, b+i cos a| |sinh b |ge |sinh b|
$$
since $|text{coth}, b|ge 1$. This leads to the rhs
$$
2frac{ e^{-pi d/h} + cosh(pi y/h)}{ |d sinh(pi d / h)|}.
$$
answered Jul 24 '18 at 8:04
Andrew
9,20711945
add a comment |
First $|tpm id-z|ge |d|$. Also for real $a$ and $b$
$$
|sin(a+ib)|=|sin a cosh b+i cos a sinh b|=|sin a, text{coth}, b+i cos a| |sinh b |ge |sinh b|
$$
since $|text{coth}, b|ge 1$. This leads to the rhs
$$
2frac{ e^{-pi d/h} + cosh(pi y/h)}{ |d sinh(pi d / h)|}.
$$
answered Jul 24 '18 at 8:04
Andrew
9,20711945
add a comment |
First $|tpm id-z|ge |d|$. Also for real $a$ and $b$
$$
|sin(a+ib)|=|sin a cosh b+i cos a sinh b|=|sin a, text{coth}, b+i cos a| |sinh b |ge |sinh b|
$$
since $|text{coth}, b|ge 1$. This leads to the rhs
$$
2frac{ e^{-pi d/h} + cosh(pi y/h)}{ |d sinh(pi d / h)|}.
$$
answered Jul 24 '18 at 8:04
Andrew
9,20711945
First $|tpm id-z|ge |d|$. Also for real $a$ and $b$
$$
|sin(a+ib)|=|sin a cosh b+i cos a sinh b|=|sin a, text{coth}, b+i cos a| |sinh b |ge |sinh b|
$$
since $|text{coth}, b|ge 1$. This leads to the rhs
$$
2frac{ e^{-pi d/h} + cosh(pi y/h)}{ |d sinh(pi d / h)|}.
$$
answered Jul 24 '18 at 8:04
Andrew
9,20711945
edited Jul 24 '18 at 9:07
answered Jul 24 '18 at 8:04
Andrew
9,20711945
answered Jul 24 '18 at 8:04
Andrew
9,20711945
answered Jul 24 '18 at 8:04
Andrew
9,20711945
9,20711945
add a comment |
add a comment |
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