On the Zeros of the Complex Error Function
$begingroup$
I'm working on something that involves the error function, except with complex inputs, i.e. erf(z) with $$zinmathbb{C}$$. In particular, at this point, I need a formula for the zeros of this function. I've found the paper "Complex Zeros of the Error Function and of the Complementary Error Function" by Fettis, Caslin, and Kramer (mathematics of computation V. 27 num 122 April 1973, pp 401-407). They provide approximate zeros, but I'm more interested in understanding the original equation they are trying to solve, because I only need a formula not actual values. In working through their paper I'm confused about a quick deduction they make, fairly off casually.
To explain further, they are working with a function denoted $$w(z)=e^{-z^2}mbox{erfc}(imath z)=e^{-z^2}(1-mbox{erf}(imath z))$$ First they point out that (sym prop 1): $$w(-x+imath y)=w^*(x+imath y)$$ and that (sym prop 2): $$w(x-imath y)=2e^{-(x-imath y)^2}-w^*(x+imath y)$$ (where * denotes complex conjugation). Then that it can be expressed as $$w(z)=frac{1}{pi}int_{-infty}^infty e^{-t^2}frac{[y+imath(x-t)]}{(x-t)^2+y^2}ds$$ which after a substitution of $$x-t=s$$ can be written in the form $$frac{2e^{-x^{2}}}{pi}Big{[}yint_0^{infty}e^{-s^{2}}frac{mbox{cosh}(2xs)}{y^2+s^2}ds+imathint_0^infty e^{-s^{2}}frac{mbox{sinh}(2xs)}{y^2+s^2}s;dsBig{]}=u(x,y)+imath(v(x,y)$$ at this point they make the following series of comments:
from which it is clear that $$u(x,y)>0$$, $$v(x,y)>0$$ for
$$x>0$$, $$y>0$$. This together with the symmetry property
(sym prop1) shows that any zeros of w(z) must lie in the lower half plane, and these according to (sym prop 2) will be
determined from the equations
$$2e^{y^2-x^2}cos(2xy) = u(x,y)$$ and $$2e^{y^2-x^2}sin(2xy)=v(x,y)$$
Then there's a footnote saying that from now on y will be positive (because in fact for the complex error functions there are zeros in the first quadrant). They continue:
Since $$u(x,y) mbox{ and } v(x,y)rightarrow 0 mbox{ for } abs{z}rightarrow 0$$ it is evident that asymptotically $$abs{y}<x$$. Further since $$u mbox{ and } v mbox{ are both positive if } x mbox{ and } y mbox{ are positive }$$ it follows that
So up to now, everything is fine. Some of its a bit above my pay grade, but ok, other references concur on these points, so ok. Here comes the hard bit. They now conclude:
$$cos(2xy)>0 mbox{ and }sin(2xy)<0$$, and hence that $$2xy=2npi-beta mbox{ for } n=1,2,3,ldots mbox{ where } 0leqbeta<frac{pi}{4}$$
This has me flumoxed. It would seem that this last equation is the equation they are going to approximate in order to find approximate roots of the complex error function. But if u(x,y) and v(x,y) are the real and imaginary parts of the solution, then the zeros should be where u and v are both simultaneously zero, which doesn't seem like it can happen, because $$cos(theta)mbox{ and }sin(theta)$$ are never simultaneously zero, and $$e^{y^2-x^2}>0;forall x,yinmathbb{R}$$. My only guess is that they are using the norm squared of $$u(x,y)+imath v(x,y)$$, but they very carefully pointed out that both functions are positive for positive x,y, although that seems to be false since sin(x) clearly can be negative for positive x. So after all that :) is that what they are doing? Taking the norm squared? And if so which of them is negative, u or v? Or am I missing something entirely...
Thanks in advance.
complex-analysis special-functions
asked Dec 13 '18 at 22:39
Michael CloudMichael Cloud
866
$endgroup$
add a comment |
$begingroup$
I'm working on something that involves the error function, except with complex inputs, i.e. erf(z) with $$zinmathbb{C}$$. In particular, at this point, I need a formula for the zeros of this function. I've found the paper "Complex Zeros of the Error Function and of the Complementary Error Function" by Fettis, Caslin, and Kramer (mathematics of computation V. 27 num 122 April 1973, pp 401-407). They provide approximate zeros, but I'm more interested in understanding the original equation they are trying to solve, because I only need a formula not actual values. In working through their paper I'm confused about a quick deduction they make, fairly off casually.
To explain further, they are working with a function denoted $$w(z)=e^{-z^2}mbox{erfc}(imath z)=e^{-z^2}(1-mbox{erf}(imath z))$$ First they point out that (sym prop 1): $$w(-x+imath y)=w^*(x+imath y)$$ and that (sym prop 2): $$w(x-imath y)=2e^{-(x-imath y)^2}-w^*(x+imath y)$$ (where * denotes complex conjugation). Then that it can be expressed as $$w(z)=frac{1}{pi}int_{-infty}^infty e^{-t^2}frac{[y+imath(x-t)]}{(x-t)^2+y^2}ds$$ which after a substitution of $$x-t=s$$ can be written in the form $$frac{2e^{-x^{2}}}{pi}Big{[}yint_0^{infty}e^{-s^{2}}frac{mbox{cosh}(2xs)}{y^2+s^2}ds+imathint_0^infty e^{-s^{2}}frac{mbox{sinh}(2xs)}{y^2+s^2}s;dsBig{]}=u(x,y)+imath(v(x,y)$$ at this point they make the following series of comments:
from which it is clear that $$u(x,y)>0$$, $$v(x,y)>0$$ for
$$x>0$$, $$y>0$$. This together with the symmetry property
(sym prop1) shows that any zeros of w(z) must lie in the lower half plane, and these according to (sym prop 2) will be
determined from the equations
$$2e^{y^2-x^2}cos(2xy) = u(x,y)$$ and $$2e^{y^2-x^2}sin(2xy)=v(x,y)$$
Then there's a footnote saying that from now on y will be positive (because in fact for the complex error functions there are zeros in the first quadrant). They continue:
Since $$u(x,y) mbox{ and } v(x,y)rightarrow 0 mbox{ for } abs{z}rightarrow 0$$ it is evident that asymptotically $$abs{y}<x$$. Further since $$u mbox{ and } v mbox{ are both positive if } x mbox{ and } y mbox{ are positive }$$ it follows that
So up to now, everything is fine. Some of its a bit above my pay grade, but ok, other references concur on these points, so ok. Here comes the hard bit. They now conclude:
$$cos(2xy)>0 mbox{ and }sin(2xy)<0$$, and hence that $$2xy=2npi-beta mbox{ for } n=1,2,3,ldots mbox{ where } 0leqbeta<frac{pi}{4}$$
This has me flumoxed. It would seem that this last equation is the equation they are going to approximate in order to find approximate roots of the complex error function. But if u(x,y) and v(x,y) are the real and imaginary parts of the solution, then the zeros should be where u and v are both simultaneously zero, which doesn't seem like it can happen, because $$cos(theta)mbox{ and }sin(theta)$$ are never simultaneously zero, and $$e^{y^2-x^2}>0;forall x,yinmathbb{R}$$. My only guess is that they are using the norm squared of $$u(x,y)+imath v(x,y)$$, but they very carefully pointed out that both functions are positive for positive x,y, although that seems to be false since sin(x) clearly can be negative for positive x. So after all that :) is that what they are doing? Taking the norm squared? And if so which of them is negative, u or v? Or am I missing something entirely...
Thanks in advance.
complex-analysis special-functions
asked Dec 13 '18 at 22:39
Michael CloudMichael Cloud
866
$endgroup$
$begingroup$
May be this helps: They also say that for $x>0, y>0$ there are no zeros, and that any zeros of $w(z)$ which exist must lie in the lower half-plane and these, according to Eq. (4)(b), will be determined from the equations ... In the paper (eq. 12) they have $$2e^{y^2-x^2}sin(2xy)=-v(x,y)$$ i.e. you missed the − sign.
$endgroup$
– gammatester
Dec 14 '18 at 10:37
$begingroup$
good catch, but I don't see how it changes anything. -0=0. As for zeros in the lower half plane, I spent a good bit of time wrestling with that. It turns out that refers to w(z) but not erf(z). Erf(z) has different symmetry properties than w(z) in particular erf(z) satisfies erf(z*)=erf*(z) AND erf(-z)=-erf(z), so that it is symmetric in all 4 quadrants, and has zeros in the 1st quadrant. That's why they let y be positive.
$endgroup$
– Michael Cloud
Dec 14 '18 at 14:25
$begingroup$
But anyway, I don't see how to solve it for any sign of y. y being positive or negative doesn't change whether sin(2xy) is positive or not (it is in both cases) nor does it change whether e^(y^2-x^2) is positive or not. It is never negative, and only zero at x= pm infinity
$endgroup$
– Michael Cloud
Dec 14 '18 at 14:26
$begingroup$
Indeed the comment in the paper that u(x,y) and v(x,y) go to zero as z=x+iy goes to infinity also seems quite off. e^(y^2-x^2) would seem to go to infinity as y goes to pm infinity. It is true that it goes to zero very quickly along the real axis, but is always positive for any finite x.
$endgroup$
– Michael Cloud
Dec 14 '18 at 16:19
add a comment |
$begingroup$
I'm working on something that involves the error function, except with complex inputs, i.e. erf(z) with $$zinmathbb{C}$$. In particular, at this point, I need a formula for the zeros of this function. I've found the paper "Complex Zeros of the Error Function and of the Complementary Error Function" by Fettis, Caslin, and Kramer (mathematics of computation V. 27 num 122 April 1973, pp 401-407). They provide approximate zeros, but I'm more interested in understanding the original equation they are trying to solve, because I only need a formula not actual values. In working through their paper I'm confused about a quick deduction they make, fairly off casually.
To explain further, they are working with a function denoted $$w(z)=e^{-z^2}mbox{erfc}(imath z)=e^{-z^2}(1-mbox{erf}(imath z))$$ First they point out that (sym prop 1): $$w(-x+imath y)=w^*(x+imath y)$$ and that (sym prop 2): $$w(x-imath y)=2e^{-(x-imath y)^2}-w^*(x+imath y)$$ (where * denotes complex conjugation). Then that it can be expressed as $$w(z)=frac{1}{pi}int_{-infty}^infty e^{-t^2}frac{[y+imath(x-t)]}{(x-t)^2+y^2}ds$$ which after a substitution of $$x-t=s$$ can be written in the form $$frac{2e^{-x^{2}}}{pi}Big{[}yint_0^{infty}e^{-s^{2}}frac{mbox{cosh}(2xs)}{y^2+s^2}ds+imathint_0^infty e^{-s^{2}}frac{mbox{sinh}(2xs)}{y^2+s^2}s;dsBig{]}=u(x,y)+imath(v(x,y)$$ at this point they make the following series of comments:
from which it is clear that $$u(x,y)>0$$, $$v(x,y)>0$$ for
$$x>0$$, $$y>0$$. This together with the symmetry property
(sym prop1) shows that any zeros of w(z) must lie in the lower half plane, and these according to (sym prop 2) will be
determined from the equations
$$2e^{y^2-x^2}cos(2xy) = u(x,y)$$ and $$2e^{y^2-x^2}sin(2xy)=v(x,y)$$
Then there's a footnote saying that from now on y will be positive (because in fact for the complex error functions there are zeros in the first quadrant). They continue:
Since $$u(x,y) mbox{ and } v(x,y)rightarrow 0 mbox{ for } abs{z}rightarrow 0$$ it is evident that asymptotically $$abs{y}<x$$. Further since $$u mbox{ and } v mbox{ are both positive if } x mbox{ and } y mbox{ are positive }$$ it follows that
So up to now, everything is fine. Some of its a bit above my pay grade, but ok, other references concur on these points, so ok. Here comes the hard bit. They now conclude:
$$cos(2xy)>0 mbox{ and }sin(2xy)<0$$, and hence that $$2xy=2npi-beta mbox{ for } n=1,2,3,ldots mbox{ where } 0leqbeta<frac{pi}{4}$$
This has me flumoxed. It would seem that this last equation is the equation they are going to approximate in order to find approximate roots of the complex error function. But if u(x,y) and v(x,y) are the real and imaginary parts of the solution, then the zeros should be where u and v are both simultaneously zero, which doesn't seem like it can happen, because $$cos(theta)mbox{ and }sin(theta)$$ are never simultaneously zero, and $$e^{y^2-x^2}>0;forall x,yinmathbb{R}$$. My only guess is that they are using the norm squared of $$u(x,y)+imath v(x,y)$$, but they very carefully pointed out that both functions are positive for positive x,y, although that seems to be false since sin(x) clearly can be negative for positive x. So after all that :) is that what they are doing? Taking the norm squared? And if so which of them is negative, u or v? Or am I missing something entirely...
Thanks in advance.
complex-analysis special-functions
asked Dec 13 '18 at 22:39
Michael CloudMichael Cloud
866
$endgroup$
I'm working on something that involves the error function, except with complex inputs, i.e. erf(z) with $$zinmathbb{C}$$. In particular, at this point, I need a formula for the zeros of this function. I've found the paper "Complex Zeros of the Error Function and of the Complementary Error Function" by Fettis, Caslin, and Kramer (mathematics of computation V. 27 num 122 April 1973, pp 401-407). They provide approximate zeros, but I'm more interested in understanding the original equation they are trying to solve, because I only need a formula not actual values. In working through their paper I'm confused about a quick deduction they make, fairly off casually.
To explain further, they are working with a function denoted $$w(z)=e^{-z^2}mbox{erfc}(imath z)=e^{-z^2}(1-mbox{erf}(imath z))$$ First they point out that (sym prop 1): $$w(-x+imath y)=w^*(x+imath y)$$ and that (sym prop 2): $$w(x-imath y)=2e^{-(x-imath y)^2}-w^*(x+imath y)$$ (where * denotes complex conjugation). Then that it can be expressed as $$w(z)=frac{1}{pi}int_{-infty}^infty e^{-t^2}frac{[y+imath(x-t)]}{(x-t)^2+y^2}ds$$ which after a substitution of $$x-t=s$$ can be written in the form $$frac{2e^{-x^{2}}}{pi}Big{[}yint_0^{infty}e^{-s^{2}}frac{mbox{cosh}(2xs)}{y^2+s^2}ds+imathint_0^infty e^{-s^{2}}frac{mbox{sinh}(2xs)}{y^2+s^2}s;dsBig{]}=u(x,y)+imath(v(x,y)$$ at this point they make the following series of comments:
from which it is clear that $$u(x,y)>0$$, $$v(x,y)>0$$ for
$$x>0$$, $$y>0$$. This together with the symmetry property
(sym prop1) shows that any zeros of w(z) must lie in the lower half plane, and these according to (sym prop 2) will be
determined from the equations
$$2e^{y^2-x^2}cos(2xy) = u(x,y)$$ and $$2e^{y^2-x^2}sin(2xy)=v(x,y)$$
Then there's a footnote saying that from now on y will be positive (because in fact for the complex error functions there are zeros in the first quadrant). They continue:
Since $$u(x,y) mbox{ and } v(x,y)rightarrow 0 mbox{ for } abs{z}rightarrow 0$$ it is evident that asymptotically $$abs{y}<x$$. Further since $$u mbox{ and } v mbox{ are both positive if } x mbox{ and } y mbox{ are positive }$$ it follows that
So up to now, everything is fine. Some of its a bit above my pay grade, but ok, other references concur on these points, so ok. Here comes the hard bit. They now conclude:
$$cos(2xy)>0 mbox{ and }sin(2xy)<0$$, and hence that $$2xy=2npi-beta mbox{ for } n=1,2,3,ldots mbox{ where } 0leqbeta<frac{pi}{4}$$
This has me flumoxed. It would seem that this last equation is the equation they are going to approximate in order to find approximate roots of the complex error function. But if u(x,y) and v(x,y) are the real and imaginary parts of the solution, then the zeros should be where u and v are both simultaneously zero, which doesn't seem like it can happen, because $$cos(theta)mbox{ and }sin(theta)$$ are never simultaneously zero, and $$e^{y^2-x^2}>0;forall x,yinmathbb{R}$$. My only guess is that they are using the norm squared of $$u(x,y)+imath v(x,y)$$, but they very carefully pointed out that both functions are positive for positive x,y, although that seems to be false since sin(x) clearly can be negative for positive x. So after all that :) is that what they are doing? Taking the norm squared? And if so which of them is negative, u or v? Or am I missing something entirely...
Thanks in advance.
complex-analysis special-functions
complex-analysis special-functions
asked Dec 13 '18 at 22:39
Michael CloudMichael Cloud
866
asked Dec 13 '18 at 22:39
Michael CloudMichael Cloud
866
asked Dec 13 '18 at 22:39
Michael CloudMichael Cloud
866
asked Dec 13 '18 at 22:39
Michael CloudMichael Cloud
866
asked Dec 13 '18 at 22:39
Michael CloudMichael Cloud
866
866
$begingroup$
May be this helps: They also say that for $x>0, y>0$ there are no zeros, and that any zeros of $w(z)$ which exist must lie in the lower half-plane and these, according to Eq. (4)(b), will be determined from the equations ... In the paper (eq. 12) they have $$2e^{y^2-x^2}sin(2xy)=-v(x,y)$$ i.e. you missed the − sign.
$endgroup$
– gammatester
Dec 14 '18 at 10:37
$begingroup$
good catch, but I don't see how it changes anything. -0=0. As for zeros in the lower half plane, I spent a good bit of time wrestling with that. It turns out that refers to w(z) but not erf(z). Erf(z) has different symmetry properties than w(z) in particular erf(z) satisfies erf(z*)=erf*(z) AND erf(-z)=-erf(z), so that it is symmetric in all 4 quadrants, and has zeros in the 1st quadrant. That's why they let y be positive.
$endgroup$
– Michael Cloud
Dec 14 '18 at 14:25
$begingroup$
But anyway, I don't see how to solve it for any sign of y. y being positive or negative doesn't change whether sin(2xy) is positive or not (it is in both cases) nor does it change whether e^(y^2-x^2) is positive or not. It is never negative, and only zero at x= pm infinity
$endgroup$
– Michael Cloud
Dec 14 '18 at 14:26
$begingroup$
Indeed the comment in the paper that u(x,y) and v(x,y) go to zero as z=x+iy goes to infinity also seems quite off. e^(y^2-x^2) would seem to go to infinity as y goes to pm infinity. It is true that it goes to zero very quickly along the real axis, but is always positive for any finite x.
$endgroup$
– Michael Cloud
Dec 14 '18 at 16:19
add a comment |
$begingroup$
May be this helps: They also say that for $x>0, y>0$ there are no zeros, and that any zeros of $w(z)$ which exist must lie in the lower half-plane and these, according to Eq. (4)(b), will be determined from the equations ... In the paper (eq. 12) they have $$2e^{y^2-x^2}sin(2xy)=-v(x,y)$$ i.e. you missed the − sign.
$endgroup$
– gammatester
Dec 14 '18 at 10:37
$begingroup$
good catch, but I don't see how it changes anything. -0=0. As for zeros in the lower half plane, I spent a good bit of time wrestling with that. It turns out that refers to w(z) but not erf(z). Erf(z) has different symmetry properties than w(z) in particular erf(z) satisfies erf(z*)=erf*(z) AND erf(-z)=-erf(z), so that it is symmetric in all 4 quadrants, and has zeros in the 1st quadrant. That's why they let y be positive.
$endgroup$
– Michael Cloud
Dec 14 '18 at 14:25
$begingroup$
But anyway, I don't see how to solve it for any sign of y. y being positive or negative doesn't change whether sin(2xy) is positive or not (it is in both cases) nor does it change whether e^(y^2-x^2) is positive or not. It is never negative, and only zero at x= pm infinity
$endgroup$
– Michael Cloud
Dec 14 '18 at 14:26
$begingroup$
Indeed the comment in the paper that u(x,y) and v(x,y) go to zero as z=x+iy goes to infinity also seems quite off. e^(y^2-x^2) would seem to go to infinity as y goes to pm infinity. It is true that it goes to zero very quickly along the real axis, but is always positive for any finite x.
$endgroup$
– Michael Cloud
Dec 14 '18 at 16:19
$begingroup$
May be this helps: They also say that for $x>0, y>0$ there are no zeros, and that any zeros of $w(z)$ which exist must lie in the lower half-plane and these, according to Eq. (4)(b), will be determined from the equations ... In the paper (eq. 12) they have $$2e^{y^2-x^2}sin(2xy)=-v(x,y)$$ i.e. you missed the − sign.
$endgroup$
– gammatester
Dec 14 '18 at 10:37
$begingroup$
May be this helps: They also say that for $x>0, y>0$ there are no zeros, and that any zeros of $w(z)$ which exist must lie in the lower half-plane and these, according to Eq. (4)(b), will be determined from the equations ... In the paper (eq. 12) they have $$2e^{y^2-x^2}sin(2xy)=-v(x,y)$$ i.e. you missed the − sign.
$endgroup$
– gammatester
Dec 14 '18 at 10:37
$begingroup$
good catch, but I don't see how it changes anything. -0=0. As for zeros in the lower half plane, I spent a good bit of time wrestling with that. It turns out that refers to w(z) but not erf(z). Erf(z) has different symmetry properties than w(z) in particular erf(z) satisfies erf(z*)=erf*(z) AND erf(-z)=-erf(z), so that it is symmetric in all 4 quadrants, and has zeros in the 1st quadrant. That's why they let y be positive.
$endgroup$
– Michael Cloud
Dec 14 '18 at 14:25
$begingroup$
good catch, but I don't see how it changes anything. -0=0. As for zeros in the lower half plane, I spent a good bit of time wrestling with that. It turns out that refers to w(z) but not erf(z). Erf(z) has different symmetry properties than w(z) in particular erf(z) satisfies erf(z*)=erf*(z) AND erf(-z)=-erf(z), so that it is symmetric in all 4 quadrants, and has zeros in the 1st quadrant. That's why they let y be positive.
$endgroup$
– Michael Cloud
Dec 14 '18 at 14:25
$begingroup$
But anyway, I don't see how to solve it for any sign of y. y being positive or negative doesn't change whether sin(2xy) is positive or not (it is in both cases) nor does it change whether e^(y^2-x^2) is positive or not. It is never negative, and only zero at x= pm infinity
$endgroup$
– Michael Cloud
Dec 14 '18 at 14:26
$begingroup$
But anyway, I don't see how to solve it for any sign of y. y being positive or negative doesn't change whether sin(2xy) is positive or not (it is in both cases) nor does it change whether e^(y^2-x^2) is positive or not. It is never negative, and only zero at x= pm infinity
$endgroup$
– Michael Cloud
Dec 14 '18 at 14:26
$begingroup$
Indeed the comment in the paper that u(x,y) and v(x,y) go to zero as z=x+iy goes to infinity also seems quite off. e^(y^2-x^2) would seem to go to infinity as y goes to pm infinity. It is true that it goes to zero very quickly along the real axis, but is always positive for any finite x.
$endgroup$
– Michael Cloud
Dec 14 '18 at 16:19
$begingroup$
Indeed the comment in the paper that u(x,y) and v(x,y) go to zero as z=x+iy goes to infinity also seems quite off. e^(y^2-x^2) would seem to go to infinity as y goes to pm infinity. It is true that it goes to zero very quickly along the real axis, but is always positive for any finite x.
$endgroup$
– Michael Cloud
Dec 14 '18 at 16:19
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038676%2fon-the-zeros-of-the-complex-error-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038676%2fon-the-zeros-of-the-complex-error-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
May be this helps: They also say that for $x>0, y>0$ there are no zeros, and that any zeros of $w(z)$ which exist must lie in the lower half-plane and these, according to Eq. (4)(b), will be determined from the equations ... In the paper (eq. 12) they have $$2e^{y^2-x^2}sin(2xy)=-v(x,y)$$ i.e. you missed the − sign.
$endgroup$
– gammatester
Dec 14 '18 at 10:37
$begingroup$
good catch, but I don't see how it changes anything. -0=0. As for zeros in the lower half plane, I spent a good bit of time wrestling with that. It turns out that refers to w(z) but not erf(z). Erf(z) has different symmetry properties than w(z) in particular erf(z) satisfies erf(z*)=erf*(z) AND erf(-z)=-erf(z), so that it is symmetric in all 4 quadrants, and has zeros in the 1st quadrant. That's why they let y be positive.
$endgroup$
– Michael Cloud
Dec 14 '18 at 14:25
$begingroup$
But anyway, I don't see how to solve it for any sign of y. y being positive or negative doesn't change whether sin(2xy) is positive or not (it is in both cases) nor does it change whether e^(y^2-x^2) is positive or not. It is never negative, and only zero at x= pm infinity
$endgroup$
– Michael Cloud
Dec 14 '18 at 14:26
$begingroup$
Indeed the comment in the paper that u(x,y) and v(x,y) go to zero as z=x+iy goes to infinity also seems quite off. e^(y^2-x^2) would seem to go to infinity as y goes to pm infinity. It is true that it goes to zero very quickly along the real axis, but is always positive for any finite x.
$endgroup$
– Michael Cloud
Dec 14 '18 at 16:19