Function which becomes just a relation on $R^{-}$?
$begingroup$
Consider the function on the variable $xin R$ and the parameter $ain R$:
$$f(x)=frac{a-sqrt{x}}{a+sqrt{x}}$$
Of course if $x>=0$ you have a simple function $f:R rightarrow R$.
If $x<0$ you can write:
$$f(x)=frac{a-sqrt{(-1)(-x)}}{a+sqrt{(-1)(-x)}}$$
$$f(x)=frac{a-[pm isqrt{(-x)}]}{a+[pm i sqrt{(-x)}]}$$
So now there are 4 possibilities according to the signs you put:
$$binom{+}{+} rightarrow f(x)=frac{a-[isqrt{(-x)}]}{a+[i sqrt{(-x)}]}=e^{itheta(x)} $$
$$binom{+}{-} rightarrow f(x)=frac{a-[+isqrt{(-x)}]}{a+[-i sqrt{(-x)}]}=1 $$
$$binom{-}{+} rightarrow f(x)=frac{a-[-isqrt{(-x)}]}{a+[+i sqrt{(-x)}]} =1$$
$$binom{-}{-} rightarrow f(x)=frac{a-[-isqrt{(-x)}]}{a+[-i sqrt{(-x)}]} e^{-itheta(x)} $$
Finally:
$$ f(x)=begin{cases}
1 \
e^{-itheta(x)}\
e^{itheta(x)}
end{cases}
$$
So I'd say that for $x<0$ $f(x)$ isn't anymore a function...Is it right?
real-analysis complex-analysis
asked Dec 29 '18 at 20:26
LandauLandau
447
$endgroup$
add a comment |
$begingroup$
Consider the function on the variable $xin R$ and the parameter $ain R$:
$$f(x)=frac{a-sqrt{x}}{a+sqrt{x}}$$
Of course if $x>=0$ you have a simple function $f:R rightarrow R$.
If $x<0$ you can write:
$$f(x)=frac{a-sqrt{(-1)(-x)}}{a+sqrt{(-1)(-x)}}$$
$$f(x)=frac{a-[pm isqrt{(-x)}]}{a+[pm i sqrt{(-x)}]}$$
So now there are 4 possibilities according to the signs you put:
$$binom{+}{+} rightarrow f(x)=frac{a-[isqrt{(-x)}]}{a+[i sqrt{(-x)}]}=e^{itheta(x)} $$
$$binom{+}{-} rightarrow f(x)=frac{a-[+isqrt{(-x)}]}{a+[-i sqrt{(-x)}]}=1 $$
$$binom{-}{+} rightarrow f(x)=frac{a-[-isqrt{(-x)}]}{a+[+i sqrt{(-x)}]} =1$$
$$binom{-}{-} rightarrow f(x)=frac{a-[-isqrt{(-x)}]}{a+[-i sqrt{(-x)}]} e^{-itheta(x)} $$
Finally:
$$ f(x)=begin{cases}
1 \
e^{-itheta(x)}\
e^{itheta(x)}
end{cases}
$$
So I'd say that for $x<0$ $f(x)$ isn't anymore a function...Is it right?
real-analysis complex-analysis
asked Dec 29 '18 at 20:26
LandauLandau
447
$endgroup$
add a comment |
$begingroup$
Consider the function on the variable $xin R$ and the parameter $ain R$:
$$f(x)=frac{a-sqrt{x}}{a+sqrt{x}}$$
Of course if $x>=0$ you have a simple function $f:R rightarrow R$.
If $x<0$ you can write:
$$f(x)=frac{a-sqrt{(-1)(-x)}}{a+sqrt{(-1)(-x)}}$$
$$f(x)=frac{a-[pm isqrt{(-x)}]}{a+[pm i sqrt{(-x)}]}$$
So now there are 4 possibilities according to the signs you put:
$$binom{+}{+} rightarrow f(x)=frac{a-[isqrt{(-x)}]}{a+[i sqrt{(-x)}]}=e^{itheta(x)} $$
$$binom{+}{-} rightarrow f(x)=frac{a-[+isqrt{(-x)}]}{a+[-i sqrt{(-x)}]}=1 $$
$$binom{-}{+} rightarrow f(x)=frac{a-[-isqrt{(-x)}]}{a+[+i sqrt{(-x)}]} =1$$
$$binom{-}{-} rightarrow f(x)=frac{a-[-isqrt{(-x)}]}{a+[-i sqrt{(-x)}]} e^{-itheta(x)} $$
Finally:
$$ f(x)=begin{cases}
1 \
e^{-itheta(x)}\
e^{itheta(x)}
end{cases}
$$
So I'd say that for $x<0$ $f(x)$ isn't anymore a function...Is it right?
real-analysis complex-analysis
asked Dec 29 '18 at 20:26
LandauLandau
447
$endgroup$
Consider the function on the variable $xin R$ and the parameter $ain R$:
$$f(x)=frac{a-sqrt{x}}{a+sqrt{x}}$$
Of course if $x>=0$ you have a simple function $f:R rightarrow R$.
If $x<0$ you can write:
$$f(x)=frac{a-sqrt{(-1)(-x)}}{a+sqrt{(-1)(-x)}}$$
$$f(x)=frac{a-[pm isqrt{(-x)}]}{a+[pm i sqrt{(-x)}]}$$
So now there are 4 possibilities according to the signs you put:
$$binom{+}{+} rightarrow f(x)=frac{a-[isqrt{(-x)}]}{a+[i sqrt{(-x)}]}=e^{itheta(x)} $$
$$binom{+}{-} rightarrow f(x)=frac{a-[+isqrt{(-x)}]}{a+[-i sqrt{(-x)}]}=1 $$
$$binom{-}{+} rightarrow f(x)=frac{a-[-isqrt{(-x)}]}{a+[+i sqrt{(-x)}]} =1$$
$$binom{-}{-} rightarrow f(x)=frac{a-[-isqrt{(-x)}]}{a+[-i sqrt{(-x)}]} e^{-itheta(x)} $$
Finally:
$$ f(x)=begin{cases}
1 \
e^{-itheta(x)}\
e^{itheta(x)}
end{cases}
$$
So I'd say that for $x<0$ $f(x)$ isn't anymore a function...Is it right?
real-analysis complex-analysis
real-analysis complex-analysis
asked Dec 29 '18 at 20:26
LandauLandau
447
asked Dec 29 '18 at 20:26
LandauLandau
447
asked Dec 29 '18 at 20:26
LandauLandau
447
asked Dec 29 '18 at 20:26
LandauLandau
447
asked Dec 29 '18 at 20:26
LandauLandau
447
447
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is not a function over the complex numbers, since the square root of a complex number is not a well-defined notion. See How do I get the square root of a complex number?
answered Dec 29 '18 at 20:34
Hans HüttelHans Hüttel
3,1972921
$endgroup$
$begingroup$
I know how to get the root of a cn, however do you confirm that my result is right?
$endgroup$
– Landau
Dec 29 '18 at 20:37
add a comment |
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%2f3056222%2ffunction-which-becomes-just-a-relation-on-r%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is not a function over the complex numbers, since the square root of a complex number is not a well-defined notion. See How do I get the square root of a complex number?
answered Dec 29 '18 at 20:34
Hans HüttelHans Hüttel
3,1972921
$endgroup$
$begingroup$
I know how to get the root of a cn, however do you confirm that my result is right?
$endgroup$
– Landau
Dec 29 '18 at 20:37
add a comment |
$begingroup$
This is not a function over the complex numbers, since the square root of a complex number is not a well-defined notion. See How do I get the square root of a complex number?
answered Dec 29 '18 at 20:34
Hans HüttelHans Hüttel
3,1972921
$endgroup$
$begingroup$
I know how to get the root of a cn, however do you confirm that my result is right?
$endgroup$
– Landau
Dec 29 '18 at 20:37
add a comment |
$begingroup$
This is not a function over the complex numbers, since the square root of a complex number is not a well-defined notion. See How do I get the square root of a complex number?
answered Dec 29 '18 at 20:34
Hans HüttelHans Hüttel
3,1972921
$endgroup$
This is not a function over the complex numbers, since the square root of a complex number is not a well-defined notion. See How do I get the square root of a complex number?
answered Dec 29 '18 at 20:34
Hans HüttelHans Hüttel
3,1972921
answered Dec 29 '18 at 20:34
Hans HüttelHans Hüttel
3,1972921
answered Dec 29 '18 at 20:34
Hans HüttelHans Hüttel
3,1972921
answered Dec 29 '18 at 20:34
Hans HüttelHans Hüttel
3,1972921
3,1972921
$begingroup$
I know how to get the root of a cn, however do you confirm that my result is right?
$endgroup$
– Landau
Dec 29 '18 at 20:37
add a comment |
$begingroup$
I know how to get the root of a cn, however do you confirm that my result is right?
$endgroup$
– Landau
Dec 29 '18 at 20:37
$begingroup$
I know how to get the root of a cn, however do you confirm that my result is right?
$endgroup$
– Landau
Dec 29 '18 at 20:37
$begingroup$
I know how to get the root of a cn, however do you confirm that my result is right?
$endgroup$
– Landau
Dec 29 '18 at 20:37
add a comment |
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%2f3056222%2ffunction-which-becomes-just-a-relation-on-r%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
