Can raising a number to an irrational power have infinite solutions?
$begingroup$
$a^{frac{1}{2}}$ is generally considered to be the positive square root of $a$, but it also makes sense (depending on context) to consider it to be multivalued, returning all square roots of $a$
Likewise with something like $a^{frac{3}{5}}$- it sometimes makes sense to consider the multivalued function which returns 5 values- the 5 complex values that, when raised to the 5th power, return $a^3$.
Extending this idea to irrational numbers, you don't get fractions directly, but you can get them using sequences of rational numbers that converges to the desired irrational. For example, to reach $pi$, you could use 3, 3.1, 3.14, 3.141, 3.1415, etc. For such a sequence, the denominator grows without bound, so raising a number to each subsequent power in the sequence increases the number of complex solutions.
Does it then follow that when considering non-integer exponentiation to be multivalued, raising a number to an irrational power yields an infinite number of complex solutions? If so, would that be every point on a circle concentric with the unit circle, or something else?
exponentiation irrational-numbers multivalued-functions
asked Nov 30 '18 at 17:01
SarcasticSullySarcasticSully
1012
$endgroup$
add a comment |
$begingroup$
$a^{frac{1}{2}}$ is generally considered to be the positive square root of $a$, but it also makes sense (depending on context) to consider it to be multivalued, returning all square roots of $a$
Likewise with something like $a^{frac{3}{5}}$- it sometimes makes sense to consider the multivalued function which returns 5 values- the 5 complex values that, when raised to the 5th power, return $a^3$.
Extending this idea to irrational numbers, you don't get fractions directly, but you can get them using sequences of rational numbers that converges to the desired irrational. For example, to reach $pi$, you could use 3, 3.1, 3.14, 3.141, 3.1415, etc. For such a sequence, the denominator grows without bound, so raising a number to each subsequent power in the sequence increases the number of complex solutions.
Does it then follow that when considering non-integer exponentiation to be multivalued, raising a number to an irrational power yields an infinite number of complex solutions? If so, would that be every point on a circle concentric with the unit circle, or something else?
exponentiation irrational-numbers multivalued-functions
asked Nov 30 '18 at 17:01
SarcasticSullySarcasticSully
1012
$endgroup$
add a comment |
$begingroup$
$a^{frac{1}{2}}$ is generally considered to be the positive square root of $a$, but it also makes sense (depending on context) to consider it to be multivalued, returning all square roots of $a$
Likewise with something like $a^{frac{3}{5}}$- it sometimes makes sense to consider the multivalued function which returns 5 values- the 5 complex values that, when raised to the 5th power, return $a^3$.
Extending this idea to irrational numbers, you don't get fractions directly, but you can get them using sequences of rational numbers that converges to the desired irrational. For example, to reach $pi$, you could use 3, 3.1, 3.14, 3.141, 3.1415, etc. For such a sequence, the denominator grows without bound, so raising a number to each subsequent power in the sequence increases the number of complex solutions.
Does it then follow that when considering non-integer exponentiation to be multivalued, raising a number to an irrational power yields an infinite number of complex solutions? If so, would that be every point on a circle concentric with the unit circle, or something else?
exponentiation irrational-numbers multivalued-functions
asked Nov 30 '18 at 17:01
SarcasticSullySarcasticSully
1012
$endgroup$
$a^{frac{1}{2}}$ is generally considered to be the positive square root of $a$, but it also makes sense (depending on context) to consider it to be multivalued, returning all square roots of $a$
Likewise with something like $a^{frac{3}{5}}$- it sometimes makes sense to consider the multivalued function which returns 5 values- the 5 complex values that, when raised to the 5th power, return $a^3$.
Extending this idea to irrational numbers, you don't get fractions directly, but you can get them using sequences of rational numbers that converges to the desired irrational. For example, to reach $pi$, you could use 3, 3.1, 3.14, 3.141, 3.1415, etc. For such a sequence, the denominator grows without bound, so raising a number to each subsequent power in the sequence increases the number of complex solutions.
Does it then follow that when considering non-integer exponentiation to be multivalued, raising a number to an irrational power yields an infinite number of complex solutions? If so, would that be every point on a circle concentric with the unit circle, or something else?
exponentiation irrational-numbers multivalued-functions
exponentiation irrational-numbers multivalued-functions
asked Nov 30 '18 at 17:01
SarcasticSullySarcasticSully
1012
asked Nov 30 '18 at 17:01
SarcasticSullySarcasticSully
1012
asked Nov 30 '18 at 17:01
SarcasticSullySarcasticSully
1012
asked Nov 30 '18 at 17:01
SarcasticSullySarcasticSully
1012
asked Nov 30 '18 at 17:01
SarcasticSullySarcasticSully
1012
1012
add a comment |
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%2f3020329%2fcan-raising-a-number-to-an-irrational-power-have-infinite-solutions%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%2f3020329%2fcan-raising-a-number-to-an-irrational-power-have-infinite-solutions%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
