Absolute Value and Exponents
$begingroup$
In my homework I've been accustomed to assuming that $|x|^a = |x^a|$ Recently however, I've begun to doubt that. Take the following example:
$$ begin{equation*}
begin{split}
|sqrt{-|x|} | &= sqrt{|-|x||} \
&= sqrt{|x|}
end{split}
end{equation*} $$
I don't, it just seems weird to me. So I took it upon myself to investigate my assumption, and prove/ disprove that $|x|^a = |x^a|$.
$$begin{equation*}
begin{split}
|x|^a &= begin{cases}
x^a, & x geq 0 \
(-x)^a, & x < 0
end{cases} \[10pt]
&= begin{cases}
x^a, & x geq 0 \
(-1)^a cdot (x)^a, & x < 0
end{cases} \
end{split}
end{equation*}$$
Whereas $$|x^a|= begin{cases}
x^a, & x geq 0 \
-x^a, & x < 0
end{cases} $$
In other words, it would seem to me that the two only equal each other when $a$ is odd.
I don't know, could you guys clear this up for me?
real-analysis real-numbers absolute-value
asked Dec 19 '18 at 16:27
user0009999999user0009999999
265
$endgroup$
add a comment |
$begingroup$
In my homework I've been accustomed to assuming that $|x|^a = |x^a|$ Recently however, I've begun to doubt that. Take the following example:
$$ begin{equation*}
begin{split}
|sqrt{-|x|} | &= sqrt{|-|x||} \
&= sqrt{|x|}
end{split}
end{equation*} $$
I don't, it just seems weird to me. So I took it upon myself to investigate my assumption, and prove/ disprove that $|x|^a = |x^a|$.
$$begin{equation*}
begin{split}
|x|^a &= begin{cases}
x^a, & x geq 0 \
(-x)^a, & x < 0
end{cases} \[10pt]
&= begin{cases}
x^a, & x geq 0 \
(-1)^a cdot (x)^a, & x < 0
end{cases} \
end{split}
end{equation*}$$
Whereas $$|x^a|= begin{cases}
x^a, & x geq 0 \
-x^a, & x < 0
end{cases} $$
In other words, it would seem to me that the two only equal each other when $a$ is odd.
I don't know, could you guys clear this up for me?
real-analysis real-numbers absolute-value
asked Dec 19 '18 at 16:27
user0009999999user0009999999
265
$endgroup$
$begingroup$
You are correct, this is not true generally. It only happens to be true when $a$ is even since $|x^a|$ is positive, while $|x|^a$ is always positive.
$endgroup$
– Don Thousand
Dec 19 '18 at 16:29
add a comment |
$begingroup$
In my homework I've been accustomed to assuming that $|x|^a = |x^a|$ Recently however, I've begun to doubt that. Take the following example:
$$ begin{equation*}
begin{split}
|sqrt{-|x|} | &= sqrt{|-|x||} \
&= sqrt{|x|}
end{split}
end{equation*} $$
I don't, it just seems weird to me. So I took it upon myself to investigate my assumption, and prove/ disprove that $|x|^a = |x^a|$.
$$begin{equation*}
begin{split}
|x|^a &= begin{cases}
x^a, & x geq 0 \
(-x)^a, & x < 0
end{cases} \[10pt]
&= begin{cases}
x^a, & x geq 0 \
(-1)^a cdot (x)^a, & x < 0
end{cases} \
end{split}
end{equation*}$$
Whereas $$|x^a|= begin{cases}
x^a, & x geq 0 \
-x^a, & x < 0
end{cases} $$
In other words, it would seem to me that the two only equal each other when $a$ is odd.
I don't know, could you guys clear this up for me?
real-analysis real-numbers absolute-value
asked Dec 19 '18 at 16:27
user0009999999user0009999999
265
$endgroup$
In my homework I've been accustomed to assuming that $|x|^a = |x^a|$ Recently however, I've begun to doubt that. Take the following example:
$$ begin{equation*}
begin{split}
|sqrt{-|x|} | &= sqrt{|-|x||} \
&= sqrt{|x|}
end{split}
end{equation*} $$
I don't, it just seems weird to me. So I took it upon myself to investigate my assumption, and prove/ disprove that $|x|^a = |x^a|$.
$$begin{equation*}
begin{split}
|x|^a &= begin{cases}
x^a, & x geq 0 \
(-x)^a, & x < 0
end{cases} \[10pt]
&= begin{cases}
x^a, & x geq 0 \
(-1)^a cdot (x)^a, & x < 0
end{cases} \
end{split}
end{equation*}$$
Whereas $$|x^a|= begin{cases}
x^a, & x geq 0 \
-x^a, & x < 0
end{cases} $$
In other words, it would seem to me that the two only equal each other when $a$ is odd.
I don't know, could you guys clear this up for me?
real-analysis real-numbers absolute-value
real-analysis real-numbers absolute-value
asked Dec 19 '18 at 16:27
user0009999999user0009999999
265
asked Dec 19 '18 at 16:27
user0009999999user0009999999
265
asked Dec 19 '18 at 16:27
user0009999999user0009999999
265
asked Dec 19 '18 at 16:27
user0009999999user0009999999
265
asked Dec 19 '18 at 16:27
user0009999999user0009999999
265
265
$begingroup$
You are correct, this is not true generally. It only happens to be true when $a$ is even since $|x^a|$ is positive, while $|x|^a$ is always positive.
$endgroup$
– Don Thousand
Dec 19 '18 at 16:29
add a comment |
$begingroup$
You are correct, this is not true generally. It only happens to be true when $a$ is even since $|x^a|$ is positive, while $|x|^a$ is always positive.
$endgroup$
– Don Thousand
Dec 19 '18 at 16:29
$begingroup$
You are correct, this is not true generally. It only happens to be true when $a$ is even since $|x^a|$ is positive, while $|x|^a$ is always positive.
$endgroup$
– Don Thousand
Dec 19 '18 at 16:29
$begingroup$
You are correct, this is not true generally. It only happens to be true when $a$ is even since $|x^a|$ is positive, while $|x|^a$ is always positive.
$endgroup$
– Don Thousand
Dec 19 '18 at 16:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is not correct that $|x^a| = -x^a$ when $x< 0$; only when $x^a < 0$.
answered Dec 19 '18 at 16:43
fkraiemfkraiem
2,8961610
$endgroup$
add a comment |
$begingroup$
Let's go a step further and look at this from the complex point of view. Any number can be written as:
$$z=rcdot e^{itheta}$$
with $rinmathbb{R}$ and $rge0$ and $thetain[0,2pi[$.
raising this number to any power $ainmathbb{R}$ gives you
$$z^a=r^acdot e^{i,a,theta}$$
The absolute value of this number is nothing more than $|z^a|=r^a$ and since $|z|=r$, you can state that $|z^a|=|z|^a$ for $ainmathbb{R}$
For a complex power $a+biinmathbb{C}$, this is not the case:
$$z^{a+bi}=r^{a+bi}cdot e^{i(a+bi)theta} = r^{a-btheta}cdot e^{i(ln r + atheta)}$$
which gives $|z^{a+bi}|=r^{a-btheta}$ while $|z|^{a+bi}=r^{a+bi}$
answered Dec 19 '18 at 16:55
kvantourkvantour
34619
$endgroup$
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%2f3046580%2fabsolute-value-and-exponents%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is not correct that $|x^a| = -x^a$ when $x< 0$; only when $x^a < 0$.
answered Dec 19 '18 at 16:43
fkraiemfkraiem
2,8961610
$endgroup$
add a comment |
$begingroup$
It is not correct that $|x^a| = -x^a$ when $x< 0$; only when $x^a < 0$.
answered Dec 19 '18 at 16:43
fkraiemfkraiem
2,8961610
$endgroup$
add a comment |
$begingroup$
It is not correct that $|x^a| = -x^a$ when $x< 0$; only when $x^a < 0$.
answered Dec 19 '18 at 16:43
fkraiemfkraiem
2,8961610
$endgroup$
It is not correct that $|x^a| = -x^a$ when $x< 0$; only when $x^a < 0$.
answered Dec 19 '18 at 16:43
fkraiemfkraiem
2,8961610
answered Dec 19 '18 at 16:43
fkraiemfkraiem
2,8961610
answered Dec 19 '18 at 16:43
fkraiemfkraiem
2,8961610
answered Dec 19 '18 at 16:43
fkraiemfkraiem
2,8961610
2,8961610
add a comment |
add a comment |
$begingroup$
Let's go a step further and look at this from the complex point of view. Any number can be written as:
$$z=rcdot e^{itheta}$$
with $rinmathbb{R}$ and $rge0$ and $thetain[0,2pi[$.
raising this number to any power $ainmathbb{R}$ gives you
$$z^a=r^acdot e^{i,a,theta}$$
The absolute value of this number is nothing more than $|z^a|=r^a$ and since $|z|=r$, you can state that $|z^a|=|z|^a$ for $ainmathbb{R}$
For a complex power $a+biinmathbb{C}$, this is not the case:
$$z^{a+bi}=r^{a+bi}cdot e^{i(a+bi)theta} = r^{a-btheta}cdot e^{i(ln r + atheta)}$$
which gives $|z^{a+bi}|=r^{a-btheta}$ while $|z|^{a+bi}=r^{a+bi}$
answered Dec 19 '18 at 16:55
kvantourkvantour
34619
$endgroup$
add a comment |
$begingroup$
Let's go a step further and look at this from the complex point of view. Any number can be written as:
$$z=rcdot e^{itheta}$$
with $rinmathbb{R}$ and $rge0$ and $thetain[0,2pi[$.
raising this number to any power $ainmathbb{R}$ gives you
$$z^a=r^acdot e^{i,a,theta}$$
The absolute value of this number is nothing more than $|z^a|=r^a$ and since $|z|=r$, you can state that $|z^a|=|z|^a$ for $ainmathbb{R}$
For a complex power $a+biinmathbb{C}$, this is not the case:
$$z^{a+bi}=r^{a+bi}cdot e^{i(a+bi)theta} = r^{a-btheta}cdot e^{i(ln r + atheta)}$$
which gives $|z^{a+bi}|=r^{a-btheta}$ while $|z|^{a+bi}=r^{a+bi}$
answered Dec 19 '18 at 16:55
kvantourkvantour
34619
$endgroup$
add a comment |
$begingroup$
Let's go a step further and look at this from the complex point of view. Any number can be written as:
$$z=rcdot e^{itheta}$$
with $rinmathbb{R}$ and $rge0$ and $thetain[0,2pi[$.
raising this number to any power $ainmathbb{R}$ gives you
$$z^a=r^acdot e^{i,a,theta}$$
The absolute value of this number is nothing more than $|z^a|=r^a$ and since $|z|=r$, you can state that $|z^a|=|z|^a$ for $ainmathbb{R}$
For a complex power $a+biinmathbb{C}$, this is not the case:
$$z^{a+bi}=r^{a+bi}cdot e^{i(a+bi)theta} = r^{a-btheta}cdot e^{i(ln r + atheta)}$$
which gives $|z^{a+bi}|=r^{a-btheta}$ while $|z|^{a+bi}=r^{a+bi}$
answered Dec 19 '18 at 16:55
kvantourkvantour
34619
$endgroup$
Let's go a step further and look at this from the complex point of view. Any number can be written as:
$$z=rcdot e^{itheta}$$
with $rinmathbb{R}$ and $rge0$ and $thetain[0,2pi[$.
raising this number to any power $ainmathbb{R}$ gives you
$$z^a=r^acdot e^{i,a,theta}$$
The absolute value of this number is nothing more than $|z^a|=r^a$ and since $|z|=r$, you can state that $|z^a|=|z|^a$ for $ainmathbb{R}$
For a complex power $a+biinmathbb{C}$, this is not the case:
$$z^{a+bi}=r^{a+bi}cdot e^{i(a+bi)theta} = r^{a-btheta}cdot e^{i(ln r + atheta)}$$
which gives $|z^{a+bi}|=r^{a-btheta}$ while $|z|^{a+bi}=r^{a+bi}$
answered Dec 19 '18 at 16:55
kvantourkvantour
34619
edited Dec 19 '18 at 17:00
answered Dec 19 '18 at 16:55
kvantourkvantour
34619
answered Dec 19 '18 at 16:55
kvantourkvantour
34619
answered Dec 19 '18 at 16:55
kvantourkvantour
34619
34619
add a comment |
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%2f3046580%2fabsolute-value-and-exponents%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
$begingroup$
You are correct, this is not true generally. It only happens to be true when $a$ is even since $|x^a|$ is positive, while $|x|^a$ is always positive.
$endgroup$
– Don Thousand
Dec 19 '18 at 16:29