Absolute Value and Exponents












0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













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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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















2












$begingroup$

It is not correct that $|x^a| = -x^a$ when $x< 0$; only when $x^a < 0$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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 $|z|^{a+bi}=r^{a+bi}$






    share|cite|improve this answer











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      2 Answers
      2






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      2 Answers
      2






      active

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      active

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      active

      oldest

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      2












      $begingroup$

      It is not correct that $|x^a| = -x^a$ when $x< 0$; only when $x^a < 0$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        It is not correct that $|x^a| = -x^a$ when $x< 0$; only when $x^a < 0$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          It is not correct that $|x^a| = -x^a$ when $x< 0$; only when $x^a < 0$.






          share|cite|improve this answer









          $endgroup$



          It is not correct that $|x^a| = -x^a$ when $x< 0$; only when $x^a < 0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 16:43









          fkraiemfkraiem

          2,8961610




          2,8961610























              0












              $begingroup$

              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 $|z|^{a+bi}=r^{a+bi}$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                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 $|z|^{a+bi}=r^{a+bi}$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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 $|z|^{a+bi}=r^{a+bi}$






                  share|cite|improve this answer











                  $endgroup$



                  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 $|z|^{a+bi}=r^{a+bi}$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 19 '18 at 17:00

























                  answered Dec 19 '18 at 16:55









                  kvantourkvantour

                  34619




                  34619






























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