Mixing
is ${x^{2/3}}$ the same as ${sqrt[3]{(x^2)}}$ or ${(sqrt[3]x)^2}$? domain?
$begingroup$
is ${x^{2/3}}$ the same as ${sqrt[3]{x^2}}$ and ${(sqrt[3]x)^2}$?
and what is domain of $x$?
Wolfram Alpha shows different results for ${x^{2/3}}$ and ${sqrt[3]{x^2}}$ representations.
Is domain dependant just on representation!?
roots wolfram-alpha
asked Nov 2 '15 at 3:43
ya_dimonya_dimon
1061
$endgroup$
add a comment |
$begingroup$
is ${x^{2/3}}$ the same as ${sqrt[3]{x^2}}$ and ${(sqrt[3]x)^2}$?
and what is domain of $x$?
Wolfram Alpha shows different results for ${x^{2/3}}$ and ${sqrt[3]{x^2}}$ representations.
Is domain dependant just on representation!?
roots wolfram-alpha
asked Nov 2 '15 at 3:43
ya_dimonya_dimon
1061
$endgroup$
$begingroup$
Try some values!
$endgroup$
– 9301293
Nov 2 '15 at 3:46
1
$begingroup$
As with all things, it depends on definitions. How do you define the three things? They all evaluate to the same thing on the non-negative reals, but then you have questions about definition on all the reals. You can define $x^{2/3}$ as being the same as the others on the reals, but some definition s do not.
$endgroup$
– Thomas Andrews
Nov 2 '15 at 3:48
$begingroup$
Well, they are different. $x^{2/3}$ isx^(2/3)while $sqrt[3]{x^2}$ is(x^2)^3. Furthermore, exponentiation is not associative, so why would we expect the same result?
$endgroup$
– Mark McClure
Nov 2 '15 at 11:42
add a comment |
$begingroup$
is ${x^{2/3}}$ the same as ${sqrt[3]{x^2}}$ and ${(sqrt[3]x)^2}$?
and what is domain of $x$?
Wolfram Alpha shows different results for ${x^{2/3}}$ and ${sqrt[3]{x^2}}$ representations.
Is domain dependant just on representation!?
roots wolfram-alpha
asked Nov 2 '15 at 3:43
ya_dimonya_dimon
1061
$endgroup$
is ${x^{2/3}}$ the same as ${sqrt[3]{x^2}}$ and ${(sqrt[3]x)^2}$?
and what is domain of $x$?
Wolfram Alpha shows different results for ${x^{2/3}}$ and ${sqrt[3]{x^2}}$ representations.
Is domain dependant just on representation!?
roots wolfram-alpha
roots wolfram-alpha
asked Nov 2 '15 at 3:43
ya_dimonya_dimon
1061
asked Nov 2 '15 at 3:43
ya_dimonya_dimon
1061
edited Nov 2 '15 at 4:10
ya_dimon
asked Nov 2 '15 at 3:43
ya_dimonya_dimon
1061
asked Nov 2 '15 at 3:43
ya_dimonya_dimon
1061
asked Nov 2 '15 at 3:43
ya_dimonya_dimon
1061
1061
$begingroup$
Try some values!
$endgroup$
– 9301293
Nov 2 '15 at 3:46
1
$begingroup$
As with all things, it depends on definitions. How do you define the three things? They all evaluate to the same thing on the non-negative reals, but then you have questions about definition on all the reals. You can define $x^{2/3}$ as being the same as the others on the reals, but some definition s do not.
$endgroup$
– Thomas Andrews
Nov 2 '15 at 3:48
$begingroup$
Well, they are different. $x^{2/3}$ isx^(2/3)while $sqrt[3]{x^2}$ is(x^2)^3. Furthermore, exponentiation is not associative, so why would we expect the same result?
$endgroup$
– Mark McClure
Nov 2 '15 at 11:42
add a comment |
$begingroup$
Try some values!
$endgroup$
– 9301293
Nov 2 '15 at 3:46
1
$begingroup$
As with all things, it depends on definitions. How do you define the three things? They all evaluate to the same thing on the non-negative reals, but then you have questions about definition on all the reals. You can define $x^{2/3}$ as being the same as the others on the reals, but some definition s do not.
$endgroup$
– Thomas Andrews
Nov 2 '15 at 3:48
$begingroup$
Well, they are different. $x^{2/3}$ isx^(2/3)while $sqrt[3]{x^2}$ is(x^2)^3. Furthermore, exponentiation is not associative, so why would we expect the same result?
$endgroup$
– Mark McClure
Nov 2 '15 at 11:42
$begingroup$
Try some values!
$endgroup$
– 9301293
Nov 2 '15 at 3:46
$begingroup$
Try some values!
$endgroup$
– 9301293
Nov 2 '15 at 3:46
1
1
$begingroup$
As with all things, it depends on definitions. How do you define the three things? They all evaluate to the same thing on the non-negative reals, but then you have questions about definition on all the reals. You can define $x^{2/3}$ as being the same as the others on the reals, but some definition s do not.
$endgroup$
– Thomas Andrews
Nov 2 '15 at 3:48
$begingroup$
As with all things, it depends on definitions. How do you define the three things? They all evaluate to the same thing on the non-negative reals, but then you have questions about definition on all the reals. You can define $x^{2/3}$ as being the same as the others on the reals, but some definition s do not.
$endgroup$
– Thomas Andrews
Nov 2 '15 at 3:48
$begingroup$
Well, they are different. $x^{2/3}$ is
x^(2/3) while $sqrt[3]{x^2}$ is (x^2)^3. Furthermore, exponentiation is not associative, so why would we expect the same result?$endgroup$
– Mark McClure
Nov 2 '15 at 11:42
$begingroup$
Well, they are different. $x^{2/3}$ is
x^(2/3) while $sqrt[3]{x^2}$ is (x^2)^3. Furthermore, exponentiation is not associative, so why would we expect the same result?$endgroup$
– Mark McClure
Nov 2 '15 at 11:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It would appear that Wolfram Alpha is secretly considering the functions in different contexts; for $x^{2/3}$ it is treating the $x$ as a complex variable, which can cause all kinds of shenanigans, whereas $sqrt[3]{x^2}$ has $x$ as a purely real variable, since there is no way that $x^2$ can be negative. Thus in the first case you must consider cube roots of $-1$, and in the second case you may always take the positive real root.
In short, the domain is NOT dependent on representation. Wolfram Alpha just likes to cut corners.
answered Nov 2 '15 at 6:08
TennoTenno
23315
$endgroup$
$begingroup$
if i can represent the complexxin ${x^{2/3}}$ as a realxin ${sqrt[3]{x^2}}$ than i can modify the context and the domain, just by another representation? And than the domain is dependent on representation? Or even after new representation, complex should be still complex and the domain will be still C?
$endgroup$
– ya_dimon
Nov 2 '15 at 19:46
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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$begingroup$
It would appear that Wolfram Alpha is secretly considering the functions in different contexts; for $x^{2/3}$ it is treating the $x$ as a complex variable, which can cause all kinds of shenanigans, whereas $sqrt[3]{x^2}$ has $x$ as a purely real variable, since there is no way that $x^2$ can be negative. Thus in the first case you must consider cube roots of $-1$, and in the second case you may always take the positive real root.
In short, the domain is NOT dependent on representation. Wolfram Alpha just likes to cut corners.
answered Nov 2 '15 at 6:08
TennoTenno
23315
$endgroup$
$begingroup$
if i can represent the complexxin ${x^{2/3}}$ as a realxin ${sqrt[3]{x^2}}$ than i can modify the context and the domain, just by another representation? And than the domain is dependent on representation? Or even after new representation, complex should be still complex and the domain will be still C?
$endgroup$
– ya_dimon
Nov 2 '15 at 19:46
add a comment |
$begingroup$
It would appear that Wolfram Alpha is secretly considering the functions in different contexts; for $x^{2/3}$ it is treating the $x$ as a complex variable, which can cause all kinds of shenanigans, whereas $sqrt[3]{x^2}$ has $x$ as a purely real variable, since there is no way that $x^2$ can be negative. Thus in the first case you must consider cube roots of $-1$, and in the second case you may always take the positive real root.
In short, the domain is NOT dependent on representation. Wolfram Alpha just likes to cut corners.
answered Nov 2 '15 at 6:08
TennoTenno
23315
$endgroup$
$begingroup$
if i can represent the complexxin ${x^{2/3}}$ as a realxin ${sqrt[3]{x^2}}$ than i can modify the context and the domain, just by another representation? And than the domain is dependent on representation? Or even after new representation, complex should be still complex and the domain will be still C?
$endgroup$
– ya_dimon
Nov 2 '15 at 19:46
add a comment |
$begingroup$
It would appear that Wolfram Alpha is secretly considering the functions in different contexts; for $x^{2/3}$ it is treating the $x$ as a complex variable, which can cause all kinds of shenanigans, whereas $sqrt[3]{x^2}$ has $x$ as a purely real variable, since there is no way that $x^2$ can be negative. Thus in the first case you must consider cube roots of $-1$, and in the second case you may always take the positive real root.
In short, the domain is NOT dependent on representation. Wolfram Alpha just likes to cut corners.
answered Nov 2 '15 at 6:08
TennoTenno
23315
$endgroup$
It would appear that Wolfram Alpha is secretly considering the functions in different contexts; for $x^{2/3}$ it is treating the $x$ as a complex variable, which can cause all kinds of shenanigans, whereas $sqrt[3]{x^2}$ has $x$ as a purely real variable, since there is no way that $x^2$ can be negative. Thus in the first case you must consider cube roots of $-1$, and in the second case you may always take the positive real root.
In short, the domain is NOT dependent on representation. Wolfram Alpha just likes to cut corners.
answered Nov 2 '15 at 6:08
TennoTenno
23315
answered Nov 2 '15 at 6:08
TennoTenno
23315
answered Nov 2 '15 at 6:08
TennoTenno
23315
answered Nov 2 '15 at 6:08
TennoTenno
23315
23315
$begingroup$
if i can represent the complexxin ${x^{2/3}}$ as a realxin ${sqrt[3]{x^2}}$ than i can modify the context and the domain, just by another representation? And than the domain is dependent on representation? Or even after new representation, complex should be still complex and the domain will be still C?
$endgroup$
– ya_dimon
Nov 2 '15 at 19:46
add a comment |
$begingroup$
if i can represent the complexxin ${x^{2/3}}$ as a realxin ${sqrt[3]{x^2}}$ than i can modify the context and the domain, just by another representation? And than the domain is dependent on representation? Or even after new representation, complex should be still complex and the domain will be still C?
$endgroup$
– ya_dimon
Nov 2 '15 at 19:46
$begingroup$
if i can represent the complex
x in ${x^{2/3}}$ as a real x in ${sqrt[3]{x^2}}$ than i can modify the context and the domain, just by another representation? And than the domain is dependent on representation? Or even after new representation, complex should be still complex and the domain will be still C?$endgroup$
– ya_dimon
Nov 2 '15 at 19:46
$begingroup$
if i can represent the complex
x in ${x^{2/3}}$ as a real x in ${sqrt[3]{x^2}}$ than i can modify the context and the domain, just by another representation? And than the domain is dependent on representation? Or even after new representation, complex should be still complex and the domain will be still C?$endgroup$
– ya_dimon
Nov 2 '15 at 19:46
add a comment |
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$begingroup$
Try some values!
$endgroup$
– 9301293
Nov 2 '15 at 3:46
1
$begingroup$
As with all things, it depends on definitions. How do you define the three things? They all evaluate to the same thing on the non-negative reals, but then you have questions about definition on all the reals. You can define $x^{2/3}$ as being the same as the others on the reals, but some definition s do not.
$endgroup$
– Thomas Andrews
Nov 2 '15 at 3:48
$begingroup$
Well, they are different. $x^{2/3}$ is
x^(2/3)while $sqrt[3]{x^2}$ is(x^2)^3. Furthermore, exponentiation is not associative, so why would we expect the same result?$endgroup$
– Mark McClure
Nov 2 '15 at 11:42