Mixing
Why is $-1=e^{pi i }=e^{2pi ifrac{1}{2}}neq (e^{2pi i})^{frac{1}{2}}=1^{frac{1}{2}}=1$ true?
$begingroup$
I thought there was this rule that $e^{xB}=(e^x)^B$?
Also what I don't understand is that $x^{frac{1}{2}}$ is defined as the square root of $x$. And because the square root may also be in $mathbb{C}$ it doesn't matter which number $x$ is because every complex number has a n-th root.
But for $1^{frac{1}{2}}$ we have the roots $-1$ and $1$, so I could also write $-1$ on the right side then the Statement above would be true and not true at the same time.
But we said the Expression in the Question is correct and to write $-1$ is false. Can somebody explain the reason we choose one Version over another altough they are equivalent?
complex-analysis complex-numbers exponentiation
asked Jan 27 at 12:05
RM777RM777
38312
$endgroup$
|
show 4 more comments
$begingroup$
I thought there was this rule that $e^{xB}=(e^x)^B$?
Also what I don't understand is that $x^{frac{1}{2}}$ is defined as the square root of $x$. And because the square root may also be in $mathbb{C}$ it doesn't matter which number $x$ is because every complex number has a n-th root.
But for $1^{frac{1}{2}}$ we have the roots $-1$ and $1$, so I could also write $-1$ on the right side then the Statement above would be true and not true at the same time.
But we said the Expression in the Question is correct and to write $-1$ is false. Can somebody explain the reason we choose one Version over another altough they are equivalent?
complex-analysis complex-numbers exponentiation
asked Jan 27 at 12:05
RM777RM777
38312
$endgroup$
3
$begingroup$
To be precise every complex number $($beside $0$$)$ has exactly $n$ distinct $n$th roots.
$endgroup$
– mrtaurho
Jan 27 at 12:07
$begingroup$
@mrtraurho Well, $0$ doesn't have distinct $n$th roots.
$endgroup$
– J.G.
Jan 27 at 12:08
1
$begingroup$
Conclusion: using the notation $z^{1/2}$ to mean a single number and simultaneously to mean a set of two numbers, can lead to chaos. Should we be surprised?
$endgroup$
– Did
Jan 27 at 12:10
1
$begingroup$
The main problem of taking a root in general is that it happens to be that the root function is a multivalued one. Thus, even within the reals you can state that $5=sqrt{25}=-5$ which is a contradiction as well.
$endgroup$
– mrtaurho
Jan 27 at 12:13
2
$begingroup$
Which rule? If your question is how to define a square root function on the complex plane, the answer is well known (and it has been hashed and rehashed on this site, I must add).
$endgroup$
– Did
Jan 27 at 12:18
|
show 4 more comments
$begingroup$
I thought there was this rule that $e^{xB}=(e^x)^B$?
Also what I don't understand is that $x^{frac{1}{2}}$ is defined as the square root of $x$. And because the square root may also be in $mathbb{C}$ it doesn't matter which number $x$ is because every complex number has a n-th root.
But for $1^{frac{1}{2}}$ we have the roots $-1$ and $1$, so I could also write $-1$ on the right side then the Statement above would be true and not true at the same time.
But we said the Expression in the Question is correct and to write $-1$ is false. Can somebody explain the reason we choose one Version over another altough they are equivalent?
complex-analysis complex-numbers exponentiation
asked Jan 27 at 12:05
RM777RM777
38312
$endgroup$
I thought there was this rule that $e^{xB}=(e^x)^B$?
Also what I don't understand is that $x^{frac{1}{2}}$ is defined as the square root of $x$. And because the square root may also be in $mathbb{C}$ it doesn't matter which number $x$ is because every complex number has a n-th root.
But for $1^{frac{1}{2}}$ we have the roots $-1$ and $1$, so I could also write $-1$ on the right side then the Statement above would be true and not true at the same time.
But we said the Expression in the Question is correct and to write $-1$ is false. Can somebody explain the reason we choose one Version over another altough they are equivalent?
complex-analysis complex-numbers exponentiation
complex-analysis complex-numbers exponentiation
asked Jan 27 at 12:05
RM777RM777
38312
asked Jan 27 at 12:05
RM777RM777
38312
edited Jan 27 at 18:06
user587192
asked Jan 27 at 12:05
RM777RM777
38312
asked Jan 27 at 12:05
RM777RM777
38312
asked Jan 27 at 12:05
RM777RM777
38312
38312
3
$begingroup$
To be precise every complex number $($beside $0$$)$ has exactly $n$ distinct $n$th roots.
$endgroup$
– mrtaurho
Jan 27 at 12:07
$begingroup$
@mrtraurho Well, $0$ doesn't have distinct $n$th roots.
$endgroup$
– J.G.
Jan 27 at 12:08
1
$begingroup$
Conclusion: using the notation $z^{1/2}$ to mean a single number and simultaneously to mean a set of two numbers, can lead to chaos. Should we be surprised?
$endgroup$
– Did
Jan 27 at 12:10
1
$begingroup$
The main problem of taking a root in general is that it happens to be that the root function is a multivalued one. Thus, even within the reals you can state that $5=sqrt{25}=-5$ which is a contradiction as well.
$endgroup$
– mrtaurho
Jan 27 at 12:13
2
$begingroup$
Which rule? If your question is how to define a square root function on the complex plane, the answer is well known (and it has been hashed and rehashed on this site, I must add).
$endgroup$
– Did
Jan 27 at 12:18
|
show 4 more comments
3
$begingroup$
To be precise every complex number $($beside $0$$)$ has exactly $n$ distinct $n$th roots.
$endgroup$
– mrtaurho
Jan 27 at 12:07
$begingroup$
@mrtraurho Well, $0$ doesn't have distinct $n$th roots.
$endgroup$
– J.G.
Jan 27 at 12:08
1
$begingroup$
Conclusion: using the notation $z^{1/2}$ to mean a single number and simultaneously to mean a set of two numbers, can lead to chaos. Should we be surprised?
$endgroup$
– Did
Jan 27 at 12:10
1
$begingroup$
The main problem of taking a root in general is that it happens to be that the root function is a multivalued one. Thus, even within the reals you can state that $5=sqrt{25}=-5$ which is a contradiction as well.
$endgroup$
– mrtaurho
Jan 27 at 12:13
2
$begingroup$
Which rule? If your question is how to define a square root function on the complex plane, the answer is well known (and it has been hashed and rehashed on this site, I must add).
$endgroup$
– Did
Jan 27 at 12:18
3
3
$begingroup$
To be precise every complex number $($beside $0$$)$ has exactly $n$ distinct $n$th roots.
$endgroup$
– mrtaurho
Jan 27 at 12:07
$begingroup$
To be precise every complex number $($beside $0$$)$ has exactly $n$ distinct $n$th roots.
$endgroup$
– mrtaurho
Jan 27 at 12:07
$begingroup$
@mrtraurho Well, $0$ doesn't have distinct $n$th roots.
$endgroup$
– J.G.
Jan 27 at 12:08
$begingroup$
@mrtraurho Well, $0$ doesn't have distinct $n$th roots.
$endgroup$
– J.G.
Jan 27 at 12:08
1
1
$begingroup$
Conclusion: using the notation $z^{1/2}$ to mean a single number and simultaneously to mean a set of two numbers, can lead to chaos. Should we be surprised?
$endgroup$
– Did
Jan 27 at 12:10
$begingroup$
Conclusion: using the notation $z^{1/2}$ to mean a single number and simultaneously to mean a set of two numbers, can lead to chaos. Should we be surprised?
$endgroup$
– Did
Jan 27 at 12:10
1
1
$begingroup$
The main problem of taking a root in general is that it happens to be that the root function is a multivalued one. Thus, even within the reals you can state that $5=sqrt{25}=-5$ which is a contradiction as well.
$endgroup$
– mrtaurho
Jan 27 at 12:13
$begingroup$
The main problem of taking a root in general is that it happens to be that the root function is a multivalued one. Thus, even within the reals you can state that $5=sqrt{25}=-5$ which is a contradiction as well.
$endgroup$
– mrtaurho
Jan 27 at 12:13
2
2
$begingroup$
Which rule? If your question is how to define a square root function on the complex plane, the answer is well known (and it has been hashed and rehashed on this site, I must add).
$endgroup$
– Did
Jan 27 at 12:18
$begingroup$
Which rule? If your question is how to define a square root function on the complex plane, the answer is well known (and it has been hashed and rehashed on this site, I must add).
$endgroup$
– Did
Jan 27 at 12:18
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If one is working in real analysis, the expression $x^{1/2}$ ($x>0$) is by definition $sqrt{x}$, which is defined to be the positive real number $y$ such that $y^2=x$. Hence one has
$$
1^{1/2}=1.
$$
If one is working in complex analysis, $1^{1/2}$ can be viewed as the multivalued function $f(z)=z^{1/2}$ evaluated at $z=1$. In this context, since $f$ is multivalued, it is incorrect to write $1^{1/2}=1$.
The identity $(e^x)^y = e^{xy}$ holds for real numbers $x$ and $y$, but assuming its truth for complex numbers leads to paradox like the one you have observed.
To quote Wikipedia:
Some identities for powers and logarithms for positive real numbers will fail for complex numbers, no matter how complex powers and complex logarithms are defined as single-valued functions.
See Failure of power and logarithm identities for more examples.
$endgroup$
add a comment |
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$begingroup$
If one is working in real analysis, the expression $x^{1/2}$ ($x>0$) is by definition $sqrt{x}$, which is defined to be the positive real number $y$ such that $y^2=x$. Hence one has
$$
1^{1/2}=1.
$$
If one is working in complex analysis, $1^{1/2}$ can be viewed as the multivalued function $f(z)=z^{1/2}$ evaluated at $z=1$. In this context, since $f$ is multivalued, it is incorrect to write $1^{1/2}=1$.
The identity $(e^x)^y = e^{xy}$ holds for real numbers $x$ and $y$, but assuming its truth for complex numbers leads to paradox like the one you have observed.
To quote Wikipedia:
Some identities for powers and logarithms for positive real numbers will fail for complex numbers, no matter how complex powers and complex logarithms are defined as single-valued functions.
See Failure of power and logarithm identities for more examples.
$endgroup$
add a comment |
$begingroup$
If one is working in real analysis, the expression $x^{1/2}$ ($x>0$) is by definition $sqrt{x}$, which is defined to be the positive real number $y$ such that $y^2=x$. Hence one has
$$
1^{1/2}=1.
$$
If one is working in complex analysis, $1^{1/2}$ can be viewed as the multivalued function $f(z)=z^{1/2}$ evaluated at $z=1$. In this context, since $f$ is multivalued, it is incorrect to write $1^{1/2}=1$.
The identity $(e^x)^y = e^{xy}$ holds for real numbers $x$ and $y$, but assuming its truth for complex numbers leads to paradox like the one you have observed.
To quote Wikipedia:
Some identities for powers and logarithms for positive real numbers will fail for complex numbers, no matter how complex powers and complex logarithms are defined as single-valued functions.
See Failure of power and logarithm identities for more examples.
$endgroup$
add a comment |
$begingroup$
If one is working in real analysis, the expression $x^{1/2}$ ($x>0$) is by definition $sqrt{x}$, which is defined to be the positive real number $y$ such that $y^2=x$. Hence one has
$$
1^{1/2}=1.
$$
If one is working in complex analysis, $1^{1/2}$ can be viewed as the multivalued function $f(z)=z^{1/2}$ evaluated at $z=1$. In this context, since $f$ is multivalued, it is incorrect to write $1^{1/2}=1$.
The identity $(e^x)^y = e^{xy}$ holds for real numbers $x$ and $y$, but assuming its truth for complex numbers leads to paradox like the one you have observed.
To quote Wikipedia:
Some identities for powers and logarithms for positive real numbers will fail for complex numbers, no matter how complex powers and complex logarithms are defined as single-valued functions.
See Failure of power and logarithm identities for more examples.
$endgroup$
If one is working in real analysis, the expression $x^{1/2}$ ($x>0$) is by definition $sqrt{x}$, which is defined to be the positive real number $y$ such that $y^2=x$. Hence one has
$$
1^{1/2}=1.
$$
If one is working in complex analysis, $1^{1/2}$ can be viewed as the multivalued function $f(z)=z^{1/2}$ evaluated at $z=1$. In this context, since $f$ is multivalued, it is incorrect to write $1^{1/2}=1$.
The identity $(e^x)^y = e^{xy}$ holds for real numbers $x$ and $y$, but assuming its truth for complex numbers leads to paradox like the one you have observed.
To quote Wikipedia:
Some identities for powers and logarithms for positive real numbers will fail for complex numbers, no matter how complex powers and complex logarithms are defined as single-valued functions.
See Failure of power and logarithm identities for more examples.
answered Jan 27 at 17:29
user587192
add a comment |
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3
$begingroup$
To be precise every complex number $($beside $0$$)$ has exactly $n$ distinct $n$th roots.
$endgroup$
– mrtaurho
Jan 27 at 12:07
$begingroup$
@mrtraurho Well, $0$ doesn't have distinct $n$th roots.
$endgroup$
– J.G.
Jan 27 at 12:08
1
$begingroup$
Conclusion: using the notation $z^{1/2}$ to mean a single number and simultaneously to mean a set of two numbers, can lead to chaos. Should we be surprised?
$endgroup$
– Did
Jan 27 at 12:10
1
$begingroup$
The main problem of taking a root in general is that it happens to be that the root function is a multivalued one. Thus, even within the reals you can state that $5=sqrt{25}=-5$ which is a contradiction as well.
$endgroup$
– mrtaurho
Jan 27 at 12:13
2
$begingroup$
Which rule? If your question is how to define a square root function on the complex plane, the answer is well known (and it has been hashed and rehashed on this site, I must add).
$endgroup$
– Did
Jan 27 at 12:18