$-1$ to the power of a irrational number
$begingroup$
According to Wolfram Alpha, $(-1)^pi approx -0.90 + 0.43i$.
But $pi$ has proven to be irrational (we can't write $pi$ in terms of a fraction $a/b$ with $a$ and $b$ integers) and, in the real numbers, a negative number raised to a fraction with even denominator is undefined*.
So why is $(-1)^pi$ a complex number, if there is no even denominator?
*Of course there are exceptions, like $(-1)^{4/2} = (-1)^2 = 1$ and $(-1)^{6/2} = (-1)^3 = -1$, but consider that the numerator is not a multiple of the denominator.
complex-numbers exponentiation pi
edited Mar 10 at 3:18
YuiTo Cheng
2,0532637
asked Mar 9 at 14:12
EnzoEnzo
662
$endgroup$
add a comment |
$begingroup$
According to Wolfram Alpha, $(-1)^pi approx -0.90 + 0.43i$.
But $pi$ has proven to be irrational (we can't write $pi$ in terms of a fraction $a/b$ with $a$ and $b$ integers) and, in the real numbers, a negative number raised to a fraction with even denominator is undefined*.
So why is $(-1)^pi$ a complex number, if there is no even denominator?
*Of course there are exceptions, like $(-1)^{4/2} = (-1)^2 = 1$ and $(-1)^{6/2} = (-1)^3 = -1$, but consider that the numerator is not a multiple of the denominator.
complex-numbers exponentiation pi
edited Mar 10 at 3:18
YuiTo Cheng
2,0532637
asked Mar 9 at 14:12
EnzoEnzo
662
$endgroup$
4
$begingroup$
Do you know how $(-1)^pi$ is defined?
$endgroup$
– James
Mar 9 at 14:13
$begingroup$
In the complex domain you don't have to have a rational exponent. But if you don't, you'looking as a function with infinitely many possible values.
$endgroup$
– Oscar Lanzi
Mar 9 at 14:15
add a comment |
$begingroup$
According to Wolfram Alpha, $(-1)^pi approx -0.90 + 0.43i$.
But $pi$ has proven to be irrational (we can't write $pi$ in terms of a fraction $a/b$ with $a$ and $b$ integers) and, in the real numbers, a negative number raised to a fraction with even denominator is undefined*.
So why is $(-1)^pi$ a complex number, if there is no even denominator?
*Of course there are exceptions, like $(-1)^{4/2} = (-1)^2 = 1$ and $(-1)^{6/2} = (-1)^3 = -1$, but consider that the numerator is not a multiple of the denominator.
complex-numbers exponentiation pi
edited Mar 10 at 3:18
YuiTo Cheng
2,0532637
asked Mar 9 at 14:12
EnzoEnzo
662
$endgroup$
According to Wolfram Alpha, $(-1)^pi approx -0.90 + 0.43i$.
But $pi$ has proven to be irrational (we can't write $pi$ in terms of a fraction $a/b$ with $a$ and $b$ integers) and, in the real numbers, a negative number raised to a fraction with even denominator is undefined*.
So why is $(-1)^pi$ a complex number, if there is no even denominator?
*Of course there are exceptions, like $(-1)^{4/2} = (-1)^2 = 1$ and $(-1)^{6/2} = (-1)^3 = -1$, but consider that the numerator is not a multiple of the denominator.
complex-numbers exponentiation pi
complex-numbers exponentiation pi
edited Mar 10 at 3:18
YuiTo Cheng
2,0532637
asked Mar 9 at 14:12
EnzoEnzo
662
edited Mar 10 at 3:18
YuiTo Cheng
2,0532637
asked Mar 9 at 14:12
EnzoEnzo
662
edited Mar 10 at 3:18
YuiTo Cheng
2,0532637
edited Mar 10 at 3:18
YuiTo Cheng
2,0532637
edited Mar 10 at 3:18
YuiTo Cheng
2,0532637
2,0532637
asked Mar 9 at 14:12
EnzoEnzo
662
asked Mar 9 at 14:12
EnzoEnzo
662
asked Mar 9 at 14:12
EnzoEnzo
662
662
4
$begingroup$
Do you know how $(-1)^pi$ is defined?
$endgroup$
– James
Mar 9 at 14:13
$begingroup$
In the complex domain you don't have to have a rational exponent. But if you don't, you'looking as a function with infinitely many possible values.
$endgroup$
– Oscar Lanzi
Mar 9 at 14:15
add a comment |
4
$begingroup$
Do you know how $(-1)^pi$ is defined?
$endgroup$
– James
Mar 9 at 14:13
$begingroup$
In the complex domain you don't have to have a rational exponent. But if you don't, you'looking as a function with infinitely many possible values.
$endgroup$
– Oscar Lanzi
Mar 9 at 14:15
4
4
$begingroup$
Do you know how $(-1)^pi$ is defined?
$endgroup$
– James
Mar 9 at 14:13
$begingroup$
Do you know how $(-1)^pi$ is defined?
$endgroup$
– James
Mar 9 at 14:13
$begingroup$
In the complex domain you don't have to have a rational exponent. But if you don't, you'looking as a function with infinitely many possible values.
$endgroup$
– Oscar Lanzi
Mar 9 at 14:15
$begingroup$
In the complex domain you don't have to have a rational exponent. But if you don't, you'looking as a function with infinitely many possible values.
$endgroup$
– Oscar Lanzi
Mar 9 at 14:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This has to do with complex numbers. $$(-1)^{pi}=left(e^{ipi}right)^{pi}=e^{ipi^2}=cos{pi^2}+isin{pi^2}$$
which is the value given by WolframAlpha.
However, since $-1=e^{(2n+1)pi i}$ for every integer $n,$ there are actually infinitely many values of $(-1)^pi.$ With complex numbers, exponentiation works differently from the way it does with real numbers.
edited Mar 10 at 1:59
J. W. Tanner
3,3351320
answered Mar 9 at 14:20
saulspatzsaulspatz
17k31435
$endgroup$
$begingroup$
You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
$endgroup$
– drhab
Mar 9 at 14:23
$begingroup$
@drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
$endgroup$
– saulspatz
Mar 9 at 14:25
2
$begingroup$
You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
$endgroup$
– user1551
Mar 9 at 22:14
2
$begingroup$
@user1551 Even when $z$ is real?
$endgroup$
– saulspatz
Mar 9 at 22:47
$begingroup$
The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
$endgroup$
– user1551
Mar 10 at 5:36
|
show 1 more comment
$begingroup$
For complex numbers, we define $a^b = exp(b log a)$. Of course $log$ is "multivalued", so $a^b$ is also multivalued. Even $(-1)^{1/3}$ has three values, two of them are non-real. In case of $(-1)^pi$, there are infinitely many values, all non-real.
Wolfram as a system of choosing a "principal value" in these cases, and it uses the logarithm with imaginary part in $(-pi,pi]$. So the principal value of $log(-1)$ is $ipi$. And the principal value of $(-1)^pi$ is $exp(pilog(-1)) =
exp(ipi^2) = cos(pi^2)+isin(pi^2)approx -0.90 - i 0.43$.
answered Mar 9 at 14:25
GEdgarGEdgar
63.1k267171
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This has to do with complex numbers. $$(-1)^{pi}=left(e^{ipi}right)^{pi}=e^{ipi^2}=cos{pi^2}+isin{pi^2}$$
which is the value given by WolframAlpha.
However, since $-1=e^{(2n+1)pi i}$ for every integer $n,$ there are actually infinitely many values of $(-1)^pi.$ With complex numbers, exponentiation works differently from the way it does with real numbers.
edited Mar 10 at 1:59
J. W. Tanner
3,3351320
answered Mar 9 at 14:20
saulspatzsaulspatz
17k31435
$endgroup$
$begingroup$
You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
$endgroup$
– drhab
Mar 9 at 14:23
$begingroup$
@drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
$endgroup$
– saulspatz
Mar 9 at 14:25
2
$begingroup$
You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
$endgroup$
– user1551
Mar 9 at 22:14
2
$begingroup$
@user1551 Even when $z$ is real?
$endgroup$
– saulspatz
Mar 9 at 22:47
$begingroup$
The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
$endgroup$
– user1551
Mar 10 at 5:36
|
show 1 more comment
$begingroup$
This has to do with complex numbers. $$(-1)^{pi}=left(e^{ipi}right)^{pi}=e^{ipi^2}=cos{pi^2}+isin{pi^2}$$
which is the value given by WolframAlpha.
However, since $-1=e^{(2n+1)pi i}$ for every integer $n,$ there are actually infinitely many values of $(-1)^pi.$ With complex numbers, exponentiation works differently from the way it does with real numbers.
edited Mar 10 at 1:59
J. W. Tanner
3,3351320
answered Mar 9 at 14:20
saulspatzsaulspatz
17k31435
$endgroup$
$begingroup$
You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
$endgroup$
– drhab
Mar 9 at 14:23
$begingroup$
@drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
$endgroup$
– saulspatz
Mar 9 at 14:25
2
$begingroup$
You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
$endgroup$
– user1551
Mar 9 at 22:14
2
$begingroup$
@user1551 Even when $z$ is real?
$endgroup$
– saulspatz
Mar 9 at 22:47
$begingroup$
The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
$endgroup$
– user1551
Mar 10 at 5:36
|
show 1 more comment
$begingroup$
This has to do with complex numbers. $$(-1)^{pi}=left(e^{ipi}right)^{pi}=e^{ipi^2}=cos{pi^2}+isin{pi^2}$$
which is the value given by WolframAlpha.
However, since $-1=e^{(2n+1)pi i}$ for every integer $n,$ there are actually infinitely many values of $(-1)^pi.$ With complex numbers, exponentiation works differently from the way it does with real numbers.
edited Mar 10 at 1:59
J. W. Tanner
3,3351320
answered Mar 9 at 14:20
saulspatzsaulspatz
17k31435
$endgroup$
This has to do with complex numbers. $$(-1)^{pi}=left(e^{ipi}right)^{pi}=e^{ipi^2}=cos{pi^2}+isin{pi^2}$$
which is the value given by WolframAlpha.
However, since $-1=e^{(2n+1)pi i}$ for every integer $n,$ there are actually infinitely many values of $(-1)^pi.$ With complex numbers, exponentiation works differently from the way it does with real numbers.
edited Mar 10 at 1:59
J. W. Tanner
3,3351320
answered Mar 9 at 14:20
saulspatzsaulspatz
17k31435
edited Mar 10 at 1:59
J. W. Tanner
3,3351320
edited Mar 10 at 1:59
J. W. Tanner
3,3351320
edited Mar 10 at 1:59
J. W. Tanner
3,3351320
3,3351320
answered Mar 9 at 14:20
saulspatzsaulspatz
17k31435
answered Mar 9 at 14:20
saulspatzsaulspatz
17k31435
answered Mar 9 at 14:20
saulspatzsaulspatz
17k31435
17k31435
$begingroup$
You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
$endgroup$
– drhab
Mar 9 at 14:23
$begingroup$
@drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
$endgroup$
– saulspatz
Mar 9 at 14:25
2
$begingroup$
You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
$endgroup$
– user1551
Mar 9 at 22:14
2
$begingroup$
@user1551 Even when $z$ is real?
$endgroup$
– saulspatz
Mar 9 at 22:47
$begingroup$
The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
$endgroup$
– user1551
Mar 10 at 5:36
|
show 1 more comment
$begingroup$
You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
$endgroup$
– drhab
Mar 9 at 14:23
$begingroup$
@drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
$endgroup$
– saulspatz
Mar 9 at 14:25
2
$begingroup$
You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
$endgroup$
– user1551
Mar 9 at 22:14
2
$begingroup$
@user1551 Even when $z$ is real?
$endgroup$
– saulspatz
Mar 9 at 22:47
$begingroup$
The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
$endgroup$
– user1551
Mar 10 at 5:36
$begingroup$
You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
$endgroup$
– drhab
Mar 9 at 14:23
$begingroup$
You mean $cdots=left(e^{ipi}right)^{pi}=cdots$ I guess.
$endgroup$
– drhab
Mar 9 at 14:23
$begingroup$
@drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
$endgroup$
– saulspatz
Mar 9 at 14:25
$begingroup$
@drhap Of course I do thanks. I struggled with the MathJax, and by the time I got it to accept anything, I wasn't looking too closely any more, I guess.
$endgroup$
– saulspatz
Mar 9 at 14:25
2
2
$begingroup$
You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
$endgroup$
– user1551
Mar 9 at 22:14
$begingroup$
You need to justify that $(e^{ipi})^pi=e^{ipi^2}$. In general, we don't have a rule $x^{yz}=(x^y)^z$ for complex numbers.
$endgroup$
– user1551
Mar 9 at 22:14
2
2
$begingroup$
@user1551 Even when $z$ is real?
$endgroup$
– saulspatz
Mar 9 at 22:47
$begingroup$
@user1551 Even when $z$ is real?
$endgroup$
– saulspatz
Mar 9 at 22:47
$begingroup$
The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
$endgroup$
– user1551
Mar 10 at 5:36
$begingroup$
The rule $(x^y)^z=x^{yz}$ fails even when $z$ is real. The problem is that neither $(x^y)^z$ nor $x^{yz}$ are unambigusously defined. In case $x>0$ and $z$ is real, both $x^y$ and $x^{yz}$ are well-defined, but $(x^y)^z$ is still not. In the OP's case, as you said, $(-1)^pi$ can assume infinitely many values. In fact, $$(-1)^pi=exp(pilog(-1))=exp(pi(ipi+2inpi))=e^{ipi^2}e^{2inpi^2}$$ but the value of $e^{2inpi^2}$ varies with $n$. You may see Marc van Leeuwen's answer in another thread about the applicability of the rule $(x^y)^z=x^{yz}$.
$endgroup$
– user1551
Mar 10 at 5:36
|
show 1 more comment
$begingroup$
For complex numbers, we define $a^b = exp(b log a)$. Of course $log$ is "multivalued", so $a^b$ is also multivalued. Even $(-1)^{1/3}$ has three values, two of them are non-real. In case of $(-1)^pi$, there are infinitely many values, all non-real.
Wolfram as a system of choosing a "principal value" in these cases, and it uses the logarithm with imaginary part in $(-pi,pi]$. So the principal value of $log(-1)$ is $ipi$. And the principal value of $(-1)^pi$ is $exp(pilog(-1)) =
exp(ipi^2) = cos(pi^2)+isin(pi^2)approx -0.90 - i 0.43$.
answered Mar 9 at 14:25
GEdgarGEdgar
63.1k267171
$endgroup$
add a comment |
$begingroup$
For complex numbers, we define $a^b = exp(b log a)$. Of course $log$ is "multivalued", so $a^b$ is also multivalued. Even $(-1)^{1/3}$ has three values, two of them are non-real. In case of $(-1)^pi$, there are infinitely many values, all non-real.
Wolfram as a system of choosing a "principal value" in these cases, and it uses the logarithm with imaginary part in $(-pi,pi]$. So the principal value of $log(-1)$ is $ipi$. And the principal value of $(-1)^pi$ is $exp(pilog(-1)) =
exp(ipi^2) = cos(pi^2)+isin(pi^2)approx -0.90 - i 0.43$.
answered Mar 9 at 14:25
GEdgarGEdgar
63.1k267171
$endgroup$
add a comment |
$begingroup$
For complex numbers, we define $a^b = exp(b log a)$. Of course $log$ is "multivalued", so $a^b$ is also multivalued. Even $(-1)^{1/3}$ has three values, two of them are non-real. In case of $(-1)^pi$, there are infinitely many values, all non-real.
Wolfram as a system of choosing a "principal value" in these cases, and it uses the logarithm with imaginary part in $(-pi,pi]$. So the principal value of $log(-1)$ is $ipi$. And the principal value of $(-1)^pi$ is $exp(pilog(-1)) =
exp(ipi^2) = cos(pi^2)+isin(pi^2)approx -0.90 - i 0.43$.
answered Mar 9 at 14:25
GEdgarGEdgar
63.1k267171
$endgroup$
For complex numbers, we define $a^b = exp(b log a)$. Of course $log$ is "multivalued", so $a^b$ is also multivalued. Even $(-1)^{1/3}$ has three values, two of them are non-real. In case of $(-1)^pi$, there are infinitely many values, all non-real.
Wolfram as a system of choosing a "principal value" in these cases, and it uses the logarithm with imaginary part in $(-pi,pi]$. So the principal value of $log(-1)$ is $ipi$. And the principal value of $(-1)^pi$ is $exp(pilog(-1)) =
exp(ipi^2) = cos(pi^2)+isin(pi^2)approx -0.90 - i 0.43$.
answered Mar 9 at 14:25
GEdgarGEdgar
63.1k267171
answered Mar 9 at 14:25
GEdgarGEdgar
63.1k267171
answered Mar 9 at 14:25
GEdgarGEdgar
63.1k267171
answered Mar 9 at 14:25
GEdgarGEdgar
63.1k267171
63.1k267171
add a comment |
add a comment |
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4
$begingroup$
Do you know how $(-1)^pi$ is defined?
$endgroup$
– James
Mar 9 at 14:13
$begingroup$
In the complex domain you don't have to have a rational exponent. But if you don't, you'looking as a function with infinitely many possible values.
$endgroup$
– Oscar Lanzi
Mar 9 at 14:15