Is there any known transcendental $b$ such that $b^b$ is also transcendental?
$begingroup$
Numbers such as $e$ and $π$ are known to be transcendental, however, $e^e$ or $π^π$ are not even known to be irrational, let alone transcendental.
There are infinitely many transcendental numbers $a$ such that $a^a$ is rational, namely the solution of every $x^x = p$ where $p$ is prime.
My question is: do we know of any transcendental number $b$ such that $b^b$ is transcendental?
transcendental-numbers
asked Jan 18 at 16:36
JanJan
541
$endgroup$
add a comment |
$begingroup$
Numbers such as $e$ and $π$ are known to be transcendental, however, $e^e$ or $π^π$ are not even known to be irrational, let alone transcendental.
There are infinitely many transcendental numbers $a$ such that $a^a$ is rational, namely the solution of every $x^x = p$ where $p$ is prime.
My question is: do we know of any transcendental number $b$ such that $b^b$ is transcendental?
transcendental-numbers
asked Jan 18 at 16:36
JanJan
541
$endgroup$
add a comment |
$begingroup$
Numbers such as $e$ and $π$ are known to be transcendental, however, $e^e$ or $π^π$ are not even known to be irrational, let alone transcendental.
There are infinitely many transcendental numbers $a$ such that $a^a$ is rational, namely the solution of every $x^x = p$ where $p$ is prime.
My question is: do we know of any transcendental number $b$ such that $b^b$ is transcendental?
transcendental-numbers
asked Jan 18 at 16:36
JanJan
541
$endgroup$
Numbers such as $e$ and $π$ are known to be transcendental, however, $e^e$ or $π^π$ are not even known to be irrational, let alone transcendental.
There are infinitely many transcendental numbers $a$ such that $a^a$ is rational, namely the solution of every $x^x = p$ where $p$ is prime.
My question is: do we know of any transcendental number $b$ such that $b^b$ is transcendental?
transcendental-numbers
transcendental-numbers
asked Jan 18 at 16:36
JanJan
541
asked Jan 18 at 16:36
JanJan
541
asked Jan 18 at 16:36
JanJan
541
asked Jan 18 at 16:36
JanJan
541
asked Jan 18 at 16:36
JanJan
541
541
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $b = 2/W(2)$, where $W$ is the Lambert W function. Note that $z = W(2)$ satisfies
$z e^z = 2$. If $z$ were algebraic, then $e^z$ would also be algebraic, but this would contradict Lindemann's theorem. Therefore $z$ is trancendental, and so is $b$. Now
$ W(2) + log(W(2)) = log(2)$, so $log(b) = log(2) - log(W(2)) = W(2)$,
and $$b^b = exp(b log(b)) = exp(2)$$
which is transcendental.
answered Jan 18 at 17:03
Robert IsraelRobert Israel
332k23222481
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1 Answer
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$begingroup$
Let $b = 2/W(2)$, where $W$ is the Lambert W function. Note that $z = W(2)$ satisfies
$z e^z = 2$. If $z$ were algebraic, then $e^z$ would also be algebraic, but this would contradict Lindemann's theorem. Therefore $z$ is trancendental, and so is $b$. Now
$ W(2) + log(W(2)) = log(2)$, so $log(b) = log(2) - log(W(2)) = W(2)$,
and $$b^b = exp(b log(b)) = exp(2)$$
which is transcendental.
answered Jan 18 at 17:03
Robert IsraelRobert Israel
332k23222481
$endgroup$
add a comment |
$begingroup$
Let $b = 2/W(2)$, where $W$ is the Lambert W function. Note that $z = W(2)$ satisfies
$z e^z = 2$. If $z$ were algebraic, then $e^z$ would also be algebraic, but this would contradict Lindemann's theorem. Therefore $z$ is trancendental, and so is $b$. Now
$ W(2) + log(W(2)) = log(2)$, so $log(b) = log(2) - log(W(2)) = W(2)$,
and $$b^b = exp(b log(b)) = exp(2)$$
which is transcendental.
answered Jan 18 at 17:03
Robert IsraelRobert Israel
332k23222481
$endgroup$
add a comment |
$begingroup$
Let $b = 2/W(2)$, where $W$ is the Lambert W function. Note that $z = W(2)$ satisfies
$z e^z = 2$. If $z$ were algebraic, then $e^z$ would also be algebraic, but this would contradict Lindemann's theorem. Therefore $z$ is trancendental, and so is $b$. Now
$ W(2) + log(W(2)) = log(2)$, so $log(b) = log(2) - log(W(2)) = W(2)$,
and $$b^b = exp(b log(b)) = exp(2)$$
which is transcendental.
answered Jan 18 at 17:03
Robert IsraelRobert Israel
332k23222481
$endgroup$
Let $b = 2/W(2)$, where $W$ is the Lambert W function. Note that $z = W(2)$ satisfies
$z e^z = 2$. If $z$ were algebraic, then $e^z$ would also be algebraic, but this would contradict Lindemann's theorem. Therefore $z$ is trancendental, and so is $b$. Now
$ W(2) + log(W(2)) = log(2)$, so $log(b) = log(2) - log(W(2)) = W(2)$,
and $$b^b = exp(b log(b)) = exp(2)$$
which is transcendental.
answered Jan 18 at 17:03
Robert IsraelRobert Israel
332k23222481
answered Jan 18 at 17:03
Robert IsraelRobert Israel
332k23222481
answered Jan 18 at 17:03
Robert IsraelRobert Israel
332k23222481
answered Jan 18 at 17:03
Robert IsraelRobert Israel
332k23222481
332k23222481
add a comment |
add a comment |
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