Mixing
Do two pairs of distinct natural numbers exist such that AGM(A,B) equal to AGM(C,D)?
0
$begingroup$
Here AGM is arithmetic-geometric mean.
Are there natural numbers A,B,C,D such that $1leq A<C<D<B$ and arithmetic-geometric mean AGM(A,B)=AGM(C,D) ?
In other words, is AGM a homomorphism of an unordered pair of natural numbers on a set of real numbers?
means natural-numbers
asked Jan 13 at 15:44
StepanStepan
468312
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add a comment |
0
$begingroup$
Here AGM is arithmetic-geometric mean.
Are there natural numbers A,B,C,D such that $1leq A<C<D<B$ and arithmetic-geometric mean AGM(A,B)=AGM(C,D) ?
In other words, is AGM a homomorphism of an unordered pair of natural numbers on a set of real numbers?
means natural-numbers
asked Jan 13 at 15:44
StepanStepan
468312
$endgroup$
$begingroup$
Isn't $A<AGM(A,B)<B?$ How could this be true?
$endgroup$
– saulspatz
Jan 13 at 15:46
$begingroup$
The questions in your title and body are different. In particular, the answer to the question in the body is trivially no: $AGM(A,B) leq B < C leq AGM(C,D)$, but the question in the title does not suffer from this problem. In particular, for any such example, we must have either $A < C < D < B$, or one of the obvious permutations of that.
$endgroup$
– user3482749
Jan 13 at 15:48
$begingroup$
Typo. Corrected
$endgroup$
– Stepan
Jan 13 at 15:49
add a comment |
0
0
0
$begingroup$
Here AGM is arithmetic-geometric mean.
Are there natural numbers A,B,C,D such that $1leq A<C<D<B$ and arithmetic-geometric mean AGM(A,B)=AGM(C,D) ?
In other words, is AGM a homomorphism of an unordered pair of natural numbers on a set of real numbers?
means natural-numbers
asked Jan 13 at 15:44
StepanStepan
468312
$endgroup$
Here AGM is arithmetic-geometric mean.
Are there natural numbers A,B,C,D such that $1leq A<C<D<B$ and arithmetic-geometric mean AGM(A,B)=AGM(C,D) ?
In other words, is AGM a homomorphism of an unordered pair of natural numbers on a set of real numbers?
means natural-numbers
means natural-numbers
asked Jan 13 at 15:44
StepanStepan
468312
asked Jan 13 at 15:44
StepanStepan
468312
edited Jan 13 at 15:49
Stepan
asked Jan 13 at 15:44
StepanStepan
468312
asked Jan 13 at 15:44
StepanStepan
468312
asked Jan 13 at 15:44
StepanStepan
468312
468312
$begingroup$
Isn't $A<AGM(A,B)<B?$ How could this be true?
$endgroup$
– saulspatz
Jan 13 at 15:46
$begingroup$
The questions in your title and body are different. In particular, the answer to the question in the body is trivially no: $AGM(A,B) leq B < C leq AGM(C,D)$, but the question in the title does not suffer from this problem. In particular, for any such example, we must have either $A < C < D < B$, or one of the obvious permutations of that.
$endgroup$
– user3482749
Jan 13 at 15:48
$begingroup$
Typo. Corrected
$endgroup$
– Stepan
Jan 13 at 15:49
add a comment |
$begingroup$
Isn't $A<AGM(A,B)<B?$ How could this be true?
$endgroup$
– saulspatz
Jan 13 at 15:46
$begingroup$
The questions in your title and body are different. In particular, the answer to the question in the body is trivially no: $AGM(A,B) leq B < C leq AGM(C,D)$, but the question in the title does not suffer from this problem. In particular, for any such example, we must have either $A < C < D < B$, or one of the obvious permutations of that.
$endgroup$
– user3482749
Jan 13 at 15:48
$begingroup$
Typo. Corrected
$endgroup$
– Stepan
Jan 13 at 15:49
$begingroup$
Isn't $A<AGM(A,B)<B?$ How could this be true?
$endgroup$
– saulspatz
Jan 13 at 15:46
$begingroup$
Isn't $A<AGM(A,B)<B?$ How could this be true?
$endgroup$
– saulspatz
Jan 13 at 15:46
$begingroup$
The questions in your title and body are different. In particular, the answer to the question in the body is trivially no: $AGM(A,B) leq B < C leq AGM(C,D)$, but the question in the title does not suffer from this problem. In particular, for any such example, we must have either $A < C < D < B$, or one of the obvious permutations of that.
$endgroup$
– user3482749
Jan 13 at 15:48
$begingroup$
The questions in your title and body are different. In particular, the answer to the question in the body is trivially no: $AGM(A,B) leq B < C leq AGM(C,D)$, but the question in the title does not suffer from this problem. In particular, for any such example, we must have either $A < C < D < B$, or one of the obvious permutations of that.
$endgroup$
– user3482749
Jan 13 at 15:48
$begingroup$
Typo. Corrected
$endgroup$
– Stepan
Jan 13 at 15:49
$begingroup$
Typo. Corrected
$endgroup$
– Stepan
Jan 13 at 15:49
add a comment |
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$begingroup$
Isn't $A<AGM(A,B)<B?$ How could this be true?
$endgroup$
– saulspatz
Jan 13 at 15:46
$begingroup$
The questions in your title and body are different. In particular, the answer to the question in the body is trivially no: $AGM(A,B) leq B < C leq AGM(C,D)$, but the question in the title does not suffer from this problem. In particular, for any such example, we must have either $A < C < D < B$, or one of the obvious permutations of that.
$endgroup$
– user3482749
Jan 13 at 15:48
$begingroup$
Typo. Corrected
$endgroup$
– Stepan
Jan 13 at 15:49