My first rebus, so I hope that it makes sense.
Given that $f(z)=\dfrac{1}{(z-A)^n}$, is it true that $A=[N,\ldots,O]$?
Hint:
$f$, $n$, and $z$ are not important
Hint:
It's about fruits
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Sign up to join this communityMy first rebus, so I hope that it makes sense.
Given that $f(z)=\dfrac{1}{(z-A)^n}$, is it true that $A=[N,\ldots,O]$?
Hint:
$f$, $n$, and $z$ are not important
Hint:
It's about fruits
The community wiki answer is
No.
Community wiki argument:
1. $A$ is a pole of the function
2. $[N..O]$ is the $N$-$O$-range, respectively the EN-O-RANGE.
3. Now the question is: IS A-POLE EQUAL TO EN-O-RANGE?
4. And the answer is: No, since you cannot compare apples to oranges.
Another try - is it?
There is no answer - i.e. it cannot be deduced
because
The rebus last line is comparing apples and oranges
as
Apples is A pole (plural as pole of degree n)
and
oranges is 0 (zero) ranges as No is the range or interval.
Is it:-
Poles apart ?
Because
$A$ is a pole of the function in variable $z$ and these letters are distant from each other in the alphabet(similar to the poles of a magnet).
It may be noted that a monopole does not exist, so the South and North Pole of a magnet are always some distance apart, howsoever small this distance might be.
So the following (ie.$A=[N\,.\!.\, O]$) is:
False, as $O$ and $N$ are not poles apart(ie. far apart) in the alphabet.
Is it:
An apple a day keeps the doctor away.
Because
Apple is A pole
Doctor away is Dr No separated
Is it:
A Polar Molecule?
So the following (ie.$A=[N..O]$) is :
True, as $[N..O]$ (read as $N$ to $O$),ie. $N_2O$ is a polar molecule.