<Comment deleted by user>
@polylogarithmique said in #8:
> What about this one: 1/x=2/x.
Not infinity, because that's no number.
So it must be 0 (!!)
But we don't know how that goes! (!!!)
Yes! But it HAS to be! (!!!!)
It's none of all the other numbers! (!!!!!!)
By process of elimination! (!¡!¡!¡!¡!¡!¡!¡!¡!¡!¡!¡!¡!¡)
> What about this one: 1/x=2/x.
Not infinity, because that's no number.
So it must be 0 (!!)
But we don't know how that goes! (!!!)
Yes! But it HAS to be! (!!!!)
It's none of all the other numbers! (!!!!!!)
By process of elimination! (!¡!¡!¡!¡!¡!¡!¡!¡!¡!¡!¡!¡!¡)
<Comment deleted by user>
@Marcin2 said in #14:
> If we consider finite numbers only, then this equation has no solutions. If we consider infinite
> numbers also, then it has three solutions: x=0, x=infinity and x=-infinity.
Infinity is not a number, it's a concept. You cannot divide 1 by infinity.
Division by infinity (given a finite numerator) is a notion that isn't well defined in ordinary arithmetic because there's no nonzero real number that, when multiplied by infinity, gives a finite number.
Similarly 1/0 is undefined in ordinary arithmetic because there is no number that, when multiplied by 0, gives 1.
In short: polylogarithmique's equation has no solutions.
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What you probably meant is that the following statements are all true:
lim_(x to +∞) 1/x = lim_(x to +∞) 2/x
lim_(x to -∞) 1/x = lim_(x to -∞) 2/x
lim_(z to 0) 1/z ––– This limit does not exist because lim_(z to 0+) 1/z = +∞ does not equal lim_(z to 0-) 1/z = -∞.
lim_(z to 0) 2/z ––– This limit does not exist because lim_(z to 0+) 2/z = +∞ does not equal lim_(z to 0-) 2/z = -∞.
In other words, the two-sided limit as z goes to 0 of either function does not exist because the meromorphic functions f(z) = 1/z and g(z) = 2/z both have a pole of order 1 at z = 0. In particular, for any function which has a pole of odd order (like order 1 in this case) the limit at said pole does not exist.
en.wikipedia.org/wiki/Zeros_and_poles
For functions which have a pole of even order however the limit at said pole does exist and is either +∞ or -∞. For example:
The function h(z) = 1/(z^2) is meromorphic and has a pole of order 2 at z = 0. Its order is even, so indeed the two-sided limit exists and takes the value:
lim_(z to 0) 1/(z^2) = +∞
Looking at the (real-valued) graphs of these functions will make it very apparent as to why the two-sided limit at the pole does (doesn't) exist for functions which have an even (odd) numbered pole:
Functions of the form R(z) = 1/(z^n), where z is a real number and n is a positive integer, have rotational symmetry with respect to the origin for odd values of n (i.e. if they have an odd numbered pole at z = 0) and are symmetric with respect to the y-axis for even values of n (i.e. if they have an even numbered pole at z = 0).
At their pole (at z = 0) they diverge to positive or negative infinity, so naturally the two-sided limit at said pole can only exist if the function is symmetric with respect to the y-axis. Only then does the function equally diverge to positive (or negative) infinity from both sides as the pole is approached. And the two-sided limit only exists when the function approaches the same value from both sides.
> If we consider finite numbers only, then this equation has no solutions. If we consider infinite
> numbers also, then it has three solutions: x=0, x=infinity and x=-infinity.
Infinity is not a number, it's a concept. You cannot divide 1 by infinity.
Division by infinity (given a finite numerator) is a notion that isn't well defined in ordinary arithmetic because there's no nonzero real number that, when multiplied by infinity, gives a finite number.
Similarly 1/0 is undefined in ordinary arithmetic because there is no number that, when multiplied by 0, gives 1.
In short: polylogarithmique's equation has no solutions.
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.
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What you probably meant is that the following statements are all true:
lim_(x to +∞) 1/x = lim_(x to +∞) 2/x
lim_(x to -∞) 1/x = lim_(x to -∞) 2/x
lim_(z to 0) 1/z ––– This limit does not exist because lim_(z to 0+) 1/z = +∞ does not equal lim_(z to 0-) 1/z = -∞.
lim_(z to 0) 2/z ––– This limit does not exist because lim_(z to 0+) 2/z = +∞ does not equal lim_(z to 0-) 2/z = -∞.
In other words, the two-sided limit as z goes to 0 of either function does not exist because the meromorphic functions f(z) = 1/z and g(z) = 2/z both have a pole of order 1 at z = 0. In particular, for any function which has a pole of odd order (like order 1 in this case) the limit at said pole does not exist.
en.wikipedia.org/wiki/Zeros_and_poles
For functions which have a pole of even order however the limit at said pole does exist and is either +∞ or -∞. For example:
The function h(z) = 1/(z^2) is meromorphic and has a pole of order 2 at z = 0. Its order is even, so indeed the two-sided limit exists and takes the value:
lim_(z to 0) 1/(z^2) = +∞
Looking at the (real-valued) graphs of these functions will make it very apparent as to why the two-sided limit at the pole does (doesn't) exist for functions which have an even (odd) numbered pole:
Functions of the form R(z) = 1/(z^n), where z is a real number and n is a positive integer, have rotational symmetry with respect to the origin for odd values of n (i.e. if they have an odd numbered pole at z = 0) and are symmetric with respect to the y-axis for even values of n (i.e. if they have an even numbered pole at z = 0).
At their pole (at z = 0) they diverge to positive or negative infinity, so naturally the two-sided limit at said pole can only exist if the function is symmetric with respect to the y-axis. Only then does the function equally diverge to positive (or negative) infinity from both sides as the pole is approached. And the two-sided limit only exists when the function approaches the same value from both sides.
Maybe x=2
Maybe I’m too drunk
Edit: I’m too drunk, already found my mistake, now I think the answer is infinity
Maybe I’m too drunk
Edit: I’m too drunk, already found my mistake, now I think the answer is infinity
<Comment deleted by user>
Don’t think it matters if you give infinity a positive or negative value in this case
@Thalassokrator said in #15:
> The function h(z) = 1/(z^2) is meromorphic and has a pole of order 2 at z = 0. Its order is even, so indeed the two-sided limit exists and takes the value:
> lim_(z to 0) 1/(z^2) = +∞
That is frankly misleading if not inaccurate. If you allow z to be a complex number, then the function z -> 1/z^2 does NOT have a limit at 0.
Proof: Assume it has a limit λ. Take z=x where x is a real number and send x to 0. Then we see that λ should be +oo. Now take z=ix where x is a real number and send x to 0. This time, we see that λ should be -oo, a contradiction.
> The function h(z) = 1/(z^2) is meromorphic and has a pole of order 2 at z = 0. Its order is even, so indeed the two-sided limit exists and takes the value:
> lim_(z to 0) 1/(z^2) = +∞
That is frankly misleading if not inaccurate. If you allow z to be a complex number, then the function z -> 1/z^2 does NOT have a limit at 0.
Proof: Assume it has a limit λ. Take z=x where x is a real number and send x to 0. Then we see that λ should be +oo. Now take z=ix where x is a real number and send x to 0. This time, we see that λ should be -oo, a contradiction.
@Marcin2 said in #17:
> There exists the extended set of reals which we define as E=R sum {-infinity, infinity}, where R is the set of real numbers
> and -infinity and infinity are different points which do not belong to R. Some operations in E are defined, for instance
> 1/0=infinity and 1/infinity=0. Of course, you are right when it comes to limits, but we can also calculate using infinite
> numbers. Some expressions are indeed undefined, for instance, infinity/infinity. If we reformulate the title problem
> posed in #1, in the following way:
> Find a number x such that x/x=2,
> then it has no solution even in E.
It doesn't matter because you could come up with a an extended set F which contains R and a special element x that satisfies x/x=2 and all other expressions involving x are left undefined.
> There exists the extended set of reals which we define as E=R sum {-infinity, infinity}, where R is the set of real numbers
> and -infinity and infinity are different points which do not belong to R. Some operations in E are defined, for instance
> 1/0=infinity and 1/infinity=0. Of course, you are right when it comes to limits, but we can also calculate using infinite
> numbers. Some expressions are indeed undefined, for instance, infinity/infinity. If we reformulate the title problem
> posed in #1, in the following way:
> Find a number x such that x/x=2,
> then it has no solution even in E.
It doesn't matter because you could come up with a an extended set F which contains R and a special element x that satisfies x/x=2 and all other expressions involving x are left undefined.
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