Googling about for negative triangular numbers doesn’t yield any responses. Yet it is easy to generate the series by summing negative integers just as you do the positive ones:
0 -1 -3 -6 -10 etc.
There is an interesting square root algorithm for perfect squares:
S^.5 = 2*((S/2)-T) Where T is the nearest triangular number less than S/2
Applied to negative triangular numbers:
-1^.5 = 2((-1/2)-0) = -1
This plots as a square in quadrant 3 with sides of -1 and an area of -1.
-25^.5 = 2*((-25/2)-10) = -5
This plots as a square in quadrant 3 with sides of -5 and an area of -25.
Seems like the kind of thing that interests mathematicians, but I don’t find any references.
The absence of negative numbers in most areas of classical number theory is largely a historical artifact of its development in ancient Greek mathematics: number-theoretic results on the natural numbers were studied long before negative numbers came into standard use.
And since most elementary number-theoretic results can be trivially extended to the negative integers, as in your example, mathematicians don’t bother with them much.
If you found out something about the properties of negative integers that is not already covered by a simple reformulation of corresponding descriptions of the natural numbers, that would definitely get mathematicians’ attention. But don’t hold your breath.
The length of sides and areas of ordinary shapes like triangles are always positive.
No real point in noting the negative of a very well known sequence.
Even the amazing The On-Line Encyclopedia of Integer Sequences doesn’t carry it. Plugging in the first few terms gives a sequence that starts off right but goes off in a different direction and then the regular triangle numbers.
If OEIS doesn’t specifically ID this, it isn’t worth studying on its own. Just make the numbers positive and go to town on generating identities.
I gotta say “Huh?” as well. You show a formula that produces the wrong answer for sqrt(-1) and then you say it “doesn’t require” the correct answer. I have no idea what you mean by that.
One could also argue that if T(0) = 0, T(1) = 1, T(2) = 3, T(n) = 0 + 1 + … + n, then when you sum from zero to a negative index you pick up a minus sign, so T(-1) = 0, T(-2) = 1, etc. (T(n) = (n^2 + n)/2 for all n in this case), and the “negative triangular numbers” are the same as the positive triangular numbers after a change of index.
Why not? In quadrant 3 there is a square with an altitude of -5 and a base of -5. Is the nature of it’s area identical to a 5 by 5 square in quadrant 1?
Using the algorithm above:
-25^.5 = 2*((-25/2)-(-10) = -5
(I am aware that this contradicts elementary math)
False arithmetic is… false. Note that the square of an integer number is never negative. Other than that, your identity seems to amount to n = 2(n^2/2 - (n^2 - n)/2) or n = 2(-n^2/2 + (n^2 + n)/2).
I don’t know much math, but I’m getting the impression that what happened here is a demonstration of the difference between ‘i’ (imaginary numbers) and what I’ll call ‘f’ (fake numbers).
One thing we do see in all this is that under some circumstances it makes sense to define the sum of f(k) as k runs from 0 to n, where n < 0, as -f(-1) - f(-2) - … -f(-n+1).
If you are expecting square roots with what you have in the first post that is why you are having problems finding information.
square sqrt(-5) == √5*ⅈ
The reason you aren’t finding anything is because you are trying to avoid imaginary numbers. Just replace the term imaginary with complex in your mind and you will find a lot.
There is nothing imaginary about i, and it is used everywhere.
Nature of it’s area? That’s a philosophical discussion of some kind.
Starting with a positive-area square in quadrant 1, we can flip it across the X axis to put it in quadrant 4. It now has a negative height and positive width, and so a negative area. Or, we can just consider that we’ve taken the mirror-image of the shape and that it has negative area for that reason.
We can then flip across the Y axis to give it a negative width. Negative width times negative height makes positive area. Alternatively, the negative-area shape mirrored goes back to being positive-area. Both ways of looking at it are consistent.
No it doesn’t . It just trivially just restores the negative to the front after taking the square root of the absolute.
What if I square root your bank balance ? If its in the +ve thats fine.
If its negative, what , do you suddenly turn a debt (you owing money) into +ive balance, you being owed money ?
Its already been said, the -ve version of your triangular series is trivial. And its trivial to have "if ( x < 0 ) sqrt_x= -1 * sqrt(-1 * x ) else sqrt_x = sqrt(x) "
Its overcomplicating it to say he needs to use complex numbers. Its still going to require an arbitrary correction for the negative case.
Also I should point out, be careful about floating point on computers.
Floating point numbers in computers are finite. Being finite numbers a large number of postulates like they are not associative nor distributive.
These rules do not necessarily hold in computer floating point arithmetic.
Associative: A*(BC) = (AB)*C
Distributive: (A+B)C = AC + BC
-25**(1/2) == -5 only in floating point.
But that .5 in the -25^.5 you have above is really significand * base ^ exponant or 510**-1
You really need to use either complex numbers, or a symbolic math function to mess around with these types of numbers on a computer.
-25^.5 == sqrt(-25) ==5⋅ⅈ
-1^.5 == sqrt(-1) == ⅈ
That last one is the identity of ⅈ and this all just works in the complex plane.
Floating point or not, many calculators and computers are going to evaluate -25**(1/2) or -25^(1/2) or -25^.5 as -5, simply because the order of operations they’re programmed to follow specifies that exponents are evaluated before the -.
