8:53 AM
-13

So, when the zeta function was created, were they aware that i^4 equals 1, therefore (by commutation) i = 1? Were the various workarounds (if they even exist) available in his time or was he just ignoring that imaginary numbers arent actually valid within standard axioms? And (apparently) the rie...

-4

With n >2, no possitive integers X,Y,Z $X^n+Y^n=Z^n$ so $Z^n-Y^n=X^n$ $Z^n-X^n=Y^n$ The little known notable product difference nth power says that $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^{2}+...+b^{n-1})$ and i have observed that $a^n-b^n=(a-b)(a^{n-1}+ab\frac{a^{n-2}-b^{n-2}}{a-b}+b^{n-1})$ an...

9:13 AM
-3

Does the sum of the squares of the unit vectors sum to 1 when there is an equation variant like the following? \begin{aligned} \mathbf{Q}\mathbf{a} &=a(\cos \theta) \mathbf{I}+(1-\cos \theta)\mathbf{A} \cdot \mathbf{a} \\ &=\mathbf{a}(\cos \theta) \mathbf{I}+(1-\cos \theta)\left[...

-3

1/16, 1/8, 3/16, 1/4, 5/16, 3/8, 7/16 ___, ___, ___

[ SmokeDetector | MS ] Link at beginning of body (42): removing i,j,k coefficients by josh.bauer2020 on math.SE

3 hours later…
12:36 PM

12:50 PM

5 hours later…
6:20 PM
@XanderHenderson The OP has edited that question a bit. He says: "I have edited it to clarify there is a question being asked".
This was mentioned in another chatroom. Since the question was pointed out by you, I thought it might be reasonable to let you know - in case you have something to add to that.

5 hours later…
11:12 PM
[ SmokeDetector | MS ] Link at beginning of body (42): Help me to determine when a series converge and the limit of an integral function by Qcc on math.SE

is it possible to delete a question if no one upvoted it and the answers just point out I was ignorant on the subject?

11:28 PM
@JorgeFernández-Hidalgo first, you should not deface the post. Second I don't think the question is that bad. I'd leave it around (unless somebody comes up with a duplicate).

Fine whatever
I never should have even asked it in the first place
deep down I knew it would just worsen my anxiety
I'll remember to stick to just answering questions

@JorgeFernández-Hidalgo if you want you can ask to be disassociated from the post. But again I think the question is not bad.

that would be awesome I would really appreciate it
if it isn't too much trouble
sorry for the trouble

@JorgeFernández-Hidalgo I'll need to ask SE to do it, thus it might take a little while.
But it is not difficult to do.

Oh, don't worry about it then

11:35 PM
@JorgeFernández-Hidalgo Alright. If you change your mind later feel free to raise a flag.