It appears that the Latex commands have been lost during the transition? So my latex math isn't showing up as it use to. Don't know if you can read my equations above.
It looks like the blackslash command doesn't work anymore - at least when I try to edit. Aha, you have to double the backslash like \\
Testing:
\[
\left( \sum_{k=1}^n a_k b_k \right)^{!!2} \leq
\left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)
\]

The Fly
It appears that the Latex commands have been lost during the transition? So my latex math isn't showing up as it use to. Don't know if you can read my equations above.

It looks like the blackslash command doesn't work anymore - at least when I try to edit. Aha, you have to double the backslash like \\

Testing:

\[

\left( \sum_{k=1}^n a_k b_k \right)^{!!2} \leq

\left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)

\]

Edited July 18th, 2019
by The Fly

The backslash command \ for Latex disappears when you re-edit a post to correct it.

The Fly
The backslash command \ for Latex disappears when you re-edit a post to correct it.

Posted July 18th, 2019
by The Fly

I found a general solution, but it isn't a single formula -- I can't figure out how to piece the data together into one.
I started with the idea that, well, -1 is just i[sup]2[/sup]:
[*]\[ \large \frac{i^m + i^2}{i^n + i^2} \][/*]
I then did a lot of unnecessary work, and eventually realized that you can just [g]modulus[/g] m or n by four because i[sup]5[/sup] is just i*i[sup]4[/sup], or put another way, it's 1*i, which is just i.
This means that m and n collapse to either 1 or 3.
I noticed that you can reduce i[sup]1[/sup] and i[sup]3[/sup] in this way:
[*]\[ \large \frac{i^3 + i^2}{i^1 + i^2} \][/*]
[*]\[ \large \frac{i^2 ( i + 1 )}{i ( i + 1 ) } \][/*]
The i+i cancels out, and you're left with some power of i.
I then looked all cases, changing 1 to i[sup]4[/sup] where appropriate:
* If m is 1 and n is 3, the answer is i[sup]-1[/sup]
* If m is 3 and n is 1, the answer is i[sup]1[/sup]
* If m is 1 and n is 1, the answer is i[sup]4[/sup]
* If m is 3 and n is 3, the answer is i[sup]4[/sup]
Given more time, I could probably reduce this into a single formula (since that's kinda been my specialty lately), but I couldn't quite figure out how to put it together.

Xhin
I found a general solution, but it isn't a single formula -- I can't figure out how to piece the data together into one.

I started with the idea that, well, -1 is just i

2:

\[ \large \frac{i^m + i^2}{i^n + i^2} \]
I then did a lot of unnecessary work, and eventually realized that you can just

modulus m or n by four because i

5 is just i*i

4, or put another way, it's 1*i, which is just i.

This means that m and n collapse to either 1 or 3.

I noticed that you can reduce i

1 and i

3 in this way:

\[ \large \frac{i^3 + i^2}{i^1 + i^2} \]
\[ \large \frac{i^2 ( i + 1 )}{i ( i + 1 ) } \]
The i+i cancels out, and you're left with some power of i.

I then looked all cases, changing 1 to i

4 where appropriate:

If m is 1 and n is 3, the answer is i-1

If m is 3 and n is 1, the answer is i1

If m is 1 and n is 1, the answer is i4

If m is 3 and n is 3, the answer is i4

Given more time, I could probably reduce this into a single formula (since that's kinda been my specialty lately), but I couldn't quite figure out how to put it together.

Edited July 18th, 2019
by Xhin

That backslash thing is extremely annoying. Will fix.

Xhin
That backslash thing is extremely annoying. Will fix.

Posted July 18th, 2019
by Xhin

Nevermind, figured it out.
[go]i[sup][sup][ (m%4)-(n%4) ]/2[/sup][/sup][/go]
Basically, this takes advantage of the fact that i[sup]0[/sup] is the same as i[sup]4[/sup]

Xhin
Nevermind, figured it out.

i[ (m%4)-(n%4) ]/2
Basically, this takes advantage of the fact that i

0 is the same as i

4
Edited July 18th, 2019
by Xhin

If n is 0 or a multiple of 4 the denominator is 0 so the fraction ((i^m)-1)/0 is meaningless; [g]not a power of [i]i[/i][/g].
Otherwise, if m is 0 or a multiple of 4 the numerator is 0 so the fraction 0/((i^n)-1) is 0. [b][o]This is not a power of [i]i[/i][/o].[/b]
Otherwise if m and n are congruent modulo 4 then the numerator and denominator are equal so the fraction has value 1. This is a power of [i]i[/i].
Otherwise you have either (i-1)/(-i-1) or (-i-1)/(i-1)
-i = i^3 so (-i-1)/(i-1)=((i^3)-1)/(i-1)=(i^2)+i+1 = -1 + i +1 = i.
(i-1)/(-i-1) is 1/i is -i. These last two values are both powers of [i]i[/i].
The fraction never evaluates to -1. That’s the only power of [i]i[/i] to which the expression does not evaluate for any m or n.
Can you calculate i^i ?
Can you prove it’s a real number?

chiarizio
If n is 0 or a multiple of 4 the denominator is 0 so the fraction ((i^m)-1)/0 is meaningless;

not a power of *i*.

Otherwise, if m is 0 or a multiple of 4 the numerator is 0 so the fraction 0/((i^n)-1) is 0.

**This is not a power of ***i*.
Otherwise if m and n are congruent modulo 4 then the numerator and denominator are equal so the fraction has value 1. This is a power of

*i*.

Otherwise you have either (i-1)/(-i-1) or (-i-1)/(i-1)

-i = i^3 so (-i-1)/(i-1)=((i^3)-1)/(i-1)=(i^2)+i+1 = -1 + i +1 = i.

(i-1)/(-i-1) is 1/i is -i. These last two values are both powers of

*i*.

The fraction never evaluates to -1. That’s the only power of

*i* to which the expression does not evaluate for any m or n.

Can you calculate i^i ?

Can you prove it’s a real number?

Edited July 25th, 2019
by chiarizio

Can you calculate i^i ?
Can you prove it’s a real number?

chiarizio
Can you calculate i^i ?

Can you prove it’s a real number?

Posted March 20th
by chiarizio