for y∈[0,1] chance of winning supporter
People who provide invalid answers are automatically disqualified
People who provide invalid answers are automatically disqualified
forum
Guess the x∈ℂ²
posted
Total Posts
44
Topic Starter
x = (i, -i)
I've completely forgotten complex numbers which I was taught more then a year ago.
I've completely forgotten complex numbers which I was taught more then a year ago.
(621, 926)
3.1415762986757277
Topic Starter
nopeRhythm32 wrote:
x = (i, -i)
I've completely forgotten complex numbers which I was taught more then a year ago.
nopeCorne2Plum3 wrote:
(621, 926)
disqualifiedPenguin wrote:
69
disqualifiedAireunaeus wrote:
3.1415762986757277
(626907, 4635891)
Jangsoodlor
Sorry im not into eldritch linguistics
the number you are looking for is the friends we made along the way abraker <3, Friendship always prevails!
Topic Starter
nopeCorne2Plum3 wrote:
(626907, 4635891)
nopez0z wrote:
(-∞-2i, ∞+2i)
Wellllll technically speaking there might be a quantum particle inside of friends made along the way bearing the normalized spin characterstic of the mystery value. Tho this does make me wonder what kind of tensor dimensionality would be needed to represent friendshipBlushing wrote:
the number you are looking for is the friends we made along the way abraker <3, Friendship always prevails!
Fix Tit-bot or shitpost on OT
Abraker chooses the latter
I hate my chud life
Abraker chooses the latter
I hate my chud life
(621, 621i)
you ask an impossible riddle. friendship cannot be so simply quantified as a quantum particle. friendship SKEWS the normal spin characteristics to always align perfectly to where the friendship is observed to be at, therefore always being the right value. Heh, you mistake my answer for simple physics, as if numbers can easily quantize the deeper understanding that is friendship. When one goes through the unending cycle of friendship they are always and forever entangled with each other, always making small waves through the other even if in the past. Memories and memories building up to the sum of the individual, building up to an infinite amount of numbers that when observed always point to the correct number. therefore... friendship is always the answer, Dr. braker.abraker wrote:
nopeCorne2Plum3 wrote:
(626907, 4635891)nopez0z wrote:
(-∞-2i, ∞+2i)Wellllll technically speaking there might be a quantum particle inside of friends made along the way bearing the normalized spin characterstic of the mystery value. Tho this does make me wonder what kind of tensor dimensionality would be needed to represent friendshipBlushing wrote:
the number you are looking for is the friends we made along the way abraker <3, Friendship always prevails!
Corne try not to make porn reference challenge.Corne2Plum3 wrote:
(621, 621i)
(0, 0)
Topic Starter
I never said friendship is a spinnor. A spinnor lacks the degrees of freedom to represent friendship. Something on the order of a very high dimensional tensor would. Also you imply friendship consists of infinite number of memories even tho the human mind is finite. It be more like a moving window of memories. You would forget the minor details the longer ago it has been. Meanwhile the individual self is a beast best described by chaos theory. That said, if you lay out the entire manifold onto which individualism can ve mapped onto, surely there are coupling points, we'll call them affinities for friendships, that intersect at various points of other such manifolds.Blushing wrote:
you ask an impossible riddle. friendship cannot be so simply quantified as a quantum particle. friendship SKEWS the normal spin characteristics to always align perfectly to where the friendship is observed to be at, therefore always being the right value. Heh, you mistake my answer for simple physics, as if numbers can easily quantize the deeper understanding that is friendship. When one goes through the unending cycle of friendship they are always and forever entangled with each other, always making small waves through the other even if in the past. Memories and memories building up to the sum of the individual, building up to an infinite amount of numbers that when observed always point to the correct number. therefore... friendship is always the answer, Dr. braker.
Corne2Plum3 wrote:
(621, 621i)
not even closeCorne2Plum3 wrote:
(0, 0)
It seems like you're asking for a value of
𝑥
∈
𝐶
2
x∈C
2
(the complex plane squared) for
𝑦
∈
[
0
,
1
]
y∈[0,1], but the problem as stated is somewhat ambiguous because the relationship between
𝑥
x and
𝑦
y isn't fully clear.
Could you provide more details or clarify the relationship between
𝑥
x and
𝑦
y? For instance, are you trying to solve an equation involving
𝑥
x and
𝑦
y, or is there a specific function or constraint you're working with? This will help in giving a more precise answer
𝑥
∈
𝐶
2
x∈C
2
(the complex plane squared) for
𝑦
∈
[
0
,
1
]
y∈[0,1], but the problem as stated is somewhat ambiguous because the relationship between
𝑥
x and
𝑦
y isn't fully clear.
Could you provide more details or clarify the relationship between
𝑥
x and
𝑦
y? For instance, are you trying to solve an equation involving
𝑥
x and
𝑦
y, or is there a specific function or constraint you're working with? This will help in giving a more precise answer
Topic Starter
disqualified for gpt'ing my braincellsLaxxer wrote:
It seems like you're asking for a value of
𝑥
∈
𝐶
2
x∈C
2
(the complex plane squared) for
𝑦
∈
[
0
,
1
]
y∈[0,1], but the problem as stated is somewhat ambiguous because the relationship between
𝑥
x and
𝑦
y isn't fully clear.
Could you provide more details or clarify the relationship between
𝑥
x and
𝑦
y? For instance, are you trying to solve an equation involving
𝑥
x and
𝑦
y, or is there a specific function or constraint you're working with? This will help in giving a more precise answer
(42, 918)
Topic Starter
nopeYoisaki Kanade wrote:
(42, 918)
nope, although this seems to be the first actual attemptcatzlucky wrote:
(sqrt(2)*(i + 1)/2, sqrt(2)*(i - 1)/2)
(69, 69i)
You're probably disqualified, abraker is looking for a pair of complex numbers, which you didn't provided the first timefluffpup wrote:
abraker you completely ignored my answer
----
(3, 5i)
Topic Starter
Serraionga wrote:
(69, 69i)
nopeCorne2Plum3 wrote:
You're probably disqualified, abraker is looking for a pair of complex numbers, which you didn't provided the first timefluffpup wrote:
abraker you completely ignored my answer
----
(3, 5i)
also if disqualified person guesses it nobody gets supporter. So if you are ignored, you def didn't get it
(1+i,1-i)
Username change?Hoshimegu Mio wrote:
(1+i,1-i)
(sqrt(2)/2 + i*sqrt(2)/2, 0)
Maths is so weird
Like some old guy in a desert just decided that he will get y∈[0,1] camels one day
Actual modern day witchcraft
Like some old guy in a desert just decided that he will get y∈[0,1] camels one day
Actual modern day witchcraft
Yes I changed it today.Corne2Plum3 wrote:
Username change?
Topic Starter
nopeHoshimegu Mio wrote:
(1+i,1-i)
also who tf are you?
nopeCorne2Plum3 wrote:
Username change?Hoshimegu Mio wrote:
(1+i,1-i)
(sqrt(2)/2 + i*sqrt(2)/2, 0)
Nooo abrak forgot about me I'm sad now...abraker wrote:
nopeHoshimegu Mio wrote:
(1+i,1-i)
also who tf are you?
pretty cool, i should change mine sometimeHoshimegu Mio wrote:
Yes I changed it today.Corne2Plum3 wrote:
Username change?
Lets all change our usernames
I remember your 古代星 name.z0z wrote:
pretty cool, i should change mine sometimeHoshimegu Mio wrote:
Yes I changed it today.Corne2Plum3 wrote:
Username change?
x = (sqrt(y)((1+i)/sqrt(2)), sqrt(1-y)((1-i)/sqrt(2))) for y∈[0,1]
(6,9)
hehey i changed mine yesterday!Laxxer wrote:
Lets all change our usernames
this just got fun!abraker wrote:
also if disqualified person guesses it nobody gets supporter.