$3x-x=7$
#help-13
sonic thistle
wraith daggerBOT
masteroogway0599
sonic thistle
?
cloud sinew
?
sonic thistle
This is your question?
earnest steppe
bro ur question makes 0 sense
cloud sinew
ye
sonic thistle
Ok
cloud sinew
bro
i will say again
question is
check which of the following is the solution of 3x-x=7
options are a (2,3) b (-1,6) c (1.-4) d (0,7)
cedar kilnBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
cloud sinew
BRO ITS NOT A HW
sonic thistle
What is it
cloud sinew
I FINISHED EXAM SO IM CHECKING ANSWER
earnest steppe
bro it's a fkn linear equation and your options are coordinates
you live in the matrix
cloud sinew
thats the question
sonic thistle
You sure there wasnt a "y" in your equation
earnest steppe
probably 3x - y = 7
cloud sinew
oh wait
sonic thistle
Yeah
cloud sinew
it was
earnest steppe
cloud sinew
3x-y=7
sonic thistle
Well yeah
earnest steppe
just substitute each option
sonic thistle
You have to replace x with the first number and y with the second number
And check if it's equal to 7
cloud sinew
i think its (1, -4)
sonic thistle
Yes
wet rock
Yea
cloud sinew
nice
sonic thistle
3×1-(-4)=7
cloud sinew
1 more queston
wet rock
@cloud sinew - + - = +
cloud sinew
the value of p(x) =5x square - 4x+3 when x = -1 is
cloud sinew
i got answer 4 is it 4
cloud sinew
ok
celest plinth
I think it's -4
wraith daggerBOT
masteroogway0599
sonic thistle
Ok
celest plinth
tbh, I still didn't get the question lol
cloud sinew
**masteroogway0599**
this is the question where x = -1
wraith daggerBOT
IamHappyLittleBoy
cloud sinew
no
cloud sinew
**masteroogway0599**
it is this
celest plinth
So can you type again the question ?
cloud sinew
where x = -1
celest plinth
it is this
But if so, what will be the result ?
cloud sinew
i will say question again
celest plinth
thx
cloud sinew
$(5x)^2 - 4x +3$ when x = -1
wraith daggerBOT
Suuuuuuuuus
cloud sinew
there are 4 options
celest plinth
ayt
cloud sinew
a 3 b -12 c 12 d 4
wet rock
Is 7 is in question or not?
cloud sinew
Is 7 is in question or not?
no
cloud sinew
no
Thats not the question
the value of p(x) = $(5x)^2 - 4x +3$ when x = -1 options = a 3 b -12 c 12 d 4
wraith daggerBOT
Suuuuuuuuus
cloud sinew
5(-1)² -4(-1) + 3
i m not getting thats why im asking
sonic thistle
Whats 5×-1
cloud sinew
Whats 5×-1
u mean 5x(-1)
sonic thistle
Yes
cloud sinew
5x
sonic thistle
No
cloud sinew
-5x
cloud sinew
**Suuuuuuuuus**
then what is this questions answer
sonic thistle
Bro
cloud sinew
I wrote × not x
-5
cloud sinew
yes
celest plinth
25
celest plinth
lol
cloud sinew
its -1 x -1 right?
wet rock
@cloud sinew is 12 is in option?
sonic thistle
What?
cloud sinew
<@910181884204564532> is 12 is in option?
ye
sonic thistle
Are you listening
cloud sinew
Yes
cloud sinew
25
wet rock
5x² - 4x+3 at x=-1
→ 5(-1)² -4(-1) + 3
= 5 + 4 + 3
cloud sinew
yes
cloud sinew
-4
sonic thistle
-4x+3
wet rock
5x² - 4x+3 at x=-1
→ 5(-1)² -4(-1) + 3
= 5 + 4 + 3
@cloud sinew u need to just put x = -1 in question
cloud sinew
4 + 3
wet rock
Yea
wet rock
@cloud sinew what's add?
sonic thistle
Tell me the answer
wet rock
Lol
sonic thistle
Wrong
wet rock
Oh!?
celest plinth
Your calcul makes (5x-1) x (5x-1) into 5x((-1) x (-1))
cloud sinew
then what is the answer
quartz scaffold
according to the question u will get 32 but that is not available in the options
cloud sinew
ye
quartz scaffold
maybe there is a mistake in the question
nova snow
I think I did something wrong but idk where?
cedar kilnBOT
I think I did something wrong but idk where?
dreamy sleet
Why do you think it’s wrong
nova snow
i checked the answer page and it shows something completely different
dreamy sleet
what does it show
nova snow
dreamy sleet
have you tried rearranging your equation
dawn junco
almost diabolic
nova snow
no i only leave it in y=mx+b form
is that what they get if they change the form
is that point slope or somethin?
dreamy sleet
is that point slope or somethin?
it is in general form
dreamy sleet
is there a specific advantage to putting it in that form instead of y=mx+b?
no, my personal opinion is that general form is the most useless
but it is the form that the books want for some reason
nova snow
yeah i find y=mx+b a lot more convenient and intuitive 🤔
dreamy sleet
yeah i find y=mx+b a lot more convenient and intuitive 🤔
yes, slope-intercept form best form
dreamy sleet
almost diabolic
Why is it diabolic?
dawn junco
555
dreamy sleet
is that what they get if they change the form
could you change the form of the answer you got?
nova snow
is that the ax+by+c=0?
dreamy sleet
I mean change the form of y = (-1/9)x + 70/3
nova snow
not sure how to do that
glossy shoal
nova snow
for this question do i first differentiate and then set the derivative equal to 0
and then solve for x there
south tundra
Yes
nova snow
it becomes -5?
nova snow
so only x=+-3 is a solution?
glossy shoal
yes
glossy shoal
dy/dx=3
nova snow
ah okay
glossy shoal
because dy/dx is the slope of tangent of y at specific x
nova snow
so its just x=0
cedar kilnBOT
cedar kilnBOT
Available help channel!
quasi plover
cedar kilnBOT
cedar kilnBOT
Show your work, and if possible, explain where you are stuck.
quasi plover
Show your work, and if possible, explain where you are stuck.
ok
So i choose this because i tried to give x a value
first substitute with 1
y = 1/4 * 1 - 2
y = -7/4
y = 3 * 1 - 7/5
y = 8/5
Second substitute with -1
y = 1/4 * -1 -2
y = -9/4
y = 3 * -1 - 7/5
y= -22/5
now i see that the first line will be in the III part and in the IV part
for the 2nd line I see that it'll be in the I part and in the III part
So in conclusion I think its in III, IV or in I,III part
which in the choice there dont have I, III
so I choose the III, IV
cedar kilnBOT
@quasi plover Has your question been resolved?
cedar kilnBOT
@quasi plover Has your question been resolved?
cedar kilnBOT
@quasi plover Has your question been resolved?
wraith basin
cedar kilnBOT
@quasi plover Has your question been resolved?
spare meadow
Can somebody help me understand this boolean algebra simplification. I don't understand how a 0 was derived in the second step:
carmine bronze
$\bar{B}B = 0$
wraith daggerBOT
Kookiemon
spare meadow
Ok, and is the (B')' = B?
carmine bronze
Yes.
spare meadow
Ok, and then why in the last step is the answer A + (B)' instead of A(B)'?
spare meadow
Ok, thanks @carmine bronze
carmine bronze
yw
spare meadow
.close
peak fog
Where do cos and sin come from when finding the solution to something like this
cedar kilnBOT
Where do cos and sin come from when finding the solution to something like this
@peak fog Has your question been resolved?
dawn junco
Where do cos and sin come from when finding the solution to something like this
your roots should be 2i and -2i btw
the basic idea is that, when you have a complex solution U(t) + iV(t) to a linear DE, then U(t) and V(t) are also solutions to the DE (U(t) and V(t) are real functions)
from that basis of complex solutions you have (i.e. e^-0x, e^-2ix, e^2ix), we can extract a basis for the real solutions
by splitting apart the real and imaginary parts of those 3 complex functions, and extracting a 3D basis of solutions from it (since the equation is 3rd order, we expect a 3D space of solutions)
this is what we get
now from those 6 solutions
- 0 is out of the question (it can't be part of a basis)
- there are other linear dependencies (cos(-2x) = cos(2x), sin(-2x) = -sin(2x)), so we can get rid of, say cos(-2x) and sin(-2x) since they're redundant
now we're pretty much left with 1, cos(2x), sin(2x)
those are linearly independent
so they form a basis of the real solutions to the DE
from that you directly get that the general real solution is of the form A(1) + B(cos(2x)) + C(sin(2x))
voilà
peak fog
so prior to all that I'm having trouble recognizing when the roots are complex which if I understand correctly is what leads us to this solution
dawn junco
yeah
if your solutions were already real, you wouldn't be asking this question hehe
@peak fog anyway still have questions ?
peak fog
Not anymore, while trying to figure out how to word my question it answered itself
thanks for the help
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
odd wadi
You are standing on planet earth. You walk one mile south, one mile west and one mile north. You end up exactly where you started. Where are you?
So there is an obvious answer to this question, which is the north pole.
There is also some not so obvious ones which are near the south pole.
Basically if you at at least a mile away from the south pole when you start, then there are various intervals which would also be valid.
Walking west round the South Pole once, walking west round the south twice, and so on, until the final answer which would basically require turning round on the spot thousands of times.
I want to know what the distances are from the South Pole that would be valid answers.
I want to know how many times you would need to walk in a circle for each distance.
I want to know how many total options there are.
This seems like impossibly hard math for me but I am really curious to figure it out.
violet flume
seems to me like there would be infinitely many
maybe 
wait
yea, i think infinitely many
because as long as you wind up on a uhh
is it latitude
,w latitude
odd wadi
Maybe if we put a realistic cutoff in place to where we said you are technically no longer walking west
wraith daggerBOT
odd wadi
like if you are spinning on the spot
violet flume
well then your cut off is going to determine the number of solutions
because you can land in a spot where the circumference at that latitude is 1 mile
or 1/2 mile
or 1/4 mile
or what 1/5 mile
any clean fraction, i think
maybe youre saying well it doesnt make sense to consider looping the pole 1064 times and 2034 times as different places
but theyre different solutions on paper
odd wadi
Yeah maybe we could cut it off at looping the pole 1000 times or something
even if the answer is technically infinite
violet flume
well then theres your answer
odd wadi
"I want to know what the distances are from the South Pole that would be valid answers.
I want to know how many times you would need to walk in a circle for each distance.
I want to know how many total options there are."
violet flume
funny you caught that, i was thinking thered be only 2
odd wadi
funny you caught that, i was thinking thered be only 2
Yeah for a long time I only thought of 2
and then it dawned on me
violet flume
hmm my spherical geometry is weak
you want to answer the question
at what latitude is the circumference about a sphere 1/n of a mile
whats the relationship between latitude and circumference?
odd wadi
whats the relationship between latitude and circumference?
I was thinking it might be something related to pie
but honestly my maths skills are not good at all
violet flume
im sure theres a fantastic clean way
my thought would be
angle to radius
radius to circumference
lets say the equator is what, 0 degrees latitude?
i dont know what the convention is
but it works in my head
if you set it that way, say the earths radius at the circumference is R
then the radius at some latitude theta degrees from the equator would be Rcos(theta)
which gives your circumference as uhh
2 pi R cos(theta)
then if we want to know where $\frac1n = 2\pi R \cos \theta$ for some integer n
wraith daggerBOT
janniku
violet flume
no
hrm
my cat is screaming and its really distracting im gonna blame that while i think for a sec
theres a unit issue
we want R in miles
right?
hmm idk id have to think about this more
cedar kilnBOT
@odd wadi Has your question been resolved?
cedar kilnBOT
@odd wadi Has your question been resolved?
cedar kilnBOT
@odd wadi Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
jolly sable
Let G be a group with identity element e. Let H and K be normal subgroups of G such that
H∩K={e}. Then what is H×K
jolly sable
10
short blade
oh i have no idea what that is lol
gilded elm
it's basically the same thing as external direct product
that's the definition, lol
some authors denote external direct products differently
not sure what help is needed here, this is a straightforward question?
cedar kilnBOT
@jolly sable Has your question been resolved?
crimson delta
well then look them up?
do you know what internal and external sums of vector spaces are? similar thing here
cedar kilnBOT
@jolly sable Has your question been resolved?
limpid plume
Let G be a group with identity element e. Let H and K be normal subgroups of G s...
H x K isomorphic to HK in this case, that's what most group theory textbooks say
The isomorphism goes like (h, k) -> hk
You can check like Dummit Foote (even though I hate this book), Herstein, Artin, or whatever there is with you, it's listed everywhere
For Groups this is how External Direct Product and Internal Direct Product relate by isomorphism
Also by the way that question 10 is bullshit
If they actually made that question number 1 is also the correct answer
flint plinth
If they actually made that question number 1 is also the correct answer
3 is perfectly valid as well
jolly sable
It is an old paper of TGT mathematics
solid juniper
are the bottom questions part of that
limpid plume
It's probably a different subject
limpid plume
jolly sable
184 and 186
gilded elm
also these questions are written so weirdly wtf
cedar kilnBOT
@jolly sable Has your question been resolved?
limpid plume
Not the direct sum being written as External Direct Product 😭😭😭
If the Groups are finite that's equivalent though
cedar kilnBOT
@jolly sable Has your question been resolved?
remote salmon
muje toh ye dono hi sawal samajh ni aaya rikuu
jee wale ho kyaa?
wraith daggerBOT
jolly sable
Here I tried this nth term will be sum (n+2)÷2^n
crimson sedge
2^(n-1)
jolly sable
Lim n tends to infinite (n+2)÷2^n
crimson sedge
nth term will be $\frac{n+1}{2^{n-1}}$
jolly sable
Here I tried this nth term will be sum (n+2)÷2^n
This is what I wrote
crimson sedge
oh my bad
jolly sable
How to solve it?
wraith daggerBOT
3998
jolly sable
Then?
crimson sedge
How to solve it?
I'm wondering the same
it seems obvious that this is converging
but I have little to no knowledge about sums 😢
solemn torrent
I don't have knowledge on the subject either but i was thinking of finding an expression for like the kth partial sum and then taking the limit of that as k tends to infinity
or some convergence test
jolly sable
I tried comparison test an+1/an
wooden geode
Let G be a group with identity element e. Let H and K be normal subgroups of G s...
is this still the question
jolly sable
is this still the question
Of course
jolly sable
also how are Q1, Q2 even related to math 💀
No
wooden geode
solemn torrent
your questions are all over the place
jolly sable
your questions are all over the place
What does it mean?
limpid plume
muje toh ye dono hi sawal samajh ni aaya rikuu
aani bhi nahi chaiye faltu ka sawaal hain
You know what, if you really want to try practicing for the love of any deity up above please find any other sample test paper
These are horrible
wraith daggerBOT
Riku
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
limpid plume
All of these is available in any introductory group theory textbook though, you can revisit that, should be helpful
limpid plume
Good luck 
jolly sable
limpid plume
But it's very complicated
try maybe some easier books? hmm
jolly sable
Please suggest
limpid plume
i usually suck at these but this sounds like combination of two infinite series idk
oh wait
is this an AP
but no that's absurd
jolly sable
It's arithmetic geometric series
jolly sable
What
elfin orchid
slate lintel
oh the numerators are going up by 1 and the denominators are doubling
vagrant elbow
$\sum_{k \geq 0} \frac{k + 2}{2^k}$
vagrant elbow
oh the numerators are going up by 1 and the denominators are doubling
Makes more sense for the denominator to be a geometric series, if it's doubling the sum doesn't converge
wraith daggerBOT
neonperseus
vagrant elbow
Ah wait they're the same in this context
slate lintel
yes 
cedar kilnBOT
@jolly sable Has your question been resolved?
jolly sable
**neonperseus**
Here
humble fiber
Here
hey , just multiply the common ratio of gp with sum and subtract it from the original sum ... you'll get your answer
jolly sable
hey , just multiply the common ratio of gp with sum and subtract it from the ori...
What??
But this is not GP
Can you write it please? I didn't understand what you mean by sum here?
humble fiber
But this is not GP
yeah it's AGP
there's the base method of solving AGP and you can even deduce a general formula by this method to solve AGP , but that formula isn't worth remembering
but this question seems easy , cause you can eliminate then options just by looking
jolly sable
What you did in 3 lines...by 1 and 2 equations? If you substrate then s/2 okay but right side 2?
Right side i guess you kept 2 and substrate next terms 3/2-1/2, 4/2^2-3/4
jolly sable
**neonperseus**
I was asking to solve the nth term by the way
jolly sable
**neonperseus**
No solution?
jolly sable
So here I did this
Sum 0 to infinity (k+2)× sum 0 to infinity 1/2^n....
2+3+4.....×1+1/2+1/2^2+....
jolly sable
<@&286206848099549185>
tacit loom
ur problem not solved ?
jolly sable
No
tropic oxide
there's a trick here
first break up the sum as $\sum_{k=0}^{\infty} \frac{k+1}{2^k} + \sum_{k=0}^{\infty} \frac{1}{2^k}$ to make this more convenient. the second sum is a simple geometric series and its sum is not hard to calculate so let's focus on the first sum
wraith daggerBOT
Ann (glomed)
tropic oxide
the reason why i split things up like this is not apparent now but it will be apparent shortly
now consider the function $f(x) = \sum_{k=0}^{\infty} (k+1)x^k$. our sum is equal to $f(1/2)$
wraith daggerBOT
Ann (glomed)
tropic oxide
$f(x) = \sum_{k=0}^{\infty} \dv{x} x^{k+1} = \dv{x} \sum_{k=0}^{\infty} x^{k+1} = \dv{x} \frac{x}{1-x}$
wraith daggerBOT
Ann (glomed)
tropic oxide
@jolly sable
jolly sable
**Ann (glomed)**
I didn't understand this one @tropic oxide
X^k in multiply how does it come?
tropic oxide
it does not "come" from anywhere
well, i could say that i rewrote the sum as $\sum_{k=0}^{\infty} (k+1) \cdot (1/2)^k$
wraith daggerBOT
Ann (glomed)
tropic oxide
and then i abstracted that 1/2 away
but really there is no need for a "formal justification" for just introducing a function like this
jolly sable
**Ann (glomed)**
Hmm understand now u put gp formula here for it
I'll need some more steps to understand more of this kind problem
tropic oxide
which part do you want me to elaborate on?
jolly sable
K+1 is disappeared
tropic oxide
$(k+1)x^k = \dv{x} x^{k+1}$
wraith daggerBOT
Ann (glomed)
jolly sable
Ohh yes right
umbral hull
Hmm you can pull the derivative out? It's sorta subtle but that needs a theorem right?
jolly sable
So after you sum series first and use infinite formula a/(1-r)
tropic oxide
Hmm you can pull the derivative out? It's sorta subtle but that needs a theorem ...
handwaving that away
tropic oxide
So after you sum series first and use infinite formula a/(1-r)
missing parentheses
tropic oxide
Hmm you can pull the derivative out? It's sorta subtle but that needs a theorem ...
it's a power series and the thing converges for |x|<1 so it's good enough for me
jolly sable
**Ann (glomed)**
So what will be the next step?
tropic oxide
our sum is equal to f(1/2)
f(x) = d/dx x/(1-x)
it should be fairly obvious what to do from here
umbral hull
Now this is gonna bother me I'm looking it up xD
jolly sable
I have a clerk job exam tomorrow early in the morning so i need to sleep. I will surely come to this. I'll read all the chat again. Thanks ann again
umbral hull
Ahh yeah it is something like MCT
Ann your explanations are always great sorry if I bud in and look at them, I use them to review and test myself xD
limpid plume
**Ann (glomed)**
ah yes generating functions
jolly sable
Hello again
jolly sable
**Ann (glomed)**
So here we need to differentiate@tropic oxide the last term?
Here I found that
(K+2)/2^n = k/2^n+2/2^n
0 to infinity series sum k/2^k =2
2/2^n =4
So the total is 6
solid juniper
Ahh yeah it is something like MCT
it’s not - idt it really has a name but you could call it “power series termwise differentiation theorem” (ok after typing that out it sounds stupid) or something and it’s usually proven using the fact that power series converge uniformly on any closed interval in their radius of convergence
tropic oxide
So here we need to differentiate<@183668144404037632> the last term?
... yes?
jolly sable
Here I found that
(K+2)/2^n = k/2^n+2/2^n
0 to infinity series sum k/2^k =2
2/2^...
@tropic oxide
Please check this
wraith daggerBOT
tropic oxide
,w sum[k=0,infty] 2/2^k
wraith daggerBOT
tropic oxide
@jolly sable there
jolly sable
Yes edited
jolly sable
**Ann (glomed)**
So what should we do next in it after differentiation x/(1-x)
tropic oxide
as i said before, $\sum_{k=0}^{\infty} \frac{k+1}{2^k} = f(1/2)$
wraith daggerBOT
Ann (glomed)
tropic oxide
maybe i should repeat it 17 more times so that you definitely hear it at least ocne
jolly sable
I got 1/(x-1)^2 as differentiation
As this is the final value after differentiation
Here we got the value of it finally 4+2 =6
0 to infinity sum of series 1/2^k =1 but multiplying 2 so it is 2
Umm wait a minute
Thank you so much by the way the differentiation is nice. I never encountered such a term @tropic oxide
You explained the difficult part in nice way
.close
topaz wind
I got 2 questions
cedar kilnBOT
I got 2 questions
cedar kilnBOT
@topaz wind Has your question been resolved?
topaz wind
<@&286206848099549185>
unkempt raft
I can't see the picture u sent clear 😔
nimble veldt
@topaz wind, think about what it means that arg (...) = -pi/2 and try to express this in terms of x and y. then expand the denominator and use the relation between x and y you found.
weary vessel
I will show you how to do it @topaz wind
But you need to focus with me
It's a bit complicated
nimble veldt
It's a bit complicated
complicated? really?
weary vessel
but its readable?
yeah, I could read!
They skipped many steps
That's why you got lost :P
topaz wind
2 sec im here in 5 mins or smth
waxen raft
,w 182398129383928482938423948293842394823942983948239489482398394823942948+394923842394832423423324234234234
wraith daggerBOT
weary vessel
wrong channel
topaz wind
They skipped many steps
yes indeed
waxen raft
,w 1379218321731298313138313873817312873831832187318791273823812739821381273982173812738912738217381273812371287381238123781273128371893712371873128937182372813718273918371897318237981237981237189371893821371283728137217321837829173218937218371293128371732987312897392138271739723718237192738127387128973918723987198738271387218378217387238273891728378127381273812738173827131928372198732897218738972138912737918+23482348238423489247288472834723894327423894234242334234329548723534752397257320523245825849285923853485394355
wraith daggerBOT
Requested by PRO Player#4534
topaz wind
arg means the angle right?
weary vessel
Yes
weary vessel
Yes
Until here, you understood?
$\frac{z-1}{z+1-4i}=\frac{[(x-1)+yi] \cdot [(x+1)-(y-4)i]}{(x+1)²+(y-4)²}$
wraith daggerBOT
LUNA
topaz wind
i dont understand why you would kwadrate the things under the line
under the fractal
i would do it because there is an i in it?
weary vessel
It's not quadrate
We multiplied by the complex conjugate of the denominator!
To remove the imaginary part
topaz wind
yes
weary vessel
What do you think we should do next?
topaz wind
just work this big line out?
topaz wind
distribution or something
topaz wind
maybe i did it wrong but i didnt get the answer
weary vessel
What did you find
topaz wind
weary vessel
$\frac{z-1}{z+1-4i}=\frac{x²-1+y²-4y + (4x+2y-4)i}{(x+1)²+(y-4)²}$
topaz wind
wraith daggerBOT
LUNA
weary vessel
Like this
weary vessel
Now, this is the final expression of $\frac{z-1}{z+1-4i}$
wraith daggerBOT
LUNA
weary vessel
We can do arg on both sides
$arg\frac{z-1}{z+1-4i} = arg \frac{x²-1+y²-4y + (4x+2y-4)i}{(x+1)²+(y-4)²}$
wraith daggerBOT
LUNA
topaz wind
the angle of this should be -90 degrees, so tan b/a should be infinity?
b = imaginair
nimble veldt
is it just the part without i?
Yes. as arg () = -pi/2 means the real part is zero.
weary vessel
No, it's okay :P
Even if you work with tan(b/a) and -oo
it works too
Wanna try it?
topaz wind
i just need a way to know this for a similar question on the exam
which one is the easiest?
weary vessel
The direct one of course, you just notice that an argument of -pi/2 means that the complex number is on the imaginary axis
topaz wind
i dont think my class worked with -infinity
weary vessel
Zero real part
weary vessel
i dont think my class worked with -infinity
$tan(-\frac{ \pi }{2}) \to -\infty$
wraith daggerBOT
LUNA
topaz wind
yes
weary vessel
But in a way or another, the infinity is what matters here
Not its sign
Still works!
weary vessel
$arg\frac{z-1}{z+1-4i} = arg \frac{x²-1+y²-4y + (4x+2y-4)i}{(x+1)²+(y-4)²}$
wraith daggerBOT
LUNA
weary vessel
Do you know the following property?
topaz wind
property?
weary vessel
$arg(\frac{a}{b})=arg(a)-arg(b)$
wraith daggerBOT
LUNA
weary vessel
This, you know it right?
topaz wind
no i dont think so
weary vessel
it's what you use here
topaz wind
whats arg (a) and (b) then?
i think i know what it is, but this doesnt help with the question
weary vessel
$arg \frac{x²-1+y²-4y + (4x+2y-4)i}{(x+1)²+(y-4)²} = arg(x²-1+y²-4y + (4x+2y-4)i) - arg((x+1)²+(y-4)²)$
wraith daggerBOT
LUNA
weary vessel
it helps xD
arg(top/bottom) = arg(top) - arg(bottom)
But bottom is purely real
So its arg is zero
topaz wind
aaaaaa
weary vessel
You will only have the top
topaz wind
arg is for imaginair
weary vessel
$arg \frac{x²-1+y²-4y + (4x+2y-4)i}{(x+1)²+(y-4)²} = arg(x²-1+y²-4y + (4x+2y-4)i)$
wraith daggerBOT
LUNA
topaz wind
arg(a+bi) = arctan(b/a)
no didnt see it in class
weary vessel
Okay never mind then
Maybe it's a bit advanced this method
I don't recommend it
Stick with the method of real part = 0
topaz wind
yes agree
weary vessel
The second method works too, it worked for me well
nimble veldt
hmmm, tan (z) = imaginary part/ real part, so if this is infinity it means real part = zero. so you again "here".
weary vessel
But it's very long
topaz wind
this is how i understand it=
you just set the function equal to 0 now?
why dont i get it
weary vessel
$arg \frac{x²-1+y²-4y + (4x+2y-4)i}{(x+1)²+(y-4)²} = - \frac{ \pi }{2}$
wraith daggerBOT
LUNA
weary vessel
Notice that real part must be zero
topaz wind
ok this i understand
topaz wind
so 4x+2y-4 = -pi/2
no wtf this is incorrect
weary vessel
No, it's arg [ (4x+2y-4)i ] which is equal to -pi/2
weary vessel
But you do not need this one as well
Real part = 0 gives you the equation of the circle
And you're done
topaz wind
nimble veldt
but what is the formula for arg?
the formula for arg (z) = arctan(b/a) where z = a+bi, but you do not need this here.
topaz wind
woooooow
REAL PART IS 0 why did i have to hear this 10000x times before understanding...
weary vessel
You're surprised how such a huge fraction
Becomes just a little equation
arg power
topaz wind
lmao
weary vessel
😛
weary vessel
topaz wind
is this correct?
for the circle
yea i think so
THANK YOU SOOOOO MUCH @weary vessel @nimble veldt
soooo soooo much
topaz wind
.close
limber notch
Hello, I just had a question that I keep forgetting but why is it that we can cancel the 2's from a fraction like the one on top and not like the one at the bottom of the pic (unless im wrong on that too)? When can I cancel the fractions from numerator and denominator
limber notch
is it cause one is a product and the other one has addition?
or can I still cancel there
cold harbor
huh
nimble veldt
you can cancel from products.
cold harbor
as long as the entire numerator and entire denominator share a factor you can cancel it
limber notch
when wouldn't I be able to cancel it besides it not being the same number, sorry if the question is confusing and a little of a no brainer....
cold harbor
think of cancelling more like dividing
well multiplying but same thing
so if you were to remove a factor of 2 youd actually be left with a 3/2
limber notch
got it, thank you and sorry if it was kind of a simple question i just wrap my mind in circles and overcomplicate some ideas
.close
stiff stratus
cedar kilnBOT
stiff stratus
Is this right? is something missing there? Have I been able to show that the function equals 1/(1+x)?
cedar kilnBOT
@stiff stratus Has your question been resolved?
stiff stratus
<@&286206848099549185>
cedar kilnBOT
@stiff stratus Has your question been resolved?
next isle
yeah that works for the y =1 case
nimble veldt
@stiff stratus:
- you are not using f(1) = 1/2, right?
- i do not understand how you get f(r)r=r+1?
- maybe the result (f(x) = 1 /(1+x) )is right - but the way to get it? I doubt.
stiff stratus
<@1085105676138844180>:
1. you are not using f(1) = 1/2, right?
2. i do not unde...
I only use f(1)=1/2 to check if i got the correct function
nimble veldt
and what about f(r) r = r+1? How do you argue this (espacially the r+1 part)?
stiff stratus
and what about f(r) r = r+1? How do you argue this (espacially the r+1 part)?
$f(y \cdot f(x)) \cdot f(x) = f(x+y) \
x \rightarrow t-1 \
y \rightarrow y \
t \geqslant 1 \
\implies f(y \cdot f(t-1)) \cdot f(t-1) = f(t-1+y) \
f(t-1) = r \
f(y \cdot r) \cdot r = r + y \
\implies f(y \cdot r) = 1 + \frac{y}{r} \cdots \$
So r + 1 comes from multiplying by y on the left side and adding y to it on the right side. In my calculation I had y = 1. However, as you can see, I just generalized it and left it as y. But there you can also see that I have to put in y =1 so that we don't have a y.
wraith daggerBOT
Gilgamesh
stiff stratus
Or do you have another idea?
nimble veldt
$f(y \cdot r) \cdot r = f(y+t-1)$
wraith daggerBOT
ThM
nimble veldt
how do you argue f(y+t-1)=r+y?
it seems you are using $f(t-1+y) = r+y$ which i do not understand.
wraith daggerBOT
ThM
stiff stratus
Because I am adding y on the right side and multiplying on the left side. So I come to the presentation.
nimble veldt
f(t-1) = r, ok thats a definition. but why should f(t-1+y) be r+y, its a kind of linearity argument which is not given i think.
stiff stratus
But the solution is right, so it means that f(t-1+y) = r +y is right. But idk how can I show you that...
nimble veldt
as i said before, the function may be right, but the way you get this solution is not (in my humble opinion).
stiff stratus
how would you do it
nimble veldt
i havent done yet. you asked if your solution is right and till now i tried to understand your solution. give me a little time.
nimble veldt
- with x = 1, y = 2 you get f(3) = 1/4
- knowing this you get f(7) = 1/8 with x = 3 and y = 4
- knowing this you can prove by induction that f(-1+2^n)= 1 / (2^n) (using y = 1/f(x))
- knowing this you can assume f(x) = 1 / (1+x) and test if this function fulfill the given conditions.
But: this is not a proof that this f(x) is the only solution.
cedar kilnBOT
@stiff stratus Has your question been resolved?
stiff stratus
1. with x = 1, y = 2 you get f(3) = 1/4
2. knowing this you get f(7) = 1/8 with ...
How can i prove that's the only one solution for f(x)?
hard merlin
cedar kilnBOT
@stiff stratus Has your question been resolved?
stiff stratus
<:devastation:898725325335298088>
💀🔫
nimble veldt
How can i prove that's the only one solution for f(x)?
As you are asking for a uniqueness-proof why are you sure there is only one solution?
stiff stratus
As you are asking for a uniqueness-proof why are you sure there is only one solu...
Do you know of another function that also describes the values e.g. for f(1) = 1/2 or f(3) = 1/4 etc.? I don't know of any other function for this...
nimble veldt
I think you need first to be clear about what you need. if there is only one such function you need no uniqueness-proof.
stiff stratus
I think you need first to be clear about what you need. if there is only one suc...
I see...
fallow steppe
a mod ping suffices, no need to ping us individually
hard merlin
he got banned?
hard merlin
a mod ping suffices, no need to ping us individually
yep ik
stiff stratus
a mod ping suffices, no need to ping us individually
Ok next time
hard merlin
banned?
stiff stratus
Same
cedar kilnBOT
@stiff stratus Has your question been resolved?
cedar kilnBOT
@stiff stratus Has your question been resolved?
glossy grail
i have a question, when i use hopital on for example (3^x-4^x)/x, i just do the derivative of the numerator and the denominator without applying the formula for D[f(x)/g(x)]?
woeful ermine
Im absolutely ass at everything here and I’m gonna need help in most of these questions 😭😭
obsidian coral
Im absolutely ass at everything here and I’m gonna need help in most of these qu...
Is this a test?
woeful ermine
Is this a test?
pre test
It’s for summer school
I have to do it in order to do assignments
obsidian coral
If I recall, pre tests are to test your prior knowledge on that material
So you should be doing that on your own
tropic oxide
^
warm bone
Im absolutely ass at everything here and I’m gonna need help in most of these qu...
I don't think estimation questions are very precise, but do you notice anything about the values?
woeful ermine
So you should be doing that on your own
rightt
but i rlly need to pass these
woeful ermine
I don't think estimation questions are very precise, but do you notice anything ...
wdym
it’s js multiplying but idk how to put like the stuff in words 😭
cosmic steppe
So you should be doing that on your own
Adding onto this, it'll be against the rules if we do help you on your pre test
warm bone
Estimation is 'as simple' as noticing that 100 x 100 is (a lot) higher than the value you need to calculate
cosmic steppe
It's a pre-test for a reason
woeful ermine
It's a pre-test for a reason
ig
woeful ermine
Estimation is 'as simple' as noticing that 100 x 100 is (a lot) higher than the ...
ohh ok gotchu gotchu
well since it’s against the rules I’ll js come back later wit the work instead of tests for explanations
i don’t rlly wanna break the rules here
obsidian coral
rightt
but i rlly need to pass these
The other thing I recall from pre tests as mentioned, it's testing your prior knowledge. Pre tests are used to see where you stand in the class and how much assistance you need from the teacher
woeful ermine
The other thing I recall from pre tests as mentioned, it's testing your prior kn...
True
The thing is I think they grade it
Like I said earlier, I don’t wanna fail because if I do I repeat the grade
I really don’t wanna do that so I’m js trying my best to get help in any way
cedar kilnBOT
@woeful ermine Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
@crimson sedge Has your question been resolved?
cedar kilnBOT
@crimson sedge Has your question been resolved?
slate lintel
expand out all the products so multiply (Ax-1)(x-1) for instane
high coyote
For method 2 you have to expand first the right-hand side
Oh, sorry, Hayley, keep explaining.
I didn't see your message
slate lintel
and then you have $4x^2 - 3x + 1 = \alpha x^2 + \beta x + \gamma$
wraith daggerBOT
Hayley
slate lintel
so you can set the x^2 coefficients equal to each other and the x coefficients equal to each other
you get a system of equations and can solve it
kinda painful, you should do it at least once though
slate lintel
I didn't see your message
no problem! I think we started typing at the same time
so like... first step is to expand and simplify the right side, put it into ( )x^2 + ( )x + ( ) form
wraith daggerBOT
picl
slate lintel
put the Bx in with the (A+1)x so you'll get
$$Ax^2 + (-A + B + 1)x + (C + 1)$$
wraith daggerBOT
Hayley
slate lintel
and we know that's equal to $4x^2 + (-3)x + 1$
wraith daggerBOT
Hayley
slate lintel
that should be Bx - B in your original
hang on let me update your equation, I think this is right but you should check me
$$Ax^2 + (-A + B - 1)x + (-B + C + 1) = 4x^2 + (-3)x + 1$$
wraith daggerBOT
Hayley
slate lintel
it's a -B constant because it comes from B(x-1)
(like I said... pain)
so now you can equate the x^2 terms together, the x terms together, and the constant terms together
like right away you can see A = 4
on the right? just so it was easier to match them up
pog
this method is sometimes necessary for more complicated functions but if you can get answers by subbing values you should do that
slate lintel
sure, you can if it makes it easier to see what's going on
i'm not entirely sure you'll be able to substitute values to find a solution here
binomial means 2 terms yeah
but if you're asserting that it has to have the form (Ax + 2)(x-2) + B(x-2) + C
then yeah it'll probably have nonzero values for A, B, and C
cedar kilnBOT
@crimson sedge Has your question been resolved?
static fern
$\cos 45^\circ = \frac{\sqrt{2}}{2},$ not just any decimal (you should remember this).
wraith daggerBOT
fireblazer8961
surreal cave
I wish to compute $\int_{0}^{\infty}\frac{t}{e^t-1}\mathrm{d}t$ by integrating around a semi-annular contour. If I let $f(z)=\frac{z}{e^z-1}$I have then:\
$\lim_{(R,r)\to(\infty,0)}\int_{-r}^{-R}f(z)\mathrm{d}z+\int_{\Gamma}f(z)\mathrm{d}z+\int_{R}^{r}f(z)\mathrm{d}z+\int_{\gamma}f(z)\mathrm{d}z$\~\
I guess my question is, is this even the correct approach?
wraith daggerBOT
XxMrFancyu2xX
upper abyss
Yeah that should work
Namely, the upper curve goes to zero as R goes to inf
Oh wait 0 to inf hmm?
<@&268886789983436800>
surreal cave
<@&268886789983436800>
upper abyss
It would be nice to get an easy way to turn that into an -inf to inf integral
Actually what contour did you have in mind?
surreal cave
ok here's a really bad representation 
of the basic idea im trying
upper abyss
Problem being this will lead to an integral over -inf to inf
surreal cave
oh wait
that's a good point lol
see this is what I mean by tarnishing my reputation 
upper abyss
Mine too I don't know many good contours
surreal cave
then do you have another suggestions for a different contour?
upper abyss
... Ehhh?...
surreal cave
mmhh a singular keyhole 
upper abyss
On second thought, does this even contain any residues
surreal cave
well that's why I tried the semiannulus contour bcz there's a hole at t=0+0i
cedar kilnBOT
@surreal cave Has your question been resolved?
surreal cave
ah and so it begins 
surreal cave
.close
amber axle
can someone help me with figuring out (a<=b and b<=a st a=b)
dreamy sleet
this looks like one of the last levels of this game https://www.ma.imperial.ac.uk/~buzzard/xena/natural_number_game/
dreamy sleet
can someone help me with figuring out (a<=b and b<=a st a=b)
what have you done so far on this question
amber axle
what have you done so far on this question
i can figure out something
like
a = b + x
b = a + y
a = (a+y) + x
but i dont know where to go after thart
amber axle
a = (a+y) + x
and i used this to prove a<= a but thats fine
i cant understand how i can use it to use to solve the asymmetricity
dreamy sleet
you are trying to prove, given that a and b in N, that at least one of a <= b and b <= a is true
amber axle
"leq" ?
dreamy sleet
a = (a+y) + x
so this is what you want to show
amber axle
so this is what you want to show
apparently yes
amber axle
oh okay
dreamy sleet
a = (a+y) + x
here, you assumed that both were true
amber axle
yes
dreamy sleet
can someone help me with figuring out (a<=b and b<=a st a=b)
a starting point might be induction on a and b
amber axle
i did use that to prove transitivity between a and b
a = b + x
c = b + y
c = (b+x) + y
dreamy sleet
What is c?
amber axle
a = b + x
c = b + y
c = (b+x) + y
and by using induction c = b + (x+y)
dreamy sleet
do you mean associativity
dreamy sleet
have you tried doing induction on b
cedar kilnBOT
@amber axle Has your question been resolved?
chilly shuttle
If we know that B = cos^-1(-5/13) which is positive, does anyone know why sinB > 0 is a rule?
thick cipher
If B is less than pi sin B >0
scarlet widget
If we know that B = cos^-1(-5/13) which is positive, does anyone know why sinB >...
sinB could be negative
rose swan
it's not a rule, but a restriction to tell you what quadrants your angle B may lie in
rose swan
how did you get that expression?
chilly shuttle
A + B = cos^-1(323/325)
so A = cos^-1(323/325) - cos^-1(-5/13)
because B = cos^-1(-5/13)
rose swan
seems kinda complicated though
like you still have to pick out the correct signs, and that's not clear from your expression
cedar kilnBOT
@chilly shuttle Has your question been resolved?
rose swan
hello, are you still there?
cedar kilnBOT
@chilly shuttle Has your question been resolved?
queen veldt
Can someone help me with this integral?
cedar kilnBOT
Can someone help me with this integral?
queen veldt
$\int{}{}\frac{(1 + ln x)^2}{x} dx$
wraith daggerBOT
totomc
upper abyss
Expand the square, split into three integrals
slate lintel
i mean you could but that's a bit of a waste
queen veldt
I need to resolve it using substitution
slate lintel
okay pick a sub-expression to set equal to u
south tundra
Tried anything so far?
queen veldt
So I got u = ln x and du = 1/x dx