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#help-39
autumn fossil
I'd try a contradiction
assume it's finite
Another hint: ||What can you say about the set of numbers such that f(n) < n? And what can you imply from this, together with the assumption?||
(ping me if you need me)
pearl pondBOT
@topaz sonnet Has your question been resolved?
autumn fossil
What is finite
It actually is finite
because there are finitely many numbers less than n
and f is injective
so it can go to each of these finitely many numbers at most once
this set actually has a cardinality smaller than (or equal to) n
is that injective?
oh, yeah, it is
but the set is still finite
Oh wait
im stupid, i completely misinterpreted the question
for some reason i thought that the RHS n would be a constant
Okay, I think this can still be done by contradiction
so let A be the finite set of n, s.t. f(n) >= n
Yep exactly
we take the largest element of A
then for n > max(A), f(n) < n
autumn fossil
I think I got an idea.
So basically, we now consider the maximum f(a) for a in A, and if this is greater than max(A), we call it m. Otherwise we just call max(A) m.
We now have that for n > m, f(n) < n. Then f(m+1) < m + 1. But since m+1 is greater than the maximum f(a), all numbers below m+1 were assigned to some numbers below m+1. Which means that m+1 itself cant be assigned to such number
the idea is there
yes, it's informal
ill try to write it more rigorously
My point is essentially this:
Imagine that we are trying to keep the set f(n) >= n finite. Then e.g.
f(0) = 0
f(1) = 4
f(3) = 7
f(4) = 5, and for all other n, f(n) < n
this means that f(0), f(1), f(2), f(3), ..., f(8) are all less than 8
wait
i probably made a mistake
damn
autumn fossil
this means that f(0), f(1), f(2), f(3), ..., f(8) are all less than 8
oh I didnt
this is the contradiction
9 f's
all less then 8
there are 8 numbers less than 8
fuck yesh
@topaz sonnet , idk if youre still here but I got it
Consider M = max({f(a) | a in A})
now look at f(0), f(1), ..., f(M)
my claim is that all of these are < M
why?
yeah, i just simplified it by quite a bit
if we stopped at f(M-1), then we wouldnt reach the desired conclusion
autumn fossil
now look at f(0), f(1), ..., f(M)
All I want is have all of these < M
then there are M+1 pigeons in M holes
exactly, except for one of them
and the one of them might cause problems
unless you decide to extensively deal with it
autumn fossil
now look at f(0), f(1), ..., f(M)
With this, you have M + 1 integers having their image in [0, M-1]. So it's a bit easier
there are M integers in [0, M-1]
autumn fossil
With this, you have M + 1 integers having their image in [0, M-1]. So it's a bit...
.
this makes it far easier
ill show you my argument:
I'll now argue that f(0), f(1), ..., f(M) are all < M
so take some n <= M
we will deal with it by cases
if n is in A, then f(n) < max({f(a) | a in A}) = M
if n is not in A, then f(n) < n <= M, lest it actually would be in A
so f(n) < M for all n <= M
and this is sufficient
M+1 pigeons, M holes
this question was really interesting
,ti physicsrocks
jolly parrotBOT
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autumn fossil
damn
current time for one helper who was just typing in this channel
at 1:20 AM lol
autumn fossil
if n is in A, then f(n) < max({f(a) | a in A}) = M
Oh youre right, this line is wrong
Just takee f's up to M+1 then
Then for n <= M+1, f(n) <= M
pearl pondBOT
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@rose sapphire Has your question been resolved?
ocean temple
pearl pondBOT
plush bramble
is that the entire graph or is there more
ocean temple
that is the entire graph
glass meadow
is that the entire graph or is there more
The function is |x-4|
plush bramble
what about f(x)
cyan lily
domain being R was accepted though, so i don't see why you would be wrong
plush bramble
is there more to the function?
glass meadow
how the heck is range wrong
You're correct, the software is bad
plush bramble
,w plot y=3
jolly parrotBOT
ocean temple
is it not constant on both fronts
plush bramble
constant functions are of the form y=k for some number k
cyan lily
constant is not increasing nor decreasing
ocean temple
so because the ENTIRE function is not a straight line
glass meadow
is it not constant on both fronts
It's linear (affine) on both intervals, not constant
plush bramble
lines of the form y = mx + b are not constant unless m=0
glass meadow
doesnt matter what parts are constant?
There's literally no part of it that is constant
plush bramble
when m = 0, then y = 0 * x + b = 0 + b = b and it is a constant function
cyan lily
which is not constant
plush bramble
it's consistently going up by 1
you misunderstand what constant means
ocean temple
apparently so
ocean temple
what would constant be
glass meadow
Horizontal
plush bramble
constant functions are of the form y=k for some number k
.
cyan lily
so when x is always y
what do you mean?
plush bramble
y=1 is a constant function yes
ocean temple
okay dope
plush bramble
,w plot y=1
jolly parrotBOT
plush bramble
it looks like that
glass meadow
Yes but it's not "x is always y"
ocean temple
also i found out the C they just wanted me to include 0 in the function
i did a parenthesis not a [
glass meadow
It's "y is always c" where c is... a constant
pearl pondBOT
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velvet wolf
When finding eigenvalues does it matter what we call Iambda_1 and what we call lambda_2 ?
tropic saddle
no
velvet wolf
no
So then when I need to determine matrices P and D such that P ^(−1)AP = D and D is a diagonal matrix.
How do I know which order to write my eigenvectors in?
plush bramble
same order as your eigenvalues
velvet wolf
So then the order of the eigenvalues matters?
plucky python
So then the order of the eigenvalues matters?
it doesn't matter how you arrange ur eigenvalues and eigenvectors, as long as they have the same order
tropic saddle
well you only need to be consistent
tropic saddle
if you name one eigenvalue lambda1 then you should also name the corresponding eigenvector as v1 and not v2 for example
tropic saddle
the second eigenvalue is wrong
velvet wolf
-4* sorry
those r the eigenvectors for λ_1
and then for λ_2 it's:
So now when determining matrix P
is that matrix gonna be in this exact order of vectors?
1, 0 ,0 then 0, 1 , 0 and then -1/2 ...
tropic saddle
if your matrix D has the diagonal 0,0,-4, yes
velvet wolf
okay and then that would work?
so there can be more than one answer for P and D here then?
tropic saddle
yes
velvet wolf
it just depends on what order u chose for your eigenvalues?
tropic saddle
they just need to fit together
velvet wolf
Also, can the fractions for λ_2 eigenvector be removed by multiplying it by 2?
Would that still be it's eigenvector?
tropic saddle
yes
pearl pondBOT
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velvet wolf
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desert dagger
pearl pondBOT
desert dagger
how do i do this
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lilac bolt
woudl this be correct
pearl pondBOT
woudl this be correct
cyan lily
doesn't seem wrong, though i would change the first part, it almost sounds like you're assuming it's composite then proving it is, maybe just put that statement lower down after you have shown the properties of r and s
lilac bolt
ah i see what you mean. I did it first because I looked up the definition of what it takes to be composite before I could even manipulate it
agile plinth
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lilac bolt
hi
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forest sun
how do i use M3, I have a table with t and v. and was able to find L6 and R5 but got confused with the M3. Is there a formula for that?
forest sun
i am estimating the distance traveled
by right and left end points
but not sure about the midpoint
plush bramble
Are we supposed to know all those M3 etc
forest sun
no just if there is a formula when calculating distance
its like a graph, but with points, where I guess we find the area of the rectangles
like in finding the area under the curve
plush bramble
I have no idea what you're talking about
forest sun
hmm
t
something like this
like there are point
and you do rectangles to calcualte the area
this one is right hand
but i am confused with how the midpoints are done
like if i have the graph, the points, do i just devide everything by half or some
cyan lily
how do i use M3, I have a table with t and v. and was able to find L6 and R5 but...
what is M3
forest sun
midpoint 3 ig
calm acorn
Sorry, I’m not sure I understand your question. Are you trying to find the area beneath that curve with like the midpoint of the rectangles?
calm acorn
There is a formula for that, I had to do this last semester. It’s called a midpoint series or something
forest sun
yea im lost with it cus L and R is easy enough with points coming from either the right or the left
but i don't understand the midpoinr
midpoint
do we take the midpoint between the know points
like this thing L6 = 15 (0+15+30+45+60+75)
R6 = 15(15+30+45+60+75+90)
calm acorn
So for midpoint it’s basically the same process. The only that’s going to change is the number of times you add beneath the curve
forest sun
like this thing L6 = 15 (0+15+30+45+60+75)
R6 = 15(15+30+45+60+75+90)
so do you perphs know how to get to that
but for M3
i found my proffesor giving this, M3 = 30(10+60+74) but don't have the clue where that came from exp the 30
calm acorn
Okay, at first I thought this was a series/ calculus problem, but I’m looking at your previous equations and they’re not at all from a series so I’m really not sure what the question is asking
forest sun
yea 😭 calc 2 here
idk
its all mixed
but how would u do the M3
i understand that the 30 came from 90/2
90/3
since there are 3 intervals
but where does the inside come from
calm acorn
Like I’m just reading through my notes from last semester, here’s the midpoint formula for the question you’re asking
forest sun
what is xi +xi?
like intial
wait i see it
it starts from 0
so in this case it would be the 0+90
+1?
over 2
calm acorn
So xi is the numbers you’re actually subbing into the equation. So in LHR you start at 0 and go up by the step size stopping befoee the upper limit. In mid, you start at your lowest than increment at double the width if I remember correctly and stop before your upper limit
Here’s the example I have on hand atm. Where f(x) = 2x^2 and n=4
forest sun
ah i see
ima try asking my prof this on monday
cus its kinda confusing
thanks for helping
.closed
calm acorn
Sry I couldn’t be of more assistance, this stuff confused me when I was covering it
calm acorn
Best wishes!
analog depot
help
analog depot
sharp vigil
,rccw
jolly parrotBOT
feral sedge
Do you know pigeonhole principle
analog depot
just basics
pearl pondBOT
@analog depot Has your question been resolved?
pearl pondBOT
@analog depot Has your question been resolved?
fossil drum
I'm struggling with this one
pearl pondBOT
I'm struggling with this one
fossil drum
I created the linear transformation that represents X1->X2 and Y1 -> Y2
T([x,y]^t) = [x+1, y+1]^t
But what should I do this information?
plush bramble
That doesn't map [1,1] to [4,5]
fossil drum
oh
says X1 to Y1 oof
so + 3
T([x,y]^t) = [x+3, y+3]^t
wait
no
thats wrong
uhh
now im really stuck
oh waity
no
its just
plush bramble
Have you learned matrix multiplication
fossil drum
T([x,y]^t) = [x+3, y+4]^t
yeah
I know that if we have a linear transformation T(X) it can be also represented by multiplying by some matrix A
plush bramble
T([x,y]^t) = [x+3, y+4]^t
This isn't a linear transformation
fossil drum
but im not sure how to get A
plush bramble
There is no matrix that does what you're describing
fossil drum
oh is the linear transformation T([x+3,y+4])?
plush bramble
Use an arbitrary matrix A [[a, b], [c, d]] with 4 unknowns and solve for a b, c, d using the maplings
fossil drum
Use an arbitrary matrix A [[a, b], [c, d]] with 4 unknowns and solve for a b, c,...
I still dont quite get it
whats the mapping?
Are you saying do something like A[] = []
multiply and then solve?
or would it be []A = []
plush bramble
Have you learned matrix multiplication
?
fossil drum
yeah we learned matrix multiplication
im going through a different book rn and I think im starting to understand
the book for my course only spends 2 pages on linear transformation
fossil drum
I'm struggling with this one
@plush bramble im just confused on if multiplication by A mean XA or AX here
since it isnt commutative in some cases
I believe its XA right?
X1A = Y1?
fossil drum
oh right
plush bramble
A is 2x2
plush bramble
Are you saying do something like A[] = []
You should have two equations of this form
With all 4 given vectors
fossil drum
With all 4 given vectors
like this?
hmm and if I multiply the first row in the second equation by 1/4
I get x + z = 5/4
and we also have x + z = 2
and so we know there doesn't exist an 2 by 2 A that transforms them?
oh dang it
I forgot to change X1->X2 to X1->Y1
fossil drum
Your conclusion is just probably wrong
the way I worded my conclusion?
plush bramble
No sorry I read your work wrong
fossil drum
oh ok
real shard
Would anyone be able to clear up how he is doing this proof? He says that hes doing the contrapositive, but it seems like he just takes the converse and negates the conclusion? Im a bit confused this is the link to the original video
summer imp
He's proving "linear independence => none of the vectors are linear combinations of the others" as
"there is a vector that is a linear combination of the others => linear dependence", which is the contrapositive
real shard
never mind i think im trippin bad
oh my gawd thank yu
are there generally some clues to help you determine when to use the contrapositive or not?
summer imp
Well I guess that in general, it's conceptually harder to prove that something NEVER holds (like NONE of the vectors being a linear combination of the others). But the negation of "none" is "there is some", which is easier to deal with, so you'd use contradiction or contrapositive
real shard
i see so when "none" or "there is some" appears in a question does that typically mean we can approach by using contradiction or contrapositive?
pearl pondBOT
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summer imp
It's just that it's easier to work with something that exists rather than proving something doesn't exist (in most cases).
If you see "none" and it seems hard to prove, try and see if the contrapositive gives something simpler
pearl pondBOT
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long steeple
I was learning Waves for my upcoming exams and came upon this derivation. They solved the differential by first assuming x = Ce^(at). Is that a standard method to solve second order DEs? If yes then where can I read more about it?
plush bramble
guessing solutions are indeed very common
just do a bunch of problems i guess. start simple
https://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
long steeple
just do a bunch of problems i guess. start simple
https://tutorial.math.lamar.e...
thankyou ill refer to it :D
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lethal vigil
How would I begin solving this?
pearl pondBOT
How would I begin solving this?
plush bramble
Find the formula for tangent plane
woven garnet
z+3=f(2,-1)(x-2)+f(2,-1)(y+1)
Assuming you already know the general form as you have this as a problem
lethal vigil
Find the formula for tangent plane
I didn't know we had a formula
My professor didn't teach it
is this this one?
I think i'll figure it out from here
thank you
.close
pearl pondBOT
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carmine zinc
Is what I have done correct?
pearl pondBOT
Is what I have done correct?
carmine zinc
Q45*
copper matrix
looks correct to me @carmine zinc
btw, do you know how many neutrinos are released in a core-collapse supernova?
carmine zinc
btw, do you know how many neutrinos are released in a core-collapse supernova?
No, sorry
happened before I started existing
carmine zinc
looks correct to me <@839109098980507659>
Thanks, I have other doubts too
copper matrix
just a fun fact
copper matrix
Thanks, I have other doubts too
👍
copper matrix
weird
scarlet tinsel
plug in values of z
carmine zinc
plug in values of z
z is a complex number
copper matrix
plug in values of z
can we assume the formula holds for all z, or just a single z?
carmine zinc
that Im not given
scarlet tinsel
yea exactly
gusty thorn
what math is that
carmine zinc
can we assume the formula holds for *all z*, or just a single z?
a single z
scarlet tinsel
u can still plug in stuff
copper matrix
a single z
yeah this is what i thought
carmine zinc
what math is that
IAT 2017
but it should hold for all z
well atleast the first one will
gusty thorn
whats iat 2017 thats the first time i heard that
scarlet tinsel
mhm
copper matrix
it has something to do with z -> 2z but i'm not sure
carmine zinc
since its just the binomial thm
carmine zinc
whats iat 2017 thats the first time i heard that
IISER Aptitude Test, 2017
Its on the same level as JEE mains in india
gusty thorn
is that like calculus 2
scarlet tinsel
if u ask me, u should try z=-2 and z=2
carmine zinc
since its just the binomial thm
but written in a different way
carmine zinc
if u ask me, u should try z=-2 and z=2
that aint a complex number tho
copper matrix
if u ask me, u should try z=-2 and z=2
see we technically have to assume this holds for any z then, both z = 2 and -2, not just a single value of z
copper matrix
i wonder if that's what the problem meant, even though the problem didn't say that
carmine zinc
see we technically have to assume this holds for any z then, both z = 2 and -2, ...
I did say it should
scarlet tinsel
im not sure if ur seeing what im seeing
copper matrix
I did say it should
yeah but that is an extra assumption. in general it doesn't have to be true
carmine zinc
but I dont have a confirmation that it will
scarlet tinsel
if u ask me, u should try z=-2 and z=2
js manipulate this
carmine zinc
yeah but that is an extra assumption. in general it doesn't have to be true
But it just looks like the binomial thm
scarlet tinsel
u should see smth familiar
copper matrix
for example. (1 + z)^2 = a0 + a1 z + a2 z^2 means a0 = 1, a1 = 2, a2 = 1 ... if this holds for all z
otherwise, a0, a1, and a2 can be anything
say it only holds for z = 1
then we could just choose a0 = 4, a1 = 0, a2 = 0
then it holds for z =1, but not all z
carmine zinc
u should see smth familiar
lemme try this
copper matrix
i think they must have meant it holds for all z, otherwise the problem seems unsolvable
so yeah if i met this question writer i might argue with them. anyway we're good, proceed
carmine zinc
well atleast mentally
I dont think itd give you anything remotely close to the sum of squares we have here
copper matrix
i do think something similar to z = 2 is relevant
carmine zinc
my best guess is that its got something to do with
carmine zinc
i do think something similar to z = 2 is relevant
something like this
then the conjugate of that, maybe
copper matrix
ahh there must be a reason z is complex
maybe z = 2 and z = 2i are relevant
or z = 2i and -2i, like conjugate like you said
idk, something like this
carmine zinc
or z = 2i and -2i, like conjugate like you said
ye
copper matrix
interesting
scarlet tinsel
and then add or subtract the results
carmine zinc
and then add or subtract the results
that would be way too tedious still
even assuming that works
you have to find a power of 15
This problem needs to be done in 3 minutes
something like that wont come up here
I think we're on the right track, but we're missing something crucial
copper matrix
This problem needs to be done in 3 minutes
with the right technique, i think it's possible it could be done in 3 mins 🪦
we have powers of 15, but, we also have a form (big sum)^2 + (other big sum)^2
the sums themselves are only squared
carmine zinc
we have powers of 15, but, we also have a form (big sum)^2 + (other big sum)^2
If it was a multiple power of two
maybe
we could use difference of squares?
copper matrix
ok the first sum
(first big sum)^2
the first big sum itself
can maybe be equal to (1 + 2i)^15 + (1 - 2i)^15 :p
(not to give away the answer, but i'm sure you'll see it yourself)
copper matrix
we could use difference of squares?
ah that'd be an interesting way too! possibly
i think that can work too
so there are multiple ways
a^2 + b^2 = (a + ib)(a - ib)
(where a and b are the big sums)
carmine zinc
a^2 + b^2 = (a + ib)(a - ib)
Thats the definition of a modulus
copper matrix
Thats the definition of a modulus
that too
but it's equivalent to a^2 - b^2 = (a + b)(a - b)
which i figure is what you were thining
here we have + instead of -, but it's equivalent, if b can be complex
carmine zinc
here we have + instead of -, but it's equivalent, if b can be complex
but it cant be
b is a real nuber by definition
copper matrix
well ig i mean
here
in the formula c^2 - d^2 = (c + d)(c - d), d can be anything
carmine zinc
thats fair
carmine zinc
hold on
copper matrix
then this implies a^2 + b^2 = (a + ib)(a - ib)
copper matrix
yeah basically
however we get to it, we can say it's true, for real (or any) a and b, that a^2 + b^2 = (a + ib)(a - ib)
carmine zinc
however we get to it, we can say it's true, for real (or any) a and b, that a^2 ...
yea that checks out
copper matrix
then a + ib can probably be equal to (1 + z)^15 for some good choice of z
carmine zinc
then a + ib can probably be equal to (1 + z)^15 for some good choice of z
guesswork seems hard
here
copper matrix
right, we did have some guesses tho, like z = 2i
carmine zinc
plus this is supposed to test actual concepts better
carmine zinc
right, we did have some guesses tho, like z = 2i
hmm
copper matrix
tbh i do think this is pretty conceptual
carmine zinc
tbh i do think this is pretty conceptual
true
copper matrix
hmm
i'd try this tho
and they use the ... dots bc a pattern will probably emerge so you don't have to calculate all of them
carmine zinc
and they use the ... dots bc a pattern will probably emerge so you don't have to...
thats because thats just the binomial thm
with a bunch of terms replaced
copper matrix
yeahh exactly
carmine zinc
a0, a1 etc arent hard to get
just check the choose fn's values for n=0 > 15
copper matrix
we can leave them as just letters/variables
but yes you're right
so for z = 2i
we can use the first formula they give, with z = 2i
and just leave a0, a1, a2, as variables/unknowns for now
copper matrix
their formula is (1 + z)^15 = a0 + a1 z + a2 z^2 + a3 z^3 + a4 z^4 + a5 z^5 + ... + a15 z^15
carmine zinc
omfg
copper matrix
we just wanna plug in z = 2i
carmine zinc
its just
copper matrix
😮
copper matrix
yess
carmine zinc
of (1+z)^15
copper matrix
modulus squared
copper matrix
just +-2i will work
i agree i think
carmine zinc
and only +-2i i think
carmine zinc
nope both are equally valid
desert tendon
is it from geometrical significance of complex numbers
carmine zinc
is it from geometrical significance of complex numbers
never heard of that
desert tendon
which standard
desert tendon
your class
carmine zinc
12th
desert tendon
complex part 2 padha hai
desert tendon
Jee ki preparation mei 2 part mei padhate hain
copper matrix
nope both are equally valid
yeah both are fine, only those two are right, no others
carmine zinc
no coaching
desert tendon
ohkk
carmine zinc
also jee is taking rationalised syllabus this year
iirc
so nothing more than 11th and 12th ncert
carmine zinc
another insight
both series's sums are equivalent
a0=a15
and so on
BT
so you can sum it up further
copper matrix
hm isn't one sum (a + ib) and the other (a - ib)
so it is modulus, (a + ib) * conjugate
carmine zinc
hm isn't one sum (a + ib) and the other (a - ib)
?
carmine zinc
so it is modulus, (a + ib) * conjugate
ye
copper matrix
but not (a + ib)^2
carmine zinc
also ye
carmine zinc
so you can sum it up further
I was wrong about thiws
copper matrix
ah k np
carmine zinc
itd be true if the powers of two didint exist
copper matrix
1 + i and 1 - i should still be different i think (like if z = i or -i)
whatev tho :p
it is interesting to think about
carmine zinc
1 + i and 1 - i should still be different i think (like if z = i or -i)
uhh
i think i got it
if z is a complex number
then
|z^2|=|z|^2
copper matrix
|z^2|=|z|^2
yeahh that is helpful
carmine zinc
the question is if it holds always
copper matrix
|z^2|=|z|^2 is always true yeah
carmine zinc
|z^2|=|z|^2
consider
|z^n|
and
|z|^n
if theyre equal
you just get your answer
copper matrix
yeah they should be equal (for real n)
carmine zinc
yeah they should be equal (for real n)
in which case
copper matrix
i think a way to prove it is to use |ab| = |a||b| always
copper matrix
in which case
|z|^n = |z^n| for any z, and real n
copper matrix
you get 5^30
that sounds almosttt right
and
i think you're basically doing it right
so
we want
(|1 + 2i|^15)^2 right
carmine zinc
the square root of this is a possible answer
carmine zinc
but
copper matrix
|1 + 2i| = sqrt(5)
carmine zinc
copper matrix
happens xd
carmine zinc
happens xd
LETS GOO
I
copper matrix
🎉
carmine zinc
IT ONLY TOOK 1 HOUR!!!
copper matrix
lool that was hard though
like that just is a hard problem
i'm happy i could even do it too lol
carmine zinc
% time lost in the exam?
60/3*100=2000%
copper matrix
🪦
carmine zinc
want the entire paper?
copper matrix
sure lol
carmine zinc
copper matrix
starts with biology :x dang
carmine zinc
you're indian right?
ye
copper matrix
you guys are too smart lol
i haven't seen tougher exams from anywhere else
it's a good thing tho
carmine zinc
i haven't seen tougher exams from anywhere else
jee advanced is like hell compared to thiss
Boy am I glad I dont have to do jee adv
copper matrix
i might'vee seen that before too and yeah it was
hollow owl
Boy am I glad I dont have to do jee adv
i do 😔
carmine zinc
one 3 hour exam is enough lol
multiple answers
copper matrix
ig the break in between is important
carmine zinc
ig the break in between is important
like 2 hours to have lunch i think
carmine zinc
hollow owl
oh yeah it's morning there
ye
hollow owl
I mean you can only give *hell incarnate* if you qualify jee mains
ill qualify jee mains with 99.9 that im confident lma
jee adv only god knows
carmine zinc
ill qualify jee mains with 99.9 that im confident lma
.close
pearl pondBOT
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Available help channel!
heavy hill
can someone help me with this:
pearl pondBOT
can someone help me with this:
plucky meadow
$n(n-1)(n-2)(n-3)(n-4)\neq (n-4)!$
jolly parrotBOT
Luke
pearl pondBOT
plucky meadow
oh
just wanted to say that if you are doing this by hand
it might be easier to keep it as (n choose 5) = 1287
and try different values of n
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swift lodge
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swift lodge
Can someone explain this to me its not a academic question but i was having a chat with gpt about simulation theory and proposed one and then asked it to try in mathematical equation
cosmic garnet
dont use ai for math things
little sundial
whats ur question here
pearl pondBOT
@swift lodge Has your question been resolved?
vernal grove
why did they give the CDF of the max function for (a)
pearl pondBOT
why did they give the CDF of the max function for (a)
vernal grove
they asked for pdf
shouldnt it have been
n(y/theta)^(n-1) (1/theta) ?
whats tripping me out is that they ended up doing that for part b
red relic
Here’s a good way to start visualizing the question. Let $n=2$ and $\theta=1$. For some $r\in[0,1]$, what is $\mathbb{P}(max{Y_1,Y_2}\leq r)$?
jolly parrotBOT
Oreo
red relic
Hint: draw the area in the Cartesian plane that represents the sample space and the area where the event is true.
From this, it’ll be a easy to generalize
pearl pondBOT
@vernal grove Has your question been resolved?
vernal grove
Here’s a good way to start visualizing the question. Let $n=2$ and $\theta=1$. F...
no I mean why did they use the CDF formula for q1
when its asking for pdf
red relic
The first question is asking about the cumulative density, because that’s what you derive first usually
red relic
no I mean why did they use the CDF formula for q1
The wording is confusing though, density usually refers to pdf. In this context though, I’d implore students to first find the CDF
hasty moss
Hey I was wondering what can I do here to find the derivative from here !
light helm
firstly you're missing () around the 64x
then, if you don't want to use product rule, distribute/simplify first
tardy oak
yes you dropped the ()
hasty moss
ok so i might be stupid but how do i do the power of 1/3 to 64 
tardy oak
recall 4³
restive briar
$x^{\frac{1}{n}}=\sqrt[n]{x}$
jolly parrotBOT
Luca M
hasty moss
ahh yeah ok ok but like uhmm i actually suck at square roots i still don't even know what to do when theres a number in front of the square root lol (talking about the n)
is it like how many times i multiply n by it self to get to an answr so like using that it would be 3 sqrt 64 so its like n * n * n = 64?
sorry if i sound stupid but yeah i never was taught how to do square roots and im in calculus :)
restive briar
yeah thats right
hasty moss
oh hell yeah !!
ok so here do i do the product rule to each one
and then would multiply whats everything with the outside thingie
tardy oak
I mean you can use the product rule but it kind of feels unnecessary
hasty moss
im afraid its the only way ive been taught lol
tardy oak
$f'(x) = u'v + v'u$ so yeah just do the product rule
jolly parrotBOT
eththorn
hasty moss
and my proffesor hates it when we do another way which is dumb i love my shortcuts but she takes points off when we do
tardy oak
im afraid its the only way ive been taught lol
I meant you can expand and use the power rule instead, if you feel like it.
hasty moss
Yeah im more used to it
tardy oak
and my proffesor hates it when we do another way which is dumb i love my shortcu...
in that case you should just use the product rule, because that would be expected
hasty moss
ok here now what do i do
do i multiply the first thing by both of them
or leave it alone lol
tardy oak
do i multiply the first thing by both of them
well you were supposed to expand before applying the power rule
hasty moss
expand?
tardy oak
because now you differentiated a product like it was a sum which is a bit dodgy
tardy oak
sure, so you multiply the first term by all inside the brackets
a(x + y) = ax + ay
just how multiplication works
hasty moss
so wait for this one can i do 3 times 4 and add the 1/4 and 1/3
tardy oak
could you show what you mean by that exactly?
tardy oak
ah yes you can do that
hasty moss
lmao im so dumb
tardy oak
it's just the distributive law
tardy oak
the thing you showed, ignoring the bottom right, yes
you do need to use index laws to simplify before you can continue differentiating though
hasty moss
would this be right?
oops ignore the last one
im not multiplying them so its not right
but the ones before it are
tardy oak
yeah and you differentiate them
hasty moss
I solved it in the end ! Thank you again tho !!!!
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Available help channel!
north talon
Let a, b be positive integers such that a^2 - b^2 = 5005, then what is the sum of all possible values of a?
north talon
i know that a - b has to be a factor of 1 x 5 x 7 x 11 x 13
is there a faster way to do this question other than just testing every single case?
dapper kraken
i dont think so but its not that much
north talon
okay
dapper kraken
maybe someone else have a diffrent way
compact ridge
note that sqrt(5005) is around 70
so since a - b < a + b that means that around 70 is the upper limit for a - b
I guess list out all the factors of 5005 less than that
1, 5, 7, 11, 13, 35, 55, 65
the next one is 77 and 77^2 > 5005
compact ridge
no worries!
north talon
.close
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Available help channel!
echo elk
can someone help with this 1?
pearl pondBOT
can someone help with this 1?
sudden badger
!status
pearl pondBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
sudden badger
Do you know what a median is
sudden badger
its like a line which bisects the angle is 2 equal parts
that’s a bisector
a median is a line that drawn from the vertex, to its opposite side
or now that I think about it, it can be an angle too
but for this one it’s something like this
sudden badger
you know how to find the equation of a line?
we have point A(3,5) which is on the line we want to find, since it goes through it
echo elk
yea but i want to know that , we have to find D first and then find the eqtn
sudden badger
yep
echo elk
but does a median divide the opposite line equally?
sudden badger
yep
jolly parrotBOT
higher's secret twin brother
oak quiver
Do you know that cosex²x - cot²x = 1
compact ridge
then you should be able to multiply top and bottom by (cos t + sin t + 1)
oak quiver
@echo elk
echo elk
then you should be able to multiply top and bottom by (cos t + sin t + 1)
yea ik that property but i just dont get this
compact ridge
yea ik that property but i just dont get this
you want to use cos^2 t + sin^2 t = 1 and various identities
,w (cot t + csc t - 1)/(cot t - csc t + 1) simplify
echo elk
what?
compact ridge
also there's a typo in the q
should be (cos t + 1)/(sin t)
that should be the RHS
echo elk
and when we find the common ratio between 2 values who are in a G.P , do we divide the first by second or the second by first?
compact ridge
and when we find the common ratio between 2 values who are in a G.P , do we divi...
second divided by first
cause for $a, ar, ar^2 \cdots$ the common ratio is $\frac{ar}{a} = \frac{ar^2}{ar} = \cdots = r$
jolly parrotBOT
higher's secret twin brother
echo elk
sm1 please tell me then answer ,i want to cross check it
compact ridge
sm1 please tell me then answer ,i want to cross check it
if n = 1 all 3 numbers of m work
if n = 4 then only m = 2 works
so just 4/9
oak quiver
!1q
pearl pondBOT
It is suggested that you limit yourself to one question per help channel, opening a new one once your original question is answered and your original channel has been closed. This is to make your channel easier to follow for potential helpers and can bring attention to the fact that your question has changed.
oak quiver
!noans
pearl pondBOT
The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.
oak quiver
You should send your calculations instead
echo elk
in this question coordinates of p will be ( x , 0) right?
compact ridge
in this question coordinates of p will be ( x , 0) right?
yes
echo elk
and the ratio will be 2:1 right?
oak quiver
Yes
compact ridge
and the ratio will be 2:1 right?
so a 2:1 ratio of the x-coordinates 3 and 4 is?
oh wait that's the ratio yes
I probably gave you the next part of the q
echo elk
i have solved this and got a , b what does factorize the expression completely mean??
echo elk
yes
oak quiver
Did you find a and b?
echo elk
yes , i know how to but didnt find them
oak quiver
Kk
echo elk
wait ill find them
echo elk
ik about quadratic , cubic isnt in my syllabus
oak quiver
Factoring cubic is the same as quadratic
Do you know any 1 root of this cubic?
Lets say the cubic is ax³+bx²+cX+d and alpha is a root
Then it will become (x-alpha)(px²+qx+r)
Then factorize the quadratic and you are done
Lol
echo elk
no i clearly remember that this wasnt taught in class
midnight haven
lol
pearl pondBOT
@echo elk Has your question been resolved?
carmine zinc
@summer gorge sry went to eat
pearl pondBOT
<@1245017023835148302> sry went to eat
summer gorge
<@1245017023835148302> sry went to eat
np.remember that if a function satisfies f(x+y)=f(x)f(y) then the function is in the form of a^x
carmine zinc
np.remember that if a function satisfies f(x+y)=f(x)f(y) then the function is in...
Ye
I did think on those lines
but
acoustic path
np.remember that if a function satisfies f(x+y)=f(x)f(y) then the function is in...
(+ differentiability at 0)
summer gorge
yes now that you know that f(x)=a^x ,this means that f'(x)=a^x log a
we know that f'(0)=1 i think with this you can find f(x)
carmine zinc
yes now that you know that f(x)=a^x ,this means that f'(x)=a^x log a
thats just loga = f'(0)
summer gorge
thats just loga = f'(0)
its a not x mb typo
summer gorge
yep e^x
summer gorge
so i think you can say how f and f' behaves
carmine zinc
summer gorge
why did this question take me literal hours
brainfreeze it happens its np
carmine zinc
brainfreeze it happens its np
fair enough Ig, i should probably stop working on math and just go to sleep
.close
pearl pondBOT
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icy narwhal
does someone know how to solve 2x =sin x
pearl pondBOT
does someone know how to solve 2x =sin x
icy narwhal
<@&286206848099549185>
cunning comet
!15min
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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
midnight haven
does someone know how to solve 2x =sin x
is this thing ur question?
icy narwhal
is this thing ur question?
yes
midnight haven
mathwa
y
x = 0
use graphing calculator
it intersects at 0
so x = 0
got it
?
red line is graph of 2x
and blue is graph of sin x
@icy narwhal
midnight haven
wait i will do
cunning comet
wait i will do
!nosols
pearl pondBOT
wait i will do
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.
@icy narwhal Has your question been resolved?
pearl pondBOT
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sterile turtle
pearl pondBOT
jolly parrotBOT
Sho
sterile turtle
What have i done wrong?
sinful nebula
What have i done wrong?
how did you get that 6-
sterile turtle
3x2
pearl pondBOT