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#help-19
modest plinth
Anyone know where lm going wrong?
odd edgeBOT
Anyone know where lm going wrong?
modest plinth
Trying tó Find A
clever fjordBOT
lean willow
Why did u write it as a negative
modest plinth
Just pretend it's negative it doesn't matter
I used radians instead of degrees by accident
Help 😦
glass trench
Anyone know where lm going wrong?
where does the negative sign come from
modest plinth
I used radians instead of degrees
either way the answer is roughly 0.98 + or -
i checked
modest plinth
and when i put arcsin in the calculator it comes out the same as the initial input
glass trench
,w 5*sin(80 degrees)/8.67
modest plinth
I did like I said
clever fjordBOT
glass trench
ok so i have .5679
glass trench
Anyone know where lm going wrong?
not 0.022
glass trench
bruh
modest plinth
I can't do math cause i keep doing rshit like this
glass trench
ok well yeah tahts what you did wrong then
but also
stick to the correct units
it happened to work this time
but it usually doesnt
modest plinth
well first I realised i had the wrong units
so I went back and tried with the correct ones
and got the same mistake
it was like a red herring in a way
so yea
google for some reason always prefers radians, should use a diff calc
thanks for the help though
glass trench
google for some reason always prefers radians, should use a diff calc
you can specify degrees
instead of radians
glass trench
then use wolfram
modest plinth
yea will do ty
glass trench
np
modest plinth
.close
odd edgeBOT
odd edgeBOT
Available help channel!
brisk dew
how are you supposed to solve this? the way i solved it is making two variables, a and b and a being the volume of the positive portions and b being the negative portions, then i setup the equations a-b=3 and a+b=13 and then solved for b and got 5, and the answer is -5 because its gotta be negative. my question is, is this the way your supposed to do it? Also why did i not get -5
odd edgeBOT
how are you supposed to solve this? the way i solved it is making two variables,...
@brisk dew Has your question been resolved?
rigid bramble
how are you supposed to solve this? the way i solved it is making two variables,...
It’s just to do with the signing conventions you chose at the beginning. By writing a - b = 3, you’re already taking into account the fact that b is negative. When you plug your value for b in afterwards, you get a - 5 = 3.
If you instead decided to set up your equations as a + b = 3 and a - b = 13, you’d get b = -5
It’s really the same thing either way, it’s just about whether you decide to take whether b is positive or negative into account before solving or not
rigid bramble
No problem :)
brisk dew
.close
languid oar
how do I do this question?
odd edgeBOT
how do I do this question?
rigid bramble
There are a couple of ways of going about it
Let me try to answer your question with a question: what do the factors tell you about the roots of the equation?
For example, if x - 3 is a factor, what information does that give us about one of the roots?
rigid bramble
Yup
And what does the root of a polynomial mean?
What would happen if you plugged a root into a polynomial?
languid oar
it'd equal to zero
languid oar
yes
rigid bramble
Yes!
rigid bramble
Another way of doing it would be dividing (x-3)(x+5) into the cubic using polynomial long division, and since you know that’ll divide evenly, you can let the remainders equal 0
It’s a bit more work but that’s another possibility
languid oar
Another way of doing it would be dividing (x-3)(x+5) into the cubic using polyno...
wouldn't that not be as nice, because you'd have to juggle having variables and numbers together as coefficients?
rigid bramble
No problem 😄
languid oar
Another way of doing it would be dividing (x-3)(x+5) into the cubic using polyno...
can you demonstrate this method
rigid bramble
Hope you can follow that
I’d usually do it neater but it’s half past midnight and I gotta sleep now lol
languid oar
I’d usually do it neater but it’s half past midnight and I gotta sleep now lol
holup u in UK too?
🤣 this is my homework due tomorrow
rigid bramble
holup u in UK too?
Ireland, but close enough lol
rigid bramble
You too 😎 👍
languid oar
.close
odd edgeBOT
languid oar
!close
odd edgeBOT
Available help channel!
ornate hare
odd edgeBOT
dim sandal
!15min
odd edgeBOT
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.
ornate hare
but please let me know if this is correct
dim sandal
Looks good to me
odd edgeBOT
@ornate hare Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
proud fog
odd edgeBOT
royal herald
What’ve u tried @proud fog
proud fog
This might sound odd
But
I know how to do it
I think
But I just need the answers coz it takes me a while
royal herald
We don’t give answers in this server unfortunately
proud fog
Ok
royal herald
If ur looking for answers, look online
proud fog
Can u confirm my working out tho?
royal herald
Yea
proud fog
A) So u make the denominators the same by multiplying by the first fraction by root2 and the second by root3
Then u add and simplify
Correct?
royal herald
No
Look at the instructions
First rationalize denominators
How would u rationalize 1/sqrt3
royal herald
Yes
Correct
If u don’t have to show work, do them in ur head and write down the answers
Stimulates your brain
odd edgeBOT
@proud fog Has your question been resolved?
cold sage
odd edgeBOT
@proud fog Has your question been resolved?
mellow notch
odd edgeBOT
mellow notch
I need help with parts i and ii
for part i, I've managed to prove than when a < 1, the sequence a_n is both bounded below and is a constantly falling sequence, hence proving that a it's convergent. But idk how to actually calculate its limit value, which i believe should be 0.
and for part ii, i have no clue how to do it.
any help is highly appreciated
open onyx
i: use the fact that a_n is given to be positive
mellow notch
yoo @open onyx thank god you're still up
mellow notch
i: use the fact that a_n is given to be positive
using this we can prove the existence of a limit
but how do we calculate it?
open onyx
you know that it’s bounded below and that it shrinks
which means regardless of the value of a_k you know a_{k+1} is a smaller positive real number (assuming sufficiently large k that the limiting behaviour is dominant)
so given that you can always find a smaller positive real number in the sequence, show that 0 is the limit
open onyx
this is where you use ε-L
mellow notch
this is where you use ε-L
is this the thing from the definition of limit?
open onyx
yes
open onyx
yes (more commonly you’ll see the limit denoted as L hence ε-L)
mellow notch
what I have in my notes: there exists Nsuch that | a_n - L | < ε for all n>=N
do we just guess L to be zero?
and plug it in?
open onyx
yup
mellow notch
or do we get to the 0 value somehow
would it work with an arbitrary sequence that isn't defined like in this case though?
open onyx
if you can show it holds for some L then L is a limit for that sequence without needing anything else
mellow notch
ok let me try doing it
odd edgeBOT
@mellow notch Has your question been resolved?
mellow notch
yo im a but stuck
so |a_n - L| < ε right
and we assume that L=0
so |a_n - 0| < ε right?
a_n < ε
this means that all the terms of a_n are smaller than ε but bigger than 0
since our sequence was positive real numbers
mellow notch
How would I find an N or ε when the thing is arbitrary though
that's what confusing me
open onyx
use the limiting behavior of your sequence
you should be able to bound it above by a geometric sequence (note you can freely bound this above because if the upper bound<ε then so does a_n)
mellow notch
geometric sequence?
open onyx
$a_n= a_0 r^n$
clever fjordBOT
IV
mellow notch
how do I know that this arbitrary sequence can be described as such? like a geometric sequence?
open onyx
specifically because the limiting behavior is a constant ratio
which is a property of geometric sequences
mellow notch
what do you mean by the limiting behaviour?
like in what way the sequence converges?
open onyx
$\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$
clever fjordBOT
IV
mellow notch
this is the limiting behaviour?
open onyx
if you always had a_{n+1}\a_n= a then it’s a geometric sequence by definition
mellow notch
but this is the limit though
open onyx
oh i should probably call it asymptotic behavior if it makes it clearer
mellow notch
oh i should probably call it asymptotic behavior if it makes it clearer
definitely doesn't xD
umm so
I'm very confused
open onyx
ok, consider this very hastily made up example of 1, 10, 100, 1000, 499, 248, 122.5, 60..
if you consider the whole sequence it’s some nonsense
but the last 4 terms (and my intention is every term afterwards as well) is bounded above by a geometric sequence 1000(0.5)^(n-4)
mellow notch
ok+
open onyx
this is the difference between being a geometric sequence and being bounded above by a geometric sequence
the since you have $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} =a<1$
clever fjordBOT
IV
open onyx
at some point your sequence starts behaving nicely enough that it’s bounded above by a geometric sequence
but the key point is this sequence doesn’t necessarily have r=a
but instead some r between a and 1 (consider any case where this convergence is coming from above
mellow notch
ok
so at a certain term onwards
the sequence starts behaving like a geometric sequence
aka bounded by a geomtric sequence
open onyx
the sequence starts behaving like a geometric sequence
this is false and will probably be -1 marks
you actually do not have any guarantee in this regard, only the bounded by geometric sequence part
but since it’s an upper bound, if you can show this geometric sequence converges to 0 then necessarily so does the original sequence
mellow notch
so the limit of our sequence is literally a geometric sequence
but the r is not constant?
open onyx
the upper bound is a geometric sequence with constant r
open onyx
a_n itself can be whatever abomination it wants to be
mellow notch
a_n itself can be whatever abomination it wants to be
ok
so we were trying to find the value of the limit of a_n, which is probably 0
but we have to actually calculate it
so we use the definition of a limit (since we know that a_n as a limit) to formulate it
but since we don't know from which N the sequence actually starts to converge, we have to describe N as a function of ε
so far right?
and now
what now
open onyx
show that a_n is bounded above by a geometric sequence with some r ∈ (a, 1)
(you might even want to use ε-N for this)
mellow notch
when you say bounded, you don't mean like a limit right?
ε-N is how we define limits right?
open onyx
nope, i just need you to show that beyond some point, every a_n is smaller than a term in a geometric sequence with the aforementioned r
this is about 2 lines using ε-N
mellow notch
ε-N is the definition of limit |a_n - a_0*r| < ε for n>=N
a_n < a_0*r + ε
and the point from which every term is smaller than the geometric sequence is ε?
open onyx
hmm, not quite
i can just show you the argument and see if you can follow it
since $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}= a, \forall \varepsilon >0 \exists N s.t. \forall n>N, \frac{a_{n+1}}{a_n} < \varepsilon$
clever fjordBOT
IV
open onyx
and here’s a fun trick
we pick $r=\frac{a+1}{2}$ and let $\varepsilon = \frac{1-a}{2}$
clever fjordBOT
IV
mellow notch
ok
open onyx
now see if you can finish the argument using the N already given as the point where the geometric sequence starts acting as an upper bound
direct computation should be sufficient but various methods work
open onyx
once you wrap your head around it a little and can write out the geometric sequence the rest is pretty quick
open onyx
i can just show you the argument and see if you can follow it
since $\lim_{n \r...
small mistake (forgot a in the last line), should be
$\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}= a, \forall \varepsilon >0 \exists N s.t. \forall n>N, |\frac{a_{n+1}}{a_n} - a| < \varepsilon$
clever fjordBOT
IV
mellow notch
a_n+1 <a_n*((a+1/2) + (1-a/2))
for all n>=N
wait
no
|a_n+1/a_n - a| < ε would become a_n+1 / a_n < ε + a
open onyx
depends on the sign inside the abs value
mellow notch
is this hard or am i just dumb
isn't the sequence always positive?
does it even matter if we have an absolute value there after we carry over the a?
open onyx
well but the limit is on the ratio
so if the ratio is smaller than a and approaches it from below then the value inside the absolute value is negative
limit of a sequence that gets bounded by a geometric sequence is pretty confusing tbh
open onyx
which?
mellow notch
I just can't wrap my head around how we need to do all this to figure out the value of the limit
like we literally already know that the limit is 0
open onyx
welcome to the land of mathematical rigor (and tbh this isn’t very long after you get used to it but yes it’s pretty annoying at the beginning)
open onyx
sure, but it’s effectively the same thing
mellow notch
oh
open onyx
show this time that it’s bounded below by another geometric sequence with r ∈ (1, a)
mellow notch
but we don't have to determine a limit of a_n as it is divergent
open onyx
and therefore show since the lower bound diverges so must the original sequence
mellow notch
open onyx
the reason why you have to use bounds here is because the original sequence is too uninformative
and the bounds let you do a lot more
mellow notch
just question
open onyx
(and it’s much easier to put everything in words.. but part of maths is figuring out how to precisely express everything you want to say)
mellow notch
for part i, I know that it has a lower bound at 0, but do I also need to show that it has an upper bound to prove its convergence?
open onyx
if you can show by ε-N that it has a limit, you do not need any bounds at all
open onyx
it’s your formal definition of a limit
mellow notch
ok i know that one
my question
how can we apply the definition of a limit to something and see if it has one or not
like take sequence 1/n for example
open onyx
you basically always solve N as a function of ε
mellow notch
|1/n - L| < ε for n>=N
this means that at some point N onwards, every 1/n will be between the limit and the epsilon value
open onyx
ok, let’s do |1/n-0|<ε
then 1/n<ε rearranging gets n > 1/ε
then immediately for any ε we pick N = 1/ε
and for all n>N we have n>1/ε which yields 1/n<ε as required
open onyx
because we know it is :p
just like how you already know yours is
mellow notch
ah ok
open onyx
and if you weren’t shown something like that, you should see how having a closed form for a_n can let you directly solve for N
which goes back to my point about getting an geometric sequence as a bound, because a geometric sequence has a closed form formula
mellow notch
geometric sequence as a bound
wait geometric sum actually
with and 0<r<1
which would mean that the sum of the sequence is a finite value
open onyx
i mean if you can show the bounding geometric sequence has infinite sum equal to finite value that’s fine too
mellow notch
bro
I give up
I'm just gonna fail this weeks assignment
Thank you so much for helping me out and being so patient. I really appreciate it
odd edgeBOT
@mellow notch Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
wispy vine
For question 1 (b), how am i supposed to find Q and R. Apparently, I’m supposed to solve the line and parabola but idk how to get the parabola
odd edgeBOT
For question 1 (b), how am i supposed to find Q and R. Apparently, I’m supposed ...
@wispy vine Has your question been resolved?
fiery ermine
Let x,y,z be non zero reals, such that x^2+y^2+z^2=1, find the minimum value of |x|+|y|+|z|
toxic monolith
$\text{Accordingly to the symmetry let's assume that: }x,y,z\ge 0\text{ then }\\\left| x \right|+\left| y \right|+\left| z \right|=x+y+z\\\text{ in addition }z=\sqrt{1-x^{2}-y^{2}}\text{ for }x^{2}+y^{2}\le 1\\\text{ Hence let's consider the following function: }\\f\left( x,y \right)=x+y+\sqrt{1-x^{2}-y^{2}}$
clever fjordBOT
Joanna Angel
toxic monolith
Now you may identify the minimum of the function f, that should not be a problem
fiery ermine
I just take the partial derivative in terms of x and y yeah?
toxic monolith
yes
and equate them to zero, both fo them
tha is the first condition , necessry one fo exisitng the extremum value
next, you need to investigate hessian
determinant consisted of secodn derivatives
toxic monolith
yes, x and y both of them must exisit
we name it :
partial derivatives
of the first order
fiery ermine
Yeah yeah just treat y as a constant or x as a constant
toxic monolith
doyou know the theory ?
when you evalaute first derivative in respect to x, then , you treat y as a consnt
fiery ermine
Briefly
toxic monolith
and when you compute derivative in respect to y, then you treat x as a consntat
fiery ermine
yeah. I got 1 - y/ sqrt (1-x^2-y^2)
toxic monolith
$\left{ \begin{array}{l}
\frac{{\partial f}}{{\partial x}}\left( {x,y} \right) = 0\
\frac{{\partial f}}{{\partial y}}\left( {x,y} \right) = 0
\end{array} \right.$
clever fjordBOT
Joanna Angel
fiery ermine
I get x=+-1 and y=0, but this can’t be true since then the square root would be zero, and our assumption about them all being nonzero would be false
Since they are nonzero reals
toxic rose
Let x,y,z be non zero reals, such that x^2+y^2+z^2=1, find the minimum value of ...
think theres a non-calc way for this
$(x+y+z)^2 \cdot (x+y+z) = 27xyz$
clever fjordBOT
ItzKraken
clever fjordBOT
ItzKraken
fiery ermine
Well yeah but we never get out the separate abs |x|+|y|+|z|
toxic rose
so $xy+yz+zx = 1 - 729x^2y^2z^2$
clever fjordBOT
ItzKraken
toxic rose
Well yeah but we never get out the separate abs |x|+|y|+|z|
assume x y and z are positive and non-zero for now
fiery ermine
well we know that xy+yz+zx>=-1/2 if they are all@reals
toxic rose
how @fiery ermine 
$1 - 729x^2y^2z^2 \geq \frac{-1}{2} \Rightarrow \frac{1}{2 \cdot 243} \geq x^2y^2z^2$
clever fjordBOT
ItzKraken
clever fjordBOT
ItzKraken
toxic rose
bruh that gives us the maximum value
we need an upper bound on xyz
oh wait
i messed up in AM-GM
fiery ermine
The upper bound is sqrt(3) via AM-QM
fiery ermine
The upper bound is sqrt(3) via AM-QM
On |x|+|y|+|z~
toxic monolith
$$\begin{array}{l}
\left{ \begin{array}{l}
1 - \frac{x}{{\sqrt {1 - {x^2} - {y^2}} }} = 0\
1 - \frac{y}{{\sqrt {1 - {x^2} - {y^2}} }} = 0
\end{array} \right. \Leftrightarrow \left{ \begin{array}{l}
1 - {x^2} - {y^2} = {x^2}\
1 - {x^2} - {y^2} = {y^2}
\end{array} \right. \Leftrightarrow \
\
\left{ \begin{array}{l}
2{x^2} + {y^2} = 1\
{x^2} + 2{y^2} = 1
\end{array} \right. \Leftrightarrow \left{ \begin{array}{l}
x = \frac{{\sqrt 3 }}{3}\
y = \frac{{\sqrt 3 }}{3}
\end{array} \right.
\end{array}$$
clever fjordBOT
Joanna Angel
toxic monolith
that calculation i s corrct plz analsye it
and now plz uinvestigate hessian
and als correct yoruprevious calculations
both , aS you can see, are positive
fiery ermine
oh I know where I messed up silly signs
toxic monolith
🙂
fiery ermine
I unfortunately don’t know what the hessians are
toxic monolith
i write it moment
fiery ermine
Thank u
toxic monolith
$\det \left[ {\begin{array}{*{20}{c}}
{\frac{{{\partial ^2}f}}{{\partial {x^2}}}\left( {{x_0},{y_0}} \right)}&{\frac{{{\partial ^2}f}}{{\partial x\partial y}}\left( {{x_0},{y_0}} \right)}\
{\frac{{{\partial ^2}f}}{{\partial x\partial y}}\left( {{x_0},{y_0}} \right)}&{\frac{{{\partial ^2}f}}{{\partial {y^2}}}\left( {{x_0},{y_0}} \right)}
\end{array}} \right]$
clever fjordBOT
Joanna Angel
toxic monolith
if this determinant is positive then f reaches extremeum value in point
if it is negatvei , there is no extremeum
toxic monolith
the point
i have foudn for you
above
so noow, you need to plug yoru pint isndie this determinant
fiery ermine
I’m sorry but I don’t know any linear algebra:(
toxic monolith
aww
fiery ermine
Can I just check if they are zeros of the second derivatives?
toxic monolith
no
toxic monolith
when it comes to the calculus-based method, we first look for the zero of the partial derivatives, i.e. the system of equations that I found for you
and then you need to examine the determinant that I wrote at this found point,
fiery ermine
Can you maybe tell me how to calculate the detriment I’ll do that
toxic monolith
$\det \left[ {\begin{array}{*{20}{c}}
a&b\
c&d
\end{array}} \right] = ad - bc$
clever fjordBOT
Joanna Angel
fiery ermine
but this will be the maximum value
toxic monolith
btw , are you student of what sort of school mayeb it helps in selectign method
fiery ermine
not the minimum
toxic monolith
if there is no minimum value inside the aera
fiery ermine
I’m a high schooler 11th grade
fiery ermine
well I know this is the maximum since you can get that using Am<=Km
or QMS*
Like quadratic mean
fiery ermine
Yeah
fiery ermine
but they are non zero
fiery ermine
and satisfy that their squares some up to 1
toxic monolith
but moment
odd edgeBOT
odd edgeBOT
Available help channel!
odd edgeBOT
glossy sparrow
odd edgeBOT
odd edgeBOT
Available help channel!
hard field
odd edgeBOT
hard field
yo
im dreading having to find the det of (A-λI) , how would I go about this with any properties ?
I dont have any knowledge that lets me do much without actually having the eigenvalues
dreamy yacht
I dont have any knowledge that lets me do much without actually having the eigen...
Are you not allowed here to look for the eigenvalues?
hard field
I can
it's just going to be quite the timesink I think
I can maybe manipulate (A-λI) to have only 1 non zero coefficient when finding det(A-λI)but it would still take centuries :((
boreal crag
Well it would take 5 minutes probably
Is the determinant of A 0?
Actually that's probably not important
hard field
Is the determinant of A 0?
no
hard field
Well it would take 5 minutes probably
howww
there are so many computations is so poop
hard field
alright
and once I get them im expecting a to have some algebraic multiplicity of eigenvalues
it then has to match the number of eigen vectors (for that eigenvalue)
boreal crag
Obviously for B choose the row/column with the 0
The number of independent eigenvectors yeah
Geometric multiplicity
hard field
ye
this shit is unbearable
I get to the end
cant factor it
mistake made somewhere on previous lines
cant find it
fuck off 
clever fjordBOT
boreal crag
did you get this
hard field
yes but 439 was on the end originally
:)
so values of 3 , -1,-1
andddd
now to find eigenvector
(s)
ok
got it
and it worked
B can actually FUCK Off
im not doing that ever again
got {[-2,0,1],[-1,1,0]}
now are those the eigen vectors?
and we say the eigen space is the space spanned by those vectors?
boreal crag
I think every eigenvalue has it's own eigenspace
Really you can probably just use wolfram
left knot
help
left knot
I know the derivative of ln(x) is 1/x
so H'(z)= 1/(the stuff inside the log that im too lazy to write out)
spiral raptor
nope
spiral raptor
I know the derivative of ln(x) is 1/x
this statement is incorrect actually
left knot
ok
left knot
oh
spiral raptor
so u have to differentiate wrt (im lazy too) in order to use that fact
spiral raptor
with respect to
left knot
ok
spiral raptor
u havent studied chain rule?
left knot
my teacher went over it like once in 5 minutes
and expects us to know it
but im not great with it
I know f'(g(x))*g'(x)
spiral raptor
ok how about i give u a example
spiral raptor
I know f'(g(x))*g'(x)
yea thats the rule
but its hard to understand in that form
so heres the example
u need to diiferentiate z wrt x
instead u can differentiate z wrt y and multiply that by the derivative of y wrt x
spiral raptor
umm ok ok
spiral raptor
u see how the du's cancel out right
left knot
yeah
spiral raptor
if u want to see y this works i'll link a vid at the end
but now lemme show how to solvve that prob
left knot
ok
left knot
ok
spiral raptor
but consider differentating wrt that whole mess inside the ln
that thing can be treated just like this du
left knot
ok
spiral raptor
ok imma try a doodle to show
left knot
so dx here is our dz
ok
spiral raptor
i dont even have a mouse sorry; there s is that big thing
so if we differentiate wrt that s
we can directly use that fact u mentioned before
d(ln(x))/dx = 1/x
left knot
i dont even have a mouse sorry; there s is that big thing
what is on the bottom of the second fraction?
spiral raptor
dz
left knot
wait the second one is also dz?
spiral raptor
the first one isnt...
the first one is ds
then with the top ds in the second fraction
it cancels out
left knot
ok
spiral raptor
but dont cancel them!!
left knot
ok
spiral raptor
they do if u want to
and see that the equation is true
but our goal here is to differentiate
left knot
ok
spiral raptor
yea show me ur final results or i'll be unsatisfied
left knot
ok
spiral raptor
💀
spiral raptor
yep
left knot
ok
spiral raptor
u can differentiate that wrt z right
spiral raptor
i might die in like 5min btw
left knot
ok
left knot
yeah
spiral raptor
what
spiral raptor
yep
left knot
or howevert its spelt
left knot
ok
spiral raptor
u okay with exponentials?
left knot
yeah
spiral raptor
and product rule?
left knot
should i write the sqrt as 1/2
left knot
and product rule?
yea
spiral raptor
quotient rule?
spiral raptor
should i write the sqrt as 1/2
yea that helps
left knot
less yeah but it should be fine
spiral raptor
if u have difficulties with the quotient rule
u can just use the product rule
u know that right
left knot
yeah
spiral raptor
okay so
the second fraction
Differential of something that looks like (a^1/2)/(b^1/2)
right?
left knot
ok
spiral raptor
and ull see what i mean
left knot
ok yeah i think i get it
spiral raptor
root of something divided by root of something else
left knot
yeah
left knot
ok
spiral raptor
if u can quotient is better
left knot
ok
spiral raptor
and while u do this using chain rule is easier
like how we used the chaos as u in this
we can stilll use the same trick if we face any more difficulty
so yea imma die now
if u need more help
!15min
odd edgeBOT
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.
spiral raptor
sleek falcon
odd edgeBOT
echo ginkgo
$(U\Sigma)^T = \Sigma^TU^T$
clever fjordBOT
aPlatypus
sleek falcon
idk wdym
sleek falcon
that's a general thing with transposes
sorry i put +
odd edgeBOT
@sleek falcon Has your question been resolved?
odd edgeBOT
@sleek falcon Has your question been resolved?
sleek falcon
right there
could you use something like I = I^2
echo ginkgo
why
sleek falcon
my prof said to do that
thin anvil
5 coins are flipped at the same time what is the probability of 3heada and 2 tails
hollow hearth
shouldn’t be too hard
do you need to draw a table or are you allowed to use other things to solve
thin anvil
I’m allowed to use other things
hollow hearth
alright
so let’s make the probability of this event occurring is the probability of 3h2t over the probability of all possible outcomes
correct?
thin anvil
Yes
hollow hearth
oops. not probability just numbers of outcomes
ok anyways
the denominator is easier to start with
how many different combinations are there?
(also are the coins distinct or no)
hollow hearth
i think it should be 2^5 instead
thin anvil
Can you please explain why?
hollow hearth
every coin has heads or tails right
thin anvil
Yes
hollow hearth
so we use the uhh
it’s like fundamental theorem of counting or smth
it’s basically you multiple the outcomes together to get the total amount
thin anvil
Oh
hollow hearth
not 1/2 just 2, we are calculating total outcomes first, and then using fractions
you could use it, but it doesn’t really make a different
difference*
anyways we have our denominator as 32 right now
thin anvil
Yes
hollow hearth
let’s say we have coins 1 2 3 4 5
thin anvil
Yes
thin anvil
Nope
hollow hearth
it will speed it up so you don’t need to count
n choose k is n!/(k!)(n-k)!
so for here we have 5 choose 3
thin anvil
Ye
hollow hearth
so plug into the formula and you will get the amount of outcomes
hollow hearth
yep
boreal wave
help
royal herald
What should be 2?
royal herald
Why would it be 2
royal herald
f(2x+1)
boreal wave
yes
royal herald
The derivative of that is
boreal wave
2
royal herald
f’(2x + 1) * (2x+1)’
royal herald
Chain rule
Derivative of the outside function times derivative of the inside function
boreal wave
outside function times inside function times derivative of inside function
but
what’s the outside function
??
royal herald
f
boreal wave
hmm
royal herald
f()
frozen stream
What does canonical mean?
I have to prove that (Z^n, •, const0) has a structure of a canonical group
frozen stream
"Let (G, ·, e) be a group, and X be a set. Show that (G^X, •, conste) with (f • f′) : x → f(x) · f′(x) forms a group. Conclude that ∏nZ has a canonical abelian group structure."
I've shown that it's an abelian group (each f has inverse f' and • ist commutative)
Only the canonical property is left.
echo ginkgo
can't you just use the first proof for that ?
"schliessen" it means "deduce" I suppose
that's what they want you to go for
echo ginkgo
it can mean a few things, the meaning here is prolly "it's a standard construction"
which is irrelevant for the proof, just show it's an abelian group
@frozen stream
and the idea is, can you write $\prod_n \mathbb Z$ as $G^X$ for some $G$ and $X$ ?
clever fjordBOT
aPlatypus
echo ginkgo
do you have a specific meaning for canonical in your class ?
frozen stream
We defined canonical projection, canonical embedding and canonical homomorphism (Z → R)
But not specifically for groups.
echo ginkgo
well if they didn't define canonical group structure just consider it's a fluff word
also don't forget the abelian part
but it's pretty easy to see why it works
shrewd cosmos
Can this be solved?
odd edgeBOT
Can this be solved?
ember oak
Can this be solved?
Do you mean simplified?
shrewd cosmos
yes
clever fjordBOT
SWR
shrewd cosmos
bruh, i just randomly wrote that, lol, didn't even notice I have 1^π.
ember oak
,w -i^(e*sqrt(-i-1))+3/i
clever fjordBOT
sleek falcon
im confused is the low rank formula just
odd edgeBOT
im confused is the low rank formula just
sleek falcon
ie for rank 1, A = u1 sigma1 and V1?
low locust
odd edgeBOT
@sleek falcon Has your question been resolved?
zenith terrace
I'm confused why is the graph going to the left
odd edgeBOT
I'm confused why is the graph going to the left
crisp wadi
I'm confused why is the graph going to the left
Because you can't plug in values of x that are greater than 4 for that function
crisp wadi
Isnt it supposed to go right because b is over 0
b is not positive in this case
zenith terrace
Because you can't plug in values of x that are greater than 4 for that function
Im confused wdym
crisp wadi
Try rewriting 4 - x in the form b(x - h)
dense thorn
physics problem; somewhere ive made an error, but all of the calculations are correct (i've triple checked)
dense thorn
sorry for the sloppy work, wasn't expecting to have to show to anyone else
odd edgeBOT
@dense thorn Has your question been resolved?
dense thorn
u forgot to convert from rpm to radians
if you look just under the calculation for angular velocity at 10s, i convert it to rad/s just underneath
i used rad/s as my units for angular vel in the linear velocity calculation
i can't see why it'd matter at what point i convert to rad/s in as long as its consistent throughout the angular velocity calculations
odd edgeBOT
@dense thorn Has your question been resolved?
dense thorn
realized it was a dumb error
in the ang vel to linear vel i used diameter not radius
.close
odd edgeBOT
hazy tartan
i need help with telling if something is a proportion because i dont understand at all
hazy tartan
i dont know if the first one is right and i still need to know how to do the other ones
<@&286206848099549185>
hazy tartan
yes
half fulcrum
are you sure?
half fulcrum
can 16/21 be reduced at all?
16 = 2x2x2x2
21 = 3x7
does anything cancel?
(2x2x2x2)/(3x7)
hazy tartan
no
half fulcrum
ok
then 16/21 does not reduce
that means 16/21 cannot be rewritten as a smaller fraction
so what would that mean then
its asking if 16/21 can be written as 6/9
hazy tartan
?
half fulcrum
can 16/21 be written as 6/9
hazy tartan
no
hazy tartan
no
half fulcrum
ok
hazy tartan
half fulcrum
we're trying to find all values of x that make the statement true
hazy tartan
ok
half fulcrum
is multiplying both sides by 12 a valid thing to do?
hazy tartan
yes
half fulcrum
ok
half fulcrum
yes
hazy tartan
i need help with these
half fulcrum
you just cross multiplied to solve the previous one right
so can you do that here too?
carmine hatch
gotta be same denominator
native yew
how do i use the squeeze theorem to find the limit to (cos(1/n)-1)/(1/n)?
native yew
i know that by the squeeze theorem -1 <= cos(1/n) <= 1
solemn jay
how do i use the squeeze theorem to find the limit to (cos(1/n)-1)/(1/n)?
limit as n approaches infinity?
native yew
and dividing by 1/n is the same as multiplying by n so im finding the limit for n(cos(1/n)-n
native yew
limit as n approaches infinity?
im assuming
i can get a screenshot of the book but it doesnt say anything
native yew
and dividing by 1/n is the same as multiplying by n so im finding the limit for ...
ive tried multiplying both sides by n and then subtracting n but it doesnt work
-n-n <= cos(1/n) <= n-n
-2n <= cos(1/n) <= 0
the answer is supposed to be 0
not sure what im doing wrong
solemn jay
it’s a sequence so yes it is infinity
native yew
-2n <= cos(1/n) <= 0
yeah but i got negative infinity is less than the sequence which is less than 0
the left side should also be 0
but idk what to do
pale scroll
P1 + P
о
cos (v - р) t cos („ + 1) t ^ + ^ ,(i_
х ?
V>x(t) = V.(t) = f(x + t) + at + b,. \L(t)eSf°, (3,1)
= 1(^ 1 ) = о,.
|[А(0 —2[А(0) + (L(—)|<<о2(/)
(х(- t) = 0 (in ^L_o>8 -()),. (3.2)
f(x + t)-2f(x) + f(x-t) = li(t)-2lK(0) + lL(-~t)==l
,(t) + l,(~t),
2tt utx
p+l l
tg /J \ 1 С /.v cos (l — r) t — cos (p 4-1) 7.1 , /p/ / 1 \
^.p(/.* ) = 5T(7+Ij } v(i ) ~ t» -еy + 0(а>2 ( t ))
о
27Т UTT
= m^uТ 5 v<> ch*
- °(cht))«
О
P ' R + x = WTv^u ??
native yew
<@&286206848099549185>
odd edgeBOT
@native yew Has your question been resolved?
odd edgeBOT
@native yew Has your question been resolved?
odd edgeBOT
@native yew Has your question been resolved?
odd edgeBOT
@native yew Has your question been resolved?
odd edgeBOT
@native yew Has your question been resolved?
wintry temple
can someone tell me the answer for the first one i don't have answer
