#help-36
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post your question first
but if i post it and you dont know how to solve it the timer is gonna run out
you can extend the timer by telling the bot your question isn't solved yet
wym
there is no timer
is it graded?
yea
we cant help with graded assignments, that violates academic integrity
wym by graded
U want us to provide you with the answers?
it wouldn't otherwise make sense that it is timed
wait u guys dont help w hw
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yoo
my question is for the equation of the generating curve (the radius) how do you know which plane thats on
for example how do i know that r(z) = y and not r(z) = x
well r(z) should only depend on z, shouldn't it
right
but whats the output
could be either y or x
neither?
wdym
we aren't doing r(z) = x or r(z) = y
wait what
we are doing x^2 + y^2 = [r(z)]^2
it's the radius of the surface at a given z
yes but that radius is represented by an output value of the radius function
yeah and we aren't graphing that output value, we're graphing an equation related to the output value
wait
its squared
so even when from the other direction
itll be the same
is that why
i'm not sure what you mean
but even then
the generating curve has to be on a particular plane
if my generating curve was in the yz plane then my radius would be z
if it was on the xy plane my radius would be x
so idk
the intersection of the surface of revolution with any plane containing the axis of revolution gives a generating curve. it doesn't matter which plane because they all give the same curve, just rotated
look
it just says "an" equation of "a" generating curve
you can pick any of them
also you should take the square root to find r(z)
true
that would just be what i had right?
well its what you have in the bottom right
yea
although it should say r(z) =
i have a question, howd you get good at math
would you say someone needs to practice since they were a kid to get to that level or you could manage that within a couple years later along the line? No
not really, you can get there eventually
we should also note that this is math that i learned years ago and have been continuing to practice since then by helping people out with it. so i'm much better at it now then i was when i was learning it, and i'm better at it than the stuff i'm learning currently
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induced emf = rate of change of flux linkage
how do they know its like the yellow one
and not the blue one
oh
its negative derivative
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im abit confused about part 3b
shouldnt tha area for A1 be [....]-2 and -4
not -2 and 4
You need to figure out when is y positive
then substract the integral of the part between -4 and 4 where y is positive with the integral of the part between -4 and 4 where y is negative
you understand?
cant i just add all 3 areas
no because when x is negative, the integral is negative
using abolute values
so changing it to positive
yea but thats just = to what i said
so is area one correct?
.
or should it be this instead
huh
im trying to figure out all 3 areas
ah okay
so
Is this how I calculate area 1 then
yea
so the answer to the question is 36.333
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you seem to consider that 6 is the only outcome of your d20
sure, but then if you know in advance that you're gonna keep rolling 6s, then I guess your strategy works.
but that's if you are 100% certain you are going to roll nothing but 6s
oh hm
that's fair. I think that this strat works with high remaining roll count
but I remember seeing this exact problem from you and a couple of helpers have given you a good strat as to when to cash out.
I think you can use the general solution to find your specific case, is that not?
I see
if you cash out do you have to reroll on the next round?
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First of all, the dx is missing everywhere
And that's the first mistake 😬
Second, yes how did the x² turn into just x?
yep
$x^2\left(\frac{1}{x}\right)dx=xdx$
Allen
so its not like differentiation where the chain rule is used for composite functions?
and that is like IBS
DI method by bprp
and +c
yes thanks
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not done
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there’s no chain rule for integration
Substitution (usually) works well if you have the derivative of some part of the integrand multiplied with the integrand itself
yep this is the inverse (kinda) of chain rule
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Ok theres some stuff here I need clarified
Where is the question?
Im Just trying to understand this concept
Well
This concept
In that idk what they mean by the word "concept" here
All I've understood so far is that under PAC, we only need the learner be such that it chooses a hypothesis which with high probability gets a low enough loss
As opposed to exact framework where the loss is expected to be 0, and hypothesis is chosen with a 100% probability
Also is this class of algos
Something which is tunable with epsilon and delta params?
(i.e., the accuracy and confidence parameters)
<@&286206848099549185>
it doesn't seem different from "subset", as far as i can tell
i'm not actually sure..
In computational learning theory, probably approximately correct (PAC) learning is a framework for mathematical analysis of machine learning. It was proposed in 1984 by Leslie Valiant.
In this framework, the learner receives samples and must select a generalization function (called the hypothesis) from a certain class of possible functions. The ...
This si the wikipedia artikel im looking at
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can someone solve 11 where the limit is in the form of x^m - a^m / x^n - a^n ?
You can factorize the numerator and the denominator
It's a little difficult but the limit gives you a big hint
I tried to solve it myself and i got 20/27 instead of 20/3
Im not sure whats wrong
I still haven’t taken that yet unfortunately
Do you know polynomial long division?
/synthetic division for polynomials
/Horner's method
not that one either
my teacher solved it by turning it from x-> -2/3 to 3x = -2 then at the end substituted a for -2 instead of -2/3 but im not sure how that’s right
I'm not sure about your teacher's method, and I might be overcomplicating this, but if you know a root r of a polynomial P(x), you can rewrite it as (x - r)(Q(x)) where Q(x) is one degree lower than P(x)
hi
hmm i’ll try to figure it out, thank youuu
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how do i know there are vertical and horzontal forces at A?
@rain sentinel Has your question been resolved?
pins always have horizontal and vertical reactions
thanks
You good with calculating torque/moment?
yes CWM-ACM
Well, theres a neat rule about torque, its that its the same everywhere, for this case, considering that the piece is static and not rotating, its 0
It makes it pretty easy to find the perpendicular components of most parts
Suppose we use the point C as reference, we know that the torque caused by vertical component of reaction A is equal and opposite to the force in B.
You can do the same from A, considereding the vertical component of the strut force
From there, you can find both the horizontal of the strut, and the horizontal of the reaction
how?
Do you know how projections of forces using cosine and sine works?
can you expand
On a different but similar example.
Lets say I apply a force at an angle you know of, would you be able to tell me the vertical and horizontal components?
yes the x and y axis
Well, yeah, but theres a formula for it
For this situation, imagine i gave you the Value of "F" and the angle "α"
You could find the vertical components anyways.
Fx = F . cos(α)
Fy = F . sin(α)
yes ik this
Well, you can also do it the other way around
Fy / sin α = F
So, if you know the vertical, and also the angle it comes from, you can know the real force
and by extension, also the horizontal.
On our problem, you can probably see that the only real force we know is purely vertical
But there is a horizontal reaction, and its because the stunt is pushing against the wall too, so the reaction has to be keeping the bar attached to the wall by tugging on the opposite side
Thats how you get the true reaction
because every action has a reaction
so because its pushing on the wall the wall pushes back leads to horizontal force
For example, we wouldnt get no horizontal reaction if the stunt was purely vertical too
On a really "stupid" sense, the horizontal reaction comes from the fact that we have to do extra work to keep the bar static, because some is wasted in pushing the bar away.
The vertical blue force coming from the bar is the one mantainind the bar from not falling
But because it comes at an angle, theres some part of the real force that is "wasted" pushing the bar away.
So the reaction at the left side must be keeping it from just flying away
normal force is the blue force?
and is a consequence of gravity
We are not considering gravity here, just the force at the right side
this is the horizontal force
and the purple force?
is a result of fv and fh?
its the combination of the two
aka, the real force which is coming at an angle and we decomposed into the horizontal and vertical
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How can i find the derivatives of sin cos and tan?
well for sin and cos it depends on how you have defined them
(assuming you're asking how to prove that sin' = cos, for example)
Yeah
there are a couple of standard approaches
one defines sin and cos as the solutions to certain differential equations
another defines sin and cos as certain power series
and in both cases you can rigorously show that sin' = cos and cos' = -sin
for tan, it's easy, just write tan = sin/cos and use the quotient rule
😂
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why is u+ υ = M?
isnt it u + υ = t
i dont understand how we went from the first region to the second with the substitution
?
can someone explain
You would have three variables, obv you wanna rid t
and how would ido that
like on the graph u and υ how do i find the new region
t runs from 0 to M
thats what i dont understand since t = u + υ
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Im having trouble understanding the surjectivity of this proof:
The existance of T should not have such importance unless its a basis right?]
thats the lemma they are talking about
neither of these are the basis so im lost as to how they made the connection
the linear map lemma guarantees that for every matrix, there is a linear map corresponding to it
so let's say we have a fixed matrix $A$, and pick fixed bases $v_1,\ldots ,v_n$ and $w_1,\ldots,w_m$. then we define a second list of vectors in $W$, $w_1',\ldots w_n'$ by
[ w_k' = \sum_{j = 1}^m A_{j,k} w_j ]
then we are saying there exists a linear map $T$ such that
[ T v_k = w_k' ]
by the linear map lemma, and by definition $A = \mathcal M(T)$
κλαουντ ☁ (cloud)
is this not only for the basis
or are we going through the linear combinations
ok nvm that makes sense
so it garenttes that for any matrix there is a lienar map corresponding to it
through linear combinations of the basis?
yes
ok we just took A to be arbrtraty here
and showed that there is a linear map to every T
from A
@formal trail
well we are saying that the map which rakes in a linear map and spits out its matrix is bijective. in this phase we are saying it is surjective because every output (matrix) has an input (linear map)
oh because the elements of the matrix are just coefients, and legit all elements and combinations in R are included here
?
well we are saying that if we use the entries of the matrix in a particular way that gives us a list of outputs w_k which, when paired with the basis v as inputs, uniquely specify a linear map (by 3.4)
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i cant seem to figure out how to evaluate this
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hi, teachers, do you know where i can find a proper proof about this equility
1
use set builder
double inclusion
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I tried it in two ways and I got different answers. I am off by one (-) sign and I can't figure out from where its seeping in into my answer.
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@cloud zephyr did you figure it out?
no still trying
why close then
but i dont know how to send in my solution because i cant send a picture of my work rn and its hard to explain what i did via text
✅
sorry isnt it physics?
physics is fine
oh mb
-# even Bio and Chem are fine
so what you want to do here is take force perunit length as lamda times g
you want to assume a pully system with mass
assum,e the pulley si the sphere with a mass
the sphere is practically irrelevant except for it being a smooth round surface
I got (λg) (1/2 - 2/pi)
ahh mb
-# So are English essay proofreads
-# we do all sciences, even political science
Ok ty
-# /j in case you couldnt tell
-# awww, I thought I can send my SAT reading here
Yes thats my issue
im curious where the minus came from
me too 
can you explain your thought process a little more?
Okay I assume a small element of length dl, mass dm, subtending a small angle dθ at an angle (θ) from the vertical .
dl = Rd(θ), dm = (λ)dl. so dm = Rλd(θ).
small gravitational force on this is dF = g* dm. The component of that force we're concerning with here is gsin(θ)dm. so basically gsin(θ)Rλd(θ). Then i integrated it
ah there it is
check your triangles
oh angle from the vertical
i need to be better at reading physics
i took my angles from the vertical
F_net - λg/2 = integration of gsin(θ)Rλd(θ) (limits pi/2 o 0)
so its just a sign thing
but the weird thing is if you integrate backwards, the idea is the same, but the sign flips
wait wait
and i cant tell you whats making the difference
if we take the angle from the horizontal right?
yeah that works
but it still doesnt fix the issue that the direction of integration flipping will change your answer
I KNOW, i cant figue it out 
I've been going crazy
wait i might just be silly
integral of sin is -cos
-cos(pi/2)-(-cos(0))=0-(-1)
which is 1
isnt the limit 0 to pi.2?
but its pi/2 (LL) to 0 (UL), which is -1
t1 is 0 adn t2 is pi/2
I think it si the reverse
right
im being extra silly
just use magnitude
you know the sphere isnt removing any forces
This is still bothering me though. I would've gone : (4) None of the above, in a high stakes situation. How would I know to take the magnitude?
is it a olympiad?
No
No no, i am prepping for some medical school entrance tests and came across this question
and couldnt resist
( T + dT = T + (dm)g \cos\theta ), ( N = (dm)g \sin\theta )
qfalcon_
is this right
$$
\int_{T_1}^{T_2} dT = R \lambda \int_0^{\pi/2} \cos \theta , d\theta
$$
qfalcon_
ok yay the converter is working
do you mind explaining a little bit?
@cloud zephyr Has your question been resolved?
the chain links pull on the rope, adding to the tension still
they dont relax the tension
thus, take magnitude
🍆
@cloud zephyr Has your question been resolved?
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ou will need to create a plot that:
• has Temperature along the X-axis and Impact Energy along the Y-axis
• Y-axis crosses X-axis at -120 deg C
Impact Energy (J)
Temperature, T (°C)
• includes three series of data – representing the lowest, mid and highest
results for Impact Energy at each temperature
• uses different styles of marker for each series, and includes a legend
however when ive tried plotting this in excel the axis looks like this
im not sure how to change it to look like this
@rain sentinel Has your question been resolved?
In the excel data put rows below for the minimum using MIN and maximum using MAX and the one in between (which tbh I can’t think of how to get easily)
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Assuming the pulleys are massless, find the time at which the 2kg Block will lift from the ground.
I had too many variables and I looked at the solution, they didnt consider net accelerations for either block. Why is that?
Like I was made the equation:
For the 2kg block:
N + 2T - 20 = 2a
And then N = 0 (condition for leaving contact)
2T - 20 = 2a
I know that T = 5t/3. and even if I make equations for the 8 kg block as well, I would have too many variables and too few equations
@cloud zephyr Has your question been resolved?
i hated physics
and now im making myself learn classical dynamics in a month and a half 😃
Maybe u forgot that for the block to just lift off, the forces have to be balanced?
Did you use string constraint?
Using that you can find the relation between the acc of the blocks
@cloud zephyr
oh hell, i confused it with a prior question
We don't need this here
Mb
No, I am reviewing that approach next. I learnt mechanics a while ago, so I've forgotten a lot of it
okay so if i want it to just lift, I just need to consider the bare minimum scenario with net force = 0 ?
Yes
You need to do a > 0
Which means Fnet > 0
So solve for Fnet = 0 to get the limiting case
yep
okay what if the Force was like 1000N (constant and huge). Wouldnt that cause a sudden acceleration when liftoff?
It would
Yeah. The second it leaves the ground, the normal force is gone, so u get a sudden net upward force.
but then wouldnt i need to give at an initial acceleration the moment the N = 0?
?
nah, a=0 is the condition for it to just start moving. If the upward force is bigger than the weight, there's a net force, and that's what causes the acceleration.
I am just a little confused that the Normal reaction if 0 the moment it leaves the ground, and it experiences an upward acceleration the moment it leaves the ground, why do i not include acceleration in my equations?
It's cuz you're solving for the tipping point. The exact moment the upward force equals its weight. At that single instant, a=0. The acceleration kicks in right after.
okay, that makes a little more sense to me
i'll try to process that
thank you so much:)
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I am getting an answer different from the options, and would like to find my mistake.
I am denoting vector quantities in bold here. i and j are unit vectors along x axis and y axis respectively.
ds is the small displacement.
ds = dx i + dy j
small work $$dW = ydx-xdy$$
$$y^2 + x^2 = a^2 $$
$$y = \sqrt{a^2 - x^2} $$ and $$ x= \sqrt{a^2 - y^2} $$
$$dW = \sqrt(a^2 - x^2) dx - \sqrt(a^2-y^2)dy $$
$$W = \int_{\a}^{\0} \sqrt{a^2 - x^2} dx - \int_{\0}^{\a}\sqrt{a^2-y^2}dy$$
$$W = (\frac{a^2}{2} - \frac{a^2 \times pi}{4}) - (\frac{a^2 \times pi }{4} - \frac{a^2}{2})$$
Adrien-Marie Legendre
Compile Error! Click the
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(You may edit your message to recompile.)
Oof
okay this is it, this is about the best i can do lmao
I am getting an answer different from the options, and would like to find my mistake.
I am denoting vector quantities in bold here. $\vb i$ and $\vb j$ are unit vectors along $x$ axis and $y$ axis respectively.
$d\vb s$ is the small displacement.
$$ d\vb s = dx \vb i + dy \vb j $$
small work $dW = ydx-xdy$
\begin{gather*}
y^2 + x^2 = a^2 \
y = \sqrt{a^2 - x^2} \quad \text{and} \quad x= \sqrt{a^2 - y^2} \
dW = \sqrt{a^2 - x^2} dx - \sqrt{a^2-y^2}dy \
W = \int_{a}^{0} \sqrt{a^2 - x^2} dx - \int_{0}^{a}\sqrt{a^2-y^2}dy \
W = \left(\frac{a^2}{2} - \frac{a^2 \times \pi}{4}\right) - \left(\frac{a^2 \times \pi }{4} - \frac{a^2}{2}\right)
\end{gather*}
κλαουντ ☁ (cloud)
since both of those essentially integrating the area under the circle for 1 quadrant they should each contribute a quarter circle area (negative)
magnitude understood, negative why?
OHHHH
I couldve gotten the -ve by analysing the direction of force and displacement
is that right?
well just noting that [ \int_0^a \sqrt{a^2 - x^2} \dd{x} ] is the area of the quarter circle; the first one has the bounds reversed and the second one is being subtracted so they both contribute negatively
κλαουντ ☁ (cloud)
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So $x'=x^{-1}$ for convention's sake.
\
$xx' =e \implies xx'(x'') = x'' \implies x(x'x'')=x'' \implies x''=x$.
\
Then $xx'=e=x'x''=x'x$
wai
what are axioms (ii) and (iii)?
It's fine, I just have to sort out part of the problem now
If I can show the left identity is the right identity then I'm done
because this works for inverses I believe
Yeah, this looks fine. I'd just write (xx')x'', parens around a single element are a bit pointless
well, subject to showing ex''=x''
I suppose I use the hint to prove that
yeah, that's a good idea
that hint actually gives u sth slightly different, but you can get ex = x from that
$ee=e\implies e(xx')=xx' \implies (ex)x'=xx' \implies ex=x$
wai
This works too I think
yep, that looks fine
cool :D. I know this is an easy problem, but kind of proud of myself for doing it
||x'xx'x'' = x'(xx')x'' = x'ex'' = x'x'' = e
x'xx'x'' = (x'x)(x'x'') = x'xe = x'x
so e = x'x||
This is what they intended to do with the hint btw
its actually not that trivial to come up with
took me like an hour when i first tried it
I had first seen this long ago
so u can be rightly proud of urself
never got it then
Cool!
I'll eat lunch now, thanks for the help again!
,ti
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I have a math uni- related question if thats okay. I am first semester and I wanna take a course that is available a few years later normally. Is it okay to take it from now? And can I also take the exams if I wanna?
Whoever answers answer in terms of experience, because I know it prolly depends
you can take the course if:
1 - your uni lets you
2 - you are ready in terms of mathematical maturity and prerequisites
it is set theory, but its for 6th semester for sum reason
Some courses require knowledge taught in previous courses, so unless you are good with these prerequisites, you would find it difficult to keep up with the pace of the course
this might mean its a grudute level course
you should check online to see if this is the case
true, but if I try and I can follow, is it possible to allow me to take the exams?
nono I checked it
this is up to your university
its a choice subject thats why its so later on
those prerequisites need not be the courses, but might also come in the form of mental juggling that gets taught during the lectures, like some specific way to handle a specific type of problem, and that comes with the practice which you might lack right now
if you are able to keep up with the pace, I dont see a reason why not?
might help you later on, if you do good in the course and if you are interested to pursue it further, in terms of finding undergrad research opportunities or thesis or any such thing
I had a lesson in something similar and was able to follow almost the whole of it, thats why Im thinking of taking set theory. it is even easier than what I attended
anyways ill ask the people here too
thank you
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✅ Original question: #help-36 message
talk to your seniors who have taken the course, if you know any (might be tough since you are just starting uni) it might help you more if you are able to find videos of lectures from the prof (might give you an idea of what the prof expects and how the stuff is going to be paced)
do you have a syllabus for that 3rd year class ? I'd assume it's much different than what you've seen about now, starting from ZFC and all that
He let me attend a similar class he had, but idk if its okay to ask to actually join it ya know? i dont wanna piss him off lol
what does syllabus mean? second language oops
and homework/exam format
Ive read a quarter of a book abt set theory but I have no idea what he will do in there
it should appear online
thats our point
I have to request online in order to see the contents of his class
its weird
the system
oh, so you know the prof? Then you might as well ask him for a meeting and talk this directly. Profs are very forthcoming when you express an interest in their topics. He would be able to help you better in that regard
okeoke, will do that then
thank you chat
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chat
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I don't get how ths is differnt from a partial order
Not sure how you defined partial order but the usual difference is that two elts of a poset need not be comparable
Ur case (i) ensures this never happens.
Think of power set of a set (with more than 2 elements) under set inclusion. That's a partial order, but not a total order
If you've ever seen a tree describing a tournament that's a partial order which is not total for ex
Oh the powerset is a good one too.
so this is a partial order
A total order is just a partial order where all elements are comparable
Yah so in this case if I give u a and b, nothing tells you a<=b or b<=a
so in a partial order a can be NOT related to b
Yah!
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(I'm using a variety of other books to study)
It's very interesting that they didn't cover this
Tbf we did do this in RA
Total order in class and partial order in the notes
It's hit or miss whether you do this.
Yeah same
Often you just start with a linear order or something that has one because delving into order theory stuff is kinda niche and unproductive if you're focused on other stuff
Order theory is cool in general. I think the first time I saw these things was set theory or discrete math.
I've to eat and then head to a class , see y'all !
Extra classes
Yeah figures
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26
!status
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
- Ig I lack in theory
Ofc😭
do you know what trace is
Yes
i don't think there's any fancy theory here
or at least i don't see anything
maybe there is some stupid bs trick but i would just bash
write sth like $A=\bmqty{a_1&a_2\a_3&-a_1}$ and similar for $B$
Ann
you want to check AB vs BA for if they are equal
But why diagonally, opposite of a_1 i.e. -a_1
trace zero.
Oh fr
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30
Assuming A to be I_4
Or all elements to be zero except 1 elements (which is one)
It can be known
But proving in a proper way I need help
,rccw
let u = [1; 1; 1; 1]
(ie u is the vector of size 4 with all coordinates 1)
then your condition is equivalent to saying Au = u
think about why that is based on what you know about multiplying matrices
$\bd{u}=\bmqty{1\1\1\1}$
Just using
Matrices
I haven't learnt vectors yet
Except vector in physics
Ann
then pretend i said 4×1 matrix.
Then?
then your condition is equivalent to saying Au = u
in fact it is a general fact that Au, with my u here, is a 4×1 matrix containing the row sums of A as entries
so you will want to look at A^10 u
think about what that must be
don't try to descend back into what'stheformulaism
A^10 should be A only, shouldn't it?
u as it is
I mean should it or not?
do you mean "be" or "become"
What difference does it make
My weak English
well i don't speak Hindi so we are a bit stuck here
& can't figure out how to say same thing again but in simpler form
Thats not a big difference
If it becomes A
It will be equal to A
What do you want to translate, lmk if i can help
Yes bro?
Pls Help me solve question no. 30
impossible because not true!
Oh there she comes
Then what's the approach then?😵
one way to solve is to take advantage of the generality of the problem
and consider an identity matrix
It's not a way to solve it that's completely hit n trial like
which one?
30
I have a question, can someone help me with absolute value functions? I don't understand and have tried many things
occupied!
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Sorry new!
i called it u
not e
just to be clear on that
That that thing I can't understand man
oh sorry
see Ae=e
multiply A by both sides
then A^2e=Ae
but Ae =e
so A^2e = e
do you see how this can be continued
Freakingly hell
now, remember that multiplying by e is effectively summing the rows
now how do you sum the columns?
do you have any idea?
try to find a vector similar to e
Now further I can do it myself
nice
Thank you🙇♂️🙂↕️
👍
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log2(x-1)+log2(x-1)=3
How to solve this?
just to clarify, your equation is $$\log_2(x-1) + \log_2(x-1) = 3$$ yes?
Ann
yep
ok then yes as @tall cape said the first step is definitely to add them as $$2 \log_2(x-1) = 3$$
Ann
chomp are you okay from here or you need hint for next step
i really do need guidance in solving it
okok
so
think about the end goal
we want
x = ???????
so look at what we have, our x is trapped in log right?
is it possible to take it out?
yes?
do you know how to solve logarithmic eqs in general
yes
ok right
so if you had something that looked like $\log_b(x) = k$ how would you rewrite it
Ann
(dont worry about your original eq yet; we will tie back to it shortly)
(i wanna see whether you know the key step)
(b)(k) = x correct me if im wrong
oh just remove the ()
Progameyen
yes
you should use ^ for exponents btw
backslash \ not forward slash /
Progameyen
okok tq
yeah so how can we get the log term on its own here
(just basic algebra, no log-law manipulation yet)
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new here
?
wait, how did we get the 2log₂?
hello and welcome to the server. this channel is currently occupied
so like
we added two of the same term
like a + a = 2a
annn need help for yr8 hw
Progameyen
go to your channel #help-29 and post your thing there.
realise that both are exact same thing
so we can just add them together
as if $a + a = 2a$, where a = $\log_2 (x - 1)$
Progameyen
@ember escarp Do i need to clarify anything more?
nonee
Progameyen
And you know that if your expression is in the form of
$$\log_a b = c$$
Then you can rewrite it as:
$$a^c = b$$
Progameyen
Where a, b, c can be constant (just number) or any variable (any expression that as x)
So far so good?
yupp
Now
If we can get our expression to...
...this form
Then we can 'take out' the x right?
cause the x is sitting inside log()
we have to take it out
Now the question is
Is it possible to arrange our expression $2 \log_2 (x - 1) = 3$ to the form of $\log_a b = c$?
Progameyen
Note: There are multiple ways to do so, we could just choose one
i think yes
Hm how would you do so then?
combining like terms?
the (x-1)
When we talk about like term__s__, we need at least two of them tho 🤔
So for your information, I cannot see where is the like terms hm
Unless you do think there are like terms, you can point it out to me
Otherwise we try other method
what is the other method?
Lets go to our goals in case we forgot
Is it possible to arrange our expression $2 \log_2 (x - 1) = 3$ to the form of $\log_a b = c$?
Progameyen
First subquestion:
Is this expression $2 \log_2 (x - 1) = 3$ already in the form of $\log_a b = c$?
Progameyen
Do tell me if you dont understand what im saying
not yet
so why is it not yet the same? What is the difference?
and a = 2 right?
yea
But dont forget the 2 that is sitting outside
it is extra
so both of them isnt the same thing yet
cause the 2 sitting outside of log
So if we want to make both same thing, we have to eliminate the extra 2 (or move it somewhere)
So any idea?
noopee
yea
So is it possible to somehow make the 2 no longer stay there?
hmmm, yea?
do tell me how
by dividing?
2
log₂ (x-1) = 3/2 ?
yup
now
Is our expression $\log_2 (x - 1) = \frac{3}{2}$ in the form of $\log_a b = c$?
Progameyen
yes
yup
a = 2
b = x - 1
c = 3/2
So we can finally apply the rules that we are waiting for!
So after apply the formula, what we will get?
2^2 = 3/2 ?
ermmmm not really tho...
do we have to add x-1?
And you know that if your expression is in the form of
$$\log_a b = c$$
Then you can rewrite it as:
$$a^c = b$$
Progameyen
In case you forgot
2^3/2 = x+1?
oh i thought we done
Almost
but not done
cause our goal is to solve it
We have to get it to x = ???
@ember escarp
hello
Have you got the answer?
Progameyen
btw its x - 1, not x + 1
we are at final step!
oh right, srry
uhm
Progameyen
how to get the expression x = ???
square root?
no need
x is at left side right?
so to get form of x = ???
we can just move that -1 to right side
$$x= 2^\frac{3}{2} + 1$$
Progameyen
oh so transpose?
errrr you just +1 from both sides
oh
$$x - 1 + 1 = 2^\frac{3}{2} + 1$$
Progameyen
so thats the answer?
Progameyen
im still kinda confused but I understand the point
here
It is a formula for logarithm
And it is basically how logarithm is defined
Wait are you asking how this formula work?
Or you are asking how to apply it?
both, bcuz this was given to us without prior knowledge huhu
I only know the basics
hmmm...
Cause like, the logarithm itself is an inverse of exponent
Like its existance is basically the "reverse" of exponent
do you know the logarithm rules?
nop
Hm...someone has to teach you about log again tho...
I wish i can do that but im about to sleep ;-;
oh alright
youll probably be taught it later in your class ig
you dont need it for this problem though
can I have the summary steps on how to solve this?
$$\log_2 (x - 1) + \log_2 (x - 1) = 3$$
$$2 \log_2 (x - 1) = 3$$
$$\log_2 (x - 1) = \frac{3}{2}$$
$$2^{\frac{3}{2}} = x - 1$$
$$x = 2^{\frac{3}{2}} + 1$$
Progameyen
- add the same term
- divide both sides by 2
- Convert logarithm form to exponential form
- put +1 both sides
is that all or did I miss any?
Progameyen
what step is that
After you divide both side by 2
anything else?
nope
is there anything else I should remember or take note of?
for now try to understand what log do
and this thing
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I don't even know how to start this, can someone break it down for me into manageable chunks?
draw a visual representation while reading the question out
use different shapes to represent a group type and write a number ( frequency ) for the group type in thoese shapes
okay, i got the total amount of combinations possible, how do i restrict it to only 3 substitutes possible?
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@blissful condor @sick roost right I'd like to formulate a thought process for this now
I'll try proving that bounded continuous functions are UC
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hold on!
you have a continuous periodic function
consider your compact period interval
show UC and bounded on that first
then, since its UC and bounded on that interval, periodicity implies UC and bounded everywhere
this is not even true btw