midnight heron
Or else we would get negative value which is not possible
#help-33
whole quiver
sup
marsh citrusBOT
@compact ether Has your question been resolved?
still temple
marsh citrusBOT
still temple
How dis wrong?
broken zealot
hearty dust
if I multiply 7560 * e^(293) * 30^0.5 and 7560, 293, and 30 are all measured values
would I round the final ans to 1 sigfig?
because the 30 is the least SF
but the 30 is a counted value so im not sure if it counts
marsh citrusBOT
if I multiply 7560 * e^(293) * 30^0.5 and 7560, 293, and 30 are all measured val...
proud ice
if I multiply 7560 * e^(293) * 30^0.5 and 7560, 293, and 30 are all measured val...
If it's counted, then no.
hearty dust
so it's based on only measured values?
proud ice
It has infinite precision
hearty dust
so if it's recorded with some type of instrument that wouldn't be infinite precision right?
great tysm
.close
exotic wolf
hi
exotic wolf
i dont get how they get this
on left side we can get
ln(y^2)
but its like they are ignoring the constant C1
if C1 = 0, i get it that y^2 = x
exotic wolf
but then we get e^C1
and thats not a "random" constant, because it cant take on all values
stoic slate
C=e^C1
exotic wolf
@stoic slate C means a random constant
right?
but e^constant cant take all values
can u get all the numbers with e^(some constant) ?
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fathom quail
what is that
hearty dust
is the second one I and III
yes
hearty dust
the first one should be right right?
I got 361.66 as well
marsh citrusBOT
fathom quail
.close
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lucid turret
Is the decimal system in the base 10 just a random picked value for our usual system or 10 is natural to us?
marsh citrusBOT
Is the decimal system in the base 10 just a random picked value for our usual sy...
lucid turret
.close
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vital tartan
some check my work
marsh citrusBOT
some check my work
still temple
correct ☑️
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upper sigil
$y(0) = 1$
$\frac{dy}{dt} + 2y = 3 \$
$\frac{dy}{dt} = 3 - 2y \$
$dy = 3-2ydt$
how else would I do this?
marsh citrusBOT
$y(0) = 1$
$\frac{dy}{dt} + 2y = 3 \\$
$\frac{dy}{dt} = 3 - 2y \\$
$dy = 3-2ydt$...
elfin berryBOT
Tomi
upper sigil
the variable of the integration is not in the function itself, so I can't really evaluate that.
velvet cedar
marsh citrusBOT
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mossy kettle
could someone explain how to do this problem, i feel like its much easier then I believe but I am lost
marsh citrusBOT
could someone explain how to do this problem, i feel like its much easier then I...
frail orbit
are you familiar with definition of the derivative?
mossy kettle
sort of, but very new
frail orbit
what is the definition of the derivative?
mossy kettle
rate change
frail orbit
yes
mossy kettle
but in function notation
mossy kettle
f(x2)-f(x1) /x2-x1
mossy kettle
could someone explain how to do this problem, i feel like its much easier then I...
so then this?
frail orbit
f(x2)-f(x1) /x2-x1
yes
now, imagine h as a 'step' away from either x2 or x1, which every point you want to find the derivative at
say x2 is h away from x1, and say x1=x
frail orbit
f(x2)-f(x1) /x2-x1
then how would you write this?
mossy kettle
so kinda like -h?
digital field
$f'(x_0) = \frac{f(x_0+h)-f(x_0)}{h}$
elfin berryBOT
Xetrov
mossy kettle
f(x2+h)-f(x1+h) / (x2+h)-(x1+h) ?
so then this is wrong?
digital field
yeah
mossy kettle
**Xetrov**
then i have to put the eq, to that format?
digital field
then i have to put the eq, to that format?
yes
mossy kettle
how exactly? do i put the f(x) eq + 0.001?
trim quest
for h in here, as 0.001?
Yes
The idea is that h should be approaching 0, but you can estimate it with a small h like 0.001
trim quest
No, $\frac{f(-3.75+0.001)-f(-3.75)}{0.001}$ where $f(x)$ is $6\cdot 0.55^x$
elfin berryBOT
tatpoj
trim quest
So $\frac{6 \cdot 0.55^{-3.75+0.001} - 6 \cdot 0.55^{-3.75}}{0.001}$
elfin berryBOT
tatpoj
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wary bluff
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polar granite
Can someone help me here idk how to identify the (x,y) here
marsh citrusBOT
Can someone help me here idk how to identify the (x,y) here
marsh citrusBOT
@polar granite Has your question been resolved?
polar granite
<@&286206848099549185>
marsh citrusBOT
@polar granite Has your question been resolved?
red nimbus
Can someone help me here idk how to identify the (x,y) here
What is this supposed to mean?
tawdry fog
How to show the rate of change of Q
marsh citrusBOT
How to show the rate of change of Q
compact crescent
Why nobody helps him?
You guys are cold blooded
It is weird because I found that sqrt(k) is attainable
I don’t know
next raft
brine blooded
marsh citrusBOT
@tawdry fog Has your question been resolved?
tawdry fog
Lol
marsh citrusBOT
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faint grove
What is area of line PQ when there's sadalikkol ABCD
And AD∥EF∥BC , AD=20 ,BC=24 while point P and Q are intersection of line EF,AC and BD
marsh citrusBOT
What is area of line PQ when there's sadalikkol ABCD
And AD∥EF∥BC , AD=20 ,BC=2...
@faint grove Has your question been resolved?
marsh citrusBOT
@faint grove Has your question been resolved?
eager tinsel
What is area of line PQ when there's sadalikkol ABCD
And AD∥EF∥BC , AD=20 ,BC=2...
what's a sadalikkol?
faint grove
사다리꼴
elfin berryBOT
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eager tinsel
huh?
faint grove
What the hell is this
eager tinsel
What the hell is this
agreed
@atomic dune where does all that come from in geometry?
atomic dune
<@163934899550683136> where does all that come from in geometry?
latex
atomic dune
<@163934899550683136> where does all that come from in geometry?
and geometry yes
atomic dune
<@163934899550683136> where does all that come from in geometry?
\documentclass{article}
\usepackage{amsmath}
\begin{document}
To find the length of the line segment (PQ) in the quadrilateral (ABCD) with the given conditions:
Given:
- (AD \parallel EF \parallel BC)
- (AD = 20), (BC = 24)
- (P) and (Q) are intersections of lines (EF), (AC), and (BD)
Solution:
Since (AD \parallel EF \parallel BC), the dominant terms for large (n):
[ a_n = \frac{2 + 4^n}{2 + 3^n} \quad \text{and} \quad b_n = \left(\frac{4}{3}\right)^n ]
Using the limit comparison test:
[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{2 + 4^n}{2 + 3^n}}{\left(\frac{4}{3}\right)^n} = \lim_{n \to \infty} \frac{(2 + 4^n) \cdot 3^n}{(2 + 3^n) \cdot 4^n} = \lim_{n \to \infty} \frac{2 \cdot 3^n + 12^n}{2 \cdot 4^n + 12^n} ]
For large (n), the dominant terms (12^n) cancel out:
[ = \lim_{n \to \infty} \frac{\left( \frac{2}{4^n} + 1 \right)}{\left( \frac{2}{3^n} + 1 \right)} = 1 ]
Since (\sum_{n=1}^{\infty} \left(\frac{4}{3}\right)^n) diverges and the limit is positive finite, the original series also diverges.
Conclusion:
The series (\sum_{n=1}^{\infty} \frac{2 + 4
elfin berryBOT
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atomic dune
<@163934899550683136> where does all that come from in geometry?
actually my bad its Calculus 2, i answer questions to different person lol
ah why it doesnt work
eager tinsel
okok
elfin berryBOT
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atomic dune
Latex generation is the best
marsh citrusBOT
@faint grove Has your question been resolved?
still temple
What is area of line PQ when there's sadalikkol ABCD
And AD∥EF∥BC , AD=20 ,BC=2...
사다리꼴의 사진 있나요? 괜찮으시면 좀 시도하겠습니다
faint grove
still temple
이건 좀 비정통적인 방법인데 이렇게하면 돼나요
marsh citrusBOT
@faint grove Has your question been resolved?
red nimbus
Why nobody helps him?
I gave a hint but seems he ignored it
frosty jay
What is area of line PQ when there's sadalikkol ABCD
And AD∥EF∥BC , AD=20 ,BC=2...
Are you still trying this?
frosty jay
이건 좀 비정통적인 방법인데 이렇게하면 돼나요
The correct answer is not 12cm.
faint grove
I am still trying
frosty jay
Actually never mind, I'm not so sure about my solution anymore
Ok I think I have an easy solution
frosty jay
I am still trying
My hint is this: can you list down all similar triangles you see?
Using the similar triangles you will be able to use length ratios of the points along EPQF to known lengths
still temple
The correct answer is not 12cm.
sorry
marsh citrusBOT
@faint grove Has your question been resolved?
indigo plume
I am needing help with my math final project. I have to take a table of data and create a function from it algebraically (I will add my data and its graph in this post). I know how to do this with smaller numbers but I would be lying if I said my head wasn't spinning trying to figure out where to start here... I tried finding the possible degree of the polynomial algebraically but I think I didn't do it right. Thank you inadvance for your help!
marsh citrusBOT
I am needing help with my math final project. I have to take a table of data and...
@indigo plume Has your question been resolved?
indigo plume
<@&286206848099549185>
brazen yoke
You could use Lagrange interpolation
spark siren
i am not sure what your question exactyl is. you have 24 datapoints (1997-2020) and what should you do now?
brazen yoke
If you want a polynomial through these points
indigo plume
I have to take those data points and find a function mathematically that is a polynomial with a degree of 3 or higher
spark siren
and you need to find the polynom or the degree of the polynom?
brazen yoke
You could use Lagrange interpolation
indigo plume
The complete polynomial
brazen yoke
https://en.m.wikipedia.org/wiki/Lagrange_polynomial
This should do the job
indigo plume
I looks like it would fit the data well!
My math course is just really particular and they havev't covered Lagrange Interpolation so I'm not sure I could use it though 😦
brazen yoke
My math course is just really particular and they havev't covered Lagrange Inter...
It’s just a formula, I think you should be fine using it
indigo plume
I can give it a shot! Im not super familar with it though, any pointers you can share to help me find my heading??
brazen yoke
But the wiki article maybe isn’t the best explenation wise, maybe look up some other websites
This looks a bit better
marsh citrusBOT
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sinful vessel
I need help
marsh citrusBOT
I need help
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sinful vessel
[
\log_2 \left( x^{3/2} \right) = 3
]
clever tangle
\[
\log_2 \left( \frac{x^{3/2}} \right) = 3
\]
huh
elfin berryBOT
zA
clever tangle
alr
sinful vessel
how do I find what x is…
clever tangle
so we need to use the rules of the logarithms
clever tangle
I have a few ideas
ah yes
sinful vessel
so we need to use the rules of the logarithms
ik which one I think
sinful vessel
**zA**
do I take the exponent and put it in front of “log”
so it becomes [
\ 3^{\log_3(x)}
]
clever tangle
yes
sinful vessel
now what shall I do?
clever tangle
we want to isolate the x
sinful vessel
hmm
clever tangle
you have a coeficcient
sinful vessel
do I get rid of it?
clever tangle
yeah
sinful vessel
3 divided by 3/2?
clever tangle
but since we're isolating the x we try to bring it to the other side
sinful vessel
but since we're isolating the x we try to bring it to the other side
yeah
clever tangle
3 divided by 3/2?
you have the correct idea here
so then it's (1/2)(log_2(x))=1
but we still have 1/2 on LHS
we still need to get rid of it
sinful vessel
times 2 on both sides?
clever tangle
yes
elfin berryBOT
zA
sinful vessel
so now we need to use another logarithmic rule
ik what it is but can u write it in the general way?
all I know it’ll be x = 2^2
sinful vessel
**
sinful vessel
**zA**
can I do is
take log2 of both sides
so I get x = 2log_2 2
which using the law
it’s
x = log_2 2^2
x = log_2 4
sinful vessel
okay I see
clever tangle
take log2 of both sides
it should be 2^(log_2(x)) = 2^2
which then 2^(log_2(x)) cancel to get x
sinful vessel
i’ll write this in latex so I can see it better
clever tangle
alr
elfin berryBOT
zA
sinful vessel
would it cancel to be just x?
clever tangle
yes
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iron meadow
what exactly is the relationship between how many times an eigenvalue will show up along the diagonal of an upper triangular representation of a linear operator and multiplicity of the factor associated with that eigenvalue in the minimal polynomial? are they always the same? i think they...aren't? but i could use some clarification on how this works
marsh citrusBOT
what exactly is the relationship between how many times an eigenvalue will show ...
limber hearth
low ruin
what exactly is the relationship between how many times an eigenvalue will show ...
Can you explain this "upper triangular representation of a linear operator "
low ruin
what exactly is the relationship between how many times an eigenvalue will show ...
If you talking about matrices then number times an element appear in a upper diagonal matrix's diagonal will determine the multiplicity
iron meadow
so given some basis for the vector space thats the domain of the linear operator, say the matrix representation of the operator is upper diagonal under that basis
then the eigenvalues appear along the diagonal right?
let me i guess explain more of why i'm asking this question, so if i'm understanding correctly, this theorem is saying that if the minimal polynomial has a factor with multiplicity of more than 1, then an operator wouldn't be diagonalizable?
iron meadow
If you talking about matrices then number times an element appear in a upper dia...
what do you mean by multiplicity in this context?
low ruin
So if (x-2)^2=0 Then the root x=2 will have multiplicity 2
iron meadow
in both the minimal and the characteristic polynomials or?
low ruin
I am familiar with term characteristic polynomial but not minimal polynomial. I was referring to the characteristic polynomial
iron meadow
i think part of my confusion is also axler presenting things in terms of minimal polynomial also and not introducing characteristic polynomial / determinants until later on in the book and not being sure about the relationship between the two, but my question is specifically about minimal polynomial
it's a bit of a different approach than most LA books
so i guess for this matrix, the minimal polyonomial would be (x-1)^2 and it isn't diagonalizable because 1 is the only eigenvalue, the dimension of the eigenspace is 1 and you would be able to tell the minimal polynomial has multiplicity 2 so it doesn't meet the necessary condition of that theorem
but this is well obviously diagonalizable, the minimal polynomial is (x-1) with multiplicity 1 so that checks out, the eigenspace for 1 has dimension 2
so i guess in conclusion, how many times the eigenvalue appears along the diagonal of an upper triangular matrix doesn't correspond with the multiplicity in the mnimal polynomial
devout mauve
you will later encounter the jordan form
and that will tell you the best relationship I suppose
the multiplicity in the min poly gives you the size of the largest jordan block for that eigenvalue
iron meadow
ah
devout mauve
you can also interpret that in terms of the generalized eigenspaces
iron meadow
so a matrix is diagonalizable only if each jordan block is of size 1?
devout mauve
yes
hushed egret
you need a basis of eigenvectors
iron meadow
so before introducing that concept, you can't really tell easily visually from just looking at the matrix what the multiplicity in the minimal polynomial will be, the way you can for characteristic polynomial?
devout mauve
no
hushed egret
but if the factor (z - λ) appears more than once in the min poly, you need to apply T - λ more than once to kill the generalised λ eigenspace
iron meadow
trying to wrap my mind around what you just said
that's using the concept of annihilator that was in the earlier axler readings?
hushed egret
i have not read axler 
iron meadow
oh sorry youre in the axler group so i thought you might have been following along
hushed egret
i havent
devout mauve
what axler group
iron meadow
so wait lemme try to wrap my mind around what it means for T- λI to "kill" an eigenspace?
hushed egret
austin's LADR reading group
devout mauve
ah
hushed egret
well say you just have a normal eigenspace with eigenvalue λ
iron meadow
hushed egret
then for each vector v in that eigenspace, Tv = λv
so (T - λ)v = 0
v gets killed by T - λ
iron meadow
ah right ok
hushed egret
the presence of (T - λ)^2 or some higher power in the min poly means that things dont quite get killed by a single T - λ
iron meadow
so if the min poly has multiplicity 2 for something, then that root is not an eigenvector?
hushed egret
it means there are eigenvectors with eigenvalue λ, but also something more
iron meadow
oh hm
those "Something more" vectors, would they be vectors in the null space of the operator?
hushed egret
they would be in the so called generalised eigenspace
ker(T - λ)^n for some large n
iron meadow
i guess that's something that is introduced later
hushed egret
probably
the point is that you need to apply T - λ more than once now to kill that subspace
iron meadow
gotcha, interesting
i've just been a bit confused by minimal polynomial because i took some LA course long ago and i remember learning about eigenvectors in terms of the characteristic polynomial but it was long ago and my recollections are pretty vague so the minimal polynomial has been vaguely familiar but also different
hushed egret
say for example (T - λ)^2 was a factor of the minpoly
then there exists a vector w such that (T - λ)w is not zero, but (T - λ)^2 w = 0
iron meadow
so does T^2(w) = λw ?
hushed egret
(T - λ)^2 = T^2 - 2λT + λ^2
iron meadow
ah
hushed egret
call v = (T - λ)w
then (T - λ)v = (T - λ)^2 w = 0
so Tv = λv
you've found a genuine eigenvector
but at the same time, v and w are linearly independent
for if you had c_1 v + c_2 w = 0, then c_1 (T - λ)v + c_2 (T - λ) w = c_2 v = 0
which forces c_2 = 0
and then c_1 needs to be 0 as well
so theres an eigenvector v of eigenvalue λ, while at the same time theres a linearly independent vector w which is only killed by (T - λ)^2 and not (T - λ)
iron meadow
i see
are these w vectors like separate from other eigenspaces associated with other eigenvalues?
iron meadow
so with these more generalized eigenspaces can you always build a basis or something?
hushed egret
yes
iron meadow
i see, sounds interesting
hushed egret
so thats the obstruction to finding a basis of eigenvectors
hence, T cannot be diagonalisable
iron meadow
(T - λ)^2 = T^2 - 2λT + λ^2
also this seems...oddly nonlinear for LA 😄
hushed egret
studying the action of a linear operator on a vector space is a lot more algebraic in flavour
somehow it boils down to understanding polynomials
iron meadow
i think axler was hinting at some point at identifying linear operators to monic polynomials? and how there was a correspondence between the two?
hushed egret
something like the action of a linear operator on a vector space is the same as a F[x] representation maybe
iron meadow
so what's the relationship between minimal polynomial and characteristic polynomial?
hushed egret
the minpoly divides the charpoly by cayley hamilton
iron meadow
i guess if something is diagonalizable then they will be the same? since multiplicity of each eigenvalue will be 1?
hushed egret
the charpoly divides the minpoly^(dimV) for other reasons
hushed egret
i guess if something is diagonalizable then they will be the same? since multipl...
only if distinct eigenvalues
iron meadow
oh right
devout mauve
the charpoly divides the minpoly^(dimV) for other reasons
is there a "nice" reason for that other than just knowing they have the same factors and then forcing the factors to be there often enough?
iron meadow
ah so minimal polynomial will just have lower degree than the dim V if something is diagonalizable but some eigenspace has dim > 1 ?
hushed egret
is there a "nice" reason for that other than just knowing they have the same fac...
hmmm not off the top of my head
devout mauve
ah sad. sounded a bit like that from the way you phrased it
hushed egret
could probably do something using structure theorem of k[x]-module but i dont want to think about it
devout mauve
understandable
hushed egret
primary decomp
iron meadow
the charpoly divides the minpoly^(dimV) for other reasons
charpoly divides minpoly ^ (max dim of any eigenspace?)
hushed egret
its more like the maximum of (dim V/dim of any eigenspace)
iron meadow
well i will get back to reading, thank you for the help snow, and denascite
hushed egret
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small stream
marsh citrusBOT
mystic ledge
what is r?
nimble prairie
try plotting the graph to z = sin(r^2)/r^2, that will help
you can work in the "rz"-plane for this
marsh citrusBOT
@small stream Has your question been resolved?
small stream
,w plot z = sin(r^2)/r^2
elfin berryBOT
small stream
hmm
nimble prairie
since r = sqrt(x^2+y^2) then there is a rotational symmetry
red nimbus
cool
nimble prairie
i.e this plot can be thought of as a slice, with each slice being identical for any "rotation" about the z-axis
nimble prairie
slice cutting from zx/zy plane?
yeah depending on the rotation so to speak
small stream
Also i'm really trying to understand this
small stream
btw can we draw this by hand
nimble prairie
btw can we draw this by hand
yeah as a rough sketch i suppose
nimble prairie
Also i'm really trying to understand this
what about it?
small stream
continuity on domain
nimble prairie
drawing pictures might help, all theyre saying is that a open subset of D is an open set which is contained in D, i.e if you take any open set in R^2, and intersect that with D (assuming D is open here) you will have have an open set contained in D
nimble prairie
continuity on domain
the thing you marked with green is not really about that
so youre asking about the first part?
small stream
i feel like i cannot understand the whole part lol
small stream
drawing pictures might help, all theyre saying is that a open subset of D is an ...
ye i cant imagine the pic in mind
nimble prairie
okay there is really seperate parts here imo
let's start with the open subset, do you know what an open set is?
nimble prairie
doesnt have to be an open disk
small stream
you mean it deosnt have to be a circle?
nimble prairie
yeah, why should it?
small stream
ye i see
nimble prairie
open sets generalizes the notion of open intervals in R
i.e (0,infty) and (0,1) etc
are open sets
why (0,1] and [0,1] etc isnt
so as it's written there
an open set E in R^2 is one where for each point p in E there exists some open disk around p which is still contained in E
basically if we go back to R
pick any point in (0,1)
notice how you can always find a (open) neighborhood around that point so that it's contained in (0,1)
this is why drawing crude examples is good
also if it helps
the complement of an open set is closed
small stream
so far i can understand
nimble prairie
ok so good, so an open subset is clear now too?
small stream
yes
nimble prairie
cool
so back to the first part, they are saying that a function f from D to R is continuous on D iff the preimage of any open subset of R is open
nimble prairie
so back to the first part, they are saying that a function f from D to R is cont...
alternatively, you can phrase it as this, if f is continues from D to R, and we let M be a subset of R, then it follows that if M is open then the preimage of M is open aswell
i.e the idea here is that continuous functions preserves open sets (through the preimage) and the same is true for closed sets
slim surge
image to illustrate that this def spot discontinuous functions (function from R to R)
nimble prairie
this is true if we're dealing with a closed set aswell
it's not the fact that you map an open set to an opet set
small stream
Then why did they keep emphsising open set all the time
nimble prairie
closed sets sort of come for free if you take the complement
and i guess it might be easier to show it for open sets sometimes(?)
frosty jay
Then why did they keep emphsising open set all the time
It is to indicate closure (i.e. remains inside) within the set family
An open set is defined with respect to being a member of a set family
nimble prairie
they dont deal with topogoly yet in that sense
small stream
can we say that a function f from D to R is continuous on D iff the preimage of any close subset of D is closed
nimble prairie
so i dont think it helps
frosty jay
You can just call it set in my_magical_set_X
nimble prairie
can we say that a function f from D to R is continuous on D iff the preimage of ...
yeah
or alt. if f is continues from D to R, and we let M be a subset of R, then it follows that if M is closed then the preimage of M is closed aswell
you get this fact for free by just considering the complement of M in the case that M is open
marsh citrusBOT
nimble prairie
can we say that a function f from D to R is continuous on D iff the preimage of ...
oh and it should be closed subset of R
not D, sorry
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lethal bridge
marsh citrusBOT
lethal bridge
how can i do this?
cunning oak
f(g(x)) is just the intersection of both functions
lethal bridge
but f(x) and g(x) dont intersect
lethal bridge
do i start by finding the domain for each functino separately
vernal forge
do i start by finding the domain for each functino separately
yes
and for f, find the range too
novel juniper
do i start by finding the domain for each functino separately
Yup!
lethal bridge
[0,9] for f(x)
[-2,2] for g(x)
vernal forge
[0,9] for f(x)
[-2,2] for g(x)
the arrow on f(x) means infinity
novel juniper
uh
vernal forge
so [0, +inf) instead
vernal forge
yes
novel juniper
yup
lethal bridge
so what should i do now
novel juniper
Find the intersection of their domains!
cunning oak
f(g(x)) is just the intersection of both functions
the basic explaination is to find f(g(x)) you first need to know both f(x) and g(x), so where are those values defined?
ive never learned about this so i might be wrong
novel juniper
You are right.
cunning oak
thanks
lethal bridge
so im finding [0,inf) U [-2,2] ?
elfin berryBOT
🏳🌈f(why am i here )= idk
brave trail
the domain of $f(g(x))$ should be, if $f:D_1\to E_1, g:D_2\to E_2$ then the domain of $f(g(x))$ is $D_2\cup (E_2\cap D_1)$
trash latex
fr
elfin berryBOT
ZeroMemory
brave trail
as first, for x, g(x) should be defined
lethal bridge
is it -2 <=x<=2?
brave trail
and then, we have output of g(x) and for this g(x)=y, f(y) should be defined
brave trail
is it -2 <=x<=2?
yes, because for $g(x)$ it is defined for $x\in[-2,2]$, and the range of g is $[0,4]$ where f is defined for here
elfin berryBOT
ZeroMemory
marsh citrusBOT
lethal bridge
thanks
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vital tartan
can someone check my work
marsh citrusBOT
vital tartan
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void elm
marsh citrusBOT
@vital tartan Has your question been resolved?
tardy fox
tardy fox
<@&286206848099549185>
marsh citrusBOT
@tardy fox Has your question been resolved?
sand fable
have you done anything with the previous advice
tardy fox
Sure?
tardy fox
label angle APB x and label angle QPB y and go from there
I have to prove that angle AQB- angle PQA = angle AOB where o is the centre right?
sand fable
I have to prove that angle AQB- angle PQA = angle AOB where o is the centre righ...
yes
tardy fox
label angle APB x and label angle QPB y and go from there
I still dont see it...
Am i missing some theorem?
sand fable
what do you have so far?
tardy fox
Being honest nothing
I am missing some theorems for sure
First of all i have no clue how to develop a relation
sand fable
so far all you need to do is label angles and do basic angle chasing to label the rest of the diagram
tardy fox
so far all you need to do is label angles and do basic angle chasing to label th...
But you will get stuck after labeling angles in triangle APQ
Theres nothing else you could
sand fable
and there's more
tardy fox
Like?
tardy fox
and there's more
I got angles around point Q
sand fable
can you show me what you have?
marsh citrusBOT
tardy fox
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sand fable
did you label AQB
tardy fox
AQB is 90+x-y
sand fable
so what's AQB - PQA?
tardy fox
2x
tardy fox
Not exactly...
How do i prove AOB is 2x?
OH
OMG
TRANK YOU
THANK YOU
I lovee you
sand fable
yw 🎉
tardy fox
Bro can i send you friend req
sand fable
yeah i don't mind
tardy fox
I mean i wont disturb you unless its a really annoying problem
Thank you so much
If i would have spotted that AOB is twice that of APB i would have maybe tried angle chasing But i didbt see motivation
Thank you so much
Sorry for being annoying
.close
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still temple
Calculate the number of quadrantal angles that exist in the interval from 15π/4 rad to 29π/4 rad.
marsh citrusBOT
Calculate the number of quadrantal angles that exist in the interval from 15π/4 ...
thorny bison
because 29π/4 - 15π/4 = 14π/4 / π/2 = 7
Yes ur correct
marsh citrusBOT
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plucky trail
marsh citrusBOT
plucky trail
I need help with section c. I've tried multiple times I keep getting it wrong
I'll do :
ln 8/5.52/0.0222
I have even tried
5.52 * 1.0222^t=8
Please @me when you see this
wise jackal
I'll do :
ln 8/5.52/0.0222
I have even tried
5.52 * 1.0222^t=8
are you including the entire thing in ln()?
because 0.0222 shouldn’t be included in that
wise jackal
well let’s write the steps
8 = 5.52 e^(0.0222t)
8/5.52 = e^(0.0222t)
ln (8/5.52) = ln (e^0.0222t)
next?
plucky trail
So the answer is 16.7?
wise jackal
yup
plucky trail
And next I have to find the equation for the doubling time
Lemme show you
D. Would it be ln 16.7^2/6.3
Or would it be ln 8^2/6.3
wise jackal
uhh what’s doubling time again?
sry it’s been a minute lol
may want to wait for someone else
plucky trail
Sure ^^
Should I @ a helper
<@&286206848099549185>
Please @ me when you see this I'll be off discord
plucky trail
Im not sure the proper equation,
red nimbus
Determine the expo function?
red nimbus
ah it's d then
plucky trail
Or would it be ln 8^2/6.3
This is what I thought the equation would be
I'm not 100% positive though
plucky trail
D. Would it be ln 16.7^2/6.3
Or this
red nimbus
5.52 and doubling time would be to find such t where you get 2 * 5.52, no?
red nimbus
to my knowledge doubling time is just looking for the such value t where the exponential growth doubled
elfin berryBOT
𝔸dωn𝓲²s
elfin berryBOT
𝔸dωn𝓲²s
red nimbus
Now you can do what next
plucky trail
Divide again
plucky trail
Ln 2/ln 1.0222?
elfin berryBOT
red nimbus
so after 31 years and a half population doubled
red nimbus
,, P(t) = 5.52e^{0.0222t}
elfin berryBOT
𝔸dωn𝓲²s
red nimbus
so that's the right one right
red nimbus
I'll do :
ln 8/5.52/0.0222
I have even tried
5.52 * 1.0222^t=8
i took it from here previously lol
anyway
,w 5.52e^(0.0222t) = 2 * 5.52
elfin berryBOT
red nimbus
31 or 31.2 idk
red nimbus
bruh
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half stump
is it possible to find p(a intersection b) with this info?
marsh citrusBOT
is it possible to find p(a intersection b) with this info?
desert dirge
you can find Pr(A and not B) which then should be easy to get just the intersection of a and b since you know Pr(A)
desert dirge
aha i have remembered the formula incorrectly, i do that a lot, sorry
thought A would be on the bottom
red nimbus
bruh
half stump
half stump
bruh
i figured it would be easier to send the first thing
but it seems thats not the case
maybe im doing something wrong
i had x = P(A and B) / y
so i needed P(A and B)
red nimbus
y = P(B) = P(A ∩ B)/x
Now the q ist can we rewrite P(A ∩ B) in terms of other information
we dont know if they are independent events so we assume they are not
I think Bayes could help
half stump
Bayes?
half stump
=0.3 P(B | A)
red nimbus
P(B | A) = [P(B) - P(B | A') * P(B')]/P(A)
Let's try again
We know some stuff so I plug it in
P(B') = 1 - P(B)
half stump
**𝔸dωn𝓲²s**
i dont really get where this came from
red nimbus
That's the Law of Total Probability
red nimbus
.
half stump
ok
half stump
xy + 0.55 - 0.55y = 0.3
half stump
xy - 0.55y = -0.25
half stump
y(x-0.55) = -0.25
red nimbus
ye
half stump
x-0.55 = -0.25/y
red nimbus
no
half stump
y= -0.25/x-0.55
half stump
y= -0.25/(x-0.55)
red nimbus
sounds good
red nimbus
,w 5/9
elfin berryBOT
red nimbus
ye
marsh citrusBOT
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proper grove
Anyone cracked jee here?
marsh citrusBOT
Anyone cracked jee here?
tawdry stag
lyric trench
U cracked jee?
you mean cracked at jee?
lyric trench
ok
i don't know about jee that much
why is your nickname "jee" anyway
just because that's what you got discord for?
proper grove
Ok
marsh citrusBOT
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stark apex
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sharp sonnet
Hello can anyone tell me how i can find the distance between these 2 points? Would it just be the circumference *2/3?
marsh citrusBOT
Hello can anyone tell me how i can find the distance between these 2 points? Wou...
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elfin berryBOT
riyobi
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@small stream Has your question been resolved?
deft apex
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@small stream Has your question been resolved?
dense beacon
So in maths for example the derivative of x^2 we just evaluate from the limit proof for a arbitrary small value h and if we just take at x= 1 and h=0.25 it just stands for all x the derivative right ?
marsh citrusBOT
So in maths for example the derivative of x^2 we just evaluate from the limit pr...
unreal oxide
if u take x = 1 and h = 0.25, u get an approximation for the derivative of x^2 at x=1
by evaluating the ratio
$f'(1)\approx\dfrac{f(1+0.25)-f(1)}{0.25}$
elfin berryBOT
SilverSoldier
marsh citrusBOT
@dense beacon Has your question been resolved?
dense beacon
by evaluating the ratio
So to proof the theorem of derivative we can evaluate it at any point we’re interested
The only prerequisite is to be a non stochastic function
Right
?
@unreal oxide
unreal oxide
$f'(1)=\lim_{h\to0}\dfrac{f(1+h)-f(1)}{h}$
elfin berryBOT
SilverSoldier
unreal oxide
if u want the derivative at 1, u evaluate the limit above
if u want the derivative at 2, u evaluate $f'(2)=\lim_{h\to0}\dfrac{f(2+h)-f(2)}{h}$
elfin berryBOT
SilverSoldier
unreal oxide
if u want it at $x$, its $f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$
elfin berryBOT
SilverSoldier
dense beacon
if u want it at $x$, its $f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$
Yes I mean I want to prove that the difference is 2x for any x.
So I just have to evaluate the function at any arbitrary point with a small displacement point right ?
@unreal oxide
I just want to derive to prove this in general
marsh citrusBOT
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novel lantern
why this set is closed ?
marsh citrusBOT
why this set is closed ?
still temple
Can you give more context? What is G? any topological group? What is a?
novel lantern
oh yeah sorry i forgot
G is a topological group, x \in X which is a topological space, a is the action of the group to X and its continuous
spice raptor
hmm..
Why Fn,m is Closed
Definition: The set Fn,m consists of elements g in G such that if x is close to x0, then a(g, x) is close to a(g, x0).
Continuity: The function a(g, x) is continuous, meaning small changes in g and x result in small changes in a(g, x).
Sequence Convergence: If a sequence (g_k) in Fn,m converges to g, the condition that defines Fn,m still holds due to the continuity of a.
Sorry if I'm wrong, it was just a quick check
still temple
Ok, I assume you mean X is a metric space. If you can use g \mapsto d(a(g,x), a(g,x0)) is a continuous map for a fixed x and x0, then the goal is the show F_{n,m}^c is open. If g is not in F_{n,m}, then there is an x with d(x,x0)<2^{-n} and d(gx,gx0)>2^{-m}. Informally speaking, small change in g doesn't change the fact that d(gx,gx0)>2^{-m}, this should be enough to find a small neighbourhood around g not in F_{n,m}.
marsh citrusBOT
spice raptor
no problem!
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ionic burrow
hello
desert dirge
yeah?
ionic burrow
but in the marking scheme it isnt
like it is not -cosu
so then the marking scheme is incorrect?
desert dirge
indeed it is
ionic burrow
man i knew it
im over here trying to reason with myself whether or not i did something wrong
smh
so then the fina answer would be uhh
1 sec
ionic burrow
oki perfecto
elfin berryBOT
desert dirge
yeah
ionic burrow
oh yh they just factored some stuff out
i c
ok thank you so much have a great day!
desert dirge
np you too
ionic burrow
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frosty minnow
probability of winning a game is 0.5. a person plays the game 3 times. what is probability that his second win occurs at 3rd game.
marsh citrusBOT
probability of winning a game is 0.5. a person plays the game 3 times. what is p...
frosty minnow
will it be 2C1 (1/2)¹×(1/2)¹ for first two matches? using binomial distribution
and 1/2 for 3rd match?
so (2×1/2×1/2)×(1/2)=1/4
first part, of only 1 win in two matches using binomial gives probability 1/2
but if question needs only 1 win in first two matches, then it could be WL or LW. and we do it manually, so it would be 1/2×1/2 (WL) + 1/2×1/2 (LW), which eventually also gives 1/2
is my interpretation of logic and interpreting cases of binomial correct here?
because I was getting so confused
i just need someone to read my message and tell me if I'm right or wrong <@&286206848099549185>
pliant inlet
second win occurs at 3rd game.
it means that there must be one winning during the first 2 games
right?
pliant inlet
i just need someone to read my message and tell me if I'm right or wrong <@&2862...
i thinks yr right.
still temple
Another argument is that there are two outcomes in our event, namely (WLW) and (LWW), and in total, there are 2^3 outcomes in the sample space. So 2/2^3 = 1/4.
frosty minnow
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elfin berryBOT
clikz
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
indigo wedge
Wait i just realized idk if they want me to find the x intercept or the y intercept
or both?
brave marsh
Yeah I'd go with both. Your method is fine
vernal forge
,w maclaurin series x^x
marsh citrusBOT
,w maclaurin series x^x
elfin berryBOT
vernal forge
why does WA give a result with logs?
still temple
because it didn't do what you asked
vernal forge
oh it says there that it's a Puiseux series. but does a maclaurin/taylor series exist for x^x?
,w taylor series x^x at x=2
elfin berryBOT
elfin berryBOT
vernal forge
interesting
ig it's because x^x is undefined within reals almost everywhere for x<0
,w taylor series x^x at x=1
elfin berryBOT
vernal forge
,w taylor series x^x at x=0.02
elfin berryBOT
vernal forge
,w taylor series x^x at x=0.00001
elfin berryBOT
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honest hemlock
Hello, I'm learning probability and got stuck on this example. I understand the function of max and min but I'm confused about their outputs and input
marsh citrusBOT
Hello, I'm learning probability and got stuck on this example. I understand the ...
honest hemlock
here is the graph it mention (fig. 1.7.)
here it says the min is 2. I understand this as the set contain minimum 2 elements. right?
so you mean that the min element value is 2?
what about A?
the max which mean no element is bigger than what?
it says m = 1, 2, 3 ,4
how?
so?
oh
so like { 1 }, { 1, 2 }, {1 , 2, 3} ...
?
but what the set looks like?
I'm sorry but what does that mean ? ordered pairs?
so far I understand that A would be like { (X, Y) ... } like if m = 1 then { (1, Y) }?
how?
oh
okay
yeah I can see that. umm I'll try and come back.
is there a limit on the channel?
oh okay I'll try to solve it and come back
thank u
is this close?
Given A = { Max(X, Y) = m };
And B = { Min(X, Y) = 2 };
We can say the value of each set as
m = 1
A = { (1, 1) } And B = { (2, 2), (2, 3), (2, 4), (3, 2), (4, 2) };
m = 2
A = { (1, 1), (1, 2), (2, 1) } And B = { (2, 2), (2, 3), (2, 4), (3, 2), (4, 2) };
m = 3
A = { (1, 1), (1, 2), (2, 1), (1, 3), (3, 1) } And B = { (2, 2), (2, 3), (2, 4), (3, 2), (4, 2) };
can you write the function?
oohh
got it
okay now I know what confused me
I misunderstood the max and min function I thought the max set a boundery like m = 3 mean 1, 2, 3. and m = 4 is 1, 2, 3 ,4
how about this
m = 1
A = { (1, 1) }
m = 2
A = { (1, 2), (2, 1) }
m = 3
A = { (1, 3), (3, 1), (2, 3), (3, 2) }
Oh same I forget them
yeah I got it
thank u again
I'll close the channel I solved it
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winged marsh
help
cunning oak
whats your problem?
winged marsh
cunning oak
what are you confused by
winged marsh
this one, i understand it a little but i just need some bit of help.
winged marsh
what are you confused by
solving part
cunning oak
well, the form for a complete square is x^2 + 2ax + a^2
take for example the first question
winged marsh
ok
cunning oak
you can see that 2a = 4
right?
(you dont have to mind the + - signs in front of ax)
winged marsh
oh