i think
#help-27
stiff crater
stiff crater
not really
solar goblet
do you know what a complex number is?
crisp tapir
imaginary part of what
stiff crater
mhm
topaz gull
A complex number is a+bi miss doritos
stiff crater
yea ik
topaz gull
So substituut that in
stiff crater
ohhh
topaz gull
You get how to continue from there?
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hushed nexus
How is 0/h equal to 0 isn't it undefined when you plug in 0 for h
feral agate
Yes, but whenever you write a limit it’s implied that h is not actually equalling zero, just getting closer and closer to it
The substitution method you’re thinking of only works because we’re abusing continuity of (most!) known functions
0/(non zero) = 0
hushed nexus
I've seen h being substituted by 0 mostly in the numerator
feral agate
Yeah, but substitution only works when the function is continuous
You will encounter cases when substitution doesn’t work
$\lim_{n\to 2}\lfloor n\rfloor$ for example
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kheerii
feral agate
Substituting n = 2 in this limit yields floor(2)=2, but the left hand limit of this is not 2 and rather its 1
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hushed nexus
Anyone???
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deep vortex
This is about \underline{dampened} electromagnetic oscillations: \begin{align*}
\ddot Q(t) + \frac{R}{L} \cdot \dot Q(t) + \frac{Q(t)}{LC} &= 0\ \lambda^2 \cdot e^{\lambda \cdot t} + \frac{R}{L} \cdot \lambda \cdot e^{\lambda \cdot t} + \frac{e^{\lambda \cdot t}}{LC} &= 0 \ e^{\lambda \cdot t} \left(\lambda^2 + \frac R L \cdot \lambda + \frac{1}{LC} \right) &= 0
\end{align*} we arrive there by setting $Q(t) = e^{\lambda t}$. Now \begin{align*}
e^{\lambda \cdot t} \left(\lambda^2 + \frac R L \cdot \lambda + \frac{1}{LC} \right) &= 0 \ \lambda^2 + \frac R L \cdot \lambda + \frac{1}{LC} &= 0
\end{align*} and there are three possibilities given by the discriminant: \begin{align*}
&1) \quad \frac{R^2}{L^2} > \frac{4}{LC}
\qquad \quad 2) \quad \frac{R^2}{L^2} = \frac{4}{LC} \qquad \quad 3) \quad\frac{R^2}{L^2} < \frac{4}{LC}
\end{align*} \underline{But why do these determine whether it's a an aperiodic limit case,} \underline{a creep case or an oscillating case?}
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This is about \underline{dampened} electromagnetic oscillations: \begin{align*}
...
acoustic leaf
the difference is in whether the quadratic has real, complex, or repeated roots
deep vortex
the difference is in whether the quadratic has real, complex, or repeated roots
Yes but why does that determine the kind of case?
Might aswell think the sign of the roots determines it
It comes a bit unmotivated
acoustic leaf
by euler's formula e^{ix} = cos(x) + i sin(x), so we expect complex roots to be oscillatory, but the same doesn't apply to real roots
deep vortex
by euler's formula e^{ix} = cos(x) + i sin(x), so we expect complex roots to be ...
Ohh
deep vortex
by euler's formula e^{ix} = cos(x) + i sin(x), so we expect complex roots to be ...
Ok so that gives us that when the discriminant is negative, it really oscillates
What about differentiating between = 0 and > 0
acoustic leaf
well we need two linearly independent solutions to fully solve the ODE, and if the root is repeated (discriminant 0), then e^{λt} is no longer able to provide both solutions with different values of λ. (we find that te^{λt} works as a second solution in this case)
deep vortex
well we need two linearly independent solutions to fully solve the ODE, and if t...
So how does that tell us whether it's the aperiodic limit case or the creep case (or however that one is called in English)
acoustic leaf
qualitatively they're not especially different, other than the fact that the t e^t factor makes the repeated roots case asymptote faster than the positive discriminant case
we can think of it this way. we start at the initial condition and grow towards the asymptote.
- positive discriminant: grow to the asymptote slowly
- 0 discriminant: grow to the asymptote as fast as possible without overshooting
- negative discriminant: grow to the asymptote so fast it overshoots the mark and starts oscillating
deep vortex
well we need two linearly independent solutions to fully solve the ODE, and if t...
Uh
then e^{λt} is no longer able to provide both solutions with different values of λ.
So what
acoustic leaf
we know that the solutions to a second order linear homogeneous ODE forms a 2-dimensional vector space, so to find a basis for that, we need two linearly independent solutions. by guessing solutions of the form e^{λt}, we can transform it into a quadratic and find two linearly independent solutions in the form of roots of the quadratic. this fails if there is a repeated root (discriminant 0)
deep vortex
we know that the solutions to a second order linear homogeneous ODE forms a 2-di...
I'm doing this for a presentation for an 11th grade so unfortunately I can't argue with linear algebra
Is there any alternative?
I even have to introduce complex numbers in the same presentation
dire forge
I'm doing this for a presentation for an 11th grade so unfortunately I can't arg...
Just say second-order DEs have two solutions
You will have two solutions, one of the form e^lambda t and one of the form t e^lambda t
The second can be obtained through the reduction of order technique, or you can just state it as fact (lambda repeats twice, so there should be two solutions involving e^lambda t)
I'd recommend watching MIT OCW's videos on differential equations
deep vortex
The second can be obtained through the reduction of order technique, or you can ...
Yeah that is called a resonance case I think
deep vortex
Just say second-order DEs have two solutions
Alright
rose girder
Sad not having linear algebra at hand
deep vortex
Sad not having linear algebra at hand
No what's worse is that not even complex numbers are at hand
dire forge
In this case the linear algebra is not very necessary
rose girder
Introducing the Jordan canonical form wouldn't be a good idea 
dire forge
If you wanted to talk about linear equations in general, it would be useful
But here, you're just solving a specific equation anyways
deep vortex
Introducing the Jordan canonical form wouldn't be a good idea <:catthimc:7320118...
In the end it's going to turn from a physics GFS to a math GFS
Probably that already happened, I spend like 20 slides on solving these DEs
dire forge
I'm just surprised you don't learn anything about complex numbers in high school in germany?
deep vortex
I'm just surprised you don't learn anything about complex numbers in high school...
Yeah we don't, sometimes at the end of 12th grade a bit
dire forge
Interesting 
that must be something German students that study at American unis really struggle with then
deep vortex
Interesting <:catthink:526983414226747393> that must be something German student...
Is that a pre-req in the US?
dire forge
I guess you don't learn the fundamental theorem of algebra either then?!
dire forge
Is that a pre-req in the US?
It's standard in US high schools
deep vortex
But I learned all these in the first weeks of linear algebra 1 at uni
dire forge
Fascinating
deep vortex
So I guess it's not a pre-req
dire forge
Linear algebra's a weird place to learn that
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deep vortex
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deep vortex
we know that the solutions to a second order linear homogeneous ODE forms a 2-di...
Wait we have two lambdas, we want to give an expression for Q(t) = e^(lambda * t)
Which one do we plug in?
Does it not matter?
acoustic leaf
well in full generality it would be [ C_1 e^{\lambda_1 t} + C_2 e^{\lambda_2 t} ] with $C_1, C_2$ determined by initial conditions (the simplest case being if one of them is 0)
woven radishBOT
cloud
deep vortex
well in full generality it would be \[ C_1 e^{\lambda_1 t} + C_2 e^{\lambda_2 t}...
Ah
deep vortex
well in full generality it would be \[ C_1 e^{\lambda_1 t} + C_2 e^{\lambda_2 t}...
How do you motivate the sum?
sand dove
How do you motivate the sum?
the set of solutions is a linear space
since we're facing a LINEAR ODE
and e^(lambda_i t) is a basis of that linear space
acoustic leaf
in linear algebra terms we're taking the span of the two basis vectors. for your purposes you can demonstrate the linearity of the ODE by showing it distributes over sums & multiplications
deep vortex
in linear algebra terms we're taking the span of the two basis vectors. for your...
Over multiplication is clear, what do you mean by over sums?
acoustic leaf
you can show that if y1 and y2 are solutions, then y1 + y2 is a solution
deep vortex
Ah
deep vortex
How do I eliminate the i
Here
In the last equation
This is what I need to get:
See how there is no i in there
frozen aurora
what is Q hat?
deep vortex
what is Q hat?
It's just a starting condition
c_1 + c_2 = Q_hat
And there is some other condition
frozen aurora
okay
deep vortex
So my c_1 and c_2 will turn into it or something
frozen aurora
recall that cosine is an even function and sine is an odd function
deep vortex
Yes
deep vortex
recall that cosine is an even function and sine is an odd function
But still, it's not like the isin(..) will cancel
In the solution, there is a sine too
frozen aurora
can the constants be imaginary?
deep vortex
can the constants be imaginary?
frozen aurora
R/2L
deep vortex
They aren't
frozen aurora
do the slides just end there?
deep vortex
i don't see Q_0 being c_1 + c_2
We know Q(0) = Q_0
lost laurel
this looks like electrical engineering
deep vortex
i don't see Q_0 being c_1 + c_2
We get 1(c_1(1 + 0)) + c_2(1 + 0)) = c_1 + c_2
frozen aurora
hmmm true
deep vortex
do the slides just end there?
I make the slides
deep vortex
hmmm true
Maybe I made a mistake somewhere? 
I also have the starting condition Q'(0) = 0
acoustic leaf
in general if we have complex roots, they will be conjugates: [ \lambda = a \pm ib ] so in general we will get solutions in the form [ e^{a t}(\cos(bt) + i\sin(bt)), \quad e^{a t}(\cos(bt) - i\sin(bt)) ] by taking linear combinations of the above, we arrive at a real-valued basis of solutions [ e^{at}\cos(bt), e^{at}\sin(bt) ]
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cloud
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deep vortex
in general if we have complex roots, they will be conjugates: \[ \lambda = a \pm...
Ok somehow I need to argue c1 = c2 and I'm done
I plugged in the condition c1 + c2 = Q_0 already
Now I guess what is left is Q'(0) = 0
But
deep vortex
in general if we have complex roots, they will be conjugates: \[ \lambda = a \pm...
So, any idea to argue c1 = c2 somehow?
acoustic leaf
for the solution to be real-valued you would require c1, c2 to be complex conjugates
proud perch
schwingungen
deep vortex
for the solution to be real-valued you would require c1, c2 to be complex conjug...
Wait nvm, actually this is the solution
Where CU_0 = Q_0
We need toget Q_0 out of the brackets first
And somehow do something about that i
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half badge
x, y and z complex numbers of modulus 1
does this mean magnitude 1
slim condor
not a math problem but a clarifying question, all trig ratios give a value of -1 < x < 1 right?
half badge
x, y and z complex numbers of modulus 1
does this mean magnitude 1
simple tundra
y are are ppl asking for help instead of helping u lol
stone stump
yes its another name for magnitude
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snow grove
Hi
idle swift
What does |X_i| mean?
cardinality?
snow grove
Because X_1, ..., X_N are random variables
which is really just a probability distribution
idle swift
|A| usually means the number of elements in set A
snow grove
no, that doesn't apply here
if X_i was really a set
then
the expected value of X_i wouldn't make sense
idle swift
the is it just the absolute value?
idle swift
uhhhh
spice raft
It's just the absolute value of the random variable
snow grove
the function's absolute value at all points is less than a_i?
spice raft
Yes
snow grove
Ah, I see
And each random variable function X_i is supposed to be between -a_i and a_i at all times in its probability distribution graph
spice raft
Yes exactly
spice raft
No
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pastel pine
3∗(m− 1\2 )∗(m+ 1\4 )=2m<2 − 3\4 m i don understand why answers are (√6\4;-√6\4)
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sturdy yew
!original
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royal laurel
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royal laurel
That’s the definition of correlation coefficient
I’m not quite sure what would it be if one expanded out the numerator
I think is nr=(x_1)’ (y_1)’ + (x_2)’ (y_2)’ +……
Can someone confirm the result?
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stable storm
yeah it would be like:
$$r=\frac{{x_1}'{y_1}'+{x_2}'{y_2}'+{x_3}'{y_3}'+...+{x_n}'{y_n}'}{n}$$
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y0shi
inner ibex
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inner ibex
what is this formula
drowsy arrow
The exponential one?
drowsy arrow
Ao is like the initial amount/what u start with
drowsy arrow
Can it only be in years
I think so but check with ur teacher to confirm
I remember having to convert my time to years when I would use the equation
inner ibex
It school holiday ))))):
drowsy arrow
So probably yea
inner ibex
ohh okok
drowsy arrow
U got the Ao part yea?
inner ibex
ya
drowsy arrow
A is what u end up with basically
inner ibex
Wats B
drowsy arrow
And base is how much it goes up by
Like if it’s doubling every time b is 2
Triples it’s 3
inner ibex
Is there a way of deriving this formula
inner ibex
Idk how to explain
Ummmmm like
I dont wanna memorise the formula I wanna see where it comes from
drowsy arrow
Like a general formula?
A = I(b)^t
I = initial amount
B = base
T = time
A = what u end up with basically/when you should get after certain time
inner ibex
Am sad
drowsy arrow
Same
inner ibex
Is not in my formula sheet 😢
drowsy arrow
Damn
What part do u not get? Maybe I can help
Like do u not understand how to put info in the equation?
drowsy arrow
As we can see here it increases by 5% every day, due to the wording being INCREASING we know it will be exponential GROWTH so we use the exponential growth equation
Since 5% in decimals is 0.05 we add that to one and get 1.05 as our base
Since in your picture we don’t convert time to years we assume the time is whatever they say it is and since it says every DAY our time is in days
Since it already started with 3m² and grows every day we can say that the initial amount is 3 bc that’s what it started with than began to grow
Then u put all that into the equation (exponential growth equation)
A = Ao(1 + r)^t
smoky gyro
BUM CHICKEN
drowsy arrow
A = outcome basically
Ao is the initial amount (in this case it is 3m² so Ao is 3)
R is rate (5% -> 0.05 plus one bc we use the growth equation so 1.05)
T is time (in days bc that’s what it said in the equation so it’s safe to assume that’s the measurement of time in this equation we will use)
Then u plug everything into the equation
A = 3( 1 + 0.05)^t
Simplifies to A = 3(1.05)^t
@inner ibex
Hope this helps I’m sorry I didn’t really get what u didn’t understand 🥲
Idk what u mean by derive the equation so I broke it up and hope it helps u
I think I’ll sleep a bit, check up in 5 minutes (:
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surreal crane
I need to make sure my math is mathing
If I price something at 14.99 and they take 12% I will get 13.19
drowsy arrow
Idk but I can try to figure it out
whole ruin
i got it
drowsy arrow
I was sleeping and waking up and saw this chat opened lol
drowsy arrow
Alright Nighty night
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pallid cosmos
hello
trail eagle
hi
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spring oak
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spring oak
- Let A1, A2,..., A20 be twenty sets each having 10 elements and B1, B2,..., B, be sets. Of which 20 each have 5 ingredients. If UA₁ = UB; = S and 1=1 j=1 If each element of S exists in exactly 5 sets of the first group and exactly 9 sets of the second group, then n is equal to -
(a) 45
(b) 54
(c) 72
(d) 50
elfin atlas
okay so firstly, we have 200 elements in A and consider the number of elements in S = k.
If UA₁ = UB; = S and 1=1 j=1 If each element of S exists in exactly 5 sets of the first group and exactly 9 sets of the second group
200 = 5k => k = 40.
B_j has 5 elements in each set so we have, 5n = 9k.
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elfin atlas
Tq very much
:D anytime!
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sweet crystal
help
sweet crystal
can i please have help in understanding
what they do for c
step by step im confused for all of it
languid root
@sweet crystal whats the formula can you write it in plain text i would b happy to help!
sweet crystal
<@684329174034677783> whats the formula can you write it in plain text i would b...
the formula for depreciation
languid root
the formula for depreciation
ah ok
sweet crystal
Future value = principle value x (1+r/100)^n
languid root
one sec
sweet crystal
okayy
languid root
you dont have any answer right?
sweet crystal
that is the awnser key
sweet crystal
thank you
languid root
sorry could not type it here without discord bugging out can i just send a ss
sweet crystal
sure thing
sweet crystal
so we find future value
which is the 26800 x AVG depreciation for 6 years
thats out FV value
and than we just substitute that with our current value of 3 years
and than we can solve for interest rate yea?
languid root
yeah!
languid root
your welcome! ping me any time you need 
sweet crystal
.close
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how do i do this?
late estuary
midpoint theorem states that PR = 2 * KL
do that for all 3 middle lines and you have that the perimeters halve after each iteration
a_n := 0.5^(n-1) * k (* 3 for perimeter)
urban harbor
!nosols
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late estuary
how do you help with this without naming the theorem that solves it automatically
apparently still not understood
trail furnace
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frigid hare
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frigid hare
i need to prove this inequality with x1,x2,x3...
hazy leaf
what is $\alpha$ and $\beta$?
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nameless individual
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frigid hare
positive real numbers
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worn totem
I have to solve for dr/du = (r-2M)/(r+2M) (light going into a black hole in Eddington-Finkelstein coordonates) but since it's nonlinear i tried to developp it into entire series but it seems all orders are zero except for the first one which is r=2M the constant solution, can anyone help ?
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I have to solve for dr/du = (r-2M)/(r+2M) (light going into a black hole in Eddi...
worn totem
Please 😭
stable storm
whatre you solving for exactly
and is M a constant
honestly if this is physics based, you should ask for help in the physics server
this server is mainly for math
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worn totem
M is a constant
I search for the function r of u
It's just a maths issue
I tried to expand r, like r = sum from zero to infinity of a_n*u^n, and I only have a_0 = 2M, and the rest is zero
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stable storm
this differential equation is separable
$\dv{r}{u}=\frac{r-2M}{r+2M} \ \
\frac{r+2M}{r-2M} dr=du \ \
\int\frac{r+2M}{r-2M} dr=\int du$
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y0shi
worn totem
You're right I did it like that but that doesnt give an expression r of u
stable storm
is it possible to solve that equation for r in terms of u?
not sure what integrating this would look like yet
havent done it
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real osprey
albert if you're still here i think you might need to use the lambert W function for that since you have r outside and inside the logarithm
dire forge
Indeed, you need to use the lambert W function
Wolfram Alpha solves it
,w dy/dx = (y-2M)/(y+2M)
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dire forge
@worn totem There's your solution ^
worn totem
Thanks ! Indeed it's a difficult one, I will try to developp the lambert W function in the first orders to see the tendancies of the solution
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restive river
Pretty simple question in theory, but my brain is fried so I can't work it out for the life of me.
Starting at a 50% chance of winning or losing a game. Upon winning, your success rate is bumped up by 1%, ie. to 51%, then 52%, then 53%, etc.. Upon losing a round, your chance of victory is reset to 50%.
All in all, what is the total chance (and formula to calculate it) of a player reaching a 100% victory chance in one run?
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icy quail
it's just 0.5 x 0.51 x 0.52.. x 0.98 x 0.99, or to put it simply which had a chance of
oh wait
wrong photo
it's the multiplication rule of probability
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restive river
it's just 0.5 x 0.51 x 0.52.. x 0.98 x 0.99, or to put it simply which had a cha...
Ah yes, thanks haha
opaque talon
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opaque talon
for the 2nd fraction
Why does the numerator just become irrelevant
I thought it would be 2ln(2x-1)
covert root
Theorem : d/dx ln(f(x)) = f'(x)/f(x)
opaque talon
ohhh
covert root
Using this theorem you get ln|2x-1|
covert root
correct
opaque talon
damn ok thank you
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opaque talon
.close
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viral skiff
how do i do 6 part i
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how do i do 6 part i
quick cedar
well from part B you have the angle BAC
and the bearing you’re given loops clockwise from North to the line AC
viral skiff
angle BAC is 82.8 degrees
ya
ya
so how do i fine b from a
find*
ohhh
i get it
quick cedar
yep
viral skiff
210-82.8
quick cedar
there you go
somber ingot
how can i do this question pls
devout snowBOT
how can i do this question pls
eager nova
how can i do this question pls
What have you tried?
Use pythagorean theorem and the area of the triangle.
eager nova
Use pythagorean theorem and the area of the triangle.
Do you know these?
somber ingot
ye duh
eager nova
Write them
eager nova
So using pyt, tell me what is a^2-c^2
somber ingot
-b^2
eager nova
This is wrong
eager nova
It is a^2=b^2+c^2
eager nova
But you will get different b now
eager nova
It is a^2=b^2+c^2
What is b^2
somber ingot
b is middle
eager nova
But we are using the data from the exercises
a is hypotenuse
So a^2=b^2+c^2
Hence a^2-c^2=?
somber ingot
OH
i get itt
ok
i was confused
ok i got the question
thx
wait canu. help with another one
pls
jovial maple
Hello, looking for web applications or Android mobile applications
to practice and prepare for SAT like exams. Problems which at most require pencil and paper, under 2 to 1 at most minutes to solve (with as fast as is humanly possible).
I am especially in need of an app that involves simplifying or solving equations with integer powers and roots
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to practice a...
pastel pasture
calculator?
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last sequoia
how 2 good at meths
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mellow talon
learn more, do exercises, challenge hard questions
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viral lynx
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viral lynx
I got pretty far on this one but im stuck
sin^2(x)cos^4(x)
(1 - cos(2x))(1 + cos(2x))^2/16
(-cos^3(2x) - cos^2(2x) + cos(2x) + 1)/16
(-cos^3(u) - cos^2(u) + cos(u) + 1)/32
Im not really sure about the next step...
eager nova
Rewrite sin^2(x) as 1-cos^2(x)
viral lynx
I did that in the first step
viral lynx
That would let (cos^4(x))(1-cos^2(x))
Im confused how you got this
Is it not (cos^4(x))(1-cos^2(2x))/2
eager nova
Rewrite sin^2(x) as 1-cos^2(x)
.
eager nova
It is pythag
viral lynx
Ok but what can I do with (cos^4(x))(1-cos^2(x))
eager nova
Expand
viral lynx
(cos^4(x)-cos^6(x))
eager nova
Integral of a-b is integral of a - integral of b
viral lynx
But what about the sin(x)
eager nova
What about that
You know have integral of cos^4x-cos^6x
That is integral of cos^4x - integral of cos^6x
viral lynx
But we cant integrate that cause of the sin(x) no?
I assume you want to use u=cos(x)
so du = -sin(x)
eager nova
What do you mean, i am not following you
viral lynx
so du = -sin(x)
I dont see what we do next because of this
Can you show me what you have in mind
eager nova
That is integral of cos^4x - integral of cos^6x
Did u understand this?
viral lynx
You know have integral of cos^4x-cos^6x
How do you integrate this without usub
viral lynx
Did u understand this?
Yes I understand up to here
eager nova
I have only 1 minute to go, look for reduction formula for cos^n(x)
viral lynx
Ok I saw a video on it I definatly havent covered it so I dont think its how my course wants me to do it...
viral lynx
Im not really sure about the next step...
I feel like I got pretty far with my method though
Does someone know the next step?
sin^2(x)cos^4(x)
(1 - cos(2x))(1 + cos(2x))^2/16
(-cos^3(2x) - cos^2(2x) + cos(2x) + 1)/16
(-cos^3(u) - cos^2(u) + cos(u) + 1)/32
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smoky gyro
sin^2(x)cos^4(x)
(1 - cos(2x))(1 + cos(2x))^2/16
(-cos^3(2x) - cos^2(2x) + cos(...
Why did you switch sin^2(x) to 1 -cos2x?
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urban harbor
i need some help with this problem from the LA done right group, i think i have an idea of how to get started how to create the function but i'm getting hung up on proving its an isomorphism
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i need some help with this problem from the LA done right group, i think i have ...
wooden veldt
what is your map?
urban harbor
ok so say v1 + U, ... vm + U is a basis for V/U
then my idea is that for some random v in V, pi(v) = c1(v1+U) + ... + cm(vm +U) is the representation of v + U in that basis
so v - (c1v1 + ... + c_m v_m) should be in U right ? since pi(that vector) = 0 ?
so lets say the map T: V -> U x V/U is
T(v) = (v - (c1v1 + ... c_m v_m), c1v1 + U + ... c_m v_m + U)
does that make any sense? trying to split the vector v into the part that's in U and what's "outside of U" in V?
I'm not really sure how to show that T is bijective
wooden veldt
i think your map is fine. i think i personally would have defined a map in the other direction as that seems easy to prove bijectivity
but it should be doable for yours
surjectivity seems easier at first glance
urban harbor
hmm
wooden veldt
for a start, if you want to hit (u,0) what should v be?
urban harbor
i guess u
wooden veldt
yea
urban harbor
oh so i guess in general for any (u, w) i would just get the representation of w = c1v1 + ... + cmvm and T(u+c1v1 + ... + cmvm) = (u,w)
so that would show its surjective?
wooden veldt
v1 + U etc but yeah
urban harbor
erm yeah
wooden veldt
you can also just do it on a basis element of V/U
if you want to hit (0,v_i + U) then v must be v_i
then linearity takes you home
ig you should prove that your map is linear lol
urban harbor
i kinda see what you mean though not sure how i'd say it "formally"
wooden veldt
oh so i guess in general for any (u, w) i would just get the representation of w...
i mean yeah its a bit of a ballache thats why we hide it behind "by linearity" lol, it would basically amount to what you did here
urban harbor
ig you should prove that your map is linear lol
ok
wooden veldt
point being if you can hit any (u,0) and any (0,v_i + U) then by linearity you can hit everything
urban harbor
so for linearity, its been a while, all i need to show is that T(u+w) = T(u) + T(w) and also T(cu) = cT(u), do i also need to show T(0) = 0 or is that taken care of by the other 2 i forget
urban harbor
point being if you can hit any (u,0) and any (0,v_i + U) then by linearity you c...
yeah that does make sense
wooden veldt
taken care of: T(0) = T(0 + 0) = T(0) + T(0) => T(0) = 0
urban harbor
aha ok
so i think of u, v are in V then we'd look at pi(u) and pi(v) and get the representations of that in V/U as something like c1v1 + U + .... + vmvm + U and d1v1 + U + .... + dmvm + U
then when we apply T, we'd end up with (c1+d1)v1 + ... (cm+dm)vm for the first one
and similar sum for the 2nd coordinate
so that would take care of the sum part?
since pi(u+v) would be the same representation in V/U
wooden veldt
thats the general idea yeah, you just follow your nose and everything just works out because of how you defined the map
urban harbor
ok so i think i get the gist of how to do sum and scalar mult, then just injective is needed
im guessing i could take an element in codomain of T and show that if T(u) and T(v) map to that element then u = v, the classic approach?
wooden veldt
that is the classic approach yeah
you could also build an inverse
which actually would have needed us not to show surjectivity explicitly
i think both ways are enlightening
wooden veldt
its linear and that both compositions with T are the identity
urban harbor
ah and that automatically takes care of both T and the inverse being bijections?
wooden veldt
yep! injective iff left inverse exists and surjective iff right inverse exists
urban harbor
this might be more educational for me since it doesn't seem as obvious
so lemme think how i'd define the inverse
wooden veldt
sure, follow your nose
urban harbor
i guess for (u, w) where u in U and w in V/U, we'd have w = c1v1 + U +... +cmvm + U
so we'd define Tinv(u, w) = u + c1v1 + ... + cmvm ?
wooden veldt
call it S for our sanity but yep that seems like a likely candidate
urban harbor
lol
aha yeah that makes so much sense
so wait have to show that TS and ST are identity uh linear maps like ST(v) = v and TS(u, w) = (u, w)
ok that seems pretty straightforward, i think i'd have it from here
wooden veldt
but now you have both maps you dont need to
sorry my internet has just died
this is the map that i initially came up with when i saw the problem, just because its a lot easier to write down, surjectivity was fine i didnt take a look at injectivity yet
and now my messages are out of order lol
urban harbor
hm maybe just on your screen
wooden veldt
so wait have to show that TS and ST are identity uh linear maps like ST(v) = v a...
yep!
you do also wanna check that S is linear but that seems pretty clear
urban harbor
but now you have both maps you dont need to <:catking:812386361297862747>
wait i don't need to show that?
or im not sure what you were referring to here
wooden veldt
i mean that now you have both maps you dont explicitly need to check inj or surj on either of them, just that they are mutually inverse
urban harbor
right right
wooden veldt
but everything you did was good sometimes it can be really hard to write down a backwards map
so inj and surj is the only way to go
urban harbor
ok thank you that was mucho helpful, really was tripping up and needed to talk that problem "out loud" in a chat at least : )
wooden veldt
no worries good stuff
urban harbor
i think i got it well enough that i can write down a solution in latex now
:p
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urban harbor
you do also wanna check that S is linear but that seems pretty clear
.reopen
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✅
urban harbor
one more question, if you show that T is linear and invertible, wouldn't that be enough without showing that S is also linear?
even though it seems like a bit of a silly point
wooden veldt
this is true for linear maps yes. heed caution with this thought though, the set map inverse of a continuous function (on a top space whatever that means) need not be continuous
and im sure a million other similar examples
linear maps are sufficiently nice though
urban harbor
but i mean if you've shown T is linear and invertible, doesn't that automatically make it an isomorphism
without needing to think about S specifically
wooden veldt
right but we didnt do injectivity
we (well really I) posited that it might be easier to just find the inverse map instead. not on good authority mind, i didnt do the injectivity proof
urban harbor
so is it really important to show that the inverse S is linear if doing the inverse map route?
wouldnt showing TS = I and ST = I be enough?
thin fern
i need some help with this problem from the LA done right group, i think i have ...
Can't you show that they have the same dimension
wooden veldt
one more question, if you show that T is linear and invertible, wouldn't that be...
yes, this is what i said in response to this
wooden veldt
Can't you show that they have the same dimension
only V/U is finite dimensional
urban harbor
Can't you show that they have the same dimension
if finite dimensional yeah it becomes a lot more trivial, but this doesn't assume that V or U are finite dim
thin fern
ah I see
urban harbor
yes, this is what i said in response to this
ok thanks again 🙏 
wooden veldt
to clarify Soosh, the two paths were either: write down T, show linear, write down S and show its a set inverse OR write down T show linear, inj,surj
we did half of the second and then sacked it off for the first lol
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gentle nymph
how do i solve c
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how do i solve c
mild sorrel
close my mouth.
gentle nymph
xD
gentle nymph
x * f(x)
frozen aurora
missing something
gentle nymph
whats that
frozen aurora
an integral sign lol
gentle nymph
oh x * integral f(X) ?
woven radishBOT
artemetra
gentle nymph
ohh okk
woven radishBOT
artemetra
frozen aurora
well
gentle nymph
like that?
nvm typo
last answer is 9/4
so how do i get var(2x - 5)
the formula is E(X^2) - (E(X))^2
so the right part i got 81/16
gentle nymph
so E(X^2) do i just repeat the process here but using x^2 * f(X) instead of just x * f(x)
@frozen aurora
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reef prism
Hi! I just want to ask, why's c=-296-999t' = 703+999t" ?
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Hi! I just want to ask, why's c=-296-999t' = 703+999t" ?
@reef prism Has your question been resolved?
celest valve
they just wanted to move from negative values to positive
so we have c= -296 + (multiple of -999)
999 is a multiple of -999: -999*(-1)
so we can rewrite that as c= -296 + (multiple of 999)
furthermore, we have c=-296 + 999t'= -296 + 999 + 999(t'-1), but since t ranges over all integers, we can just rename (t'-1) as t''.
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reef prism
so we can rewrite that as c= -296 + (multiple of 999)
still kinda stuck as to why we added 999 on -296…
celest valve
just to have it positive
unborn atlas
Question 19 what am I doing wrong
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Question 19 what am I doing wrong
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hasty flame
this correct?
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this correct?
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@hasty flame Has your question been resolved?
olive snow
Yes
restive river
help
restive river
so theres two cases right
you can either choose M first in which case you cant sell any tickets because you cant give change
you can choose N first and then give the next person who will be an M
so that is £1 from N, and then £1 given to M as change but you end up keeping a £2 so you cant give change to the next M
so it ends at 2
so i thought that the probability would be 2/(1+m)
obviously this is wrong
please help
zenith sequoia
you can either choose M first in which case you cant sell any tickets because yo...
You start with no change, so as soon as someone comes with a £2 coin, you have to stop. You seem to have this backwards
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zenith sequoia
Sorry that's wrong, I ignored the fact that you gain change after selling a ticket to someone with a £1 coin
restive river
yeah bur
but
that change you get only lasts for 1 person
because everyone after that first person has a £2 as n = 1
zenith sequoia
You only need to give change to people with £2 coins, so as long as the single person with a £2 coin isn't first, you can sell a ticket to everyone
restive river
but you cant
because
so first person gives me £1
2nd person gives me £2, and i give them £1 change
i then have a £2 coin which i cant give as change to the 3rd person
do you see what i mean?
zenith sequoia
Ok I see what you're saying
Only one person gives you a £2 coin though
You don't have to give change to the third person because they'll give you a £1 coin
zenith sequoia
It happens haha, np
zenith sequoia
This is the alternate form for arsinh
restive river
i shouldnt use that
hyperbolic functions are not apart of my cours
this is the full question
this is my working thus far
wait its not improving my maths skills by asking
,close
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zenith sequoia
wait its not improving my maths skills by asking
I'd suggest looking at the answer, you might be able to figure it out from there
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jaunty abyss
This is not a right triangle, right? Two slopes would have to be negative reciprocals?
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This is not a right triangle, right? Two slopes would have to be negative recipr...
jaunty abyss
Oh ok
jaunty abyss
wrote some numbers wrong
Could you give me a hint?
dense jay
AC
dense jay
if youre consistent in which point is 1 and 2, no
jaunty abyss
Uhm I got 1/7
jaunty abyss
Okay! So it’s not a right triangle correct?
None of the slopes are negative reciprocals
dense jay
are they not?
dense jay
-2/6=?
dense jay
si
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oak jasper
Wondering if this is correct and if not where I went wrong? I think the fraction messed me up so I don’t understand this with fractions.
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Wondering if this is correct and if not where I went wrong? I think the fraction...
oak jasper
4.5
woven radishBOT
faiyrose
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subtle oar
Which polynomial represents the total area in square meters?
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Which polynomial represents the total area in square meters?
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
vast drift
Which polynomial represents the total area in square meters?
which part are u stuck on
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void knot
$\lim_{x\to\frac\pi2}\frac{(1-\sqrt{\sin x})(1-\sqrt[3]{\sin x})\cdots(1-\sqrt[n]{\sin x})}{(1-\sin x)^{n-1}}.$
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$\lim_{x\to\frac\pi2}\frac{(1-\sqrt{\sin x})(1-\sqrt[3]{\sin x})\cdots(1-\sqrt[n...
woven radishBOT
nino
void knot
Status 1
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vast drift
Status 1
im assuming theres meant to be actual techniques for this but im limiteed in my math knowledge so heres my attempt
split them into matching fractions iwth (1-sinx) as the denominator and an excess term from the numerator aside
every single fraction of the type (1-(sinx)^1/i) / (1-sinx)^n-1 should approach infinite
u are left with an excess 1 - (sinx)^1/i term in your numerator
as x approaches pi/2 this excess numerator term approaches 0 (1-(1))
so your overall limit should approach 0
its not rigorous by any means but thats all i can think of
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tiny panther
The limit does not approach 0, as substituting n=2 shows
hint: use difference of powers formula somewhere
void knot
This?
tiny panther
yes
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restive river
hello I need help!
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hello I need help!
restive river
I'm trying to use chat gpt to solve a simple problem but i don't know the name of it
the closest thing i got is
Columnar intersections
where you change every x value to 0 to find y
or f(x) =0 to find x
(c,0) or (c,k)
restive river
where you change every x value to 0 to find y
example
x^4- 1= (0,-1)
wicked turtle
!nogpt
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Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).
stable storm
i believe you're trying to find x and y intercepts
stable storm
Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they ...
but dont use chatgpt like said here
restive river
but dont use chatgpt like said here
what's the alternative?
stable storm
wolfram alpha is usually correct
edgy nacelle
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edgy nacelle
could i get help with c please
this is what i have done and ive done the questions before
im not too sure where to go from here as it is only 2 marks
ive now tried to reason with what i did before
im also stuck on the next question
gentle nymph
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stable storm
basically you want to find P(x>k)=0.5
what is the center of this distribution
the center divides the distribution exactly in half
also this distribution isnt binomial, so you dont need to make it P(x<=169)
P(x<=170) works just fine
stable storm
yeah
gentle nymph
basically you want to find P(x>k)=0.5
so i have to do inverse normal?
gentle nymph
yeah i got the same
stable storm
since the normal distribution is symmetrical
gentle nymph
oh so thats the concept?
stable storm
mhm
gentle nymph
mean will always be the middle
stable storm
mean is the center
gentle nymph
thanks bro
gentle nymph
so k is just mean?
stable storm
mhm
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meager terrace
Can anyone help me factorise this?
devout snowBOT
Can anyone help me factorise this?
meager terrace
Pretty sure its meant to look like this but i think i have some + and _ the wrong way round
zenith vortex
Can anyone help me factorise this?
in the end why is it sin(theta)) why double )) ?
meager terrace
i think its just a mistake
zenith vortex
Can anyone help me factorise this?
Let sqrt(sin^2(theta)-2lambda)-sin(theta)=y, then we can write ycos^2(theta)-sqrt(sin^2(theta)-2lambda)(-y)sin(theta)
factorize the y now
meager terrace
alright ill give it a try
Wait how do u factorize it out? isnt the y only in the first term'
zenith vortex
Wait how do u factorize it out? isnt the y only in the first term'
-y=-sqrt(sin^2(theta)-2lambda)+sin(theta)
meager terrace
-y=-sqrt(sin^2(theta)-2lambda)+sin(theta)
I see, like that?
zenith vortex
y=sqrt(sin^2(theta)-2lambda)-sin(theta)
this means -y=sin(theta)-sqrt(sin^2(theta)-2lambda)
itll look like ycos^2(theta)-sqrt(sin^2(theta)-2lambda)(-y)sin(theta)
=y(cos^2(theta)+sqrt(sin^2(theta)-2lambda)sin(theta))
zenith vortex
Pretty sure its meant to look like this but i think i have some + and _ the wron...
yeah probably you wrote the terms incorrectly