1,4,11,26
#help-49
woeful turret
that as well
midnight plankBOT
woeful turret
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btw theres another method for doing this.. by finding an expression in terms of n but that is a little tricky
manic bison
btw theres another method for doing this.. by finding an expression in terms of ...
I thought of that originally but there were 3 terms in n so it was tough
woeful turret
yeah
midnight plankBOT
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manic bison
I can find the sum this way but I need a_23-2a_22 and the calculations will get really large by then
wary scroll
===== Pg9 - Challenge and Thrill Of Pre-college Mathematics =====
"By definition the sum of two complex no. is a complex number and
the product of two complex no. is a complex no."
This wouldn't be the case if one of the complex numbers was the
complex conjugate of the other, right? For example:
(3 + 2i)(3 - 2i) = 3^2 + 2^2 = 13
We know that 13 is real.
So whats going on???
===== Pg10 - Challenge and Thrill Of Pre-college Mathematics =====
Definition 3: |2 + 3i| = 5
Shouldn't it be equal to sqrt(14)??
carmine sigil
!help @wary scroll
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manic bison
yes do the calculation man
I ran it through chatgpt and the values are 8million and 4 million roughly
wary scroll
!help <@1176480264235532288>
thanks i didnt know i had to do that
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spring latch
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spring latch
i know this is false but i dont know how to prove it
lavish venture
you need a counterexample
spring latch
you need a counterexample
yes but how do i prove the counterexample is a counterexample
like
like sin(pi*x)+sin(x) is not periodic
lavish venture
what did you have in mind?
spring latch
but how do i prove thjat
lavish venture
well if it were periodic then the ratio of the periods would have to be rational
can you prove that?
like
spring latch
nope my brain isnt big enough for that
intuitively like obviously
the period is the lcm
spring latch
and if its irrational you cant
lavish venture
if you can prove "if f is p periodic and g is q periodic and p/q is rational then f + g is periodic" then the contrapositive gives you f + g not periodic implies p/q is irrational
lyric charm
when both pix and x are multiples of 2p
right so there must exist integers $m$ and $n$ such that $\pi x = 2\pi n$ and $x = 2\pi m$
grand pondBOT
Ann
spring latch
right so there must exist integers $m$ and $n$ such that $\pi x = 2\pi n$ and $x...
damn thats clever
spring latch
thats basically the same thing but yea
spring latch
i would not have thought of that
alright just 14 more exercises to go until im done with subspaces for now
i hate linear algebra
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midnight plankBOT
lavish venture
lol
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molten bay
Derivative of log|logx|
midnight plankBOT
Derivative of log|logx|
midnight snow
nova yoke
also beware of x=1
molten bay
Yes
Domain would be 0<x<1 and x>1
1/xlogx so here what is the problem x=/1
It seems good
modest star
you get 1/0
nova yoke
you mean x=1
?
the issue is the abs val, to get rid of it you can consider separately the cases 0 < x < 1 and x > 1
in the first case, |log(x)| becomes - log(x)
long dagger
how
modest star
what part
long dagger
every part
modest star
$H_n = \sum_{i=1}^n \f1i$ by definition
grand pondBOT
long dagger
i know what H_n and H_n^(m) are
modest star
so you can say
long dagger
no that's where i'm stuck
the author broke the sum
that's fine, but then why subtract H_n^(2)
modest star
$\sum_{i=1}^n \left(\sum_{j=1}^i \f{1}{ij}+ \sum_{j=i}^n \f{1}{ij}\right) - \sum_{i=1}^n \f{1}{i^2}$
grand pondBOT
modest star
so the reasonn theres a minus 1/i^2 is because we overcount in the middle
do you see the first sum goes from j = 1 to i
then the second part starts from i to n
we count i twice
grand pondBOT
modest star
and the key observation is that
$$
\sum_{i=1}^n \sum_{j=1}^{,i} \frac{1}{i,j}
;=;
\sum_{i=1}^n \sum_{j=i}^{,n} \frac{1}{i,j}
$$
long dagger
cool
modest star
lmao the engineer verification
long dagger
lol i'll think abt it for a sec
wait
ohhhhhhhhhhhhhhhhhhhhhhhh
thanks @modest star
modest star
np
long dagger
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tidal turret
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lyric charm
!status
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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
tall lantern
Also would be great if we could have a translated version
lyric charm
i can do that
Let B = {v1, v2, v3} be a basis for a vector space V and let f: V -> V be a linear transformation such that [see image].
a) Find all k ∈ R for which f is injective (or, as the text says, a monomorphism).
b) For each k ∈ R, find ker(f).
lyric charm
ok
if V was R^3 and {v1, v2, v3} was the canonical basis, would you know what to do?
at least the line of attack
tidal turret
i wanted to ask what is a mono
lyric charm
it means injective
tidal turret
Let B = {v1, v2, v3} be a basis for a vector space V and let f: V -> V be a line...
what does injective mean?
tidal turret
,w injective
grand pondBOT
lyric charm
inyectiva
burnt flame
makes sense
lyric charm
basically the same word 
tidal turret
yeah xd
lyric charm
latin go brrrr
tidal turret
ker(f) is trivial
lyric charm
anyway yeah injective generally means f(x)=f(y) implies x=y i.e. no two inputs get the same output
tidal turret
basically mono <=> ker is trivial
lyric charm
and yes for linear maps it's equivalent to trivial kernel
thats a pretty important thing to keep in mind generally when doing LA
grand pondBOT
tidal turret
k = R-{3,-1}
@burnt flame did you ghost pinged me?
or was something deleted or no?
burnt flame
basically mono <=> ker is trivial
just deleted something i was just stating something obvious like how you can show this
tidal turret
thats a pretty important thing to keep in mind generally when doing LA
basically, if v1 ∈ ker(f) then f(v1)=0 same for the other ones, v2, v3, so we need to find a k such that they are all linearly independent vectors right,
basically the image must be full
and since kernel is trivial
the nullity is zero
lyric charm
basically, if v1 ∈ ker(f) then f(v1)=0
what you said is true but pointless
tidal turret
yeah im just stupid
lyric charm
,w det {{1,2,0},{1,k,3},{0,1,k}} = 0
this is good
tidal turret
the rank nullity approach is correct i think tho
lyric charm
so yes for all k other than -1 or 3 you have that the kernel is trivial and so the map is injective
and now for b) you need to work out the kernel when k=-1 and when k=3
tidal turret
and now for b) you need to work out the kernel when k=-1 and when k=3
is it okay if I map in ny brain that mono = trivial kernerl
lyric charm
eh
yeah why not
"monomorphism" is specialized to abstract algebra anyway
probably no harm
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humble siren
could i get insight on how to start and a few steps, i dont need the answer just the process on how to do this
midnight plankBOT
could i get insight on how to start and a few steps, i dont need the answer just...
lyric charm
ok step 0 is to crop away the transparent region that's 4 times taller than the image itself
there is no "process" really as there are a great many functions that exist
easiest is to draw a graph
outer quartz
Hi i am unsure how to simplify this to get to the next line of working where the arrow is. I have tried but I get different values. any help is appreciated
midnight plankBOT
Hi i am unsure how to simplify this to get to the next line of working where the...
gaunt jetty
Hi i am unsure how to simplify this to get to the next line of working where the...
Recall that 1/j = -j
This would make the process more straightforward for you in this case
gaunt jetty
i'll try that
Also, just a tip, addition/subtraction is easier done in Cartesian form, whilst division/multiplication is easier done in Polar form
You could convert between the two to do those computations if you may
midnight plankBOT
@outer quartz Has your question been resolved?
gaunt jetty
Do you still need help? @outer quartz
outer quartz
Do you still need help? <@1177488097949266043>
yes please i still dont understand how my teacher got 10.88 + j6.4
when i do it by hand and also by calculator i get 0.1603 + 0.457j
unique juniper
it would help if you'd realise that $\frac{1}{Z}=\frac{Z^*}{\abs{Z}^2}$ 🤔
grand pondBOT
parabolicinsanity
outer quartz
yes i did use the complex conjugate
gaunt jetty
when i do it by hand and also by calculator i get 0.1603 + 0.457j
You seem to be right yeah. Seems like a mistake from your professor's end
outer quartz
You seem to be right yeah. Seems like a mistake from your professor's end
oh ok yea i'll send him an email. thanks for the help i appreciate it
gaunt jetty
oh ok yea i'll send him an email. thanks for the help i appreciate it
Btw is this the equivalent impedance of an inductive motor or something
outer quartz
yep
gaunt jetty
Oh nice
outer quartz
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humble prairie
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subtle blaze
Why does it say 2p instead of a
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humble prairie
.reopen
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tranquil zinc
hi
astral canyon
multiply A by itself n times
fallow scarab
how can i calculate a matrice A of the power n
!original
midnight plankBOT
how can i calculate a matrice A of the power n
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
tranquil zinc
multiply A by itself n times
not like that we do that when the n is defined 2 3 4 ect
ivory saddle
not like that we do that when the n is defined 2 3 4 ect
brev what
astral canyon
Please show the original problem, **exactly as it was stated to you**, with the ...
You can diagonalize the matrix... refer to this please
livid lynx
not like that we do that when the n is defined 2 3 4 ect
Are you looking for some properties?
midnight plankBOT
@tranquil zinc Has your question been resolved?
limber plaza
I think I'm being slow here. He says that the difference between the two pictures is that once we've changed the y-axis to be multiplicative (e.g., 10 to 100), "we can say that we are now plotting the logarithm of the total number of cases."
midnight plankBOT
I think I'm being slow here. He says that the difference between the two picture...
limber plaza
He doesn't say that the function has changed, so I don't see how it is plotting the logarithm. E.g., if f(3) = 5, then on both graphs it will be 5. And if f(4) = 50, then on both graphs it will be 50. But, since we have rescaled our y-axis, those data points won't just shoot off into an exponential curve, and we get more visually interesting data.
I don't see how rescaling it changes what we're plotting.
blissful pier
I imagine he's explaining that it helps us get a better idea of how the data is evolving: that is, if the data appears linear of hte log chat then we can say that log(Y)~aX+b meaning Y~e^(aX+b) which lets us say things like "the data appears to grow exponentially"
also I've seen it certain sciences where log scales will be used to better display data that may be exponential by nature
an example of the top of my head eludes me at the moment but I saw it multiple times in my bio and chem classes in secondary school
limber plaza
No, I believe this. That is the purpose of using the logarithmic scale.
I'm confused by the wording "we are plotting the logarithm." I do not know what he means by this.
dusty portal
No, I believe this. That is the purpose of using the logarithmic scale.
I'm con...
I think he means that we’re just plotting log(y) against x.
Or rather 10^y, but you get my point.
coral turret
can someone explain the whole of 8’th question
midnight plankBOT
can someone explain the whole of 8’th question
coral turret
i kinda understand the first one but i got no clue how to do the 2nd part
also for the first one qhy cant i do 100-21 why 1-0.21
fallow scarab
also for the first one qhy cant i do 100-21 why 1-0.21
What answer would you get
fallow scarab
They're not the same thing
fallow scarab
If n=1, what's 79^n * N
coral turret
79N
oops
i see
but then for the second one why did they do 100-x then?
if its x%
should it be 1 - x/100
oh shit
it is that
lol
but why did they equate it to the previous one
arent they unrelated
one is 21% another is x%
coral turret
If n=1, what's 79^n * N
Help!!
fallow scarab
What
coral turret
I dont understand why they equated 1-x/100 to 0.79^n
they are completely unrelated right?
fallow scarab
What are "they"
fallow scarab
fallow scarab
I dont understand why they equated 1-x/100 to 0.79^n
This follows from the information given in the questions
n = 20
coral turret
n = 20
but the previous one disinfects at 21% per minute and this one is x% per minute so their 2 different things right?
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inland patio
Consider some real $2$-dimensional inner product space $\mathsf V$. If $\det(\mathsf T)=1$, let $\beta$ be an orthonormal basis for $\mathsf V$. We then have $\det([\mathsf T]\beta)=1$ and hence the map $\mathsf L_A(x)=Ax$ with $A=[\mathsf T]\beta$ for $x\in \mathsf R^2$ is a rotation. Now the map $\phi_\beta$ that sends a vector $v\in\mathsf V$ to its coordinate vector $x\in\mathsf R^2$ with respect to $\beta$, preserves the inner product (i.e. distances and angles). Does it follow from this that $\mathsf T$ is a rotation whenever $\mathsf L_A$ is? I don't know how to rigorously motivate this.
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Consider some real $2$-dimensional inner product space $\mathsf V$. If $\det(\ma...
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inland patio
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viral dagger
find all functions $f:\mathbb{R}\to\mathbb{R}$ such that
$$\forall x,y\in\mathbb{R}, f(x)+f(y)+1\geq f(x+y)\geq f(x)+f(y)$$
$$\forall x \in [0,1), f(0)\geq f(x)$$
$$-f(-1)=f(1)=1$$
midnight plankBOT
find all functions $f:\mathbb{R}\to\mathbb{R}$ such that
$$\forall x,y\in\mathb...
grand pondBOT
skissue.in.a.teacup
viral dagger
status 1 (well well well)
<@&268886789983436800>
P(x,y)=P(1,0)
2+f(0)>=1>=1+f(0)
1>=-f(0)>=0
-1<=f(0)<=0
this implies theres a discontinuity at x=1?
midnight plankBOT
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carmine sigil
I would imagine we can start by pinning down certain values, or at least placing limits on them
Considering f(0), we have f(0) = f(-1 + 1) so f(-1) + f(1) + 1 ≥ f(0) ≥ f(-1) + f(1) giving 1 ≥ f(0) ≥ 0
Then we can use f(0)'s bounds to place bounds on f(1/2) and vice versa.
f(1) = f(1/2 + 1/2) => f(1/2) + f(1/2) + 1 ≥ f(1) ≥ f(1/2) + f(1/2)
2f(1/2) + 1 ≥ 1 => f(1/2) ≥ 0
1 ≥ 2f(1/2) => f(1/2) ≤ 1/2
Similarly, we can find any dyadic rational q/2^n in (0, 1) will have the bounds q/2^n ≥ f(q/2^n) ≥ 0
Now consider f(1/2) = f(1/2 + 0)
f(1/2) + f(0) + 1 ≥ f(1/2) ≥ f(1/2) + f(0)
Considering only f(1/2) ≥ f(1/2) + f(0) we have f(0) ≤ 0
But we also have f(0) ≥ 0, so f(0) = 0, and because of f(0) ≥ f(x) for all x in [0, 1) we have f(q/2^n) = 0 for all q/2^n in [0, 1)
See if you can take it from here pinning down more values!
viral dagger
oh sorry i didnt see this
but this is terrible timing, its kinda late rn :((
.close ill reopen a new one tomorrow, sorry!
sacred gate
i need heeelp
midnight plankBOT
i need heeelp
sacred gate
i dont understand how do i find a
cause when i get a
without derivating
i get 0 in the deominator but a is 0 cause of multiplying
when i do derivativr the denominator is 1
or does that mean that a is 0
im stupid
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rancid vigil
[I_n = \int_1^e \ln(x)^n] is there a way to find $I_{n+1}$ as $I_n$ plus some other constant
midnight plankBOT
\[I_n = \int_1^e \ln(x)^n\] is there a way to find $I_{n+1}$ as $I_n$ plus some ...
grand pondBOT
<rajel />
rancid vigil
if i try to integrate I_n+1 with IBP i dont get anywhere
,, I_{n+1} = \int_1^e \ln(x)^{n+1} \dd{x}
grand pondBOT
<rajel />
sharp coral
you should be able to IBP a recursion formula relating I_n+1 and I_n
rancid vigil
i couldn't 
sharp coral
what did you select for your u and dv?
frozen talon
maybe define (I_n(x)=\int_1^x{(\ln t)}^n dt)?
rancid vigil
,, I_{n+1} = [\ln(x) (n+1) \cdot \frac{1}{x} \cdot \ln(x)^n]_1^e - \int_1^e (x\ln(x) - x) \cdot (n+1) \frac{1}{x} \ln(x)^n \dd{x}
rancid vigil
ther first term is 0
nvm
anyways i dont care about it , since its a constant
,, I_{n+1} = c - n\int_1^e (\ln(x) - 1) \cdot \ln(x)^{n-1} \dd{x}
sharp coral
**<rajel />**
i'm not entirely sure how you got this
rancid vigil
by making ln^n+1 =ln*ln^n
glass lantern
hmm
rancid vigil
oh i made a mistake
grand pondBOT
<rajel />
rancid vigil
c is that term
sharp coral
i would instead use the fact that f(x) = f(x) * 1
rancid vigil
huh
glass lantern
can the answer have
do i have to learn to use this bot now 🌚
[I_n = \int_1^e \ln(x)^n] is there a way to find $I_{n+1}$ as constant * $I_n$ plus some other constant
grand pondBOT
ChinChin
glass lantern
copy pasted from yours :>
rancid vigil
i would instead use the fact that f(x) = f(x) * 1
this would make it way more easier
thx
rancid vigil
copy pasted from yours :>
and whats the question
glass lantern
read that
like
does it have to be
integral n+1 = integral n + c
or can it be
integral n+1 = k * integral n + c
rancid vigil
it should just be in this form $I_{n+1} = \frac{2n}{e^2} I_n +...$
grand pondBOT
<rajel />
rancid vigil
just an example
glass lantern
got it
wait
IBP on integral (lnx)^n+1 dx
=[xln(x)^(n+1)][evaluated from 1 to e]
−∫x⋅(n+1)ln(x)^n * 1/x dx
1st part becomes e,
and the 2nd part is (n+1)*In
it should just be in this form $I_{n+1} = {n+1} I_n + e$
grand pondBOT
ChinChin
rancid vigil
lol
grand pondBOT
ChinChin
rancid vigil
yeah yeah , the thing is i was taking the wrong u ,dv
cloud already told me above
but yeah thx
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dusty portal
Only did the n=2 part so far
midnight plankBOT
Only did the n=2 part so far
dusty portal
Don’t know how to do inductive step
lavish venture
try dividing the right by |z_1|
dusty portal
I think
You’re cooking
$n=k+1\implies (k+1)\frac{|z_2|^k}{|z_1|^{k-1}}<\frac{1}{|z_1|-|z_2|}$
Ok now what
I put n instead of k
🤦♂️
grand pondBOT
;(
lavish venture
you’re trying to prove $n \biggr| \frac{z_2}{z_1}\biggr|^{n-1} < \frac{1}{1 - \biggr|\frac{z_2}{z_1}\biggr|}$
grand pondBOT
knief
jolly pike
[|z2|/|z1|]^n is decreasing function
dusty portal
What the freak
lavish venture
Infinite series
ding ding ding
dusty portal
I didn’t think we could invoke that lmao
lavish venture
why not?
dusty portal
$n<\infty$
lavish venture
the modulus makes it real
grand pondBOT
;(
lavish venture
and |z_2| < |z_1| means their ratio is < 1
lavish venture
see if you can prove n |z_2/z_1|^{n-1} < infinite sum
jolly pike
take derivative of the geometric sequence
dusty portal
No calculus ❌
lavish venture
no need
dusty portal
see if you can prove n |z_2/z_1|^{n-1} < infinite sum
Yeah I think I can just use the nz^{n-1} infinite series closed form I derived in the last exercise
And compare to the n-th term
lavish venture
what book is this?
dusty portal
what book is this?
Howie’s Complex Analysis
This copy had a freaking IP address as its URL
lavish venture
never heard of it
lavish venture
yea no
dusty portal
Yeah I think I can just use the nz^{n-1} infinite series closed form I derived i...
Valid?
lavish venture
i’m not sure if you mean the closed form for the series nz^{n-1}?
isn’t it simpler to just say something like
jolly pike
x^n become nx^n-1 x is ratio of z2 and z1 we know the sum of all nx^n-1 is 1/(1-x)^2 which gives z1^2/(z1-z2)^2 which is less than z1/z1-z2 i didn’t write absolute sign sorry
lavish venture
|z_2/z_1|^(n-1) < |z_2/z_1|^(n-2) < … < |z_2/z_1|) + 1 and we have n terms here so n |z_2/z_1|^(n-1) < partial sum < infinite series?
dusty portal
|z_2/z_1|^(n-1) < |z_2/z_1|^(n-2) < … < |z_2/z_1|) + 1 and we have n terms here ...
oh
Interesting
jolly pike
so sum s> any term in the sequence z1/z1-z2> n[z1/z2]^n-1
lavish venture
1 + |z_2/z_1| + … + |z_2/z_1|^(n-1) < |z_2/z_1|^(n-1) + … n times
lavish venture
i’m not sure if you mean the closed form for the series nz^{n-1}?
because we aren’t summing n |z_2/z_1|^(n-1)
dusty portal
|z_2/z_1|^(n-1) < |z_2/z_1|^(n-2) < … < |z_2/z_1|) + 1 and we have n terms here ...
Why is there a plus 1
lavish venture
Why is there a plus 1
to get n terms
night hawk
-# serry
dusty portal
to get n terms
lavish venture
it’s just making the partial sum
dusty portal
Ok
jolly pike
try my way
dusty portal
Mkay
lavish venture
why are you doing complex analysis before real analysis btw
dusty portal
$$\frac{|z_1|}{|z_1|-|z_2|}=\frac1{1-\frac{|z_2|}{|z_1|}}=\sum_{n=0}^{\infty}\bigg\vert\frac{z_2}{z_1}\bigg\vert^n>1+\frac{|z_2|}{|z_1|}+\bigg\vert\frac{z_2}{z_1}\bigg\vert^2\ldots\bigg\vert\frac{z_2}{z_1}\bigg\vert^{n-1}>n\bigg\vert\frac{z_2}{z_1}\bigg\vert^{n-1},$$
grand pondBOT
;(
dusty portal
why are you doing complex analysis before real analysis btw
Real analysis sucks
lavish venture
💀
lavish venture
you’re welcome
dusty portal
try my way
Yeah, makes sense (factorization stuff and decreasing things)
dusty portal
2.3 to 2.11 were all ez
dusty portal
did this one take you hours?
A full hour yeah
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fading ore
My abstract algebra homework is having me make a group of symmetries on a circle
midnight plankBOT
My abstract algebra homework is having me make a group of symmetries on a circle
fading ore
I decided to just use the horizontal reflection matrix and the general rotation matrix in R2
But idk how I properly denote this group
main current
That's a way to denote it!
fading ore
That's a way to denote it!
Well i moreso mean
Idk what to write down mathematically to demonstrate that this is what I mean
main current
What you just said is perfect
fading ore
Hm
main current
I don't know if your class might use a symbol for the reflection and rotation matricies
fading ore
I thought i would need to say that they are generators
fading ore
I don't know if your class might use a symbol for the reflection and rotation ma...
No this was my own doing
I could've used any group thats isomorphic probably
main current
Oh yeah, sure. I did kind of stick "these are the generators" in there. you're right you should say that
fading ore
At first I just said the complex numbers with magnitude 1 but thats only symmetric to the rotation group
main current
Or, sorry. I did just think of a better way to notate this group
This is similar to a group you may have put on a triangle, a square, etc
This is a dihedral group
fading ore
This is a dihedral group
Yeah the homework was for the dihedral group of a circle
But I of course can't use the standard notation since the rotations are by an arbitrary angle
midnight plankBOT
@fading ore Has your question been resolved?
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twilit field
really silly but I feel I'm missing smthg here
ST(v)=av
TS(ST(v))=aTS(v)
I want to get the RHS to be ST(v), I think?
midnight plankBOT
really silly but I feel I'm missing smthg here
ST(v)=av
TS(ST(v))=aTS(v)
I want...
sharp coral
note that they have the same eigenvalues, not necessarily the same eigenvectors
twilit field
note that they have the same eigenvalues, not necessarily the same eigenvectors
well, I need to start from sat ST(v) and somehow from that get to TS(u), right
sharp coral
yeah
twilit field
ST(v)=av
TS(ST(v))=aTS(v)
I want to get the RHS to be ST(v), I think?
so something like this
lyric charm
maybe try looking at $TST$ instead of $TS^2T$ as you are rn
grand pondBOT
Ann
midnight plankBOT
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twilit field
Let $dim(V)=n$
Let $v_1,v_2,\dots v_m$ be all the $m$ distinct eigenvectors $0≤m≤n$ as eigenvectors are Linearly independent.
\
We then have
Let $T(v_i)=u_i$
\
Let $\sum_{i=1}^{n} a_iv_i≠ 0$
so $\sum_{i=1}^{m}{a_iv_i} = \sum_{i=1}^{m} a_i u_i ≠0$
\
Only dim(Range(T)) vectors in the image are LI
midnight plankBOT
Let $dim(V)=n$
Let $v_1,v_2,\dots v_m$ be all the $m$ distinct eigenvectors $0...
twilit field
and here I;m stuck
grand pondBOT
wai
flat spire
wai
twilit field
wai
hi
I feel like I'm half-way there but missing something simple
Like my idea is as eigenvectors are LI we can start with that and extend that to form a basis
twilit field
Let $dim(V)=n$
Let $v_1,v_2,\dots v_m$ be all the $m$ distinct eigenvectors $0...
like this would mean dim ( range (T))+1 eigenvectors, no?
.Close
.close
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viral dagger
the amount of mappings $f:{1,2,3,4,5}\to{1,2,3,4,5}$ such that $\forall x\in{1,2,3,4,5}\quad f(f(x))\in {1,2}$
midnight plankBOT
the amount of mappings $f:\{1,2,3,4,5\}\to\{1,2,3,4,5\}$ such that $\forall x\in...
grand pondBOT
skissue.in.a.teacup
visual tiger
the amount of mappings $f:\{1,2,3,4,5\}\to\{1,2,3,4,5\}$ such that $\forall x\in...
!status
midnight plankBOT
the amount of mappings $f:\{1,2,3,4,5\}\to\{1,2,3,4,5\}$ such that $\forall x\in...
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
is there a better way than casework
visual tiger
I mean you can get rid of some casework
but I'm not sure if you can get rid of all
do you know what f(1) and/or f(2) should be?
viral dagger
casework as in there is only 1 domain that points to {1,2}, 2 domain that points to {1,2} ect
viral dagger
do you know what f(1) and/or f(2) should be?
f(1) and f(2) should be in the {1,2}
visual tiger
f(1) and f(2) should be in the {1,2}
not necessarily both
viral dagger
ehh
visual tiger
as I said it's hard to get rid of casework
viral dagger
casework as in there is only 1 domain that points to {1,2}, 2 domain that points...
is this the best way to split the cases or is there a better way
visual tiger
what does "1 domain" and "2 domain" mean
viral dagger
what does "1 domain" and "2 domain" mean
like there are 2 elements in the domain that point to {1,2}
visual tiger
like there are 2 elements in the domain that point to {1,2}
this is a bit more far fetched than what I had in mind
and I think there would still be some casework INSIDE each of those
rendering this kind of split kinda useless
viral dagger
whats a better split then
visual tiger
I would have started with "both 1 and 2 are sent to {1,2}" or only one of them
show that none being sent to 1,2 is impossible
and if only 1, for example, is sent to {1,2}, what to say about f(1)
midnight plankBOT
@viral dagger Has your question been resolved?
tawdry laurel
reminds me of aime question i could swear
hard shard
seems like it could be considering the size of the solution instinctually feels around three digits
viral dagger
show that none being sent to 1,2 is impossible
if neither f(1) nor f(2) is sent to {1,2}, then the thing that sends to {1,2} wont go to {1,2}
if only one of them say f(1) goes to {1,2} like you wrote, then it has to go to {1} as if it were to go to {2} then f(f(1)) wouldnt go to {1,2} as f(2) doesent go to {1,2}
vice versa with f(2)
so either f(1)=1, or f(2)=2, or f(1) and f(2) both go to {1,2}
for the f(1)=1, if the range 1 only has 1 coresponding domain then there is 1 possibility, if there is 2 then there is 4c1×2^3, if there is 3 then there is 4c2×3^2, if there is 4 then there is 4c3×4^1, if there is 5 then there is 1, the sum is 103 and since its the same for both f(1)=1 and f(2)=1 it will be 103×2
so now, whats next is f(1)=1 and f(2)=2, and f(1)=2 and f(2)=1, they are also the same so you can think about one of them
if there is 2 thrn there is 3c0×2^3
if there is 3 then there is 3c1×3^2
if there is 4 then there is 3c2×4^1
if there is 5 then there is 1
48×2=96
so total is 302?
visual tiger
for the f(1)=1, if the range 1 only has 1 coresponding domain then there is 1 po...
don't know how you got this
viral dagger
huh
visual tiger
suppose you have f(1) = 1 and f(2) not 1 or 2
then pick f(2), which is either 3,4 or 5
and then we must have f(f(2)) = 1 no matter what (why?)
viral dagger
ah
visual tiger
suppose wlog now that f(2) = 3, f(3) = 1
we're only left to choose f(4) and f(5)
f(4) = 1,3 or 5
if f(4) = 5 then f(5) = 1
otherwise f(5) = 1 or 3
so in total, 2 * 3 * (1 + 4) (this is false, see remark below)
if exactly one of 1,2 is sent to 1,2
oh wait
if f(4) = 1 then we also have f(5) = 4 that becomes a possibility
so in total, 2 * 3 * (1 + 4+1)
and now we're good to go I believe
in the case where both 1,2 are sent to {1,2}
now splitting into how many elements outside are directly sent to {1,2} makes sense
viral dagger
oo
visual tiger
choosing f(1) and f(2) gives you then 2^2 choices
if only one outside is sent to {1,2}, choose that element and its image (3C1 * 2)
because then the other two are sent exactly to that element
if two outside are sent to {1,2}, choose them and their image (3C2 * 2^2)
and then two choices for the image of the remaining element
if all are sent to {1,2}, choose all their images (2^3)
so
2 * 3 * 6 + 2^2(3C1 * 2 + 3C2 * 2^2 * 2 + 2^3)
188
sorry I went a bit too much into details, I can erase some of those if you want to work it out
viral dagger
whats an image btw? is it like what the element corresponds to from the function?
visual tiger
f(x) is the image of x by f
viral dagger
ok i understand
midnight plankBOT
@viral dagger Has your question been resolved?
midnight plankBOT
viral dagger
bruh sorry it failed to send lmao
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lofty chasm
help
wheat patrol
hint they have said r is dependent on time t, so you use chainrule
inland sparrow
Yeah use chain rule, you have clues as to 3 different differentials.
inland sparrow
isnt it 2pi x r + pi x l
No, l is dependent on r. It's not a constant.
inland sparrow
Use this: dS/dt = dS/dr * dr/dt
wheat patrol
but before you do the second question, have you done the first part?
inland sparrow
well you found an expression for l in terms of r, right?
wheat patrol
presenting l in terms of r?
inland sparrow
but before you do the second question, have you done the first part?
that's a fair point.
inland sparrow
then you need to substitute that into the expression for the surface area
inland sparrow
then expand, collect like terms and diff.
wheat patrol
this is why mathematicians love the blackboard, helping or doing math over chat is tough
inland sparrow
l is root 25 + r2
is the whole thing square rooted or just 25?
inland sparrow
this is why mathematicians love the blackboard, helping or doing math over chat ...
very true.
lofty chasm
ohhh
lofty chasm
is the whole thing square rooted or just 25?
whole thing
ok so then i use chain rule ??
wheat patrol
yes but only when its applicable ie there is r
inland sparrow
whole thing
then you might have an issue, cause if it is you wont be able to do much
or no its fine
inland sparrow
ok so then i use chain rule ??
yes
when you sub in l you should get S = pi(r^2) + pi(r)(25+r^2)^1/2
then you differentiate that expression to get dS/dr
lofty chasm
yuh
ok wait
i got htis
i forgot the pi in the second and third addition
i got tit
thanksssss
inland sparrow
hold on.
inland sparrow
Try to make your writing more legible, after the second plus on the last line it looks like you put an equals sign
lofty chasm
Try to make your writing more legible, after the second plus on the last line it...
okaies
inland sparrow
An examiner might get very confused and mark you down for it, so just watch out for that
lofty chasm
ofc
lofty chasm
and for dv/dt i sub the V
lofty chasm
to get dv/dh i use chain rule yea
i got this 1/
inland sparrow
and for dv/dt i sub the V
yes
lofty chasm
idk where i went in simplyifing it
inland sparrow
i got this 1/
first expand the V expression given, then use the expanded version to subsitute into dV/dt
lofty chasm
is this sufficient
no wait
i messed up
the last part
llollll
its supposed to be 6--3h
60-3h
inland sparrow
yeah I would say it is sufficient working.
inland sparrow
But personally after reviewing loads of mark schemes, to be on the safe side state what your derivatives are rather than subsituting in straight away.
There are usually marks for that stuff on mark schemes.
inland sparrow
Well I am sure you got full marks, reminder if you don't need help anymore, please close the channel to allow other people to use it.
lofty chasm
how to get the partial fraction
lofty chasm
help
lyric charm
are you in an exam right now?
gaunt nimbus
wait
D lies on ${l}$ and therefore is a scalar multiple of ${\vec{AB}}$. Thus,
[ \vec{AD} = k\vec{AB}. ]
Using the cross product definition for area, we have that [A_{CAB} = | \vec{AB} \times \vec{AC} |.]
Note also that for scalar ${c}$,
[ (c\vec{u}) \times \vec{v} = c(\vec{u} \times \vec{v}). ]
grand pondBOT
k
gaunt nimbus
@lofty chasm
grand pondBOT
k
midnight plankBOT
@lofty chasm Has your question been resolved?
narrow dragon
Can someone help with this question
midnight plankBOT
Can someone help with this question
quaint shore
Can someone help with this question
What is the question? This is only the game rules
midnight plankBOT
@narrow dragon Has your question been resolved?
cursive swan
lol
lofty chasm
help
green arrow
Hopefully you're not currently in an exam
desert siren
you can just use the formula for integration of 1/a^2-y^2
lofty chasm
would copy from chatgpt n sleep
desert siren
you can just use the formula for integration of 1/a^2-y^2
a here is 2
lofty chasm
you can just use the formula for integration of 1/a^2-y^2
whts this frmula
desert siren
whts this frmula
integration of 1/a^2-x^2 = 1/2a * ln mod (a+x/a-x)
green arrow
It's a "hence" question so they don't get any points if they don't use the formula from the first part
desert siren
oh sorry didn't see the first part
green arrow
He has to do it manually
desert siren
well then firstly use a^2-y^2 = (a+y)(a-y)
then you can split that into A/a+y + B/a-y, where A and B are constants
so we get 1/a^2-y^2 = A/a+y + B/a-y
Now multiply both sides by (a+y)(a-y)
Then you can just find A and B
lofty chasm
alrr
desert siren
Lmk when you're done till here
Then we'll do the integration part
(btw in the explanation above a would be 2 since that's what's given in the first part)
green arrow
He got the partial fractions right did he not
lofty chasm
yeah
desert siren
ok so you're till there already, the mistake is you took 2 across but wrote it as ln2
desert siren
?? how did you get ln (-4)
desert siren
notice there's a mod on the ln argument
desert siren
just take all the other terms except ln sec x to one side
lofty chasm
desert siren
take ln 2 to the other side so it becomes ln (4-y^2/2)
lofty chasm
yes
desert siren
got g(x)?
lofty chasm
take ln 2 to the other side so it becomes ln (4-y^2/2)
i rmove the ln and square it ?
desert siren
what's ms?
lofty chasm
apparently the answer here is the answer
lofty chasm
what's ms?
marking scheme
desert siren
apparently the answer here is the answer
the answer you wrote? how? you wrote 2 as ln2 in that
desert siren
why has your teacher written y=8+4y/2-y?? Aren't you supposed to find g(y)?
desert siren
i rmove the ln and square it ?
did you do this tho
lofty chasm
yes
desert siren
idk what your teacher's written tell them to cross check
lofty chasm
i need help wit his too
why is it integrate theta
desert siren
why is it integrate theta
the parameter can be anything it doesn't matter, change it all to x if you feel more comfortable with it
integrating theta basically means finding the area of the given function from one value of theta to another (here, from 0 to pi/2)
desert siren
theres a theta before the function
At first glance I'd say it's just integration by parts
desert siren
so like 1/theta ^ 2
? integration of theta is (theta)^2/2
lofty chasm
desert siren
yes, good progress
now the first integration can just easily be done
for the second integration you need to use by parts
lofty chasm
how to integrate that 1/2 theta
desert siren
just think of theta as x, and now tell me what the answer would be
lofty chasm
desert siren
see, the integration sign is blind to your parameter as long as its the one being used (eg, dx for x, and similarly dtheta for theta)
lofty chasm
yasss
desert siren
perfect
desert siren
wdym
desert siren
oh damn lmao
not much will change tho except the by parts
you know the integration of cos2x and sin2x right?
desert siren
cool, still got it then
desert siren
sin is -cos
wb sin2x
desert siren
coolio
lofty chasm
😰
desert siren
nice
desert siren
np 
midnight plankBOT
@lofty chasm Has your question been resolved?
rancid vigil
whats a good algorithm to find the particular solutions for the type of equations:
[
\alpha x + \gamma y = \theta
]
midnight plankBOT
whats a good algorithm to find the particular solutions for the type of equation...
grand pondBOT
<rajel />
rancid vigil
i.e 5x +3y = 12
rancid vigil
,, x,y \in \mathbb{Z}^2
grand pondBOT
<rajel />
desert siren
I don't think there's a specific algorithm since this is just one equation two variables
brute forcing
rancid vigil
I don't think there's a specific algorithm since this is just one equation two v...
yep seen many and im a bit confused
desert siren
unless there's advanced techniques i don't know about
rigid cypress
extended euclidean algoeithm no?
desert siren
you can take over lmao idk that
rigid cypress
yeah I'd use the algorithm i mwnfioned. it finds integers x and y such that alpha x + gamma y = gcd of alpha and gamma
if that gcd divides theta just scale the solution by theta/gcd and you jave an integer solution
midnight plankBOT
queen herald
whats a good algorithm to find the particular solutions for the type of equation...
It's enough to find just one solution (x0, y0)
After that all solutions are just (x0, y0) + (-γ/d, α/d)k, where d = gcd(γ, α)
rancid vigil
It's enough to find just one solution (x0, y0)
After that all solutions are just...
thats what i meant by particular solutions
queen herald
Ah ok
Euclidean algorithm allows to compute u, v such that uα + vγ = d
From here it's possible to contruct an algorithm for finding x0, y0
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rancid vigil
,, a|b , c|b \implies ac|b
grand pondBOT
<rajel />
rancid vigil
is this always true
midnight plankBOT
is this always true
rancid vigil
i tried for a bunch of numbers and it seems to be working
last slate
**<rajel />**
only when a and c are coprime
last slate
i tried for a bunch of numbers and it seems to be working
try 6,4 and 12
a = 6
c = 4
and b = 12
rancid vigil
i see
last slate
wait
coprime is a strong word
it doesn't need to be coprime tho
2^2 | 2^64 and 2^4 | 2^64 and 2^6 | 2^64
but the sum of powers of primes in the prime factorization of a and c <= power of the corresponding prime in the prime factorization of b
@rancid vigil
queen herald
lcm(a, c) | b
rancid vigil
lcm(a, c) | b
if this than ac|b ?
queen herald
No, I meant that if a | b and c | b then lcm(a, c) | b
rancid vigil
ah
last slate
No, I meant that if a | b and c | b then lcm(a, c) | b
yeah
midnight plankBOT
sharp coral
<@&268886789983436800>
lyric charm
<@&268886789983436800> scam
pearl idol
.close
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inland patio
I have a basic question. Consider $(A-2I)x=b$ where $$A=\begin{pmatrix}
2&-1&0\
0&2&-2\
0&0&2
\end{pmatrix}\text{ and }b=e_1.$$What is the solution to the above system? We get the equations $x_2=-1$ and $-2x_3=0$, but what about $x_1$? Do we assign it a parameter or $0$? Why?
grand pondBOT
psie
lyric charm
it's a free variable
inland patio
it's a free variable
right, so is the solution x = t(1,-1/t,0) for t \neq 0 and x = (0,0,0) for t = 0?
midnight plankBOT
inland patio
.reopen
midnight plankBOT
✅
inland patio
x = t(1,0,0) + (0,-1,0)
Isn't it kind of strange that the basis of the solution space is {(1,0,0),(0,-1,0)}? Doesn't this say that (1,0,0) should satisfy the equation, though v = (1,0,0) is an eigenvector to A, so (A-2I)v = 0, which isn't equal to v. Where am I reasoning incorrectly?
lyric charm
Isn't it kind of strange that the basis of the solution space is {(1,0,0),(0,-1...
no it's not
grim vector
The point isn't a vector of the basis
lyric charm
the solution space of (A-2I)x = e1 is not even a subspace of R^3 at all
nova yoke
Isn't it kind of strange that the basis of the solution space is {(1,0,0),(0,-1...
if v is an eigenvector with eigenvalue 2, then Av = 2v, which is the same as (A - 2I)v = 0, not (A - 2I)v = v
inland patio
if v is an eigenvector with eigenvalue 2, then Av = 2v, which is the same as (A ...
right, but the equation is (A-2I)x = e1, and e1 was found to be one vector in the basis of the solution space, though I suspect that {(1,0,0),(0,-1,0)} is not a basis for the solution space?
nova yoke
but e1 is not a solution to (A-2I)x = e1
inland patio
Right, so somewhere I'm not thinking correctly 😅
nova yoke
what was your reasoning to say that it was a solution?
btw, as Ann said, the solution set to Ax = b is not even a subspace when b is nonzero, so it doesn't make sense to talk about a basis
inland patio
what was your reasoning to say that it was a solution?
well, we found the solution to be x = t(1,0,0) + (0,-1,0). Isn't this a line in R^3?
nova yoke
sure, but it doesn't pass through the origin
so it's an affine subspace, not a linear subspace
in particular, it's formed by taking the null space of A and translating it by (0,-1,0)
(1,0,0) is in the null space of A
inland patio
in particular, it's formed by taking the null space of A and translating it by (...
Ah, ok. The reason why I asked the question from the beginning was because I'm trying to find a Jordan canonical form of A, the matrix above. It has eigenvalue 2 with multiplicity 3, and e1 is basis for the eigenspace. So in a "cycle/chain of generalized eigenvectors", this is our first vector, since (A-2I)e1 = 0. Now, the second vector v, shouldn't it be the solution to (A-2I)v = e1?
nova yoke
Ah, ok. The reason why I asked the question from the beginning was because I'm t...
yea that looks right, but note that the second vector v is only a generalized eigenvector, not an actual eigenvector, maybe that was causing confusion?
inland patio
yea that looks right, but note that the second vector v is only a generalized ei...
yeah, so since v = t(1,0,0) + (0,-1,0), do we pick v to be a vector on this line?
nova yoke
yeah, so since v = t(1,0,0) + (0,-1,0), do we pick v to be a vector on this line...
yep, any vector on this line
midnight plankBOT
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Available help channel!
viral dagger
fins the polynomial $p(x)$ such that for every $x$, $$p(x^2)=x^{2019}(x+1)p(x)$$, and $p(\frac{1}{2})=-1$
midnight plankBOT
fins the polynomial $p(x)$ such that for every $x$, $$p(x^2)=x^{2019}(x+1)p(x)$$...
grand pondBOT
skissue.in.a.teacup
viral dagger
not to sure what to do here
p(x) degree is 2020 i think
i thought of p(x)=x^2020-x^2019 but thst doesent satisfy the second condition
fossil knot
i thought of p(x)=x^2020-x^2019 but thst doesent satisfy the second condition
I think multiplying this by a constant still keeps the first equality
midnight plankBOT
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nova pike
Hiii. I just want to check whether or not what I am saying is mathematically right, or not, just to check my understanding. The main focus I’m looking into are points on f(x). A turning point is classified as a point where the derivative of that point has an isolated zero and changes signs (or direction) at that point. A stationary point is just where the derivative at that point is 0 and where the function stops increasing/decreasing. A point of inflection is where the curvature changes sign and the second derivative is either 0 or undefined. However a point of undulation is where the second derivative vanishes but has no change in sign. Now, all turning points are stationary points but not all stationary points are turning points (like the ones where the derivative becomes + then 0 then + as seen in f(x) = x^3 at the point (0,0 in which case this is a stationary point of inflection). If the derivative at the point is 0 and concavity changes, then it’s a stationary point of inflection. Now onto a problem: What are the possible number of turning points of a polynomial, p(x) of degree n. So what I did for this problem was break it into n odd and n even and examine the polynomial at ±∞ by examining the term of the highest degree, which we will call ax^n, which we can do because for large |x|, p(x) is very close in some sense to ax^n (is there any rigorous proof of this?). So for when n is even we must have the sign at x→∞ the same as the sign as x→−∞ because this is true for ax^n (again looking for a rigorous proof). Now because turning points are points where direction changes, we can clearly see that there must be an odd number of turning points. Similarly for odd n, the sign at x→∞ is opposite the sign as x→−∞ so there clearly has to be an even number of turning points.
midnight plankBOT
Hiii. I just want to check whether or not what I am saying is mathematically rig...
nova pike
Hiii. I just want to check whether or not what I am saying is mathematically rig...
sorry for the long message, I’ll break it down if I have to one by one
lyric charm
big wall of text that is for sure
coarse zinc
nova pike
big wall of text that is for sure
I’m so sorry, I tried to fit everything in one.
desert siren
Hiii. I just want to check whether or not what I am saying is mathematically rig...
turning point needs the derivative to have an isolated zero??
nova pike
turning point needs the derivative to have an isolated zero??
Yes, I’m pretty sure that’s right. Is it not? Sorry, how would you say it?
lavish venture
Hiii. I just want to check whether or not what I am saying is mathematically rig...
just n - 1 because that’s the degree of p’?
lavish venture
Yes, I’m pretty sure that’s right. Is it not? Sorry, how would you say it?
y = |x|
desert siren
hm, in a function with no cases that might be true, but it is not universal. Here's an example:
if f(x) is defined as x^2 for x<=0 and as -x^2 for x>0, x=0 can still be a turning pt even if the zero is not isolated (f'(x) can be zero at many points depending on how we take smoothing into consideration like gsc and stuff
lavish venture
doesn’t have to be a zero
desert siren
but that's not applicable here
anyway, im getting off topic let me read the original question
lavish venture
the question is "what are the possible number of turning points of a polynomial p(x) of degree n"
if p has degree n then p’ has degree n - 1 and so it has at most n - 1 turning points
desert siren
yes i agree it's n-1 then
desert siren
y = |x|
also im gonna have to amend my statement because of this lol, it is not certainly true for a function with a single case either
lavish venture
single case?