alright, what about E?
#help-17
fossil pagoda
is it two lines?
fossil pagoda
x and y axis
fossil pagoda
it does go through 0,0
because they have a 0,t) or vice versa
vector addition is true because t is any real number and any real number added to a real number is a real number
peak matrix
vector addition is true because t is any real number and any real number added t...
you sure?
if you take any 2 vectors in E and add them, will you also get a vector in E?
it will be in E only if 1 of its components is 0, will it always be?
fossil pagoda
[0,t] + [0,Real number] = [0,Real number]?
peak matrix
[0,t] + [0,Real number] = [0,Real number]?
yes
fossil pagoda
yes because when doing vector addition
peak matrix
but what if we chose another 2 vectors
fossil pagoda
you have to add by a vector also in E
fossil pagoda
and it would not be in E because you end up with [real number, real number]
peak matrix
yes, where we have no guarantee that one of these is 0
so condition i) fails
what about ii), just for fun
fossil pagoda
yes it does work
peak matrix
yes, condition ii works
fossil pagoda
because any scalar multiplied by a vector in E
gives you 0,T or vice versa
because that scalar * 0 =0
peak matrix
exactly
it either gives [0, t] or [t, 0]
for some t
alright, now F
can you draw this?
fossil pagoda
its a vector passing through 3,1?
fossil pagoda
yes
peak matrix
yes, it's a line and it does pass through 3, 1
but there are many lines with that property
fossil pagoda
T multiplies that point
so its a line going through 3,1 depending on T as if it were a scalar
fossil pagoda
0,0
peak matrix
yes
so now we have 2 points and that uniquely determines a line
so what about condition i)?
fossil pagoda
yes
fossil pagoda
because any real numbers in E added to 3,1 give us another point on F
peak matrix
T multiplies that point
It's mainly because all vectors in E are scalar multiples of [3, 1], as you said. And if we add 2 scalar multiples of [3,1], we get another scalar multiple of [3,1]
a[3, 1] + b[3, 1] = (a+b)[3, 1], which is a scalar multiple of [3, 1]
and so is in E
fossil pagoda
right
peak matrix
what about condition ii)?
fossil pagoda
yes because its just another scalar
peak matrix
yes
if we just add another scalar in front of it, it's still gonna be a scalar multiple of [3,1]
t*a[3, 1] is a scalar multiple of [3, 1]
so is it a subspace?
fossil pagoda
yes
fossil pagoda
its a line going through that point and 0
peak matrix
yep
so it's a really similar situation to what we had in F
except instead of [3, 1], there is [1, 1]
Can we skip directly to H?
fossil pagoda
yes
peak matrix
alright
this one looks a bit tricky
it actually is tricky, very tricky
this doesnt have a definite answer, as it depends on u and v
fossil pagoda
kind of a trick question?
fossil pagoda
everywhere?
peak matrix
yes, it's whole R^2
fossil pagoda
gotcha
peak matrix
but sometimes, we may get something like this:
span{ [2, 0], [1, 0] }
what would this be
fossil pagoda
the same
peak matrix
now it's not everwhere
peak matrix
yep
fossil pagoda
thought it was 2,0 0,1
peak matrix
so it's either everywhere, or it can be a line
there is one last situation
can you figure it out?
it's a very special case
peak matrix
nvm idk lol
it has sth to do with [0, 0]
fossil pagoda
its not within the span?
peak matrix
the last special case is
span{ [0, 0], [0, 0] }
fossil pagoda
ic
fossil pagoda
right
peak matrix
and all of those are subspaces
fossil pagoda
right
peak matrix
in general, all spans are subspaces
because spans were designed to be closed under vector additon and scalar multiplication
So H is a subspace of R^2 as well
fossil pagoda
gotcha
fossil pagoda
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heavy yoke
which part are you having trouble with?
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brazen sluice
how could i prove that $f(2n+1) < 4n+2$
vocal sleetBOT
how could i prove that $f(2n+1) < 4n+2$
twin meteorBOT
General_Jacob
brazen sluice
f(x) returns the sum of all factors of x
for example f(12) = 1 + 2 + 3 + 4 + 6 + 12
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dense eagle
how could i prove that $f(2n+1) < 4n+2$
is this the original question?
dense eagle
**General\_Jacob**
i'm pretty sure this is an open question
bcus if we could prove f(2n+1) < 4n+2
that means there are no odd perfect numbers
(odd perfect number => f(2n+1) = 4n+2)
dense eagle
that means there are no odd perfect numbers
but no odd perfect numbers is currently unproven
dark kiln
nice
dense eagle
how could i prove that $f(2n+1) < 4n+2$
well if u manage to finish this problem, please let me know how to do it cus i'd love to get a fields medal
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inland egret
hi
inland egret
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left portal
Let
$A = \begin{pmatrix} 1 & -5 & 1 \ k & -k & 1 \ 5 & -1 & k \end{pmatrix}$
and
$\mathbf{b} = \begin{pmatrix} -11 \ 1 \ k^2 \end{pmatrix}$.
Find all $k \in \mathbb{R}$ for which
$\begin{pmatrix} 3 \ 3 \ 1 \end{pmatrix}$
is one of the infinite solutions of the system $A \mathbf{x} = \mathbf{b}$. For each of the values found, determine all solutions of the system.
twin meteorBOT
ඞඞඞ
bitter pilot
Let
$A = \begin{pmatrix} 1 & -5 & 1 \\ k & -k & 1 \\ 5 & -1 & k \end{pmatrix}$ ...
(3,3,1) is a solution for Ax=b
left portal
,w rref {{1,-5,1,-11},{k,-k,1,1},{5,-1,k,k^2}}
twin meteorBOT
left portal
,, \begin{cases} \frac{k^2+3k-4}{4(k-2)} =3\ \frac{k^2+7k-20}{4(k-2)} =3 \ \frac{k^2-3k-2}{k-2} =1 \end{cases}
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ඞඞඞ
bitter pilot
,w rref {{1,-5,1,-11},{k,-k,1,1},{5,-1,k,k^2}}
hmm
left portal
i might have messed something up
bitter pilot
I thought of doing [ A\mathbf{x} = \mathbf{b} \Leftrightarrow \begin{pmatrix} 1 & -5 & 1 \ k & -k & 1 \ 5 & -1 & k \end{pmatrix}\begin{pmatrix} 3 \ 3 \ 1 \end{pmatrix} = \begin{pmatrix} -11 \ 1 \ k^2 \end{pmatrix} ] and then find such $k \in \mathbb{R}$ so that $A\mathbf{x}$ has one zero row (inducing infinite many solutions).
twin meteorBOT
bacc the sigma😔🤞
left portal
yeah sure
we can perform matrix multiplication
,w {{1,-5,1},{k,-k,1},{5,-1,k}}×{{3},{3},{1}}
twin meteorBOT
left portal
what
bitter pilot
Ax produces a vector
left portal
k=-12
bitter pilot
k+12 = k²
bitter pilot
**bacc the sigma😔🤞**
I noticed the last sentence with the zero row didnt make sense 
left portal
,w solve k^2 -k+12=0
bitter pilot
,w k+12 = k^2
twin meteorBOT
bitter pilot
,w rref {{1,-5,1,-11},{-3,-(-3),1,1},{5,-1,-3,(-3)^2}}
twin meteorBOT
twin meteorBOT
left portal
ah because determinant
bitter pilot
yea or you look at the rows
bitter pilot
here it's fine gives exactly one solution
bitter pilot
here infinite cause zero row
left portal
yeah one row full of zeros means one variable is free
bitter pilot
yes exactly
left portal
k=4 and thats it
bitter pilot
k=-3
left portal
bitter pilot
k=4
bitter pilot
yesss
left portal
with k=4
bitter pilot
you can check
left portal
.solved
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main jolt
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main jolt
how do I find i1 and i2?
oblique kettle
silk osprey
(kirchhoffs junction law)
main jolt
how does that look?
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winged briar
How do I graph for the function f(x) = [[x-c]] given c = -1, 1, and 3?
winged briar
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dawn spindle
why's this the case? (last sentence)
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why's this the case? (last sentence)
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karmic forge
Hello, can someone help me with this?
vocal sleetBOT
Hello, can someone help me with this?
pine hull
full ferry
could you translate some stuff?
full ferry
doesnt it say a > 1?
karmic forge
The first question says to prove that : a^n+1....
karmic forge
doesnt it say a > 1?
Yes yes sorry
full ferry
alright
karmic forge
I can prove that it's correct for n=1. But when I get to the second step, I get stuck
hybrid flicker
I can prove that it's correct for n=1. But when I get to the second step, I get ...
the idea is
look at a^(n+1) + 1/(a^(n+1)) + a^(n-1) + 1/(a^(n-1))
(there is no need for recurrence if you're asking)
hybrid flicker
look at a^(n+1) + 1/(a^(n+1)) + a^(n-1) + 1/(a^(n-1))
looking at this, group together the a^..., and group together the 1/a^...
and factor appropriately
karmic forge
(there is no need for recurrence if you're asking)
I have to use that method
hybrid flicker
I have to use that method
were you told to use that method during the exercise at some point, or for EVERY question?
because I would understand if it were for this
karmic forge
were you told to use that method during the exercise at some point, or for EVERY...
No, not really. The exercise came following that paragraph. So I tried to prove it using the recurrence although it can be done differently.
dense eagle
ur talking about 1). right? show that for all n \in N (recurrence)
karmic forge
Both of the questions actually, but as Rafilou said, 1 can be proven without it.
karmic forge
Are you talking about the 2nd question ?
karmic forge
first one
But for the first one (if I do it by recurrence) I will need to prove it for a^(n+2)+1/a^(n+2)
karmic forge
once u've done the first one, the 2nd one is really obvious by induction
Really? It's not obvious to me — can you elaborate?
dense eagle
But for the first one (if I do it by recurrence) I will need to prove it for a^(...
i'm not sure what u mean?
$(a^n + \frac{1}{a^n})(a+\frac{1}{a}) = a^{n+1} + \frac{1}{a^{n+1}} + a^{n-1} + \frac{1}{a^{n-1}}$
twin meteorBOT
LY
karmic forge
i'm not sure what u mean?
P(n) =>P(n+1)
dense eagle
$\implies a^{n+1} + \frac{1}{a^{n+1}} = 3(a^n + \frac{1}{a^n}) - (a^{n-1} + \frac{1}{a^{n-1}})$
twin meteorBOT
LY
dense eagle
P(n) =>P(n+1)
but you only need that for an induction?
dense eagle
**LY**
so by this recurrence, a^2 + 1/a^2 is an integer
and so using the recurrence again, a^3 + 1/a^3 is an integer etc.
karmic forge
I get it. Thank you so much for your time.
dense eagle
nw!
manic lance
https://math.stackexchange.com/questions/2027832/l-notation-in-linear-algebra
what other commentators?
hard atlas
mild flower
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warped egret
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twin meteorBOT
ren
hybrid flicker
and why was c and d introduced xd
warped egret
thats just what the question said
hybrid flicker
then the question is dumb xd
warped egret
ill get rid of the c-d part later
i should be able to do smth like this right?
or is there a significantly easier way
hybrid flicker
oh you could but induction is significantly easier
hybrid flicker
induction is pretty easy:
you want to prove some property P(m) holds true for all m in Z+ (or N)
show P(0) or P(1) holds, whichever is the first case
then, show that if P(k) holds, then P(k+1) holds
warped egret
its kinda like a recursive loop in coding right
hybrid flicker
kinda
I think of it as dominoes
I want to show every domino is flipped
All I have to do is flip the first domino
and make sure every domino flipped makes the next one flip
warped egret
i dont think we should be using induction for this cuz we barely learned it monday
hybrid flicker
i dont think we should be using induction for this cuz we barely learned it mond...
if you learnt it i highly advise using it here
warped egret
we did it for a series but thats it
modern ledge
we did it for a series but thats it
the structure of the question leads to induction
well at least it leads nicely to it
more often than not a direct proof rather htan induction is harder
warped egret
ig i can try it rq
ugh ive been doing this hw for like a significant number of hours. its 5 am
so is base case checking m = 1?
hybrid flicker
I'm only asking because sometimes Z+ and N have different meanings
specifically to include 0 or not
warped egret
well Z+ is positive integers
hybrid flicker
positive in the english sense means > 0
warped egret
yeah so starts at 1 right?
hybrid flicker
yes
warped egret
so i gotta show any work for this part or can i just put a is equivalent to b modn and show thats the same as one of the conditions?
hybrid flicker
a ≡ b mod n because we supposed so
warped egret
wow thats a fance congruence sign
lemme type smth up rq
so smth like this?
i feel like im doing this wrong
should i be checking divides or smth instead
yk what im gonna take a power nap
hybrid flicker
you're not supposed to say this
your goal is a^1 equivalent to b^1 mod n
in a proof, you always start off from what you know, to end up on what you want
warped egret
o
hybrid flicker
"Suppose m = 1. Then, since it is known that a = b mod n, we then naturally have a^1 = b^1 mod n. Thus the base case holds"
warped egret
ok i put that in my own words
im unsure what to do for the actual inductive step tho. for the series we were able to set it equal to smth, but can i really do that on this?
all i got but from what we did before im unsure how to precede. again, we barely have done any induction.
so i appreciate u bearing with me
hybrid flicker
all i got but from what we did before im unsure how to precede. again, we barely...
ok so have you proved the ac = bd mod n thing?
warped egret
yeah
tho i ended up deleting it cuz for some reason the prof wrote ac=bc instead
basic idea was just rearranging so we already have the ac on one side and then subbing the other stuff in then factoring out an n
hybrid flicker
ok so
if you already proved that
use interesting values of c and d
to prove the induction step
warped egret
well its not on the paper anymore, but we do have it as prior knowledge so i can use it anywya
when u say that do u mean setting c equal to a and d equal to b or smth
cuz thatd just produce the i+1 naturally right?
does this work?
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modern ledge
modern ledge
.reopen
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✅
modern ledge
am stuck at part (v) now 💀
modern ledge
<@&286206848099549185>
🙏 manifesting solution to come into my mind
So according to question, i should have that $Re(\alpha^{2^{n+1}})= a_{n+1}-b_{n+1}\sqrt{3}$ which is equivelant to $Re((\alpha^{2^{n}})^{2})$. Then i tried letting $\alpha ^{2^{n}} = x+ iy$ but that doesnt really work since idk what $y$ is since $Re(\alpha ^{2^{n+1}}) = x^{2}-y^{2}$ according to the set up i have (placed into 1 message for readability)
twin meteorBOT
Dootud
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modern ledge
put the$ $ around your masterpiece
ur tex needs to be in math mode to be recognised by bot
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flat whale
<@&268886789983436800> troll
hybrid flicker
viscious term
knotty crystal
Image
Limit at orgin?. My orginial thoughts were no
so tried to show the limit is not unique considering seqeunces which were a subset of R^2
Using f(0, 1/k) i got limit f is zero
now struggling to find another two seqeunces which returns a finite limit which isnt zero to show that the limit is not unique i.e it does not exist
Any thoughts?
vocal sleetBOT
Image
Limit at orgin?. My orginial thoughts were no
so tried to show the limit i...
knotty crystal
bitter pilot
Image
Limit at orgin?. My orginial thoughts were no
so tried to show the limit i...
I think this one's exist actually
knotty crystal
So lost with it man?
If it exist i guess i go about proving it with the formal defonition so ill try that first
bitter pilot
Yea or use polar coords + squeeze theorem
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jovial parcel
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jovial parcel
The answer is supposed to be ln(50/27) but I'm getting ln(100/81).. I can't tell what I did wrong. I'm correct up till the point where I put double ticks.
nvm I got it
.close
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craggy hamlet
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outer warren
yiu over reduced it with the
2k > 0
note that you gave the condition that k>=3
trim ice
feel dumb, but i cannot for the love of god solve this, can anyone help?
trim ice
not sure how i can approach it (never done any modular arithmetic before, which i am guessing this would require)
dense eagle
feel dumb, but i cannot for the love of god solve this, can anyone help?
so a) is meant to be an 'exploration'
so try n=3, n=5
do they work?
trim ice
do they work?
i mean 3 and 9 do work, i can see that, but this is intended to be solved within quite a short timeframe, how am i exactly supposed to "explore" wether odd n > 1 is a factor of 2023^n - 1?
dense eagle
the fastest way would indeed be modular arithmetic
but you can do part a without
i'll think abt how u can do part b
dense eagle
feel dumb, but i cannot for the love of god solve this, can anyone help?
but factoring for a would work for n=3
trim ice
i think you can do part b with induction
dense eagle
i think you can do part b with induction
yep
lifting the exponent here would be the fastest fastest way to do b but it's kinda an obscure-ish theorem
dense eagle
i mean 3 and 9 do work, i can see that, but this is intended to be solved within...
but yeah notice that powers of 3 work then induct
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dense eagle
thanks i will try
nw!
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ruby scarab
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ruby scarab
someone help me pls
cyan shadow
!status
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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
3
feral coral
what have you got
ruby scarab
65.9 and 45.2
ruby scarab
idk
ive tried to answer it and it was wrong
then i tried apps then it was wrong aswell
i just need to get this done
ruby scarab
its the last question i have
feral coral
to be fair sparx is pretty bugged
ruby scarab
for the edges i was confused
for the faces i did 17 squared -8 squared all square root
then i used that height and did bxh/2
feral coral
then i used that height and did bxh/2
but you're finding the angle between AOB and COD
feral coral
you will need to use trig here
ruby scarab
oh no
feral coral
you have got the height of AOB and COD here
ruby scarab
non right angle trig confuses me
feral coral
non right angle trig confuses me
you can make it a right angle triangle here don't worry
ruby scarab
ok
feral coral
you see the triangles here are all isosceles
ruby scarab
ok
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jolly plover
I need help using the shell method to find the volume V generated by rotating the region bounded by the given curves about y = 8. The equations are 27y=x^3, y=0, and x=6. I made a graph, but I need help construcitng the integral.
vocal sleetBOT
I need help using the shell method to find the volume V generated by rotating th...
full ferry
27y =^3?
jolly plover
sorry, 27y=x^3
full ferry
i would plug y = 8 to find the other x coord
jolly plover
wdym other x cord?
jolly plover
what?
jolly plover
bro idk
finding the radius and the height of the cylidrical shell are the two things i need help
also wouldnt shifting the curve 8 units down mess with my bounds??
full ferry
no, if you had the bounds of the x coordinates
jolly plover
no, if you had the bounds of the x coordinates
but its being rotated around y=8, so i would have the bounds on the y-axis
bro did you read my question😭
<@&286206848099549185>
full ferry
if you move it down by 8, then you can change the question to being rotated around the y axis
then just find the y coord for x = 6
jolly plover
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full ferry
if you move it down by 8, then you can change the question to being rotated arou...
*x axis
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lofty knot
4x+9=12
lofty knot
inland sail
pale seal
4x+9=12
You want to isolate the term with x , i.e the LHS should contain only the x term and the RHS will have a number
lofty knot
You want to isolate the term with x , i.e the LHS should contain only the x term...
please just give me the answer
pale seal
!noans
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iron quiver
i want to know , what is correct
vocal sleetBOT
i want to know , what is correct
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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.
pine hull
<@&268886789983436800>
iron quiver
Can you give me your reasons?
pine hull
Can you give me your reasons?
Better show original question, sometime things gets confusing dealing with fraction/fraction
iron quiver
Better show original question, sometime things gets confusing dealing with fract...
thank you
iron quiver
thank you
pine hull
No problem
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regal zephyr
the answer is tan y= -1/2 but how do i get negative?
vocal sleetBOT
the answer is tan y= -1/2 but how do i get negative?
silver bloom
Its in Q2 so it's negative
uneven flicker
!done
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hushed cosmos
why cant you just get rid of the 4x by moving it to the other side
hushed cosmos
why cant you just get rid of the 4x by moving it to the other side
they do it heree so i dont understand why they dont just do that with the one above
heady vine
why cant you just get rid of the 4x by moving it to the other side
you just lost a whole solution bcz of this
modern ledge
they do it heree so i dont understand why they dont just do that with the one ab...
division by x versus 6 is different
hushed cosmos
oh is it cuz theres a 6 there
modern ledge
if x = 0
then u are dividing by 0
which is no bueno
whereas division by 6 on both sides does not affect solution
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@hushed cosmos Has your question been resolved?
rare sedge
Hello, I wouldnt mind some help with this. For some reason I am having some issue just plugging in values... One specific question I have is how arccos (-1/2) is actually possible if the range of arccos is only from 0 to pi/2.. Maybe i have some kind of conceptual misunderstanding...
vocal sleetBOT
Hello, I wouldnt mind some help with this. For some reason I am having some issu...
rare sedge
for arccos(pi/2) do we just reflect it over to the positive side?
This is the answer fyi
thanks in advance for help
torn timber
Hello, I wouldnt mind some help with this. For some reason I am having some issu...
its from 0 to pi not 0 to pi/2
torn timber
for what?
rare sedge
for x
torn timber
arccos can have negative values as its inputs
rare sedge
ok..
torn timber
Hello, I wouldnt mind some help with this. For some reason I am having some issu...
also your third step is wrong arccos(-1/2)=2pi/3 also why is there an extra 1/2 multiplied with it?
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elder rock
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feral coral
the vertical lines surrounding 3x-1 basically turns everything inside them positive
elder rock
so would it be -|3x+1|?
elder rock
okay
feral coral
okay how would you do it if it is like this
3x - 1 ≤ ±41
elder rock
oh
sorry i didnt realize it was on dnd
uhm
i would move 1 to the other side by adding one to both sides
then i would divide 3 to both sides
which leaves me with x<= +-41+1/3
feral coral
yes
but ± here suggests that it could be positive and negative
so ((±41) +1) /3) has two cases: (+41 + 1)/3 and (-41+1)/3
feral coral
yea kinda
feral coral
you have got the first one here
elder rock
so i got it partially right im just missing the second part?
feral coral
yes
feral coral
kinda realised I might have make this a bit complicated
this is also the same as saying
$-41\le3x-1\le41$
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VjFaU1dsb3hjRVZhTW1SUFVqQXhibGRY
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elder rock
Im still confused
vocal sleetBOT
heavy yoke
,rccw
twin meteorBOT
heavy yoke
looks like that's cramer's rule
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grim leaf
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grim leaf
does anyone know how to solve this?
i was told to use a calculator (ti84 plus ce) but i cant afford one so is there a way to do it like without one?
blazing wigeon
What should you do before even trying the quadratic formula or factoring or completing the square
grim leaf
in case i ever get a paper one or anything?
grim leaf
What should you do before even trying the quadratic formula or factoring or comp...
i was trying to like factorise it
but i wasnt rlly sure how to break up 18
blazing wigeon
First try dividing everything by ...?
grim leaf
First try dividing everything by ...?
3?!
blazing wigeon
What is the largest number that can divide each of the coefficients
It is larger than 3
blazing wigeon
Yes
blazing wigeon
$n^2-3n-4=0$
twin meteorBOT
mari
grim leaf
yup
blazing wigeon
Can this factor evenly?
grim leaf
and i broke the -3n to .... +n - 4n
blazing wigeon
Sounds good
grim leaf
ok wait im a bit stuck here
blazing wigeon
$n^2+n-4n-4=0$
twin meteorBOT
mari
grim leaf
could it be (n + 1)(n +4)?
blazing wigeon
No
dull bear
could it be (n + 1)(n +4)?
(you were very close 
it's almost this, but check again 
)
blazing wigeon
What can you factor out of n^2 and n
blazing wigeon
Yes
vast shale
Is there anyway to prove something by not using induction?
So for example instead of assuming that n=k is true and prove for n=k+1 by using given identity instead we use this identity to prove only n=k case. Or it just not possible?
vocal sleetBOT
Is there anyway to prove something by not using induction?
So for example instea...
brisk moss
what?
ancient knoll
if you’re asking what i think you’re trying to say, then the answer is you’ll often have to prove an infinite number of cases instead of 1 or 2
because if you want to show something holds for all n>1
instead of proving holding at n=k implies at n=k+1 and n=2, you need to prove n=2, n=3, …, and prove by (physical) exhaustion
vocal sleetBOT
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vast shale
if you’re asking what i think you’re trying to say, then the answer is you’ll of...
I actually have the example but it just too long
So the point is to prove divisibility of certain polynomial by other polynomial after specific operation performed n times
So instead of just using proof by induction I tried to use different definitions to prove base case for n=k instead of assuming it being true . So I actually got curious what induction actually gives you in more formal way.
I just think proving for n=k is just like proving for any other integer while proving for all integers should utilise induction step. So basically I was doing wrong thing all along, since only what I was asked for is to proved that operation is valid for consecutive case and n=1 case.. Ahh sometimes is so confusing where is induction and where isn't
Thanks
outer warren
non induction proofs would be stuff like direct proofs
quiet river
i got up to here, how what? (should also be parallel to the line)
vocal sleetBOT
i got up to here, how what? (should also be parallel to the line)
@quiet river Has your question been resolved?
cobalt arch
need help solving this problem, "Describe a sequence of transformations that will map DEF onto GHJ
vocal sleetBOT
need help solving this problem, "Describe a sequence of transformations that wil...
cobalt arch
<@&286206848099549185>
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vast shale
why do we convert an ODE into a set of linear first order ODEs?
vocal sleetBOT
why do we convert an ODE into a set of linear first order ODEs?
vast shale
what advantage does that have?
vocal sleetBOT
@vast shale Has your question been resolved?
pallid zenith
@vast shale you can use methods that apply to first order odes
isnt that good enough 
vast shale
<@456226577798135808> you can use methods that apply to first order odes
what methodss?
vast shale
how about rk4
whats rk4?
pallid zenith
vast shale
ohh okay! so mostly when we approximate solutions via iterations it might be helpful to have it as a system?
pallid zenith
well, we have access to pretty good methods for first order differentia equations
vast shale
Does it have any other advantages?
pallid zenith
why should we have to give them up
vast shale
hmm but theyre still coupled
pallid zenith
thats fine
vast shale
thats what kind of confuses me
does it have any other advantages?
besides that it looks cooler
pallid zenith
IDK if id count looking cooler, lol
pallid zenith
fair enough
vast shale
thanks tho, i dont think there are more advantages
vast shale
tell me
pallid zenith
but we can make pretty firm statements about simultaneous equations
this would be my thought
but, im sure there are more advantages
just not a lot coming to mind at 9pm on a thursday
anyways
pallid zenith
idk what that is
vast shale
snapchat ai
-
Simplified Analysis: First-order ODEs are generally easier to analyze than higher-order equations. You can use established methods for first-order equations, which can save time.
-
Numerical Methods: Many numerical methods, like Euler's method or Runge-Kutta methods, are designed for first-order ODEs. By converting to first-order, you can apply these methods directly.
-
Initial Conditions: It's often easier to specify initial conditions for first-order systems, which can be crucial for solving ODEs in real-world applications.
-
Phase Space Representation: Converting to a system allows for a better representation in phase space, helping visualize the behavior of the system over time.
-
Stability Analysis: It can facilitate stability analysis, as you can examine the behavior of each first-order equation separately to understand the overall system's stability.
gave me this
pallid zenith
ah, sure, stability
you know, i think there are other properties based on spectra
where we'd like it in a system
i did not feel very focused during this course though
if you are curious
youll find stability in chapter 4 and spectral methods towards the end
vast shale
he gave me 10 more 😭
vast shale
youll find stability in chapter 4 and spectral methods towards the end
im currently watching a nice series about them but was just curious what advantage it brings
pallid zenith
to me
it is that we may just use plain ol first order method
everything else is gravy
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patent nymph
If 0 < p/n < 1 where p and n are integers with n being odd, and p/n is in lowest terms, then do p/n and 1/n necessarily have shortest repetends of the same length in their binary expansion?
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If 0 < p/n < 1 where p and n are integers with n being odd, and p/n is in lowest...
@patent nymph Has your question been resolved?
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@patent nymph Has your question been resolved?
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@patent nymph Has your question been resolved?
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@patent nymph Has your question been resolved?
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@patent nymph Has your question been resolved?
graceful girder
If 0 < p/n < 1 where p and n are integers with n being odd, and p/n is in lowest...
is this still available?
patent nymph
is this still available?
My question still hasn’t been answered
graceful girder
My question still hasn’t been answered
Let me try lol hahaha
The answer is yes! FOR ME MY OPINION FOR ME
Both fractions will behave similarly in binary because "n" is odd. When you divide by an odd number, the way the digits repeat in binary will be consistent.
Since "p/n" is just a scaled version of "1/n", their repeating parts will match in length.
So, if p/n and 1/n are in the right form, they will have the same length of repeating digits in their binary forms!
When we have two fractions, like p/n and 1/n
=>This is the part of the fraction that keeps repeating when you turn it into a decimal. For example, in 1/3 =>0.333..., the "3" keeps repeating
=>This means turning the fraction into a base-2 (binary) number instead of the usual base-10 (decimal).
Now, for your question:
=>if we have p/n and n is an odd number (which means it can't be divided evenly by 2), and p/n is in its simplest form like 1/3 is simpler than 2/6 we want to know if both p/n and 1/n will have repeating parts in their binary form that are the same length.
@patent nymphIt's just my opinion, so it's 50/50. I don't even trust myself that it's correct lol
patent nymph
The answer is yes! FOR ME MY OPINION FOR ME
>Both fractions will behave similarl...
I hope you haven’t used chatgpt lol
patent nymph
The answer is yes! FOR ME MY OPINION FOR ME
>Both fractions will behave similarl...
For your second point: Consider 1/63 and 7/63 in binary
graceful girder
I hope you haven’t used chatgpt lol
the fuck is chatgpt? new messanger?
patent nymph
The first one has a repetend of length 6, but the second one has a repetend of length 3
So they don’t match in length, although they are multiples
patent nymph
the fuck is chatgpt? new messanger?
It’s just that your message seemed a bit like an ai generated one
like the format and how you tell me something I did know already
graceful girder
It’s just that your message seemed a bit like an ai generated one
like gemini type shit
patent nymph
and how you’re a bit vague
and you just rephrased the question at the end
but I suppose it could just be you understanding the question
so forgive me if I’m wrong
graceful girder
and how you’re a bit vague
it true i am bit vague because i dont want to want to failed you or assume this is correct maybe it right but anyone can just say something super mathematics that little convincing and you assume it is correct hehe
graceful girder
so forgive me if I’m wrong
no no no forgive me
graceful girder
For your second point: Consider 1/63 and 7/63 in binary
hmmmm it seemsssssssssssss some parts of my chat or explanation is incorrect thank you broooo for pointed that out! God bless you
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@patent nymph Has your question been resolved?
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@patent nymph Has your question been resolved?
half imp
If 0 < p/n < 1 where p and n are integers with n being odd, and p/n is in lowest...
Yes. The length of the shortest repetend is the smallest number k such that 10^k = 1 mod n, regardless of what the numerator is (so long as it's coprime to the denominator)
If n has factors of 5 or 2 (ik you said it's even but if it wasn't this would still work) remove them first
patent nymph
Yes. The length of the shortest repetend is the smallest number k such that 10^k...
I see, how do you prove that?
patent nymph
Yes. The length of the shortest repetend is the smallest number k such that 10^k...
also should that be 2^k = p since the expansion is binary
patent nymph
oh right
patent nymph
Yes. The length of the shortest repetend is the smallest number k such that 10^k...
how do you prove that?
half imp
one way is examining the process of long division
the other is you know that if you have, say, 0.1100110011001100...
no there's an easier way
hmm
multiplying a fraction by an integer can't increase the length of the repetend
so you can multiply it by whatever integer is p^-1 mod n
but yea examine the process of long division or use the fact that if you have say 0.[1100] where the 1100 repeats forever
that's 1100/1111 = 12/15 = 4/5
so looking at 4/5 we see that k=4 is the first k such that 2^k = 1 mod 5
where was i
anyways the fact that 4/5 has a repetend of k=4 digits is a consequence of the fact that it can be written in the form a/(2^k-1)
which is in turn a consequence of the fact that k=4 is the first number for which n=5 divides 2^k-1
which is another way to say that 2^k = 1 mod n
patent nymph
multiplying a fraction by an integer can't *increase* the length of the repetend
I don’t see how to show that it doesn’t decrease the length of the repetend though
does that just follow from them being coprime?
half imp
I don’t see how to show that it doesn’t decrease the length of the repetend thou...
that's not part of the final proof i ended up settling on
but yes
if we have an irreducible fraction like 14/27
we can multiply it by 14^-1 mod 27 (in this case 2) to get a number with fractional part 1/27
patent nymph
and that can’t increase the repetend length right
I think I see
Thanks for helping!! @half imp
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half imp
np
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vast shale
I am not sure how to proceed from here. I don't know how to go from there:
twin meteorBOT
forMaths
hybrid flicker
I am not sure how to proceed from here. I don't know how to go from there:
v stands for y?
well you gotta find the value of C
plugging some known values
and then solve for x when y = 0
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hexed kestrel
For sequence a(n)=[2a(n-1)-1]/3, a(0)=4x-3, Rewrite a general formula
hexed kestrel
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quartz lodge
I would try to expand the sequence by doing something like
$a(1) = p a(0) + q$ where p = 2/3 and q = -1/3
twin meteorBOT
delta
quartz lodge
$$a(2) = p a(1) + q$$
$$a(2) = p (p a(0) + q) + q$$
$$a(2) = p^2 a(0) + q(1 + p)$$
twin meteorBOT
delta
quartz lodge
$$a(3) = p^3 a(0) + q (1+p+p^2)$$
twin meteorBOT
delta
quartz lodge
once you start to see a pattern and introduce the values for p = 2/3 and q=-1/3 and a(0) = 4x-3 it should be straightforward to solve
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delta
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hollow thorn
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hollow thorn
How to do this ?
hybrid flicker
How to do this ?
substitute first
hollow thorn
substitute first
which one 、
hybrid flicker
this
we don't know anything about integral of f(3x+1)dx
but we do know about integral of f(x)dx (or f(y)dy or replacing with any other mute variable)
mossy finch
just change the inside
3x+1 = a
@hybrid flicker can u try to solve a Variational inequality because i'm not the brightest at it
mossy finch
Yeah it is multi choices question
so 1/3 part
i just threw it at the end
for easier calculation
hollow thorn
but we do know about integral of f(x)dx (or f(y)dy or replacing with any other m...
pull the - sign, and turn upside down the limit ?
mossy finch
???
hollow thorn
owh whats this
mossy finch
Lemme check my old highschool book
so i can explain it better
but the main part is that u consider 3x+1 = a
And that pic is wrong btw
it should be 1/3 with that pic
hollow thorn
And that pic is wrong btw
alright
twin meteorBOT
rafilou is not not born in 2003
hollow thorn
never seen u sub?
you mean u substitution ?
hybrid flicker
yes
mossy finch
is Differentiable a word in math @hybrid flicker ?
hollow thorn
ohh yea i do knew that
hybrid flicker
ohh yea i do knew that
alright so do some u substitution on f(3x+1)
to get some f(u)
and use the other integrals you know
hollow thorn
to get some f(u)
okay how to combine with those 2 integral function which is 21 and -15 ?
mossy finch
okay how to combine with those 2 integral function which is 21 and -15 ?
find the f(u) first
then u can see that how u should combine the first 2 integral
vast shale
Substitute
a = 3x+1
da = 3 dx
dx = da/3
Figure the limits
when x is 1, u=3(1) + 1 = 4
when x is 2, u=3(2) + 1 = 7
Substract the other two integrals given the value of them
-15-21=-36
Now we substitute
1/3 * (-36) = -12
So the answer is B
mossy finch
Substitute
a = 3x+1
da = 3 dx
dx = da/3
Figure the limits
when x is 1, u=3(1) ...
chill bro
he is learning
cosmic cloud
don't give answers
cosmic cloud
!noans
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The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.
hybrid flicker
I just explained how to do it
yeah but you didn't let the guy work through it, you handed it like an "already finished and ready to be copied" solution
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ruby heart
using x^2+y^2+z^2=1 I have to minimise and maximise x+4y+8z
I think the minimum is -8 but I couldn't prove it I tried to prove it homogenising RHS I need some hints/help
vocal sleetBOT
using x^2+y^2+z^2=1 I have to minimise and maximise x+4y+8z
I think the minimum ...
ruby heart
x,y,z is real btw
magic wasp
It's not -8
magic wasp
x^2+y^2+z^2=1 is a sphere (radius 1, center origin)
x+4y+8z=a is a plane, your goal is basically to find the extremum values for a, and that happens when the plane is tangent to the sphere
Think about the normal to the plane
ruby heart
x^2+y^2+z^2=1 is a sphere (radius 1, center origin)
x+4y+8z=a is a plane, your g...
uh I haven't had this I only had like basic inequalities, but it's fine, wait so like I have to find when it is tangent from "above" and "below" the sphere? and it's probably like similar to 2d so I can try doing it
magic wasp
Yes it's somewhat similar
ruby heart
okay thank you
mossy finch
The largest is easier to find btw
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ruby heart
so the normal vector to this plane is (1,4,8)?
magic wasp
Yes
magic wasp
Yes
ruby heart
yayy thanks a lot
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ruby heart
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plucky ferry
guyz help
twin meteorBOT
altaccountinthespotline
plucky ferry
rearrange equation to find x
Substitute your found value back into the equation to see if you arrive at a new result.
terse forum
!original
vocal sleetBOT
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.
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prisma linden
claim
vast shale
split this
vast shale
k * 2 ^ (k+2) + 2 * 2 ^ (k+2)
prisma linden
alright I'll give it a try brb
nevermind I did not cook
it burned
where do we go from here?
vast shale
there are 2 same terms
prisma linden
k*2^(k+2)?
vast shale
add them
prisma linden
like this?
prisma linden
does this work?
vast shale
you can also simplify this
prisma linden
oh yeah
vast shale
write 2+1 as 3 please
prisma linden
prisma linden
now you will see 2 same terms again
do I add them together like before
vast shale
yes
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vast shale
not k+4
vocal sleetBOT
prisma linden
oh alright tysm