-9/5 x^-4
#help-0
tender anchor
surreal meadow
mhm, can you try writing it as a fraction?
ocean sealBOT
maximofs
surreal meadow
what would this be if we changed that exponent to a positive one?
think back to the first rule we applied here
tender anchor
-9/5 * 1/x^4
surreal meadow
as in, how can we write this with the x in the denominator
perfect
same thing as
$\frac{-9}{5x^4}$
ocean sealBOT
maximofs
surreal meadow
but yes that’s the idea for this problem
lone heartBOT
keen furnace
What is an increasing function?
keen furnace
Try by proving x^2+1/x^2>=2
I will try that
lone heartBOT
@rocky escarp Has your question been resolved?
rocky escarp
i still can't figure it out
tacit arch
https://tutorial.math.lamar.edu/classes/calciii/TangentPlanes.aspx
which part of this was confusing?
vocal palm
rocky escarp
ok i appreciate it
have fun
does this look correct? and how can i find the normal line?
lone heartBOT
@rocky escarp Has your question been resolved?
rocky escarp
.close
rocky escarp
lone heartBOT
lone heartBOT
@rocky escarp Has your question been resolved?
lone heartBOT
@rocky escarp Has your question been resolved?
lone heartBOT
@rocky escarp Has your question been resolved?
trim heron
consider your function as $z - ln (x^2+y^2) = 0$
ocean sealBOT
ertansinansahin
trim heron
normal line?
trim heron
your normal vector is (2/e, 0, -1) in component notation. This is a direction vector for your normal line.
trim heron
put z part to the right hand side
rocky escarp
ok
trim heron
check this
rocky escarp
0 = .... (2-z)
trim heron
now you need to write this
trim heron
0 = .... (2-z)
yes, z with -1 as a coefficient.
V is the direction vector, which is your normal vector
P_0 is the initial vector, for your initial point, which is that point where you have this normal.
rocky escarp
check this
this really sums it up
lone heartBOT
@rocky escarp Has your question been resolved?
cloud pollen
alpine sable
lone heartBOT
lone heartBOT
@alpine sable Has your question been resolved?
lone heartBOT
alpine sable
.reopen
lone heartBOT
✅
upbeat hornet
did you consider that the two volumes overlap
alpine sable
did you consider that the two volumes overlap
Yeah I subtracted them
lone heartBOT
@alpine sable Has your question been resolved?
upbeat hornet
Yeah I subtracted them
that doesn't solve the problem
lone heartBOT
lone heartBOT
Available help channel!
granite rapids
Hi
granite rapids
To prove this can't we just say a=2 and b=2
2*2=4 which is perfect square
gcd(2,2)>1
Thus we proved this?
remote heron
You can't prove by example
upbeat hornet
for any integers a and b
remote heron
You can only disprove by example
remote heron
how do you know?
granite rapids
Just don't know how to prove it
remote heron
I guess thats the first question, are you sure youre convinced that its true or false
granite rapids
Because logically speaking just do the calculations
upbeat hornet
Because logically speaking just do the calculations
but you can’t do the calculations for every case
remote heron
even order group dont leave i might need your help 
upbeat hornet
Because logically speaking just do the calculations
but you need to give a proof
granite rapids
Okay
upbeat hornet
even order group dont leave i might need your help <:blobsweat:58450623966111337...
same
granite rapids
So how would I do this
remote heron
I'd start with definition, personally
specifically perfect square
see if you can write out something meaningful in math language
upbeat hornet
are you allowed to use fundamental theorem of arithmetic
granite rapids
so how would I do this?
remote heron
you just asked us that
granite rapids
are you allowed to use fundamental theorem of arithmetic
wym by this?
upbeat hornet
so how would I do this?
what is the definition of perfect square?
granite rapids
oh ok
so how is that gonnna get me the answer
bc i dont need to provide that in the answer
upbeat hornet
one step at a time
remote heron
no offense but your attitude is not gonna help you with proof based problems lol
granite rapids
its literally grade 8 stuff
upbeat hornet
its literally grade 8 stuff
you are supposed to provide a proof, so you have to start from the definitions and build up from there
granite rapids
oh nm i dont need ur help. u guys where so based on the little stuff that i found a=4 and b=9
remote heron
it doesnt matter what grade it is, proofs are hard no matter what the subject material is
dull cedar
Hello I am having trouble solving this geometry question. I've looked over many thereoms to use but I'm just stuck because I'm not sure which one to use or which will help me answer it.
dull cedar
My first guess was that X = 15 and Y = 30 but then I noticed the triangle was a little lopsided so I assumed those wouldn't be the solutions.
upbeat hornet
You should be able to use similar triangles if AC is parallel to A’C’
dull cedar
I'll try that out, would it be possible for you to keep this channel open in the case I have another question?
upbeat hornet
you can always open another channel to ask
dull cedar
Ok thank you
upbeat hornet
have you solved it?
dull cedar
I'm working on the problem but I think i'll be able to solve it
lone heartBOT
Available help channel!
normal summit
how do i began this question
lone heartBOT
how do i began this question
upbeat hornet
have you sketched the line?
vale wigeon
oh what unfortunate punctuation at the beginning
tardy stag
~~oh what unfortunate punctuation at the beginning~~
oh i hate that
normal summit
have you sketched the line?
yes
tardy stag
i hate that very much
normal summit
so the labels are right ?
it’s a bit of guess
then i just did the Pythagorean theorem for 5
vale wigeon
yes your side length labels are in order
now just need to sketch the angle
remember where those are measured from
normal summit
ok
lone heartBOT
@normal summit Has your question been resolved?
outer kelp
What is the answer to the sum of cubes problem? But I don’t have any work written down
ruby current
what problem?
outer kelp
A^3 + b^3 = c^3
lone heartBOT
@outer kelp Has your question been resolved?
outer kelp
or are you here to troll
I’m not trolling. I’m actually trying to get the answer
So then it’s actually solved
vale wigeon
okay so can you post the statement of the problem in full? if you don't, i'll maliciously comply with your incomplete statement.
the equation a^3 + b^3 = c^3 has infinitely many real solutions, among them, for example, (0,0,0).
@outer kelp
also (1, 1, 2^(1/3))
among many many others, i must add!
lone heartBOT
@outer kelp Has your question been resolved?
digital moss
1/2/2/2/2/2/2/2/2/2/1=?
lone heartBOT
1/2/2/2/2/2/2/2/2/2/1=?
gusty gorge
,w 1/2/2/2/2/2/2/2/2/2/1
ocean sealBOT
zenith leaf
1/2/2/2/2/2/2/2/2/2/1=?
just multiply all the denominators which will get u 512 then put 1 over it
digital moss
(0+2)+(1+2)+(2+2)+(3+2)+(4+2)/2=(y-312)/2 y=?
gusty gorge
,w (0+2)+(1+2)+(2+2)+(3+2)+(4+2)/2=
ocean sealBOT
gusty gorge
not sure if you're asking for a calculator or if you want a neat trick to compute it
lone heartBOT
@digital moss Has your question been resolved?
lone heartBOT
@digital moss Has your question been resolved?
digital moss
(0+2)+(1+2)+(2+2)+(3+2)+(4+2)/2=(y-312)/2 y=?
gusty gorge
!status
lone heartBOT
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
@digital moss Has your question been resolved?
alpine sable
Is this correct?
lone heartBOT
young finch
yes
lone heartBOT
Available help channel!
scenic wing
what am i doing wrong
lone heartBOT
what am i doing wrong
young finch
needs to be squared
scenic wing
needs to be squared
so just make the sin theta on the bottom squared ?
young finch
yep
young finch
where is it saying its wrong
scenic wing
where is it saying its wrong
do i need to make it cot^2
young finch
try it
young finch
ong
fallen verge
She sunny on my klaus till i violet
lone heartBOT
@scenic wing Has your question been resolved?
fierce panther
can you please help me with the hence part !! I have exams tomorrow.
safe tartan
in terms of b and c
wait i might be wrong but a quadratic is written by
x^2-(a+b)x+ab
where a and and b are the roots
safe tartan
so whats its asking is what is the quadratic equation
x^2-(a^3+1/b^3)x+ (a^3+1/b^3)(B^3+1/a^3)
fierce panther
ah yes but in the Answers for that question they have mentioned this method is wrong
safe tartan
but writing the roots in terms of a and b
cloud chasm
Can you explain what's happening in the hence part ?
c^3x^2+(b^3-3b*c)(c^3+1)x+c^3+1
fierce panther
wi8 I'll send the answers so I think then you can explain wuts happening
this is the Answers for that question
cloud chasm
fierce panther
this is the Answers for that question
i want to know why they are doing like that ?
cloud chasm
i want to know why they are doing like that ?
this is my answer
fierce panther
this is my answer
uh !! thx I'm looking at it
cloud chasm
uh !! thx I'm looking at it
it's a good question
i think
a good question about Vieta theorem
fierce panther
ok but why are you subtracting the roots from X and then multiplying with the other roots which has been subtracted from the same X ?
cloud chasm
ok but why are you subtracting the roots from X and then multiplying with the ot...
because the question asked us to construct a quadratic equation which the roots is that two
that is a common trick to construct a quadratic equation that has roots we already knew
fierce panther
Do you mean that if the roots of a quadratic eqn is a , b
the eqn can be written as X^2 - (a+b) + ab right ?
cloud chasm
Do you mean that if the roots of a quadratic eqn is a , b
the eqn can be writte...
yes
nonono
cloud chasm
Do you mean that if the roots of a quadratic eqn is a , b
the eqn can be writte...
(x-a)(x-b)
u missed an x
u may typed wrongly
fierce panther
yh sorry !
i missed the x
that's how I did it in the first place then checked the marking scheme but they have mentions that's wrong. !
because they have mentioned "you have to find the eqn for the new roots of the quadratic eqn using the previous eqn"
cloud chasm
because they have mentioned "you have to find the eqn for the new roots of the q...
i think they may means "u have to use b and c to complete this quadratic eqn"
u can use α and β at first and then use vieta theorem to replace them
just like what i has written
fierce panther
uhh !! thanks
but can you also tell wuts happening in the picture I have sent ? the second sums ?
cloud chasm
uhh !! thanks
but can you also tell wuts happening in the picture I have sent ?...
can u take a screenshot? idk what eqn you're referred to
fierce panther
from that 05 at the top
cloud chasm
from that 05 at the top
to the end???
fierce panther
upto the 4th 05
cloud chasm
upto the 4th 05
will,i can tell u what they're doing but i cant tell u clearly why they do it like that
fierce panther
umm okay No problem
cloud chasm
the first three eqn are serving for the conclusion of the first part
u already knew that its roots is α^3 and β^3
these there are aimed to find a value that can replace α^3 and β^3
fierce panther
ohh 😯
what is that new " Y " variable then ?
is that the root of new Quadratic eqn ?
cloud chasm
and it's actually is
cloud chasm
is that the root of new Quadratic eqn ?
it's a.....coreference?
idk how to discribe it
they set up the Y just for expression convenience
fierce panther
🤣 My countries education !! what to do !!
cloud chasm
just like they want to forcely apply the conclusion of the first part
cloud chasm
🤣 My countries education !! what to do !!
can i know where are u from?
cloud chasm
my true name?
fierce panther
yh
cloud chasm
no
cloud chasm
that is vital personal info
cloud chasm
Are you from Japan ?
is japan a socialism country?
fierce panther
idk
🙁
Anyway thank you so much for giving your time and helping me !!
thanks brother ❤️
safe tartan
how would i solve this using reverse chain>
lone heartBOT
how would i solve this using reverse chain>
strong compass
What’s the derivative of tan?
safe tartan
um sorry dont understand
weary lake
silk lintel
.close
scarlet drum
help
scarlet drum
Alrighty
mystic mulch
lone heartBOT
@scarlet drum Has your question been resolved?
lone heartBOT
@scarlet drum Has your question been resolved?
tender anchor
derivative of f(t) = e^(t+5)
lone heartBOT
derivative of f(t) = e^(t+5)
nocturne robin
compare 1 and \sqrt{\frac{3}}$ -1
lone heartBOT
compare 1 and \sqrt{\frac{3}}$ -1
brave cave
Hi
mellow grail
compare 1 and \sqrt{\frac{3}}$ -1
send a photo 👍
last ether
What's inside the square root
ocean sealBOT
mehdi_moulati
ocean sealBOT
umbraleviathan
nocturne robin
Oh are you meant to put an inequality in between them?
yes
last ether
Okay well you should be familiar with the fact that as the radicand increases, the value of the square root also increases
In this case, if you were to consider a value inside the square root other than 2, so:
$$1 = \sqrt{x} - 1$$
What would $x$ be
ocean sealBOT
umbraleviathan
nocturne robin
$\sqrt{2}$
ocean sealBOT
felixyn_ellington
last ether
I just said to consider a value other than 2
nocturne robin
hm
last ether
**umbraleviathan**
What would x be such that 1 = sqrt(x) - 1
nocturne robin
idk?
last ether
Think about it
If you did anything with algebra, you would know that 1 = sqrt(x) - 1 implies that 2 = sqrt(x)
tacit arch
but the only answer is sqrt(2), is there any another answer?
x=2 does not solve the equation umbra gave you
nocturne robin
no?
tacit arch
If you did anything with algebra, you would know that 1 = sqrt(x) - 1 implies th...
$\sqrt{x} = 2$
ocean sealBOT
rie.mann
tacit arch
what happens when you plug in $x = \sqrt{2}$ into that equation
ocean sealBOT
rie.mann
last ether
Yeah
nocturne robin
i'm dumb wtf.-.
last ether
Okay well you should be familiar with the fact that as the radicand increases, t...
Now because of this statement, 2 < 4. What does this imply
tacit arch
,w define radicand
ocean sealBOT
nocturne robin
okay
nocturne robin
Now because of this statement, 2 < 4. What does this imply
you lost me here
what do you mean?
so the radicand is 2
and 2<4?
last ether
Okay well you should be familiar with the fact that as the radicand increases, t...
This statement implies that sqrt(2) is less than sqrt(4), right?
nocturne robin
2<4 so sqrt(2) < sqrt(4)
nocturne robin
Okay and we know that if the radicand is 4, they will be equal
it's kinda like an answer i found on gg, i didn't get that so that's why i'm here
last ether
Which means that sqrt(2) - 1 must be (what) than 1
_ _
I'll give you a hint. Replace the standalone 1 with "sqrt(4) - 1"
Now:
1 ? sqrt(2) - 1 ==> sqrt(4) - 1 ? sqrt(2) - 1
What must ? be then
last ether
Mmhm
scarlet drum
nvm figured it out
Answer was -19. Thanks!
lone heartBOT
Answer was -19. Thanks!
scarlet drum
.close
shell cedar
Quick question, how do i get a list of numbers quickly (no brute force) where no number subtracted from another number in the list equals a number in the list
marsh rapids
You should add constraints to that question
shell cedar
i mean like for example:
marsh rapids
Arbitrarily large sets ? But with elements not too large ?
shell cedar
As small numbers as possible
but apart from that any number. preferably under 255 but larger is possible too
tardy stag
that's equivalent to saying the sum of any two numbers in the list is not in the list btw
marsh rapids
If you say 255 you have a target number of elements in mind
tardy stag
that might be easier to work with
shell cedar
that might be easier to work with
true lol
marsh rapids
A simple case is if min+min > max
shell cedar
not really, as long as possible would be nice. but again prefably with values under 255
what would max be in this example?
255?
marsh rapids
{128, ..., 255} trivially works
lone heartBOT
Available help channel!
hidden surge
I need to solve for x, sorry for bad quality
lone heartBOT
I need to solve for x, sorry for bad quality
split mantle
What is something special with a triangle that has two equivalent sides?
hidden surge
i don’t know im sorry
split mantle
My bad, I forgot to add the "w".
But, when it comes to a triangle with two equivalent sides, the angles that they don't intercept, are the same.
For instance, $$\angle CBD = \angle BDC = x$$
ocean sealBOT
very3good
lone heartBOT
@hidden surge Has your question been resolved?
hidden surge
My bad, I forgot to add the "w".
But, when it comes to a triangle with two equ...
okay i see it, then?
split mantle
That's the principle you'll be using.
alpine sable
Consider a set of polynomials $P(x) = \sum_{k=0}^{n} a_kx^k$ where $a_k \in \mathbb{R}$ for $0 \leq k \leq n$. Let $P(x)$ have $n$ distinct real roots $x_1, x_2, \ldots, xn$. Prove that the minimum value of $\sum{i=1}^{n} \left(\frac{1}{xi}\right)^2$ is attained when $P(x)$ is the unique polynomial $P(x) = \sum{k=0}^{n} \frac{(-1)^{k+1}}{x^k}$.
ocean sealBOT
scientia0623
gusty gorge
$\sum\limits_{i=1}^{n} \left(\frac{1}{x_i}\right)^2$
$P(x) = \sum\limits_{k=0}^{n} \frac{(-1)^{k+1}}{x^k}$
ocean sealBOT
Saccharine
plain summit
Whats stopping you from making all the roots really large
ocean sealBOT
Saccharine
gusty gorge
I'm trying to fix the LaTeX to understand the question
P(x) isn't a polynomial and I'm not sure
what's going on
but yeah
vale crag
Consider a set of polynomials $P(x) = \sum_{k=0}^{n} a_kx^k$ where $a_k \in \mathbb{R}$ for $0 \leq k \leq n$. Let $P(x)$ have $n$ distinct real roots $x_1, x_2, \ldots, x_n$. Prove that the minimum value of $\sum_{i=1}^{n} \left(\frac{1}{x_i}\right)^2$ is attained when $P(x)$ is the unique polynomial $P(x) = \sum_{k=0}^{n} \frac{(-1)^{k+1}}{x^k}$.
ocean sealBOT
_aplatypus
gusty gorge
but the unique "polynomial" isn't a polynomial is it?
desert island
am i using the alternating series test correctly?
lone heartBOT
am i using the alternating series test correctly?
last tendon
Seems fine to me
desert island
Seems fine to me
can u also plug in n+1 and compare it to the original series?
last tendon
I'm not sure I understand what your proposed strategy is there
desert island
This is my note from class
does taking the derivative and substituting n+1 both show if the the series is decreasing?
last tendon
The idea is that f(x) agrees with f(n) for all n in the set {0,1,2,....}. By considering f(x) and taking the derivative, you are showing it is decreasing for all values of x, this includes all values of n. There is really no need to do anything like plugging in n+1.
If you wanted to beat it to an inch of a dead horse, you can write that since f(x) is decreasing for all x in R^+, f(x0) < f(x1) whenever x0 < x1 and x0, x1 in R^+. In particular, this means that f(n) < f(n+1) for all n. i.e., it satisfies the first hypothesis of the alternating series test.
last tendon
@lofty flare I misunderstood your question, but yes those are two different techniques. You can just take the derivative in the case of 1/n or 1/sqrt(n) instead if you wanted.
last tendon
.close
lone heartBOT
lone heartBOT
Available help channel!
azure bison
If I have a subspace x - 3y = 0 with z = 0, how can I find orthonormal bases for this?
azure bison
So I have to find two bases, as far as I understand they need to be orthogonal and have their vectors be 1
The solution for one of the bases is that we take (3y, y, 0) = y(3, 1, 0) so an orthonormal base is (3/sqrt(10), 1/sqrt(10), 0)
but I dont understand what happened
azure bison
correct
vale crag
yeah so given these equations you can simplify quite a lot what your vectors look like
first we have z = 0
so our subspace is {(x, y, 0) | x-3y = 0}
and now x-3y = 0 means that x=3y
so our subspace is {(3y, y, 0) | y is a real number}
that's where the (3y, y, 0) comes from
azure bison
Ah I see
vale crag
so our subspace is {(3y, y, 0) | y is a real number}
it's a line in R^3, 1 dimensional subspace
azure bison
And how is this used?
vale crag
well then it's very easy to get a basis of that
exactly what they did in your correction
each vector in this subspace can be written as some multiple of (3,1,0)
azure bison
True
vale crag
and ofc this way of writing each vector is unique
so { (3, 1, 0) } is a basis of that subspace
azure bison
Right, thats a base but not orthonormal yet right?
azure bison
I cant recall what I had to do
vale crag
what does it mean for a set of vectors to be orthogonal ?
azure bison
I just remember that I could check if two vectors were orthogonal by multiplying them and summing and if its 0 they are
vale crag
yeah that works
azure bison
that they are perpendicular, or that one is 90 degree from the other
vale crag
there's only one vector though here
azure bison
ye
vale crag
there's no 2 different vectors to dot prod here
so there's nothing to check
and now there's the normal part
azure bison
so I just skip the orthogonal part?
vale crag
so I just skip the orthogonal part?
it's trivially true
you only have one vector in the basis
azure bison
I think I understand
if there were two vectors then I would have to do something else I suppose
vale crag
yeah you'd have to check their dot prod is 0 indeed
azure bison
So for the normal part, I just have to see if they are normal by dot producting itself and see if its 1
and if its not, then divide the vector by its norm
I think
vale crag
yes
azure bison
the resulting vector should be an orthonormal base
I see, but then, how could I get another one?
just multiplying the base by whatever?
vale crag
nah
multiplying
but not by whatever
do you know what the homogeneity of the norm is ?
(how the norm behaves when you scale a vector)
azure bison
I think so
vale crag
yeah
azure bison
But im unsure how I use that here?
vale crag
well you need your norm to still be equal to 1
otherwise your 2nd basis will not be orthonormal
let's suppose we have a vector v with norm(v) = 1
what would norm(k*v) be ?
(k is a real number)
vale crag
nah
not exactly
well i'll be spoiling you the answer here
if you scale a vector by -1, does its norm change ?
azure bison
not really, would just flip its direction no?
vale crag
yeah
so if you flip the vector in your basis
it still has norm 1
so your basis is still orthonormal
azure bison
So I can just flip my original basis? the one with (3, 1, 0)?
Just for reference I'll send the solution I'm looking at, it's in spanish but you might be able to understand what I'm looking at
azure bison
sorry Ill send the full thing
yes
So it wants me to find two orthonormal basis B1 and B2 for V, and its orthogonal complement(?), then it says that with these basis obtain an orthonormal basis B of R3
azure bison
Oh I see damn im dumb, so its 2, one for each respectively
I still don't really understand what I'm looking at then though XD
I understood the orthonormal part, but the other I dont really get
Cl I think means linear combination just in case
azure bison
yeah so the perpendicular equivalent to the point of the vector no?
I think, I remember it was something like that with projections
vale crag
well it has to do a bit with projections
azure bison
the orthogonal complement would be the vector that goes down from the ending of the projection to the plane
vale crag
but here we're talking about a whole vector space being ^perp'ed
vale crag
the orthogonal complement would be the vector that goes down from the ending of ...
yea well here it's the orthogonal complement of V
ie the vectors which are orthogonal to everything in V
azure bison
its kinda hard to visualize but I think I understand yes
vale crag
V^perp = {x in R^3 | <x, v> = 0 for all v in V}
so yeah anyway it's sufficient to find the x's that are orthogonal to a basis of V
it's your typical "if it works for a basis, it works for the whole space" kinda deal you find everywhere in lin alg
azure bison
I see, I just dont understand why this is so hard to grasp for me
At least the solution kinda is
Like I dont get where 1, 0, 1 is coming from and whats happening with it afterwards
vale crag
Like I dont get where 1, 0, 1 is coming from and whats happening with it afterwa...
yea it seems like they typo'd here
they meant the 3, 1, 0 we found earlier
vale crag
so yeah anyway it's sufficient to find the x's that are orthogonal to a basis of...
so yeah what they're doing is exactly what I suggest
finding the vectors (x, y, z) such that <(3,1,0), (x,y,z)> = 0
azure bison
I see thank you so much, just one last thing, to get an orthonormal basis of R3 using these two found basis what must be done?
Also sorry for the maybe confusing questions I just don't know how some things translate directly
vale crag
they just combined them
you have an orthonormal basis of V and one of V^perp
so the 3 vectors they found have norm 1
the vectors in the basis for V are orthogonal to each other
azure bison
Oh damn I think I see what happened
vale crag
the vectors in the basis for V^perp are orthogonal to each other
and all the vectors in V^perp are orthogonal to the vectors in V
it's also true for the basis vectors we found
azure bison
1,-3, 0 isn't normal so they had to normalize it
same with the other one but it would be divided by 1 so its stays the same
I dont know why I suck so much at this even with the solution in my face XD, I appreciate this server so much
Thank you very much @vale crag , you'll probably see me around later lmfao
vale crag
I dont know why I suck so much at this even with the solution in my face XD, I a...
takes a bit of time to get used to the language of lin alg yea
azure bison
takes a bit of time to get used to the language of lin alg yea
I appreciate you, I have other things I was unsure about but can't remember, I'll post them later if I stumble upon em
thanks a lot
oh btw, just one last thing @vale crag, if they ask me to calculate [(1, 0, -1)}_B, I just have to make the dot product of this against all of the basis vectors that I just found?
vale crag
yeah
azure bison
What was this called again? the coordinates?
vale crag
you have an orthonormal basis so it's very easy to find the coords in basis B
it's not the case for a general basis
azure bison
I cant recall if in general it was something like this:
but I dont have any variables in this case so Im unsure
azure bison
wdym no variables ?
like to change the values of the variables I have with the coordinates im given
azure bison
not sure what you mean still
No problem I don't think I do either xD
I'll try to figure it out
Thanks again 🙏🏼
.close
wet sorrel
lone heartBOT
wet sorrel
hey guys i'm trying to find a closed form for this expression.
so the second one with even powers of jw i was able to simplify it down to a geometric series
so i was thinking i could do the same for the other expression using geometric series but i'm having a hard time seeing it.
this is where i'm at atm:
is it even possible to make this into a nice clean form with geometric series? or is there something else that i can do that i'm not aware of. thanks
vale crag
it's pretty much the same as the second one
wet sorrel
i figured but i wasn't seeing it so i wanted to get confirmation on my guess
vale crag
if you factor 1/2 e^jw it's easier to see yea
ah yeah nvm you already did it, didn't look at the 2nd pic
I guess we're done here lol
vale crag
$$\frac12 e^{j\omega}\sum_{k=0}^{\infty} \left(\frac14 e^{2j\omega}\right)^k$$
ocean sealBOT
_aplatypus
vale crag
@wet sorrel if we want to be formal
wet sorrel
ah that makes it easy to see now. thank you
silent crane
how do you find the max/min points for a trig function?
lone heartBOT
how do you find the max/min points for a trig function?
silent crane
what i've done so far is find the derivative
y = sin2x - cosx - 4x
y' = 2(cos2x) + sinx - 4
carmine reef
you could use a double angle identity next
silent crane
whats the identity? also are you expected to memorize these identities
also if it helps, the example shows how they went about finding critical numbers
queen harbor
what i've done so far is find the derivative
y = sin2x - cosx - 4x
y' = 2(cos2x...
Look a little closer and notice that y' can never reach 0
carmine reef
ah, that works
queen harbor
Since the interval is open there are no maxima or minima
silent crane
Look a little closer and notice that y' can never reach 0
what are you seeing that tells you this
queen harbor
Maybe it's just poorly written
queen harbor
what are you seeing that tells you this
You have a -4 at the end and then three trigs
2*cosine of something and a sine
cosine and sine are at most 1 and at least -1
so anything you do with 2*cos(s) + sin(r) will always be between -3 and 3
no matter the s or r
silent crane
wow
youre like. brilliant. tf. ive been trying to factor this all day and you got it at first glance
how do i express that there is no minimum value
the graph
queen harbor
normally when you have a closed bounded domain [0, 2pi] the next step is to evaluate f in the bounds of the domain and choose the greatest value
silent crane
i thought i was just really stupid for not seeing the max/min
silent crane
normally when you have a closed bounded domain [0, 2pi] the next step is to eval...
how?
queen harbor
how do i express that there is no minimum value
The usual way to go about these questions is:
1- Find all points where derivative is 0 or not defined and evaluate your function there. Save the results.
2- Evaluate your function in the bounds of the domain whenever it is a closed domain and save the results.
3- Pick the greatest, lowest value from the results from before.
So we found that first step returns nothing because derivative is always nonzero
Let's just pretend for a moment that the question asks us to optimize for 0 <= x <= 2pi
so second step would be acually evaluating at 0 and 2pi
And those will be your maximum and minimum
silent crane
The usual way to go about these questions is:
1- Find all points where derivativ...
this might be a stupid question but what is a closed domain? youre referring to 0 < x < 2pi right?
rigid prawn
what i've done so far is find the derivative
y = sin2x - cosx - 4x
y' = 2(cos2x...
y'= 2(cos2x-1) +(sinx-1) - 1 so the derivative is always negative since the parenthesis are always negative or 0
queen harbor
the way it is written suggests that the borders are not included in the domain. It states that x must be strictly less than 2pi and strictly greater than 0. I think that is not what it meant to say though.
silent crane
im looking at the derivative and both minimums occur between 0 and 2pi, do you think it was intentional?
also i know that these questions r really annoying but im so dumb, bear with me
silent crane
The usual way to go about these questions is:
1- Find all points where derivativ...
what did you mean by evaluate the function?
queen harbor
evaluate function at a given value: plug that value in your function
silent crane
the steps we take to find the max/min is through a second derivative test or through intervals, since the second derivative is never 0, should i look at the third derivative?
silent crane
evaluate function at a given value: plug that value in your function
i see! thank you for clarifying
queen harbor
im looking at the derivative and both minimums occur between 0 and 2pi, do you t...
You wanted to find maxima and minima for your function, not its derivative
silent crane
y'= 2(cos2x-1) +(sinx-1) - 1 so the derivative is always negative since the pare...
youre right, the whole derivative is below 0. i didnt know you could rearrange the function to show parenthesis like that
silent crane
You wanted to find maxima and minima for your function, not its derivative
yes
i thought you needed the derivative to find max/mins
queen harbor
Yes, but it's not necessary to find maxima and minima of the derivative
We are only interested in its zeroes
queen harbor
the steps we take to find the max/min is through a second derivative test or thr...
It's not necessary here because the domain is closed and bounded. It would be a different story if it asked us to find maxima and minima over all the reals.
silent crane
for the entire unit we've been using derivatives to find max/min points so im completely drawing blanks on how you do that without derivatives
when i sub in 0 or 2pi wouldnt it give me the two maximums?
how would i find the minimums?
queen harbor
im looking at the derivative and both minimums occur between 0 and 2pi, do you t...
blue curve here is the derivative of your function
we only want its zeroes, which it doesn't have
so just ignore that part. What it's saying it's that the function won't change direction
it will always go up or always go down
silent crane
right
queen harbor
Also, this is what the graph should look like
Very strong emphasis in the fact that they asked us to optimize over a bounded domain (and I'm very sure that the borders should be included, so it's also closed)
So the strategy is kind of brute forcing it
Look at all the points where it might change direction (a.k.a. critical points, the points at which the derivative is zero or undefined)
And also look at the borders
queen harbor
Also, this is what the graph should look like
Looking at this it should be clear why it's necessary to look at the borders
Because that's where the highest and lowest points are
When I say "look at [something]" I mean plug that value in your function and save it
at the very end pick the highest and lowest values of that list
silent crane
oh god that makes so much more sense, thank you for clarifying that. i was looking at the derivative thinking damn, why do the borders matter again? but completely forgot to look at the original function
do these points have any significance
im still confused on what teach wants me to do tbh
queen harbor
Not for what they asked us to do
silent crane
hypothetically
if you were me
how would you answer this question
this is hypothetical so even if its wrong you arent responsible for my mental failings at all
queen harbor
Those are the points in which the derivative reaches maxima and/or minima. Those are the inflection points
They are determined by the changes of concavity of your function
The zeroes of the first derivative determine where your function changes its direction (up or down)
And the zeroes of the second derivative determine where your function changes its "acceleration"
tardy stag
inflection points are where your function starts putting on the brakes / stepping on the gas
silent crane
i see! thank you both
even if this question is unsolvable i learned a lot from chatting with you guys lmao
cant even ask my teacher anything because it's all asynchronous and online
tldr: there is no "max/min" because the derivative has no zeroes so the function doesn't change direction, so instead we brute force it and plug in 0 and 2pi and say and (0, y) is the max and (2pi, y) is the min?
queen harbor
It's brute-forcing from the very beggining, when we notice that the domain is closed and bounded. Evaluate at all critical points (see here? not ignoring the derivative) and borders and pick the highest or lowest from that list.
tardy stag
there are no local maxima/minima because the derivative has no zeros, so the function doesn't change direction, so we just look at the endpoints (we'd have to do this anyway)
silent crane
It's brute-forcing from the very beggining, when we notice that the domain is cl...
what do critical points include? i know i need to evaluate the endpoint/borders but is there anything else i'm missing
queen harbor
Points where derivative is 0 or undefined
silent crane
OHH right
alright thank you!!
is there a formal way to thank you like through a point system?
queen harbor
Consider this as an example for undefined derivative. x=0 is a minimum and I will notice that even without the graph because f' is not defined at 0, so I will include that in my list.
queen harbor
is there a formal way to thank you like through a point system?
I'm sure there isn't
silent crane
Consider this as an example for undefined derivative. x=0 is a minimum and I wil...
i see^^ thank you for showing an example of an undefined minimum
silent crane
I'm sure there isn't
oof that sucks. you guys need more appreciation. or maybe youre insane and actually enjoy math
queen harbor
I'll go with the second one 😅
azure bison
does anybody know why these are the basis?
lone heartBOT
does anybody know why these are the basis?
queen harbor
Parece que están al revés, no?
vague coral
eigenvectors of a matrix form a basis since they are linearly independant I guess
azure bison
eigenvectors of a matrix form a basis since they are linearly independant I gues...
I dont get it tho
How do I get (1, 0, 0) from that resulting matrix
or those two from the second one
queen harbor
How do I get (1, 0, 0) from that resulting matrix
It's a basis for the kernel of that matrix
vague coral
^ basically
azure bison
what was that again?
vague coral
$\operatorname{Ker} (A-\lambda I_3) = { v \in E | Av = \lambda v }$
ocean sealBOT
herels
vague coral
but in general, the kernel of a linear application f (or a matrix A) is the set that contains all the vectors v such that f(v) = 0 (or Av = 0)
azure bison
I think I understand now thanks a lot
Just another question
If I have this
can I make I try to make it diagonal and THEN subtract lambda from the diagonal?
I feel like that would make the determinant calculation easier
real gazelle
Sorry what's the question, are you trying to find the eigenvalues of this?
azure bison
yea
real gazelle
What do you mean by try to make it diagonal, do you mean like row reduction?
real gazelle
Oh okay, that won't work because row reduction changes the eigenvalues/eigenvectors of a matrix
azure bison
I see
real gazelle
I think the easiest way is to calculate det(lambda I - A) unfortunately
You'll have to expand it out but it won't be terrible because it's a 3x3 matrix
jagged pagoda
i'm back with some more trigonometry!
c = sqrt( a^2 + b^2 - 2ab * cos(C) )
cos(C) = ( a^2 + c^2 - b^2 ) / 2ac
i'm trying to make these two into a single simplified function... for an hour now. i literally forgot all maths and had to relearn on the fly lmfao
i started with
C = acos ( ( a^2 + c^2 - b^2 ) / 2ac)
i reduced c and got
C = acos ( a/2c + c/2a - (b^2 / 2ac) )
i tried mixing everything back together and ended up with
C = acos ( 2(a^3)(b^3)-2b^2 / 3(a^2)(c^2) )
i started going insane, i didnt know where i was going with this or if it was even correct up to this point. i tried substituting c for sqrt( a^2 + b^2 - 2ab * cos(C) ) so maybe the a's and b's would cancel out or something and got this:
C = acos (2(a^3)*(a^2+b^2-2ab*cos(C))^2-2b^2 / 3a^2*(a^2+b^2-2ab*cos(C))
jagged pagoda
i literally dont know what im doing anymore so i stopped to ask for help here
am i on the right track?
proven leaf
i'm back with some more trigonometry!
c = sqrt( a^2 + b^2 - 2ab * cos(C) )
cos(...
is your final end goal to isolate big C?
proven leaf
$b^2=a^2+c^2-2ac\cos\beta\implies\cos\beta=\frac{a^2+c^2-b^2}{2ac}\implies\beta=\arccos\left(\frac{a^2+c^2-b^@}{2ac}\right)$
that's what you got so far by the looks of it, right?
proven leaf
dw that's fine :)
jagged pagoda
but yeah thats what i started with
im trying to turn it into something potentially useful if possible
ok one sec
proven leaf
im trying to turn it into something potentially useful if possible
so you're saying that given the sides of a triangle
you want a formula to compute one of the angles
jagged pagoda
why is it b^2-(a^2+c^2) and not (a^2+c^2)-b^2?
do they switch places when cos turns into arccos?
proven leaf
why is it b^2-(a^2+c^2) and not (a^2+c^2)-b^2?
I subtracted (a^2+c^2) to the b^2 side
so I could isolate 2ac*cos(B) :)
lol that should be right now
ocean sealBOT
XxMrFancyu2xX
proven leaf
ugh there's an @ sign in the exponent, supposed to be a 2 you get the idea 
jagged pagoda
so from what i understand, if cos() goes on the other side it becomes an arccos. is this everything that happens or does more stuff change?
jagged pagoda
so from what i understand, if cos() goes on the other side it becomes an arccos....
^
B = arccos ( c^2 + a^2 − b^2 ) / 2ca
proven leaf
so from what i understand, if cos() goes on the other side it becomes an arccos....
what is the range of arccos(x)?
jagged pagoda
what is the range of arccos(x)?
wdym range?
proven leaf
essentially what are the possible values it can output?
jagged pagoda
btw i know barely anything about trigonometry, let alone cosines and stuff
jagged pagoda
essentially what are the possible values it can output?
no idea
proven leaf
alright that's fine, but I'll tell you the range of arccos is from 0 to pi
think about it this way arccos(1)=0
and arccos(-1)=pi
jagged pagoda
but we know
- the value of a,
- the value of b,
- the value of cos(C)
- and that c = sqrt( a^2 + b^2 - 2ab * cos(C) )
proven leaf
so from what i understand, if cos() goes on the other side it becomes an arccos....
this is your main question, right? Does anything change when taking arccos?
jagged pagoda
no, just a side question
proven leaf
oh LOL 
I was trying to answer that im sorry
jagged pagoda
the main question is whether i can simplify the stuff
jagged pagoda
oh LOL <:KEK:586240877358350341> I was trying to answer that im sorry
xD
proven leaf
,w arccos(((a^2+c^2)-b^2)/(2ac))
ocean sealBOT
proven leaf
appears not :) and ig that makes sense
vapid shuttle
lone heartBOT
vapid shuttle
Trying to find all x where this converges absolutely
I did the ratio test
and this is what I am getting
vapid shuttle
hm?
the final limit I am getting is
$\lim_{n \to \infty} x^{(n+1)} \cdot n + x^{(n+1)}$
as n-> infinity
ocean sealBOT
austinu
tardy stag
I'd be somewhat surprised if this isn't |x|<1
fallen verge
!show please
lone heartBOT
Show your work, and if possible, explain where you are stuck.
fallen verge
Should be 0 RoC
vapid shuttle
$|\frac{(n+1)!x^{(n+1)!}}{n!x^{n!}}|=|(n+1)x^{(n+1)}|$
ocean sealBOT
austinu
vapid shuttle
?
fallen verge
Exponents sjbtract when you divide
^^
fallen verge
(n+1)!-(n)! != n+1
worn fox
!=
vapid shuttle
does not equal
fallen verge
!= as is not equal lol
worn fox
unfortunate doubling up of exclamation marks lol
fallen verge
couldve gona with =/= tbf
torn elk
≠ use this from now on xd
wary stream
!= as is not equal lol
~=
ocean sealBOT
austinu
vapid shuttle
better?
vapid shuttle
my b
fallen verge
All good
vapid shuttle
and then of course we are taking lim n to infinity
and we want to find when absolute value of this is less than 1
I mean if x>1 this has to diverge right
no chance it doesn't
fallen verge
That parts the easy part
fallen verge
I think so
vapid shuttle
x<-1 then I'd say it still diverges
because it will alternate
so thus far |x|<1
yea?
fallen verge
Wdym it will alternate
fallen verge
Yeah
vapid shuttle
then it will alternate
fallen verge
Alternate and grow absolutely
worn fox
im not even sure if the ratio test can be applied here bc its not in the form a_n(x-a)^n ?
vapid shuttle
we haven't shown that that necessarily converges yet
fallen verge
(n+1) grows linearly while
x^(n!×n) decays exponentially
vapid shuttle
oh also we can't have x=0
worn fox
oh also we can't have x=0
?
vapid shuttle
well because then ratio test wouldn't apply
but apparenty it just never applies to this anyways
lol
real gazelle
when x=0 it trivially converges
worn fox
right right, and that should send alarm bells bc it definitely converges for x = 0
real gazelle
It's not that it sounds alarm bells
The ratio test never tells you that 0+0+0+0+... converges
keen plinth
whats wrong with ratio test on successive nonzero terms
vapid shuttle
yeah the terms have to be all nonzero
keen plinth
we arent doing ratio test on the zero terms
vapid shuttle
for ratio test
real gazelle
Oh, fair
keen plinth
those terms dont really exist here
worn fox
point being if trying to apply the ratio test tells you x=0 does not converge then you probably shouldnt be applying the ratio test
real gazelle
That's not right, the ratio test is never applicable when x=0
vapid shuttle
point being if trying to apply the ratio test tells you x=0 does not converge th...
no, it just tells me I can't apply the ratio test when x=0. I see what you mean though
real gazelle
That doesn't mean it's not useful