rocky wraith
Never mind I think I can just do basic manipulation - we can close this now, thanks for the help anyways!
#help-28
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uncut marlin
Need help with linear equations
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Need help with linear equations
uncut marlin
keen vector
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uncut marlin
<@&286206848099549185> can anyone help me
placid sapphire
YE
uncut marlin
mgm
mhm
how do you solve this
i made the right fraction postive because it had two negatives
so it would be -5/8 + 1/6
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uncut marlin
@floral scrollcan someone just guide through this whole thing
gritty rose
so it would be -5/8 + 1/6
Find a common denominator
uncut marlin
alredy did
gritty rose
So what are you stuck on
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torpid perch
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torpid perch
f(1,1,1,1)=0
f(3,1,2,2) = (1,5,1,0)
f(0,0,1,0)=(1,1,1,1)
f(0,0,0,1)=(1,5,1,0)
,w det {{1,1,1,1},{3,1,2,2},{0,0,1,0},{0,0,0,1}}
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torpid perch
,w rank {{1,1,1,1},{1,5,1,0},{3,1,2,2}}
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torpid perch
.solved
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novel lava
can someone show me how they'd draw the rectangles for part a)
torn jolt
ocean cedar
The red rectangles are the lower ones
The green ones are the upper ones
You can distinguish the upper and the lower based on their position with the graph of your function
novel lava
hmm
ocean cedar
Although I wonder what people would think is the right one when the function is decreasing ...
Here, I think the exercise has been loose on the terms, in a way that they've let themwelves write "lower" because the top of the rectangles are just below the curve, and "upper" because the top of the rectangles are above the curve
novel lava
hm
i see
in question b
it says
of equal width
but isn't it already supposed to be of equal width?
novel lava
no like in part a
novel lava
yes but it doesn't specifically mention that in part a
but it then says "equal width" in part b
ocean cedar
Yeah probably forgot about it
novel lava
k
ocean cedar
Or else there would be different possible bounds than the ones asked to find
For that area
novel lava
how would you answer this
ocean cedar
Look at how the area of the rectangles bound the value of your integral
novel lava
yeah they get closer
but I need to give a specific quantitative value
but idk how they get this
ocean cedar
Yeah I don't know either without actually knowing the theory around integration
obsidian cedar
Please help, what is the value of the yellow shaded area? Is it twice as large as angle A (x1 + x2) or x2?
Here we have a triangle ABC inscribed in a circle, and ABEC is a quadrilateral. x1 = x2 and y2 = y1, because AE is the bisector. Triangles ABE and ACE are similar.
Now, I want to examine quadrilateral OBED and prove that a circle can be circumscribed around it. To do this, the sum of angle O and angle E (y2) should equal 180 degrees
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rotund ibex
Could someone check my solution to this problem? Everything is in the screenshot. Thanks!
gritty rose
,w solve y''-y' = x e^x
glossy valveBOT
gritty rose
looks right
rotund ibex
Thanks.
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torn jolt
how do u know when to switch the direction of a vector
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how do u know when to switch the direction of a vector
torn jolt
@ancient folio
torn jolt
Use this instead
how does this one work?
shell adder
Can someone help me with geometry
torn jolt
Can someone help me with geometry
make a channel
shell adder
How
ancient folio
Kinda
torn jolt
is it not the tail to tip method
torn jolt
my question is why did they make the u vector the opposite direction
ancient folio
I mean when u actually do that math, its the same thing
ancient folio
my question is why did they make the u vector the opposite direction
huh?
U mean it points downward?
torn jolt
how do u know when to switch the direction of a vector
look at the bottom
u is pointing down
ancient folio
U mean it points downward?
Well
Its all about the direction that u choose
U can see the problem original coord system
torn jolt
yes
ancient folio
It doesnt provide which direction is positive/negative
Since the arrows are appear on both side
So, it basically tell u that you can choose your own
torn jolt
oh, interesting
ancient folio
In the example, ur teacher choose downward as positive
But u can choose the other side
Usually, the upward is positive
But its just convention/standard
ancient folio
Different countries/region use different standard
ancient folio
But when u convert them into math then consider the dir, u still get the same result
torn jolt
why did they subtract 90 from 180
ancient folio
Look at this triangle
torn jolt
that’s not a 90 degree tho
ancient folio
They're trying to get the green angle
Its 180 - 2 others angle
35 is one of them
The other one is made up of 90 and 15
torn jolt
oh
ancient folio
So its 180 - 35 - (90 + 15)
ancient folio
Well
By the way they writing it
They probably trying to get the angle using something else and not the triangle
torn jolt
oh
ancient folio
if i had to guess it probably this
The total angle of 1 side of a line = 180
The green angle = 180 - 90 - 15 - 35
ancient folio
But its a bit overcomplicated though, u could just use the triangle
torn jolt
But its a bit overcomplicated though, u could just use the triangle
could u check if i did this one right when ur available
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buoyant hinge
can someone explain the last line?
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can someone explain the last line?
fast peak
what do you not understand about it
buoyant hinge
what is it trying to say?
i mean i kind of get what it says i just don't understand how
does it just say that every T in L(V, W) has a unique A = T[B_1, B_2]
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void shore
I have a qusetion
Why can't we use Cauchy integrals in the ellipse? Isn't it considered a closed curve?
charred raft
I have a qusetion
Why can't we use Cauchy integrals in the ellipse? Isn't it con...
depends on holomorphicity of the function inside the ellipse
ya abuuudeeeh
void shore
One sec I'm sending the image 😅
charred raft
okeh
void shore
charred raft
,rccw
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void shore
Here, we cannot say zero because there are no singular points?
Here too cant we just say 0 ?
charred raft
,rccw
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charred raft
ur aware ur function isnt holomorphic anywhere @void shore ?
void shore
ur aware ur function isnt holomorphic anywhere <@920328827786387517> ?
how do I know if its or not ? Cauchy-Riemann conditions?
void shore
oki 😁
charred raft
let me know if there is good competitive programming question kekw good luck
void shore
its in my uni I have an exam
charred raft
sheeesh, u been slackin off, whats the other question?
look if ur function isnt holomorphic over the connected region, dont use cauchy-integral ur violating the first condition
charred raft
sure go ahead
void shore
In real analysis, there are constants, meaning a function is differentiable if it is a polynomial or something similar. Are there constants in complex analysis that allow me to immediately conclude that a function is analytic?
I mean holomorphic
Here for example how did we instantly know that the function is holomorphic ?
charred raft
because the singularities lies inside the contour @void shore
charred raft
In real analysis, there are constants, meaning a function is differentiable if i...
practice alot
charred raft
bet bet free syria!!
void shore
because the singularities lies inside the contour <@920328827786387517>
we divided two holomorphic functions by each other, and the singularities lie inside the contour. Therefore, the resulting function is holomorphic everywhere on the contour except at the points where the singularities occur.
got it ❤️
charred raft
we divided two holomorphic functions by each other, and the singularities lie in...
correct
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halcyon canopy
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halcyon canopy
Quick question is the zero there because of the free variable?
frozen fiber
whatever the question is lol
sand escarp
which zero and where
halcyon canopy
the small green arrow xd
halcyon canopy
thanks
it just felt wierd to just forward the zero
but i guess its a bit redundant to have a forth matrix that is multiplied by zero?
or is there a more interresting take?
sand escarp
I usually just think of it as any number multiplied by 0 is equal to 0
sand escarp
np!
halcyon canopy
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prisma cradle
variance?
cause range is independent of freq
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variance?
cause range is independent of freq
prisma cradle
ping if replied
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strong phoenix
variance?
cause range is independent of freq
250th? bro what test are you doing
prisma cradle
its a question bank
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craggy tapir
$X\oplus\ker T = Y\oplus\ker S$
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$X\oplus\ker T = Y\oplus\ker S$
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craggy tapir
if $\ker S\subseteq\ker T$, does this tell us anything about the relationship between $X$ and $Y$?
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torn jolt
Hi, can I ask you guys something?
hollow herald
Did you try something with this?
craggy tapir
i'm not sure how should i verify this for now
hollow herald
You can start with applying the definition always. The space is decomposed in two different ways: \ $x + k_T = y + k_S , x \in X, y \in Y, k_T \in ker T, k_S \in ker S$
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craggy tapir
but it's a direct sum. each vector in V can only be written in one way using x and kT. likewise for y and kS
if kT is in ker(T) but not in ker(S), then x = kS and y = kT
i think?
but the vector only X and kerS share is the zero vector
so in the case of kT in ker(T) but not in ker(S), v must be in Y cap ker T
original problem btw
wait i just had an idea
hollow herald
The forward direction is trivial. Are you stuck with the backward direction?
Construct a linear map E: S(V) -> W as follows ||- For each w in S(V), pick v in V with S(v) = W. Let E(w) = T(v).|| Show that this is well defined and extend it.
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topaz valley
$X\oplus\ker T = Y\oplus\ker S$
i was gonna feed you the next part of the proof
topaz valley
you're supposed to define T' : X -> W, T'(x) = T(x), and apply your lemma again to T'
to get X = ker(T') (+) Y and V = ker(S) (+) ker(T') (+) Y
craggy tapir
.reopen
topaz valley
then prove that ker(T) = ker(S) (+) ker(T')
topaz valley
The forward direction is trivial. Are you stuck with the backward direction?
no, the forward direction is the highly non-trivial one
topaz valley
i'm not on my pc rn, i'll read this later
what a coincidence, neither am i
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hollow herald
no, the forward direction is the highly non-trivial one
Sv = 0 => Tv = ESv = E(0) = 0?
topaz valley
that's the backwards direction
topaz valley
no
hollow herald
Don't you first show the "if" and then the "only if" ?
topaz valley
that's not what people mean when they say "forward"
hollow herald
nomenclature, alr got it 
topaz valley
A <=> B, A => B is forward
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buoyant hinge
and every A has a unique T
cool, thank you 👍
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cool, thank you 👍
buoyant hinge
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smoky cipher
how do i do this
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how do i do this
torn jolt
plug y
gritty flax
You can sub the value of y
smoky cipher
oh yeah
thanks
wait a seond
ah
wait bro
how do i do this
plug y?
nvm got it
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thorny shuttle
what does it mean for it to be that
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what does it mean for it to be that
thorny shuttle
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viral wraith
question : show that if a decreasing sequence converges to zero, it is positive
viral wraith
hey
silver topaz
question : show that if a decreasing sequence converges to zero, it is positive
what would happen if there was a single negative term?
hint: ||all successive terms will be smaller than or equal to that negative number, which in turn is smaller than 0||
viral wraith
thnks bro
woven grove
What are some reasons i would get a worse approximation using Gauss hermite integration with more nodes? I have a task given to compute it for 9 given points and weights and 18 nodes and weights. Shouldn't the 18 nodes give a better approximation? I know the integral analytically computes to 1 but for 9 nodes its about 0.95 and for 18 nodes its 0.8 ish. Why is this?
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velvet sedge
how are you distributing your nodes
definitely runge phenomenon
try using chebyshev nodes
@woven grove
wait
woven grove
how are you distributing your nodes
Both nodes and weights are given in a table so i can't choose them
velvet sedge
yea
woven grove
Imma triple check so i didn't miss enter any weight or node to be super super sure
Already did it once
but the weights dont sum to sqrt(pi) which i saw online they should?
They do for the 9 nodes and weights
frail hound
haughty plover
well isn't it easier to just divide 29 by 1.4?
which results in 20.71 disc cases could fit in
so basically 20
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frail hound
so basically 20
yes i divided numbers but its bounds
i divided and still didbnt work
lemem put 20
wait i dont got ti open now
versed lily
let f : $\mathbb{R} \rightarrow \mathbb{R}$ be differentiable twice, bounded form above, and has a global minimum point. prove that there eixsts an x $\in \mathbb{R}$ such that f''(x) = 0
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jason2D
viral star
I have a problem I need help with
versed lily
this channel is taken
viral star
Where do I go to get help for a problem
versed lily
go to an empty help channel
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versed lily
<@&286206848099549185>
viral jasper
let f : $\mathbb{R} \rightarrow \mathbb{R}$ be differentiable twice, bounded for...
!status
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let f : $\mathbb{R} \rightarrow \mathbb{R}$ be differentiable twice, bounded for...
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
Ngl, i thought it wanted to show f'(x)=0, and was confused why this problem seemed so trivial
It just got more interesting
versed lily
interesting isnt the word id use lol
viral jasper
I'm guessing you have learned about mean value theorem. And i am thinking that is what should be used here
versed lily
i have
viral jasper
What can you say about f' and f" at the global min?
versed lily
one is 0 one is negative
viral jasper
Negative? You sure?
versed lily
isnt it?
viral jasper
$f(x)=x^2$ has a global min at $x=0$. What is f''(0)$?
glossy valveBOT
SWR
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
viral jasper
I guess, at a global min, it is also possible the f"=0, but then we're done, so let's ignore that case
It is also possible the derivative does not exist at a global min, but the function is twice differentiable, so we can rule that out
versed lily
yeah
viral jasper
Here's one idea using intermediate value theorem: $f">0$ somewhere. If we can show that $f"<0$ somewhere else, then $f"=0$ follows from IVT
glossy valveBOT
SWR
viral jasper
You can also use MVT if you can find two spots where f' would be equal
Essentially, they're very similar proofs, so take your pic
versed lily
yeah i thought of that but i coudnt find a point like that
viral jasper
this is where you need to use the upper boundedness of f
what class is this for btw? What level of rigor are they wanting?
versed lily
this is where you need to use the upper boundedness of f
im not sure how to use that
viral jasper
that's good, because I can only think of epsilon-delta limit definitions to prove this
versed lily
we dont know anything about the upper bound
versed lily
so how can i use that
viral jasper
There are two cases you need to consider
You know f''>0 somewhere. If f''<0, then IVT tells you that there is some f''=0 somewhere. That make sense?
versed lily
yes
what im doing now is i assumed that there isnt another extrema point and im trying to reacha contradiction
viral jasper
Okay, so now the task is to prove that $f''<0$ somewhere. To do this, we can try to use MVT. If we can find some $x_1, x_2\in\bR$ such that $x_1<x_2$ and $f'(x_2)>f'(x_1)$, then MVT will tell us that $f''(x)$ for some $x\in[x_1, x_2]$
glossy valveBOT
SWR
viral jasper
Oh. There's one other way to find $f''=0$ that I want to touch on, before we move forward
glossy valveBOT
SWR
viral jasper
We know $f'=0$ at the global min. Suppose $f'=0$ somewhere else, then MVT will tell you that there is some $f''=0$ somewhere inbetween. That make sense?
glossy valveBOT
SWR
versed lily
Okay, so now the task is to prove that $f''<0$ somewhere. To do this, we can try...
you mean f'(x2) < f'(x1)
viral jasper
you mean f'(x2) < f'(x1)
Oh you're right. Typo'd the inequality sign
viral jasper
yes
So, if f'=0 somewhere, we are done. So, now we're left with the final case where f'=0 nowhere else except at the global min
versed lily
yes
viral jasper
So, in our last case, here are the facts:
- f is twice differentiable everywhere
- f is bounded above
- f has a global minimum
- f''>0 at the global minimum
- f'=0 only for one specific x
And here is what we want to find
- f''<0 somewhere
If we find that, then the rest is ezpz. But finding f''<0 is not so ezpz.
Can you think of some kind of example function that has all of the above facts that I shared? And is there something you can see from that function which helps you see why f'' would be negative somewhere?
versed lily
a function where f' = 0 only at a single point?
that means the the function would have to be monotone increasing after the min and decreasing before the min
right so i have to reach a contradiction form that
but how
viral jasper
Ask yourself this one question:\
\
What is $\lim_{x\to\pm\infty}f(x)$?
glossy valveBOT
SWR
viral jasper
Anyway, I'll be back in like an hour
this one question should help you a lot I hope
versed lily
Anyway, I'll be back in like an hour
no problem thanks a lot!
versed lily
0
versed lily
why
if f has one extrema which is a global minimum
and f is bounded
then at +inf f approaches some M, and at -inf too
which means f' approaches 0
fringe silo
sin(x) ?
versed lily
what about it
fringe silo
well sin does have a global minimum, is bouded, but it's derivative cos has no limits
versed lily
it has more than 1 extrema
fringe silo
let f : $\mathbb{R} \rightarrow \mathbb{R}$ be differentiable twice, bounded form above, and has a global minimum point. prove that there eixsts an x $\in \mathbb{R}$ such that f''(x) = 0
glossy valveBOT
Lebrown
fringe silo
to prove it cant be unique
what ?
versed lily
thats the original question
we split it into 2 cases
one where it has more than 1 extrema point, which means theres differnet point at which f' is 0, so were done
and the other case is where there is 1 extrema point
fringe silo
oh kk
versed lily
and in this case we need a contradiction
fringe silo
why do you need a contradiction though ? Just need to find two points where f'(x) = f('y)
charred raft
well sin does have a global minimum, is bouded, but it's derivative cos has no l...
f(x)=sin(x) does not have a global minimum in the strict sense required by the problem. While sin(x) achieves its minimum value of −1 at infinitely many points, it does not settle into a single global minimum point because it oscillates indefinitely. Thus, f(x)=sin(x) does not satisfy the third assumption of the problem
charred raft
A global minimum requires the function to achieve its lowest value at a single point (or a finite set of points) in R.
versed lily
sorry, i was replying to lebrown
charred raft
okey just follow what SWR suggested
since f as has global minimum at c
then f'(c) = 0
and since its bounded, then f'(x) -> 0 as x-> plus/minus infty
so f grows from 0 at x=c to 0 again as x -> p/m infty
so it has to reach a local maximum/minmum somewhere , call it x0
then by IVT
we get f''(x0) =0
versed lily
oh my god i completely forgot we have a point where f' is 0
that sounds stupid now that i say it outloud
thanks
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viral jasper
so it has to reach a local maximum/minmum somewhere , call it x0
@versed lily @charred raft, the proof is actually a bit more subtle than this. It is not necessarily true the f has a global max. This is why I was asking how formal your proof had to be
charred raft
<@1263213304755650600> <@597465547650760723>, the proof is actually a bit more s...
f?
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charred raft
in the conditions they said f has global min
viral jasper
,w plot -e^(-x^2)
glossy valveBOT
versed lily
<@1263213304755650600> <@597465547650760723>, the proof is actually a bit more s...
yeah but the proof doesn't require that
viral jasper
in the conditions they said f has global min
yes, but a global max is not guaranteed
versed lily
f' is 0 at some point, and it approaches 0
viral jasper
so it has to reach a local maximum/minmum somewhere , call it x0
problem is that you were using this idea that there was a global max somewhere
versed lily
if its constant 0 were done
if its not, then theres a point where f' < 0
because it approaches 0 that means this point is a local min which mean f'' at this point is 0
charred raft
,w plot -e^(-x^2)
this function has a point where f'' = 0
so what are u trying to counter here?
viral jasper
this function has a point where f'' = 0
It has no global max.
charred raft
why would that be relevant
viral jasper
Your argument was that it will have a max somewhere, and therefore f''=0 somewhere by ivt
charred raft
min/max
one of them is guaranteed to exist
doesnt have to be both to make the proof fail, u see what i mean?
viral jasper
this logic doesn't feel rigorous enough
It's not wrong, but the rigor isn't too great
it's why I was asking jason what level of rigor he needed
charred raft
i never made a proof for someone in this server since i joined
im just sketching ideas
which apparently seems to some degree probabilistically correct to me
viral jasper
which apparently seems to some degree probabilistically correct to me
the idea is correct. It's a good idea for jason to use. But if I were his grader, I wouldn't give it full points
Not trying to tell you that you are wrong, just that the proof might need more substance
But that's based on what level of proving jason needs
charred raft
yeah for sure, we are just givin him hints, not spoon feeding him the solution, he has the tools we had given him, he can sketch a proof for himself
viral jasper
@versed lily, if that all feels good to you, and you feel confident to finish the proof from this, then go ahead and close
viral jasper
Yeah, I thought you needed like an epsilon-delta proof 
versed lily
i do but i know how to continue from here
viral jasper
.close
ripe stream
im obviously doing something dubiously stupid but i cant see what or why
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fervent owl
im obviously doing something dubiously stupid but i cant see what or why
lol practising for STEP?
anyway the answer is you've assumed that there exists some function T such that T(x+y) = t((x+y)/(1-xy))
we don't know if there does exist such a function
(for example, if t(x) = x, then can you find a function F such that F(x+y) = (x+y)/(1-xy))
ripe stream
lol practising for STEP?
Yeah….
ripe stream
anyway the answer is you've assumed that there exists some function T such that ...
hmm but the rest of the q hints towards assuming it does though
after a bit of messing around I realised if u sub T(x) = t(tanx) it works tho
fervent owl
well basically we want to massage our condition t(x) + t(y) = t(z) to become what we have before
fervent owl
after a bit of messing around I realised if u sub T(x) = t(tanx) it works tho
yeah basically it's you look at x+y/(1-xy) and realise it's the thingy for tan
ripe stream
you get t(x) = karctanx but im still somewhat confused why my initial approach was wrong D:
fervent owl
nw!
fervent owl
Ok thank you
if u've gotten a cambridge offer then GL for step!
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cobalt atlas
would the radius be 3 here ?
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would the radius be 3 here ?
vestal moat
No, 2
cobalt atlas
why
cobalt atlas
No, 2
is it cause thats the amount of lines it takes to reach each point from the center ?
vestal moat
Think in terms of diameter
cobalt atlas
Think in terms of diameter
still don't get it
cobalt atlas
3 counting the line in the center if that doesn't count then 2
vestal moat
Do you know what a diameter is ?
sand escarp
The horizontal distance between points (-4,2) and (0,2) is 0-(-4) = 4
And so the radius is 2
cobalt atlas
The horizontal distance between points (-4,2) and (0,2) is 0-(-4) = 4
why do you have to divide by 2 or something ?
sand escarp
yes
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wanton basin
(k+1)! + (k+1)! * (k+1) = (k+2) * (k+1)! how do I expand this?
wanton basin
or rather simplify it so I can show both sides are equal
I got to this point but can't figure out how to progress further
ocean lark
factor (k+1)! out of the the left hand side
dusky oxide
factor out the factorial
dusky oxide
(k+1)!(1+(k+1))?
wanton basin
oh that's it
huh
because the k+2 on the right side is * (k+1)!
so they are equal after that step
man the one thing I hate about all my math courses is I never actually get better
I just learn new things
anyways thanks for the help
I got it
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modest obsidian
need help solving this cause i don't even know where to begin 😭
young oracle
i got 70
modest obsidian
can you explain how please
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wanton basin
1 sec
so I solved a) and b)
but for c I don't know how I would indicate letters
I know it is m, n, o, p
part of the alphabet
steel solar
Think about their positioning.
steel solar
so I would imagine
Hmm, well...
steel solar
Think about their positioning.
^
steel solar
Yes.
wanton basin
so 13 <= x <= 16
steel solar
You could have $x\in\text{Position in alphabet}$
wanton basin
but then what about x is an element of alphabet
glossy valveBOT
;(
wanton basin
ah position in alphabet
wanton basin
ok well thanks, but does the other way I gave work?
if I say letter in alphabet
where x is any given letter
just trying to figure out what's acceptable
steel solar
where x is any given letter
Well, we need numbers in there somewhat.
Let me think.
I don't know, it feels strange.
wanton basin
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dreamy spear
ive got a circle defined by 3 points, a start, middle, and end; and a parametric equation that draws that circle https://www.desmos.com/calculator/z47o7prgup.
you can see in the circle that it starts at a_0 (angle of the start point) and ends at a_2(angle of the end point). the problem is, this isnt always the case. if a_2 is less than a_0, it wont draw. other times (and i cant tell what causes it for the life of me), it draws but the visible section of the circle doesnt intersect the middle point.
im adding 4 pictures, the first 2 are examples of what i want, the last 2 are examples of whats going on that i dont want.
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ive got a circle defined by 3 points, a start, middle, and end; and a parametric...
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dreamy spear
<@&286206848099549185>
dreamy spear
E<=t<=S+2pi and E<=t<=S define the different sections of the circle at every combination ive tried, so thats helpful. the only hurdle now is writing something that knows which one to choose
torn jolt
Have you tried to sustitute t in your parametrization by t-a_0 and take t in the interval [0,a_2-a_0]?
dreamy spear
I have, but i get the same issues
torn jolt
If we consider t = a_2*x + a_0 with x in [0,1] that should fix the problem when a_2<a_0
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torn jolt
Sorry, I made a mistake take t = (a_2-a_0)/a_0 * x + a_0 and x in [0,a]
shy edge
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shy edge
can x have nnegative value..?
sharp lagoon
can x have nnegative value..?
They say it converges at x = -3, so yes.
shy edge
but usually we write it as abs(x) < 3 right
dont we usually take the smaller value in this case so it guarantee converges
hot lily
but usually we write it as abs(x) < 3 right
Is this supposed to be the ratio test?
shy edge
i dont think so.?
vestal moat
I think everything except a and d
shy edge
wait it doesnt make sense abs(x) < -3
so can radius of convergence have a negative value?
hot lily
I think everything except a and d
!nosols let them work through the problem
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I think everything except a and d
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
shy edge
sooo..?
hot lily
sooo..?
In general, there is a limited interval where a power series converges
So you can be suer that $\mathrm S$ converges if $x \in [-3, 4]$
glossy valveBOT
King Leo
shy edge
oh
thats what it meant
i thought it was saying abs (x) <4 and abs(x) <-3
so anything falling within the domain -3<x<4 are all value that converges?
hot lily
so anything falling within the domain -3<x<4 are all value that converges?
Should be < but yes
hot lily
in ratio test we do <1 for it to converge
Yes, but
- we generally test the endpoinst individually after that
- its given that the power series converges if
x = -3, 4
shy edge
Yes, but
- we generally test the endpoinst individually after that
- its given t...
oh so its based oon question
okok got it
thank you very much
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shy edge
!occupied
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shy edge
not thread, channel
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mild flicker
Sorry.. Layman.. Might come off as a rube/shitpost but its been bugging me today.. I've always read that "0.9999...=1" without any caveats.. but also that 1/0 is undefined..
But whenever I look at why 0.999..=1 it always seems to based on limits to infinity...
So if I follow that line of reasoning...
0.999999... = lim(n→∞) [1 - 1/n] = 1
0.00000... = lim(n→∞) [1/n] = 0
Which means if you can substitute 0.999.. for 1 (and vice versa), then could you substitute 0 for 0.0000.. and write 1/0 as 1/0.0000.. and now 1/0=∞?
minor crater
the reason that 1/0 is undefined is because if you take 1/0.0000.. -> +infinity and 1/(-0.0000) -> -infinity, even though in the limit both +0.0000 and -0.0000 are 0
minor crater
Sorry.. Layman.. Might come off as a rube/shitpost but its been bugging me today...
this is just a simplified version of a more rigorous argument but i hope this is understandable
indeed, $\lim_{x\to 0^+} \frac{1}{x} \to +\infty$
glossy valveBOT
artemetra
mild flicker
I read about the issue with the discontinuity.. that it can't work because of it. I just can't tell why following the same line of reasoning that leads to 0.9999.. being valid can also lead to 0.000.. being invalid, particularly since it seems like you need to invoke the 1/inf to be able to do 0.9999..=1
I mean it got a bit more sensible if I limited to 0+
since I guess that solves the discontinuity
minor crater
mhm
minor crater
I read about the issue with the discontinuity.. that it can't work because of it...
moreover here you are never dividing by zero
mild flicker
yeah thats the point.. I can "pretend" not to divide by zero by dividing by 0.0000...
minor crater
0.999999 goes to 1 but also 1.000000....1 goes to 1
mild flicker
and if 0.9999..=1 then 0.000..=0
minor crater
0.999999 goes to 1 but also 1.000000....1 goes to 1
you can go from both sides
minor crater
the reason that 1/0 is undefined is because if you take 1/0.0000.. -> +infinity ...
but as i say here, you can't go from both sides here
both 0.0000 and -0.0000 go to zero, while 1/0.0000 goes to positive infinity and 1/-0.0000 goes to negative infinity
,calc 1/0.000001
glossy valveBOT
Result:
1e+6
,calc 1/(-0.000001)
glossy valveBOT
Result:
-1e+6
deep gale
Sorry.. Layman.. Might come off as a rube/shitpost but its been bugging me today...
It would suffice to know what the expression 0.999... means, i would consider the limit of something that will look like $x_{i+1} = x_i(10^{-i} + 1)$
glossy valveBOT
Prelude to archbishop
minor crater
graph of 1/x if that helps
deep gale
It would suffice to know what the expression 0.999... means, i would consider th...
A real sequence defined recursively*
if given initial terms of course,
Preferably can choose a better sequence
minor crater
It would suffice to know what the expression 0.999... means, i would consider th...
i don't see a reason for a recursive definition tbh, isn't it just 1-1/n okay?
deep gale
I've heard of lower and upper decimal approximations of real numbers
deep gale
i don't see a reason for a recursive definition tbh, isn't it just 1-1/n okay?
Or that
Whichever sequence works better
But still makes sense*
mild flicker
I think in the proof I saw it expanded out to an infinite series.. but it seems roughly equivalent to do 1-1/n as n approaches infinity..
fast peak
but the point is that 0.9999... is not the limit of 1-1/n, it stands for a different limit (which also happens to equal 1)
so they arent the same situation
the specific notation 0.9999... means something
so if you want to apply that to the notation 0.0000... you also have to use that same meaning
and 1/n is just something else
deep gale
In order to properly define decimal approximations of a real number
fast peak
and while 1/x->infty is true, there are good reasons to let 1/0 not be a thing in the real numbers
deep gale
It might perhaps suffice to consider irrational real numbers
mild flicker
hm alright thanks - I think the issue is in how I'm formulating 0.9999..?
is that where I'm ending up with the incorrect premise?
eg. the whole limit 0.999 = 1-1/inf isn't a valid representation of the infinite series?
fast peak
whenever you are writing inf as if it was a real number you are already writing wrong things
mild flicker
yeah sorry, I'm not familiar with latex.. I mean as a limit as n approaches infinity then 1-1/n
fast peak
while thats not wrong, its more accidental as just both things equal 1. but strictly speaking 0.9999... is a different limit
deep gale
eg. the whole limit 0.999 = 1-1/inf isn't a valid representation of the infinite...
The limits of two sequences may be equal, but the sequences aren't equivalent to each other
In the sense that there exists an i in N, so that ai != bi, where ai is the i'th term of the (infinite) upper decimal expansion of 1, and bi is 1 - 1/i
fast peak
the bigger problem is just that you want to replace 1/0 by its limit which just isnt a real number
and then you just run into issues with infinity
a/b = lim a/b_n where b_n -> b is true, unless b=0
or b=infty
in both cases the lhs just isnt defined
for good reasons
(in the real numbers)
mild flicker
hmm okay thanks. I think I need to go learn how convergence and sequences actually work. I think I'm taking way too much liberty and misunderstanding WHY 0.9999..=1
deep gale
hmm okay thanks. I think I need to go learn how convergence and sequences actual...
What else would the expression 0.9999.. mean?
It is defined in math as an infinite decimal expansion
mild flicker
well.. the limity thing I had earlier, which converges to 1
well - thanks, I know where I'm messing this up; I think I know what I need to go read more of - I will stop making silly statements
and go read until I get it
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stuck fiber
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stuck fiber
Im not really sure what I did wrong here
First I subbed in e^-3 into f(x) to calculate the y coordinate of the tangent, that gave me -4
Then I got the derivative of f(x), which was 2/x
Then I subbed in y=-4, x=e^-3 and m = 2/x into y=mx+c
slate violet
the slope $a$ should be $f'(e^{-3})$
glossy valveBOT
south
slate violet
First I subbed in e^-3 into f(x) to calculate the y coordinate of the tangent, t...
so that's not correct, this is not the slope of the graph at x = e^(-3)
stuck fiber
Ah I see
slate violet
Then I got the derivative of f(x), which was 2/x
you carry on from here and then sub in e^(-3) in here
and then you use point-slope form, so you need y when x = e^(-3)
stuck fiber
I will 1 moment I will try again to solve it
slate violet
$y - y_1 = m(x - x_1)$
glossy valveBOT
south
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stuck fiber
nice to see you again odie!
And you aswelllllll ❤️ ❤️
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signal creek
Can someone help
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quiet zodiac
anyone who can help with a physics problem help 8 pls
alpine pecan
partial fractions
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signal creek
partial fractions
Yea
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C&rroter
final wing
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final wing
I dont really get why the integral goes from a to a?
does it have to do that g(x) = integral a to x and g(a) will say g(x) is integral a to a?
cant be that easy right?
rapid rain
it is that easy
hearty plover
It is that easy yes
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distant quiver
@limpid scroll
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distant quiver
@limpid scroll bro you are maths genius , please help us we all are noobs
Hi guys?
slate violet
<@715766571540414464> bro you are maths genius , please help us we all are noob...
just ask your question
distant quiver
just ask your question
Ya dude why you pissed of
slate violet
Ya dude why you pissed of
I'm not
I'm telling you that people aren't helping cause of this
cause there's no question yet
distant quiver
I'm telling you that people aren't helping cause of this
Help me with this
slate violet
Help me with this
oh, use the symmetry of Pascal's triangle
,w 3r - 1 = 39 - (r^2 - 1), 3r = 39 - r^2
distant quiver
oh, use the symmetry of Pascal's triangle
??
slate violet
hmmm wait
distant quiver
,w 3r - 1 = 39 - (r^2 - 1), 3r = 39 - r^2
no way we can do that?
slate violet
basically use the fact that 39 choose k = 39 choose (39 - k)
glossy valveBOT
distant quiver
basically use the fact that 39 choose k = 39 choose (39 - k)
Ohh I get it now thanks bro
slate violet
I don't have the right equation but it must be close to this
arghhh
it's okay other people will come given enough time
distant quiver
Nahh your right
long rock
Help me with this
is this chemistry problem? how 39 got up there?
slate violet
yikes
distant quiver
is this chemistry problem? how 39 got up there?
Algebra
long rock
oh thanks I see
distant quiver
no it's combinatorics
What career prospects are there after undergraduate in maths?
slate violet
What career prospects are there after undergraduate in maths?
search up "maths careers" on Google
little badge
Hi! every one can anyone help me to solve this assignment i shall be very greatful to you all for this help. Thanks
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tulip badge
ok i have 2 problems i need help with
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ok i have 2 problems i need help with
stiff musk
probably do row and column manipulations to transform the first matrix into the second
sharp flame
you can use determinant properties
frail lance
do cross product and then solve
stiff musk
i would probably take the transpose first since the xyz stuff is in col 1 of the first matrix and row 1 of the second, etc
then just do row/col operations until you get the second matrix
rapid rain
If you haven't found yet: first thing you can do is linear operations to to clear out the original matrix (for example factoring columns/rows, adding/subtracting multiples of one row to another...)
then once this is done, see what further operations (switching rows/columns, transpose, etc...) need be done to obtain the matrix you want
tulip badge
ok so
i got to
---|x y-2z z|
18 |r s-2v t|
|u v-2u w|
ok ye
where to go from here
how to remove the -2 things
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ashen garnet
hi zap
today I can't help you, I'm studying for my exam
but use these properties of the determinant (when you do elementary row operations the determinant changes):
row interchange: $R_1 \leftrightarrow R_2 \implies |A| = -|A|$
glossy valveBOT
nico.alesi
ashen garnet
row scalar multiplication: $kR_1 \rightarrow R_1 \implies |A| = \frac{1}{k}|A|$
glossy valveBOT
nico.alesi
ashen garnet
addition of different rows (and their multiples): $R_1 + kR_2 \rightarrow R_1 \implies |A| = |A|$
glossy valveBOT
nico.alesi
ashen garnet
last one basically does nothing
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NICO
tulip badge
today I can't help you, I'm studying for my exam
goodluck man
ashen garnet
thanks : )
ashen garnet
use these properties and you'll find the answer 100%
tulip badge
---|x y-2z z|
18 |r s-2v t|
|u v-2u w|
i got to this point
hm ok il ltry again
tysm
ashen garnet
I can't see what you've done
but quickly looking at the exercise
I would also add
before using elementary row operations
do the transpose of the matrix
$|A| = |A^T|$ so you can do it without affecting the determinant
glossy valveBOT
nico.alesi
tulip badge
i did it
ashen garnet
oh ok so just try man
tulip badge
.
tulip badge
---|x y-2z z|
18 |r s-2v t|
|u v-2u w|
.
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solid ruin
(1)תכפיל שורה ראשונה ב1/6
(2)תכפיל עמודה ראשונה ב1/3
(3)תוסיף לשורה ראשונה את -2שורה שנייה
(4)תכפיל עמודה ראשונה ב-1
(5)תכפיל שורה ראשונה ב1
(6)תעשה טרנספוז
תראה בכל שלב מה צריך לשנות ב-180
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(1)תכפיל שורה ראשונה ב1/6
(2)תכפיל עמודה ראשונה ב1/3
(3)תוסיף לשורה ראשונה את -2...
tysm
torn jolt
hello
torn jolt
i have been thinking about this limit for a while
to no success
i will write it down
lim x->0 of (sqrt^4(x+1) - sqrt^3(2x+1))/x
the sqrt^3(2x+1) can be solved easily through adding and subtracting an x for the first and second sqrt
then using the formula sqrt^3(x) - sqrt^3(y) = x-y /sqrt^3(x^2)+sqrt^3(xy)+sqrt^3(y^2)
but sqrt^4 what do i do with that
placid vale
$\lim_{x\to 0} \left(\frac{\sqrt[4]{x+1} -\sqrt[3]{2x+1}}{x}\right)$
placid vale
is that the limit?
torn jolt
no
placid vale
then?
glossy valveBOT
General_Jacob
restive geyser
Again, this is a 0/0 limit
We can say that, if this does have a limit, then by l'Hopital it should be the limit of "the derivative of the numerator over the derivative of the denominator"
restive geyser
ngl I think that's a little overkill
torn jolt
We can say that, if this _does_ have a limit, then by l'Hopital it should be the...
it does not end well..
i tried it
placid vale
taylor series?
torn jolt
u can't get away of x^2
torn jolt
taylor series?
i dont know that
ancient folio
LHopital?
placid vale
$\sqrt[4]{x+1} \approx 1 + \frac{1}{4}x$
glossy valveBOT
General_Jacob
torn jolt
**General\_Jacob**
this feels like cheating..
ancient folio
that does not end well
Hows come?
placid vale
$\sqrt[3]{2x+1} \approx 1 + \frac{2}{3}x$
glossy valveBOT
General_Jacob
torn jolt
Hows come?
differentiate it and you will find out
placid vale
$1+\frac{1}{4}x - 1 - \frac{2}{3}x = -\frac{5}{12}x$
glossy valveBOT
General_Jacob
placid vale
$\frac{-\frac{5}{12}x}{x} = -\frac{5}{12}$
glossy valveBOT
General_Jacob
torn jolt
oh
that's cool but suppose i dont have that
binomial approximation
how do i ddo
it
torn jolt
wait
restive geyser
There's no quotient rule at play
restive geyser
That's quotient rule yeah