Has to be differentiable in (a,b)
#help-27
faint gorge
spring oasis
Well
Like remember that f(x,y) = P(x,y) + R(x,y)
Basically if f(x,y) - P(x,y) = R(x,y)
Then lim (x,y)->(a,b) R(x,y)/||(x-a,y-b)||=0
Because it feels like impossible to continue
There's too many terms
And I don't think I have enough space in my page to expand all that
Also feels like brute forcing the answer
zenith spoke
ok so there is something that you can do to simplify things quickly
the denominator is (x-1)^2+(y+2)^2
so if you can get multiples of this expression in the numerator for example then you can simplify right?
spring oasis
Yes
zenith spoke
say you get something like $\frac{4(y+2)^2+4(x-1)^2-3}{(x-1)^2+(y+2)^2}$ then that would simplify to $4-\frac 3{(x-1)^2+(y+2)^2}$
woven radishBOT
ali yassine
zenith spoke
so now the goal is to do things like this
spring oasis
(x-1)^2 / ||(x-1,y+2)|| -> 0
spring oasis
Yes I can prove it
spring oasis
also where did this come from
Things we can use for simplifying
zenith spoke
ohhh i see
zenith spoke
(x-1)^2 / \||(x-1,y+2)\|| -> 0
also remember that the denominator is not exactly this
spring oasis
Also (y+2)^2 / ||(x-1,y+2)|| -> 0
zenith spoke
its actually ||(x-1,y+2)||^2
zenith spoke
your statements are true
zenith spoke
its actually \||(x-1,y+2)||^2
but with this they are false
spring oasis
Wait
spring oasis
but with this they are false
No
zenith spoke
yes they are
if you convert to polar coordinates via the substitutions x-1=rcosθ and y+2=rsinθ then the limit will become a limit when r->0
spring oasis
Ah it's not 0, it's 1?
zenith spoke
it doesnt exist
zenith spoke
if you convert to polar coordinates via the substitutions x-1=rcosθ and y+2=rsin...
when you do this, the value of the new limit will depend on θ
so it doesnt exist
do you know about this thing with polar coordinates?
if no then i recommend that you pick it up because it is sometimes useful
zenith spoke
so if you can get multiples of this expression in the numerator for example then...
so lets get back to the beginning and try to evaluate the original limit using this idea
you dont have to overthink and try to get things that may be useful in the process
just begin and you can figure out what to do along the way
spring oasis
just begin and you can figure out what to do along the way
Ok
Sorry
Can we start from scratch then?
zenith spoke
no need to say sorry you didnt do anything wrong. I am just telling you this because it may save you some time and not mess things up for you
spring oasis
Shit got messy
zenith spoke
Can we start from scratch then?
yes thats what we will do
zenith spoke
so if you can get multiples of this expression in the numerator for example then...
remember that you want this
zenith spoke
alright then lets do this
zenith spoke
remember that you want this
since you want this
notice that there is already a term 4(x-1)^2 in the numerator
so its probably better to not touch that for now right?
spring oasis
Ye
zenith spoke
ok so keep this term as it is for now
now subsitute P_3 instead of f
and simplify things
from there we will proceed
spring oasis
now subsitute P_3 instead of f
I'm trying
zenith spoke
you did it correctly here btw
zenith spoke
notice that there is already a term 4(x-1)^2 in the numerator
but now you shouldnt expand this
zenith spoke
I'm trying
np take your time
zenith spoke
yes
zenith spoke
thats right, nice!
ok so now is the tricky part
now we want to play around with the terms in order to factorize things nicely
so first notice the 4y^2+16y that you have
we want everything that has y to somehow have the form of k(y+2)^2 and everything with x to have the form n(x-1)^2 where k and n are some constants right?
because then we can somehow simplify things with the denominator
so can you make 4y^2+16y into this form somehow?
its missing a constant term there
which constant is it missing?
zenith spoke
yes
spring oasis
+4
zenith spoke
ah mb, when i wrote that i didnt want you to answer right after it. I was just trying to show you the picture better
spring oasis
4Y^2 + 16y + 16 = 4(y+2)^2
zenith spoke
right
spring oasis
so can you make 4y^2+16y into this form somehow?
add + 16
zenith spoke
exactly
spring oasis
And subtract -16
zenith spoke
nice
spring oasis
What's your point?
zenith spoke
the point is that now we have 4(y+2)^2+4(x-1)^2 for now
there are other terms too
but we can separate the fractions
and then simplify
zenith spoke
you can even write things in a nicer way
but first lets do this
then we will continue
zenith spoke
alright, now the fraction on the left simplifies to?
spring oasis
4
spring oasis
zenith spoke
(also small note that the denominator of the right fraction is ||(x-1,y+2)||^2 and ik you meant to write that but just so you dont find it weird if you return to this later)
spring oasis
(also small note that the denominator of the right fraction is \||(x-1,y+2)||^2 ...
Mb
spring oasis
zenith spoke
lets try to factorize in a nice way
spring oasis
(x-1)^2 = x^2 -2x + 1
9(x-1)^2 = 9x^2 -18x + 9
zenith spoke
so here let me tell you the idea
since there is (x-1)^2+(y+2)^2 in the denominator
we know that it becomes something nice if we convert to polar coordinates with the right substitutions right?
spring oasis
Sure
zenith spoke
exactly so we would like to use that if it makes things easier
but if we convert to polar coordinates rn the numerator will be a mess
so its better to write the numerator in a nicer way where using polar coordinates gives a nice form without tedious calculations
so if we want to convert to polar coordinates here what would be the most natural substitution?
spring oasis
but if we convert to polar coordinates rn the numerator will be a mess
Yes
spring oasis
so its better to write the numerator in a nicer way where using polar coordinate...
Care to elaborate on that
zenith spoke
i think you will know what i mean soon
zenith spoke
so if we want to convert to polar coordinates here what would be the most natura...
think about this, if you have (x-1)^2+(y+2)^2 and you want to convert to polar coordinates
what would be the substitution
zenith spoke
Care to elaborate on that
after you know the answer to this question you will know what i meant by this
spring oasis
think about this, if you have (x-1)^2+(y+2)^2 and you want to convert to polar c...
X-1 = rcostheta -> 0
Y-1 = rsintheta -> 0
zenith spoke
Y-1 = rsintheta -> 0
be careful
spring oasis
As r -> 0 for some theta
zenith spoke
Y-1 = rsintheta -> 0
this is not correct
spring oasis
Y+2
zenith spoke
exactly
ok nice
so we want to substitute x-1=rcosθ (and y+2=rsinθ) at the end of the day
so what i really meant by "write the numerator in a nicer way" is to have something of the form k(x-1)^n for some k and n or something like that since then converting to polar coordinates would become much easier
spring oasis
(x-1)^2 = x^2 -2x + 1
9(x-1)^2 = 9x^2 -18x + 9
zenith spoke
hmmm but then what will you do with this?
zenith spoke
if you dont know how to proceed then maybe start by trying to put similar things together
spring oasis
hmmm but then what will you do with this?
Idk. I'm stuck
zenith spoke
for example -3x^3 and 3 have the same coefficients (up to a sign) . also 9x^2 and -9x have this similarity too
so maybe if you factorize each of these alone a common factor between them will appear?
so try to factorize each of -3x^2+3 and 9x^2-9x then see what you get
zenith spoke
9(x^2 -x)
you meant -x not -2x right
ok can these expressions be factorized more?
or is this it?
zenith spoke
remember here it is -3x^3+3 not -3x^2+3
spring oasis
remember here it is -3x^3+3 not -3x^2+3
Wdym
zenith spoke
-3(x^2 - 1)
i mean that this should be -3(x^3-1) not -3(x^2-1)
because the numerator has -3x^3 and doesnt have -3x^2
spring oasis
so try to factorize each of -3x^2+3 and 9x^2-9x then see what you get
Ok one second
Well
zenith spoke
right, so what is the next step?
spring oasis
Well
spring oasis
right, so what is the next step?
Isnt it overkill to use difference of cubes
Like clearly you want me to factorize (a-b) in the numerator
Correct?
zenith spoke
right, but you most probably should show this step in the exam
personally i dont mind if you skip it and just take the common factor and simplify everything in one step as long as you do that correctly and you know what you are doing
spring oasis
Ok
zenith spoke
also i wouldnt call it overkill because its essentially what you are doing even if you dont explicitly show the step
rain summit
im confused about $4 + \lim_{(x, y) \to (1, -2)} \frac{-3(x^3 - 1) + 9x(x - 1)}{\abs{\abs{(x - 1, y + 2)}}^2}$
woven radishBOT
1 divided by 0 equals Infinity
rain summit
but why $(x - 1, y + 2)$?
woven radishBOT
1 divided by 0 equals Infinity
rain summit
and the denominator consists of addition of 2 squares which is hard to work with
zenith spoke
$\lVert (x-a,y-b)\rVert=\sqrt{(x-a)^2+(y-b)^2}$ so $\lVert (x-1,y+2)\rVert ^2=(x-1)^2+(y+2)^2$
woven radishBOT
ali yassine
rain summit
.
spring oasis
(x^3-1) = (x-1)(x^2 + x + 1)
zenith spoke
right
spring oasis
manic onyx
it does
spring oasis
How so?
manic onyx
expand -3(...) + 9x and simplify it again
spring oasis
-3(x-1)^2
manic onyx
yes -> so the lim is -3(x-1)^2 / [ (x-1)^2 + (y+2)^2 ]
sorry idk how to write plain latex lol
zenith spoke
(x-1)/|(x-1,y+2)| -> 1
no
manic onyx
yes -> so the lim is -3(x-1)^2 / [ (x-1)^2 + (y+2)^2 ]
this is the form that you want for all of these questions regarding taylor polynomials
manic onyx
because you can use a variable change u = x-1 ; v = y+2
spring oasis
zenith spoke
no
manic onyx
because you can use a variable change u = x-1 ; v = y+2
and the limit instead of converging in the ugly (1,-2) that you can do nothing with
zenith spoke
-3(x-1)^3
manic onyx
it converges in (0,0)
zenith spoke
remember that you had a factor of (x-1)
zenith spoke
because you can use a variable change u = x-1 ; v = y+2
right but even then there is no immediate way to see if the limit exists or no
spring oasis
it converges in (0,0)
No spoilers
zenith spoke
you would need something like polar coordinates
which would do everything with one hit
manic onyx
you would need something like polar coordinates
yess
in polar coordinates the denominator is trivial
zenith spoke
so now is the time for what i told you earlier @spring oasis
manic onyx
so you can solve it
zenith spoke
so now you have $\lim_{(x,y)\to (1,-2)}\frac{-3(x-1)^3}{(x-1)^2+(y+2)^2}$
woven radishBOT
ali yassine
zenith spoke
we agreed that the substitution we would do is x-1=rcosθ and y+2=rsinθ right?
in that case the limit will become a limit in 1 variable which is r
we want to know where r goes
zenith spoke
we agreed that the substitution we would do is x-1=rcosθ and y+2=rsinθ right?
if you let (x,y)->(1,-2) in this pair of equations and hold θ as a constant, where would r go?
@spring oasis
spring oasis
zenith spoke
ohhhh great job!
one last question
what if the value of the last limit depended on θ?
spring oasis
Then it dne
zenith spoke
i will give you a limit to compute using polar coordinates
because there is a trap that i want you to be aware of and avoid if you encounter it later on
$\lim_{(x,y)\to (0,0)}\frac{x^2}y$
woven radishBOT
ali yassine
zenith spoke
evaluate this limit using polar coordinates (if it exists)
spring oasis
$\lim_{(x,y)\to (0,0)}\frac{x^2}y$
Wait a second let me try path y = x and x = y
Wat about y = x^2
Aha!
spring oasis
$\lim_{(x,y)\to (0,0)}\frac{x^2}y$
Does not exist
Path y = x^2 gives 1
Path y = x gives 0
zenith spoke
you cheated!
zenith spoke
rcos^2(@)/sin(@)
so from here if you let r->0, the expression should go to 0 right?
so the limit should exist and be 0
but it clearly doesnt exist
you just gave 2 paths with different limits
spring oasis
so from here if you let r->0, the expression should go to 0 right?
I mean if sin is 0 for some theta like 2pi we are fucked
zenith spoke
hahaha great job!
so there is no problem in the polar coordinate way
its not wrong or something like that
but you should be careful
zenith spoke
I mean if sin is 0 for some theta like 2pi we are fucked
and thats the trap i was talking about
zenith spoke
I mean if sin is 0 for some theta like 2pi we are fucked
so the limit does depend on theta if you change to polar coordinates
spring oasis
Yeah I guess
zenith spoke
nice job 
zenith spoke
Yeah I guess
if you are done type please type .close
spring oasis
.close
devout snowBOT
spring oasis
nice job <:catking:812386361297862747>
Thanks for the help
zenith spoke
np, you did the whole work!
errant harbor
Send your problem
copper harbor
Things seem to have worked out?
vital edge
Sorry for the tag lance
copper harbor
This doesn't seem like a math problem 
vital edge
Brother did you really open a channel to ask that
errant harbor
The moderators tag made this so crowded
vital edge
I agree
errant harbor
Xd
vital edge
Neither of these are problems
...are you doing a test
Well given the amount of context you've given us
There is very little we can say
Bruh you've given us stage 1
Nothing else
And asked what happens in stage 5
We aren't clairvoyant
Also stop deleting your messages, it makes it very very annoying to try to help you
How did bot not auto close this channel
.close Original message deleted, OP seems to have disappeared
devout snowBOT
devout snowBOT
Available help channel!
mighty crypt
i need help with d) this is genuinly impossible
devout snowBOT
i need help with d) this is genuinly impossible
uneven salmon
idk if im allowed to help but the domain is between the 2nd and 3rd quadrants
its asking for
$sin \theta = \frac[-1][2]$
shitt
aight basically
soft umbra
$\sin \theta = \frac{-1}{2}$
woven radishBOT
🌙 ЅκψΑиdΝιɡħτ
uneven salmon
ty lol
soft umbra
uneven salmon
ur base angle is going to be pi/6
cause
sin pi/6 = 1/2
and its asking for -1/2
which means its in the 3rd quadrant
so u add pi to pi/6
therefore ur answer is 7pi/6
uneven salmon
so bascially theta is my beta basically
i dont understand what u mean
mighty crypt
theta is suppsoed to be terminal arm and beta is supposed to be my corresponding angle
but in this situation it switches around
devout snowBOT
@mighty crypt Has your question been resolved?
royal laurel
Number seems wrong
devout snowBOT
Number seems wrong
royal laurel
Check please
uneven salmon
dont do long division
untold ravine
!xy
devout snowBOT
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.
uneven salmon
?
royal laurel
dont do long division
Stop
uneven salmon
what 😭
uneven salmon
u just showed me a video of u trying to do long division im not sure what u want help with
misty crest
i mean your calculation is correct
uneven salmon
you are trying to divide 0.27 by 0.8 then
so u should be dividing 2.7/8
u need to move ur decimal point 1 place back
royal laurel
I think i got it now
uneven salmon
ok
proven girder
uneven salmon
just go to ur profile
shut meteor
i saw this hard problem online and i would like to know the answer ((x^{3}+y^{3}+z^{3}=k)): Find integers (x,y,z) for any integer (k). For (k=33) and (k=42),
woven radishBOT
eido
shut meteor
really i did not know that
uneven salmon
i recommend u look at andrew sutherland and andrew bookers works
uneven salmon
really i did not know that
i saw it on yt shorts
pseudo basin
i saw this hard problem online and i would like to know the answer (\(x^{3}+y^{3...
!occupied
devout snowBOT
i saw this hard problem online and i would like to know the answer (\(x^{3}+y^{3...
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
uneven salmon
its famous, unsolvable for decades
pseudo basin
!redir
devout snowBOT
This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.
uneven salmon
sorry
shut meteor
ok ty
devout snowBOT
@royal laurel Has your question been resolved?
hexed wedge
devout snowBOT
hexed wedge
Can someone give example
faint zinc
,rccw
woven radishBOT
faint zinc
This is very chem.
hexed wedge
yeah apologies but chem server doesn’t have much people to get instant help
and I kinda need to know as well
faint zinc
So water will always make some H+ ions and OH- ions, which is why neutral pH is 7.
If you add acid to water there are more H+ ions and correspondingly fewer OH- ions, the reaction reaches a new equilibrium
It's not the best example, but it's the one I thought of in the moment.
normal bolt
Consider N2(g) + 3H2(g) <--- ---> 2NH3(g)
at t= 0 it has some initial concentration and when you make disturbance in it or say add reactant it will try to achieve equilibrium (moles, concentration etc with time becomes constant) by shifting that moles from left to right
hexed wedge
CO2 + H2O <-> H2 CO3
Okay but don’t I have to write two reactions
faint zinc
We force the concentration of CO2 in the water as high as we can, which drives the reaction forward.
Then when you open the can the concentration of CO2 is significantly less, so CO2 bubbles out
(which is the reaction going in reverse)
hexed wedge
I’ll take both for convenience
hexed wedge
Consider N2(g) + 3H2(g) <--- ---> 2NH3(g)
at t= 0 it has some initial concen...
is t in Celsius
Can’t be kelvin
normal bolt
is t in Celsius
It depends what the author of question gives
It sometimes give celcius or K
celsius*
hexed wedge
It sometimes give celcius or K
0 K can’t be happening right?
normal bolt
It may happen
hexed wedge
Consider N2(g) + 3H2(g) <--- ---> 2NH3(g)
at t= 0 it has some initial concen...
Oh so the reaction goes forward as we add some concentration to the reactants
What about vice versa
Let’s say we add to the products
AND
The place where there are more moles
Like in the left side it’s 4
On the right it’s two
SO
Will the reaction go in the direction where there are less moles
normal bolt
Yes to achieve equilibrium state
devout snowBOT
@hexed wedge Has your question been resolved?
devout snowBOT
devout snowBOT
Available help channel!
spring oasis
devout snowBOT
woven radishBOT
Renato
fossil locust
you should be able to figure out what to compare a) to, using the limit comparison test
fossil locust
like, $\frac{x^3 + 5}{x^4 + 4}$ is similar enough to $\frac{x^3}{x^4 + 4}$
woven radishBOT
south
spring oasis
fossil locust
fossil locust
even to x^3/x^4 tbh
I didn't want to give it away but yes
spring oasis
what am I supposed to do with LCT for improper integral
And how do I know if it even applies here
we are skipping steps here
@fossil locust
fossil locust
I should just give you the entire document then: https://websites.umich.edu/~mconger/dhsp/lct.pdf
spring oasis
so basically I need to find a lower bound
spring oasis
like, $\frac{x^3 + 5}{x^4 + 4}$ is similar enough to $\frac{x^3}{x^4 + 4}$
this
fossil locust
so basically I need to find a lower bound
yeah, direct comparison also works
spring oasis
but. . . how do I even know if comparison test even applies here, what are the previous conditions that need to happen in order to apply
fossil locust
you can use 1/(2x) as a lower bound, given that you have something like "for all x >= 10"
spring oasis
is x in the inverval [1, infty+)?
can I receieve more handholding? @fossil locust idk if you guys are rushing through this or being handwavy
supose x > 0, then this whole $$\frac{x^3 + 5}{x^4 + 4} \geq \frac{x^3}{x^4 + 4}$$
woven radishBOT
Renato
fossil locust
so it is true that $\frac{1}{2x} \le \frac{x^3 + 5}{x^4 + 4}$ for all real $x \ge N$
spring oasis
the second une is g(x)
woven radishBOT
south
fossil locust
where N is finite: I'd encourage you to try estimating N
by rearranging the inequality
fossil locust
the idea is that it's approximately (x^3)/(x^4) = 1/x
but 1/x is too big
so you can multiply 1/x by some smaller constant
spring oasis
this handwavy
fossil locust
it's the right intuition for these problems
spring oasis
we dont have x^3 / x^4 tho
fossil locust
like, you'd do a similar thing to find the horizontal asymptote of (x^3 + 5)/(x^4 + 4)
pseudo basin
i wouldnt say it is too big
rather you should be thinking in somewhat looser terms as far as comparisons go
"looks like 1/x" means bounded above/below (as necessary) by k/x with a constant k which may need to be pinned down
well i mean ok like, "too big" means k<1 in this case
fossil locust
do you remember $\lim_{x \to \infty} \frac{x^3 (1 + 5/x^3)}{x^3 (x + 4/x^3)} = \lim_{x \to \infty} \frac{(1 + 0)}{(x + 0)} = \frac{1}{x}$
pseudo basin
just saying that it may help to open yourself up to a slight bend in the argumentation road
woven radishBOT
south
hasty zenith
You are using the comparison test to determine whether or not the sequences converge
fossil locust
that the integral of (x^3 + 5)/(x^4 + 4) behaves like the integral of 1/x
spring oasis
right
fossil locust
because you're considering the behaviour of the integrals as $x \to \infty$
woven radishBOT
south
fossil locust
so they have a finite ratio as x to infinity
and you can pull constants out of the integral; that doesn't affect convergence or divergence
spring oasis
**south**
so we got by simple intuition that the integral diverges, how to show it formally
fossil locust
I think the approach is to use the integral test to approximate those as summations
then use the limit comparison test for summations
both of these have proofs
spring oasis
comparison tests work for me
I just need to find a suitable lower bound and call it a day
spring oasis
I think the approach is to use the integral test to approximate those as summati...
surely we can do this more formally but comparison test is the simplest way to go
just this, simplest approach, what would it be
?
help me find a suitable lower bound @fossil locust
hasty zenith
You have lower bounds in the question you were given no?
spring oasis
even to x^3/x^4 tbh
is this argument true or false $$\frac{x^3 + 5}{x^4 + 4} \geq \frac{x^3}{x^4 + 4}$$ and then $$\frac{x^3 + 5}{x^4 + 4} \geq \frac{x^3}{x^4} \geq \frac{x^3}{x^4 + 4}$$ for $x \in [1, \infty+)$ \\ And finally $$\frac{x^3 + 5}{x^4 + 4} \geq \frac{x^3}{x^4}$$
spring oasis
You have lower bounds in the question you were given no?
what do you mean?
I dont think so, no
spring oasis
You have lower bounds in the question you were given no?
its a different type of lower bound what I mean, is not the integral bounds but like, a striclty increasing/decreasing sequence that is always lesser than the one we have
hasty zenith
Right I see
woven radishBOT
Renato
spring oasis
lmfao I always mess up with this bounding arguments
spring oasis
so it is true that $\frac{1}{2x} \le \frac{x^3 + 5}{x^4 + 4}$ for all real $x \g...
where does the 1/2x come from
seems out of the blue
pseudo basin
we say $+\infty$ btw and not $\infty+$
woven radishBOT
Ann
spring oasis
the problem is, this bound SOUTH found 1/2x seems out of the blue
spring oasis
"looks like 1/x" means bounded above/below (as necessary) by k/x with a constant...
what do you mean by that
ah I think I see what you mean
pseudo basin
i mean that "f(x) looks like 1/x" should mean "f(x) ≥ k/x [or ≤ k/x if that's what we need] with some constant k that we'll figure out now"
spring oasis
let me clarify things, so we have that $x \in [1, +\infty)$
having said that, we also have that
$$\lim_{x \to \infty} \frac{x^3 (1 + 5/x^3)}{x^3 (x + 4/x^3)} = \lim_{x \to \infty} \frac{1 + 5/x^3}{x + 4/x^3} = \lim_{x \to \infty} \frac{1}{x}$$
so $$\int_1^{+\infty} \frac{x^3 + 5}{x^4 + 4} \ dx \approx \int_1^{+\infty} \frac{1}{x} \ dx$$ and now before direct comparison $$\frac{1}{x} \geq \frac{1}{2x}$$ when $x \in [1, +\infty)$ \\ having said that, since $$\int_1^{+\infty} \frac{1}{2x} \ dx = \frac{1}{2} \int_1^{+\infty} \frac{1}{x} \ dx$$ clearly diverges and using direct comparison: $\frac{x^3 +5}{x^4 + 4} \geq \frac{1}{2x}$ then we have that $$\int_1^{+\infty} \frac{x^3 + 5}{x^4 + 4} \ dx $$ diverges
@pseudo basin @fossil locust something among that lines for the argument?
pseudo basin
mmm
something along those lines i guess, sure. but what you've written is nonrigorous
it's an unpleasant mix between formal proof and intuition
spring oasis
which parts are handwavy 😂
sorry this is my first improper integral exercise
i guess the part where I get that x^3(..)/x^4(....) = 1/x
spring oasis
I should just give you the entire document then: https://websites.umich.edu/~mco...
also, we are not expected to be fully rigurous, take for example this arguments from this pdf, theres a lot of handwavyness imo
pseudo basin
so i'd say for a formal proof you just dont put any intuition in writing
it's your rough work
spring oasis
sure
spring oasis
mb
woven radishBOT
Renato
devout snowBOT
@spring oasis Has your question been resolved?
faint gorge
whats ur q
spring oasis
please how would you do it I need more handhold
faint gorge
we would like to find a smaller integrand, so make the denominator greater and numerator smaller
spring oasis
we would like to find a smaller integrand, so make the denominator greater and n...
how do I find that
woven radishBOT
faint gorge
because the estimation would probably not be valid for all x in [1, +oo)
spring oasis
how so?
faint gorge
and it's no big deal since 1/x diverges in [x0, +oo)
faint gorge
then x0=0 but the point is sometimes estimations dont hold for all x, but sometimes for sufficiently large values of x
for example x^2 >= x is not true for all x, for example in (0,1)
so we would then consider the interval [1, +oo)
spring oasis
so we would then consider the interval [1, +oo)
dude
it is appearant that $\frac{x^3+5}{x^4 + 4} \ge \frac{1}{2x}$ correct?
woven radishBOT
Renato
faint gorge
because you wanted your work checked
and imo your justification was not good
so do it rigorously
spring oasis
where is it not good
faint gorge
the limit thingy
spring oasis
my point is, why are you splitting the integrals
spring oasis
the limit thingy
I see
yeah that part was very handwavy but was proposed by another helper I think
faint gorge
my point is, why are you splitting the integrals
because i am interested in the integral we integrate over where the estimation holds
spring oasis
I see
faint gorge
the other one is just some finite number in the end which doesnt make a difference
full geode
**Renato**
This isn’t difficult to show, following universe’s advice here is good
woven radishBOT
spring oasis
ok, do you mind we start from scratch
yeah in this case x >= 1
we have that advantadge
faint gorge
u just have to find C such that x^4+4 =< Cx^4
might indeed just take 2, and see for which x that inequality holds
spring oasis
u just have to find C such that x^4+4 =< Cx^4
what happened to x^3 + 5?
spring oasis
what a moment, I want to prove $\frac{x^3+5}{x^4 + 4} \ge \frac{1}{2x}$ how do you go to $x^4 + 4 \leq Cx^4$
woven radishBOT
Renato
spring oasis
C is 2? no
faint gorge
C can be greater than 0
woven radishBOT
faint gorge
u forgot, we wanted to find a C > 0 so that we can estimate the integrand down, so do that
faint gorge
ok show it
faint gorge
no
sqrt(2) =< |x| which implies x =< -sqrt(2) or x >= sqrt(2)
but since x >= 1 we only care about the very latter
winged pumice
Hii
spring oasis
what about the numerator now? @faint gorge
faint gorge
think
winged pumice
brodie you gonna say something related to q?
Bro I need question to solve where is the question I can find to solve my self
calm blade
. @winged pumice
faint gorge
**Renato**
woven radishBOT
spring oasis
,, \frac{x^3 + 5}{x^4 + 4} \ge \frac{Cx^3}{x^4 + 4} \implies x^3 + 5 \ge Cx^3
woven radishBOT
Renato
faint gorge
thats overkill
you can just omit constants, in the numerator
x^3+5 >= x^3 immediately since it's equivalent to 5 >= 0
worn plover
31
spring oasis
x^3+5 >= x^3 immediately since it's equivalent to 5 >= 0
yeah so C1 = 1
and C2 = 2
where C1 is for the numerator and C2 is for the denominator
something like dat
faint gorge
ok sure
spring oasis
I see
I appreciate the help, but this should be pretty much it for the first exercise
devout snowBOT
@spring oasis Has your question been resolved?
faint gorge
I appreciate the help, but this should be pretty much it for the first exercise
indeed
the second one would use a similar approach
using |sin(x)| =< |x|
spring oasis
-1 <= Sin(x) <= 1
0 <= Sin^2(x) <= ?
spring oasis
using |sin(x)| =< |x|
Need more hunts
full geode
-1 <= Sin(x) <= 1
0 <= Sin^2(x) <= ?
1
For any real x, if -1 <= x <= 1 then 0 <= x^2 <= 1
faint gorge
Need more hunts
holds for all x btw
spring oasis
holds for all x btw
We don't have absolute value in the integrand
spring oasis
using |sin(x)| =< |x|
Are you saying it converges
faint gorge
We don't have absolute value in the integrand
so what does my suggestion imply
faint gorge
Are you saying it converges
you are skipping many steps, do it one by one
spring oasis
The problem is we don't have absolute value
faint gorge
that's not a problem
spring oasis
Please elaborate
faint gorge
if |sin(x)| =< |x| but also sin(x) =< |sin(x)| then what do you get by transitivity?
spring oasis
Sin(x) <= |x| doesn't make any sense
faint gorge
Sin(x) <= |x| doesn't make any sense
plot it on paper
anyway by sin(x) =< |x| you can get sin(x) =< x, given that x >= 0
spring oasis
,w sin(1/2)
woven radishBOT
full geode
faint gorge
yea (you may try to prove it some other time for yourself)
faint gorge
so?
faint gorge
yea okay
faint gorge
ye
spring oasis
2/2-3/2
full geode
yea (you may try to prove it some other time for yourself)
Use Taylor series
(If you are going to do this Renato)
faint gorge
too good 🥀
nah i am having a break from work
hexed wedge
I think I’ll ask for help as well
faint gorge
2/sqrt(x) >= (...)
you dont even need the >
spring oasis
you dont even need the >
?
spring oasis
They are not equal
faint gorge
,w 2x/(x^{3/2})
woven radishBOT
spring oasis
,w 2x/(x^{3/2})
This is greater equal to the b) integrand
spring oasis
What now?
spring oasis
well dont skip steps
Ok
spring oasis
what do you have now?
2/sqrt(x) >= (integrand)
faint gorge
you should document your work well not just write the end result, thats what i mean
spring oasis
Ok
faint gorge
2/sqrt(x) >= (integrand)
yes
spring oasis
Sorry I'm in the toilet
faint gorge
consider the integral of 2/sqrt(x) from 0 to 1
spring oasis
consider the integral of 2/sqrt(x) from 0 to 1
It's fucked when x = 0
faint gorge
so take the limit
spring oasis
What limit
faint gorge
instead of fucking x=0
spring oasis
X -> 0
spring oasis
It's infinity, unrefined
spring oasis
1/x as x-> 0
faint gorge
???
spring oasis
Is infinity
harsh stream
it's like saying 1 >= 1
Thats a true statement
harsh stream
Whats the question btw
faint gorge
pin
faint gorge
The lateral limits dont coincide
what do you mean??
spring oasis
The limit doesn't exist when x approaches 0
faint gorge
,, \dots \le \lim_{a\to0}\int_a^1 \f{2}{\sqrt{x}} , dx
woven radishBOT
faint gorge
The limit doesn't exist when x approaches 0
prove it
spring oasis
,w limit of 2/sqrt(x) as x goes to 0
woven radishBOT
spring oasis
,, \dots \le \lim_{a\to0}\int_a^1 \f{2}{\sqrt{x}} \, dx
Why you doing this
ebon coyote
you're taking the limit of the whole integral, not of the integrand
faint gorge
you're taking the limit of the _whole integral_, not of the _integrand_
he would have known it, if i didnt handhold too much
spring oasis
South took the limit of the integrand earlier
faint gorge
thats why he needs to write the stuff down
faint gorge
read your theorem
faint gorge
This one is when one of the bounds is unbounded
use the substitution u=1/x
improper integrals are not just bounds with inf
spring oasis
Wdym
faint gorge
look at your work again
spring oasis
use the substitution u=1/x
This I don't get it
faint gorge
to show your theorem works still
spring oasis
But how to substitute
faint gorge
by substituting
spring oasis
faint gorge
Why is there $\lim_{a\to0} \int_a^1 2\sqrt{u} , du$??
woven radishBOT
spring oasis
,, \dots \le \lim_{a\to0}\int_a^1 \f{2}{\sqrt{x}} \, dx
Is this after substituting
faint gorge
where did you state a substitution? did you check the bounds?
i said to apply the substitution for the theorem
but okay anyway, you need to check the bounds
they dont make sense that way
spring oasis
.close
devout snowBOT
devout snowBOT
Available help channel!
little sonnet
Right so ive just learned integrals. This is all what ive understood => integral ( y ) dx . Integral symbol stands for sum of y . dx which is formula of area for rectangles. Lets take for example x =1; were adding an infinitesimal value to 1 making dx as one of the sides. Thus also adding an infitesimal value to y ( dy) making another side of that rectangle dy. Adding every possible x value gives us the area of the whole function. Now here’s my question: why does the area of a derivative give us its original equation? What does the area of a derivative have to do with its original equation?
devout snowBOT
Right so ive just learned integrals. This is all what ive understood => integral...
supple knot
were you given a proof of one of the fundamental theorems of calculus
little sonnet
What do u mean ?
pure kelp
What do u mean ?
if you really just want to understand the intuition behind why we define the process of integration as the reverse process of taking the derivative, then you might want to recall that a derivative is the slope of the tangent to the curve (our function) defined at all points on the curve if our function is continuous and diffetiable. so its kinda like breaking up the curve into very very tiny points of tangency?
now when you add all these slopes up (integrate it), you get your curve back!
other helpers please correct my language if ive added anything incorrectly
devout snowBOT
@little sonnet Has your question been resolved?
little sonnet
if you really just want to understand the intuition behind why we define the pro...
Thank u this was what i was looking for
river urchin
Hello im new here i got my o'levels exam next really need some help in maths
supple knot
yea just post your question
river urchin
I need to finish the whole probability chapter can we start from the beginning
supple knot
yea just post your question
river urchin
basically need to finiash the whole thing
can you explain it to me i dont understand anything
supple knot
it's expected you do some reading and come up with a math question to use help channels
supple knot
see
supple knot
they used sum of all probabilities equals 1 to solve for x, then plugged x into 3x + 0.2
devout snowBOT
@river urchin Has your question been resolved?
devout snowBOT
devout snowBOT
Available help channel!
junior rose
I need some help
devout snowBOT
I need some help
junior rose
Let be pull it up
I am very confused on how they got this number
I have a 3.2 for P
on my question, but the example question (Image) is confusing me, how did they simplify the last step
lunar harbor
Multiply both sides by $32$. $32 \cdot \frac{1}{4}=8$.
woven radishBOT
Civil Service Pigeon
woven radishBOT
lunar harbor
rip
devout snowBOT
If you are done with this channel, please mark your problem as solved by typing .close
junior rose
.close
devout snowBOT
devout snowBOT
Available help channel!
lilac crescent
,, z^{2n} = -1
devout snowBOT
,, z^{2n} = -1
woven radishBOT
lilac crescent
Ik how to solve z^n = -1
But this one is a bit confusing
I'm trying to write the value of Z_k
,, z_k = e^{\frac{\pi}{2n} + \frac{k\pi}{n}}
woven radishBOT
supple knot
use sqrt(-1) = exp(i pi / 4)
lilac crescent
Is it correct?
lilac crescent
you can check by plugging in your z_k into the first equation
I get exponential of n(k +π)
,, e^{i(n\pi + k\pi)}
woven radishBOT
woven radishBOT
supple knot
yea that looks right
proud perch
Ramn!
fringe patio
devout snowBOT
fringe patio
Am I dumb or did I do it wrong
woven vale
,rccw
woven radishBOT
pure kelp
did you draw a rough diagram?
winter patrol
this is a bit of a trick question
pure kelp
do you know section formula?
winter patrol
note that point P doesn't necessarily have to be between point A and B
pure kelp
here the ratio AP/BP>1
fringe patio
note that point P doesn't necessarily have to be between point A and B
it says p is a point on AB
pure kelp
P divides externally
winter patrol
AB refers to a line passing through A and B extending infinitely
it would be different if they said
P lies on segment AB
pure kelp
it says p is a point on AB
you can use section formula for external division or find the ratio AB/BP with some simple manipulations and apply the formula for internal division
fringe patio
is it B then?
winter patrol
yes
fringe patio
yes
why doesnt ab refer to just joining the 2 points?
or does produce have anything to do with it
pure kelp
read your question carefully again
it says, "P is a point on AB produced"
which could only mean that P does not lie between A and B but lies somewhere else and AB has been produced to P
devout snowBOT
pure kelp
what does produced mean here?
extended
devout snowBOT
Available help channel!
fringe garnet
sec = 1/cos x, I want to isolate the cos x but I forgot how.
devout snowBOT
sec = 1/cos x, I want to isolate the cos x but I forgot how.
fringe garnet
sec x=$\frac{1}{cos x}$
eager nova
What do you mean, like cosx = 1/secx?
fringe garnet
What do you mean, like cosx = 1/secx?
How'd you isolate cos x?
woven radishBOT
Gold/Train
fringe garnet
Can you show it to me?
eager nova
left side = right side
1 / left side = 1 / right side
left side = sec x
right side = cos x
1 / sec x = 1 / (1 / cosx) = cos x
fringe garnet
1 / sec x = 1 / (1 / cosx) = cos x
Proof of 1/sec x = cos x?
fringe garnet
You don't want to isolate cosx
I do.
fringe garnet
you want the proof that cosx = 1/secx?
Well yeah, since I don't really know where it comes from. Is it part of trig identities or?
fringe garnet
Ok you want to veryfy the identity ok
ye...
young spade
$$\sec x = \frac 1{\cos x}$$
$$(\sec x)^{-1} = \left(\frac 1{\cos x} \right)^{-1}$$
$$\frac 1{\sec x} = \cos x.$$
This is also sort of the definition for multiplicative inverses.
eager nova
xD
woven radishBOT
young spade
THERE we go
young spade
no single clue why would you do this tbh
eager nova
Write the definition of cos(x) and sec(x)
fringe garnet
Ok so, gold
Is $$\frac 1{\sec x} = (\cos x)$$ a trig idenity?
woven radishBOT
Gold/Train
fringe garnet
First, give me the definition of cos(x)
Cos (x) is the ratio of the length of the adjacent side over the hypotenuse.
fringe garnet
and the definition of secx?
sec(x) = hyp / adj
fringe garnet
That gives 1 / sec(x)
eager nova
but sec x = hyp / adj
i want you to do
1 / (hyp/adj)
without using the word secx
do you remember how to operate with fractions?
fringe garnet
do you remember how to operate with fractions?
1 / (hyp / adj) = 1 * adj / hyp = adj / hyp = cos(x)
eager nova
nice
eager nova
what you did was using the definition of sec x and cosx to verify that 1/secx = cosx
fringe garnet
what you did was using the definition of sec x and cosx to verify that 1/secx = ...
So Trig verification, right?
eager nova
you can call like that if you want
fringe garnet
you can call like that if you want
Ok, can you help me with my Chaos theory problem?
eager nova
Chaos? i am not a physicist xD
fringe garnet
It's not physics, It's math Lol
eager nova
if it's non related, is better you close this channel which solved your doubt
and open a new one
young spade
just send
fringe garnet
if it's non related, is better you close this channel which solved your doubt
Well, it is actually so my bad.
young spade
Greens Theorem
fringe garnet
Greens Theorem
That's why you're green? Who made that theorem?
eager nova
that's why he is green
young spade
Greens
eager nova
young spade
Its the precessor to stokes theorem
fringe garnet
Its the precessor to stokes theorem
Okay, George Green
young spade
Basically, the integral over a closed curve in a field is equal to the integral over the whole region of the curl of said field one dimension above
Sooo, it relates a 2d curve to a surface in 3d
Stokes Theorem is the generalized form of that for any n-1 and nth dimension manifold(s)
for most multivariable calculus courses, it just feels like a neat trick, meanwhile, its arguably one of the most important identities in calculus, lmao
fringe garnet
Basically, the integral over a closed curve in a field is equal to the integral ...
Here is my first problem.
young spade
okay imma just ask, if you were struggling with the inverse multiplicative relation between cos and sec
Why the hell are you doing this
fringe garnet
Why the hell are you doing this
Dw.
Bro, why does everyone keep dissapearing everytime I'm putting those Chaos Theory problems??????
devout snowBOT
@fringe garnet Has your question been resolved?
viral lynx
Exercise 62
vital edge
They didn't even show themselves lmfao
viral lynx
Anyways, this is the answer that they gave, I got the derivative but why is it >= 0
Why is it not just equal to 0
drifting sierra
It is sometimes equal to 0
vital edge
Oh well they haven't exactly explained what's happening here
viral lynx
I thought the method was to find the derivative and set it equal to 0
drifting sierra
You understand that cos(2x) varies from 0 to 1, right?
vital edge
The derivative is sometimes zero but that doesn't necessarily make that point an extrema
Have you seen the graph of x³
viral lynx
You understand that cos(2x) varies from 0 to 1, right?
Ummm
vital edge
The derivative is 0 at x, but the point is very much not an extrema
viral lynx
Have you seen the graph of x³
Yes
drifting sierra
Sorry, -1 to 1
viral lynx
Ahh oke haha
vital edge
To be an extrema, you need the first derivative to be zero and the second to be non zero
If not, you get weird stuff happening
viral lynx
To be an extrema, you need the first derivative to be zero and the second to be ...
Why non 0
vital edge
Essentially cuz you want the graph to sort of turn around at that point
drifting sierra
If not you can get inflection points
viral lynx
And if it doesnt then there is no extrema?
viral lynx
If not you can get inflection points
Ah ok
Because it will only be increasing or decreasing, right?