seeing that it's a quadratic is probably the hardest part there
#help-39
hazy crypt
hazy crypt
no problemo
lethal plover
.close !
worldly glacier
pearl pondBOT
worldly glacier
hi
pearl pondBOT
worldly glacier
sorry just closed the other one
I knwo that z_o is an eigen value
that's the inner proudc btw
.close
pearl pondBOT
pearl pondBOT
Available help channel!
neat cave
could someone explain how this works?
dont give me the answer I just wanna know why we're allowed do operations like this on a set
cobalt monolith
Think about $\mathbb{R}^2$ for instance
jolly parrotBOT
Cain
neat cave
what does that do
cobalt monolith
Aren't you familiar with the real plane?
neat cave
like the coordinate axis?
cobalt monolith
Yes
jolly parrotBOT
Cain
cobalt monolith
exactly
neat cave
so then is it the exact same concept with C^3
cobalt monolith
yes
neat cave
just instead theres 3 values
cobalt monolith
how many in C^2?
cobalt monolith
no
cobalt monolith
You have (0,0) in the plane right?
neat cave
yea
cobalt monolith
(a,a),(b,b) are also there
pearl pondBOT
pearl pondBOT
Available help channel!
burnt storm
Hi there! If I want to prove that m and n (as defined in the screenshot) are divisible by 17, can I solve it like this (treating it kinda like a regular equation)? If not, what’s the reason/proof/intuition behind it? Is there a good resource with guidelines in relation to stuff like this? I searched online and I couldn't find anything on solving congruences like this.
I was also able to prove it using algebra, but wanted to do so with congruences just to test my knowledge. Thank you! :)
pearl pondBOT
Hi there! If I want to prove that m and n (as defined in the screenshot) are div...
@burnt storm Has your question been resolved?
pearl pondBOT
@burnt storm Has your question been resolved?
burnt storm
<@&286206848099549185>
pearl pondBOT
@burnt storm Has your question been resolved?
haughty pagoda
could someone explain how this works?
dont give me the answer I just wanna know ...
(a,b)^3
fallen cobalt
Hi there! If I want to prove that m and n (as defined in the screenshot) are div...
Is it that you know mn = m^2 + n^2 (mod 17) and we want to show m = 0 (mod 17) AND n = 0 (mod 17)?
burnt storm
yes
fallen cobalt
OK let me play around with this for a few minutes then
burnt storm
mn = 0 (mod 17) and m^2 + n^2 = 0 (mod 17) too actually
(sorry i'm kinda new to this stuff)
fallen cobalt
Ok whats the original question then lol
burnt storm
a simplified version of this where i let c and d = 0 just to experiment before tackling it fully
and i let a = 7m and b = 7n
fallen cobalt
mn = 0 (mod 17) and m^2 + n^2 = 0 (mod 17) too actually
How do you know this? Im not sure thats obvious to me
burnt storm
a = 7m, b = 7n, c = 0, d = 0
(i) 7m + 7n = 2023
=> m + n = 289
(ii) 49mn/2023 = k (k is a natural number)
=> 7mn/289 = k
=> 7mn is divisible by 289
=> mn is divisible by 289
Since 289 = 17^2, mn must also be divisible by 17 i.e. mn = 0 (mod 17)
(iii) (49m^2 + 49n^2)/2023 = l (l is a natural number)
=> (7)(m^2 + n^2)/289 = l
=> m^2 + n^2 is divisible by 289
Since 289 = 17^2, m^2 + n^2 must also be divisible by 17 i.e. m^2 + n^2 = 0 (mod 17)
I might've made an assumption/mistake somewhere but that was my logic
fallen cobalt
You assumed c and d are zero which may not be true
burnt storm
I know. As I said, I was just tackling the case where c = 0 and d = 0 to mess around with the problem a bit before coming to a full solution
burnt storm
Hi there! If I want to prove that m and n (as defined in the screenshot) are div...
It was just about whether you're allowed to treat a congruence equation like this where you let the factors of each side equal each other
But yeah I get what you mean about the flaw that n could have more prime factors
I watched a bit of your stream
fallen cobalt
and was able to show that a^2 + b^2 = 0 mod 2023 and c^2+d^2 = 0 mod 2023
I think if you use the divisibility by 7 here and use prime factorizations then you don't run into the n^2 issue
fallen cobalt
I watched a bit of your stream
oh thank you 
I feel like i was close but was on a bit of a time crunch
burnt storm
a = 7m, b = 7n, c = 0, d = 0
(i) 7m + 7n = 2023
=> m + n = 289
(ii) 49mn/2023 ...
This is the algebra way that I was able to prove that both m and n are divisible by 17. Then I tried using the congruences, and got the same answer that m and n must be divisible by 17, but I wasn't not sure if that was just coincidence or a rigorous solution
fallen cobalt
yes i think this is a good argument
n^2 must have two copies of 17
yeah thats usually how I think about stuff. I like prime power factorizations for trying to figure out where the missing factors come from
I gotta run though, sorry I couldnt be of more help
burnt storm
No problem, you helped me to think about it and I appreciate the ideas you put forward 😄
pearl pondBOT
@burnt storm Has your question been resolved?
pearl pondBOT
@burnt storm Has your question been resolved?
burnt storm
.close
tired zealot
How do you do 5(ii)?
pearl pondBOT
How do you do 5(ii)?
tired zealot
, rotate
jolly parrotBOT
tired zealot
not sure if you can sub in if its cut the curve at 2 distinct points i only know that if its tangent to the curve you are able to sub in y
cloud zephyr
not sure if you can sub in if its cut the curve at 2 distinct points i only know...
yes, you can sub, and you'll get 2 distinct points if it's the case of 2 distinct intersections
cloud zephyr
hmm, lemme reread
tired zealot
i did get c^2-2c>-97 not sure if this shows it
cloud zephyr
did you sub the y in the line into the quadratic before finding the discriminant?
tired zealot
yes i did
tired zealot
oh you are supposed to use c=3 for part (ii) too?
cloud zephyr
let's see
tired zealot
,rotate
jolly parrotBOT
tired zealot
not sure if this shows it or did i mess up somewhere
cloud zephyr
i get it now
tired zealot
oh?
cloud zephyr
you'll have to use completing square
which actually
it's just in the form of
(-c+1)²+96
since (-c+1)² is always ≥0
so (-c+1)²+96>0
discriminant>0
therefore we can conclude it has 2 distinct roots
tired zealot
ohhh
is such things like completeing a square like how do u figure that you are required to use it
the only thing i know is that if there is some weird value of x^2 like 0.01x^2 then i wld complete the square and +c+ b those constants then i would use discirminant but well here c^2 had a coefficent of 1
cloud zephyr
is such things like completeing a square like how do u figure that you are requi...
usually, when we need to prove something is always greater than (or equal to) 0, we will use completing square method to make the quadratic become the form of
(something)²+postive number
tired zealot
discriminant>0
therefore we can conclude it has 2 distinct roots
wait but isint (c-1)^2>-96 doesnt that mean as long as discriminant>-96 there would be 2 distinct roots or must it always be >0
tired zealot
usually, when we need to prove something is always greater than (or equal to) 0,...
ohh
cloud zephyr
well, the discriminant is
(c-1)²+96, but not (c-1)²
tired zealot
so wouldnt that mean in this case if discrimiant >-96 in this scenario there would be 2 distinct roots?
cloud zephyr
yea, (c-1)² can be zero.
but (c-1)²+96 is 96
tired zealot
ohhhhhh
cloud zephyr
in the case c=1
pearl pondBOT
cloud zephyr
cheers!
pearl pondBOT
Available help channel!
opaque hearth
[\text{Find equation of the line tangent to}]
[x=2cos(t),y=2sin(t),\text{ at t=}\frac{\pi}{4}]
opaque hearth
To solve would I first need to convert to parametric equation?
Or can I solve with x and y being seperate equations?
jolly parrotBOT
dopediscorduser
merry carbon
You don't need to put text inside an equation environment, you can just do it like
[
x = 2\cos(t), y = 2\sin(t)
]
jolly parrotBOT
@merry carbon
merry carbon
You can keep them as parametric equations for the time being sure, making use of the chain rule ⛓️
opaque hearth
You can keep them as parametric equations for the time being sure, making use of...
[x' = -2\sin t]
[y' = 2\cos t]
[m = -\frac{\cos \frac{\pi}{4}}{\sin \frac{\pi}{4}} = -1]
[x(\frac{\pi}{4})= 2cos(\frac{\pi}{4}) = \sqrt{2}]
[y(\frac{\pi}{4}) = \sqrt{2}]
[y -\sqrt{2} = -(x-\sqrt{2})]
Have I gone wrong here?
merry carbon
That’s not too bad - you need to put in t=pi/4 into the m you have though 
opaque hearth
That’s not too bad - you need to put in t=pi/4 into the m you have though <:RooP...
pi/4 would be m?
Oh oh
Nvm
merry carbon
jolly parrotBOT
dopediscorduser
opaque hearth
That’s not too bad - you need to put in t=pi/4 into the m you have though <:RooP...
Like this you mean?
$y = 2\sqrt{2} -x$
jolly parrotBOT
dopediscorduser
merry carbon
Like this you mean?
yep like that!
quaint crown
How many variations can there be if the length of the line is from 1 to 100, and the symbols that can be used are all the letters of the English language and all the numbers(from 0 to 9)?
It's a 1^36+2^36+3^36+...+100^36?
vestal tapir
pearl pondBOT
pearl pondBOT
Available help channel!
waxen smelt
when does a doubel integral give area and when does it give volume?
waxen smelt
i am having difficulty in understanding that
like double integral can give volume of a 3d surface
but so can tripple integral
but double integral also gives area?
how are they all different?
pearl pondBOT
@waxen smelt Has your question been resolved?
warm current
when does a doubel integral give area and when does it give volume?
a double integral is just an abstract concept. What it gives you depends on thje units you are choosing.
a double integral can give you anything.
waxen smelt
a double integral can give you anything.
Yeaa thats what i am having a difficult time understand. when does it give what?
warm current
depends on your units
waxen smelt
i am a physics student so the way i imagine it is like
i take a small square
of length and height
dx dy
warm current
Then you should know dimensional analysis
waxen smelt
and we sum it all over
warm current
Use that to get your answer
waxen smelt
so an integral sign comes over
wanton steeple
pearl pondBOT
wanton steeple
i have no clue how to even proceed
woven matrix
a point of the curve has coordinates (x, y) which satisfies the equation
if it intersects the y-axis, it means x = 0
so it's a system of two eq
x^3+y^3 = xy+x+y
x = 0
substitution, done
modern fjord
and tangent as well
wanton steeple
so it's a system of two eq
x^3+y^3 = xy+x+y
x = 0
Okay
woven matrix
your 3 points of intersection are the solutions and then you write the equations for the tangents at these 3 points
late sun
My task is to do this by calculation
this is my workings but somehow I've ended up with the wrong statement and I'm not sure how
$$ x\in(S\T)\cap (S\U) \iff x\in {z:X | z \in s \wedge z \notin T} \cap {z:X | z\in S \wedge z \notin U $$
nimble lily
what did you do so far
late sun
what did you do so far
i didnt mean to send it, im trying to type my tex sorry
nimble lily
no worries, slightly odd that it cut off the tex haha
nimble lily
I also got an agda/lean vibe from that
late sun
is this lean or smth
This is the written part, but yeh i am assessed partly in lean, its a weird module
Sorry for the delay, I'll have to learn more latex lol
late sun
Because i was copying the question
shrewd basin
so where does the proof start?
shrewd basin
\[x \in (S\setminus T) \cap (S \setminus U) \iff x \in \{z : (z \in S \land z\notin T ) \} \cap \{z : (z \in S \land z\notin U ) \} \]
\[ \iff x \in \{z : (z \in S \land z \notin T) \lor (z \in S \land z \notin U) \} \]
\[ \iff x \in \{z : z \in S \land (z \notin T \lor z \notin U) \}
\]
idk if i made a mistake
i did
watever
omd
okay let me try to fix this one sec
jolly parrotBOT
Pure
late sun
thanks for taking the time to fix the very awkward tex, how come on the second line, the $\cap$ turns into a $\lor$ and not $\land$
jolly parrotBOT
lewis_f04
shrewd basin
yeah idk im so not fimiliar writing set proof like this one sec lemme just think
[x \in (S\setminus T) \cap (S \setminus U) \iff (x \in S \land x \notin T) \land (x \in S \land x \notin U) ]
right
jolly parrotBOT
Pure
late sun
yeh makes sense
shrewd basin
,, (x \in S \land x \notin T) \land (x \in S \land x \in T) \iff (x \in S) \land (x \notin T \land x \notin U)
jolly parrotBOT
Pure
shrewd basin
,, (x \in S) \land (x \notin T \land x \notin U) \iff (x \in S) \land (x \notin T \cap U)
jolly parrotBOT
Pure
shrewd basin
,, (x \in S) \land (x \notin T \cap U) \iff (x \in S) \land (x \in (T \cap U)^c)
yey
jolly parrotBOT
Pure
shrewd basin
now use de morgans law
,, (x \in S) \land (x \notin T \cap U) \iff (x \in S) \land (x \in T \cup U)
jolly parrotBOT
Pure
late sun
and the in goes to not in
late sun
ive seen demorgans law applied to sets, let me check the lecture notes for the form
shrewd basin
brb give me 1 min
late sun
ok
shrewd basin
okay this is not gonna work
lemme just write down in words
let's take x in S\ (TUV)
then x is in S but not in the union of T and V
so x is in S but not in either T or V
then x is in S but not in T and x is in S but not in V
so S\ (TUV) subsets (S\ T) cap (S\ V)
late sun
yeh i think we started this all wrong, ive got to the answer with de morgans
$\iff x\in S \land \neg (x\inT \land x\in U)$
$\iff x\inS \land x\notin T \lor x\notin U$
shrewd basin
u have to seperate \in T
jolly parrotBOT
lewis_f04
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
shrewd basin
$\iff x\in S \land \neg (x\in T \land x\in U)$
$\iff x\in S \land x\notin T \lor x\notin U$
late sun
stupid tex
jolly parrotBOT
Pure
late sun
i think that argument holds tho right? but im not sure what is wrong with my previous method of using the definitions
shrewd basin
well i think if x is in S and notin T AND x is in S and notin U then it follows that x is in S and x is not in either T and U.
yeah thats it
if x is not in the interserction of T and U then that means x is not in one of them or both
so it becomes x notin T or x notin U
late sun
yeh, i guess it has to go to that by this much shorter way of working
pearl pondBOT
@late sun Has your question been resolved?
terse epoch
can you assume a formula you are trying to prove is polynomial in your proof? If you can't, can you prove a function(which is the formula in this case) polynomial or rational function if all you know is that it is function of x and f(x)-f(x-1)=P(X), P(X) is another polynomial function?
terse epoch
I know that difference of two polynom is definitely polynomial but what if there is non polynomial expression in the function "f" but non polynomial part cancels each other? is this impossible
pearl pondBOT
@terse epoch Has your question been resolved?
pearl pondBOT
@terse epoch Has your question been resolved?
grizzled lodge
Saw this earlier today
pearl pondBOT
Saw this earlier today
hallow kite
Saw this earlier today
This is third time i see that
grizzled lodge
modern fjord
I’ve seen this as well lmao I have no idea how to approach it 💀
grizzled lodge
I think it’s some kind of Olympaid q
grizzled lodge
Not really that just gives f =0 as a solu I think
cinder flower
pure will know how to do it
grizzled lodge
meager trellis
...are we given any conditions like continuity or just that equation? since if it's just that equation i'm pretty sure there's going to be loads of bizarre solutions that do random nonsense at irrationals
grizzled lodge
It’s just this I’m pretty sure
shrewd basin
Trivially, \( f(x) = 0 \) is a solution. Now assume \(\forall, f(x) \neq 0\).
Swapping \(y\) and \(x\) immediately gives us \( f(-x) = -f(x) \).
Now setting \( y=-x \), we have
\begin{align*}
-f(x)f(x)f(2x) &= -x^2f(x) - x^2f(x) \\
f(2x)f(x) &= 2x^2 \quad (f(x) \neq 0 \text{so division is fine}) \\
\end{align*}
Putting $2x$ and $x$ we have
\begin{align*}
f(2x)f(x)f(x) &= 4x^2f(x) - x^2f(2x) \\
f(2x)f(x)^2 &= x^2(4f(x) - f(2x)) \\
2x^2f(x) &= x^2(4f(x) - f(2x)) \\
2f(x) &= 4f(x) - f(2x) \\
f(2x) &= 2f(x)
\end{align*}
Pure
shrewd basin
from the two equations we have its pretty easy to deduce $f(x) = \pm x$ are the only solutions besides f being identically 0
jolly parrotBOT
Pure
modern fjord
🐐
grizzled lodge
Hm yeah but what if like we have f(x) = x for some x then for other x’s f(x) = -x
shrewd basin
suppose f(x1) = x1 and f(x2) = -x2 then you can show x1 = x2 = 0
cinder flower
pure i love you
grizzled lodge
.close
pearl pondBOT
grizzled lodge
Wat
pearl pondBOT
Available help channel!
pearl pondBOT
pearl pondBOT
Available help channel!
chrome crescent
How do I get B
pearl pondBOT
How do I get B
chrome crescent
pearl pondBOT
@chrome crescent Has your question been resolved?
bold yarrow
how do you find the angle between 2 lines (given the equation of the lines)
regal herald
in what form are said lines given
bold yarrow
just the normal y int form
unkempt yacht
in what form are said lines given
i assume they both are linear equations
bold yarrow
wait ill send pic of the question
jolly parrotBOT
ed
bold yarrow
I’m just stuck on the last part of it
unkempt yacht
I’m just stuck on the last part of it
find the coordinates of A and B first, then find the linear equation of the tangent line at A and B
i assume you already did this?
bold yarrow
yes
unkempt yacht
if yes, then the formula for the angle between two y=mx+b equations is $\tan\theta = \frac{m_1 - m_2}{1 + m_1m_2}$
jolly parrotBOT
FungusDesu
bold yarrow
i got the 2 equations
unkempt yacht
where m1 and m2 are the slopes of two equations
bold yarrow
ill just sub in the gradients to find angle right
unkempt yacht
correct
unkempt yacht
i would have thought 0
same, numerator < denominator so therefore 0
regal herald
no its 1/x^2
last summit
pearl pondBOT
unkempt yacht
for limits like this, a way to solve them fast is to simply remember a rule
- numerator's highest factor > denominator's highest factor -> infinity
- NHF = DHF -> NHF coefficient/DHF coefficient
- NHF < DHF -> 0
pearl pondBOT
Available help channel!
fossil drum
i’m having trouble figuring out the abs max and min and rel max and min
midnight haven
M is the absolute maximum if M = f(c) for some c in D, and f(x) ≤ M for all other x in D.
Basically highest x value
fossil drum
and if there are two highest x value is there an absolute max
midnight haven
and if there are two highest x value is there an absolute max
There can't be two highest x values
one will be greater
fossil drum
i. trying to figure out the question marks in the image whether they are neither or max or min
midnight haven
like the one on ur pic
midnight haven
i. trying to figure out the question marks in the image whether they are neither...
Oh because
first is open circle
fossil drum
If we have a closed interval I know the end points can be considered as a max or min but what if we have (x, x]
do we consider just the right end point
or neither
fossil drum
yeah same
midnight haven
Is this for AP Calc?
fossil drum
yeah basically taking calc 1 at a uni though
midnight haven
Have you learned critical values
fossil drum
yeah
midnight haven
ok so if ur endpoints aren't max or min
they are critical points
Or where f'(x)=0 or undefined
midnight haven
i’m having trouble figuring out the abs max and min and rel max and min
Also endpoints cant be max or min
Because it needs to be higher on both sides
if it makes sense
Like look at the (4,39) dot
Its the highest x value from right and left
but endpoints we don't know the other side
if it makes sense
pearl pondBOT
@fossil drum Has your question been resolved?
hollow saffron
does anyone which part of my answer is wrong?
pearl pondBOT
does anyone which part of my answer is wrong?
violet tundra
does anyone which part of my answer is wrong?
can you send your working
civic crane
does anyone which part of my answer is wrong?
constant and second term
hollow saffron
can you send your working
i did the antiderivative
F(x) = 4x^(3/2) + 2x^(5/2) + C
then i did
9 = 4(1)^(3/2) + 2(1)^(5/2) + C
9 = 6 + C
3 = C
plush bramble
i did the antiderivative
F(x) = 4x^(3/2) + 2x^(5/2) + C
then i did
9 = 4(1)^(3/...
Read lonelys comment
hollow saffron
Does that get you 10x^(3/2)?
i got 5x^(3/2) after differentiated
hollow cobalt
Right
So what needs to be changed about the coefficient?
What would make it 10x^(3/2) after differentiated
pearl pondBOT
@hollow saffron Has your question been resolved?
pearl pondBOT
pearl pondBOT
Available help channel!
fluid lark
Hi
hollow cobalt
With what?
nimble lily
isn't this the generalised schrodinger equation or something? i.e. why are you trying to interact with it
nimble lily
yeah, and the conjugate of psi makes me think quantum
hollow cobalt
Probably some part from the standard model equation
@fluid lark Anyway do you have a question to post?
midnight haven
Quite trivial if I'm being honest.
rustic gate
,,\mathcal{L}{G W S}=\sum_f\left(\bar{\Psi}f\left(i \gamma^\mu \partial \mu-m_f\right) \Psi_f-e Q_f \bar{\Psi}f \gamma^\mu \Psi_f A\mu\right)+ \
+\frac{g}{\sqrt{2}} \sum_i\left(\bar{a}L^i \gamma^\mu b_L^i W\mu^{+}+\bar{b}L^i \gamma^\mu a_L^i W\mu^{-}\right)+\frac{g}{2 c_w} \sum_f \bar{\Psi}f \gamma^\mu\left(I_f^3-2 s_w^2 Q_f-I_f^3 \gamma_5\right) \Psi_f Z\mu+ \
-\frac{1}{4}\left|\partial\mu A\nu-\partial_\nu A_\mu-i e\left(W_\mu^{-} W_\nu^{+}-W_\mu^{+} W_\nu^{-}\right)\right|^2-\frac{1}{2} \mid \partial_\mu W_\nu^{+}-\partial_\nu W_\mu^{+}+ \
-i e\left(W_\mu^{+} A_\nu-W_\nu^{+} A_\mu\right)+i g^{\prime} c_w\left(W_\mu^{+} Z_\nu-\left.W_\nu^{+} Z_\mu\right|^2+\right. \
-\frac{1}{4}\left|\partial_\mu Z_\nu-\partial_\nu Z_\mu+i g^{\prime} c_w\left(W_\mu^{-} W_\nu^{+}-W_\mu^{+} W_\nu^{-}\right)\right|^2+ \
-\frac{1}{2} M_\eta^2 \eta^2-\frac{g M_\eta^2}{8 M_W} \eta^3-\frac{g^{\prime 2} M_\eta^2}{32 M_W} \eta^4+\left|M_W W_\mu^{+}+\frac{g}{2} \eta W_\mu^{+}\right|^2+ \
+\frac{1}{2}\left|\partial_\mu \eta+i M_Z Z_\mu+\frac{i g}{2 c_w} \eta Z_\mu\right|^2-\sum_f \frac{g}{2} \frac{m_f}{M_W} \bar{\Psi}_f \Psi_f \eta
jolly parrotBOT
fast zealot
I dont know if I should be scared or impressed tbh
plain spade
please provide more context for this problem
midnight haven
I dont know if I should be scared or impressed tbh
yeah same
I'd be scared if I was writing something that long and decide to write \frac{g}{2}
rustic gate
midnight haven
yeah man
tawny delta
rustic gate
tawny delta
pearl pondBOT
@fluid lark Has your question been resolved?
fluid lark
- posts part of standard model
True
pearl pondBOT
fluid lark
,,\mathcal{L}_{G W S}=\sum_f\left(\bar{\Psi}_f\left(i \gamma^\mu \partial \mu-m_...
Mf what
Nahhhh
These nerds bro
To smart
pearl pondBOT
Available help channel!
vernal siren
pearl pondBOT
vernal siren
i used a u substitiution of u=x-1 and worked through and ended up getting 7/2 + 2ln6 so not sure if went wrong somewhere
grave thistle
show work
stone python
can anyone help me with this question
pearl pondBOT
can anyone help me with this question
stone python
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copper sierra
.
the intensity of a sound wave increases by 1000 w/m2 what is this increase equal to in decibels
copper sierra
what formula do i apply
the 10log I/Io
?.
um i applied to btw but my answer was wrong
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pearl pondBOT
grizzled lodge
71?
feral sedge
Horizontal asymptotes is end behavior
grizzled lodge
divide the top and bottom thing by e^x
feral sedge
What happens when x goes to infinity or negative infinity
grizzled lodge
why would u do ln
no
you would get -1 when x goes to inf
but something else when x goes to -inf
no
what happens when x is a very big negative number
e^x yeah
so youd just have...
no where did the 2 go
its (8 - something small)/(2+something small)
okay
..
thats only useful for x -> inf
which we already know is -1
because it doesnt help when x goes to -inf
the original expression already works
like u said you get weird inf\inf stuff
inderterminate
well i think you would have to show the function is not gonna go above or below that by taking derivative
i guess solving the equation y = -1 and y=4
show theres no root?
and then use the fact that y is continuous
then that means it doesnt cross
(or go above or below them)
,, \frac{8-e^x}{2+e^x} = 4
jolly parrotBOT
Faq
grizzled lodge
show therers no such x that satisfies this
idk this might become a lil circular 
yeah if youre fine with the range of e^x
e^x > 0 for all x
which is easy to show because exp(x) = exp(x/2 + x/2) = exp(x/2)^2 >=0
and then
since we have exp(x) = exp(k)exp(x-k) if exp(k) is 0 then exp(x) is zero for all x but thats clearly not the case
hazy pilot
pearl pondBOT
@midnight haven Has your question been resolved?
hybrid basin
Hello
hybrid basin
$\int 6x^3e^{6x^2}dx$
jolly parrotBOT
LE SSERAFIM
hallow kite
**LE SSERAFIM**
DI method
hybrid basin
Haven't learned that yet
hybrid basin
I started using IBP
But it looks like a continuation of IBP
hallow kite
Well it's same like you use integration by part , but better organise
Let me send video
hybrid basin
And for u-sub I don't know what to sub in for u
hybrid basin
Let me send video
Video?
hallow kite
YouTube
Check this
@hybrid basin it's same rule of how to integral function
But he write it in Better way
hallow kite
but we haven't learned it in class
I think you will never learn it in class 👀
Well let me think about way
You know u-sub? @hybrid basin
Can u use it?
hybrid basin
Yes
hybrid basin
Can u use it?
We can
hybrid basin
Or u need help?
I do need help
hallow kite
I do need help
Well , first u need to select best value for your u= something
By checking your function we can select u = x² because it's best one for exponential
Now find your derivative u
Du= x*dx/2
2du = xdx
Now replace those value in your function
hallow kite
Du= x*dx/2
2du = xdx
Now replace those value in your function
Can you do that? @hybrid basin
hybrid basin
But u = x^2
hallow kite
Yea
So we will replace x² with u
But what about second x? 
Try to replace your dx with du and you will find the answer
hallow kite
6 $\int \frac{ue^{6u}}{2}du$
jolly parrotBOT
Μήλιος
hallow kite
@hybrid basin
So first i take 6 outside
Then i replace e^6x² by e^(6u)
Then for rest i have x³dx
I write it like x² * x dx
This equal to u * 0.5du
@hybrid basin
Now you see how i write the x³?
hallow kite
Then for rest i have x³dx
I write it like x² * x dx
This equal to u * 0.5du
All those replace them in your Integral
hybrid basin
Uhhh
hallow kite
I'm confused
Where?
hallow kite
I just thought you were saying that I am wrong
Oh , i just try to write the explain and i didn't check your message
So now let integral it
hallow kite
Do I just ibp now?
Yes
hybrid basin
Yes
hallow kite
Yes
Well about DI it's something like ibp
It's just that you put your integral and derivatives into table which make the work very easy
hybrid basin
I know, it's just I'm not sure if our prof will allow us using DI method
And so just to be safe
$3\int ue^{6u}=u\frac{1}{6}e^{6u}-\int e^{6u}du$
hybrid basin
Take f= u and g'=e^(6u)
Like this?
jolly parrotBOT
LE SSERAFIM
hybrid basin
$$3\int ue^{6u}=u\frac{1}{6}e^{6u}-\frac{1}{6}e^{6u}+C$$
$$3\int ue^{6u}=\frac{1}{6}e^{6u}\left(u-1\right)+C$$
jolly parrotBOT
LE SSERAFIM
hybrid basin
Yay?
hallow kite
**LE SSERAFIM**
Here -integral of e^(6u)/6 du
Because you g(x) = e^(6u)/6
This multiple f'(x) = 1
So it's= g(x)
hallow kite
**LE SSERAFIM**
So you will have small edit here and write -1e^(6u)/36
hybrid basin
Why is it negative?
hallow kite
And some calc and u will get:
=$\frac{ue^{6u}}{2}$ - $\frac{e^{6u}}{12}$
hybrid basin
Huh?
hallow kite
Negative int e^(6u) du
hallow kite
**LE SSERAFIM**
From here
jolly parrotBOT
Μήλιος
pearl pondBOT
pearl pondBOT
Available help channel!
lean dome
Hello! I need help with Solving Inverse Trig. Functions. My goal here is to isolate X without a calculator and using the unit circle (limited to [0,2π)).
lean dome
I have made it all the way through sin^-1(-2), though I don't know where to proceed from there. Even desmos says its undefined.
worldly quail
Desmos is right
sin(x) takes values between -1 and 1, so sin(x) + 2 takes values between 1 and 3
sin(x) + 2 cannot be zero
lean dome
sin(x) takes values between -1 and 1, so sin(x) + 2 takes values between 1 and 3
That is very clever I did not think of it that way
Thanks a lot I can sleep peacefully now
🙏
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pearl pondBOT
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inland nacelle
pearl pondBOT
inland nacelle
need help with this question
cannot seem to get the right answer
i will post my work in a bit
if someone could take a look at my work and see where i made a mistake that would be greatly appreciated
i just looked at my answer again and i just missed the negative on the quadratic equation
i put it in it is good my bad
lol should have looked at it more carefully before putting it in
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heady elm
hi i am stuck on this question
pearl pondBOT
hi i am stuck on this question
wooden merlin
$(2n-1)^2 \neq 4n^2 -1$
jolly parrotBOT
Good
heady elm
oh
rigid junco
need help
near haven
yes
near haven
second derivative test
unkempt yacht
$3^{\frac{3+\sqrt{37-8x}}2} + 3^{\frac{3-\sqrt{37-8x}}2} = 12$
jolly parrotBOT
FungusDesu
unkempt yacht
when pressing this into my calculator it gives me $x = \frac92$
but i want to know the step by step solution
jolly parrotBOT
FungusDesu
pearl pondBOT
@unkempt yacht Has your question been resolved?
pearl pondBOT
pearl pondBOT
@midnight haven Has your question been resolved?
summer imp
First off it's 4n, not 4^n.
Maybe notice that for large n, sqrt(n^3+4n) = sqrt(n^3).
For large n, 4n is insignificant compared to n^3.
So 4n+n^3 is essentially n^3.
You could do this limit by checking the leading coefficients tbf
Sorry I didn't mean coefficients, rather then leading terms' degrees
Here, the top has degree 2, the bottom has (effectively) degree 3/2
Yes
forest vine
Question regarding conrvergent series. Need to prove the convergence of these 2 series using the cauchy criteria but having a hard time applying it.
pearl pondBOT
Question regarding conrvergent series. Need to prove the convergence of these 2 ...
@forest vine Has your question been resolved?
pearl pondBOT
@forest vine Has your question been resolved?
frosty wigeon
for the first one, I tried using this inequality $\frac{1}{k^2} \leq \frac{1}{k^2-k} \leq \frac{1}{k} - \frac{1}{k-1}$
jolly parrotBOT
Macacofonico
frosty wigeon
so when you do $|a_m - a_n | = | \sum_{k=m}^{n} \frac{1}{k^2} |$
jolly parrotBOT
Macacofonico
frosty wigeon
you have some telescopic terms that cancel out
for the second I would try something like this (suppose that $m\geq n-1)$
jolly parrotBOT
Macacofonico
frosty wigeon
$|b_m - b_n | = | \sum{k=m}^{n} \frac{(-1)^n}{k!} | \leq \sum{k=m}^{n} \frac{1}{k!} = \frac{1}{m!} + \frac{1}{(m+1)!+\cdots + \frac{1}{n!}} \leq \frac{m-n}{m!} \leq \frac{m}{m!}$
jolly parrotBOT
Macacofonico
frosty wigeon
and you can make this arbitrarly small for large m
forest vine
hold up
so we want to show that for this partial sum, this inequality is valid
for n>=m>=n_ε
this inequality means: the absolute value of the partial sum of the sequence a_k from the mth to the nth term
is smaller than ε
so we need to calculate the partial sum of all the terms to the nth term, and subract the partial sum of all the terms upto the mth term from it
So | s_n - s_m | < ε
and that would be this
which for our sequence would be this
I honestly don't know how to evaluate a series for convergence using the cauchy criterion.
normally, we would just evaluate the sequence in the partial sum, and if the sequence converges to 0, then we would know that the series is also convergant. but idk how to use the cauchy criteria
<@&286206848099549185>
Do I need to plug in actual values for m and n and n_ε for the test?
<@&286206848099549185> anyone available for a real analysis question?
if anyone sees this and can help please write me a dm. Much appreciated
pearl pondBOT
@forest vine Has your question been resolved?
dusk dagger
for the first one, I tried using this inequality $\frac{1}{k^2} \leq \frac{1}{k...
You messed up the signs here but the idea seems right
For the second part, the trick is to note that the sequence is an oscillating sum of terms of strictly decreasing magnitude, and thus can easily be bounded.
dusk dagger
so we want to show that for this partial sum, this inequality is valid
To be exact, you need to show for any $\epsilon$ there is a m such that this is true for all n. Which is the same as showing it for $n = \infty$.
You do not need to calculate the entire sum of the sequence; that automatically shows it is convergent and would make the test redundant.
Instead, the point of the Cauchy convergence test is that you can pick a large enough m such that showing the sum = 0 will be easy. You never need to calculate the earlier terms and certainly do not have to calculate the whole sum.
jolly parrotBOT
chencking
forest vine
To be exact, you need to show for any $\epsilon$ there is a m such that this is ...
I watched a video where the guy used n=m+1 so that when he'd be evaluating the inequality, it would turn out to a_n > ε
But I've never used the cauchy criterion to evaluate a series for convergence before.
could you possibly use a simple example to show me how it works?
dusk dagger
Consider the sequence $a_n = \frac{\sin(n)}{2^n}$.
I have no idea how to evaluate the sum, but for any $\epsilon > 0$ there is a $k > 0$ such that $\epsilon > 2^{-k}$.
For any $m > 2k$, $|\sum_{n = 2k}^m a_n| \leq \sum_{n = 2k}^m |a_n| \leq
\sum_{n = 2k}^m 2^{-n} < 2^{1 - 2k} < 2^{-k} < \epsilon.$
Remark: Note taking the limit as m goes to infinity gives that $| \sum_{n = 2k}^\infty a_n| = 0$, which automatically gives that the sequence converges (since the sum of the first m terms is always finite).
An equivalent but more direct argument would have been: $|\sum_{n = 2k}^\infty a_n| \leq \sum_{n = 2k}^\infty |a_n| \leq \sum_{n = 2k}^\infty 2^{-n} < 2^{1 - 2k} < 2^{-k} < \epsilon.$
jolly parrotBOT
chencking
dusk dagger
I watched a video where the guy used n=m+1 so that when he'd be evaluating the i...
Yes, that case is for proving the sequence does not converge. If $a_n$ never goes to zero, then it automatically diverges.
I might have phrased my earlier comment about it being the same as $m = \infty$ a bit poorly (there could be edge cases I do not see right now) but nevertheless the intuition still holds. The cauchy convergence test is a formalization of the idea that a sum converges if and only if the later terms eventually become arbitrarily small.
jolly parrotBOT
chencking
pearl pondBOT
forest vine
.reopen
pearl pondBOT
✅
forest vine
so, my goal is to show that the lim |s_n - s_m| approaches 0
that would mean that the next term that would be summed up in the partial sum, is always getting smaller, therefore the series is convergent.
that make sense?
frosty wigeon
If I understood correctly, your statement doesn't work for the partial sum $\sum\limits_{n=0}^{k} \frac{1}{n}$ because the next term $\frac{1}{k+1}$ that is summed up is getting smaller, but the (harmonic) series $\sum\limits_{n=0}^{\infty} \frac{1}{n}$ diverges
jolly parrotBOT
Macacofonico
pearl pondBOT
@forest vine Has your question been resolved?
forest vine
If I understood correctly, your statement doesn't work for the partial sum $\sum...
yo are you still there?
the problem is that this stuff is still really new to me. I literally learned them yesterday. so I don't really know how to apply them properly
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pearl pondBOT
pearl pondBOT
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lucid frost
pearl pondBOT
lucid frost
Also I have a question about this- it is all good when they are both positive, but what happens if the top one is -9. If it was +9 we would have said the interval from 5 to 9. But when it is (-) do we say 5 to -9. But then it doesn't make sense to go from a positive value to a (-) value.
faint prism
👋 You can switch the limits of integration, but you have to change the sign of the integral, like this:
jolly parrotBOT
Nonna
lucid frost
So after I get my final answer do I just -(answer)
faint prism
Intuitively I guess it makes sense because it's a bit like you are integrating from right to left, so your tiny dxs are negative
Applying the fundamental theorem of calculus, you just end up finding that those values are the same
lucid frost
Intuitively I guess it makes sense because it's a bit like you are integrating f...
So -(answer)
faint prism
I don't know what you mean by answer, but if you want to apply this, you have to switch the limits of integration too
Otherwise you can just apply the formula normally and ignore that b < a, it's the same
Because if you use the fundamental theorem of calculus on both sides, you find that
F(b) - F(a) = - (F(a) - F(b))
F(b) - F(a) = F(b) - F(a)
Do you see what I mean?
lucid frost
So at the end I don't need to worry about signs or the answer itself I get the same result as +9
faint prism
I don't know what you mean by answer, but if you want to apply this, you have to...
Are you referring to that function you posted? In this case, that function is defined only for values bigger than 4^(1/3), so it won't work
But you can't change the signs of the integration limits and always expect to get the same result
Just apply the fundamental theorem of calculus and you'll get the right result, even if b < a (← corrected)
If you want to switch the two limits of integration, you can do this, but you are just switching the two limits and the sign of the integral, the signs of the integration limits stay the same!
lucid frost
<:thonk:406575732563902485> Are you referring to that function you posted? In th...
What do you say "even if b>a" for- bc that's the normal equation
faint prism
oh mb, you are right! I meant a > b
pearl pondBOT
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brittle pebble
S is nonempty, a function, and a partial order, find a set S that satisfies this or show why such a set doesn't exist
brittle pebble
!status
pearl pondBOT
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
1
dim zodiac
If you start with a base set X, both a partial order and a function can be viewed as a subset of $X\times X$
jolly parrotBOT
Willow
dim zodiac
In particular, what property’s does this subset of $X\times X$ need to be considered a function
jolly parrotBOT
Willow
dim zodiac
And what properties are needed for it to be a partial order?
brittle pebble
function should be many to one/one to one
partial order properties are anti-symmetric, reflexive and transitive
would f(x) = x work where S is the natural numbers
since it would be reflexive
dim zodiac
Yep!
In general the identity map should work over any space, and also should be the only such possible relation
brittle pebble
identity map?
silver spoke
pearl pondBOT
silver spoke
So I'm thinking that in Z5, that [2] and [3] satisfy part a. That seems too simple, just want a second opinion
lament dagger
[2] + [3] = [0]
merry carbon
Examples don't need to be elaborate all of the time 
silver spoke
Kk. Ty
Yeah. However where a majority are elaborate, when something seems too simple, misreading may be at fault. Just wanted to make sure
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broken mist
why is inequality bigger than zero?
pearl pondBOT
why is inequality bigger than zero?
vale meadow
anything within log must be > 0
ancient bolt
what are wavefronts
pearl pondBOT
pearl pondBOT
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lethal harbor
im kinda confused with the concept of inital conditions in multivariable calculus, since i cant imagine this as anything other than just like a point on a graph
pearl pondBOT
im kinda confused with the concept of inital conditions in multivariable calculu...
@lethal harbor Has your question been resolved?
pearl pondBOT
@lethal harbor Has your question been resolved?
lethal harbor
<@&286206848099549185>
pearl pondBOT
@lethal harbor Has your question been resolved?
iron arch
im kinda confused with the concept of inital conditions in multivariable calculu...
Initial conditions are indeed a point
However these points will change the resulting function
Take the function a+b for exemple
At inutial condition 0,1 you’ll have 1 but at initial condition 2,4 you’ll have 6
Si you found the function for all of its domain using a point
lethal harbor
ohhh ok thank u
so to get the trajectory curve i should plug in the inital points into the 2 functions? ( we were given a pair of equations)
pearl pondBOT
@lethal harbor Has your question been resolved?
silk osprey
pearl pondBOT
inland lantern
dont advertise your channel in others
inland lantern
Well honestly just d, I dont know what to do with that x^2
note that g(m) has a minimum at m=4 (as per part c)
therefore we want the input to g(x) to be 4
inland lantern
assuming you got x = 2, then yes
silk osprey
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pearl pondBOT
pearl pondBOT
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old shuttle
I am trying to solve an equation in physics about pressure, sorry if that is not what this server is intended for.
The question is about breaking bricks with your hands and whatever whatever, the information I'm given is that the length of the brick is 25cm and the width is 12cm. It can also endure 30MPa. Approximate how much force the hand has to endure with such a hit
I got this but it feels very wrong, and I'm not even sure what I'm doing.
midnight haven
I am trying to solve an equation in physics about pressure, sorry if that is not...
The method is correct.
I am not checking the calculations.
old shuttle
The method is correct.
I am not checking the calculations.
Ok, thank you!! Your pfp is awesome, awesome cat 
I'll call this solved if the method is correct 🙂
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pearl pondBOT
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native bolt
I found a very interesting function and am curious about a few properties
native bolt
${frac{f(x,y)}{g(x,y)}} \leq \alpha$, where $0 < \alpha < 1$
jolly parrotBOT
7aman
native bolt
For linear $f(x,y)$ and $g(x,y)$, what ratio of the cartesian plane is shaded for a given $\alpha$?
jolly parrotBOT
7aman
unborn abyss
surely it depends on f and g right?
as a silly example take f(x, y) = 0
then the whole plane is shaded
native bolt
surely it depends on f and g right?
Yeah
For
$f(x,y) = f_1 x + f_2 y + f_3 \
g(x,y) = g_1 x + g_2 y + g_3$
jolly parrotBOT
7aman
native bolt
Is there a way to create a closed form function for alpha based on these six parameters?
pearl pondBOT
@native bolt Has your question been resolved?
pearl pondBOT
@native bolt Has your question been resolved?
pearl pondBOT
@native bolt Has your question been resolved?
native bolt
${\frac{f(x,y)}{g(x,y)}} \leq \alpha$, where $0 < \alpha < 1$
Realized fractional part is curly brackets sorry
jolly parrotBOT
7aman
subtle ginkgo
no need to use fractional part, then your function becomes liek a constant
nothign interestign in it then
native bolt
nothign interestign in it then
Gives it a nice upper bound
Also graphs very nicely
subtle ginkgo
yes graph is nice, reminds me some bifurcations taken from PDE
pearl pondBOT
@native bolt Has your question been resolved?
pearl pondBOT
@native bolt Has your question been resolved?
pearl pondBOT
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torn willow
Is my answer to this question correct?
pearl pondBOT
Is my answer to this question correct?
pearl pondBOT
@torn willow Has your question been resolved?
pearl pondBOT
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stone sage
my textbook says:
pearl pondBOT
my textbook says:
hallow kite
Take x/x⁴ = 1/x³ = 1/inf = 0
hallow kite
As i think:
Consider the ln function graph and also the function that they give it go u
We know that this function that they give it to u go to 0 when x->inf
That mean when x=inf then y=0
Now let put this function into ln
Ln (y) = x
Ln (0) = something
Since ln (0) is undefined we calc the limit so
limit Ln (0) = -inf
<@&286206848099549185> sorry for mention but is that true 👀
hallow kite
Ln of 0 is not - inf
I mean limit
open nimbus
Oh yeah then it is
open nimbus
Keep pinging me
nocturne plover
hold up
nocturne plover
$\lim_{x \to 0} \ln(x) \neq -\infty$
jolly parrotBOT
Katharine
nocturne plover
$\lim_{x \to 0^{+}} \ln(x) = -\infty$ and $\lim_{x \to 0^{-}} \ln(x) = $ undefined
jolly parrotBOT
Katharine
nocturne plover
this means that $\lim_{x \to 0} \ln(x) = $ undefined
hallow kite
So here this limit = $0^{+}$
jolly parrotBOT
Μήλιος
hallow kite
Cuz we will go to calc for $\lim_{x \to 0^{+}}$ ln(x)
jolly parrotBOT
Μήλιος
pearl pondBOT
@stone sage Has your question been resolved?
pearl pondBOT
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foggy dome
hey guys how can i start to solve a derivative if they have 3 ()
$(u^2 + u) (u^2 + 1) (u^2 - u + 1)$
jolly parrotBOT
odokawa
foggy dome
btw i dont know the name of something that is inside of a parenthesis
midnight haven
product rule
eternal tulip
you can mutliply/foil out, then apply product rule to the remaining two
midnight haven
you can do product of 3 functions
eternal tulip
yes, you can do this as well
eternal tulip
if you have f(x)g(x)h(x), then you can just treat 2 functions as one, for example g(x)h(x)=j(x)