look over this please
#help-39
iron stream
iron stream
so that you can understand
Thanks 💎
iron stream
interesting
Nah you made a mistake 🥲
dusty heart
but the problem here is that
we didnt substitute the bounds for the integral
so it shouldnt be 0 to 2pi
bc we are using tg2x
it should be from 0 to 0
which is why we are getting 0
iron stream
so it shouldnt be 0 to 2pi
Do it like, indefinite, at last we shall substitute tan2x in place of z, but that too would yield same results
dusty heart
i think that in this exercise we shouldnt use tg substitutions
iron stream
i think that in this exercise we shouldnt use tg substitutions
So this
dusty heart
but
do you agree that as we did it now
the integral should be from 0 to 0?
bc we did the t substitution
iron stream
the integral should be from 0 to 0?
I agree
iron stream
the integral should be from 0 to 0?
But you must note sin and cosine are periodic functions
That assume 0 at multiple points
dusty heart
int (a;a) = 0 no matter what
iron stream
That assume 0 at multiple points
So we're facing issues
dusty heart
yea
dusty heart
int (a;a) = 0 no matter what
so if this is correct, then we should avoid tan substitutions
iron stream
Consider the integration of sinx from 0 to pi
iron stream
Consider the integration of sinx from 0 to pi
If you substitute sinx with x
You get 0-0
iron stream
Consider the integration of sinx from 0 to pi
But the actual result is 2
iron stream
so if this is correct, then we should avoid tan substitutions
tan sub is the easiest way to solve the integral, we can use it, just don't put the limits
iron stream
2√2 pi will be the answer 😭
iron stream
,calc 2√2
jolly parrotBOT
The following error occured while calculating:
Error: Syntax error in part "√2" (char 2)
dusty heart
let me write it
iron stream
,calc 2 * sqrt(2)
jolly parrotBOT
Result:
2.8284271247462
smth like this
i forgot
but like
we can write 2pi as 2 time the int from 0 to pi
and so on
like
iron stream
we can write 2pi as 2 time the int from 0 to pi
Yeah
dusty heart
pi = pi/2
iron stream
In this case we can write 4× int 0 to pi/2
dusty heart
and then pi/4
iron stream
and then pi/4
Yeah pj/4 works too
dusty heart
so
can you tell me how the actual thing had to be
like, what was the condition for that to work
i forgot
iron stream
and then pi/4
But that was irrelevant, if I were able to explain the intuition i applied.
iron stream
i forgot
So see
If interested 0 to a f(x) dx = int 0 to a f(2a-x) we can pull out that 2
dusty heart
If interested 0 to a f(x) dx = int 0 to a f(2a-x) we can pull out that 2
you meant 0 to 2a
right?
dusty heart
ok but
iron stream
You know the rule
Int 0 to 2a fx dx = int 0 to a f(x) + a to 2a (f(2a-x))dx
Now for sin and cosine this pi range is interchangeable kinda so we can say from this
dusty heart
one sec cuz i dont think we got the right answer
iron stream
one sec cuz i dont think we got the right answer
Huh ?
2√2 pi
No way bro !! 😭 It must be the correct one
iron stream
Ok
iron stream
Feeling sleepy bro, it's 1:38 am
iron stream
Yeahhhh
iron stream
tan pi/2 = ∞
dusty heart
we would have pi/4/ sqrt 2
iron stream
Again tan^-1 ∞ = pi/2
dusty heart
which would be pi sqrt2/8
iron stream
no cuz we got arctg t/sqrt 2
That t /√2 doesn't matter
iron stream
from 0 to pi/4
No you didn't change the limits
dusty heart
Feeling sleepy bro, it's 1:38 am
pls, im so close to understanding
dusty heart
No you didn't change the limits
OMG
iron stream
T = ∞
dusty heart
NO IT WOULD BE SQRT2/2/SQRT2
iron stream
SO IT WOULD BE FROM 0 TO SQRT 2/2
Nah from 0 to inf
iron stream
See tan2x = t
At x = 0, t = 0
At x= pi/4 , tan pi/2 = t = ∞
dusty heart
OMG CUZ WE USED FUCK TAN2X
iron stream
we got pi/4
We have tan" 2 "x
dusty heart
MOTHERFUCKER
i think i may be able to figure it out from here
thanks a lot
you should sleep
iron stream
Goodbye Comrade
iron stream
You will win the battle !
pearl pondBOT
@dusty heart Has your question been resolved?
dusty heart
You will win the battle !
i figured it out, may god bless you with the best head
pearl pondBOT
i figured it out, may god bless you with the best head
dusty heart
.close
pearl pondBOT
vast wadi
min is better
pearl pondBOT
pearl pondBOT
Available help channel!
pearl pondBOT
rancid depot
<@&268886789983436800> asking for help with test
radiant terrace
.close
brittle onyx
RTP: if a^n -1 is prime then, a = 2
can i take the contraposiitve and attempt to prove it by contradicting it
so the contrapositive would be if a ≠ 2, then a^n -1 is composite
and i would prove the contrapositive by assuming a = 2, and show how does it doesn't work by using counterexample (n=1)
i feel like i'm definitely doing something wrong here
dapper kraken
do you have to use contrapositive?
rancid depot
im assuming theres a restriction on n, n>=2
if so, then contrapositive would be a great way to do this problem
dapper kraken
can you not factor a-1?
pearl pondBOT
@brittle onyx Has your question been resolved?
pearl pondBOT
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sterile turtle
pearl pondBOT
low crystal
When you can use integration by parts technique? for example can I use it for arcsinx * 1/x^3
light helm
try and see what happens
stoic imp
pearl pondBOT
thorny ridge
lol
pearl pondBOT
@stoic imp Has your question been resolved?
sharp vigil
how could you translate between the standard basis and the given basis?
stoic imp
Wdym?
a(2,2,-1)+b(2,5,-2)+c(-1,-2,2)=(a,b,c)
a((2,2,-1)-(1,0,0))+b((2,5,-2)-(0,1,0))+c((-1,-2,2)-(0,0,1))=(0,0,0)
a(1,2,-1)+b(2,4,-2)+c(-1,-2,1)=(0,0,0)
(a+2b-c,2a+4b-2c,-a-2b+c)=(0,0,0)
stoic imp
how could you translate between the standard basis and the given basis?
Can you help
stoic imp
(a+2b-c,2a+4b-2c,-a-2b+c)=(0,0,0)
Like i need to do something with this
Unless i am tripping
sharp vigil
we have methods in linear algebra which give us a formula for translating from one basis to the other
and we also have the concept of "this vector is the same after transforming it"
stoic imp
Wdym?
sharp vigil
if we had a matrix A, then how could we find all the vectors which remain the same after multiplying by A?
stoic imp
Idk
sharp vigil
so we have Ax = 1x
sharp vigil
it doesn't have to be true for all vectors, just some certain special vectors
the special vectors for which multiplying by A is the same as multiplying by the scalar 1
sharp vigil
are you at all familiar with the equation [ Ax = \lambda x ] for vector $x$, matrix $A$, scalar $\lambda$?
jolly parrotBOT
cloud
sharp vigil
it is exactly the definition of an eigenvector
stoic imp
And the scalar is eigenvalue
What about it?
We were talking about coordinates
How did we ended up here
sharp vigil
but we can find a matrix which translates between coordinate systems
sharp vigil
yes
stoic imp
How is that related to eigenvectors
sharp vigil
because if the change-of-basis matrix is A, then the vectors which have the same coordinates in each basis satisfy Ax = x
stoic imp
How do I do that for my situation I am working with
sharp vigil
(a+2b-c,2a+4b-2c,-a-2b+c)=(0,0,0)
to go back to what you were working with before, can you turn this into a system of equations?
stoic imp
,w rref {{1,2,-1,0},{2,4,-2,0},{-1,-2,1,0}}
jolly parrotBOT
stoic imp
a+2b-c=0
sharp vigil
the fact that the vectors must be in S gives an additional restriction
stoic imp
Yeah
a=-2b+c
(a,b,c)=(-2b+c,b,c)=b(-2,1,0)+c(1,0,1)
x1+x2+2x3=0
-2b+c+b+2c=0
b=3c
(-6c+c,3c,c)=c(-5,3,1)
sharp vigil
being in the subspace is just another equation, you already have a system of equations
stoic imp
Imma be honest, I am getting better at this algebraic thinking, tho is hard to explain but is really good for the mind
pearl pondBOT
pearl pondBOT
Available help channel!
sharp tendon
pearl pondBOT
sharp tendon
You just multiply numerator and deno with √(1+z²) + z
and what about the left part sqrt1+z^2
hyperbolic trig sub
trig sub
iron stream
and what about the left part sqrt1+z^2
Nah
There's a formula for it
Else you can just derive using that tan sub, buth there's a direct formula
sharp tendon
only if i remembered
iron stream
only if i remembered
Ok √a² + x² = 1/2 [x√(x²+a²) + a²ln(x+√x+a))]
iron stream
how do you remember such big formulas
Idk !
iron stream
how do you remember such big formulas
Maybe coz there are formulas bigger than this one 😭😭
surreal relic
💀
inland ivy
Then the integrand becomes cosh^2(t) which is just (1+cosh(2t))/2
sharp tendon
Then the integrand becomes cosh^2(t) which is just (1+cosh(2t))/2
ohh yes
almost did byparts at that step for no reason
inland ivy
Hyperbolic trig sub makes integrals with terms like sqrt(1+x^2) and sqrt(x^2-1) much easier than if you use trig sub
You just gotta get used to the identities
They're similar to the trig ones but there are sign differences
sharp tendon
and remember formulas of hyperbolic inverses
inland ivy
and remember formulas of hyperbolic inverses
If you want to write it in terms of log then sure
I think just arcsinh(x) and stuff like that suffices too
iron stream
Hyperbolic trig sub makes integrals with terms like sqrt(1+x^2) and sqrt(x^2-1) ...
hyperbolic subs are surely good !
midnight haven
hi
midnight haven
here is my work so far
i am stuck on the final part about showing that inequaality
autumn trellis
i would rather use the formula $x^{N} - 1 = (x-1)(x^{N-1}+x^{N-2}+...+1)$
autumn trellis
and here you know how to integrate each term via the previous part of the question
jolly parrotBOT
4573r01d|)d357r0y3r 45²
midnight haven
and here you know how to integrate each term via the previous part of the questi...
thankyou
pearl pondBOT
Available help channel!
pearl pondBOT
vast wadi
you could use general form of fibinaci
jolly parrotBOT
bruhmoment
vast wadi
you could look at even and odd terms seperately
vast wadi
you could look at even and odd terms seperately
if it helps
pearl pondBOT
pearl pondBOT
Available help channel!
sharp smelt
Define $\lim_{m,n \to \infty} =a$ to mean that for all $\varepsilon >0 \exists N \in \N st , m,n >N \implies \abs{a_[m,n}-a}<\varepsilon $
\
Let $a_{m,n} = \frac{1}{m+n}$ . Does the limit exist?
\
The limit is 0. We thus have $\abs{\frac{1}{m+n}}< \varepsilon$., or $m+n > \frac{1}{\varepsilon}$
jolly parrotBOT
math_rocks
Define $\lim_{m,n \to \infty} =a$ to mean that for all $\varepsilon >0 \exists N \in \N st , m,n >N \implies \abs{a_[m,n}-a}<\varepsilon $
\\
Let $a_{m,n} = \frac{1}{m+n}$ . Does the limit exist?
\\
The limit is 0. We thus have $\abs{\frac{1}{m+n}}< \varepsilon$., or $m+n > \frac{1}{\varepsilon}$
```Compilation error:```! Extra }, or forgotten $.
l.1422 ... \N st , m,n >N \implies \abs{a_[m,n}-a}
<\varepsilon $
I've deleted a group-closing symbol because it seems to be
spurious, as in `$x}$'. But perhaps the } is legitimate and
you forgot something else, as in `\hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.
Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]```
sharp smelt
I feel that even if I say N> m+n, I won't get anywhere
jolly parrotBOT
PinkPurpleBlue
fiery charm
whenever $m,n > N$
jolly parrotBOT
PinkPurpleBlue
fiery charm
cant we pick $N(e)$ as ceiling of $1/e$?
jolly parrotBOT
PinkPurpleBlue
fiery charm
then $m,n > N(e) = ceil(1/e) \ge 1/e$
jolly parrotBOT
PinkPurpleBlue
fiery charm
therefore $m+n > N(e)$
jolly parrotBOT
PinkPurpleBlue
sharp smelt
hmm
warm current
Okay, so you are saying that $(a){m, n} :\bN\times\bN\to\bR$ is a multidimensional sequence. Specifically,
$$a{m, n}=\frac1{m+n}$$
Naturally, for some fixed $m_0\in\bN$, we can define the one-dimensional sequence $(b)n :\bN\to\bR$ simply by
$$b_n=a{m_0, n}=\frac{1}{m_0+n}$$
If we fix the one-dimensional case, then $\lim_{n\to\infty} b_n=0$ is easy to find. We could then define the sequence $(c)m :\bN\to\bR$ by $$c_m=\lim{n\to\infty} a_{m, n}=0$$
Then, very clearly, $\lim_{m\to\infty} c_m=0$.\
\
Intuitively, this is all obvious. The real question here is going to be if whether your $\lim_{m,n\to\infty}a_{m,n}$ definition will give us the same thing.
jolly parrotBOT
SWR
sharp smelt
**PinkPurpleBlue**
I don't think so
fiery charm
I don't think so
what have I got wrong?
warm current
@sharp smelt you got some tex errors btw
sharp smelt
N(e) seems to imply N is a function of epsilon
brave sluice
N=1/e seems to work
warm current
Are you trying to say
$$\lim_{m,n\to\infty}a_{m,n}=a\Leftrightarrow\forall(\varepsilon>0)\exists(N\in\bN)(m,n>N\to\abs{a_{m,n}-a}<\varepsilon)$$
jolly parrotBOT
SWR
fiery charm
I mean its okay no?
brave sluice
seems ok to me
warm current
N(e) seems to imply N is a function of epsilon
It usually is, fyi. Even in normal limits
fiery charm
I mean for every epsilon there is exists $N$ is sorta the same as saying that we can pick an $N$ dependent on epsilon.
jolly parrotBOT
PinkPurpleBlue
sharp smelt
Define $\lim_{m,n \to \infty} =a$ to mean that for all $\varepsilon >0 \exists N \in \N st , m,n >N \implies \abs{a_{m,n} - a}<\varepsilon $
\
Let $a_{m,n} = \frac{1}{m+n}$ . Does the limit exist?
\
The limit is 0. We thus have $\abs{\frac{1}{m+n}}< \varepsilon$., or $m+n > \frac{1}{\varepsilon}$ ( with TeX. fixed)
jolly parrotBOT
math_rocks
brave sluice
m,n > ceil(1/e)
m,n > 1/e
|1/(m+n)| < |1/(1/e + 1/e)| = e/2 < e
brave sluice
basically
that shows that for all e>0, if you let N=ceil(1/e), then m,n>N implies |a_m,n| < e
which fits the definition so it means the lim exists and is equal to 0
sharp smelt
I see
so I start by finding a N that works for the one dimensional cases
and then combine them?
warm current
As you wrote, you basically want to satisfy $m+n>\frac1{\varepsilon}$ for all $m,n>N$. Now ask yourself, if $m,n>N$, then what inequality conclusion can you make about $m+n$ and $N$?
jolly parrotBOT
SWR
brave sluice
once you finish a limit proof it's a little easier to follow
but when you're looking for delta or N it feels like you're working backwards and it's kind of mind-bending
at least for me 
sharp smelt
at least for me <:opencry:586078614865444865>
yeah, same here
warm current
if m,n>N,then 1/m<1/N, 1/n<1/N
I'm asking about what conclusion you could draw between m+n and N
sharp smelt
m+n>2N
sharp smelt
N= \eps/2
sharp smelt
hmm
warm current
Hint: consider $\varepsilon=1$
jolly parrotBOT
SWR
sharp smelt
so N=1
warm current
mhm
warm current
You wrote $N=\frac{\varepsilon}2$. But I asked you what $N$ would be if $\varepsilon=1$, and you said $N=1$, which is not consistent with your first statement. From your first statement, I would infer that you would have chosen $N=\frac12$.
jolly parrotBOT
SWR
sharp smelt
yeah, but N=1/2 seems sus
jolly parrotBOT
SWR
warm current
You're close to the right answer, but not there yet
sharp smelt
yeah, I'm lost
warm current
yeah, but N=1/2 seems sus
Okay, let's start here. Why does N=1/2 feel sus to you?
sharp smelt
because, say we chose ,m=n=1
warm current
what about it?
warm current
we then have 1/2<1/2
where'd you get this?
warm current
so LHS, $\frac{1}{m+n}$ would be $\frac12$, yeah. But how is RHS also $\frac12$?
jolly parrotBOT
SWR
warm current
indeed it is
That wasn't the exact issue I had with your $N=\frac{\varepsilon}2$ attempt.
jolly parrotBOT
SWR
warm current
Another hint: we require $N\in\bN$
jolly parrotBOT
SWR
jolly parrotBOT
math_rocks
jolly parrotBOT
SWR
sharp smelt
1
jolly parrotBOT
SWR
warm current
And with all that, we still have one more thing to fix 
sharp smelt
hmm?
jolly parrotBOT
SWR
warm current
So we are good
warm current
sharp smelt
Can I close this now
warm current
yes
brave sluice
blessed cat
warm current
as far as I can see, your solution is all the way correct
pearl pondBOT
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nocturne grail
If f(x) =5x + b was an inverse function to r(x)=cx+4
Find cb
oblique river
if they're inverses tthen f(r(x))=x
glacial sequoia
5(cx + 4) + b = x needs to be independent of x
oblique river
You could write it as 5cx+(20+b) and compare coefficients with x+0
nocturne grail
You could write it as 5cx+(20+b) and compare coefficients with x+0
I dont understand
I dont study in english
So explaining right away will be appreciated
pearl pondBOT
@nocturne grail Has your question been resolved?
sacred hollow
Find the derivative of the function 3|x+2| at x = -3
pearl pondBOT
Find the derivative of the function 3|x+2| at x = -3
sacred hollow
Please help 🙏
rancid depot
absolute value is defined as a piecewise function
what is |x+2| defined as when x+2 < 0?
sacred hollow
-(X+2)
rancid depot
so use that, find the derivative, then plug in x=-3 if necessary
sacred hollow
Ohkk
So we can differentiate mod functions?
As long as they continuous?
In this case it is not continuous at x = -2 right?
.close
pearl pondBOT
rancid depot
In this case it is not continuous at x = -2 right?
yes
glacial sequoia
In this case it is not continuous at x = -2 right?
*differentiable
sacred hollow
.reopen
pearl pondBOT
✅
sacred hollow
*differentiable
Is it also not not continuous?
glacial sequoia
it's continuous
rancid depot
*differentiable
oh right my bad
spare lark
It is continious
sacred hollow
How?
glacial sequoia
,w plot y = 3|x + 2|
jolly parrotBOT
sacred hollow
Like is mod of smtng equal to 0 not not continuous?
glacial sequoia
I can draw the graph without lifting my pen up
glacial sequoia
Like is mod of smtng equal to 0 not not continuous?
is the not not meant to be a double negation or is it a typo
spare lark
is the not not meant to be a double negation or is it a typo
Bro is doing boolean xd
sacred hollow
Yea double negative
glacial sequoia
Is it also not not continuous?
i'm asking because you did it here too and i thought it was a typo
sacred hollow
i'm asking because you did it here too and i thought it was a typo
Sorry TwT
spare lark
Then yeah its not not continous
sacred hollow
So it is continuous? But mod of 0 is not?
But this is mod of 0 cause -2+2 = 0?????
spare lark
Mod of 0 is abs(x) ?
sacred hollow
I mean like mod of x where x = 0 is not continuous?
spare lark
Its always continious
sacred hollow
Wait mod is always continuous?
spare lark
As long as no division yeah
sacred hollow
Wait mod of x is = |x| ?
spare lark
Yeah
sacred hollow
And it is continuous?
spare lark
Yes
sacred hollow
My whole last year of highschool was a lie
spare lark
,w abs(x) graph
sacred hollow
Sorry I was only taught in words and not much graphs soo
jolly parrotBOT
glacial sequoia
at 0 you can have a tangent either x = y or x = -y
spare lark
Even 0
sacred hollow
y= -x and x at 0?
spare lark
Kekw
sacred hollow
But if it's like mod of -2 then it is differentiable?
spare lark
No cuz it will be the same thing just on a different part of the graph
,w plot y = 3|x + 2|
sacred hollow
So not differentiable?
jolly parrotBOT
sacred hollow
I mean like |x| where x =-2
spare lark
Ah
sacred hollow
It is not differentiable?
sacred hollow
But y = 2 for both x= 2 and -2
sacred hollow
Oh is |x| not differentiable cause both values coincide while being different and hence not possible?
spare lark
Abs is not differentiable where what inside abs is 0
sacred hollow
Oh is |x| not differentiable cause both values coincide while being different an...
When 0 I mean
I c I c
Thank u soo much for helping with that
pearl pondBOT
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Available help channel!
shadow cargo
How do I turn (a+b)²+2(ac+bc) to a power
pearl pondBOT
How do I turn (a+b)²+2(ac+bc) to a power
calm wing
How do I turn (a+b)²+2(ac+bc) to a power
wdym to a power
you can factor this
if that's what you mean
tawny harness
How do I turn (a+b)²+2(ac+bc) to a power
add c² and remove c²
shadow cargo
thats what the exercise says, i assume it means factor the make it into a parenthesis thats in a power of 2
tawny harness
becomes difference of squares
use the identity:
(a+b+c)² = a² + b² + c² + 2ab + 2ac + 2bc
shadow cargo
use the identity:
(a+b+c)² = a² + b² + c² + 2ab + 2ac + 2bc
oh
that makes sense
thank you!
.close
pearl pondBOT
tawny harness
this wont give you a power of two though
so i dont think there exists a power expression for what youve given
pearl pondBOT
Available help channel!
ancient lotus
Hi
sharp ivy
Is there any particular exemption to the question?
ancient lotus
Nope ,
sharp ivy
wait i am being tripping
ancient lotus
Should I substitute lim x->0 (e^x-1)/x
sharp ivy
use lhospital rule , differetiate num and denom and then solve
This
plain cedar
can anyone give solution to this question
ancient lotus
use lhospital rule , differetiate num and denom and then solve
I got the answer using this , thnks
plain cedar
answer my question please
ancient lotus
Idk
Dude , you asked this question already in other channels ??!!
.closed
.closed
Uhh
.close
pearl pondBOT
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balmy scaffold
lets differentiate this?
pearl pondBOT
lets differentiate this?
fringe raft
oh my bad, must be a multiplication
anyway, you are right, you can define u(x) = x^3-2x
balmy scaffold
v is basically (x^3-2x)^(1/2) so we need to differentiate that to get v'
fringe raft
then what is v(y), so that v(u(x)) becomes the whole term?
balmy scaffold
i assume it is 1/2 * (x^3-2x) ^-1/2
balmy scaffold
then what is v(y), so that v(u(x)) becomes the whole term?
so basically ?
do you have experience doing these?
well next one is this
fringe raft
yeah this is right
balmy scaffold
here we have to use 3 rules i guess
it reminds me of a recursive function at this point lol
the rule for fractions is
$$\frac {u' \cdot v - u \cdot v'}{v^2}$$
fringe raft
not quite
jolly parrotBOT
Jaeger
fringe raft
you can use that, or directly use -1/2 as exponent
balmy scaffold
ok i guess we start with the bottom part first
well lemme try to show you my solution first and then we discuss it
fringe raft
not that I think they differ too much in difficulty
balmy scaffold
not that I think they differ too much in difficulty
ok no i am missing something
i assume it was just left divided by right
wolfram says its this
fringe raft
almost got it, just forgot the minus
balmy scaffold
almost got it, just forgot the minus
something is off i don't get it
fringe raft
the u' v just becomes 0, then don't forget the remaining minus
the denominator you got $\left(\sqrt{x-1}\right)^2$
jolly parrotBOT
rbit
fringe raft
yes, first get rid of the square root
balmy scaffold
basically (x-1) ^1/2
fringe raft
and the (x-1) on top has a negative exponent, so what does that mean?
balmy scaffold
that you can put it into bottom?
fringe raft
yes
fringe raft
$\left((x-1)^{\frac{1}{2}}\right)^2\cdot (x-1)^{\frac{1}{2}}$
jolly parrotBOT
rbit
balmy scaffold
yeah that becomes (x-1) ^3 i would say
fringe raft
you sure?
balmy scaffold
oh damn sec
jolly parrotBOT
rbit
fringe raft
yes
balmy scaffold
was not hard after all
just needed some accuracy
@fringe rafthave you done these before? sin, cos, arctan always scared me
fringe raft
you can do these with the product rule and chain rule once again
balmy scaffold
sin x * cos x
would be $$sin x \cdot cos x ' + sin x' \cdot cos x$$
jolly parrotBOT
Jaeger
balmy scaffold
well that would be rule 1
to get cosx' and sinx' we need to use chain rule i guess
fringe raft
=> sin(x) * cos(x)' + sin(x)' * cos(x)
balmy scaffold
x' * (cosx)'
fringe raft
please don't use chain rule here, it's way simpler
you said the derivative of sin(x) is cos(x), which is correct
what is the derivative of cos(x)?
fringe raft
yes
balmy scaffold
kinda odd with these
fringe raft
just like x^2 ' = 2x, we have sin(x)' = cos(x)
balmy scaffold
ohhh
makes sense then
well it gets complicated with stuff like
sin(2x)
sin(x^2)
sin(x^3 -2x^2)
i guess
then you have to use chain rule?
balmy scaffold
but if its as you said, in our case it just means
fringe raft
yes
fringe raft
if you really wanted to you can simplify it
balmy scaffold
here chain rule i guess
fringe raft
yes
fringe raft
v(u(x)) what is v and what is u?
balmy scaffold
?
fringe raft
right
fringe raft
cos times anything'
balmy scaffold
yep
this nightmare though
chain rule i guess since i think its $$arctan(sqrt(x))$$
jolly parrotBOT
Jaeger
fringe raft
yes
jolly parrotBOT
Jaeger
fringe raft
then arctan'?
fringe raft
it does
fringe raft
well you either just remember it or derive it using the inverse rule
balmy scaffold
not that hard to remember
but i guess its $$1/2 \cdot x^{-1/2} \cdot\frac {1}{1+x^2} \cdot x^{1/2}$$
jolly parrotBOT
Jaeger
fringe raft
not quite
balmy scaffold
this is what wolfram says
fringe raft
you plug x^(1/2) into the 1/(1+x^2) term
balmy scaffold
you plug x^(1/2) into the 1/(1+x^2) term
where
fringe raft
$\frac{1}{1+\left({x^{\frac{1}{2}}}\right)^2}$
jolly parrotBOT
rbit
balmy scaffold
$\frac{1}{1+\left({x^{\frac{1}{2}}}\right)^2}$
something is off
doesn't add up
like you would get a sqrt in bottom but also on top
and they would cancel eachother out
fringe raft
nono, you just don't multiply by the square root, you plug it in
you get 1/2 x^(-1/2) times arctan'(x^(1/2))
balmy scaffold
you get 1/2 x^(-1/2) times arctan'(x^(1/2))
the squareroot gets plugged in where the x is?
fringe raft
yeah
balmy scaffold
ah interesting
so if it was 4x it would be
(4x)^2
ye i guess that's it
well some loud people entered the library so i got distracted for a bit xD
fringe raft
right, then simplify
jolly parrotBOT
Jaeger
fringe raft
just correct
balmy scaffold
or dunno tis is their solution
ohh i get what they did here
they just multiplied it by 2
leaving the sqrt(x) on the left
so 2 * this basically
nice the more i deal with these rules the more i understand them intuitevely
this is the endboss i guess
what the heck is arccos
does this mean cos^-1
fringe raft
yes
fringe raft
no
balmy scaffold
damnit this time intution does not work
fringe raft
what makes e^x so special
balmy scaffold
what makes e^x so special
i assume that and log have their own rules when it comes to derivatives
fringe raft
yes
balmy scaffold
$$e^x \cdot cos^-1'$$
jolly parrotBOT
Jaeger
fringe raft
$e^x \cdot \cos^{-1}'(e^x)$
jolly parrotBOT
rbit
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
balmy scaffold
y/7 -4/7 = x
that's how it should be wtf
why do they casually swap it
f(x) = cos^-1
f^-1 = jesus
@fringe raftis this something that you just have to remember
or is it something that you can use logic for
it says inverse of cos is just cos
fringe raft
the inverse of cos is arccos
the derivative should be -1/sqrt(1-x^2) or something
,w arccos(x)'
jolly parrotBOT
balmy scaffold
looks like they are similar
@fringe rafti'll just remember these 3
not that hard to memorize
thx for the help dude i solved the 1st wave of exercises
there are 12 waves in total and they get exponentially harder, i will open new channels for later waves etc
but thx for helping me solving the 1st wave
pearl pondBOT
Available help channel!
lavish blade
this is a wierd exercise, i need help
pearl pondBOT
this is a wierd exercise, i need help
desert timber
what is F(x)
quasi sundial
I think you're gonna need information about F(x) to solve the question
pine raptor
Translate the question to english?
lavish blade
the text doesnt say much
calm wing
F is primitive of f right
lavish blade
just that there exists f and f(0) = -2
lavish blade
is f(x) the derivative of F(x)?
yea
quasi sundial
Oh right that makes sense
lavish blade
but its undefined
quasi sundial
Write f(x) as dF(x)/dx rearrange the terms and I think you could get something integrable
calm wing
this is a wierd exercise, i need help
problem is there are infinitely many F's out there
lavish blade
problem is there are infinitely many F's out there
for any x sure
we gotta find it for f(-1)
calm wing
no i meant that if F(x) is a primitive of f, then F(x)+C is also a primitive of f
lavish blade
no i meant that if F(x) is a primitive of f, then F(x)+C is also a primitive of ...
mhm
also the answers all have e
so thats wierd
lavish blade
They should've given F(0) instead of f(0) I think
this was at a uni entrance exam
im 100 % sure the problem is not wrong
quasi sundial
Wait I think you could substitute 0 into the original equation then get F(0)
quasi sundial
Did you solve the differential equation?
calm wing
this is a wierd exercise, i need help
Let $f:\mathbb{R}\to\mathbb{R}$ be a function, with $f(0) = -2$, that has a primitive $F$. Given that $f(x) + 2F(x) = 3$ holds for all $x\in\mathbb{R}$, calculate $f(-1)$
is that a good translation?
lavish blade
perfect
calm wing
great
lavish blade
did that help u?
jolly parrotBOT
artemetra
quasi sundial
$\frac{dF\left(x\right)}{dx}+2F\left(x\right)=3$
jolly parrotBOT
Photon
calm wing
,w y' + 2y = 3
lavish blade
i tried that too
jolly parrotBOT
quasi sundial
Obtain F(x) then you could get f(x)
lavish blade
howww
quasi sundial
f(x) is the derivative of F(x)
lavish blade
i know
quasi sundial
Do you have trouble with solving the differential equation?
calm wing
this is a wierd exercise, i need help
it's option d
sharp ivy
Can't we solve the DE's to obtain F(x) after that we can find f(x) by deriving F(x), it should left the integration constant with sufficient information provided
lavish blade
Do you have trouble with solving the differential equation?
i dont know what i should do
calm wing
from this you get that $F(x) = c e^{-2x} + \frac{3}{2}$ which becomes $f(x) = -2c e^{-2x}$ and using the fact that $f(0) = -2$ we get that $c = 1$ so $f(x) = -2e^{-2x}$
jolly parrotBOT
artemetra
lavish blade
??
calm wing
**artemetra**
okay this is terribly explained
quasi sundial
$$$
\frac{dF\left(x\right)}{dx}+2F\left(x\right)=3
\frac{dF\left(x\right)}{dx}=3-2F\left(x\right)
\frac{dF}{3-2F}=3dx
$$$
Integrate both sides
jolly parrotBOT
Photon
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
quasi sundial
welp
lavish blade
so i get that F'(x)=3-2 F(x) right?
that what you get
btw this ' means derivative
calm wing
so i get that F'(x)=3-2 F(x) right?
yes
quasi sundial
$$\frac{dF\left(x\right)}{dx}+2F\left(x\right)=3$$
$$\frac{dF\left(x\right)}{dx}=3-2F\left(x\right)$$
$$\frac{dF}{3-2F}=3dx$$
jolly parrotBOT
Photon
quasi sundial
Finally
lavish blade
i still dont get it
calm wing
which part confuses you
lavish blade
the last line from the last pic
calm wing
how familiar are you with differential equations
lavish blade
extremely
i just dont have the same notations
we use diff notations so thats why im getting confused
quasi sundial
What notation are you familiar with
calm wing
we use diff notations so thats why im getting confused
are you able to solve y' + 2y - 3 = 0?
lavish blade
so after the second line, you just derivate both sides again?
calm wing
if this is the notation you use
quasi sundial
so after the second line, you just derivate both sides again?
Nah, it's just simple rearranging
lavish blade
if this is the notation you use
yes
quasi sundial
So we can integrate easily
quasi sundial
I mean yeah
calm wing
$$\int \frac{dF}{3-2F}= \int 3dx$$
jolly parrotBOT
artemetra
quasi sundial
This would give you a general solution for F(x)
We can get an exact solution by getting a value for the constant of integration
From the original equation with the help of the value provided
If you have F(x) then you can obtain f(x)
lavish blade
We can get an exact solution by getting a value for the constant of integration
thats the first thing i did when i tried this exercise
and i remember get that c= -3/2
it might be wrong tho
quasi sundial
It seems you got the opposite sign
calm wing
and i remember get that c= -3/2
which c are you talking about?
quasi sundial
wait nvm
lavish blade
which c are you talking about?
there are multiple??
calm wing
eh no it's more of a language ambiguity
can you just show us your solution
like your attempt
lavish blade
i dont have a solution
i am stuck at F'(x)= 3-2F(x)
had i known just the next step
maybe i could solve the problem
did you integrate after that?
and what did you rearrange?
did you divide by 3-2F(x)
quasi sundial
Yes
So on the LHS the integration is wrt to F and on the RHS wrt x
Simple variable separation
lavish blade
and then u integrated
quasi sundial
Yes
quasi sundial
Oh shit I wrote it wrong
Shouldn't be 3, should be 1
My bad
$$\int \frac{dF}{3-2F}= \int dx$$
jolly parrotBOT
Photon
lavish blade
and also, why is there dF on LHS and dx on the right
quasi sundial
We have both F and it's derivative of F in the equation so we have dF
lavish blade
i get that part yes
quasi sundial
Are you familiar with the variable separation method?
Maybe you refer to it with another name
Because that's what we're doing if that rings any bell
lavish blade
u mustve done some sort of substitution bc otherwise idk how u got dF straight up
i feel like im kind of wasting your time at this point
quasi sundial
Yeah, then you could rearrange the terms and integrate and voila
pearl pondBOT
pearl pondBOT
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pearl pondBOT
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pure jackal
Can someone explain the solution (attached on the right)? In particular, I'm confused on why they broke the problem into subgroups of 3 and 6, and where they got the 3 ways and 6 ways from.
pure jackal
Shouldn't there be more groups? If 0 pairs with 2, 1 pairs with 3, doesn't this form a group of 4?
rancid depot
that red, blue, green solution is for part 3 where there are groups of 3, not pairs
pure jackal
that red, blue, green solution is for part 3 where there are groups of 3, not pa...
Ohh I see, but where did the two groups of 3 and 6 come from?
rancid depot
you can either group 3 people who are in adjacent time zones (green) or leave a gap (blue) that must be filled by the next group (red)
a group on the left side of blue cannot reach the gap, so it must be reached from the right
pure jackal
you can either group 3 people who are in adjacent time zones (green) or leave a ...
why can't you leave two gaps? Like 0 pairs with 3 and 6?
rancid depot
then 0 and 6 will be 6 apart
pure jackal
then 0 and 6 will be 6 apart
i see, one other question: why couldn't we have flipped the second left blue and the second right-most red?
wait nvm, that doesn't work
pure jackal
i see, so the solution's configuration is the only possible solution for a group of 6?
rancid depot
yeah
technically you could also have the case with 3,3, but you count that already and you dont want to overcount
pure jackal
Ok I see, thanks so much
Is there a particular name for these types of problem/techniques that I can look up?
pearl pondBOT
@pure jackal Has your question been resolved?
pearl pondBOT
pearl pondBOT
Available help channel!
tawny portal
I do not know how to find the iroc at -3 without using derivatives
tawny portal
cunning comet
draw the tangents at this points.
elfin summit
or you could use the vertex and roots to find the quadratic
tawny portal
draw the tangents at this points.
just a estimation of a tangent line?
cunning comet
for example: its really obvious that the tangent at x = -5 is a horizontal line.
tawny portal
.close
pearl pondBOT
pearl pondBOT
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rancid steppe
we complete the square now$2(x^2 -8x +16 -16 +17)+1 = 2(x-4)^2 +2$ but it is supposed to be +3 i don't see where it is wrong
jolly parrotBOT
Simon James B
lofty berry
you dropped the +1 on the outside of the parenthesis