sorry if that's a bit anticlimactic btw
I should've picked the a, b, and c to make it have nice roots
#help-41
craggy sundial
civic spindle
sorry if that's a bit anticlimactic btw
I should've picked the a, b, and c to ma...
bro wanted it to look like this
craggy sundial
you can try the same steps with x^2 + (4/2) x - (6/2) = 0 and x^2 + (4/2) x = 6/2if you want
craggy sundial
bro wanted it to look like this
haha,
no, just making the roots integers instead of some irrational number
civic spindle
real
but
which one was it suppose to be?
was it 10/2 or 5/2 NVM IT WAS 10
im blind
why did it end out in that value specefically?
oh those were the roots erm
still
why was 10/2 the root??
craggy sundial
it wasn't
civic spindle
so that was just random or
craggy sundial
that 3.81174... was the root
civic spindle
oh
craggy sundial
one of the roots ig
civic spindle
but my original question is still unanswered
well not exactly unanswered i didnt finish it
craggy sundial
sure, continue
civic spindle
you can try the same steps with ```
x^2 + (4/2) x - (6/2) = 0
and
x^2 + (4/2) x ...
i was still thinking about that
how subtracting that would be equivalent
thinking: \ | /
civic spindle
you can try the same steps with ```
x^2 + (4/2) x - (6/2) = 0
and
x^2 + (4/2) x ...
5+5=X
5=X-5
'I1:'
So from what im getting from this is that even if the quadratic is not exactly equal to the prior state/structure it had the root is still required for x^2 + (4/2) to equal 6/2 otherwise it would be false and that if it were at zero prior they simply moved it up a bit and if they wanted to they could undo it by just adding it again.```
is this correct?
like how
5=X-5
shows that 5 is half of X
which
although not the same in structure as 5+5=X
in that that would be 10 whereas the other one shows 5 is half of X
youd still solve them out to be 10
in that they would be the same root expressed differently and still solving to the same
is that the answer?
craggy sundial
well, that's for you to decide
but all of what you said is correct
craggy sundial
shows that 5 is half of X
although I would express that as 5 is 5 less than X
or something like that
civic spindle
craggy sundial
anw, feel free to ping me if you have a related question in the future
craggy sundial
well, idk
are you still confused about it?
civic spindle
i mean a bit but it seems to make sense despite abstract
its annoying because i kept saying the structure would be unequal and i was right and wrong simutaneously
because it didnt matter if it was it would still technically be equal as what mattered was the root not how it was expressed.
civic spindle
well, that's for you to decide
but all of what you said is correct
oh you meant if that was the answer
well alr
tyy
craggy sundial
well, idk
are you still confused about it?
(tbh, a game is playing in the background
and the music sounded inspirational so my mind went for that wording)
civic spindle
https://www.youtube.com/watch?v=fAIqvJ1Ebr0
2:18
craggy sundial
and yea
you are right
those equations are not the same
just that you can imply one's truthiness from the other's truthiness
by ways that was agreed to preserve the truthiness
or by ways that can be reached from the ways that were agreed to preserve truthiness
I'm describing axioms btw
civic spindle
or by ways that can be derived from the ways that can be reached from the paths that were agreed to preserve the truthiness by the truthiness of the truthestness.
craggy sundial
yea I'm not sure what I've just said tbh
it makes sense in my head though
anw, time to close the channel before I said something like that again haha
amber waspBOT
craggy sundial
cya
amber waspBOT
Available help channel!
languid bough
hello, I am currently doing Descartes rule of signs and I'm stuck can someone please help me?
languid bough
I'm using a video for a guide and idk if I should put 2 in the positive or 3 bc there's 3 sign changes
I'm a bit confused, so if anyone can I rlly appreciate it
strange field
what does the descartes sign rule say?
languid bough
number of sign change is equal to number of positive or negative real root
that's how I understood it
strange field
not exactly (the number of positive roots cant be more than the number of sign changes)
so what should you put?
2 or 3?
languid bough
3
strange field
yes
languid bough
okay
thank u
I tried solving for f(-x)
there were no sign changes so
would that make my negative real root and imaginary root 0?
just for clarification 😓
strange field
0 negative real roots yes
languid bough
okay2, same for imaginary?
strange field
no
languid bough
ohh may i ask why?
strange field
sign change in f(x) or f(-x) only tells you about positive or negative real roots
not imaginary
its up for you to deduce that
like in ur question u have max 3 positive real roots and 0 negative real roots
as you know, imaginary roots are always in pair
so you could have 1 positive real root and 2 imaginary roots
but if u have 3 positive real roots, then there's no room for imaginary roots (since the degree of the polynomial is 3 here so it can have max 3 roots)
languid bough
ohhh
okay2
i understand now
okay2 thank you
if thats the case btw
do i have to solve for x or its not necessary anymore?
strange field
do i have to solve for x or its not necessary anymore?
the descartes sign rule just tells you about the max possible number of negative/positive real roots
it doesnt tell if thats the exact number of roots
so it depends on the question
if it asks like "max number of positive real roots?" no need to find the roots
but if it asks "number of positive real roots?" then you will have to find the roots
languid bough
part c says "write the number of positive and negative real roots"
so like ill have to find the roots right?
languid bough
oh okay
so like, ill only have roots for positive?
im really sorry if i sound dumb we js started to tackle abt this lesson 😭
strange field
so like, ill only have roots for positive?
maybe imaginary, but no negative
also maybe what the c part of the question is asking is just the results of descartes sign rule
but its unclear
languid bough
ohh
alr
its says in the instructions in my book to identify the number of positive and/or negative real roots for each polynomial function.
and then write the number of positive and negative real roots
so maybe
like for num. 3 it has 3 pos real roots and 0 negative real roots?
amber waspBOT
@languid bough Has your question been resolved?
amber waspBOT
@languid bough Has your question been resolved?
jaunty shale
like for num. 3 it has 3 pos real roots and 0 negative real roots?
it has 0 negative real roots for sure
but you cant say for sure that it has 3 positive real roots
the rule predicts 3 or 1
in case it had 3 positive roots, then it wouldnt have any complex roots
and in case it had 1 positive root, then it would have one pair of complex-conjugate roots
jaunty shale
its says in the instructions in my book to identify the number of positive and/o...
you cant find the roots to know which is the case here, its cubic
im pretty theyre just asking for the results of the sign rule
aka to list all the possible scenarios
languid bough
ohh okay thank u so much!!!!
amber waspBOT
@languid bough Has your question been resolved?
amber waspBOT
amber waspBOT
Available help channel!
oblique oak
im kind of confused
amber waspBOT
im kind of confused
south stag
which bit
south stag
that would be the squared (sum of the deviations)
as opposed to the sum of (the deviations squared)
south stag
no thats E
wait sorry LOL i was indeed looking at E
oblique oak
so D is right?
amber waspBOT
@oblique oak Has your question been resolved?
edgy cave
amber waspBOT
edgy cave
Can somebody give me pointers on how to approach this problem
south stag
mhm
edgy cave
So P is just P(t)
south stag
yea
edgy cave
But how would I keep track of the final results unit?
south stag
ummm i'm not sure how you do it
but P [J/s] is given
and integrating over time removes the [/s]
so you get U [J]
edgy cave
Interesting
Thank you
So when I set ip my integral it would look like this correct?
south stag
that checks out
amber waspBOT
@edgy cave Has your question been resolved?
edgy cave
Does this process seem correct?
edgy cave
<@&286206848099549185>
amber waspBOT
@edgy cave Has your question been resolved?
amber waspBOT
@edgy cave Has your question been resolved?
amber waspBOT
amber waspBOT
Available help channel!
spark iris
https://cdn.discordapp.com/attachments/1018700472988749904/1283288980770521099/IMG_1666.jpg?ex=66e31c3f&is=66e1cabf&hm=a07983a7d48054aa48f6931f9ad75776f754ae8a679064a5a100f463dd4f3ffb& how do we prove the arc angle is 60 degrees?
spark iris
https://cdn.discordapp.com/attachments/1018700472988749904/1283313840318320692/Screenshot_2024-09-11_at_6.29.00_pm.png?ex=66e33366&is=66e1e1e6&hm=8169414cfe95b3aa4b22b271f19a47be9bf2674ad275ebc6def4566d5007107b& i got something like this from somebody else but i am unsure how the 2 red angles are 90 degrees
amber waspBOT
@spark iris Has your question been resolved?
spark iris
<@&286206848099549185> ?
misty nimbus
<@&286206848099549185> ?
what is the number youve written at the corner of each triangle
spark iris
what is the number youve written at the corner of each triangle
the radius length
its a 20
misty nimbus
alright so do you see that the red triangles are equilateral?
spark iris
alright so do you see that the red triangles are equilateral?
yeah
so the angles are 60 each
dapper verge
The length of the rope is 240+40pi
spark iris
The length of the rope is 240+40pi
im not rlly looking for the total rope, just tryna understand why its 90 there
or if you know anything else that proves the arc is 60 there
dapper verge
yeap 2 (90's and 2 60's) so total of 300
spark iris
yeap 2 (90's and 2 60's) so total of 300
why is it 2 90's there?
dapper verge
fact: tangent to center is 90 degree
spark iris
fact: tangent to center is 90 degree
but wouldnt the 90 degree be where the black angle is?
dapper verge
so two tangent touching each circle and two equilateral
spark iris
but wouldnt the 90 degree be where the black angle is?
???
spark iris
so two tangent touching each circle and two equilateral
can u explain this bit
dapper verge
yeap
we can see angle at p is 90 and q is 90 and see since each circle is congruent angle QAB=90 and angle PAF= 90 and angle FAB=120( two equilaterals) thus angle QAP=60 as well
ivory sorrel
https://cdn.discordapp.com/attachments/1018700472988749904/1283313840318320692/S...
Why wouldn't they be?
I didn't look at any other messages btw
Also, this is a chemistry thing btw 😉
I mean, would be useful there
spark iris
Why wouldn't they be?
im not sayings it not im kinda saying like why is it
ivory sorrel
im not sayings it not im kinda saying like why is it
It is because that's how it's drawn
They drew a 90⁰ to the normal line
spark iris
It is because that's how it's drawn
the red angle isnt given btw its something that could be true it given that it is drawn 90 degrees in the question
ivory sorrel
the red angle isnt given btw its something that could be true it given that it i...
I think there's a clear rectangle Q R A B and a bunch others in the figure
It must be a rectangle by the tangent normal property of circles
R on the right of p *
keen pawn
amber waspBOT
@keen pawn Has your question been resolved?
coral wigeon
hi kepe
kindred lark
Is this equality correct or not? I thought maybe it can be true because they will end up having an extremely big end value for that sigma in either cases
kindred lark
but wait infinity is not a number so I guess it's wrong?
I believe it's wrong because we are being talk about infinity = infinity problem? which is obiviously uncertain
lusty saffron
Is this equality correct or not? I thought maybe it can be true because they wi...
yeah i think that's true
lusty saffron
I believe it's wrong because we are being talk about infinity = infinity problem...
i have no idea what either of these two messages mean
kindred lark
i have no idea what either of these two messages mean
But isn't applying limit on sigma uncertain?
I mean we dont know how the function behaves
lusty saffron
...as in we don't know what f is, or...?
kindred lark
but wait thats not important i guess
lusty saffron
I mean we dont know how the function behaves
also this logic just fundamentally doesn't seem to make sense?
you can't object to someone solving a problem on the grounds that we don't know the answer to the problem
kindred lark
so do I not need to write two limits for that sigma as long as all I care about is making a sigma which is summing f(i) to infinity?
lusty saffron
...i have no idea what you mean by "two limits"
kindred lark
...i have no idea what you mean by "two limits"
there are two limits on the first expression
kindred lark
you can't object to someone solving a problem on the grounds that we don't know ...
what I didnt mean that
lusty saffron
the usual notation for this is to just write it as $\sum_{i=0}^\infty f(i)$
grizzled pagodaBOT
bee [it/its]
lusty saffron
which is usually defined as $\lim_{n \to \infty} \sum_{i=0}^n f(i)$
grizzled pagodaBOT
bee [it/its]
kindred lark
Also I want to ask, Is infinity / 0 considered undefined?
or just infinite
I'm talking about positive infinity
@lusty saffron
lusty saffron
well ``$\frac \infty 0$'' doesn't make much sense reading it literally because $\infty$ isn't a number and also you can't divide by $0$
grizzled pagodaBOT
bee [it/its]
lusty saffron
but assuming you mean what's $\lim_{a \to \infty, b \to 0} \frac ab$
grizzled pagodaBOT
bee [it/its]
kindred lark
yeah
lusty saffron
...oh wait hang on i was wrong earlier actually, because this will depend on the sign of $b$
grizzled pagodaBOT
bee [it/its]
lusty saffron
something like $\frac{1000000000}{-0.0000000001}$ ends up being a big negative number
grizzled pagodaBOT
bee [it/its]
kindred lark
let's say It's ->+ 0
lusty saffron
same reason that $\lim_{x \to 0} \frac 1x$ doesn't exist, because from one side it's $\infty$ and from the other side it's $-\infty$
grizzled pagodaBOT
bee [it/its]
kindred lark
because that's like + 0.000000000000000001 right?
lusty saffron
let's say It's ->+ 0
yeah in that case you get +infty
amber waspBOT
kindred lark
.reopen
amber waspBOT
✅
lusty saffron
or i think normally the way people write it is $0^+$
grizzled pagodaBOT
bee [it/its]
kindred lark
does that mean our number is getting closer to zero from the positive side?
lusty saffron
**bee [it/its]**
because something like $\lim_{x \to 4^-}$, approaching $4$ from below, is very different to $\lim_{x \to -4}$, approaching the number $-4$
grizzled pagodaBOT
bee [it/its]
lusty saffron
does that mean our number is getting closer to zero from the positive side?
yeah
kindred lark
.close
maiden turret
Heya, i sadly had to stop going to school in 7th grade im supposed to be 11th grade currently, i was pretty good back then (98 average all subjects) im lost on what to study to get back into math as im gonna start school again at 9th grade
timid vault
See Khan Academy if you haven't yet. It's nice because it has a curriculum.
maiden turret
See Khan Academy if you haven't yet. It's nice because it has a curriculum.
its progressing too slow for my likings
spiral zealot
some things can't be rushed
ensure that you get good foundations
maiden turret
some things can't be rushed
ensure that you get good foundations
is there other resources you guys can recommend?
spiral zealot
organic chem tutor channel
Paul's online notes
prof leonard channel
come highly recommended
they go into more deptg
sinful crag
amber waspBOT
sinful crag
Idk what i did wrong but its wrong. Pleas help.
BTW. Im Bad in english and subtraction
amber waspBOT
@sinful crag Has your question been resolved?
sinful crag
No
neat tusk
simplify (6+27^1/2)-(3+3^1/2)+(1-2*3^1/2)
amber waspBOT
simplify (6+27^1/2)-(3+3^1/2)+(1-2*3^1/2)
neat tusk
can someone help me to simplify this question
frank zealot
a^½ = square root of a
neat tusk
yes
frank zealot
evaluate those square roots
neat tusk
in question it is square root only i cannot type them i did this
it is already in form of squt(a)
winter estuary
you need to group like terms to get to for form a+b\sqrt(b)
winter estuary
make the things under the root the same and then use the distributive law
true jackal
x + 4x = 5x (instead of x you'll have √3 for instance)
true jackal
,w simplify (6+27^1/2)-(3+3^1/2)+(1-2*3^1/2)
grizzled pagodaBOT
true jackal
Yep
amber waspBOT
amber waspBOT
Available help channel!
split sail
.reopen
gritty holly
help i beg
amber waspBOT
help i beg
sullen beacon
SOMEONE PLS I HAVE TEST TOMORROW
good luck
shrewd elm
!nogpt
amber waspBOT
Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).
gritty holly
i just need help from a human :((
gritty holly
in the time you spent writing to me you could have been devising a solution to my problem u know
you say u have no time but u do have a little
sullen beacon
it easy
gritty holly
ive already found the solutions for when the depth = 2
ive tried - 105/60 but no working
sullen beacon
i just can explain it in vietnamese
shrewd elm
can you show the rest of the problem? there's missing information and it's not clear how the depth of the water affects this
shrewd elm
ah ok good
gritty holly
so ive got t =
4.124
7.875
16.125
19.875
the first wavecycle was used in a 'previous' part
like we used the 7.875 for another problem before this
ill let u work it out
@shrewd elm
im sorry to ping but i need to go to sleep very soon
does no one want to try this question is it too hard xD
sullen beacon
does no one want to try this question is it too hard xD
wwait me
sullen beacon
does no one want to try this question is it too hard xD
The latest time the ferry can leave Main Beach is 2 hours and 55 minutes after 3 a.m., which is 5:55 a.m.
gritty holly
My textbook said the answer is 6:12pm
sullen beacon
My textbook said the answer is 6:12pm
idk
gritty holly
<@&286206848099549185>
I’m terribly sorry for my third ping
It’s 11:55pm
I have test tomorrow
I’d like to sleep in the near future
sullen beacon
I’m terribly sorry for my third ping
why you dont study early
gritty holly
I’m in high school
sullen beacon
I had work after school Tuesday Wednesday and today
mid term test or what
gritty holly
Original question for newcomers…?
gritty holly
This too
gritty holly
mid term test or what
No just a circular functions test
gritty holly
Thx
sullen beacon
No just a circular functions test
if you dont prepare ,you get low score
sullen beacon
But I fear I’ll do bad because I haven’t had time lol
bro you dont trust yourself
sullen beacon
But I fear I’ll do bad because I haven’t had time lol
trust yourself
you can do it
you are lebob
not lebron james
gritty holly
Thank u bro I’ll try
amber waspBOT
@gritty holly Has your question been resolved?
sacred hawk
what is the value of theta here?
amber waspBOT
what is the value of theta here?
quick spoke
i don't think there's enough information
sacred hawk
unsolvable with just this?
quick spoke
yeah
amber waspBOT
amber waspBOT
Available help channel!
jagged mist
We need to prove this
amber waspBOT
We need to prove this
celest cove
amber waspBOT
@jagged mist Has your question been resolved?
keen pawn
amber waspBOT
keen pawn
so to start the defn of a limit at infty is
$\forall \varepsilon >0 \exists N | x > N \implies |f(x)-L|<\varepsilon$
proven vapor
indeed the definition is $
grizzled pagodaBOT
Veni, vidi, perii
proven vapor
probably add a \in\bR in there
keen pawn
0
proven vapor
so write down what you want to prove
keen pawn
proven vapor
To prove: $\forall\epsilon >0\exists N>0(x>N\implies \left|\frac1{2x+3}\right|<\epsilon)$
grizzled pagodaBOT
kheerii
keen pawn
No idea
proven vapor
you found it already
proven vapor
actually you partially found it
keen pawn
which is?
proven vapor
N must also be positive
keen pawn
I mean, yeah
keen pawn
but you can just set N = (...)
N= 1-3eps/2eps?
proven vapor
again, not quite
you actually need to set $N=\max\left(\frac{1-3\varepsilon}{2\epsilon},1\right)$
grizzled pagodaBOT
kheerii
proven vapor
(I chose the 1 arbitrarily, it can be any positive number)
amber waspBOT
amber waspBOT
Available help channel!
dawn token
isn't f(x)=sin^-1(sinx) => f(x)=x?
amber waspBOT
isn't f(x)=sin^-1(sinx) => f(x)=x?
full marsh
u know the graph of this?
dawn token
Which gives f'(x)=x^2/2
tropic wadi
isn't f(x)=sin^-1(sinx) => f(x)=x?
no it is not
full marsh
sin(arcsin(x))=x but not arcsin(sinx)=x
pseudo crescent
sin(arcsin(x))=x but not arcsin(sinx)=x
specifically, sin(arcsin(x)) = x for -1 <= x <= 1
that's domain, not range
knotty reef
no it is not
Damn 
mb
pseudo crescent
Should work both ways
Try thinking about arcsin(sin(2pi)), do you get 2pi?
knotty reef
Which gives f'(x)=x^2/2
Wat derivative of x is x²/2 
full marsh
u know the graph of this?
best if u know the graph
dawn token
Try thinking about arcsin(sin(2pi)), do you get 2pi?
but arcsin(sin(x))=x 0<x<1
dawn token
pseudo crescent
what calculator is that?
dawn token
desmos
weak zinc
Desmos smoking 🚬
pseudo crescent
what is sin(2pi)?
full marsh
knotty reef
Don't trust calculators
Mine shows 700! Is infinity calculator companies are truly trying to revolutionize math
pseudo crescent
this has a domain and range [-1,1]
Well, that's true, but here we care about sin^-1(sin(x))
dawn token
right
pseudo crescent
My question is: What is sin^-1(sin(2pi))? And as a hint: compute sin(2pi) first
dawn token
0?
knotty reef
My question is: What is sin^-1(sin(2pi))? And as a hint: compute sin(2pi) first
Sin(2pi) is def not || 0 ||
dawn token
pseudo crescent
Do you have any questions related to this pic?
pseudo crescent
note that sin^-1(sin(2pi)) = 0, but 2pi certainly isn't 0
pseudo crescent
but that says sin^-1(sin(x)) = x
This is true only for -1 <= x <= 1
knotty reef
but that says sin^-1(sin(x)) = x
Who is that?
pseudo crescent
This is true only for -1 <= x <= 1
Wait, actually this is wrong
it's true also for -pi/2 <= x <= pi/2, sorry
knotty reef
Who is that?
If that is desmos then desmos confirms this
pseudo crescent
it's true also for -pi/2 <= x <= pi/2, sorry
Between -pi/2 and pi/2, sin(x) looks like this
pseudo crescent
Between -pi/2 and pi/2, sin(x) looks like this
and arcsin(x) is the inverse of this
and so between -pi/2 and pi/2, arcsin(sin(x)) = x
dawn token
isn't f(x)=sin^-1(sinx) => f(x)=x?
okay that means i can not simplify f(x) to x since this is R(x)=[-infinity, infinity] and the other one is R(x)=[-pi/2, pi/2]
pseudo crescent
okay that means i can not simplify f(x) to x since this is R(x)=[-infinity, infi...
You can't do it mainly because arcsin(x) isn't a precise inverse of sin(x). (Because sin(x) is not injective, so it doesnt even have a full inverse). Arcsin(x) is actually an inverse of sin(x), with domain restricted to [-pi/2, pi/2]
dawn token
right
pseudo crescent
but arcsin(sin(x)) itself is defined over all of R, because sin(x) ranges from -1 to 1, and that's also the domain of arcsin(x)
okay, so now let's think about arcsin(sin(x)) for say pi/2 < x < 3pi/2
dawn token
isn't f(x)=sin^-1(sinx) => f(x)=x?
but how would I start with simplifying this?
pseudo crescent
but how would I start with simplifying this?
I think that the best approach would be trying to think about what does the graph of it look like, and possibly trying to plug in some values, for which you can compute sin(x) easily
the name "sawtooth function" is also quite hinting
dawn token
okay, so now let's think about arcsin(sin(x)) for say pi/2 < x < 3pi/2
in the exam we wont have any type of devices
no graph calculators
dawn token
pseudo crescent
Yes, that's what it looks like
note that sin(x) is periodic with period of 2pi
so arcsin(sin(x)) should be periodic too, with period (at most) 2pi
dawn token
f(x)=sin(x)
R(f) = [-1,1]
g(x)=sin^-1(sinx)
R(g)=[sin^-1(-1),sin^-1(1)]
pseudo crescent
so arcsin(sin(x)) should be periodic too, with period (at most) 2pi
So it would be actually sufficient to find its values on e.g. the interval [-pi/2, 3pi/2], and then fill in the rest
dawn token
3pi/2?
pseudo crescent
Yes, that would work
that's essentially one of the tooths
we know that it's periodic with a period 2pi, so we can fill in the rest easily
and to find the values of arcsin(sin(x)) between pi/2 and 3pi/2, note that arcsin(sin(x)) with pi/2 < x < 3pi/2 = arcsin(sin(u + pi)), where u = x - pi (and so -pi/2 < u < pi/2). But then arcsin(sin(u + pi)) = arcsin(-sin(u)) = -arcsin(sin(u)) = -u = pi - x
This description is a bit technical, but you can make a good guess by just plugging in pi, 3pi/2, and maybe something like 5pi/4
pseudo crescent
You would get something like this if you plugged in some values
and then you can do some extrapolation and just draw the line
dawn token
how would go about to simplify it to cosx/sqrt(cos^2(x))
pseudo crescent
how would go about to simplify it to cosx/sqrt(cos^2(x))
sqrt(u^2) = |u|
apply that first
dawn token
oh sorry I meant, how would I even get f'(x)?
pseudo crescent
how would go about to simplify it to cosx/sqrt(cos^2(x))
so this seems correct
grizzled pagodaBOT
MæthIsAlwaysRight
pseudo crescent
This is what you did, right?
pseudo crescent
oh, I see
apply chain rule
arcsin(sin(x)) is a composition of arcsin(x) and sin(x)
dawn token
I got this
pseudo crescent
(f(g(x)))' = f'(g(x)) * g'(x)
dawn token
oh i did primitive
pseudo crescent
I got this
I think you confused sin^-1(x) with 1/sin(x)
full marsh
but I cant get there
maybe ure not taking |cosx| on the root cos^2x
pseudo crescent
I think you confused sin^-1(x) with 1/sin(x)
arcsin(x) and 1/sin(x) are very different
dawn token
right
pseudo crescent
do you know the derivative of arcsin(x)?
full marsh
**MæthIsAlwaysRight**
did u get till here?
pseudo crescent
but I cant get there
.
full marsh
ah ok
dawn token
do you know the derivative of arcsin(x)?
no
pseudo crescent
no
That's something you should probably remember, there is an derivation for that (by implicit differentiation), but it's too long to do it everytime you need to recall the derivaitve of arcsin(x)
the derivative is 1/sqrt(1-x^2)
y = arcsin(x)
sin(y) = x
d/dx sin(y) = d/dx x
dy/dx cos(y) = 1
dy/dx = 1/cos(y) = 1/cos(arcsin(x)) = 1/sqrt(1-x^2)
pseudo crescent
y = arcsin(x)
sin(y) = x
d/dx sin(y) = d/dx x
dy/dx cos(y) = 1
dy/dx = 1/cos(y) ...
this is the derivation in case you're interested
dawn token
got it
pseudo crescent
Ok, try using it to get the derivative of arcsin(sin(x))
by chain rule
(f(g(x)))' = f'(g(x)) * g'(x)
f(u) = arcsin(u)
g(x) = sin(x)
dawn token
is the derivitive of arccos the same?
pseudo crescent
(f(g(x)))' = f'(g(x)) * g'(x)
Do you know this formula?
dawn token
I know it
pseudo crescent
like i mean the f'(u)
f'(g(x))?
pseudo crescent
f(x) = arcsin(x)
f'(x) = 1/(1-x^2)
g(x) = sin(x)
What is f'(g(x))?
dawn token
f(x) = arcsin(x)
f'(x) = 1/(1-x^2)
g(x) = sin(x)
What is f'(g(x))?
AHAAA
cosx/sqrt(1-(sinx)^2)
the full answer
pseudo crescent
the full answer
Yes
couple of simplifications can be made though
1-sin^2(x) = cos^2(x)
that gives
cos(x) / sqrt(cos^2(x))
sqrt(cos^2(x)) = |cos(x)|
that gives cos(x) / |cos(x)|
pseudo crescent
that gives cos(x) / |cos(x)|
And this is very interesting actually
if cos(x) is negative, then cos(x) / |cos(x)| simplifies to cos(x) / -cos(x) = -1
if cos(x) is positive, then cos(x) / |cos(x)| simplifies to cos(x) / cos(x) = 1
and this is another way to draw the graph
dawn token
those graphs look interesting
thank you though
will reopen if i stumble acroos anything else
.close
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worn vault
I need help finding the volume of the space in the unit cube that is above the surface defined by $z=\dfrac{\ln X}{\ln(xy)}$ where $0 \leq X < 1$
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The د
sterile nymph
are X and x different?
worn vault
yeah X is just a constant
worn vault
if so is X a random variable?
yup
worn vault
ok sure ignore the equality
this would be 1 - the volume under the surface from (0, 0) to (1,1) but im not so sure how to get that
sterile nymph
This inequality seems to be equivalent to xy^z > X
sterile nymph
Is this about the 3b1b twitter post?
worn vault
lol yes
sterile nymph
lol
worn vault
the volume should represent the cdf of the the random variable
if my logic was correct
sterile nymph
I think your logic is indeed correct.
neat tusk
@sterile nymph just a small dought pls tell me 3*squt(5)=15 is this correct
sterile nymph
!occupied
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sterile nymph
I'll catch you in #serious-discussion
tough mica
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tough mica
how do I determine if this sets are subspaces
shadow stump
use the subspace test (contains 0, closed under addition, closed under scalar multiplication)
worthy mountain
tough mica
okay
for a)
,, x_1 - x_2 + 2x_3 = 0 \textbf{ contains zero } \implies 0 - 0 +2(0) = 0
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tough mica
my problem is with addition
shadow stump
for context this is chapter 4.2 in anton
but the idea is that we have two vectors in our subspace [ \vb x = (x_1,x_2,x_3), \quad \vb y = (y_1,y_2,y_3) ] where they both individually satisfy the requirement: [ x_1-x_2+2x_3=0, \quad y_1-y_2+2y_3 = 0] and we want to know whether their sum [ \vb x + \vb y = (x_1+y_1,x_2+y_2,x_3+y_3) ] is in the subspace. in other words, is it true that [ (x_1+y_1)-(x_2+y_2)+2(x_3+y_3)=0? ]
grizzled pagodaBOT
cloud
tough mica
yes is it is true because we have the zero vector aswell
so addition should be cleared, about scalar multiplication though
shadow stump
scalar multiplication is very similar. take a vector in the subspace, say x, and some scalar, say k, then is kx in the subspace?
shadow stump
yes is it is true because we have the zero vector aswell
well we should be able to justify it using only the information in that picture
shadow stump
**cloud**
you need to justify whether x + y satisfies the equation using the fact that x and y satisfy it individually
tough mica
ok
now, regarding scalar multiplication, how would that go?
progress:
$x_1 - x_2 + 2x_3 = 0 \ u_1 - u_2 + 2u_3 = 0 \ u = k \cdot x \ kx_1 - kx_2 +2kx_3 = 0 \ k(x_1 - x_2 + 2x_3) = 0$
xd
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shadow stump
use the first equation
tough mica
,, \lambda \in \mathbb{R} \land \textbf{v} \in \mathbb{W}\ \lambda \cdot \textbf{v} = \left(\lambda \cdot v_1, \lambda \cdot v_2, \lambda \cdot v_3\right) \ \textbf{v} \in \mathbb{W} \implies \begin{cases} x_1 = \lambda \cdot v_1 \ x_2 = \lambda \cdot v_2 \ x_3 = \lambda \cdot v_3 \end{cases} \ x_1 - x_2 + 2x_3 = 0 \ \lambda v_1 - (\lambda v_2) + 2(\lambda v_3) = 0 \ \lambda v_1 - \lambda (v_2) + 2\lambda(v_3) = 0 \ \lambda(v_1-v_2+ 2v_3) = 0 \implies \lambda = 0
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shadow stump
given:
$v$ is in W, so
\begin{equation}
v_1 -v_2 +2v_3 = 0
\end{equation}
show that, \emph{for all} $\lambda \in \R$
\begin{equation}
(\lambda v_1) -(\lambda v_2) + 2(\lambda v_3) \qeq 0
\end{equation}
you should \emph{use} equation (1) to prove equation (2)
grizzled pagodaBOT
cloud
tough mica
I am trying
but I cannot do it
,, v_1 -v_2 +2v_3 = 0 \ (\lambda v_1) -(\lambda v_2) +2(\lambda v_3) = 0 \ \lambda(v_1 -v_2+2v_3)=0 \ \implies \lambda = 0
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harsh sable
i want to know if my proof is sound
amber waspBOT
i want to know if my proof is sound
harsh sable
i feel like i was just changing semantics and is not confident that the manipulations are correct
thats the powerset
hybrid sedge
whats the method for converting an integrals upper and lower limits from respect to one variable to another?
indigo cloud
hybrid sedge
i had a feeling thats what it was but i know how to do u substitution it mightve been related to upper or lower limits being functions themselves
i think maybe a function of ‘u’ inside an integral with upper limit of a function of x
this is what reminded me of it because i thought i remembered there was a method i forgot about to go from e to 1 for the upper limit
pallid canopy
,rotate
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eternal breach
how to write an short response?
amber waspBOT
how to write an short response?
shrewd elm
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open ruin
So I am trying to get the Fourier series for the ode as written in purple
open ruin
I’m at the stage where I am finding one of the coefficients but I don’t know how to integrate a product of trigonometric functions
clear meteor
I’m at the stage where I am finding one of the coefficients but I don’t know how...
this seems like it might be over my head but could you use a trig identity here?
specifically product to sum
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rain harness
i did y = 6x-5 and this and it keeps saying my answer is wrong
rain harness
does anyone know why
rain harness
i used the y-y1=m(x-x1) formula
slender blaze
yup that’s point slope
ivory sorrel
i did y = 6x-5 and this and it keeps saying my answer is wrong
Looks right. Why did you keep 5 on the RHS
I think that's the only issue
rain harness
i also wrote it as y=6x-5
slender blaze
standard form is equaled to 0
rain harness
ohh
ivory sorrel
i also wrote it as y=6x-5
Standard form means RHS must be 0
ivory sorrel
Yes
rain harness
ok it worked ty!
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fallen vigil
diameter: 12 cm
OA Is a right angle
BA = 14 cm
Find the area of OAB
final matrix
Wouldn't diameter be 12 given radius is 6
fallen vigil
Wouldn't diameter be 12 given radius is 6
My bad
fallen vigil
can i use this channel for a question?
Its used
final matrix
can i use this channel for a question?
Go to an open help channel
final matrix
You know O to beta is also 6cm
Since its also radius
And I think you could use the properties of that isosceles triangle to find out further information about angle AOB
Which would let you solve for the triangle
Not 100% sure tho
misty nimbus
diameter: 12 cm
OA Is a right angle
BA = 14 cm
Find the area of OAB
i might be misunderstanding the problem
but since A is a right angle
woudlnt the area be 6*14/2 = 42?
weak zinc
woudlnt the area be 6*14/2 = 42?
Diagram is a bit misleading, assuming it’s meant to mean the sector of the circle
misty nimbus
Diagram is a bit misleading, assuming it’s meant to mean the sector of the circl...
oh i see
weak zinc
but like ik how to find OB with soh cah toa
And, also, it’s not sohcahtoa that’ll get you OB right now, unless you take an intermediate step first, but assuming it’s the sector of the circle you want (is it?)
…then you can find the angle that’s at the centre of the circle, that time, with sohcahtoa 
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lime jasper
Does (a-b)(a+b)+1 = (a-b)(a+b+1)?
amber waspBOT
Does (a-b)(a+b)+1 = (a-b)(a+b+1)?
hearty ledge
no
final matrix
you can try expanding it out to check
hearty ledge
You know how to multiply and open up these?
lime jasper
Wait then how does (a+b)(a-b)+(a-b) = (a-b)(a+b+1)
hearty ledge
Wait then how does (a+b)(a-b)+(a-b) = (a-b)(a+b+1)
Do you know how to expand them?
lime jasper
Yea
hearty ledge
If you just multiply and expand them, you can compare the both sides
What is (a+b)(a-b)+(a-b)?
Try expanding it
lime jasper
(a^2-b^2)+(a-b)
hearty ledge
Yep
hearty ledge
(a^2-b^2)+(a-b)
And now add these
amber waspBOT
@lime jasper Has your question been resolved?
lime jasper
And now add these
Uh idk how
hearty ledge
It just becomes:
$a^2-b^2+a-b$
grizzled pagodaBOT
moaforlife
hearty ledge
Ok?
lime jasper
Oh
hearty ledge
Now can you also multiply (a-b)(a+b+1)?
lime jasper
Now can you also multiply (a-b)(a+b+1)?
a^2+a-b^2-b
lime jasper
Yea but
hearty ledge
You just have to multiply and check whether they are equal or not... Thats the key to solve this kind of problems
lime jasper
(a+b)(a-b)+(a-b
Yea but I wanna know how this
hearty ledge
RIght!
Yeah it
is
Wait let me tell you how to...
Do you know this property?:
$xa+xa=x(a+b)$
grizzled pagodaBOT
moaforlife
lime jasper
Where did the b come from
hearty ledge
No these are random variables...
I am just asking do you know this property of multiplication?
like if x is common in several things, you can just factor it out...
lime jasper
Yea ik that
hearty ledge
(a-b)(a+b)+1
No.
If you factor (a-b) it should be multiplied with both a+b and 1
like this:
(a-b)(a+b)+(a-b)(1)=(a-b)(a+b+1)
hearty ledge
**moaforlife**
Look at this property...
@lime jasper do you get what you did wrong?
when we bring something to common, it's then multiplied to both the numbers, not by only one...
lime jasper
(a-b)(a+b)+(a-b)(1)=(a-b)(a+b+1)
I don’t understand how 1 gets inside the parentheses
hearty ledge
**moaforlife**
Don't you see this property?
lime jasper
Reverse distributive
hearty ledge
Its like this:
(a-b)(a+b)+(a-b)(1)=(a-b)(a+b+1)
So, when we are taking a-b in common
a+b and 1 will be summed
and then multiplied with a-b
lime jasper
Oh
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proper umbra
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proper umbra
How did they transform n(y^n+1)/n+1 to n/n+1
Its probably something obvious since the prof in the vid didnt explain but idk my slow ass cant tell
gloomy tide
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solemn scarab
can someone explain how we got zero?
amber waspBOT
can someone explain how we got zero?
lavish wind
can someone explain how we got zero?
dividing top and bottom through by the highest order of x
in this case, x^2
and plugging in x = infinity
taking the limit of each term
because it's very easy to evaluate lim x->inf c/x^n, where c is a constant and n>=1
lavish wind
$\lim_{x \to \infty} \frac{-\frac{1}{x} + \frac{2}{x^2}}{1 + \frac{2}{x} + \frac{2}{x^2}}$
grizzled pagodaBOT
Melvin Eugene Punymier
lavish wind
would it be something along the lines of this
no
lavish wind
**Melvin Eugene Punymier**
plug infinity into this
(you can't really "plug" infinity in because we can't reach it, but we can take the limit of every term)
...effectively what we would expect if we did plug in infinity
dividing by a power of infinity greater than or equal to 1 gives us zero
so all of the terms with powers of x in the denominator go to zero
all of the terms in the top go to zero...
(and all but the constant term in the bottom go to zero)
so the answer is zero
I want to caution you
this problem began
where if you just plugged in infinity right away
you would have infinity in the top and infinity in the bottom
this would be "indeterminant"
your work in these kinds of limit problems requires you to reformulate what you are given so you can avoid an "indeterminant form"
that's why you divide by the highest order of x you have
to avoid that problem
lavish wind
np
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mighty wasp
I'm trying to prove for all positive continous random variables, the expected value $E(X)$ can be calculated as $E(X) = \int_{0}^{\infty}{(1-F_x(X))dx}$. I attempted to derive the equation by using the definition of the expected value and solve the inetegral $E(X) = \int_{-\infty}^{\infty}{g(x)f_x(x)dx}$
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Delusional J
mighty wasp
Using integrals by parts and the fact that we are dealing with a positive continous random variable, I get $E(X) = \int_{0}^{\infty}{g(x)f_x(x)dx} = g(x) -\int_{0}^{\infty}{g'(x)dx}$
grizzled pagodaBOT
Delusional J
mighty wasp
This is just wrong though. I'm pretty sure that $\int_{0}^{\infty}{f_x(x)dx} = 1$
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Delusional J
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dreamy slate
someone help me what's 2x^5x(x)7x>6x+5x
amber waspBOT
someone help me what's 2x^5x(x)7x>6x+5x
sterile nymph
[
2x^{5x} \times 7x > 6x + 5x
]
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OmnipotentEntity
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toxic canopy
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toxic canopy
am i missing something here?
knotty reef
am i missing something here?
Yes you are missing the answer
eager lintel
CBD is the same angle as A by similar triangles/parallel lines. Call it x. Using that BF bisects ABD I think you can write an equation down and solve for x.
knotty reef
We got some jokesters in here
|| here's a hint can you find angle dbf given that bd|ae
Since bf is the angle bisector of angle dba can you find angle dba and correspondingly angle DBC ||
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haughty gust
Can someone help me w this?
amber waspBOT
Can someone help me w this?
haughty gust
In the first picture it’s easy to make the tables since we just put the values as per f(x) and then root it
But
How would I do it for the second graph without any information about the function
Oh nvm
I got it
Sorry thanks
.close
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safe yarrow
hi
safe yarrow
I need Help! I need f`(x) in this picture
tulip tapir
use the quotient rule
or chain rule jeez
i just think for beginners quotient is easier here
continue chartbit
i swear y'all gonna incinerate me one day
weak zinc
i swear y'all gonna incinerate me one day
weak zinc
i just think for beginners quotient is easier here
Either works to be fair, though personally I disagree (at least for me, quotient rule was taught as a result of chain and product rule!)
safe yarrow
idk but it is f´(x) = sin(x) / 1 ???
tulip tapir
Either works to be fair, though personally I disagree (at least for me, quotient...
really? we were just taught it, had to prove it myself
weak zinc
really? we were just taught it, had to prove it myself
Yea at least in one course, when chain and product rule were given, the quotient rule was introduced by working with f(x) * (g(x))^{-1} 
in any case, "most important thing first is that it works" 
safe yarrow
it is f´(x) = sin(x) / cos(x)^2 ???
tulip tapir
Yea at least in one course, when chain and product rule were given, the quotient...
ah
yes
good job
safe yarrow
My mistake f´(x) = ( u´ * v - u * v´ )/ u^2
But this is right: f´(x) = ( u´ * v - u * v´ )/ v^2
thx
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ancient raft
How do I show that 1+2i is a root of -3+4i?
amber waspBOT
How do I show that 1+2i is a root of -3+4i?
ancient raft
The argument of -3+4i doesn't give the common angles. And $$w^n=z$$ doesn't seem to help at all.
grizzled pagodaBOT
Good
pallid canopy
,w (-3+4i) / (1+2i)
grizzled pagodaBOT
pallid canopy
wow look at that
ancient raft
use this with n=2
What does the result tell, or the logic behind? I understand that n=2, but not that z is divided by the root, and the result tells you something.
pallid canopy
if w^2 = z, then z/w = w
ancient raft
So the other root is also 2i+1?
ancient raft
The formula for roots is $$w_{k}=w_{0}\cdot (e^{i 2\pi/n})^k$$
grizzled pagodaBOT
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