okayy
#help-36
glass wren
mint orbit
$x^2(ax+b)+(ax+b)$
soft zealotBOT
jan Niku
mint orbit
can you see how factor this now?
mint orbit
no im giving out of pocket help haha
glass wren
HAHHHAHHA
mint orbit
i think so
we have a common term right
glass wren
Yes
mint orbit
its not clear with your problem to be honest like 
so we can make assumptions about m
and find roots
we can just write down all the roots
maybe they want something more specific
glass wren
They doooo
mint orbit
like classify where there are roots?
glass wren
ill just give you the entire thing
c) Determine the real number m, knowing that the sum of two roots of the polynomial f is equal to 1.
mint orbit
man these annoying sum of roots problems
do you have any results you can use already?
glass wren
HAHHAHHA u can leave it if ud like
mint orbit
no I'm just wondering how far back we have to start
glass wren
My friend went to sleep soi cant ask him for hints
glass wren
do you have any results you can use already?
nope
glass wren
no I'm just wondering how far back we have to start
Frol the very start ๐ข
mint orbit
okay
lets say this thing has 3 roots
lets call them r1 r2 and r3
can you write the factored form of f(x)?
mint orbit
๐
mint orbit
no its alright
glass wren
https://cdn.discordapp.com/attachments/908076393198419988/1444505315042263264/33...
-7x
mint orbit
lets say we have a quadratic
glass wren
yes
mint orbit
it looks like x^2 + bx + c
and we know it has roots r1, r2
then we can write x^2 + bx + c = (x-r1)(x-r2)
glass wren
mhm
mint orbit
so, same thing in your problem
we have $f(x) = 2\qty(x^3 -2x^2-\frac 72 x+\frac m2)$
and r1, r2 and r3 are roots
soft zealotBOT
jan Niku
glass wren
Yes
mint orbit
then we can write x^2 + bx + c = (x-r1)(x-r2)
can you extend this?
to the cubic case we have
mint orbit
i just mean, you agreed on the quadratic case
so can you guess how it should work out in the cubic
mint orbit
do you get why if r1 and r2 are roots of the quadratic then we can write (x-r1)(x-r2)
like, we assume we can factor it right
so something times something
glass wren
i think so
glass wren
Hold on 3 minutes
mint orbit
okay
mint orbit
so we want 2 roots
2 places its equal to 0
so where x=r1, we want it to be 0
and where x=r2, we want it to be 0
we can do this by (x-r1)(x-r2)
now its 0 times something at r1, and at r2
glass wren
and where x=r2, we want it to be 0
Yea
mint orbit
so we can do the same thing for a cubic
glass wren
okay
mint orbit
say that cubic looks like $2\qty(x^3-2x-\frac72 x + \frac m2)$ and its 0 at $r_1, r_2, r_3$.
glass wren
Hahhha
soft zealotBOT
jan Niku
mint orbit
then $2\qty(x^3-2x^2-\frac72 x + \frac m2) = 2 (x-r_1)(x-r_2)(x-r_3)$
soft zealotBOT
jan Niku
mint orbit
seem reasonable?
mint orbit
why what?
glass wren
Like why are there only 3 of r
mint orbit
this is just something we have in general
if you have a polynomial of degree n
it can only have at most n roots
so a line can only cross the x axis once
glass wren
ohhh
mint orbit
a quadratic twice
glass wren
Ohh yez
glass wren
Sure
mint orbit
c) Determine the real number m, knowing that the sum of two roots of the polynom...
but the way this is phrased makes it sound like, yea, we have three
and the sum of some 2 is 1
glass wren
mint orbit
if they said "the sum of the two roots**"
although who knows what the hell they mean
anyways
glass wren
Hahaahhhhahhhahhha
mint orbit
so we can do some fancy things
glass wren
lets do them
mint orbit
**jan Niku**
we can expand the right hand side
mint orbit
yea just foiling like normal
glass wren
mhmmm
glass wren
wowwwww
mint orbit
kind of
glass wren
Hahhhahhahhha
soft zealotBOT
mint orbit
okay
glass wren
wolfie
mint orbit
$2\qty( x^3 - 2x^2 - \frac 72 x + \frac m2 ) = 2\qty( x^3 - [r_1 + r_2 + r_3] x^2 + [r_1(r_2 + r_3) + r_2r_3] x - r_1 r_2 r_3 )$
soft zealotBOT
jan Niku
glass wren
damn
mint orbit
but, we know some things
we know the some of two roots is 1
so lets call it, $r_1 + r_2 = 1$
soft zealotBOT
jan Niku
mint orbit
do you see what should happen then?
glass wren
My bad for the dumb questions
mint orbit
no, youre actually right
glass wren
no, youre actually right
Oh :)
glass wren
HAHAHAHA nahh its cool
mint orbit
these things have to be the same
glass wren
these things have to be the same
Why
mint orbit
fancy name is 'equating coefficients'
glass wren
ohhh so thats just fixed theres no choice
mint orbit
Why
its just how it works out, if you have two polynomials on either side of an =
the part attached to the x^2 has to equal on both sides
glass wren
its just how it works out, if you have two polynomials on either side of an =
Sure also why that part
OH
WAIT
IM SEEING IT
IM SEEING IT
OK OK
mint orbit
youre right to doubt it but yea maybe you can convince yourself
so we figure
$-2 = -(r_1 + r_2 + r_3)$
soft zealotBOT
jan Niku
soft zealotBOT
jan Niku
glass wren
youre right to doubt it but yea maybe you can convince yourself
Iwhat do you mean
glass wren
but we know r1 + r2 = 1
RIGHT
mint orbit
Iwhat do you mean
about the equating coefficients
like lets say you had idk
$mx+b = nx+c$
and this was true for all x
heck we can even make it more so
soft zealotBOT
jan Niku
glass wren
Sure
mint orbit
**jan Niku**
this has to be true at x=0
is true for all x
so then $m\cdot 0 + b = 0 \cdot n + c$
soft zealotBOT
jan Niku
glass wren
Mhm
soft zealotBOT
jan Niku
soft zealotBOT
jan Niku
soft zealotBOT
jan Niku
glass wren
hahhahha
mint orbit
like, if we do this process one time to a quadratic, we end up with the linear case
we let x=0, every term except the constants falls away
theyre equal, so we subtract them
this leaves a common x factor, we divide it out
rinse and repeat
mint orbit
so were back here
we know that the x^2 coefficients must be equal
and we know that r1+r2=1
glass wren
Yr
mint orbit
so $-2 = -(r_1 + r_2 + r_3)$
soft zealotBOT
jan Niku
mint orbit
or $2 = 1 + r_3$
soft zealotBOT
jan Niku
glass wren
hahhhah omgg
mint orbit
you dont want the answer so
mint orbit
**jan Niku**
maybe you can go forward from this
knowing the coefficients are equal blah blah
glass wren
maybe you can go forward from this
I think i can use that
mint orbit
glass wren
knowing the coefficients are equal blah blah
THANKS
Thanks a lotttttt you really helped me i was so lost
THANK YOU
glass wren
thanks so much
thannkkss!! Goodbyee
Byee
Id give you 5 stars on yiur service if i could
byee!!
bold dust
help pls
plucky rover
bold dust
casework but it's too tedious
safe ruin
i understand it now
plucky rover
What stage of study are you in
Are you familiar with random variables and distributions
bold dust
mathcounts
bold dust
Are you familiar with random variables and distributions
yeah kind of
plucky rover
Actually nvm we can just use symmetry
plucky rover
You have 37 tosses
plucky rover
try heads>=19
Overcomplicating it
bold dust
yeah like 37 choose 19 = 37 choose 18, etc
plucky rover
You have 37 tosses
Let's say h heads and t tails
polar agate
yeah like 37 choose 19 = 37 choose 18, etc
oh nononono pls no
plucky rover
With me so far?
safe ruin
oh nononono pls no
why?
plucky rover
Because there are way less computational ways of doing this
plucky rover
Void you aren't the helpee
bold dust
With me so far?
si
plucky rover
Cool
So now h and t are positive integers
Well non negative integers
Do you agree with me that at least of them is always โฅ 19
(and more importantly, do you see why)
bold dust
yes heads cause its greater
plucky rover
No
I mean in every case
Not just the favourable ones
h + t = 37
No matter what happens, one of them is always >= 19
bold dust
yes because for them to add up to 37 one of them have to be greater than half of 37 or basically 19 since they're integers
plucky rover
Yup
Now, how many total cases are there
How many total possibilities
Accounting for ordering
polar agate
(btw once you're done, I'll reveal a nice fact about this problem)
bold dust
18 cases?
plucky rover
(I have a feeling I know which fact it is)
bold dust
wait no
bold dust
2^37 total possibilty
plucky rover
Yes
bold dust
sry "cases" threw me off
plucky rover
Now if you pick an outcome
It's either favourable or not
Those are the only two possibilities
Now this next idea might be a bit weird so take it in
If you take an outcome, and reverse every coin in it
It changes it from favourable to unfavorable and vice versa
bold dust
cause the number of heads and tails switch?
plucky rover
If you take an outcome, and reverse every coin in it
Now notice that
If you repeat this you get the original back
Do you see where I'm going with this
bold dust
So half the outcomes work while the other half doesn't?
plucky rover
Exactly
You can pair them up
And only one of each pair works
So what's the probability
bold dust
1/2
plucky rover
Good
bold dust
that makes a lot more sense thanks
plucky rover
Note that this only works cuz the total tosses is odd
If it was even it would be a bit more involved
But still the same idea
plucky rover
(btw once you're done, I'll reveal a nice fact about this problem)
๐
plucky rover
LMAO
polar agate
but anyway the fact is that this works for any odd number of tosses
plucky rover
(I have a feeling I know which fact it is)
Called it 
manic helm
why does this line hold true
final saddleBOT
why does this line hold true
vital crag
Do you know any properties of measures
manic helm
Do you know any properties of measures
yes
vital crag
Show the ones you learned
manic helm
both measures and outer measures?
vital crag
Yea 12.16 iii
manic helm
Yea 12.16 iii
that's for outer measures tho
the red underlined part is using the measure on R(S)
vital crag
12.12 iv with some cleverly chosen sets
manic helm
12.12 iv with some cleverly chosen sets
but R(S) isnt a sigma algebra
nor is it a sigma ring (i think)
swear it's the finite disjoint unions of sets in S
vital crag
Then where's the definition of R(S)
vital crag
Yea I don't know then if R(S) isn't a sigma algebra
manic helm
rip
final saddleBOT
@manic helm Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
exotic shale
show that the 3 lines intercept at 1 point if and only if a+b+c=0
final saddleBOT
show that the 3 lines intercept at 1 point if and only if a+b+c=0
iron mist
show that the 3 lines intercept at 1 point if and only if a+b+c=0
what have you tried
exotic shale
what have you tried
i tried using the determinant with variables being a,b,c
and gaussian elimination but that takes a lot of cases
iron mist
hmm my first instinct is to notice the symmetry
iron mist
each of the variables (and 1) takes turns being multiplied to each constant
exotic shale
oh wait i just noticed something
if we add all the equations
we get (a+b+c)(x+y+1)=0
iron mist
yes
iron mist
yeah you can't really use the a+b+c=0 there
exotic shale
wait though
iron mist
but what happens if a+b+c is not zero
exotic shale
im i seeing this wrong?
if a+b+c=0
then x,y are free variables
so the solution is R^2
well not R^2
iron mist
no not necessarily
iron mist
but if a+b+c is nonzero, what does that force x+y+1 to be
exotic shale
not line like basically the span of 2 independent vectors
exotic shale
no not necessarily
how so if its 0 then x,y can be anything
oh wait no
x is dependent on y
exotic shale
so if a+b+c= 0 it has infinetely many common points?
iron mist
not necessarily
exotic shale
can you explain that
iron mist
but if a+b+c is not 0, then there are infinitely many (x,y) that work
exotic shale
oh nvmv yea i see this (a+b+c)(x+y+1)=0 just becomes useless if its 0
iron mist
show that the 3 lines intercept at 1 point if and only if a+b+c=0
we are proving two directions
iron mist
yes
exotic shale
ok so if a+b+c=/0
iron mist
but if a+b+c is not 0, then there are infinitely many (x,y) that work
but can you see how this proves one direction?
exotic shale
x+y+1=0
exotic shale
but can you see how this proves one direction?
idk what that means
iron mist
ok, when we say "p if and only if q", we need to prove two things: "if p then q" and "if q then p"
iron mist
there's also such a thing as the contrapositive
exotic shale
yea ik what it means nvm
i thought you were talking about the direction of the line
iron mist
no that's irrelevant lol
exotic shale
ok no i dont see it yet how it proovevs it
also the excersice says to show that they intercept
not that they intercept at only one point
if it says they intercept at one point does that also mean they cant have a second point
im not sure if its supposed to say only to mean that or yea not sure how to interpret it
exotic shale
no that's irrelevant lol
so the problem means exactly at 1 point
not minimum at 1 point
exotic shale
but if a+b+c is not 0, then there are infinitely many (x,y) that work
yea can you explain this one
iron mist
we get (a+b+c)(x+y+1)=0
from here, if a+b+c is nonzero, then x+y+1=0, or equivalently, x+y=-1, which is a line with infinitely many points
exotic shale
i see that it proves the backwards direcction but i dont see why there are infinite
exotic shale
from here, if a+b+c is nonzero, then x+y+1=0, or equivalently, x+y=-1, which is ...
yea but you have the other 3 equations
aswell
iron mist
we got this equation by adding the three others
iron mist
wait
exotic shale
dont we need to sub back y=-x-1
and check that they are all true for any x
to say this is a line
iron mist
dont we need to sub back y=-x-1
ok so you can do this if you want for all three equations, which yields three different forms of "x"
exotic shale
ok so you can do this if you want for all three equations, which yields three di...
how else do i know that it verifies all 3 for any x
iron mist
since it's a system, that "x" has to satisfy them all, so we can equate them, and it yields an equation $a^2+b^2+c^2=ab+bc+ca$, which only holds if $a=b=c$
soft zealotBOT
CST (reply ping for help)
iron mist
but since the x's had the form (b-c)/(a-b), that's indeterminate
exotic shale
**CST (reply ping for help)**
this is (a+b+c)^2 = 0
iron mist
and if you plugged it back into the equations, you'd just get the line x+y+1=0 after dividing out the nonzero constants which yeah infinitely many solutions
iron mist
not what we're trying to prove
exotic shale
what are we trying to proe
like when we got y+x+1=0 how do we know thats the solution we only used 1 equation
iron mist
but that equation came from adding all of the equations in the system
exotic shale
but that equation came from adding all of the equations in the system
and what does this mean
is this always true?
iron mist
if we add all the equations
you added them lol
Gaussian elimination works under the same principle, right?
adding multiples of a row to another row
exotic shale
no
iron mist
yes
exotic shale
because like that will only change the top line
the other 2 equations at the buttom will stay
youd have that equation and the buttom 2 in gaussian elimination
actually yea idk
iron mist
if you think of the first three equations as vectors
the fourth equation is a linear combination of the first three
so it must exist in the same vector space
and so anything that works for all of the first three equations
must also work for the fourth one
exotic shale
if you think of the first three equations as vectors
like addding equations can give you a relation between 2 variables but plugging them back in to an equation can give you a alue no?
exotic shale
and so anything that works for all of the first three equations
yes ik that
but is this also true in the other direction
if anything works for the 4th then it works for the first three
like adding 3 equatioons you can lose information
seems like it has no solutions unless a=b=c
iron mist
yeah it's either a=b=c and infinitely many solutions, or at least one pair of them is not equal and we don't have any solutions
but yeah the tldr is: if a+b+c is nonzero, then it does not intersect at exactly one point
thus the contrapositive: if they intersect at exactly one point, then a+b+c=0. So that's one direction
NOW we assume that a+b+c=0 and try to get that (x,y) is unique
exotic shale
yea but i didnt understand this part
how do we do it
like am i allowed to take the det
een thouhg i have x y and 1 with 1 being a scalar
iron mist
i mean you can try
exotic shale
i mean you can try
so do you know how we move one from here
like how do i show that it either has 0 or infinite
but never 1
iron mist
the issue with this determinant approach is that we'll get -a^3-b^3-c^3+3abc, which factors into -(a+b+c)(a^2+b^2+c^2-ab-bc-ac). Obviously if a+b+c=0, then our determinant will be zero and so the matrix is noninvertible
so it's important that 1 does not become a "variable"
exotic shale
so how do i solve this problem
man it cant be this complicated
it takes 1.5 points/10 in a 2 hour exam
this supposed to be like a 15 min problem
iron mist
you're overthinking it that's why
anyway
other way: if a+b+c=0, what value of (x,y) will obviously satisfy the three equations
exotic shale
other way: if a+b+c=0, what value of (x,y) will obviously satisfy the three equa...
1
iron mist
yeah, (1,1)
iron mist
so now it's time to show that no other (x,y) will work for all possible a,b,c where a+b+c=0
so perhaps one equation at a time: if ax+by+c=0, and a+b+c=0, then try eliminating the c?
iron mist
do this for the other two equations also
tulip willow
ya sin pidora
exotic shale
b(x-y) + a(1-y) =0
final saddleBOT
final saddleBOT
Available help channel!
delicate fiber
final saddleBOT
delicate fiber
its saying my answer is only partially correct
wicked robin
I mean ur w is simply not even in the domain of f
U want to find a w in the domain of f
delicate fiber
ive tried -3-i and -3+i as well and they are also marked as partially correct
wicked robin
Whatโs ur definition of log_alpha
This seems sketchy because the usual definition is analytic everywhere in its defined domain unless Iโm missing something
I.e there is no such w
delicate fiber
wicked robin
You can deduce what alpha must be here
delicate fiber
You can deduce what alpha must be here
5pi?
wicked robin
4pi no?
delicate fiber
oh because thats when its not defined?
does that mean that the arg(z) = 4pi
which is just 0?
wicked robin
which is just 0?
Sorry was away, well it depends on ur definition
Ive seen definitions that are such that arg_alpha must lie in (alpha, alpha + 2pi]
But maybe urs is different?
delicate fiber
uhh wha
delicate fiber
thats what they wrote here no?
wicked robin
Right, well I mean in some sense this makes it so we can sort of ignore what alpha is
Anyways, log_alpha(z) is log|z| + i arg_alpha(z) right?
delicate fiber
yes
wicked robin
Iโm assuming the assignment want you pick a w on a certain ray then
If we ignore f for a moment
We can pick say z = e^i(alpha)
This lies on the ray where log_alpha(z) is not analytic
(If we allow a bigger domain; in most cases this isnโt defined tho, but Iโm guessing ur assignment is using the defined domain of arg instead)
delicate fiber
yep
wicked robin
Now to choose w accordingly for ur f, do u see what to do?
delicate fiber
so whenever f outputs a value on the pos real axis?
oh yeah thats where i mustve got confused
normally its (-pi, Pi]
so in that case it would not be defined on the neg real axis
wicked robin
so whenever f outputs a value on the pos real axis?
log_alpha(z) wonโt be analytic for zโs on the positive line
Is the main takeaway
delicate fiber
yeah thats what i meant
wicked robin
Now choose w accordingly to work for f
delicate fiber
iw - 3 + 3i = k where k is real and >0
wicked robin
Yes
wicked robin
Ur welcome!
final saddleBOT
@delicate fiber Has your question been resolved?
old quarry
final saddleBOT
tired walrus
your work is kinda messy, and you're suffering from a severe case of fraction-bar shrivel-itis, but it looks like you got the final answer as 4 sqrt(2), which is correct and desmos agrees
where did you get it from that you've gone wrong?
old quarry
oh
one sec let me send the whole q
for a i got 1/8
and b = got 4sqrt(2)
which is not matching any option
worldly spruce
I swear this writing follows y = -2^x
old quarry
I swear this writing follows y = -2^x
hm?
tired walrus
,w lim[x->0] (sqrt(1+sqrt(1+x^4))-sqrt(2))/x^4
soft zealotBOT
tired walrus
looks like it was a that you made a mistake in
old quarry
looks like it was _a_ that you made a mistake in
OH YEA got it
final saddleBOT
old quarry
thanks
final saddleBOT
Available help channel!
leaden moon
I'm needing help with inequalities again. I know that $(a, b) \in (\mathbb R^+)^2$ and $a + b = 9$. I need to find the minimum of $P = a^3 + b^4$. I need to guess the landing point for this one, how one should do that?
final saddleBOT
I'm needing help with inequalities again. I know that $(a, b) \in (\mathbb R^+)^...
soft zealotBOT
1 divided by 0 equals Infinity
barren pebble
Use UCT:
(aยณ+216+216)+(bโด+81+81+81)โฅ108(a+b)=972
=>Pโฅ297
barren pebble
Undefined coefficients technique
barren pebble
Sure
Equality happens at a=x, b=y
So x+y=9
aยณ+2xยณโฅ3xยฒa
bโด+3yโดโฅ4yยณb
You want 3xยฒ=4yยณ
From there you find x and y and just plot it in, done
leaden moon
Equality happens at a=x, b=y
that's our assumptions?
leaden moon
alr
leaden moon
aยณ+2xยณโฅ3xยฒa
bโด+3yโดโฅ4yยณb
like at this part, why do we specifically add $2x^3$ and $3y^4$?
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
can i just add in $x^3$ and $y^4$ alone?
soft zealotBOT
1 divided by 0 equals Infinity
leaden moon
so the 2s are to make it clean
barren pebble
Not just clean, to make it the correct power
aยณ+xยณ+xยณโฅ3xยฒa (AM-GM)
Here a is degree 1
Because you're given a+b=9
leaden moon
$a^3 + 2x^3 = a^3 + x^3 + x^3 \geq 3\sqrt[3]{a^3x^3x^3} = 3ax^2$
soft zealotBOT
1 divided by 0 equals Infinity
barren pebble
Yes
leaden moon
so i have to do some manipulations just to make this clean
and same degree as what they gave me
barren pebble
What manipulations?
barren pebble
and same degree as what they gave me
This, yes
leaden moon
What manipulations?
like adding $2x^3$ in
soft zealotBOT
1 divided by 0 equals Infinity
barren pebble
Yeah that's just the technique
leaden moon
aight
lemme try that out rq
thank god wolfram alpha comes to play
landing point at $x = 6$ and $y = 3$
soft zealotBOT
1 divided by 0 equals Infinity
final saddleBOT
final saddleBOT
Available help channel!
final saddleBOT
tired walrus
.open
this is not a command
in fact, starting your message with . will prevent channel opening
but you sent an image so that opened the channel anyway
so,
!status
final saddleBOT
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
I'm at step 1
tired walrus
do you know how to take derivatives
pine night
Yeah
tired walrus
then roadblock is?
pine night
I just can't generate the expression. I can't make sense of it.
pine night
The one given in the question
tired walrus
congratulations that's like the least precise answer possible
in fact it's pessimally uninformative 
i have no idea wtf you mean
but all you need is to find dy/dx and find d^2y/dx^2 and plug in. simple!
pine night
but all you need is to find dy/dx and find d^2y/dx^2 and plug in. simple!
Yeah
I did that
tired walrus
ok so roadblock is???
pine night
ok so **roadblock is???**
I can't get the answer. I differentiated it twice, and it looks nowhere near that stuff
tired walrus
tired walrus
I can't get the answer. I differentiated it twice, and it looks nowhere near tha...
"it" is?
"that stuff" is?
pine night
"it" is?
"that stuff" is?
The expression in the question
tired walrus
for someone who is literally nicknamed the "CEO of pedantry" you sure are forcing me to be a pedant myself.
ok last attempt: show your work up to now.
pine night
for someone who is literally nicknamed the "CEO of pedantry" you sure are forcin...
Quite the underhanded compliment. Should I be honored? Or beg your pardon
tired walrus
it wasn't a compliment at all.
pine night
Fair enough
tired walrus
show your work please.
tired walrus
ok, looks like your dy/dx is correct and you have found that dy/dx = ay + b e^(ax) cos(bx)
tired walrus
and then you did that again and found
d^2y/dx^2 = a dy/dx + ab e^(ax) cos(bx) - b^2 y
from which, if we're going to continue with this idea, i would put dy/dx = ay + b e^(ax) cos(bx) back in there
and you would get
d^2y/dx^2 = a^2 y + 2ab e^(ax) cos(bx) - b^2 y
assuming there is no algebraic mistake that escaped me
tired walrus
did you mean this truthfully or were you just being polite?
pine night
I meant it, trust me!
The exclamation mark should be proof enough for my gratefulness
tired walrus
ok
pine night
.close
waxen bear
Where did the 1023 come from?
final saddleBOT
Where did the 1023 come from?
tired walrus
simplify 2^5 - 2^(-5) to get 1023/32
worldly spruce
Any ideas
tired walrus
but oh well
worldly spruce
๐
waxen bear
simplify 2^5 - 2^(-5) to get 1023/32
how i see it become is
32/(32ln2)
tired walrus
ok first off
let's factor out that 1/ln(2)
so you get $\frac{1}{\ln(2)} (2^5 - 2^{-5})$
soft zealotBOT
Ann
tired walrus
do you agree or disagree thus far?
waxen bear
agree
tired walrus
ok
now simplify 2^5 - 2^(-5) for me please.
do it on paper.
show me how you get 32/32 [a.k.a. 1] and i will tell you where you are screwing up.
final saddleBOT
final saddleBOT
Available help channel!
gritty mesa
Task: Compute the limit of this series as x approaches 1. No differentiation or L'Hopital allowed.
Intuitively, I'd say the limit is n/m. But I cannot find a way to prove nor disprove it. Perhaps complete induction?
final saddleBOT
Task: Compute the limit of this series as x approaches 1. No differentiation or ...
blissful meadow
Is L'Hopital's Rule out of the picture?
gritty mesa
I tried to but failed
tired walrus
you tried to prove $\lim_{x \to 1} \frac{x^n-1}{x-1} = n$ but could not do it?
soft zealotBOT
Ann
gritty mesa
correct
tired walrus
is induction permitted?
gritty mesa
not prohibited at least
tired walrus
is the geometric series sum formula permitted
final saddleBOT
final saddleBOT
Available help channel!
gritty mesa
Task: Find the minimum value of this set. In Latex, you can see my argumentation. I'm curious if there exists an easier way - or perhaps more correct?
final saddleBOT
Task: Find the minimum value of this set. In Latex, you can see my argumentation...
gritty mesa
$B(3)=-13/10$.
Assume $B\ge -13/10$ for all $k\in\mathbb{Z}$:
[
\frac{8k+11}{k^2+1}\ge -\frac{13}{10}.
]
Find all zeros of $13k^2+80k+123$:
[
x_1=-\frac{41}{13},\quad x_2=-3.
]
Since the interval $\left(-\frac{41}{13},-3\right)\cap\mathbb{Z}=\emptyset$
the inequality is true and therefore the minimum is $-13/10$.
soft zealotBOT
qrpd
gritty mesa
Not allowed to use differentiation
tired walrus
Task: Find the minimum value of this set. In Latex, you can see my argumentation...
do you mean minimum or infimum
also your notation's kind of dodgy
final saddleBOT
@gritty mesa Has your question been resolved?
gritty mesa
do you mean minimum or infimum
well, if the infimum is part of the set of all values, then its a minimum right? So the task is to find the minimum (and infimum) if they exist.
gritty mesa
also your notation's kind of dodgy
yeah, sorry, I didnt want to write too much on Discord. My notation is much longer, but you probably dont care about how to compute the zeros, i thought
tired walrus
no, i mean the fact that you use B to mean first a function and then an element of set B and all the while it is actually a set and neither of these things
but like... mm. i would have liked to see how you get to the quadratic in the first place
final saddleBOT
final saddleBOT
Available help channel!
quartz flame
so for this one
final saddleBOT
so for this one
quartz flame
could i do it where its f(x,g(x,z),z ?
since you can project D onto the Z plane?
like this
well i did it wrong
lol
formal trail
from the perspective of the yz-plane the cylinder is there twice (once above once below) which would necessitate splitting into two integrals
formal trail
well it would double the work required to get the answer
formal trail
could i do it where its f(x,g(x,z),z ?
this approach would be for projecting onto the xz-plane which might be less work
formal trail
the homework wants me to use divergence theorem
you can do that as well
quartz flame
so complicated ๐ฉ
quartz flame
this approach would be for projecting onto the xz-plane which might be less work
oh thats what i meant to do
i wasnt thinking when i drew it
evidently
final saddleBOT
@quartz flame Has your question been resolved?
gentle zephyr
final saddleBOT
robust mulch
are you even trying the problems before sending them
gentle zephyr
are you even trying the problems before sending them
are you trying to help with sending this comment?
robust mulch
yes, show your work or say youre stuck
gentle zephyr
what if I don't know how to begin
robust mulch
say that then
gentle zephyr
yes, show your work or say youre stuck
so how do we start dis
gentle zephyr
how to find critical points in the boundary
robust mulch
you pick a restriction and set it
e.g. the first part of the restriction is 0<=x so set 0=x
gentle zephyr
let me see if I understand you are telling me to parametrize this function to make it a single variable optimization problem 3 or 4 times, like find 3 or 4 different parametrisations?
robust mulch
sure you can call it parametrization
gentle zephyr
if its not a parametrization then what is it then
robust mulch
idk its kinda just a substitution
gentle zephyr
either way, we can do this, but all my classmates are using lagrange multipliers, should be pretty similar, I think(?
main mirage
either way, we can do this, but all my classmates are using lagrange multipliers...
Right so, have you figured out what your function "g" should be? (I.e what the constraint function is?)
gentle zephyr
Right so, have you figured out what your function "g" should be? (I.e what the c...
our constraints are inequalities
so we need to, work with the interior of the region, the pairwise intersections of the boundaries of the region and the boundaries of the region themselves
main mirage
Right. The Lagrange multiplier method is specifically detailed to optimising the function on the specific boundary, right?
gentle zephyr
or use something like karush kuhn tucker
gentle zephyr
Right. The Lagrange multiplier method is specifically detailed to optimising the...
yes
main mirage
yes
So, if you want to figure out what you need to use for g, you should probably figure out what the 'boundaries' of this region are.
That's what the inequality is there to tell you.
gentle zephyr
how do I do dat
main mirage
how do I do dat
Well for starters, did you draw the region A?
gentle zephyr
I don't need necessarily to draw it to find the absolute extremas
main mirage
But you are trying to find the boundary lines right?
Of course, you don't really need to draw the region to find the boundary. I'm just saying it might be useful
main mirage
Of course, you don't really need to draw the region to find the boundary. I'm ju...
It's similar to what the other guy was saying: iirc Lagrange Multipliers is used particularly when the boundary aren't really convenient to parameterise.
main mirage
Here it kind of is, because you have pretty simple boundaries (i.e you got three constant boundaries and one parabola).
gentle zephyr
But you are trying to find the boundary lines right?
well, I am just trying to find the maximums and minimums along the boundary of the region, if that makes any sense
main mirage
Yeah I know
gentle zephyr
It's similar to what the other guy was saying: iirc Lagrange Multipliers is used...
who is this other guy you are referring to
main mirage
I'm saying that there's two methods to do that. You either use Lagrange multipliers to find a function g that represents the boundaries of the region A.
Or, you manually parameterise the function using the boundaries, and treat this question as a single variable optimisation question.
main mirage
sure you can call it parametrization
:)
gentle zephyr
so you want to do 4 different parametrizations for the boundary
main mirage
Yes
gentle zephyr
and use single variable calculus
main mirage
Yep.
gentle zephyr
I'm saying that there's two methods to do that. You either use Lagrange multipli...
I am not sure if I am allowed to do this for the exam, even though if that's the simplest way of doing this, it was never taught in class, but lagrange multipliers was taught, also, in this case the function is easy to parametrize but in case I get other function this method is not guaranteed to work flawlessly is more like a shortcut or trick in order to avoid lagrange
main mirage
Yeah
I mean that's why I poised the two methods here - obviously not every boundary will be simple to sketch. Especially if the region is a 3D space rather than a 2D space which is what we have here.
main mirage
I am not sure if I am allowed to do this for the exam, even though if that's the...
I mean I don't see any reason why you wouldn't be allowed to use the parameterisation method. It usually gets taught hand-in-hand with Lagrange, but oh well.
In any case, if you are going to use Lagrange then you just have to figure out what the constraint function "g" is for each of the boundaries.
Otherwise you can use that KKT method you mentioned (although I'm not sure how you would properly use it here, you can ask someone else if you need further clarification).
gentle zephyr
no lets forget about kkt idk why I mentioned it
main mirage
o
gentle zephyr
I mean I don't see any reason why you wouldn't be allowed to use the parameteris...
Actually I did solved last exam with this parametrization method, luckily the region I got was simple, but results are coming out tomorrow morning so we will see if I got it correct or they don't like it
having said that, still, all of my classmates use lagrange multipliers for this
I just never really understood lagrange multipliers by when the exam was due
main mirage
having said that, still, all of my classmates use lagrange multipliers for this
๐คท
Why use a more complex formula over a nicer strategy
Well it's not really more complex but it's just a little more tedious in calculations
gentle zephyr
is it?
main mirage
I would think so, because with Lagrange, you have to calculate all the partial derivatives of both f and g, then solve a bunch of simultaneous equations.
With the parameterisation, you just plug it in and take the derivative normally.
But eh, that's my own opinion.
gentle zephyr
can we just do both and then compare them?
or if we have to do just 1 if we can do lagrange multipliers method
main mirage
can we just do both and then compare them?
Outside of the exam sure, but during the exam it probably won't be wise
Use just one method, and in this case that would be using Lagrange.
gentle zephyr
0 = x
x = sqrt5
0 = y
y = 5 - x^2
main mirage
0 = x
x = sqrt5
0 = y
y = 5 - x^2
Good! So,
We want to now think of a function g for each of these boundaries such that
g(x, y) = 0.
What would be the function for g for each of these equations? (I.e g1, g2, g3 and g4).
gentle zephyr
We want to now think of a function *g* for each of these boundaries such that
g(...
wdym?
care to explain what you mean
main mirage
care to explain what you mean
This is the method of lagrange multiplier right?
main mirage
0 = x
x = sqrt5
0 = y
y = 5 - x^2
Each of these equations represent a specific constraint, right?
main mirage
0 = x
x = sqrt5
0 = y
y = 5 - x^2
My question is, what is the 'constraint function' for each of these four equations?
So as an example, the constraint for x = 0 is just g(x, y, z) = x, because when you are solving g(x, y, z) = 0, you get x = 0.
gentle zephyr
My question is, what is the 'constraint function' for each of these four equatio...
oh I see
gentle zephyr
So as an example, the constraint for x = 0 is just g(x, y, z) = x, because when ...
well this is the part that is fuzzy
wait
g1 = x
g2 = x - sqrt5
g3 = y
g4 = 5 - x^2 - y
@main mirage
main mirage
g4 = 5 - x^2 - y
Perfect!
So now you just use Lagrange multipliers with all of these, separately.
gentle zephyr
ok
gentle zephyr
So now you just use Lagrange multipliers with all of these, separately.
the lagrangian
we need the lagrangian first
main mirage
I mean
You get the lagrangian by solving the actual lagrange multiplier equations no?
o wait
that's a different thing
Eh? Isn't the lagrangian 0 in this case?
gentle zephyr
no what I mean is
we get the lagrangian first, then differentiate wrt x,y, and lambda
and then equate those to zero
and we get a system of equqtions
gentle zephyr
Eh? Isn't the lagrangian 0 in this case?
the lagrangian for the first constraint is L1 = f + ฮปg1
gentle zephyr
we differentiate L1 wrt x, y, lambda
main mirage
Yeah
gentle zephyr
we equate those expressions to zero we get a system
main mirage
I always just remember Grad(f) = lambda * Grad(g1)
gentle zephyr
I always just remember Grad(f) = lambda * Grad(g1)
wdym?
main mirage
wdym?
That's what you are saying
Like
You find the lagrangian
Then you take the derivative (the partials) of the lagrangian, then set each result equal to 0.
main mirage
This is the method of lagrange multiplier right?
What you are saying is basically this, but you're starting at the very beginning.
gentle zephyr
I always just remember Grad(f) = lambda * Grad(g1)
how is it your way of doing it? maybe I am mistaken
forget everything of what I said
main mirage
Nah you're good
What you are saying is literally what I'm saying
Wait, perhaps if I show it to you visually, that might make more sense
gentle zephyr
beautiful hand writing
gentle zephyr
This is the method of lagrange multiplier right?
but the picture says other stuff
main mirage
but the picture says other stuff
Oh yeah don't worry about that
When I was taught Lagrange, we were taught to set k = 0 always
So your constraint function will always be in the form g(x, y) = 0
But it doesn't really matter whether it's 0 or k (because it will disappear when you take the partial derivative)
Also the actual value of lambda is irrelevant
Like
It could be positive or negative
So the sign kinda gets absorbed
I can redo my working out with that into consideration
gentle zephyr
When I was taught Lagrange, we were taught to set k = 0 always
oh interesting
there is a problem
gentle zephyr
0 = x
x = sqrt5
0 = y
y = 5 - x^2
we are not considering the pairwise intersections
those are also candidates for extremas
main mirage
True. Idk how you fix this with Lagrange, I'd personally just substitute in the points of intersection and take that as an 'extra point' to consider.
gentle zephyr
True. Idk how you fix this with Lagrange, I'd personally just substitute in the ...
yes
main mirage
So for example, the points of intersection should be (0, 0), (sqrt(5), 0) and (0, 5).
Whoops, the region should have three pairwise intersections.
gentle zephyr
True. Idk how you fix this with Lagrange, I'd personally just substitute in the ...
are you from america?
main mirage
Nope, I'm an Australian student :)
gentle zephyr
So for example, the points of intersection should be (0, 0), (sqrt(5), 0) and (0...
how to find this
main mirage
how to find this
Drawing the region is sort of the best strategy
That way you can figure out how they intersect
main mirage
is not necessary
๐ญ fine
gentle zephyr
๐ญ fine
let's do the drawing method
and then check with algebra
u here? @main mirage
main mirage
Yep I'm alive
main mirage
I need some handhold with the drawing
ahhh that's okay
main mirage
0 = x
x = sqrt5
0 = y
y = 5 - x^2
0 = x
x = sqrt5
0 = y
y = 5 - x^2
Remember, these are the constraints that we have for this function.
On a xy plane, do you know how to sketch these functions?
Like for example, how would you sketch x = 0, x = sqrt(5)?
gentle zephyr
Like for example, how would you sketch x = 0, x = sqrt(5)?
is hard
main mirage
hmmmm
gentle zephyr
main mirage
What kind of graph is x = 0? Is it a line, is it a parabola..?
main mirage
Good! So those are where the points are.
What sort of graph do these equations form though ..?
gentle zephyr
What sort of graph do these equations form though ..?
idk a line segment
gentle zephyr
What kind of graph is x = 0? Is it a line, is it a parabola..?
a line
is the y axis
main mirage
a line
Good! That's much more accurate :)
The graph x = 0 is a line, in fact - it's the y axis :)
How about the graph x = sqrt(5)? What sort of graph does that make?
gentle zephyr
How about the graph x = sqrt(5)? What sort of graph does that make?
idk
a line that passes through sqrt5
and has direction of y axis
gentle zephyr
help
gentle zephyr
idk how to draw it
gentle zephyr
Literally a straight line
OK
main mirage
||Straight line :)||
gentle zephyr
||Straight line :)||
OK
main mirage
but how to put all of this together
Oh you gotta like
Literally just draw them all together
On the same plane
gentle zephyr
||Straight line :)||
this?
main mirage
O I mean
That's only x = 0, x = sqrt(5)
You now need to draw y = 0 and y = 5 - x^2 :)
gentle zephyr
You now need to draw y = 0 and y = 5 - x^2 :)
last one is hard
help
it's a parabola and shit
in the xy plane
gentle zephyr
0
main mirage
Well a vertex is a point right?
main mirage
the vertex is (-b/2a, f(-b/2a))?
yep, which is?
main mirage
wait a second... why is there a backslash
gentle zephyr
wait a second... why is there a backslash
its such that dude
main mirage
Oh
Where I come from, a backslash means 'exclude'
Okay good thing it doesn't mean that
gentle zephyr
Where I come from, a backslash means 'exclude'
inside a set builder aswell?
main mirage
Yeah
main mirage
What's b here?