well it seems you're off by a factor of two
#help-0
buoyant kayak
also the triangle on the right doesn't make sense
hypotenuse is the longest side
what info are you directly given
sand dagger
oh actually nvm
i kinda did an error when counting it
so its all good now
.close
covert agate
suppose we have a paremetric curve defined as \
$\begin{cases}
x = 3 \sin t \
y = 2 \cos t
\end{cases}$
ocean sealBOT
Chromium
covert agate
and we'd like to eliminate the paramter
why would $t = \arcsin \frac{x}{3}$ followed by $y = 2 \cos (\arcsin \frac{x}{3})$ be incorrect?
little drum
screams
covert agate
smaercs
little drum
quietly leaves
covert agate
cries
little drum
hopes he's realized what he's written
ocean sealBOT
Chromium
little drum
lol
covert agate
**Chromium**
apparently this is wrong?
i know we have to like use $\sin^2 + \cos^2 = 1$ but why won't this method work
covert agate
we end up with uh
little drum
well ahaha
covert agate
not an ellipse
little drum
to start with
this thing right here
doesn't have a
umm
bound on "t"
but the moment you say
t = arcsin(whatever)
it's kinda.. umm yk
won't give you the entire thing
covert agate
$\sin t = x \not\implies t = \arcsin x$ 
ocean sealBOT
Chromium
covert agate
how do you write not implies ;-;
ocean sealBOT
little drum
lol
covert agate
i like to eat milk
little drum
$\nrightarrow$
ocean sealBOT
little drum
$\centernot \implies$
covert agate
end times
ocean sealBOT
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
covert agate
$\not \implies$
ocean sealBOT
Chromium
little drum
yeah let's make do with that
little drum
**Chromium**
and yes that's true lol
little drum
exactly the same thing you enjoyed discussing with Shuri
covert agate
we are young life is fun
little drum
.-.
covert agate
(new question!)
little drum
10 more yrs and you're already too old for shi
covert agate
.close
vale wigeon
no
that's not what symmetric means, and also math is case sensitive
do not UPPERCASE letters that should be lowercase
anyway, one of your conditions does not belong
and also the entries of the matrix are lowercase letters and it is wrong to uppercase them especially a
symmetry means that A^T = A, and the only non-tautological equality you get from that is c = b
math is case sensitive!!!!!!!!!!!!!!
a and A aren't the same thing!
you KNOW? then why don't you DO it?
this is a written conversation. you have no excuse not to respect letter case for mathematical notation
that's all i ask for
now
$A = \bmqty{a&b \ c&d}, A^T = \bmqty{a&c\b&d}$
ocean sealBOT
Ann
vale wigeon
do you understand this?
you have never been introduced to the concept of transpose of a matrix?
what the actual fuck
i mean ok like
your misuse of notation aside
the discriminant of λ^2 - (a+d)λ + ad-bc
is (a+d)^2 - 4(ad-bc)
are you saying you are unable to simplify (a+d)^2 - 4(ad-bc) into (a-d)^2 + 4bc?
answer my question: are you or are you not able to do the simplification i said?
....
okay so
first off
let's make sure you understand that you are NOT asked to show (a-d)^2 + 4bc >= 0
you are asked to show that if (a-d)^2 + 4bc >= 0 then the eigenvalues of A are real
do you understand that this is your goal?
@alpine sable
lone heartBOT
@alpine sable Has your question been resolved?
vale wigeon
you have this equation here
its roots are the eigenvalues of your matrix
do you understand this, yes or no
the discriminant of λ^2 - (a+d)λ + ad-bc
is (a+d)^2 - 4(ad-bc)
do you understand this, yes or no
no
do not use the letters a, b and c for something other than what they're used for in the problem
you shot yourself in the foot by saying anything more than a "yes"
you could have and should have simply said yes
and left it at that
so
a quadratic equation has real solutions if and only if its discriminant is nonnegative
do you understand this, yes or no
so your matrix has real eigenvalues if and only if (a+d)^2 - 4(ad-bc) >= 0
do you understand this, yes or no
(a+d)^2 - 4(ad-bc) can be simplified into (a-d)^2 + 4bc
do you understand how this happens, yes or no
do you want me to take you through the simplification step by step or just tell it to you all at once?
(a+d)^2 - 4(ad-bc) = a^2 + d^2 + 2ad - 4ad + 4bc = a^2 + d^2 - 2ad + 4bc
so we have now shown that your matrix has real eigenvalues if and only if (a-d)^2 + 4bc >= 0, as required
vale wigeon
(a+d)^2 - 4(ad-bc) = a^2 + d^2 + 2ad - 4ad + 4bc = a^2 + d^2 - 2ad + 4bc
i literally just demonstrated to you here that (a+d)^2-4(ad-bc)^2>=0 and (a-d)^2+4bc>=0 are the same inequality
do you think there is something left unexplained still?
have you been reading what i have been saying
vale wigeon
so your matrix has real eigenvalues if and only if (a+d)^2 - 4(ad-bc) >= 0
.
i asked whether you understood this and you said yes
.......
vale wigeon
you are asked to show that **if** (a-d)^2 + 4bc >= 0 **then** the eigenvalues of...
.
you are asked to show that
IF (a-d)^2 + 4bc >= 0
THEN the eigenvalues of A are real
why is this only coming up now
have you never proved any if-then statements before?
little drum
Given (a - d)² + 4bc ≥ 0, you need to show discriminant of this quadratic: (a + d)² - 4(ad - bc) ≥ 0 holds true [because only then is it certain that the eigenvalues are real]
So your problem reduces to: Given (a - d)² + 4bc ≥ 0, show that (a + d)² - 4(ad - bc) ≥ 0.
Are you unable to do that?
ait, do it first then we'll come back to what the question be asking
"why the eigenvalues of A (a - d)² + 4bc ≥ 0" doesn't make any sense
can you rephrase it better?
vale wigeon
well it sounds like they did not teach you what a transpose is or what a symmetric matrix is...
so you cannot be expected to be able to do this problem without the requisite knowledge
no, i'm specifically expressing frustration not at you
but at whoever teaches you less than what you need to do the exam
little drum
A symmetric matrix A has $A^t = A$
ocean sealBOT
little drum
.wdym you get it (@_@;) were you taught this in class?
.that $A^t = A$ for symmetric matrices?
ocean sealBOT
lone heartBOT
alpine sable
Holy
little drum
sad :o make sure you talk to the teacher about this ^^"
robust dagger
i have a simple equation 100/(10-0.05x)^2, i looked up on mathway the deriviative, it game me 80000/(-x+200)^3, whilst i got 4000/(-x+200)^3, im not sure what i got wrong, i started by * my original equation by 20 to make x an integer, which gave me 2000/(200-x)^2, i then brought the bottom part up as 2000(200-x)^-2, i then did the chain rule and it came out as -4000(200-x)^-3 * -1, which then became 4000(200-x)^-3 and finally 4000/(200x)^3, what did i do wrong?
vale wigeon
started by * my original equation by 20 to make x an integer, which gave me 2000/(200-x)^2
there are several things wrong with this statement
first off, there is a difference between x itself and the number next to x, i.e. its coefficient.
it's the latter that you want to make an integer
second, you're not multiplying your equation by 20, you're multiplying the numerator and denominator by 20 each. that way, the value of your expression remains unchanged as it's supposed to.
third, $20 \cdot (10 - 0.05x)^2 \neq (200 - x)^2$.
ocean sealBOT
Ann
robust dagger
oh yeah your right
vale wigeon
you want to multiply the num and denom by 20**^2**, not just 20
robust dagger
so basically my mistake is just the integerisation
ok i will fix this and see what i come back with
so basically because the 0.05x is affected by a power, i have to affect everything else by that power too
vale wigeon
you're overthinking it.
robust dagger
no i was just confirming it to make sure i have the idea right
ok so i simplified it as 40000/(40000-400x-x^2), this is before actually deriving
robust dagger
i think ive made a mistake tho because the end result has a ^-3
vale wigeon
(200-x)^2 does not expand to 40000 - 400x - x^2
and there is also no need to actually expand it
so you've done something unnecessary and fucked up while doing it
robust dagger
(200-x)^2?
vale wigeon
......
robust dagger
i had (10-0.05x)^2 on the bottom, i * it by 20^2 which is 400, so 400(10-0.05x)^2
ah wait
ait
ye
and then i expanded the brackets
and then x by 400
vale wigeon
you did not need to expand anything
robust dagger
i agree
vale wigeon
you did not even need to write 20^2 as 400
robust dagger
but where does 200-x come from
vale wigeon
$20^2 \cdot (10-0.05x)^2 = [20 \cdot (10 - 0.05x)]^2$...
ocean sealBOT
Ann
robust dagger
ok so you just put a second set of brackets
vale wigeon
overthinking it again
robust dagger
but anyway i see where (200-x)^2 is from now
vale wigeon
overfocusing on the precise symbols i write instead of the underlying idea
and now you're omitting parentheses where you shouldn't
that's the other extreme
there we go
robust dagger
well thank you for your help
i derived and got the right answer
i think thats all, but just to confirm, its only when the two things are to the same power that you can multiply them like that i assume
because in the end the power has the same effect
anyway
thank you ann im going to close this now
very appreciative
.close
rough acorn
how do you mathematically show that the vectors are parallel? can you do cross product but just assume j for both vectors are 0?
rough acorn
finite flax
if two vectors are ||, cross is 0
idk how to set that up in LaTeX
|i j k|
|6 0 1|
|6 0 -1|
so you can do what i was saying?
finite flax
oh idk, I didn't read what you were saying 😛
I'm like the worst one here
but I still get there
yes
lone heartBOT
lone heartBOT
Available help channel!
cold cedar
How do I draw a function based on this? Can anyone refer me to a video/source, where I can learn about it?
harsh swallow
Well this seems like you have to follow the yellow lines in the direction that theyre pointing while going through point P
cold cedar
Well this seems like you have to follow the yellow lines in the direction that t...
Yup
harsh swallow
In other words, if i start at P and go a bit to the right, what direction does the yellow lines point to
cold cedar
Then you can find the equation too
I just determine three points, and find the equation, right?
Lemme try to see if I can do so
harsh swallow
Although
I think im wrong
The reason i think that im wrong about the parabola bit is because it seems like the line y = x is a asymptote to the function
harsh swallow
I think it might actually be a hyperbola, those can have diagonal asymptotes
cold cedar
How do I draw a function based on this? Can anyone refer me to a video/source, w...
How would you draw it then?
Now I'm confused xD
harsh swallow
Sorry!
I think if youre only given that first picture
Then you only really have to draw the correct function
cold cedar
I'm only given the picture, and told that it crosses the point
Lemme double-check
I'll try to ask my teacher, might just be confused at the task itself
Appreciate the help nonetheless @harsh swallow
.close
fallow pagoda
how would i find all the zeros of
P(x) = x^3 − 4x^2 + x + 6
lone heartBOT
how would i find all the zeros of
P(x) = x^3 − 4x^2 + x + 6
fallow pagoda
is it just subbing random numbers into it and hoping i get zero?
hardy moth
Looks like a polynomial division problem to me, so i would try plugging in some integers and then do polynomial division
fallow pagoda
okay thanks
lone heartBOT
Available help channel!
weak prairie
Let $p>0$ be a prime number. Let $q$ be a natural number such that $\phi(p^2) = \phi(q^2).$ Show $p=q$ where $\phi$ denotes the Euler Totient function
ocean sealBOT
duckduckdo
weak prairie
I know that a proof by contradiction is the way to go, i.e. start with $p\not| q^2$
ocean sealBOT
duckduckdo
weak prairie
we know $\phi(p^2) = p(p-1) = \phi(q^2)$
ocean sealBOT
duckduckdo
weak prairie
by def of $\phi$ function
ocean sealBOT
duckduckdo
weak prairie
so $\phi(q^2)$ in its prime fac $=\phi({p_{1}}^{k_1})\phi({p_{2}}^{k_2})....\phi({p_{r}}^{k_r})$
ocean sealBOT
duckduckdo
oak perch
f(x)=x(x-1) is monotonic when x is greater than 1
weak prairie
by def of $\phi$ being multiplicative with co prime numbers
ocean sealBOT
duckduckdo
oak perch
Oh sorry didn’t notice that q is just a natural number…
weak prairie
f(x)=x(x-1) is monotonic when x is greater than 1
bro do u know what the $\phi$ funciton is
ocean sealBOT
duckduckdo
weak prairie
Oh sorry didn’t notice that q is just a natural number…
im sorry but you're way off
<@&286206848099549185>
oak perch
let q=Πp_j^r_j then p(p-1)=φ(p^2)=φ(q^2)=Π(p_j-1)p_j^(2r_j -1)
If p doesn’t equal any p_j then p divides some p_j -1
Then p<p_j
p(p-1) is smaller than (p_j-1)p_j^(2r_j-1)
Which is impossible
So p=p_j for some j
Then p(p-1)<=(p_j-1)p_j^(2r_j-1)
weak prairie
**duckduckdo**
your notation is it $p_j -1 or p_{j-1}$?
ocean sealBOT
duckduckdo
oak perch
Other terms are 1 so r_j=1 in order to make that inequation an. Equation
p_j -1
I will use p_(j-1) otherwise
weak prairie
so you expand $\phi(q^2)$ into its prime fac right first
oak perch
Therefore q=p_j=p
ocean sealBOT
duckduckdo
weak prairie
yeah i know youve applied def of phi function yeah using $\phi(p^n) = p^{n}-p^{n-1}$ ?
ocean sealBOT
duckduckdo
weak prairie
on prime fac of $q^2$
ocean sealBOT
duckduckdo
weak prairie
so i understand $p(p-1) <= (p_j -1) {p_j}^{2r_{j}-1}$
ocean sealBOT
duckduckdo
weak prairie
but where from there
oak perch
You need to make that inequation an equation
So there must be just one term, one j
And r_j need to be 1
weak prairie
ahh okay then that implies p is a factor
oak perch
If p doesn’t equal any p_j then p divides some p_j -1
From here
That’s why p has to be equal to one of the p_j
little drum
huh :o
weak prairie
if p is equal to one p_j that means p is a prime fac of q^2 ?
little drum
Given $$\phi(p^2) = \phi(q^2)$$ Let $q = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$, then the given equation is equivalent to $$p(p-1) = \prod_{i=1}^{r} p_i^{2k_i - 1} (p_i - 1)$$
ocean sealBOT
oak perch
Π(p_j-1)p_j^(2r_j-1)=p(p-1)<=(p_j-1)p_j^(2r_j-1)
You want to make this an equation
So there has to be just one j and r_j=1
weak prairie
yeah so $p(p-1) = (p_{1} - 1)p_{1}$
ocean sealBOT
duckduckdo
weak prairie
right ?
little drum
excuse me? how'd you conclude i = 1?
weak prairie
p is prime
little drum
so? for some prime $p_i$, $p_i , | , p(p - 1) \implies p_i , | , (p - 1)$ or $p = p_i$
ocean sealBOT
little drum
how'd you reject p_i divides (p - 1)?
oak perch
Yes
oak perch
yeah so $p(p-1) = (p_{1} - 1)p_{1}$
I mean, yes
weak prairie
how'd you reject p_i divides (p - 1)?
because we say p_{j} = p
oak perch
_excuse me? how'd you conclude i = 1?_
You are already smaller than or equal to one term on your right hand side
weak prairie
for some
oak perch
So you can’t have another terms greater than 1 on your right hand side
oak perch
Yeah
weak prairie
in the uk ?
proof is finished after that right ? as youve shown p is the only factor of q^2 ?
@oak perch
oak perch
No Japan right now
First year graduate student
Anyway I better not occupy the help channel
weak prairie
is that the end ?
of the proof yeah
ah damn we never did any number theory first year
ありがとうございます
あなたは数学が得意ですか。
あなたは数学が得意です
no か
ohh grad student
nicee
how i end channel
.close
main cedar
[Type Theory] What did Bertrand Russell do when developing type theory to avoid the Russell's Paradox he discovered in set theory? (I know little to nothing about formal logic, but I have to do a research paper on this guy)
valid lake
you can google it for the detail
main cedar
I did.
valid lake
ok then.
lone heartBOT
@main cedar Has your question been resolved?
lone heartBOT
@main cedar Has your question been resolved?
main cedar
.close
jade star
lone heartBOT
oak perch
Do you know how to calculate Σj^k j from 1 to n?
oak perch
Two ways to calculate, first you can use (t+1)^(k+1)-t^(k+1)= ΣC(k+1,p)t^p then take sum in terms of t from 1 to n you have that (n+1)^k-1=ΣC(k+1,p)S_p when you can find S_k inductively if you have all smaller S_p
Another method is
Given a sequence { f(n): n>=0} you define Ef(n)=f(n+1) Df(n)=f(n+1)-f(n) If(n)=f(n)
So three maps D E I mapping sequences to sequences satisfying D=E-I
Then Σf(k)=ΣE^kf(0)=Σ(D+I)^kf(0)
jade star
I think i got it
oak perch
=ΣC(n+1,q+1)D^qf(0)
jade star
is it possible if u can confirm my answer?
jade star
for q.7 i got 4n
for q.9 i got B
oak perch
strange I got 4/3 for the first question I need to check it what’s wrong…
jade star
strange I got 4/3 for the first question I need to check it what’s wrong…
strange
oak perch
And the second… I think they all do😂
The first is the limit of n(n+1)(2n+1)/6n^3 + n(n+1)/n^2 that’s why I got 1/3+1=4/3
spice cedar
How can I find y? How can I move 2 in other side? this is the final part for minimum average total cost
noble sinew
y? Aren’t those a’s or what?
spice cedar
I have to find y =
noble sinew
y isn’t in what you posted
slender gull
Can you rewrite what you've written?
noble sinew
Suggest you post a full question
spice cedar
I need to find minimum average total costs. I have C(Y)=(2/25)Y^2+10Y+2.
ACT= (2/25)y^2+10y+2 / y = (2/25)y+10+2/y
MC (derivate)= 4/25y+10
ACT=MC
2/25y+10+2/y = 4/25+10
I remove 10 from both side. I remove 2/25y in left side e transform 4/25 in 2/25 in right side.
Final I have 2/y=2/25y
spice cedar
Wait I will do a photo
noble sinew
Oh you forgot a y, but there is no solution to ACT=MC. Are you sure you should equate this?
Finding minimum is taking derivative and setting derivative equal to 0
Why are you setting them equal?
noble sinew
What
spice cedar
Can you show me how can i resolve this with derivate equal 0?
noble sinew
The 1st thing you said isn’t a thing
You have derivative
Set it equal to 0
Solve for y
spice cedar
How?
noble sinew
?
noble sinew
You have derivative
.
noble sinew
Set it equal to 0
.
noble sinew
Solve for y
.
noble sinew
Yes that is what I said
noble sinew
Then you wrote the question down wrong
Suggest you take a picture of the question as its written in your book
spice cedar
If a firm's total cost curve is given by C (Y) = (2/25) Y ^ 2 + 10Y + 2, the average total cost is minimal for Y equal to:
noble sinew
What is Y?
Because what it sounds like is Y is units and you are asked for the number Y such that C(Y)/Y is minimal
That is different than what you have been doing (and saying)
noble sinew
You can’t just write it how you want and work on that. Where does 1st line come from? That is not what you have
(And why do you write ‘y’ as ‘a’?)
noble sinew
Sure if that is what the question asks
spice cedar
is mathematically correct?
noble sinew
Yes
sudden sphinx
So I've the following function and Fourier series, and I want to calculate the value for b1
sudden sphinx
I know how to calculate Bn generally, with the formula
but I´m getting the wrong answer, So there must be something wrong with my calculations
lone heartBOT
@sudden sphinx Has your question been resolved?
sudden sphinx
<@&286206848099549185>
tacit arch
but I´m getting the wrong answer, So there must be something wrong with my calcu...
Show your work
tacit arch
The antiderivative of sin is not cos
sudden sphinx
-cos
i know, corrected it after the pic
Been looking at this for 1 hour, my brain is dead, sorry
it aint even hard, did a lot of these yesterday
dunno why im stuck today
tacit arch
Share your updated work
sudden sphinx
sure, one sec
Here
@tacit arch
Wow, I get my mistake
cos(pi/2) is zero lol
fuk me
solved
raven rover
.close if appropriate 🙂
stone dove
hello, i need to find the value of this complex number
lone heartBOT
hello, i need to find the value of this complex number
stone dove
tacit arch
hello, i need to find the value of this complex number
where is $n$ ?
stone dove
in the form of x+yi
ocean sealBOT
riemann
pale kestrel
Show the full question
stone dove
that's all
pale kestrel
what on earth is that n = 5 lol
tacit arch
well we're not your instructors who wrote the problem, so go back to your instructor and ask
noble silo
Forgot how to do this one. You need to do a system of equations but forgot everything else. Thanks
candid torrent
Forgot how to do this one. You need to do a system of equations but forgot every...
do you know what a linear combination is?
raven rover
,rotate
ocean sealBOT
candid torrent
basically you have two vectors (-3, 1) and (-1, 2)
raven rover
Basically just scale u and v by some real numbers so their sum yields (-3, -4)
candid torrent
you want two constants that ield (-3, -4)
@raven rover let me help
i got here first
😠
so set up an equation for this
noble silo
Linear comb is just i and j
candid torrent
what
noble silo
You asked if I knew what a linear comb is
noble silo
i is the component in the x and j is the component in the y
candid torrent
ok yeah but you don't have to do that in this case
let x = (-3, 1) and y = (-1, 2)
let a, b \in R
so a and b are real numbers
ax + by = (-3, 4)
this is the equation
do you understand
noble silo
Yes. So what next?
candid torrent
you tell me
what do you want to do before you do anything elese
(convert x and y in that euqation back into vectors)
noble silo
Ok.
noble silo
a(-3,1)+b(-1,2)= (-3,4)
noble silo
We need another equation
candid torrent
no you don't
expand a and b into the vectors
trust me, you'll see you don't need another equation
noble silo
Into I and j?
noble silo
Never mind. Got it
candid torrent
I and J have nothing to do with this
noble silo
(-3a-b,a+2b)
noble silo
Factor out a and b?
noble silo
Yes
candid torrent
now you can solve this two variable two equation system
👍
do you need help with that or are you alright?
noble silo
I got b as 9/5
noble silo
2/5
dusty grove
Agua best teacheeeeeeer
noble silo
Then distribute and then it’s finished?
candid torrent
Then distribute and then it’s finished?
no don't distribute
you're already done
you found a and b
noble silo
Ok. But that’s not the final answer
candid torrent
@noble silo wdym
noble silo
It’s 2u-3v
candid torrent
what is u and v?
noble silo
The original vectors
candid torrent
do you mean (1,0) and (0, 1)
noble silo
u=(-3,1) v=(-1,2)
candid torrent
uhhhh
actually wait
no they're right
there are multiple answers to these questions
your answer should also be correct
check on a calculator
remote heron
remote heron
as a and b values
i get a=2, b=-3
noble silo
We didn’t learn that
candid torrent
but is our answer still a correct linear combination?
@remote heron no
we made 2 variable system
and solved
remote heron
is this for algebra?
noble silo
Precalc
remote heron
so
have you ever tried elimination techniques?
sometimes you see those in algebra 2
like adding equations
im using fancy notation but this is just fancy elimination
you want to look at this
notice if you multiply the top equation by 2, and add it to the bottom equation, a variable cancels
dusty grove
we got 9/5, 2/5
I think you copied the text of the exercise wrongly
Because I did too and I got your same result
remote heron
this is sufficient to solve the system
since you can then just solve for a
and plug into the other eqn
or either eqn really
noble silo
@dusty grove I just took a pic
remote heron
@noble silo does this technique make sense
or you could do like substitution instead, thats usually the first technique in algebra i think
but its slower
bright remnant
lone heartBOT
bright remnant
So this is a question I really do not understand how to do, what I have done so far is find r=3. From there, I did 1.2^3 to get 1.728. After, I divided 200,000 by 1.728. Then, I arranged everything in terms of factorials and multiplied by 3! (Which is 6) to get 694444.44, leaving me with n(n-1)(n-2) = 694444.44 but I then get as my final formula as n^3-3n^2+2n-694444.44.... I'm graphing it out and then I get 89.5 as my x intercept however the answer is around or is exactly 27, where did I go wrong?
Please dm me immediately if you have a proof because I am about to go to bed
little drum
So this is a question I really do not understand how to do, what I have done so ...
uhh
bright remnant
Yes
little drum
the 3rd term would perhaps be $$\binom{n}{3} (x^2)^3 (1.2)^{n-3}$$
ocean sealBOT
little drum
so, the coefficient would be $\binom{n}{3} (1.2)^{n-3}$
ocean sealBOT
bright remnant
No it would not
bright remnant
The format is
(nCr) (a)^n-r * (b)^r
little drum
Yes.
vale wigeon
the 3rd term if you index from zero
which you should unless you like impaling yourself on fenceposts
the \textbf{term} is $\binom{n}{3} (1.2)^{n-3} (x^2)^3$
ocean sealBOT
Ann
vale wigeon
the \textbf{coefficient} is $\binom{n}{3} (1.2)^{n-3}$
ocean sealBOT
Ann
vale wigeon
@bright remnant do you object to this? Y/N
bright remnant
Yes because it doesnt make sense bro
bright remnant
x^2 is in the (a)^n-r position
bright remnant
do not call me bro.
sorry
vale wigeon
...oh, so you're insistent on placing the terms strictly as written
and rejecting the commutative law of addition
bright remnant
ummm
vale wigeon
so writing (x^2 + 1.2)^n as (1.2 + x^2)^n, and then taking the x^6 term from that as the 3rd term, is something you refuse to do?
do i understand you correctly?
i mean you can take the (n-3)rd term if you really insist on rejecting the commutative law of addition, and its coefficient will still be (nC3) * (1.2)^(n-3) because nC(n-3) = nC3 anyway
bright remnant
so writing (x^2 + 1.2)^n as (1.2 + x^2)^n, and then taking the x^6 term from tha...
from what I was taught, how it is written matters because it could determine if the co-efficent is n-r or just r
vale wigeon
you mean just the binomial coefficient?
the nCr thing?
$\binom{n}{r} = \binom{n}{n-r}$ for all $n$ and $r$, you know...
ocean sealBOT
Ann
vale wigeon
or did your teacher FORBID you from EVER using that fact under threat of ZERO ON YOUR ENTIRE ASSIGNMENT
bright remnant
i mean you can take the (n-3)rd term if you really insist on rejecting the commu...
Oh dang I understand that
little drum
$(x^2 + 1.2)^n = \sum_{r=0}^n \binom{n}{r} (x^2)^r (1.2)^{n-r}$ \ $= \sum_{r=0}^n \binom{n}{r} (1.2)^r (x^2)^{n-r} = (1.2 + x^2)^n$
ocean sealBOT
bright remnant
or did your teacher FORBID you from EVER using that fact under threat of ZERO ON...
nah
little drum
choice of "a" and "b" doesn't really matter, you choose accordingly to whichever is more efficient
bright remnant
can we continue?
bright remnant
<@!287899997251436545> do you object to this? Y/N
this does not solve my problem 100% though
what do I do after this
vale wigeon
you need to find the smallest value of n such that that coefficient is greater than 200,000
bright remnant
so I must equate nC3(1.2)^n-3 to 200,000 correct?
how do I solve a question with a number that has n-3 as a power
little drum
$\binom{n}{3} (1.2)^{n-3} > 200000$
ocean sealBOT
bright remnant
yep..
tacit arch
guess and check / binary search i guess
bright remnant
but I would loose marks if I dont show working out, this would be like a 7 mark question
amber urchin
but I would loose marks if I dont show working out, this would be like a 7 mark ...
Are you allowed calculators?
bright remnant
yes
little drum
perfect.
amber urchin
I wouls say changing 200 000 to (1.2)^x and divide both sides by (1.2)^(n-3)
Then at least you move all n's from exponents to one side and keep n in the binominal, but I don't know if it gives any progress
little drum
not quite.
bright remnant
Then at least you move all n's from exponents to one side and keep n in the bino...
it doesnt do anything for me
im just left with nC3 > 1^x-n-3
little drum
me just patiently waiting for Ann
bright remnant
i dont think she wants to speak with me anymore
little drum
little drum
but how do I actually solve for n?
should probably ping Ann and give with appropriate reasoning why and where you're stuck
stoic hound
There's no elementary way to do this, you need a computer
bright remnant
what? 😭
little drum
calculator is allowed and there's a fine solution to this, please ignore
little drum
I know a way out but I really want to know what Ann suggests :c
little drum
well, mine might be less efficient.. which does matter cause efficiency with solutions is utmost priority
altho, the first priority is being able to solve (@_@;)
lone heartBOT
@bright remnant Has your question been resolved?
mellow vapor
I need to find the formula for this on my midterm review did I do impartial fractions correctly ?
oak perch
How about solving the inverse of a matrix
mellow vapor
?
oak perch
You can factor your denominator as polynomial with coefficients in C, I know you will have some complex functions, but they always appear in conjugate pairs
So eventually when you add up everything you will still get real function
mellow vapor
My professor wants me to just set it up in terms of a b c over part of the denominator
He doesn’t want me to actually solve it
I just don’t know if I did it correctly here
oak perch
Then in my picture you take off those integral symbols in those I_k and in b
glass lichen
if you have a quadratic factor in the denominator, you guess a linear term in the numerator
@mellow vapor
mellow vapor
So would it be cx+ d
ocean sealBOT
Mosh
oak perch
So you replace I_k with x^k/Q(x), integral dx/(x-a)^j with 1/(x-a)^j
glass lichen
@oak perch I suggest not overcomplicating solutions if you want to help
glass lichen
Yeah, you just follow the rules depending on the factor(s) in the denominator
mellow vapor
oak perch
I mean it’s a solution,what he’s looking for is linear combination of components of x, x=A^-1b components of b are of the form 1/(x-a)^j so what he’s looking for is linear combination of functions of the form 1/(x-a)^j solving inverse of A
glass lichen
Yep, it's unfortunately that ugly
mellow vapor
oak perch
Surely a can be a complex number but conjugate of a also appears so when he finally add up everything he will have a factorization of real functions
mellow vapor
Hopefully it isn’t this bad on the exam
mellow vapor
I don’t know how to actually solve it tho and the exam is tonight
I know the next step is to multiply both sides by the denominator but I I get lost from there
glass lichen
you'd put everything back onto a common denomiator then expand everything out
then compare coefficients and get a system of equations to solve for the constants
mellow vapor
Compare the coefficients to what
glass lichen
the cubic in the original numerator
mellow vapor
No wonder he said not to solve this
Gawd damn
Thx for teaching me the method to figure out how to write these
Gonna do some more practice ones and memorize formulas
glass lichen
yep
mellow vapor
.close
fervent cloak
lone heartBOT
buoyant kayak
zero product property
buoyant kayak
yeah
fervent cloak
oh okay i get it
i’m dumb i was confused cuz i thought we had to subtract it by 0
thanks
.close
lone heartBOT
lone heartBOT
Available help channel!
vivid mango
Can someone help me with this problem?
lone heartBOT
Can someone help me with this problem?
buoyant kayak
what have you tried
vivid mango
I've tried to write cos2x as cos^2(x) - sin^2(x).
Then we have: t = sin^2(x) - cos^2(x).
=> (2/3)^t + (2/3)^-t = 13/6
Then, (2/3)^t = u
u + 1/u = 13/6
u = 2/3, u = 3/2
=> t = -1, t = 1
1)
sin^2(x) - cos^2(x) = 1
(1-cos^2(x)) - cos^2(x) = 1
1-cos^2(x)-cos^2(x) = 1
-2cos^2(x) = 0
cos^2(x) = 0
x = π/2, x = 3π/2
sin^2(x) - cos^2(x) = -1
(1-cos^2(x)) - cos^2(x) = -1
1-2cos^2(x) = -1
cos^2(x) = 1
x = 0, x = π
That's 4 solutions, but it says that there should be 4+
oak perch
cos2x=1 has 3 solutions and cos2x=-1 has 2
The shape of image of cos2x between 0 and 2π looks like letter W right
3 up 2 down
lone heartBOT
@vivid mango Has your question been resolved?
vivid mango
I see, when I graph this function I can see that it has 3 solutions, and the other has 2. But the solutions are required to be in the interval [0, 2π]
vivid mango
So for the 2), we have x = 0, x = π, x = 2π, right?
tall zealot
hiii, im struggling to integrate x/1+x, ive tried substitution and integration by parts but i cant find the correct answer
buoyant kayak
what sub did you try?
pine kettle
$\int ! \frac{x}{1+x} , \dd x$
ocean sealBOT
sills
tall zealot
x+1=u
pine kettle
did you solve for x in terms of u?
tall zealot
im sorry, how do you mean?
buoyant kayak
if you were to make that sub, you'd get $\int\frac{x}{u}\dd u$
ocean sealBOT
a disappointing son
tall zealot
yea, thats why it didnt work
tall zealot
yea
buoyant kayak
so you can replace that x with something in terms of u
tall zealot
x=u-1?
buoyant kayak
mhm
tall zealot
makes sense, but in that case i should do a substitution again?
buoyant kayak
no need, split the fraction over a common denominator
lone heartBOT
lone heartBOT
Available help channel!
lavish ether
how would i show that the improper integral $\int_{-\pi}^{\pi}x^n\sin(mx)dx=\infty$ for $n\in{-3,-5,-7,\dots}$ (i.e. $n$ is a negative odd integer) and $m\in{1,2,3,4,\dots}$?
ocean sealBOT
Beous
lavish ether
wait maybe i can show that there is an interval $[0,\epsilon]$ such that $\frac{1}{2}mx\leq\sin(mx)$ and then use the fact that $x^n\left(\frac{1}{2}mx\right)\leq x^n\sin(mx)$ and the fact that $\int_{-\pi}^{\pi}x^n\left(\frac{1}{2}mx\right)dx$ diverges to show the integral diverges
ocean sealBOT
Beous
lavish ether
i know that $\int_{-\pi}^{\pi}x^n\sin(mx)dx=2\int_{0}^{\pi}x^n\sin(mx)dx$ since $x^n\sin(mx)$ is even
ocean sealBOT
Beous
lavish ether
but how do i show that there is an interval $[0,\epsilon]$ such that $\frac{1}{2}mx\leq\sin(mx)$?
ocean sealBOT
Beous
pale kestrel
standard limit
bitter cliff
When defining P(A n B), in the independent case is it true that P(A n B) = P(A) * P(B) = P(A|B)P(B) = P(B|A)P(A)?
Or is it strictly just P(A) * P(B)?
I suppose that question becomes does P(A) = P(A|B) and P(B) = P(B|A) when events A and B are independent?
Furthermore, if this is true, does that imply that P(A n B) = P(B n A) only when the events are independent?
lone heartBOT
When defining P(A n B), in the independent case is it true that P(A n B) = P(A) ...
@bitter cliff Has your question been resolved?
bitter cliff
<@&286206848099549185>
tacit arch
read this
https://stats.libretexts.org/Bookshelves/Introductory_Statistics/Book%3A_Introductory_Statistics_(Shafer_and_Zhang)/03%3A_Basic_Concepts_of_Probability/3.03%3A_Conditional_Probability_and_Independent_Events#:~:text=Two events A and B,B)%20of%20their%20individual%20probabilities.
lone heartBOT
@bitter cliff Has your question been resolved?
dry echo
lone heartBOT
dry echo
How can I do this?
We've learned about inverse trig derivatives/integrals today
But I can't manipulate this in any way to use those formulas
dry echo
if I do u = sqrt(7x)
That's a bad idea hold on
Oh
I see
u = (1 + 7x)
NO
u = 7x?
buoyant kayak
you were closest with the first one lol
buoyant kayak
that's why i normally go for a search and replace route instead of explicitly solving for dx
dry echo
OH
buoyant kayak
y'see it?
dry echo
YEAH
jesuhs
I looked for it at first but didn't see it
its cuz I always look for it in the numerator
ok let me continue 🙏
Did I do it
ember mirage
How do i do this?
lone heartBOT
How do i do this?
placid zinc
Strikes me as similar triangles
severe sluice
there's also a general formula for this. see inverse pythagorean theorem.
lone heartBOT
Available help channel!
cinder sundial
lone heartBOT
versed thunder
looks right
lone heartBOT
@cinder sundial Has your question been resolved?
lone heartBOT
lone heartBOT
Available help channel!
loud oyster
lone heartBOT
loud oyster
I found that matrix A has two pivots, so Col(A) basis has two vectors and null(A) has two vectors
so why is null A subspace of R^4?
isn't its dimension two with two basis vectors?
lone heartBOT
@loud oyster Has your question been resolved?
lone heartBOT
@loud oyster Has your question been resolved?
amber iron
isn't its dimension two with two basis vectors?
i think youre thinking of Col A
lone heartBOT
lone heartBOT
Available help channel!
sterile vale
Hello, how would I go about solving this limit?
lone heartBOT
Hello, how would I go about solving this limit?
sterile vale
Where E(x) is the floor function
I tried to go about it using the squeeze theorem but failed :-(
rigid smelt
usually if it were to be a similar limit but without the floor function, you would want to multiply by the conjugate. So here you can do something similarly, it will just be a bit different when evaluating the limit with the floor function
sterile vale
usually if it were to be a similar limit but without the floor function, you wou...
Ok I will try
I am asked to solve this with the squeeze theorem but I will see
rigid smelt
yeah you will need to use a squeeze theorem as an intermediate step
lone heartBOT
@sterile vale Has your question been resolved?
sterile vale
I’m working on it
sterile vale
yeah you will need to use a squeeze theorem as an intermediate step
How is it looking
sterile vale
I forgor 💀 to change the y here for the first line of 4th bullet point
rigid smelt
yeah overall it looks good
rigid smelt
i would say you dont need this line
simply because x-1<floor(x)<=x is already a known identity of the function
or well its trivial enough to see so
sterile vale
Right, thank you
rigid smelt
the rest looks fine to me
sterile vale
I was trying to copy another proof we did in class
rigid smelt
hmm ok
sterile vale
But this is another case I guess where it’s more important to write it I think?
Anyway regardless thanks so much for helping me I really had no idea wtf I was doing at the beginning there
.close
lapis gyro
lone heartBOT
lapis gyro
how do they get to 14?
because even after flipping the sign i am still at 28
do i simplify 54/2 and 28/2
but then what happens to the stuff under the radical
jovial breach
sterile vale
It looks like you’re going too quickly
lapis gyro
no
sterile vale
You skipped the multiplication step and did it in your head
lapis gyro
just havnt done this in 4 years
lapis gyro
and have been working all day