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
#help-49
midnight plankBOT
placid spoke
sure lol
twilit field
$bc=ak_1$.
\
We now divide across by $c$
\
$b = \frac{ak_1}{c}$
grand pondBOT
A dense set
twilit field
Have to figure out how gcd(a,b)=1 is relevant here
placid spoke
what is the proposition mentioned?
strong lava
what does the proposition on page 152 say
placid spoke
oh perfect
twilit field
so 1=ak+bl
placid spoke
here's the hint that makes everything fall together: multiply both sides by c
twilit field
$1=ak+\frac{ak_1l}{c}$
grand pondBOT
A dense set
grand pondBOT
A dense set
twilit field
Which proves $a \mid c$
grand pondBOT
A dense set
twilit field
wish I had thought of this
bezouts would have never struck me
$ab=pk_1 \implies b= \frac{p}{a}k_1$
grand pondBOT
A dense set
twilit field
wait
oops
I was accidentally proving a\mid p and b\mid p
We prove the contrapositive, if $p \nmid a$ and $ p \nmid b$ then $p \nmid ab$
grand pondBOT
A dense set
twilit field
$a=pq_1+r_1; 0<r_1<p$
\
$b=pq_2+r_2;0<r_2<p$
\
It thus follows that $ab= p^2q_1q_2+pq_2r_1+pq_1r_2+r_1r_2; 0<r_1r_2<p^2$
$r_1r_2 \neq p$ as that would require one of $r_1$ or $r_2$ to be p
\
It thus follows that $ p \nmid ab$
How does this look?
grand pondBOT
A dense set
jaunty canopy
**A dense set**
doesn't work
so you've found an expression for ab
ab = p*stuff + r1r2
but you've said that in order for r1r2 to be divisible by p, that would mean r1 or r2 is divisible by p
which is circular cus that's exactly what we're trying to prove
paper inlet
bro just stop if you keep swimming in lead rivers you will turn into stone
twilit field
but you've said that in order for r1r2 to be divisible by p, that would mean r1 ...
How come
twilit field
which is circular cus that's exactly what we're trying to prove
Oh nuts
jaunty canopy
why does r1r2 = p?
why is it not the case r1r2 = 2p?
btw i've got my galois theory supervision now
jaunty canopy
btw i've got my galois theory supervision now
so try to see if another helper can help u
also <@&268886789983436800> please can u deal with this
if you don't have anything useful to add please stop spamming the chat
twilit field
why is it not the case r1r2 = 2p?
Ah, good point
twilit field
why is it not the case r1r2 = 2p?
That would require one of the reminders to be $p$, which isn't possible
grand pondBOT
A dense set
twilit field
<@&286206848099549185>
twilit field
no
twilit field
This
twilit field
I know
shy marsh
ok, so you want to prove this elementary right (without prime factorization)?
or what do you need exactly?
ok you want to continue your proof i guess
twilit field
yes
shy marsh
you got p|r_1r_2
the idea is now that you can keep reducing mod p, do you have an idea how?
like you know r_1,r_2<p also
and assume that r_1r_2>p
then consider the minimal k such that r_1*k>p
twilit field
I would like to understand why my proof is wrong
grand pondBOT
A dense set
shy marsh
yeah but you are using here the lemma which you try to prove
shy marsh
like why cant it be that $$r_1\cdot r_2 = 2p$$ with $r_1,r_2 \neq p$
grand pondBOT
cosmic
twilit field
again, that would require r_2 =p
shy marsh
why, you are using here that the Z is a UFD, which we havent proven yet cuz this proof relies on this lemma
like we dont know that the primefactorization is unique yet
twilit field
oo
shy marsh
like there are rings where this is false
so you have to still prove this for the integers
twilit field
like there are rings where this is false
Aaa, rings
I literally had to prove soemthing relating to dual sided ideals today lol
shy marsh
Aaa, rings
yeah like Z[sqrt(-5)] 6=2*3 and 6=(1+sqrt(-5))(1-sqrt(-5)), so there your conclusion wouldnt hold
so you have to use something specific to integers, and that are often size arguments
i have to leave now unfortunatelly, sorry hope it helped a bit at least
twilit field
Thanks!
midnight plankBOT
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twilit field
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midnight plankBOT
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obtuse totem
midnight plankBOT
obtuse totem
how did they find the general solution?
cause we learnt a method in which constants are infront of y”,y’ and y
what do you do if it’s not a constant in front?
thorny coyote
Trial solution of y=X^n
Its just something u remember, when it has x infront u just try that
P much it
its also like in the screenshot u sent, u go from the equation to the characteristic equation by using a trial soln of y=e^rx
you just suggest a trial solution
midnight plankBOT
@obtuse totem Has your question been resolved?
obtuse totem
Trial solution of y=X^n
y=x^n?
obtuse totem
n as in the highest degree of x?
thorny coyote
No just a constant
when u sub in the trial soln y=x^n into that first image equation, youll find an equation which leads to n=2 and n=-2 which leads to ur general soln
Try it out trust its easier to understand
midnight plankBOT
@obtuse totem Has your question been resolved?
thorny coyote
u got it?
midnight plankBOT
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Available help channel!
dreamy tapir
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Available help channel!
potent fractal
midnight plankBOT
potent fractal
I know how to do this question but I don't understand why my method doesn't work
I tried doing this to get
1/244sin(60) for 4sqrt3
I also got 60/360pi*4^2 =8/3 pi for the sector
Do when we consider a quarter of a circle, it is 4pi
so the shaded area for 1/4s of a circle is
4pi-(8pi/3+(8pi/3-4sqrt3))
4pi-16pi/3+4sqrt3
=4pi-16pi/3+4sqrt3
so in the whole circle,
16pi-64/3pi+16sqrt3
= -16pi/3 + 16sqrt3
which is correct
nvm I played myself
.close
midnight plankBOT
potent fractal
.reopen
midnight plankBOT
✅
potent fractal
the answer is indeed -16pi/3 + 16sqrt3 but I did a step wrong
potent fractal
Do when we consider a quarter of a circle, it is 4pi
so the shaded area for 1/4...
I should have minused the 4sqrt3 instead of added it but why does that guve me a correct answer in the end
.close
midnight plankBOT
midnight plankBOT
Available help channel!
oak rain
Help me with parabolas
midnight plankBOT
Help me with parabolas
oak rain
midnight plankBOT
@oak rain Has your question been resolved?
oak rain
No it hasn’t but im uses to it so I might leave the server
midnight plankBOT
@oak rain Has your question been resolved?
oak rain
Still no, no one fucking helps me
daring bison
can u take another pic of the questions
midnight plankBOT
@oak rain Has your question been resolved?
livid seal
Guys help my homework 😭
midnight plankBOT
Guys help my homework 😭
keen seal
midnight plankBOT
midnight plankBOT
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Available help channel!
twilit field
midnight plankBOT
grand pondBOT
A dense set
twilit field
Let the nullity of $T$ be $r$.Let $dim(V)=n$ Let the basis of $null(T)$ be ${e_1,e_2,\dots,e_r}$. This basis can be extended to one of $V$ , giving $\beta ={e_1,\dots,e_r,f_1,\dots, f_{n-r}}$.
\
We now show that $dim(range(T)) = n-r$. To do so we consider an arbitrary vector, spanned by $\beta$.
\
$v = a_1e_1+ \dots a_re_r+b_1f_1+b_2f_2+ \dots b_{n-r}f_{n-r}$
\
$T(v) = T(\sum_{i=1}^{n} a_ie_i + \sum_{i=1}^{n-r} b_iT(f_i)$
\
$T(v)= \sum_{i=1}^{n-r} b_iT(f_i)$
Let's suppose these vector weren't linearly independent.
\
in that case there exist non-zero scalars such that
\
$T(\sum_{i=1}^{n-r} b_iT(f_i))=0$
\
This would imply $\sum_{i=1}^{n-r} b_iT(f_i)$ lies in $null(T)$.
\
This is obviously absurd, as the vectors in the span of $f_i$ don't belong to null(T). we've thus arrived at a contradiction.
\
\
From that, we can conclude that $dim(rnage(T))= n-r$
\
From this the desired result is evident
grand pondBOT
A dense set
twilit field
I feel this is missing something though
Let the nullity of $T$ be $r$.Let $dim(V)=n$ Let the basis of $null(T)$ be ${e_1,e_2,\dots,e_r}$. This basis can be extended to one of $V$ , giving $\beta ={e_1,\dots,e_r,f_1,\dots, f_{n-r}}$.
\
We now show that $dim(range(T)) = n-r$. To do so we consider an arbitrary vector, spanned by $\beta$.
\
$v = a_1e_1+ \dots a_re_r+b_1f_1+b_2f_2+ \dots b_{n-r}f_{n-r}$
\
$T(v) = T(\sum_{i=1}^{n} a_ie_i + \sum_{i=1}^{n-r} b_iT(f_i)$
\
$T(v)= \sum_{i=1}^{n-r} b_iT(f_i)$
Let's suppose these vector weren't linearly independent.
\
in that case there exist non-zero scalars such that
\
$T(\sum_{i=1}^{n-r} b_iT(f_i))=0$
\
This would imply $\sum_{i=1}^{n-r} b_iT(f_i)$ lies in $null(T)$.
\
This is obviously absurd, as the vectors in the span of $f_i$ don't belong to null(T). we've thus arrived at a contradiction.
\
\
From that, we can conclude that $dim(rnage(T))= n-r$
\
From this the desired result is evident
grand pondBOT
A dense set
midnight plankBOT
midnight plankBOT
Available help channel!
coarse agate
does anyone know how the ford-fulkerson algorithm works under graph theory? 🥹
midnight plankBOT
does anyone know how the ford-fulkerson algorithm works under graph theory? 🥹
@coarse agate Has your question been resolved?
coarse agate
<@&286206848099549185>
here's the iterations I've done but I'm not sure if I did something wrong or is this okay already 🥹
midnight plankBOT
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coarse agate
.close
midnight plankBOT
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sacred musk
hey guys I have a calc 3 midterm worth 50% of my grade tmr and the professor didnt explain this topic ive spent like 5 hours trying to understand can someone help me pls?
I need to be able to change the order of integration to dydzdx and dydxdz:
midnight plankBOT
hey guys I have a calc 3 midterm worth 50% of my grade tmr and the professor did...
sacred musk
it gets super weird and the answer doesnt make sense to me
a GSI basically said its gonna be on the test tho and im kinda panicking
better picture of the integral
(ive drawn many pictures)
tall lotus
better picture of the integral
This is just normal integration but 3 times. First step would be integrating $\int_{0}^{1-y}dz$.
grand pondBOT
deepfriedrich
sacred musk
This is just normal integration but 3 times. First step would be integrating $\i...
the problem isnt integrating its changing the bounds to dydxdz
tall lotus
the problem isnt integrating its changing the bounds to dydxdz
As far as I know, you're supposed to do integrate with dz first, then dx and then dy.
$\int_{0}^{1}(\int_{0}^{1-y^2}(\int_{0}^{1-y}dz)dx)dy$
grand pondBOT
deepfriedrich
sacred musk
As far as I know, you're supposed to do integrate with dz first, then dx and the...
this is the full question on my practice midterm
it has nothing to do with integrating its just showing you know how to manipulate bounds
tall lotus
Oh, my bad I did not see that.
sacred musk
but the only ones I struggle with are dy[dxdz,dzdx]
getting dy first and satisfying the inequalities proves tricky
midnight plankBOT
@sacred musk Has your question been resolved?
last slate
can u draw the region
midnight plankBOT
@sacred musk Has your question been resolved?
midnight plankBOT
midnight plankBOT
Available help channel!
civic lagoon
The problem is here
midnight plankBOT
The problem is here
tall lotus
what's a "chain"?
civic lagoon
what's a "chain"?
Chain rule
civic lagoon
Yes
civic lagoon
Is this the entire question?
oh wait that's the answer
tall lotus
Happens.
civic lagoon
The problem is here
The problem is here
tall lotus
Ah, I'm not good at reading handwriting but I can try.
civic lagoon
Ok
tall lotus
The problem is here
There are some simplifications you can do
nothing more than that.
tall lotus
Like cancel out x from numerator and denominator
civic lagoon
so the x^2 from the numerator will be gone?
tall lotus
The problem is here
Look here, it's not possible to cancel out x^2.
civic lagoon
So how do I get the second form?
civic lagoon
In this
tall lotus
So how do I get the second form?
Take x common from numerator
tall lotus
No.
Like this
$x(x(2e^{2x}-2e^{-2x})+2e^{2x}+2e^{-2x})$
You can do differentiation, this shouldn't be a big deal, right?
civic lagoon
You can do differentiation, this shouldn't be a big deal, right?
Differentiation of?
tall lotus
Differentiation of?
Differentiation is another name for derivatives.
tall lotus
I missed the 2, my bad.
grand pondBOT
deepfriedrich
civic lagoon
so the x^2 is the x(x?
tall lotus
$\frac{x(x(2e^{2x}-2e^{-2x})+2e^{2x}+2e^{-2x})}{x^4}$\
\
$\frac{2xe^{2x}-2xe^{-2x}+2e^{2x}+2e^{-2x}}{x^3}$
grand pondBOT
deepfriedrich
civic lagoon
Ohhh
tall lotus
Then you take 2x common and simplify further.
civic lagoon
ok ok
tall lotus
$\frac{2(e^{2x}(1+x)+e^{-2x}(1-x))}{x^3}$
grand pondBOT
deepfriedrich
tall lotus
This is after clubbing e^{2x} and e^{-2x} terms together.
civic lagoon
Oh okk
civic lagoon
**deepfriedrich**
Wait how?
tall lotus
See what terms have e^{2x} multiplied to them
2xe^{2x} and 2e^{2x}
Then take e^{2x} common
e^{2x}(2x+2)
2e^{2x}(x+1)
civic lagoon
I still don't get it, I'm sorryyy.
tall lotus
I can't spoon-feed harder than this </3
civic lagoon
Okkk
grand pondBOT
deepfriedrich
civic lagoon
Yes
tall lotus
e^{2x}(2x+2)
This and the example I sent are the same on a fundamental level.
tall lotus
**deepfriedrich**
Just imagine e^{2x} instead of "b" here.
barren onyx
midnight plankBOT
steep hinge
it is like the number of positioning 6 object in 6 different places
divided by the number of symmetries that the cube has
barren onyx
divided by the number of symmetries that the cube has
so 6!/3!?
barren onyx
But there are more than 6 rotational symmetries of a cube
how
storm spindle
Well each of the six sides can be up top
But for each side being up top you can rotate the cube along its vertical axis
4 times
barren onyx
Well each of the six sides can be up top
how do u even calculate it
steep hinge
A position of the cube is defined by bottom face and a rotation of the z-axis
storm spindle
With then different faces being in the front each time
steep hinge
so it seems like 6*4
barren onyx
yea
storm spindle
how do u even calculate it
Like I described above. For any of the possible 6 sides on the top there are 4 rotations. Meaning 6*4 rotations in total
barren onyx
Like I described above. For any of the possible 6 sides on the top there are 4 ...
oh
so every side is symmetrical to other
then how is it not 6*6
storm spindle
then how is it not 6*6
With a given side up top, how would you rotate the cube six more times?
barren onyx
With a given side up top, how would you rotate the cube six more times?
oh so i get it rotational symmetry is when u rotate an object and still get the same shape
After rotating
storm spindle
in our solution it is given as 5*3!
Sure, that’s the same as 6!/(6*4)
barren onyx
Sure, that’s the same as 6!/(6*4)
after painting the first side(be any) take the opposite side and you have 5 ways to do it and the rest is circular combination (n-1)! I didn't get their method
barren onyx
Sure, that’s the same as 6!/(6*4)
this is what it says
storm spindle
this is what it says
This is basically the comprised version of the 6!/24 explanation.
We say we start with the first color. For the opposite face there are 5 colors and for the other for there are 3! Ways to do it. Since the starting color doesn’t restrict our options, 5*3! Must be the solution
I don’t really like that explanation though, I find it too unconvincing
barren onyx
This is basically the comprised version of the 6!/24 explanation.
We say we st...
shouldn't the first side have 6!
how is it 1
storm spindle
You start with the first color, not the first side
barren onyx
I don’t really like that explanation though, I find it too unconvincing
unfortunately that's what i have to write in my exam
barren onyx
You start with the first color, not the first side
oh
storm spindle
divided by the number of symmetries that the cube has
Shouldn’t this be fine too?
barren onyx
Shouldn’t this be fine too?
my teacher is a .... he will probably cut my marks
Doesn't allows to use different methods
barren onyx
🤡
storm spindle
Yw
barren onyx
have a good day
storm spindle
You too
barren onyx
.close
midnight plankBOT
midnight plankBOT
Available help channel!
twilit field
I want to prove the following lemma
midnight plankBOT
I want to prove the following lemma
storm spindle
I actually have a nice proof of this at hand
Well actually, only in one direction i suppose
twilit field
As in cgcd(a,b) \geq gcd(ca,cb)
grim vector
I think I can use bezouts?
Yea sure
twilit field
or something like that
storm spindle
As in cgcd(a,b) \geq gcd(ca,cb)
g(ac,bc) | c*gcd(a,b)
May be a bit restricted though, depending on what you’re allowed to assume
visual tiger
take $d = gcd(a,b)$ and $D = gcd(ca,cb)$
grand pondBOT
rafilou is not not born in 2003
twilit field
Yeah, I know how to do that I think
twilit field
For bezouts, I was thinking $k_1a+k_2b = gcd(a,b)$
grand pondBOT
A dense set
twilit field
we now multiply across by $c$ to obtain
grand pondBOT
A dense set
twilit field
$k_1ca+k_2cb=cgcd(a,b)$
grand pondBOT
A dense set
twilit field
but k_1(ca)+k_2(cb)=gcd(ca,cb)$ for some choice of $k_1,k_2$
grand pondBOT
A dense set
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
twilit field
therefore $gcd(ca,cb)=cgcd(a,b)$
grand pondBOT
A dense set
twilit field
the problem we run into is proving that the K_1, k_2 are the same
visual tiger
alternatively, you could use the fact that the gcd is the only common divisor that can be written as a linear combination
for d >= 0, d = gcd(x,y) <=> d|x, d|y and d = ux+vy
twilit field
Hmm
twilit field
This one is the other direction
Thanks!
twilit field
**rafilou is not not born in 2003**
thanks
storm spindle
This one is the other direction
One thing I always like to point out here is that the 4th line follows only from the cancelability of c (I find that part is hard to spot)
So c can’t be 0 here
twilit field
Let me try now :
Let gcd(a,b)=d
$d \mid a \land d\mid b$ by definition
\
As proven earlier $ca \mid k \land cb \mid k$
\
It also follows that $ca \mid ck \land cb \mid ck$.
\
\
It thus follows that $gcd(ca,cb) \mid cgcd(a,b) \implies gcd(ca,cb) \geq cgcd(a,b)$
grand pondBOT
A dense set
twilit field
We now attempt to prove $cgcd(a,b)\mid cgcd(ca,cb)$
grand pondBOT
A dense set
storm spindle
$d \mid a \land d\mid b$ by definition
\\
As proven earlier $ca \mid k \land cb ...
What’s k here?
midnight plankBOT
@twilit field Has your question been resolved?
twilit field
What’s k here?
My bad, meant to write $d$
grand pondBOT
A dense set
twilit field
Let $gcd(ca,cb)=k$ . By definition $k \mid ca \land k \mid cb$. We let $gcd(a,b)= l$. We wish to show that $ cl \mid k$
visual tiger
not sure this is what you want
because proving what you wrote would imply gcd(ca,cb) | gcd(a,b)
grand pondBOT
A dense set
twilit field
We prove this via contradiction
We assume that $ k \nmid cl$
so $k = clq_1 + r; 0<r<cl$
grand pondBOT
A dense set
sinful trout
this is still going?
twilit field
I was eating dinner until a while ago
We start by dividing across by $c$ . This gives us $\frac{k}{c} = lq_1 + \frac{r}{c}$
grand pondBOT
A dense set
twilit field
We know that $c \mid k$. It thus follows that $r$ is a multiple of $c$ or is 0
grand pondBOT
A dense set
twilit field
Okay, now what
<@&286206848099549185>
Let $gcd(ca,cb)=k$ . By definition $k \mid ca \land k \mid cb$. We let $gcd(a,b)= l$. We wish to show that $ cl \mid k$
grand pondBOT
A dense set
twilit field
From this we can conclude that $ l\ mid a \land l\mod b$
so $ca=kq_1; cb = kq_2$
\
We also know that $a=lq_3; b=lq_4$
grand pondBOT
A dense set
twilit field
we know d|a and d|b, so cd|ca and cd|cb
ooh
okay
thanks
so just to be sure we let d be the gcd of (a,b)
so $d \mid a\land d \mid b$ . We then let $ D \mid ca$ and $D \mid cb$ . It also follows that $cd \mid ca \land cd \mid cb$
grand pondBOT
A dense set
twilit field
so just to be sure we let d be the gcd of (a,b)
so $ca=dk_1;c b=dk_2$ . We also know that $Dk_3=ca; Dk_4=cb$
grand pondBOT
A dense set
twilit field
It thus follows that $dk_1=Dk_3$; $dk_2=Dk_4$
grand pondBOT
A dense set
grand pondBOT
A dense set
twilit field
Let's assume that $cgcd(a,b) \nmid gcd(ca,cb)$
grand pondBOT
A dense set
twilit field
$cl \nmid d$
grand pondBOT
A dense set
twilit field
so $d = clq_1+r$
grand pondBOT
A dense set
twilit field
dividing across by $c$, we find thet $r=0$ or $r$ is a multiple of $c$
grand pondBOT
A dense set
twilit field
.close
twilit field
let $\alpha >1$. Show if $|f(x) \leq |x|^{\alpha}$, prove $f$ is differentiable at $0$
midnight plankBOT
let $\alpha >1$. Show if $|f(x) \leq |x|^{\alpha}$, prove $f$ is differentiable ...
grand pondBOT
A dense set
twilit field
So here, I'll have to prove continuity first, do I have to use $\varepsilon-\delta$
grand pondBOT
A dense set
zealous schooner
twilit field
LHL=RHL = F(x)
fallow scarab
Why do you need to prove continuity first?
twilit field
Only then can I even consider it being differentiable
fallow scarab
I don't see why
twilit field
I don't see why
I need continuity to obtain $f(0)$ too
grand pondBOT
A dense set
grand pondBOT
A dense set
twilit field
that gives me $f(0)=0$
grand pondBOT
A dense set
merry pewter
Just apply the definition of differentiability
twilit field
Yeah
realised now
We have $|f(0+h)-f(0)| \leq |h|^\alpha$
\
We thus have $\abs{\frac{f(0+h)-0}{h}} \leq |h|^{\alpha -1}$
grand pondBOT
A dense set
twilit field
We now take the limit of both sides as h tends to $0$
grand pondBOT
A dense set
twilit field
$\lim_{h \to 0} \abs{\frac{f(0+h)-0}{h}} \leq \lim_{h \to 0} |h|^{\alpha -1}$
grand pondBOT
A dense set
twilit field
It's apparent from this that the lhe derivative is $0$, and that the function is differentiable
grand pondBOT
A dense set
twilit field
Is this enough
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last slate
mfs i need help
last slate
Maple Grove High School is designing an L-shaped walking path for their campus. The longer straight section of the path is 4y+7 meters, and the shorter section is 3y−2 meters, where y represents a variable length. The width of the path is consistent at y+1 meters throughout. Write a simplified polynomial expression to represent the perimeter of this L-shaped walking path.
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long dagger
A sphere of solid material of specific gravity 8 has a concentric spherical cavity and just sinks in water. What must be the ratio of radius of cavity to that of outer radius of the sphere?
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A sphere of solid material of specific gravity 8 has a concentric spherical cavi...
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last slate
ok
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marsh veldt
What did I do wrong here to get the incorrect solution pi/2 + npi instead of pi/2 + 2npi (more specifically, why can't n be an odd integer in npi)?
Given: cscx + cotx = 1
1/sinx + cosx/sinx = 1
cosx + 1 = sinx
cos^2 x + 2cosx + 1 = sin^2 x
2cos^2 x + 2cosx = 0
2cosx(cosx + 1) = 0
x = pi/2 + npi, pi + 2npi
midnight plankBOT
What did I do wrong here to get the incorrect solution pi/2 + npi instead of pi/...
blissful talon
When you square both sides, you introduce new answers to the question. That means that you have to check the answers you get.
For example, say I have (x-1)=4
(x-1)^2 = 16
When I factor and solve, I get x=5 or x=-3
but -3 was not an answer to the original question
marsh veldt
When you square both sides, you introduce new answers to the question. That mean...
how can I get the right solution instead
fallow scarab
where even is your problem
blissful talon
how can I get the right solution instead
What you did is good. You just need to check your answers.
blissful talon
where even is your problem
cscx + cotx = 1
fallow scarab
they say that's a given?
marsh veldt
That’s the problem
fallow scarab
that's an equation. what are you supposed to calculate
marsh veldt
x
blissful talon
I know it's not the best way to pose a problem, but you can tell from context
marsh veldt
yea mb
blissful talon
unless you're not actually asking and you're just trying to get GoldenTrash to format better
marsh veldt
I just listed my steps afterwards
dusk owl
What did I do wrong here to get the incorrect solution pi/2 + npi instead of pi/...
U applied the wrong formula
By comparing with Cos theta = Cos Alfa
We get, theta = 2nπ + alfa
marsh veldt
U applied the wrong formula
wait sry where
dusk owl
Check ur last step
dusk owl
By comparing with Cos theta = Cos Alfa
We get, theta = 2nπ + alfa
By comparing with Cos x = Cos Alfa
We get, x = 2nπ + alfa
marsh veldt
what is alfa in this case
marsh veldt
but 0 comes every pi interval
marsh veldt
I don’t understand why to compare cos theta with cos alfa
dusk owl
Replace theta with x
dusk owl
I don’t understand why to compare cos theta with cos alfa
It's a formula
R u from India?
marsh veldt
no united states
dusk owl
Ohh so ig ur textbook can have different formula
marsh veldt
It's a formula
but 0 comes every pi interval, so why 2npi
marsh veldt
no
dusk owl
Any textbook?
marsh veldt
no
dusk owl
but 0 comes every pi interval, so why 2npi
n belongs to integer
dusk owl
Yess
marsh veldt
why doesnt pi/2 + npi work for the equation if the solution for 2cosx = 0 is pi/2 + npi though
dusk owl
Is π/2 + nπ correct answer
marsh veldt
no
marsh veldt
no worries, but the answer is pi/2 + 2npi
I was confused on how to obtain that answer
dusk owl
Wait I am sending u the solution
In my country, it's completely based on formula
Please replace theta with x cuz we use theta mostly so it's in my habit 😅
marsh veldt
but the correct answer is pi/2 + 2npi
dusk owl
but the correct answer is pi/2 + 2npi
R u sure?
dusk owl
Ig now I am unable to solve it
dusk owl
you can graph it
Can u send me the graph
dusk owl
Do u know the concept behind this
dusk owl
Ig I'll take time to understand this 😅
marsh veldt
ok ty for your help
dusk owl
Sorry sorry
I am unable to help
Please reopen ur question
But thanks a lot u helped me clearing my concept ✨
marsh veldt
yea np ig
marsh veldt
What did I do wrong here to get the incorrect solution pi/2 + npi instead of pi/...
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mental bone
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mental bone
How do I find where the function is increasing and decreasing?
I found the derivative:
dawn dagger
you can also use |x| and think about how the transformations affected f(x)
mental bone
you can also use |x| and think about how the transformations affected f(x)
So in a sense, I don't rlly need the derivative?
dawn dagger
nope
mental bone
Hmm... how would I determine them using the derivative tho
mental bone
Yeah I know but I'm just having trouble
Because how I do it is i find the critical point then determine how the function behaves using nearby points
runic hamlet
its so much easier to do the derivative by cases instead of this inconvenient thing
mental bone
its so much easier to do the derivative by cases instead of this inconvenient th...
What do you mean by cases?
runic hamlet
just do cases depending on whether the inside of the absolute value is >=0 or not
mental bone
ok
grand pondBOT
bacc (unhelpful)
mental bone
Ahh I see
runic hamlet
also your derivative isnt correct
mental bone
okay let me try
mental bone
also your derivative isnt correct
Let me work on that
floral apex
Would this be the derivative?
they are telling you to take the deriv of a linear function
you need to apply the piecewise defn of absolute value
|y| = y if y>= 0
mental bone
ohhh i see
floral apex
|y| = -y if y < 0
dawn dagger
Would this be the derivative?
it's correct
you neglect the denominator since at x = 3 is not differentiable anyway so you really solve -2(x-3) >= 0 and -2(x-3) < 0
neat canyon
midnight plankBOT
violet lion
$\sin^2 x = 1 - \cos^2 x$
grand pondBOT
timuko
neat canyon
for the rhs?
violet lion
for the rhs?
Yes
neat canyon
is it easier from there
violet lion
Maybe try using complex numbers
neat canyon
i have not learnt them
violet lion
$\cos(3x)=\cos(x+2x)=\cos(x)\cos(2x) - \sin(x)\sin(2x) = \cos x(2\cos^2 x - 1) - 2\sin^2 x \cos x = 2\cos^3 x - \cos x - 2\cos x(1 - \cos^2 x) = 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x = 4\cos^3 x - 3\cos x$
grand pondBOT
timuko
last slate
You could do this
Or
You could use the property
CosA +CosB = 1/2 Cos(A+B/2) Cos(A-B/2)
A being 3x and B as 5x
violet lion
You could use the property
CosA +CosB = 1/2 Cos(A+B/2) Cos(A-B/2)
Wow, I completely forgot about this property
last slate
You could then use Cos(2x) = 2cos^2(x) -1
last slate
i can take out the negative sign
Yes
last slate
but its a difference
It's a sum
violet lion
do they cancel out with the 2
$-\frac{1}{2}(\cos 3x + \cos 5x)$
grand pondBOT
timuko
violet lion
cos(x) * (1-cos(4x))
It reminds me of half-angle identity of $\sin \frac{\theta}{2}$
grand pondBOT
timuko
violet lion
Like $\sin 2x = \pm \sqrt{\frac{1-\cos 4x}{2}}$
grand pondBOT
timuko
neat canyon
yea i see it
violet lion
Here is a cheat-sheet:
neat canyon
i know them
last slate
You could then use Cos(2x) = 2cos^2(x) -1
.
last slate
First your x would be 2x,if that made any sense XD
neat canyon
2cos^2(2x) - 1
last slate
2cos^2(2x) - 1
Yeah
neat canyon
Yeah
does the minus sign affect both
like
cosx*(1 - 2cos^(2x) - 1)
or cosx * (1 - 2cos^2(2x) + 1)
neat canyon
honestly
neat canyon
or cosx * (1 - 2cos^2(2x) + 1)
this
last slate
You are right
neat canyon
1+1 right?
last slate
Good
neat canyon
so
last slate
Go on
neat canyon
do i use
cos(2x) = cos^2(x) - sen^2(x)
or ig i could take the 2 out
2(1-cos^2(2x)
and change that to sin^2(2x)
last slate
cos(x) * (1-cos(4x))
Rather than using cos2x equals 2cosx minus 1
Using cos2x equals 1- 2sin^2 x eliminates the 1
neat canyon
oh yeah
but we get to the same thing anyways
or not
i think not
ok
cos(x) *(1 - 1 + 2sin^(2x))
oh we do
cos(x)*(2sin^2(2x)
last slate
cos(x)*(2sin^2(2x)
Sin2x can be expanded to?
neat canyon
cos(x)*(2.2sin^2(x)cos^2(x)
last slate
Yep
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vital vault
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vital vault
This is the one issue I have is finding the asymptotes, I tried y = 900, y = 0 and was wrong? not sure what I am missing
hearty rune
you have written the equation of a vertical line
what happens as t tends to infinity
specifically to the e^(-0.1692t)
oh actually
i see, did you actually write 'y'?
try p
vital vault
Originally I did because I was listing the asymptots at y = 900
I did t = kinda as just a shot in the dark
vital vault
ohh
so would it be p(t) = and then where the asymtots cordinate is?
like p(t) = 900 yeah?
hearty rune
you can skip the t but yeah
vital vault
ohh ok ok
hearty rune
p=900, p=0 should be okay
wait no
the 0 is fine, the 900 not
sorry ignore that, got confused
youre good
vital vault
So my original listing of 0 and 900 was right I was just missing p=?
hearty rune
yup
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meager plover
hi can someone explain how for problem c, it’s 42 seconds? i dont get it 🥲🥲🥲
sleek cloud
Why did you label the second peak 42
midnight plankBOT
@meager plover Has your question been resolved?
meager plover
Why did you label the second peak 42
that’s the x-value
sleek cloud
How?
meager plover
for c or just on my graph
sleek cloud
Your graph
meager plover
the max distance is 25 yds, and michael reached 25 yds in 21 seconds
since the graph is in a cycle, the period is 21 seconds. so he completed 25 yds for the second time in the timeframe of 0-42 seconds
meager plover
Isn't the second time when he returns back?
the period is the distance between the peaks
meager plover
When the distance is back to 0?
that’s just when he completed his first lap
sleek cloud
The 42 should be when the distance returns to 0 first time
The period can also be from 0 to 0
Which is 42
meager plover
The 42 should be when the distance returns to 0 first time
no bc after 21 seconds, he reached maximum distance
after another 21 seconds, he reached the max distand again
21+21=42
sleek cloud
Max distance?
meager plover
pool is 25 yds long
sleek cloud
Yeah it takes 21 seconds to swim 25 yards
And then 21 seconds again to return from the other side to starting point
meager plover
Hm
because the answer key said point D is 84 seconds
Wait
Ohhhhhhhhhhh
no youre right
i get it now
sorry i thought for whatever reason, 42+21=84 😅😅😅
Wait but i dont get why it takes 42 seconds to get from point A to point C
sleek cloud
Because that's a period
meager plover
Oh so the period is actualy 42 seconds
😭😭😭
Ok my brain just doesn’t wanna function simple math
ok ty
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grand canopy
help me
grand canopy
first part is just solve for roots and factorise over the real field noting cos4π/5 = -cosπ/5
how do i find cosπ/5 and cos2π/5
midnight plankBOT
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 tried SOR and POR of the quadratics but that doen't really help me
grand canopy
with the 2nd quadratic
cis2π/5 + cis(-2π/5) = 2cos2π/5
but that's just a true statement cuz LHS is 2Re(z)
similar with POR
small jasper
Consider the sum of the roots from all three factors. What must this sum be equal to (||consider the left hand side||)?
grand canopy
ohh
1 + cis(2π/5) + cis(-2π/5) + cis(4π/5) + cis(-4π/5) = 0
1 + 2cos(2π/5) + 2cos(4π/5) = 0
cos2π/5 - cosπ/5 = -1/2
then do i use POR ?
grand canopy
oh so i do cos(2π/5) = 2cos^2(π/5) - 1
sub into the other equation and get a quadratic
solve then choose whichever solution is right?
ok i got it now
THANK you
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leaden matrix
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leaden matrix
could someone explain how to go from that top right equation to the bottom one?
sharp coral
lavish venture
$\frac{y-3}{y-3} + \frac{3}{y-3}$
grand pondBOT
knief
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pseudo moat
The perimeter of a rectangle is 80dm. The width is 1 more than 1/2 of the length. Find the length and the width of the rectangle.
midnight plankBOT
The perimeter of a rectangle is 80dm. The width is 1 more than 1/2 of the length...
pseudo moat
do I use substitution
lavish venture
what’s the formula for perimeter of a rectangle
pseudo moat
P = 2(L + W)
stable dawn
half of length
lavish venture
P = 2(L + W)
yes then express W in terms for L
pseudo moat
how
weary nest
L/2 is half the length
^^^
pseudo moat
L/2?
lavish venture
yes L divided by 2
weary nest
Length divided by 2
lavish venture
which is half the length
pseudo moat
can you show it, i don't understand in terms of text
weary nest
Ok what do you not understand
pseudo moat
all of it, can you start again, im lost
stable dawn
all of it, can you start again, im lost
If length was 4(say) then width will be (half of 4) + 1
But length is a variable here which we have to find so we denote it with L.
weary nest
Because we have already established that W is equal to L/2 +1
Then because you multiple everything by 2
You get this
Then you add them together
pseudo moat
but it's only 2L
weary nest
Do you know how to expand brackets?
Look at the very top line in the photo
Everything in between the brackets have to be multiplied by the number directly next to the brackets
So multiply everything by 2
pseudo moat
yes
pseudo moat
just got a lil confused with this, but I understand it
pseudo moat
so what's the length and width
midnight plankBOT
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pseudo moat
where did 3L + 2 = 80 came from
this will become 3L = 80-2 which will become 3L=78 then divide both sides by 3 to get L which will be 3L/3 = 78/3 which will be L=26
is that correct
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nova pike
hello, i dont know how to approach this following problem: in graph theory, let us describe a simple graph of n vertices and m edges assigned an integer value from 0 onwards. in each operation, when you chose a pair of connected vertices, one minuses 1 from one vertice's value and adds 1 to the other. call the task winnable if it is impossible after applying any number of iterations no two linked vertices are both 0, and let the smallest sum across all the n vertices be k. find all possible values of k, given m and n
midnight plankBOT
hello, i dont know how to approach this following problem: in graph theory, let ...
nova pike
hello, i dont know how to approach this following problem: in graph theory, let ...
i tested it for small values of n like 2 and 3, but for greater than 3 testing becomes difficult and ideas need to be come acrossed
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nova pike
hello, i dont know how to approach this following problem: in graph theory, let ...
<@&286206848099549185> Anyone good at graph theory or know any techniques for solving this problem, it would be would be much appreciated if you could help me
merry pewter
hello, i dont know how to approach this following problem: in graph theory, let ...
so you have a graph with n vertices each numbered from 0 to n-1, and at each step you can take two connected vertices and subtract 1 from one vertex and add 1 to the other?
hexed pilot
hello, i dont know how to approach this following problem: in graph theory, let ...
You will need to iterate through all the vertices of the graph making sure no connected vertices (having same parent) are both 0
Disjoint Set with specific optimization can help you with that traversal
merry pewter
well it seems like if the initial sum is fixed, then the sum is always the same since subing 1 and adding 1 keeps the sum constant
nova pike
so you have a graph with n vertices each numbered from 0 to n-1, and at each ste...
Sorry I did not mean this. We can have values of each vertex any integer from 0 and above. For example with four vertices, we could have values 0, 1, 3, 1, etc. also I am not the best at graph theory, so feel free to correct me anytime
nova pike
well it seems like if the initial sum is fixed, then the sum is always the same ...
Yes, exactly
merry pewter
Sorry I did not mean this. We can have values of each vertex any integer from 0 ...
i see
nova pike
You will need to iterate through all the vertices of the graph making sure no co...
I’m sorry, how would we do this?
merry pewter
i think he means a search with a computer
hexed pilot
yes
I can write a program in Java that can help you solve this
but the problem is I don't understand the point of this problem?
What is it trying to achieve ultimately??
merry pewter
Sorry I did not mean this. We can have values of each vertex any integer from 0 ...
well then it seems like the graph part of the problem is not relevant, all that you need is the initial sum of the vertices
also fyi, a forest is a set of disjoint trees, which i assume is not what you are talking about
hexed pilot
yes, and a tree can also be classified as a type of graph
merry pewter
yes, and a tree can also be classified as a type of graph
well im saying that it seems like he is ignoring the forest part of the statement
nova pike
I can write a program in Java that can help you solve this
That would be great. I don’t know how to script things
merry pewter
well then it seems like the graph part of the problem is not relevant, all that ...
so then the minimum sum is 0, and the maximum is n(n-1)
nova pike
also fyi, a forest is a set of disjoint trees, which i assume is not what you ar...
I think this is what I am talking about. The reason why I am not that great with this problem is because I tried to convert another problem in terms of graph theory which become this problem. Due to the original problem conditions, I figured that they only allowed ‘forests’ as graphs
merry pewter
I think this is what I am talking about. The reason why I am not that great with...
so it's important that the graph be a set of disjoint trees?
nova pike
so it's important that the graph be a set of disjoint trees?
Yes. The original problem had that each edge is connected to some two vertices and each pair of vertices is connected by one edge at most
merry pewter
Yes. The original problem had that each edge is connected to some two vertices a...
that's just a simple graph
not necessarily a tree
or a forest
simple means that there is at most one edge between vertices and each edge is between exactly two vertices like you said
nova pike
simple means that there is at most one edge between vertices and each edge is be...
Ah, I’m so sorry, then I meant only simple graphs work for this problem, not forests
merry pewter
Ah, I’m so sorry, then I meant only simple graphs work for this problem, not for...
ok, then if i understand correctly, you need to find all possible k for all winnable graphs with n vertices and m edges
then it's a bit tricky because for some values of m, there are no winnable graphs
merry pewter
namely if m = n(n-1)/2, so that the graph is a complete graph
then, you can always make two of the vertices 0
im not sure if there will be a closed form
nova pike
then, you can always make two of the vertices 0
Yeah
nova pike
im not sure if there will be a closed form
I tried working with smaller n and building my way up, but it becomes hard to search for n>3
What happens if we fix a m and n? Then would we be able to compute all possible values of k easily?
grizzled pivot
Is the following graph winnable? (0)—(1)—(0)
merry pewter
wait but now it seems like any graph with three vertices that are connected in a row is unwinnable
merry pewter
Is the following graph winnable? (0)—(1)—(0)
no, as you can move the 1 to the end
nova pike
no, as you can move the 1 to the end
Yes
merry pewter
ok, then it seems like there are actually few values of m so that a graph with n vertices is winnable
nova pike
Is the following graph winnable? (0)—(1)—(0)
No, it’s not winnable which bair explains why exactly
grizzled pivot
wait but now it seems like any graph with three vertices that are connected in a...
okay then yeah i agree with this
yeye you got it
nova pike
ok, then it seems like there are actually few values of m so that a graph with n...
Interesting. How exactly?
merry pewter
so i think when m < n/2, there are n(n-1) possible values of k, and 0 otherwise
merry pewter
Interesting. How exactly?
"few" meaning not all of them, only for m < n/2
because when m > n/2, then there three vertices that are connected
which means that the graph is unwinnable
nova pike
so if there was a graph with 100 vertices and 1000 edges, then there does not exist such graph?
merry pewter
so if there was a graph with 100 vertices and 1000 edges, then there does not ex...
right, because there are three vertices that form a connected component
and therefore, if the numbers are x-y-z, then you can make (0)-(0)-(x+y+z)
actually a funny edge case is when m = 0, then you can have n(n-1)+1 values of k
nova pike
and therefore, if the numbers are x-y-z, then you can make (0)-(0)-(x+y+z)
Wait, how is this possible?
nova pike
actually a funny edge case is when m = 0, then you can have n(n-1)+1 values of k
Interesting, how is that so?
merry pewter
Wait, how is this possible?
well can't you first move (0)-(x+y)-z and then (0)-(0)-(x+y+z)
merry pewter
Interesting, how is that so?
when m = 0, no vertices are connected, so you can make every vertex have a value of 0
but if there is at least one edge, then the sum has to be at least 1
nova pike
Oh, I think I made also a terrible mistake in what I was trying to say which makes the problem much more complicated. The person that is making each iteration wants there to be a 0-0 but the vertices move in a way that always try to avoid this
merry pewter
Oh, I think I made also a terrible mistake in what I was trying to say which mak...
the vertices move??
how
do you mean the vertices are arranged so that the graph is unwinnable?
nova pike
do you mean the vertices are arranged so that the graph is unwinnable?
Yes, I mean this.
Im so sorry, I really wanted to convert the original problem into a graph theory one
merry pewter
Yes, I mean this.
but then the problem is even simpler, then if m>=2, then there is always an unwinnable graph
so for m>=2, the only value of k is 0
and when m=1 there are n(n-1) values of k
and when m=0 there are n(n-1)+1 values of k
merry pewter
Im so sorry, I really wanted to convert the original problem into a graph theory...
can you send the original problem, maybe it will help makes things clear
nova pike
can you send the original problem, maybe it will help makes things clear
Yes, I’ll send it
nova pike
I thought replacing 100 with n and 1000 with m would help solve the problem in general
merry pewter
oh i see, so merlin gets to chose the edge, but not the direction in which the wise men move
this is quite different to the earlier statement
nova pike
oh i see, so merlin gets to chose the edge, but not the direction in which the w...
Merlin does not matter. It’s just like a person choosing a corridor and asks one of the people from one of two rooms to go through the corridor into the other room
nova pike
this is quite different to the earlier statement
I understand, I’m so sorry. I tried to rephrase the problem in terms of graph theory, but to no avail
merry pewter
Merlin does not matter. It’s just like a person choosing a corridor and asks one...
yes, but the point is that we cannot have the situation from before, when (x)-(y) became (0)-(x+y), since the wise men could just keep moving back and forth
nova pike
yes, but the point is that we cannot have the situation from before, when (x)-(y...
Yeah, I was wanting to tell you this but couldn’t express it properly
merry pewter
hmm, then it is a tricky problem, let me think
nova pike
hmm, then it is a tricky problem, let me think
alright, thank you for your time and patience
merry pewter
well im very stuck, it seems like it highly depends on the structure of the castle
one idea is to alternate the parity of every room
since then, the parity of each adjacent room will swap, hence not both can be 0
but this is not always possible, depending on castle layout
you would need a bipartite coloring
nova pike
since then, the parity of each adjacent room will swap, hence not both can be 0
wait, wdym by this?
merry pewter
wait, wdym by this?
like if i have (even)-(odd), then no matter which way the wise men go, we will get (odd)-(even)
wait but then the parity will be the same on the other side
so this doesn't work
nova pike
like if i have (even)-(odd), then no matter which way the wise men go, we will g...
yes, I get this, but why different parity?
merry pewter
yes, I get this, but why different parity?
don't worry, it doesn't make sense
ok, it is giving me a lot of trouble, i would suggest pinging helpers again, or asking in #discrete-math , or asking again when more people are awake
nova pike
ok, it is giving me a lot of trouble, i would suggest pinging helpers again, or ...
Oh ok, thank you for your time and help. I was also thinking of converting this into a string of numbers followed by suitable operations, but idk where that would lead to
nova pike
ok, it is giving me a lot of trouble, i would suggest pinging helpers again, or ...
Thank you so much, and sorry for causing such a big inconvenience
merry pewter
Thank you so much, and sorry for causing such a big inconvenience
np, next time i would just suggest sending the original problem first
merry pewter
and then saying how you reformulated the problem
nova pike
and then saying how you reformulated the problem
Oh ok
midnight plankBOT
@nova pike Has your question been resolved?
nova pike
<@&286206848099549185> here is the problem. instead of 100 rooms, i tried breaking it down into 2, 3 etc. but from >3, it becomes hard finding the best and minimal possible construction
fervent ember
i have two questions:
- whats the meaning of "every minute" as there is no other time relevant information in it? would it be the same if its just "every thirty seconds" or "every hour"?
- if merlin has choosen a corridor and a wise man psssed along it, would it be a simple strategy if the wise man will go back?
nova pike
i have two questions:
1. whats the meaning of "every minute" as there is no othe...
for the first question, yes it would be the same. it would just be considered every single iteration. for the second question, that is if merlin has chosen the same corridor again, then yes, a wise man can go back
fervent ember
does this mean the wise men can only move after a choice of merlin?
nova pike
does this mean the wise men can only move after a choice of merlin?
yes, and only one wise man each iteration
fervent ember
ok, then i do understand nothing. whats the meaning of "the wise men needs to scatter across the rooms in a manner they wish"?
if only one man can move and only in the corridor which is defined by merlin i cant see a "a manner they wish"?
nova pike
ok, then i do understand nothing. whats the meaning of "the wise men needs to sc...
they do this btw before merlin starts picking a corridor
nova pike
ok, then i do understand nothing. whats the meaning of "the wise men needs to sc...
its like saying if you have a_i be the number of wise men in the ith room, then you have the sum of all a_i for i = 1 to n be the number of wise men in total
nova pike
<@&286206848099549185> if someone could help with this problem (which looks like theres some graph theory in disguise), that would be much appreciated
ocean dune
Does Merlin win if there's no wise man in a room but there's one in the other
nova pike
Does Merlin win if there's no wise man in a room but there's one in the other
Not necessarily. Merlin wins if at the end of any iteration there exists a corridor connecting two rooms both with 0 people in them