A <==> B iff B <==> A
#help-13
north wyvern
so if i prove A <==> C, then i also proved C <==> A
now, very important that we cant use rules that only say A ==> B
because then B ==> A isnt necessarily true
so like, substitution doesnt necessarily work backwards
sometimes it can, but not in general
but if i add something to both sides of an equation, then it holds both ways
if that makes it clearer
anyways, just prove this from axioms and i believe it should be good
glossy fractal
cedar kilnBOT
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solid patrol
is this the correct way to use chain rule?
cedar kilnBOT
is this the correct way to use chain rule?
swift moth
hexed vortex
is this the correct way to use chain rule?
yeah thats right but how did you get $(x^2 - 4)^{\frac{1}{3}}$ in the final result
wraith daggerBOT
Branshi
solid patrol
swift moth
ya that is correct
hexed vortex
yeah
solid patrol
i realize i thought i can time x^1 with the (x^2-4)^-2/3
swift moth
not really...
unless you kinda put it into brackets
hexed vortex
oh no you cant do that becaue they dont have the same base
solid patrol
the question ask how to find the horizontal or vertial tangent lines?
i plugged in f'(0) and got 0
i guess x can't be 2 for vertical tangent lines right?
i know that 3(4-4)^2/3 would be 0 , anything / 0 would be undefined
swift moth
the question ask how to find the horizontal or vertial tangent lines?
i plugged...
yes x cannot be 2
but to find horizontal tangent you actually try solving x such that g'(x) = 0
swift moth
yup
solid patrol
does that mean that at (0,0) there is a horizontal line and at x = 2,-2 , there is a vertical tangent lines?
swift moth
does that mean that at (0,0) there is a horizontal line and at x = 2,-2 , there ...
erm when x = 0, y isnt 0
solid patrol
erm when x = 0, y isnt 0
oh ok so just x = 0 there is a horizontal line and x = -2,2 there is a VTL
swift moth
yea
swift moth
yes
calm notch
hi i am new whatshould i startwith
solid patrol
hi i am new whatshould i startwith
if you have any math question just send a picture , problem to the #help-3
solid patrol
......
calm notch
he he eh
cedar kilnBOT
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calm notch
i dont understand what u mean
cedar kilnBOT
i dont understand what u mean
uneven dome
that's Collatz. what exactly do you want to know about it?
calm notch
explain plzzzzzzzz
supple coral
What is there to explain
supple coral
Take f(5)
calm notch
how to do it
supple coral
Just put the number
calm notch
how it works
supple coral
how to do it
f(5) will give 3*5 +1
16 which is even
so n/2 will be applicable here
keep doing that
calm notch
oooooooh
supple coral
16/2=8
calm notch
👍
calm notch
thNKS
supple coral
If done .close
calm notch
THANKS
uneven dome
!done
cedar kilnBOT
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fossil dawn
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calm notch
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small elbow
cedar kilnBOT
small elbow
This seems so random but i guess its eventually used for determinants
and this too ig
mental trail
Well it'll come together once you talk about the determinant
if you name e1 = (1,0,...,0), e2 = (0,1,0,....,0), etc...
The determinant is the only multilinear alternating form on Mn(F) with D(e1,...,en) = 1
small elbow
kk
oh i gotta remind myself D maps to a scalar
and v1, .., vn are row vectors of some matrix?
mental trail
well you can just view v1,...,vn as vectors on their own
ofc they can always assemble to make a matrix
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crimson sedge
hi
crimson sedge
I need help
uneven dome
question please
crimson sedge
tropic oxide
progress?
uneven dome
what do you understand by solving for a variable?
wicked mantle
tropic oxide
you factored?
crimson sedge
then I divided by 1/3a
crimson sedge
by factoring?
tropic oxide
??
uneven dome
there's only one term here though
tropic oxide
you're saying $\frac13ar^2$ became $r \cdot \frac13a$ ??
wraith daggerBOT
Ann
simple epoch
Where's the square...
tropic oxide
again how are you going from ar^2 to ra
tropic oxide
there is no need nor even possibility to factorize
multiplying by 3 was a good idea tho
ar^2 = 3C
crimson sedge
ok
tropic oxide
how do you undo a square?
crimson sedge
I don't know
tropic oxide
do the words "square root" ring a bell?
crimson sedge
kinda
tropic oxide
do you know what a square root is
tropic oxide
...i was gone for like 3 min
anyway, you can either tell me what you understand "square root" to mean,
tropic oxide
the things that make a number
"things that make a number" is quite vague
the square root of 144 is 12 because...?
crimson sedge
12x12=144
tropic oxide
12x12=144
ok right.
or rather 12**^2** = 144.
so in fact, sqrt is precisely the operation that undoes squaring.
in a similar way that division undoes multiplication, or subtraction undoes addition.
cedar kilnBOT
@quiet pier Has your question been resolved?
ebon cloud
Any idea how to approach this? Question is to find an integral function F for f that is continous when x = 2 and to F(0) = - 3
cedar kilnBOT
Any idea how to approach this? Question is to find an integral function F for f ...
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
2
cedar kilnBOT
Show your work, and if possible, explain where you are stuck.
@ebon cloud Has your question been resolved?
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formal crypt
If it's -1<x<3 then write the expression in the absolute value
A= |x-3| + |-x-1|
can someone tell me how i can go about solving this?
tropic oxide
do you know the piecewise-function defn of absolute value?
lavish vine
What they mean by piecewise function is breaking this into multiple other functions
Which do not contain absolute values
formal crypt
What they mean by piecewise function is breaking this into multiple other functi...
ohh then i don't think so
lavish vine
For example when we consider the function f(x)=|x|
How do we write this function in a simpler form
formal crypt
idkkk 😭
tropic oxide
I'm not that well educated in the English math terms 😔
whats your native language
formal crypt
whats your native language
greek
tropic oxide
aight nevermind
ok let me just write it out
$|x| = \begin{cases} x & x \geq 0 \ -x & x < 0\end{cases}$
wraith daggerBOT
Ann
tropic oxide
that ring a bell?
formal crypt
Yess
If x is either bigger or equal to 0 then the absolute value is x
and if it is smaller then the absolute value is -x
languid charm
If it's -1<x<3 then write the expression in the absolute value
A= |x-3| + |-x-1...
Are you trying to write A in terms of x(without the absolute value)?
formal crypt
i guess I'm solving for a
formal crypt
If it's -1<x<3 then write the expression in the absolute value
A= |x-3| + |-x-1...
that's what the math problem says
languid charm
Ok ima just guess so then
So we have x<3
What would happen if we minus 3 on both sides
formal crypt
So we have x<3
What would happen if we minus 3 on both sides
|x-3|-3 + |-x-1|-3
this is what you mean?
languid charm
No i meant for the inequality
languid charm
this is what you mean?
we would get x-3<3-3 right?
cedar kilnBOT
@formal crypt Has your question been resolved?
languid charm
Do you still not understand it?
cedar kilnBOT
@formal crypt Has your question been resolved?
formal crypt
Do you still not understand it?
i don't no
languid charm
i don't no
Ok let me ask this
If x>0 then is x+1>1
formal crypt
Ok let me ask this
If x>0 then is x+1>1
i guess yea
languid charm
i guess yea
If x<2 then is x-1<1
formal crypt
ye
languid charm
Then what would |x-3| equal
formal crypt
Then what would |x-3| equal
something negative i suppose
languid charm
something negative i suppose
Why negative?
languid charm
That symbol means the absolute value of x-3 by the way
formal crypt
how the fuck do i determine if it's positive or negative if it has x in it
dire geode
|x - a| = x - a if x > a and
|x - a| = a - x if x < a
languid charm
how the fuck do i determine if it's positive or negative if it has x in it
x-3<0 since x<3(given info)
formal crypt
|x - a| = x - a if x > a and
|x - a| = a - x if x < a
x<3 so it's the second one?
languid charm
Yeah
languid charm
If it's -1<x<3 then write the expression in the absolute value
A= |x-3| + |-x-1...
Ok so we've found that |x-3|=-x+3
Now what do you think we should start with for |-x-1|
formal crypt
uhh
i was re-reading the caption of the problem
If –1 < x < 3, then write the expressions without the absolute value:
a. A = |x-3| + |x-1
it said without the absolute value
does this change anything?
cedar kilnBOT
@formal crypt Has your question been resolved?
formal crypt
I'm giving up
bronze briar
i need to find the stationary points of this equation
cedar kilnBOT
i need to find the stationary points of this equation
bronze briar
first i found the derivative: Cw + b = 0 -> Cw = -b
boreal lichen
hi i need help about a question can I send it here?
bronze briar
so then i did
2 1 -1
1 3 -1
and reduced that matrix to
1 0 -.04
0 1 -.2
so if i understand this correctly, does that mean that w must be
-.4
-.2?
flint plinth
assuming you meant -0.4 instead of -.04 in the reduced matrix then yes
bronze briar
assuming you meant -0.4 instead of -.04 in the reduced matrix then yes
yes, thank you. i'm also supposed to plot this equation (to determine if the point is a saddle, min, or max). Do you know how to plot it on desmos?
like the other equations were easy to plot, like a previous question was just g(w) = wtanh(w)
flint plinth
sorry, i don't really use desmos, you could try asking in #computing-software
cedar kilnBOT
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sacred quest
I dont fully understand how to do this can someone explain how im supposed to do it please?
y³-y²-6y-4 divided by y+1
clear ember
like polynomial division?
sacred quest
Yeah
clear ember
can you show where you're currently stuck?
sacred quest
Its once it gets to after I dived y³-y² by y³-y²
sacred anchor
I dont fully understand how to do this can someone explain how im supposed to do...
you can try remainder theorem
clear ember
(y + 1)(y^2) = y^3 + y^2
like regular long division you would write what you get after multiplication
then decide to subtract the highest powers and then iterate through
that way yeah
so in your case y^3 - y^3 = 0
but -y^2 - y^2 isn't 0, right?
the -6y and -4 part is fine
sacred quest
Ohh okay
clear ember
btw
do you have to use polynomial long division
if not then you can equate coefficients
if you feel more comfortable that way
idk
something like $y^3 - y^2 - 6y -4 = (y + 1)(y^2 + bx + c)$
wraith daggerBOT
mmmm7
sacred quest
if not then you can equate coefficients
Sorry I havent learned that yet 😭
sacred quest
No im not sure how to do it sorry
clear ember
okie this (poly division) is fine then
scenic jolt
cedar kilnBOT
neon void
Please advise, thansk
what does the 50'44'' mean
are those angles even right? the big one looks like 270...
scenic jolt
Don't worry about that. The angle is 244.845
Degrees
And the other angle is 92.078333deg
scenic jolt
are those angles even right? the big one looks like 270...
Image is not representative
Just a sketch
Angles shown is what is important
Please advise if anyone can help
Tha ks
Also
I should also mention
0deg starts at positive y axis
Sorry about that for not mentioning earlier
Please advise thabks
neon void
dafuq
scenic jolt
If you don't understand then that's fine
This is advances math
Not everyone will get it
neon void
ok why is 244 deg angle touching the x axis though...
neon void
thats... worse
if is starting from the y axis (0 deg) it should stop just before the negative x axis
digital cliff
scenic jolt
There you go
But yea still trying to find angle x
This is advances math if anyone is curious
Please advise thanks
lusty pumice
scenic jolt
I'll wait any a few longer in case there any Geometry experts in here
Thabks
Got the answer
Thanks everyone
.done
cedar kilnBOT
@scenic jolt Has your question been resolved?
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foggy trout
Suppose that the sequence $c_k$ does not converge to 0. Show that the radius of convergence of $\sum_{k\geq 0}c_k(z-z_0)^k$ is at most 1
wraith daggerBOT
Alrik
foggy trout
I dont really know where to start to show this
If c_k doesnt converge to 0, it can either diverge or converge a complex number L that is not 0
If c_k diverges i dont think its meaningful to speak about a radius of convergence. If c_k converges to some complex number we should be able to use the root test or the ratio test, but I'm struggling to find an upper bound for these limits
dire geode
did you try using radius of convergence formula
sand cradle
If c_k diverges i dont think its meaningful to speak about a radius of convergen...
If c_k does not converge to 0 you can always claim something
Well, there are two cases: c_k -> L and c_k does not exist
In any case, you can pick a subsequence (c_(k_j)) for which for arbitrary epsilon, (c_(k_j)) >= epsilon for j >= N (just from the fact that it does not converge to 0)
\begin{proof} Let $(c_k)$ be a sequence that does not converge to $0$. Either it converges to $L$ or it does not converge. Pick a $\varepsilon > 0$ that is arbitrary. In any case, we can pick a subsequence $(c_{k_j})$ so that {\bf for all} $j \in \mathbb N$, we have $c_{k_j} \geq \varepsilon$. And then surely you can say [R=\frac{1}{\limsup {k\to \infty }|c{k}|^{1/k}} \leq \frac{1}{\limsup {j\to \infty }|c{k_j}|^{1/{k_j}}} \leq \frac{1}{\limsup_{j\to \infty } \varepsilon^{1/{k_j}}} = 1.] \end{proof}
And this gives you the answer, you can directly evaluate that last limit
(What is it?)
sand cradle
Yeah, epsilon^(1/k_j) -> 1
foggy trout
I see what you did
for sufficiently large N it will be less than epsilon, so you can use that to find something greater than c_k IF it converges to a L
That's very very clever
sand cradle
If c_k converges to 0, then all subsequences of c_k will converge to 0.
foggy trout
yes!
sand cradle
This is true, you can prove it and probably did (trivially)
foggy trout
And if it diverges its meaningless to talk abt radius of convergence
sand cradle
Wait, actually we can make it cleaner:
wraith daggerBOT
Kepe
sand cradle
The subsequnce bit is just to get rid of having to pick an N
Surely you could do it without that
foggy trout
Is it not the same thing..?
sand cradle
The alternative could go like this:
\begin{proof} Let $(c_k)$ be a sequence that does not converge to $0$. Either it converges to $L$ or it does not converge. Pick a $\varepsilon > 0$ that is arbitrary. In any case, we can pick a $N \in \mathbb N$ so that for all $n > N$, we have that $c_{n} \geq \varepsilon$. And then we have that [R=\frac{1}{\limsup {k\to \infty }|c{k}|^{1/k}} \leq \frac{1}{\limsup_{k\to \infty } \varepsilon^{1/{k}}} = 1] since $k \to \infty$ will exceed $N$ eventually \end{proof}
wraith daggerBOT
Kepe
sand cradle
So yes, it is basically the same thing
foggy trout
Okay yes that feels more like something ive seen before
the subsequence is just a part of the sequence?
sand cradle
Yes
foggy trout
Im not sure ive heard that term before, but it could be a language thing
okay yes!
thank you very very much
sand cradle
np
cedar kilnBOT
foggy trout
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✅ Original question: #help-13 message
sand cradle
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The definition of non-convergence is this: [\exists\ \epsilon > 0,\ \forall\ N \in \mathbb R\ \exists\ \mathbb N \ni n > N : |x_n - l| \ge \epsilon]
wraith daggerBOT
Kepe
sand cradle
So actually, we can't just say 'Pick varepsilon > 0' as I did at the beginning
Instead, we consider the varepsilon that is given by the non-convergence
foggy trout
so we'd split the proof in two parts? one for converging but not to zero and one for not converging?
sand cradle
Now we pick an arbitrary N in the first step. For this N, we can find an n_1 > N that makes the sequence greater epsilon. Then we consider N_2 > n_1 and do the same. And so on. And in the end, we really need a subsequence..
We can't get around that unfortunately
One sec, let me write it down
foggy trout
okay!
sand cradle
\begin{proof} Let $(c_k)$ be a sequence that does not converge to $0$. Per negation of the definition, we can find a $\varepsilon > 0$ so that for all $N \in \mathbb N$, there exists an $n > N$ so that $c_n \geq \varepsilon$. So, first pick $N_1 = 1$, then we find an $n_1 > N$ with $c_{n_1} \geq \varepsilon$. Then we pick $N_2 = n_1$, then find $n_2 > N_2 = n_1$ with $c_{n_2} \geq \varepsilon$ and so on. This gives us a subsequence $c_{k_j}$ with $c_{k_j} \geq \varepsilon$ for all $j \in \mathbb N$. And then we have that [R=\frac{1}{\limsup {k\to \infty }|c{k}|^{1/k}} \leq \frac{1}{\limsup {j\to \infty }|c{k_j}|^{1/{k_j}}} \leq \frac{1}{\limsup_{j\to \infty } \varepsilon^{1/{k_j}}} = 1] \end{proof}
wraith daggerBOT
Kepe
sand cradle
Now, it's right
\li \ The definition of convergence to $0$ would be that [\forall_{\varepsilon > 0} \exists_{N \in \mathbb R} \forall_{n > N}: |c_n - 0| < \varepsilon.] So the negation is [\exists_{\varepsilon > 0} \forall_{N \in \mathbb R} \exists_{n > N}: |c_n - 0| \geq \varepsilon.]
wraith daggerBOT
Kepe
sand cradle
(We need to switch all exists to forall and forall to exists and negate the last statement, that's how it's obtained)
foggy trout
Okay!
sand cradle
So yes, sorry about before, I was a little distracted - we really do need a subsequence
foggy trout
No worries! Thank youuu
sand cradle
My first proof is correct but without all the details that I gave now
The second one would be wrong where I tried to leave out subsequences as I did not correctly negate convergence
foggy trout
Got it
sand cradle
np
foggy trout
.close
pseudo merlin
cedar kilnBOT
pseudo merlin
how isthat the
normal
when its jus
OH
THE
formnulka
r dot n
i get it
ok thanks
bye guys
love you
cutie pies
sacred iron
Boy i leave seeing lightbulb moments’
pseudo merlin
cedar kilnBOT
dire geode
let w = z^3 then find w^(1/3)
pseudo merlin
let w = z^3 then find w^(1/3)
wait
is q supposde to be the power
but then isnt that the solutions to z^(1/q) according to the formula
dire geode
wherever you found this should define q for you
dire geode
¯_(ツ)_/¯
pseudo merlin
wait what is q
what the flip
im just gonna act like the 1/q is q isntead on the z
and then uset he formula normally
cedar kilnBOT
@pseudo merlin Has your question been resolved?
hollow trail
solving z^3 = 1 is equivalent to finding the value of 1^(1/3) isn't it
pastel vault
wait what is q
it's 3
wraith daggerBOT
south
pastel vault
(keep adding 2pi to the angle)
and then you just take the cube root of 1 for the magnitude
and divide the angles all by 3, to find z
zinc quarry
The problem is asking if these are consistent, but I don’t even know where to start
zinc quarry
When we did it in class, we made a truth table but there’s too many propositions and variables here that I think it would get extremely large, so how can I do this problem not using a truth table?
last apex
You would start piecing what are the truth values of p,q,r,s
And see if there are any contradictions
You know s is false
So (not p) must be false by 4
So p must be true
zinc quarry
So (not p) must be false by 4
What do you mean false by 4
last apex
4 tells you (not p) implies s
zinc quarry
which means 4 cant be true because it would lead to a false conclusion?
wait
i get what youre saying
im gonna back up a second
night wedge
4 can be true if the not(p) is false
zinc quarry
4 can be true if the not(p) is false
yeah ok that makes sense
night wedge
False -> false is okay
night wedge
Yeah so you can find what p is
zinc quarry
alright so by that same logic, does that mean that (not q) is also F because 3 is (not q) > s
last apex
Yes
zinc quarry
ok sweet so
now on 2, not p is F, so it doesnt matter what r is
and same for 1
right?
becuase the hypothesis isnt going to be T so it cant possibly make the conditional false
so therefore its consistent
last apex
1 is true becuase its F-->T
last apex
r will be the same as (not p)
zinc quarry
oh wait
yeah
and not p is F, so r is F, and 2 Fs in a biconditional make it T
and because all statements agree with each other, the system is consistent
right?
cedar kilnBOT
@zinc quarry Has your question been resolved?
zinc quarry
I think so
last apex
Yes
idle sky
how do i do the show that Pn is a polynomial of degree n part. i tried induction but goes no where, as it works for n = 1
green coral
binomial ?
sand cradle
how do i do the show that Pn is a polynomial of degree n part. i tried induction...
Let's call $\operatorname{deg}$ the term of a polynomial with the highest degree. By the binomial theorem, surely [\operatorname{deg}\l((x^2 - 1)^n\r) = \deg\l(x^{2n}\r). \tag{$$}] You can see that [\frac{\dd}{\dd x} \operatorname{deg} (P(x)) = \operatorname{deg}\l(\frac{\dd}{\dd x} P(x)\r)] easily (write out $P(x) = a_0 + a_1x + \cdots + a_nx^n$ for this and verify). Now with this notation, we can say: [\deg \l(\frac{\dd^n}{\dd x^n} (x^2 - 1)^n\r) = \frac{\dd^n}{\dd x^n} \operatorname{deg}\l((x^2 - 1)^n\r) \overset{()}= \frac{\dd^n}{\dd x^n} \operatorname{deg}\l(x^{2n}\r) = \operatorname{deg}\l(\frac{\dd^n}{\dd x^n} x^{2n}\r)]
idle sky
oh right can i justify by saying before differentiating the polynomial is x^2n, when u differentiate n times the p_n(x) becomes x^n degree
is this university content? im preparing for a university admissions test so havent seen it but think i get it
sand cradle
This is probably more rigorous than you need to be then
wraith daggerBOT
Kepe
sand cradle
interesting
And now for that last bit, you just care about differentiating x^(2n) n times
Which will give you 2n(2n-1)(2n-2)...(n+1) x^n
wraith daggerBOT
Kepe
sand cradle
And now well, you are done
sand cradle
appreciate it
Less rigorously, you can say this:
(x^2 - 1)^n is of degree 2n (binomial theorem)
Each time you take the derivative, the degree of a polynomial decreases by 1
So after taking the derivative n times, you will be at degree n
This is what I wanted to make rigorous in the above, but if this is not expected then you can just argue like that
idle sky
okok ty
sand cradle
np
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void sand
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sacred iron
cerulean sail
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wide elm
What is going on here?
cedar kilnBOT
What is going on here?
wicked mantle
.solved
cedar kilnBOT
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✅ Original question: #help-13 message
wicked mantle
wide elm
Logarithm's constituents cannot be separated in this manner
So, how is the second equation achieved?
cedar kilnBOT
@wide elm Has your question been resolved?
cedar kilnBOT
@wide elm Has your question been resolved?
dire geode
that's some awful formatting
sand cradle
How did they arrive at the second equality?
That's just pulling out the 1/h
\li [{\scriptstyle \lim _{h\rightarrow 0}\frac{\log _{\sec (-\frac{\pi }{3}+2h)}\left|\cos 4\left(-\frac{\pi }{6}+h\right)\right|+\left|\sin \left(-\frac{\pi }{6}+h\right)\right|-\left(\log _{\sec (-\frac{\pi }{3})}\left|\cos \left(-\frac{2\pi }{3}\right)\right|-\left|\sin \left(-\frac{\pi }{6}\right)\right|\right)}{h}}] And this becomes [\scriptstyle \lim _{h\rightarrow 0}\frac{1}{h}\left(\log _{\sec (-\frac{\pi }{3}+2h)}\left|\cos \left(-\frac{2\pi }{3}+4h\right)\right|-\log _{\sec (-\frac{\pi }{3})}\left|\cos \left(-\frac{2\pi }{3}\right)\right|\right)+\lim _{h\rightarrow 0}\frac{1}{h}\left(\left|\sin \left(-\frac{\pi }{6}+h\right)\right|-\left|\sin \left(-\frac{\pi }{6}\right)\right|\right)]
cedar kilnBOT
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cedar kilnBOT
@wide elm Has your question been resolved?
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pseudo merlin
How would I do this
cedar kilnBOT
How would I do this
hollow trail
so if you expand it out into x = ... and y = ... you have a system of equations involving x, y, and lambda
the goal is to combine into a single equation which does not involve lambda
pseudo merlin
oh
chrome elk
pseudo merlin
ok thanks
proud pawn
guys i have my maths exam next week
proud pawn
bruh
muted bear
omg i had four last week
proud pawn
same to samw
tropic oxide
guys i have my maths exam next week
!occupied
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muted bear
best of luck on your exams
proud pawn
yeah i need help for my maths exam
pseudo merlin
i started studying yesterday
tropic oxide
yeah i need help for my maths exam
!occupied
cedar kilnBOT
yeah i need help for my maths exam
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muted bear
get an open help channel please
pseudo merlin
.close
cedar kilnBOT
proud pawn
what
hollow trail
!redir
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This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.
livid hound
people are willing to assist if you follow the guidelines of the server
muted bear
you have not read a single direction we have given you
proud pawn
then what is use of this channel
hollow trail
this channel is specifically for bum chicken's problem. you can open your own channel if you want to get your own help
tropic oxide
then what is use of this channel
this was for somebody else's question and had somebody else's name on it
you need to claim your own channel
proud pawn
got it
tropic oxide
our bot right now is just telling you the same info in a condensed format
tropic oxide
read #❓how-to-get-help
proud pawn
idk\
chrome elk
I tagged you in an available channel
cedar kilnBOT
Available help channel!
dim tundra
please help with this question
cedar kilnBOT
please help with this question
tropic oxide
do you know what each of those words in the answer options mean
tropic oxide
ok then go study these.
tropic oxide
we can't really give you any pointers without giving out the answer
but you're saying you do not even know "rotation"?
tropic oxide
ok
dim tundra
and reflection
tropic oxide
can you look at the shapes and tell me whether these look like rotated versions of each other
dim tundra
just translation and dialation
dim tundra
can you look at the shapes and tell me whether these look like rotated versions ...
it does
oh
?
vague burrow
please help with this question
translation means you move the object without changing its directions
so in this case , when you translate A , the pointy end will always point in the same direction , even though you move it
rotation means you rotate the object
so in this case , when you rotate A , you can change the direction of the pointy end
reflection means you take a mirror image of the object
dilation means you increase or decrease the size of the object , keeping the proportions
which is like zoom in and zoom out
in this case , even after dilation , the pointy end of A , will always point in the same direction
so the question is, which operation you need to perform on A , so that it looks like B
dim tundra
OHHHH
dim tundra
translation means you move the object without changing its directions
so in this...
TYSM!!!
so this must be rotation
vague burrow
YUP !
dim tundra
ty!!!
vague burrow
welcome !
torn marsh
cedar kilnBOT
oblique flare
you didn't even complete number 2 and it looks like you are going in the right direction
for number 3 you multiplied by a negative wrong
tropic oxide
this is called going in circles.
you had 2x - 3 = (NUMBER)
you need to break out of samsara, solve for x, and achieve enlightenment
torn marsh
you had `2x - 3 = (NUMBER)`
That number is log base 10 of 666
tropic oxide
yes sure
it is some log shit
that doesnt stop it being a NUMBER.
you solve this equation the same way you would solve 2x - 3 = 42069
torn marsh
ohh
let me work on it
That number is an irrational
Should I include all of the digits
oblique flare
Should I include all of the digits
Sure but you won’t finish the question doing that
torn marsh
Should I include all of the digits
Actually, I know the answer to this.
tropic oxide
and what is this answer you say you know
torn marsh
and what is this answer you say you know
here you go
tropic oxide
the question said to round to 2dp?
torn marsh
yes
tropic oxide
ok
torn marsh
.close
crimson sedge
cedar kilnBOT
crimson sedge
Is the proof right for the -v is (-1)v
last apex
You say v + (-1)v = 0... unless you already proved this you should begin from the axioms
cedar kilnBOT
@crimson sedge Has your question been resolved?
crimson sedge
You say v + (-1)v = 0... unless you already proved this you should begin from th...
Ahhhhhh
last apex
1v + (-1)v = (1 + (-1))v = 0v = 0
quasi oriole
Can anyone check what i got wrong?
cedar kilnBOT
Can anyone check what i got wrong?
rich iron
Can anyone check what i got wrong?
wait
north wyvern
case 2 counts cases in case 1 i think
quasi oriole
The "choose the required length" line alr counted for that .... For all possible positions of a line length a, choose two distinct positions (That is the (6-x)A2)
It couldn't match the same space for both hor and ver
So a rectangle would be impossible
north wyvern
<@&268886789983436800> scammers
cedar kilnBOT
@quasi oriole Has your question been resolved?
cedar kilnBOT
north wyvern
.reopen
cedar kilnBOT
✅ Original question: #help-13 message
north wyvern
cases 1 and 2 work, its definitely a problem with case 3
cedar kilnBOT
@quasi oriole Has your question been resolved?
north wyvern
oh i see the issue
north wyvern
i think the second part in case three is overcounting 4 times, as opposed to once
hmm but that still doesnt give 2159
nocturne spear
cedar kilnBOT
nocturne spear
how is the adjacent side -3/2
for e
this is the acceleration vector
so the 3d vector would look like this from the side?
so to find the angle it makes with positive j, we would do arccos(0.5 / hyp of a) + 90 ?
nocturne spear
cobalt iron
Well its 2D representation
Well the angle is the supplementary of what u have taken
Theta should be in between resultant line u have drawn and the y axis
nocturne spear
thanks
crimson sedge
Can someone draw 17? I dont understand what does “on different sides from line of action of vector F” mean
crimson sedge
wicked mantle
tropic oxide
crimson sedge
Is “but on different sides from the line of action of vector F” supposed to mean anything?
onyx cove
You can transpose F1 to the end of F2 and use Pythagorean theorem
crimson sedge
You can transpose F1 to the end of F2 and use Pythagorean theorem
I did that but i don’t really have side lengths to use pythagorean theorem, idk what to do next
languid charm
I did that but i don’t really have side lengths to use pythagorean theorem, idk ...
Youre comparing F1 to F2, so you can use a trig function with the angle given
onyx cove
First we have:
(F1)^2+(F2)^2=F (Using the theorem)
So change F for his equivalent using (F1) and (F2)
crimson sedge
Youre comparing F1 to F2, so you can use a trig function with the angle given
okay yeah i think i got it
I think its F1 = 2/rt 3 F2
alrriight thank you smmm every1
.close
cedar kilnBOT
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Available help channel!
nocturne spear
cedar kilnBOT
tropic oxide
one moment, this is best explained with a diagram
nocturne spear
i tried to draw one but i am horrendous at drawing 3d vectors
tropic oxide
so this is like, not remotely to scale or anything but it will do fine to illustrate my pt
you decompose vector $\ora{BA}$ (red) into the sum of two vectors: one parallel to $\ora{BC}$ (green) and one perpendicular to it (blue).
then you negate the blue part, add it back to the green, and that's your $\ora{BD}$.
wraith daggerBOT
Ann
tropic oxide
mathematically you can write down ig something like this:
let $\ora{BA} = \lambda \ora{BC} + \bd{p}$, where $\lambda \in \bR$ and $\bd{p} \cdot \ora{BC} = 0$. (there exists exactly one value of $\lambda$ that makes this work; find it.)
then $\ora{BD} = \lambda \ora{BC} - \bd{p}$.
wraith daggerBOT
Ann
nocturne spear
you decompose vector $\ora{BA}$ (red) into the sum of two vectors: one parallel ...
this makes sense
tropic oxide
btw this also doesnt give a shit how many dimensions youre working with
all thats required is that you know how to take dot products
cedar kilnBOT
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Available help channel!
cloud mango
why is x from 0 to 2?
cedar kilnBOT
why is x from 0 to 2?
jade blaze
why is x from 0 to 2?
its the first octant
so z will be greater than 0
and so will be x and y
slender ginkgo
jade blaze
so you can out those equations greater than 0
and you will get values of x and y in some range
cloud mango
doesnt that mean that both x and y will be a range from 0 to 2?
jade blaze
doesnt that mean that both x and y will be a range from 0 to 2?
no only one will be
strange venture
jade blaze
and then we will take other as variable of another
cloud mango
and then we will take other as variable of another
i see.. so i can technically say x = 2-y and y ranges from 0 to 2?
jade blaze
yess
cedar kilnBOT
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Available help channel!
queen stirrup
Determine all polynomials P(x) with real coefficients satisfying (P(x))^n = P(x^n) for all x ∈ R, where n > 1 is a fixed integer.
queen stirrup
I know that the leading coefficient is 1,-1
It cant have any roots other than 0,1,-1
And I think the form of the polynomial solution is -1,0,1,x^k,-x^k
basically I need to prove that all the coefficients are 0 other than the leading one
mental trail
It cant have any roots other than 0,1,-1
Not necessarily, real coefficients doesn't imply real roots
queen stirrup
Not necessarily, real coefficients doesn't imply real roots
Yeah I meant it as if it has any real roots then its gotta be one of those
mental trail
Yeah I meant it as if it has any real roots then its gotta be one of those
and I'm telling you you're missing some xd
queen stirrup
Oh and I mean real roots
mental trail
well, ok, the only real roots can be 0,-1,1
that doesn't tell you much about the polynomial still
mental trail
roots of unity, yes. Not necessarily n-th roots of unity (at least we can't tell yet), but roots of unity
or 0 of course
queen stirrup
basically a^n = a
mental trail
now the idea to pursue is to show there are no roots of unity that are roots of our polynomial
queen stirrup
now the idea to pursue is to show there are no roots of unity that are roots of ...
Not sure how I would go about doing that
Are you saying any root of unity?
mental trail
I'm finding out that there's an easier way
wait brb
Ok so
There might be a way to reason with root multiplicity if we were factoring
mental trail
yep
queen stirrup
all roots as we know are roots of unity
perhaps that helps
n-1 roots of unity
any factored terms that represent a root of 0 cancel out
mental trail
ok I think I found a way, using principal root
queen stirrup
Whats a principal root?
mental trail
like, looking for roots of a complex number
you take the principal root = the only root with argument in some specific range
queen stirrup
|√ω| ?
mental trail
√ω only
|√ω| would be its module
which we don't really need bc it's always 1 here
queen stirrup
you take the principal root = the only root with argument in some specific range
oh and how do we define that range
mental trail
oh and how do we define that range
well for example to find the square roots of $\omega$
wraith daggerBOT
Raphaelisius Maximus MMIII
mental trail
one root will have argument in $(-\frac \pi 2,\frac \pi 2]$ and the other will have argument outside that
wraith daggerBOT
Raphaelisius Maximus MMIII
mental trail
you just split the trig circle exactly in half
queen stirrup
all right
mental trail
and you're guaranteed that one root is inside, the other outside
(because the other root has argument pi + the first one)
we can do the same for k-th roots, etc...
queen stirrup
ok I get it now
wraith daggerBOT
Raphaelisius Maximus MMIII
queen stirrup
yes
mental trail
then $P$ is divisible by $(x-\omega)$
wraith daggerBOT
Raphaelisius Maximus MMIII
mental trail
so $P(x^n)$ is divisible by $(x^n-\omega)$
wraith daggerBOT
Raphaelisius Maximus MMIII
mental trail
but that's (P(x))^n
so any n-th root of omega is also a root of P
If omega = 1, take any one of those and call it omega_1
otherwise, using principal root, you can find one, call it $\omega_1$, such that $0 < Arg(\omega_1) < Arg(\omega)$
queen stirrup
yeah i know what argument is
mental trail
sorry
wraith daggerBOT
Raphaelisius Maximus MMIII
queen stirrup
alright
mental trail
repeat all we did with omega, this time with omega_1
mental trail
well, i wouldn't call it like that
queen stirrup
so the only roots P(x) has is 0 with multiplicity
or P is constant
That was pretty good
queen stirrup
so any n-th root of omega is also a root of P
but let me just try to understand this better
queen stirrup
but that's (P(x))^n
actually
can we use this given that the relation is only true for all real x
queen stirrup
Determine all polynomials P(x) with real coefficients satisfying (P(x))^n = P(x^...
original q said that its true for all real x
mental trail
can we use this given that the relation is only true for all real x
exercise : show that it's also gonna be true for all complex x
queen stirrup
uh
mental trail
(hint : the polynomials Q(x) = (P(x))^n and R(x) = P(x^n) are congruent on an infinite number of points)
queen stirrup
:| it was really so simple
both have degree nk
yeah so Q(x) is congruent to R(x)
if I need to be more formal Q(x)-R(x) has infinite roots
but have degree <=nk
last apex
exercise : show that it's also gonna be true for all complex x
I thought of the identity theorem for holomorphic functions 
night wedge
What’s wrong with using that
last apex
Overkill
queen stirrup
I tried to find a solution with galois theory to the og question
did you find one
last apex
No
night wedge
Any complex roots gotta be roots of unity
I still can’t see why this is the case
queen stirrup
I still can’t see why this is the case
right
night wedge
Oh nvm
queen stirrup
you saw it?
night wedge
If z is a root z^n is and then you get an infinite chain
queen stirrup
exactly
queen stirrup
so any n-th root of omega is also a root of P
is that because any nth root of omega is a root of P(x^n)
yeah sorry i got it
thanks
but I'm pretty sure theres a solution that is not so involved in complex numbers
.solved
cedar kilnBOT
mental trail
but I'm pretty sure theres a solution that is not so involved in complex numbers
Probably differentiating the equality
but it's messy
queen stirrup
Probably differentiating the equality
i did that
didnt see anything myself at least
mental trail
If P(0) is not 0, it's about showing x coefficient = 0, x^2 coefficient = 0, etc...
mental trail
say deg(P) <= k, with k > 0
queen stirrup
.reopen
cedar kilnBOT
✅ Original question: #help-13 message
mental trail
and P(0) is not 0
queen stirrup
so $P(x) = \sum_{i=0}^{k}a_ix^i$
wraith daggerBOT
CherryMan
night wedge
a_0\ne0 and a_k\ne0
queen stirrup
we'll show that it is impossible
that P(0) is not 0??
mental trail
that P(0) is not 0??
that deg(P) > 0 when P(0) is not 0
(btw, in that case P(0) = 1 or (-1)^n)
so, nx^(n-1) P'(x^n) = n(P(x))^(n-1) P'(x)
queen stirrup
$\frac{d}{dx}[P(x)]^n = \frac{d}{dx}P(x^n) \implies nP(x)^{n-1}\cdot P'(x) = \sum_{i=1}^{k}nia_ix^{in-1}$
oops
queen stirrup
so, nx^(n-1) P'(x^n) = n(P(x))^(n-1) P'(x)
is that chain rule application correct
nvm sorry
queen stirrup
my mistake again
wraith daggerBOT
CherryMan
queen stirrup
it was correct the firs ttime...
night wedge
And the right should be nx^(n-1)P’(x^n)? yeah use a sum
Maybe
Unless you want to do this sum thing
Tbh idk how my rhs is helpful
queen stirrup
~~And the right should be nx^(n-1)P’(x^n)?~~ yeah use a sum
whyd you strike that
night wedge
It’s not wrong (i think) just not useful
mental trail
so, nx^(n-1) P'(x^n) = n(P(x))^(n-1) P'(x)
If you want to continue from there, just realize that LHS is 0 and RHS is (constant)P'(0), when x = 0
queen stirrup
If you want to continue from there, just realize that LHS is 0 and RHS is (const...
P'(0) = 0 is true for any polynomial
mental trail
P'(0) = 0 is true for any polynomial
???
fallow drift
cedar kilnBOT
fallow drift
can someone please check if i used the comparison theorem correctly
night wedge
At the start of step 2 you put = but they aren’t equal
(For 5a)
Also you are integrating over $(0,\infty)$ not $(1,\infty)$ so that doesn’t work
Also, $$\int_{0}^{\infty}\frac{1}{x^{2}}=\infty$$
night wedge
So we need a different upper bound on 1/(x^2+1) for (0,1)
cedar kilnBOT
@fallow drift Has your question been resolved?
fallow drift
Also you are integrating over $(0,\infty)$ not $(1,\infty)$ so that doesn’t work
it does not???
fallow drift
Also, $$\int_{0}^{\infty}\frac{1}{x^{2}}=\infty$$
😭
how should i solve that question..
night wedge
On (0,1) just let the upper bounding function be 1
Then you can use your 1,infinity = converges then add 1
fallow drift
On (0,1) just let the upper bounding function be 1
sorry im so confused
so i break the interval into 2? 0-1 and 1-inf
night wedge
Yeah
cedar kilnBOT
fallow drift
.reopen
cedar kilnBOT
✅ Original question: #help-13 message
fallow drift
@night wedge
can you explain
what you mean
cuz like
would i get
a definite number for 0-1
then an indefinite number for 1-inf