im just so confused
#help-19
midnight mauve
i just noticed the denominators are different too
can someone please explain this whole problem to me from start to finish. and ignore the worked solution because i found that online and im not sure its correct bc its answer doesnt match the textbook solution
odd edgeBOT
@midnight mauve Has your question been resolved?
midnight mauve
do i need to do partial fractions
odd edgeBOT
@midnight mauve Has your question been resolved?
odd edgeBOT
@midnight mauve Has your question been resolved?
brittle drift
Im stuck on how to start this problem
odd edgeBOT
Im stuck on how to start this problem
@brittle drift Has your question been resolved?
brittle drift
<@&286206848099549185> Any help is appreciated C:
midnight mauve
i believe you implicity differentiate the given y and just plug in 3
brittle drift
let me try that
midnight mauve
i tried it like this see if u work it out the same way
brittle drift
i tried it out and it worked out the same! I got 1 as my answer
midnight mauve
yes i got 1 for part a and 25 for part b
brittle drift
for part b, did you the same process but with x?
upper onyx
$\frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}$
clever fjordBOT
ThM
midnight mauve
for part b, did you the same process but with x?
i deleted my work but yeah basically i solved for x first and then continued implicit differentiating the same way
clever fjordBOT
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brittle drift
$y'x = 1/2(2x+1)^-1/2*2$
clever fjordBOT
nicheo
midnight mauve
did you solve for x first?
i dont know how to use texit but it should be x = ( 1/2 )( y^2 - 1 )
brittle drift
oohhhhhhh
midnight mauve
and then u can differentiate
i would write it out for u but my ipad is charging now 😔
brittle drift
so i before diff i need to rework the x to be in the y's place icic
midnight mauve
yup that would make it a lot easier
brittle drift
implicit diffs are difficult :C let me try that!
midnight mauve
wait i might be setting u up for failure give me 1 second LMFAO
midnight mauve
how did you get remove the x from the square root?
you square both sides of the equation
squaring gets rid of the sq root
brittle drift
on the second line, what's the process to getting there? im trying to see how you got it but im stumped
midnight mauve
the second line like where x' starts?
brittle drift
the one above that one ^
midnight mauve
oh ok ok ok
brittle drift
the x = 1/2(y^2-1)
yayayya C:
OH
I FIGURED IT OUT !
let me diff and see what i get
midnight mauve
okiee
odd edgeBOT
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brittle drift
I’m currently stuck on part b of this question
odd edgeBOT
I’m currently stuck on part b of this question
brittle drift
heres what i have so far
ember oak
heres what i have so far
Looks good so far. Write your exponents in terms of radicals and fractions and try to factor
sullen ferry
note critical values can also be when the function has derivative that is undefined
brittle drift
oooo ok let me try and rewrite eq
odd edgeBOT
@brittle drift Has your question been resolved?
brittle drift
ok i converted it into this, would factoring out 1/2 be a good idea?
i had some extra stuff to do so i had to take a break from math for a lil bit 😓
.close
odd edgeBOT
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thorny ivy
BC is BC is a diameter
GA = AC
ABC is blocked in the circle
Triangle GAB=Triangle CAB
I need to find that there is 2 Degrees that equal to each other in GBC and GAD
thorny ivy
i already found that Degree BGC of triangle GDA is equal to Degree BGC of triangle GCB
because they share it
although i cant find a way to get another degree
<@&286206848099549185>
ruby nacelle
i already found that Degree BGC of triangle GDA is equal to Degree BGC of triang...
use isosceles triangle and you find AG = AD
thorny ivy
how do you know that
thorny ivy
wait
i dont see
oh
the line thing equals to the third line
i forget the sentence
ruby nacelle
with AG = AD, you know AD = AC, and use eq. chords, eq. chords and you know angle GBA = angle ABC
thorny ivy
with AG = AD, you know AD = AC, and use eq. chords, eq. chords and you know angl...
there is nothing like that
how do you know AG=AD
there is no equal degrees for it to be a isosceles triabgle
<@&286206848099549185>
ruby nacelle
i already found that Degree BGC of triangle GDA is equal to Degree BGC of triang...
bro u said BGC = GDA
thorny ivy
i already found that Degree BGC of triangle GDA is equal to Degree BGC of triang...
i meant GCB
odd edgeBOT
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simple oar
i have 6 matching pairs of shoes in how many different combinations could i select 2 pairs of not matching shoes from the 6 pairs
simple oar
<@&286206848099549185>
odd edgeBOT
@simple oar Has your question been resolved?
spare kelp
i have 6 matching pairs of shoes in how many different combinations could i sele...
I am assuming there is no distinction of left or right shoes. Choose a pair among all of the pairs and pick a shoe (this is 6). Then choose a different pair amongst the remaining pairs and complete a non matching pair (this is 5). Here, we have 6x5 = 30. Then, do the same thing again. There no pairs that are completely gone, (and assuming the is no distinction in a pair of shoes) it is still 6x5 = 30. So, that makes 30x30 = 900
spare kelp
but the answer greatly changes depending on the initial conditions of the question.
.close
odd edgeBOT
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sleek moth
Can anyone tell me if the reasoning is correct?
odd edgeBOT
Can anyone tell me if the reasoning is correct?
gray whale
Hey there
sleek moth
Hey
sleek moth
Sure, thanks
sleek moth
yea
sleek moth
Yes
sleek moth
On case 1 i only counted digits
gray whale
Mhm
sleek moth
This questions is bricking my brain
sleek moth
What number did u got?
gray whale
129024
gray whale
Np
sleek moth
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odd edgeBOT
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urban gale
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proven saffron
thats what i did
fickle wasp
is the following correct?
$\int{}^{}\sqrt{x^2-4x+3}dx=\int{}^{}\sqrt{(x-2)^2-1}dx = \int{}^{}\sqrt{u^2-1}du=u/(arccosh(u)) + C = (x - 2)/(arccosh(x-2))+C$ since $\int{}^{}1/(\sqrt{x^2-1})dx = (\int{}^{}1dx)/(\int{}^{}\sqrt{x^2-1})dx = arccosh(x)+C \Rightarrow \int{}^{}\sqrt{x^2-1}dx = x/(arccosh(x)) + C$
clever fjordBOT
erxuanli
vivid girder
I don't recall this statement to be true, check again just in case
fickle wasp
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odd edgeBOT
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hearty cape
odd edgeBOT
hearty cape
what does 1/x represent ?
random sigil
notice that if the whole fraction is x then what I circled is also x
boreal crag
they did
mystic saffron
yeah my bad
random sigil
cause its infinite
boreal crag
ah wait nvm
iron bear
no don't nvm
dawn scarab
notice that if the whole fraction is x then what I circled is also x
I‘d say look at this and make sure you get it
hearty cape
I do not
dawn scarab
If you consider 1 + fraction as x, then you‘ll notice that there is precisely 1+fraction within x
(The circled part)
Maybe let‘s do another example that might be easier to see
$\sqrt{x \sqrt{x \sqrt{\dots}}}$
clever fjordBOT
ℑμΤ𝛄𝛗θ
dawn scarab
If you call the whole thing, say, "y", you‘ll notice that y is inside of itself. Therefore you‘d have $y = \sqrt{xy}$
clever fjordBOT
ℑμΤ𝛄𝛗θ
hearty cape
yes
dawn scarab
Great! Well, this is what‘s happening with your fraction
You just have to see what part is repeated and give it a name
hearty cape
if the whole thing can be denoted as x why put it as 1/x
odd edgeBOT
@hearty cape Has your question been resolved?
hearty cape
\
winter spruce
I need to check for what values of a and b this function is continuous
winter spruce
I'm stuck here because I'm not sure how to approach this limit here
mystic saffron
I need to check for what values of a and b this function is continuous
$\lim_{x \to 0^+} a\left(\sin^2(x)\right)^{\frac{1}{2\log{x}}$?
clever fjordBOT
45
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
odd edgeBOT
@winter spruce Has your question been resolved?
winter spruce
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odd edgeBOT
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limber sequoia
How do I simplify: $square root(1+cos3pi/8)/2)$
odd edgeBOT
How do I simplify: $square root(1+cos3pi/8)/2)$
clever fjordBOT
Zexarium
mystic saffron
How do I simplify: $square root(1+cos3pi/8)/2)$
cos(2a)=2cos^2(a)-1
warm nacelle
cos(2a)=2cos^2(a)-1
cos(2a)=2cos^2(a)-1
mystic saffron
oh mb
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jagged sable
can someone send me the formulas for like apothem??
odd edgeBOT
can someone send me the formulas for like apothem??
jagged sable
like idk how to ask the qs but like
i need the formulas for radius of circumscribed circle
and apothem of hexagon
idk bro
i'll send a qs
wait nvm sorry guys
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safe flower
odd edgeBOT
safe flower
This more of a calculator qwestion
What formula do I input to get this in a graphing calculator
odd edgeBOT
@safe flower Has your question been resolved?
safe flower
I figured our the question by doing it manually, but I'd like to know the formula for doing this on a graph calculator
odd edgeBOT
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odd edgeBOT
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limber willow
List all derangements of {1,2,3,4,5} where the numbers are 1, 2, and 3 all occur (in some order) before 4 or 5,that is, 1, 2, and 3 are the first 3 numbers in the derangement.
limber willow
I got (2, 3, 1, 5, 4) & (3, 1, 2, 5, 4). Is this correct?
odd edgeBOT
@limber willow Has your question been resolved?
wintry yew
Looks correct to me
calm raft
I dont really understand where the C comes from... From what I know the likelihood function is just the joined pmf for all values of x given a parameter theta
calm raft
I get that C means LHS and RHS are proportional, but idk why
green elm
it's a definition, it's saying that the scale factor doesn't matter as long as it's positive
any positive scalar multiple of a pmf/pdf is a valid likelihood function
the reason the scale factor doesn't matter is that you're generally looking to maximize it, and the maximum will occur at the same point regardless of scale factor
green elm
if you multiply it by a negative number, then the max becomes a min
which is ok as long as you remember to minimize the result instead of maximizing it
but since your goal is to "maximize" the "likelihood" you don't want a negative multiple, otherwise the names don't make sense anymore
calm raft
I see
its more convenient that way
so technically we can call it a "minimum likelihood estimation" if the scale factor C is smaller than 0?
green elm
it wouldn't really make sense to call it "likelihood" in that case, some people call it a "loss function" if your goal is to minimize it
green elm
it's all just wording, the math doesn't care
calm raft
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zinc mantle
odd edgeBOT
zinc mantle
what did I get wrong here
mystic saffron
I believe 3
mystic saffron
What is a linear pair?
mystic saffron
Nope has to be 180, because 180 degrees makes a straight line(ar)
So yep, always true
mystic saffron
np
odd edgeBOT
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dawn tiger
!15mins
odd edgeBOT
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
dawn tiger
for each digit, you have 10 choices: the digits 0 through 9
since you can repeat the digits, you'll have 10 choice every time
so you have 10 passwords of length 1
yeah
10^4 passwords of length 4
yeah 10000 different passwords
yes
yes
1234 is 1 of the 10000 possible passwords, that's one way to think about it
the other way is that for each digit, you have a 1/10 chance of getting the right digit to match the password
so (1/10)^4
yes
yes
odd edgeBOT
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stray ivy
need more help
odd edgeBOT
need more help
stray ivy
can someone check this?
@dawn tiger could u help me again?
are there any other answers for c?
<@&286206848099549185>
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odd edgeBOT
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pastel orbit
odd edgeBOT
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
1
I literally have no idea where to even begin constructing such a set
any suggestions are appreciated 
green gorge
there's a really cursed way to do this
using cauchy's functional equation
but their hint gives you a pretty good starting point i think
pastel orbit
using cauchy's functional equation
unfortunately, I don't know what that is and Spivak hasn't yet said anything of it
green gorge
try just picking one point from each of these fractions of the square
pastel orbit
okay, I'm seeing that if we're gonna have [0,1] X [0,1] as our boundary, we have to have a point in every single smaller square
if we don't, then the boundary will never be the whole square
green gorge
yea
pastel orbit
but this doesn't give me a systematic way of constructing A, does it? 
green gorge
idk how to convey this without just telling you outright sorry that im being vague
green gorge
in this case the limit doesn't really work
after all what is the limit of a sequence of sets
pastel orbit
I was thinking more like a sequence of partitions
but idk how to make that rigourous either 
so... maybe not
green gorge
a good way to work with infinite families of sets
either take their union or intersection
because these are easily defined :D
can u construct a family of sets such that
- each subsection with length 1/2^k is hit eventually
- no two points have an identical component
pastel orbit
green gorge
if we have like a family of sets A_n then all we have to do is ensure that A_k doesn't "conflict" (share any points on the same line) with any of the sets from A_1 to A_k-1
and then the infinite union of all A_n won't share any coordinates
so the goal really is a family of sets such that
- each subsection with length 1/2^k is hit eventually
- for any set A_k in this family, A_k doesn't conflict with any of the sets A_1 to A_k-1
i think this is a pretty good hint
pastel orbit
wait, what do you mean by “hit eventually”?
green gorge
like
we divide it into quarters
then sixteenths
then 64ths etc.
(each has sidelength 1/2^k which is why i said that)
pastel orbit
(each has sidelength 1/2^k which is why i said that)
right
green gorge
and we need to ensure that in each of these there's a point
so by "hit" i mean "make sure there's a point from our family in the set"
pastel orbit
ah, I see what you mean now
I’m gonna have to take some time to mull this over though…
pastel orbit
thanks, and thank you for the hints! 
I’ll close this channel now, and try it myself
.close
odd edgeBOT
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quaint urchin
context
w = wheelbase
t = track width
x = center steer angle
I began experimenting with the ackermann formula arctan((w*tan(x))/(w+t*tan(x))) because it suffered a problem in which it only spanned from -pi to pi on the y axis. This caused an inversion past certain angles that I wished to get rid of.
In an attempt to remedy this, I added pi*floor((x+arctan(w/t))/pi) this makes the function continuous, but seems very computationally expensive. I would like to ask if there's a way I could simplify this expression, since the use of floor seems to prevent any simplification.
odd edgeBOT
```
context
w = wheelbase
t = track width
x = center steer angle
```
I began ex...
@quaint urchin Has your question been resolved?
odd edgeBOT
@quaint urchin Has your question been resolved?
odd edgeBOT
@quaint urchin Has your question been resolved?
odd edgeBOT
@quaint urchin Has your question been resolved?
gleaming schooner
In order to construct a rectangular box, it costs 5$ per square foot to construct an xy face, 1$ per square foot to construct a yz face, and 3$ per square foot to construct an xz face. Total budget = 180$.
Whats the maximum volume possible with this budget?
gleaming schooner
so basically i have the answer and solution, this question was asked in a midterm and i just wanna know if I answered it correctly
modern gyro
hm
clever fjordBOT
TheRuleOfEngineering
gleaming schooner
just wanna know if there's anything wrong in here
coz i wrote that answer in the midterm and im scared lmao
i appreciate anyone who reads through my solution to find flaws
gleaming schooner
ember oak
All your algebra is fine
gleaming schooner
our textbook solves problems considering both cases where lambda = 0 and non-zero
in this case, from the nature of the equations, lambda=0 would mean either x,y,z = 0 which cannot happen in a rectangle
gleaming schooner
our textbook solves problems considering both cases where lambda = 0 and non-zer...
i realized this after i wrote the midterm so idk if the prof will take away credit for not mentioning this
ember oak
our textbook solves problems considering both cases where lambda = 0 and non-zer...
It would just be a damn cheap box
gleaming schooner
lmao
gleaming schooner
All your algebra is fine
woah u got the Advanced Math role so ur above undergrad i see
in that case i guess its enough verification
.close
odd edgeBOT
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elfin bear
what can I do to determine whether this expression is always true for every sets A, B, C?
elfin bear
I think that just trying to think up edge cases where one of then is false (as in {1,2...}) is not what I should be doing
weary pelican
what can I do to determine whether this expression is always true for every sets...
If A is true, then what does the expression give?
If A is false, same question
odd edgeBOT
@elfin bear Has your question been resolved?
elfin bear
If A is true, then what does the expression give?
If A is false, same question
those are sets, not expressions
modern gyro
The statement is quite easily visualised
It would be better if you approach the proof using Venn diagram maybe
elfin bear
It would be better if you approach the proof using Venn diagram maybe
I tried, not working
is there a trick to do here?
or any simplification I can do
modern gyro
Im looking for it
Given that AUB = AUC
⇒ (AUB) ∩ C = (AUC) ∩C
⇒ (A∩C) U (B∩C) = C [ ∴(AUC)∩C = C ]
⇒ (A∩B) U (B∩C) = C ..........(1) [ ∴(A∩C) = A∩B ]
Again AUB = AUC
(AUB) ∩ B = (AUC) ∩ B
B = (A∩B) U (C∩B)
B = (A∩B) U (B∩C) ...........(2)
From 1 & 2 we get
B = C
elfin bear
Given that AUB = AUC
⇒ (AUB) ∩ C = (AUC) ∩C
⇒ (A∩C) U (B∩C) = C [ ∴(AUC)∩C = C ]...
$Given that AUB = AUC
⇒ (AUB) ∩ C = (AUC) ∩C
⇒ (A∩C) U (B∩C) = C [ ∴(AUC)∩C = C ]
⇒ (A∩B) U (B∩C) = C ..........(1) [ ∴(A∩C) = A∩B ]
Again AUB = AUC
(AUB) ∩ B = (AUC) ∩ B
B = (A∩B) U (C∩B)
B = (A∩B) U (B∩C) ...........(2)
From 1 & 2 we get
B = C$
clever fjordBOT
Ayanokoji
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elfin bear
$Given that AUB = AUC
⇒ (AUB) ∩ C = (AUC) ∩C
⇒ (A∩C) U (B∩C) = C [ ∴(AUC)∩C = C ...
I cant read this
modern gyro
It wont work like that i think
modern gyro
Given that AUB = AUC
⇒ (AUB) ∩ C = (AUC) ∩C
⇒ (A∩C) U (B∩C) = C [ ∴(AUC)∩C = C ]...
This?
elfin bear
also I don't know the triple dots u r using
modern gyro
its therfore sign
elfin bear
oh I see
elfin bear
Given that AUB = AUC
⇒ (AUB) ∩ C = (AUC) ∩C
⇒ (A∩C) U (B∩C) = C [ ∴(AUC)∩C = C ]...
I don't understand this
what is this proof exactly doing/assuming?
elfin bear
We re just generalising
generalizing what
modern gyro
Like the value of B and C
elfin bear
Like the value of B and C
assuming what is true pre determined exactly?
modern gyro
We do ΠB and ΠC just to find B and C values exclusively
modern gyro
We do ΠB and ΠC just to find B and C values exclusively
And we see that both equal to the same rhs
So B=C we conclude , the way of proof is quite trivial
elfin bear
So B=C we conclude , the way of proof is quite trivial
I think I get get your intuition, I'll reread it now
modern gyro
Yeah
elfin bear
Yeah
this proof feels all over the place I cant understand it
I get what you're trying to do but there seems to be a mistake
odd edgeBOT
@elfin bear Has your question been resolved?
mystic sable
toppology
frigid canopy
what about it?
mystic sable
his set of Computer Networks Multiple Choice Questions & Answers (MCQs) focuses on “Network Topology”.
- Physical or logical arrangement of network is __________
a) Topology
b) Routing
c) Networking
d) Control
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ionic falcon
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coral linden
@zenith jungle I’m awake now if you want to reopen a channel
coral linden
.close
odd edgeBOT
static totem
odd edgeBOT
✅
coral linden
I'm pretty sure this is just a supposed rule of inference written in a different notation
For example Modus ponens would be ${p, p \implies q} \vDash q$ in your notation
clever fjordBOT
Existentialistic
zenith jungle
ok and how do i find if the semantically is correct
i need to have a row of trues in the truth table?>
coral linden
There's different ways
I found it to be incorrect via a counterexample
p = True, q = False
zenith jungle
using the truth table
coral linden
yeah
zenith jungle
how can i see it
coral linden
there's a faster way though
zenith jungle
but i need to use the truth table sadly
coral linden
wait why lol
zenith jungle
it is an exercise in my exam
coral linden
that's overly tedious
that's really dumb but ok
This is how I did it without a truth table
you start with labelling p and q as T and F, and work from there
I guess with a truth table we have to build up each expression
zenith jungle
ye ahaahah i aldo need to learn the Method of analytic tableaux
zenith jungle
you start with labelling p and q as T and F, and work from there
how can u see if there is a semmantic from there
coral linden
well $\neg p \lor q$ is clearly false
clever fjordBOT
Existentialistic
coral linden
can you see that?
for p True and q False
in fact it's the only case where the conclusion is false
zenith jungle
isnt that or
zenith jungle
yes
coral linden
if it's single letter propositision on the left and right
ok
bad picture
so we have the only case with False on the right side
in all 3 other cases the right side is true, so the entire inference is true.
zenith jungle
ok so this is semmantical?
coral linden
since both $(F \implies T) \Leftrightarrow T$ \
and $(T \implies T) \Leftrightarrow T$
zenith jungle
dw ahahah
coral linden
there we go
coral linden
ok so this is semmantical?
what do you mean by "semantical"?
zenith jungle
semmantically
clever fjordBOT
Existentialistic
zenith jungle
⊨
coral linden
so like
I'm pretty sure $\vDash$ and $\implies$ have the same meaing, just different connotations
clever fjordBOT
Existentialistic
zenith jungle
that is implies on wikipedia
coral linden
$\vDash$ signifies that it's the overall conclusion on the right
clever fjordBOT
Existentialistic
coral linden
it's just like putting a box around your final answer
logically they should be equivalent
so we can treat it like logical implication
and use the table for the material conditional
this table #help-19 message
zenith jungle
ok so what im trying to understand is that if in at least in a row for multiple columns is true then it can be simenticaly
coral linden
Also to simplify your expression, "$q \lor \neg q$" is always true, so the first proposition is always true.
zenith jungle
taking this as an example
clever fjordBOT
Existentialistic
coral linden
ok so what im trying to understand is that if in at least in a row for multiple ...
Oh I see what you're asking, it's true semantically if it's true for all possible values of the propositions.
zenith jungle
if it is like this then it can be simenticaly
coral linden
There's four possibilities
here
p True, q True
p True, q False
p False, q True
p False, q False
zenith jungle
dont i need to look at this part?
coral linden
we want to look at
$(\neg p \lor q \lor \neg q) \land (\neg p \implies (q \implies p)) \implies (\neg p \lor q)$
clever fjordBOT
Existentialistic
coral linden
I should add parentheses
$((\neg p \lor q \lor \neg q) \land (\neg p \implies (q \implies p))) \implies (\neg p \lor q)$
clever fjordBOT
Existentialistic
coral linden
if this is True in all four cases, it's semantically true
zenith jungle
arent the cases 3?
zenith jungle
oh u were talking about rows
zenith jungle
so if it isnt true for all the rows in all the 3 columns it isnt semantically true
here is false
zenith jungle
only 1 column?
coral linden
well we should "build up" to it
to make it easier
the point of drawing out truth table like that is you can build up to larger expressions
from smaller expressions
zenith jungle
here is false
w this one i cant prove if it is simantically trye
coral linden
I mean I guess we could?
zenith jungle
all i want to know is looking at fully built truth table what i need to look at to see if it is simantically true
this one should have everything
for what i understood i need to look at the last 3 columns (in this case) and if in a row they are all true it is simmentically true
coral linden
This column at the end is not true for all possibilities, so the statement is not semantically true
zenith jungle
why did u do this?
clever fjordBOT
Existentialistic
clever fjordBOT
Existentialistic
coral linden
In English: "Given A and B, we have sufficient information to demonstrate C"
zenith jungle
i usually put A1 A2 A3 below the columns and do not in the truth table but below {A1,A2}⊨A3
coral linden
oh ok that works to then
ok then we're looking for cases where A1 and A2 are True but A3 is false
that would be the only possible counterexample
if there are no counterexamples, the statement is semantically true.
If and only if there are no counterexamples
zenith jungle
so on the table what do i look for a row full of trues?
or even if i have a row full of trues but in other rows A1 and A2 are True but A3 is false it isnt semantically
coral linden
a row where the right column is false, and all the other columns are true
that's a counter-example
zenith jungle
so in this case it isnt simmentically true
coral linden
right
because there's a possibility for the premises to be true, but the conclusion false
zenith jungle
so the answer to this whole thing is{A1,A2}⊭ A3
coral linden
yes
In general if we have ${A1, A2, A3, ... An} \vDash C$, then it's semantically false if and only if there's at least one case where A1 ... An are all true but C is false.
clever fjordBOT
Existentialistic
zenith jungle
so this case is false again
coral linden
what's the original statement?
zenith jungle
coral linden
we need $\neg p \land q$ somewhere
clever fjordBOT
Existentialistic
zenith jungle
why? isnt it there
coral linden
I don't see it
clever fjordBOT
Existentialistic
coral linden
this is going to be semantically false by counterexample when p is False and q is True
zenith jungle
so besically what i need to look at is this
if a1 a2 true but a3 false in a single row it is simmentically false
.
coral linden
I think it's "semantically contingent" technically
because it depends on the values of p and q
coral linden
yes
it can be both
so it's not semantically true
semantically true means it's always true
coral linden
all rows
zenith jungle
so i should answer it can be simmentically true
coral linden
no it's not semantically true
it's semantically contingent
There's no such thing as "can be semantically true"
semantically true already means it's true in all possible worlds
zenith jungle
can i just answer w a1 a2 ⊨ a3 ?
zenith jungle
semantically true already means it's true in all possible worlds
ok
coral linden
yeah since there are two premises here and one conclusion that works
I was writing a more general case with n premises
This is for n premises
zenith jungle
and if i had a table where not even in a single row they were all true then it is a1 a2 ⊭ a3
coral linden
well
if a1 and a2 are always false
then logical implication says ${A1 \land A2 } \vDash A3$
clever fjordBOT
Existentialistic
coral linden
it's like saying
"If 1 = 2 then pigs can fly"
it's sort of vacuously true
since 1 never equals 2
zenith jungle
what i meant was t f t
f f t
f f t
t t f
zenith jungle
not always false
coral linden
"The story goes that Bertrand Russell, in a lecture on logic, mentioned that in the sense of material implication, a false proposition implies any proposition.
A student raised his hand and said ”In that case, given that 1 = 0, prove that you are the Pope.”
Russell immediately replied, ”Add 1 to both sides of the equation: then we have 2 = 1. The set containing just me and the Pope has 2 members. But 2 = 1, so it has only 1 member; therefore, I am the Pope.”
Classic example
t t f should be a counterexample
The first two premises are true, but the conclusion is false
zenith jungle
t t t
t t f
f t t
t f t
a1 a2 ⊨ a3
zenith jungle
we would need t t f
so if there is no t t f i can type a1 a2 ⊨ a3
zenith jungle
basically only when i have t t f a1 a2 ⊭ a3
coral linden
We're looking for cases where the premises are true but the conclusion is false
coral linden
basically only when i have t t f a1 a2 ⊭ a3
yes
coral linden
ok :)
zenith jungle
do u know anything about gentzen and the analytic tableaux
xD
if u want we can take a break
coral linden
I don't think so
I'm not familiar with the terminology at least
but I might have seen it before and just not know the name
coral linden
ok so we're turning ^'s into commas and splitting V's
makes sense
I haven't seen it before but it makes sense
coral linden
You're welcome
odd edgeBOT
@coral linden Has your question been resolved?
viral forum
!help
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neat jasper
multiply by 10 to get rid of 0.1
viral forum
oh ok
mystic sable
k
odd edgeBOT
@viral forum Has your question been resolved?
viral forum
help
is it just me or math isn't intuitive at all
like you need to see someone else do the steps first
fallow osprey
#problem 2)
#solve the heat eq on a du/dt = 2 nabla^2 u on a circular cylinder of height 0 <= z <= 6
#subject to the initial condition u(r,theta,z,0)=r^2 z and
#boundary conditions: u(r,theta,0,t)=0, u(r,theta,6,t)=0, u(4,theta,z,t)=0,
I am having trouble with the algebra when it comes to speration constants
fallow osprey
Here is where I am at. From the cylindrical laplacian and the assumption of seperation of variables and after some algebra $\frac{T'}{T}= 2[\frac{1}{r} \frac{\partial}{\partial r}\frac{rR'}{R} + \frac{1}{r^2}\frac{\Theta ''}{\Theta}+ \frac{Z''}{Z}$
clever fjordBOT
Tavin
fallow osprey
Then applying seperation constants by the assumption that $f(x)=g(y)$ for a constant value. Then $\frac{T''}{T} = -\lambda$. From here I don't understand the justification but I know I should say $\frac{Z''}{Z}=-\mu$ and $\frac{1}{r} \frac{\partial}{\partial r}\frac{rR'}{R} + \frac{1}{r^2}\frac{\Theta ''}{\Theta} = \mu - \lambda$ and I am very confused about why this step is allowed: $\frac{\Theta ''}{\Theta} = -\gamma$ then I can solve for the time, height and angular component but I don't understand why the $\frac{1}{r^2}$ term 'goes away' in the angular component and I don't understand how to then apply a constant of seperation to the radial part.
clever fjordBOT
Tavin
odd edgeBOT
sand horizon
the function is even
so ig it would be with a cos
instead of a sin
it's something like 4cos(b(x-h)+k
4 due to the amplitude
nah more like -2sin(b(x-h)+k
sand horizon
no it's 2pi/2pi so it's just 1 no ?
2pi over the period length which is 2pi
it has to be something like -2sin(x+ 8/5) + 1
ruby sigil
ohh right
sand horizon
so to let u know how to find it, you search for k = (min + max)/2
then |a| = (max-min)/2
b = 2pi / the period (the distance between two maximum or two minimum)
and h is the distance between the function you have, and the function cos(x) or sin(x)
therefore with all of this you can find the equation @hidden vault
odd edgeBOT
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dim igloo
find area
frigid canopy
this is composed of two figures, do you notice that?
dim igloo
yeah i dunno how to find the internal angle
brittle notch
yeah i dunno how to find the internal angle
Doesn't it look like semi circle??
dim igloo
yeah
brittle notch
Yea so what's the problem?
dim igloo
ik that semi circle area is 9.42 but i dont know how to find the sector
brittle notch
Look u said the semicircle area is 9.42
Leftover part is a triangle right?
What sector?
dim igloo
oh is it
brittle notch
Yea
dim igloo
oh
brittle notch
The red line is diameter
dim igloo
ohh ok
brittle notch
Above that is semicircle
dim igloo
thanks
dim igloo
.close
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opal temple
solving equations in mod
odd edgeBOT
solving equations in mod
opal temple
so lets say im solving x^2 = 2 in Z mod 3
how can i prove that theres no x that satisfied that for any 2 + 3n?
and if there is a solution, for example:
x^2 = 4 (mod 5)
how can i find all of the solutions?
becuase here 2 is a solution, and so is 8
and 3
well for a small mod its easy, but lets say its mod 57
.close
odd edgeBOT
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sacred yoke
hello do you know statistics
odd edgeBOT
hello do you know statistics
sacred yoke
can someone elaborate how to prove their similarity
wanton bison
Ist das Deutsch
wanton bison
Verstehst du nicht die Zeilen oder was
sacred yoke
ich kann mir ja merken dass die dinge gleich sind aber ich verstehe nicht wie die dazu gekommen ist diese Gleichheit zu pruefen
wanton bison
Lass mich schauen
Das gelbe ist ein unabhängiger Faktor
den darf mam rausziehen
Beim blauen wird der gleiche Werte mit sich selbe n mal aufsummiert
Also kann man schreiben auch
n • (x²/n)
Der vorletzte Schritt ist tricky
Die Summe die da steht über alle x_i ist ja das aritmethische Mittel
also würde man
,,2\bar x^2
clever fjordBOT
she opened my eyes
wanton bison
Erhalten
Ok jetzt machts dann sinn
Vorletzte Zeile
,,\frac{1}{n} \sum_{i=1}^n (x_i^2) - 2\bar x^2 + \bar x^2
clever fjordBOT
she opened my eyes
clever fjordBOT
she opened my eyes
wanton bison
Hoffe das macht jetzt mehr Sinn
sacred yoke
Hoffe das macht jetzt mehr Sinn
ich kann definitiv damit weiter arbeiten, dankee dir !!
odd edgeBOT
Available help channel!
vocal quest
Hey! I solved this by looking at a graph, but is there a way to solve it analytically?
odd edgeBOT
Hey! I solved this by looking at a graph, but is there a way to solve it analyti...
@vocal quest Has your question been resolved?
vocal quest
<@&286206848099549185>
long tinsel
Even if you did you'd mentally have to draw a graph anyways
vocal quest
Y-yeah.. But how to get the actual values?
thin geyser
regardless of the sign of the quadratic term, it's just a factored cubic and you know the roots come in conjugate pairs
vocal quest
yeah, the roots are 0, 1, and 2
thin geyser
Actually I misread the question playing a video game and didn't think about it carefully enough
It's a bit more complicated than that
vocal quest
I drew this
And this is from Desmos
The vertex of the parabola in the abs function is clearly =1.5 (mean of its roots)
And playing with the graph in desmos it can be seen that these two bumps have the same maximas
But I don't know how to "prove" it?
olive sparrow
And playing with the graph in desmos it can be seen that these two bumps have th...
find the derivative of the function (without abs) and set it equal to 0
odd edgeBOT
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brittle steeple
Q. Kendra wants to arrange and divide the letters in her name to create 3 strings, each of which must
contain at least one letter. In how many ways can she do so, assuming that the order in which she
creates the strings is irrelevant, and she uses each letter exactly once?
Is it possible to solve this with complimentary counting?
My work:
First find the total number of ways to arrange them on the strings ignoring the constraint that every string must have at least 1 letter. Then subtract the number of ways that the strings have no letters at all.
Why is this wrong?
The answer is 1200, but I'm getting 729-189=540
odd edgeBOT
Q. Kendra wants to arrange and divide the letters in her name to create 3 string...
@brittle steeple Has your question been resolved?
brittle steeple
<@&286206848099549185>
vivid girder
how did you arrive at 729 - 189 ?
brittle steeple
3^6 ways to distribute the letters to the strings not considering each string has to have at least 1 letter
then it is 2^6 ways that at least one of the strings has 0 letters
but then
i used Principles of inclusion exclusion
to reach 64+64+64-1-1-1 ways that there are none on some strings
which is 189
vivid girder
Am I right to assume that the 3 in 3^6 comes from 3 strings.
vivid girder
hmmm, 3^6 I can interpret combinatorically as choosing 1 of the 3 strings 6 times, with repeating strings.
I find it hard to interpret it as 'distributing letters'
brittle steeple
hmmm, 3^6 I can interpret combinatorically as choosing 1 of the 3 strings 6 time...
repeating strings?
vivid girder
yes, like you have 3 marbles in a bowl. You pick one and return it, then pick another and return it, and so on 6 times.
brittle steeple
yes, like you have 3 marbles in a bowl. You pick one and return it, then pick an...
but is it w/ replacement??
late dust
3^6 does distribute 6 letters into 3 strings, but you are not taking into account the ordering
late dust
Basically yeah
brittle steeple
gotcha
do u mind walking me through it tho
just to see if what im doing is correct
brittle steeple
well the solution said this,
but i dont get why they do 5 pick 2 too split the letters
so i chose a different solution
vivid girder
Imagine you have an ordering of the letters
you can place some dividers in the string separating it into substrings
brittle steeple
yea ik, but why 5 pick 2?
vivid girder
for 6 characters there are 5 'slots' and you want to place dividers in 2 of them
then you get 3 groups
brittle steeple
this wouldnt work with stars and bars tho right...?
brittle steeple
for 6 characters there are 5 'slots' and you want to place dividers in 2 of them
but why are there 5 objects though
vivid girder
see image
brittle steeple
ok
late dust
That's not really it though, the solution only considers the 3 remaining letters
late dust
<https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)#Theorem_two_proof>
It's this but with n=k=3
brittle steeple
It's this but with n=k=3
so wait
solve it then with stars and bars pls?
i think if i see a full solution id get it
[
\binom{3 + 3 - 1}{3 - 1}
]
and then what
[
\binom{3 + 3 - 1}{3 - 1} = \binom{5}{2}
]
clever fjordBOT
sdatta
late dust
Ye
late dust
brittle steeple
.close
odd edgeBOT
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last citrus
how did we go from |z|²=1 to zż=1
odd edgeBOT
how did we go from |z|²=1 to zż=1
last citrus
|z| is the modulus
odd edgeBOT
@last citrus Has your question been resolved?
dusk fossil
odd edgeBOT
dusk fossil
@gritty oar I have an idea: Let $q_1, q_2 \dots$ be an enumeration of rationals in $[0, 1]$. Consider $\bigcup\limits_{i} \left (q_i - 3^i, q_i + 3^i\right ) = \mathcal{F}$. Then $\mathcal{F}$ contains all the rationals, is open and $|\mathcal{F}| = \sum\limits_{j=1}^\infty \frac{1}{3^j} = \frac{1}{2}$. Thus, $\mathcal{F}^c$ is closed, contains only irrationals and has a measure greater than or equal to $\frac{1}{2}$. We replace $3$ with suitable constants.
clever fjordBOT
szpol
gritty oar
yep was the idea i had
gritty oar
.close
mystic saffron
odd edgeBOT
spice geyser
if it is non-trivial it must be surjective, my first thought would be trying to figure out whats in the kernel.
f(aaa)=f(a)+f(a)+f(a)=0
things like that
mystic saffron
if it is non-trivial it must be surjective, my first thought would be trying to ...
Why surjective ? Am I to dumb to understand 😅
spice geyser
Why surjective ? Am I to dumb to understand 😅
a group homomorphism must map to a subgroup of Z/3Z
there are 2 subgroups of Z/3Z, one is trivial
mystic saffron
O
spice geyser
there are 2 subgroups of Z/3Z, one is trivial
if it maps to the trivial group the homo is trivial
spice geyser
There order must be multiple of 2
ok so what is f((123))=f((12)(23))
similarly what is f((132))
mystic saffron
= f(12)f(23) = 0
so everything is 0
mystic saffron
Got it
spice geyser
so some kernel bashing might work for these types of problems
mystic saffron
Thank You
spice geyser
you're welcome
mystic saffron
so some kernel bashing might work for these types of problems
wait
For which types of problems?
@spice geyser
spice geyser
For which types of problems?
um for questions like "does there exist a non-trivial homomorphism"
mystic saffron
.close
odd edgeBOT
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amber current
hey guys just need some quick help, what happens in the second row? it says "with integrating the series above we get: "
amber current
how and why?
slate onyx
you integrate each summand and then you use the sum of geometric sequence
spice geyser
how and why?
you can interchange sum and integeral under some condition, in this case integral of the infinite sum equals the infinite sum of integral of each term
in the radius of convergence
amber current
so in the above one I integrate each summand?
I just dont understand why he gets x + x there
amber current
nono im sure he wouldnt mess up
slate onyx
probably a mistake because the last = equals to 1 + x + x^2 + ...
mystic saffron
You can't use sum of infinite g.p , if there were 2 x
spice geyser
so in the above one I integrate each summand?
they forgot a +C
mystic saffron
1
mystic saffron
There's a 1
spice geyser
There's a 1
its not +C
mystic saffron
Not sure where that came from though
spice geyser
Not sure where that came from though
1+C and +C are practically the same
pretty sure they just forgor the +C
mystic saffron
its not +C
They perssumed it to be 1?
mystic saffron
Yes
spice geyser
You can't use sum of infinite g.p , if there were 2 x
whats g.p
spice geyser
They should be giving the condition amplitude x <1
i remember them calling it amplitude in hs
but yeah x<1
spice geyser
nono im sure he wouldnt mess up
thats for sure a typo
you should tell your teacher they forgot an C and put an extra x
amber current
you should tell your teacher they forgot an C and put an extra x
he wrote after that he chose C = 1
thats why theres a 1
but ig the x + x is a mistake
I'll say to him
spice geyser
he wrote after that he chose C = 1
i see
odd edgeBOT
Available help channel!
gritty oar
youink