i understand that the limit of r/r^2 is 0 but i dont understand why it's insufficient to say the whole limit is equal to 0 without showing that (cost + sint)/(1+sintcost) is bounded
#help-39
urban jacinth
or could you please give me a counterexample where 1/r goes to 0 but the rest of the function is unbounded so the whole limit doesn't go to 0?
compact ridge
then the limit would vary between r/r^2 and r^2/r^2 or something
a constant multiple of a limit to 0 is also a limit to 0
but not something like $\lim_{x \to \infty} x \cdot \frac{1}{x}$
jolly parrotBOT
south
compact ridge
cause x is unbounded
urban jacinth
this is using that one thing that goes like
If $f$ is bounded, and $\lim_{x \to a} g(x) = 0$, then $\lim_{x \to a} f(x)g(x) = 0$
that?
jolly parrotBOT
Moonful
urban jacinth
is this what's being used here
urban jacinth
**south**
ahh of course
so the goal here is to show that $|\frac{\cos t + \sin t}{1 + \sin t \ cos t}| <= k$
jolly parrotBOT
Moonful
pearl pondBOT
compact ridge
If $f$ is bounded, and $\lim_{x \to a} g(x) = 0$, then $\lim_{x \to a} f(x)g(x) ...
yeah!
no worries
oh yeah it's one of the algebraic limit theorems, f(x) <= M and then you have lim M * f(x) = M * lim f(x)
and then M * 0 = 0 for any M
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slim vigil
∆ABC has circumcenter O, orthocenter H. M, N are midpoints of BC, OM, resp. The line through A perpendicular to BN intersects CH at P. E,F are the reflections of A through C,B resp. Prove that FP ⊥ HE.
slim vigil
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silent rock
can someone remind me how this is supposed to be solved?
pearl pondBOT
can someone remind me how this is supposed to be solved?
silent rock
i dont remember any way to remove one of the variables
pulsar flax
generally you want to isolate one variable
another option would be to add/subtract the equations
silent rock
another option would be to add/subtract the equations
adding them just makes another double variable equasion
pulsar flax
but a linear one
silent rock
and i dont think theres any way to add to either or both of the lines to make them cancel something out
naive zinc
You just add up, you can replace one variable with an expression solely in another variable
silent rock
but a linear one
adding them would make 2x + 8y = 2, or x + 4y = 1, if i represented x as x=1-4y, does that apply to the original 2 equasions?
pulsar flax
adding them would make 2x + 8y = 2, or x + 4y = 1, if i represented x as x=1-4y,...
yes
silent rock
.....since when
spiral coyote
.....since when
-# forever
pearl pondBOT
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tired grove
A function is defined as e^{ 2|x|+x}. Let x1>0 and x2<0 be two points on the curve such that the tangents at these points is perpendicular.
The number of such pairs (x1,x2) are
(a)0
(b)1
(C)2
(D) More than 2
steady vapor
Rewrite the equation so that x2 is on one side
Then check whether the expression can ever be negative when x1 > 0
toxic lichen
$f(x) = e^{2|x| + x} = \begin{cases} e^{3x} & x \geq 0 \ e^{-x} & x < 0\end{cases}$
jolly parrotBOT
Ann
toxic lichen
I don't know what to do after this
consider the range of f'(x) and how it can overlap with the range of -1/f'(x)
tired grove
3x1>0 -x2>0 so no such pairs?
pearl pondBOT
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drifting valley
A school sells adult tickets at ₹65 and child tickets at ₹40. The number of child tickets sold was 15 more than adult tickets. Total collection was ₹3,025.
A) Let number of adult tickets be x. Form an equation.
B) Solve to find number of tickets sold.
C) If child ticket price increases by ₹7, what will be the new total collection?
tulip marsh
!status
pearl pondBOT
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
All
toxic lichen
ok so let's start with A
the number of adult tickets is x as the question directly tells you.
then what's the number of child tickets?
(answer as an expression in terms of x)
drifting valley
40(x+15)
toxic lichen
if you need help on questions like this then you should be going about this in a structured way
yes, x+15.
so
x adult tickets sold for Rs. 65 each
(x+15) child tickets sold for Rs. 40 each
can you tell me the total money collected in terms of x
drifting valley
40(x + 15)
toxic lichen
no, that's from the child tickets only
i want you to tell me the total collection. from child and adult tickets, combined.
in terms of x.
drifting valley
Rs. 3025
toxic lichen
no, not what i am asking
3025 will appear on the right side of your equation
i am asking about the left side
drifting valley
65+ 40
toxic lichen
no, that's the cost of one adult and one child ticket.
40(x+15)
is the money collected from the sale of (x+15) child tickets
this is only "half" of what you need here.
the other "half" is the money collected from selling x adult tickets.
drifting valley
65x + 40×x+15
drifting valley
Oh
toxic lichen
65x + 40(x+15) would've been correct.
so now, can you write the equation from this, now using the question's given value for the total collection?
drifting valley
65x+ 40x+ 600?
toxic lichen
i said equation.
equations always have an equals sign.
it is even in the name.
you can expand 40(x+15) as 40x + 600 if you want but for part A this is optional.
toxic lichen
no, because you did not actually write down the equation as i asked you to.
so we are not done with A.
drifting valley
65x+40(x+15)
toxic lichen
the equation
please reread what i told you.
equations always have an equals sign.
using the question's given value for the total collection
65x+40(x+15)
there is no=here
drifting valley
65x+40(x+15)= ₹3025
toxic lichen
the rupee symbol is going to get in the way a bit
but ok i'll take it
65x + 40(x+15) = 3025
NOW we are done with A, and NOW we can move on to B.
can you solve this equation and find the value of x?
a) "Yes. Let me go and do that."
b) "I am not sure. Let me try."
c) "No, I definitely can't."
d) "Something else."
drifting valley
Yes i can do it on my own
toxic lichen
ok
do you need any more help from this point onward, or are you good on your own until the end?
drifting valley
No thank you!
drifting valley
.close
eager jewel
pearl pondBOT
eager jewel
i took cases
7 even 0 odd--> 5^7
5 even 2 odd-->5^7
3 even 4 odd-->5^7
1 even 6 odd--->5^7
so 4 x 5^7
and then discount cases where 0 is the first digit
6 even 0 odd-->5^6
4 even 2 odd-->5^6
2 even 4 odd-->5^6
0 even 6 odd-->5^6
so 4x5^7 - 4x5^6 = 4 x 5^6 x 4
why is this wrong?
toxic lichen
is 4 * 4 * 5^6 = 16 * 10^3? 
toxic lichen
what is your actual final answer for m+n
oh hold on
they ask for m * n * 10^n?
impossible as written (if the exponent is INTENDED to be the same as the n factor)
eager jewel
yeah
toxic lichen
the actual count seems to be 250000 and you calculated it correctly
is the answer listed as 10?
eager jewel
is the answer listed as 10?
14
toxic lichen
eager jewel
but i ignored their logic and wanted to correct mine
toxic lichen
i just realized your mistake
sorry it took me so long
you did not account for where the digits go in the number
you're counting 2461111 and 1214161 as the same thing
eager jewel
huh meaning i need to multiply by 7!?
toxic lichen
so you would need to multiply by some binomial coefficients
not by 7!, no
but by the number of ways to arrange the corresponding number of E's and O's in a string in each case
eager jewel
but by the number of ways to arrange the corresponding number of E's and O's in ...
oh
so like for 2 even case
i multiply by 7 C 2?
verbal whale
<@&268886789983436800>
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robust nexus
lowkey lost for 67 d
pearl pondBOT
lowkey lost for 67 d
robust nexus
i obvs change to dx dy
now for fixed y, i need bounds for x
in terms of y
BUT
when i draw the thing out
the curves are symmetric between the intersection
cuz yk how for x limits, we look at the leftmost curve going to the rightmost curve
that technique doesnt work
plush bramble
sure it does. you just need to work to find the functions of y
robust nexus
sure it does. you just need to work to find the functions of y
x = sqrt(y)
x = sqrt(2-y)
lucid anvil
you might need to break it up into the sum of two integrals
robust nexus
what to do after that
lucid anvil
one where y goes from 0 to 1 and the other when y goes from 1 to 2
robust nexus
you might need to break it up into the sum of two integrals
x = -sqrt(y) to x = -sqrt(2-y) ?
then
x = sqrt(y) to x = sqrt(2-y)
robust nexus
one where y goes from 0 to 1 and the other when y goes from 1 to 2
im fine for the y bounds i think, its the x thats troubling me
frosty rampart
i was thinking same as @lucid anvil
robust nexus
i was thinking same as <@325044604212871179>
yep i get that, breaking it up seems good
i dont understand the x bounds
frosty rampart
cuz how r u gonna create a function for your x
ok
then you have the same function
but negative
lucid anvil
x = -sqrt(y) to x = -sqrt(2-y) ?
then
x = sqrt(y) to x = sqrt(2-y)
not quite
try fixing some y
robust nexus
oh
frosty rampart
x=sqrt(2-y)
x=-sqrt(2-y)
lucid anvil
^
robust nexus
x=sqrt(2-y)
x=-sqrt(2-y)
wait how did you get this
frosty rampart
that's one of the two double integrals
cuz like qian said
you should probably do one from 0 to 1 and another from 1 to 2 (for y)
for y = 0 to 1, you use the red curve as x bounds
and for y = 1 to 2, you use the blue curve as x bounds
robust nexus
and for y = 1 to 2, you use the blue curve as x bounds
wait ok
so when i do y = 0 to 1 i use the red curve
that would get me the area with the y-axis woulnd it
oh yea it would kk
ok i think i get what you mean
frosty rampart
bet
robust nexus
oh shit ye i see it
frosty rampart
i probably explained like shit
robust nexus
thanks guys
pearl pondBOT
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swift spindle
pearl pondBOT
swift spindle
i get +- but how do i know that its one of either
i.e. which one do i reject, or is my working out just flat out wrong
viscid shale
To check, plug both in into the equation of the derivative.
swift spindle
To check, plug both in into the equation of the derivative.
i see
and one of them gives 2 other gives -2, so i use theo ne that gave 2
for
wait so does this mean
if i h ave x^2 = a
and i do x = a^(1/2), the a^(1/2) has to be +-?
viscid shale
and one of them gives 2 other gives -2, so i use theo ne that gave 2
Are you really sure?
swift spindle
viscid shale
the rule where $\left(a^b\right)^{\frac1b} = a$ only applies for $a \geq 0$
jolly parrotBOT
viscid shale
Try to decompose it entirely and then solve.
viscid shale
yes, $x^{\frac23} = \sqrt[3]{x^2}$
jolly parrotBOT
viscid shale
Also, notation wise, you missed where the minus sign goes.
viscid shale
oh yeah, mb
pearl pondBOT
swift spindle
.reopen
pearl pondBOT
✅ Original question: #help-39 message
swift spindle
oh yeah, mb
wait so is this wrong?
mortal epoch
wait so is this wrong?
No, that is correct
Both the positive and negative solutions are valid
viscid shale
You can have more than 1 solution
fresh olive
Domain of $n$th roots: (For real numbers)
$$\text{If }n\text{ is even: }[0,\infty)\text{ (Postive numbers and zero)}$$
$$\text{If }n\text{ is odd: }\mathbb{R}\text{ (Every number)}$$
jolly parrotBOT
BBMaths
pearl pondBOT
@swift spindle Has your question been resolved?
swift spindle
Both the positive and negative solutions are valid
look at my working where did i go wrong
swift spindle
i found dy/dx of the negative version is -2
pearl pondBOT
mortal epoch
.reopen
fresh olive
.reopen
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✅ Original question: #help-39 message
@swift spindle Has your question been resolved?
pearl pondBOT
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void estuary
pearl pondBOT
void estuary
how to solve this type of limits?
hard crystal
What's ur status?
eager jewel
what is thisss bruh 😭
hard crystal
Yep
viral axle
Start by solving the summation
void estuary
alrightt
hard crystal
After that just take n common
pearl pondBOT
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Available help channel!
timber zinc
ok good
toxic lichen
you should send your question(s) as the first message btw
timber zinc
Theorem 1.4.3. (Mader 1972):
$k\in\mathbb{N}$ (non-zero) and $G$ a graph with $d(G)\geq 4k$. Then there exists a k-connected subgraph H with $d(H)>d(G)-2k\geq 2k$
jolly parrotBOT
Affe
timber zinc
you should send your question(s) as the first message btw
i know but i didnt want anyone to take this channel
question inc now
now diestel uses this theorem to deduce that given $d(G)\geq 10r^2$, then $\epsilon(H)>\epsilon(G)-r^2\geq 4r^2$ in his proof of theorem 7.2.3
jolly parrotBOT
Affe
jolly parrotBOT
Affe
timber zinc
shouldnt it be $\epsilon(H)>\epsilon(G)-r^2\geq r^2$?
jolly parrotBOT
Affe
timber zinc
havent read 7.2.3. completely yet
tired
Th 3.5.3 (Thomas&Wollan 2005) states that every 2k connected graph with $\epsilon(G)\geq 8k$ is k-linked
jolly parrotBOT
Affe
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
jolly parrotBOT
Affe
timber zinc
which is equivalent to $\epsilon(G)\geq 5r^2$
jolly parrotBOT
Affe
timber zinc
.close
pearl pondBOT
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void estuary
how to easily differentiate this ?
pearl pondBOT
how to easily differentiate this ?
void estuary
will any substitution work here?
unborn abyss
i'd be tempted to just try taking d/dx of both sides and see where you end up
feral sedge
pearl pondBOT
@void estuary Has your question been resolved?
pearl pondBOT
void iris
bruh
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Available help channel!
light pine
$Let $\pi$ be the plane given by the equation $x+2y+z=0$. Define $F:\mathbb{R}^3\to\mathbb{R}^3$ by
[
F(x)=\operatorname{proj}_{\pi} x,\qquad x\in\mathbb{R}^3.
]
Use the formula for the orthogonal projection onto a plane through the origin to show that $F$ is a linear transformation and to determine the standard matrix of $F$.$
jolly parrotBOT
Quantie
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light pine
dope
cobalt hinge
(capital Pi maybe?)
light pine
ive proven its a linear transformation by F(ax + by) = aF(x) + bF(y)
and to get the standard matrix for F i put in the base vectors e1 e2 e3 and i was done
but in my professors solution, he does a loot of stuff with orthonormal bases, and wrote "to determine F´s standard matrix we determine an explicit ortogonal base B to pi"
and gets this, which is easy steps
but he doesnt use it for anything and gets the standard matrix and just goes on
maybe he missed something in the exam problem? or am i missing something
id send the full solution but its in swedish
or no he uses the new base to calculate the F(e)
is it just a more rigorous method, rather than just
jolly parrotBOT
Quantie
Compile Error! Click the reaction for more information.
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light pine
cobalt hinge
ive proven its a linear transformation by F(a**x** + b**y**) = aF(**x**) + bF(**...
I mean yours is fine...what's your actual question though?
light pine
I mean yours is fine...what's your actual question though?
im just confused over his solution lol im trying to understand it
why is an orthogonal base so important that he highlighted it
ok i think its just
different solving methods
alr alr
.close
pearl pondBOT
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whole jolt
pearl pondBOT
pearl pondBOT
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
whole jolt
which question
all
burnt copper
nCr/nCr-1 is (n-r+1)/r
warped violet
Why is part of this redacted?
burnt copper
Idk if this gonna help tho
warped violet
Is this classified work for the government
burnt copper
Looks like it will
burnt copper
Is this classified work for the government
Bro can u help w my question
whole jolt
nCr/nCr-1 is (n-r+1)/r
tried it
whole jolt
Is this classified work for the government
lol no
it was some random comment frm teach
which question we doing first
whole jolt
nCr/nCr-1 is (n-r+1)/r
wht do i do with the n
autumn fossil
1 feels weird, shouldnt the n.o. terms be always odd
whole jolt
idk trinomial expansion
whole jolt
1 feels weird, shouldnt the n.o. terms be always odd
nope, the n in first is 6
tht ik
whole jolt
But nCn is 1 right
yeah
autumn fossil
,w expand (1-2/x+4/x^2)^6
jolly parrotBOT
autumn fossil
thats 13 terms
burnt copper
Yo i think this is a hp in n
whole jolt
this is purely binomial ;-;
autumn fossil
the number of terms is 2n+1
burnt copper
Binomial suck tbh
whole jolt
ik
compact ridge
what does $C_n$ even mean
jolly parrotBOT
south
whole jolt
idk
autumn fossil
is it like $^{15}C_n$
jolly parrotBOT
MathIsAlwaysRight
compact ridge
like I've seen $C(n, r), ^nC_r, C^n_r$ for binomial coefficient
jolly parrotBOT
south
compact ridge
but the denominators are all $C_0, C_1, C_2 \cdots$ which doesn't make sense
jolly parrotBOT
south
autumn fossil
so that's 2 vague questions... nice
do u have notes / textbook or anything where that notation is used?
burnt copper
Apply that formula n-r+1/r
burnt copper
Why don't u just write 1
whole jolt
wht do i put in place of n
whole jolt
thts just for the sigma
whole jolt
doesnt it mean n \* Cn
yeess
burnt copper
Toh n and r are equal to n so it's basically sigma 1 to 15 1/n
autumn fossil
It was the standard
if it was that, what'd the denominators mean?
C_13? What does that mean?
whole jolt
whole jolt
What does Cn even mean
whole jolt
whole jolt
google doesnt know
poor us
iron basin
yeah
autumn fossil
okay, yeah, this matches my result
plush bramble
amazing how much clearer things are when you don't redact important information
burnt copper
natural numbers
Fr
autumn fossil
how are we using it here
$\frac{i \cdot ^iC_2}{^{15}C_{i-1}}$
warped violet
random comment from teach 😭
jolly parrotBOT
MathIsAlwaysRight
whole jolt
daym
autumn fossil
**MathIsAlwaysRight**
so this is how the terms actually look like
whole jolt
oh
autumn fossil
try doing some algebra and simplifying that
whole jolt
i will
omg i hate these questions leave them ill go make ramen chill and then come back to them
iron basin
divide and multiply with n! to start
burnt copper
It's so easy it's e^3
whole jolt
It's so easy it's e^3
how
whole jolt
damnnnn
burnt copper
Idk e^3 looked the best
whole jolt
whole jolt
Idk e^3 looked the best
how old are you ;-;
burnt copper
Im 41
whole jolt
cap
burnt copper
No cap gng
whole jolt
no 41 year old talks like tht
burnt copper
I got 2 kids
burnt copper
no 41 year old talks like tht
My kids taught me
whole jolt
cap ;-;
burnt copper
I only speak 📠 no 🧢
autumn fossil
try simplifying the inner sum first
whole jolt
try simplifying the inner sum first
ogk
autumn fossil
note that both n and r is fixed, so you can factor out the first 2 things and you're left with sum of rCt 4^t
autumn fossil
nah, we dont hand out answers + im lazy to do it all
burnt copper
What are u even preparing for mate
burnt copper
Who tf is asking that shit
whole jolt
Who tf is asking that shit
school?
autumn fossil
what i mean is start by simplifying the red sum
and you can ignore the blue stuff, because it doesnt depend on t. So its just a constant
whole jolt
what i mean is start by simplifying the red sum
leave it;-; one question wont hurt me ;-;
whole jolt
and you can ignore the blue stuff, because it doesnt depend on t. So its just a ...
ohh
burnt copper
Ts too much for high school exams
whole jolt
nope
burnt copper
leave it;-; one question wont hurt me ;-;
It's gonna be on the exam
whole jolt
no its not like tht
i just doing previous year questions for practice
for the finals
and pre annuals
burnt copper
no its not like tht
No I'm telling you the question u think won't hurt much to leave will show up
Ok the exam
whole jolt
they give ipad to first ranker
burnt copper
they give ipad to first ranker
Crazy
whole jolt
No I'm telling you the question u think won't hurt much to leave will show up
i doing it off the web;-;
burnt copper
i doing it off the web;-;
I mean u never know
burnt copper
What like a entrence typa shi
autumn fossil
is the answer 5 btw?
whole jolt
yep
burnt copper
is the answer 5 btw?
U didn't just solve it like in 5 minutes
whole jolt
What like a entrence typa shi
JEE ;-;
whole jolt
U didn't just solve it like in 5 minutes
legit
whole jolt
is the answer 5 btw?
show wht u did
autumn fossil
U didn't just solve it like in 5 minutes
its not difficult, its tedious. Typical JEE. It only looks difficult
autumn fossil
what i mean is start by simplifying the red sum
well once we ignore the blue sum, its just sum of rCt 4^t which is (1+4)^r by binomial, but the sum is only up to r-1, so we subtract 4^r (the last would-be term)
burnt copper
its not difficult, its tedious. Typical JEE. It only looks difficult
I just don't know where to begin
autumn fossil
so the inner sum is just blue * (5^r - 4^r)
autumn fossil
I just don't know where to begin
begin at the inner sum
burnt copper
No in general i think jee math is too much
autumn fossil
its literally the same trick applied 3 times
whole jolt
well once we ignore the blue sum, its just sum of rCt 4^t which is (1+4)^r by bi...
go tit
whole jolt
its literally the same trick applied 3 times
got it
burnt copper
so u are a kid
Nah this is my 32rn drop year
whole jolt
did it
whole jolt
Nah this is my 32rn drop year
uh sure
pearl pondBOT
Available help channel!
cold scaffold
$\int_{0}^{1} \frac{\sin(\arctan x + \pi)}{1 + x^2} , dx$
okay, this is a call for help too, i genuinely hate feeling like this, im forced into college against my will by my family and got threatened to keep studying while i wanted to pursue something else. i hate how i can understand basic intuition behind everything but whenever a curveball gets thrown at me in form of a problem i dont recognize i genuinely get stuck
i hate feeling behind so much, please throw whatever resources you have, books you have, hard problem sets, ill do whatever it takes to catch up and i want to fullclear all kind of exams, this is probably really easy to you guys but i genuinely dont know where to go from here, and im also scared ill forget this after a year
anyway, what i observed so far was sine shifting when its +π, making it int0,1 [ [-sin(arctan(x))]/1+x^2 ], thats as far as my intuition goes, and for i know i could be wrong too, im scared to push forward
jolly parrotBOT
Sylvaline
cold scaffold
there, thanks everyone in advance
viscid shale
id advice you delete the previous image and simply put the text on its own
cold scaffold
done, sorry
viscid shale
Well, to start with, you will prob want a trig identity table at hand at all times
cold scaffold
Well, to start with, you will prob want a trig identity table at hand at all tim...
got from google, good resource?
viscid shale
Apart from some basic knowledge on trig functions
viscid shale
got from google, good resource?
I myself can only recommend wikipedia's list
https://en.wikipedia.org/wiki/List_of_trigonometric_identities
there are some obscure identities that relatively frequently come at hand
you can usually find them on your own but make some things faster
cold scaffold
wikipedia never crossed my mind because i always went to images but oh wow thats really useful
i might make a note compiling them
viscid shale
**Sylvaline**
anyways, just to start, you should see a really easy simplification on sin( arctan x + \pi)
cold scaffold
yes, i saw +pi shifting the function
it makes for a -sin(arctan x + \pi), for all i know, this could be wrong too
cold scaffold
oh right, it has already shifted the function
viscid shale
you can go two ways from here.
The crazy way > Which is using the table for trig(inverse trig) identities.
Or theres a really easy sub.
viscid shale
kind of the reverse, but yeah
you prob want to learn the derivatives of all the main trig functions + arctan.
arcsin and arccos rarely appear too.
And honestly ive never seen arccsc, arcsec and arccot being used.
cold scaffold
arcsin and arccos rarely appear too.
right, i only saw them in one problem
ever, or maybe my palette is small
viscid shale
In HS/Uni youll rarely see them, and depending on what kind of work youre doing, it becomes even rarer past that
cold scaffold
by the way, what might this crazy inv trig table route be? would it help me bruteforce problems in exams?
viscid shale
Theres an identity for every composition of trig function and another inverse trig function
sin(arctan) is there
viscid shale
the problem would then become $-\int_0^1 \frac{x}{(1+x^2)^{\frac 32}}dx$
jolly parrotBOT
viscid shale
which you prob have seen already
viscid shale
its also a u sub, but it just requires for you to know off the top of your head the result of sin(arctan x)
cold scaffold
the problem would then become $-\int_0^1 \frac{x}{(1+x^2)^{\frac 32}}dx$
oh my god
say u = 1+x^2
du = 2x
viscid shale
yep, you get 1/2du = x dx
cold scaffold
oh mmyyy goddddd
viscid shale
and then you simply do 1/u^{3/2} which is just power rule.
cold scaffold
i think youre the best
viscid shale
also, in the wikipedia page i sent you can check up on the forgotten trig functions
viscid shale
anyways, any other question?
cold scaffold
not at the moment, so all is clear now
how may i close?
thank you so much im goonna cry
viscid shale
how may i close?
,.close when youre done
cold scaffold
.close
pearl pondBOT
cold scaffold
thank you
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Available help channel!
carmine juniper
Why is this like that?
pearl pondBOT
Why is this like that?
carmine juniper
If sqrt(2)^2 =4
Then why is that equation in the picture bigger than that if it’s just plus
north talon
Then why is that equation in the picture bigger than that if it’s just plus
what do you mean by this?
lavish stream
If sqrt(2)^2 =4
i think sqrt(2)^2=2
robust lynx
If sqrt(2)^2 =4
(√2)² ≠ 4, it equals 2
carmine juniper
I mean sqrt(2) times sqrt(2)
lavish stream
yup it is equal to 2
reef rampart
Then why is that equation in the picture bigger than that if it’s just plus
You mean why is it not 2? And why 2root2
robust lynx
Why is this like that?
It’s like that because that’s they simplified using multiplication.
𝑎 + 𝑎 = 2𝑎
robust lynx
I mean sqrt(2) times sqrt(2)
It would still equal 2.
lavish stream
sqrt(2) is basically 2 raised 1/2
so then when you multiply two same numbers with powers
their powers add up
therefore 1/2 plus 1/2 equals to 1
hence its 2
proper nova
Why is this like that?
but this is addition
proper nova
It’s like that because that’s they simplified using multiplication.
𝑎 + 𝑎 = 2...
so this explanation makes sense the most
cinder flower
proper nova
Why is this like that?
the thing is that you are not multiplying 2 sqrt(2)s, you are adding 2 sqrt(2)s
proper nova
Why is this like that?
the operator is +, not *
lavish stream
so this explanation makes sense the most
yup but she said that sqrt(2)2^2 is equal to 4
proper nova
yup but she said that sqrt(2)2^2 is equal to 4
hell naw
proper nova
If sqrt(2)^2 =4
also $\sqrt{2}^2 = 2$
lavish stream
If sqrt(2)^2 =4
^
jolly parrotBOT
1 divided by 0 equals Infinity
carmine juniper
It would still equal 2.
It isn’t…
robust lynx
…you proved that it is yourself. 😭
proper nova
It’s like that because that’s they simplified using multiplication.
𝑎 + 𝑎 = 2...
look at this explanation
robust lynx
It’s like that because that’s they simplified using multiplication.
𝑎 + 𝑎 = 2...
Again, for the picture you attached, the answer is this. They have used the definition of multiplication to transform √2 + √2 into 2√2.
proper nova
Ok blue
now tell us, if there's ANYTHING that you can't understand
please TELL us that EXACT confusion
pearl pondBOT
dusk raven
lmao
lavish stream
cooked
carmine juniper
Ik these
lavish stream
it’s fine though
dusk raven
its ok, don't say that
robust lynx
Let’s not.
carmine juniper
Im aware that root square is just number under the root
I needed help with the first one
proper nova
Im aware that root square is just number under the root
squaring and square roots cancel each other
carmine juniper
Ik that…
dusk raven
I needed help with the first one
yeah its ok, you understood right
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fading ledge
pearl pondBOT
crystal dew
,rccw
jolly parrotBOT
ivory basin
This ain't maths
rigid mist
is this a test
pearl pondBOT
@fading ledge Has your question been resolved?
fading ledge
It is a reasoning question
toxic lichen
ok well what do you think the answer is
we can't really say much without spoiling it
proper nova
Provide a translation please
toxic lichen
Provide a translation please
there's an English version there
proper nova
there's an English version there
My bad
bronze heath
This feels like social science
toxic lichen
and op is gone
pearl pondBOT
@fading ledge Has your question been resolved?
still hamlet
I've no clue where to begin
pearl pondBOT
I've no clue where to begin
still hamlet
M(T) is the matrix of the linear map T
waxen agate
I've no clue where to begin
well, what does dim range T = 1 mean?
still hamlet
well, what does dim range T = 1 mean?
a single basis vector of W can span the range
still hamlet
forward
waxen agate
a single basis vector of W can span the range
so every output points in the same direction w_0, the whole problem becomes 1) choose a basis of W starting with that direction w_0 and 2) choose a basis of V so that the scalar multiplier of w_0 is 1 on every basis vector
gilded vortex
Hi y'all
crystal dew
Hi y'all
hi, welcome to the server! to chat, please head to #discussion or #chill.
copper matrix
welcome
still hamlet
so every output points in the same direction w_0, the whole problem becomes 1) c...
for 1), we can extend w_0 to a basis, I'm not sure about 2) though
do i take a v_0 that is mapped to w_0 and then extend the basis?
waxen agate
you do take v_1 with T(v_1) = w_0, but instead of just extending it arbitrarily, you extend it by adding vectors from ker(T) to it.. (we know by rank nulity that dim ker(T) = dim V - 1)
still hamlet
wouldn't that force all coeffs to be zero
because if Tv_k=0,
=> 0=a_0 w_0+...+a_m w_m and it is linearly independent because it's a basis so each a_n = 0
waxen agate
where did that linear combination come from?
diamond$ds T\p(sum a_i v_i) = sum a_i T(v_i)$.
jolly parrotBOT
Oléagineux Distilliànus VIVII
still hamlet
waxen agate
I don't understand your confusion
rank 1 automatically forces all rows except the first to be zero once you pick the right basis of W
still hamlet
but the question says we need all entries as 1
waxen agate
oh I somehow misread that
still hamlet
my thinking is if Tv_k = 0,
0 = A_1,k * w_0 +..+ A_m,k w_(m-1)
but if this is 0, every A_p,q has to be zero
still hamlet
oh I somehow misread that
oh 😭
waxen agate
one sec
pearl pondBOT
@still hamlet Has your question been resolved?
waxen agate
Choose u in V such that T(u) = w_0 and set u = v_1, then you let k_2,...,k_i be a basis of kerT and define v_i = v_1 + k_i then T(v_i) = w_0..
and (v_1,...,v_n) is a basis of V
still hamlet
how do u know every w_m apart from w_0 will also have coeff 1
waxen agate
choose a basis of W such that w_1 + ... + w_m = w_0
waxen agate
still hamlet
.close
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wind lagoon
Problem. Determine $E(X + Y \mid X > Y)$ for $X,Y$ i.i.d. $N(0,1)$. \
I've read somewhere you can simply relabel $X, Y$ so that $X$ is almost surely greater than $Y$ without changing what you're taking the expectation of (and $P(X=Y) = 0$), so $$E(X + Y \mid X > Y) = E(X+Y) = E(X) + E(Y) = 0.$$ I don't understand this reasoning. $A={X>Y}$ is an event. So $$E(X + Y \mid X > Y)=\frac{E[(X+Y)\mathbf{1}_A]}{P(A)}.$$How do we go from here?
jolly parrotBOT
psie
zenith mural
how do I solve this?
pearl pondBOT
how do I solve this?
zenith mural
i honestly havent a clue
turbid summit
Firstly, what can we find
We know it is a hemisphere and we know from the bottom centre of the cuboid to the top left corner is the radius of the hemisphere, right?
turbid summit
Now from here, what can you tell from these 2 lines?
bronze heath
hint: congruent triangles
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wind lagoon
If $\phi$ is the standard normal density, i.e. $(2\pi)^{-1/2}\exp(-x^2/2)$, how can I compute $$\int_c^\infty x^2 \phi(x) , dx ?$$
jolly parrotBOT
psie
wind lagoon
Here c >= 0.
jolly parrotBOT
random ermine
yup
wind lagoon
Ah, ok. Shame.
potent light
Can't you integrate by parts? With u = x and v' = x*psi?
random ermine
smth prolly goes wrong in ur new integral
cursive wraith
Can't you integrate by parts? With u = x and v' = x*psi?
That's how you get the error function
erf(e) is oddly close to 1 but isn't expressible by standard functions
potent light
I didn't see the lower and upper bounds. 😂
pearl pondBOT
If $\phi$ is the standard normal density, i.e. $(2\pi)^{-1/2}\exp(-x^2/2)$, how ...
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
random ermine
I didn't see the lower and upper bounds. 😂
yeah if it was -infty to infty it would be doable
cursive wraith
erf(e) is oddly close to 1 but isn't expressible by standard functions
Well, not 'oddly' close, since e is pretty big we approach 1
erf(infty) is 1
wind lagoon
!xy
We know if $X$ and $Y$ are jointly normally distributed then they are independent iff they are uncorrelated. Now, let $X\sim N(0,1)$ and $c\geq0$. Define $Y_c$ as follows, $$Y_c=\begin{cases} X,&\text{for }|X|\leq c\ -X&\text{for }|X|> c\end{cases}.$$
(a) Show that $Y_c\sim N(0,1)$.\
(b) Show that $X$ ad $Y_c$ are not jointly normal.\
(c) Show that $\operatorname{Cov}(X,Y_0)=-1$ and that $\operatorname{Cov}(X,Y_c)\to 1$ as $c\to\infty$. Show that there exists $c_0$ such that $\operatorname{Cov}(X,Y_{c_0})=0$.\
(d) Show that $X$ and $Y_c$ are not independent (when $c=c_0$).
jolly parrotBOT
psie
wind lagoon
I'm stuck on (c). Computing $\operatorname{Cov}(X, Y_c) = E[X Y_c]$ (since $E[X] = E[Y_c] = 0$), we have
\begin{align*}E[X Y_c] &= E[X Y_c \cdot \mathbf{1}{|X| \leq c}] + E[X Y_c \cdot \mathbf{1}{|X| > c}] \
&= E[X \cdot X \cdot \mathbf{1}{|X| \leq c}] + E[X \cdot (-X) \cdot \mathbf{1}{|X| > c}] \
&= E[X^2 \mathbf{1}{|X| \leq c}] - E[X^2 \mathbf{1}{|X| > c}] \
&= E[X^2] - 2 E[X^2 \mathbf{1}{|X| > c}] = 1 - 2 E[X^2 \mathbf{1}{|X| > c}].\end{align*}
Let $f(x) = x^2 \phi(x)$ where $\phi$ is the standard normal density. Then
$$E[X^2 \mathbf{1}_{|X| > c}] = 2 \int_c^\infty x^2 \phi(x) , dx.$$This is where I'm stuck.
jolly parrotBOT
psie
pine jay
I'm stuck on (c). Computing $\operatorname{Cov}(X, Y_c) = E[X Y_c]$ (since $E[X]...
What happens as c approaches infty here ?
In the last few lines
wind lagoon
What happens as c approaches infty here ?
I'd say the integral vanishes, or? But I can't prove it. After all, $x^2$ gets very large, though $\phi(x)$ gets very small.
jolly parrotBOT
psie
pine jay
Yes it vanishes
Just need to show the integral is convergent (for every c>= 0)
Which is a lot more doable
wind lagoon
Which is a lot more doable
Hmm, ok. How should one approach this if the integral can not be solved?
wind lagoon
Hmm.
pine jay
Have u never had to show a certain integral converges before?
Seems like a prerequisite for this problem
wind lagoon
Have u never had to show a certain integral converges before?
It was a long time ago. 😄 I'm a bit rusty with integral convergence. I'm better with series convergence.
wind lagoon
We can go through it together
We need to show perhaps that it is bounded above by something that does converge?
pine jay
Mhm
wind lagoon
I notice that the integrand is nonnegative.
pine jay
Mhm
Also we can safely ignore most of the constants or factors
The meat of the integrand is x^2 e^(-x^2)
pine jay
I notice that the integrand is nonnegative.
I mean
Yes this is a good observation
You can probably see the other more trivial approach to this
wind lagoon
Also we can safely ignore most of the constants or factors
Yeah. Perhaps this would do it. $$\int_c^\infty x^2\phi(x),dx\leq \int_{-\infty}^\infty x^2\phi(x),dx=E[X^2].$$ Now, $\operatorname{Var}(X)=E[X^2]=0$, since $X\sim N(0,1)$. Is this correct?
jolly parrotBOT
psie
potent light
Var(X) is not 0, it is 1
pine jay
Oops yeah but the main point is its finite
potent light
The argument is still valid
wind lagoon
Var(X) is not 0, it is 1
Yeah, typo. 😄
wind lagoon
Mhm
Ok. So we know $\int_{c}^\infty x^2\phi(x),dx$ is finite for all $c$. Why would it vanish for $c\to\infty$? Also, for $c=0$, why is the integral $-1$, and for which finite $c$ is the integral $0$?
jolly parrotBOT
psie
wind lagoon
I'd expect the integral to equal 1/2 for c = 0. I'm mistaken here I think.
pine jay
Ok. So we know $\int_{c}^\infty x^2\phi(x)\,dx$ is finite for all $c$. Why would...
The last one follows by intermediate value theorem if im not missing anything
For c = 0, what have u tried?
The first one is almost by definition
Try for urself
wind lagoon
For c = 0, what have u tried?
Well, I was thinking $$1=\int_{-\infty}^\infty x^2\phi(x),dx=2\int_{0}^\infty x^2\phi(x)\implies I=\frac12$$where $I$ is the integral, isn't it?
jolly parrotBOT
psie
pine jay
Mhm
pine jay
I'm stuck on (c). Computing $\operatorname{Cov}(X, Y_c) = E[X Y_c]$ (since $E[X]...
Okay I think I see a mistake
Why simplify from X^2 1_{|X| <= c} to X^2?
If c = 0
Then this part must vanish
Hence you get -1
But the argument from before still works when we let c -> infty
It’s only then we can simplify it to X^2
wind lagoon
Why simplify from X^2 1_{|X| <= c} to X^2?
Hmm, ok. I will look at my attempt again. Thanks for the comments. Will get back. 🙂
wind lagoon
It’s only *then* we can simplify it to X^2
Ok. Yeah, I probably simplified too much. It's not necessary to do as much as I did. One can simply read off the case c = 0 implying -1 and c tending to infty implying 1 from the expressions earlier. But regarding there existing a c0 such that Cov(X, Y_c0) = 0, you mentioned the intermediate value theorem. Does this not apply to continuous functions on intervals [a,b]?
I'm looking at Theorem 4.23 in Rudin, and there he requires continuity and intervals of the form [a,b]. 🤓 I'm not sure which are the continuous functions we are looking at here.
pine jay
Consider f(c) = Cov(X, Y_c) as a function of c
We can choice our own interval
Since we know f(c) > 0 for big enough c
And f(c) = -1 for c = 0
So an interval like [0, 1337]
Should work
Or at least we know there’s an endpoint that works
Only have to show f is continuous
Which I would say follows by elementary results from analysis
(Integral of continuous function is C^1 and so in particular continuous)
wind lagoon
(Integral of continuous function is C^1 and so in particular continuous)
Ah ok. You mean the simplification I did way up above, where I expressed Cov(X,Y_c) as a constant minus an integral?
pine jay
Which I would say follows by elementary results from analysis
An application of the fundamental theorem of calculus
wind lagoon
Nice. Ok. 👍 Only (d) left then. 😄
wind lagoon
Yeah
I figured it out. 🤓 Thanks for your help above! Appreciate it. 
I'm closing here.
.close
pearl pondBOT
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quick venture
Hi,guys I have statics problem
pearl pondBOT
Hi,guys I have statics problem
quick venture
im not sure how to do this could someone help please
honest berry
im not sure how to do this could someone help please
ok imma get on it rn
So,between the beam carries a uniformly distributed load, and the shear force diagram there is a straight sloping line.
betwenn b and c
shear at b is 19.8,at c is -10.2 and lenght bc is 6,so change in shear is shear at b - shear at c is30kn so uniformly distributed load is change in shear/length so answer to no 19 is 5.0KN/Mˆ-1
Then for 20
so for mamb it occurs where shear force is minimum or 0,so look for where that is,we can see that shear drops from 19.8 to -10.2,so distance from b to 0 shear is
$x=(19.8/30)*6=3.96m$
jolly parrotBOT
LUC1DV1B3S_010
honest berry
then bending moment is area under the sfd from b to that point. that area is a triangle
and normal area is
quick venture
So,between the beam carries a uniformly distributed load, and the shear force di...
isnt it s slope line between b and c
honest berry
$A=1/219.83.96=39.2KNM$
jolly parrotBOT
LUC1DV1B3S_010
honest berry
isnt it s slope line between b and c
Yes it is
quick venture
i dont understand this bit isnt moment = to area under SFD
honest berry
i meant constant udl
honest berry
**LUC1DV1B3S\_010**
but its in the option
honest berry
i dont understand this bit isnt moment = to area under SFD
yes its equal but that area rrly means the bending moment area under the shear force diagram from a reference point up to that point
you get?
anyways ive given you the answers and solutions
honest berry
is this high school statics for you?
and it isnt statistics
its engineering/applied mechnics
quick venture
its engineering/applied mechnics
in high school?
quick venture
**LUC1DV1B3S\_010**
im assuming you used proportional triangles for this
pearl pondBOT
@quick venture Has your question been resolved?
pearl pondBOT
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rare field
where do i even start 
pearl pondBOT
where do i even start <:catthink:526983414226747393>
rare field
i was thinking of differentiating but... can i even
cursive wraith
Do you know the Maclaurin series for exp(z)
rare field
ive only learnt how it works for reals
tho i assume it generalises quite nicely
cursive wraith
It does
rare field
same thing?
cursive wraith
it is in fact the general definition of exp for complex numbers
rare field
huh neat
tropic saddle
i was thinking of differentiating but... can i even
why shouldnt you be able to?
rare field
i have no experience with differentiating complex things
tropic saddle
works the same as differentiating e^(2x)
cursive wraith
Well it is a real variable
rare field
its the root of unity thats bugging me
cursive wraith
omega^k is a constant like any other
rare field
sooo i can differentiate per se exp(ix) without something going wrong for smth like that?
and the derivative is iexp(ix)
tropic saddle
yes
rare field
huh oke
pearl pondBOT
@rare field Has your question been resolved?
pearl pondBOT
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proud frost
So I’m taking another stab at trying to understand Axiom of Choice: #help-8 message
And my understanding is:
Why is AC is needed: ZF set theory alone does not allow us in general to somehow “choose” an element from each set of a collection of sets. That is in general there is no way to specify a “general rule” for choosing elements for any arbitrary collection of sets.
I.e. the statement “pick any element from each set in the collection all at once” is NOT allowed in ZF set theory since we cannot create such a set from the axioms alone and there is no deterministic/well defined function?
Well-ordering: Hence this is why we need well-ordering to be assumed. Since then it provides the necessary structure to create a choice function (based upon a well ordering)
Would this be correct? Are there any misunderstandings I have and/or missing pieces?
cursive wraith
Why is AC is needed: ZF set theory alone does not allow us in general to somehow “choose” an element from each set of a collection of sets.
Yes
That is in general there is no way to specify a “general rule” for choosing elements for any arbitrary collection of sets.
In ZF without AC yes
I.e. the statement “pick any element from each set in the collection all at once” is NOT allowed in ZF set theory since we cannot create such a set from the axioms alone and there is no deterministic/well defined function?
Yes
Well-ordering: Hence this is why we need well-ordering to be assumed. Since then it provides the necessary structure to create a choice function (based upon a well ordering)
"Well ordering" (as in Well-ordering Theorem/axiom, stating that any set can be given a well-ordering) is equivalent to AC. So if you want to be able to use the axiom of choice, just assume AC directly, or the well-ordering theorem, given that when you assume one of them you immediately get the other as well
proud frost
“Well ordering" (as in Well-ordering Theorem/axiom, stating that any set can be given a well-ordering) is equivalent to AC. So if you want to be able to use the axiom of choice, just assume AC directly, or the well-ordering theorem, given that when you assume one of them you immediately get the other as well
Yes sorry that is what I meant. The “Well Ordering” as an any non empty set X has a well ordering. And trying to show the equivalence between Well Ordering and AC.
I.e. the statement “pick any element from each set in the collection all at once” is NOT allowed in ZF set theory since we cannot create such a set from the axioms alone and there is no deterministic/well defined function?
Regarding this statement, I think I’m still kind of shaky as to why this cannot be proved using ZF alone. Since the definition of a “rule” seems a bit arbitrary and I’m not sure why “pick any element from each set” would not be a valid rule
cursive wraith
> “Well ordering" (as in Well-ordering Theorem/axiom, stating that any set can b...
Regarding this statement, I think I’m still kind of shaky as to why this cannot be proved using ZF alone. Since the definition of a “rule” seems a bit arbitrary and I’m not sure why “pick any element from each set” would not be a valid rule
I mean, it's been an axiomatic/philosophical question for a while whether you're allowed to 'pick any element from each set' for an infinite number of sets, and how big the infinity can be such that you can still choose. From a human perspective, we can only accomplish a finite amount of things because we need time to accomplish each thing, so the idea of "choosing randomly an element in this set" requires you to almost "physically" grab an element in each set, and that for an infinite number of sets, while you only have a finite amount of time.
Also, I'm sure you're aware of the kinds of things we can prove once we allow ourselves to magically use the axiom of choice (like the Banach Tarski paradox)
Have you seen the Veritasium video on the AC subject? It's very entertaining and I think is well written so you can learn a thing or two from it, at least I recommend it
proud frost
> Regarding this statement, I think I’m still kind of shaky as to why this canno...
I mean, it's been an axiomatic/philosophical question for a while whether you're allowed to 'pick any element from each set' for an infinite number of sets, and how big the infinity can be such that you can still choose. From a human perspective, we can only accomplish a finite amount of things because we need time to accomplish each thing, so the idea of "choosing randomly an element in this set" requires you to almost "physically" grab an element in each set, and that for an infinite number of sets, while you only have a finite amount of time.
Ah okay. So like “pick any element from each set” for an infinite number of sets is more philosophical than mathematical? As in we just define set theory as in saying that not a rule? Since then I’m not even sure what can be a rule then/why it’s not a rule
Also, I'm sure you're aware of the kinds of things we can prove once we allow ourselves to magically use the axiom of choice (like the Banach Tarski paradox)
Yeah I remember reading a proof (Hahn-Banach Extension Theorem) that used Zorn’s Lemma and then began down the rabbit hole 😭
cursive wraith
> I mean, it's been an axiomatic/philosophical question for a while whether you'...
Math axioms always have this philosophical aspect to it. The purpose of axioms is that you need to assume the most basic and fundamental things in order to build stuff on top of it that is less basic/fundamental. So an axiom will always be subject to questions like "Why do we need it?" or "Is it really needed?" all the way up to "Isn't that too strong of an axiom" and "Can't we assume a much weaker axiom instead?"
toxic fractal
remember that the whole point of defining axioms in the first place is just establishing your avaliable "computer instructions"
and everything that isnt there, or isnt deduced from there, cant be used
crystal dew
maybe OP can think about axioms as being pre-defined rules of a game, and any theorems based on those axioms as 'strategies' that players develop using the axioms/rules
proud frost
maybe OP can think about axioms as being pre-defined rules of a game, and any th...
I see yeah. I’ve been going off of axioms are just rules I must play by/must be true and then theorems + lemmas etc build off from there
toxic fractal
basically
in terms of programming languages, it would be the reserved keywords, for example
proud frost
Math axioms always have this philosophical aspect to it. The purpose of axioms i...
Yeah ig more so where in ZF does it disallow statements like “choose any element from each set”? I think that’s kinda my main question now, as this seems to be a valid rule
cursive wraith
Yeah ig more so where in ZF does it disallow statements like “choose any element...
using the language above, you haven't defined that you can do that, “choose any element from each set”
toxic fractal
well, i guess some do disallow things, like the collection of all sets not being a set
imagine two very basic programs in pseudocode:
Program 1:
ADD(2, 3)
error: ADD is not defined
Program 2:
#define ADD(a, b):
return a+b
ADD(2,3)
5
your axiom would be the #define here
you have defined a thing that you can do
proud frost
using the language above, you haven't defined that you can do that, “choose any ...
Oh like here using the ZF axioms alone I can’t do that or
proud frost
an axiom just defines a thing that you can do
Yeah but I thought ZF allows us to use that statement to create a set
toxic fractal
the axiom of choice is not needed for finite sets, it's just for allowing some stuff on infinite sets
proud frost
I think am missing smthing you guys are saying but idk what
crystal dew
I don't think you need an axiom to create a set
proud frost
Yeah
crystal dew
but you do need axioms to assume certain properties about the set you create
proud frost
I just don’t really understand why the rule “choose any element from each set” can’t be used in ZF
Or like proven ig?
Idk how to say, like it seems I’m not allowed to do this in ZF alone (otherwise AC would be meaningless) but I’m not sure why
cursive wraith
Idk how to say, like it seems I’m not allowed to do this in ZF alone (otherwise ...
I guess the first instance of choosing you get is related to the quantifier $\exists$
jolly parrotBOT
Rafilouyear2026
cursive wraith
to say I can choose something in set $A$ is "applying" the statement "$\exists x (x\in A)$"
glass meadow
You might want to realize that set theory (and in particular ZF) is built from first-order logic
jolly parrotBOT
Rafilouyear2026
cursive wraith
from first order logic, the rule you have is that you can choose one element from one set
glass meadow
I recommend reading this: https://math.stackexchange.com/a/1083058
cursive wraith
So of course, repeating this operation many times in a row, it's ok to choose one element each for a finite number of sets
But just like you can't give a computer an infinite amount of instructions
you can't, from this rule only, choose one element each for an infinite number of sets
crystal dew
maybe one close analogy is mathematical induction and how it works for any finite set, but does not imply the validity of the infinite case
proud frost
You might want to realize that set theory (and in particular ZF) is built from f...
Wait sorry, how is first order logic related to this/to the statement? I’m not too familiar with it
toxic fractal
ZF and ZFC are defined after first order logic
cursive wraith
Oh like here using the ZF axioms alone I can’t do that or
All the axioms you see written here are using first order logic to be formulated
glass meadow
Wait sorry, how is first order logic related to this/to the statement? I’m not t...
Rafilou hinted at it. ∃x, p(x) means you can choose an x such that p(x)
proud frost
you can't, from this rule only, choose one element each for an infinite number o...
ooh wait. So are you saying my statement/rule "choose an element from each set from all sets in the collection" is the same as saying for all sets S in the collection, $\exists x(x \in S)$? But isn't this a valid rule? Like this is choosing one lement each for an infintie number of sets
jolly parrotBOT
LXDL
proud frost
I recommend reading this: <https://math.stackexchange.com/a/1083058>
I think what Rafilout is saying is similar to what Nel shared here
but isn't for all set $S$ in the collection ... bascially syaing we can apply all at once?
jolly parrotBOT
LXDL
crystal dew
first-order logic cannot quantify over undefined sets, iirc
let alone an infinite number of them
glass meadow
Yes, and on top of that the resulting thing you would get from "choosing an element from each set in the collection" might not qualify to be a set
Like you can't tell unless you prove it's a valid set by some other means (see Russel's paradox for an example of an invalid set)
proud frost
first-order logic cannot quantify over undefined sets, iirc
ohhh okay woudl you happen to know fo any resoruces that teach first order logic from basics? ngl i am kinda nooby at this
and pretty lacking in that department but woudl like a solid foundation in it
crystal dew
https://forallx.openlogicproject.org/
this is open-source (or rather, free to distribute).
pearl pondBOT
@proud frost Has your question been resolved?
pearl pondBOT
pearl pondBOT
Available help channel!
quasi patrol
im trying partial fractions but i cant work it out or find a pattern
