wait that actually makes so much sense
#help-27
frosty crescent
becuase its a property for all sets
transfinite induction is basically epsilon induct specifially for ordinals
cerulean ruin
\textbf{Induction Scheme of Theorems:}\textbf{ Let $\varphi(u,z_1,\dots, z_k)$ be any L-formula \\
Theorem:\ $\forall z_1 \dots \forall z_k \Big(
\big[
\varphi(0, z_1, \dots, z_k)
;\land;
\forall n \big(
\varphi(n, z_1, \dots, z_k)
\implies
\varphi(S(n), z_1, \dots, z_k)
\big)
\big]
;\implies;
\forall n, \varphi(n, z_1, \dots, z_k)
\Big).$}
is it this
woven radishBOT
toast
cerulean ruin
rip formatting
cerulean ruin
wait i am not sure we learned epsilon induction
frosty crescent
if a statement is true for empty set
frosty crescent
if a statement is true for empty set
and for any set if its true for its elements implies its true for it
then the formula is true for all sets
frosty crescent
transfinite induction is induction + unreachable ordinals
frosty crescent
limit ordinals
frosty crescent
is epsilon-indctuion covered in these ^
basically
but for ordinals
would post theorem but typing on phone 🙁
cerulean ruin
do we need to do induction actually
we could consider just the set of all ordinals that are contained within d
frosty crescent
we could consider just the set of all ordinals that are contained within d
how do you know that exists
Ord is a class 🙁
frosty crescent
so you cant schema specify
frosty crescent
if d is a set
wait oh
cerulean ruin
if there is a subset of ordianls in d that isnt a set, then d cant be a set
frosty crescent
do the set of elements of d such that they are in Ord
cerulean ruin
like separation scheme
frosty crescent
thats the construction oops
frosty crescent
do the set of elements of d such that they are in Ord
and this must have a smallest element right
cerulean ruin
yes
bc set of well orders
well
set of ordinals must be well ordered
so there exists a least element
frosty crescent
wait
i just realized
its not the least element of d intersect Ord
its the least element thats NOT in it
@cerulean ruin
i think you have to take infinitary union over d intersect Ord
call d intersect Ord = x
least element of S(infinitary union x) \ x
cerulean ruin
i think you have to take infinitary union over d intersect Ord
Isn’t this just separation scheme applied to d
frosty crescent
d intersect Ord is {x ε d : x ε Ord}
cerulean ruin
so take the Union of this set ?
cerulean ruin
hmmm
I think this will make sense once I write it down
Immhust just walking around rn
frosty crescent
find min of S(inf union x) \ x
S is successor and \ is complement
btw inf union x is an ordinal
since its taking the max of the ordinals
in x
cerulean ruin
X = {0,1,2,4,6}
U X = {0,1,2,3,4,5}
S (U X) = 6
5 \ X = {0,1,2,3,4,5} \ {0,1,2,4,6}
= {3,5}
min = 3
hmm
I think the intuition behind S(U x) is to achieve the biggest ordinal and put it into its definitional form@frosty crescent
Then you eliminate the common elements of with X and take the minimum and you get alpha
frosty crescent
I think the intuition behind S(U x) is to achieve the biggest ordinal and put it...
yeah
cerulean ruin
\textbf{c) Let $d$ be any set. Prove there is an ordinal $\alpha$ such that $\alpha \notin d$, yet every $\beta < \alpha$ is an element of $d$.}
\\
let $d$ be any set. By separation scheme let $X = {x \in d : x \text{ is an ordinal}}$. Then $X$ is a set of ordinals. Observe that $\bigcup X$ is a set by the Union Axiom. We can take $S (\bigcup X) \backslash
X$ which is a set by a theorem. Note because this is still a set of ordinals, we can take the minimum of this set since it is a well-ordered set. We will argue that $\alpha = \min S(\bigcup X) \backslash X \notin d$, but every element $\beta < \alpha$ will be in $d$.
is this set up fine
woven radishBOT
toast
cerulean ruin
its kinda hard to formalize it
cuz it seems like im defininig a random set
even though i really just want to take the biggest ordinal of X in its definition form, then set difference with X itself
but that sounds really unrigorous idk lol
cerulean ruin
\textbf{c) Let $d$ be any set. Prove there is an ordinal $\alpha$ such that $\alpha \notin d$, yet every $\beta < \alpha$ is an element of $d$.}
\\
let $d$ be any set. By separation scheme let $X = {x \in d : x \text{ is an ordinal}}$. Then $X$ is a set of ordinals. Observe that $\bigcup X$ is a set by the Union Axiom. We can take $S (\bigcup X) \backslash
X$ which is a set by a theorem. Note because this is still a set of ordinals, we can take the minimum of this set since it is a well-ordered set. We will argue that $\alpha = \min S(\bigcup X) \backslash X \notin d$, but every element $\beta < \alpha$ will be in $d$.
\\
First, assume by way of contradiction that $\alpha \in d$. So $\alpha$ is an ordinal, thus $\alpha \in X$, a set of all ordinals in $d$. But by definition, $\alpha \in S(\bigcup X) \backslash X \implies \alpha \in S(\bigcup X)$ and $\alpha \notin X$ hence a contradiction.
\\
Now suppose by way of contradiction there exists a $\beta < \alpha$, but $\beta \notin d$. Because $\beta \notin d$, then $\beta \in S(\bigcup X) \backslash X$. But this contradicts $\alpha$ being the $\in$-least element of $S(\bigcup X) \backslash X$.
\\
Thus we have found our ordinal such that $\alpha \notin d$, but every $\beta < \alpha$ is an element of $d$ $\blacksquare$
@frosty crescent
woven radishBOT
toast
cerulean ruin
$\beta \in S(\bigcup X) \backslash X$
woven radishBOT
toast
cerulean ruin
idk if i can assume it
logically beta < alpha and b ecause alpha is in this set, shouldnt beta clearly be in this set, since we're doing the union operator?
well ok evidently, $\beta \notin X$ since $X \subseteq d$
woven radishBOT
toast
cerulean ruin
and $\alpha \in S(\bigcup X)$, and $\beta \in \alpha$
woven radishBOT
toast
cerulean ruin
thus by transitivity, $\beta \in S(\bigcup X)$
woven radishBOT
toast
cerulean ruin
\textbf{c) Let $d$ be any set. Prove there is an ordinal $\alpha$ such that $\alpha \notin d$, yet every $\beta < \alpha$ is an element of $d$.}
\\
let $d$ be any set. By separation scheme let $X = {x \in d : x \text{ is an ordinal}}$. Then $X$ is a set of ordinals. Observe that $\bigcup X$ is a set by the Union Axiom. We can take $S (\bigcup X) \backslash
X$ which is a set by a theorem. Note because this is still a set of ordinals, we can take the minimum of this set since it is a well-ordered set. We will argue that $\alpha = \min S(\bigcup X) \backslash X \notin d$, but every element $\beta < \alpha$ will be in $d$.
\\
First, assume by way of contradiction that $\alpha \in d$. So $\alpha$ is an ordinal, thus $\alpha \in X$, a set of all ordinals in $d$. But by definition, $\alpha \in S(\bigcup X) \backslash X \implies \alpha \in S(\bigcup X)$ and $\alpha \notin X$ hence a contradiction.
\\
Now suppose by way of contradiction there exists a $\beta < \alpha$, but $\beta \notin d$. Because $\beta \notin d$, then $\beta \notin X$, since $X \subseteq d$. Now observe that $\bigcup X$ is an ordinal if $X$ is a set an ordinal by a proposition. By another proposition $S(\gamma)$ is a ordinal for any ordinal $\gamma$. Thus $S(\bigcup X)$ is an ordinal. Therefore, $\beta, \alpha, S(\bigcup X)$ are ordinals and $\beta \in \alpha$ and $\alpha \in S(\bigcup X)$ by definition. By transitivity of ordinals, then $\beta \in S(\bigcup X)$. Because $\beta \notin X$, then $\beta \in S(\bigcup X) \backslash X$ and $\beta < \alpha$. But this contradicts $\alpha$ being the $\in$-least element of $S(\bigcup X) \backslash X$.
\\
Thus we have found our ordinal such that $\alpha \notin d$, but every $\beta < \alpha$ is an element of $d$ $\blacksquare$
ok more formal
woven radishBOT
toast
cerulean ruin
@frosty crescent ur acc goated for that hint though
😭
idk how u came up with that so fast
cerulean ruin
wow nice solution!
Thx! Honestly tho I really needed that hint where I had to consider the minimum ordinal not in d
I think I would be still stuck on this problem without it
😭
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cerulean ruin
.close
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Available help channel!
random badge
Cot(x)>= -1
devout snowBOT
Cot(x)>= -1
hollow vine
yes?
random badge
I solved for the solution, but Im having trouble interpreting the points in relation to the inequality
what to do after finding them
as In how do I know
the intervals and wether they are above or below
hollow vine
cotangent decrease, try using arccot to find it, or if you don't know that bring everything to one side and solve for when it is =
random badge
im sorry what
what do you mean by bring everything to one side and solve for when it is =
I feel like thats what I did when I said I solved for the solutions
random badge
Im confused as to the intervals
its a inequality
how do I tell when the interval increases or decreases
hollow vine
find it's first order derivative
hollow vine
i'm not sure then
random badge
<@&286206848099549185>
hollow vine
usually, you find it's derivative, negative in this case, and since it is then it's before the solution point
try graphing it in desmos
random badge
I am not learning derivatives by tomorrow
potent dirge
I am not learning derivatives by tomorrow
There are some derivative rules you could learn by tomorrow
Power rule being the main one
potent dirge
Im learning precalculus not calculus
Can you give an example of one of the questions?
potent dirge
solve for the domain of [0,2pi)
So, what angles between 0 and 2pi result in cot >= -1
random badge
I have no clue
for these problems
I think unit circle
is the easiest way to solve it
problem is idk where to start imaginging it
mortal tendon
a graph might help
random badge
cotangent and tangent are a bit harder to imagine
than sin and cosine
cause sine is just simple above below x axis
and cosine for y axis
I know the pattern of tangent
potent dirge
cause sine is just simple above below x axis
But you really only need sine and cosine
potent dirge
But you really only need sine and cosine
Because every other ratio can be described as sine and cosine
random badge
the problem is after that
we have like different properties
of the different graphs
random badge
like cotangent has asymptotes at 0 and pi
likewise secant has asymptotes at like pi/2 and 3pi/2 right
potent dirge
like cotangent has asymptotes at 0 and pi
And you can figure that out because:
cotθ = cosθ/sinθ
sinθ can't equal 0, because it would cause a division by 0. And some of the values where that happens are:
θ = 0, π
random badge
OH
OHHHH
that makes SENSE
ok
Cot(x)>= -1
so i got the solutions 3pi/4
and
7pi/4
and tangent is 0 at 0 and pi
and 2pi
potent dirge
Cot(x)>= -1
First tip, make the right side equal to 0
random badge
First tip, make the right side equal to 0
no I solved the inequality already
now
how do I know if cotangent increaes or decreases within 0, 3pi/4, pi, 7pi/4, 2pi?
how can I imagine this?
random badge
than k yoiu
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potent dirge
than k yoiu
Almost. Sorry bro
random badge
ur chill
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wet robin
I need help
devout snowBOT
I need help
wet robin
i do not understand why p = 0, p>0
there isn't enough data to write that down
im confused
topaz axle
p > 0 is what they asked you to do
willow helm
these shits easy
What was the point of stating this
devout snowBOT
last parrot
What was the point of stating this
Dude thought dudes tuff😂
sour ibex
@spice jetty
willow helm
. what happened here
runic dew
Am struggling with this question on my homework, cyclic permutation notation confused the hell out of me so i dont even know if i got part (a) right, did i?
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cerulean ruin
\textbf{b) Use Transfinite Induction to prove that for all ordinals $\alpha$, $V_\alpha$ is transitive and $V_\beta \subseteq V_\alpha$ for all $\beta < \alpha$}
\\
Let $()\alpha$ be the statement "$V\alpha$ is transitive and $(\forall \beta < \alpha), V_\beta \subseteq V_\alpha$. I will show $\forall \alpha(\alpha$ ordinal $\implies ()\alpha)$ by Transfinite Induction.
\\
I will first show $(*)\alpha$ holds. Observe that there are no such elements where $\beta < 0$. Thus $V_\beta \subseteq V_\alpha$ is vacuously true. $V_0$ is also vacuously transitive because we cannot pick any element of $V_0$. Thus $()_0$ holds
\\
Now I will assume $()\alpha$ holds and prove $(*){S(\alpha)}$. So for any ordinal, $V_\alpha$ is transitive and for all $\beta < \alpha$, $V_\beta \subseteq V_\alpha$. By definition, $V_{S(\alpha)} = \mathcal{P}(V_\alpha)$. Because $V_\alpha$ is transitive, and from part a), we know that $\mathcal{P}(V_\alpha)$ must be transitive as well. Then $V_{S(\alpha)}$ is transitive. Also by part a), $V_\alpha \subseteq \mathcal{P}(V_\alpha) = V_{S(\alpha)}$. For any $\beta < \alpha$, $V_\beta \subseteq V_\alpha$. But $V_\alpha \subseteq V_{S(\alpha)}$. So $V_\beta \subseteq V_{S(\alpha)}$ for any $\beta < S(\alpha)$. Therefore $()_\alpha \implies ()_{S(\alpha)}$
devout snowBOT
\textbf{b) Use Transfinite Induction to prove that for all ordinals $\alpha$, $V...
woven radishBOT
toast
cerulean ruin
Now, I will assume $\alpha$ is a non-zero limit ordinal and assume $()_\beta$ holds for all $\beta < \alpha$. I will show $()\alpha$ holds. Observe that for a non-zero limit ordinal, $V\alpha = \bigcup{V_\beta : \beta <\in \alpha}$. Let $\gamma \in \bigcup {V\beta : \beta <\in \alpha} \iff \exists V\beta^* \in {V_\beta : \beta < \alpha}$ such that $\gamma \in V_\beta^$. But $V_\beta^$ is transitive by $()\beta$, so $\gamma \subseteq V\beta^$. We also know that $V_\beta^* \subseteq V_\alpha$ by our hypothesis. Therefore $\gamma \subseteq \bigcup{V_\beta : \beta < \in \alpha}$. So $V\alpha$ is transitive.
\\
Now we want to show for all $\gamma < \alpha$, $V_\gamma \subseteq V_\alpha$. We know that $V_\gamma \subseteq V_\beta$ by $(*)\beta$, and $V\beta \subseteq V_\alpha$. Thus $V_\gamma \subseteq V_\alpha$.
\\
Therefore by Transfinite Induction, for all ordinals $\alpha$, $V_\alpha$ is transitive and for all $\beta < \alpha$, $V_\beta \subseteq V_\alpha$ $\blacksquare$
woven radishBOT
toast
cerulean ruin
Can someone review my proof :)
This is the first time doing transfinite induction so idk if its rihgt
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main hornet
I need help on my exam review!!!
I didnt take enough notes my last two classes so now im lost and the exam is tmmr 
these are the questions I could not figure out on my own using my notes
devout snowBOT
I need help on my exam review!!!
I didnt take enough notes my last two classes s...
main hornet
not sure how quickly someone will respond so in the meantime ill be working on #6
faint gorge
not sure how quickly someone will respond so in the meantime ill be working on #...
Sure what questions have you not figured
main hornet
id like to go in order so #5,, i cant find any notes similar to it so im clueless
untold ravine
id like to go in order so #5,, i cant find any notes similar to it so im clueles...
do you know how to graph an arbitrary function like say x^2 (when it is not given using cases)
main hornet
yea, thats the U shaped one i think
untold ravine
yeah
so here
if we go step by step
it first says that f(x) = 2 for x<-1
that means we draw the graph for the function that is just 2
but in the given range
so we only draw it till x=-1
main hornet
i dont understand 😓
untold ravine
right, if you were told to graph f(x)=2 (no further instruction) how would you do it
main hornet
straight horizontal line up 2
untold ravine
yes
so if the question was graph f(x) = 2, x<-1
then you have the same graph
but cut it off at x=-1
main hornet
OH🤦 ok yea
main hornet
so id write it as (-inf,-1)
untold ravine
the question doesnt actually ask for that, but yes that is the domain in which f(x)=2
the next line says f(x) = x+1, -1<=x<3
it is similarly interpreted
main hornet
its an open dot for less/greater than or equal to when graphing right?
main hornet
untold ravine
well the standard instruction is that they are reversed
like the filled dot is the "or equal to" end and the open dot is the other
otherwise this is perfectly correct
main hornet
and for the second part it would be increasing (-1,3)
untold ravine
increasing [-1,3)
main hornet
ok
untold ravine
[ -> including that end, ( -> not including that end
main hornet
right
untold ravine
and then the third part is 5-x^2, x>=3
if you want, i can share what the final graph for f(x) should look like
main hornet
yes please
untold ravine
main hornet
i forgot it would be all one graph lol
untold ravine
yeah, all three pieces combined give f(x). thats the meaning of this curly brace notation
main hornet
ok so it would be that graph and
constant (-inf,-1)
increasing [-1,3)
decreasing [5,inf)
for final answer?
main hornet
ill have to look at a few similar questions but i get it now ty
ok now #6 if you still have time?
main hornet
i thought it was y=mx+b, is that a different thing?
untold ravine
c,b are both names standing in for a number
so they are the same basically
right
main hornet
ok just making sure
untold ravine
so m is for slope
so if we have to find a line parallel to the given one
we have to make sure their slopes are same
that is we use the same m value
so the first step would be to write the given equation in y=mx+c form
main hornet
untold ravine
everythings correct except the last line
-3/-2 = 3/2
and -8/-2 = 4
(the -1's cancel out)
main hornet
so would that be the final answer?
untold ravine
no, this is only writing the given equation in standard form
from this we extract that the slope of the line we want to find is 3/2
and they already told us that the y intercept is -4
so b=-4
so we get that the new line's equation is y = 3/2 x - 4
which is the solution, maybe it would be safer to write it in the form they gave for the previous line
main hornet
3x-2y = 2
im pretty sure my prof posted an answer key with the assignment let me go look
untold ravine
you should recheck your working
main hornet
found it
main hornet
it could be a typo, same thing happened last time
ill ask my professor about it tmmr
i need help with this one, my notes just made me confused
supple knot
i need help with this one, my notes just made me confused
it's better to think of $(f\circ g)(x)$ as $f(g(x))$ so plug $g(x)$ in for $x$ in $f(x) = x^2 - 4$
woven radishBOT
riemann
supple knot
first line would be $f(g(x)) = (g(x))^2 - 4$
woven radishBOT
riemann
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sand fable
I lowk need help w smth
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I lowk need help w smth
trail eagle
sand fable
<:helpjustask:371662163489718273>
Not a prob if it's 9th grade math ? 😭
Im waiting for my friend to answer me so im checking here if i can get help
dim ridge
Ask i can help too
trail eagle
Just post it
sand fable
Need help with the last question
dim ridge
It's in french idk how to translate
Jsuis francais tkt frro
trail eagle
En gros tu dois montrer que DE = 0
dim ridge
Jsp c quoi 'confondus'
Ils sont au mm endroit
sand fable
En gros tu dois montrer que DE = 0
Aaaaa
dim ridge
C sa que tu dois montrer
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blazing wadi
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solid osprey
are there any constraints on a,b,c?
solid osprey
are a,b,c positive? integers? or what
solid osprey
2 -27 is -25 right
yes
blazing wadi
all positive
solid osprey
you can always pick (1,99999999999999999999999999,1) and get a really small value
solid osprey
you can pick b to be really large and a and c to be really small
blazing wadi
it says find the smallest amount
solid osprey
blazing wadi
-25 is big
oh nooo
-1 is big
+1 is small
2a -3b solution is gonna be negative?
if it says to be the smallest
reef orbit
wait would this be smallest as in least or smallest as in closest to 0?
blazing wadi
bigger the negative number the smaller it is
bigger the positive number the bigger it is
so away from 0 on negative side
reef orbit
so -999999999999 is smaller than -1 in this context?
reef orbit
-1 is biggest +1 is smallest
wait what?
reef orbit
ok
yeah unless there are other restrictions on a, b, c this equation has no lower bound
blazing wadi
but +2 is bigger than -1 right
im confused enough already and asking these in english makes me even more confused
reef orbit
yeah there's definitely a language barrier
blazing wadi
the answer has to be the smallest number
after doing the solution of 2a-3b+3c
and the answer is here
but im confused on why
if it was -25 after 2a-3b
why not make it smaller by making C 8
oh then it would be -1
if it was 3b-3c then C would have been 8
does this mean 3b is gonna be a negative number
or do u just calculate 2a minus 3b
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left cliff
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left cliff
hello guy, im back, how about this
twilit comet
hint: AC is a diameter. what does that say about the triangle ABC
soft umbra
Do you have any idea?
last parrot
then DBC = half of DOC
left cliff
right angle i know
last parrot
use theorems of angles isncribing
thick schooner
OAB is isosceles
left cliff
is abc and dab the same?
candid maple
actually, I may have a better idea, if I may. (spoiler): ||DOB is a straight line. what does that make angle COB? from there, what does that make angle x?||
twilit comet
bina's idea works too, it's shorter
last parrot
I would use other approach: ||What can you say about triangle COB? And what is the relationship of angles DOC and OBC?||
Or: ||What is angle AOB compare to DOC? What is the property of triangle AOB? The sum of angles in a triangle is?||
tepid stag
is it the straight line is 180 and 180-110?
some of angles on one side of straight lines, yes
left cliff
i think i got it
last parrot
great!
twilit comet
what's your answer, then
left cliff
is aob and boc the same?
twilit comet
uh
no
twilit comet
BOC is half of aob
you mean the angles?
that's clearly not true
last parrot
DBC
last parrot
you mean the angles?
that's clearly not true
i mistyped 
twilit comet
lol is fine
thick schooner
i mean instead of doing all that
u could've also just noted that <DOC = <BOA
and OAB is isosceles so you're done
candid maple
well, you can do every single listed approach to check.
left cliff
so if the line is a diameter, then the triangle in the diameter is 100% isoceles right?
candid maple
every single approach should result in the exact same answer.
left cliff
like the aob in this situation
tepid stag
you coulve also noticed angle bound by arc DC + arc BC at the centre would be 180, so angle bound by arc BC would be 70 and since A is the angle of BC from circle's circumference it would be 70/2=35 lol
left cliff
70+70 =140
ohh yeah my bad
candid maple
so if the line is a diameter, then the triangle in the diameter is 100% isoceles...
no. notice that the full length of the diameter is not creating the triangle you mentioned (AOB).
left cliff
no. notice that the full length of the diameter is not creating the triangle you...
is aob not isoceles? i though it is both 35 degree plus 110 degree
candid maple
an inscribed triangle is isosceles iff two of its arms are radii. (in this case, OA and OB.)
thick schooner
it is because AO = OB = r
candid maple
is aob not isoceles? i though it is both 35 degree plus 110 degree
AOB is isosceles but not for the reason mentioned.
left cliff
ohhhh
what about this?
my brain is even more confused
does parallel mean they have the same degree?
candid maple
they here meaning which two angles?
mark the two angles you think have the same measure.
sour ibex
I need help
left cliff
I need help
cont create multiple channel
candid maple
I need help
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sour ibex
!occupied, and please don't spam help requests in the help channels of others. p...
Nobody is coming
left cliff
Nobody is coming
you havent even post your question
sour ibex
you havent even post your question
Nobody is coming that's why
left cliff
Nobody is coming that's why
what do you expect them to answer
candid maple
you've just opened your channel without a question. please be patient, and send your question in your own help channel.
left cliff
tell you the answer before you ask?
sour ibex
what do you expect them to answer
I'm here
candid maple
plus, the way help channels work, you're supposed to lead with the question.
sour ibex
they are not paid so dont expect an instant reply
one should help for humanity
left cliff
one should help for humanity
yes but dont take it as granted
you have to wait
anyways, i have to continue my brainstorm with the question
sour ibex
No offence but we study this in grade 3 btw
left cliff
No offence but we study this in grade 3 btw
yeah i know
left cliff
No offence but we study this in grade 3 btw
im dumb if that the answer you expect
tepid stag
ragebait peak lowk
sour ibex
im dumb if that the answer you expect
I'm not saying that but I'm asking are you in grade 3 or else
left cliff
I'm not saying that but I'm asking are you in grade 3 or else
it ok gng, just stay in your help channel and discuss for answer
green crypt
cook brother 🔥🔥
sour ibex
i just want to do my math
We all forget the basic sometimes dw gng
candid maple
<@&268886789983436800> I'm sorry for the ping, but this is getting a little unproductive, in my opinion.
left cliff
is the angle of x same as c?
faint gorge
this is a good reference
you are a good reference
last parrot
Just kick this dawg guy😭
tender cobalt
is the angle of x same as c?
yea, you gotta name why its true though
left cliff
yea, you gotta name why its true though
you mean to proof it?
tender cobalt
here you just gotta name what the reason would be
btw proof is a noun
you want to prove it, you use "prove"
for now, doing a proof just means you list down your steps and why theyre true
left cliff
abc is isoceles right?
tender cobalt
wait wait you havent named the reason yet
tender cobalt
abc is isoceles right?
(if youre good at finding reasons, you dont need to ask questions like this)
left cliff
😭
tender cobalt
(this is how you can answer them on your own)
tender cobalt
wait wait you havent named the reason yet
lets look at these reasons
we have this big list of reasons
each row is one
left cliff
it because they are parallel
tender cobalt
thats not a specific enough reason
you can see all of these have parallel lines in them
but which one?
left cliff
alternate exterior angle
tender cobalt
youre saying its this one?
left cliff
yeah
tender cobalt
for this?
left cliff
for this?
yeah i think so
i cant find anything is similar
or is it the first one?
yeah i think it is alternate interior angles
so that is the reaon?
reason
candid maple
correct. x and the marked angle are alternate (interior) angles.
candid maple
correct.
left cliff
yes
last parrot
Then BCA=?
left cliff
180-42-90
candid maple
oh ABC is not isosceles; I'm really stupid here. sorry.
left cliff
shit yeah i forgot not every triangle are isoceles
candid maple
no, I'm not. that was a simple error.
last parrot
Eh okay, everyone make a mistake👍
left cliff
no, I'm not. that was a simple error.
you really help me alot
last parrot
Then you find BCA, because BC // OD, then x = BCA
left cliff
thank everyone
last parrot
*thanks
devout snowBOT
left cliff
im back again
this time should be faster
a should be 90-19
and since b and a are radii
they have the same length which mean this time they should be isoceles right?
so a is 71 and b is 71 too?
and x is 38 degree
am i correct
candid maple
.reopen
devout snowBOT
✅ Original question: #help-27 message
last parrot
No
left cliff
noooooooo
last parrot
I dont know how to call CAB in english (i know viet), but CAB is half of AOB
CAB in this case i call Góc chắn dây cung và tiếp tuyến let me translate
left cliff
ok
last parrot
Angle subtended by chords and tangents.
This is half of the angle from centre subtend same arc
last parrot
CAO is 90 degree, yes, but it WONT associate with the angles in this question
you only need CAB and AOB
devout snowBOT
@left cliff Has your question been resolved?
left cliff
so aob is double of cab?
acoustic onyx
What's the problem can you send it again?
left cliff
last parrot
Correct
acoustic onyx
Now if angle CAO is 90 degrees then this is easy but....
acoustic onyx
It's not given that angle CAO is 90 degrees nor there is any indication
left cliff
It's not given that angle CAO is 90 degrees nor there is any indication
but isnt it a tangent
acoustic onyx
but isnt it a tangent
Are you sure?
left cliff
Are you sure?
i dont know
acoustic onyx
Again nothing is mentioned
So we have to make some assumptions
We do know for a fact that triangle OAB is isosceles triangle
left cliff
yeah
left cliff
yes
acoustic onyx
Now if AC is a part of the tangent at A then angle CAO is 90 degrees
acoustic onyx
And if angle CAO is 90 degrees and CAB is 19 degrees then BAO is 90-19 = 71
acoustic onyx
If BAO is 71 then ABO is also 71 due to isosceles triangle property
left cliff
see this image
same as this?
last parrot
Correct
left cliff
If BAO is 71 then ABO is also 71 due to isosceles triangle property
and 71 divided by 2 right?
acoustic onyx
and 71 divided by 2 right?
Bro no
last parrot
No, 19*2
acoustic onyx
and 71 divided by 2 right?
You do know in isosceles triangle we have two equal sides and two equal angles
left cliff
You do know in isosceles triangle we have two equal sides and two equal angles
ohhh yeah
sry
somber ibex
same as this?
What's the answer ?
acoustic onyx
however, this is the answer
Wrong answer bro
Both minh's explanation and my explanation results in the answer 38
left cliff
yeah i learn 2 theory
but i just got confused by the answer
my original answer are correct
acoustic onyx
but i just got confused by the answer
The answer key is wrong
Man most of these online questions have wrong answer keys, I have seen in this help section many times now
left cliff
Man most of these online questions have wrong answer keys, I have seen in this h...
it is from cambridge though
devout snowBOT
left cliff
anyways, thank everyone
acoustic onyx
What the F is this?!
acoustic onyx
anyways, thank everyone
You are welcome
left cliff
What the F is this?!
lol
devout snowBOT
Available help channel!
keen sundial
what is this slide saying?
devout snowBOT
what is this slide saying?
rain summit
Basically it talks about the side you're approaching $x$ to
woven radishBOT
1 divided by 0 equals Infinity
tepid stag
what is this slide saying?
when we say x tends to 0, there are two things, either x is very very very slightly greater than 0 or very very very slightly less than 0 like 0.00000......01 or -0.00000...01
tepid stag
when we say x tends to 0, there are two things, either x is very very very sligh...
it is defined that if we want to say limit x->a is defined, both limit as x-> a+ and x->a- should be defined AND equal
a+ means, slightly more than a and a- means slightly less than a generally in calc a+=a+h and a-=a-h where h is like a very small number
tepid stag
what is this slide saying?
for limit of f(x)= sqrtx, at x=0, if x=0+ then it would be 0, cuz sqrt (0+) would be 0 but at x=0- it is not defined because sqare root of negative would be imaginary right
@keen sundial do you understand this
?
tepid stag
it is defined that if we want to say limit x->a is defined, both limit as x-> a+...
on the number line x->a+ means x is slighly more than a and almost equal to a so it is like approaching a from the right side
for x->a- it means x->a- meaning it is like barely smaller than a and hence on the number line it seems like x is going towards a from the left side
devout snowBOT
@keen sundial Has your question been resolved?
keen sundial
for limit of f(x)= sqrtx, at x=0, if x=0+ then it would be 0, cuz sqrt (0+) woul...
but its 0-, which is still 0
or are we reffferring to very close values of of 0 from LHS?
tepid stag
or are we reffferring to very close values of of 0 from LHS?
yes
keen sundial
Isee
tepid stag
what is this slide saying?
from this slide
keen sundial
any other point that is bothering you
yes
but trust me its is very hard to understand through text
are you able to vc?
since there is a whiteboard
tepid stag
are you able to vc?
no sorry
tepid stag
when we say x tends to 0, there are two things, either x is very very very sligh...
..??
keen sundial
I understand that
tepid stag
then which point is bothering you?
keen sundial
defintion of a limit
tepid stag
defintion of a limit
for some cases like say lim x+3 as x->3 would just be 6
for some cases like say lim [x] as x tends to 1 where [] denotes greatest integer function
do you know greatest integer function?
@keen sundial
keen sundial
No
keen sundial
for some cases like say lim x+3 as x->3 would just be 6
Why?
When do we need to sub the limit in the function?
tepid stag
When do we need to sub the limit in the function?
for most of the cases we can if the value we get is not like undefined
if we are getting undefined, we should furhter simplify or do some manipulations
and we can also check this with the x->3+ and x->3- aswell
keen sundial
Oh I see
keen sundial
and we can also check this with the x->3+ and x->3- aswell
Wdym?
tepid stag
and we can also check this with the x->3+ and x->3- aswell
both of them will be equal to 6 aswell
tepid stag
Wdym?
ok wait lets say lim x+3 as x->3+ this means x is like barely greater than 3 so the limit would be 6+ which is barely more than 6 or equal to 6
for 3- you can do the same
tepid stag
its basically like just rounding off to the nearest least integer
like for example g.i.f of 0.5 would be 0 because 0 is the nearest smaller integer less than 0,5
nvm ig that is out of context
tepid stag
for some cases like say lim [x] as x tends to 1 where [] denotes greatest intege...
so for this example to find the left hand limit say x=1- meaning it is barely less than 1 or x=0.9999999999.... so the greatest integer function would say this is equal to 0
but for x->1+ meaning it is barely more than 1 or x=1.0000000....1 so the greatest integer function say this is equal to 1
0 and 1 are no where near to being equal, hence limit of [x] as x tends to 1 is NOT defined or does NOT exist
@keen sundial u there ;-;?
keen sundial
Yeah I’m trying to understand
devout snowBOT
@keen sundial Has your question been resolved?
devout snowBOT
devout snowBOT
Available help channel!
silver breach
so this is the solution to a hw problem by my prof. but im still not sure why the directions are the way it is in the z-plane and the w-plane.
devout snowBOT
so this is the solution to a hw problem by my prof. but im still not sure why th...
last parrot
The hyperbola is facing vertically because you were given c1 is negative...
When you map, which w = z^2 (that is the real part of w (called u) is x^2 -y^2. and since this equal c1, then u= c1
silver breach
ohhh but why is the little arrows going the way it is?
silver breach
yeah
last parrot
When the angle theta increase, then the angle of image 2theta also increase
And to do this you do counter clockwise around the origin (move from right to left)
To make the angle increase (like the arrow direction)
And on the vertical line u=c1, where c1 is negative, so its to the left of v-axis, and increase angle also mean moving upward
For the red one, hmm maybe c2 is negative and y = c2/2x, this means x and y are in different signs, placing this hyperbola in 2nd and 4th quadrants
And following the arrows here u notice that moving from top left to x-axis, other way, is clockwise
Which implies theta is decreasing
silver breach
for the red one, so as the angle is decreasing on the z-plane since its either going close to 0, or its getting a bigger negative angle, so thats why its decreasing on the w-plane and thats going to the left with the left arrow
devout snowBOT
last parrot
No worries have a good one!
devout snowBOT
Available help channel!
mellow willow
How do I multiply cos and sin together?
devout snowBOT
How do I multiply cos and sin together?
north roost
,rccw
woven radishBOT
north roost
u js did
mellow willow
Wdym?
north roost
oh u want to simplify
mellow willow
Yea
north roost
view ${\sin ^2 \theta = \sin \theta \times \sin \theta}$
woven radishBOT
k
north roost
see if anything cancels
mellow willow
Ohhh
ebon coyote
also, careful:
mellow willow
I would get tan but I’m trying to get this
ebon coyote
!xy
devout snowBOT
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.
mossy berry
I would get tan but I’m trying to get this
Can you show your steps that got you here, something is going wrong
mellow willow
pine minnow
you forgot the square on cos in the 3rd-4th step
restive river
mellow willow
Ohhh
pine minnow
the < but 90 degree clockwise, cant find it :(
edit: thanks:)
devout snowBOT
@mellow willow Has your question been resolved?
devout snowBOT
devout snowBOT
Available help channel!
ocean quest
I have this; I'm not sure how to properly get to the last line from what I have up to that point. I would think that I need to make a sequence of $\epsilon_n$ where ${\epsilon_n} \to 0$
devout snowBOT
I have this; I'm not sure how to properly get to the last line from what I have ...
woven radishBOT
Mr. BananaHead
devout snowBOT
@ocean quest Has your question been resolved?
ocean quest
Would I just be able to set $\epsilon_n > \sup{x^nS(x)}$ and then use the fact that $x^nS(x)$ converges pointwise to 0 to let $\epsilon_n \to 0$?
woven radishBOT
Mr. BananaHead
ocean quest
I have this; I'm not sure how to properly get to the last line from what I have ...
Updated to be reflect that these epsilon values are for fixed n so I can build a bounding sequence out of them
supple knot
this part's a little confusing. delta1 depends on eps1 and delta2 depends on eps2. you let eps = max(eps1, M*eps2), how can you tell which delta is smaller?
supple knot
either way setting delta1 = delta2 seems wrong
ocean quest
I need to be able to cover the whole interval [0,1]
I don't see another way to do that
supple knot
you use eps1-delta1 for convergence on [delta1, 1]. you can fix this first, then take a smaller epsilon in your eps-N proof for the remaining interval [0, delta1]. you don't need delta2 at all
supple knot
the overall proof depends on both [0, delta1] and [delta1, 1], but one interval's convergence can depend on the variables in the other interval
supple knot
this should probably be min
ocean quest
I thought it needed to be the maximum to bound the function over the total interval.
if you use min instead of max, could you be bounding by a constant where the function is actually larger than it on one of the intervals?
ocean quest
the overall proof depends on both [0, delta1] and [delta1, 1], but one interval'...
so because I have bounds where epsilon can go to 0 on either interval, I'm just fixing delta to a point where one interval's epsilon will always be larger than the other's?
supple knot
yea that's confusing
you need one epsilon that works for both. usually that's the min of the two epsilons
ocean quest
hmm... now, does it matter that the same delta might not work for every n in the sequence?
ocean quest
you need one epsilon that works for both. usually that's the min of the two epsi...
I'm pretty sure it would still need to be the max here so that it bounds everything. Even if one is always smaller than the other, you need to pick the bigger one to bound both intervals
ocean quest
The big problem is that once I bound by epsilon_n, how do I make it go to 0 as n->infty?
devout snowBOT
@ocean quest Has your question been resolved?
devout snowBOT
@ocean quest Has your question been resolved?
ocean quest
you use eps1-delta1 for convergence on [delta1, 1]. you can fix this first, then...
I don’t know how to show the bound goes to 0 as n goes to infinity
karmic ferry
is this?
ocean quest
The big problem is that once I bound by epsilon_n, how do I make it go to 0 as n...
<@&286206848099549185>
hollow vine
<@&286206848099549185>
this is a fairly advanced question, consider asking in another channel
most people (including me) doesn't know this stuff
ocean quest
this is a fairly advanced question, consider asking in another channel
Oh, I didn’t know that was an option
Would it be in #real-complex-analysis ?
void fox
you use eps1-delta1 for convergence on [delta1, 1]. you can fix this first, then...
This is also what I told you last time
ocean quest
.close
devout snowBOT
devout snowBOT
Available help channel!
fervent helm
how should I approach this question? First of all I tried to figure out if there is some way to find out "A" matrix from adj(A) but I couldn't. That is what I tried.
devout snowBOT
how should I approach this question? First of all I tried to figure out if there...
fervent helm
.close
devout snowBOT
last parrot
Can you recall definition of adj
restive river
how should I approach this question? First of all I tried to figure out if there...
X = A^-1 B
devout snowBOT
Available help channel!
robust bobcat
why is it on the open interval instead of the closed interval?
devout snowBOT
why is it on the open interval instead of the closed interval?
wicked rover
why is it on the open interval instead of the closed interval?
we typically avoid defining derivatives at endpoints of the domain
robust bobcat
we typically avoid defining derivatives at endpoints of the domain
why is that? also wouldn't this limit us?
one more thing I have a couple of questions regarding how he formulates the theorem
why does he say function that is integrable on [a,x] where x in [a,b] why not just say on [a,b] for all x in [a,b]?
and for c. isn't a<= c <= b the same as c in [a,b]? why one choice over the other?
void fox
why is that? also wouldn't this limit us?
at the end points you can, at best, define the "one-sided" derivatives
faint gorge
the question is how much they are useful
void fox
and for c. isn't a<= c <= b the same as c in [a,b]? why one choice over the othe...
it's just style
robust bobcat
it's just style
is it style for both statements I questioned?
robust bobcat
at the end points you can, at best, define the "one-sided" derivatives
so normal derivative is defined by a standard limit which can be approacheed from sides. The endpoints can only be approached from one side. So like you said it is one sided derivative not the standard one we think of
robust bobcat
the question is how much they are useful
?
void fox
is it style for both statements I questioned?
Well, if f is integrable on an interval [a, b] then it is also integrable on any subinterval
but I don't know if you have proved this fact already
robust bobcat
Well, if f is integrable on an interval [a, b] then it is also integrable on any...
ok. so your saying that if we haven't proven this fact we need to state our assumptions in such a way that shows that each subinterval is integrable
devout snowBOT
@robust bobcat Has your question been resolved?
frozen aurora
<@&286206848099549185>
devout snowBOT
<@&286206848099549185>
frozen aurora
Bro I need help😭
viral kernel
what questions
pure cedar
using seperate sheets seems like a waste
restive river
devout snowBOT
@frozen aurora Has your question been resolved?
restive river
Q1 : They're asking you to evaluate the anti-... of f(x) at some value x.
Q2 : You won't really be able to directly integrate sin^3xcosx directly, can you use a special property to split the sin^3 x?
Q3 : As the denominator tends to 4 from the left side, meaning 3.9, 3.999, 3.9999,... it will be a division by zero. Ask yourself : Is it a case of L'Hopitals? If yes, perform L'Hopitals. If not, then what is something divided by something that becomes infinitely large? (Note the positive & negative signs.)
Q4. Sketch the graphs (Isolate as a function of y or as a function of x. Decide it on your own.)
Q5. Read Q3.
Q6. Sketch.
Q7. Type of rate question (Think perhaps antiderivative?)
Q8. Find the point of discontinuity then set a value of x to ensure continuity.
Q9. The gradient of the function (m) at a certain point is the anti-___?
Some pointers that maybe can help you start. Attempt and im sure other helpers will check and guide you further. (Also this is a very quick glance at the questions, I did not attempt it.)
<@&268886789983436800>
potent compass
could any of u teach me trig function basics pls
devout snowBOT
could any of u teach me trig function basics pls
potent compass
(I’m gone for a while to shower)
lucid compass
hi
polar chasm
could any of u teach me trig function basics pls
https://www.khanacademy.org/math/trigonometry
Have you tried this?
lucid compass
https://www.khanacademy.org/math/trigonometry
Have you tried this?
anyone help me with maths ? exam comming in few days
polar chasm
anyone help me with maths ? exam comming in few days
you can open your own help channel, e.g. #help-36 is currently free (read #❓how-to-get-help for more details)
restive river
also og chemistry tutor
pure cedar
anyone help me with maths ? exam comming in few days
!occupied
devout snowBOT
anyone help me with maths ? exam comming in few days
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
@potent compass Has your question been resolved?
potent compass
https://www.khanacademy.org/math/trigonometry
Have you tried this?
No
I’ll try it
rapid gulch
well, can this be closed?
rapid gulch
right, but where is @potent compass
potent compass
I’m here
harsh stream
I’m here
I have a website recco but its for uk exams
Do u do uk gcse or a level
Theres a website which literally teaches everything
potent compass
I do ieb
harsh stream
Its very good
potent compass
Oh ok
harsh stream
Idk what that is
potent compass
Can I have that pls
harsh stream
But sure
potent compass
A diff system
harsh stream
Maths genie
potent compass
I used to do gcse
harsh stream
Maths genie
He has videos too
potent compass
Ty
pure cedar
thats fine
ill still recommend sl loney for problems
and proofs
its not a calc book tho
potent compass
its not a calc book tho
Wdym
pure cedar
it doesnt relate trig to calc
potent compass
What’s calc
pure cedar
its a “doing trig for trig sake, who cares about applications” kinda book
pure cedar
What’s calc
calculus
potent compass
its a “doing trig for trig sake, who cares about applications” kinda book
Oh
I dont think I’ll understand it
blissful comet
I am confused in 5. I got my normal vector as <3,-2,2> but I am confused what to do next. I thought we would take dot product, but we only have one vector and a point
devout snowBOT
I am confused in 5. I got my normal vector as <3,-2,2> but I am confused what to...
acoustic leaf
did you do the previous questions?
blissful comet
i have to do only 3 and 5. i m done with 3
our prof assigns practice problems
do u wanna help
mortal tendon
the line is in the direction of the normal vector
blissful comet
how do u know that its same direction
vital sedge
"perpendicular"
vital sedge
perpendicular means orthogonal to the plane
blissful comet
yeah the line is orthogonal to the plane
orthogonal basically is perpendicular right
vital sedge
pretty much
vital sedge
try to actually visualize
blissful comet
there is a plane
vital sedge
wow you beat me to it
blissful comet
and theres a line
vital sedge
think of a sheet of paper
blissful comet
okay
vital sedge
and an arrow going perpendicular off of the plane
blissful comet
okay
vital sedge
that's the direction the line you're searching for is
blissful comet
can i say that the normal vector and the point are on the same line?
that is perpendicular to the plane?
pure cedar
I am confused in 5. I got my normal vector as <3,-2,2> but I am confused what to...
when you got the normal vector as <3,-2,2> what is the doubt
simple use the point and direction vector
to describe the line
blissful comet
when you got the normal vector as <3,-2,2> what is the doubt
I am confused how is it parallel
pure cedar
parallel to what
blissful comet
like why r we using the scalar multiple
derived formula
to get our line
when the direction vector is not parallel to the line
pure cedar
you are confusing plane and line
blissful comet
oh
blissful comet
okay
pure cedar
i.e. the direction vector = normal vector
blissful comet
okay
blissful comet
yes that i knoq
mortal tendon
normal direction=line direction
blissful comet
oh so when both line and normal vector are perp to plane
mortal tendon
take a plane
and a line going perpendicular to it
and now visualise the normal vector (a vector perpendicular to the plane)
do u see they're parallel
blissful comet
line and normal vector are parallel so we use (x -xo, y -y0, z -z0> = t<a,b,c>
t is the scalar multiple?
am i thinkng correctly
pure cedar
the result that is obtained from the line being perpendicular to the plane
is that the direction vector of the line
and normal vector of the plane
are same
blissful comet
arent they parallel
mortal tendon
i thought i got it
did u get why normal vector and line are parallel
blissful comet
yep
vital sedge
you see how that's the case?
blissful comet
it extends in x y z directions
vital sedge
yes
blissful comet
i know this is the general eq of a plane
blissful comet
i also got normal vector <3,-2,2>