When I solve these does the numerator matter that much
#help-28
torn jolt
So like
5-x > 0
-x > -5
x < 5
And if I were to write it would it be like (-∞,5] ?
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torn jolt
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clear owl
When I solve these does the numerator matter that much
the denominator is what matters and you would not not have a ] for the 5 since ] is only used if its included in the domain
if the symbol was less than or equal too you would use ], but since it is only > you would use ) for both sides
torn jolt
if the symbol was less than or equal too you would use ], but since it is only >...
So what if it was x ≤ 5
clear owl
So what if it was x ≤ 5
then the domain would be (-∞,5] , if its x<5 its (-∞,5)
torn jolt
For square roots in the denominator it's not x ≥ 0? And in this case ^5-x
clear owl
For square roots in the denominator it's not x ≥ 0? And in this case ^5-x
for square roots in the denominator it will always be x>0, since the denominator cannot be 0 you would not include it
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lusty steeple
I need help on this problem
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I need help on this problem
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lusty steeple
I need help on part b
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I need help on part b
wild sleet
t is 8
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solemn rampart
someone help I beg
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scenic summit
can someone check my proof please? i wasn’t entirely sure if it’s 4 or 3 because i couldn’t figure out how to make a valid table for n=4 (which makes me skeptical of if the largest n can be is 4)
plush mural
I'm have 0 clue what ur doing, just here to tell u that ur handwriting is beautiful
scenic summit
LMAO aww thank you!!
twilit leaf
if you start with
WW
W
in your square, you should be able to find it
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crimson tiger
What do I do next
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What do I do next
crimson tiger
dull seal
What is that 2?
crimson tiger
What is that 2?
If you mean the top it’s just the question no.
hardy snow
@crimson tiger do you see the substitution or no
crimson tiger
I don’t need to use any integration formulae right?
hardy snow
idk what you reallyh mean by that but u substitution is what you need to use 👍
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torn jolt
Why is all x epsilon Set A P(x) equivalent to all x ( p(x) if x is part of set A) equivalent
torn jolt
The truth tables of the former would have a truth table which has F T as false
fast peak
do you mean $\forall x\in A: P(x)$ and $\forall x: x\in A\implies P(x)$ ?
glossy valveBOT
Denascite
torn jolt
do you mean $\forall x\in A: P(x)$ and $\forall x: x\in A\implies P(x)$ ?
What language is this btw i need to learn it it’ll be easier
fast peak
latex
torn jolt
Oh
fast peak
The truth tables of the former would have a truth table which has F T as false
not sure what truth table you mean
I actually cant properly put it into words why these statements are the same
torn jolt
not sure what truth table you mean
For A:p(x) wouldn’t it be A and P
fast peak
uhh
you split that up very weirdly
A is a set, it's not a statement
the first option says, if we take an element x from A, then P(x) has to be true
the second one says "let's take an element x. if x is not in A, then the implication is true anyway, so we dont care. if x is in A, then P(x) has to be true for the implication to be true"
so thats really just the same as before
if x is an element of A, then P(x) has to be true
snow mulch
Which expression is equivalent to 9x2 – 16y2?
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Which expression is equivalent to 9x2 – 16y2?
faint karma
9 times 2 minus 16 y 2 ?
snow mulch
i believe so.
snow mulch
yeah
Which expression is equivalent to 9x2 – 16y2?
(3x – 4y) (3x – 4y)
(3x + 4y) (3x + 4y)
(3x + 4y) (3x – 4y)
(3x – 4y)2
(the four are the options)
torn jolt
well
faint karma
ok
snow mulch
i thought so too.
torn jolt
.close if you're done
faint karma
it would have been clearer if you wrote your squares properly
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peak estuary
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peak estuary
answer sheet says something about reversing the order of integration
but the bounds change
so how would i do that
snow mulch
one moment let me solve the problem
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As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
snow mulch
yes.
peak estuary
y=root x/3
how would i sketch this
snow mulch
so.
let me get a rough sketch rq
i have an app
this is a rough copy
not sure though
peak estuary
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willow fable
The greatest common divisor of a and b is k. a/k = c, b/k = d. Can c+d be relatively prime to k? Please provide evidence why this cannot be the case
willow fable
Oh and k != 1.
also c is even, and d is odd
k is obviously odd, since a is even and b is odd
<@&286206848099549185>
quasi oak
use chatgpt
willow fable
a is even b is odd
quasi oak
bro, im not an engineer
robust slate
Doesn’t a=10 and b=5 work as a counterexample
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brittle parcel
I can give this a shot, but the only thing which confuses me is - why is dx under the radical? Would it matter if I solved the integral as if it wasn't under it?
brittle parcel
would make more sense to me in this scenario
narrow path
I can give this a shot, but the only thing which confuses me is - why is dx unde...
Haha this is the free wifi integral
brittle parcel
yep
narrow path
yes if you want to integrate this you need to integrate with the dx outside the radical
brittle parcel
I haven't seen the answer nor want to (yet)
narrow path
probably just a type error
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brittle cobalt
Hello, as a part of self-studying meteorology, I'm currently calculating Rossby wavelengths. I reproduced the sample application. My outcome is 7001 and not 6990 though. Which one is correct?
keen agate
They rounded intermittently and assumed pi as 3.14
@gritty rose What do you mean by that reaction?
gritty rose
oh nothing you did
brittle cobalt
They rounded intermittently and assumed pi as 3.14
Alright, thank you very much for your answer!
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vapid knot
why are sines and cosines of a certain angle constants why dont they depend on side lengths of the triangle
vapid knot
yea
lime ether
given a certain angle, the ratios of the side lengths is fixed despite it being possible to have different side lengths
vapid knot
not the side lengths themselves
they make up the ratio tho
lime ether
they make up the ratio tho
1/2 = 2/4
vapid knot
oh
lime ether
consider the common 3 4 5 triangle
there is also the 6 8 10 which is really a multiple of the 3 4 5 you could say
the angles are the same
and the ratios are the same
but the 6 8 10 is obviously bigger
lime ether
wdym
lime ether
you’re welcome
vapid knot
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hollow cove
can someone help me out on this problem
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can someone help me out on this problem
hollow cove
i think I know a) which is closed but im unsure of all the other ones.
b) i have no clue
c) i think its closed bc its bounded?
d) neither open or closed?
I did a) by showing that A^c is open. A^c being the compliment of the set A. if the set A is whats defined for a, then A^c would be A_1 union A_2 where A_1 is x^2 + y^2 <1 and A_2 x^2 + y^2 > 2
by a theorem, if A_1 and A_2 are open/closed then their union is open/closed. thus A^c is open. then A is closed.
we know that those A_1 and A_2 are open because they are [0,1) U (2, infinity) Im actually not sure about the first one because its [,) so idk if its closed or open. but thats my reasoning
edit: the way im defining the domain is in like polar coordinates so its r [0,1) U (2, infinity). I dont know if that is still allowed to solve these problems.
placid oak
To show a subset $S \subseteq R$ is open, you need to show for all elements $a \in S$ there exists a $\delta$ such that $(a - \delta, a + \delta) \subset S$. Have you done that for $A_1$ and $A_2$?
glossy valveBOT
Shuba
hollow cove
I dont think thats a definition i have learned in class
placid oak
What definition did you learn?
hollow cove
We have learned the one using complement, the limit point one and the infinite sequence that converges??
Something like that
placid oak
You mean, a subset is closed if it contains all the limits of every convergent sequence within that subset?
placid oak
I mean, you are right. A is closed. I'm not sure you've got a rigerous-enough proof for it.
we know that those A_1 and A_2 are open because they are [0,1) U (2, infinity)
It's not[0,1), it's the ball{(x,y) : x^2 + y^2 < 1}.
We have learned the one using complement
Just to clarify, that's "a subset is closed if it's complement is open", or "a subset is open if it's complement is closed"?
hollow cove
> We have learned the one using complement
Just to clarify, that's "a subset is ...
either way should work its arbitrary which one is which
bc at the end of the day A^c^c is just A
placid oak
Ok. So how have you shown {(x,y) in R : x^2 + y^2 < 1} is open?
hollow cove
the interval is from 0 to 1 right? the thing is idk if the fact that it is [0,1) changes everything, because then it might be neither open or closed
placid oak
It's not [0,1)
hollow cove
how come?
placid oak
it's in 2 dimensions for a start
placid oak
In polar, it's equally (-1,+1)
hollow cove
thats true
i didnt think of that
oh okay
so -1,1 is open
the intersection of that with 2 to infinty is also open
then A^c is open
A is closed
placid oak
But that's only for r. What about the angle t?
hollow cove
well why would that matter?
t = arctan (y/x) so as long as x=! 0 and y/x =! +-pi/2
it should be ok
placid oak
Ah, so there's a hole at x = 0?
placid oak
How do you know (-1,+1) is open in R?
hollow cove
it containts all of its limit points
placid oak
prove it
hollow cove
call that set D,
let x be a limit point of D
assume x e/ D
by def of limit point every ball of radius r>0 must contain points in D other than x.
this is not possible because x e/D thus
x must be in D
unless x is a boundary point but thats something else xd
placid oak
It doesn't contain the limit of the sequence a_n = 1 - 1/n, right?
hollow cove
i suppose not because the limit of that sequecne should be 1 and thats not in D
placid oak
So I don't confuse you any further, have you been shown a subset can be both open and closed?
hollow cove
R^n and the empty but i only understand that they are both open and closed by the complement definition. not really because it makes sense
placid oak
The reals are nice because it's Hausdorff, meaning every subset is either open or closed. But this isn't true for every set.
placid oak
yep, point-set topology did that to me too XD
hollow cove
im in calc 3 ODE and Lin alg tho😭😭😭
hollow cove
i mean this unit IS topology according to my professor but thats not the whole emphasis of the class
we are basing the class off this book advanced calculus of several variables by c.h. edwards
placid oak
Ok, so I'm assuming the way your prof would want you to show (-1,+1) is open is by showing it's complement is closed?
ignore that last XD
placid oak
I'm trying to gauge if you can assume a subset (excluding the empty set and the entire set) is either open or closed, but not both.
You can with subsets of the reals, but I'm not sure if your prof will allow you to use this property.
hollow cove
i think the only ones we say are both open and close are R^n and the empty set
placid oak
Ok, makes life MUCH easier XD
hollow cove
i mean this problems shouldnt be THAT hard bc i have 10 total and this one is one of the short ones
placid oak
So, all your prof want to show a subset is not open is to find a sequence whose limit is not in the subset?
let me try that sentence again
hollow cove
yeah so a subset A of R^n is closed IFF A contains all its limit points
thats the theorem we learend in class
learned^
placid oak
So all your prof wants, to show if a subset is not closed (i.e. open), is to give a sequence whose limit is not in the subset.
hollow cove
i suppose yeah
placid oak
So, how would you show a subset IS closed?
hollow cove
well i define x to be a limit point and i would show that for all x in R^n, x e D where D is the subset of R^n im trying to show is closed
hollow cove
call that set D,
let x be a limit point of D
assume x e/ D
by def of limit p...
which is this sorta
i think
placid oak
Right, I think I get what he wants.
- A subset is open if there exists a sequence whose limit is not in the subset.
- A subset is closed if it's complement is open.
Then, clearly, (-1,+1) is open because the limit of a_n = 1 - 1/n is not in the subset.
hollow cove
and because 1 - 1/n is in the subset
placid oak
and because 1 - 1/n is in the subset
for all n > 0.
hollow cove
yes okay
i feel like i get the general idea of a) and even though my proof might not be as rigurous as it should, i still feel good with that tone
one^
placid oak
Yeah, fine 🙂
hollow cove
but the rest have my lost
b) i have no clue how to even think of it
bc the complement way doesnt seem like a smart way of doing it
but i cant figure out how to approach it using the lim point definition
placid oak
With B, I would look at Q in R, instead of Q^2 in R^2.
Because, if Q in R is open/closed, then Q^2 in R^2 is open/closed.
hollow cove
yes I tried that too because we also proved in class that the cartesian product holds closedness
but im still unsure
placid oak
Can you think of a sequence of rationals whose limit is irrational?
hollow cove
didnt we just do 1 - 1/n, doesnt that work?
placid oak
1 - 1/n is rational for all n.
placid oak
Can you think of a rational sequence converging to e?
hollow cove
i would just do e - 1/n
but thats irrational for all n and converges to an irrational number
placid oak
yep, because e - 1/n is irrational, and we want a rational sequence.
Do you know the definition of e?
hollow cove
perhaps?
placid oak
which is?
hollow cove
isnt it something like the lim of (1 + 1/n ) ^n
placid oak
$$e := \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n$$
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Shuba
placid oak
you're absolutely right XD
hollow cove
okay so every one of them is rational and it converges to an irrational number
i see
placid oak
So, clearly, Q in R is NOT CLOSED because that sequence of rationals converges to the irrational e.
Can you show Q in R is NOT OPEN?
hollow cove
but i still need to show that the sequence is in Q right?
placid oak
but i still need to show that the sequence is in Q right?
you can show, for any finite n, the limitand is rational.
hollow cove
okay... then to show that Q is NOT open in R, i would try to show it IS open
which by the defition i have written down in my notebook from class says:
A subset U of R^n is an open subset if for all p e U, there exists epsilon>0 such that the ball of radius epsilon centered at p is a subset of U (is completely contained in U)
placid oak
okay... then to show that Q is NOT open in R, i would try to show it IS open
wh...
we've just shown it's not open.
hollow cove
didnt we show it was not closed?
placid oak
yes, my bad
hollow cove
all g
placid oak
ok good. my train of though does make sense. small worry there XD
hollow cove
lol
hollow cove
okay... then to show that Q is NOT open in R, i would try to show it IS open
wh...
so what i said kinda works or no
bc then i just let x be in Q and say that there are no open balls fully contained in Q centered at x.
i just dont know how to really show thats the casae
case^
placid oak
A subset U of R^n is an open subset if for all p e U, there exists epsilon>0 such that the ball of radius epsilon centered at p is a subset of U (is completely contained in U)
this is correct
placid oak
okay... then to show that Q is NOT open in R, i would try to show it IS open
wh...
you mean, show it's complement is not closed?
hollow cove
i just dont think going down the complement route is smart because I dont know how to figure out what R^n/Q is
for a) it was easy to define the complement, for this one idk
it must contain AT LEAST one irrational number
placid oak
you don't need to know exactly what it is, you only need to be able to choose a sequence whose elements are in R\Q
i.e. you could choose a sequence of irrationals
hollow cove
hm so by showing that the complement is not closed we showed that Q is not closed and Q^c is not closed. thus its neither closed or open?
placid oak
Yeah, I think we'd need to use the "open ball" definition to show a subset is not open.
i.e. it contains at least one boundary point
No, I'm being silly again. We CAN use sequences. We want a sequence of irrationals that converges to a rational. (i.e. the subset isn't open if it's complement isn't closed). I think that's what your prof is after?
hollow cove
yeah thats kind of what i was thinking. using the "same method" to prove that neither of them are closed
hollow cove
now i need to figure out a sequence that converges to a rational made up of irratinoals
dw its all g
placid oak
Yep
hollow cove
do u already have one in mind?
placid oak
I do, but there's an infinite selection you could choose from
hollow cove
okok dont tell me im trying to think
so irrationals are stuff like " e, pi, sqrt 2" but what else
placid oak
yep
hollow cove
are all sqrt irrational? (ofcourse avoiding numbers like 4, 9, 16...)
placid oak
all sqrts of primes are
hollow cove
so yeah bc for any n as long as its not a square already i can factor it to its primes and then get m*sqrt p where m is an integer and p is a prime.
i suppose that m * sqrt p is always irrational tho
placid oak
you're on the right lines
sqrt p also diverges to infinity, and we want it to converge to a rational.
Can you do it using only sqrt 2?
hollow cove
wdym
placid oak
Can you think of a sequence in Q[sqrt 2] which converges to a rational?
Q[sqrt 2] means "the result of any algebraic expression using only rationals and sqrt 2".
hollow cove
placid oak
I say that, because it's easy to prove sqrt 2 is irrational, but a little more work to show sqrt p is irrational for all primes p.
hollow cove
does this kinda work?
placid oak
does this kinda work?
$$1 - \frac{2^{\frac{1}{2}(e^{\frac{1}{4}})}}{\sqrt{2}}$$
glossy valveBOT
Shuba
placid oak
That?
hollow cove
that 4 is an "n" sorry my hadnwritting on the computer isnt the best
hollow cove
well but that is for sure not rational like, just by looking at it you can tell
the only way for that to be rational is if n was infinity
which is the whole point
bc the exponent of the 2 on top will never be 1/2 unless that happens
placid oak
Maybe not. It's still up for question whether pi^e is irrational. We don't even know if it's not an integer XD
placid oak
strange things happen when you take powers of irrationals
hollow cove
placid oak
I'll give you a hint.
||a rational times an irrational is irrational||
hollow cove
right so basically the m sqrt p that we had but instead of p its just sqrt 2
placid oak
yeah, that would work.
hollow cove
so how about 1 - sqrt2 /n
placid oak
now, just make it converge instead of diverging
perfect
for all finite n, 1 - sqrt{2} / n is irrational, and it's 1 when n tends to infinity.
hollow cove
okay and i fs do NOT need to prove thaat sqrt 2 is irrational (right??)
placid oak
nah XD
if you wanted to be really rigerous, you could use "1 - r / n where r is any irrational"
then theres no need to prove r is irrational, because you've just stated it's irrational XD
That should solve question B for you.
Question D is very similar to B.
hollow cove
oh so i can just say that the number is irrational to start with
thats nicer
okay okay
placid oak
yep. gotta love the axiom of choice
hollow cove
okay that sounds good i will work on that later bc my brain is fried rn
thank you so so so so much for your help
placid oak
❤️
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iron flame
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iron flame
how do I do this
narrow ermine
The derivative is defined like $f'(-1) = \lim_{h\to 0} \frac{f(-1+h) - f(-1)}{h}$.
In this problem, they want you to approximate it by taking h = 0.1 instead of taking the limit, so you're essentially approximating it as the slope of the secant line through points (-1, f(-1)) and (-1 + 0.1, f(-1 + 0.1))
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Azyrashacorki
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unique sluice
is the work for this just g(f(0)) then g(2) then 2->4?
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is the work for this just g(f(0)) then g(2) then 2->4?
narrow ermine
Yes.
nimble crane
yup!
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edgy dome
is there any chance i could hop in a vc with someone for help with some of my absolute value stuff? i can barely even understand it and my test is tomorrow
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is there any chance i could hop in a vc with someone for help with some of my ab...
@edgy dome Has your question been resolved?
left bone
that seems unlikely to happen, but if you have a specific problem i can go over it
edgy dome
so i already have all the answers to my study guide, but i still can't figure out how to actually get them myself, the main thing that's tripping me up is where do i figure out that i should use the (-∞ and ∞) respectively
violet bolt
what's the biggest number you can think of that is greater than or equal to 5?
edgy dome
that's fair
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torn jolt
still need?
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crisp fjord
How does example 4-5 work? The house thing is jointly proportional to days and people and it goes 5 over 12 but then it flips to 12 over 5 at the bottom??
crisp fjord
I’m pretty sure days and people are inversely proportional and houses is jointly proportional with both but idk how to use that
My main question I guess is why does 12 flip to the top and 5 flip to the bottom
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nova basin
How does example 4-5 work? The house thing is jointly proportional to days and p...
3 days, 4 people, 5 houses
1 day, 4 people, 5/3 houses
1 day, 2 people, 5/6 houses
6 / (6/5) = 36/5 days, 2 people, 6 houses
Just like that
crisp fjord
Ok thanks
torn jolt
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limpid moat
torn jolt
i tried to make some zeroes and then simplifying the determinant then
but
i got stuck
limpid moat
i tried to make some zeroes and then simplifying the determinant then
what is the meaning of your nickname?
torn jolt
that my balls are sweaty?
limpid moat
is it supposed to be funny ?
torn jolt
dude can you actually help me pls 😑
limpid moat
now I have to go ... you can eventually ping the helper if nobody come here
torn jolt
oh okay
torn jolt
Do you know how to compute determinants?
yes
should i just solve the determinant
Yes
It will probably give you something nice
aight hold on lemme try
,rotate
glossy valveBOT
torn jolt
ts definitely not nice bro 😭😭🙏💀
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torn jolt
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torn jolt
hello, for these kind of question i know that i need to differentiate, but im not sure once or twice base on the informations given by the question?
torn jolt
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rough shadow
you need to differentiate once i think
torn jolt
could you please tell me why? i dont quite understand why its differentiating once not twice
rough shadow
cuz you want to find out in which direction it is moving, if towards or away from the point O
So if the distance increases the derivative is positive it moves away from O
and otherwise it comes closer and the derivative is negative
So you want to find out the sign of the derivative on the intervalls
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A fabric when washed: "Its length shrinks by
1/3, and it stretches; its width; in 1/5. How many
meters should be purchased? For after
washed, 240 m2 is available. Knowing that the
original width is 60m.
a) 6m b) 15m c) 5m d) 10m e) 20m
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edgy siren
feels like im doing bogus make-believe math rn
this is my work so far
ive somehow proved it is injective by evaluating odd n and even n cases (which yields + ceil(n/2) and - ceil(n/2))
then surjective i tried same approach separating into odd even
odd gives me
ceil(n/2) = b
let n = 2k+1
ceil( (2k+1) / 2) = b
ceil ( k+0.5 ) = b
k + 1 = b
k = b - 1
n = 2b - 1
mild correction, b is limited to >= 2 instead of >=1 as i wrote on my paper
i think idk
if someone could help me with this it would be amazing
im already 3hrs into this problem alone and i dont know what else to do
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@edgy siren Has your question been resolved?
edgy siren
<@&286206848099549185>
rough shadow
ok yeah, so have you proven injective yet?
edgy siren
yea injective is the first half of my first page
i dont know if its a sound proof
my conclusion is kinda
idk 😐
edgy siren
surjective i tried separating to odd and even
for odd n, i get n=2b-1 which must be positive
and this function shows there is a mapping from every n to every b
which i think is the definition of surjective?
so now im trying to do with even but i get n = -2b
and that dont make sense cuz for b=1 i get n=-2 which is not in the domain of natural numbers
rough shadow
and this function shows there is a mapping from every n to every b
for b positive yeah
edgy siren
tbh i probably violated the laws of the universe 12 times in my work
rough shadow
so now im trying to do with even but i get n = -2b
what do you mean by that
edgy siren
odd n:
let n = 2k+1, k is a member of Natural Numbers
(unsure how to define b)
ceil(n/2) = ceil( (2k+1) /2) = b
ceil (k+0.5) = b
k+1 = b
k = b-1
Back to n definition
n = 2(b-1) + 1
n = 2b -1
rough shadow
yeah this works, so you proved that for each b>1 there exists such an n
cuz you simply choose n=2b-1
edgy siren
even n:
Let n = 2k, k is a member of Natural Numbers
(unsure how to define b)
-ceil(n/2) = - ceil(2k/2) = b
-ceil (k) = b
floor(-k) = b
-k = b
k = -b
n = -2b
(No idea what to do from here)
-ceil because of (-1)^(n+1) so even n gives -1
rough shadow
No you proof is complete, cuz for b<=0 you can choose n=-2b and you have f(n)=b
edgy siren
so b doesnt have any restrictions in particular to make this work?
rough shadow
No surjective is defined as follows: Let $f:A\to B$ be function. We call $f$ \emph{surjective} if for every $b\in B$ we can find an $a\in A$ such that $$f(a)=b$$
edgy siren
if i do b = 0 though i get n = 0
glossy valveBOT
cosmic
edgy siren
so in this case f(n) = b
rough shadow
and in your case you always found such an $n$ for every $b$ be giving a concrete constrcutuin
glossy valveBOT
cosmic
edgy siren
i see
rough shadow
if i do b = 0 though i get n = 0
yeah sometimes $0\in \mathbb{N}$
glossy valveBOT
cosmic
rough shadow
depends on context (e.g. country from which this problem is from)
edgy siren
any ideas on how i can set up f-1 then?
im thinking of the following:
let f(n) = y
n = (-1)^(y+1) * ceil(y/2)
this this a good setup?
cuz if so
my first problem is how do i isolate y in ^(y+1)
rough shadow
yeah you do that like this i think
edgy siren
using log wont work cuz i get (y+1) * log(-1) which is undefined
wait
nvm log(-1) isnt undefined
rough shadow
yeah log is also bad because of the ceil function
rough shadow
to log this isnt such a good idea i think cuz we are in a discrete context
like whole numbers
edgy siren
right
i guess i can separate this one as a piecewise function?
n = ( ... if y even
... if y odd)
still need to do smth about the ceil tho 😐
edgy siren
my setup is
n =
- ceil(y/2) for y=2k+1
- -ceil(y/2) for y=2k
ig i can just sub in 2k+1 and 2k
rough shadow
yes exactly, remember that y is now natural and n is the integer
rough shadow
hmm i think you made a mistake somewhere, a minus went missing i think
rough shadow
oh wait
you still need to write the inverse map down
like now you have $$f(2k)=-k$$ and $$f(2k+1)=k+1$$
glossy valveBOT
cosmic
rough shadow
But you want to find $f^{-1}$
glossy valveBOT
cosmic
rough shadow
So you have to rearragne a little bit
cuz you want to express y in dependence of n and you have it currently the other way round
but simplyfied
rough shadow
yeah kinda but it is like this $f^{-1}(-k)=2k$ and $f^{-1}(k+1)=2k+1$
glossy valveBOT
cosmic
rough shadow
maybe you meant this and i just misunderstood
rough shadow
**cosmic**
for k>=0
rough shadow
the inverse function maps the non-positive numbers to the even ones and positive numbers to the odd ones
cuz the original function maps the even integers to the negative integers and the odd integers to the positive integers
edgy siren
right
edgy siren
so original function:
negative to even
positive to odd
inverse function
positive to even
negative to odd
?
rough shadow
i think we just mean the same thing
rough shadow
**cosmic**
i thought your inverse function was defined like this
edgy siren
probably, im just trying to confirm i understood what you said by rewriting it my breaking it down
uhh
the inverse function just says f inverse
where f(n) is the original function
(-1)^(n+1) * ceil(n/2)
rough shadow
yeah maybe just tell me your inverse function if i give you an integer y, where does this get mapped to?
rough shadow
the inverse function just says f inverse
yeah thats why i was also confused
rough shadow
yeah maybe just tell me your inverse function if i give you an integer y, where ...
sorry if i also confused you, just forget what i said and just tell me your inverse cuz then I know you mean the right thing
edgy siren
since y=f(n)
f^-1 of y is just f^-1(f(n)) which is n
rough shadow
no i mean what is this n, like as a set of operations which I can perform that gives me n in dependence of y
edgy siren
n for y%2==0 gives k+1 (idk what k is supposed to be though)
n for y%2!=0 gives -k (again idk what k is)
rough shadow
yeah thats kind of the problem
rough shadow
xD
rough shadow
n for y%2==0 gives k+1 (idk what k is supposed to be though)
n for y%2!=0 gives...
but when you want to find out $f^{-1}(y)$ you dont look at parity of y
glossy valveBOT
cosmic
rough shadow
when evaluating $f(y)$ you would be looking at the parity of y
glossy valveBOT
cosmic
rough shadow
you are mixing these two things up
edgy siren
ok so if i understand correctly
the og function does this in a very abstract form:
- pick n
- get f(n)
and now we want the inverse function which says
- pick f(n)
- get n
is this correct to begin with
edgy siren
f(n) -> n
is
f^(-1)(n) right
rough shadow
yeah
edgy siren
so therefore, is my input n or f(n) ?
rough shadow
i dont like the notation very much i would write y=f(n) cuz you dont know n yet
but your imput would be y, yeah
edgy siren
pick y
get n
rough shadow
yeah exactly
rough shadow
no
rough shadow
that is what the original function does
edgy siren
😔
rough shadow
im trying my best
yeah np, i also explained this pretty bad
edgy siren
whether u explained it good or bad, im smoothbrained as fk
my teachers always repeat shit 3 times in office hours and i sit there not having understood a word
rough shadow
whether u explained it good or bad, im smoothbrained as fk
my teachers always r...
idt so, also he is just a bad teacher then
wait so let me explain what you just did
edgy siren
can i say y is k or something 😭
rough shadow
we have a function $$f(n)=(-1)^{n+1}\left\lceil \frac n2 \right\rceil$$
Now you split up into cases odd and even
glossy valveBOT
cosmic
rough shadow
And you wrote $n=2k+1$ for odd and found out that $$f(2k+1)=k+1$$
glossy valveBOT
cosmic
glossy valveBOT
cosmic
edgy siren
so still the non-inverse function is
f(2k+1) -> k+1
and
f(2k) -> -k
rough shadow
yeah exactly
edgy siren
Ok
rough shadow
you have previously like mixed up y and k and n and all that made it very confusing
so we try to avoid that now
everything else was correct
rough shadow
so still the non-inverse function is
f(2k+1) -> k+1
and
f(2k) -> -k
also we should remember that $k\in \mathbb N$ right cuz $n\in \mathbb N$
glossy valveBOT
cosmic
rough shadow
right?
edgy siren
yuh
rough shadow
ok and now to the inverse
edgy siren
f^-1(2k+1)
glossy valveBOT
cosmic
rough shadow
and you know that f^-1 exists cuz you have proven that f is bijective
edgy siren
and second
2k = f^-1(-k)
rough shadow
yeah what does this mean?
edgy siren
1 sec
rough shadow
ofc
edgy siren
problem is im tryna invert k+1 and -k but i usualy do this with y x
so now im brain is exploding
and i only see one variable k
so my brain double explodin
rough shadow
XD
Maybe we look at the first equation
$$f^{-1}(k+1)=2k+1$$
We can write this in your form $$f^{-1}(x)=y$$ with $x=k+1$ and $y=2k+1$
edgy siren
first half gives 2k+1 = f^(-1) (k+1)
wat happened to this
rough shadow
oh wait i fucked up xD
edgy siren
Ok np
rough shadow
sorry let me fix this quick
glossy valveBOT
cosmic
rough shadow
for the second one $$f^{-1}(-k)=2k$$ and $$f^{-1}(x)=y$$ with $x=-k$ and $y=2k$
glossy valveBOT
cosmic
rough shadow
So if I give you an arbitary $x$ what $f^{-1}$ does it checks first if $x$ can be written as $x=k+1$ or as $x=-k$
glossy valveBOT
cosmic
rough shadow
and then computes as our equations say, so it maps k+1 to 2k+1 and -k to 2k
rough shadow
swapping x and y is kinda weird tho
yeah it can be kinda confusing
rough shadow
yeah we know that $k\ge 0$, so for positive $x$, we first write $x=k+1$ (which we can do) then we know that $$f^{-1}(x)=f^{-1}(k+1)=2k+1=y$$
glossy valveBOT
cosmic
rough shadow
2k+1 doesn't change
wdym by that?
edgy siren
like uh
our operation is f^-1(x) = y
y is y
i aint doin anything with that
but im inverting x
using f^-1
rough shadow
you want to find y in terms of x right?
edgy siren
uh
not sure
idk if i need x in terms of y or y in terms of x
i guess y in terms of x
but terms of x are k
rough shadow
our operation is f^-1(x) = y
in this case you want y in terms of x cuz x is the variable inputted so you know x
rough shadow
but terms of x are k
yeah k is in term of x
rough shadow
yeah exactly almost everything is right but remember that k is positive, so you only can do if x-1 is bigger or equal to 0 right?
So if you can do that everything is fine
rough shadow
yeah exactly almost everything is right but remember that k is positive, so you ...
cuz 0<k=x-1
Maybe do the same for the second equation then
what you just did and then we will talk about k>=0
edgy siren
but that wasnt the inverse right
we only wrote in terms of x
no inversion took place right
rough shadow
yes you calculated the inverse for positive inputs, so we now have if $x>$ we know that $$f^{-1}(x)=2x-1$$
glossy valveBOT
cosmic
edgy siren
oh
rough shadow
but yeah we didnt apply the inverse anymore we are just simplifing
rough shadow
I think you have the wrong idea of an inverse in general
edgy siren
probably
im just following the definition rn
we have f-1 (x) so whatever y ends up being is the f-1
because f-1(x) = y
so just find y
rough shadow
yeah exactly
rough shadow
yeah
rough shadow
yes do that now
edgy siren
ye i did
er
i had f(-1)(-k) = 2k
f(-1)(x) = y
x = -k
y = 2k
k = -x
y = -2x
f(-1)(x) = -2x
rough shadow
yay exactly we know have got $$f^{-1}(x)=\begin{cases}2x-1 & \text{for } x>0 \ -2x & \text{for } x\le 0\end{cases}$$
glossy valveBOT
cosmic
edgy siren
Oh
Uh
that last bit
x>0 and x<=0
need to think about that
idk how we got there
rough shadow
yeah we have to talk about this still xD
edgy siren
ik you mentioned k earlier
k is in natural numbers
so x>0 ensures k stays positive
and x<=0 ensures k stays positive
?
edgy siren
last q
rough shadow
of course
edgy siren
how do i reach x>0 for the first one
like
my input was f-1(k+1) which gave 2k+1
so why is 0 not inlcuded there
if its 0 i get 1
rough shadow
yeah but wrote x=k+1
edgy siren
i see
rough shadow
if x=0 then k=-1
which doesnt work
cuz k is natural as we said earlier
you get this?
edgy siren
ye
if natural is defined as 1,2,3 + only
not including 0
then does it change to x>=1
rough shadow
k can also be zero sorry
rough shadow
if natural is defined as 1,2,3 + only
but yeah if natural would mean this then this would be true
edgy siren
but then i want x>=1 cuz that gives me k=0
actually i want x>1
forcing x to 2
and k's first value to 1
rough shadow
no k can also be zero
edgy siren
then x>=1 is correct?
rough shadow
yeah
rough shadow
when i said k natural i meant k>=0
rough shadow
x>0 literally means x>=1
yeah xD
rough shadow
np
rough shadow
it was a lot at a time but i hoped it helped
edgy siren
helped a lot
now imagine this 2 times
thats how long i solve 1 problem on average
and sometimes i give up
💀
edgy siren
yesterday i did 5hrs on 3 questions
Nahh i wont do that i'll just create another channel in here
rough shadow
ok sure np
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Available help channel!
maiden owl
can someone explain this question to me
full forumBOT
can someone explain this question to me
lime ether
what’s this for btw
narrow path
1-??
maiden owl
its school work
lime ether
show the full screen
lime ether
homework
lime ether
yeah you need to fill in the blank
yeah you need to fill in the blank
maiden owl
yeah you can say that
narrow path
yeah you need to fill in the blank
do you understand
maiden owl
so i am not exactly understanding what i need to do so first i take the derivative of y = x^1/x
done that whats next
lime ether
do you understand
do you understand
narrow path
no
glossy valveBOT
knief
lime ether
then use the product rule and implicit differentiation
maiden owl
can you show me the working ?
narrow path
,tex .diff rules
glossy valveBOT
Steel
lime ether
can you show me the working ?
you don’t know implicit?
maiden owl
i am a little rusty right now
lime ether
ok fine
$\frac{1}{y} y’ = \frac{1}{x} \cdot \frac{1}{x} + \ln(x) \cdot \frac{-1}{x^2}$
glossy valveBOT
knief
maiden owl
so it would become
y' = y (1 - In(x) / x^2)
glossy valveBOT
knief
maiden owl
so here f(x) = 1-In(x)
lime ether
$1-\frac{\ln(x)}{x^2}$
glossy valveBOT
knief
lime ether
not that
lime ether
so here f(x) = 1-In(x)
yuh
lime ether
yes
full forumBOT
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Available help channel!
torn jolt
so when converting 300 degrees to radians i have an issue coming up with the answer in radians in terms of pi. I get the final decimal which the fraction value is (in the actual answer 5pi/3 = 1.6666), i get the 1.6, and when i press 2nd and the button to convert to decimal on my calculator it gives 1 weird u symbol and 3/5. Is there a fast way to convert this 1u3/5 to 5pi/3?
torn jolt
in other words im asking is there a fast way to convert 1u3/5 (mixed number) to the fraction 5pi/3 (or just 5/3, and then add the obvious pi for my answer)
gritty rose
what does 1u3 mean
torn jolt
The "u" symbol on the TI-30XIIS calculator simply represents a separation between the whole number and the fractional part in a mixed number format. It's not a mathematical symbol but rather a way for the calculator to indicate the structure of a mixed number.
chat gpt'd that cuz i didnt know myself xd
gritty rose
$1 \frac{3}{5} = 1 + \frac{3}{5} = \frac{5}{5} + \frac{3}{5} =?$
glossy valveBOT
riemann
topaz valley
😒
shrewd hamlet
For some reason, I feel like Blake is unamused.
torn jolt
ooooo
gritty rose
$\frac{a}{a} = 1$ for any nonzero numbers a
glossy valveBOT
riemann
torn jolt
got it
so ya 8/5s
then what
end result needs to be 5pi/3
or just 5/3
and ill know where to go from there
like how in the flying fuck do you just instantaneously get 5pi/3
1 3/5 = 5/5 + 3/5 = 8/5
idk where to go from there
ima just skip it if i fail it on the quiz whatevs
.close
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compact lantern
i need help on finding out is 6 and 7 are “and” or “or”, i attached a pic and the notes
torn jolt
,rotate
glossy valveBOT
glossy valveBOT
torn jolt
right
torn jolt
well it says "and" for 7. @compact lantern
compact lantern
wdym
torn jolt
i mean that 4 < 18 is true axiomatically so its addition into the question doesn't really change anything
compact lantern
so how do i find if its and or or
torn jolt
it asks you to solve the compound inequality
compact lantern
yes but theres two ways to do it, and or or
compact lantern
torn jolt
I misread the question, super sorry I'm high right now
it's an and inequality, to answer your question, because 7c + 4 is greater than or equal to -3 AND less than 18
for 6
torn jolt
well it says "and" for 7. <@798959292018851862>
i thought they were trying to represent the logical connective of "OR" with a + sign
which makes no sense in hindsight lmao
anyway
you want to solve $-3 \leq 7c + 4$ AND $7c + 4 < 18$ then find the intersection of these two intervals
glossy valveBOT
maxim
compact lantern
so we solve them first and then see if its and or or?
for and ones do the signs go the samw way
and for or they go opposite ways? @torn jolt
torn jolt
so we solve them first and then see if its and or or?
no
solve those two for now basically and tell me what you get
i will guide you from there
compact lantern
can we do 5 first
also if the thing like 3p+7 is in the middle/sandwitched does THAT mean its an and problem?
@torn jolt there
torn jolt
also if the thing like 3p+7 is in the middle/sandwitched does THAT mean its an a...
ye[p
yes
full forumBOT
@compact lantern Has your question been resolved?
compact lantern
<@&286206848099549185>
in addition 10 says its an and problem but when i checked online its an or
<@&286206848099549185>
obtuse marten
I got 141 is that correct if not lmk what to do
full forumBOT
I got 141 is that correct if not lmk what to do
digital shard
I got 141 is that correct if not lmk what to do
its right
obtuse marten
yayaya
thanks
the problem I’m on currently is a bit more complicated
could you help with that?
urban imp
nut
obtuse marten
