#help-17
vocal sleetBOT
warm rapids
Basically, I get the idea.
but yeah
I don't know if it's correct, but basically what I did was move the -x/a to the y/b side and get
that
prisma rain
you could just plug x = a and solve for y (which is 0) and plug in x = 0 and solve for y (which is b)
warm rapids
what do you mean?
acoustic monolith
that
you cna put it in the canonical form y=mx+p
find m
m = (b-0)/(0-a)
m= -b/a
warm rapids
yes, i did that
acoustic monolith
you cna put it in the canonical form y=mx+p
now solve for p
and then put it in that form
y=-(b/a )x + p
by putting (x,y)=(a,0) we get
0 = (-b/a)*a +p
p = b
then we get y=-(b/a )x + b
prisma rain
what do you mean?
if when x is a, y is 0 and when x is 0, y is b, then it passes through the points (a, 0) and (0, b)
acoustic monolith
then we get y=-(b/a )x + b
then you multiply it out and you get your answer
acoustic monolith
remember to cloe the channel
warm rapids
.close
vocal sleetBOT
vestal dagger
vocal sleetBOT
vestal dagger
i think i know the answers but not the formula
I've already answered no. 2 but I can't figure out the rest
vocal sleetBOT
@vestal dagger Has your question been resolved?
warm vine
I've already answered no. 2 but I can't figure out the rest
check for patterns
in 1. its 8, -24 -> 8*-3
then 72, so -24*-3
vestal dagger
yes but there should be a formula
warm vine
what formula
vestal dagger
like
warm vine
just keep multiplying with -3
vestal dagger
arithmetic i think?
warm vine
so $a_n = 8(-3)^{n-1}$
twin meteorBOT
Astar777
warm vine
what?
vestal dagger
this is no. 1
warm vine
why is it a_n
vestal dagger
cause a_1 should be 8
vestal dagger
**Astar777**
if I do it like this, the answer is 24 for a_1
vestal dagger
the calculator told me so😔
warm vine
8(-3)^(1-1) = 8
vestal dagger
what
warm vine
**Astar777**
put n=1 here
vestal dagger
no. 2
warm vine
seems right
vestal dagger
vestal dagger
- 1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8
ruby heart
For (\varepsilon > 0), we define the set
[
S_{\varepsilon} = \bigcup_{k \in \mathbb{Z}} (k - \varepsilon, k + \varepsilon).
]
Is it true that for any (\varepsilon > 0), the set of real numbers (\mathbb{R}) can be covered by a finite number of sets of the form (a S_{\varepsilon} = {a x : x \in S_{\varepsilon}}), where (a \in \mathbb{R})?
vocal sleetBOT
For \(\varepsilon > 0\), we define the set
\[
S_{\varepsilon} = \bigcup_{k \in \...
twin meteorBOT
ω
ruby heart
I need some help with this, it seems very difficult
hybrid flicker
oh I get it now
wraith mist
Interesting question
The answer is likely no
Without loss of generality we can assume a \in (0, 1]
hybrid flicker
not sure why you can assume a <= 1
wraith mist
If a finite cover with such sets exists, we can choose the set with the largest a and rescale the entire cover by 1/a
hybrid flicker
ok
wraith mist
If we manage to show that the segment [0, 1] can't be covered then we won
wraith mist
Yes, perhaps
hybrid flicker
ok yes
vestal dagger
look at every step
kinda complicated but I understand
I've also finished answering that item lol but thanks
hybrid flicker
alright I get it I think
tidal dock
**ω**
can't you pick something like a = 2/eps
you would get that you need only one aS_eps to cover all R
hybrid flicker
can't you pick something like a = 2/eps
not really
the "integer" references are scaled too
so the holes in between the intervals still exist
tidal dock
are they though?
hybrid flicker
if epsilon = 1/3 for example
S_epsilon = .... U (-4/3, -2/3) U (-1/3,1/3) U (2/3,4/3) U ...
well 1/2 doesn't belong to that set for example
so if S_epsilon is scaled by a
a/2 won't belong to the scaled version
because everything is scaled by a
including the points that don't belong
tidal dock
right okay
vocal sleetBOT
@ruby heart Has your question been resolved?
tidal dock
including the points that don't belong
what if we do something like $3 S_{\varepsilon} \cup \pi S_{\varepsilon}$
(hear me out)
twin meteorBOT
artemetra
wraith mist
There will be holes
tidal dock
the thing is that if we have something like nS_eps union mS_eps where n and m are integers then there will be holes at lcm(n,m) but here because pi is irrational the same argument doesn't apply
yk what idk what i am talking about lmao
hybrid flicker
it means $(a_kS_\varepsilon)_{1\leq k\leq n}$ is a finite open cover of the compact $[0,1]$
twin meteorBOT
rafilou is not not born in 2003
wraith mist
the thing is that if we have something like nS_eps union mS_eps where n and m ar...
We can assume a <= 1:
#help-17 message
tidal dock
ok
hybrid flicker
thus there exists some $\delta$ such that $(x-\delta,x+\delta)\cap [0,1]\in a_{k_x}S_\varepsilon$ for all $x\in [0,1]$
twin meteorBOT
rafilou is not not born in 2003
wraith mist
Yes
hybrid flicker
my hunch is that epsilon = 1/3 doesn't work
so taking this value for example
mmmh
wraith mist
It becomes hard when a gets very small
We start covering half of [0,1] with just one aS_eps
hybrid flicker
thus there exists some $\delta$ such that $(x-\delta,x+\delta)\cap [0,1]\in a_{k...
but by this thing a can't be smaller than delta/epsilon
otherwise the set can be removed from the cover
and won't change coverage
hybrid flicker
It becomes hard when a gets very small
We start covering half of [0,1] with just...
is it really half of [0,1] btw?
yeah I guess for epsilon = 1/3 yes
wraith mist
is it really half of [0,1] btw?
Well, for a=1/2^n and proper eps it's exactly half
vocal sleetBOT
ruby heart
.reopen
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wraith mist
If the center of an interval is at x, then the radius of that interval is xε, so if we want to cover the region near ε, the furthest center point should be at such x that x - xε = ε, so x = ε / (1-ε), and the furthest covered point isn't greater than (1+ε) ε / (1-ε)
hybrid flicker
I had an idea if the topic is limits
Instead of thinking about x in the union of scaled versions of S_epsilon
thinking about the existence of d1,...,dn such that dkx always in S_epsilon for some k
so we can create f(x) = inf{d>=1, dx in S_epsilon}
wraith mist
Isn't it 0?
hybrid flicker
take d >=1
wraith mist
Ah okay
ruby heart
We can assume a <= 1:
https://discord.com/channels/268882317391429632/9034633534...
okay I get that
hybrid flicker
Well, for a=1/2^n and proper eps it's exactly half
ok wait to go back on this
I have an idea
instead of looking for a finite cover of [0,1] with infinite union (ak - aepsilon, ak + aepsilon)
since this last union only has a finite number of intervals within it that intersect [0,1]
we can instead look for a finite cover of [0,1] with intervals (ak-aepsilon,ak+aepsilon)
with epsilon fixed and some a1,...,a_n <= 1, k1,...,kn integers
then
we first notice that to optimize the number of intervals chosen
we can start by picking an interval that covers the number 1 for example
and wlog, we always pick the interval that contains 1 and covers the most out of [0,1]
and repeat, always trying to cover sup(what's left to cover) this way
with a <= 1, the interval that contains 1 and covers the most of [0,1] is (1-epsilon,1+epsilon) (left to prove)
actually let's do it the other way around if you don't mind
start by covering 0
then cover inf(what's left to cover)
etc...
so cover 0 with interval (-epsilon,epsilon) (to prove)
then, we must cover the point epsilon
ruby heart
then cover inf(what's left to cover)
you mean sup(of whats left) right?
or Im misunderstanding
hybrid flicker
you mean sup(of whats left) right?
if we start by covering 0 with some interval
the sup of that interval, call it b, is not covered
ruby heart
yes
ruby heart
oh okay I get it
hybrid flicker
so the idea to cover the most amount of what's around b, here b = epsilon
is to scale down (1-epsilon,1+epsilon)
by some factor a1
ideally the lower bound a1(1-epsilon) is "just enough" to the point we have a1(1-epsilon) = epsilon
so if we take a1 = epsilon/(1-epsilon), we cover less than [0,(1+epsilon)epsilon/(1-epsilon)) in two intervals
to make our life easier we could take the "closed" versions of S_epsilon
meaning our optimized solution covers [0,(1+epsilon)epsilon/(1-epsilon)] in two (closed) intervals
And I just realized that there is a finite cover of [0,1] with that 😭
ruby heart
damn
hybrid flicker
ok so @wraith mist trying to show there is no finite cover of [0,1] is probably a wrong lead
vocal sleetBOT
@ruby heart Has your question been resolved?
vocal sleetBOT
ruby heart
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ruby heart
hmm okay Ill think about it
hybrid flicker
the ks are integers tho
there's a finite cover for a different reason
the intervals you are allowed to use are a*(k-epsilon,k+epsilon) with a in (0,1] and k in Z
the largest k?
if you don't wanna talk about it ok
just that this is false
and the problem 'restricted to [0,1]' fails for a different cover
and now if you wanna talk about this
hybrid flicker
the largest k?
what do you mean by "the largest k"?
vocal sleetBOT
@ruby heart Has your question been resolved?
vocal sleetBOT
@ruby heart Has your question been resolved?
wraith mist
ok so <@301295716066787332> trying to show there is no finite cover of [0,1] is ...
Indeed, I also thought about that when I was offline
It seems that we can indeed find a finite cover of [0, 1] for any eps but that doesn't guarantee that we will cover [1, 2], for example
vocal sleetBOT
Indeed, I also thought about that when I was offline
It seems that we can indeed...
wraith mist
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vocal sleetBOT
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untold arrow
Which statements hold?
Btw I think 1 is incorrect, but What kind of extra conditions would make this statement correct?
hybrid flicker
Which statements hold?
Btw I think 1 is incorrect, but What kind of extra condi...
think about what happens when D is not connex
untold arrow
what's connex?
hybrid flicker
connected
untold arrow
But if D is not connected, then can f"(x,y) still be continuous?
hybrid flicker
But if D is not connected, then can f"(x,y) still be continuous?
why couldn't it be
f(x) = 1/x is continuous on R\{0} isn't it?
untold arrow
Aah i see
untold arrow
think about what happens when D is not connex
Hmm but then what will happen? 
hybrid flicker
idk, try to find counterexamples
is it necessary for f(x,y) to only depend on x when f_y(x,y) = 0?
for example when D is two disjoint balls on top of each other?
untold arrow
is it necessary for f(x,y) to only depend on x when f_y(x,y) = 0?
No ig, what if the y is a constant function
But idk how is this related to disjoint D
untold arrow
for example when D is two disjoint balls on top of each other?
Uhh idk topological stuff..
wraith venture
But idk how is this related to disjoint D
you're trying to generalize the theorem in R
"if f : (a, b) -> R has 0 derivative everywhere, it is constant"
Notice that ||intervals are the connected sets in R||
So this should naturally generalize to ||being constant on connected components.||
hybrid flicker
Uhh idk topological stuff..<:blobsweat:584506239661113371>
imagine f being equal to some constant in the first ball
and f equal to some OTHER constant on the other ball
can you write f(x,y) = phi(x,y) for some function phi?
and prove why/why not
untold arrow
Prove what
untold arrow
and f equal to some OTHER constant on the other ball
And why did we do this? Why do we want the constant different?
hybrid flicker
And why did we do this? Why do we want the constant different?
didn't you want a counter example?
if the constant is the same
then f(x,y) = constant = phi(x)
untold arrow
Why is that? What I mean is that the function only related to y is constant, like phi(y)
Since fy=0, not fx=0
hybrid flicker
?
you want to show 1) is false
so find f
such that fy = 0
and f(x,y) isn't phi(x)
if you took the same constant for the two disjoint balls I told you to consider
meaning
f(x,y) = c on first ball
f(x,y) = c on second ball
then f(x,y) = phi(x) with phi(x) = c
untold arrow
and f(x,y) isn't phi(x)
Ooh! Darn i get it now😂
But what i understood is that fy=0, so assume m(y) is a constant function like m(x)=1 on x>0; m(x)=-1 on x<0, so f(x,y) will be like phi(x) on x>0 and (-1)phi(x) on x<0
So f(x,y) isn't phi(x)
But phi(x) doesn't have to be c if c is a constant, cuz fx != 0
Not sure if I understand correctly tho
hybrid flicker
still makes little sense
I think you don't understand what we're asking
we want f
such that fy = 0
and f(x,y) isn't phi(x)
so we don't care about fx if we wanna disprove "f(x,y) = phi(x)"
if you want to disprove f(x,y) = phi(x)
show that there is some x
and two different y1, y2
such that f(x,y1) different from f(x,y2)
which is impossible since they're both supposed to be phi(x)
untold arrow
But what i understood is that fy=0, so assume m(y) is a constant function like m...
Hmmm could you maybe point out that what's wrong with this reasoning?
hybrid flicker
you write "m(y) constant function" and then you write 'm(x) = ...'
hybrid flicker
so first of all typo, and then second of all, is it constant if it outputs multiple values?
untold arrow
m is the function only related to y
untold arrow
so first of all typo, and then second of all, is it constant if it outputs multi...
Hmm yes..? It outputs 1 and -1 depends on which interval it's in
hybrid flicker
😑
hybrid flicker
So sgn(x) is not constant function?
no
hybrid flicker
But its derivative is also 0 in D 😭
again
just because a function has 0 derivative
doesn't mean it's constant
it's constant ON ITS connected components (thanks for edit)
wraith venture
(connected components)
untold arrow
But Isn't sgn(x) constant function on its connected components?😭
hybrid flicker
But Isn't sgn(x) constant function on its connected components?😭
there is only one connected component, R
untold arrow
But its derivative is also 0 in D 😭
I didn't mean that it's constant function bc of derivative=0 here lol, i mean it fits the statement in the q
hybrid flicker
like
for example
|x| is constant on {-2,2}
does this mean |x| is constant on R?
untold arrow
No
hybrid flicker
because you can find two points
that have different images
so the function is not equal to a CONSTANT
similarly
if I had
wraith venture
patiently screaming be like
untold arrow
Sryyyyy, really unfamiliar with these
untold arrow
No
hybrid flicker
no
because
sgn(-1) is not equal to sgn(1)
so not all outputs are equal
and the function isn't equal to a constant
compared to:
sgn is constant on (-infinity,0)
because when the DOMAIN = (-infinity,0)
sgn = -1 on that domain
so sgn is constant ON (-infinity,0)
now
let D = the union of two disjoint balls on top of each other
for example
D = B_(-2) U B_2
where B_x is the unit (open) ball centered at x
and I took f(x,y) = 1 if (x,y) is in B_2
and f(x,y) = -1 if (x,y) is in B_(-2)
is f constant on each ball separately?
untold arrow
Yes
untold arrow
Uh no..?
hybrid flicker
exactly, why?
untold arrow
Cuz it's separately constant in distinct connected components?
hybrid flicker
?
like the fact it's constant in each connected component isn't bad
why, when we take the union of those components, f is no longer constant?
hybrid flicker
so not all outputs are equal
recall what a constant function is
untold arrow
it has two distinct values (1 and -1) depending on the region within D
A constant function is a function that has the same value for all inputs within its domain
But in this case f has different values in the domain
hybrid flicker
yep
untold arrow
I think I get the overall idea of 1 and 2, but how to think of 3?
hybrid flicker
I think I get the overall idea of 1 and 2, but how to think of 3?
- means "locally independent of x"
untold arrow
It looks similar with 1
hybrid flicker
not really
- said that f(x,y) is globally independent of y when fy = 0
- says that it's locally independent of x when fx = 0
meaning:
if fx = 0, then around any point (x0,y0), if I don't venture out too far, then f(x,y) doesn't depend on x
intuitively, what happens is that we're staying on a single connected component
untold arrow
Sounds like 3 is correct then
Because if it is narrowed down to a small area, then we don't have to consider the issue of disconnection ig
@hybrid flicker Thank you so much for your patience😭
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vocal sleetBOT
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tame wadi
Find limit to the left and right side of the origin.
Where are you stuck?
Okay mb
Converting to polar coordinates helps
I think it depends on theta
so it's not differentiable
untold arrow
Ty!
marble fjord
Those Orange things are Rand switches in a game called RUST. When powered and set they have two functions, on and off (0 and 1). When the power goes on, all the switches flip at once, and have a 50% chance of being on or off. I'm trying to find the probability of only the #1 switch going on, with the rest all being off.
vocal sleetBOT
Those Orange things are Rand switches in a game called RUST. When powered and s...
@marble fjord Has your question been resolved?
knotty lynx
Two easy ways to go about this
You could use the product rule. The probability of any switch being either on or off is 1/2 and that is independent of the state of any other switches.
So you could just multiply the probabilities for 1 0 0 0 (#1 on, #2 off #3 off, #4 off)
vocal sleetBOT
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marble fjord
appreciate it
knotty lynx
No problem
marble fjord
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vocal sleetBOT
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real swallow
Prove that f(x) is < or equal to 1/4
vocal sleetBOT
Prove that f(x) is < or equal to 1/4
gloomy wadi
I would study $f(x)-1/4$ and I would try to use derivatives to find a maximim point $x_M$
twin meteorBOT
cristorenzo99
real swallow
We still didn't study the derivatives. Is there a way to do it without it ?
gloomy wadi
Then try to solve $f(x) \leq 1/4$ and check that the solutions are $x \in \mathbb{R}$
twin meteorBOT
cristorenzo99
real swallow
ok thank you I'm gonna try!
gloomy wadi
Ok! I'm gonna try too to check
real swallow
thanks a lot for your help anyway !
gloomy wadi
thanks a lot for your help anyway !
You're welcome!
I don't know if this works since you get a 4th grade degree equation
real swallow
yes i got the same thing when we'll do the derivatives, i'll try doing this exercise again
gloomy wadi
Or maybe there is a way
Solving the equality, you would get: $\frac{x(1-x)^2}{(1+x^2)^2} \leq \frac{1}{4} \Leftrightarrow x^4-4x^3+10x^2-4x+1 \geq 0$
twin meteorBOT
cristorenzo99
gloomy wadi
But: $x^4-4x^3+10x^2-4x+1 = (x^4-4x^3+6x^2-4x+1)+4x^2=(x-1)^4+4x^2$
This is a sum between two non negative things, hence the whole expression is $\geq 0$
twin meteorBOT
cristorenzo99
vocal sleetBOT
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plush summit
Hi I wanted to check if my solutions are correct for this exercise:
Let a real function f(x) be given. The Taylorpolynomial T1(x; x0) is determined for them at the development point x0.
a) What similarities exist between T1(x; x0) and the tangent g(x) to f(x)
at point x0?
b) What similarities do T1(x; x0) have in common with the properties of the given function at position x0?
My answer:
similarities a)
- Tangent equation g(x) is exactly the same as the first degree Taylor polynomial T1(x;x0)
- Therefore, the slope of g(x) and T1(x;x0) is the same as f(x) at x0
- For a location x that lies in a small region around x0, g(x) and T1(x;x0) both provide a linear approximation to f(x) around x0
similarities b)
- the Taylor polynomial T1(x;x0) has the same function value as f(x) at position x0
- T1(x;x0) is a linear approximation of f(x) around x0
flat whale
do you mean tangent line g(x) ?
plush summit
do you mean `tangent line g(x) ` ?
yup
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thick niche
vocal sleetBOT
thick niche
why is it not root3 -1
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vast shale
if they are equal then their square must also be equal
the square of the first number is $4 - \sqrt{12}$
twin meteorBOT
lunaflower
vast shale
find the squares of each of the other answers
thick niche
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vast shale
how do you get from the first equation to the second form of that equation?
drifting jackal
how do you get from the first equation to the second form of that equation?
Complete the square
vast shale
Complete the square
Yes but im talking about the method used, which is algebraical, not a visual
drifting jackal
Not sure what exactly you are asking. You asked how to go from the first line to the second, you complete the square
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brave wasp
Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
vocal sleetBOT
Michael walks at the rate of $5$ feet per second on a long straight path. Trash ...
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Jonathan Xu
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jade verge
is it possible to prove that the two white triangles are congruent?
kind light
is there any info at all?
brittle cipher
without any info, I'd probably just say the fact that I can rotate the picture ||180 degrees and it looks the same means they're congruent.||
well idk, maybe that's technically giving the answer away sorry
silent sorrel
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fervent sphinx
<@&286206848099549185>
hi
fervent sphinx
so what problems
silent sorrel
i dont know how to do any of them
silent sorrel
mhm
fervent sphinx
mhm
what would her distance be if she bikes 10 miles per hour
i'll help you out
20 minutes is 1/3 of an hour
so
so it's just 1/3 of 10
or 3.33 miles
so
silent sorrel
oh okay
fervent sphinx
so the way we'd derive this is by multiplying the rate(10 mph) by the time
so, D=rate * time
so finish the equation
silent sorrel
so 3.33 * 20 minutes ?
fervent sphinx
basically, we don't knwo the distance, but we do know the time. So, D= R*T, or D=1/3 R
silent sorrel
so the rate is 13 miles per hour right
fervent sphinx
so the rate is 13 miles per hour right
yeah
then plug it in the equation
tell me what u get
silent sorrel
4.3
fervent sphinx
4.3
that's right! but in algebra we ususally use fractions if we can, so a better answer would be 13/3
or 4 1/3
silent sorrel
that's right! but in algebra we ususally use fractions if we can, so a better an...
ohh okay, i understand thank you
fervent sphinx
ohh okay, i understand thank you
how long does the person go in the first 5 minutes
(remember the time is different)
silent sorrel
1.5 ?
fervent sphinx
i did 60 / 5
remember, the unit is miles
and the time is in minutes
so we need to convert those minutes to miles
silent sorrel
covert 5 min to hours
as a fraction ?
fervent sphinx
as a fraction ?
yeah
silent sorrel
1/12
so thats the rate?
i think i’ll do it laterr or try to review my nottees morre cs its due next week
vocal sleetBOT
fervent sphinx
yw
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dusty cypress
How to show this set is convex?
vocal sleetBOT
How to show this set is convex?
hybrid flicker
How to show this set is convex?
triangular inequality + homogeneity?
dusty cypress
triangular inequality + homogeneity?
Whats homogeneity
hybrid flicker
||cz|| = |c|*||z||
dusty cypress
I think we can ignore abs
hybrid flicker
c real number
dusty cypress
because a is in 0, 1 right
hybrid flicker
\|\|cz\|\| = |c|*\|\|z\|\|
sure but the original property is this
dusty cypress
wait i think i know how to do it now
is it like |ax + (1-a)y| <= a |x| + (1-a) |y|
hybrid flicker
yes
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jagged cargo
vocal sleetBOT
jagged cargo
im not sure how to find the inverse of f
im applying the formula $(f^{-1}(x))' = \frac1{f'(f^{-1}(x))}$
twin meteorBOT
vast shale
what about g(f(x))= x where g is f^-1(x)?
jagged cargo
what about it?
vast shale
We don't have to find the inverse function to find its derivative
heavy yoke
the formula above is what you get if you follow that logic to its conclusion
vast shale
Oh
jagged cargo
in the end i still need to find the inverse
and im sure the inverse of x^3 + x isnt pretty at all
,w inverse of x^3 + x
twin meteorBOT
vocal sleetBOT
@jagged cargo Has your question been resolved?
acoustic nest
vocal sleetBOT
acoustic nest
these are both the same axiom in set theory, why are they worded differently? What are the differences?
empty frigate
well they're using different variable names, and one of them is written entirely in symbols while the other one has more words in it, but neither of those really affect what's actually being said, it matters about as much as which font the text is in
aside from that the only difference is that the second version uses "iff", so it also includes the claim that, if two sets are equal, they have the same elements
this can already be proven with no axioms just by the meaning of =
acoustic nest
Thank you bee
empty frigate
...if you want to say something you could just say it here?
acoustic nest
I feel like I understand the gist of it, but not fully
And also, if you had experience learning set theory, do you know any good resources
sinful magnet
is the final matrix the farest we can go?
vocal sleetBOT
is the final matrix the farest we can go?
sinful magnet
like we cannot make 0's anymore?
drifting jackal
You can
sinful magnet
L1-2L2?
drifting jackal
But from that last matrix, you can solve for the unknowns
drifting jackal
Depends on the instructions
drifting jackal
Yeah it should
drifting jackal
You can just reduce it and see if you result in the same solution set
vocal sleetBOT
vocal sleetBOT
Available help channel!
desert apex
I got f(x) = 2070 + (x/4 - 2070/4)
vocal sleetBOT
I got f(x) = 2070 + (x/4 - 2070/4)
gaunt sparrow
You're including the money spent on salaries
gaunt sparrow
Indeed, But they ask for the money spend on monthly activities, not salaries
desert apex
and 2070 represents the salary
gaunt sparrow
So that's just it, no need to add 2070
desert apex
so are you insisting it is d?
gaunt sparrow
Well I mean your original guess is none of the choices
gaunt sparrow
Either the key is wrong, or the question is severely ambiguous
desert apex
yea
ma boi gpt saying its not A but D instead
but anyways I can ask my teacher tmrw
TYSM!!!!
God Bless
gaunt sparrow
ma boi gpt saying its not A but D instead
I would personally go D as well but yea I suppose jsut ask your professor.
desert apex
👍 👍 👍 👍 👍
naive imp
Im doing a question about proofing using induction 5^n + 9^n + 2, where n is an integer im stuck on the step after applying k+1
vocal sleetBOT
Im doing a question about proofing using induction 5^n + 9^n + 2, where n is an ...
gaunt sparrow
What are you proving?
steep crater
whats the problem statemen
naive imp
after that
steep crater
what
steep crater
so your stuck on showing $5^{k+1} + 9^{k+1} + 2$ is a multiple of 4
naive imp
kind of
steep crater
ah
hmmmmm
you know $5^{k} + 9^{k} + 2$ is a multiple of 4 by induction assumption, then i would consider subtracting $5^{k} + 9^{k} + 2$ from $5^{k+1} + 9^{k+1} + 2$
steep crater
to get $5^{k+1} + 9^{k+1} - 5^k - 9^k$ and we are done if this is a multiple of 4
steep crater
so we have to consider mod 4, $5^{k+1} + 9^{k+1}- 5^k - 9^k$ is equal to $1^{k+1} + 1^{k+1} - 1^k - 1^k$ (mod 4).
steep crater
and $1^{k+1} + 1^{k+1} - 1^k - 1^k = 0$, so $5^{k+1} + 9^{k+1}- 5^k - 9^k$ is divisible by 4.
steep crater
@naive imp
gaunt sparrow
I mean this works, but it's hardly using induction if you could've just done mod 4 on 5^n + 9^k + 2 to begin with.
steep crater
I mean this works, but it's hardly using induction if you could've just done mod...
but there is essence of inductive step
even though you could so mod 4 in the beginning
brittle cipher
I think you either use induction or mod, not both on this
steep crater
I think you either use induction or mod, not both on this
oh
vocal sleetBOT
@naive imp Has your question been resolved?
tired ruin
unkempt spire
i need help with annuities please.
vocal sleetBOT
i need help with annuities please.
vocal sleetBOT
@unkempt spire Has your question been resolved?
vocal sleetBOT
@unkempt spire Has your question been resolved?
sharp dove
vocal sleetBOT
sharp dove
which one is true? red or blue?
and why?
i am trying to find the power series of ln(4-x) via differentiation
i know ln(1+(3-x)) is easier but im trying something else here
cyan shadow
red and blue are the same, assuming the +C is from indefinite integration
incorrect, doesn't work over an infinite sum, see clouds answer
sharp dove
yes it is the same but if i arbitrairly choose C=3 its different
or are they should not be the same number?
heavy yoke
the +C should be outside the sum
you just add one constant for the whole integration, it doesn't add to every single term
sharp dove
that makes sense logically speaking. because the blue colored one will have infinity*c, is this correct?
heavy yoke
it would diverge if you added any constant other than 0 to the terms, yes
sharp dove
the +C should be outside the sum
if the upper bound of the sum is not infinity, lets say its 10, would this still hold?
heavy yoke
if the upper bound was finite, then it still "should" be outside the sum, but if it was inside the sum then you end up adding 10*C, (which is itself an arbitrary constant), so it's not a problem in the same way
sharp dove
ah i see because both inside and outside are constants... this makes a lot of sense
i kinda find it weird that the solution makes the +c pass through the summation but because they are esentially arbitrary constants, that is saying the same thing
okay i understand now. thankyou very much everyone!
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
vast shale
,calculate 10/(sqrt)2
twin meteorBOT
The following error occured while calculating:
Error: Undefined symbol ulate
vocal sleetBOT
,calculate 10/(sqrt)2
vast shale
hmm...
torn oracle
,calc 10/(sqrt)2
twin meteorBOT
The following error occured while calculating:
Error: Unexpected type of argument in function multiplyScalar (expected: number or Complex or BigNumber or bigint or Fraction or Unit or string or boolean, actual: function, index: 0)
vast shale
,calculate 10 / (sqrt)2
twin meteorBOT
The following error occured while calculating:
Error: Undefined symbol ulate
fervent wasp
,calculate 10/(sqrt)2
twin meteorBOT
The following error occured while calculating:
Error: Undefined symbol ulate
fervent wasp
,calculate 10/sqrt2
twin meteorBOT
The following error occured while calculating:
Error: Undefined symbol ulate
fervent wasp
Bruh
fervent wasp
,calculate 10/(sqrt)2
What’s your question?
brisk moss
,calc 10/sqrt(2)
twin meteorBOT
Result:
7.0710678118655
I kinda solved it already
Should I make a new one for another question or should I paste it here?
inner osprey
standard protocol dictates you should claim a new channel
in practice, it will likely not matter much
arctic mantle
but it doesn't really matter
vast shale
Kiomi says that It's either I get a warning for pasting a different subject question or a warning for making many channels
arctic mantle
if it's not maths we most likely can't help
and we just recommend that you keep one channel OPEN
instead of yk opening 3 for the same question
it's perfectly fine to post another maths question here though
vocal sleetBOT
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
lyric sluice
Do you know the side length of the first square
tame wadi
She solved it already she is stuck on last row ig
lyric sluice
Ah
vast shale
She solved it already she is stuck on last row ig
Yeah thanks
I still don't know how to solve for the last row column
tame wadi
See,
I'm taking an example here.
2,4,8,16,....
It's a geometric series.
here, a = 2 and r = 4.
let's say I wish to find partial sum upto n digits.
Do you know the formula?
Sn=a(1-r^n)/(1-r)
vast shale
Yeah
tame wadi
Okay so if I wish to find S5.
I'll be finding the partial sum upto 5 digits.
Which is (2+4+8+16+32)
vast shale
2(1-4^2)/(1-4)?
tame wadi
2(1-4^2)/(1-4)?
r=2
vast shale
See,
I'm taking an example here.
2,4,8,16,....
It's a geometric series.
here, a ...
?
a = 2 and r = 4?
Did you do it backwards?
vast shale
a = 4 and r = 2?
tame wadi
a is the first term which is 2.
vast shale
2(1-4^2)/(1-4)?
So what I did is correct
vast shale
a = 2 and r = 4
Sn=a(1-r^n)/(1-r)
Sn = a(2)(1-r(4)^n(2)/(1-r(4))
Okay so a = 4 and r = 2
Sn=4(1-2^4)/(1-2)
tame wadi
Okay so a = 4 and r = 2
no, a is 2 and r is also 2.
vast shale
a = 2 and r = 4?
then what is this?
So you just confused me for no absolute reason okay
a = 2 and r = 2
tame wadi
a = 2 and r = 2
yes
vast shale
Sn=2(1-2^2)/(1-2)
tame wadi
Sn=2(1-2^2)/(1-2)
Provided you want to find partial sum upto two terms in the series
vast shale
Sn=2(1-2^2)/(1-2)
Sn = 6
tame wadi
Yes
Which is similar to (2+4), the first two terms of the series.
So partial sum upto two terms of the series is 6.
What would be partial sum upto five digits?
vast shale
36
tame wadi
Sn=a(1-r^n)/(1-r)
Put n=5 in this formula.
tame wadi
Correct.
vast shale
How about this one?
tame wadi
Now let's get back to your example.
You found perimeter of squares which makes geometric series like..
40,40/√2,20,20/√2,...
Correct?
row 3 in your question
tame wadi
Now if we want to know partial sum upto two terms in this geometric series, we'd do something like
(40+40/√2), right?
vast shale
Wait wait
tame wadi
or we could do
sn=40(1-(1/√2)^2)/(1-(1/√2))
tame wadi
Now if we want to know partial sum upto two terms in this geometric series, we'd...
Since we know the series and we want to find just upto n=5. I think it'd be best approach.
vast shale
The thing is my answer... is... kinda wrong
My perimeter of nth square is 40,40/√2,20,10, 10/√2
vocal sleetBOT
@vast shale Has your question been resolved?
vast shale
<@&286206848099549185>
vocal sleetBOT
@vast shale Has your question been resolved?
vast shale
<@&286206848099549185>
strong isle
what have you tried, or what are you up to?
vast shale
what have you tried, or what are you up to?
!status
vocal sleetBOT
what have you tried, or what are you up to?
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
n is 5?
vast shale
what?
strong isle
what?
oh you said last column of the table, so i thought you meant you did not know how to fill in the n=5 column
vast shale
Idk.. how to solve for it
strong isle
Idk.. how to solve for it
isnt it just using the previous values?
vast shale
isnt it just using the previous values?
I do not know
strong isle
I do not know
did you do the n=1 column?
vast shale
did you do the n=1 column?
Everything... is n=1
strong isle
alright, whats the side length of the first square?
vast shale
10
vast shale
yes, now how would you find the side length of the second square
the answer is 10/(sqrt)2
Oh-
Erm
strong isle
the answer is 10/(sqrt)2
this is correct
strong isle
yes, do you see what you are doing each time to the previous number?
strong isle
the answer is 10/(sqrt)2
also this is an interesting way to put it, i never really thought of that
vast shale
also this is an interesting way to put it, i never really thought of that
what were you thinking?
strong isle
what were you thinking?
i was thinking you would do sqrt( (10/2)^2 + (10/2)^2) = sqrt ( 2 * (10/2)^2 ) = (10/2) * sqrt(2) = (10sqrt(2))/2 which is still the same as your one because 1/sqrt(2) is the same as sqrt(2)/2
strong isle
then 2.5
anyways you got the side lengths for all the 5 squares im pretty sure, so then now i think you should be able to fill in the rest of the rows and even find recursive formulas after
vast shale
How do I do that
strong isle
so you know how to do side length of the nth square divided by 2 and the perimeter of the nth square, right?
vast shale
Yes
vast shale
Yes
strong isle
now do you know how to find the nth partial sum of perimeters?
vast shale
no
strong isle
no
its basically adding the nth square to the (n-1)th partial sum of perimeters
so you just keep adding the perimeter of the smaller square each time
vast shale
Something like geometric series or sequence?
strong isle
Something like geometric series or sequence?
yeah, should be a geometric series
strong isle
Something like geometric series or sequence?
this will also help you in finding the nth partial sum of perimeters, no matter how large the n im pretty sure
vast shale
Oh Sn=a1/1-r?
strong isle
Oh Sn=a1/1-r?
yes, thats for an infinite geometric series
vast shale
Oh then Sn=a1(1-r^2)/1-r?
strong isle
Oh then Sn=a1(1-r^2)/1-r?
thats for when n=2
in general its S_n = (a_1)(1-r^n)/(1-r)
strong isle
Oh then Sn=a1(1-r^2)/1-r?
here you did it for n=2, aka sum of the first 2 terms in the sequence
vast shale
What's r?
strong isle
What's r?
r is the ratio between 2 consecutive terms. basically r is (a_n)/(a_(n-1)) where a_n is the nth term of the sequence/series
vast shale
So r is 0?
strong isle
So r is 0?
no
vast shale
I don't get it
strong isle
I don't get it
what sequence are you looking for firstly?
vast shale
A1
vast shale
uhh
strong isle
i think you mean you are looking for the nth partial sum of perimeters sequence
strong isle
uhh
im sorry if i was not much help, but i have to go now, and i hope someone else could properly help you soon
vast shale
i think you mean you are looking for the nth partial sum of perimeters sequence
Okay
<@&286206848099549185>
vast shale
Yes
tame wadi
Okay that's a geometric series right?
vast shale
Yes
tame wadi
Since, n=1 represents first inner square
We'd add the perimeter of outer square and first inner square to get partial sum for n=1.
vast shale
I don't get it
tame wadi
Do you know perimeter of outer square?
vast shale
No
tame wadi
You did calculate it in the third row.
It's four times the length of the outer square.
vast shale
You mean the perimeter of nth square?
tame wadi
Perimeter of the square with length of 10.
vast shale
You did calculate it in the third row.
Yes
tame wadi
Can I see it?
tame wadi
vast shale
?
tame wadi
I think you should interchange these two terms.
It's wrong as
if you multiply the first row by 4 you get the third row.
Start from beginning,
If you multiply 10 by 4 you get 40.
tame wadi
No.
20/2 = 10
20/√2 != 10
(!= not equal to)
tame wadi
Your 3rd should be changed to 40,40/√2,20,20/√2,10.
Then we'll start focusing on fourth row.
vocal sleetBOT
vocal sleetBOT
Available help channel!
cunning olive
Can someone confirm if these two answers are correct for both of these questions? thanks
vocal sleetBOT
Can someone confirm if these two answers are correct for both of these questions...
@cunning olive Has your question been resolved?
vocal sleetBOT
@cunning olive Has your question been resolved?
cunning olive
<@&286206848099549185>
vocal sleetBOT
@cunning olive Has your question been resolved?
cunning olive
<@&286206848099549185>
cunning olive
<@&286206848099549185>
hybrid citrus
Hdlooo
barren jolt
where do i start with taking the derivative of this 2x(5x-8)-x^4 *4
barren jolt
can i jus multiply parenthesis with 2x?
barren jolt
prisma rain
2x * 5x = ?
barren jolt
barren jolt
Product rule?
❓
barren jolt
first one
imagine youre using latex
i didnt use {4*4}
That was my thought process while writing it
prisma rain
so then -16x^3?
yeah
prisma rain
nope
barren jolt
yes
prisma rain
its -16x³
twin meteorBOT
kaue
twin meteorBOT
Luh Roub
prisma rain
turn the square root into a power and simplify
prisma rain
$\sqrt{p} = p^{1/2}$
twin meteorBOT
kaue
barren jolt
i know
barren jolt
$2x^{2} \cdot 7y^{-\frac{2}{4}}$ , correct?
twin meteorBOT
Luh Roub
prisma rain
you mean x instead of y
prisma rain
!occupied
vocal sleetBOT
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).
mystic orchid
barren jolt
4/2 is 2 therefore x^2
prisma rain
**Luh Roub**
now rearrange constant with constant, and x power with x power, then add the powers and multiply the constants
barren jolt
now rearrange constant with constant, and x power with x power, then add the pow...
lots of constant in 1 sentence
prisma rain
join the 2 and the 7, and the x² with x^(-2/4)
prisma rain
$x^a \cdot x^b = x^{a + b}$
twin meteorBOT
kaue
prisma rain
its all a product there
and you can rearrange terms in product bc order doesnt matter
prisma rain
isnt x 2x and 2nd x 7x?
what does that mean
prisma rain
its double of x²
barren jolt
so then $14x^{\frac{3}{2}}$
twin meteorBOT
Luh Roub
prisma rain
yep
prisma rain
yes
barren jolt
got it tysm