Where does the + 1 come from?
#help-4
terse chasm
pure mirage
From the (k+1)(k+2) summand
terse chasm
Oh right
pure mirage
vale dockBOT
pure mirage
(you can just pull a 1/3 out of the last bracket now, and multiply what's inside of it with 3 to even that 'pulling out' out, giving you k + 3 inside the bracket and a 1/3 factor infront)
And you are there
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mellow iris
Consider a circle with symbol AB and center O. Let points C and D differ on this circle. The line CD intersects AB at point M (with MB < MA and MD < MC). Let K be the intersection of the intercepting circles (different from O) on the triangular circles AOC and DOB. Prove that <)MKO = 90
mellow iris
cant solve this
wild bobcat
Consider a circle with symbol AB and center O. Let points C and D differ on this...
what have you done so far
did you get the figure
mellow iris
i tried to solve this with power of point but failed
prof told us we need to use spiral similarity
wild bobcat
could you favor me the diagram
mellow iris
im out rn
exotic grove
hw for hw
mellow iris
you mean the drawing?
vale dockBOT
@mellow iris Has your question been resolved?
wild bobcat
yeah
mellow iris
not on me
wild bobcat
i could help you if you get me the fig
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@mellow iris Has your question been resolved?
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sudden forum
hi kings and queen š how this wrong inflection is between min max am i missing something
waxen scarab
The function isn't differential able at x = -4
Also you can think of infliction point as that a function was concaving before being convexing
sudden forum
Also you can think of infliction point as that a function was concaving before b...
4
waxen scarab
Yeah and what else
waxen scarab
What about 0
sudden forum
wait
waxen scarab
there's only 1 min
It doesn't have anything to do with mins
waxen scarab
Yeah
waxen scarab
I don't know, it is supposed to be that
short moth
what in the freaky
sudden forum
what in the freaky
pls daddy toji help
u so much aura help me
pls keep typing
i go on two knees
short moth
lil bro
short moth
0 and 4 are def correct idk ab 6 tho and i do not wanna pretend like i know
sudden forum
<@&286206848099549185> daddys please
why u wilt me start helping
š š
i got it
@short moth apparently 2 is a point inflection as well why that
.close
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ivory tree
how do i solve this?
vale dockBOT
how do i solve this?
ivory tree
i know i can simplify the fraction to (1+x_n)/(1+x_n+x_n^2) -1
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@ivory tree Has your question been resolved?
midnight pier
how do i solve this?
can u send a translated version+
midnight pier
Is it 67?
6 7 š„
ans is 86 ig
@wraith quarry u got 1/x1+1 as the ans right?
xn=1/xn+1-1/x(n+1)
wraith quarry
xn=1/xn+1-1/x(n+1)
Ive changed the terms for the telescopic so idk wht u talking boutš„²
midnight pier
yea i made telescopic 3
wraith quarry
xn=1/xn+1-1/x(n+1)
Mann, can u pls write it in like math form, i cant understand if n is up or down
midnight pier
wait
wraith quarry
Oh yeah got it,
Mine was 1/((1/x1)+1)
The answer i mean
Its impossible to write the telescopic in chatš¶āš«ļø
Moreover the mods r gonna freak out if i send a snap anymore
midnight pier
yea
vale dockBOT
@ivory tree Has your question been resolved?
wraith quarry
@ivory tree should i send u the solution and explain the thing?
ivory tree
<@740215914502291507> should i send u the solution and explain the thing?
yeah sure
ivory tree
can u send a translated version+
it's just a recurrence relation, but basically just find the sum of all the terms of that series from 1 to infinity
wraith quarry
wraith quarry
1st term comes from simple rearrangement of the question
vale dockBOT
@ivory tree Has your question been resolved?
idle void
<@740215914502291507> should i send u the solution and explain the thing?
Hi as a helper your role is to guide helpees to find a solution themself, not just give them the solution
Next time you do this it will incur a time out
graceful saddle
Given a subspace X = <0,1] U {2} U <3,4> of an euclidean space (R, d2), describe euclidean balls in the subspace X.
pine prairie
a subspace?
graceful saddle
yes
pine prairie
what does <>, [], and {} mean
graceful saddle
open interval, segment and a set
notation
I don't get which points should I include in a ball
I know that I have to check the counstruction of balls for x element of <0,1] then x = 1 then x = 2 and then x element of <3,4>
.close
vale dockBOT
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vivid plover
I have a question regarding limits
vale dockBOT
I have a question regarding limits
vivid plover
I know the idea behind limits and the epslion-delta definition this is fine.
I dont even know, if my question is more philosophical. Sorry, I am not sure how to use Latex on discord
Now the epsilon-delta definition explicitly states that lim x->a f(x) = L if and only if for every (\epsilon >0), there exists a (\delta >0) such that if (0<|x-a|<\delta ), then (|f(x)-L|<\epsilon ).
This implies the distance between L and f(x) is never 0, which fits with the general intuition about points around our point a, but we can get arbitrarily close.
Assume we have a function f(x)=x which is undefined at x=3
Assume we have a second function g(x)=x^2
Why are we allowed to calculate lim x->3 g(f(3))? lim x->3 f(x) will be 3, since our function approaches 3 as x approaches 3, but we dont evaluate at the point exactly, just arbitrarily close.
So why is it fine then to plug in the limit L into g(x)? Because 3 is actually not defined in f(x)?
rocky lotusBOT
Mathmatix
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vale dockBOT
@vivid plover Has your question been resolved?
silent dune
We can show that if f(x) -> L as x -> a and g(x) is continuous, then g(f(x)) -> g(L) as x -> a
If g say wasnāt continuous, then it would not necessarily be true in your example that g(f(x)) -> g(f(3))
Try and find an example of that
It might be fruitful
Also nothing stops us from computing the limit of g(f(x)) in general (with a priori no relation to what g(L) is, since they donāt have to be equal)
If it exists
Since g(f(x)) implicitly defines a function, say h
vivid plover
Okay, mabe let me rephrase the question a bit.
Lets take the definition of the derivative lim h-> 0 f(x+h)-f(x)/h.
For f(x)=x^2 we get lim h->0 (x^2+2xh+h^2-x^2)/h thus we end up with
lim h->0 (h(2x+h))/h, canceling the h's leads to. We were able to cancel h, because it is not 0, just super close.
lim h->0 (2x+h) Taking the limit yields 2x.
Now we say, to calculate the instantanous rate of change we can use f'(x)=2x. At x=2 it would be 2(2). But why is this indeed the derivative? Because we never really set h equal to 0? Why is lim h->0 (2x+h) = 2x valid? Even tho we get super close to 0, there is a non 0 distance that we throw away? For me it sounds like we implicitly do some 'rounding' by ignoring super small numbers but they are still there? Where is my misconception?
silent dune
The derivative and limit are the same
Thatās what defined the derivative, that limit
Okay I see ur question now
thereās different ways of thinking about it, for 2x + h
Weāre for fixed x using the fact that the limit preserves sums
I.e lim of a+b is lim of a + lim of b
Whenever both sides make sense
silent dune
We can show that if f(x) -> L as x -> a and g(x) is continuous, then g(f(x)) -> ...
Or we could also use this fact as well
For fixed x, lim h->0 2x is just 2x (a constant relative to h)
And lim h->0 h =0
silent dune
Or we could also use this fact as well
Using this, our f(x) is say f(h) = h, and g(h) = 2x + h, for fixed x. As g is continuous, then lim of g(f(h)) is jut g(0) = 2x
Here x and h have swapped roles since thatās what is varying
x is just some constant
For clarity
vivid plover
Okay wait, take the function f(x)=1/x as x approaches 0. At f(0) it is undefined. However, lim x->0 f(x) is 0. This simply states that for every epsilon > 0 we can get so close, that |L-f(x)| < epsilon right? So we are really not evaluating the point itself just the surroundings correct?
silent dune
That limit is undefined
silent dune
Using this, our f(x) is say f(h) = h, and g(h) = 2x + h, for fixed x. As g is co...
Does this make sense btw?
I think this way of thinking might be helpful
Since continuity often times is sort of obvious and intuitive
vivid plover
Using this, our f(x) is say f(h) = h, and g(h) = 2x + h, for fixed x. As g is co...
Yeah, algebraically it makes sense
silent dune
Do you know the defining property of continuity?
It preserves limits
I.e if g is continuous, then lim(g(x)) = g(lim(x))
vivid plover
Correct
silent dune
We can show that if f(x) -> L as x -> a and g(x) is continuous, then g(f(x)) -> ...
So thatās why this should also make intuitive sense
vivid plover
Indeed it makes sense
silent dune
Okay great
vivid plover
But I think the problem I had was a bit different, even tho you cleared the first part with the function composition up
silent dune
I mean it explained lim h->0 2x + h as well
This could be seen as a function composition too
The other route would be to say use the sum rule for limits and what not
silent dune
Using this, our f(x) is say f(h) = h, and g(h) = 2x + h, for fixed x. As g is co...
But this approach is a bit more general
vivid plover
I think my issue is less formal and more intuitive or philosophical
If we say, a function f(x) is undefined at a, but we take the limit for x->a f(x) and the limit exists, then it means, that we can get as close as we want to L when approaching, but we still keep a distance delta > 0 around the point a and therefore not actually evaluate the point itself right?
silent dune
Yes
silent dune
I think my issue is less formal and more intuitive or philosophical
If we say,...
Also btw my intention was that it could be seen as intuitive
The way I explained it before, but Iāll let u continue
vivid plover
Also btw my intention was that it could be seen as intuitive
I know, and it helped a lot! Already thankful for that
I think my underlying issue is that when using limits we always are talking about points AROUND the limit and not the point at the limit itself. But then in following calculations we use the limit as if it was a value that was evaluated AT that point you know?
Like in the 2x+h example h was replaced with 0, in the composition we plug the limit 3 into g(x)
Because of delta > 0 and epsilon > 0, for me there is always some microscopic distance, that we just ignore or truncate but even tho it is super small, it is still there
silent dune
Well I think continuity might help explain this phenomenon; in that points close enough are exactly the same as plugging it in given some circumstances
vivid plover
For continuous functions sure
silent dune
Well in all ur examples we can rigorously replace things with a continuous function of some sort
Unless u have an example where this is not the case?
Continuity plays nice with limits in essence
Thatās what youāre witnessing
vivid plover
But f(x) in our case wasnt continuous since I defined it as f(x)=x with x=3 being undefined
silent dune
Yes but thatās not what Iām arguing for, where are u plugging things in solely with f(x)?
Also btw f is continuous in this case
But that doesnāt matter
silent dune
I think my underlying issue is that when using limits we always are talking abou...
Iād love to give u an example where this intuition fails btw
silent dune
We can show that if f(x) -> L as x -> a and g(x) is continuous, then g(f(x)) -> ...
It ties into this
vivid plover
True that, maybe I need to take a step back, and the issue will resolve itself. Or I will come back when I have a better concrete example
Thanks for that awesome help tho!
silent dune
I could maybe try and rephrase myself
silent dune
Thanks for that awesome help tho!
Ur welcome
vivid plover
Right now, I dont even know, what exactly I could ask š
silent dune
Well I think I understand ur issue
In that you think itās weird how we can sometimes just plug in what we are trying to approach?
vivid plover
More like, how we can use the limit of a non defined point in further calculations, because in my head, there is still an error margin.
Plugging the Limit into an continuous function like g(x) is not the issue, since the function is continiouos. The issue is more about the limit of f(x) itself. Taking the limit for an undefined point? Thats cool, we just look which value we approach as we get arbitrarily close to the point without reaching the point exactly itself. But then using this result L and plugging it into further calculations, makes the subsequent claculations in my head not exact anymore and more like ā insted of a = because we never reached the limit as the point was undefined
silent dune
I see, well if it helps I can provide an example where itās not okay to just plug in L for a certain limit; could that be something worth looking at?
vivid plover
Sure, any view angle is appreciated
silent dune
Okay awesome, letās suppose we had f(x) as u defined it before. Now say we wanted to compute the limit of g(f(x) as x -> 3 where g(x) = -1, x <= 3 & 1, when x > 3
Then g(lim f(x)) = g(3) = -1, but lim of g(f(x)) as x->3 is the same as computing the limit of g(x) as x -> 3 since weāre not looking at x = 3
Do u see the problem?
vivid plover
Okay awesome, letās suppose we had f(x) as u defined it before. Now say we wante...
x <= 3 & 1, when x >= 3 I think there is an issues, one of the equal signs has to go right?
silent dune
Oops x > 3 on the right one
vivid plover
The problem is g is not continuous right?
silent dune
Yeah sure, but moreover does the limit of g(x) as x ->3 even exist ?
vivid plover
No becasue left and right side of the limit are different
vivid plover
Exactly
silent dune
Your question boils down to, is g(x) continuous or not
As we can always rephrase a computation as such
vivid plover
But thats an issue about g(x) itself since it is not continous. Your issue regarding the limit of g(x) has nothing to do with f(x) itself. You could also just have purely taken lim x->3 g(x) and arrive at the same issue. I am talking about using the Limit of any function at an undefined point in further calculations introduces a margin of error (in my head). Because we are just approaching the undefined point in f(x)
silent dune
The fact that it reduced to lim x -> 3 g(x) was because it happened to be the case that f(x) = x when x is not 3
āFurther calculationsā here is supposed to be represented by g(x)
And Iām assuming most of the āfurther calculationsā youād be doing is simple combinations of common operations?
Well guess what that is a continuous function
Put differently, putting in the limit of f(x), call it L, in further calculations is the same as computing g(L) for some function g that encapsulates our āfurther calculationsā
vivid plover
Ahhh the last sentence was a good one!
I think that helped to clear a lot of the confusion. I will take a break now. If I will have any questions left I would ping you, if thats cool
silent dune
I think that helped to clear a lot of the confusion. I will take a break now. If...
Yes that sounds good
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@vivid plover Has your question been resolved?
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low yoke
here I have a question, given a sequence $u_(n+1) - u_n = 2n -1$, identify succession u_n
rocky lotusBOT
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red tulip
You mean find the general term of Un in terms of n?
red tulip
Oh just a graph
Then in the given domain follow the steps of the formula for each term
So u1 is 7
red tulip
It says u1=7
low yoke
where
pliant nova
isn't that a different question
low yoke
isn't that a different question
oh, a very cool person joined the chat
red tulip
Bit
low yoke
anywho
yeah it doesn-t say the first term
I-ll say how I did it, because it-s correct
and I just don't think
I got there the right way
pliant nova
ok
low yoke
give me just a sec to write it out
red tulip
The increase with each term is 2n-1
Which is never below 0 for n>0
So it always increases
low yoke
$u_(n+1) - u_n = 2n -1 \Leftrightarrow u_(n+1) = 2n -1 + u_n \Leftrightarrow u_n = (2(n-1)-1)*2$
Alright
rocky lotusBOT
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pliant nova
is it linear?
red tulip
It wouldnt be linear though?
low yoke
I got there by thinking it out
pliant nova
yeah it isnt linear
low yoke
but
low yoke
It wouldnt be linear though?
look at the options
red tulip
The rate its increasing is increasing as well
low yoke
all the options are parabolas tho
red tulip
Yeah those options arent linear
pliant nova
it's a quadratic
you're adding 2n - 1, n times essentially
low yoke
ya that's the issue
red tulip
Linear is where they make a line
pliant nova
can you explain where you get the *2
red tulip
Remember n increases by each term
low yoke
can you explain where you get the *2
I did this a few hours ago, but I think if was to change the term to u_n intead of u_(n+1)
low yoke
it's incorrect
let me see
essentially I think
how it's supposed to be solved is
you figure out the sequence and you plot it into the graphical calculator
red tulip
the difference equation essentially states that the difference between any 2 consecutive terms is 2*(the first index in question) -1
With each term n increases
Therefor the rate is increasing
Which makes it quadratic
low yoke
yeah, but my calculate (I think) do not allows me to represent terms based on previous terms, I'd have to find some way of defining it without relying on previous terms, I think the real question I have is, how to transform the given expression in the question to a sequence
which I can plug into the calculator and see
how do I do that
red tulip
Thats the thing with difference equations
Do you know how to solve these using the complementary and particular solutions?
low yoke
Do you know how to solve these using the complementary and particular solutions?
ummm, it might be the language but I don't understand the term
low yoke
Thats the thing with difference equations
this is called difference equations in english? it ways geometric and arithmetic progressions tho
(litterally translating)
low yoke
Do you know how to solve these using the complementary and particular solutions?
you can give an example
red tulip
Alr for this example the first thing im talking about is pretending the 2n-1 bit is just 0
low yoke
ok
red tulip
So $U_{n+1}-U_n=0$
rocky lotusBOT
ImOakley
low yoke
ok
red tulip
Then you imagine $U_{n+1}$ is x and $U_n$ is 1 and find the root(s) of the equation
rocky lotusBOT
ImOakley
red tulip
So $x-1=0$ and $x=1$
rocky lotusBOT
ImOakley
red tulip
And you put the x value into the general term equation which is $U_n=Ax^n+$(particular solution)
rocky lotusBOT
ImOakley
red tulip
Then you find what the particular solution is, which involves the 2n-1 bit
Since its linear we say the particular solution is an+b
And pretend for a sec that thats the sequence and put it into the original equation, using that to solve a and b
low yoke
Since its linear we say the particular solution is an+b
it's not linear no
Ćt's a parabola in the solution
red tulip
I mean the 2n-1 part
low yoke
oh yeah
that's linear
that's sounds overcomplicated though, I don't remember anything like that in my lectores
red tulip
Alr then ignore what i said lol
low yoke
no
that's not what I mean
amazing thinking
and I believe you this is the most elegant way
red tulip
Its probably not im just rehashing what im doing in app maths which is this but with U_{n+2} terms
low yoke
ok
red tulip
The 2n-1 bit messes things up a bit thats why i went there
low yoke
hmm
then again, maybe they never asked me to "determine" because they knew I couldn't show my reasoning in that wat, they asked to identify, maybe I just solved it how they wanted to, by using a table and thinking the terms
sorry, I think this was all pointless, kind of unsatisfying resolution
red tulip
The simplest thing to do here might be writing out the first few terms without actually solving them if you know what i mean
Do we know what the first term is?
low yoke
no
however
we can start with u_(n-1)
but we also don't know that
yeah
and if I try to get u_n, it will require me to get u_(n+1) first, and I cannot start at infinity
damn
red tulip
Start with some first term a
U2=2(1)-1+a
U3=2(2)-1+2(1)-1+a
U4=2(3)-1+2(2)-1+2(1)-1+a
Its like an arithmetic series - n + the starting term
vale dockBOT
@low yoke Has your question been resolved?
dull cipher
3/2|x-1| + 11 = 5x - 7
vale dockBOT
3/2|x-1| + 11 = 5x - 7
dull cipher
3/2|x-1| = 5x - 18
quasi valve
is |x-1| in the denominator?
dull cipher
|x-1| = 5x/3/2 -12
dull cipher
is |x-1| in the denominator?
no
dull cipher
|x-1| = 5x/3/2 -12
how do i proceed from this
quasi valve
the most straightforward way is to consider two cases
x >= 1 and x < 1
and see what solutions occur in each case
dull cipher
can i do |x-1| = -(5x/3/2 - 12) and |x-1| = 5x/3/2 - 12
quasi valve
no, your equation is the second one
you can't just introduce a new minus sign on the right
quasi valve
if x >= 1 then |x-1| becomes x-1
and if x < 1 then |x-1| becomes -(x-1)
this way you can get rid of the abs value but at the cost of dealing with two cases
dull cipher
-(x-1) = 5x/3/2 - 12 and (x-1) = 5x/3/2
@quasi valve this ok?
dull cipher
mb x-1
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granite trellis
hi! When I'm proving that two sets have the same cardinality, I've to come up with a function that is bijective but im kinda stuck when thinking of a funciton.
The recent problem i tried was |(0,1)| = |R|
granite trellis
here for LHS since its a subset of i know that the cardinality is |(0,1)| <= |R|
but for the other side, im confused as to how to identify a function
river shale
I NEED HELP WITH 1+1
Donāt troll.
left fox
I NEED HELP WITH 1+1
here gng
keen tundra
but for the other side, im confused as to how to identify a function
try crafting something with functions you know
reciprocals, exponents, exponential functions, etc
granite trellis
reciprocals, exponents, exponential functions, etc
i thought of arctan and using piecewise function when x < 0 and when x > 0, but that would be a closed interval. How can i convert it into an open interval?
left fox
Don't encourage trolling
sorry
granite trellis
it gives you a bijection between R and (-pi/2, pi/2)
and if i multiply that by 2/pi and -2/pi,would that make the range (0,1)
granite trellis
i meant in the function
keen tundra
yeah so you get 2/pi * arctan(x)
but notice this is a bijection between (-1, 1) and R
not (0,1) and R
granite trellis
for that i would make it a piece wise with +- 2/pi * arctan(x)
keen tundra
but then injectivity will fail
as this will be an even function, if understand you correctly
granite trellis
oh i didnt consider that
keen tundra
something like this you meant right?
granite trellis
yes like that
keen tundra
so you can see its not injective
keen tundra
but notice this is a bijection between (-1, 1) and R
this is really close
try to think how to modify this slighly to get what you want
granite trellis
i think the graph i need would have an e
keen tundra
hint: ||try to get a bijection (-1,1) --> (0,1), so you can compose it with your original bijection to get a bijection between R and (0,1)||
keen tundra
i think the graph i need would have an e
there is something related to that! just slightly different then what we are going for with arctan
granite trellis
to remove the -ve 1,. adding +ve 1 would be helpful but i need to make sure that +ve 1 changes according to the value of x right
keen tundra
ve?
granite trellis
positive and negative, i wrote that as -ve and +ve
keen tundra
yeah so adding 1 after the original bijection gives a bijection between R and (0,2)
whats the next step?
granite trellis
divide the whole thing by 2 again?
keen tundra
yes!
now, can you give the expression for the bijection we got between R and (0,1)
just for completeness sake
granite trellis
1/pi * arctan(x) + 1/2
keen tundra
nice
keen tundra
there is something related to that! just slightly different then what we are goi...
now the other bijection you can build is $\frac{1}{1+e^{-x}}$ which is called the sigmoid function
rocky lotusBOT
ExpertEsquieESQUIE
granite trellis
so sigmoid can be used for other types of proof as well?
lets say the range was (0,2)
keen tundra
yeah so multiplying sigmoid by 2 works
keen tundra
this is one of infinitely many functions that work for this kind of thing
granite trellis
and if it were any other number instead of 0 i could do: n + (sigmoid function)
keen tundra
something like that yeah
you have bijections between any two open intervals (a,b) and (c,d) given by linear functions f(x)=qx+p for some numbers q and p
granite trellis
im thinking of some examples, just a sec
so the p and q are going to change according to x and (c,d) right?
it wont be the same p and q for the whole interval?
keen tundra
according to a,b,c,d
for example between (-1,1) and (0,1) you had the linear function f(x) = (x+1)/2 that gave you a bijection
so there q = p = 1/2
granite trellis
oh i understand it now
so ive thought of this, if i have to create a bijection between any sets N, R or Q, ill try to simplify it to get a small sample and then use a linear equation to get the set I want
is that a good approach to figuring out the functions?
keen tundra
generally your bijections will be very very very ugly
for example a bijection between Q^4 and N
so instead of finding a direct bijection you would use other methods to show the existence of one
granite trellis
for example a bijection between Q^4 and N
that would be a lot of variables
keen tundra
ugly yeah
granite trellis
is that where you use theorems like cantor?
keen tundra
cantor tells you a method to show that two sets do not have any bijections between them
but there is the CantorāSchrƶderāBernstein theorem which tells that if |A| <= |B| and |B| <= |A| then |A| = |B|
which is a very useful result for these kinds of questions
granite trellis
yes thats the one i was looking for
keen tundra
this one is very useful
granite trellis
but even then i would need to show an injection between Q ^4 ->N
keen tundra
that isn't too difficult
using some prime numbers
Also you can do this by first getting an injection Q^4 -> Q
Which might be easier
granite trellis
yes i was looking at something similar where Q -> N, by mapping it to Z x Z+
keen tundra
The point is that there are plenty of creative ways to get such injections
But bijections are harder to build explicitly
granite trellis
but there is the CantorāSchrƶderāBernstein theorem which tells that if |A| <= |B...
if i go with this i could create two different functions for each injection right?
granite trellis
i understand it alot better now, Thank you! will give few more questions a try
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vast minnow
Can someone help me with exercise 5-16?
vale dockBOT
Can someone help me with exercise 5-16?
vast minnow
I did it on the right, but Iām sure thereās a faster way to do it
I just did brute force casework but I donāt think thatās what the author intended
And I donāt wanna look at the solution either because I wanna think through it
hazy pivot
Are you familiar with ā¤_8
vast minnow
Uhh
I know thatās the integers modulo 8
But idrk what that means tbh
And also the book is meant for middle schoolers so
hazy pivot
Now look at the image of x ā¦ xĀ² in that
hazy pivot
And also the book is meant for middle schoolers so
Oh in that case this is prolly the best you can do lmao
vast minnow
Oh okay
I still kinda feel like it was overkill though
Well ig thereās not really another way to approach it
Thanks š
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spark zenith
How do I do Ļ (n) ā¤ Hn +ln (Hn)eHn
vale dockBOT
How do I do Ļ (n) ā¤ Hn +ln (Hn)eHn
river sapphire
Huh?
river sapphire
No it's not
spark zenith
It is
spark zenith
=
river sapphire
How do I do Ļ (n) ā¤ Hn +ln (Hn)eHn
This isn't an equation. But anyways. What is sigma(n) and H_n
spark zenith
This isn't an equation. But anyways. What is sigma(n) and H_n
Sum of divisors
Nth harmonic number
vivid fiber
How do I do Ļ (n) ā¤ Hn +ln (Hn)eHn
Isn't this an open problem bruh š„
A restatement of the riemmann hypothesis
@spark zenith , I don't think you will be getting the answer to your problem any time soon, but maybe it would be better if you talk about this open problem in #math-discussion
hazy pivot
.close We ain't gonna solve the Riemann Hypothesis on a discord server mate
spark zenith
<@1147986854969352374> , I don't think you will be getting the answer to your pr...
Itās on my math homework look
vale dockBOT
Itās on my math homework look
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gleaming kindle
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fluid hemlock
Itās on my math homework look
dude thats hard wtf
vale dockBOT
dude thats hard wtf
river shale
.solved
vale dockBOT
vivid fiber
Itās on my math homework look
2nd question isn't even math
quasi valve
2nd question isn't even math
for shame, asking biology questions on a math server š¤£
vivid fiber
Is that even biology
I would see that as an economy problem
Or maybe even politics
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bronze copper
for this double integral question how do i consider the <= part?
bronze copper
like what if it doesnt even intesect with the circle?
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@bronze copper Has your question been resolved?
thorn storm
guys how do you calculate the area of the green plane encompassed by the red cone? š
distant galleon
guys how do you calculate the area of the green plane encompassed by the red con...
!occupied
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guys how do you calculate the area of the green plane encompassed by the red con...
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thorn storm
shoot my bad
ivory stag
No clue how to start on this
vale dockBOT
No clue how to start on this
ivory stag
Just started doing calculus again
Would i use the quotient rule to start? or convert it into a different form
vale dockBOT
@ivory stag Has your question been resolved?
quasi valve
this form seems promising
ivory stag
Nvm i been in the lab with gpt
ivory stag
this form seems promising
Yea it was the answer key but it was split into 2 steps so i didnt know how to do the inbetween steps
west wharf
vale dockBOT
west wharf
how can I identify the arctan of this.
stark wedge
do you really mean specifically arctan(-1-i)?
like the extension of the arctan function to complex numbers?
or do you MAYBE mean arg(-1-i) instead
west wharf
Yeah, its polar form
stark wedge
ok then NO you're not talking about taking the literal arctan of the number -1-i
anyway this number's in the 3rd quadrant.
you can use the arctan(Im/Re) thing with the necessary adjustment for that.
west wharf
you can use the `arctan(Im/Re)` thing with the necessary adjustment for that.
Yeah like
I would like to know how to do the adjustment.
stark wedge
ok for this we will need to know sth important
in your class are arguments of complex numbers taken from 0 to 2pi, or from -pi to pi
west wharf
Beats me.
stark wedge
ok then we'll just have to assume [0,2pi) and then maybe find out that we have to do a last minute correction
so you know that arctan() only ever outputs angles between -pi/2 and +pi/2 yes?
devout mist
ok then we'll just have to assume [0,2pi) and then maybe find out that we have t...
Generally it's taken (-pi,pi) right
So why assume [0,2pi)
west wharf
so you know that arctan() only ever outputs angles between -pi/2 and +pi/2 yes?
Nope, I didn't know that.
stark wedge
Generally it's taken (-pi,pi) right
both conventions exist
stark wedge
Nope, I didn't know that.
ok wait do you even know what the arctan function does
west wharf
ok wait do you even know what the arctan function does
right?
devout mist
both conventions exist
Both do exist but we were taught the -pi to pi
devout mist
It returns angles given argument.
What angles does it give
Like any random angles or anything else
west wharf
If I do arctan 1, I get Ļ/3
stark wedge
It returns angles given argument.
too vague tbh
stark wedge
If I do arctan 1, I get Ļ/3
nope, it's pi/4. aka 45Ā°
arctan(x) answers the question "tan of what angle gives x"
anyway the adjustments i was talking about are adding or subtracting pi depending on quadrant
it probably isn't going to make a lot of sense if i try to explain just the rule
vale dockBOT
@west wharf Has your question been resolved?
vale dockBOT
hollow rune
The help channels are usually meant for math problems
neat igloo
generally when we translate english statements into logic, are we allowed to transform the logic to a different, albeit equivalent, formation?
neat igloo
these two statements are equivalent
but can i write the (f)'s formation as answer for (g)?
shadow moss
idk if this is the right term, but its like de morgan for existential and universal statements
keen tundra
The point is to try to translate as accurate as possible
shadow moss
generally when we translate english statements into logic, are we allowed to tra...
in general, no
usually you save the modifying until after translation
neat igloo
i see
shadow moss
yeah you can
pastel siren
Itās just when u do it directly u might loose readers
shadow moss
hence, save it for later
tight lake
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merry crystal
find two more vectors that will form a linearly independent set
tight lake
Then what
ruby sleet
Then you'll have a collection of 4 linearly independent vectors in R^4
... so a basis
ruby sleet
find two more vectors that will form a linearly independent set
To be clearer, you must find 2 more vectors which make, in addition to v1 and v2, a linearly independent collection
tight lake
Could the two other vectors be (1,0,0,0) or (0,1,0,0)
ruby sleet
See if they work
tight lake
They do work but I think my teacher wants a solution like this
Would I just replace one of the 2 vectors with (a,b,c,d)?
glad prism
Can I have help of the part B?
vale dockBOT
Can I have help of the part B?
short moth
glad prism
i done the part a but I do t have any ideas of the part b
vale dockBOT
@glad prism Has your question been resolved?
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heavy girder
Does anybody know the Pythagorean theorem?
vale dockBOT
Does anybody know the Pythagorean theorem?
lyric atlas
what about it
heavy girder
I kind of understand it but not all that good
heavy girder
Uh so a and b the faces two faces of the triangle must make c, so for example a=5 b=6, so you know that you times 5x5 and 6x6 and then you add the answers up by addition and then you square it and thatās your answer right?
lyric atlas
not necessarily
it only holds when ABC s a right angled triangle
and c (or AB) is the hypotenuse
heavy girder
Oh ok
lyric atlas
now do you know what the hypotenuse is
heavy girder
Yeah, thank you
coral wren
WHATS 1+1 ?
gusty jasper
11
lyric atlas
what
vale dockBOT
@heavy girder Has your question been resolved?
glad rivet
vale dockBOT
eternal osprey
Yes, but did you get the steps right?
glad rivet
i examined the points from left to right; -3, 1, 7/3
and saw where they satisfied the inequality
1 sec
might be a bit hard to read sorry for that
also i have some other questions, can i just post them in this thread?
eternal osprey
Yeah, sorry. It was hard to read.
toxic ridge
can i post the other question here?
yeah you can
vale dockBOT
@glad rivet Has your question been resolved?
glad rivet
it will take a while until i post the new questions so i think i should just close this thread for now
main kraken
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main kraken
I can seem to figure out how to continue to reduce the equation i get after
i have e^x-e^-x(2+e^x)= 0
how do i expand the bracket?
ebon glade
how do you expand a-b(c+d) ?
main kraken
does e work diffrently?
cuz by my understand i would get 2e^-x-e^2x
which is incorrect
ebon glade
you seem to be messing up the exponent laws
not the expanding
after expanding you get $e^x - e^{-x}\cdot 2 - e^{-x}\cdot e^x$, yes?
rocky lotusBOT
Denascite
torn gazelle
main kraken
go to
ebon glade
are you ok with $e^x - e^{-x}\cdot(2+e^x) = e^x - e^{-x}\cdot 2 - e^{-x}\cdot e^x$ ?
main kraken
"how to get help"
rocky lotusBOT
Denascite
ebon glade
main kraken
**Denascite**
no sorry
main kraken
$ a-bc-bd $
lyric sundial
Then do the same here š š¤
ebon glade
are you ok with $e^x - e^{-x}\cdot(2+e^x) = e^x - e^{-x}\cdot 2 - e^{-x}\cdot e^...
this is the exact same
lyric sundial
If you're doing linear algebra, you should know exponents and expansion really well, right?
main kraken
yeah
why does the last e^x positive ?
wait im stupid
omg
i get it
but lke how do i do the rest
ebon glade
**Denascite**
so you are fine with this?
what do you get when you simplify this with exponent laws?
rocky lotusBOT
levy
ebon glade
no
ebon glade
the tex itself is already wrong
rocky lotusBOT
levy
ebon glade
we have three terms
main kraken
the tex itself is already wrong
yeah im not used to this
rocky lotusBOT
Denascite
ebon glade
write .close
main kraken
.close
lapis marsh
hey
lapis marsh
a function f has the derivative function f'(x) = 4x + 6cos(x/2)
determine whether the function can ever have a minimum
or a maximum
I tried setting f'(x) = 0 but couldn't solve for x without a calculator
tawny moon
then it might be time to consider the wording of the question
lapis marsh
precisely what I did
the precondition for a point of maximum or a minimum is
the derivative = 0
the second derivative = 0
tawny moon
that you were unable to find an x that resolves the derivative to 0 suggests something
prime sail
can someone help?
tawny moon
can someone help?
!occupied
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can someone help?
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lapis marsh
what does it suggest
tawny moon
look at your question and its wording
specifically, this question didn't ask you to find the max or min of f
this question asked if f can ever have a max or min
that's the biggest bombshell hint
lapis marsh
yeah you're right but
I wanted to set it to 0 to establish whether a max or a min exists at all
lapis marsh
yes
tawny moon
then it suffices to consider a range of x from -3/2 to 3/2. if there are no answers in this range, you're never going to get one
lapis marsh
okey
tawny moon
(but also I hope you know why we can limit x to this range)
lapis marsh
why
lapis marsh
yes
lapis marsh
any x out of what range
lapis marsh
where did you get -3/2 to 3/2
tawny moon
you have -2x, and -2x cannot be out of the range -3 < -2x < 3
so just divide the entire range by -2
lapis marsh
oh right
that makes sense
okey but I can't test every individual x to see whether the equations match
that'll take too much time
within that range
lyric sundial
That's why you're not asked to find the exact value
You're asked to show whether it exists or not
lyric sundial
this question asked if f **can ever have a max or min**
@lapis marsh
tawny moon
I mean, you could also consider some extreme values
tawny moon
at x = 0, the whole RHS becomes 3, but the LHS is 0
at x = 3/2, the LHS becomes -3, but cos only switches signs at pi/2, which unfortunately is a greater number than 3/2
you can do the same analysis for the negative side, keeping in mind that cos is strictly increasing from [-pi/2, 0]
tawny moon
well you want to first determine if there are even any points where the first deriv is 0
lapis marsh
yes
tawny moon
you can do the same analysis for the negative side, keeping in mind that cos is ...
hence
lapis marsh
so we set the derivative to zero
tawny moon
I mean I kinda solved half the problem for you already
lapis marsh
in the key sheet
they skip past the first derivative
and take a look at the second derivative
tawny moon
idk why they would choose to do that, and idk why you're bringing up this without showing us what the answer key is doing
but technically you don't need the second deriv at all
lapis marsh
the way I traditionally solve these problems
is I set the derivative to zero
if I find an x-coordinate that satisfies that equation
I input that x-coordinate into the second derivative
lyric sundial
Again...
tawny moon
ok sure, that's the right path to follow. but here's the thing then - have you concluded that there are points where the derivative is 0?
tawny moon
at x = 0, the whole RHS becomes 3, but the LHS is 0
at x = 3/2, the LHS becomes ...
like I mean, I already did some analysis over here, and you can kinda copy this and do the negative part
lapis marsh
that's above my pay grade
lyric sundial
how to solve it alberto
By using Bolzano's theorem, for example
(which is a corollary of IVT)
tawny moon
you don't need to be an undergrad to see that though
in fact, nothing of what I used in my analysis requires any specific tools other than to just, use the definitions of the two sides
tawny moon
at x = 0, the whole RHS becomes 3, but the LHS is 0
at x = 3/2, the LHS becomes ...
like look at this, it's just sign analysis and observing the behavior of cos
no fancy theorems, no heavy machinery
lyric sundial
And also, most important, if you've been assigned this exercise, it means you've already been taught the suitable math instruments for being able to do it
lapis marsh
how about
I send a picture of the key
and we can discuss it
and I'll tell you why I'm skeptical
lyric sundial
Sure
lapis marsh
okey forgive me it's a bit blurry
they set the f''(x) = 0
determine that f''(x) varies between 1 and 7
hence no max
which to me is crazy considering we haven't established whether the first derivative can be zero or not
tawny moon
it's technically not crazy, it's just that they might be using a conclusion you're not allowed to use
hushed aspen
can anyone help me with kinematic 3d motion?
tawny moon
it is true that if f''(x) > 0 for all x in an interval (in this case the whole domain), then the graph is concave up in the whole interval
hushed aspen
2d*
lyric sundial
can anyone help me with kinematic 3d motion?
!occupied
vale dockBOT
can anyone help me with kinematic 3d motion?
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tawny moon
but also, the key only addressed maximums, while you're asked for max/min
so even though I kinda see the point, I don't like the key's solution
lapis marsh
oh okey
i'm fine with
ruling out a max because the second derivative is always positive
tawny moon
but ruling out a max is all you can do using this method, because if you do the sign analysis I did on the negative side, you will find something unexpected (or expected)
vale dockBOT
@lapis marsh Has your question been resolved?
narrow arrow
i know this is absurdly easy but i dont know whats wrong with me but i keep getting e^2/(1-e^2) which is not a option š
dreamy scroll
i know this is absurdly easy but i dont know whats wrong with me but i keep gett...
another physics question dang
narrow arrow
i just did max height befor ecollission is d+h
and after collission is h
max height directly proportional to velocity square
h+d directly proportional to v^2
h directly proportional to e^2v^2
then just took ratio
narrow arrow
i get the problem now
a
dreamy scroll
i just did max height befor ecollission is d+h
and after collission is h
max hei...
v^2 is proportional to h not h+d
dreamy scroll
and e^2v^2 is proportional to h-d
narrow arrow
huh
dreamy scroll
i considered v the velocity before collission
look at it after the first collision instead of looking at it from the start
it's max height is h
as the question states it is now h above the next step
so the previous height it had must be h-d
narrow arrow
oh
the question just badly worded it never said it was purely vertical motion
i thought it would fall onto the next stair
and then have that max height
dreamy scroll
the question just badly worded it never said it was purely vertical motion
ye you just have to assume it ig
dreamy scroll
i thought it would fall onto the next stair
me too
narrow arrow
.close
vale dockBOT
narrow arrow
wait @dreamy scroll
i did get the answer now but
the question states that the ball is
"bouncing down a set of stairs"
so its not even bad wording
dreamy scroll
yes
narrow arrow
its just wrong
dreamy scroll
its just wrong
not rly actually since the velocity in the x direcn should not be effected by the collision
since it's always a head on collision
though yes it is poorly worded
narrow arrow
not rly actually since the velocity in the x direcn should not be effected by th...
wdym
nono
the question explicitly said
"bouncing DOWN a set of stairs"
which implies its going DOWN the steps of the stair
dreamy scroll
wdym
during a collision the velocity affected is only the component perpendicular to the surface
narrow arrow
during a collision the velocity affected is only the component perpendicular to ...
yeah i know
narrow arrow
the v i stated in my first response
dreamy scroll
.reopen
narrow arrow
.reopen
vale dockBOT
ā Original question: #help-4 message
narrow arrow
was v in y direction
dreamy scroll
yes
narrow arrow
as x remain constant
dreamy scroll
yes
dreamy scroll
we like assumed its on one step
yes bcs otherwise we wont rly get a use of h
dreamy scroll
ok
dreamy scroll
lemme try to make a image
you cant rly use the starting value of velocity since we dont know the initial height it started on
we just know it reaches h after collision
so i just took the first case after a collision happened
narrow arrow
no i meant like what if this is initial
and 1 sec lemme maek final
this final
like in the previous collission
to the one we are studying
it already got to h for it
now its falling the entire d+h
to the next step
do yk wim
dreamy scroll
this final
not rly it says h over the next step which it is now assumingly going towards
dreamy scroll
this final
and these cases you have taken here are during diffrent collisions
dreamy scroll
no i meant like what if this is initial
ts case here is after collision from step 1
narrow arrow
yeah
dreamy scroll
this final
ts here is after collision from step 2
narrow arrow
and its at rest at height h
i am starting the study
from AFTER the collission from step 1
its at entire height h+d from step 3 at that point
if it did have some velocity in x lets say
it would fall on step 3
instead of step 2
so its velocity squared would be prportional to h+d
and after that it will go up height h from step 3
so proportional to h
u feel what im sasying now
we kinda assumed it stays on 2
to solve it
dreamy scroll
it would fall on step 3
i dont think we have to do that
dreamy scroll
i know this is absurdly easy but i dont know whats wrong with me but i keep gett...
one step each bounce
narrow arrow
the question said its going DOWN the stairs
yeah
if u assume one step each bounce
u get the answer im saying
dreamy scroll
you cant rly miss a stair with one step each bounce
narrow arrow
wait
dreamy scroll
u get the answer im saying
not rly
vale dockBOT
dreamy scroll
thanks for being patient
no problem
vale dockBOT
Available help channel!
hollow mica
Hey! I've been trying to prove that the odd numbers in (n+1)! divide 1x3x5x..x2n-1. Though I'm having some trouble
hollow mica
Up to now I have:
Let us consider $(n+1)!$, and write it as
[(n+1)! = 2^k \cdot (\text{odd part}),]
where $k$ is the number of factors of $2$ in $(n+1)!$, and the odd part is the product of all odd numbers appearing in $(n+1)!$ after factoring out powers of $2$ from each term.
We consider the odd part of $(n+1)!$. This can be organized as a series of sequences, where the $i$-th sequence corresponds to the odd numbers remaining after factoring out $2^i$ from each term. The lengths of these sequences are
[\left\lceil \frac{n+1}{2} \right\rceil, \quad
\left\lceil \frac{n+1}{4} \right\rceil, \quad
\left\lceil \frac{n+1}{8} \right\rceil, \dots]
and each sequence consists of odd numbers starting from $1$.
The idea is then to show that the product
[1 \cdot 3 \cdot 5 \cdots (2n-1)]
contains enough factors to cover all the odd sequences of $(n+1)!$. In particular, for each new sequence of length $\left\lceil \frac{n+1}{2^i} \right\rceil$, as I don't see any other way currently, I want to check one by one (and then try to generalise in some way), that each sequence (from /2 to /2^i) divides the product.
Anyways, the way I wanted to prove that is by showing that for each "sequence", what we have in 1x3x...x2n-1 is twice as long if that makes sense
Though I'm having some trouble getting there, so not really looking for the answer or something, just trying to know if this is a way to do it or if I'm on the wrong track?
this is better š
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@hollow mica Has your question been resolved?
vale dockBOT
@hollow mica Has your question been resolved?
hollow mica
.close
vale dockBOT
light saddle
Up to now I have:
Let us consider $(n+1)!$, and write it as
[(n+1)! = 2^k \cdot...
are you looking for a more formal way of phrasing it?
or can you just "do the calculation"?
(what I mean is, are you required to prove it formally)
hollow mica
not sure what formally means? š
Im not really required to do anything, Im just doing this one for fun haha
light saddle
oh ok
hollow mica
though I'd like a proof that holds and I'm having trouble currently haha
light saddle
you can write 1 * 3 * ... * 2n-1 as (2n)! / ((2^n) * n!)
hollow mica
ah yeah
light saddle
uhhh wait