Show that if $lim (a_n)=a, then lim(ca_n)=ca$
\
$ We have $ |a_n-a| < \varepsilon \implies \abs{c} \abs{ a_n -a} < |c| \varepsilon$
\
$\abs{ca_n - ca} < c \varepsilon$
#help-49
twilit field
midnight plankBOT
Show that if $lim (a_n)=a, then lim(ca_n)=ca$
\\
$ We have $ |a_n-a| < \varepsi...
grand pondBOT
ƒ( wai ina teacup)= I don't know
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runic hamlet
what about all the quantifiers
twilit field
what about all the quantifiers
Will introduce them now
grand pondBOT
ƒ( wai ina teacup)= I don't know
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twilit field
Show that if $\lim (a_n) = a$, then $\lim (c a_n) = c a$.
We have $\abs{a_n - a} < \varepsilon \implies \abs{c} \abs{a_n - a} < \abs{c} \varepsilon$.
Thus, $\abs{c a_n - c a} < \abs{c} \varepsilon$.
We wish to show the following: $\forall \varepsilon > 0, \exists N \in \mathbb{N}$ such that $n \geq N \implies \abs{ca_n -c a} < \varepsilon$.
We have $\abs{ca_n - ca} < \varepsilon \implies \abs{a_n - a} < \frac{ \varepsilon}{\abs{c}}$.
\
As ${a+n}$ coverges , we can chose an appropriate value of $N$. Thus the seqeunce converges
grand pondBOT
ƒ( wai ina teacup)= I don't know
twilit field
Show that if $\lim (a_n) = a$, then $\lim (c a_n) = c a$.
We have $\abs{a_n - a} < \varepsilon \implies \abs{c} \abs{a_n - a} < \abs{c} \varepsilon$.
Thus, $\abs{c a_n - c a} < \abs{c} \varepsilon$.
We wish to show the following: $\forall \varepsilon > 0, \exists N \in \mathbb{N}$ such that $n \geq N \implies \abs{ca_n -c a} < \varepsilon$.
We have $\abs{ca_n - ca} < \varepsilon \implies \abs{a_n - a} < \frac{ \varepsilon}{\abs{c}}$.
\
As ${a_n}$ coverges , we can chose an appropriate value of $N$. Thus the seqeunce converges
grand pondBOT
ƒ( wai ina teacup)= I don't know
quiet hinge
As a formal proof this just has false and incomplete statements
twilit field
Suppose that
twilit field
quiet hinge
"We have that |ca_n -ca| < epsilon => ..." just doesnt make sense
You didnt explain what n is here
twilit field
N or n?
quiet hinge
n
twilit field
n is the index
quiet hinge
I can see that
That explains where n is
Not what n is
n is the index so it makes sense to write a_n
That doesnt answer what n is
quiet hinge
n is an index
n = eps/|c|
quiet hinge
>n is an index
>n = eps/|c|
Do you see a problem with the two statements written here
twilit field
yes
runic hamlet
you need to start writing chains of (in)equalities
grand pondBOT
Asteroid destroyer 45²
runic hamlet
you are messing up with all those implications and in which order you are doing them
quiet hinge
**Asteroid destroyer 45²**
Note how the existence of N is justified, along with what n acutally is is written. In particular n is any number > N
twilit field
you are messing up with all those implications and in which order you are doing ...
Yeah, I realise now
runic hamlet
**Asteroid destroyer 45²**
(I hope you arent suggesting that this is a full proof)
runic hamlet
the whole proof should be reordered
twilit field
So I start off with let eps >0
quiet hinge
- 0 case needs to be added
runic hamlet
you can include it if you just dont divide by |c|
and instead divide by something else
twilit field
We wish to show that if $\lim(a_n)=a$ then $\lim(ca_n)= ca$
\
We thus wish to find a value of $N$, such that if $n \geq N$, then $\abs{ca_n-ca} < \varepsilon$
\
Let $\varepsilon >0$ and $c \neq 0$
\
sinc $a_n$ converges, we have that there exists a $N$ , such that $\forall n >N$, $\abs{a_n-a} < \frac{\varepsilon}{\abs{c}} \implies $\abs{ca_n -ca}< \varepsilon$ as desired
grand pondBOT
ƒ( wai ina teacup)= I don't know
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midnight plankBOT
quiet hinge
I hope you know it isnt complete but the remaining bit is easy to completr
twilit field
.reopen
midnight plankBOT
✅
twilit field
Yeah, c =0. is left
quiet hinge
you can include it if you just dont divide by |c|
You can also try this
twilit field
But showing that seqeunce converges isn't too hard
runic hamlet
**ƒ( wai ina teacup)= I don't know**
Compile Error! Click the <:errors:504266556...
the fact that you used epsilon before defining it should immediately tell you that this is the wrong order
the last implication should be written out more
twilit field
the fact that you used epsilon before defining it should immediately tell you th...
What do you mean by this
quiet hinge
the fact that you used epsilon before defining it should immediately tell you th...
Oh yes my bad
quiet hinge
What do you mean by this
You used eps without explaining what it is in the 2nd line
twilit field
epsilon is a value such that ${a_n}$ converges
grand pondBOT
ƒ( wai ina teacup)= I don't know
quiet hinge
Thats wrong and also doesnt make sense
twilit field
yeah, i realise that now
quiet hinge
Convergence (of a sequence) isnt a property of a single number
twilit field
I don't know how to frame what O want to sat
twilit field
You used eps without explaining what it is in the 2nd line
so a simple let $\varepsilon >0$ would suffice?
grand pondBOT
ƒ( wai ina teacup)= I don't know
quiet hinge
Let eps > 0 where?
quiet hinge
**ƒ( wai ina teacup)= I don't know**
Compile Error! Click the <:errors:504266556...
You did say let eps > 0 in the 3rd line here
runic hamlet
you need to show that something holds for all eps>0
so you take an arbitrary eps>0 and show that for this fixed eps the claim holds
and because eps was arbitrary you have therefore shown it for all eps
twilit field
Let eps > 0 where?
before the second line
quiet hinge
Idk you tell ne
twilit field
you need to show that something holds for all eps>0
Okay
quiet hinge
Idk you tell ne
This is a logical issue and has nothing to do with analysis so you should be able to tell where what statements have to go for it to be logically coherent
twilit field
Let $\varepsilon >0$. As ${a_n}$ converges , it follows that we can chose a $N$, such if if $n >N$, then $\abs{a_n-a}< \varepsilon$ and consequently, $\abs{a_n-a} < \frac{ \varepsilon}{\abs{c}}$. From this it follows that for the same $N$, $\abs{ca_n - ca} < \varepsilon$ , thus proving it converges
twilit field
This is a _logical_ issue and has nothing to do with analysis so you should be a...
That should be the first line
quiet hinge
We just had a discussion about saying what a variable is when you introduce it
twilit field
I introduced eps in the first line
quiet hinge
Ok
twilit field
isn't that logically coherent
quiet hinge
It is
quiet hinge
Bringing in a variable eps_0 without any explanation isnt logically coherent
And moreover you dont need to
grand pondBOT
ƒ( wai ina teacup)= I don't know
dreamy lichen
how does |an - a| < eps imply |an - a| < eps / |c|?
quiet hinge
|an-a|<eps does not imply |an-a|<eps/c
twilit field
how does |an - a| < eps imply |an - a| < eps / |c|?
as the stament is true for all epsilon
we can chose an number >0, eps/0
dreamy lichen
as the stament is true for all epsilon
whats the point of stating |an - a| < eps if you dont use it then?
quiet hinge
The way you written it makes it seem like the same N that works for eps works for eps/|c|
twilit field
I'm trying to figure out how to use it
dreamy lichen
as the stament is true for all epsilon
you are right with this tho
you could indeed chose N such that |an - a| < eps / |c| whatever your eps is
twilit field
so my proof holds?
twilit field
Let $\varepsilon >0$. As ${a_n}$ converges , it follows that we can chose a $N$, such if if $n >N$,$\abs{a_n-a} < \frac{ \varepsilon}{\abs{c}}$which implies , $\abs{ca_n - ca} < \varepsilon$ , thus proving it converges
dreamy lichen
way better
grand pondBOT
ƒ( wai ina teacup)= I don't know
dreamy lichen
Let $\varepsilon >0$. As ${a_n}$ converges, it follows that we can chose a $N$, such if if $n >N$, $\abs{a_n-a} < \frac{ \varepsilon}{\abs{c}}$ which implies $\abs{ca_n - ca} < \varepsilon$, thus proving ${ca_n}$ converges.
twilit field
Thanks
dreamy lichen
just fixed the typography
but it looks pretty good
I assume that the goal is stated in theorem, so you dont necessarily need to repeat it in the proof
maybe you could at least replace "it" with (ca_n)
grand pondBOT
MæthIsAlwaysRight
dreamy lichen
are you used to denoting sequences by a_n or (a_n) btw?
a_n is commonly used for a single term
(a_n) is used to denote whole sequence
but that might depend on your book
twilit field
I hadn't given it much thought until now, but I think I'll stick to ${a_n}$ from now
grand pondBOT
ƒ( wai ina teacup)= I don't know
dreamy lichen
**ƒ( wai ina teacup)= I don't know**
this could be confused with a set
dreamy lichen
and one which wasnt even offered
unfortunately its pretty common to use that one
at least in HS
(a_n) is a better notation to use imo
dreamy lichen
so when you wanna say that a sequence converges, you'd write "(a_n) converges" instead of "a_n converges"
runic hamlet
unfortunately its pretty common to use that one
still the worst one and one that you didnt offer in your selection of options
dreamy lichen
agreed
runic hamlet
I still dont like how the end of the proof just claims |ca_n-ca| < eps
dreamy lichen
still the worst one and one that you didnt offer in your selection of options
btw what do you think of the proof, does it look good to you?
runic hamlet
what are you struggling with
twilit field
what do you mean by write down a chain of inequalities
dreamy lichen
|ca_n - ca| = [intermediate steps] = epsilon
(where some of the ='s will be replaced with <)
the goal of this is to explain why does |a_n - a| < eps / |c| imply that conclusion
dreamy lichen
the goal of this is to explain why does |a_n - a| < eps / |c| imply that conclus...
so the intermediate steps should be chosen such that they are either simple identitites, or easily follow from |a_n - a| < eps / |c|
twilit field
$\abs{a_n-a}< \frac{\varepsilon}}{abs{c}} = \abs {c} \abs{a_n-a} < \varepsilon = \abs{ca_n-ca} < \varepsilon$
runic hamlet
no implications
grand pondBOT
ƒ( wai ina teacup)= I don't know
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runic hamlet
by that I didnt mean to just repeat teh same thing but without the arrows
runic hamlet
|ca_n - ca| = [intermediate steps] = epsilon
(where some of the ='s will be repl...
here was a template for you
it already gave you the first and last thing
twilit field
Isn't that the opposite order I want to follow
like I want to show that (a_n) converges \implies (ca_n) converges
dreamy lichen
we are not changing the whole proof
the only part that should change is this one
and the change is just expanding on it and explaining that implication using a chain of inequalitites
runic hamlet
what book of RA are you reading
twilit field
Abbott
runic hamlet
chains of inequalities should not be news to you
twilit field
I know, yeah
twilit field
the only part that should change is this one
yeah, so we want to show $\abs{a_n-a} < \frac{\varepsilon}{\abs{c}} \implies \abs{ca_n-ca} < \varepsilon$
grand pondBOT
ƒ( wai ina teacup)= I don't know
dreamy lichen
abbott actually uses exactly the same chain
twilit field
I'm royally confused
dreamy lichen
**ƒ( wai ina teacup)= I don't know**
you are trying to explain this implication
and chain of inequalities is one way to do that
quiet hinge
multiplying by |c| and writing an implication is a perfectly fine explanation
runic hamlet
multiplying by |c| and writing an implication is a perfectly fine explanation
thats really not the point
you cannot succeed in RA if you cant even write chains of inequalities
quiet hinge
sure
runic hamlet
frankly wai you jumped into doing exercises too early
you need to read a few more proofs before that
to absorb what they are doing
and then trying to copy that
dreamy lichen
unfortunately, this is the first proof in the book that uses chain of inequalitites
runic hamlet
have you even taking your break yet
dreamy lichen
it's not an exercise though
twilit field
have you even taking your break yet
I'm travelling from monday
runic hamlet
there are chains of inequalities before that
I saw that while scrolling through
for example th 1.4.5
dreamy lichen
oh, yeah
dreamy lichen
btw that thing you are trying to prove is not an exercise, right?
It's a theorem with a proof given
runic hamlet
too early
twilit field
Bad idea probably
quiet hinge
learn the writing style
runic hamlet
you literally just jumped into the topic
you dont know yet how real analysis proofs go
dreamy lichen
You could try to think about the proof before reading it, but only spend like 5-10 mins doing that. After that, compare your thinking to the proof given by author
twilit field
Got it, will do that in the future
thanks everyone
I'll close this for now then and go though the proof I suppose
dreamy lichen
make sure to look at the chain
midnight plankBOT
midnight plankBOT
Available help channel!
crude galleon
Find min(x^x).
midnight plankBOT
Find min(x^x).
thorn sorrel
use derivatives
crude galleon
I don't k calculus
lethal path
I don't k calculus
bruh
thorn sorrel
idk how else to do it
keen herald
If you have a graphing calculator it can find it for you, but otherwise the methods to solve this are either trial and error or calculus
crude galleon
Note: Don't type if you don't know how to find min(x^x) without calculus or are just inexperienced.
keen herald
Keeping everyone accountable today, love it
crude galleon
If you have a graphing calculator it can find it for you, but otherwise the meth...
Well I could use wolfram alpha for that purpose too
keen herald
You certainly could
thorn sorrel
calculus is the messiah
lethal path
Note: Don't type if you don't know how to find min(x^x) without calculus or are ...
why do you want to do it without calculus
keen herald
Note that it's defined on (0,infty)
crude galleon
Because I don't know it
lethal path
then learn it first
crude galleon
And it's a normal function
crude galleon
Note that it's defined on (0,infty)
yeah
keen herald
Calculus is the study of "normal" functions
thorn sorrel
wym normal function
lethal path
otherwise you're limited to finding the turning points of quadratics
and a few other functions like x + 1/x
or ones with classical inequalities
lethal path
or ones with classical inequalities
these can't save you here
hell even classical inequalities use the tangent line method
which is just calculus
crude galleon
How did people before the existence of calculus calculate it?
lethal path
approximation
keen herald
Trial and error and approximations
crude galleon
Yeah analysis simply
keen herald
There is another way to write the function that may help but I don't know how it was derived, may as well have been via analysis
crude galleon
Analysis is a very broad term
keen herald
My intended meaning being real analysis
midnight plankBOT
keen herald
Try playing with e^(xln(x))
midnight plankBOT
Available help channel!
fossil saffron
i have to find the limit sup and inf
midnight plankBOT
i have to find the limit sup and inf
lethal path
well, ${{n + 1} \choose 2} = \frac{(n + 1)(n + 1 - 1)}{2}$
grand pondBOT
south
lethal path
also you know that $-1 \le \sin(\text{anything}) \le 1$
grand pondBOT
south
midnight plankBOT
@fossil saffron Has your question been resolved?
grave kernel
Let
f (x) be a periodic function with period 2 , and g(x) = integral from 0 to x (f(t)dt)
midnight plankBOT
Let
f (x) be a periodic function with period 2 , and g(x) = integral from 0 to x...
grave kernel
need to comment about both their parity, even or odd
g'(x) is just f(x)
cant figure out what f(x) will be
modern ice
need to comment about both their parity, even or odd
What do you get if you evaluate g(-x)?
grave kernel
g(-x) = g(x)?
then g(x) is even
and g'(x) is odd which is f(x)
right?
wait nvm i got it
.close
midnight plankBOT
midnight plankBOT
Available help channel!
chilly cobalt
Let $P(x) = x^3 + 9x^2 +27x + 24$ be a polynomial, and $a_0, a_1,...$ are real numbers with $a_0 = 0$ and $a_i = P(a_{i-1})$ for all integers $i \geq 1$ Find the minimum positive integer n such that $1000 | a_n$
midnight plankBOT
Let $P(x) = x^3 + 9x^2 +27x + 24$ be a polynomial, and $a_0, a_1,...$ are real n...
grand pondBOT
Copter
chilly cobalt
i translated it correctly this time :3
P(x) = (x+3)^3 -3
so a_i+1 = (a_i + 3)^3 -3 for i>=0
is there like a way to generalize a_n from here? ive never seen stuff like this before
midnight plankBOT
@chilly cobalt Has your question been resolved?
chilly cobalt
<@&286206848099549185>
modern ice
I'm unfamiliar with that kind of notation. Does the last line mean a_n should be divisible by 1000?
modern ice
Is this problem from a course on elementary number theory? I don't know much about number theory so I'm pretty much just asking stuff cuz I'm curious
chilly cobalt
no, its from a middle school competition
modern ice
Wow..
chilly cobalt
but the competition isnt serious though
i just think the problems seem interesting
cedar coral
seems like you need to use modular arithmetic
modern ice
You do have that P(P(x)) = x^6
If you could find a general expression for P^n then that might be helpful
I messed up tho, gimme a sec
P(P(x)) = (x+3)^6 - 3
anyway it's not straightforward no. I suddenly got a little busy with something else
modern ice
If you could find a general expression for P^n then that might be helpful
Ok this is actually very manageable. Have you been able to find this @chilly cobalt ?
midnight plankBOT
@chilly cobalt Has your question been resolved?
modern ice
Well I'm just gonna show what I have found so far since the channel is closing soon
The general form for a_n is a_n = 3^(3^n) - 3
And uh finding patterns when dealing with that large numbers wasn't easy
And uh according to this problem it's possible to write 3^n as 4k + 1 for some natural number k
For the correct value of n
radiant roost
that doesn't seem right
radiant roost
3 = 4*1 - 1
27 = 4*7 - 1
modern ice
Oh
No you're using -
I'm saying +. And these will be candidates for the solution. I.e. a necessary condition the solution must have
I might be overcomplicating it
radiant jacinth
midnight plankBOT
radiant jacinth
u see for integration. what would you do in a situation where the curve dips below the x axis. like this. when integrating this, would the area under the curve be included in the calculation when taking the limit of 0 - 5? and since its negative wouldn't it make it smaller?
uneven sandal
you want to find the area?
radiant jacinth
yh but im confiused on what happens when interfrating the curve
worthy wing
And what do you think you have to put?
radiant jacinth
no, im asking if the part below the x axis is included in the calculation of intergrating the curve with the limit of 0 - 5
is it?
worthy wing
!original
midnight plankBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
worthy wing
I dont see the text that asks you what to do
radiant jacinth
!original
im not asking how to do the question but a question about intergrating a curve
last slate
Here's how you can do it:
-
find the area between y = 2x between the origin (0, 0) and point A completely ignoring curve since this area is only bounded by the straight line and the x-axis
-
find the area bounded between the curve and the y = 2x line between A and B. This is just integral of (top line - bottom curve)
add the two up and you would have the shaded area
TLDR: Till point A, only consider the area bounded by the line and the x-axis
radiant jacinth
Here's how you can do it:
1. find the area between y = 2x between the origin (...
ah i see
but lets just say, you was to intergrate the curve, what will happen?
last slate
just the curve?
radiant jacinth
yes
last slate
and between what limits?
uneven sandal
what do you mean with integrate the curve
radiant jacinth
cuz it dips below x axis
radiant jacinth
and between what limits?
0 - 5
last slate
,w integeral between 0 and 5 of x(x - 3)
grand pondBOT
last slate
You would get this area 👆
uneven sandal
I think he means
radiant jacinth
so is it including the bit under the x axis as a negative
uneven sandal
,w integeral between 0 and 5 of (2x-x(x - 3))
grand pondBOT
last slate
ahhhhh okkay mb thanks @uneven sandal
uneven sandal
which could be considered as the enclosed area, but its not
midnight plankBOT
last slate
,w integral between 0 and 3 of x(x - 3)
grand pondBOT
last slate
yeah it would come out to be negative @radiant jacinth
midnight plankBOT
Available help channel!
safe oriole
i want to prove that the sequence ${u_n}$ defined by $u_n=-\ln(n)+\sum_{k=1}^n \frac 1k$ is convergent , i proved that ${u_n}$ is decreasing but i still need a lower bound
midnight plankBOT
i want to prove that the sequence $\{u_n\}$ defined by $u_n=-\ln(n)+\sum_{k=1}^n...
grand pondBOT
pirateking0723
safe oriole
so how can i find a lower bound to {u_n}
fallow scarab
Try plotting the difference
safe oriole
isnt -ln(n) not bounded below?
fallow scarab
i.e. plot ln(x) and 1/x
And the differences between the two from 1 to n are your sequence
safe oriole
last slate
Also note that
$$H_n = \sum_{k=1}^{n} \frac{1}{k} \approx \ln(n) + \gamma, \quad \text{where} \quad \gamma \approx 0.57721$$
grand pondBOT
Edmund Cloudsley
safe oriole
**Edmund Cloudsley**
thats not something that i can use
fallow scarab
i.e. plot ln(x) and 1/x
Sorry not ln(x)
Just 1/x and its lower Riemann sum
Area of 1/x will give ln(x)
safe oriole
Area of 1/x will give ln(x)
but it didnt give ln(x)
ohh wait
i plotted the wrong thing
now i see
but this is obvious graphically
how can i prove this
that $-\ln(n)>-\sum_{k=1}^n \frac 1k$
grand pondBOT
pirateking0723
midnight plankBOT
@safe oriole Has your question been resolved?
safe oriole
<@&286206848099549185>
gray prism
how can i prove this
there's many ways, you could use the definition of ln for example
safe oriole
which is ?
safe oriole
there's many ways, you could use the definition of ln for example
which definition
ln as an integral ?
gray prism
yes
safe oriole
i cant do that
gray prism
well what can you do?
safe oriole
i am taking analysis and still in limits and continuity
i can use monotone convergence theorem for example
in fact i proved that u_n is decreasing
i need a lower bound
gray prism
yes here 0 as the lower bound should be easiest
another way iirc is some telescoping series
safe oriole
how can i do this
safe oriole
yes here 0 as the lower bound should be easiest
how can i reach this lower bound
gray prism
you can see this by plotting your sequence no?
safe oriole
yes but i cant just say that it is clear by plotting the curves
i need to prove that this is a lower bound
gray prism
yes yes
well can you use anything related to integrals/differentiation?
or any established results?
safe oriole
well can you use anything related to integrals/differentiation?
no
safe oriole
or any established results?
established results like what exactly
fallow scarab
but it didnt give ln(x)
Area from 1 to n, not 0 to n
write log(n) = integral of 1/x dx
gray prism
then how do you define ln?
safe oriole
as the inverse function of e^x
gray prism
established results like what exactly
maybe something like this?
gray prism
another way iirc is some telescoping series
then it should be pretty straightforward with this
gray prism
you couldve mentioned it earlier..
safe oriole
you couldve mentioned it earlier..
mb
gray prism
i feel like this is weird and not very useful.. do you have any other results?
fallow scarab
this is the only hint ive got
Show the entire original question
safe oriole
sage helm
euler-mascheroni lol
afaik you write ln(n) as some sort of telescoping sum and then bound with integrals
safe oriole
maybe something like this?
i mean doesnt this do the job
safe oriole
ln(n+1)-ln(n)=ln(1+1/n)
safe oriole
then use this
gray prism
yeah thats it
gray prism
haha np
safe oriole
in fact we should use -ln(n)+ln(n+1)>=-1/n
gray prism
why flip the equation around?
safe oriole
$\ln(n)-\ln(n-1)\leq\frac 1{n-1},\ln(n-1)-ln(n-2)\leq\frac 1{n-2}\cdot\cdot\cdot$ so $\ln(n)-\ln(n-1)+\ln(n-1)-\ln(n-2)+\cdot\cdot\cdot+\ln(2)-\ln(1)=\ln(n)\leq\sum_{k=1}^{n-1}\frac 1k\implies -\ln(n)\geq\sum_{k=1}^{n-1}\frac 1k\implies\sum_{k=1}^n\frac 1k-ln(n)\geq\frac 1n+\sum_{k=1}^{n-1}\frac 1k-\sum_{k=1}^{n-1}\frac 1k=\frac 1n>0$
grand pondBOT
pirateking0723
safe oriole
this works right
gray prism
the first thing on the 3rd line is missing a - sign or smth, otherwise yes
midnight plankBOT
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Available help channel!
edgy knot
midnight plankBOT
edgy knot
i cant see the proof that y = -x (according to answer sheet it is)
lethal path
do you get how when you set x' = 0
that 2x(x - y) = 1 for this to happen?
then the next step is 2y(x - y) = -1 for y'
so adding both equations, (2x + 2y)(x - y) = 0 or 2(x + y)(x - y) = 0
ah but if x - y = 0, when you sub into x' and y' you don't get 0
or you don't get 2x (x - y) = 1, or similar
edgy knot
im not quite following, can send the answer sheet my notes are kinda hard to follow here
lethal path
well, e^(anything) can never be 0 right
lethal path
yeah then what don't you get about this working
edgy knot
i cant find where it is calculated that y = -x
lethal path
it comes from (x + y)(x - y) = 0
but then x - y = 0 is impossible if you sub in x = y into the original expressions for x' and y'
edgy knot
so we know that x = y is impossible because it gives us 2x(0) which is not = 1. we want only x terms so we replace y with x. (y = -x) and we get the expression 2x(x-(-x)). i figured that now. how is x-y = 0 turned into y = - x is the last detail i cant figure out
lethal path
so we know that x = y is impossible because it gives us 2x(0) which is not = 1....
not x - y = 0, but x + y = 0
if x + y = 0 just subtract x from both sides
you get y = -x
midnight plankBOT
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tidal turret
How many distinct prime factors does the sum of the positive divisors of $400$ have?
midnight plankBOT
How many distinct prime factors does the sum of the positive divisors of $400$ h...
grand pondBOT
938c2cc0dcc05f2b68c4287040cfcf71
feral sedge
tidal turret
,calc 50/2
grand pondBOT
Result:
25
Result:
400
400 has (4+1)(2+1) factors
400 has 5×3=15 factors
factors of 400: 2^0,2^1,2^2,2^3,2^4,5^1,5^2, 2×5, 4×5, 8x5, 16x5, 2×25, 4x25,8x25,16x25
,calc 16*25
grand pondBOT
Result:
400
1,2,4,8,16,5,25,10,20,40,50,100,200,400
,calc 1+2+4+16+5+25+10+20+40+50+100+200+400
grand pondBOT
Result:
873
,calc 291/3
grand pondBOT
Result:
97
Result:
873
,calc 31*31
grand pondBOT
Result:
961
,calc 2^0 + 2^1 + 2^2+2^3+2^4+5+5^2+25+45+85+165+225+425+825+1625
grand pondBOT
Result:
961
,calc 961/13
grand pondBOT
Result:
73.923076923077
,w prime factorization of 961
grand pondBOT
tidal turret
.close
manic bison
midnight plankBOT
manic bison
How do I begin?
cerulean oyster
How do I begin?
What are we supposed to find here?
manic bison
Sum of infinite terms
It's obviously gonna be a converging series
That won't help with the question though
north vector
How do I begin?
Can you write this in sigma notation and recognize a series of some function you may know?
You'd want to find a pattern
A pattern based on the terms
north vector
Tn = Sigma ( n +2019)^2/(n-1)!
Expand e^x and differentiate
manic bison
Taylor expansions of e^x comes close
north vector
Try that.
manic bison
Expand e^x and differentiate
Are you sure?
north vector
Taylor expansions of e^x comes close
I think you've got it
north vector
Are you sure?
Yes. ,and you'd multiply by x both sides and differentiate after that
I hope you can see it now
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tidal turret
The sum of the positive divisors of a positive integer of the form $2^i3^j$ is equal to $600$. What is $i + j$?
midnight plankBOT
The sum of the positive divisors of a positive integer of the form $2^i3^j$ is e...
grand pondBOT
938c2cc0dcc05f2b68c4287040cfcf71
blissful totem
The sum of the positive divisors of a positive integer of the form $2^i3^j$ is e...
wat have u tried
tidal turret
there is a formula for the sum of positive divisors of a function, is the sum of its prime factors, it uses a geometric series I think no?
blissful totem
there is a formula for the sum of positive divisors of a function, is the sum of...
i think there is a formula yes
tidal turret
blissful totem
but the logic essnetially is that every integer has an unique prime factorisation
so u can find some combination of 2s and 3s
up until u exhaust every combo
tidal turret
how do I do that
blissful totem
what context of math is this
tidal turret
info I have gathered:
manic bison
so u can find some combination of 2s and 3s
Using the sum of divisors will make this process easier
tidal turret
what context of math is this
pre university math
tidal turret
Using the sum of divisors will make this process easier
how
up until u exhaust every combo
tidal turret
yes, divisor function is multiplicative
you are multiplying the sum of divisors of 4 with the sum of divisors of 3
manic bison
That's the principle
manic bison
If we have a number 12
12 = 3x2^2
Sum of divisors=( 2^0 + 2^1 + 2^2)(3^0 + 3^1)
Use this till i and j
2^0 + 2^1....+2^i
It's a GP
tidal turret
sigma(2^i) = 2^0 + ... + 2^i
sigma(3^j) = 3^0 + ... + 3^j
sigma(2^i x 3^j) = sigma(2^i) x sigma(3^j)
manic bison
Yep
manic bison
Do you know the sum of a geometric progression?
tidal turret
if |r| < 1 for a geometric series the sum converges to
1/(1-r)
ar^n is the geometric series
manic bison
What about for terms up to n
manic bison
if |r| < 1 for a geometric series the sum converges to
1/(1-r)
This will be valid for a series with infinite terms
tidal turret
This will be valid for a series with infinite terms
manic bison
It's a(r^n-1)/r-1
tidal turret
ok
manic bison
Even if you can't get the answer from there using algebra, it should be easier to intuitively guess and test the values of i and j
tidal turret
I need more hints
manic bison
I'll brb in few minutes till then try that
tidal turret
a1 = 2^0 and 3^0
for sigma(2^i) , r = 2, r > 1
for sigma(2^i), a = 2^0=1
,, \sigma(2^i) = \left(\frac{2^{i+1}-1}{2-1}\right) \ \sigma(3^j) = \left( \frac{3^{j+1} -1}{3-1} \right)
grand pondBOT
938c2cc0dcc05f2b68c4287040cfcf71
tidal turret
,, \sigma(2^i) = \left(\frac{2^{i+1}-1}{2-1}\right) \ \sigma(3^j) = \left( \frac{3^{j+1} -1}{3-1} \right)
grand pondBOT
938c2cc0dcc05f2b68c4287040cfcf71
tidal turret
,, \left(\frac{2^{i+1}-1}{2-1}\right)\left( \frac{3^{j+1} -1}{3-1} \right) = 600
grand pondBOT
938c2cc0dcc05f2b68c4287040cfcf71
last slate
**938c2cc0dcc05f2b68c4287040cfcf71**
you have missed out multiplying first term (a) in each summation
no it;sright
my bad
tidal turret
a = 2^0
a = 3^0
tidal turret
you have missed out multiplying first term (a) in each summation
yeah
,, \left(\frac{2^{i+1}-1}{2-1}\right)\left( \frac{3^{j+1} -1}{3-1} \right) = 600 \\ (2^{i+1}-1)(3^{j+1}-1) = 1200
grand pondBOT
938c2cc0dcc05f2b68c4287040cfcf71
tidal turret
,calc 1200/2
grand pondBOT
Result:
600
,calc 600/2
grand pondBOT
Result:
300
,calc 300/2
grand pondBOT
Result:
150
,calc 150/2
grand pondBOT
Result:
75
,calc 75/5
grand pondBOT
Result:
15
,calc 15/3
grand pondBOT
Result:
5
5^2
tidal turret
yeah
grand pondBOT
Result:
1200
glossy valve
I have a question on triangles,
tidal turret
I have a question on triangles,
other channel
last slate
!occupied
midnight plankBOT
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).
tidal turret
I have a question on triangles,
open other channel
tidal turret
I have a question on triangles,
open other channel
glossy valve
thx
tidal turret
factors of 1200
1)2^0
2) 2^1
3) 2^2
4)2^3
5)2^4
6)5
7)3
8)2 x 5
9) 2 x3
10) 2^2 x 5
11) 2^3 x 5
12) 2^4 x 5
13)2^2 x 3
14) 2^3 x 3
15) 2^4 x 3
16) 2 x 3 x 5
17) 2x3x5^2
18) 2x5^2
19) 2^2 x 5^2
20) 2^3 x 5^2
21) 2^4 x 5^2
22) 2^2×3×5
23) 2^3x3x5
24)2^4x3x5
25) 2x3x5^2
26)2^2x3x5^2
27)2^3x3x5^2
28)2^4x3x5^2
I am missing some of them
last slate
you didn't have to list them
tidal turret
manic bison
Back
last slate
there's no need of finding each factor
manic bison
factors of 1200
1)2^0
2) 2^1
3) 2^2
4)2^3
5)2^4
6)5
7)3
8)2 x 5
9) 2 x3
10) 2^2...
What 😭
last slate
just factorise you see 1200= 2^4 *3^1 *5^2
manic bison
You should first see that the left one will be an odd factor and the right one even
last slate
now you need to use brute force where you start by putting i=1 and so on till 4 maximum...because 2 has power of 4...then you find corresponding value for j...we need to make sure j is a positive integer
manic bison
now you need to use brute force where you start by putting i=1 and so on till 4...
Yeah this works too
manic bison
now you need to use brute force where you start by putting i=1 and so on till 4...
Wait actually
tidal turret
1200= 2^4 *3^1 *5^2
last slate
ig it should be max 4 because 2 power is 4 but i believe i need to rethink
manic bison
Try 3
tidal turret
cause 3+1 = 4
2^(i+1)
manic bison
That's pretty probable
tidal turret
15(3^(j+1) - 1 ) = 2^4 x 3 x 5^2
last slate
That's pretty probable
yeah!! that works
manic bison
i = 3 and j =3 satisfies it
last slate
i = 3 and j =3 satisfies it
j=4*
manic bison
It's 3^(j+1) too
tidal turret
3^(j+1) -1 = 2^4 x 5
last slate
oh ya
tidal turret
,calc 3^4-1 -2^4*5
manic bison
In these kinds of questions, you really need to have a strong intuitive grasp
grand pondBOT
Result:
0
j+i=6
manic bison
factors of 1200
1)2^0
2) 2^1
3) 2^2
4)2^3
5)2^4
6)5
7)3
8)2 x 5
9) 2 x3
10) 2^2...
There is no way to power through and test that many cases, it's only your intuition which will help
last slate
In these kinds of questions, you really need to have a strong intuitive grasp
true instead of trying out each possibility it becomes much easier with some calculative trials and errors
manic bison
true instead of trying out each possibility it becomes much easier with some cal...
Yeah
manic bison
Welcome :))
tidal turret
.close
midnight plankBOT
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Available help channel!
manic bison
midnight plankBOT
manic bison
This is an entirely new type of question to me
last slate
is that 25 or 75
manic bison
75
manic bison
It's 2
last slate
calculate a3
noble holly
square and make an inequality
might work
We get (An)^2 - (An-1)^2 >= 2
this is telescopic
manic bison
Oh damn
manic bison
We get (An)^2 - (An-1)^2 >= 2
I can solve from here , but how exactly did you get this inequality?
noble holly
square both sides and transpose An-1^2, since Ak is alwasys +ve we get this
manic bison
By squaring
An^2 = an-1^2 + 1/ an-1 ^2
noble holly
+2 also
midnight plankBOT
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Available help channel!
sharp dragon
I want to make a calculation sheet about my income from job.
Let’s say I get $17/h, and every minutes counts.
Let’s say that day I worked exactly 7 hours and 43 minutes. How on the sheet would I be able to do in order to calculate how much I earned that day?
midnight plankBOT
I want to make a calculation sheet about my income from job.
Let’s say I get $1...
visual tiger
43 minutes = 43/60 hours
modern ice
well well well
visual tiger
so you've worked (7 + 43/60) hours
sharp dragon
so you've worked (7 + 43/60) hours
How to do that on Google spreadsheets?
<@&286206848099549185>
spring wave
=17*((A1)+(A2)*(1/60))
if A1 is where you write down how many hrs
and A2 is where you write down how many miuntes
sharp dragon
Did you call me a hoe…
sharp dragon
so you've worked (7 + 43/60) hours
How to do that in equation?
$17 * 7 hours * 0,716 minutes?
visual tiger
Did you call me a hoe…
they wanted to say how
visual tiger
How to do that in equation?
$17 * 7 hours * 0,716 minutes?
no
$17/hour * (7 + 0.716)hours
grand pondBOT
Result:
85.283333333333
Result:
85.283333333333
<@&286206848099549185>
spring wave
Did you call me a hoe…
typoed
sharp dragon
Why not work
@spring wave
spring wave
should be
,calc 17*(7+(43/60))
grand pondBOT
The following error occured while calculating:
Error: Syntax error in part "\*(7+43/60)" (char 3)
sharp dragon
,calc 17*(7+43/60)
spring wave
you can set the rounding in spreadsheets
spring wave
17/60
sharp dragon
,calc 17/60
grand pondBOT
Result:
0.28333333333333
Result:
38.36
Wow I only get 38 kr…
midnight plankBOT
@sharp dragon Has your question been resolved?
midnight plankBOT
@sharp dragon Has your question been resolved?
sharp dragon
.close
frank osprey
Hey
frank osprey
Can anyone help? I am trying to solve this by using transformations. I tried reflection for a long time but i just could not do anything with it, and i dont see how others transformations would help
I know about translations, reflections, rotations, glide reflections, homothety and inversion
spring wave
Try 180 degree rotation
frank osprey
I thought about it, but i thought it would not help lol
I should try more things when i think about them
I'll try thx
spring wave
BTW, what book is this?
Use rotation by n degrees + translation = rotation by n degrees
glass dome
answer that may or may not work:
||if they are not concurrent, then they would form a small triangle at the center of the hexagon, with three vertices, each the intersection of two of the diagonals. the hexagon plus its diagonals should be symmetric under 180 degree rotation; ie. all the lines in the diagram should be exactly where they started.
however, no triangle remains in the same position when rotated 180 degrees unless it is degenerate. therefore the triangle is degenerate and all the diagonals are concurrent.||
i'm curious as to whether this is actually a valid proof
obviously don't click this unless you're willing to just read a (possible) answer
spring wave
You have to show you can rotate the hexagon 180 degrees to itself
But otherwise it is correct
midnight plankBOT
@frank osprey Has your question been resolved?
midnight plankBOT
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Available help channel!
hybrid grail
Can someone help me with this question?
midnight plankBOT
Can someone help me with this question?
@hybrid grail Has your question been resolved?
midnight plankBOT
@hybrid grail Has your question been resolved?
young topaz
Can someone help me with this question?
first i would look at how many critical points each equation has and where. then how can you find what direction the arrows would be moving to the left and right of the critical points?
midnight plankBOT
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hybrid grail
first i would look at how many critical points each equation has and where. then...
can you do and walk through the first one so that i get a feel of how to do it? i would like to then try the other ones myself
midnight plankBOT
can you do and walk through the first one so that i get a feel of how to do it? ...
@hybrid grail Has your question been resolved?
hybrid grail
.close
midnight plankBOT
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humble prairie
h'(0) = 3 right? not 0
midnight plankBOT
h'(0) = 3 right? not 0
tribal temple
Yea think they meant 3 there 
midnight plankBOT
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edgy knot
midnight plankBOT
fallow scarab
What even is the question
edgy knot
find the biggest and smalest value in the equation on top left and restrictions are in middle to top right
fallow scarab
Did you learn Lagrange multipliers
edgy knot
yes
but we dont ever use lagrange on these questions there is a seperate question where lagrange is used.
this was solved accordingly and i cant understand how he did it
carmine sigil
for an inequality, you can find the extrema on the edges by treating the inequality as an equality and using LM. And then you use regular derivatives to find the extrema on the interior.
fallow scarab
https://tutorial.math.lamar.edu/classes/calciii/lagrangemultipliers.aspx
Example 4 for an example of interior extrema
edgy knot
for an inequality, you can find the extrema on the edges by treating the inequal...
alright i see
thank you both of you
.close
midnight plankBOT
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steady trail
$\int x^2e^x\sin(x)dx$
midnight plankBOT
$\int x^2*e^x*\sin(x)dx$
grand pondBOT
Task Bot
steady trail
How to solve this ?
grim vector
Ibp and good luck, at first sight
steady trail
I'll try
grim vector
Setting dv as x²
midnight plankBOT
@steady trail Has your question been resolved?
paper inlet
Tabular method cooks so hard here lol
kindred lagoon
Is my construction of the logical 'or' correct as a categorical diagram? The diagram commutes, but I'm not sure if I'm missing something. I can't find any other abstraction of it online so I'm not sure how to check my answer. Here's the diagram:
midnight plankBOT
Is my construction of the logical 'or' correct as a categorical diagram? The dia...
@kindred lagoon Has your question been resolved?
midnight plankBOT
@kindred lagoon Has your question been resolved?
kindred lagoon
.close heading out
midnight plankBOT
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Available help channel!
modern shard
ik, its highest/highest
midnight plankBOT
ik, its highest/highest
modern shard
but i am getting confused on taking coefficients
midnight plankBOT
@modern shard Has your question been resolved?
sharp wave
Can you show what you have done
modern shard
uhm i cann tell you whatt i have done
since its highest/highest, thats how we solve limit
we take out 2(root2x)^10*x^5 as numerator
midnight plankBOT
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Available help channel!
blissful drum
midnight plankBOT
blissful drum
this is physics but im hoping someone can help me
i rly struggling on figuring how how to identify which valye is u, u', and v
midnight plankBOT
@blissful drum Has your question been resolved?
last slate
You can pick either of them
to be either of them
This is IB physics right?
I was literally studying this exact topic yesterday
usually $u$ si the variable for the velocity of the object which is also our frame of reference
so in this case
grand pondBOT
Edmund Cloudsley
last slate
$$u_\text{velocity of particle A with respect to B} = \frac{v_B - v_A}{1 - \frac{v_Bv_A}{c^2}}$$
grand pondBOT
Edmund Cloudsley
blissful drum
This is IB physics right?
yea how did you know lmao
blissful drum
but yeah this is it
thanks
im just really strugglng on
assigning variables
when solving realtivity questions
last slate
yea how did you know lmao
bceause I am also doing IB physics HL
blissful drum
bceause I am also doing IB physics HL
nice nice
last slate
and I am LITERALLY studying this topic as we speak
I usually like to come and help others on this server
blissful drum
are you '25?
last slate
yeah
blissful drum
nice
wait so for this questoin, everything measured in the frame of the ground are varuables without prime
last slate
no not exactly
blissful drum
hm is that so?
last slate
so we have 3 variables for such questions right?
u
x
and t
- u is teh velocity of an object as measured in one inertial reference frame
- u prime is the velocity of the SAME object as measured in another referential frame
same case for x and t
to make it easier
last slate
if a question was framed as "Find <quantity> of Object A with respect to Object B"
- u would be velocity of object A with respect to ground
- u prime would be velocity of the same object with respect to object B
- v would be velocity of object B
last slate
but doesnt one have to be a stationary frame and the other moving
nope that's not entirely true
one has to be not accelerating
not the same thing as not moving
Which book are you using for your IB physics prep?
blissful drum
Which book are you using for your IB physics prep?
oxford
blissful drum
cambridge has good q's
blissful drum
yea i have that one
last slate
Why so?
blissful drum
i use textbooks mainly for understanding and learning concepts
and solve questions early on in the learning stage
blissful drum
yea
last slate
but yeah your call
blissful drum
to build foundation
last slate
whatever you like
last slate
gotcha
blissful drum
so fricking confusing
last slate
I have 0 intuition for relativity
last slate
A4 was Rigid body mechanics right?
blissful drum
A4 was Rigid body mechanics right?
yea
last slate
Do you do revision villlage as well?>
blissful drum
uhh not rly
last slate
you can get the fake version of it on r/pirateIB
last slate
wait really? first every IB student I know and now you as well? 😆
blissful drum
bro i know like every p*rated site
last slate
Alrighty man good luck
last slate
Have a good day
tender vector
midnight plankBOT
tender vector
no clue what to do here
dense ruin
you are given f'(a) = 2f(a) + 2a
tender vector
yeah I got that but not sure where to go from there
flat moat
yeah I got that but not sure where to go from there
$f'(a) = 2f(a) + 2a$ implicitly means $f'(x) = 2f(x) + 2x$ so it is going to be $\frac{dy}{dx} = 2y + 2x$
grand pondBOT
raxdius
flat moat
sorta first order ode
tender vector
.close
midnight plankBOT
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molten socket
how to find the limit of [(e^(cosh))-e]/h as h approaches 0?
midnight plankBOT
how to find the limit of [(e^(cosh))-e]/h as h approaches 0?
gaunt nimbus
${\lim_{h \to 0}\frac{e^{\cos h} - e}{h} = \lim_{h \to 0}\frac{e^{\cos h} - e^{\cos 0}}{h - 0}}$
grand pondBOT
k
molten socket
ok ty
edgy schooner
how to find the limit of [(e^(cosh))-e]/h as h approaches 0?
how silly of me to read cos(h) as hyperbolic cosine
ember marsh
just the derivative of e^(cos(x)) at zero
peak herald
whats the limit definition of a derivative
peak herald
whats x in that expression
molten socket
im more interested in solving the limit itself
midnight plankBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
ember marsh
im more interested in solving the limit itself
this limit is eqivalent of the derivative of e^cos(x) at 0
ember marsh
wet bridge
bottom is -, not ->
ember marsh
yeah i found the image on internet
this is a better one
consider f(x)=e^cos(x)
a=0
this is same as ur problem
molten socket
im even more cofused now
ember marsh
emm
rain wasp
i could be wrong, but i think the fact that this is the derivative of e^cos(x) doesn't mean it's how the limit should be solved
wet bridge
ye
rain wasp
more like it's a l'hopital deterrent
ember marsh
i could be wrong, but i think the fact that this is the derivative of e^cos(x) d...
just a way of simpifying the problem:)
wet bridge
more like it's a l'hopital deterrent
i dont think he knows what a derivative is
molten socket
lets say im doing a limits exam where i cannot use any differentation techniques
yeah i dont
ember marsh
emm