to some extent
#help-26
prisma mesa
most modern ai doesn't use logic like this
current AI is based more on statistics
chatGPT has analyzed millions of pages of text during its training and what it does now is basically predicting what word comes next based on what it saw during its training
hybrid pasture
so it has some coefficient attached to words?
prisma mesa
3blue1brown has quite a cool vid on that if you wanna learn more, my explanation is just the absolute basics
hybrid pasture
definitely I'll watch that
modern AI's are like aliens in my eyes
compared to just ~3 years ago I wouldn't have expected them to do the things they can
prisma mesa
https://www.youtube.com/watch?v=aircAruvnKk&list=PLZHQObOWTQDNU6R1_67000Dx_ZCJB-...
and this is a series about neural networks
neural networks are much more fundamental concept, they underline basically every AI we currently have
hybrid pasture
I mean we are the product of hundred of millions of years of evolution
makes sense that our structure is fairly good to base things off
prisma mesa
yeah, but in NN's it's simplified a lot
while our neural system is an utter mess basically (the nerves are connected more or less randomly), NNs neurons are connected in a structured manner in layers
hybrid pasture
I think there is probably some key insight there
our random nerve connections might be responsible for our intelligence
structure could impede fast thinking
but also probably impedes large computation capacity (our brain)
prisma mesa
Could be. But the problem is that it's hard to replicate that complicated structure in a way it would work
neural network is basically an extremely large function that takes many inputs and converts them into many outputs. It also has some parameters which change how exactly the function looks (sth like a, b, c in standard quadratic function). Then the training is done by giving the network some inputs, making it compute the outputs and then rate how good the outputs are and essentially compute some "error". The next step is slightly changing the parameters, such that the error gets smaller. And by repeating this many times, we get a small enough error and the network is trained. After that, when it gets inputs it has never seen, it's often able to guess the correct output quite accurately
but ofc there are limitations to that
hybrid pasture
the limitations seem to be fairly distant
I saw the o3 benchmarks and they were impressive
prisma mesa
I actually think that with how current AI works we cant get that much better
we can make the models larger, more capable but nothing too inovative will happen imo
current AI is usually only good at one task
and it can only do stuff that's similar to something it has done before
hybrid pasture
and it can only do stuff that's similar to something it has done before
I think this is honestly fine
I repeat things all the time in my life
But I tend to waste time repeating them
If I had an AI that could repeat them for me, I would save a lot of time
To think about harder stuff
prisma mesa
Yeah, but it's a limitation. Current AI woudldnt be a very good researcher e.g.
hybrid pasture
Yeah, but it's a limitation. Current AI woudldnt be a very good researcher e.g.
For sure
prisma mesa
it's a great tool for researchers, but not a researcher itself
prisma mesa
anyway, ill have to go now
prisma mesa
indeed, cya
hybrid pasture
.close
topaz sinewBOT
topaz sinewBOT
Available help channel!
terse depot
Given differential equation
x y''(x)-2y'(x)=3+x
Please write a function of two variables (a and b) that solves the boundary value problem
y(0)=a
y(1)=b
using the finite difference method by dividing the interval [0,1] until reaching the absolute error of order 1e-5.
The values of y(x) at the intermediate points are used to approximate the function by the least squares method by a polynomial of degree 3.
Ax^3+Bx^2+Cx+D
To solve the least squares problem, the program must use the Cholesky decomposition.
Define the function that returns A as a function of a and b.
For a fixed value of a (a=1) determine the solution of equation A(1,b)=0 using the Newton-Raphson method and calculating the derivative numerically.
The initial approximation for b is b0=1.
terse depot
i've made a code to solve it. But my professor keeps saying that is wrong, this, that and at this point i'm so damn confused i don't even know what to do anymore
i can send the code here if i may
topaz sinewBOT
@terse depot Has your question been resolved?
topaz sinewBOT
@terse depot Has your question been resolved?
terse depot
can someone help me?
agile harness
can someone help me?
for starters no one probably wants to click that but you’re unlikely to get help for this in these channels
try here
terse depot
thx buddy
terse depot
please..
topaz sinewBOT
@terse depot Has your question been resolved?
neon iron
@terse depot What have you tried? Ping me when you reply
terse depot
<@373541179276656640> What have you tried? Ping me when you reply
the code is above. I've tried solving it using numpy libraries
but my professor keeps saying something's wrong
i cant figure what
neon iron
Does it run?
terse depot
yep
neon iron
What did prof say is wrong
terse depot
and it gives a result
neon iron
Specifically
terse depot
What did prof say is wrong
THAT'S THE WHOLE DAMN POINT
HE JUST SAY THAT IS WRONG
HE DOESNT SAY WHAT IS WRONG
EVEN THOUGH I SOLVED THE QUESTION
neon iron
Did you?
terse depot
yes i did
i'm so confused
because he gave me some insights
on error calculations and stuff
especialy on h*
for central differences
and using central differences in newton raphson
which you can see i did correctly
finite differences is correct
cholensky is correct
i believe that everything is correct
but still, he said something's wrong.
neon iron
Fr?
terse depot
hope you don't mind some portuguese words
neon iron
Lol
terse depot
YEAH
i was looking for a discord server
cause i was so confused
my professor is ukranian
he thinks he's teaching me something by saying "you're wrong."
come on
can you notice any mistakes?
terse depot
i think so
i'll do it your way 🙂 if you don't mind, i'll hit u up to give some feedback k?
neon iron
Ye no problem
Actually looking at the documentation it says the opposite
So what you did would be right
topaz sinewBOT
Available help channel!
limpid pulsar
For the chain rule, if we have (2x+1)^8 why can’t I just use like the constant multiple rule and then subtract one from the exponent?
topaz sinewBOT
For the chain rule, if we have (2x+1)^8 why can’t I just use like the constant m...
limpid pulsar
I know this question is kind of stupid I’m just trying to understand the chain rule better
ashen pecan
For the chain rule, if we have (2x+1)^8 why can’t I just use like the constant m...
Can you elaborate? What would you think the derivative would be instead?
pallid zenith
For the chain rule, if we have (2x+1)^8 why can’t I just use like the constant m...
recall the definition of the chain rule
limpid pulsar
Can you elaborate? What would you think the derivative would be instead?
I was just wondering why you can’t use the power rule
limpid pulsar
recall the definition of the chain rule
F’(g(x))*g’(x)
ashen pecan
It's just bc that function is a composite function
So you can most conveniently write (2x+1)^8 as a f(g(x)) where g(x)=2x+1 and f(x)=x^8
You can avoid the chain rule by foiling out (2x+1)^8 into an explicit degree 8 polynomial
And you just apply the power rule to what you get when you expand
pallid zenith
I was just wondering why you can’t use the power rule
its just an extended version; for (ax+b)^n we have that its derivative is a(ax+b)^(n-1)
same for something like (x^2+1)^2
ashen pecan
And you just apply the power rule to what you get when you expand
But that's not a viable strat at higher powers, for example (2x+1)^100
limpid pulsar
So you can most conveniently write (2x+1)^8 as a f(g(x)) where g(x)=2x+1 and f(x...
Okay this makes sense
limpid pulsar
its just an extended version; for (ax+b)^n we have that its derivative is a(ax+b...
Okok
Thank you guys
I understand the chain rule better now
.close
vapid notch
how do i do 22
topaz sinewBOT
how do i do 22
vapid notch
since tan pi/2 is undefined should i just say the answer is undefined?
or how do i simplify it into cotx
loud oasis
you can put tan in terms of sin and cos
pallid zenith
since tan pi/2 is undefined should i just say the answer is undefined?
think about a shift if you want to prove it visually
note that cot=1/tan and that cot(pi/2)=0
for the algebraic proof
topaz sinewBOT
@vapid notch Has your question been resolved?
minor sand
How to calculate returns on investment with fixed rate but increasing monthly contribution every certain amount of years?
topaz sinewBOT
How to calculate returns on investment with fixed rate but increasing monthly co...
minor sand
If I set aside 500 each month, and my investment grows annually by 7%, and if I increase the monthly payment to 1,000 from the third year, to 2,000 from the fifth year, to 20,000 from the seventh year, and to 50,000 from the twelfth year until the thirty-fifth year, how much would I end up with?
pallid zenith
If I set aside 500 each month, and my investment grows annually by 7%, and if I ...
sum it individually
i dont think there would be a continuous formula for this
minor sand
But dosent it give a different answer? What I mean is, lets say I have 500 usd that compounds for 2 years and gets to 1000. If I wait another 2 years, it will grow to 4000 (just an example). But if I take 1000 and invest it now and wait 2 years, wouldn’t it be lower than if it was 500 that compounded for 4 years?
I hope I can be understood
minor sand
Oh didn’t even notice ur username😂, was just sad that my question couldn’t be answered
pallid zenith
oh lol
the best i can answer here is partitions
dont really have much knowledge of this subject tbh
topaz sinewBOT
@minor sand Has your question been resolved?
topaz sinewBOT
@minor sand Has your question been resolved?
odd flower
Hello
odd flower
In this question I have a doubt with my work
i got the correct answer but want to make sure
topaz sinewBOT
@odd flower Has your question been resolved?
ancient valve
In this question I have a doubt with my work
Can you show the work?
topaz sinewBOT
@odd flower Has your question been resolved?
mortal thunder
i got the correct answer but want to make sure
!show
topaz sinewBOT
i got the correct answer but want to make sure
Show your work, and if possible, explain where you are stuck.
mortal thunder
Without showing your work, are you expecting us to solve the Q for you n then verify your answer with ours? Is it feasible? Is it not convenient for the both of us if you share ur soln along with the Q
brisk ice
What is the correct way to solve this? I cant find any solutions online.
brisk ice
I tried it and I got I = 2 so I think something went really wrong in my solution.
rigid star
I think if you split the big integral into 3 parts, you can find explicit expressions for f on each interval
brisk ice
yea i tried that but it didn’t work
from 0 to 1 it doesn’t matter whether it’s t-x or x-t because the answer for x-1/2 or 1/2-x is 0 either way
here i meant from -1 to 0
then x-1/2 from 1 to 2 so it’s 1
and i got 2
idk what i did wrong
topaz sinewBOT
@brisk ice Has your question been resolved?
brisk ice
<@&286206848099549185>
mortal thunder
$\int_{-1}^{2} \int_{0}^{1} |x - t| \dd t \dd x$
thorny flameBOT
Arya
twin pendant
You have to split it like this
mortal thunder
$= \int_{-1}^{0} \int_{0}^{1} (t-x) dt dx + \int_{0}^{1} \left[\int_{0}^{x} (x-t) dt + \int_{x}^{1} (t-x)dt \right] dx + \int_{1}^{2} \int_{0}^{1} (x-t) dt dx$
thorny flameBOT
Arya
brisk ice
.close
topaz sinewBOT
topaz sinewBOT
Available help channel!
maiden owl
In an international competition, participants must answer 18 questions numbered from 1 to 18. An answer to each question is either right or wrong. Each participant obtains a result corresponding to the sum of these two calculations:
- The number of successful questions;
- The sum of the numbers of the successful questions, divided by a thousand.
For example, Christiano passed questions 3 and 5. His result is: 2 + (3 + 5) ÷ 1000 = 2.008
How many different results are there in this competition?
This question is so hard. You have to consider that you can't use a calculator and this is an 8 grade question. Also suppose you got 2 questions right, n7 and n3, it will give the same result if you got n4 and n6, hindering the possible number of results. I tried one way of finding the max/min of possible results for each number of questions that someone got correct but it doesn't consider the example I just gave. I don't know what to do now
topaz sinewBOT
In an international competition, participants must answer 18 questions numbered ...
@maiden owl Has your question been resolved?
wispy pier
There are a lot of calculations still !
mortal thunder
Consider two different participants with same results. $P_1$ got a total of i questions right, which were $S_1 = {a_1, a_2, \cdots, a_i}$ and similarly for $P_2$ who got $S_2 = {b_1, b_2, \cdots, b_j}$ questions right. Now, for the results to be same, we must have $i + \frac{\sum_{k=1}^{i} a_k}{1000} = j + \frac{\sum_{k=1}^{j} b_k}{1000}\ $ Note that max of $\sum_{k=1}^{i} a_k < 1000$ so you must have $i = j$ and hence, $\sum_{k=1}^{i} a_k = \sum_{k=1}^{j} b_k$
thorny flameBOT
Arya
mortal thunder
Now, it's a matter of counting combinations for i = 1 to 9
maiden owl
i cant bro 😭
wispy pier
What's the answer ?
maiden owl
i dont know man
storm flume
I think you should indeed approach it by finding the possible result per each number for correct questions as you already did, but you only need to consider the possible sum of the number of succesful questions. E.g. if you get 2 questions right, the sum could range from 3 (by getting n1 and n2 right) to 35 (by getting n17 and n18 right). Any sum between 3 and 35 is possible, so that means getting 2 questions right gives you 33 possible results
You also don’t need to worry about whether a person who gets 1 question right has the same result as someone who gets 2 questions right, since the result for getting 1 question right will definitely lie between 1-2, and the result for getting 2 questions right will lie between 2-3
@maiden owl
mortal thunder
Question is indeed difficult, but still doable with some brute force
topaz sinewBOT
@maiden owl Has your question been resolved?
maiden owl
I think you should indeed approach it by finding the possible result per each nu...
so you are doing the min/max
ok lemem try it again
storm flume
so you are doing the min/max
Yupp, it doesn’t matter if the sum is 10 by adding 6+4 or 7+3, you only care about the endresult
storm flume
yay!!!!
wait let me calculate too
okay i got 997 but i might have messed up somewhere
it's close enough
xD
ohh wait +1 because 0 answers right is also a result
okay it's 988!
topaz sinewBOT
@maiden owl Has your question been resolved?
maiden owl
okay it's 988!
wait wait
what
no
@storm flume
@storm flume
bro plz tell me ur lying
it has to be 987
@storm flume
little pine
lemme brute force
storm flume
it has to be 987
yeess you're right it's 987 if you count all the possible results from getting 1 to 18 questions right, and then one more result because it's also possible to get zero questions right and your result would be 0
little pine
lemme brute force
yup, got 988 too
little pine
<@595185421743554570>
feel free to check my python code
R = set()
for i in range(0, 2**18):
S = (0, 0)
for b in range(0, 18):
if (i >> b)&1 == 1:
S = (S[0]+1, S[1]+b+1)
R.add(S)
print(len(R))```
I'm using the binary of a number to represent the questions that is right (1) or wrong (0)did you check where the sum = 0 like they said?
little pine
no its 1-18
what if they answered 0 questions right?
little pine
yea?
little pine
yes
maiden owl
.close
topaz sinewBOT
topaz sinewBOT
Available help channel!
topaz sinewBOT
neon iron
Help pls
neon iron
simplify it, you start to see that most of the terms get cancelled
What do the … mean?
do you see a pattern in the terms?
round island
What do the … mean?
It means that it is continuous depending on the pattern up to (1-1/100)
So the itll be (1-1/7)(1-1/8)(1-1/9) up to (1-1/100)
topaz sinewBOT
@neon iron Has your question been resolved?
topaz sinewBOT
topaz sinewBOT
Available help channel!
errant hamlet
Find the smallest nonzero natural number, a multiple of both 41 and 73, that has all its digits identical.
topaz sinewBOT
Find the smallest nonzero natural number, a multiple of both 41 and 73, that has...
errant hamlet
n=M41, n=M73, gcd(41,73)=1
n=M41x73=M2993
n=2993k
we easily show that n>9999
but i dont think we can limit it until we find the number
because i think n has about 30-40 digits
and im stuck here
rigid ivy
,w lcm 41, 73
thorny flameBOT
rigid ivy
,w sum a*10^k for k=0 to n
thorny flameBOT
rigid ivy
Yeah, not sure if that same-digit summation formula would help much
Actually, you can try every a from 1 to 9, and see if you could solve for n. But that may be a poor idea
craggy haven
start by finding a multiple of 41 with all digits same
or characterize all such multiples
errant hamlet
11111
upbeat musk
topaz sinewBOT
https://media.discordapp.net/attachments/1015759484708208792/1045835008012714017...
upbeat musk
So the method they are proposing is to divide 5x by 2 then square that result to gain a number whos factors would multiply to make the original number, yes?
pallid zenith
So the method they are proposing is to divide 5x by 2 then square that result to...
no
also, specify which “number” you are talking about
meager dawn
not quite. they aren’t actually dividing 5x by 2. they’re doing stuff with the 5 because it’s the term with the x attached to it.
pallid zenith
"(x+a/2)^2 formula"
the whole “factors” thing is irrelevant her e
it’s more of “create a trinomial such that you have it as a perfect square trinomial” and adjust the constant according to that
upbeat musk
how is it irrelevent the entire operation is a factorization operation
pallid zenith
how is it irrelevent the entire operation is a factorization operation
this is completing the square, not factorization.
upbeat musk
where you forge the middle term via factors of x^2 and well the third term didnt have a factor so they deleted it and made 25/4 out of 5
pallid zenith
where you forge the middle term via factors of x^2 and well the third term didnt...
eh?
upbeat musk
thats what they did yes?
they couldnt forge 5 within the resulting factorization bc no factors were available to forge 5
so they deleted the 3rd term
bright trail
What does forge mean?
upbeat musk
and made a new one via 5
pallid zenith
please don’t use words such as “forge” and “delete”, this is confusing me
upbeat musk
alr
pallid zenith
try to use mathematically sound words
upbeat musk
by forge I basically mean that they used factors to recreate the middle term since it isnt actually present in the resulting binomials being squared
but x and the third term typically are in factored form
recreating the middle term
upbeat musk
where I get completely lost is when they divded and moved 8 to the other side
pallid zenith
could you take a screenshot or something please?
upbeat musk
i get that youd be unable to factor properly with the 3 and divide but then he moved 8 because he couldnt get factors of 8 multiplied by x to get 15
but then
he started doing nonsense
pallid zenith
where I get completely lost is when they divded and moved 8 to the other side
so it was asking for 3x^2+15x-8=0, right?
i’ll explain this one
pallid zenith
i get that youd be unable to factor properly with the 3 and divide but then he m...
the first part is correct; it is necessary to divide by 3.
upbeat musk
he started doing nonsense
because what i dont understand is after he divded and moved to the other side he divded 5 by 2 then squared it to make a number whos factors would generate 5 again when multiplied by x
(perfect square)
pallid zenith
so we get x^2+5x-8/3=0, to get a cleaner result
now we add to isolate the x^2+5x term
so x^2+5x=8/3
is this where you are confused?
upbeat musk
now what would be really tragic is if this is all just a general method the entire time and is a self-evidential truth but i really hope that isnt it but thatd explain alot because those always torture me
ok so
bright trail
is this where you are confused?
It would be good to answer this fwiw
Hard to tell where you are getting lost without you answering it
upbeat musk
i know but it takes me like 5 minutes to read any form of mathematical writing
upbeat musk
so we get x^2+5x-8/3=0, to get a cleaner result
I really dont understand but like.
pallid zenith
i know but it takes me like 5 minutes to read any form of mathematical writing
do you want me to latex it?
upbeat musk
so basically you are just saying they moved 8/3 over because it didnt have a factor thatd multiply to make 5?
pallid zenith
so basically you are just saying they moved 8/3 over because it didnt have a fac...
well, to be precise, 8/3 is not the a^2/4 term that we want
so i just moved it over to create the a^2/4 term
i do it for any problem, really
upbeat musk
a
pallid zenith
well, do you want me to continue, or can you finish it?
pallid zenith
you said “(x-a/2)^2 factorization” earlier
upbeat musk
i dont know what that is either
i just copy pasted that
because someone said it was what i was trying to do
pallid zenith
so you legitimately don’t know what completing the square is?
upbeat musk
not at all
upbeat musk
the funny thing
is that like
if you told me what it was
i would probably know
but then youd go in context and id know nothing
pallid zenith
well, it’s most commonly known as completing the square
what country do you live in?
upbeat musk
USA EST
pallid zenith
same...
upbeat musk
can you elaborate on completing the square?
pallid zenith
k
upbeat musk
what grade ??
idk / N/A / not applicable
pallid zenith
so imagine you have $\qty(x-\frac{a}{2})^2$
thorny flameBOT
;(
pallid zenith
dang you typed that fast
experience
anyways this expands to $x^2-ax+\frac{a^2}{4}$
thorny flameBOT
;(
pallid zenith
so that’s what you’ve been doing the whole time basically
trying to find that a^2/4 term, adding it, then making the trinomial a perfect square trinomial
sorry my keyboard turned to español
in fact you can use completing the square to prove the quadratic formula
i did it a while back
@upbeat musk if you would like once we are done i could show you
upbeat musk
._.
bright trail
It's a good proof to see fwiw
upbeat musk
I dont know what a^2/4 is nor do i understand the a/2 term.
pallid zenith
> I dont know what a^2/4 is nor do i understand the a/2 term.
ah.
damn, you watched the video too, right?
upbeat musk
yes
pallid zenith
upbeat musk
I was just talking to you earlier about how he divided because the 3 term next to X since thatd stop any factorization without removing it
But then I told you I didnt understand at all because creating a new term as 5/2 powered by 2 was nonsense to me
and so long as 5/2 pow 2 isnt explained nothing else can be
pallid zenith
ah, okay
upbeat musk
i swear this is going to end up being a generalization concept smh cuz those are agony (nigh impossible due to false principles)
pallid zenith
well, based on what i told you earlier, what do you think “completing the square” means intuitively?
upbeat musk
if a quadratic is a set of operations/functions done unto X and you want to solve for what value of X would generate zero given those operations (root)
factoring would be when you factor the problem into smaller binomials that multiply to create the whole to solve for individual parts
but completing the square would be when you isolate X and solve for X by having two factors that are the same allowing you to break the multiplication between both binomials isolating X alone and solving for it linearly.
Completing the square also would entail manipulating parts of the equation to have factors that are equal so you can undo the squaring operation (multiplication of binomials)
pallid zenith
yeah
another way to think of it is finding a trinomial perfect square (i hope you know what this means) such that you do the operations that you have said earlier
hmm
let’s take a quadratic if the form x^2+ax+b
the way to get that nice perfect square trinomial is to add the constant a^2/4
this is because. as shown earlier, x^2+ax+a^2/4 in and of itself is a perfect square ((x+a/2)^2 is its equivalent)
upbeat musk
yes its my favorite song i use it for work
pallid zenith
upbeat musk
pallid zenith
i thought you might have accidentally put in the wrong link lmfao
anyways do you understand?
upbeat musk
the way to get that nice perfect square trinomial is to add the constant a^2/4
see the problem is you are out of the scope
you are using terms to paint a whole i can never see because i dont have the constituents first
basically if i dont know what a^2/4 means or other such things its impossible for me to figure this out
pallid zenith
which is why i’m asking if you know the vocab
pallid zenith
ok
a perfect square trinomial is one in which it can be favored into some (a+b)^2 where a and b are TBD
btw, (a+b)^2=a^2+2ab+b^2
this is the backbone of completing the square
so let’s say we plug in a for x
since those are the kinds of problems we are dealing with
you follow?
@upbeat musk
hearty cave
Hello, can someone check my work for me. I boxed the answers and the way I got them is all written on the respective questions. Thank you I really appreciate it!
topaz sinewBOT
Hello, can someone check my work for me. I boxed the answers and the way I got t...
@hearty cave Has your question been resolved?
topaz sinewBOT
@hearty cave Has your question been resolved?
pallid zenith
Hello, can someone check my work for me. I boxed the answers and the way I got t...
yeah
dont know why it took 2 hrs to get help but it looks good
topaz sinewBOT
topaz sinewBOT
Available help channel!
neon iron
topaz sinewBOT
crisp raptor
if you divide by a fraction its the same thing as multiplying with the fraction being flipped, that's probably where you want to start, because multiplying is easier then dividing
neon iron
alright but what is the colon for
crisp raptor
what does colon mean here Im sorry my english is to bad for that
sterile finch
It signifies a ratio.
For instance, if you have nine fruits and there is a ratio of two apples to one orange, 2 : 1, you can determine that you have six apples and three oranges.
neon iron
For instance, if you have nine fruits and there is a ratio of two apples to one ...
yea but this is an expression
it isnt a ratio
sterile finch
Are you sure it isn't a ratio? What subject are you currently covering?
loud loom
Its the first term divided by the second
neon iron
its a university entrance exam
neon iron
Its the first term divided by the second
so left over right?
loud loom
Yes
neon iron
hold on let me try
loud loom
Aight lmk
loud loom
yea well done 😄
neon iron
thanks bro
loud loom
Gl on the rest !
neon iron
.close
topaz sinewBOT
topaz sinewBOT
Available help channel!
cobalt oyster
this gonna sound really stupid but ive been cramming calc 1 for my finals lately and now i cant even remember whether im supposed to change the bounds when replacing the variable like this
topaz sinewBOT
this gonna sound really stupid but ive been cramming calc 1 for my finals lately...
thorny flameBOT
Edmund Cloudsley
cobalt oyster
missing an x after 12
thorny flameBOT
Edmund Cloudsley
neon iron
so this is your origiunal question
cobalt oyster
ya
round island
Let u = 1+ x²
neon iron
you take the 1 + x^2 as your substitution
and then you re-evaluate the limits based on the substitution
$$\int_{0}^{\sqrt{8}} 12x \cdot \sqrt{1 + x^2} dx$$
$$u = 1 + x^2$$
$$\frac{du}{dx} = 2x$$
$$dx = \frac{du}{2x}$$
Re-evaluating the limits
$$a_1 = {1 + \sqrt{8}^2} = 9$$
$$a_2 = {1 + {0}^2} = 1$$
Put the new limits back in. Put the substition of $u$ back in. And voila! You're done
thorny flameBOT
Edmund Cloudsley
cobalt oyster
104 alr thanks
neon iron
,w evaluate integral from 0 to sqrt{8} of 12x * sqrt{1 + x^2}
thorny flameBOT
cobalt oyster
just had brainfart i remembered the arc length formula but forgot how to integrate lmao
cobalt oyster
.close
topaz sinewBOT
topaz sinewBOT
Available help channel!
neon iron
topaz sinewBOT
neon iron
can someone translate to normal english
wispy pier
can someone translate to normal english
?
wispy pier
P_1 = 3x²-x+4
P_2 = x³+x²
P_3 = x²-1
wispy pier
The other part of the question reads
x³+4x²-x+4
3x⁴-x³+x²+x-4
x⁴-2x²+1
x³+1
(x²)/(x-1)
wispy pier
They want you to perform operations between P_1,P_2 and P_3 so that you get these results
wispy pier
i divide the equation with the polynomials
You can
Add, subtract, divide and multiply
That is, you can perform all elementary operations
neon iron
Add, subtract, divide and multiply
but only the polynomials right?
wispy pier
Yes
neon iron
oh fun
mental basin
In general, try thinking about what happens to the degrees of the polynomias as you add/subtract/multiply/divide them.
topaz sinewBOT
topaz sinewBOT
Available help channel!
elfin parcel
could someone help me understand how to do this
topaz sinewBOT
could someone help me understand how to do this
smoky sparrow
could someone help me understand how to do this
what don't you understand about the solution?
elfin parcel
what don't you understand about the solution?
The first step
smoky sparrow
the trick is say $\frac{1}{3} - \frac{1}{5} = \frac{5 - 3}{3 \cdot 5}$
thorny flameBOT
south
smoky sparrow
The first step
have you seen other examples of summations that cancel out like this?
telescoping sums?
the solution to "how do you even know you need to do this" is to practice, look at more worked examples and practice some others yourself
elfin parcel
have you seen other examples of summations that cancel out like this?
I've seen a bit but not much though
smoky sparrow
yeah that makes sense
elfin parcel
I'm a bit confused on how you split the fractions into two
smoky sparrow
I'm a bit confused on how you split the fractions into two
so what they did is to multiply by 2 then divide by 2 outside
so $\frac{1}{2} \cdot \frac{2}{35} = \frac{1}{2} \cdot \frac{7 - 5}{7 \cdot 5}$
thorny flameBOT
south
smoky sparrow
then $\frac{7}{7 \cdot 5} - \frac{5}{7 \cdot 5} = \frac{1}{5} - \frac{1}{7}$
thorny flameBOT
south
elfin parcel
so what they did is to multiply by 2 then divide by 2 outside
so $\frac{1}{2} \...
Is there an easy way to know how we split up the fractions
smoky sparrow
Is there an easy way to know how we split up the fractions
the trick is that 2/3, 2/15, 2/35 and so on all simplify really nicely
again, you learn you need to do that through practice
elfin parcel
again, you learn you need to do that through practice
Alright thanks for the help!
.close
vital moth
given a sequence ${u_n}$ defined by the recurrence relation $u_{n+1}=(1-u_n)^2$ with $u_0=\frac 12$ and in a previous part i proved that $0<u_n<1\ \forall n\in\mathbb{N}$ \ Study the convergence of ${u_n}$
topaz sinewBOT
given a sequence $\{u_n\}$ defined by the recurrence relation $u_{n+1}=(1-u_n)^2...
thorny flameBOT
vital moth
this sequence isnt monotone since u_0<u_2 while u_0>u_1 for example
but after writing out some terms i noticed that the subsequence of the even numbered terms is increasing and that of the odd numbered terms is decreasing
so i thought about proving that both of them are convergent and converge to the same limit
this should suffice to prove that {u_n} converges right
because if both of these subsequences converge then every subsequence of them converge
and all subsequences of {u_n} consist of terms of these 2 subsequences
i began with {u_{2n}}
but i am not able to show that it is increasing
topaz sinewBOT
@vital moth Has your question been resolved?
vital moth
<@&286206848099549185>
vital moth
for the base case , $u_2>u_0$. now suppose that $u_{2n}>u_{2n-2}\implies u_{2n}^2>u_{2n-2}^2$ but $2-u_{2n}<2-u_{2n-2}\implies (2-u_{2n})^2<(2-u_{2n-2})^2$ so that doesnt tell us anything about $u_{2n+2}-u_{2n}$ since $u_{2n+2}=(1-(1-u_{2n})^2)^2=u_{2n}^2(2-u_{2n})^2$
thorny flameBOT
vital moth
so what can i do
vital moth
<@&286206848099549185>
sudden spade
It's a quartic but already factorised, you should be able to show that if $y=x^2(2-x)^2$ then for $x\in[a,b]$ that $y>x$, which is equivalent to monotonically increasing.
But yeah once you show they both are monotone, as the sequence is bounded then it must converge, the only suitable limit must be the limit.
vital moth
wdym by cubic isnt it quartic
sudden spade
wdym by cubic isnt it quartic
You saw nothing 😑😂
thorny flameBOT
Max
vital moth
thats what i am trying to show
how will i do that
also this must not be always true
sudden spade
Plot it
sudden spade
**Max**
I used [a,b] not [0,1] because you are right, they both can't be increasing, you'll find that either side of the limit you get y>x and y<x
Which you could prove by solving y>x with the y I provided above
vital moth
Which you could prove by solving y>x with the y I provided above
the roots of y-x are 0,1,(3+sqrt(5))/2 and (3-sqrt(5))/2
(3+sqrt(5))/2 is not relevant in the specific problem that i asked about
since 0<u_n<1
now $u_{2n+2}-u_{2n}=u_{2n}(u_{2n}-1)(u_{2n}-\frac{3-\sqrt 5}2)(u_{2n}-\frac{3+\sqrt 5}2)$
thorny flameBOT
vital moth
the first term is > 0
the 2nd and the 4th are < 0
what varies is the 3rd
so $u_{2n+2}-u_{2n}<0$ when $0<u_{2n}<\frac{3-\sqrt 5}2$ and $u_{2n+2}-u_{2n}>0$ when $\frac{3-\sqrt 5}2<u_{2n}<1$
thorny flameBOT
vital moth
Which you could prove by solving y>x with the y I provided above
but now what
how does this help
btw $u_{2n+2}$ and $u_{2n}$ can be replaced by $u_{n+2}$ and $u_n$ respectively in all of what i wrote above
thorny flameBOT
sudden spade
-
Write $u_{n+2}$ in terms of $u_n$.
-
Let $x=u_n$ and $y=u_{n+2}$, plot the graph for $x\in[0,1]$
-
When is y>x? When is y<x?
-
Look at $u_1$ and $u_2$, the. Relate this to the graph
thorny flameBOT
Max
vital moth
i cnt do anything with the graph
other than observation
but that doesnt help me prove anything
sudden spade
Plot the graph and plot y=x, then show me
vital moth
but didnt i solve y>x some moments ago ?
sudden spade
It helps understand what you NEED to show, if we can understand informally what is going on it will help us create a formal argument
You did, plot it tho, we are visual creatures, it's good to see things
You won't always need to plot (personally I wouldn't plot it) but for your first few a plot helps
vital moth
alright here you go
vital moth
You won't always need to plot (personally I wouldn't plot it) but for your first...
i dont prefer plotting too xD
vital moth
alright here you go
the shaded parts are the regions of solution to the inequality y>x
so what now
sudden spade
$x\in[0,1]$ so zoom in on that part
thorny flameBOT
Max
sudden spade
And plot y=x on the same axis, (desmos is a better tool btw)
vital moth
vital moth
And plot y=x on the same axis, (desmos is a better tool btw)
this is y=x
by y you mean x^4-4x^3+4x^2 right
sudden spade
Well yes but I wouldn't have expanded it, we don't need to
It's easier to plot in factorised form
sudden spade
this is y=x
I don't mean a y, x axis
I mean plot y=x
On that graph
Because you can see where y>x clearly
And vice versa
vital moth
I mean plot y=x
sudden spade
vital moth
oh what i did was to plot f(f(x))-x
vital moth
ok so now what
sudden spade
oh what i did was to plot f(f(x))-x
Ah okay
sudden spade
ok so now what
Well we see that when x is less than that point of intersection yeah time we put x in we get a smaller y value out
Whereas when x is on the right of the intersection we see that each x put in gets a value of y out that is bigger
So what is happening in relation to the sequence?
$u_0=1/2$ which is on the right of the intersection. So we expect ,$u_2>u_0$ and hence this will continue.
Whereas $u_1=1/4$, which is on the left of the intersection so we expect $u_3<u_1$ and so this will continue
Ie. The even sequence is monotonically increasing from 1/2
And the odd sequence is monotonically decreasing from 1/4
Both these sequences must converge due to Cauchy BUT they clearly converge to different limits so the overall sequence does not converge.
It's likely they converge to 0 and 1 respectively
thorny flameBOT
Max
vital moth
this much is evident from here
vital moth
and here
vital moth
**Max**
but how do i prove what is written here about the 2 subsequences consisting of even numbered terms and odd numbered terms
so how do i prove that the first is increasing and the second is decreasing
i can do that by proving that the point of intersection is an upper bound of the odd numbered subsequence and a lower bound of the even numbered subsequence
which can be done by ?
topaz sinewBOT
@vital moth Has your question been resolved?
vital moth
<@&286206848099549185>
weak ridge
vital moth
this is the given
rigid ivy
@vital moth hello
rigid ivy
Doin fine
vital moth
good to hear that
rigid ivy
I have two ways to finish your proof
vital moth
the one from yesterday ?
rigid ivy
Yeah
vital moth
tysm
rigid ivy
Where did we leave off?
vital moth
yesterday we left off trying to prove that both the odd and even numbered sequences converge to the same limit
we proved that both converge but we still needed this to prove that they are convergent
although it may not necessarily converge since the question was to study convergence in the first place
rigid ivy
Okay. We had a bound for the even subsequence. What was it?
vital moth
Okay. We had a bound for the even subsequence. What was it?
upper or lower ?
rigid ivy
We had $u_{2n}$ in some interval
thorny flameBOT
SWR
vital moth
ah yes give me a second to write it down
$u_{2n}\in (\frac{1-\varepsilon+\sqrt{(\varepsilon-1)(\varepsilon-5)}}{2},\infty)$
thorny flameBOT
vital moth
actually i shouldve said one second to copy it 
vital moth
now that i think about it cant we replace infinity by 3 since u_n=< 3 for all n in N
rigid ivy
now that i think about it cant we replace infinity by 3 since u_n=< 3 for all n ...
We're gonna do even better
Let's leave the even subsequence alone for now and move to the odd
rigid ivy
Let's do the work similar to this, but for odd one
vital moth
ok
${u_{2n+1}}$ converges so it is cauchy and thus $|u_{2n+1}-u_{2n-1}|<\varepsilon\implies u_{2n+1}<u_{2n-1}+\varepsilon\implies 2-\frac 1{u_{2n-1}+1}<u_{2n-1}+\varepsilon\implies u_{2n-1}^2+(\varepsilon-1)u_{2n-1}+\varepsilon-1>0$
thorny flameBOT
rigid ivy
So far so good
rigid ivy
so the first thing written there was wrong ?
I missed whatever it was
vital moth
the last inequality was < instead of >
rigid ivy
The inequality here is correct i think
But some of the algebra that follows May be wrong
Maybe double check
vital moth
i will do that because this should be the same inequality that we had for u_{2n} right ?
rigid ivy
Brb
vital moth
i am done checking and it looks to me like this is correct
but why should it not be the same as that of u_{2n}
after all $u_{n+2}=2-\frac 1{u_n+1}\ \forall n\in\mathbb{N}$
thorny flameBOT
rigid ivy
I am back
rigid ivy
but why should it not be the same as that of u_{2n}
Because one is increasing and the other is decreasing
vital moth
but still both of them are positive
rigid ivy
but their difference of sequential terms are not
vital moth
so where does the inequality have to flip in a different way
rigid ivy
but their difference of sequential terms are not
it flips because of this
vital moth
but their difference of sequential terms are not
but where is the difference
where does this difference appear in the inequality
rigid ivy
$|u_{2n+1}-u_{2n-1}|$
thorny flameBOT
SWR
rigid ivy
(although, you should do $2n+1$ and $2n+3$...)
thorny flameBOT
SWR
rigid ivy
yes. One sequence is decreasing and the other is increasing
one moment
could you show me the work you did for the even subsequence?
vital moth
$2-\frac 1{u_{2n}+1}<u_{2n}+\varepsilon\implies\frac 1{u_{2n}+1}>2-u_{2n}-\varepsilon\implies (u_{2n}+1)(2-u_{2n}-\varepsilon)-1>0\implies u_{2n}^2+(\varepsilon-1)u_{2n}+\varepsilon-1<0$
thorny flameBOT
vital moth
but when i went back to this message after you asked me about the interval
i thought this was wrong
rigid ivy
$2-\frac 1{u_{2n}+1}<u_{2n}+\varepsilon\implies\frac 1{u_{2n}+1}>2-u_{2n}-\varep...
further back than this. How did we get here?
vital moth
ahh wait
ok i see what you are getting at
we started at $|u_{2n+2}-u_{2n}|<\varepsilon$
thorny flameBOT
vital moth
there is a mistake here from the second inequality to the third
so the last inequality should be $u_{2n}^2+(\varepsilon-1)u_{2n}+\varepsilon-1>0$
thorny flameBOT
rigid ivy
the inqualities are tricky
I do them backwards a lot
And they're so dangerous and squares and absolute values
rigid ivy
this is correct for even
vital moth
as for the odd subsequence we have $|u_{2n+3}-u_{2n+1}|<\varepsilon\implies u_{2n+1}-u_{2n+3}<\varepsilon\implies u_{2n+1}-2+\frac 1{u_{2n+1}+1}<\varepsilon\implies\frac 1{u_{2n+1}+1}<2-u_{2n+1}+\varepsilon\implies (u_{2n+1}+1)(2-u_{2n+1}+\varepsilon)>1\implies 2u_{2n+1}-u_{2n+1}^2+u_{2n+1}\varepsilon+2-u_{2n+1}+\varepsilon-1>0\implies u_{2n+1}^2-(\varepsilon+1)u_{2n+1}-(\varepsilon+1)<0$
thorny flameBOT
rigid ivy
yeah this is correct
vital moth
but also almost all coefficients are different
vital moth
ok but before that
since the inequality of the even subsequence changed
isnt the interval of u_{2n} different now
vital moth
it should be between the 2 roots if i am not mistaken
rigid ivy
isnt the interval of u_{2n} different now
From what I checked, the last interval we had was from the correct equation, but you can double check. It can't hurt
vital moth
ah yes you are right
the inequality was written incorrectly but the solution was that of the correct one
vital moth
now from this $u_{2n+1}\in (\frac{\varepsilon+1-\sqrt{(\varepsilon+1)(\varepsilon+5)}}2,\frac{\varepsilon+1+\sqrt{(\varepsilon+1)(\varepsilon+5)}}2)$
thorny flameBOT
rigid ivy
Okay, we only care about the upper bound.
rigid ivy
We already know that every $u_{2n+1}>\frac12$, so the lower bound here isn't helpful for sufficiently small $\varepsilon>0$
thorny flameBOT
SWR
rigid ivy
Okay, we have some good bounds. Now is when the tricky part is gonna start.
Before we continue, do you have all of your work so far? How we're going to progress is going to depend on some other things which may or may not have been proven already. So I want to know what you've proven so far.
rigid ivy
exactly we can replace that bound by 1/2
(not always. It will depend on epsilon. But for small enough epsilon, then yea)
It's why we just discard it overall. We won't care about it
rigid ivy
although it is always < 0
what is always less than zero?
vital moth
the lower bound here
vital moth
so we can always discard it
vital moth
we can always put the lower bound 1/2 too since u_n>=1/2 so no need for values <1/2
vital moth
my own proof missed that
ohh
its probably the first thing you missed in ages 
so what we (you) did so far was find an interval that u_{2n} must lie in
the same thing for u_{2n+1}
we also proved that both of these converge where one increases and the other decreases
and established upper and lower bounds to u_n in general
did i miss anything
thorny flameBOT
SWR
vital moth
why would we want to do that
rigid ivy
because we're gonna want this
vital moth
i still dont see why we need this too but i will probably see in a moment
rigid ivy
because we're gonna want this
this must be true for any epsilon
so it squeezes the series to a single value
vital moth
it squeezes it between 2 different values right
ah no
nvm
now i see
ok lets do it
rigid ivy
Show me how you proved that your even subsequence was increasing
vital moth
Show me how you proved that your even subsequence was increasing
i searched in the channel and it turns out that we didnt prove that
rigid ivy
Yeah I didn't think we did
rigid ivy
I had trouble proving it, so I would've been surprised if you did it easily
vital moth
in that case then i would be surprised if i did it easily too
rigid ivy
I gotta take a ShoWeR
vital moth
in that case then i would be surprised if i did it easily too
you can even remove easily
rigid ivy
I'll bbl
vital moth
I gotta take a ShoWeR
ok take your time
vital moth
Show me how you proved that your even subsequence was increasing
in this time i will try
i am getting that {u_{2n}} is decreasing wth
i am trying induction
the base case is true since u_2=4/3>1/2=u_0
now assume that u_{2n}>u_{2n-2}
then $-u_{2n}^2<-u_{2n-2}^2$ and $\frac 1{u_{2n}+1}<\frac 1{u_{2n-2}+1}\implies -\frac{u_{2n}^2}{u_{2n}+1}<-\frac{u_{2n-2}^2}{u_{2n-2}+1}\implies 2-\frac{u_{2n}^2}{u_{2n}+1}<2-\frac{u_{2n-2}^2}{u_{2n-2}+1}\implies u_{2n+2}<u_{2n}$
thorny flameBOT
vital moth
<:hmmcat:831508801646755850>
i have the same expression tbh
i have to be wrong somewhere
vital moth
but am i
rigid ivy
your flaw is here
$$-\frac{u_{2n}^2}{u_{2n}+1}<-\frac{u_{2n-2}^2}{u_{2n-2}+1}$$
when you multiply by negatives, it flips the inequality direction
thorny flameBOT
SWR
vital moth
when you multiply by negatives, it flips the inequality direction
but i already flipped
originally u_{2n}>u_{2n-2} by assumption
vital moth
originally u_{2n}>u_{2n-2} by assumption
since f(x)=x^2 is increasing for all x>=0 then squaring here doesnt change the inequality
now multiplying by -1 flips
and u_{2n}+1>u_{2n-2} so inverting both sides wrt to multiplication flips the inequality since both are >0
then multiply this inequality with the inequality of squares
vital moth
yooo
if you find this long then you might be more surprised when you know that from here onward is a continuation of something that began yesterday
vital moth
is this a 6hr long help thread
.
vital moth
idk now
idk too i mean the conclusion i reached is definitely wrong
since u_2>u_0 which automatically contradicts my result
@rigid ivy
nvm
i think that i am a bot
like literally
maybe even worse
so thats where i made the mistake
there is no u_{2n}^2
idk why i even included such a thing
so $u_{2n+2}=2-\frac 1{u_{2n}+1}$
thorny flameBOT
vital moth
so $u_2>u_0$ and now assume that $u_{2n}>u_{2n-2}$ then $\frac 1{u_{2n}+1}<\frac 1{u_{2n-2}+1}\implies u_{2n+2}=2-\frac 1{u_{2n}+1}>2-\frac 1{u_{2n-2}+1}=u_{2n}$ and thus ${u_{2n}}$ is increasing
thorny flameBOT
vital moth
now in a similar way, $u_3<u_1$, assume that $u_{2n+1}<u_{2n-1}$ then $\frac 1{u_{2n+1}+1}>\frac 1{u_{2n-1}+1}\implies u_{2n+3}=2-\frac 1{u_{2n+1}+1}<2-\frac 1{u_{2n-1}+1}=u_{2n+1}$ thus ${u_{2n+1}}$ is decreasing
thorny flameBOT
vital moth
**SWR**
i am sorry for doing something so stupid
things were going smoothly until i did that stupid thing
so from here
vital moth
slight correction here , the base case is u_3<u_1 not u_1<u_0
rigid ivy
there is no u_{2n}^2
Ohh right. Haha
Sorry for delay. I was getting food
@vital moth
Okay we have proven increasing and decreasing
Now another crucial step
We must prove $u_{2n}<u_{2n+1}$ for all $n\in\bN$
thorny flameBOT
SWR
vital moth
Sorry for delay. I was getting food
np you can come and go whenever you want
vital moth
more induction
ah yes it follows easily by induction @rigid ivy
vital moth
because we're gonna want this
after that these bounds become clear too
and then the limit also follows since the bounds here are in terms of ε for all ε>0
thus by squeeze theorem they have the same limit which is (1+sqrt(5))/2 and hence {u_n} converges to (1+sqrt(5))/2
this is all right ?
we are done like this
vital moth
if you want we can do this some other time
you probably have things to do
personally, i have nothing
but you probably have
sorry for troubling you for a long time @rigid ivy
rigid ivy
we are done like this
yeah we're looking good
rigid ivy
if you want we can do this some other time
We can make this easy by proving something else first
vital moth
so to sum up the solution of the first problem
first i bound u_n from above and below then consider the odd and even subsequences
i prove that they are monotonic so they converge
next i find the intervals that these subsequences lie in using the fact that they are cauchy by solving the inequality that comes from this fact
after that i prove that the terms of the even subsequence are less than the terms of the odd subsequence
finally i bound these 2 subsequences by bounds that converge to the same value to use squeeze theorem
is this right ?
thats the whole procedure that we did to solve the first problem
is anything missing here
@rigid ivy
rigid ivy
is this right ?
yeah that all looks good
rigid ivy
Not sure if you are allowed to use continuity yet. But it works regardless
vital moth
that f(a_n)→L is clear
since for all ε>0, |a_{n+1}-L|<ε for sufficiently large n
so that a_{n+1}=f(a_n)→L
vital moth
Not sure if you are allowed to use continuity yet. But it works regardless
i probably can but i am not pretty sure
so continuity was given but idk if this exercise assumes this knowledge
anyway assuming continuity f(a_n)→f(L) since a_n→L
vital moth
so that a_{n+1}=f(a_n)→L
and then with this the result that f(L)=L follows
given a sequence ${u_n}$ defined by the recurrence relation $u_{n+1}=(1-u_n)^2$ with $u_0=\frac 12$ and in a previous part i proved that $0<u_n<1\ \forall n\in\mathbb{N}$ \ Study the convergence of ${u_n}$
thorny flameBOT
vital moth
@rigid ivy so thats the question
i copy-pasted itso that it is easier to refer to
so here it is possible to start in the same way as before
consider the even and odd subsequences
so u_1<u_0 and u_2>u_0
so consider the even subsequence
prove that it is increasing by induction
$u_{2n+2}=(1-u_{2n+1})^2=(1-(1-u_{2n})^2)^2$ and $u_2>u_0$, now assume that $u_{2n}>u_{2n-2}$ then $u_{2n}>u_{2n-2}\implies (1-u_{2n})^2<(1-u_{2n-2})^2$ (since $1-u_n<0\ \forall n\in\mathbb{N}$ and $f(x)=x^2$ is increasing $\forall x>0)\implies 1-(1-u_{2n})^2>1-(1-u_{2n})^2\implies (1-(1-u_{2n})^2)^2>(1-(1-u_{2n-2})^2)^2\implies u_{2n+2}>u_{2n}$. Hence, ${u_{2n}}$ is increasing
thorny flameBOT
vital moth
basically the same process with every inequality flipped proves that {u_{2n+1}} is decreasing
along with u_1<u_0
this with 0<u_n<1 for all n in N imply that the odd and even subsequences converge
it remains to check if their limits are the same or no
let $f$ be a function defined as $f:(0,1)\mapsto (0,1),f(u_n)=u_{n+1}=(1-u_n)^2$ and let $l_1,l_2$ be the limits of the even and odd subsequences respectively
thorny flameBOT
vital moth
first of all i need to prove that f is continuous over its domain
hmmm
how do i do that
i am a bit confused
@rigid ivy
how to prove continuity over a whole interval
i take an arbitrary point in the interval
and work with the usual definition of continuity ?
cedar wagon
Continious on each point of the interval
vital moth
ik that but i was asking about how to do that
vital moth
surely you wont be checking at each point separately XD
cedar wagon
Some crazy man propose me this once, now i take the preshot xd
vital moth
that crazy man is still working on that problem
vital moth
that crazy man is still working on that problem
he wants to show you how it is done but if he is in big trouble if the domain was taken to be an infinite set
vital moth
So f(un) = un+1
yes
vital moth
Here f(x) = (1-x)²
and yes
vital moth
can i just say that
cedar wagon
Yes you can
vital moth
Or you not allowed to say that ?
good question
the answer is probably yes
ok that makes it easier
i will try proving continuity when i am done with the question
that will serve as a little practice
cedar wagon
Sure
vital moth
so now f is continuous
vital moth
from here consider {u_{2n}}
then $f(l_1)=l_1\implies (1-l_1)^2=l_1\implies l_1^2-3l_1+1=0\implies l_1=\frac {3+\sqrt{5}}2$
thorny flameBOT
vital moth
i didnt take the other root since it is < 0
cedar wagon
Sure
vital moth
and 0<u_n<1
cedar wagon
,calc (3+sqrt(5))/2
thorny flameBOT
Result:
2.6180339887499
Kekw