hey
#help-28
tame notch
tame notch
sorry
quaint prawn
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pearl hinge
hey
pearl hinge
In the expansion of [1 + x² – x³ + x⁴]¹⁰, the sum of all coefficients of powers that are multiples of 3 is
a) 114. b) 228. c) 342. d) 456. e) 570.
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@pearl hinge Has your question been resolved?
hushed briar
verbal musk
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verbal musk
I tried drawing lines from A to the top corners of the square and found its corresponding angle
but im stuck
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old basalt
f
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modern zodiac
modern zodiac
idk why they're dividing by 2! 2!
I just did 6P6 for this
treating all the e 's as one entity
weak iron
you good?
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pale dome
<@&268886789983436800>, just realized this guy spammed this emoji across multiple channels
calm nacelle
<@&268886789983436800>, just realized this guy spammed this emoji across multipl...
dealt with
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mortal sigil
I already know the answer but is there an easier way than doing it brute force?
spiral vigil
not really because carries
hot herald
apart from the issue with using variables to represent different things,
carrying doesn't prevent that from working
rough plaza
apart from the issue with using variables to represent different things,
carryin...
my bad, lemme change
6(1000x+y)=(1000y+x)
mortal sigil
6(1000a+b)=(1000b+a)
a__>__2 then the thing get's ruined
mortal sigil
a__>__2 then the thing get's ruined
wait a minute
let's count a and b as separate digits first
6(10a+b)=10b+a
59a=4b
nvm
I suck at math
mortal sigil
6(1000x+y)=(1000y+x)
You misplaced the y and x
mortal sigil
apart from the issue with using variables to represent different things,
carryin...
what if like y * 6>1000?
rough plaza
(editing is so much fun!)
mortal sigil
since y is three digits
mortal sigil
since y is three digits
so like over 166 then it can't work, right?
oh wait
it still works
🤦♂️
rough plaza
so like over 166 then it can't work, right?
that's a good spot out
that's the reason why we can write that out
like that
because when x is > 166, the term will become 7 digits
mortal sigil
that's the reason why we can write that out
oh y3ah
hot herald
doesn't stop you from setting up that equation, you can put additional restrictions of x,y being 3 digit integers
mortal sigil
doesn't stop you from setting up that equation, you can put additional restricti...
Oh my god
I overthink everything
I really shouldn't brute force things
thanks ya'll
hot herald
prime factorisation would be helpful in identifying such integer solutions
mortal sigil
prime factorisation would be helpful in identifying such integer solutions
if there's a restriction then youonly have tof ind the number that fits the bill
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thorny shuttle
shouldnt h be -ve?
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shouldnt h be -ve?
stiff musk
looks like you're doing a change of variables, h = x + 3, so x = h - 3
stiff musk
x = h - 3, so |x| = |h - 3|
thorny shuttle
umm not sure ill figure it out later thanks anyway
thorny shuttle
h is approaching to 0 from negative side
h is small positive number
so it must be substracted from -3 to reach its lhl right?
thorny shuttle
forget it
narrow helm
so h is obv negative but too small
loud sapphire
need help
loud sapphire
@ornate bobcat
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@loud sapphire Has your question been resolved?
tribal oxide
!show
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Show your work, and if possible, explain where you are stuck.
@loud sapphire Has your question been resolved?
mortal sigil
There are 30 problems in a mathematics competition. The scores of each problem are allocated in the following ways: 5 marks will be given for a correct answer, 0 marks will be given for a blank answer or a wrong answer. Find the minimum number of candidate(s) to ensure that 6 candidates will have the same scores in the competition.
is it 5*30+6-1?
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There are 30 problems in a mathematics competition. The scores of each problem a...
@mortal sigil Has your question been resolved?
summer echo
No, first of all there are 5 * 31 possible scores
and then use pigeonhole principle
Try out edge cases where everyone has scores as different as possible
mortal sigil
No, first of all there are 5 * 31 possible scores
so 5*30+6-1?
summer echo
No, but 5 * 31 + 6 - 1 also wouldn't be correct
with bad luck there could be 5 * 31 * 5 people all divided into groups of 5 with exactly the same scores
mortal sigil
with bad luck there could be 5 * 31 * 5 people all divided into groups of 5 with...
so
is it
156
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ornate bobcat
I think we need probabilities to be able to solve this question.
The probability of scores can be done by the number of ways that someone can solve the test to get a certifician score.
The order of the questions doesn't affect the probability
We can take that number of ways and devid it by the total number of ways to solve the question
The question can be solved true or wrong(or kept blank)
Which gives us 2³⁰ ways to solve the exam
Since we will use combination to solve this question we know the the highest probability that could occur is to solve 15 question out of 30
So the probability will be
15C30 / 2³⁰
$$\frac{\binom{30}{15}}{2^{30}}$$
glossy valveBOT
Sherif Player
ornate bobcat
So this will give us approximately
$14.45%$
glossy valveBOT
Sherif Player
ornate bobcat
So from 100 people there are 14.45 people who will solve 15 question in the exam which is the highest probability in the test
So to ensure that 6 people will be able to score the same score we will divide
6 / 14.45% and approximate this to the nearest integer
Which will give us 42 people needed to make at least 6 participates get the same score
All right ? @mortal sigil
mortal sigil
All right ? <@910426153217581057>
uh
it seems to be pigeonhole principle
imagine a hole
there needs to be at least 6 for each one
so what's max
before 6
5
in each hole
however it's also a trick question, cuz there's also one without hole
I mean
ornate bobcat
there needs to be at least 6 for each one
Umm I don't think I understand this
mortal sigil
Umm I don't think I understand this
there needs to be at least 5
yk what I already understand but it's not statistics dependent
thanks anyways
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ornate bobcat
That answer really doesn't satisfy me
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torn jolt
how do i expand $\sum_{n=k}^{\infty} \frac{1}{n-k!}$
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how do i expand $\sum_{n=k}^{\infty} \frac{1}{n-k!}$
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bearcw
spiral vigil
$\sum_{n=0}^\y \f1{n!} = e$ is a well-known equality
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hayley!
spiral vigil
and if you start writing out terms of $\sum_{n=k}^\y \f1{(n-k)!}$ you'll see that they match what you hav
glossy valveBOT
hayley!
torn jolt
**hayley!**
in this if i give n=1 then "k" is also 1 then the series become 1 + 1 + 1 + ..
or am i wrong
?
torn jolt
then what is the point of "=" sign under summation ?
spiral vigil
it tells you the starting point
torn jolt
oh
spiral vigil
just like in $\sum_{n=0}^\y \f1{n!}$ it tells you that $n$ starts at 0 and goes up to $\y$
glossy valveBOT
hayley!
normal thicket
150x=35mod31
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150x=35mod31
normal thicket
,w 14k+11=47x
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,w 14k+11=47x
glossy valveBOT
normal thicket
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vast fossil
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#latex-testing or #bots next time
normal thicket
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✅
normal thicket
i am doing an old quesiton so this is the question
47x=11mod249
so here i did this
x=k+(14k+11)/47
what should I do next? i have put some random values of k but couldn't get
vast fossil
Just try solving 14k + 11 = 0 mod 47
normal thicket
i got it
vast fossil
I am sure finding the inverse of 14 mod 47 is must easier than finding the inverse of 47 mod 249
normal thicket
it works for 16
but it takes time
i got it 16 more easy way
let me show you
the kast term was 14k+11/47
14x=-11mod47
7x=18m47
x=47k+18/7
x=6k+(5k+18)/7
so here i got 16 easily
wow i solved this question two types
u there bean?
@vast fossil
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austere breach
I’ve been looking into the integral of the function x^x, or e^(xln(x)). It can’t be represented by any elementary functions, so I was looking at the expanded form of the MacLaurin series I got from Wolfram Alpha to see how much of the pattern of the series I can deduce.
The series is just the integration of the e^x MacLaurin series, but with xln(x) substituted in.
So, I’ve been able to figure out that every term contains x^(n+1)/(n+1)^(n+1). It’s also obviously still being divided by n!.
This gives me the exact constant divisor for even inputs, but there’s something strange going on with odd inputs. The divisors are being multiplied by a series of constants that follows no obvious pattern. I have the first several that I’ve found shown below:
- n=1 » 1
- n=3 » 2
- n=5 » 24
- n=7 » 16
- n=9 » 640
- n=11 » 20,736
- n=13 » 7,168
- n=15 » 2,048
- n=17 » 23,887,872
- n=19 » 8,192,000
So, I have no idea what kind of pattern these multipliers are following. The only thing I’ve noticed is that they all seem to be extremely composite. The fact that they aren’t constantly increasing or decreasing and seem to almost jump about randomly and at extreme intervals is a bit disturbing. I would love for anyone to shed some light on where these are coming from and how they would be represented in a summation.
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@austere breach Has your question been resolved?
austere breach
Try oeis.org
Nothing
The closest thing is that the first 4 terms are also the first 4 in the expansion of (1-x)^(-1/x)/e
spiral vigil
try it with the even ones in there as well
austere breach
Ooh ok so that’s interesting
It’s the GCD of (n-1)! and n^n
But it has more terms than there are shown I think. I need to look at this more
So it’s this series but every other term gets thrown out and changed to 1
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slender lynx
Hey I'm a bit stuck on vector projections in three dimensions where two dimensional vectors do not share the same dimensional scope, and I can't find anything online that points me in the right direction for this
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torn jolt
what is mean by " i < j = 1 " ?
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what is mean by " i < j = 1 " ?
torn jolt
shrewd hamlet
4 • -5 = -20
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ripe shale
https://flexbooks.ck12.org/cbook/ck-12-elementary-intermediate-college-algebra/s...
Is it just me or there is a typo. 11(20) is not factor of ac=4(5)=20
ripe shale
why -20?
Basically what it is saying is; a =4, b=8, and c=-5 Multiply ac=4(-5) =-20 and find all factors of -20
-20 = -1(20) or 1(-20) and -20 = 2(-10) or -2(10) and so on
Now you add the factors to match the middle term +8. The only factor that matches the middle term is -2(10) because -2+10 = 8
torn jolt
what about this step, how can I do it?
what about this step
ripe shale
So now you want to rewrite: 4x^2 +8x -5 = 4x^2 +10x -2x -5 Now you want to group
torn jolt
yes the group step is confusing
how it change from the format 4x^2 +8x -5 = 4x^2 +10x -2x -5 to the new format () ()
ripe shale
yes the group step is confusing
You are correct. You have to factor out 2x+5. Let me group. (4x^2 +10x) -(2x +5) now you factor out 2x and you get 2x(2x+5)-1(2x+5) Notice 2x+5 can be factor out.
Let 2x+5 be Y. then you have 2x(2x+5)-1(2x+5) = 2xY-1Y It is more easier to see that you can factor out Y. Now you get Y(2x-1) = (2x+5)(2x-1)
Is that make sense?
torn jolt
yes, I have last question
ripe shale
yes, I have last question
sure
torn jolt
$6x^{2} - 13x -5$
glossy valveBOT
Chocolate
torn jolt
if multiply it should equal -30, and if added it should be -13 right?
ripe shale
if multiply it should equal -30, and if added it should be -13 right?
correct
torn jolt
is it -15 and +2?
ripe shale
is it -15 and +2?
correct
torn jolt
thank you sir
ripe shale
You are more than welcome!
torn jolt
6x^{2} - 15x + 2x -5
torn jolt
$3x(2x-5)$
glossy valveBOT
Chocolate
ripe shale
**Chocolate**
look good
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sacred thorn
Hello, i want some help with a problem i came up with, the set up is in the picture below. There are 2 questions i have about this,
Inside a collection of K objects:
1)what's the least amount of types of orderly techniques can you use to create any orderly techniques?
2) Can you create ALL shuffle techniques out of just orderly techniques? What's the least amount of types do you need to reach all shuffle techniques?
sacred thorn
I first converted the problem into a matrix thingy
After that, i tried to create a T3 with T2 and T4 shuffles and got this
Then i tried to make a T4 with T2 and T5,
I saw a pattern so i tried to make a T5 with just T2 and T6,
So i got this formula from that
Though i don't know if this is true for all n
I don't know how to prove it
Can somebody help me?
zenith kernel
There is a way to select two generators for all n, but I am not sure whether it’s fair not to reveal it to you. Generally it’s either proven or given as a hint in textbooks
People in these channels don’t like giving direct answers , but I feel it’s somehow necessary to give it here, and let you prove it yourself. Because most textbooks directly give this:
Hint: consider ||(1->2,2->3,…,n-1->n, n->1|| and ||1->2, 2->1, anything else remains||
sacred thorn
But i already thought of that, and i got that formula, or are you saying the hint's about something else?
zenith kernel
You need two “shuffle techniques” to generate all “shuffle techniques” right
sacred thorn
That's what i'm trying to prove, yes
zenith kernel
I thought you wanted to create all of them using only (12…n)
So you need to prove any shuffle is product of (12…n) and (12)
sacred thorn
I think i'm confused about your notations, what do you mean by (12...n)?
zenith kernel
1->2,2->3,…,n-1->n, n->1
sacred thorn
Oh
sacred thorn
So you need to prove any shuffle is product of (12…n) and (12)
And yes
Do i need induction?
zenith kernel
Induction might not work here. Do you know that { (i,j) (switching only i and j) : i doesn’t equal j } generate all shuffle?
sacred thorn
So you need a shuffler that shuffles all objects? It doesn't matter where the objects end up?
zenith kernel
I mean any shuffle is product (composition) of shuffles of the form (i,j)
Where (i,j) means the shuffle that only switch i and j, and doesn’t change any other elements
sacred thorn
But that only shuffles 2 elements
zenith kernel
Product of them
sacred thorn
Oh i think i get what you mean
zenith kernel
Like you do (12) first, then do (23) later, in the end 1->3, 2->1, 3->2
sacred thorn
Ok, but what i'm asking is can you use only two types of shuffle techs
zenith kernel
Induction might not work here. Do you know that { (i,j) (switching only i and j)...
Yes. But do you know that, for now, this is true
sacred thorn
Yes
zenith kernel
Okay
Now I use (23)(12) to denote the composition of these two shuffles, (doing (12) first , doing (23) later)
And as a result, {(i,i+1): i from 1 to n-1} generate al shuffles
sacred thorn
Ok, yes that seems to be true
zenith kernel
So now the question becomes
Proving that (i,i+1) can be written as composition of (12…n) and (12)
If you show this, then the proof is complete
sacred thorn
Ok i'll think about this for a sec
zenith kernel
Okay
sacred thorn
Ok i get it now, so if i prove that any shuffle techniques of the form (i,i+1) can be made from (12..n) and (12), i can just chain those to get all the orderly techniques
Right?
zenith kernel
Yeah
sacred thorn
So how do i go about proving this?
sacred thorn
Oh yeah
zenith kernel
So you use (12…n) to “rotate” i,i+1 to be on positions 1,2
Switching 1 and 2 then becomes switching i and i+1
Then “rotate” back
sacred thorn
Why'd you tell me the solution? The rotation part already gave me the answer i needed
zenith kernel
Oh my bad
zenith kernel
That’s answer of both I think
First it answered the second, any shuffle of n elements can be generated by (12…n) and (12). And what you called orderly something, (12…m) for m<=n, are also shuffles
zenith kernel
Shuffling m elements is also a shuffle of n elements, m<=n
Where n-m elements are mapped to themselves
Any shuffle of n elements can be generated by (12…n),(12), so any shuffle of m elements ,m<=n, (viewed as a shuffle of n elements ) can also be generated by (12,…,n),(12)
zenith kernel
Great
sacred thorn
Thanks a ton
zenith kernel
Np
sacred thorn
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summer beacon
I kinda need help on all of them but especially #29 and #30
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I kinda need help on all of them but especially #29 and #30
summer beacon
yeah like they go through each other or something
summer beacon
so would intersect just be like the line is going through the other one but it isnt into 2 equal parts?
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fallow pecan
alr whatd i get wrong
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alr whatd i get wrong
torn jolt
Your vertical asymptote is at -6
fallow pecan
SHEIT
torn jolt
Also, as x approaches -infinity, y approaches infinity
As your values of y become greater as x moves to the left
You made the same mistake again
As x approaches infinity, y approaches negative infinity
fallow pecan
You made the same mistake again
like this?
this is end behavior right
i should look up khan academy
torn jolt
Y approaches positive infinity as x approaches negative infinity
And the domain is (-6, infinity)
torn jolt
<:soobig:726878809202229380>
It is decreasing on its domain
fallow pecan
It is decreasing on its domain
besides that am i all set
torn jolt
I think so.
shadow badger
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shadow badger
just want someone to see if i’m doing these correctly since i’ve spent a while doing them & don’t want the work to be for naught
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ripe shale
B) looks good. C) I got -98, not 64. It could be me.
shadow badger
oh ur right
all the terms are negative
so i’m wrong
i put 81 as positive
-81+1=80 and + 18 is 98
ripe shale
d) I got 171 for constant.
shadow badger
oh ur right again
i fucked up the simple stuff
ty a lot
have a good one @ripe shale
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torn jolt
I need help with b?
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I need help with b?
torn jolt
And if someone could check this for me that would be great
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@torn jolt Has your question been resolved?
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torn jolt
What’s the average rate of change formula? Would it be 42.24-28.89/25-10? I forgot it sorta so I just wanna see if what I put in is correct. This is for** 1 a**
normal thicket
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spiral vigil
i think all options are wrong
x = 0 would result in the identity matrix
which has a determinant of 1
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normal thicket
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tacit rapids
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tacit rapids
why is it saying do not distribute?
it just says simplify
is it wrong if you distribute the denominator?
i really wanna distribute and write it as my final answer
but its saying do not distribute
why?..
torn jolt
you can do whatever you want
tacit rapids
but i just want to know why it's written do not distribute
is it just teacher saying
i am the god
just follow my rules
torn jolt
yeah, kinda
sharp nebula
Why is 1. (a) not
Positive, positive, neg, neg ?
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Why is 1. (a) not
Positive, positive, neg, neg ?
sharp nebula
Ans:
onyx glen
you got your concavity backwards i think
sharp nebula
Above x-axis = pos
Below x-axis = neg
?
onyx glen
...oh ok no i see the issue
you're listing the values of f itself and not f'', the second derivative, as the problem asks.
sharp nebula
Above x-axis = pos
Below x-axis = neg
?
Oh this is not f'' ?
onyx glen
look at concavity!
sharp nebula
Neg cubic graph
sharp nebula
look at concavity!
Oh shit I'm dumb-
sharp nebula
you got your concavity backwards i think
It just hit me what this meant
Thanks so much for waking me Ann :)
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rare quartz
hey friends. this isnt exactly homework but i'm just wondering how you would calculate the sine or cosine of an angle without a calculator
rare quartz
like if i wanted to know sin(48) (degrees) how could i do this?
torn jolt
Sometimes there are some nice formulas that you can use to calculate it
Sometimes there isn't going to be
rare quartz
yea i have unit circle memorized
rare quartz
oh really?
rare quartz
well that's good news. you said there's some formulas that might be handy?
torn jolt
No
rare quartz
no derivation comes to mind. i'd have no idea where to start approximating
torn jolt
I said sometimes those formulas work, sometimes they don't really help you. Your case, they don't really help
sharp nebula
well that's good news. you said there's some formulas that might be handy?
U might be asking for this?
sharp nebula
For example sin(75)
Can be further broken down into sin(45 + 30)
rare quartz
like what if i wanted sin(pi/52)
rare quartz
For example sin(75)
Can be further broken down into sin(45 + 30)
ah this is really interesting
torn jolt
It doesn't matter
Sin(48) is irrational and doesn't have a nice closed form
Whether in degrees or radians
rare quartz
but lets say i wanted just a few degits
sharp nebula
U might be asking for this?
Ye u might be looking for this
torn jolt
You can do Taylor series of sin(x) and calculate up to like idk
Second or third order
If you wanted to approximate
fervent lava
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fervent lava
I tried this rubbish but it did not work
onyx glen
well of course it didn't work -- you didn't plug the thing in correctly
why's the denominator 1/x - 1 ?
fervent lava
do I not do the function thing there?
onyx glen
it's x - 1 not f(x) - 1 ...
fervent lava
but if I had f(x), then f(1/x). I would replace everything with x
onyx glen
but if I had f(x), then f(1/x)
bad notation + definitely not wht you meant...
look, this is simple
f(x) = 1/x
therefore $\frac{f(x) - 1}{x - 1} = \frac{\frac{1}{x} - 1}{x - 1}$. there's no $f$ in the denominator so you don't touch it. understand?
glossy valveBOT
Ann
onyx glen
@fervent lava
fervent lava
just trying to solve it
onyx glen
do you understand what i'm saying here, yes or no?
fervent lava
yes
onyx glen
great
fervent lava
I only replace where it says f(
onyx glen
do you see how to proceed with simplifying $\frac{\frac1x - 1}{x - 1}$?
glossy valveBOT
Ann
fervent lava
I tried making the top x into x^-1, but don't know what to do next
onyx glen
possible but not very helpful.
try multiplying the outer fraction's num and denom by x instead.
fervent lava
do you mean to factorise it?
onyx glen
no
i mean exactly what i said.
go from $\frac{\frac{1}{x} - 1}{x - 1}$ to $\frac{x(\frac{1}{x} - 1)}{x(x-1)}$.
glossy valveBOT
Ann
fervent lava
$1-1x/x^2-1x$
glossy valveBOT
Timsaay
onyx glen
bad!
do it on paper if you do not know LaTeX.
also don't expand x(x-1), it won't help.
fervent lava
oh okay
I take the minus out and cancel them
what I don't understand is why it does not work if i bring x to denominator
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distant trellis
Use the results from the last task a to solve
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Use the results from the last task a to solve
tranquil cave
,rccw
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vast fossil
!status
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What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
3
vast fossil
Could you show your answers and possibly your working as well?
distant trellis
split swift
,rotate
glossy valveBOT
split swift
you've omitted the square roots from the middle part, but the question says you can use part a to do this
so $\sqrt{20}+\sqrt{180}=2\sqrt{5}+6\sqrt{5}$
glossy valveBOT
Desync
tribal oxide
Use question a to resolve it
distant trellis
you've omitted the square roots from the middle part, but the question says you ...
Oh I understand
8V5 right?
split swift
yes
distant trellis
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pale plank
Can someone help me understand this question? I used the simulation to find that it was in the 2nd percentile, but i dont get part b, i guessed that it came from a different population but i dont have any reasoning behind it
light sonnet
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pale plank
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untold vessel
Hii
untold vessel
I need to write the truth tables for this
I already understand how the truth tables work, but idk how it works with this
torn jolt
P = True or P = False
simple
p is true, so p -> p is true and (p -> p) -> p is true
p is false, so p -> p is true and (p -> p) -> p is false
untold vessel
Right
untold vessel
what is the "og"?
"and"
for which one of these can we set biimplication arrows and which ones only implication arrows
my own guess would be that for 1, 2, 3, 4 we can set biimplication arrows
and for x = -1 V x = 5, we can send implication arrows:
x => -1 v x => 5
torn jolt
are you sure with the biimplication between 1 and 2?
torn jolt
and for x = -1 V x = 5, we can send implication arrows:
x => -1 v x => 5
I think they are misunderstanding the question. Ehd, x => -1 doesn't make any sense
x isn't a statement and neither is -1
torn jolt
The implications and biimplications you should be looking for are between equations
lol
untold vessel
oh
untold vessel
not 100% sure what you mean
torn jolt
not 100% sure what you mean
Does the solution x=-1 satisfy equation 1?
torn jolt
its x-2 => 3
Do you understand what => means?
untold vessel
Do you understand what => means?
yeah it means that P implies Q
torn jolt
yeah it means that P implies Q
And P and Q are "..."?
torn jolt
Now explain to me exactly what x-2 is stating
untold vessel
idk why x-2 = 3 isnt a statement
untold vessel
okay
torn jolt
As I've said earlier the arrows are between equations
Take equation 1 and 2 for example
untold vessel
(x-2)^2 and 3^2 are not statements either
torn jolt
Yep
torn jolt
Mhm
torn jolt
so no arrow
Not inside of the statement, for sure
Well uh no that's misleading statements can have arrows inside of them just not this one
So let's take equation 1 and 2
Which are both statements
What kind of implication is there between the 2? Is it just an implication? A biimplication?
untold vessel
what implication there is between equation 1 and 2?
none
since you are squaring the second one
torn jolt
none
There seems to be a misunderstanding
Here's a clearer question, is this statement true:
x-2=3 => (x-2)^2=3^2
untold vessel
Here's a clearer question, is this statement true:
x-2=3 => (x-2)^2=3^2
i dont actually fully understand what implies means in the case of equations.
torn jolt
i dont actually fully understand what implies means in the case of equations.
Think of it this way:
eq A => eq B means that any solution to eq A is also a solution to eq B
torn jolt
No arrow?
torn jolt
Yes but are all solutions to x-2=3 also solutions to (x-2)^2=3^2
torn jolt
so does it go both ways?
You still haven't actually figured out if it even goes one way
untold vessel
all solutions from x-2=3 are solutions to (x-2)^2=3^2
but not all solutiosn to (x-2)^2=3^2 are for x-2=3
untold vessel
aha
torn jolt
There's a good reason this stuff is teached
untold vessel
but there is a biimplication between 2 and 3 then
torn jolt
If between all your equations there were biimplications then you know all the solutions you get at the end are correct
untold vessel
and same goes for 3 and 4
last line here is just a solution
4 to 5 has no biimplication or even implication, right?
torn jolt
But if what you get is a mix of biimplications and implications between an equation and the one that comes just before it then you know all the solutions to your first equation are there but you don't know if all the solutions you get are solutions to your first equation
torn jolt
4 to 5 has no biimplication or even implication, right?
Is what makes you say that that statement 5 does not look like your typical equation?
untold vessel
Is what makes you say that that statement 5 does not look like your typical equa...
well its just a solution
i dont see why 4 to 5 should have arrows
i guess solutions are the same?
torn jolt
well its just a solution
It's also a statement about x
untold vessel
It's also a statement about x
that it can be -1 and 5
torn jolt
What they all have in common is that they're all statements about x
So all of them can potentially have implication or biimplication arrows between themselves
untold vessel
all of line 5 is 1 statement right? not two ?
torn jolt
Does statement 4 imply statement 5 and vice versa?
torn jolt
all of line 5 is 1 statement right? not two ?
Yes
untold vessel
Does statement 4 imply statement 5 and vice versa?
well, yes, kinda
torn jolt
If A is a statement and B is a statement then A or B is a statement
torn jolt
well, yes, kinda
"kinda"?
This isn't the continuum hypothesis
Either 4 implies 5 or it doesn't and vice versa
untold vessel
Either 4 implies 5 or it doesn't and vice versa
im not 100% sure
torn jolt
im not 100% sure
What are all the values of x that satisfy statement 4?
untold vessel
What are all the values of x that satisfy statement 4?
x=-1 and x=5
so the ones in statement 5
torn jolt
so the ones in statement 5
Yep, they have the exact same solutions
untold vessel
so its a double arrow
torn jolt
Yep
If x is a solution to statement 4 then it is a solution to statement 5 and vice versa
So there's a biimplication between them
So we have:
1 => 2 <=> 3 <=> 4 <=> 5
Which means 1 => 5
torn jolt
It's => not -> but yeah
torn jolt
Yes
torn jolt
If you get <= you know something went wrong
Think of it this way:
=> sometimes adds solutions (this is where the x=-1 came from here)
<=> keeps all the solutions
<= sometimes removes solutions
untold vessel
untold vessel
Think of it this way:
=> sometimes adds solutions (this is where the x=-1 came f...
gonna save this in my notes
torn jolt
If you get <= you know something went wrong
Correction: I guess <= is fine but you might miss solutions
untold vessel
in the one i just posted, there is <=> for all
becasue x = 5 is the only solution, even to the last one
right?
torn jolt
I was gonna say 2 to 3 is an implication because of the squaring but that's not the case here
torn jolt
right?
right
Typically squaring adds solutions because (-x)^2=x^2 but if the thing being squared has to be 0 then it's fine
(-0)^2=0^2 isn't an issue since -0=0
So yes
<=> everywhere
=> sometimes adds solutions
This is why you always have to check that the solutions you got are all valid
@untold vessel Anything else?
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pale shadow
okey so on the right side the angle all the way down
i got all the other angles right
nvm just got it
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lavish knoll
tribal gust
Linear algebra question:
Prove that the following system in the unknowns x_i and the real coefficients ai,bi,ci,di has a non trivial solution using an argument about linear independence in R^3.
a1x1 + b1x2 + c1x3 + d1x4 = 0
a2x1 + b2x2 + c2x3 + d2x4 = 0
a3x1+b3x2+c3x3+d3x4 = 0
tribal gust
I genuinly have no idea how to go about this. Since in the system AX = B with A the coefficient matrix. Then since B is the zero vector it is definitly an element of the column space and so there exists a solution X so that AX = B. Idk how to show that that X is not the zero vector tho
onyx glen
consider this system as the equation $x_1 \bd{a} + x_2 \bd{b} + x_3 \bd{c} + x_4 \bd{d} = \bd{0}$, where $\bd{a} = (a_1, a_2, a_3)$ and likewise for the other letters
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Ann
tribal gust
consider this system as the equation $x_1 \bd{a} + x_2 \bd{b} + x_3 \bd{c} + x_4...
Im honestly not sure how to continue, my brain is a bit overwhelmed rn.
I get that i should check for linear independance of a,b,c and d but idk how
onyx glen
is there a reason why you are doing math in a bad brain state
tribal gust
bcs exam friday haha
tribal gust
Yeah haha, uni doesnt gaf sadly enough
onyx glen
I get that i should check for linear independance of a,b,c and d but idk how
they are vectors in R^3 and there's 4 of them
tribal gust
oh
That makes alot of sense haha
Slightly related question, given a set of vectors (like here a,b,c,d), is there an algorithm to check if they are linearly independant
onyx glen
yes it's called gaussian elimination
tribal gust
Ahh okey i thought that too but it doesnt work here, im guessing because a,b,c and d are unspecified?
onyx glen
you don't need any computational algorithm here
tribal gust
Yeah sure but imagine i applied the usual algorithm, we need coordinates for that right?
onyx glen
yes
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barren iris
For finding the area between two cruves,
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barren iris
I'm sort of confused by how you select the upper and lower curve or the right or left
timid girder
Yeah
barren iris
Especially for when I need to do this for the y axis.
I'm also confused about setting the bounds for the definite integral.
sleek apex
this comes with practice
just do examples and review a solution to make sure you did it correctly
post problems here and we can walk you through it if you do not understand
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barren iris
post problems here and we can walk you through it if you do not understand
I'll post an example
I think I did the second page wrong. But I'm fine with going over the first page first.
At least for the bounds, I was told to set the equations equal to eachother to see where they intersect. Does that sound right so far?
proper hawk
For the last part shouldn't it be f^-1(y)?
barren iris
For the last part shouldn't it be f^-1(y)?
I'm assuming you're looking at page 2?
proper hawk
Yes
proper hawk
y=f(x) so for y integral x=f^-1(y)
proper hawk
Both
proper hawk
I'm talking about the last part
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distant halo
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distant halo
I tried to get the x value of the x axis
by finding the midpoint of a and c
but cant i also use the x value of b
.x
buoyant wadi
Only if A and C have the same Y value, which they don't in this case
distant halo
What do you mean
buoyant wadi
Well if A and C have different Y values, the line between them isn't horizontal
So the line between O and B won't be vertical
buoyant wadi
Well the midpoint can still be found, but it won't be vertical to B
distant halo
it seems so
void nova
Have you drawn it?
buoyant wadi
I'm not sure it is, note that C is at (4,-1) not (4,1)
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feral widget
The definition of an injective function is : {x1, x2} element of A, f(x1) = f(x2) implies that x1 = x2.
My teacher say that x^2 in an injective fct BUT if
x1 = -1 and x2 = 1
=>
f(x1) = 1 and f(x2) = 1
=>
f(x1)=f(x2) BUT x1 ≠ x2 so x^2 inst a injective fct ?
scenic sundial
what's the question?
feral widget
what's the question?
look
scenic sundial
f(x) = x^2 is not injective using that definition, which looks right
unless perhaps the domain was restricted?
feral widget
yes but check the internet they say x^2 is injectiv
scenic sundial
can you show me a place that says that?
scenic sundial
yes, they did
feral widget
but on my note x^2 say that x^2 is injectiv so my note are false ?
scenic sundial
f(x) = x^2 for all x >= 0 is injective
pseudo cape
The definition of an injective function is : {x1, x2} element of A, f(x1) = f(x...
[f: \mathbb R \rightarrow \mathbb R, \hspace{1cm} x \mapsto x^2] is not injective, that's correct, and most people don't say that, see https://math.stackexchange.com/a/1919990 for example.
We could make it injective if we restrict the domain to $[0; \infty]$, for example.
scenic sundial
but on my note x^2 say that x^2 is injectiv so my note are false ?
yeah that's false
pseudo cape
and if we restrick the domain to R+ to R+ the fct is bijective
If we restrict the domain to R^+, yes
feral widget
and if we dont restrick the domain, i can already say that x^2 is a surjective
pseudo cape
You don't say "R^+ to R^+" when talking about the domain
scenic sundial
and if we dont restrick the domain, i can already say that x^2 is a surjective
that sounds wrong
surjective means the domain is the same as the codomain, right?
pseudo cape
surjective means the domain is the same as the codomain, right?
image is the same as the codomain
scenic sundial
image?
pseudo cape
range/image, yes
pseudo cape
what is the codomain ?
If we have a function [f: \mathbb N \rightarrow \mathbb R] for example, then $\mathbb R$ is the codomain. It's a set containing the image.
If the codomain is the same as the image, then the function is surjective
(So the function takes on all of the elements in the codomain)
pseudo cape
so if all the element in R have one or more pre-image in N by f ?
Sounds right, yeah.
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pseudo cape
so if all the element in R have one or more pre-image in N by f ?
Though can you specify what you mean with "all the elements in R"? Do you mean them as a set?
Or do you mean them as single elements
feral widget
single like -1 ; 2 ; 12 ; sqrt4
pseudo cape
single like -1 ; 2 ; 12 ; sqrt4
For the pre-image, you need a set though
pseudo cape
single like -1 ; 2 ; 12 ; sqrt4
I guess you could create sets with only single elements
feral widget
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pseudo cape
I guess you could create sets with only single elements
What you said should be correct, then.
The pre-image of a set $M$ is defined as [f^{-1}(M) = {x \in A \mid f(x) \in M},] for $f(x)$ to be in $M$, it needs to take on that single value.
feral widget
i mean x^2 is surjective coz the element 1 in the arrived set have 2 image in the starter set wich are -1 and 1
i m not sure i understand the question
pseudo cape
i mean x^2 is surjective coz the element 1 in the arrived set have 2 image in th...
Yes, all elements in the codomain need to get mapped to by atleast one element from the domain for the function to be surjective, that's correct.
pseudo cape
i m not sure i understand the question
I was talking about your pre-image definition of surjectivity, which should also be fine.
feral widget
and injective in the inverse all elements in the codomain need to get mapped to 0 or 1 element from the domain
feral widget
I was talking about your pre-image definition of surjectivity, which should also...
okayyy
pseudo cape
and injective in the inverse all elements in the codomain need to get mapped to ...
Yes
At maximum 1 element from the domain
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brave blaze
what have you tried ? 🤔
queen pebble
Using the table and the quadratic function to find a and b
As and a and b of the function
queen pebble
what have you tried ? 🤔
I know that c is -3
brave blaze
I know that c is -3
yes
queen pebble
And b= -2a
brave blaze
hmm you sure ?
queen pebble
I think
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sharp fable
Is this right?
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Is this right?
light saddle
No?
sharp fable
what is wrong?
vast fossil
The amplitude is 3 indeed
sharp fable
The amplitude is 3 indeed
thank you.
light saddle
Ok but still the period
vast fossil
But yeah you gotta find the period
sharp fable
how do I find the period?
light saddle
You know what a period is right
sharp fable
yes.
light saddle
So you look at a convenient start and end for one period
And then you look how long it is
Look at the lowest and highest points in terms of x you have
Does it make a period
sharp fable
Does it make a period
no.
light saddle
How
sharp fable
the function doesn't repeat at the lowest and highest points, if that's what you're asking.
light saddle
No a period is like when it has the same curve over and over again
sharp fable
right.
sharp fable
Yes, i understand the graph repeats.. but how do we find the period?
vast fossil
Alternatively you could just look at the distance between the peaks
sharp fable
We're only given 5 points.
light saddle
Look at the blue part
sharp fable
Alternatively you could just look at the distance between the peaks
But there aren't any values in the peaks.
light saddle
Good
vast fossil
But there aren't any values in the peaks.
Ah I thought that desmos graph was yours nvm
sharp fable
-0.642 and 0.142 right?
light saddle
Close but not quite
The period ends after 0.142
If not it doesn’t make the whole curvy thing
The tip of the blue part is on the biggest x you have for a point
light saddle
The one that starts the blue part and the one that ends the blue part on this
sharp fable
The one that starts the blue part and the one that ends the blue part on this
but that isn't a full period?
light saddle
It is
sharp fable
isn't it trough to trough?
light saddle
Wdym
sharp fable
like peak to peak.
light saddle
It can start anywhere as long as you finish at a point that begins an other curve like the one you just made
light saddle
0.142 isn’t far enough
It doesn’t make the whole curve
You’ll be missing the part under y=0
sharp fable
is this not a repetition?
light saddle
-0.642 isn’t the first circle you made