imaginary error function i believe
#help-39
quiet tendon
daring wedge
What was that infinitesimal length you were talking abt
quiet tendon
it's just a special name for the antiderivative that we've come up with since this kind of integral shows up a lot
daring wedge
.
quiet tendon
What was that infinitesimal length you were talking abt
what infinitesimal length?
oh
daring wedge
It was the other guy
quiet tendon
well the idea is you approximate the area under the curve using shapes like rectangles or trapezoids, right?
daring wedge
Yeah
quiet tendon
the fewer the rectangles and the wider they are, the less accurate
daring wedge
Yup
quiet tendon
but increase the number and make them thinner, you get better approximation
daring wedge
So the anti derivative method uses very thin shaoea
quiet tendon
in theory, if you could have infinitely many rectangles with infinitely thin width, you'd have a perfect approximation
daring wedge
Shapes?
quiet tendon
not exactly, the antiderivative approach works because of the fundamental theorem of calculus
daring wedge
not exactly, the antiderivative approach works because of the fundamental theore...
What does that say
quiet tendon
very thin shapes work in the case of e^(x^2) even though that doesn't have an elementary antiderivative
quiet tendon
What does that say
that to evaluate a definite integral from a to b of f(x) dx, it's equal to F(b) - F(a) where F(x) is the antiderivative of f(x)?
daring wedge
that to evaluate a definite integral from a to b of f(x) dx, it's equal to F(b) ...
Oh
Ok
I skipped that theorem
Hehe
Anyways thanks, I think my question has been answered
.close
quaint sluice
how did they simplify the denominator ?
pearl pondBOT
how did they simplify the denominator ?
quaint sluice
woops.
light helm
where they multiplied numerator and denominator by sqrt(1-y^2)
for what's happening with the red arrow you drew, they just wrote an additional y' =
lucid moth
pearl pondBOT
lucid moth
I dont really understand what I did wrong
3sin(x/3) differentiates to cos(x/3)
cos(x/3) = 0
x/3 = cos^-1(0)
x/3 = pi/2 + pi(k) and 3pi/2 + pi(k)
x = 3pi/2 + 3pi(k) and 9pi/2 + 3pi(k)
The main operator in your answer is not an equal sign Implies that there is only 1 value for x, so im guessing its just x/3 = pi/2 + pi(k) , but I dont see why
light helm
see what happens when you combine
3pi/2 + 3pi(k)
into a single fraction
lucid moth
(3pi + 6pi(k))/2
lucid moth
see what happens when you combine
3pi/2 + 3pi(k)
into a single fraction
Like this?
lucid moth
and combine the other into a single fraction too
(9pi + 6pi(k))/2
light helm
can you list out the values you get for each one for k=0,1,2,34 (should be enough)
lucid moth
(3pi + 6pi(0))/2 = 3pi/2
(3pi + 6pi(1))/2 = 9pi/2
(3pi + 6pi(2))/2 = 15pi/2
(3pi + 6pi(3))/2 = 21pi/2
(3pi + 6pi(4))/2 = 28pi/2
lucid moth
can you list out the values you get for each one for k=0,1,2,34 (should be enoug...
Like that?
light helm
yeh and for the other one too
lucid moth
(9pi + 6pi(0))/2 = 9pi/2
(9pi + 6pi(1))/2 = 15pi/2
(9pi + 6pi(2))/2 = 21pi/2
(9pi + 6pi(3))/2 = 28pi/2
(9pi + 6pi(4))/2 = 33pi/2
lucid moth
yeh and for the other one too
I see a trend but idk what it means
light helm
notice the values are already present in the first list
so these describe the same set of values
so you don't need to write both
lucid moth
Ohhh
I think I get your point
Would it be different if the question was with sin instead of cos?
Or does this also apply to sin
lucid moth
Hmm ok
I will try one with sin now cause im curious lol
Thank you!
β€οΈ
.close
pearl pondBOT
unkempt cliff
That is my working out and you can see my current answer down the very bottom. I'm probably going to simplify it a bit more once I find the mistake, but thats what I'm sticking with for now
And that gets the same answer as my calculator, except mine is negative when the calculator's is positive. I have no idea why this is the case and any help would be very much appreciated
pearl pondBOT
unkempt cliff
i've got no idea what happened there, i didn't delete the message :/
open ivy
you forgot to reevaluate the bounds of integration when you did the substitution
unkempt cliff
so would that need to happen when i put the integral back in terms of x? I've done this same method for other integrals and gotten the right answer so i'm not really sure what you mean by reevaluate the bounds
pearl pondBOT
Available help channel!
unkempt cliff
**kneeecaps**
im just gonna reply to this so it says the channel is still claimed
open ivy
in a definite integral, when doing a substitution you need to reevaluate the bounds, originally the integration was on $-5\leq x\leq -3.75$, when you substituted $x=5\sin\theta$, and you want to integrate wrt $\theta$, you can't have $-5$ and $-3.75$ as bounds. More specifically you have $-5\leq5\sin\theta\leq -3.75 \leftrightarrow \sin^{-1}-1\leq \theta \leq \sin^{-1}-0.75$ as your new integration bound.
jolly parrotBOT
Crystopher
open ivy
but I see, as you said, you replaced it back later so the bounds should be -5 and -3.75 at the end.
unkempt cliff
so is that part right then since i put it back in terms of x?
or should i have reevaluated the bounds anyway?
open ivy
If we are a bit strict/pedantic, these steps would be wrong since you state that the new integration has bounds which are incorrect as per the substitution, but you seem to replace back later so the answer should not be affected by that.
unkempt cliff
ok, so next time I'll make sure to do those bounds better to make it a bit clearer
but since they don't affect the final answer after I put it back, can you see how I managed to get the negative of the right answer?
open ivy
I think I found it
you thought this probably
$$x=5\sin\theta $$
$$\frac{x^2}{25} = \sin^2\theta$$
$$\frac{x^2}{25} = 1-\cos^2\theta$$
$$1-\frac{x^2}{25} = \cos^2\theta$$
$$\frac{25-x^2}{25} = \cos^2\theta$$
$$\cos\theta = \frac{\sqrt{25-x^2}}{5}$$
jolly parrotBOT
Crystopher
open ivy
but we actually have $\cos\theta = \pm\frac{\sqrt{25-x^2}}{5}$
jolly parrotBOT
Crystopher
open ivy
the error is that it should be $\cos\theta = -\frac{\sqrt{25-x^2}}{5}$
jolly parrotBOT
Crystopher
open ivy
since $y=\sqrt{25-x^2}$ describes the upper half of a circle centered in origin with radius 5.
jolly parrotBOT
Crystopher
open ivy
if $-5\leq x\leq -3.75$ then the range of $\theta$ lies in the second quadrant, where $-1\leq\cos\theta\leq 0$
jolly parrotBOT
Crystopher
unkempt cliff
I get what you're saying there, that makes sense but that still gets an answer different to what I'm getting on the calculator
pearl pondBOT
@unkempt cliff Has your question been resolved?
unkempt cliff
unless, would the $\sin^{-1}$ be negative too? since that gets the right answer, but I can't figure out where the negative comes from there
jolly parrotBOT
kneeecaps
open ivy
hmmm, the expression $\frac{25}{2}[\sin^{-1}(\frac{-3.75}{5})-\frac{3.75\sqrt{25-3.75}}{25}-\sin^{-1}(-1)]$ gives the correct value.
jolly parrotBOT
Crystopher
open ivy
I think here you evaluated $F(-5)-F(-3.75)$ instead of $F(-3.75)-F(-5)$
jolly parrotBOT
Crystopher
open ivy
**Crystopher**
Which would've given this.
unkempt cliff
yep, that seems to be exactly what I've done. I think I must have confused myself by looking at 5 and thinking it is higher from forgetting about the negative
thank you so much for the help here
.close
pearl pondBOT
pearl pondBOT
Available help channel!
wild fable
does path A imply it was launched horizontally with no angle?
pearl pondBOT
wild fable
.reopen
pearl pondBOT
β
wild fable
> When launched horizontally, the golf ball follows Path A
im not sure if i have enough info to solve it
i mean i know sx = 111m
sy = -16m
but thats about it
timid spindle
you know how fast things fall under gravity, and you know how far it had to fall and how far it had to travel horizontally
so you should be able to work out how fast it had to be travelling
wild fable
is this a speed = d/t
timid spindle
is what a speed
timid spindle
right sure
wild fable
cuz i dont think the suvat stuff will help
timid spindle
this is kinematics suvat will certainly help
wild fable
well for this part at least
timid spindle
figure out how long it will take the ball to hit the floor if it started 16m above the floor
wild fable
figure out how long it will take the ball to hit the floor if it started 16m abo...
im trying to use s = ut + 1/2 at^2 but i dont have u and if i go with other formulas i need v which i dont have (this is for y component)
i get same problem with x component
timid spindle
what is the initial vertical velocity of the ball if you hit it horizontally
wild fable
oh zero?
timid spindle
yea
timid spindle
yep
timid spindle
youre done with ux
wild fable
i am?
timid spindle
what is it youre looking for
wild fable
wait so was it just 0 for part a is that it lol
timid spindle
no it was the total velocity
but from the start you knew Uy was 0
and youve just found Ux
wild fable
rite
My time is going to be different from part A right
because its being launched at an angle
timid spindle
yes, now the vertical initial velocity will not be 0
wild fable
alright
timid spindle
althought this is now harder because you dont have theta
wild fable
yeah
timid spindle
but you do have the height of its peak
timid spindle
yep!
so you could try splitting the total time up into 2 times which should be easy to calculate
eh hm not knowing theta is still a pain for the first part
ah one of the other suvats will help
timid spindle
yeah that works, you could have also used s = vt - 1/2at^2
youre currently kinda doing part C, since now you'll be able to get theta
but part B doesnt require you to know theta
wild fable
o yeah
wait so for part b to find the total flight time
do i have to divide the projectile motion into two parts or something?
timid spindle
so you could try splitting the total time up into 2 times which should be easy t...
yes
t_up and t_down for instance
timid spindle
it does yeah
pearl pondBOT
pearl pondBOT
Available help channel!
vast berry
pearl pondBOT
vast berry
then map it
i have to find A^-1 right
burnt dust
no
vast berry
why not
burnt dust
just use A
vast berry
why?
burnt dust
any triangle should work
pick whatever is convenient
there might be a celver way of doing this with detrminants
if you want to look into that
that would just be for theoretical interest though
burnt dust
at this point you know how to do it I think
the rest is just computation
draw diagrams to make sure you are computing the areas of the triangles correctly
burnt dust
sorry I'm not going to check the numbers
maybe somone else can if they want to
just be careful that using A you are mapping area d to 30
vast berry
yep
15=2x, 4=x+y
if i let 15 and 4 be the baes and height of the new triangle
but answer says d = 15
burnt dust
you are probably assuming that the preimage of your triangle is right triangle at the origin
which is not true in general
burnt dust
15=2x, 4=x+y
the way you are using these equations is incorrect too
you are asking what the preimage of (15,4) is
you need to actually draw diagrams
a triangle has 2 points that need the mapping applied to assuming the third point is the origin
vast berry
Oh
ohhh
i get it
so assuming one of the points is the origin, i gotta do A on two points
thanks
.close
pearl pondBOT
pearl pondBOT
Available help channel!
naive dirge
I have a really interesting geometric question.
pearl pondBOT
I have a really interesting geometric question.
naive dirge
In a coordinate plane, there's A(4,3), and an movable point P upon the x-axis, and another movable point Q on the line y=x. What is the minimum value of lineAP + line PQ + lineQA.
Hereβs a simple diagram that I drawn
thick dagger
ur drawing point A on the x=y
naive dirge
WHAT
naive dirge
ur drawing point A on the x=y
what
thick dagger
point a is not supposed to be on the y=x ;-;
naive dirge
I see
thick dagger
its A (4;3)
naive dirge
lets ignore the mistake that I made
thick dagger
anyway i guess the answer would bve
naive dirge
naive dirge
good guess
thick dagger
line ap would be min when ap is perpedicular to x
thick dagger
so is QP
naive dirge
what if these 3 points have to form a triangle?
thick dagger
but I have another question doggy
πΆ πΏ
thick dagger
what if these 3 points have to form a triangle?
the question says "have to" ;-; ?
naive dirge
no, but i just wonder how we find the min then.
then my method would be ideally fit in the question
thick dagger
then my method would be ideally fit in the question
what is it
naive dirge
So I assume these lengths of lines as a,b, and c.
thick dagger
umhm
naive dirge
and a is actually dependent on b,c and the theta.
thick dagger
true
naive dirge
b,c, has the min when the triangle looks like the one in picture 2.
naive dirge
also cos (theta) would be 0
thick dagger
b,c, has the min when the triangle looks like the one in picture 2.
how do u know π€π€
naive dirge
I imagine the situation in my mind and find the possible min.
thick dagger
but can u prove it is min in htat position ;)
thick dagger
looks correct though..
thick dagger
ye in that case u find the min of AP but not the other 2
or aq can be perpendicular to y=x and ap perpendicular to x axis
that d give the min of aq and ap
but not the last
so im not so sure
or maybe its unsolvable π
naive dirge
or maybe its unsolvable π
My heart would sink to my stomach then.
thick dagger
π
naive dirge
<@&286206848099549185>
thick dagger
sometimes things just dont possess a min bro ToT
naive dirge
the triangle does not has a maximum
but it is obvious that it would def possess a min
thick dagger
wait a triangle can still arguable be smallest when it is a straight line
thick dagger
RAGHH
@naive dirge ITS COMMING OUT
just my guess btw
Call aq = a; ap = b ; pq = c
=> c+ b > a
=> a + b + c > 2a
And a is min when aq is perpendicular to y = x
Find aq when that and ull get a + b + c > sth. Not the min but at least we have sth
(This is assuming that a is the smallest side in the 3)
@naive dirge ITS COMING OUT
naive dirge
I'm depressed
thick dagger
Bro why
naive dirge
a p q is a straight line
cause even a p q forms a straight line, the min would not happens
thick dagger
cause even a p q forms a straight line, the min would not happens
Waterflip thats not possible
It should be
thick dagger
How can u tell
naive dirge
5(sqrt2)
thick dagger
And how
naive dirge
I actually been taught some method to find the min tho
but I just forgot its existence.
it says to draw some symmetric point
thick dagger
π
Dang we failed education
No jobs no life living from paycheck to paycheck as dc mods ToT
naive dirge
i;m so sad
thick dagger
i;m so sad
So am i πππ
pearl pondBOT
@naive dirge Has your question been resolved?
thick dagger
@naive dirge dang u was correct
indeed the min as in ur text book would be the 2nd image
naive dirge
You mean the book is correct
thick dagger
.
naive dirge
Thatβs correct?
pearl pondBOT
thick dagger
PLS TELL ME WHAT THE METHOD IS BE4 MY INTERNET DIED @naive dirge
pearl pondBOT
Available help channel!
mellow bluff
I have to find the min and max value of this this function on the interval [1,2].when i do the table with critical points 1 and 2 i get that the min value is e^2 but if I take separately 1<=x<=2 and construct the function from here the min value becomes -2e. The min value should be 0.
spare lark
I have to find the min and max value of this this function on the interval [1,2...
Can you show your work ?
mellow bluff
the table is pretty simple. the function increases until 1, at 3e, decrease to 2, at e^2, and then increases to inf.
1<=x^2<=4
-10<= -5x <=-5
-9<=x^2-5x<=-1
-2<=x^2-5x+7<=6
e<=e^x<=e^2
-2e<=f(x)<=6e^2
spare lark
Ima try
mellow bluff
ty
spare lark
1<=x^2<=4
-10<= -5x <=-5
-9<=x^2-5x<=-1
-2<=x^2-5x+7<=6
e<=e^x<=e^2
-2e<=f(x)<...
You cant add the third line like this
If there is a trouble this is here
Or yeah you can i dont remember
pearl pondBOT
pearl pondBOT
Available help channel!
gilded hollow
In how many ways can the word MISSISSIPPI be arranged such that two S aren't next to each other?
gilded hollow
I have 5 spots where i can place the letters MIIIIPP where i have 1 M, 4 I and 2 P
The total amount of ways that the word MIIIIPP can be arranged would be $\frac{7!}{4!2!}$
jolly parrotBOT
Merineth
gilded hollow
How do i continue?
astral rapids
sorry i cant be of much help today this is what i found
summer imp
Please don't put people's questions in chat gpt
astral rapids
is 3am rn my head on a roll
astral rapids
Please don't put people's questions in chat gpt
ye wont do again but shes been asking for a while so i figured id try it
but i wont do it again
sudden heath
I have 5 spots where i can place the letters `MIIIIPP` where i have 1 M, 4 I and...
ig count the ways the s's can be arranged and multiply
summer imp
I understand. The thing is chat gpt is not made for computation, and makes logical errors almost always, so it's completely unreliable when it comes to maths
astral rapids
I understand. The thing is chat gpt is not made for computation, and makes logic...
ye i gotchu thing is the answer on the back of her book was the same as chatgpt's answer so i thought it would somewhat help
summer imp
ig count the ways the s's can be arranged and multiply
That would be a good approach, find the number of ways the S's can be placed such that no two are next to eachother.
Then you multiply that by the number of arrangements of MIIIIP.
sudden heath
yeah
gilded hollow
That would be a good approach, find the number of ways the S's can be placed suc...
So in this case is 8 spots where it can be placed?
and i'm choosing 4 spots
ooooooh so thaaaat's why we get 8 over 4????
Damn i would love your thought process on how you figured that out
Could this question be solved with Incl/excl ?
sudden heath
ooooooh so thaaaat's why we get 8 over 4????
do you get 8 over 4?
gilded hollow
Yeah the solution is like you said, the total places of where S can be placed (8 over 4) multiplied by the amount of arrangements that the word MIIIPPI can create
That would be combinations multiplied by permutations where the combinations is having a total of 8 but choosing 4, meanwhile the permutation would be 7! / 4!2! where 7 is total divided by 4 I's and 2 P's
summer imp
In any placement of MIIIIPI, you'll get 8 spaces possible between the letters, so you just have to pick 4 of those for the S's. This is why you have 8C4.
sudden heath
In any placement of MIIIIPI, you'll get 8 spaces possible between the letters, s...
ah right
gilded hollow
I still don't understand how you think because i'm getting stuck or getting every single question wrong so far
For example:
I got 10 people, where 7 are men and 3 are women.
a) Make a council of 4 out of them with no restrictions
Shouldn't this just be
$\binom{10}{4}$
jolly parrotBOT
Merineth
summer imp
Yes
gilded hollow
I would like to think so but the answer in the book says 10*9*8'7
5040 is the answer apparently
gilded hollow
$\binom{10}{4} = C(10 4) = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!}$
jolly parrotBOT
Merineth
gilded hollow
This is what i was thinking?
summer imp
If you pick 4 identical balls out of a bag of 10, then yes, 10 C 4
No yeah
It's 10C4
for the council
5040 is too large
Unless the council has a hierarchy
In the sense that the council (Alex, Beatrice, Charlie, Damien) is the same as the council (Beatric, Damien, Alex, Charlie)
So 10*9*8*7 is overcounting
summer imp
It is
gilded hollow
It is wrong or it is 5040?
gilded hollow
The add more conditions after it
"In how many ways can they make a concil if at least one woman is on it"
warm sonnet
i had a similar word problem to this
if there's 3 woman, and there has to be 1, you gotta consider all 3 variations
summer imp
First, pick the necessary woman, and then pick the rest of the council from the remaining people
summer imp
Well pick the woman first, so 4C1, then 9C3.
summer imp
`3*9*8*7`
is the answer apparently
This overcounts again...
gilded hollow
let me post the entire question + answers
i might have made something wrong in translation
summer imp
Ok
gilded hollow
A small association consists of ten people, three of whom are women. At the annual meeting, the association must appoint a board consisting of a chairman, vice-chairman, secretary and treasurer.
a) In how many ways can an association appoint a board?
b) In how many ways can the association appoint a board if a woman is to be chairman?
a answer) 10*9*8*7 = 5040
b answer) 3*9*8*7 = 1512
summer imp
Ah
That's what I meant by hierarchy
Then the council
(Alex, Beatrice, Charlie, Damien) is different from the council
(Beatrice, Damien, Charlie, Alex).
gilded hollow
oooh..........
I see what you mean
So in fact it should be
we are choosing between 4 women
to take 1 spot
that should stil be = 4?
summer imp
Well for a) it should be fairly simple. 10 choices for chairman, 9 for vice, and so on
gilded hollow
makes sense
summer imp
For b), there are 3 women to choose from for chairman, then 9 people for vice, and so on
summer imp
Mandela effect I thought that too hahah
gilded hollow
7 men 3 women lol
Okay i see what you mean now
damn it's so hard to decide when to use the formulas or not
c) it must have only 1 woman
So if i am thikning right
3 women to choose from on 4 spots right?
and then 9 over 3?
summer imp
Well the woman has to be in one of the positions. There are 3 ways to pick that woman, and 4 positions for her.
Then the remaining positions can be dispatched throughout the men
gilded hollow
_ _ _ _ I have 4 spots and i need to choose between 3 women
hmm
this is SOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO confusing
holy shit
It's quite honestly impossible to rationalize
summer imp
You know there is exactly one woman.
She has to occupy one position on a council, and there are 4 possibilities for that.
So if you have 3 women, to pick from
and a position to attribute to her
You have 3 * 4 ways of doing that right?
gilded hollow
Yes
And this isn't by using a formula?
Permutations or kombinations?
This is just rationalized 3 * 4
summer imp
Yeah mostly
gilded hollow
Ok
gilded hollow
7*6*5
Which is the right answer but it's actually so hard to rationalize this
I'm so used to applying formulas
summer imp
Exactly. So once once you've chosen the woman, there are 7*6*5 ways to fit the remaining men.
That yields 12*7*6*5 yeah?
gilded hollow
yeaa
summer imp
Okok
summer imp
I'm so used to applying formulas
Yeah the thing is as soon as things get a bit more complicated then counting balls in a bag or ways of sitting on a bench, you have to do some things without formulae
gilded hollow
d) at least one woman
So first i choose one woman which is 3*4 and then i choose the remaining people (9) on the 3 seats? 9*8*7 ?
gilded hollow
Yeah the thing is as soon as things get a bit more complicated then counting bal...
Yeah you are right, i'll just have to get used to it
summer imp
One way you can help doing this is thinking as the "appointer" and try to fill in an empty council ____.
Then every time you make a choice, count how many people you're choosing from and so on.
summer imp
d) at least one woman
So first i choose one woman which is `3*4` and then i cho...
Yep that's it
summer imp
Oh yeah sorry I think my method here overcounts this time
Because if there is two women for instance, we count the situation twice, one from the POV of every woman getting picked.
Sorry
gilded hollow
Total number of ways - number of ways with only men?
summer imp
Yes that's the best approach
gilded hollow
I hate this 
summer imp
I know :C
gilded hollow
I canβt rationalize it
summer imp
I gtg right now, hopefully with more exercises it will become easier. Hope I didn't confuse you more than anything
pearl pondBOT
pearl pondBOT
Available help channel!
gilded hollow
In how many ways can 8 identical balls be distributed among 4 baskets if
a) each basket must have at least one ball
gilded hollow
Is there no logic behind combinatorics?
night scaffold
search stars and bars
gilded hollow
And how do i know if i should use that, combinations, permutations and so on?
π
pearl pondBOT
@gilded hollow Has your question been resolved?
night scaffold
just build some intuition i guess
gilded hollow
Okay so i just have to memeorize that placing items into a basket is stars and bars?
and memorize the formula for it
to be able to solve problems?
vestal tapir
yes
gilded hollow
Okiii
I do have a quick question tho
In how many ways can 8 identical balls be distributed among 4 baskets if
b) the last basket has an odd number of balls
quiet tendon
https://sites.gatech.edu/math3012openresources/lecture-videos/lecture-3/
if this makes it a bit easier to understand, check out the end of lecture 2 and the beginning videos of lecture 3
it should help you understand how to solve every single one of those questions
gilded hollow
I think i have it figured out but i'm not getting the same as answer as in the book
gilded hollow
https://sites.gatech.edu/math3012openresources/lecture-videos/lecture-3/
if thi...
I'll look at it after i've solved this problem
Either way
If the condition is that the last basket should have an odd number of balls
Does that mean that we have 5 different scenarios
1,3,5,7 balls in the last basket
and the sum of those 4 scenarios is my answer?
quiet tendon
isn't that 4 scenarios
quiet tendon
but yeah you'd consider each scenario case by case, if there's 1 ball in the last basket then you'd count the number of ways to distribute 7 balls among the other 3 baskets
if there's 3 balls in the last basket, count the number of ways to distribute 5 balls in the 3 baskets
etc
gilded hollow
$n = 7 k = 3 \ \
\binom{n+k-1}{k-1} = \binom{7+3-1}{3-1} = \binom{9}{2} = 9C2$
This was my initial guess
jolly parrotBOT
Merineth
gilded hollow
But it's wrong apparently
Should be 9 over 7
and i dont know why
Shouldn't it be k-1 in the bottom part? or n-1
if i use n-1 i get the right answer as in the book but not with k-1
gilded hollow
it should help you understand how to solve every single one of those questions
Hmm I watched it and I canβt really say I understand. This problem isnβt related to mine is it?
quiet tendon
are you sure you watched the right videos? it has to do with distributing objects into bins based on various conditions
and it shows how that can be seen as identical to counting integer solutions to some equation
and it kind of rederives stars and bars
gilded hollow
I dont really understand what he is talking about
"More on Distributing Objects: Good = All β Bad"
is that what you mean?
quiet tendon
https://sites.gatech.edu/math3012openresources/lecture-videos/lecture-2/
the 6th video here, enumerating distributions
https://sites.gatech.edu/math3012openresources/lecture-videos/lecture-3/
and then videos 1 through 3 here
the main idea is the concept of introducing false objects to subtract them off at the end
this is kind of where stars and bars comes from
gilded hollow
Like i dont understand what he is talking about.
I know what the stars and bars concept is, but i can't solve any problems
quiet tendon
okay, consider the case where the last basket has 1 ball
then you have 7 balls left to distribute among the other 3 baskets, where baskets are allowed to be empty
we know that the number of ways to distribute m identical objects into n distinct bins where empty bins are allowed is (m+n-1) C (n - 1)
so this case becomes (7 + 3 - 1) C (3 - 1) which is 9C2
and you have 9C2 different distributions for the case when there's 1 ball in the last basket
when there's 3 balls in the last basket, you'd repeat this but you're distributing 5 balls into the other 3 baskets with empty bins allowed so you get (5+3-1)C(3-1) = 7C2 distributions in this case
repeat this for when there's 5 balls in the last basket and 7 balls in the last basket, and adding up your totals should give you your answer
gilded hollow
we know that the number of ways to distribute m identical objects into n distinc...
over n-1
should be the formula?
gilded hollow
yes
quiet tendon
wait
your original question said that each basket must have at least one ball
is that also true here
gilded hollow
Let me translate entire thing, one moment
In how many ways can eight identical balls be distributed in four different boxes if... a) each box must contain at least one ball b) the fourth box must contain an odd number of balls
so the only condition is b
quiet tendon
a and b are separate problems? or is this one problem where both the conditions in a and b apply?
quiet tendon
your answer key is wrong
the correct answer to a is 7C3
which probably suggests that the answer key is wrong for b too
gilded hollow
**Merineth**
Okay good so this is correc tthen?
quiet tendon
9C2 is correct for the case when there's 1 ball in the last basket
but you also need to consider all cases where there's an odd number in the last basket
so 1, 3, 5, and 7 in the last basket
gilded hollow
Yeah i just apply the same method to 3 5 and 7
quiet tendon
great then that should work afaik
gilded hollow
Tysm π«Ά
Ohwell i'll just pick this up tomorrow since math is suicide inducing
Thanks so much for the help tho π«Ά
.close
pearl pondBOT
random stirrup
the correct answer to a is 7C3
7C4 and 7C3 are the same @gilded hollow
Instead of choosing where the bars you, you are choosing where the stars go
easier to see when you use the factorial definitions
random stirrup
the answer to a) 7C4
the answer to b) 9C7 + 7C5 + 5C3 + 3C1
and the answer for b is the same as 9C2 + 7C2 + 5C2 + 3C2
for the same reasons as above
${n \choose k} = {n \choose {n-k}}$
jolly parrotBOT
vehnil
pearl pondBOT
Available help channel!
signal kernel
pearl pondBOT
signal kernel
can someone please help with this
im starting the hyperbolic functions chapter, and so far all theyve taught us is the exponential definition of sinhx coshx and tanhx
jolly parrotBOT
Dork9399
pearl pondBOT
@signal kernel Has your question been resolved?
fickle schooner
Do you at least know this identity
$$ \cosh^2 (x) - \sinh^2 (x) = 1 $$
jolly parrotBOT
StrangeQuarkAL
signal kernel
Do you at least know this identity
$$ \cosh^2 (x) - \sinh^2 (x) = 1 $$
No, but we're meant to solve it without any identities
pearl pondBOT
No, but we're meant to solve it without any identities
crystal venture
No, but we're meant to solve it without any identities
are you sure? because it would be the easiest way
pearl pondBOT
@signal kernel Has your question been resolved?
pearl pondBOT
pearl pondBOT
Available help channel!
chrome flame
pearl pondBOT
cold oriole
pearl pondBOT
@chrome flame Has your question been resolved?
spiral pivot
Looking at the definition of sigma, I don't see any obvious patterns. There's one orbit of length 2 and one of length 8. You obviously can show this by just working out the 45 different cases, but surely there ought to be a better way.
chrome flame
Looking at the definition of sigma, I don't see any obvious patterns. There's on...
i see
thank you
pearl pondBOT
Available help channel!
ruby trail
Why is it wrong?
pearl pondBOT
Why is it wrong?
ruby trail
I just based my answer on this question. I did not write a proper solution
summer imp
Yeah well in that problem the function is a step function.
In the one you're doing it's the line 1-t
ruby trail
isnt it a piecewise function as well?
1-t from 0 to 1
0 from 1 to +inf
summer imp
Piecewise doesn't mean they're the same
summer imp
I just based my answer on this question. I did not write a proper solution
Try to graph that
summer imp
Why is it wrong?
It doesn't look like that at all
ruby trail
summer imp
I just based my answer on this question. I did not write a proper solution
It's this one you have to graph
If you compute the Laplace transform of a function and then that of a totally different function you'd expect the result to vary.
pearl pondBOT
@ruby trail Has your question been resolved?
midnight haven
Can someone explain law of proportion thing
pearl pondBOT
Can someone explain law of proportion thing
@midnight haven Has your question been resolved?
severe quarry
\begin{problem} Given the subspace $U \coloneqq \l { \mrm{x \ x \ y \ y} ,\middle \vert , x, y \in K \r } \subseteq K^4$, find a subspace $W \subseteq K^4$ with $K^4 = U \oplus W$. \end{problem}
severe quarry
Is there a general method for this?
What if I pick $W = \l { \mrm{0 \ z \ w \ 0} , \middle \vert , z, w \in K\r }$?
limber oasis
Do you think it works ?
severe quarry
Do you think it works ?
From the definition I read on this, we need to show $\mrm{1 \ 0 \ 0 \ 0}$, $\mrm{0 \ 1 \ 0 \ 0}$, $\mrm{0 \ 0 \ 1 \ 0}$, $\mrm{0 \ 0 \ 0 \ 1} \in U + W$ first
jolly parrotBOT
Kepe
severe quarry
Afterwards, we need to show that each v in V can be written uniquely as u + w
So, $\mrm{1 \ 0 \ 0 \ 0}$ is included if we pick $x = 1$, $z = -1$, $y = 0$, $w = 0$
jolly parrotBOT
Kepe
severe quarry
Afterwards, we need to show that each v in V can be written uniquely as u + w
Now this;
Intuitively yes it also fulfills this. For a rigorous argument maybe argue by contradiction?
severe quarry
you don't seem to be struggling
Yeah I'm pretty much just asking because maybe somewhere I need a theorem I haven't seen yet (I'm still not on page 75)
limber oasis
doesn't hurt to prove things straight from the definitions anyway
it is very much so doable
severe quarry
it is very much so doable
Does this look fine?
limber oasis
looks alright to me
pearl pondBOT
limber oasis
do you think you could have done this any other way?
severe quarry
.reopen
pearl pondBOT
β
severe quarry
do you think you could have done this any other way?
Maybe write k with coordinates and then compare
limber oasis
for the second part?
severe quarry
Yes
limber oasis
yes
severe quarry
uniqueness
limber oasis
for the first?
limber oasis
**Kepe**
btw
U n W = {0} is equivalent to the sum being direct
so all the rest was unnecessary once you said this
severe quarry
Write U + W in set-builder notation and mess with it until it's a * (1, 0, 0, 0)+ b * (0, 1, 0, 0) + c * (0, 0, 1, 0), d * (0, 0, 0, 1)
severe quarry
btw
U n W = {0} is equivalent to the sum being direct
Oh
limber oasis
(exercise)
limber oasis
Write U + W in set-builder notation and mess with it until it's a * (1, 0, 0, 0)...
yeah
Here's a theorem you could use to skip either of the steps:
Do you know Grassman's formula?
severe quarry
btw
U n W = {0} is equivalent to the sum being direct
Ok yeah this makes sense becasue then there is no way we can write some element in K^4 in two different ways
severe quarry
Here's a theorem you could use to skip either of the steps:
Do you know Grassman...
No
limber oasis
dim(U+W) = dim(U) + dim(W) - dim(U n W)
Just like set cardinals
Except this one is harder to show
limber oasis
btw
U n W = {0} is equivalent to the sum being direct
by this lemma, you get
U, W are in direct sum iff dim(U+W) = dim(U) + dim(W)
And this holds in general for finitely many spaces
hence you can skip either proof here with this fact
severe quarry
So really we can just say dim K^4 = 4, dim U = 2, dim W = 2 and so we are done?
limber oasis
either:
show the sum is K^4 by saying (x, z, t, y) is given by (x, x, y, y) + (0, z-x, t-y, 0) in U+W
Then sum dim is 4 = 2 + 2 the sum is direct
Or show U n W = {0} and therefore the sum is direct, hence dim(U+W) = 2+2 = 4 and the only 4D subspace fo K^4 is K^4 itself hence it's a complement
conclusion: keep reading them theorems
severe quarry
Thanks!
limber oasis
also feel free to ping me for these
severe quarry
conclusion: keep reading them theorems
Yeah I need to finally get to the chapter these exercises are about 
Thanks!
.close
pearl pondBOT
pearl pondBOT
Available help channel!
knotty urchin
Hi. I solved this question using laws of exponents and then using a u substitution. However I came across a far better method which just involves me substituting x = 0 into the equation. I am not sure I understand how this works. Could somebody please explain?
hollow cobalt
The equation supposedly should work for all values of x, meaning you can plug x = 0 in
In fact, letting x = -1 here would be a simpler approach
knotty urchin
How can we infer that?
hollow cobalt
Ah, I misunderstood the question, nvm; Just divide both sides by 5^x and the x's vanish
knotty urchin
Ah, I misunderstood the question, nvm; Just divide both sides by 5^x and the x's...
yes, that is what I did
However it takes a bit more time than using the substituting an x value trick. I would like to use this approach for all further problems like this (1 equation with 2 variables) but im not sure when to generalise
hollow cobalt
Did they plug in x = 0 without mentioning any justification?
knotty urchin
Did they plug in x = 0 without mentioning any justification?
indeed
The justification is what im trying to understand
hollow cobalt
Well the quickest way you could justify that any value of x works is that all appearances of x disappear after dividing both sides by 5^x, this is why I found the method weird
knotty urchin
ok thanks
pearl pondBOT
@knotty urchin Has your question been resolved?
agile ridge
pearl pondBOT
agile ridge
!status
pearl pondBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
tropic saddle
do you know that you can see row/column operations as the multiplication with elementary matrices?
agile ridge
yes
tropic saddle
then try rephrasing the question in terms of row/column operations
pearl pondBOT
@agile ridge Has your question been resolved?
cursive wraith
how?
You know that you have three standard row/column operations
scaling (Ri <- c*Ri, c different from 0)
swapping (Ri <-> Rj)
and transvecting (Ri <- Ri + a*Rj, a any scalar)
now picture yourself this
if A is the matrix before an operation
and B is the new matrix after some operation (say scaling row i by c)
find the invertible matrix P such that B = PA or B = AP
(you might know them as elementary matrices)
agile ridge
ok so we get the final matrix above by doing row/column operations P and Q to A?
cursive wraith
yes
So Scaling operation is represented by P = Diag(1,1,....,1,c,1,....,1), with c on the i-th column/row
and so if it's a row operation, B = PA
if it's a column operation, B = AP
(it's the same principle for other row/column operations)
Transvection is represented by P = Identity + a*E_ij, where E_ij is the matrix with a 1 on the i-th row, j-th column and 0s elsewhere
And swapping is represented by this
agile ridge
i see....but we dont know what the original matrix A looks like, how can we know what kind of operations it has gone through
cursive wraith
i see....but we dont know what the original matrix A looks like, how can we know...
well
just write A = (a_ij)
agile ridge
ok so we might go through transvection with row operations to turn lower rows into zero and we might go through scaling to turn every a_ij to I
cursive wraith
ok so we might go through transvection with row operations to turn lower rows in...
you know how to do row-echelon reduction right?
after that the matrix is "triangular"
and you transform all the "diagonal" coefficients into 1 by scaling
and then you get rid of the coefficients that aren't on the "diagonal" with row and column operations
agile ridge
ok so we might go through transvection with row operations to turn lower rows in...
so we mainly use transvection and scaling , right?
but I still don't know how to know the specific operations that were done to A
cursive wraith
So Scaling operation is represented by P = Diag(1,1,....,1,c,1,....,1), with c o...
we just went through it
if you do scaling on the ith column
the new matrix is AP for example
say for example I take a 3*3 matrix A at the start
and to row-echelon reduce it
I do:
- Row swapping 1 and 2
- Row 1 scaling by 5
- Column 3 scaling by 1/2
- Column 2 add 5*Column 1
- Row 2 add -3*Row 3
let's name the matrices associated to those operations P1,P2,P3,P4,P5 in the order they've been done
and the original matrix is A
So I start with : Row swapping 1 and 2 represented by P1
since it's a row operation, the new matrix is P1*A
now row 1 scaling by 5, represented by P2
the new matrix is P2*(P1*A)
next is a column operation
so new matrix is (P2*P1*A)*P3
then column
(P2*P1*A*P3)*P4
then row
P5*(P2*P1*A*P3*P4)
And I can check that every matrix is invertible
agile ridge
yeah I know this but how can we use this to solve this q then?
cursive wraith
yeah I know this but how can we use this to solve this q then?
literally what we said
you know there exists an algorithm that takes your m*n matrix A
does row and column operations
cursive wraith
and outputs the (Ir 0, 0 0) matrix where r is the rank of A
cursive wraith
P5\*(P2\*P1\*A\*P3*P4)
but each row and column operation consists of multiplying to the left/right by an invertible m*m/n*n matrix
cursive wraith
so it's done?
you've seen the row/column operations algorithm before right?
agile ridge
yep
btw just, i have another q @cursive wraith
do you still remember you said
#help-21ο½μ리μ€ν¨μΆ©1 message
why do we need to choose any constant between 0 and l that isn't 0?
Are you talking about this theorem?
But even we don't choose l/2, we can still get $x\in (x_0,x_0+\delta)$, $f(x)-f(x_0) > 0$, isn't it?
jolly parrotBOT
nino
cursive wraith
https://discord.com/channels/268882317391429632/903481102591750184/1261338364217...
I'm just using the definition of the limit:
$\forall \varepsilon > 0, \exists \delta > 0, \forall x\in (x_0-\delta,x_0+\delta) \setminus {x_0}, \left|f'(x_0) - \frac{f(x)-f(x_0)}{x-x_0}\right| \leq \varepsilon$
jolly parrotBOT
rafilou2003
cursive wraith
meaning
$\forall \varepsilon > 0, \exists \delta > 0, \forall x\in (x_0-\delta,x_0+\delta) \setminus {x_0}, f'(x_0)-\varepsilon\leq\frac{f(x)-f(x_0)}{x-x_0}\leq f'(x_0)+\varepsilon$
jolly parrotBOT
rafilou2003
cursive wraith
we want $\frac{f(x)-f(x_0)}{x-x_0} > 0$
jolly parrotBOT
rafilou2003
cursive wraith
so we need $f'(x_0) - \varepsilon > 0$
jolly parrotBOT
rafilou2003
cursive wraith
so we choose an epsilon > 0 (that doesn't change)
such that epsilon < f'(x_0)
plenty of epsilon will work
but as an example
the arithmetic mean of 0 and f'(x_0) will
i.e f'(x_0)/2 = l/2
pearl pondBOT
@agile ridge Has your question been resolved?
agile ridge
why do we make the absolute difference between ( f'(x_0) ) and the difference quotient ( \frac{f(x)-f(x_0)}{x-x_0} ) is less than or equal to \varepsilon ?
Isn't the definition of limit $\forall \varepsilon > 0, \exists \delta > 0 \text{ such that if } 0 < |x - a| < \delta \text{ then } |f(x) - L| < \varepsilon where\lim_{x \to a} f(x) = L$?
jolly parrotBOT
nino
agile ridge
**rafilou2003**
Why here we make f(x) as f'(x)?
pearl pondBOT
@agile ridge Has your question been resolved?
cursive wraith
Why here we make f(x) as f'(x)?
?
the limit is L = f'(x_0)
the function we study is g(x) = (f(x)-f(x_0))/(x-x_0)
so |g(x)-L| < epsilon
replace L by its value, g(x) by its value
gilded hollow
In a communication system there are 40 different symbols to send messages. The same symbol may be used several times in a message.
a) How many different messages are there where 25 signs are used?
gilded hollow
Would this be combinations? 40C20?
frosty sedge
40^25?
gilded hollow
yeah 40^25 is right.... how did you know that?
you choose between 40 and then 40 again .. 25 times?
frosty sedge
40x40x40β¦.x40 25 times
gilded hollow
And what would happen if the conditions:
"How many messages with 25 signs are there if 10 of the sylbols are only allowed to be first or last in the message but the other 30 may be everywhere"
frosty sedge
Give me a minute
gilded hollow
2*10^2 + 23^40?
frosty sedge
100 x 10^23
gilded hollow
no wait i'm thnking wrong
quiet tendon
Would this be combinations? 40C20?
for future reference, make sure to remember the difference between a string, a combination, and a permutation
gilded hollow
for future reference, make sure to remember the difference between a string, a c...
Is that not combinations?
quiet tendon
Is that not combinations?
frosty sedge
Is 100x30^23 right?
quiet tendon
this isn't a combination because you can reuse the symbols, and the order of the symbols matter
gilded hollow
I unironically never have understood the difference between Permutations and Combinations and it's been over a year trying to learn discrete math
All i know is that i use combinations when i want to choose/select while permutation is when i want to order/rearrange something?
quiet tendon
both count the number of ways to arrange objects
combinations are used when the objects are indistinguishable
permutations are used when the objects are distinct
for example, i have a club with 100 members and i need to select an executive board of 3 members
in this case, you can't tell the difference between the three positions on the executive board
so if i selected persons A, B, and C to be on the board, it doesn't matter whether i happened to select B, C, A, or C, B, A, or any of that
since you can't distinguish between the executive board members, there are 100 C 3 ways to select the executive board
but let's say i specifically had roles, like president, vice president, and treasurer, then the order that i pick the three people matters
so there are 100 P 3 ways to select the executive board here
gilded hollow
Is 100x30^23 right?
$40^2*30^{23}$
jolly parrotBOT
Merineth
gilded hollow
I still dont get why it's multiplied betwen them
40^2 is for the first and last symbol which makes sense. That gives me 10 less to choose from the other 23
oh and neil btw
quiet tendon
you know how you multiply when counting the number of ways things can be simultaneously chosen?
frosty sedge
Oh yeah I get it thanks
gilded hollow
yeah i think i know what you mean
quiet tendon
in a simpler example, if i had a string of two characters with 40 possible letters for the first character and 40 possible letters for the second, i have 40 * 40 different possible strings
so you multiply
same concept here
40 choices for the first and last letter, 30 choices for the remaining 23
gilded hollow
permutations are used when the objects are distinct
Is it by this reasoning that Permutations are used on words? For example if i had the word algebra which is 7 letters and i want to figure out how many 3 letter words i can make of these it would be 7P3 since alg and agl are unique?
But if we wanted to see how many total combinations where the same letters aren't in the same 3 letter group we'd use 7C3 ?
quiet tendon
i'm not sure if the answer is exactly 7P3, since the A can be repeated and you can't tell the different between aag and aag
i'm trying to remember how that one is done
oh imagine if the As were labelled A1 and A2 so you could distinguish them, then there would be 7P3 possible 3 letter words made from the letters
but since you can't tell the difference between A1A2 and A2A1, you also need to divide by the number of ways to arrange A1 and A2 which is 2!
so i believe it should be 7P3 / 2!
gilded hollow
the answre should be 135
i'm not getting it tho
oooooh
i completely forgot
it should be 7!/2! right?
It's not even permutations?
nvm
quiet tendon
i'm not sure about that, i feel like it's still permutations
7! counts the number of ways to make a 7 letter word out of those 7 letters, assuming the A's were distinct
gilded hollow
7P3 makes a 3 letter word out of those 7 characters
however we have 2 A's like you mentioned
i would imagine it's 7P3 / 2!
but that isn't right
quiet tendon
could this be a case where the answer key is wrong?
hasn't it been wrong before?
assume the A's are distinct, then you have 7 choices for the first letter, 6 choices for the second, and 5 choices for the third, so 7 * 6 * 5 = 7P3 different permutations of length 3
but since the A's aren't distinct, you divide by 2! to avoid double counting the duplicates
gilded hollow
Yeah that's what i was taught
I'd imagine the answer key is wrong again. tchh...
Is it normal for it to be wrong so many times?
quiet tendon
Is it normal for it to be wrong so many times?
depends where it's from lol
if it's just some random internet source or an old textbook then probably
gilded hollow
No okay i see what they did
Yeye gpt is bad blabla lol
but it also got 135
The numbers of 3 letter words using A L G B R E
quiet tendon
wait that explanation lowkey makes sense
either you have all 3 letters distinct or the A has been repeated twice
so 6P3 ways to make it with distinct letters
and then if the As are repeated, you have 5 choices for the unique letter and 3 different positions to place it
hm
gilded hollow
I really really despice combinatorics because everytime there is a question there is something new to learn. Why can't ie be straightforward like everything else?
If i wanted to find out the number of words we can create when we have all 7 letters we get 7!/2! becasue of the two A's.
I fail to see how this can't be applied to 7P3 and then divide by 2! for the same result
It's unironically mind boggeling how complicated this has to be
quiet tendon
i agree lol never liked combinatorics
but then i feel like i could make an argument that there are 75 ways
when you have just one A, you have 3 different places to place the A and 5*4 ways to fill up the other 2 places, so 3 *5 *4
gilded hollow
7P3 would give all combinations possible where double A's are included, right?
quiet tendon
when you have 2 As, you have 3 different ways to place the As, and 5 choices for the remaining letter, so 3 * 5
so why not 3 * 5 * 4 + 3 * 5 = 75
at this point i'm not sure π
gilded hollow
7P3 is 210
but
is it possible to count when the two A's are the same?
What i mean is there a permutation that counts all the
AAL , AAG, AAB, AAR, AAE, ALA, AGA ... and so on
i think so
we have 5 letters
and 3 spots
5P3 ?
no
5^3 = 125
210 - 125 = 85
How is this wrong?
jesus chirst
Or rather 7P3 counts all different arrangements of the letter algebra including the double a's
Can someone help?
"How many 3 letter arrangements can i make of the word algebra " ?
gilded hollow
Yeah but I donβt get where the 15 comes from
gilded hollow
But a wasnβt allowed to be included?
two Aβs
Three letter word where a isnβt allowed to repeat