#discrete-math

1 messages ยท Page 133 of 1

pale epoch
#

so it's not a tree

#

and i don't think this is even important for the question

brazen tendon
#

I feel like one class can have 2 prerequisites though

#

like

#

For Differential Equations you might need to complete

#

Physics 2 and Math 2

#

together

pale epoch
#

yes, but this is allowed in this solution

brazen tendon
#

what

pale epoch
#

what do you think disallows it?

brazen tendon
#

Oh I am just talking about the persons answer

#

I sent

#

so it can be

#

Directed Multigraph

pale epoch
#

what is a multigraph according to you?

brazen tendon
#

Like a tree but not really

#

disconnected tree lol

pale epoch
#

that's a forest

brazen tendon
#

:&&&

#

Yes

#

I will make my graph a forest

pale epoch
#

a multigraph means there can be multiple edges between 2 vertices

#

i don't think this would make sense for this problem though

brazen tendon
#

Oh you mean like

#

this=?

pale epoch
#

yes

brazen tendon
#

ok that would make no sense

#

So it is a simple directed graph

pale epoch
#

ye

#

and it's cycle-free, which makes it a forest

brazen tendon
#

lemme write my answer and post it if u can wait

#

is an "arrow" a correct term to write

#

in a directed graph

#

like

#

"If a vertex doesn't have any arrows pointing to it"

#

@pale epoch

pale epoch
#

don't think i ever heard that term

#

but it's fine i guess

#

alternatively this statement is equivalent to a vertex having indegree 0 (if you know what indegree is)

brazen tendon
#

I know a degree

pale epoch
#

yeah, so in a directed graph you have both indegree and outdegree

#

outdegree is the number of edges "pointing away", indegree "pointing to" a vertex

brazen tendon
#

I wrote this

pale epoch
#

seems fine i guess

brazen tendon
#

whole answer

pale epoch
#

i'd write (indegree 0) and (outdegree 0)

#

because thats what it is you explained

brazen tendon
pale epoch
#

i meant the thing in brackets

brazen tendon
#

oh

pale epoch
#

no arrows pointing at it is equivalent to having indegree 0

brazen tendon
pale epoch
#

ye

brazen tendon
#

ok

#

or should I just say

#

u and v

#

to vertexes

#

instead of using "other course" "this course"

pale epoch
#

seems like a language problem, not a math one

#

i.e. it doesn't matter

brazen tendon
#

๐Ÿ˜„

#

I will keep it like this I guess

#

I only have 1 question left to finish my exam, about probability, can you help me with that as well?

pale epoch
#

you can ask in the appropriate channel i guess, someone will help

brazen tendon
#

I asked but no answers rip

pallid lintel
#

does anyone wanna see a picture i made of a star hypergraph? its 2 coloured ๐Ÿ™‚

#

i havnt put vertices in it actually

last sigil
#

Cool!

jaunty carbon
#

What does the x operator stands for in discrete maths. For example if G1 and G2 are groups, then what is G1 x G2. Are it all the couples (a,b) with a an element from G1 and b an element from G2?

last sigil
#

x is usually the Cartesian product

#

But for groups, A x B is called the direct product of A and B(which is really just a generalization of the Cartesian product under a different name)

#

You got the idea of what elements are in it correct (a,b), a in A, b in B, but also consider what the binary operation is now

sleek swallow
#

@jaunty carbon

errant bear
#

am I missing smth, because the area of these squares is less than half of the actual area

#

so how is this an accurate approximation

#

like, we get 4 squares from the lattice points

#

actually the approximation is worse, in this case its less than 1/4 of the aactual integral

tulip ravine
#

so here n=8 right? For the integral, I get 16.6, for the sum I get 20

#

that's not too far off

errant bear
#

wait

#

I though it was counting the area of the lattice points squares

tulip ravine
#

yeah, so the tau sum is exactly the area of the lattice point squares

#

and the integral approximates taht

errant bear
#

uh

#

how

#

thers only 4 lattice points

tulip ravine
#

there are 20

errant bear
#

wait its counting all of them

#

not just for xy = 8

tulip ravine
#

all of the below the curve, but not on the axes

#

that's correct

errant bear
#

bruh

tulip ravine
#

the ones ON xy=8 are counting tau(8)

#

which is 4

#

but if we count the points (x,y) such that xy <= 8, then we pick up not just tau(8), but tau of every number less than 8.

errant bear
#

yeah, I though thats what it was talking about when it say "area of this region"

#

yeah I get that

tulip ravine
#

yeah, so we're approximating the sum by the integral

errant bear
#

yeah

#

and itl get better as k increases for xy=k

#

so they are asymptotic

#

got it

tulip ravine
#

yeah, the problem is at the boundary

#

and the boundary is of lower order of magnitude than the area

#

roughly

errant bear
#

yes

tulip ravine
#

like the square corresponding to the point (1,7) is almost entirely lost, for example

errant bear
#

mmhm

tulip ravine
#

btw @errant bear, I think the #advanced-number-theory channel is perfectly appropriate for your analytic number theory inquiries, if you have been hesitant to use it.

errant bear
#

ah yeah, I just don't know like when the "bridge" into higher NT happens

tulip ravine
#

I think it definitely counts. But even literally the phrase "elementary number theory" means number theory that doesn't use analysis. So analytic number theory is definitionally not elementary haha.

#

Here I think it mostly just a matter of level, but you're definitely at the higher level.

errant bear
#

hmmm alright, and I'm taking my first analysis class this upcoming semester so I'm not aware of like, what specifically counts as analysis lol

tulip ravine
#

analysis is just what mathematicians call calculus to feel better about themselves

errant bear
#

lol

tulip ravine
#

in an honest world, the real numbers would be called the hard numbers, the complex numbers would be called the easy numbers, and then we'd have calculus, easy calculus, and hard calculus as the names of the coruses

errant bear
#

hmm

errant bear
#

just watched your vid btw, and realized I did a bunch of the log stuff the other day lol

#

also, at times it was kind of slow ish ngl but it was good

tulip ravine
#

thanks! Yeah, I'm still finding the exact level/speed. But I definitely want "people with PhDs in non-math technical fields" to not feel left out.

#

and given you've been reading about this exact topic, you definitely have a leg up on understanding it quickly!

errant bear
#

true, I would definitely need more time to process things if I wasn't already doing stuff similar to it rn lol

tulip ravine
#

also, just generally, I was always intimidated by analytic number theory, so I really errored on the side of convincing students they shouldn't be.

vapid light
#

are you a prof zeta

tulip ravine
#

I am, yeh

vapid light
#

wow

#

thats really cool

tulip ravine
#

thanks!

vapid light
#

what do you research

weary tiger
#

A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?

#

how can i do this problem

stray reef
#

tree diagram, probably

#

wait

#

hang on

#

this is more complicated than i thought

#

or maybe it isn't.

weary tiger
#

@stray reef ??

sleek swallow
#

Two cases to consider; first case is where there are no green chips drawn. Second case is where there's one green chip drawn. The first case will be easy, the second case will be a little difficult but still doable.

#

@weary tiger

stray reef
#

hold on, lemme try to format this cleanly

#

that may prove challenging

#

bah

#

i have an idea but i can't seem to phrase it in a way that would both satisfy me and be understandable to others

halcyon ledge
#

I liked the tree idea

sleek swallow
#

Urm are you trying to do this elegantly, without breaking it down into cases?

stray reef
#

i'm trying to consider all the possible states of the bag throughout the process basically

#

basically as a markov chain but w/o explicitly calling it a markov chain

sleek swallow
#

uhm i'm not sure that's necessary for this problem but if it gives an elegant solution, then that'd be nice

halcyon ledge
#

thats why i liked the tree idea

#

the set is small enough that it can be used to prove it and it shows the solution clearly

stray reef
#

here

#

each circle represents a possible state of the bag

#

the numbers above the arrows represent the transition probabilities (and all unmarked transition probabilities are 0)

weary tiger
#

@stray reef wait what

#

is this the only way to do it? why is it complicated??

halcyon ledge
#

its not the only way but it shows you whats happening

sleek swallow
#

you can just break it down into cases, like i described above

stray reef
#

i tried my best here

sleek swallow
#

ann's diagram is a nice way of looking at it

stray reef
#

i'm trying not to use any big words here

#

but essentially what i'm doing is starting out with a bag containing 3 reds and 2 greens and then simulating drawing the chips from it one by one

#

so each circle represents a possible state of the bag (ie its contents) and the blue numbers are the probability of ending up with a bag with that content after 0, 1, 2, etc draws

#

blanks for 0

#

@weary tiger does this make any sense whatsoever

weary tiger
#

i guess so

#

@stray reef

#

it's just a lot more than i expected for this problem

#

like

#

how could you solve this on an exam

stray reef
#

ยฏ_(ใƒ„)_/ยฏ

#

i'd probably end up doing sth similar to this diagram

#

maybe in not as much detail but still

halcyon ledge
#

in an exam you just use your head

#

you have 5 stones

#

two green and 3 red

#

so you have $\binom{5}{2}$ possible combinations

vital dewBOT
stray reef
#

combinations of what

#

of two stones remaining?

#

you can't guarantee it'll always be two stones at the end

halcyon ledge
#

total patterns at the end

#

no thats why you divide

#

thats your total

stray reef
#

i'm not convinced

#

drawing two green stones ends the process before you can draw a third, and drawing all but one of each still has you draw another stone and only THEN stop

halcyon ledge
#

$\binom{4}{3}$ thats the amount of patterns with three green and one red

vital dewBOT
stray reef
#

what are your patterns bruh

halcyon ledge
#

those are the patterns we want

#

red and green stones bro

stray reef
#

can you list out all of the patterns

#

one by one

#

cause until i see that list i'm really not convinced at all

halcyon ledge
#

red red red green green

#

red red green red green

#

red red green green red

#

red green green red red

#

red green red green red

#

red green red red green

#

green green red red red

#

green red red red green

#

green red green red red

#

green red red green red

sleek swallow
#

Okay, this isn't that complicated. There are two cases here.

  1. You draw 0 green chips.

So, that's the case, then the process ends in 3 turns. Hence, the probability is just:

$P_1 = \frac{3}{5} \cdot \frac{2}{4} \cdot \frac{1}{3} = \frac{1}{10}$

The probability works out well in that case. Now, suppose you took out 1 green chip. So, you have to make 4 draws in total AND one of them has to be green. If you draw 3 red chips, then the process ends and we go back to the case above.
With this in mind, we can split this into 3 further cases:

Case 1: You drew the green chip first.

$P_2 = \frac{2}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{10}$

Case 2: You drew the green chip second

$P_3 = \frac{3}{5} \cdot \frac{2}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{10}$

Case 3: You drew the green chip third

$P_4 = \frac{3}{5} \cdot \frac{2}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{10}$

We've exhausted all cases at this point. So, $P_1+P_2+P_3+P_4 = \frac{2}{5}$

halcyon ledge
#

red gren red green red

stray reef
#

ok aight @halcyon ledge i think i see your point

vital dewBOT
halcyon ledge
#

@stray reef rofl

weary tiger
#

ngl i am lost rn

sleek swallow
#

Just look at my solution above and ask questions if you have any

stray reef
#

you're presented with three different approaches

#

well

#

two and a half

halcyon ledge
#

๐Ÿ˜ฆ

stray reef
#

to expand on deekaan's idea

#

you basically ignore the "stop drawing once one color runs out" requirement and just keep drawing until the bag is empty and record the order in which you drew your stones

halcyon ledge
#

$\frac{\binom{4}{3}}{\binom{5}{2}} = \frac{2}{5}$

vital dewBOT
halcyon ledge
#

its sexy

stray reef
#

and then count how many of your 5C2 draw sequences satisfy the "red run out before greens" requirement"

sleek swallow
#

@weary tiger I know you're lost but please, at the very least, say something or ask questions.

weary tiger
#

I don't understand deekan's approach at all

#

and in your solution

#

@sleek swallow your like

#

hmm

#

well i guess your solution makes sense

sleek swallow
#

i'll let deekan explain what he's doing

#

If you have any questions about what I've said in that solution, then go ahead and ask.

halcyon ledge
#

you just take the five chips and arrange them in all the possible patterns

#

so thats $\binom{5}{2}$ patterns

vital dewBOT
halcyon ledge
#

now a lot of these patterns aren't legal because of the restriction

#

so we want to find all the legal patterns

#

which is where the $\binom{4}{3}$ comes from

vital dewBOT
weary tiger
#

what does 4c3 say

#

@halcyon ledge

halcyon ledge
#

combining three red chips with one green

weary tiger
#

what

#

but you don't have to have one green

#

you can have 0 green

halcyon ledge
#

why red green red red

#

is possible

weary tiger
#

yes

halcyon ledge
#

red red green red as well

weary tiger
#

and so is red red red

halcyon ledge
#

yes

weary tiger
#

with no green at all

halcyon ledge
#

thats four

#

red red red green

weary tiger
#

no???

halcyon ledge
#

you just ignore the green one

weary tiger
#

you stop drawing after you get the red red red

halcyon ledge
#

just ignore it if it's to abstract

#

pretend the green stone doesn't exist

weary tiger
halcyon ledge
#

it doesn't matter you're interested in legal patterns

#

once you have your three reds youre done, drawing more doesn't change the outcome

weary tiger
#

ngl i'm not sure how you can just "ignore" the green stone

#

lol

#

@stray reef pls help

stray reef
#

the way i understood deekaan's idea, he draws all five stones and simply records the point in each draw sequence where he would've stopped had he observed the "draw until one color runs out" requirement

halcyon ledge
#

yes ๐Ÿ™‚

weary tiger
#

yeah??? why is that ok

halcyon ledge
#

because i like trees

stray reef
#

i mean... it's certainly possible to do things that way, especially given that after one color runs out you only have one color left in the bag

#

and so the subsequent outcomes are predetermined

# 
RRR|GG
RRGR|G
RRGG|R
RGRR|G
RGRG|R
RGG|RR
GRRR|G
GRRG|R
GRG|RR
GG|RRR
hardy remnant
#

i found a pretty cool discrete math series thats helped me a lot since i am self studying

#

she has a like and share at the end so i am sharing it here incase anyone else is struggling too

fading spear
#

could anyone help me understand what the lines about R and Q v P are?

#

also what is the delta symbol?

weary tiger
#

@stray reef can you explain something to me

stray reef
#

maybe

#

depends on what it is

weary tiger
#

if you have 5 shirts and 6 pants, and you want to choose 2 pants and 2 shirts, then you can do 5c2 * 6c2, but would this include order between picking them? for example, you could pick shirt pants shirt pants or shirt shirt pants pants or anything like that?

stray reef
#

for example, you could pick shirt pants shirt pants or shirt shirt pants pants or anything like that?
doesn't matter

#

you're picking the shirts and the pants independently of one another

weary tiger
#

??? yes it does

#

ok i think

stray reef
#

no it doesn't

weary tiger
#

i don't understand independence in counting then

stray reef
#

picking shirt 2, shirt 4, pants 1 and pants 6 is the same as picking shirt 2, pants 6, shirt 4 and pants 1

weary tiger
#

how tho

#

clearly different order

stray reef
#

you still end up with {pants1, pants6, shirt2, shirt4} as your selection

weary tiger
#

lol

#

that doesn't make sense

#

for example

#

if you had to arrange 2 math books and 2 history books

#

you have

#

4!

#

you still end up with {math1, math2, history1, history2}

stray reef
#

no

weary tiger
#

yes

stray reef
#

you're ARRANGING them

weary tiger
stray reef
#

while your pants/shirts problem is one where the particular arrangement doesn't matter

weary tiger
#

yes it doesn't matter you are right

#

but how do we know we aren't including order tho

stray reef
#

wdym "including order"

#

theres nothing to include

weary tiger
#

as in

#

how do you know we don't accidentally create order

#

by multiplication

#

when we multiply them together

stray reef
#

you're doing the equivalent of wondering whether to count the cards in a deck as 13 * 4 or as 4 * 13

#

or whether to calculate the number of, say, license plate IDs consisting of a letter and two digits as 26 * 10 * 10 or as 10 * 10 * 26 or as 10 * 26 * 10

weary tiger
#

yeah

#

how do we know

#

for real

#

with more complicated examples

stray reef
#

...

#

it sounds like you're questioning the commutative law of multiplication for integers, if anything.

weary tiger
#

oh god

#

i am stressing out atm

stray reef
#

go take a break

weary tiger
#

ann

#

can you explain independent choices tho

#

lol

#

for real

#

omg

stray reef
#

two choices are independent if which option you pick in one does not affect the number of options in the other

weary tiger
#

hmm

#

so

#

the deck example

#

4 * 13 and 13 * 4

#

can you elaborate

stray reef
#

4 * 13 is if you count suits first then ranks within each suit

#

13 * 4 is if you count ranks first then suits within each rank

weary tiger
#

yeah so

#

why are these the same thing

#

i almost want to divide by 2

stray reef
#

are you seriously asking me why 4 * 13 = 13 * 4

weary tiger
stray reef
#

bruh

weary tiger
#

@stray reef we need abstract algebra to demystify counting the number of cards in a deck

tulip ravine
#

I'm rather fond of saying that a deck of cards is the Cartesian product of the suit set and the rank set

weary tiger
#

that actually makes more sense

#

@stray reef this is what happens when your confusion confuses your confusion

#

and youre not really sure what is confusing anymore

stray reef
#

you really should take a break

whole shard
gleaming zephyr
#

the function one is the most clear for me

#

im a hs student

#

@whole shard

whole shard
#

oh u mean the bottom?

gleaming zephyr
#

yea

#

stating this is easy enough id guess

whole shard
#

I dont understand brother

#

why is there division

vapid light
#

I think the best way is to look at an example

#

Let's say you have a deck of cards, and you want to find the number of unique sets of $k$ cards without replacement. Then what we have is that there are $\frac{n!}{(n-k)!}$ ways to select $k$ cards without replacement, but for example if $k=2$ then $(1,2),(2,1)$ are different and both counted. However, we want it so that these two are counted only once. This means we want to account for the permutations of the set of $k$ cards, which is $k!$. So if $B$ is the set of subsets of $k$ cards that have unique elements and $A$ is the set of all subsets of size $k$, then $|B|=|A|/d$, where $d=k!$.

vital dewBOT
vapid light
#

@whole shard

whole shard
#

@vapid light hey thank u for this but i have a question

vapid light
#

Yeah go ahead

whole shard
#

Whats the difference of this method, and the subtraction rule

#

Isnt it taking out the "duplicate" way?

vapid light
#

The subtraction rule is that you count all possibilities and then subtract the ones that aren't what you're looking for

whole shard
#

It says its also called the principle of inclusion-exclusion

vapid light
#

Yeah

whole shard
#

But for division rule, I'm dividing what I dont want?

vapid light
#

You're dividing the number of duplicates

whole shard
#

I know this is crazy to ask but why do you want to divide? Because I understand the subtraction rule

vapid light
#

Because sometimes like in that example I showed what you have is an exact multiple of what you're looking for

#

So we can just map duplicates to the same thing in the other set

#

If there are d duplicates in A, then |B|=|A|/d

#

Per item

whole shard
#

Hmm i think I'm getting some of it, maybe it will show up in permutations?

#

for your example, what does n! represents? Is it the sets of cards in the deck?

#

OH i got it. thank u gupple

vapid light
#

Yep

#

That's how n choose k works

whole shard
#

AH so basically each (1,2) and (2,1) is the same

#

thats why u divide it by 2

vapid light
#

Yeah

whole shard
#

cuz theres two elements

#

but considered 1

vapid light
#

Yeah

whole shard
#

I see , that was hard to understand

vapid light
#

Yeah I bet, but over time it'll get easier

whole shard
#

Everytime my textbook say "ways" I remember knuckles

vapid light
#

Is there a downvote emote

whole shard
#

omg that was a joke please dont get mad

pallid lintel
#

i've come full circle in math. i started colouring in triangles in primary school, now im colouring them in again for graphs

lyric pumice
#

I color lines to keep track of walks.

vapid light
#

I'm just kidding @whole shard

pallid lintel
#

this question, when it says its asking for different kinds of simple graphs for n vertices, is that just asking for different ways of arranging edges for all the different numbers of edges it can have?

#

so a simple graph with 3 vertices can have 0 edges, or 1 edge, or 2 edges, or 3 edges. and we figure out arrangements for each of those numbers?

#

i think i have to multiply spanning tree count by cycle count of complete graph

whole shard
west silo
#

hey

whole shard
#

hey

west silo
#

${p^n \choose k} $ is divisible by $p$ right ? (p is prime)

vital dewBOT
whole shard
#

ur telling me its supposed to cancel n?

#

to 1

#

cuz i know how it became a factorial

#

idk where the denominator came from

west silo
#

i think this follows from induction and ${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k} $

vital dewBOT
west silo
#

induction on n

whole shard
#

oh, damn i skipped that chapter

west silo
#

yo i dont understand your questions

whole shard
#

im asking how they just came up writring that denominator

#

(n-r)!

#

what does it mean

#

because I know the numerrator is from the factorial

obtuse lance
#

you can compute the power of p dividing n! directly with Legendre's formula

#

$v_p(n!) = \frac{n-s_p(n)}{p-1}$

vital dewBOT
obtuse lance
#

v_p(n!) is the power of p dividing n!, s_p(n) is the sum of digits of n when written in base p

#

so together you can combine this to get the exact answer to your question

near prawn
#

@pallid lintel did u figure it out?

vital dewBOT
pallid lintel
#

+1 for the k hole

#

@near prawn

#

because how many different graphs on n vertices can there be? there is the graph with 0 vertices. that's just one graph. Then there is the graph with 1 edge. thats NC1, then the graphs with 2 edges. That's NC2, ect...

near prawn
#

uhh

#

not quite

pallid lintel
#

wait, wrong number

near prawn
#

im not following your reasoning either

pallid lintel
#

$\sum_{t=0}^{n} {\binom{n}{0}}$

#

that better?

near prawn
#

is that meant to be n choose 0

#

i mean no its not

vital dewBOT
near prawn
#

t instead of 0

pallid lintel
#

i thought the question asking for all different graphs on n vertices meant we trying to find all the different ways of putting edges on n

near prawn
#

yes

#

given n vertices

pallid lintel
#

so for K3, we can have 0 edge. that's 3C0=1 type. for 1 edges, that's 3C1=3 ways of arranging. for 2 edge thats 3C2=3 ways of arranging.

#

ect

near prawn
#

how many different simple graphs can you make

pallid lintel
#

1+3+3+1 for K3 right? 8 different ways?

near prawn
#

K3?

pallid lintel
#

er

near prawn
#

why are you only looking at complete graphs

pallid lintel
#

not k3

#

for G3

near prawn
#

oh

pallid lintel
#

sorry, tired, ha

#

do you agree?

near prawn
#

its fine

#

and yea thats right

#

can you generalise it now?

pallid lintel
#

what do you mean? the question was asking for all different simple graphs for graph with n vertices. the summation gives the answer right?

near prawn
#

no

#

that summation doesnt even make sense

#

theres no t in the expression

pallid lintel
#

$\sum_{t=0}^{n} {\binom{n}{t}}$

vital dewBOT
pallid lintel
#

make sense now?

near prawn
#

better

#

another way to think about

#

is that if you have n vertices

#

max no of edges is nC2

#

ie a complete graph

#

and for every possible edge u can either add it or remove it

#

so you have 2^(nC2) possible total graphs

pallid lintel
#

ahh yeah. thats a good way to think about it

hasty glade
#

Why on that linear relation an is not being multiplyed by any constant c?

gleaming zephyr
#

context ?

hasty glade
#

Homogeneous linear recurrence ratios

near prawn
#

yea but what exactly is happening here

#

what is that general form in relation to

hasty glade
#

They are trying to show which are the roots of that polinom

#

Homogeneous = When replacing all ai F(an-1, an-2, ..., n-k) = 0 ?

near prawn
#

if an has a coefficient just divide through by that coeffcient it doesnt really matter

hasty glade
#

$a_{n} = c_{1}a_{n-1} + c_{2}a_{n-2} + ... + c_{k}a_{n-k} $

vital dewBOT
hasty glade
#

If I were doing it that would be my definition

#

And the I will say that if an = 0 for all n then f(an) = 0

#

I dont know why all terms are moved to the left

#

On the original definition

#

$a_{n} + c_{1}a_{n-1} + c_{2}a_{n-2} + ... + c_{k}a_{n-k} = 0 $

vital dewBOT
near prawn
#

doesnt matter

#

the constants would just differ by a factor of -1

hasty glade
#

But my question is why that definition is done that way

near prawn
#

who knows

hasty glade
#

why the sum of all that terms = 0 ?

#

They must be the same....

#

But that = does not happen all the time

novel quartz
#

is this a bad question or

#

All I can think of is the additional dimension of the vector

hasty glade
#

Could someonle plis give me a definition of an homogeneus recurrence?

#

I think I might be confused

novel quartz
#

Do you mean homogeneous recurrence or or representation?

whole shard
#

when doing permutations and combinations, what are some tips that the order does and does not matter

last sigil
#

uhhhhhh, I would say mostly common sense; you can check by thinking do objects a, b, c = objects a, c, b

whole shard
#

so lets try this question. How many ways are there to distribute six different toys to three different children such that each child gets a least one toy?

#

When I read it, it seems like I would use 6C3 = C(6,3)

#

Because if one kid got 1 toy, then theres only 5 left to choose from

last sigil
#

You also have to account for the fact each toy is different

whole shard
#

poco what do you mean by that? The way I see it is the problem uses 6! for the toys

hasty glade
#

Could somenonle help me with recurrence ?

#

I m stuck on second order recurrency

last sigil
#

I am trying to think of a way to explain it with less brute force work

#

So let's take the case of 3 toys and 2 children

stray reef
#

it matters who gets what toys, no?

last sigil
#

yeah, ann's point is the thing to think about

whole shard
#

How I solved it is like this C(6, 3) = 6! / 3! * (6-3)!

last sigil
#

But why

whole shard
#

because it goes 6 * 5 * 4 / 3!

last sigil
#

Each child can pick more than 1 toy

whole shard
#

OH

#

at least one toy

#

is that a keyword too?

stray reef
#

if you want it to be

whole shard
#

Now maybe this is not a premutation and combination problem?

last sigil
#

Sure, but this doesn't give you a "direct" formula

stray reef
#

if "keyword" means "a word ignoring which changes the entire problem" then yes at least is a keyword

whole shard
#

yes I didnt think of a child getting two or more toys

hasty glade
#

Sorry for interrupting but is it allowed to make questions here ? or they MUST be made on any question channel ?

last sigil
#

Nah questions here are fine

hasty glade
#

Oh ok ! thanks I will wait and make my question here

last sigil
#

Its just that if you asked here right now, you'd be interupting our conversation with yo mama

hasty glade
#

Oh okey I m so sorry

whole shard
#

Sorry now I'm lost with how to proceed with this question, maybe I was wrong using the C(n, r) formula

sleek swallow
#

Its just that if you asked here right now, you'd be interupting our conversation with yo mama
yes, talking to yo mama is extremely important

whole shard
#

NO lol, Im just a regular dumb student ๐Ÿ˜›

#

jack please send ur question, I might be able to help

hasty glade
#

First of all consider that I m not an english speacker so I will try to do my best

#

Translating math is quite difficult

#

Consider an LINEAR - HOMOGENEOUS - SECOND ORDER - CONSTANT COEFFICIENTS. If I want to get its general equation I would have to find its polinom (taking a^ n-k as a commun factor) and then find the roots for P(x) in order to know which general equation use, if the one for a single x or the one for two different values for x. My question is, where are those formulas taken from ?

#

That formulas

vapid light
#

isnt that question just stars and bars

#

for the toy problem

last sigil
#

Mostly stars and bars yeah

vapid light
#

so 5C2

stray reef
#

is it stars and bars thonk

#

the toys are distinct

vapid light
#

hmmmm

#

I have terrible reading

#

I was thinking of then multiplying 5C2 with the number of permutations of the toy set but thats gonna overcount

#

hmm

near prawn
#

its just the no of all onto functions from a set with 6 elements to a set with 3@whole shard

whole shard
#

yes I udnerstand, some guy pointed out the multimnomial theorem

#

i have to read it somehow, because C(6,3) is apparently wrong

vapid light
#

Oh that's a cool approach

#

Dang

last sigil
#

@stray reef well I was thinking stars and bars treating the toys as identical first, then, multiplying by the number of toys to get permutations

#

Bad wording but hopefully you see what I am saying

vapid light
#

some1's approach is a lot better, each toy has 3 choices, child 1, 2, or 3, so 3^6 total possibilities, minus the possibilities where one person doesnt get toys, which is 6

#

so 3^6-6

stray reef
#

@last sigil i'm not sure that holds up tho

#

you might be overcounting

vapid light
#

Ok my figure is gigantic

#

So idk if I'm right

last sigil
#

Wait are you

#

I can't think rn

vapid light
#

I feel like the logic holds up

last sigil
#

Idk why you -6

vapid light
#

3C1 describes the number of mappings where the toys only go to one kid, 3C2 with two kids

#

So -6

last sigil
#

???

vapid light
#

Wdym

last sigil
#

Even if you select one kid to not get presents, there are many ways to distribute the 2 boxes amongst the others

#

let one kid not get presents

#

Then you need to divy up 6 presents amongst the last 2

#

Which can be (1,5), (2,4), etc.

#

And take into account distinct presents

vapid light
#

You're right

last sigil
#

Plus which kid (a, b, or c) doesn't get any

vapid light
#

(3C2*2^6-2)

#

I think

#

3C2 picks the two kids the presents go to, 2^6 are the number of total possibilities, and 2 is the number of possibilities where it goes to only one kid

#

I hate counting

last sigil
#

Haha yeah

vapid light
#

Gotta love inc-excl

last sigil
#

Via my method, I get 60

vapid light
#

I still get a big number

#

Let's go

#

Gg

last sigil
#

?

vapid light
#

I'm probably wrong

last sigil
#

Gn

#

Seems the answer is 540?

vapid light
#

Yeah

#

That's what I got

#

Let's go

last sigil
#

Hmm wonder how my method fails

vapid light
#

I did 3^6 - 3C1 - 3C2*(2^6-2)

fossil pewter
#

Another way is to divide the toys in 3 groups and distribute them

vapid light
#

How would that account for the distinctness of the toys

fossil pewter
#

I can show you if you want

vapid light
#

I think I get it actually

#

But can I see it anyway

#

Lol

fossil pewter
#

3 ways to distribute 6 toys in group of 3 with atleast 1 in each group
1,1,4
1,2,3
2,2,2

vapid light
#

Oh so counting partitions sorta

fossil pewter
#

Number of ways to distribute for case 1 is $ \frac{6!}{1!1!4!} $

vital dewBOT
fossil pewter
#

Multiple that with 3 To distribute to kids

#

Same for 2nd case

#

For 3rd case number of ways to distribute the groups to kids would be 1 since they are same

vapid light
#

Hold on, why is it the same for the second case

fossil pewter
#

$ \frac{6!}{1!1!4!} \cdot \frac{3!}{2!} + \frac{6!}{1!2!3!} \cdot 3! + \frac{6!}{(2!)^3} $

vital dewBOT
fossil pewter
#

I meant in similar way

vapid light
#

And where did you get 6!/4! anyway

fossil pewter
#

Dividing 6 in group of 1,1,4

vapid light
#

Let me think about this for a sec

#

Ok, I see

#

Yeah this makes sense

#

Thanks

#

I wouldn't have thought of this

vital dewBOT
weary tiger
#

@stray reef ive made a proof

#

for the thing we were talking about earlier

stray reef
#

what thing

weary tiger
#

why 4 * 13 and 13 * 4 both count the same thing: number of cards in a deck

#

though

#

it can be extended to more complicated things

#

but i mean in general independent events/things

stray reef
#

why would that need a proof beyond acknowledging the commutative law of multiplication or presenting the most obvious explicit bijection between $A \times B$ and $B \times A$ for arbitrary sets $A$ and $B$?

vital dewBOT
stray reef
#

@weary tiger

weary tiger
#

@stray reef define obvious

stray reef
#

i'm specifically talking about the bijection $(a,b) \mapsto (b,a)$

vital dewBOT
stray reef
#

A ร— B -> B ร— A

weary tiger
#

i mean

#

i guess you can say it's obvious

#

but it's not very rigorous to do so

stray reef
#

do you want me to write out in full on 100% rigorous formal notation why the function (a,b) โ†ฆ (b,a) : Aร—B -> Bร—A is actually a bijection?

weary tiger
#

that would be nice

stray reef
#

i was kinda hoping you'd say no

#

but fine...

#

call this function f

#

i.e. f: Aร—B -> Bร—A is defined by f(a,b) = (b,a) for all a โˆˆ A, b โˆˆ B

#

define g: Bร—A -> Aร—B by g(b,a) := (a,b)

#

observe immediately that f . g is the identity on Aร—B and that g . f is the identity on Bร—A

#

therefore g is the inverse of f

weary tiger
#

see

#

now this is intuitive why 4 * 13 = 13 * 4

#

are counting the same thing

#

๐Ÿ™‚

jaunty carbon
#

Given a abelian group (G,+) with a finite order and a an element of group G. If -a=a, then what is the order of element a?

#

I really dont know how to solve this question, could someone help me?

gleaming zephyr
#

do you know the definitions

#

what all those mean?

jaunty carbon
#

Yes

gleaming zephyr
#

okay so

#

what element is its own inverse ig

jaunty carbon
#

do you mean the zero element

left gate
#

Yy

#

For ur question add a on both sides

#

Order is either 1 or 2
If 1 then a=0
If not then order of a is 2

jaunty carbon
#

Tnx I understand

left gate
#

Those coloured namescatBruh

jaunty carbon
#

If I have a finite field $Z[x]/(x^4+x^2+x+1)$

vital dewBOT
stray reef
#

i don't think that's finite or even a field the way it's currently written

jaunty carbon
#

Ow imma type it again

#

If I have a finite field $Z_3[x]/(x^4+x^2+x+1)$ and need to construct a subfield with 9 elements. Can I then for example take Z_3[x]/(x^2+2^x+2)?

stray reef
vital dewBOT
jaunty carbon
#

Z_3[x]/(x^4+x^2+x+1) is a finite field with 81 elements

#

It are all polynomials with a maximum degree of 3 and numbers modulo 3 as coefficients

stray reef
#

2^x thonk

#

anyway... no you can't

hasty glade
#

Hello could somenonle help me please please please with mathematical induction

#

I m really stuck

halcyon ledge
#

whats the induction?

#

and whats the problem?

hasty glade
#

Prove that a property P(n) is valid for all n e N

#

First making sure P(1) is true

#

Then doing P(n+1)

halcyon ledge
#

yeah but what is P?

#

rofl

hasty glade
#

P is a property

faint narwhal
#

I don't really understand what you're asking

#

Or what you're stuck on

#

Just the idea of mathematical induction?

hasty glade
#

No, now that you know what I m talking about I will make the quesiton

#

I wanted to make sure you knew what I was talking about becuase I dont speack english so I dont know enligsh math vocabulary

#

Well my question is this:

#

If I have to prove P(n+1) is true

halcyon ledge
#

ah thats actually nice

hasty glade
#

Do I have to represent the equality with my general formula ?

#

Or can I do it getting two equals polinoms p(n) = p(n)

#

Please ask if it is not clear, I really need help

halcyon ledge
#

could you give an example problem

hasty glade
#

Of course

halcyon ledge
#

like something you're stuck on

hasty glade
#

I got a proof which is this

halcyon ledge
#

okay

#

you see that the left hand side is a sum?

#

$\sum_{k=1}^{n} k^2$

vital dewBOT
hasty glade
#

I m giving a proof for p(n+1)

halcyon ledge
#

yes

weary tiger
#

ok

halcyon ledge
#

but you have to build up to that

hasty glade
#

The thing is that I m not expressing it with its general fornm

weary tiger
#

thank you, very cool

hasty glade
#

but you have to build up to that
@halcyon ledge Build up ?

halcyon ledge
#

you have to prove the base case

hasty glade
#

I already done it

#

For p(1)

#

Its on other page

halcyon ledge
#

$1^2 = \frac{1(1 +1)(2 + 1)}{6}$

vital dewBOT
halcyon ledge
#

ah okay

#

next you want to use summation

#

$\sum_{k=1}^{n + 1} k^2$

vital dewBOT
hasty glade
#

This is the solution the book is giving

#

Ignore the spanish just look at the operation

#

Can you see that it is expressed in other way different than mine

#

Is that an issue or not ?

halcyon ledge
#

i still would prefer summation but alright

hasty glade
#

Yeah I know what you mean using that sumation for proving it

halcyon ledge
#

yes you basically transform the part that you know (1^2 + 2^2 + ... + n^2)

#

into the fraction

hasty glade
#

So what I did has to be good

#

I mean I would recive all points

#

For the exercise

halcyon ledge
#

yeah I mean you got the general idea its just a little confusing the way you wrote it up

hasty glade
#

Yeah I know lol

halcyon ledge
#

$\frac{n(n +1)(n + 1)}{6} + (n + 1)^2$

#

that you have to transform into P(n + 1)

vital dewBOT
hasty glade
#

But I was thinking was... If I prove that P(1) is true and P(n+1) is true then it is done

#

that you have to transform into P(n + 1)
@halcyon ledge Oh so do I have to ?

halcyon ledge
#

your intial assumption is that P(n) holds for a given n

hasty glade
#

yep

halcyon ledge
#

you want to use that to prove P(n + 1) is true

hasty glade
#

yep

halcyon ledge
#

good

hasty glade
#

I also know that p(1) is true

#

Continue..

halcyon ledge
#

thats your P(n) basically

#

the base case

#

from there you run your induction

#

see because if it holds for 1 and n + 1 then it holds for 2 and because it holds for n + 1 also for 3 and so on

#

this way you show that it holds for all n

hasty glade
#

Yep

#

But did you understand my question ?

#

Maybe you didnt because I dont express myself very well lol

#

For a spanish speacker this is pretty hard

halcyon ledge
#

tbh i dont really understand whats up, you seem to have a pretty good handle on induction

hasty glade
#

Okay mmm....

#

See

halcyon ledge
#

your write up just isn't all that well structured

hasty glade
#

1- P(1) was proved to be true

#

2- I assume that it works for P(n)

#

3- Then I try to see if it is true for P(n+1)

#

The thing is: On step 3, Does it matter the way I prove the equality ?

#

Because the way I did and the way the book did are different, but both of them give a prove to the P(n+1)

halcyon ledge
#

ahh now I get it

#

you have to start at the point where your base is

#

atm you're walking backwards

#

basically assuming what you want to prove is true and walking to your base case

#

that's all right in the conceptual stage and is actually a good way to figure out how to prove it

#

but in the actual proof you have to walk the right way around

#

the way it's shown in your book

hasty glade
#

So what you mean is: I gave a proff that it is true. But you have to express it using the base case

#

Replacing all n for n+1

halcyon ledge
#

yes I think thats it

#

at least if i understand you correctly

hasty glade
#

P(n) = 1,2,3,4, ..., n = n(n-1)/2

#

So in order to give a complete proff do I have to express it in this way?

#

P(n+1) = (n+1)((n+1)-1)/2 = (n+1)((n+1)-1)/2

halcyon ledge
#

$\sum_{k=1}^{n + 1} k^2 = \sum_{k=1}^{n} k^2 + (k + 1)^2 \ \overset{\text{induction}}{=} \frac{n(n +1)(2n + 1)}{6} + (n + 1)^2$

hasty glade
#

The exercise just says "prove"

halcyon ledge
#

and from there you transform the term step by step into the one with (n + 1)

vital dewBOT
halcyon ledge
#

if you don't like summation just use the other stile

#

but this is your starting point

hasty glade
#

but this is your starting point
@halcyon ledge That is the key word

halcyon ledge
#

basically the way you've written it down already only the other way araound

hasty glade
#

Do I have to express the proff using the starting point ?

halcyon ledge
#

yes you start with your base case

#

that's the thing you've proven

hasty glade
#

Start with base case ---> finish with base case but with n = n+1

#

?

halcyon ledge
#

something like that

hasty glade
#

I think I understood

#

Thanks deekaan

#

That makes induction pretty hard....

#

If I could express the proof other way It would have been pretty easy

halcyon ledge
#

no you basically can still solve it that way you did you just write it up backwards

#

and pretend you did it the right way around

hasty glade
#

Oh...

#

So what I did was right..

halcyon ledge
#

yes

hasty glade
#

AHHH !!!!!

#

I think I dont understand lol...

halcyon ledge
#

its pretty strange at first

#

but you got the right idea

#

its just the form thats a little off

#

@hasty glade

hasty glade
#

What that men made was to express p(n) with p(n+1)

#

After I got the prove do I have to do that ?

#

What I did is not p(n+1)

#

Now I have to express it getting roots of the polinom...

#

In order to get p(n)

halcyon ledge
#

you're doing polynoms bro

#

and you're not looking for roots

#

P(n) is just an expression

hasty glade
#

Do I have to arrive to n(n+1)(2n+1)/6

#

Or that is not necessary

halcyon ledge
#

((n + 1)(n + 2)(2n + 3))/6

#

thats your target

hasty glade
#

So the objetive will be to arrive to that expression ?

#

What I did is not right then

weary tiger
#

Hi is there anyone here that knows computational modeling? Stuff like turing machines, reductions, etc.
Im looking for a tutor. Im willing to pay a decent amount!

vapid light
#

Just ask here lol

weary tiger
#

I'd prefer dm

vapid light
#

...

weary tiger
#

sorry if thats not okay with you

whole shard
#

Hello please explain

#

is this only possible when the constant is negative

dark forge
#

Do you guys have any resources to help learn discrete math?

#

like a textbook with good practice problems

hardy remnant
grand elm
#

if Juan doesn't have a smartphone, then you can't apply the statement, but everything works out better, if you define it to be true when the premise is false

hardy remnant
#

so in a if then statement, the if has to be true, and then it doesnt matter if 2+3 is false or not?

#

i see i may be confusing this with the truth table earlier

grand elm
#

look at it like this: "if X is a cat, then X is an animal."

#

this is obviously true right?

hardy remnant
#

yes

pallid lintel
#

what are some interesting topics in ramsey theory for graphs? I just started studying it

hardy remnant
#

i see

grand elm
#

but if you take X = me, then both the premise and the consequence are false

#

but you still want the statement to be true

#

you can also look at it like this: "if false is true, then anything is possible"

hardy remnant
#

gonna need a bit more thinking, but i should be good

#

thanks

grand elm
#

๐Ÿ‘

last sigil
#

Consider the statement: "If A, then B", where "A" and "B" are some statements

#

If A is false, you can't say anything about whether A being true implies the truthfulness of B

hardy remnant
#

alrighty

last sigil
#

So B is vacuously true

hardy remnant
#

so its basically just this

if(statement = true){
Then do this}

else{
do that}

last sigil
#

Thats CS which is vaguely similar, but not really

hardy remnant
#

i get what you mean tho

lyric pumice
#

Let j=0 and k=n.

last sigil
#

what the fuck is this

lyric pumice
#

Algebraic demonstration, abridged.

last sigil
#

?????

lyric pumice
#

Pascal's rule is used.

last sigil
#

Well, there is a simpler solution

#

Consider the binomial theorem @whole shard

dusky pond
#

how would i solve a recurrence relation of

#

$$t(n^2) = t(n^2/2) + \Theta(n)$$

vital dewBOT
vapid light
#

Isn't this just Theta(sqrt(n))

#

Correct me if I'm wrong

jaunty carbon
# 
4 = 0+4 = 1+3 = 2+2 = 3+1 = 4+0 (so there are 5 ways to write 4 as an ordered sum).
Calculate the generating function f_j for the row numbers {a_i,j} with i = 0 ->infinity```
#

I dont know if there is anyone here who can solve this, but I've been trying to solve this for a few hours now, maybe someone could help me. I can't find a recursive formula for a_i,j

vapid light
#

This is stars and bars pretty sure

#

So, if you have $i$ and you want to find the number of ways you can sum up $j$ natural numbers to get it, you can think of placing some $j-1$ dividers between $i$ balls, and the number of balls between each divider is the number of items in one term of the sum.

vital dewBOT
desert ferry
vapid light
#

However, you can have 0 balls between dividers, so we need to find the number of spots in which we can put dividers

#

You should thoroughly read the Wikipedia page for stars and bars @jaunty carbon

#

It has a useful drawing that can explain the concept

desert ferry
jaunty carbon
#

tnx

vapid light
#

What's the definition of A-B @desert ferry

desert ferry
#

Wym?

#

They're sets

#

A-B is equal to the remainder of A

#

and A-C is equal to the remainder of A after substracting C

#

i just dont understand the answer of (A-B) - (A-C) although i know that it's incorrect but with a different answer

fossil pewter
#

It should have been (A-C)-(B-C)

#

Instead of (A-C)-(A-B)

desert ferry
#

u mean for it to be correct it should be (A-C) - (B-C)?

stray reef
#

(A-C) - (B-C) correctly describes the shaded area

#

er

#

hold on

#

no, no it doesn't

desert ferry
#

yeah it doesnt

stray reef
#

set diff is a bit inconvenient to work with

desert ferry
#

it is

stray reef
#

$A \cap C' \cap (B \cap C')' \ = A \cap C' \cap (B' \cup C) \ = A \cap [(C' \cap B') \cup (C' \cap C)] \ = A \cap [(C' \cap B') \cup \rien] \ = A \cap B' \cap C'$

vital dewBOT
desert ferry
#

either way the question didnt mention to correct it, and it is incorrect so ill just move on and solve other problems

#

Thanks anyway guys

#

and gals if there's any

fossil pewter
#

no, no it doesn't
what

#

It's correct

desert ferry
#

it's correct for the first shaded part not the second one

vapid light
#

Right, what I meant was using set builder notation to understand what was going on @desert ferry

jaunty carbon
#

Could it be that the greatest common divisor of 52 and 72 is 1 over R[x]?

stray reef
#

yes because 52 and 72 are both units in R[x]

jaunty carbon
#

damn

#

tnx

languid flint
#

I was wondering if finding shortest path from a vertex to all other vertices is same as finding the minimum spanning tree, is it correct?

faint narwhal
#

no

languid flint
#

umm why not?

faint narwhal
#

why would it be?

languid flint
#

it seemed so, can you give a counter example?

faint narwhal
#

The vertex A on the right side, if you wanted to find the shortest path from A to D

#

you would take the bottom edge of length 2

#

but the only minimum spanning tree of this graph is to take all three top edges of length 1

languid flint
#

oh

#

right, thanks.

stray reef
#

here's an interesting question

#

the number of 3-point subsets of a 6-point set is 6C3 = 20.
is it possible to color the vertices of an icosahedron in six colors in such a way that every face has a different set of colors at the vertices it's incident to?

#

if not, what's the maximum possible number of unique color sets on the faces?

hasty glade
#

I dont know how to get the general equation

#

I was thinking about doing telescopical sums but I have N on the right side

jaunty carbon
#

What is meant with a singularity of a function. I thought it were just the poles of the denominator but apparently it isn't that simple. For example for the function: $1/((e^x-e)*(x-2)^2)$ the dominant singularity is 2 and not 1

vital dewBOT
jaunty carbon
#

@hasty glade do you want to get a formula for a_n? If so, you could do it easily using generating functions.

hasty glade
#

Generating functions ?

#

I dont know to do that. What I was trying to do was to find a patter for an-n = a0 and then get the general formula

#

If you could please explain what that is you would help me a lot ! ๐Ÿ˜„

obtuse lance
#

nah

#

just do it as a telescoping sum

#

the sum on the RHS over n has a well known trick

#

the trick is usually explained as take the sum, write it forwards and backwards

#

1+2 + ... + (n-1) +n
n + (n-1) + ... + 1

#

then add the terms

#

(n+1) + (n+1) + ... + (n+1) + (n+1)

#

there are exactly n terms here

#

n*(n+1)

#

we double counted by adding it to itself so

#

n(n+1)/2

#

= 1+2+... + n

#

probably not easy to read in discord but you can probably find the same explanation somewhere else

hasty glade
#

I did it

#

But I did no use telescoping sum

#

I did reverse substitution

#

I dont know why I could not figure out this

#

an-a0 = ?

#

The right side of the sum

hasty glade
#

$\sum_{k = 0}^{n} \frac{1}{a^{k}} = ?$

vital dewBOT
hasty glade
#

Could someonle please give me that formula ?

obtuse lance
#

geometric series

hasty glade
#

Could someonle give me tips on finding general forms

#

It is being pretty hard for me

hasty glade
#

I have been figuring them out

#

For 4 hours

#

Please any type of help would save me lol

#

This is the last subject I need to learn for the test

urban olive
#

Hey guys I have a quick question

#

I need to write the negation of: โˆ€xฦŽy(xy โ‰ฅ y)

#

And I think I already did it

#

ฦŽxโˆ€y(xy โ‰ค y)

#

Is this right? Like when we are writing the negation the โ‰ฅ from the original one changes to โ‰ค ?

vapid light
#

The negation of greater than or equal to is just less than

urban olive
#

Ok so ฦŽxโˆ€y(xy < y)

#

Ty

whole shard
#

Guys

#

When u say repetition is allowed, does it mean the size of element is still the same?

#

like 7 options, then u still have 7 options for next

robust mango
#

Yes, the number of options stay the same.

dark forge
#

@neon thorn thank you for that recommendation. Does this resource also work for CS people as well? or is discrete math a bit different in CS?

muted tree
#

hi guys, how do I go about this problem? : Set f(x) = x3 โˆ’ 3x2 + 2x โˆ’ 4. Prove that there exists a real number r such that 2 <r< 3 and f(r) = 0.

obtuse lance
#

look at f(2) and f(3)

#

f is a polynomial, so it's continuous

#

what does that tell you

loud copper
#

damn it i was googling IVT and someone already answered it angerysad

stray reef
#

do you mean x^3 - 3x^2 + 2x - 4

obtuse lance
#

@loud copper whoops sorry

loud copper
#

fishthonk its ok

muted tree
#

@obtuse lance I honestly don't know sadly.

#

@stray reef yes

stray reef
#

does "intermediate value theorem" ring any bells

obtuse lance
#

what is f(2) and f(3) let's start there

muted tree
#

I guess they're functions f(r) with r being the real numbers 2 and 3?