help-28

[spice orchid]

might be easier to work with idk

[tribal gust]

yeah i did that aswell

i only noticed this graphically tho so thanks!

[spice orchid]

nice!

[tribal gust]

How would i go about showing that y = -1-x is an oblique asymptote?

I thought of showing that the distance between both graphs for every (y,x) tends to 0 as x tends to infinity

but then i didnt show that its an oblique asymptote but just an asymptote

[full forum]

@tribal gust Has your question been resolved?

[full forum]

@tribal gust Has your question been resolved?

[heavy geyser]

is this Dutch?

[tribal gust]

Yeah

[heavy geyser]

cool

[tribal gust]

Lol

[heavy geyser]

my friend goes to Leiden

[tribal gust]

Im not from the netherlands, thats cool tho

.close

[full forum]

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[warm flare]

hEY

[vapid barn]

Hey

[warm flare]

I want to remove module

[vapid barn]

Consider the expression inside the logarithm

[warm flare]

yea

[vapid barn]

What are you trying to do first of all

[warm flare]

splitting it

[vapid barn]

So you wish to make this a piecewise function?

Or?

[warm flare]

yea like this

[vapid barn]

Ok

Let us first consider ln(z)

Trust me on this.

When is ln(z) negative or positive?

[warm flare]

ln(z) is negative below 1

from 0 to 1

[vapid barn]

That is right

Now do you agree that the argument of the logarithm is (x+1)^2

[warm flare]

yes

I did that formula to find the zeros

and it has a double zero

on -1

[vapid barn]

So now we must find:

(x+1)^2 < 1

Good?

[warm flare]

yes

in x = 0 and x = 2

according to my calculations

so like from ]0;2[

@vapid barn

[vapid barn]

x+1 = 1, x =0
x+1=-1, x = -2

(-2,0) is the negative interval

[warm flare]

x^2 + 2x + 1 = 1

right?

[vapid barn]

Yep

[warm flare]

but then

is

[vapid barn]

x^2+2x = 0

[warm flare]

x(x+2) = 0

x = 0

x = -2 ahh

right

thanks

[vapid barn]

Ye keep track of those signs

(x+2) is a negative root

[warm flare]

~ye

and now

[vapid barn]

So this function is defined on the entirety of the real line

Except for x = -1

Agreed?

[warm flare]

Domain is IR \ -1

yes

[vapid barn]

And it is negative on the interval (-2,0)

[warm flare]

yes

[vapid barn]

So for x in (-2,0) we want to return -ln(x²+2x+1)

[warm flare]

why

[vapid barn]

Because the function is negative over that interval

So we take the negative of the negative to get the absolute value

[warm flare]

ok

[vapid barn]

Make sense?

[warm flare]

yes

[vapid barn]

Now what do we do outside of (-2,0)?

[warm flare]

outside we let it positive

[vapid barn]

Yes

All good?

[warm flare]

i will show you

its right

thank you

but my teacher has

or / -1 < x <= 0

[vapid barn]

Yeah

[warm flare]

why or

[vapid barn]

I guess you do have to split

Because function is not defined at -1

[warm flare]

yea

that's genius

thanksss dude ❤️

[vapid barn]

👍

[full forum]

@warm flare Has your question been resolved?

[full forum]

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[patent meadow]

so im trying to do this and i solved for x, then did the solution for vx but my solution wasnt any of the options

[viral jasper]

show your work

[austere dome]

@patent meadow Can you show your work and what are the possible answers?

[patent meadow]

alr one sec

so these are the solutins

[austere dome]

One of those indeed is the correct answer. Can you show your work?

[patent meadow]

ok

so i did (5x+2)=(3x-8)

and that gets us x=-5

[austere dome]

That is where your error.

[viral jasper]

How'd you arrive at this?

[viral jasper]

This would give you a negative length

[patent meadow]

since the lines in the triangle were congruent i assumed that ST and VX were equivalent

yeah

so i knew that was wrong from the start

[viral jasper]

The triangle are not congruent

[patent meadow]

i mean

[viral jasper]

They are similar

[austere dome]

They are similar.

[patent meadow]

oh

[viral jasper]

So the corresponding sides are not equal

[austere dome]

Though, there should be a theorem you've seen that will help here.

[viral jasper]

But their ratios are

[austere dome]

Or if you are learning similarity now, then use that.

[patent meadow]

alr hold on

yeah idk how to do this

i can find how to use the ratios on similar equivalent triangles

but this i just dont get

[austere dome]

There is a theorem that states what the ratios between the two segments is, and also states that they are parallel. Does this ring a bell?

[patent meadow]

oh is that the proportionality theorem?

[austere dome]

Can you state which theorem you are referring to?

[patent meadow]

the basic proportionality theorem

isnt that the name of it

[austere dome]

Can you state what the theorem says?

[austere dome]

Ak. I meant to say can you state what the theorem you are referring to says.

[patent meadow]

that if a line is drawn parallel to one of the triangle sides and intersects the other two sides of the triangle at a certain point, then the other sides are divided with the same ratio

[austere dome]

Not what I was looking for. Do midsegments ring a bell?

[patent meadow]

yeah

oh are you talking about the triangle midsegment theorem?

[austere dome]

Yes.

[patent meadow]

where the midsegment that connects the midpoints of two of the sides is parallel to the third side

[austere dome]

What does that theorem state?

[patent meadow]

and that its half the length of the third side

[austere dome]

Precisely.

How can you apply it here?

[patent meadow]

by solving for x again but using half of VX instead of the full thing?

or is that wrong

[patent meadow]

am i right

[austere dome]

Yes

Wait

Half of ST

Or double VT.

[patent meadow]

OH ok

lemme try thatagain

so x=18?

[austere dome]

Yes. So now, what is VX?

[patent meadow]

46

[austere dome]

✅

[patent meadow]

awesome

thank you

[austere dome]

No problem.

[patent meadow]

.close

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[kindred viper]

how can i prove, that (A~B)&&(C~D) -> (A x D) ~ (B x C)?

[spice orchid]

What is ~ and &&?

[kindred viper]

uh, i don't know how to translate the name of this ~

[devout valley]

Any more context than that?

[kindred viper]

it is a set theory problem

[spice orchid]

Probably a relation

[devout valley]

Equivalence relation(?)

[spice orchid]

But we need more information

[kindred viper]

oh, it sounds like it

in polish it is równoliczność

[kindred viper]

yeah, it is it

i think

[spice orchid]

But what actually is ~? Like has it been defined? Because you're using the same relation symbol for two different relations

[kindred viper]

it is something that means, that two sets have the same number of elements

[spice orchid]

Ahh okay, that makes much more sense

Are A,B,C,D finite sets?

[kindred viper]

those are any unempty sets

[spice orchid]

Have you seen something like A~B if and only if there is a bijection f : A -> B?

[kindred viper]

yes, i've seen it

[spice orchid]

So your goal will be to make a bijection h : AxD -> BxC, using the maps f : A -> B, g : C -> D

[kindred viper]

hmm, okay

i've been thinking about f: A -> B and g: D -> C

[kindred viper]

but how can i create this h function

[full forum]

@kindred viper Has your question been resolved?

[kindred viper]

i've generally managed to do accomplish this

B x C = f(A) x g(D)

but i don't really know what can i do next

anyone? <@&286206848099549185>

[kindred viper]

:((((

ok, just to sum up, we have to prove this

~ means, that two sets have the same number of elements

and now
becasue A ~ B there exist a bijection f: A -> B, and because D ~ C there exist a bijection g: D -> C
Then B x C = f(A) x g(D)

when can i do next? <@&286206848099549185>

[austere dome]

@kindred viper You need to find a function that maps pairs (a,d) in A x D to pairs (b,c) in B x C.

For a hint, think in terms of components of the pairs.

[kindred viper]

hmm

i still don't really know what to do

[austere dome]

Alright. Do you know what the elements of A x D look like?

[kindred viper]

yes

those are pairs (a, d)

[austere dome]

And what about B x C?

[kindred viper]

same, pairs (b, c)

[austere dome]

So, you want to transform a pair (a,d) into some pair (b,c). How do you think that can be accomplished?

Consider the two functions you have, f: A -> B and g: D -> C.

[kindred viper]

hmm

ok, so, i know that (b, c) = (f(a), g(d))

but it isn't relevent i think

[austere dome]

Right. Maybe you can define your function somewhat like that?

[austere dome]

(It is relevant by the way)

[kindred viper]

oh, ok

so

let h(a, d) be equal to (f(a), g(d))

h: AxD -> BxC

[austere dome]

Correct.

Now, you just have to show this is a bijection, which is a straightforward task using the fact that f and g are bijections themselves.

I believe you should be able to take it from here.

[kindred viper]

well

i believe you might be surprised @austere dome

[austere dome]

Do you know how to prove a function is bijective?

[kindred viper]

not really

i mean

i can prove it is surjection

and

injection

and then i know it is bijective

but i don't know if this is a good way in this situation

no, wait

[austere dome]

That is basically how you proce a function is bijective. Bijective functions by definition are injective and surjective.

@kindred viper

[kindred viper]

yes, i mean

now i see this

[austere dome]

(I'm heading off to dinner now, so I won't be able to respond. I think you can do this on your own now.)

[kindred viper]

kk

[full forum]

@kindred viper Has your question been resolved?

[kindred viper]

.close

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[kindred viper]

ok, and how can i solve this one?

[kindred viper]

~ means, that two sets have the same number of elements

because A ~ B, there exist a bijection g: A -> B

[devout mason]

Is this in the real numbers?

[kindred viper]

Let G: A^C -> B^C be given by G(f) = gof

and i don't know how to prove that G is a bijection

[devout mason]

Trenek is this in the real numbers?

[glass crystal]

why would it matter?

[devout mason]

Like what space is it in

A complement means nothing

unless you have a set it is contained in

[devout valley]

Doesn't matter

[glass crystal]

well lets calll E the bigger set

[kindred viper]

those are any unempty sets

[glass crystal]

w/e

[kindred viper]

sooo, anyone have any ideas how to do it?

[devout mason]

G isn't well defined

So you said g is the bijection from A to B, right? Then you said G is the same function, but from E - A to E - B?

I think I have a counter example

[devout valley]

And is f an element, or a map?

[devout mason]

Trenek, I don't think it is true

[kindred viper]

G(f) = g  ∘  f

now it's prettier

[devout mason]

So if f is in E - A, then how do we know f is a function?

I'm assuming this is composition of functions

[kindred viper]

yes, it is

[kindred viper]

yyy

[devout mason]

Is our space E a set of functions?

[kindred viper]

actually

i don't understand where does E come from

tbh

[devout mason]

It is important because otherwise we have no idea of what A complement should be

If E were the real numbers, and A were a subset of the real numbers, A^c is usually the set of all real numbers not in A

[devout valley]

That's like your "universe", in that A, B, A^c, B^c are all subsets of that

[kindred viper]

C is also a set

[devout mason]

C isn't complement?

ohhh

I thought c was complement, sorry

[devout valley]

Same here! Very poor choice of lettering

[devout mason]

what is A^C defined as

[devout valley]

Think that's all functions from A to C?

[devout mason]

Okay

[kindred viper]

it is a set of all functions f: C -> A

[devout mason]

I'll see if I can think of something

[devout valley]

In that case, because A and B have a bijection between them, you have an invertible function $f: A \to B$ (and $f^{-1} : B \to A$)

[glossy valve]

chartbit

[devout valley]

So I think that would be the idea

[devout mason]

Did you prove it was a surjection or injectino yet?

[kindred viper]

no

i don't know how

[devout valley]

Oh yeah you've done that already so far. fair enough

$f$ is a function $f : C \to A$, and then $g : A \to B$ is your bijection between the sets, so then $G(f) = g \circ f : C \to B$

[glossy valve]

chartbit

[devout valley]

Assume you have two functions $f_1, f_2 : C \to A$ such that $G(f_1) = G(f_2)$ and work with that (injectivity)

[glossy valve]

chartbit

[devout mason]

Are you going to have to use the axiom of chioce? for surjectivity

[devout valley]

For surjectivity, assume you have $h : C \to B$ and try and create an $f$ such that $G(f) = h$

[glossy valve]

chartbit

[full forum]

@kindred viper Has your question been resolved?

[kindred viper]

ok, i did this

[kindred viper]

now i have to do this one

[kindred viper]

huh, i don't really understand this one

[devout mason]

I want to quickly check what you did for the proof it is injective

Did you say G(f1) = G(f2)

Then let c be an element of C

We have g(f1(c)) = g(f2(c)), so by the injectivity of g, f1(c)=f2(c)

Thus, f1 = f2

[kindred viper]

yes

[devout mason]

Perfect

[devout valley]

I'm happy to take that

[devout mason]

The proof for surjectivity is a bit more difficult. When I tried it, it requires the axiom of choice

Do you understand the axiom of chioce

[kindred viper]

i'm not sure if i know what it is

[devout mason]

Do you know what a choice function is

[kindred viper]

not really

[devout mason]

I'm going to be afk for a few minutes. You should search up what a choice function is

[kindred viper]

kk

[devout mason]

@kindred viper I'm back

[kindred viper]

ok

now i know a bit what it is

[devout mason]

Do you have an idea why we would need the choice function?

i.e. the axiom of choice

[kindred viper]

but it provides nonconstructive proofs

so, isn't there a better way to do it? 🤔

[devout mason]

I am not sure

The way I found uses the axiom of choice but I am not qualified to say you can't prove it without the axiom of choice

[kindred viper]

not really

[devout mason]

Do you want to have an attempt at the proof or do you want me to guide you through the process

[kindred viper]

i would prefer you to guide me tbh

[devout mason]

okay

So we have a function in B^C and we look for a function in A^C where, when we apply G, we find our function in B^C

So let h be a function from C to B

If we have an element c of C, then h(c) is an element of B

and since g is surjective and function from A to B, we have some element a of A where g(a) = h(c)

So what we want to do is find a function f that goes through this process: taking c in C, and mapping it to a in A where g(a) = h(c) (so that f(c) = a so G(f) = g(f(c)) = h(c))

Does this make sense

The problem is that there may not be a unique a where g(a) = h(c)

So we use a choice function to choose a specific a

[devout valley]

g is injective too, no?

[devout mason]

I'm unsure of how that would be useful

ohhhhhh

so we don't need the axiom of choice

[devout valley]

Yea, I was gonna say why not just take $f = g^{-1} \circ h$ as is, thought I would have been talking shit

[devout mason]

and we just let f be this function and we'll find G(f) = h

[glossy valve]

chartbit

[devout mason]

Yeah that works

[kindred viper]

ok, i get it now

thanks guys

.close

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[kindred viper]

how to prove, that f is injective iff

[kindred viper]

A and B are sets

[stiff musk]

to see that it's not necessarily true when f is not injective, try a three-point domain {1,2,3} with A = {1,2} and B = {2,3}, and give f some appropriate values so that the desired result doesn't hold

[sharp cipher]

For starters, suppose $f$ is injective, and let $A, B$ be non-empty sets. Then, ...

[glossy valve]

DavidL1450

[sharp cipher]

To prove sets are equal , prove that each one is a subset of the other.

[austere dome]

@kindred viper Are you studying Category theory by any chance? I would be a bit surprised if you are taught exponentials in any introductory course on set theory/proof writing.

[kindred viper]

no, it's introductory course

it's my first semester

[austere dome]

Very strange. Well, I suppose going over the set of functions between two sets is fine in an introductory course.

[kindred viper]

it is "Logika i teoria mnogości" which translates to "Logic and Set Theory"

[austere dome]

I see. What do you have so far for this problem?

[kindred viper]

well, i see that this equation doesn't hold up if f is not injective

and it means that if

[austere dome]

Can you prove that for all sets and functions f?

Or only for specific examples?

[kindred viper]

this equation is true, than the function is injective

but i don't really know how to prove to other way around

how to prove that is the function is injective, then this identity occures

[austere dome]

I presume A and B are subsets of the domain of f, right?

[kindred viper]

yes

unempty subsets

[stiff musk]

one containment is true even if f is not assumed injective. do you know which one?

[austere dome]

Are the sets A and B fixed? Or is the statement quantifying over all non empty subsets?

[kindred viper]

what does it mean that those sets are fixed?

i mean, this is basically a problem

i will translate

[austere dome]

Dla dowolnych
I presume that means for all.

[kindred viper]

Let f: X -> Y be a function. Prove that following conditions are equivalent:

(a) f is injective
(b) [equation] for any [the_rest]

[austere dome]

So the sets are not fixed then.

So, it should be simple to show that if f is not injective, then (b) does not hold.

To negate (b), you only need a single counterexample. See if you can construct one.

[kindred viper]

and it means that if b holds, than a is also true

[kindred viper]

I can, let A = {1, 2} and B = {2, 3}

[austere dome]

Yes. That is the contraposirive.

A and B have to be defined in terms of X and f.

[kindred viper]

let f(A) = {0, 2} and f(B) = {2, 0}

[stiff musk]

this is the right idea, but you need to specify exactly what are f(1), f(2), f(3)

[austere dome]

How can you generalize that? A and B are subsets of X, which might not contain any natural numbers.

[stiff musk]

you only need one counterexample to show that the formula isn't true in general

[kindred viper]

oh, ok, so f(1) = 0, f(2) = 2, f(3) = 0

[stiff musk]

yep, that'll do it

now, i claim that one of the containments $$f(A \cap B) \subseteq f(A) \cap f(B)$$ and $$f(A \cap B) \supseteq f(A) \cap f(B)$$ is true in general even if f is not assumed injective, do you know which one?

[glossy valve]

Bungo

[austere dome]

Well, yes, though part of me says to be more general and not fix a concrete X.

[kindred viper]

oh, ok, so the first one is true

[stiff musk]

i think you need an X with at least 3 points (and a codomain with at least 2 points), so this seems like the simplest counterexample. For a larger X, you could embed this example inside that X and ignore all except 3 points of the domain

[stiff musk]

right

so now try the other direction and see where you need injectivity

[kindred viper]

if you ask about this, then i need it on the right hand side, i have one element of a set too much

[austere dome]

2 points suffice. Take A and B to be disjoint singleton sets. f maps these to the same value. LHS is the empty set, right hand side is a set with one element.

[stiff musk]

oops misread, one moment

[austere dome]

Their intersection need not be non-empty.

[stiff musk]

right, good point

yep that works

[austere dome]

And the claim is vacuously true if some sets are empty.

[stiff musk]

seems right, surely the image of an empty set is empty

[austere dome]

By definition of the image. Yes.

[stiff musk]

can you elaborate on this?

[kindred viper]

well

given the previous example

we have {2} and {2, 0}

[stiff musk]

right

[kindred viper]

and this 0 element is there because the function is not injective

[stiff musk]

what i meant above was:

if you assume that f is injective, how does that fact needed in order to show that $$f(A) \cap f(B) \subseteq f(A \cap B)$$

[glossy valve]

Bungo

[stiff musk]

like, carry out the proof and see at what point you need injectivity to proceed

[kindred viper]

🤔

[stiff musk]

sounds like you need a nudge in the right direction to get started...?

to show that one set is a subset of another, show that an arbitrary element in the first set is also an element of the second set

so in this case:

suppose $y \in f(A) \cap f(B)$

[glossy valve]

Bungo

[stiff musk]

the goal is to show that $y \in f(A \cap B)$

[glossy valve]

Bungo

[stiff musk]

now what does it mean for $y$ to be in $f(A) \cap f(B)$?

[glossy valve]

Bungo

[kindred viper]

ok, so $y \in f(A) \wedge y \in f(B)$

[glossy valve]

Trenek

[kindred viper]

and i'm not sure how to continue

[stiff musk]

$y \in f(A)$ means that there is some $x \in A$ such that $y = f(x)$

[glossy valve]

Bungo

[stiff musk]

$y \in f(B)$ means that there is some $z \in B$ such that $y = f(z)$

[glossy valve]

Bungo

[stiff musk]

agree?

[kindred viper]

oh, ok

ok, so $x \in A \wedge z \in B$

[glossy valve]

Trenek

[stiff musk]

yep

and f sends both x and z to the same output y

[kindred viper]

indeed

what means that

f is not injective

[stiff musk]

no... we are assuming that f is injective

[kindred viper]

then

x and z must the equal

[stiff musk]

yes exactly

[kindred viper]

ok, i see it

[stiff musk]

because an injective function cannot send two different inputs to the same output

therefore since x and z are the same point we can just call it x

and it's in both A and B

[kindred viper]

sorry, i'm a bit sleepy XD

[stiff musk]

no worries!

[kindred viper]

ok, now i can see it

yes, now the rest seems fairly easy

thanks for help

[stiff musk]

sure, enjoy

[austere dome]

Dobranoc! Its nearly midnight here. Good luck!

[kindred viper]

Dobrej nocy, pięknych snów 🙂

[austere dome]

Dziękuję!

[kindred viper]

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[tardy lotus]

How to solve it?

[wise widget]

Question, what does exp() mean?

[torn jolt]

Just exponential function e^(whatever in parentheses)

Anyways

[tardy lotus]

exp() function in R Language is used to calculate the power of e i.e. e^y or we can say exponential of y

[torn jolt]

I think you need to integrate by parts + trig sub

But you have to clean it up a bit first

[tardy lotus]

How to do that?

[gritty rose]

Replace all trig functions in terms of sin, cos and simplify

Use as many double angle formulas as it takes

[tardy lotus]

I see

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[onyx glen]

@topaz valley @wise wyvern my idea about that summation came to fruition. pic inbound.

[glossy valve]

Ann

[topaz valley]

amazing

if you do the opposite inequality its bounded by above by $2\sqrt {99}$

[onyx glen]

bounded from below you mean?

[topaz valley]

,w 2sqrt(99)

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[onyx glen]

anyway what im doing here is approximating our sum by one that lends itself better to calculation

and then putting a bound on how far off i was

[topaz valley]

2sqrt(99) doesnt seem to be good enough

ig your bound works

[alpine girder]

👏

[torn jolt]

Oh wow approximating that from 1 to 99

[onyx glen]

i mean the galaxy-brain move would be to say that $\int_1^{99} \frac{\dd{x}}{x^{5/2}} \leq \int_1^{+\infty} \frac{\dd{x}}{x^{5/2}}$

[glossy valve]

Ann

[topaz valley]

so no integration required

[severe basin]

Isn't this the problem you helped me to solve before, snow?

[topaz valley]

yeah it is lol

its like the 4th or 5th time ive seen it

[severe basin]

Why is this question so famous?

[topaz valley]

[severe basin]

#help-20 message

[onyx glen]

if nobody has anything else to say i'll close this

[topaz valley]

[torn jolt]

i would like to pitch in with something

https://tenor.com/view/this-cannot-continue-stinkbugs11number1smeck-fan-cat-cat-in-water-water-cat-gif-25921319

[onyx glen]

yes?

[torn jolt]

that was it

thank you for your attention

[topaz valley]

[onyx glen]

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[stark folio]

Is it true?

[tight glen]

Yes

[stark folio]

thank you

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[tribal gust]

I have a convergent, decreasing sequence that converges to 0. I've proved that its bounded (below) but how do i show that its bound = 0

[fast peak]

well clearly 0 is a bound

and if there was a bigger bound then it wouldnt converge to 0

[tribal gust]

A sequence can have multiple bounds right?

Like for example -1 here

[fast peak]

I assume you meant greatest lower bound

[tribal gust]

Oh yeah i wasnt giving a counter point

[tribal gust]

this was just another question haha

[fast peak]

yes

-1 is another lower bound

[tribal gust]

Awesome, thank fuck we dont use supremum/infinum in this calc class

Thanks!

[fast peak]

greatest lower bound is literally the infimum

[tribal gust]

Yeah i know but we dont consider it in this class, all the question asked for was a bound

we do in proofs nd reasoning tho

Anyway, thanks again

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[tribal gust]

How do i show that the (alternating) series given by pi^n/2 cos(npi) is divergent?

[tribal gust]

If i take the limit of the series as n tends to infinity i end up having to evaluate cos(npi) but im pretty sure that limit doesnt exist

In absolute value however the series tends to infinity or atleast a non zero value

So the terms keep getting bigger and bigger

[sharp vine]

to show that the limit DNE you can try to find two different subsequences that tends to the same result and show that values of cos(npi) for them are different

[tribal gust]

Im not sure how id do that, could you maybe show me an example?

[sharp vine]

ok, for examle let's say we have limit x goes to inf of sin(x)

and we want to show it DNE

let's take two subsequences

(both divergent to inf)

xn1 = pi, 2pi, 3pi, 4pi, 5pi, 6pi, ...

and xn2 = pi/2, pi/2 + 2pi, pi/2 + 4pi, pi/2 + 6pi, ...

now it's enough to show that

f(xn1) and f(xn2) goes to different numbers

[tribal gust]

I dont really see that these are subsequences

[sharp vine]

sin(xn1) = sin(pi), sin(2pi), sin(3pi), ... = 0, 0, 0, 0 ...

so it tends to 0

[tribal gust]

oh wait, a subsequence can just be the sum of random members of the series its defined from?

I thought it had to be 'in order'

[sharp vine]

sin(xn2) = sin(pi/2), sin(pi/2 + 2pi), sin(pi/2 + 4pi) ... = 1, 1, 1, ...

so it tends to 1

1 != 0

limit DNE

[tribal gust]

so for example sin(x) from n = 0 to 10 would be a subsequence, while sin(x) for n = 0,2,4,6,.. wouldnt be

[sharp vine]

it refers to Heine definition of limit

ee maybe I said it wrong, it can be just sequence haha

[tribal gust]

😅 thanks but im not sure i get it then

[sharp vine]

just two any sequences

[tribal gust]

how do they relate, the two sequences i mean

Like how should they relate

[sharp vine]

** for all sequences **

we found 2 sequences which gives other results, so limit does not exist by definition

[tribal gust]

Sorry im still not sure i get it. I thought you wanted to show that two subseries of the series had different limits and therefore the series has no limit but i suppose thats not what you're doing?

[tribal gust]

I havent really learned about this and im trying to make do with what we saw in class

[sharp vine]

Sorry if I made you confused, I've just read you wanted to calculate the limit of cos(npi) where n goes to infinity

and you thought limits does not exist, so I suggested a way how to prove it

[tribal gust]

no of pi^n/2cos(npi)

[sharp vine]

but you ended with what?

[tribal gust]

Well i thought of just calculating the limit, since its an alternating series the limit should equal zero, if not then its divergent

The issue is that i dont know how to calculate that limit

[sharp vine]

in fact you can try same thing as I did, I guess it doesn't exist, just other sequences for this one (take limit of exponent)

[tribal gust]

okey thank you!

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[timber hatch]

Hi, could someone help me by double checking whether this is correct or not. Online calculators are giving me different answers so I'm not sure. The goal is to find the series both converges and absolutely converges.

I think it does not absolutely converge, nor converges normally in this case. Am I correct ?

[gritty rose]

Don't even need ratio test

The terms don't even converge to zero. So the series diverges by the limit test

[timber hatch]

Right, but the question asks to check for absolute convergence
A series can't technically diverge, and still absolutely converge ?

[gritty rose]

If a series diverges, it also doesn't absolutely converge

[timber hatch]

In that case I'm a bit confused about the inconclusive result of the ratio test.
If the result is 1, in which case it's inconclusive, wouldn't that mean that the series could either converge or diverge ?

[gritty rose]

Yes.

,w define inconclusive

[glossy valve]

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[timber hatch]

Oooh, right. No you're absolutely right. So any time a series diverges, the result of the ratio test could either be > 1 or = 1, but not < 1 which is the only case where it absolutely converges.

So here, doing the limit test first would've led me to the conclusion that the ratio test's result would either be = 1 or > 1, which meant that it could not be absolutely convergent

Thanks you for explaining ^^

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[thick dove]

Find the set possible values of random variable X and draw a table to show each probability

1a) a box contains 3 redmarbles, 5 green. Two are taken at random without replacement and X is the number of green marbles obtained

[thick dove]

Can someone draw a table of

x | 0,1,2
P(X=x)| no clue

And explain it pls

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[amber dust]

Paper plates cost 8$ per package and plastic utensils cost 5$ per package your supplier delivers 15 packages for 90$ if x=paper plate package and y=utensil package that would mean that there are 5 paper plate packages and 10 utensil packages how do i make a system of equations to get the coordinate pair to (5,10)

[short siren]

$x + y = 15 \
8x + 5y = 90$

[glossy valve]

VulcanOne

[short siren]

Solve both equations together

👍

[amber dust]

Gracias bro I was missing something and this was it

[wise widget]

There are two things you know about the delivery: number of packets and price of packages. And given x and y, you can write the number in terms of x and y and write the price in terms of x and y. Hence the two equations above

[amber dust]

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[barren iris]

Could anyone assist me with doing the proof for #12?

[barren iris]

From what I've been helped with, once I use the hint given and apply it to the lhs. I can start to simplify the lhs.

I understand that (z_i - zbar) will give me 0.

But what about (zbar -d)?

How do I approach that?

[ivory cairn]

z_i - zbar is not always 0

[barren iris]

If you summate z_i it will be zbar.

So, zbar - zbar = 0

[ivory cairn]

ah, yeah, if you sum and average, then yes

then you're done? the middle term 0's out
zbar - d is constant

[barren iris]

Yeah , what do you do with that?

The zbar - d?

[ivory cairn]

the 2(zbar - d)(z_i - zbar) is 0 after you apply the sum
the (zbar - d)^2 is constant, so when you apply the sum and 1/n you just get it back.

[barren iris]

Sorry, I just get what back? (Zbar -d)^2 ?

[ivory cairn]

yes

[barren iris]

(z_i - zbar)^2 would also come out to 0 right?

[ivory cairn]

no

[barren iris]

Why not?

After you summate?

And divide by n

[ivory cairn]

it's squared, you can't bring in the multiplication of 1/n into it to get (zbar - zbar)^2 like you did for the other one

[barren iris]

Hmm

[ivory cairn]

and you can't bring the sum into the quadratic either

[barren iris]

So on the lhs, I'm left with (z_i - zbar)^2 + 2(zbar - d) + (zbar - d)^2

[barren iris]

Noted

[ivory cairn]

the middle term is 0

[barren iris]

?

Wait how though

You said (zbar - d) just returns the same thing

[ivory cairn]

it's multiplied by this

[barren iris]

Oh

Derp

0 * (zbar - d)

So that's 0

Wow

So now I'm finally left with the same on the rhs

[ivory cairn]

yep

[barren iris]

So I'm finished with what I wanted to prove

Thank you so much man

That little detail was messing me up

[ivory cairn]

np

[barren iris]

Well anyway, appreciate the help. Have a good one.

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[earnest frigate]

hey

[earnest frigate]

do u think info is insufficient

[vast fossil]

c = 3
8a - 4b = -1

Well this is what can be implied

[earnest frigate]

yea

[vast fossil]

The local maximum of y would be at the point where y' = 0

So 3ax^2 + 2bx + 9 = 0 for some x

Hmm

Ah wait it's asking for the local maximum value

Knowing this x might not be necessary

[earnest frigate]

[vast fossil]

y = ax^3 + (8a + 1)x^2/4 + 3x + 5

[earnest frigate]

but then u get maxima with containg a

[vast fossil]

Maybe

[torn jolt]

doesn't "touch" mean root of multiplicity ≥2?

[vast fossil]

Oh damn I didn't see that

Then it's way easier

[earnest frigate]

is it always implied?

[vast fossil]

So y(x) = a(x + 5/4)(x + 2)^2?

[vast fossil]

Yeah it's be "intersects" otherwise

[earnest frigate]

oh

i see

in that case

lemme see

[vast fossil]

And now just pick a such that y'(0) = 3

[earnest frigate]

i see

wow

great

thanks toby n bean

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[torn jolt]

I am unsure on what the differences are between a cut set and an edge set?

[torn jolt]

like according to the definitions i am reading, they are different, but they seem to be interchangeable in a lot of situations for some reason?

[fast peak]

can you give the exact definitions you are using?

[torn jolt]

Sure

[rose rain]

oh its you again

[torn jolt]

A cut set of a graph $G$ induced by a partition of $G$'s vertices into sets $X$ and $Y$ is the set of all edges with one endpoint in $X$ and another endpoint in $Y$

\vspace{3 mm}
An edge cut of a connected graph $G$ is a set $S$ of $G$'s edges such that $G-S $ is disconected and $G-S'$ is connected for any proper subset $S'$ of $S$

[glossy valve]

♡Lex♡

[rose rain]

what's disconnected here

intersection being empty?

[fast peak]

disconnected graph. quite standard

you can't get from every vertex to every other

[rose rain]

welp time to learn graph theory

[torn jolt]

what i am getting is that every cut set is not necessarily an edge cut, because like, if $X$ itself wasn't connected, then you would need more than one edge to reconnect $G$

but i am really not sure

[glossy valve]

♡Lex♡

[fast peak]

yes

[torn jolt]

I see

that makes sense

oh yeah i suppose cut sets can be defined even for disconnected graphs i guess?

which is also another distinction

[fast peak]

that aswell, yes

[torn jolt]

alright thanks for the clear up!

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[torn jolt]

Oh yeah just another quick clear up, but is null graph a subgraph of every graph?

[torn jolt]

According to what i am reading, a subgraph can be defined as ""A graph G is a subgraph of graph H if the nodes of G are a subset of the nodes of H, and the edges of G are a subset of the edges of H""

because a null graph just means empty sets for the vertices and edges, which is just like, phi

[fast peak]

well the empty set of nodes is clearly a subset of the nodes of G

same for the edges

[torn jolt]

yeah thats fair thanks again!

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[torn jolt]

oh but from what i read, some people tend to say that the set of vertices need to be nonempty

i guess it doesnt matter either way?

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[fast peak]

well empty stuff is often not interesting

or it might ruin theorems

[torn jolt]

yeah fairly isn't haha

this made me think

this is just a question i came up with, but how would we figure out how many induced subgraphs are there of a graph of n vertices?

[fast peak]

number of choices of subsets of vertices

[torn jolt]

oh yeah i just realised how non-general my question is

since i am supposing if the graph has more than n vertices, we cant really know can we

wait no

uh

i am confusing myself

[torn jolt]

yeah that does make sense in my head too

ty

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[fervent mirage]

Can anyone explain how this came pls ping

[clear lily]

multiply by 12 on each side of the equation

[golden furnace]

@fervent mirage

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[mortal sinew]

bit confused on this i can do the basics of inverse functions

[golden furnace]

I assume that notation is composition

then remember than when
f: A → B and g: B → C, then (f o g): A → B

[onyx glen]

no

it's g o f : A -> B

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[mortal sinew]

Hi sorry had to do something, i am still a bit confused on how to answer this..

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[fervent mirage]

Not that mann

That a/80-56 thing

[fervent mirage]

Yes ?

[golden furnace]

you said pls ping

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[fossil temple]

This is the bernoulli inequality right

Can x be a negative number?

I need to show that this converges with the bernoulli inequality and sandwich

[devout valley]

Needs to be at least -1

[fossil temple]

ahh alright

[devout valley]

Just in case my answer wasn’t clear, x can be negative, provided it’s at least -1

[fossil temple]

yeah I think it's always >= -1 because n => 1

thanks

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[hidden saffron]

Determine the intercepts of the graphs of the following polynomial functions:

1. y = x^3 + x^7 - 12x
2. y = (x-2)(x - 1)(x + 3)

[hidden saffron]

Im tryna do this but im still confused I keep getting the wrong answers

[silver pasture]

Workings?

[vast fossil]

Do x3 and x7 mean x * 3 and x * 7 ?

[silver pasture]

Surely not

Why write x3 then 12x

[hidden saffron]

Oh x raised to 3 like tht

Lemme rewrite it

[vast fossil]

x^3 and x^7 then

[hidden saffron]

Yup

[vast fossil]

Show your work

[hidden saffron]

Still learning it so yea

All I know no.1 is wrong

[vast fossil]

You need to find x-intercepts too I think

[hidden saffron]

Its supposed to be 0

[vast fossil]

Btw 12x = 0 when x = 0

[zenith frost]

Hey you have made a mistake in last literal of question-1

put x=0

[hidden saffron]

Oh

So its gonna be 12(0)??

[vast fossil]

Yes

[hidden saffron]

Ahh alright

[zenith frost]

for question-2, when you put y=0 it gives you x intercept but you have written as Y intercept in the last line

[hidden saffron]

Oh

I need to write it as x-intercept then

?

[zenith frost]

y intercept in question-2 will be 6

[zenith frost]

Yeah mate

[hidden saffron]

Oh alr tyyy

Well now I just have to correct my mistakes then

Ty for the help:D

[zenith frost]

Glad to help mate

Keep learning

[hidden saffron]

Aight

I will

[zenith frost]

Reach out to me if you have any doubt

[hidden saffron]

Will do

[silver pasture]

Wow and i was trying to find where those two meet

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[upbeat root]

I need help with this please

[rocky vale]

Have you learned the hinge theorem?

[full forum]

@upbeat root Has your question been resolved?

[upbeat root]

Yes

[rocky vale]

Do you see how it might be helpful?

[upbeat root]

kindaaa

srry i took that long to write that

lol

[rocky vale]

Try drawing a parallelogram ABCD

with the diagonals

[rocky vale]

so

which of your angles are the smaller ones?

[upbeat root]

the adjacent angles?

oh wait

lets say A

[rocky vale]

In your picture, it looks more like B and C are the smaller acute angles

and A and D are obtuse

It might help to make the parallelogram more slanty so the difference is more obvious

but either way

compare triangles ABD and CDB

[upbeat root]

they both have congruent sides

oh wait

Im going to say A and D are the smaller angles and B and C are the larger ones

that means B is going to be bigger than C

So from angle B its diagonal is longer than the diagonal from angle C?

[rocky vale]

angle B and C share the same diagonal, don't they?

That is, BC is a diagonal

in your picture

[upbeat root]

yeaa

The diagonal AD >CB, <B would be larger than <D because ABD had two small angles and CDB had two large angles.

Idk if this is right

Srry for my bad drawing

[rocky vale]

What you're saying doesn't really match the picture

In the picture, B and C are the small angles

They are acute

A and D are obtuse

[upbeat root]

Do I have the right idea though?

[rocky vale]

not exactly

[upbeat root]

dang it

[rocky vale]

To use the hinge theorem, you want to use two triangles, and first show that those two triangles have two congruent sides

so triangles ABD and CDB

do you see any congruent sides?

[upbeat root]

congruent sides are from the diagonals

[rocky vale]

hm

No, the diagonals are different lengths.

In fact, that's what we're trying to prove

that BC (joining the smaller angles) is longer than AD (joining the bigger angles)

the congruent sides are going to be from the sides of the parallelogram

[upbeat root]

AB and CD are congruent

AC and BD are congruent

[rocky vale]

Yes, that's good

[rocky vale]

this is true, but AC is not a side of either triangle we're looking at

Look specifically at triangles ABD and CDB

Notice that BD is a side of both triangles

[upbeat root]

oh reflexive property

[rocky vale]

right

So you have AB cong to CD
and BD cong to DB

Now do you see how we can use the hinge theorem?

[upbeat root]

If I said D was the smaller angle and angle B is the larger angle then AD>CB

[rocky vale]

yes

[upbeat root]

Can I write the proof out on paper and you check it?

Last thing

[rocky vale]

yeah sure

i might be off for a few minutes, go ahead and ping me

[upbeat root]

I know how to prove this but im having a hard time trying to write it like a proof

@rocky vale

Should I include that <A and <D are the smaller angles and <B and <C are the larger angles in my given statement

[rocky vale]

yes, but again that's the opposite of what you drew

I'd do the other way around

[upbeat root]

Idk when I should be talking about the triangles

Where do I put it

After that I just use the hinge theorem

[rocky vale]

Yes, I'd say something like "hinge theorem on triangles ABD and CDB"

[upbeat root]

Ok thanks I got it now

Sorry it took so long

Thank you for your help

[rocky vale]

np 👍

[upbeat root]

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[sleek brook]

can you use completing square on sin and cosine?

[fast peak]

can you give an example of when you would want to apply it?

you can use completing the square whenever you have something of the form a^2+a*(something). whether it's useful is another question

[sleek brook]

can i use complete square to solve this?

instead of the way he did it

after completing the square then finish it with arctan

[fast peak]

arctan?

isn't the sign wrong for that?

[sleek brook]

if we complete the square

[rotund birch]

yes it seems possible

[sleek brook]

of the dom

I can continue but my question is can i use complete square woth trig?

with

[fast peak]

yes

trig doesn't matter

[sleek brook]

alright thank you

[fast peak]

but still, the sign is wrong

[sleek brook]

which sign

[fast peak]

of the -1/4

[rotund birch]

well you need to know how to integrate 1/(1-x^2)

if you are gonna do it that way

[sleek brook]

yes thats with the arctan

integration

[rotund birch]

you can just multiply -1 to get it inthat form

[fast peak]

in the end it will turn into essentially the same thing as what is done in the picture

[rotund birch]

no thats not arctan

arctan has 1 + x^2 in denominator

on this we have 1-x^2 on denominator

its entirely different

[sleek brook]

i mean

[rotund birch]

let me check what it is rq

[sleek brook]

tanh^-1

the hyperbolic

form of arctan

[rotund birch]

oh yeah

[sleek brook]

yep

alright thank you all

[rotund birch]

sign is not really an issue here i believe

since you can just multiply by -1

[sleek brook]

yes we can add a negative outside

yes

so i believe it’s possible

[rotund birch]

also integral 1/(a^2 - x^2) is also equal to [ ln|(x-a)/(x+a)|/2a ] + c

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[fathom void]

its asking for domain of the function i dont even know where to begin 😭 how do u do this

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@fathom void Has your question been resolved?

[fathom void]

@floral scroll

damn mb

<@&286206848099549185>