help-27

[spice linden]

and If yes how would I prove this?

[rich summit]

Decreasing doesn't necessarily mean the limit is 0

[spice linden]

how do I check if it's bounded

[spice linden]

decreasing and never being negative?

[rich summit]

1, 1 - 1/4, 1 - 1/4 - 1/8, 1 - 1/4 - 1/8 - 1/16 is also decreasing, but doesn't converge to 0

[spice linden]

?

I think you calculated the first terms incorrectly

[rich summit]

no, this is another sequence to demonstrate the principle

Have you proven a_n is decreasing btw?

[spice linden]

no

that's what I'm asking

how do I prove it

i can see that it is decreasing

but I don't know how to prove it

[wooden veldt]

This is fine you just need it to be eventually decreasing

[spice linden]

it seems that it is.

but how do I prove it though

[wooden veldt]

Induction

[spice linden]

ahh ok

lemme try it

but

what would I use as my induction basis?

i thought I have to use the smallest/first term

[spice linden]

and it doesn't really follow the decreasing pattern in the first few terms

[rich summit]

You can start anywhere with your induction basis, but you will only prove it's decreasing for numbers larger than your induction basis

[spice linden]

which is all I need

dope

[wooden veldt]

Yeah when doing your induction you will probably need n to be bigger than something

So your base case will whatever that is +1

[spice linden]

it would be for n>3

the decreasing pattern holds from the 3 term onwards

[rich summit]

Yeah, so for the inductive step assume that a_(n + 2) < a_(n + 1) < a_n

[spice linden]

why not just a_n+1<a_n?

wait nvm

its just the two steps in one

[rich summit]

yeah, this way you can do induction in one go, otherwise there might things like odd and even separately

[spice linden]

ok i'm working on it

[spice linden]

bro im struggly with the induction part

where do I go from here? have I done something wrong?

[rich summit]

The trick is to rewrite the recursion this way $$a_{n+2}<\frac{1}{3}a_{n}+\frac{1}{2}a_{n}$$

[woven radish]

Jelle

[spice linden]

$$a_{n+2}<\frac{1}{3}a_{n}+\frac{1}{2}a_{n-1}$$

[woven radish]

AzerimA

[rich summit]

You leave the 1/2a_n unchanged

[spice linden]

oh

is it because we know that a_n is larger than a_n+1 so if we replace it with something bigger the equation still holds?

[rich summit]

Yeah, a_n > a_n+1

[spice linden]

and that's valid cuz it's our induction assumption right?

[rich summit]

Yeah, I think the best way is to do strong induction actually and assume the sequence is decreasing up to a_(n+2)

[rich summit]

because there is a gap between a_(n + 2) and a_n

[spice linden]

what is strong induction?

[rich summit]

You assume the inductive hypothesis is true for all k <= n and then prove it will also be true for n + 1

[spice linden]

is this the only difference?

and we can just assume that?

no problem?

[rich summit]

Yeah, well N_0 <= k <= n at least

where you started at N_0 with your induction basis

but in this case assuming the previous 2 values also works

[spice linden]

my N_0 is 3 in my case

but it doesn't tho

a_3 is larger than a_2

and a_2 is larger than a_1 and it is larger than a_0

wait nvm

[rich summit]

just assume the previous 2 values, that's easier

[spice linden]

a_0 < a_1 and a_1 > a_2 and a_3 > a_2

[spice linden]

which two values?

u mean a_0 and a_1?

[rich summit]

assume it's true for a_n and a_n+1, then it will be true for a_n+2, but also still for a_n+1

[spice linden]

I don't understand what you mean by that

[rich summit]

You know it's true for a_n and a_n+1, then you want to prove it's also true for a_n+1 and a_n+2, you keep shifting them a place

[spice linden]

yes

[rich summit]

I just noticed that my way only proves a_n+2 < a_n for all n, not a_n+1 < a_n, sorry

[spice linden]

bro I don't even understand the problem u r dealing with rn

that's bad

[rich summit]

I think you have to do induction on 2 bounds at the same time

I suspect you can prove a_n+1 > 0 and a_n+1 < 5/6a_n or something like that

[spice linden]

what does that mean

[spice linden]

bro I'm gonna go to sleep. thanks for you help.

[rich summit]

You can open a new channel if you want tomorrow, hopefully somebody else can help

[spice linden]

if you found out how it is supposed to be solved, drop me a dm

if u end up working on it

[rich summit]

yeah

[spice linden]

ok thanks for your time bro

gn

!close

[rich summit]

.close

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[humble fractal]

Can someone help me with part b please?

[humble fractal]

<@&286206848099549185>

[dense jay]

whats up?

[humble fractal]

I got stuck on part b

Can you help me please?
@dense jay

[dense jay]

have you drawn a diagram of some kind for the scenario?

[humble fractal]

yeah

[dense jay]

may i see

try take a pivot at c

[humble fractal]

my calculation and my diagram

@dense jay

[dense jay]

you didnt multiply Td by its distance from c

[humble fractal]

oh okay I see

Can I dm to ask some maths question please, if you think that is okay

[dense jay]

i dont do dm

[humble fractal]

oh okay

Can you help me with this question please?
https://discord.com/channels/268882317391429632/1174450473944748142

@dense jay

Thank you

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[heady coral]

Hi, I'm trying to understand this proof. My question is: If I'm using contraposition, how should I write the statement? For example, if 'x' is a negative number and 'n' is an even number, then 'x^n' is a negative number, the contraposition will be: if x^n is not a negative number, then x is not a negative number or n is not an even number

[upper schooner]

Well just "plainly", it'd be "if there isn't at least one box with at least ceil{N/k} objects, then there aren't N objects in the k boxes"

[heady coral]

:(

Is there another form to show it? contraposition is weird :((

[upper schooner]

It can be a bit weird at times you could e.g. phrase this one as a proof by contradiction, which basically does the exact same thing

[patent marsh]

was going to be my suggestion.

[upper schooner]

In fact, the proof is pretty much identical

[heady coral]

[upper schooner]

you're not happy with proof by contradiction?

[heady coral]

Yes, but if is similar to the other then I still without understand it

[upper schooner]

Well basically for by contradiction, you'd assume that N objects are in the k boxes, but that you don't have at least one box that has ceil{N/k} objects

Everything they said basically then gets you to that within the k boxes, you must have less than N objects

But that contradicts what they've shown

It is a bit annoying here in this case negating the statements

[heady coral]

Why I'm adding and resting the one in the yellow part?, and the light blue part is definition of ceil function?, how that inequality shows me that I have less N objects? I still without understand why that is a contradiction

[upper schooner]

For the blue part, if N/k is an integer then you know that ceil{N/k} = N/k, and that’s strictly less than (N/k) + 1

If N/k is not an integer then you know that because ceil{N/k} is the next integer greater than N/k, you have ceil{N/k} - N/k < 1 (because the distance between any two integers is at most 1)

You’ve basically assumed that no boxes have at least ceil{N/k} objects, so all boxes must have less than ceil{N/k} objects, and cause the quantity of objects is integer you then would need to have ceil{N/k} - 1 objects or less

The “largest” possibility is that all of those k boxes have the ceil{N/k} - 1 objects, which is how they get this:

[upper schooner]

They then basically replace the ceil{N/k} with the blue part you’ve mentioned, which is how the yellow ends up appearing

If any of that makes it clear? (I’m assuming not )

[heady coral]

I'm reading, english is not my native language

[upper schooner]

do let me know if i need to explain anything

[heady coral]

If N/k is not an integer why do you say that the distance between any two integers is at most 1?

[upper schooner]

N/k is in between two integers basically, so like the worst case you could have is something like this (one second)

Excuse the bad drawing, on mobile

But like say N/k isn’t an integer, then the ceiling is you rounding it up

But that green distance has to be less than 1, otherwise you’d be able to find another integer that would be closer to N/k

If that makes it any clearer?

[heady coral]

yay! yes I understand it :D

[upper schooner]

perfect

[heady coral]

How can I replace it if I isn't a equality?

[upper schooner]

Because it’s an inequality and k is positive, you just keep the < as is

[heady coral]

yes yes, sorry for being slow

[upper schooner]

It happens don’t be sorry, we’re all here to help you!

[heady coral]

So I get that my total number of objects is < than my N objects, and that is the contradiction?

[upper schooner]

Yep, you assumed that there were N objects to begin with, but then conclude that there aren’t N objects

But that’s impossible

[heady coral]

Oooh

https://tenor.com/view/excited-so-drums-gif-11982656

I understand!!!!!!!!!!!!

[upper schooner]

perfect

[heady coral]

thank you so much <<<333

[upper schooner]

Always a pleasure to help you

[heady coral]

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[raven swift]

let P be a polynomial fuction and p(x)=x^4-mx^3+4x^2-6x+8 if (x-2) is a factor what is value of m?
help?

[tribal sierra]

what have u tried

if x-2 is a factor what do u know about p(x)

[crisp niche]

then due to Bezout Theorem P(2) = 0

[raven swift]

i plug in x for 2

and get 28-8m=0

solve for m?

which is 7/2

[tribal sierra]

yes

[raven swift]

do i write -7/2

or pos 7/2

[tribal sierra]

7/2

[raven swift]

if it was negative

would the negatives cancel out

[tribal sierra]

bezouts theorem ?

[raven swift]

and it would be plus?

[tribal sierra]

wat

[raven swift]

like if it was negative 7/2

[crisp niche]

yes

[tribal sierra]

just replace m with 7/2

[raven swift]

it would be -(-7/2)

[tribal sierra]

if its -m its -7/2

[raven swift]

which is 7/2

[raven swift]

is m 7/2

or neg 7/2

[tribal sierra]

isnt it factor theorem ?

[tribal sierra]

u just solved for m and its 7/2

so m = 7/2

[crisp niche]

in english literature it is caleld Polynomial remainder theorem

[raven swift]

what if m was -7/2?

would it be -(-7/2)

which is +7/2

[tribal sierra]

yes

[raven swift]

so that is the method for those types of questions?

[tribal sierra]

idk factor theorem ?

[crisp niche]

In algebra, the polynomial remainder theorem or little Bézout's theorem

[raven swift]

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[tribal sierra]

oh little Bézout's theorem is diff to bezouts theorem probs

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[jaunty mantle]

What’s the definition of a factorial?

[trail trench]

idk but this was the answer \

i dont know how it got that

[jaunty mantle]

Well how can you simplify something if you don’t know what it means

What does ! Mean

[trail trench]

idk how to explain it with a variable but i know 5! is 1 * 2 * 3 * 4 * 5

[tribal sierra]

write (2n+1)! in terms of (2n)! * something

[trail trench]

like replace n with a number?

[jaunty mantle]

So what would (n)! Mean

[trail trench]

idk i guess n * n ?

[tribal sierra]

is 5! = 5 * 4 * 3 * 2 * 1

[trail trench]

im lost cause ive never done a factorial with a variabl;e

[tribal sierra]

n! = ?

[jaunty mantle]

What’s 3!

[trail trench]

123

oops

1 * 2 * 3

[jaunty mantle]

What’s 4!

[trail trench]

1 * 2 * 3 * 4

[jaunty mantle]

What’s 10!

Let me answer this

[trail trench]

lmao

[jaunty mantle]

I would say 1 * 2 * 3 * … * 9 * 10

[trail trench]

mhm

[jaunty mantle]

(There’s a reason)

Now what would n! be?

[trail trench]

i have no clue im so lost

[jaunty mantle]

Or perhaps what would 20! Look like?

[jaunty mantle]

Knowing this

[trail trench]

n * n *n ... n ?

[jaunty mantle]

Try 20! First

[trail trench]

like write it out or simplify it

[tribal sierra]

if n=5 then you are saying 5! = 5 * 5 * 5 * 5 * 5

[jaunty mantle]

Do what I did here for 10!

[trail trench]

i thought that would be 5^5

[tribal sierra]

it would be

[jaunty mantle]

Let’s try 20! First

[tribal sierra]

im saying based on ur definition here (which is why its wrong)

[trail trench]

oh ok

um ok

so

1 * 2 * 3 ... 19 * 20

[jaunty mantle]

What about 25!

[trail trench]

1 * 2 * 3 ... 24 * 25

[jaunty mantle]

So a cool thing you might notice is you always start with 1, 2, 3, etc

[trail trench]

mhm?

[jaunty mantle]

(Unless you’re asked 2! In which case it ends at 1 * 2)

[trail trench]

mhm

[jaunty mantle]

So in general we might want to say

n! = 1 * 2 * 3 * … * (n-1) * n

Does that make sense?

[trail trench]

why is it n -1

[jaunty mantle]

Why is it 9

[trail trench]

oh ok

[jaunty mantle]

Why is it 19

[trail trench]

i get it

[jaunty mantle]

Why is it 24

Yeah right

So what would (n+1)! Look like?

[trail trench]

so n is the final number

um

(n-1) * n * (n+1) ?

[jaunty mantle]

With the 1 * 2 * 3 * … at the front

Now we might notice the first n terms just make n!

So then (n+1)! = n! * (n+1)

[trail trench]

they ar ethe same?

[jaunty mantle]

Well (n+1)! = 1 * 2 * 3 * … * (n-1) * n * (n+1)

Do you agree?

[trail trench]

yeah

[jaunty mantle]

Well that first part is just n!

[trail trench]

oh so going back to this i get it now

[jaunty mantle]

Yep!

All you needed was to know what the ! Meant

[trail trench]

okok i get it now

yeah i see ok thanks

.close

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[royal laurel]

So the teacher was trying to find the solution of x^3=1

Yet, it so singular that it seems like he forecasted the solution

I mean how does he knew k=1,2 are the solutions?

[royal laurel]

It is impossible

[crisp niche]

the root of n-th degree is a set of n eleemnts in comlpelx number set

so cube root is a set consisted of three elements

[royal laurel]

So the “k” would always be integer, I suppose

[crisp niche]

yes, we start from k = 0

so if we have three elements, you write

k=0, k = 1, k = 2

you use , of course, the formula for elements of nht-root, that can be also derived fm de Moivre theoerm but to compute, you may use ready formula

it can also be interpreted as the vertices n of a regular polygon that lie on a circle with a radius equal to the arithmetic root n of this degree

so if oyu have cube root, you take 360 degrees and you divide by 3, so roots lie on 0, 120, 240, degress

i just simpklify my explanaiton

[royal laurel]

Why it is a regular polygon?

[crisp niche]

to have all angles equal

[royal laurel]

I can’t understand why they have to be equivalent

Is there any property of nth-root that attributes to this phenomenon

[crisp niche]

i wouild start explanations from de Moivre theorem, do you know it ?

[royal laurel]

I do, it basically just many of Z being multiplying together IMO

[royal laurel]

which we know how to compute the result

[crisp niche]

then if you replace n with 1/n, you get for z^(1/n) right ?

[royal laurel]

yes

[crisp niche]

and then you can derive the formula for comlpelx roots

[royal laurel]

emm

let me try

[crisp niche]

arguemnt of cosne and sine wil be (ϕ + 2k𝝅) / n, where k = 0, , 2, , ... n - 1

[crisp niche]

and ϕ = arg z, arguemtn fo comlelx number z

yes exactly

but you need to rememebr ab the period

of trig fucntions

[royal laurel]

im quite famillar with them

But is it related to find the nth-root

[crisp niche]

yes

[royal laurel]

I have no clue, I still can't understand why these "angles have to be equivalent"

which result in a regular polygon

[crisp niche]

let's start this frm : let x = Re(z^1/n) and y = Im(z^1/n)

then you can derive the eqautiuon of the circle

[royal laurel]

Re? lm?

[crisp niche]

yes real part and imagjanry part

[royal laurel]

i see

[crisp niche]

and take sum fo thier squaRES

then you get eqaution of circle

[royal laurel]

squaRES?

[crisp niche]

squares

i meant

[royal laurel]

i see

[crisp niche]

x^2 + y^2 = R^2

where radius R is arithmetic root

[royal laurel]

that is the distance from it to the origin

[crisp niche]

yes

that shwos oyu relation

to this circle

where vertices of polygon lie

in other words, nicer words, every nth root of comexl number is a vertice

of such polygon

but due to tri periods they rep[eat itself

how to say, wioth sme angle disdtance

repeat every same number of degrees

to be exact thsi angle distance is

2kPi / n

look

cos (( θ + 2kPi)/n) , the period

si what i wrote above

[royal laurel]

im not sure what do you mean by "take sum for their squares"

[royal laurel]

is it what you mean, which presented on the paper

[crisp niche]

no

[royal laurel]

then i suppose you mean |Z|^2?

[crisp niche]

you wrote de moivre

it s good

adn now let x be your real part of you wrote

and y be an imaginary part

of what you wrote

and then do this x^2 + y^2

sum of sqaures

[royal laurel]

I see

[crisp niche]

to be faster i wrote what you shud get =

x^2 + y^2 = R^2, whhere R is the nth root of z

[royal laurel]

that is (a+b)(a-b)

[crisp niche]

no ) i msut make pdf

[royal laurel]

ohhh

[royal laurel]

<@&286206848099549185> can someone help continue the proving process

[royal laurel]

we're trying to prove that it is true

[crisp niche]

this is the part that shows that the root lies on a circle of radius R

and now, from the period of trigonometric functions, we see that the roots are the vertices of a regular polygon

the base period is equal T = 2𝝅 / n

[royal laurel]

I see, as x^2 + y^2 = R will form a circle on the plane.

It is terrific

[crisp niche]

trig idenity si also used among calcualtiosn but i omit those clear calcualtions

[royal laurel]

I see

Thank you so much for the proof of it

It is clear and intuitive

[crisp niche]

but the conclusion you shdu take from it, is that all roots of comlelx number lies on this circle

[royal laurel]

And the distances between roots are all equivalent on the plane

[crisp niche]

yes yes

[royal laurel]

Which might result in a line, triangle, or polygon

[crisp niche]

thai is conclusion trival one

if you can see all circle in your mind, then you can compute those roots

jsut in mind

no need formula

[royal laurel]

“Formula” what specific formula is it

[crisp niche]

in my pdf, first line

[royal laurel]

I see

I think the geometry perspective is much better and clearer

[crisp niche]

yes but soem pl prefer to computer on numbers )

[royal laurel]

Thank you for explaining those

[crisp niche]

that depends on yoru audience, students

[royal laurel]

I will add them to my note, thank you

[crisp niche]

yvw)

[royal laurel]

.close

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[bleak tiger]

is there any way that we can get a range of 70 quantile when mean = 80 sd = 10 in standard normal distribution

[brisk totem]

might want a better word than range

and the answer is z table

[bleak tiger]

not using a z table

im so confused

Can we get a "range of X Random varibable" w/o using a ztable?

[brisk totem]

eh, you’re supposed to memorize 1/2/3 stds away and in general it’s intractable do to by hand

[bleak tiger]

im so sorry for bothering but could you show me. an example?

[devout snow]

@bleak tiger Has your question been resolved?

[devout snow]

@bleak tiger Has your question been resolved?

[devout snow]

@bleak tiger Has your question been resolved?

[devout snow]

@bleak tiger Has your question been resolved?

[devout snow]

@bleak tiger Has your question been resolved?

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[stone fjord]

Hey I need help with proving this:

Let 𝐴 ⊆ ℝ be non-empty and upper bounded. Show that there then exists a sequence (𝑎_𝑛) with 𝑎_𝑛 ∈ 𝐴 for all 𝑛 ∈ ℕ and lim𝑛→∞ 𝑎_𝑛 = sup𝐴.

I tried to define a sequence (a_n) which is defined on A and is corresponding to the Monotonicity criterion so that a_n < a_(n+1)

But I think I'm missing something and doing it completly wrong

[stone fjord]

I did a bit of research and I think i need to work with the upper bounds of A. So that E contains all upper bounds of A so that, a_n < I (with I ∈ E)

and we also have sup A = s < I

ok

I had it in a lecture but i think i didnt got it very well

i think i got it

you explained it well enough to get me further

that's what i needed

thank you!

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[summer edge]

This look right?

I did it in my head

Here ill do it on paper rq

[tribal sierra]

.rrcw

[dry oxide]

,rccw

[dry oxide]

,w derivative of \frac{2}{x-2}

[woven radish]

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[dry oxide]

@summer edge

[summer edge]

Thx

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[astral steppe]

Discrete math and transitive help her.

Let's say we have the relation {<1,2>,<2,3>,<1,3>,<4,5>}. If we only had {<1,2>,<2,3>,<1,3>} I know it's transitive. But is it still transitive with the <4,5>?

[sonic smelt]

Yeah, the condition of transitivity is not contradicted by including <4, 5>

[wooden veldt]

Yes, <4,5> does not produce any new chains like <x,y> AND <y,z>

[astral steppe]

ok

But let's say we add <5,6> in there

is it not transitive now?

Since we don't have the <4,6>

@wooden veldt @sonic smelt

[wooden veldt]

Correct

[sonic smelt]

Indeed

[astral steppe]

Ok thanks for clearing that up for me!

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[astral steppe]

.reopen

[devout snow]

✅

[astral steppe]

{<1,1>,<2,2>,<1,2>,<3,3>,<4,4>,<3,4>} Why is this one transitive @wooden veldt @sonic smelt ?

[sonic smelt]

Why should it not be transitive?

[astral steppe]

the definition is if x related to y and y related to z, then x must be related to z right?

[sonic smelt]

Right

[astral steppe]

So it's just because y is not related to any z?

[lethal walrus]

hello

[sonic smelt]

What do you mean by that?

[lethal walrus]

can someone help me on math homework

[sonic smelt]

Open your own channel, #❓how-to-get-help

[lethal walrus]

oh sorry

[astral steppe]

well 1 is related to 2, and 2 is not related to anything else then 2

[sonic smelt]

Yeah

[astral steppe]

therefor it's transitive ?

[sonic smelt]

If you have x = y or y = z, then the condition of transitivity becomes trivially true

[astral steppe]

ok

ty

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[dim prawn]

Taken

[dim prawn]

How do i know if this function is even or odd?

[supple knot]

,rotate

[supple knot]

find f(-x) and see if it equals f(x), which would make it even, or if it equals -f(x), which would make it odd

[dim prawn]

How do i do this in piecewise funcs?

[supple knot]

wherever you see an x, replace it with -x

[dim prawn]

,rotate

This way?

[dim prawn]

,rotate

[supple knot]

yup

[dim prawn]

So it’s even?

[supple knot]

simplify the inequalities

[dim prawn]

Oh

[supple knot]

which one of these does f(-x) equal?

[supple knot]

,rotate

[supple knot]

looks good

[dim prawn]

Thanku

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[keen crane]

Is it right that limit of 1/+inf = 1/-inf = 0?

[drifting sierra]

1/+inf and 1/-inf are not numbers

[sonic smelt]

If you meant this
[ \lim_{x\to\infty}\frac1x = \lim_{x\to-\infty}\frac1x = 0 ]
Then, yes, this is right

[woven radish]

A Lonely Bean

[keen crane]

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[lime dome]

f(n) = 2npi+pi/2

[lime dome]

why is this function one to one and onto?

sinx=1

x=2npi+pi/2

sin(2npi+pi/2)=1

thats what im supposed to find

but

idk why its one to one or onto

[lavish sigil]

Assuming the domain is integers, what's your codomain

Technically you can restrict the codomain of any function and that would make it a bijection

[lime dome]

it'd be irrationals ?

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[delicate wind]

What are these type of equations called? When I look up adding fractions on khan its not the same as these

[winged hazel]

That’s just adding, subtracting, and multiplying numbers

It’s not really a type of equation, it’s just addition subtraction and multiplication

[delicate wind]

How do you go about doing it without a calculator? For the first addition I added each row individually and got the correct number but the second one was way off

[winged hazel]

I mean there’s lots of ways to go about it, there’s probably people better suited to answer your question than me, I just kinda do it so I don’t have a great explanation of how

[topaz axle]

each row individually sounds like the right way and the wrong way at the same time

but also "the wrong way" should work too

[delicate wind]

Well the first one I added each row from the right So 1+7+1 9 then 7+1+1 9 then 5+2+1 and 899

but the second one doing that I got like 181824

[topaz axle]

yeah that's the right way, each column

[winged hazel]

Are you remembering that when something adds up to say 11, you need to carry a 1 to the next decimal place and add it there?

[delicate wind]

So the 2nd one I star far right 5+8+2+9 = 24

[topaz axle]

yeah, you should get 2 digits, you write the first digit above first row

[delicate wind]

Carry the 2?

[topaz axle]

yes

[delicate wind]

To the next row or first row?

[topaz axle]

4 goes in the answer

[delicate wind]

okay

okay perfect I got it

so it the subrtracting the same

is*

anyone here

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[atomic shale]

Hello

[atomic shale]

nvm solved it myself lmao

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[restive river]

I'm asked to determine period, amplitude and the eq. of the axis of the trig function

then to write equation of function

I've determined the following so far

period: 4pi
k: 1/2
q: -2

how do I determine if this is a sine, cosine or tan graph?

or if it's been shifted

[devout snow]

@restive river Has your question been resolved?

[restive river]

<@&286206848099549185>

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[lavish cypress]

Is this right correct me if I’m not

[stark sinew]

Looks good!

[devout snow]

@lavish cypress Has your question been resolved?

[lavish cypress]

Would it be 15 or 30

[stark sinew]

what have you tried? Also your previous question was resolved so would have been better to open a new channel

[lavish cypress]

oh i think its 30

cause inscribed

[devout snow]

@lavish cypress Has your question been resolved?

[devout snow]

@lavish cypress Has your question been resolved?

[lavish cypress]

https://media.discordapp.net/attachments/1147269436869451856/1174917535041343498/IMG_8572.jpg?ex=65695639&is=6556e139&hm=0aa3a1879d7739f9f1eb136857cab44559042fc30ce2af3a763776f9d0c4b58f&

[devout snow]

@lavish cypress Has your question been resolved?

[restive river]

@lavish cypress I think the answer is second option

[woven radish]

Aditya gamer

[restive river]

@lavish cypress

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[devout snow]

@atomic rivet Has your question been resolved?

[atomic rivet]

<@&286206848099549185>

[devout snow]

@atomic rivet Has your question been resolved?

[atomic rivet]

<@&286206848099549185>

[devout snow]

@atomic rivet Has your question been resolved?

[atomic rivet]

<@&286206848099549185>

[frozen aurora]

looks like symmetric polynomials to me

[atomic rivet]

whts tht?

i think we find

x^2y^2z^2 using the equation

[frozen aurora]

or actually nvm

[atomic rivet]

yeah

[frozen aurora]

you should get $xyz(x-y-z)=0$ and $xy-xz-yz=0$

[woven radish]

artemetra

[frozen aurora]

maybe vieta should work somehow

[atomic rivet]

but how?

[frozen aurora]

let me think

[atomic rivet]

@frozen aurora have u cooked the solution

[atomic rivet]

<@&286206848099549185>

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[fervent sundial]

https://cdn.discordapp.com/attachments/1081837589780234241/1175047510872563803/image.png?ex=6569cf45&is=65575a45&hm=535e4dde068a7ee35495a19c7b82c5bfae71f1753c3b8ee89aed89ee1da1c3fb&

[fervent sundial]

Do you just have to have the triplets memorized or something? Lmao

or have euclid's proof for generating triplets memorized?

[winter patrol]

well you should know the most basic one (and you could scale that up)
you don't need to consider that much since 24 is so low

3,4,5
5,12,13
are ones you should remember

[fervent sundial]

what if i didn't remember them?

[wraith horizon]

mb, I misread the question

[fervent sundial]

lol

[winter patrol]

if you can't remember 3,4,5
you're stuffed

[fervent sundial]

i know 3,4, and 5

i don't see why i need 5,12, and 13 though

not so common

[winter patrol]

it comes up quite frequently in trig/geo questions

so worth remembering too

generating some takes a bit more work, but isn't actually that bad

[fervent sundial]

hmm sure

thanks

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[real topaz]

Btw
Area of isosceles triangle is
1/2 base x height

But here why do they
Take height = base ?

[real topaz]

I am referring to the last solution in the second pic

[devout snow]

@real topaz Has your question been resolved?

[near trout]

it's isoceles, meaning two sides are equal

but it's a right-angled triangle too

so the two short sides must be the equal ones

because the hypotenuse is gonna be longer

so base = height

@real topaz

[real topaz]

Ah ok

[real topaz]

I was a bit confused because they didn't equal the base and height in other question that was similar to this one

[near trout]

right

[real topaz]

This

It's not equal here

[real topaz]

.

[drifting sierra]

I think you're confused about what the "base" is

[real topaz]

Hm?

[drifting sierra]

The formula 1/2 base * height works for any triangle, not just isosceles

[real topaz]

Yes

[drifting sierra]

Ok here is an isosceles triangle

And here is the same but on its side

The "base" of an isosceles triangle can refer to the side that is not equal to the other two, but here the base is just for calculating area

It can be any side, as long as you pick the corresponding height (which is perpendicular)

[real topaz]

Oh ok

I understand it now :]

[drifting sierra]

And if the triangle is also a right triangle, then you can take the base as one of the legs and the height as the other leg

(it should be obvious that the area of a isosceles right triangle is half the area of a square, with the legs being two sides of that square)

[real topaz]

Right. Thx

Will be closing now cause I understand the situation now :>

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[winged imp]

can anyone help me with Divide Remainder Theorem

[winged imp]

<@&286206848099549185>

[pastel dome]

don't ping helper

before 15 min

[winged imp]

sorry

uhmm anyone please help me

[pastel dome]

what is the question

asking you to do

[winged imp]

this one

[pastel dome]

no what is the original question

is it asking you to simplify?

[winged imp]

It just says here to Divide

[winged imp]

maybe

[deep vortex]

This is polynomial long division

[winged imp]

yeah how do you solve it?

i need help ;D

[pastel dome]

use synthetic division

[deep vortex]

Well, use the procedure of polynomial long division.

Divide the term of highest order in the dividend through the highest order term in the divisor (p^4/p)

You get p^3

[pastel dome]

$p^3−p^2−5p+5$ with a remainder of -1

[deep vortex]

Multiply p^3 with (p + 6) and subtract that from the dividend

Repeat

[woven radish]

Pure2

[pastel dome]

,w synthetic division

[woven radish]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by johnny1230851

[pastel dome]

look into that

[winged imp]

I just need a guide

even though if i read that i'll still won't get it

[pastel dome]

no search up a youtube video on that

i don't have a paper rn and it's hard tos how

[winged imp]

those this work?

P(-6) = (-6)^4 + 5(-6)^3 - 11(-6)^2 - 25(-6) + 29
= 1296 - 1080 - 396 + 150 + 29
= -1

hello?

[pastel dome]

yes

the remainder is -1

as specified

but this won't yield the quotient

[winged imp]

oh but can i still use

Remainder Theorem?

[pastel dome]

usel long division/modular arithmetic/synthetic division to get the quotient

synthetic division will be the easiest

[winged imp]

idk how though huhu

is this the one?

is this what you trying to mean?

[pastel dome]

yes

it is this one

[winged imp]

can i ask?

how did it to 6?

[pastel dome]

that's just

$p^3*6$

[woven radish]

Pure2

[pastel dome]

search up a youtube video

[winged imp]

alright

.close

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[edgy jasper]

how can i solve these two?

[edgy jasper]

<@&286206848099549185>

[devout snow]

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[limpid dome]

hi

[limpid dome]

i get 64 but there is no 64

Number of cubes along the length = 8cm / 2cm =4

Number of cubes along the width = 8cm / 2cm = 4

Number of cubes along the height = 8cm / 2cm = 4

[drifting sierra]

That would be how many are in the box, not touching the box

[limpid dome]

4x4x4=64

[final scarab]

oh shit my bad.

didnt go thru the q

[limpid dome]

sjdsjdjsd

what should i do

[restive river]

8^3/2^3

İt is this rigth?

[drifting sierra]

Read the question

[restive river]

Ok

[drifting sierra]

You have four layers, how many cubes in the top layer are touching the box?

[restive river]

Ohh

Ok

[limpid dome]

Number of cubes along the length = 8cm / 2cm =4
Number of cubes along the width = 8cm / 2cm = 4
Number of cubes along the height = 8cm / 2cm = 4
4x4x4=64

i do this

what is the problem

[drifting sierra]

.

[restive river]

İsnt İt be all the boxes?

Oh ok

İ read it wrong

[limpid dome]

bro think i cant do this

[drifting sierra]

?

You came here for help, if you can do it then do it

[limpid dome]

i write i can`t😂

[drifting sierra]

Uh ok, language barrier I guess

[limpid dome]

😂

[restive river]

That doesnt works nvm

[devout snow]

@limpid dome Has your question been resolved?

[limpid dome]

finish

[supple knot]

.close

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[real cove]

How can I integrate this through partial fraction decomposition? Seems like I'm back where I started

[devout snow]

@real cove Has your question been resolved?

[supple knot]

you should be doing a trig sub here

[real cove]

Like what?

[devout snow]

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[supple knot]

$\int \frac{du}{u^2+1} = \arctan(u) + C$

[woven radish]

riemann

[real cove]

I dont have the du

.reopen

[devout snow]

✅

[supple knot]

yes

you have to try different trig subs

[real cove]

I did a trig sub on the other half

I don't know how to do it here since it is squared

[bold shale]

Actually you don't have to use partial fraction decomposition

[real cove]

What do I use then?

[bold shale]

You can write it as $\int ( \frac{n}{(n^2+1)^2} + \frac{1}{(n^2+1)^2} ) dx$
Then the first one will be a simple u-substitution and the the second one trig substitution

[woven radish]

Alireza

You can write it as $\int \( \frac{n}{(n^2+1)^2} + \frac{1}{(n^2+1)^2} ) dx$
Then the first one will be a simple u-substitution and the the second one trig substitution
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l.57 You can write it as $\int \(
                                  \frac{n}{(n^2+1)^2} + \frac{1}{(n^2+1)^2} ...

Your command was ignored.
Type  I <command> <return>  to replace it with another command,```

[real cove]

That's what I did, I got the first part right

The second one I could do n = tanx and I end up with sec^4(x) ?

[bold shale]

Did you substitute in dn too?

[real cove]

Fck

Well played. Couldn't do this on an exam and it was bugging me thank you

[bold shale]

Any time 😁

[real cove]

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[devout snow]

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[steady violet]

I got stuck.. what do I do next? or is my answer already wrong? (note: we still didn't take L'hopital) ||(also.. the answer is -9/50 showed in this booklet)||

[steady violet]

this is the referance

[upper schooner]

How'd you get this?

[don't forget the limits btw]

[steady violet]

I rationalized 1-cos(3h)

[steady violet]

oh yeah mbmb

[upper schooner]

Yep, figured, and what happened when you did that?

[steady violet]

oh yeah speaking about it I forgot to put another sin(3h)

wait I think I can solve it now

[upper schooner]

What I was hinting at

[steady violet]

lool I thought of multiplying (1/h)^2 both sides.. but I couldn't do it since there's only one sin in the top

[upper schooner]

Welllllll also another comment too, assumedly you got those sin(5h) through Pythagoras, but you'd introduce a negative sign when that happens

[upper schooner]

And how do you mean?

[steady violet]

ohh

[steady violet]

this

[upper schooner]

So I think you should have something like $\lim_{h\to 0}\pqty{ -\frac{1}{1 + \cos(3h)} \frac{\sin^2(3h)}{\sin^2(5h)} }$

[woven radish]

@upper schooner

[upper schooner]

And that's a decent idea, it works

[steady violet]

yeah but I still don't get the negative part

[upper schooner]

Alternatively multiply by h^2/h^2

[upper schooner]

Because $1 - \cos^2(5h) = \sin^2(5h)$, but you have $\cos^2(5h) - 1$, so you have to multiply by -1

[woven radish]

@upper schooner

[upper schooner]

Like $\cos^2(5h) - 1 = -(1 - \cos^2(5h)) = -\sin^2(5h)$

[woven radish]

@upper schooner

[steady violet]

ohhhh yeah

I forgot about that

thanks so much!

[upper schooner]

no problem

[steady violet]

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[sleek panther]

I moved to equal = x^2 + 2x-3

factor = (x-1)(x+3)

so i put in

[-3,1]

but it say no good 4 the answer

[prime egret]

x^2 + 2x -3 >= 0 tho

[pastel dome]

not equal****

[prime egret]

Think about the shape of quadratic

[sleek panther]

i mean between x intercepts is below zero

[prime egret]

Right

[pastel dome]

draw the thing out with x intercepts at 1 and -3

[prime egret]

Listen to my student

[sleek panther]

0 <= x^2+2x-3, i think x^2 +2x -3 => 0 is equivelent

so i had wrong

because its opposite

[pastel dome]

what

=

[sleek panther]

yes is true

[sleek panther]

im programmer so this notation makes sense to me

[main marlin]

=>

[sleek panther]

oh yeah = matters

.close

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[pastel dome]

@prime egret .

[prime egret]

[dry oxide]

@prime egret sir, can i be your student too?

[prime egret]

@pastel dome

What do you think

[dry oxide]

pwease

i beg you

[prime egret]

@pastel dome

[pastel dome]

hmm

he can be apprentice...

[dry oxide]

oh

ok..💔

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[lethal meadow]

I’m not too sure how to set this equation up. As this is a harmonic oscillator it has to he a second order ode and idk how to make that

[devout snow]

@lethal meadow Has your question been resolved?

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[floral badger]

Order f(x) f(0)
0 7sin5x 0
1 35cos5x 35
2 -175sin5x 0
3 -875cos5x -875
4 4375sin5x 0

[floral badger]

im stuck

[timber pebble]

oh thats nice of them to provide you all this information

[floral badger]

i did 5 derivs

[timber pebble]

are they asking you to find it to a certain order

or just find the infinite one

[floral badger]

but i cant see the way to express the pattern

[floral badger]

infinite

[timber pebble]

well there are a few ways

[floral badger]

from n=0

[timber pebble]

one is to know the series

for sinx i mean

so i guess thats out lol

[timber pebble]

you can use this but its not going to be fun

you can alternatively work out the maclaurin series for $e^{ix}$

[woven radish]

jan Niku

[floral badger]

idk any other ways

[timber pebble]

because $e^{ix} = \cos x + i \sin x$

[woven radish]

jan Niku

[timber pebble]

you just take every piece of the series with an i attached to it

and the series for e^(ix) is very easy

[timber pebble]

try this:

work out the series for sinx

just sinx

using $f(x) = \sum _{n=0} ^\infty \frac{ f^{(n)} (0) }{ n! } x^n$

[woven radish]

jan Niku

[timber pebble]

do you think you could do this?

[floral badger]

like what i did with the whole thing

but just sinx

?

[timber pebble]

yea

i mean we can kind of walk through it, if you want

but i dont wanna just start rattling it off i fyou want the practice

[floral badger]

no i dont really understand it to well so

its more learning atm

[timber pebble]

okay

well we can walk through it

[timber pebble]

so the scary piece here is $f^{(n)} (0)$

[woven radish]

jan Niku

[timber pebble]

but it doesnt end up being so bad

f here is sine

whats the zeroth derivative of sine evaluated at 0?

(no derivative)

[floral badger]

sine

[timber pebble]

yea

and at x=0, its just 0

[floral badger]

ye

[timber pebble]

so, for the n=0 term, it actually vanishes

you found something like this in your table

we can maybe notice that like

the second derivative of sin(x) is -sin(x)

which is also 0 at 0

so it seems like every other term is going to vanish

start with n=0,2,4,...

now we gotta do the odd derivatives

whats the first derivative of sine? what is it evaluated at 0?

[floral badger]

cose

then 0

[timber pebble]

no

cos(0) = 1

yea?

[floral badger]

ye

[timber pebble]

alright

we know the next derivative up is -sin(x) and thats 0

whats the derivative of -sin(x)?

(were on term n=3, now)

[floral badger]

-cosx

[timber pebble]

at 0?

[floral badger]

-1

[timber pebble]

yup

so, do you think you can write out $f^{(n)} (0)$ for any n?

it seems like 0, 1, 0, -1, ...

[woven radish]

jan Niku

[floral badger]

this is where i get stuck

[timber pebble]

well, lets start with a guess

[floral badger]

am i looking at the answers or derivs

[timber pebble]

were trying to work out $f^{(n)} (0)$, in general

[woven radish]

jan Niku

[timber pebble]

heres something that might help

$f(x) = \sum _{n=0} ^\infty \frac{ f^{(n)} (0) }{ n! } x^n$

[woven radish]

jan Niku

[timber pebble]

can you remember which terms are 0?

n=0,2,4,... right?

[floral badger]

evens

[timber pebble]

yea

so lets change our index to just have odds

[floral badger]

f(n+1)

[timber pebble]

say like... n=2m+1 for m=0,1,2,3,....

[timber pebble]

i wouldnt worry about it, sine and cosine are probably the hardest series to work out at first

[timber pebble]

do you believe that this gives up only the odd indeces?

[floral badger]

ye 2 makes it even +1 makes it odd

[timber pebble]

alright

so we can replace n with 2m+1 here

and let m=0,1,2,3,... instead

[floral badger]

ok

[timber pebble]

$f(x) = \sum _{m=0} ^\infty \frac{ f^{(2m+1)} (0) }{ (2m+1)! } x^{2m+1}$

okay, that term in the numerator is asking for the odd terms of the derivative evaluated at 0

before we reindexed, we had 0, 1, 0, -1, 0, 1, .....

so if we just pull out the odd terms, what are they?

[floral badger]

0,0,0

[timber pebble]

well, thats not right

we are looking for the odd terms

it just starts at 0

[floral badger]

1, -1, 1

[timber pebble]

yea