help-0

[solemn grove]

cuz it’s easier to say

[chilly tangle]

Thanks. How do I finish this help section?

[earnest mortar]

a module can be a norm...

[solemn grove]

.close

[earnest mortar]

that's confusing

[solemn grove]

bro, it’s a single |

probs just convention

but I learnt single lines as det of matrix

[earnest mortar]

yep but i never denoted |M| the det of M so allow me to be confused 🙂

[solemn grove]

and double for norm

[chilly tangle]

How do I get the vector from this matrix, for instance?

[solemn grove]

yeah, probs convention as I said

[chilly tangle]

It's a brazilian book.

[solemn grove]

… I still don’t know what that has to do with the question

but yeah

[chilly tangle]

How do I transform a vector from a matrix to a vector form?

.close

[lone heart]

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[wispy cipher]

I have an AM-GM inequality question

[wispy cipher]

if a^3 + b^3 = c^3

find P such that a^2 + b^2 - c^2 >= P(c-a)(c-b)

I found that (a^2/(c-b)) = (b^2 + bc + c^2)/a

and (b^2/(c-a)) = (c^2 + ac +a^2)/b

but I don't really know how do i transform this to the wanted inequality

[earnest mortar]

$$a^2 + b^2 - c^2 \geq P(c-a)(c-b)$$
is equivalent to
$$a^2 + b^2 -Pab \geq c(Pc + c - a - b)$$
Notice that
$$(a+b-c)^3 = a^3 + b^3 - c^3 +3a^2 (b-c) + 3 c^2 (b+a) + 3b^2 (-c + a) - 6abc$$ $$(a+b-c)^3 = 3a^2 (b-c) + 3 c^2 (b+a) + 3b^2 (-c + a) - 6abc$$
Maybe you can try using GM-AM on $(a+b-c)^3$

That's the idea i have, not sure if it works

[ocean seal]

Silfer

[earnest mortar]

nvm with -c it doesnt work

[wispy cipher]

i tried this way and also with (a+b)^3, (c-b)^3

but they don't work out

<@&286206848099549185>

[earnest mortar]

are a b and c positive integers ?

[wispy cipher]

not indicated

[earnest mortar]

mmh

AM-GM only works with positive numbers

[wispy cipher]

positive real numbers, not integers

[earnest mortar]

so we'd have to use it on a² b² and c²

if a b and c can be negative

[wispy cipher]

i just looked at it again, a,b,c are positive real numbers

[earnest mortar]

oh ok

huh i'm dumb they cant be integers anyways

[wispy cipher]

well they can but if 0 is included

[earnest mortar]

yes

[wispy cipher]

to clarify, P is an integer

not a term including a, b or c

[earnest mortar]

intuitively i would say P = 2 or 1

[wispy cipher]

from this i found $$c(a^2 + b^2 -c^2) \geq 2(c-a)(c-b) \frac{ab}{c}$$

sorry \geq

[ocean seal]

DFGrasped

[earnest mortar]

well

i think i'll just give up

[wispy cipher]

😄

[earnest mortar]

i'm bad at those things

[wispy cipher]

this is the first time i see cubes iwith am-gm

so i am stuck

whole evening spent on this 😄

[earnest mortar]

i never really practiced am-gm

[wispy cipher]

i had to for some diophantine equations, they are helpful sometimes

[earnest mortar]

you might find it soon. recently i was thinking about a problem for the entire night. Then I woke up and had the solution in mind

the hardest problem i ever solved

[wispy cipher]

i hope so

and i am curious about urs as well

[earnest mortar]

what's your level ?

[wispy cipher]

highschool, calculus

some linear algebra but no topology or vector spaces

[earnest mortar]

i'm sorry but it's kinda useless to give it now... I mean you can understand it maybe if you know what a bijection is but it's never something that would be given in HS

but if you want to see it i can send it to you

[wispy cipher]

isn't it if the input and the output sets of functions are one-to one

[earnest mortar]

yeah kind of. The rigorous way to define it is about uniting injectivity and surjectivity

[wispy cipher]

i don't know the english part but I had a class on that

[earnest mortar]

wait where r u from ?

[wispy cipher]

Turkey, but not at highschool

at olympiads instructions

[earnest mortar]

I see

[wispy cipher]

I didin't join IMO, but at regional levels

to some extent

[earnest mortar]

i never participated to such competitions

oh, I did it once in France

[wispy cipher]

I started maths a little late 😄 so it was also hard

[earnest mortar]

I never studied maths outside of HS when I went into it

anyways, sorry for not being able to help

[wispy cipher]

no problem, nice chat though

[lone heart]

@wispy cipher Has your question been resolved?

[lone heart]

@wispy cipher Has your question been resolved?

[lone heart]

@wispy cipher Has your question been resolved?

[lone heart]

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[fossil basin]

Howdy! Can you help me figuring out the steps to this...?
2^(x+1)+2^(x-1)= 40
5. 2^(x-1)=40

[fossil basin]

I know x=4 but im struggling with the process 😦

[placid zinc]

Note that 2^(x+1) = 4(2)^(x-1)

[fossil basin]

sorry i don't get it

[placid zinc]

If I let u = 2^(x-1), then it becomes:
4u + u = 40

[lone heart]

@fossil basin Has your question been resolved?

[alpine sable]

@fossil basin we're trying to get equal powers so we can add the terms

[fossil basin]

thanks mate!

[lone heart]

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[nimble frost]

im using this as a reference

[tacit arch]

what is your question

[nimble frost]

to find the power series of the function 1/3-x by comparing it to the power series from 1/1-x

[glass lichen]

yeah that looks right

also 1/(3-x) and 1/(1-x) for typing

oh wait

centered at 2

[nimble frost]

where does the a=2 centering come in

yeah

isnt it usually (x-2)^n?

[glass lichen]

yeah

[nimble frost]

so would that be the entire series

[glass lichen]

so you want to make $\frac{1}{a-(x-2)}$ appear

[ocean seal]

Mosh

[nimble frost]

oooo

[glass lichen]

so $\frac{1}{1-(x-2)}$

[ocean seal]

Mosh

[nimble frost]

bruh

you are that mf

[glass lichen]

then just give interval of convergence

[nimble frost]

(1,3) ?

2,4

[glass lichen]

|x-2|<1

-1<x-2<1
1<x<3

[nimble frost]

ok so it is (1,3)

[glass lichen]

yes

[nimble frost]

thank you for your help

[glass lichen]

then do the endpoints specifically

but they should also diverge I believe

[nimble frost]

im not sure why i couldnt visualize it in that way with 1-x-2

.close

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[nocturne carbon]

i need help with 2 and 3 stepp equations like how to do them

[nocturne carbon]

like in general

[nimble frost]

do you have any examples

[mental agate]

help me pls

im in year 7

[nimble frost]

res, someone else is using this channel. find another

[wary stream]

#❓how-to-get-help

[nocturne carbon]

yes

ill get soem

actually i need help with that and more but its all algebra

ones like these

[nimble frost]

okay. lets look at ex8

what is the opposite of an exponent

like how do you get $x^2$ back to x

[ocean seal]

axap

[nocturne carbon]

how

idk how

idk how to solve any of these

[nimble frost]

ok what have you learned in math so far

[nocturne carbon]

man i just fell asleep for one class

[nocturne carbon]

algebra

i know the basics

[nimble frost]

ok this is algebra

[wary stream]

https://www.youtube.com/watch?v=7DPWeBszNSM&t=77s&ab_channel=TheOrganicChemistryTutor

[nimble frost]

anything you do to one side you have to do to the other

[nocturne carbon]

i know how you do thoe

but the ones with fractions and exponents r hard

[nimble frost]

so if 5x = 10, you would divide both sides by 5 to undo the multiplication

[nocturne carbon]

yes

[nimble frost]

same thing with a fraction

[nocturne carbon]

how do i do it

[wary stream]

You apply the opposite operation

[nocturne carbon]

i get so confused with fractions and exponents

[nimble frost]

just multiply by the inverse of the fraction, turn it upside down

[nocturne carbon]

i multiply the fraction??

[wary stream]

Because a fraction is division

[nocturne carbon]

ohh

[nimble frost]

maybe write it as 2x/3 instead

that could help

[nocturne carbon]

how do i do x^2

= 144

[nimble frost]

what does the exponent do

[nocturne carbon]

okay

multiply

[nimble frost]

multiply what

[nocturne carbon]

it multiplies x

[nimble frost]

by what

[nocturne carbon]

2

x x s

X x X

[nimble frost]

it multiplies x by itself

x * x

have you learned about square roots?

[wary stream]

For exponents, it's applying exponent rules, you want to have a power of 1, aka $x^1$. So $x^{p^{\frac{1}{p}}}= x^{p \cdot \frac{1}{p}} = x^1 = x$

[nocturne carbon]

yea but i forgot what it is

if i know a bit i can ptobably catch on fast

[nimble frost]

just look up what a square root does

its like the opposite of an exponent

[nocturne carbon]

it devides??

divides

[nimble frost]

so if x * x = 144

what is 144

[wary stream]

No

[nocturne carbon]

the square root of x

[nimble frost]

yes

since there are two sides to the equation you need to square root BOTH sides

[nocturne carbon]

what

[wary stream]

You don't divide

[nimble frost]

so the square root of x^2 will give you x

and that will be equal to the square root of 144

[ocean seal]

dldh06

For exponents, it's applying exponent rules, you want to have a power of 1, aka $x^1$. So $x^{p}^{\frac{1}{p}}= x^{p \cdot \frac{1}{p}} = x^1 = x$
```Compilation error:```! Double superscript.
l.55 ...o have a power of 1, aka $x^1$. So $x^{p}^
                                                  {\frac{1}{p}}= x^{p \cdot ...
I treat `x^1^2' essentially like `x^1{}^2'.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2020/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]
(./485903210272391178.aux) ) 
Here is how much of TeX's memory you used:
 18267 strings out of 479485
 359583 string characters out of 5871962```

[nocturne carbon]

how do i find a square root again

[wary stream]

A number times itself

[nocturne carbon]

144 x 144??

[nimble frost]

not quite

what number times itself equals 144

what you wrote would be 144 squared

[nocturne carbon]

thats what idk how to find

[wary stream]

Knowing your squares would be good

[nimble frost]

yes knowing squares is helpful

what is 10 times 10

[nocturne carbon]

100

[nimble frost]

ok

so if 10^2 = 100

144 is going to be more than that

[nocturne carbon]

is it 144 devided by 14??

[nimble frost]

whats 14*14

[nocturne carbon]

nvm

196

[nimble frost]

right, its too high

but you know its between 10 and 14 now

[nocturne carbon]

yes

12??

[nimble frost]

yes

12*12 =144

so what is x

[nocturne carbon]

12

now thats cleared thanks

wait let me write this down

wait

[nocturne carbon]

how do i writw this dwn

[nimble frost]

so this is how we got there

[nocturne carbon]

okay thanks

[nimble frost]

does it make sense?

its the square root of both sides

see how the square root of x squared is just x?

[nocturne carbon]

yes

[trim plinth]

not exactly

[nimble frost]

yeah

[nocturne carbon]

wait

[nimble frost]

it could be negative

[trim plinth]

think of the square of a negative number

[nocturne carbon]

how

[nimble frost]

two negative numbers multiply to a positive number

[nocturne carbon]

yea

[nimble frost]

so -12*-12 would also be 144

so technically x could be -12 or 12

or you can leave it as x= square root of 144

[nocturne carbon]

but dont u always automaticly assume that its a possitive numgebr

[nimble frost]

no

[nocturne carbon]

if no sign is shwon

[nimble frost]

not in this case

[trim plinth]

√ usually denotes the principal square root

[nimble frost]

we know the RESULT (144) is positive

[trim plinth]

so if I say √4 I usually mean 2

[nimble frost]

but we dont know if the numbers that went in are positive

[trim plinth]

but if I write sqrt(4) it is usually +2 and -2

[nimble frost]

is that really a thing

i didnt know that

[trim plinth]

yeah

[wary stream]

Yeah, that's a thing

[nimble frost]

xo, does the problem make sense to you now?

[wary stream]

If you write the root yourself, it's $\pm$ but if the root was already there, it's just the positive

[ocean seal]

dldh06

[nocturne carbon]

i get it now for this one

[nimble frost]

cool

[wary stream]

And also, for exponents, it's applying exponent rules, you want to have a power of 1, aka $x^1$. So $$x^{p^{\frac{1}{p}}}= x^{p \cdot \frac{1}{p}} = x^1 = x$$

[nimble frost]

lets look at problem 7

try writing it as 2x divided by 3 on the left

[nocturne carbon]

but for this one i had to make it does that mean its negatibe

[ocean seal]

dldh06

[wary stream]

It's both positive and negative

[nocturne carbon]

how\

im so confused

[wary stream]

Because $(-12) \cdot (-12) = 144$

[ocean seal]

dldh06

[wary stream]

And $12 \cdot 12 = 144$

[ocean seal]

dldh06

[nocturne carbon]

YES

yes\

i get this one now

ok now to the fractions ones

[wary stream]

Recall that you can rewrite fractions like $$\frac{4x}{3} = \frac{4}{3} \cdot x$$

[ocean seal]

dldh06

[nocturne carbon]

yea

[nimble frost]

see how multiplying both sides by 4 makes it so theres no fractions

[nocturne carbon]

how did 2 4s apear

[nimble frost]

i multiplied both sides by 4

youre allowed to do any operation to one side as long as you also do it to the other side

[nocturne carbon]

okay ill try remembering ut

wait let me write this down

[nimble frost]

do my examples make sense to you?

[wary stream]

I am going to suggest that you should ask your teacher for help or watch videos because it seems like you are lacking basic algebra knowledge

[nocturne carbon]

not really but ill try understanding it

[nimble frost]

since x is being multiplied by 3 and divided by 4, i do the opposite of those to get rid of them

[nocturne carbon]

ill ask her too

im still confused man

[nimble frost]

how would you solve this

[wary stream]

Then ask your teacher for more assistance

[trim plinth]

x/3=10

[nocturne carbon]

ngl idk

[trim plinth]

how do you go from $\frac{x}{3}$ to $x$?

[ocean seal]

sunflame

[nocturne carbon]

i barely have any knowledge of secondary level math

[nimble frost]

that's basic multiplication/division

i would review your arithmetic

[wary stream]

That sounds like a problem where you should look up resources and ask your teacher for help

[nocturne carbon]

i know a fraction is divison

[wary stream]

Apply the opposite operation that cancels the division

[nocturne carbon]

multiplication

how do i find x for questions like this tho

vvvvvvvvvvv

[trim plinth]

divide both sides by $\frac{2}{3}$

[ocean seal]

sunflame

[trim plinth]

$x \div \frac{a}{b} = x \times \frac{b}{a}$

[ocean seal]

sunflame

[trim plinth]

$x \divide \frac{a}{b} = x \times \frac{b}{a}$

[ocean seal]

sunflame

$x \divide \frac{a}{b} = x \times \frac{b}{a}$
```Compilation error:```! You can't use `\relax' after \divide.
\frac ->\protect 
                 \frac  
l.55 $x \divide \frac
                     {a}{b} = x \times \frac{b}{a}$
I'm forgetting what you said and not changing anything.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2020/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]
(./892725065127444480.aux) ) 
Here is how much of TeX's memory you used:```

[wary stream]

Just do another frac

[trim plinth]

$x ÷ \frac{a}{b} = x \times \frac{b}{a}$

[ocean seal]

sunflame

[trim plinth]

………………………………

[wary stream]

$\frac{x}{\frac{a}{b}}= x \times \frac{b}{a}$

[ocean seal]

dldh06

[nocturne carbon]

OKAY

I GET IT NOW

is x 18

[wary stream]

Plug it back in and check

[lone heart]

@nocturne carbon Has your question been resolved?

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[undone herald]

is this right, and if so, how do I express this properly?

[merry depot]

it's not right

$e^{-n} \neq e^{1/n}$

[ocean seal]

Zybikron

[undone herald]

ohh oh my god what was I thinking

embarrassing lol

[merry depot]

is this $\lim_{n\to\infty}\frac{n^{e^{-n}}}{n!}$ or $\lim_{n\to\infty}\frac{ne^{-n}}{n!}$?

[ocean seal]

Zybikron

[undone herald]

first one

[merry depot]

interesting. Alright

[undone herald]

ok wait lol I'm redoing the question real quick

but I still think it'll be a problem

cause u end up with infinity^0 right?

like $\frac{{\infty}^0}{n!}$

[merry depot]

that could be a problem yeah

[undone herald]

$\frac{{\infty}^0}{\infty}$

[ocean seal]

RealLifeGamer

[undone herald]

so it's not l hoptial

is it more like I need to rearrange stuff first?

[merry depot]

do you know about comparison tests or anything?

[undone herald]

yeah I do but I only just learnt them

so I'm still rusty

is the aim to use a series to prove that n! will be bigger than the numerator?

like ratio test?

or a different one

[merry depot]

direct comparison test could be nice here, with the right comparison

[lone heart]

@undone herald Has your question been resolved?

[undone herald]

what would I be comparing though?

like do I split the fraction into two series or sth?

[merry depot]

there's a nice comparison function written on your board in the first picture

[undone herald]

don't neither of the terms converge though?

isn't that a requirement for comparison test?

wait hang on

no I think the top term does converge?

[merry depot]

????

nothing about part of the series terms converging or not

[undone herald]

so I'm comparing the numerator and denominator right?

[merry depot]

no

comparison test says to find a different series where each term of the other series is larger (or smaller) than the original, then if the new series converges (or diverges) then the original series also converges (or diverges)

[lone heart]

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[unborn valve]

Hi I would like to know how the sigma here is simplified

I couldn't really see a pattern and I have no idea where to even begin

close

.close

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[mild raft]

how would you solve this?

[severe sluice]

a for loop?

[merry depot]

while loop?

definitely a loop

[lone heart]

@mild raft Has your question been resolved?

[mild raft]

does anyone know how i could know what for instance N = 4 results in using this formula?

i mean mathematically this way i can see if my answer is correct

[lone heart]

@mild raft Has your question been resolved?

[tawny schooner]

You should substitute N(n) = A*r^(n)

Where A is a constant

Then you can determine the resulting quadratic equation's roots for r

The resulting quadratic will look like this after factoring out common terms: r^2 = r+2

after you find the roots r1 and r2 of this quadratic, your solution will be a linear combination of the two solns: N(n) = Ar1^(n) +Br2^(n)

Determine A and B using initial conditions of the problem

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@mild raft Has your question been resolved?

[lone heart]

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[grave robin]

I'm new to the unit circle and I've got a question to solve an equation using it but I'm not sure how
|sin x| = 1/sqrt(2)
the given answers are x = 45 degrees, 135 degrees or x = 225 degrees, 315 degrees

[lone heart]

@grave robin Has your question been resolved?

[grave robin]

<@&286206848099549185>

[silver adder]

@grave robin can you compute sin based on this triangle?

[grave robin]

its just 1/sqrt(2) right?

[silver adder]

yup

the hypotenuse comes from pythagoras: sqrt(2) = sqrt( 1^2 + 1^2 )

[grave robin]

okay, where do I go from there?

[silver adder]

,w plot x^2+y^2=1, y = 1/sqrt(2)

[ocean seal]

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[silver adder]

sin is the y coordinate over time, both points here satisfy y = sin(45 degees) = sin(135 degrees) = 1/sqrt(2)

[grave robin]

how do I get the 45 and 135 though?

[jovial breach]

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:trig/x2ec2f6f830c9fb89:unit-circle/a/trig-unit-circle-review

You can look it up

[grave robin]

I can see they're the answer, but I don't get how to tell from this

[silver adder]

consider only the y coordinates

take |y|

[grave robin]

Y being sin X?

[silver adder]

uhh best to use a different variable while considering the unit circle

y = sin(t)

x = cos(t)

the point p = (cos(t),sin(t))

[grave robin]

Okay, I can see that 1/sqrt(2) is 0.7something which is the y axis co-ordinate but from there I'm lost

[silver adder]

,w plot sin(x) = 1/sqrt(2)

[ocean seal]

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[silver adder]

lost on what specifically?

[grave robin]

How to get the 45 and 135

[silver adder]

well sin peaks at 90 degrees

and then the point comes down in the 2nd quadrant just like it went up in the 1st quadrant

by symmetry sin(45 degrees) = sin(90 + 45 degrees)

[grave robin]

Oh that makes sense now!

Thanks!!

[silver adder]

hmm

https://twitter.com/freyaholmer/status/1173376419168247810

Freya Holmér (@FreyaHolmer)

working on a vector rendering library for Unity, for making stuff like this~✨

#madewithunity

Likes

10911

Retweets

2378

▶ Play video

Twitter · Sep 15, 2019, 11:21 PM

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[alpine sable]

Good day! Got a polynomials problem here

[alpine sable]

p(𝓍) =a𝓍³-3𝓍²-5a𝓍-9

(𝓍-a) is a factor of p(𝓍)

What are the possible values of a?

What I did:

1. tried finding p(a), can't solve it
2. tried dividing by x-a, too complicated for the question

[gray isle]

1. tried finding p(a), can't solve it
   show work

[alpine sable]

a⁴-8a²-9=0

[gray isle]

did you try anything after getting that equation?

[alpine sable]

Yeah, using any two terms and considering a difference between two squares, taking a common factor, and having two sides and rooting both.

Ok I can let b = a² and solve it quadratically normally

[gray isle]

note that this is a quadratic equation in a^2,
doing a substitution like u=a^2 may make it cleaer

[alpine sable]

yeah

didn't occur to me until now

thanks

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[flat holly]

I don't know how to find X with this equation

[earnest mortar]

wait sec = 1/cos right ?

[flat holly]

yes

[earnest mortar]

here you try to find where the derivative is 0

but is it the value of x where the curve reaches the y axis ?

[flat holly]

oh

right

so if i find the coordinates of the point where the curve intersects the y-axis, how can i find the gradient of the tangent?

[earnest mortar]

this gives you a coordinate x of the point where it touches the y axis. Just plug this x into the derivative

and x can easily be found here...

[flat holly]

?

wouldn't that x just be 0?

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[lapis sparrow]

.close

[prisma prawn]

is this allowed?

[drowsy lagoon]

no

[prisma prawn]

thanks

[wanton hare]

Thats... cursed

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[cedar vault]

how would i do number c

the total possible combinations is 6 * 6 * 6

216

i would rather not list all the possible combinations that add up to 10... it just seems really inefficient. what other way can i solve it?

https://hi-static.z-dn.net/files/da4/fd9a4bdb16ffd48a19c7ba02d753e0ff.jpg

it seems the answer is on google

why does the order matter here

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@cedar vault Has your question been resolved?

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[alpine sable]

so how do I go about finding the n? @tacit arch

[alpine sable]

based on the pdf you linked ,it seems I'm going to have to test it for each n, create a tble

[olive oar]

yea thats what youre gonna need to do

you can ball park at the start and look around there

[alpine sable]

alright but umm

if the purpose of this question is to minimize the error, then theoretically a function which represents the original function the best would have the least amount of error

so can I just go with n = 3? because then according to the formula, you would have f^4(c) which is sin x

another question: What exactly does the x represent on the last line?

is that the very point on the x-axis that minimizes error?

[lone heart]

@alpine sable Has your question been resolved?

[merry depot]

no

[alpine sable]

why not?

I guess the part that im stuck on is that I dno how to calculate the numerical value for error

[merry depot]

You took 4 derivatives, you can see that each of them is $\pm \cos(x)$ or $\pm \sin (x)$ and it will just cycle through those 4 no matter how many derivatives you take. So you can bound the n-th derivative.

[ocean seal]

Zybikron

[merry depot]

you found an x that makes your set up work. But x is in the interval [-pi,pi]. So I can just pick x = pi and break your equality.
This should work for all x in your interval.

[alpine sable]

so..

guess that means it's integration time?

[merry depot]

why?

[merry depot]

you want $|R_n|<\frac1{1000}$. You know $|R_n|<\left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right|$. So if you can find $n$ such that $\left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right|<\frac1{1000}$ you're good.

[ocean seal]

Zybikron

[alpine sable]

but you said the x thing only works at a specific point and not the interval

[merry depot]

yeah, so you need a larger value of n until it works for every x in the interval

the more terms in P_n, the better approximation it is, and the smaller R_n is.

[alpine sable]

so how do I find the n that minimizes the error given an interval?

[merry depot]

taking it a piece at a time and figuring out how to max/min it

Do you know what $|f^{(n+1)}(x)|$ looks like?

[ocean seal]

Zybikron

[alpine sable]

it's going to be plus or minus sin or cos

without any constants

[merry depot]

Can you bound that?

[alpine sable]

So I dno what that exactly means

I get the impression you're trying to find the maximum value

[merry depot]

what is the maximum/minimum value of sin(x) and cos(x)

[alpine sable]

plus or minus 1

0 to 1 if u abs value it

[merry depot]

right, so $|f^{(n+1)}(x)|\leq 1$ and $\left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right|<\left|\frac{1}{(n+1)!}x^{n+1}\right|$

[ocean seal]

Zybikron

[merry depot]

what is the maximum value that x can be?

[alpine sable]

1

no

uhh

I dno. x in your equation cancels out

[merry depot]

x is in an interval

[alpine sable]

Ok I'm lost

I'm completely lost

[merry depot]

What interval are you working on?

[alpine sable]

minus pi to pos pi

[merry depot]

that is the interval for x, yes?

[alpine sable]

yes

[merry depot]

so what is the maximum value for x?

[alpine sable]

1

in that interval

[merry depot]

is pi larger or smaller than 1?

[alpine sable]

you mean in terms of x?

it depends on whether it's cos or sin

[merry depot]

no, just x

if you're working on the interval $[-\pi, \pi]$ then $-\pi\leq x\leq \pi$

[ocean seal]

Zybikron

[alpine sable]

right

[merry depot]

so x is at most pi

[alpine sable]

yes

[merry depot]

So $x^{n+1}\leq \pi^{n+1}$ and you have $$\left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right|<\left|\frac{1}{(n+1)!}x^{n+1}\right|<\left|\frac{\pi^{n+1}}{(n+1)!}\right|$$

[ocean seal]

Zybikron

[alpine sable]

oh

wait no

that makes no sense because f(n+1)(c) is a y value no?

not x?

like I put 1 there because that's the maximum y value the fn can have

[merry depot]

yes

[alpine sable]

I don't understand where the pi came from

We're trying to find the optimal n right?

[merry depot]

your question says this is on the interval [-pi,pi]

[alpine sable]

yeah

that is true

[merry depot]

that's bounds on x

[alpine sable]

mhmm'

[merry depot]

so the largest x can be is pi

[alpine sable]

for an x value yes

[merry depot]

that's all we're doing with this right now

we already found a bound for the derivative

[alpine sable]

What does it mean by bounds on x?

[merry depot]

a min/max value

[alpine sable]

like min/max value as in the lowest/highest y value given an x value?

[merry depot]

[-pi,pi] is an interval for x. That means that -pi < x < pi. So you have a lowest and highest value of x.

You also know the derivative is either sin(x) or cos(x). You know that -1 < sin(x) < 1 for any values x (and the same for cos(x)). So you have a lowest and highest value for the derivative.

[alpine sable]

ok

yeah I understood that

[merry depot]

since you have a highest value for x and for the derivative, you can create an inequality that bounds R_n

[alpine sable]

mhmm

[merry depot]

That inequality is this

[alpine sable]

oh my God, I missed the part where pi gets subbed into x and not f(n+1)(c)

I was like.. how did a function turn into its domain

Ok

what do I do next?

[merry depot]

Now you can find an $n$ such that $$\left|\frac{\pi^{n+1}}{(n+1)!}\right|<\frac{1}{1000}$$

[ocean seal]

Zybikron

[alpine sable]

Oh

Ohhhh

Ok so

step 1) Identify the x at which you have the highest y value for a given function of an interval [a,b]
step 2) replace the x with the x value you found in step 1 into Rn
Step 3) solve for n

is that correct?

[merry depot]

almost
The x that maximizes the nth derivative may be different than the x that maximizes (x-a).

[alpine sable]

Right

and that's why they gave a =0

what should I do if a is given?

[merry depot]

a should be given

a taylor series needs a center

[alpine sable]

oh sorry I meant if a is not 0

[merry depot]

then maximize (x-a) instead of just x

[alpine sable]

ok makes sense

thank you so much

I appreciate your patience and time

.close

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[west hearth]

too late

[west hearth]

so i'm not sure how to proceed with lagrange optimization

when you have inequalities

I tried treating it as a function but it's not helping ://

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[west hearth]

<@&286206848099549185>

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[finite flax]

f looks like a hyperbola of one sheet and g looks like an ellipsoid. So the max's and min's will be level curves (ellipses)

I got the maximum...dunno how I missed the minimum

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[oak perch]

Replace y with (y/2) you got 5x^2+6xy+5y^2<=64, find the maximum and minimum of x^2+y^2

Then x=(1/sqrt(2))(u-v), y=(1/sqrt(2))(u+v), it turns into given 8u^2+2v^2<=64, find maximum and minimal of u^2+v^2

8u^2+2v^2<=64 is a region bounded by an ellipse u^2/8+v^2/32=1

So clearly maximum is 32, minimum 0

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[rigid flower]

How long does it take the minute hand of a clock to traverse an angle of 55°?

[merry depot]

what have you tried?

[rigid flower]

5 mins = 30 degrees
55 – 30 = 25
30/5 = 25/x
x = 25/6
5 + 25/6 = 55/6
55/6(60)=550secs
answer is 550 secs

here, I'm not confident about it. I was just practicing these types of probs recently

[merry depot]

that seems to work

550 seconds is correct

[vale wigeon]

overcomplicated but seems correct

[rigid flower]

Thank you!

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[iron parcel]

guys how do i know if i triangle is sss sas or asa without the lines telling u so?

[merry depot]

??? sss, asa, and sas refer to congruence of two triangles. I'm not sure what you mean about a single triangle being one of these?

[gray isle]

do you have an example?

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[clear frost]

1:200 means 1 cm in drawing = 200cm in real life

convert 60m into cm

then just find how long is the train in drawing

@dim jackal

[dim jackal]

Ok thanks 🙏

.close

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[compact cedar]

https://i.insider.com/6064c04cdaf0f10018f994a4?width=1000&format=jpeg&auto=webp

Sub in 1800 for A, and you're trying to find P. Once P is found, you could subtract it from A, and divide that by the T

@swift shore correct?

[spare mango]

so

would it look like

1800 = P ( 1 + 12.5% )^2

[swift shore]

Almost

12.5% * 2, since it’s 2 years

[spare mango]

Oh

[compact cedar]

The T is 2 years, since it's time. It's a flat interest rate, so that should ring a bell in your head that you don't need to use exponents (that's for compound interest)

[spare mango]

Ohhh I see

So would it be

[compact cedar]

yep

[spare mango]

1800 = P ( 1 + 12.5% * 2 )

[compact cedar]

Yes. I assume repayments means the "additional" payments due to the interest. So solve for P, and see if you can figure out the rest

[spare mango]

So do i do

1800 DIVIDED by 1.25P

Wait nvm i got it thanks

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[queen raven]

i get it until the third line

[gray isle]

the part where they said
$$\therefore 2^{n-1} = 64 = 2^6$$?

[ocean seal]

ℝamonov

[gray isle]

that's sort of a leap and ideally you shouldn't jump to that conclusion like that

64 can be expressed with a base of 2,
similarly 729 can be expressed with a base of 3

$\frac{729}{64} = \br{\frac32}^{n-1} \ \
\frac{3^6}{2^6} = \br{\frac32}^6 = \br{\frac32}^{n-1}$

[ocean seal]

ℝamonov

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@queen raven Has your question been resolved?

[queen raven]

that is a massive jump

but i see now

okay thank you

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[karmic falcon]

hi

[karmic falcon]

can someone explain me why f(x)*g(x) < (g(x))^2 isn't equivalent to f(x)<g(x)

i guess it has something to do with the rule that you can't divide with a function that can get 0 value right?

[tall hearth]

I think it has more to do with the fact that you dont know the sign of g(x)

so you won't know if the sign remains the same or switches

[unborn verge]

That's it

[tall hearth]

f(x)<g(x) this is true at g(x) >= 0

[unborn verge]

And, quick side note : g(x) can't be zero, otherwise the inequality wouldn't hold

[tall hearth]

right that also

[karmic falcon]

but i guess there is a rule which states that you can't add or divide or multiply or subtract any function g(x) which may get the value 0 over r in an eqaution

or sth like this

[tall hearth]

yes you cant

division by zero error

[karmic falcon]

then can you help me with this

the 3 and 4 equation

i can't spot the difference

[tall hearth]

the -1 and +1

[karmic falcon]

so the 3 can be equal to 0/0 which makes no sense

right?

and the 4 not

[tall hearth]

I don't really understand what you mean

[karmic falcon]

if you choose x=1 or x=-1

the third equation will be equal to 0/0

which makes no sense

[tall hearth]

then you mutiply 0 by f(x) which evaluates to 0

[karmic falcon]

but i already have the answers

and it says that 3 4 are not true

i'm sorry 4 is true

but 3 not

as i said

[tall hearth]

hmm yeah for 3

you do get a 0/0

but for 4 (x^2 + 1) can never be 0

[karmic falcon]

well 0 over anything is 0

so 4 is true

[tall hearth]

yes

[karmic falcon]

and 3 is not trye because 0/0 is not 0

it does not exists

[tall hearth]

yes

only at x is a part of {1, -1}

[karmic falcon]

the same logic stands for 8

right?

[tall hearth]

yes

that is also false?

[karmic falcon]

yep

and 7 is true

by the same logic

but i am a bit confused at 5

it says it is true

[tall hearth]

yeah (x^4 + 1) can never be 0 so you wont get a 0/0 ig thats the reasoning?

[karmic falcon]

yep

but what about 5?

it says true

[tall hearth]

im guessing thats true?

[karmic falcon]

wait

sqrt(0) exists right?

[tall hearth]

yes

it's 0

[karmic falcon]

then yeah it's true

thanks for your time

.close

how do i close this channel

[tall hearth]

send this again

[karmic falcon]

.close

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[dim kettle]

hi

[lone heart]

Please don't occupy multiple help channels.

[fluid eagle]

hi

could you translate this

[dim kettle]

i dont understand this division

yes it says finding the quotient

and rest

[tall hearth]

remainder?

[fluid eagle]

do you know how polynomial division works? the result is correct

[dim kettle]

im not sure

[tall hearth]

https://www.youtube.com/watch?v=3eowXt_LNbg&t=21s

I think this should be enough for your specific problem

[dim kettle]

so i have to simplify the first polynomial?

ruffini?

[tall hearth]

no

watch the part on long division

[dim kettle]

i honestly dont understand what he wrote here

[tall hearth]

uh not sure what this means

[dim kettle]

its a theorem we use to somplify polynomial higher than 2nd grade sometimes

[tall hearth]

hes dividing (x^2 + 3x + 2) by (x + 1)

[tall hearth]

oh

[dim kettle]

why did he wrote is as a multiplication?

[tall hearth]

umm thats the conventional sign for division

this

long division

[dim kettle]

i never saw divisions like this

so this is confusing me

[wheat elm]

I just got into a new section and has no idea about it

Hyelp

[tall hearth]

#❓how-to-get-help

[tall hearth]

oh damn

[dim kettle]

can we vc?

[tall hearth]

im sorry i dont think i can really help you with this

[dim kettle]

oh ok

[tall hearth]

no sorry

maybe you wan to close this open a new channel

or wait for someone to help

[dim kettle]

i can try layce

@fluid eaglehi are you free

.close

[lone heart]

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[feral frost]

Anyone willing to help out with a simple example? If for a joint distribution of discrete random variables (X and Y) we know the values of P(Y=y, X=x), how would one go about determining P(X=x, Y=y)?

[feral frost]

Are they just equal?

[lone heart]

@feral frost Has your question been resolved?

[feral frost]

<@&286206848099549185>

[noble sinew]

And opperator commutes yes

[feral frost]

sick az thanks

.close

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[south cedar]

Hello everyone, ive got 10k rows of these, the yellow is the service time (you only see 1's on the screenshot but it goes up), and the red are the waiting times. Does anyone know how I can create a graph like I have above ?I dont get how I can calculate the percentiles on the x axis like on the screenshot

[lone heart]

@south cedar Has your question been resolved?

[lone heart]

@south cedar Has your question been resolved?

[lone heart]

@south cedar Has your question been resolved?

[lone heart]

@south cedar Has your question been resolved?

[woven sphinx]

well do you know matplotlib

personally what i would do is i would import this into python (maybe using csv or smth), sort it from least to greatest, calculate the percentile for each, make a bunch of tuples consisting of (percentile, value) and get matplotlib to plot that

if you don't know python/matplotlib thats admittedly a bit unhelpful

also i dont think this really is a maths question ur better off asking in a different server i think

actually there's an easier way if you want to stick with google sheets/excel now that i think of it, but it is assuming that all of the data sets are the same length (are they?)

[lone heart]

@south cedar Has your question been resolved?

[lone heart]

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[balmy plank]

Jesse has 4 friends and 5 coworkers.

Each day in the year 2021, Jesse talked to some number of friends: always at least one, but never all 4 in the same day.

Each day in 2021, Jesse also talked to some number of coworkers, but never more than 3 coworkers in one day.

Prove that there were two days in 2021 in which Jesse talked to the same set of friends and the same set of coworkers.

[balmy plank]

<@&286206848099549185>

[remote heron]

must be pigeonhole right

[balmy plank]

i think so

[remote heron]

so lets see

you got how many possible pairings out of your friends

$\sum _1 ^3 \binom 4 i$

right?

[ocean seal]

jan Niku (Shuri for Honorable)

[balmy plank]

i being the like friends pairing, right?

yeah i think so

[remote heron]

i could just write it

$\binom 4 1 + \binom 4 2 + \binom 4 3$

[ocean seal]

jan Niku (Shuri for Honorable)

[remote heron]

ways to choose 1, ways to choose 2, ways to choose 3

how many is this?

looks like 14?

[balmy plank]

like should I simplify the expression, right?

[remote heron]

just solve

coworkers is the same, but with 5

you get 25

so uhh

i mean i dont want to get too hung up on that part

but if you had to calculate this do you think you couldL?

does it make sense how you might do this and do those numbers seem reasonable

[balmy plank]

1 would be like 1 person and coworker

[remote heron]

were not mixing friends and coworkers yet

[balmy plank]

sorry I'm trying to wrap my head around it

[remote heron]

say you only knew she had 4 friends okay

she has 4 friends

and she tells you

i talked to 2 of my friends

how many distinct groups of people could she be talking about

[balmy plank]

in 4 days, right?

[remote heron]

no just once

[balmy plank]

oh ah

[remote heron]

do you get like

[balmy plank]

then it could be one or 2

[remote heron]

nCr

this sort of thing

[balmy plank]

ye ye

[remote heron]

so one day

one time

she talked to 2 of her 4 friends

how many possible groupings

[balmy plank]

2

[remote heron]

no

[balmy plank]

but she only talks to one friend per day

[remote heron]

👀

[balmy plank]

or would it imply that it's the same person?

[remote heron]

I'm trying to tell you

she has 4 friends

assume she talked to 2 people

how many groupings could that have been

how many different groups of 2 people could she be talking about

[balmy plank]

only one right

[remote heron]

no

[balmy plank]

sorry

[remote heron]

its okay

you just need to do this problem 6 times and you can prove this problem

[balmy plank]

groupings of coworkers and friends or ...?

[remote heron]

ignore the question all together

[balmy plank]

ah okie

[remote heron]

all im asking is if you can use binomial

you have a blue, green, red, and yellow ball

[balmy plank]

ah i see

[remote heron]

you draw 2 balls

how many different possibilities are there

https://www.youtube.com/watch?v=Pcgvv6T_bD8&t=57s

he solves the exact problem i am asking you

[balmy plank]

isn't it like (n 2)2+(n 2)2!

ah okie let me watch that real quick

ah so it would just be (4 2)

[remote heron]

,w 4 choose 2

[ocean seal]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by jan Niku#2119

[balmy plank]

ah okie

[remote heron]

this is the tool you need

[balmy plank]

okie

[remote heron]

heres my thought in general, okay?

for your problem

[balmy plank]

okay

[remote heron]

you need to do the nitty gritty

say for example

Each day in the year 2021, Jesse talked to some number of friends: always at least one, but never all 4 in the same day.

here you need to find how many ways all of these pairs can happen

but generally

[balmy plank]

so that would be the (4 1)

[remote heron]

you enumerate all the different ways that she could have talked to her friends

all the different groups

(there are 14) but youll need to find that

then youll enumerate the ways she could have talked to her coworkers

(there are 25, but find it)

then you need to see

how many pairings can we make here

one from a group of 14

one from a group 25

[balmy plank]

ah I see

[remote heron]

how many unique pairings can we make before we have to start repeating?

is it enough?

im gonna go cook

but uhh

feel free to ping

[balmy plank]

ah okie

thank you so much

[lone heart]

@balmy plank Has your question been resolved?

[remote heron]

you get it?

[lone heart]

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[lone heart]

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• Show your work, and if possible, explain where you are stuck.
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[arctic mortar]

i have the function f(x)=(x+3)^2
i need to find the inverse function of this function
im lost and i have no idea how to do it

[naive valley]

on what domain?

[arctic mortar]

wdym?

[naive valley]

it doesn't have an inverse function if the domain of $f$ is all of $\mathbb R$

[ocean seal]

OurBelovedBungo

[naive valley]

because it's not one-to-one

[arctic mortar]

the answer key says f^-1(x)=+-sqrt(x)-3

[naive valley]

+-

so f^-1 is not a function

[arctic mortar]

oh, yeah i guess it isnt
so the inverse relation?

[naive valley]

yes

[arctic mortar]

i see, so how would I find that

[naive valley]

basically given y, find all x such that f(x) = y

[arctic mortar]

im confused, my teacher told me to swap the x and y's and then rearrange, so
f(x)=(x+3)^2
y=(x+3)^2
x=(y+3)^2
then rearrange for y but im not sure how to

[glass lichen]

you want to get y=(stuff)

[arctic mortar]

yeah

[glass lichen]

what would the first step be to get y=(stuff)

Hint: You can get y+3=stuff

[arctic mortar]

ohhh,
+-sqrt(x)=y+3?
then
+-sqrt(x)-3=y

[glass lichen]

yes

[arctic mortar]

ahh, ok
thanks alot!

.close

[lone heart]

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[fringe yoke]

X

[glass lichen]

Y

[tall hearth]

Z?