help-10

[broken magnet]

Yes

[velvet spoke]

we can take x^r/2 as x^z/2

[broken magnet]

where?

I got my mistake

It’s in the first step itself

Thanks👍

[velvet spoke]

sorry if i confused you a bit in the end...

i have to review my binomial as well

[broken magnet]

nah you helped me thanks 🙏

[velvet spoke]

https://media.discordapp.net/attachments/913018237783003136/1222581471173279914/Kokomi_Sip.gif

wlc

[obtuse pebble]

@broken magnet Has your question been resolved?

[obtuse pebble]

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[molten cosmos]

did this example just choose to leave out the other orbits

on 2 to 5 because they are the same as 1

nvm, we just dont count them because then X/<sigma> would have repeat elements which doesnt make sense

[obtuse pebble]

@molten cosmos Has your question been resolved?

[obtuse pebble]

@molten cosmos Has your question been resolved?

[obtuse pebble]

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[proven karma]

I need help with an induction problem:

[proven karma]

I proved that it's true for n = 1.

if we plug 1 in, we just get r+s = r+s which holds true

For the inductive step, I want to try to show that p(n-1) implies that p(n) is true

I tried plugging n-1 into the right hand side and i just got (r-s)/(r-s) , which is just 1.

I'm really struggling with induction as a whole, as you can probably tell

<@&286206848099549185>

[restive gorge]

I would try to write it as a sum

[warm shale]

bacc (unhelpful)

[restive gorge]

Assuming this for some n in N you wanna show

[warm shale]

bacc (unhelpful)

[proven karma]

it would be helpful to me if we did it with n-1 implying n

i know its the same thing but this is the way i'm used to doing it

[restive gorge]

if you now manipulate the left side it's simple

factor an r

and make the sum go from k=0 to n

then use the assumption

the rest is algebra

[proven karma]

I'll work on this then. Do i keep the channel open?

[restive gorge]

yes

also whether n -> n+1 or n-1 -> n doesnt matter

you can still apply the same strategy when you solved it

[restive gorge]

Also it wouldn't work with k = 0 to n-1 because in what you wanna prove you would miss the n-th term

[obtuse pebble]

@proven karma Has your question been resolved?

[proven karma]

why exactly

[proven karma]

do we wanna show this?

Aren't we supposed to assume this is true, and then have that imply the n case ?

i know i'm probably sounding silly

[restive gorge]

After the base case you assume for some natural number this is tru

[restive gorge]

The induction step would be to show that it also holds the consecutive n+1-th

[proven karma]

and use that to imply that the n+1 case would be true?

[restive gorge]

yes

thats why you wanna make the sum go from k=0 to n

using manipulations

[proven karma]

alright i've gotten this far, and i think this isn't bad

my thinking in this is that since we're choosing to sum to n instead of n+1 now, we're supposed to add a term to the new sum to compensate for this

i'm just not sure what this term is supposed to look like

@restive gorge i also pulled the r^1 out of the sum, that seemed reasonable since it doesn't depend on n. that's allowed, no?

[restive gorge]

the n+1-th term

which would be also be multiplied by r

you look at the expression in the sum

and plug in

k=n+1

thats what you need to add manually if you reduce the sum from n+1 to n

[proven karma]

isnt that achieved by factoring out my r^1 and having that multiply into the whole addition?

[restive gorge]

yes

so if you now write the last term manually it also needs the r term

since it was part of the sum

[proven karma]

the last term for me looks like this now: r^(n+1-k)*s^k

if we multiply the r into that we get: r^(n+2-k)*s^k

[restive gorge]

the last term has no k

k = n+1 in the last term

[proven karma]

ah right

this is slowly starting to look like what we want

[restive gorge]

,, r \cdot \sum_{k=0}^{n+1} a_k = r \cdot \sum_{k=0}^n a_k + r \cdot a_{n+1}

[warm shale]

bacc (unhelpful)

[restive gorge]

that's the logic

[proven karma]

yes we agree

now i have

[restive gorge]

,texsp so in your box you would have [ r \cdot r^{n-(n+1)} s^{n+1} ]

[warm shale]

bacc (unhelpful)

[restive gorge]

yes

now simplify

r * r^(n-(n+1)) vanishes

[proven karma]

i have the same fraction now but with n+2 in the r's exponent. and we just add s^n+1 to all of that

[restive gorge]

yes

multiply and divide by r-s the s^(n+1) term

[proven karma]

common denominator, do the addition

[restive gorge]

ye

[proven karma]

and this should give us the fraction seen here?

[restive gorge]

the right hand side of this

what we wanted to show

using our assumption that A(n) is true

[proven karma]

that's right, yes

[unborn sentinel]

can anyone help me with my algebra 1

[restive gorge]

it's a common strategy to rewrite it as a sum, manipulate in a way you can use the assumption and basically hope it works out

[unborn sentinel]

can someone help mer

[restive gorge]

!occupied

[obtuse pebble]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[restive gorge]

grab a channel, post the q and people will come by

[restive gorge]

you forgot the r

[proven karma]

the s^n+1 being multiplied by r is causing a jam in things

[restive gorge]

you didnt multiply the whole numerator by r

in 2nd step

it's

[proven karma]

jesus christ you're right

[restive gorge]

r^(n+2) - rs^(n+1)

[proven karma]

i should stop doing math when i'm tired

that solves the problem

i need to practice this a lot

the problem is each question just looks so different

thank you bacc

i'll close the channel now

.close

[obtuse pebble]

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[obtuse pebble]

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[runic crest]

guys help: calculate the length of the height of a straight circular cylinder with a base radius of 10 cm if the area of ​​the base section is equal to the area of ​​the base

[runic crest]

i need help with this beacuse text book wont show answer

[timid silo]

base section and base are two different things?

[runic crest]

yes

i used translate sorry

i dont know math english

[edgy mango]

Hi

[runic crest]

hello

cab you hel

p

[edgy mango]

?

[runic crest]

bruh

this server is useles

s

[edgy mango]

In confused on how to get help

[runic crest]

just og into the available channels

math help available category

idk why nobodys helping me

usually people take time ig idk

[edgy mango]

What type of math u need help with

[runic crest]

.

i used translate because i dont know how to write it in english

[edgy mango]

O so the hight of a cylinder

What's the volume or area of the rectangle that would be around the circle

[runic crest]

what rectangle

nono the tbing inside the shape

whatever

its called

idk

oh its that nvm

i dont speak englsj

sorry

[edgy mango]

So cylinders are 2 circles and a rectangle l

What language u speak

[runic crest]

yes'

i speak serbian

i understand a bit of eng

[timid silo]

your question is not clear

[runic crest]

can you pls help

well fuck how do i translate it to math english

[timid silo]

"base section" and "base" represent the same

the bottom circle of a cylinder

[timid silo]

also a cylinder is not 2 circles and a rectangle

[edgy mango]

откуцајте свој проблем на српском

[runic crest]

this maybe

[edgy mango]

When finding area that's what you do

[runic crest]

Izracunati duzinu visine pravog kruznog valjka poluprecnika osnove 10cm ako je povrsina osnog preseka jednaka povrsini osnove

I have no idea man

im fucking daed

[edgy mango]

Idk man just use Google I found the same problem but since I'm only in 8th grade have no fuckin clue on what anything means

[runic crest]

i tried nothing showed u

p

[obtuse pebble]

@runic crest Has your question been resolved?

[obtuse pebble]

@runic crest Has your question been resolved?

[obtuse pebble]

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[green sphinx]

is this correct because the internet says otherwise

[wooden cipher]

you get to x^2 = -1/3

so you square root both sides, you dont square both sides

when you do that what happens and what does that tell you about the function?

[green sphinx]

imaginary

?

like cannot square root a negative number

[wooden cipher]

yes, the values of x where f'=0 are imaginary

so what does that mean for the graph?

[green sphinx]

there are no horizontal tangent lines?

[obtuse pebble]

@green sphinx Has your question been resolved?

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[timid silo]

I need help

[timid silo]

@dusk widget eheh help me out rq

[ember frost]

!noping

[obtuse pebble]

Please do not ping individual helpers unprompted.

[timid silo]

I think it's a not a one to one function because it has 2 points??

[rocky pelican]

do u know vertical line test

[timid silo]

Nope

[open nacelle]

draw a vertical line and if 2 pts are on it then its not 1:1

[orchid wind]

no

horiontztazl line test

to test if one to one

vertical line test if its a function

[timid silo]

I don't get it.?

[open nacelle]

o right i mixed em up

[fallow wharf]

look at the y axis, that's all the points where x=0. Notice that line hits the curve at 3 places. So it's 1 to 3 there, not 1 to 1

as in, one value of x goes to three values of y

[timid silo]

OHHHH

[dusk widget]

isn't that to check well-definedness?

injectivity is the other way around

[timid silo]

Okay what is this-

I guess I can explain that it's a 1 to 1 function

[dusk widget]

the function happened to fail both line tests though, so it's okay

[fallow wharf]

I don't follow what you mean

[timid silo]

It just looks like they are in 1 the same position?? for x and y?

[dusk widget]

you said for one input (x = 0), we got three outputs

[fallow wharf]

one to one means it maps each single point to its own single point

[dusk widget]

but that's not what injectivity is, no?

[ember frost]

if the graph is not a function then injectivity wouldnt make sense anyway

[timid silo]

So it has to be 1?

[fallow wharf]

prove it

;P

it is what injectivity is, this is equivalent

[dusk widget]

[timid silo]

Guys change topic!

[dusk widget]

let me get back to you on this later

to Kiomi: I can't exactly help rn, I gotta go finish my own hmw

sorry

[timid silo]

It's fine!

@fallow wharf Help me out

rq

[ember frost]

apply the definition of injective function

[timid silo]

I dont get it

[ember frost]

that is, $x_1\neq x_2 \implies f(x_1) \neq f(x_2)$

[rocky pelican]

is this asking for the inverse

[ember frost]

oops, i thought its asking to verify if its injective

[timid silo]

OHHH SO LIKE

1. f(x)= 7x - 5
   skip to
   x = 7y - 5
   x + 5 = 7y
   x = 7y/5??

[rocky pelican]

yes

[timid silo]

I feel like im wrong here

[fallow wharf]

last step

wrong

[timid silo]

Isn't it suppose to be looking for y?

not x?

[rocky pelican]

ur supposed to find y

[timid silo]

Thank youu

1. f(x)= 7x - 5
   skip to
   x = 7y - 5
   x + 5 = 7y
   x + 5/7 = y ??

[fallow wharf]

close but without parenthesis it's wrong

7 divides everything not just 5

[timid silo]

Something like

1. f(x)= 7x - 5
   skip to
   x = 7y - 5
   x + 5 = 7y
   (x + 5)/7 = y ??

[fallow wharf]

yeah it's perfect now

[timid silo]

YES

Wait dang it now I don't know how to do number 2

All I did is

x = 1/3y + 4/5
x - 4/5 = 1/3y

Idk what to do else

[rocky pelican]

multiply by 1/3 just means divide by 3

so x - 4/5 = y/3

[timid silo]

So something like
All I did is

x = 1/3y + 4/5
x - 4/5 = 1/3y
x - (4/5)/3 = 1/3y/3

x - (4/5)/3 = y?

[rocky pelican]

you multiply by 3

not divide

because its y/3

[timid silo]

x - 4/15 = y?

[rocky pelican]

multiply everything by 3

3(x + 4/5)

[timid silo]

Oh okau

3x + 12/5 = y

Correct?

[rocky pelican]

yes

wait no

[timid silo]

Okay

What did I do wrong?

<@&286206848099549185>

[rocky pelican]

its - 12/5

[timid silo]

AHHH BECAUSE -4/5

[rocky pelican]

minus

[timid silo]

3x - 12/5

Got it

x = 4y + 2/3y - 2
x - 4y + 2 = 3y - 2?

[rocky pelican]

you need to multiply both sides by 3y-2 first

[timid silo]

x = 4y + 2/3y - 2
3xy - 2x = 4y + 2?

[rocky pelican]

yes

and move all the ys to one side, and all the x to another side

[timid silo]

Uhh

which side is one

and which side is another

[rocky pelican]

it doesn't really matter

y on left side lets say

[timid silo]

x = 4y + 2/3y - 2
3xy - 2x = 4y + 2
3xy - 4y = 2x + 2?

[rocky pelican]

yes

[timid silo]

x = 4y + 2/3y - 2
3xy - 2x = 4y + 2
3xy - 4y = 2x + 2
-x = 2x + 2?
whats next?

[rocky pelican]

wait what did u do

[timid silo]

3 - 4 is -1

y and y is gone

[rocky pelican]

you can't minus 3xy by 4y

also ur trying to find y, so making y disappear is not a good idea lol

[timid silo]

x = 4y + 2/3y - 2
3xy - 2x = 4y + 2
x = 4y + 2?

[rocky pelican]

3xy and 4y both have y, so u can factor out the y

[timid silo]

I don't get it

[rocky pelican]

3xy - 4y = 3x * y - 4 * y

= y(3x - 4)

[timid silo]

x = 4y + 2/3y - 2
3xy - 2x = 4y + 2
3xy - 4y = 2x + 2
3x * y - 4 * y = 2x + 2
y(3x - 4) = 2x + 2

[rocky pelican]

yes

[timid silo]

Okay then...

put y to the other side?

[rocky pelican]

keep y on left side

now its y * (3x - 4)

one last step

[timid silo]

x = 4y + 2/3y - 2
3xy - 2x = 4y + 2
3xy - 4y = 2x + 2
3x * y - 4 * y = 2x + 2
y * (3x - 4) = 2x + 2

[rocky pelican]

how do u make y the subject on the left side

[timid silo]

You put 3x - 4 on the other side

[rocky pelican]

yes

and how would u do that

[timid silo]

x = 4y + 2/3y - 2
3xy - 2x = 4y + 2
3xy - 4y = 2x + 2
3x * y - 4 * y = 2x + 2
y * (3x - 4) = 2x + 2
y = 2x + 2 (3x - 4)

[fallow wharf]

You must put parenthesis when writing fractions, like this: x = (4y + 2) / (3y-2)

Everyone who ignores this is doing you a disservice

[rocky pelican]

you divide (3x-4)

[dusk widget]

nvm I don’t feel like talking about this anymore @fallow wharf

[fallow wharf]

lol well for your own sake you should know what injectivity means

[rocky pelican]

because its y * (3x-4), dividing by (3x-4) will cancel the (3x-4) on the left side

[fallow wharf]

"one to one" means literally what it means, one thing is mapping to one thing

[timid silo]

x = 4y + 2/3y - 2
3xy - 2x = 4y + 2
3xy - 4y = 2x + 2
3x * y - 4 * y = 2x + 2
y * (3x - 4) = 2x + 2
y = 2x + 2/ (3x - 4)
y = 7x^2 + 4x - 8

correct?

[rocky pelican]

no

[fallow wharf]

x = 4y + 2/3y - 2 is incorrect

[rocky pelican]

y = (2x+2)/(3x-4)

[dusk widget]

I'm just confused because it seemed to me that what you said was injectivity was reversed from the definition I know

[fallow wharf]

f(a)=f(b) means a=b

[dusk widget]

namely, f(x) neq f(y) => x neq y

[timid silo]

x = 4y + 2/3y - 2
3xy - 2x = 4y + 2
3xy - 4y = 2x + 2
3x * y - 4 * y = 2x + 2
y * (3x - 4) = 2x + 2
y = 2x + 2/ (3x - 4)

[dusk widget]

but you said that f(0) = {0, 1, 3}

that's not f(a) = f(b) not => a = b

[rocky pelican]

u don't need to do anything else after u get this: y = (2x+2)/(3x-4)

just leave it as it is

[dusk widget]

cause we didn't start with f(a) = f(b), no?

[timid silo]

x = 4y + 2/3y - 2
3xy - 2x = 4y + 2
3xy - 4y = 2x + 2
3x * y - 4 * y = 2x + 2
y * (3x - 4) = 2x + 2
y = (2x + 2)/ (3x - 4)

this is the answer?

[rocky pelican]

parentheses for 2x+2

[fallow wharf]

you're working with the contrapositive, maybe that's your confusion?

[dusk widget]

.

even from the forwards direction, I don't see how what you said is equiv

[timid silo]

Should I divide them?

[rocky pelican]

no

just put parentheses around 2x+2

[timid silo]

subtract?

x = 4y + 2/3y - 2
3xy - 2x = 4y + 2
3xy - 4y = 2x + 2
3x * y - 4 * y = 2x + 2
y * (3x - 4) = 2x + 2
y = (2x + 2)/ (3x - 4)

this is the answer?

[rocky pelican]

yes

[dusk widget]

because if f isn't injec. (which it wasn't), we should be able to find an pair (f(x), f(y)) st f(x) = f(y) and yet x neq y

but you didn't find a pair of equal outputs, so far as I could tell?

(sorry Kiomi)

[timid silo]

Number 4 is kinda easy

Its okieee

x = 3^y+1
x = 3y + 3
x - 3/3 = y?

[rocky pelican]

its 3 to the power of y+1

do u know logs

[timid silo]

No

[rocky pelican]

oh no

[timid silo]

I ONLY HAVE AN HOUR AND 14 MINS

[rocky pelican]

i'll just tell u the answer for this one then

[timid silo]

WAIT

[fallow wharf]

please stop enabling this user

[timid silo]

COME ONNN

SOMEONE HELP MEEE

<@&286206848099549185>

[rocky pelican]

ok

[fallow wharf]

yeah you're right, I'm thinking backwards, not sure why I'm saying that sorry!

[timid silo]

DANG IT WAIT LEMME JUST IM SO SORRY

[dusk widget]

no worries, I had suspected you just went backwards haha

[fallow wharf]

yeah weird LOL brb getting my brain checked

[dusk widget]

you really do need logs for to calculate the inverse function of i(x) = 3^(x + 1)

[formal sierra]

That should be regularly done (I am not stating your mathematical ability)

[dusk widget]

there's no getting around it

[formal sierra]

Log to the base 3 to be precise

[dusk widget]

yeah Mero is usually very sharp with math, so I thought that I had messed up the def of injectivity

[dusk widget]

indeed

well.. you don't need a base of 3, but it helps

[formal sierra]

Do every honourable need to be extremely smart

[dusk widget]

no

counterexample: me

[formal sierra]

I would argue otherwise

[fallow wharf]

I think I'm just used to things being functions, then you already know it's never "many to one" so it can only ever be "one to ???" and so injectivity on top of that forces one to one.

I probably just sort of applied this shortcut in thinking without really thinking it through.

good thing I'm in these question channels, learning I gotta remember to slow down too I guess

[dusk widget]

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[dusk widget]

you need to use the natural logarithm ln(x) here

it is the inverse of the exponential function e^x

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

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[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stark moon]

4 sin thetha - 5 = 3 and the answer is 30 degrees and 150 degrees how

[stark moon]

How does this work????

am i dumb or...?

[deft magnet]

that shouldn't have an answer

unless im mistaken

sin(theta) = 2 should have no solutions

[stark moon]

that's exactly what I am thinking

[frigid prawn]

sin(θ-5) or sin(θ)-5

[stark moon]

I am so lost at this current moment

here let me send a picture of the exact question

[deft magnet]

are you sure it's not 4sin(theta)+5

or something

actually nvm

[stark moon]

nope one second

[deft magnet]

sin(30) and sin(150) are positive values of sine in the unit circle with answer 1/2

[frigid prawn]

that's possible?

[stark moon]

yes apparently and I am very, very confused

wut?

[frigid prawn]

NOPE

nvm

[deft magnet]

ran it through wolfram and got these solutions

[glacial verge]

those angles will result to -3
maybe a typo?

[deft magnet]

given this fact that indicates that 4sin(theta)=2 so it would work with -5 and -3?

[frigid prawn]

yea it's literally a complex value of θ

[stark moon]

bruh what the heck

and one last question before I move on

[frigid prawn]

probably meant to be a +5 🤔 holy crap why are there so many

[stark moon]

7 cot thetha - 4 = 6 cot thetha - 5 one of the answers would be 3pi/4 rad right?

[stark moon]

Yeah I am guessing so

[stark moon]

cuz cot thetha = -1

[glacial verge]

I'm pretty sure it's a typo
if it's 4sin(theta) - 5 = 3
30 and 150 degrees is the answer

[stark moon]

ok thank you

[frigid prawn]

double neg is correct

[stark moon]

ya

That's just the original equation

[deft magnet]

even then the answers are for positive 1/2

[glacial verge]

oh wait, no, = -3

[deft magnet]

ya

[deft magnet]

. 4sin(theta)-5=-3

or 5-4sin(theta)=3

[stark moon]

ok ty

[frigid prawn]

yes

i am stupid

[stark moon]

yippers

ty

all that helped

I will close this channel down

I forgor how to close

I'll just say close

uhh

[deft magnet]

.close

[glacial verge]

.close

[deft magnet]

@stark moon

[stark moon]

.close

lol

[obtuse pebble]

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[obtuse pebble]

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[unkempt cargo]

i've done this problem over 3 times, help pls

[patent osprey]

time to do it a 4th time

😄

what have you done so far

can you show me ur work

[unkempt cargo]

i got to -2x=sqrt(25-x^2)

[patent osprey]

what di dyou get as ur average rate of change?

@unkempt cargo

[unkempt cargo]

1/2

[patent osprey]

good good

what did you get as f'(x)

[patent osprey]

this checks out

what did you get as

x

[unkempt cargo]

-x/(sqrt(25-x^2))

[patent osprey]

yeah that matches with mine

[unkempt cargo]

that's what i got as f'(x)

[patent osprey]

ur just solving wrong then

rip

[unkempt cargo]

bro can you compute it for me rq

[patent osprey]

what 😭

[unkempt cargo]

i already tried like 5 times 😿

[patent osprey]

just tell me what you got

and ill verify

[patent osprey]

remmeber what einstein said

[unkempt cargo]

-sqrt(5)

[patent osprey]

what about sqrt(5)

oh yeah nvm that does'nt work

[unkempt cargo]

so it is -sqrt(5)

?

[patent osprey]

what is the decimal of -sqrt(5)

[unkempt cargo]

-2.236

i think i might just have to email my professor :(

[patent osprey]

yeah it looks good @unkempt cargo

sqrt(5)

[obtuse pebble]

@unkempt cargo Has your question been resolved?

[obtuse pebble]

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[magic geyser]

Person A and Person B are taking the same 5 classes, a person takes a total of 8 classes. Assume that there are any 3 periods a class can take place in, ex.

1. Math, History, English
2. Science
3. History
4. English
5. Math, Science
6. Science, English
7. History
8. Math

and so on(thats just an example of what could happen). Calculate the probability that Person A and Person B will have atleast 1 class together in the same period.

[magic geyser]

Is this possible

my friend and I are wondering if its possible to be in the same class next yr

[obtuse pebble]

@magic geyser Has your question been resolved?

[polar fossil]

the easiest way to do this is probably simulation

[magic geyser]

what is that?

[polar fossil]

like write a program to generate a class assignment and then run it thousands of times to get an idea of how likely it is

[magic geyser]

oh

[empty cypress]

i can probably write this quickly

[magic geyser]

ohh okay

[empty cypress]

although I dont really understand - each person has 8 classes over 8 periods, but they each share 5 of the same classes

and each is taking 3 the other person isnt?

@magic geyser

[magic geyser]

well heres the thing

each class is present 3 times

in 2022 say calculus

it was period 1,4,8

in 2023

it was period 2,5,7

so its random

[empty cypress]

okay

[magic geyser]

it gets changed up each year

[empty cypress]

its surely not random but yes

[polar fossil]

i think it's like...
he and his friend are both taking Accounting, Biology, Calculus, Danish, and Economics

[magic geyser]

yeah something like that

[polar fossil]

but he's taking French, German, and Hungarian while his friend isnt

[empty cypress]

is this the same as assigning both schedules at random and seeing if theres overlap

[magic geyser]

yes

[empty cypress]

like does the 3 timeslots actually matter

[magic geyser]

i mean the 3 timeslots are the same for everyone, like for everyone

theyre either in periods 2, 4, 6 for history

but they dont choose that and cant do something like period 5 history

[empty cypress]

i think since i dont know them id have to randomly generate them anyways so i can just ignore i think

[magic geyser]

if that makes sense

[polar fossil]

doesn't feel like you can ignore them but maybe

[empty cypress]

it looks like its about half if you dont ignore it

i mean do ignore it

47.5%

[magic geyser]

😭 dang

okay thank you bunny

.close

[obtuse pebble]

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[dim drum]

hello can anyone help teach me how to do cubic spline interpolation?

[dim drum]

this is my set of data and im ask to find the value of y when x is 5.8

[obtuse pebble]

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[obtuse pebble]

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[hoary kindle]

how do i use triple angle formulae for 1a?

[hoary kindle]

.close

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[timid silo]

Is it possible for me to describe an explicit sequence for a pattern of 1, -2, 6, -12

[timid silo]

This is actually kind of an XY question so I will provide the context as well

[rain sluice]

Oh my a dr du student?

[hoary kindle]

wassup

[timid silo]

I am given a linear, and time invariant transformation described by the following equation: [
\7yn + 2\7y{n-1} = \7xn + 2\7x{n-2}
]which is initially at rest $(y[n] = 0 \tss{for} n < 0)$. I am looking to find the response under $\7xn = \7\delta n$

[timid silo]

i basically just checked the first 4 values of n

you get h[0] = 1, h[1] = -2, h[2] = 6, h[3] = -12

but how do i describe this non-recursively

Does [
\7hn = (-1)^n \2 2^{n-1}\2 \floor{\4n2}!
]
work

[restive gorge]

h[0] = 0.5

[timid silo]

Yeah right

[restive gorge]

You are near tho

nvm

[timid silo]

what the heck

this is so dumb

i think you are bound to have it recursively

my recurise formulation is currently
[
\7hn =(-1)^{n}\7h{n-1} \2 3 \q \t{If } n = 2 \
\7hn = (-1)^n \7h{n-1} \2 2 \q \t{otherwise for } n >2
]
with initial conditions $\7h0 = 1$ and $\7h1 = -2$

i think this makes sense?

[timid silo]

ok maybe i should just include the n =2 case as an initial condition as well? lol

.close

[obtuse pebble]

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[restive gorge]

I even wonder how you came up with this lol, it's clear that's alternating but the rest

[obtuse pebble]

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[neon crane]

Given $0 < a < 180$ degrees satisfies $2 * \tan(a) + 2 * \cot(a) + 5 = 0$. Calculate $P = 4 \sqrt{5} \cdot \sin(a) - 16$ with $\tan a > \cot a$

[warm shale]

cffex

[neon crane]

dunno how to approach this

[ionic oar]

do you know how to find tan(a) from the given equation?

[neon crane]

no not really

[ionic oar]

let tan(a) = A for some real number A, then cot(a) = 1/A, from the equation you'd have
2A + 2/A + 5 = 0

[neon crane]

then you would have A + 1/A = -5/2, (A^2 + 1)/A = -5/2

[ionic oar]

and tan(a) > cot(a) so A> 1/A -> A^2 > 1

[neon crane]

right

go on

[ionic oar]

id multiply both sides by A in the original equation

[neon crane]

do A^2 + 1 = -5A/2

so*

and then what do I do

[ionic oar]

A^2 + 5A/2 + 1 = 0

you could factorize, or use quadratic formula

[neon crane]

oh quadratics

ohhh so you find A and then find sin

[ionic oar]

yep

you'd get 2 values of A though, which is why tan a > cot a was given to choose only 1 value

[neon crane]

-0.5 > 1/-0.5 = -2 false
-2 > 1/-2 = -0.5 true

so its -2

ok then $\tan a = 2$

[warm shale]

cffex

[neon crane]

$\frac{\sin a}{\cos a} = 2$

[warm shale]

cffex

[neon crane]

$\sin a = 2 \cos a$

[warm shale]

cffex

[neon crane]

$\sin^{2} a + \cos^{2} a = 1$

[warm shale]

cffex

[neon crane]

these two eqs are simul eqs

then its possible to find sin

[ionic oar]

i'd use the definition of tan but this also works

by the way, tan(a) = -2

[neon crane]

since $0 < a < 180$ degrees which means $\sin a > 0 \forall a$

[warm shale]

cffex

[neon crane]

oh yeah

which means $cos a < 0 \forall a$

[warm shale]

cffex

[ionic oar]

cos(pi/2) = 1/sqrt(2) > 0

[neon crane]

oh shoot

i mistaken

nevermind

[ionic oar]

cos can be either negative or positive in this range

[neon crane]

i just realized

yes

[ionic oar]

but you only need sin(a)

[neon crane]

yea

so find sin

then sub to P

then im done

ok thanks

.close

[obtuse pebble]

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[ionic oar]

yw

[obtuse pebble]

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[neon crane]

just realized i was wrong

[ionic oar]

lol

but you came to the right conclusion

[obtuse pebble]

@neon crane Has your question been resolved?

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[heady turtle]

I would like a hint to start thinking

[alpine bison]

taylor expansion could be useful

[obtuse pebble]

@heady turtle Has your question been resolved?

[obtuse pebble]

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[remote bison]

Lets take a function that is piecewise defined. In order to find the derivative of the function, do we need to check the continuity of the function at the point where the definition changes? Or does the LHD-RHD takes care of that?

[fossil crag]

if the function is not continuous, then one of the LHD-RHD doesn't exist

[remote bison]

Okay so I can directly jump at LHD-RHD, right?

[fossil crag]

yes, but you have to revert to the definition of LHD-RHD

[remote bison]

I dont have to worry whether the function is continous or not

[fossil crag]

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

with 0+ and 0-

[remote bison]

Got it. Thanks

One more thing

What is the value of the limit if we get something like (tends to 0) x (tends to infinity)

[fossil crag]

0*infinity is an undefined state, meaning it could tend to infinity, something finite, 0, or nothing at all

you have to simplify stuff to find its limit

for example limit of sin(x) * 1/x as x goes to 0

it looks like (tends to 0) * (tends to infinity)

and yet it goes to 1

another example:

[knotty dome]

L'H :trollface:

[fossil crag]

xsin(x) * sin(1/x) as x goes to infinity

this looks like (tends to some infinity) * (tends to 0)

and actually has no limit

since it keeps oscillating between -1 and 1

[remote bison]

The examples made it clear. I was wondering if it will be 0 like in the case of 0 x infinity

[fossil crag]

it can

x^2 * 1/x when x-> 0

[remote bison]

Is this applicable to (tends to 0) / (tends to infinity) as well?

[fossil crag]

0/infinity is 0

0/0 is undefined form

infinity - infinity is undefined form

etc...

[remote bison]

And what about (tends to infinity)/(tends to 0)?

[fossil crag]

depends on the sign

(tends to +infinity)/(tends to 0+) is +infinity

(tends to +infinity)/(tends to 0-) is -infinity

[remote bison]

In that case. limit doesnt exist, right?

[fossil crag]

it can exist

here's an example

$\lim_{x\to +\infty}\sqrt{x^2+x}-x$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

should go to 1/2

[remote bison]

How?

[fossil crag]

$\sqrt{x^2+x}-x = (\sqrt{x^2+x}-x)\frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

$\sqrt{x^2+x}-x = \frac{x^2+x-x^2}{\sqrt{x^2+x}+x} = \frac{x}{\sqrt{x^2+x}+x}$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

you can forcefully factor the denominator by x:

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

[warm shale]

rafilou is not not born in 2003

[fossil crag]

$\sqrt{x^2+x}-x = \frac{1}{\sqrt{1+\frac 1x} + 1} \xrightarrow{x\to+\infty}\frac{1}{\sqrt{1}+1} = \frac 12$

[warm shale]

rafilou is not not born in 2003

[remote bison]

Thank you so much for the deatiled explanation

[fossil crag]

easier example but

(x+1) - x when x to +infinity

[remote bison]

I had to confirm one more thing about differentiability

Suppose LHD = RHD at a point. Is that condition enough to show that the fn in differentiable at that point?

Or is there some other tests?

Or conditions?

[fossil crag]

if both LHD and RHD exist, with THIS exact definition

and they're equal

then f is differentiable at that point

[remote bison]

all 3 of them has to be equal

[fossil crag]

if LHD = RHD, then automatically D = LHD = RHD

[remote bison]

Ohh I see. Thanks a lot

[fossil crag]

btw here's an example of why using this definition is important

$f(x) = \begin{cases}-x, & x < 0\ -x+1, & x \geq 0\end{cases}$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

you could "differentiate by sight" both things by being like "derivative of -x is -1, derivative of -x+1 is -1, well they're equal"

except the piece "-x" never actually reaches x = 0

[remote bison]

yes, I would have

[fossil crag]

so this method doesn't work

you have to go through this for LHD :

$\lim_{h\to 0^-} \frac{f(h)-f(0)}{h} = \lim_{h\to 0^-} \frac{-h-1}{h}$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

and suddenly f is not LHD when you use the definition

all because the piece "-x" is not reaching x = 0

[remote bison]

This doesnt match with the formula my tecaher gave

[warm shale]

Crystal
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[remote bison]

$\lim_{h\to 0^-} \frac{f(0)-f(h)}{h}

I am sorry

[fossil crag]

this is the opposite of the derivative

dollar signs surround your formula from both sides

$like this$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

$not like this

[remote bison]

Idk how to use that tool

[fossil crag]

no problem

[remote bison]

Okayy

[fossil crag]

but the derivative has always been $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ hasn't it?

[warm shale]

rafilou is not not born in 2003

[remote bison]

$\lim_{h\to 0^-} \frac{f(h)-f(0)}{h} $

[fossil crag]

oh and

[remote bison]

is it?

[fossil crag]

dollar signs next to your formula

$ this doesn't work $

$this does$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

you can edit

[remote bison]

we were given 2 slightly different formulas for RHD and LHD

[fossil crag]

well yeah, unless your teacher is weird

[remote bison]

Mannn Idk he said the bigger one has to come first and smaller one gets substracted

[fossil crag]

derivative : $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ $\$
LHD : $\lim_{h\to 0^-}\frac{f(x+h)-f(x)}{h}$ $\$
RHD : $\lim_{h\to 0^+}\frac{f(x+h)-f(x)}{h}$

[warm shale]

rafilou is not not born in 2003

[fossil crag]

woooow no

[fossil crag]

I think you'll find this to be easier

just remember the derivative formula, which should never change

LHD is just we take h < 0

RHD is just we take h > 0

[remote bison]

I will be using this one from now then

[remote bison]

The RHD is same but the LHD is kind of the opposite

[remote bison]

This is easier to remember

[fossil crag]

waitwaitwait

is your formula

LHD: $\lim_{h\to 0^+}\frac{f(x)-f(x-h)}{h}$?

[warm shale]

rafilou is not not born in 2003

[fossil crag]

which is unnecessarily complicated

and h goes to 0+

[remote bison]

Yes but it was just "h tends to 0" not ".... 0+"

[fossil crag]

then it makes even less sense!!!!

because if h just tends to 0 then you allow h to go from under 0

[remote bison]

Idk I am just saying what we were taught in class

[fossil crag]

yeah your teacher is onto some substance

[fossil crag]

I would personally stick to this

[remote bison]

yes, we can say that ig

Thats all! Thank you

.close

[obtuse pebble]

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[random berry]

can anyone explain to me whats going on here

[red ice]

so due to the rank-nullity theorem, the nullity of A is 0

that means that Au and Av will never be zero, unless u, v are the zero vectors

then they did the inner product on Au and Av, which is another way of saying the dot product

the norm is a way of measuring the 'size' of a linear transformation

[random berry]

what do you mean by size?

and where did vector u and v come from are they some arbitary vectors or what?

[brazen viper]

(technically, the inner product is a generalization of the dot product which can be applied to other things as well, but for the case of linear algebra and matrices they are the exact same thing)

[red ice]

yes

[brazen viper]

Norm measures the length of a vector for some particular notion of a length.

[random berry]

am familiar with that idea but the idea of size of a transformation is new to me

[brazen viper]

In normal Euclidean space, the most common notion of length is based on the Pythagorean theorem

[red ice]

hopefully from $||u||^2 = u \cdot u$ you can see why the norm is like that

[warm shale]

south's secret twin brother

[red ice]

you just then square root both sides

[red ice]

well I mean, the determinant is just the area / generalised volume scale factor

[random berry]

wait so i then square root Au . Av and i get the "size" of the linear transformation

[red ice]

so if the determinant in R^2 is 3, that means the square formed by the basis vectors (1, 0) and (0, 1) becomes 3 times larger under the matrix transformation

it changes shape ofc

[random berry]

oh wait nvm

[red ice]

yeah

pretty much

[random berry]

wait so the size is the determinant

[red ice]

it's another concept of size unrelated to this one

I recommend watching 3Blue1Brown's videos for more detail

[random berry]

I watched his playlist but am still lost like i understand that the determinant shows how much the area/volume gets scaled by but i just dont understand where A comes from and the vectors u and v

[brazen viper]

Essentially, this is examining what happens to the basis vector (0, 1) under the linear transformation given by the matrix A

The determinant of A is 1. (Given by 1 * 1 - 2 * 0), but space is still stretched and squished by the transformation

So a determinant isn't the full story

This A and u,v pair is just for the sake of an example

They were chosen to illustrate the ideas

Specifically, the inner product in this case.

@random berry does that make sense? Or are you still a little bit unsatisfied?

If unsatisfied, do you have a particular question? Or are you just not seeing the point or purpose?

[random berry]

wait so tell me if i understood it right so the determinant tells us by how much the area was scaled but it doesn't tell us whether the area itself changed shape? so then this inner product of a matrix and a vector tells us? sorry if am being slow and thanks for your effort for helping

[brazen viper]

Yes, and it can but doesn't have to

It might be possible to select a vector whose length doesn't before and after the transformation incidentally.

(Depending on the nature of the transformation.)

My guess is this is leading up to eigenvectors and eigenvalues

Which does tell you exactly how a matrix warps space

[random berry]

you are correct the set of lectures note right after this is eigenvectors

[brazen viper]

However, the inner product is still useful, because it's an easier way to tell how any particular direction is affected by the transformation

So if you only care about a single direction (such as perhaps a special relativity calculation where you only care about the direction (inc time) of your spaceship) then the inner product gets you the information you want with less work

[random berry]

Thank you very much for your help and i appreciate your time and effort

https://tenor.com/view/bear-byebear-cute-bye-wave-gif-344809869674681480

.close

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[primal talon]

(x-2) (x-1) (x+3) should i let x = 0 so it can be (0,-2) (0,-1) (0,3) ? same topic RRT

[tardy epoch]

!original

[obtuse pebble]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[restive gorge]

i wish i could read minds

[crimson pike]

same

[primal talon]

that is the only problem our teacher showed us 😭

she said we should practice that kind of problem 😭

[timid silo]

you haven't shown what the question is to begin with

so we can't help if we don't know what is being asked for

[obtuse pebble]

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[nocturne anvil]

can any1 explain this to me?

[nocturne anvil]

<@&286206848099549185>

[void holly]

!15mins

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[nocturne anvil]

.close

[obtuse pebble]

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[ocean stone]

;rotate

:rotate

,rotate

[ocean stone]

alright so I need help with exercise 1

Now I am stuck

And no Cezaro Stolz can't work because it isn't 0/0

[obtuse pebble]

@ocean stone Has your question been resolved?

[ocean stone]

<@&286206848099549185>

[timid silo]

Yes

[ocean stone]

What is the continuation or well

[ocean stone]

Any hints?

[twin sandal]

define x_n = 1/a_n. then apply Cesaro convergence

to x_n

[ocean stone]

I will attempt

[ocean stone]

If I write it as 1/a_n it will go down with the denominator right?

[ocean stone]

Like, I don't see another way to write it as

[twin sandal]

let's do this step by step, what does x_n converge to?

[ocean stone]

Well 1/n converges to zero

[obtuse pebble]

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[spice chasm]

how to find the component of a vector in the direction of another vector

[spice chasm]

A ship is sailing in the direction of $\langle3,4\rangle$ when wind pushes it in the direction $\langle-9,5\rangle$. Use vector projection to determine if the wind helps the ship or goes against the ship.

[warm shale]

Jash

[obtuse pebble]

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[stoic bloom]

Can someone help me finish number 2?

[stoic bloom]

pls

[obtuse pebble]

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[upbeat sapphire]

Hi. Does anyone here know how to use a TI-84 plus to calculate x and/or coefficients if its a quadratic function? (Pattern: 16; x; 25)

[rocky pelican]

U mean nsolve?

[upbeat sapphire]

I haven't used TI calculators for 15 years so I'm a bit rusty. I'm afraid nsolve rings a bell but not much

[patent osprey]

do you mean given a couple points

you wanna fit a quadratic?

[upbeat sapphire]

Yeah

[patent osprey]

do you got the nice screen ti 84

or the older one

[upbeat sapphire]

As in what would X be and what would the pattern look like

[patent osprey]

what are your points

I wanna use my ti 84

[upbeat sapphire]

Ti 84 plus (non cas)

[patent osprey]

but ilke quadratic

like you have a couple points

[upbeat sapphire]

The patten goes 16...x...25

It's up to me to figure the pattern out, but i am allowed to use my ti 84 plus

[patent osprey]

hmm

can I see the original problem pls

[upbeat sapphire]

So basically n1=16, n2=x, n3=25

Ok gimme a sec

[patent osprey]

yayy

[patent osprey]

shoot 😭

[upbeat sapphire]

It's in swedish but the important parts: a1=16, a3=25

(1) - give examples of what a2 could be if it is a quadratic function.
(2) - solve a for any n.

[patent osprey]

oh swedish

i love the fish

hmm okay

did you solve a through b btw?

[upbeat sapphire]

Yeah

Ezpz 😉

[patent osprey]

what do you know about quadratic sequences

[upbeat sapphire]

I can use A B C formula manually or with the TI84 to find coefficients for a pattern with 3 values

[patent osprey]

let's do it manually shall we?

[upbeat sapphire]

I try and fiddle with n^2 to find a reasonable pattern

[patent osprey]

can you give me a general formula

for a_n

in terms of some constants a, b, and letter of ur choice

[upbeat sapphire]

In the case I dont know hoe to proceed but I can give u the start with 2 values

[patent osprey]

like I mean

hmm

kyle give me ur fav number

[upbeat sapphire]

A+B+C=16
4A+2B+C=X
9A+3B+C=25

Oh

13 or 21

7 the I suppose if it's single digits;)

[patent osprey]

A+B+C=16

9A+3B+C=25

this is what we know

we can subtract the equations to get

8A + 2B = 9

so 8A = 9 -2B

I'm sorry if this is hard to follow because of the language disconnect

[upbeat sapphire]

So far I'm onboard

[patent osprey]

God morgon

only swedish word ik

[upbeat sapphire]

I'm also an English teacher so hopefully language isn't my major issue 🙏

[patent osprey]

let's solve for 2B here

2B = 9 - 8A

[upbeat sapphire]

Still understandable

[patent osprey]

B = (9-8A)/2

plug it back into A+ B + C = 16

to get C also in terms of A

kyle are we still following

[upbeat sapphire]

Gimme 1m to absorb

Why plug it back in to A B C = 16 and not 25. Can I choose either?

[patent osprey]

you can choose either

[upbeat sapphire]

Alright then I follow

A+ (9-8A)/2+C=16

[patent osprey]

solve for C

and then ur done

[upbeat sapphire]

C=3A+11.5 ?

[patent osprey]

should be something like that

[upbeat sapphire]

I'm sorry but I cannot proceed with that. Is this usable to solve a2?

I have to head to work now but thanks for your time and effort ❤️

[obtuse pebble]

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[true heron]

Is anyone able to help with my mathematica fit? I've been trying for ages and i simply cannot get it to work, thanks.

[late elbow]

It depends on the math topic and if I learned it at school 🥲

I’m out of school for like 2 years now ‘cause I already graduated from college

[true heron]

Its just a log plot its nothing crazy, but it keeps coming out with the wrong magnitude and its starting to drive me up the wall

whenever i overlay them the fit just doesnt get displayed at all, which i assume is because of the magnitude difference

<@&286206848099549185>

So I have a .dat file that I need to log plot and then add a fit to

but the fit is coming out completely wrong eveyrtime

(Importing the data)
mylist = Import["C:\Users\liamm\Documents\Wolfram\file1.dat"];

(Creating filtered list and log list)
filteredList = Select[mylist, #[[1]] > 400 &];
logFilteredList = {#[[1]], Log[10, #[[2]]]} & /@ filteredList;

(Defining quadratic function for the fit)
fitFunc = a + bx + cx^2;

(Creating fit)
fit = NonlinearModelFit[
logFilteredList,
fitFunc,
{a, b, c},
x
];

(Plotting everything)
ListLogPlot[
filteredList,
PlotStyle -> Blue,
PlotLabel -> "Quadratic Fit with 95% Confidence Bands",
AxesLabel -> {"x", "y"},
PlotRange -> All
]

Plot[
fit[x],
{x, 400, Max[filteredList[[All, 1]]]},
PlotStyle -> {Red, Thick},
PlotRange -> All
]

Plot[
1 Evaluate[fit["MeanPredictionBands"] /. fit["BestFitParameters"]],
{x, 400, Max[filteredList[[All, 1]]]},
PlotStyle -> {{Red, Dashed}},
PlotRange -> All
]

thats the mathematica code ive used

huh?

any help is much appreciated

ive been on it for hours

thats the .dat,

only use values with x>400

[obtuse pebble]

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[karmic hinge]

is it true that Making the trig ratio positive always returns the reference angle itself

[karmic hinge]

thats what chatgpt said

and if its not can u show an example

thanks

[obtuse pebble]

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[obtuse pebble]

@dire violet Has your question been resolved?

[dire violet]

<@&286206848099549185>

[obtuse pebble]

@dire violet Has your question been resolved?