help-0

[rocky grove]

Do you know what a fixed point is?

[empty forge]

yes, so f(x)=x is a fixed point

[rocky grove]

Okay

Let p be the solutions to x = ax^2 + bx + c, which makes it the fixed points of Q(x)

[empty forge]

okay, I am not sure how that will help.

[rocky grove]

That's what the question tells you to do

Then the question wants you to prove some cases for p

If you plug p into Q'(x) and it gave 1, then prove that it is not attracting when you use fixed-point iterative method.

[empty forge]

to confirm, fixed-point iterative method would mean I would show |Q'(p)|>1

[rocky grove]

Hmm

In the question, it tells you to prove that when Q'(p) = -1, then it is attracting and for Q'(p) = 1, then it is not attracting.

[empty forge]

@rocky grove thanks. I'll think about what you said.

.close

[lone heart]

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[alpine sable]

what does something mean when it is "lying strictly between x and y"

[rocky grove]

$x\le \text{something} \le y$

[alpine sable]

oh thank you

[naive valley]

no

[fallen verge]

No

STOP

[naive valley]

strict inequality is <

not <=

[alpine sable]

tysm

[naive valley]

also, it's possible that y < x

[alpine sable]

im trying to prove that for x, y in set Q there exists some number that is strictly between x and y (when x not equal to y

[alpine sable]

wdym by this?

[naive valley]

Q = rationals?

[alpine sable]

like y is before y

yes

[naive valley]

"between x and y" doesn't necessarily mean x < y

[alpine sable]

oh sorry, more specificially its lying strictly between

[naive valley]

what's the simplest rational you can think of that's guaranteed to be between x and y?

assuming x and y are themselves rational

[alpine sable]

x/y?

[fallen verge]

Not quite

[naive valley]

no, that won't work, suppose x = 1 and y = 2

1/2 is not between them

[alpine sable]

oh wait thats a good strategy

[rocky grove]

Oops

Sorry for misleading

[naive valley]

can you think of a rational number between 1 and 2?

[alpine sable]

like 1.5

[naive valley]

yeah perfect

[alpine sable]

so

3/2

[naive valley]

yea

can you express that in terms of 1 and 2?

like a formula

[alpine sable]

well if x = 1 and y = 2, thats specifically x + y / y

[naive valley]

yea in this particular case (x+y)/y works

what if y wasn't 2 though

3/2 is midway between 1 and 2

how would you find a number midway between x and y

for arbitrary numbers x and y

[alpine sable]

like x + y / 2

[naive valley]

yep

(x+y)/2

parentheses matter here

[alpine sable]

i need to use the definition of a rational number

which i know is two integers expressed as a / b where b not eq to 0

[naive valley]

yea, can you explain why (x+y)/2 is rational if x and y are?

[alpine sable]

a rational number divided by any number is still a rational number

[naive valley]

by any rational number

[alpine sable]

yes

[naive valley]

so what reasoning can you use here

[alpine sable]

that explains why it lies strictly between x and y

[naive valley]

but specifically why is (x+y)/2 rational

2 is rational

what about x+y

[alpine sable]

because they are already rational numbers

defined

like id be like fix rational numbers x, y e Q right

[naive valley]

x and y are

what about their sum

[alpine sable]

because ive defined it

[naive valley]

x+y

[alpine sable]

isnt that closed

[naive valley]

is the sum of two rationals still rational?

i could be off base but i suspect that whoever is asking this will want you to prove this carefully

[alpine sable]

the answer is yes, but i dont know how to prove this

i appreciate suggestions (im new to this type of mth)

math

[naive valley]

i would go back to the definition

if x is rational then i can write it as a/b where a,b are integers

if y is rational then i can write it as c/d where c,d are integers

now find x+y as a function of a,b,c,d

see if i can express it as a ratio of two integers

[alpine sable]

ahh the goal is to show that it over all ends up being sum thing / another thing right

i think it comes out to like (ad) + (cb) / 2 (bd)

[naive valley]

yes!

and the numerator and denominator are integers

ad + bc = integer

2bd = integer

so when you divide them you get a rational number

[alpine sable]

thank you so much

i appreciate it

[naive valley]

glad to help

cheers

[alpine sable]

.close

[lone heart]

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[eternal kestrel]

Find the point P on the line y=5x that is closest to the point (52,0) . What is the least distance between P and ​(52,0)?

[alpine sable]

WHY IS THE WORK NEGATIVE

[eternal kestrel]

its asking for the closest point please help

[rocky grove]

!help

[lone heart]

Please read #❓how-to-get-help

[alpine sable]

i dont understand

[rocky grove]

Please use a new channel:
#help-13

[eternal kestrel]

ill dip

.close

[lone heart]

Channel closed

Closed by @eternal kestrel

Use .reopen if this was a mistake.

[rocky grove]

Wha

[eternal kestrel]

it got drowned below the rubble

gotta refresh the feed

.close

[rocky grove]

Okay lol

[lone heart]

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[strong geyser]

−3x+5>x−19 is not (6, infinity) ?

[serene junco]

Remember to flip the > sign if you divide by a negative number

[lone heart]

@strong geyser Has your question been resolved?

[lone heart]

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[sudden gale]

p(p^2+3)=12+p^3

[wary stream]

What are you asking?

[sudden gale]

I need to solve this question

my problem is i dont know how to factor p into the parenthesis

[wary stream]

Distribute

[sudden gale]

so distribute looks like p^3+3p=12+p^3

Can i cancel out the p^3 on both sides?

[alpine sable]

yes

[sudden gale]

okay so then it turns into 3p=12

[alpine sable]

now divide

[sudden gale]

then divide by 3 on both sides and get p=4?

[wary stream]

Yes

[alpine sable]

yeah

[wary stream]

That's it

[sudden gale]

idk why math solvers are making it so complicated

it says the answer to the problem is 3p^3

my brain was exploding

ty !

[wary stream]

Then you typed something wrong

[sudden gale]

.close

[lone heart]

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[ornate condor]

i donut get this example :c

mainly from the Any f in X hat g must have... onwards

[desert crane]

Just to make sure - what's X tilde g here

[limpid turret]

I believe it is the permuted elements of X tilde

[carmine reef]

that seems like the hard part

wouldnt the permuted elements just be X tilde still

[limpid turret]

I suppose if it's all functions, then it must be

[carmine reef]

maybe the subscript has some special group theoretical meaning I'm not aware of?

[desert crane]

I'm 99% sure it's fixed points

[carmine reef]

it seems weird they don't introduce it

[desert crane]

I use different notation personally lol

[carmine reef]

oh well that would do it, wouldn't it

[desert crane]

Okay wait yeah I get what this is saying lmao this has to be for some burnsides theorem thing later on

[carmine reef]

or wait maybe not idk

[desert crane]

Are you comfortable with the proof of 14.21

[raven haven]

why didnt they just call it stabG(x) lmao

[ornate condor]

NEIN

[keen plinth]

[desert crane]

Okay wait let me try to write this out then

[carmine reef]

ye so a fixed point of g is a function where "applying g to the domain" doesn't change the outputs

hence f(2)=f(4)=f(5)

since if we apply g, f(2) becomes f(4)

[desert crane]

This is such a bizarre formulation to introduce an entirely new group rather than just describing how the action of G on X extends to functions f:X -> Y

[carmine reef]

with the first equality holding specifically because f is a stabilizer

[keen plinth]

[lone heart]

@ornate condor Has your question been resolved?

[lone heart]

@ornate condor Has your question been resolved?

[lone heart]

@ornate condor Has your question been resolved?

[carmine reef]

this is why any f must have the same value on each cycle

[lone heart]

@ornate condor Has your question been resolved?

[ornate condor]

ok i shld look at it now

ok so um

the cycles have 4 unique values, each cycle has a unique value

and there are 3 options

but why 3^4

what does that represent

[keen plinth]

there are 3 choices for the value of f on each cycle

4 cycles

so 3^4

[ornate condor]

hm

is there a more

visual example

idk relate a square or triangle, with3 colors coloring mayb

prev eg they gave was a square w 2 colors, D4

is that related? na?

[keen plinth]

you can draw the functions

[carmine reef]

let A be a set of 3 colors

a way to color vertices of a square can be thought of as an ordered quadruple: (color 1, color 2, color 3, color 4)

then, valid colorings are elements of A x A x A x A, which has a cardinality of |A|*|A|*|A|*|A|, which is 3^4

[ornate condor]

wait

what happened to permutation groups and cycles

why did they take the order of um idk n how does it relate to ur eg

cuz for them they used some cycles thing right

[bleak coyote]

Hey I was wondering if I could get help with some programming in python!

[rocky grove]

!help

[lone heart]

Please read #❓how-to-get-help

[rocky grove]

And please check out #old-network and computer science server

[bleak coyote]

Thank you

[lone heart]

@ornate condor Has your question been resolved?

[ornate condor]

ok nvm

i dont get a lot of things

this is another ill pass on

lol

thanks all

thanks

snow

moni

lyra

vulcan

activemental

i love you snow

.close

[lone heart]

Channel closed

Closed by @ornate condor

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[ornate condor]

and swr

hey u saw it

[lone heart]

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[cerulean goblet]

can anyone help please, this was a question in my recent mock exam and i got 3/5 marks for it

[amber loom]

So first I believe you have to calculate the angle at DCB

[cerulean goblet]

yeah, i’ve calculated that to be 30

[amber loom]

yes that's correct

[cerulean goblet]

i can send the working out i did on the test that got me 3 marks

[amber loom]

go for it

[cerulean goblet]

but it’s messy

[amber loom]

Is this GCSE?

[cerulean goblet]

yeah

[amber loom]

alright sick I can help with this

[cerulean goblet]

you from the uk too?

[amber loom]

yes I'm doing a levels now

got a 9 last year

[cerulean goblet]

oh damn

[amber loom]

do you see what you did wrong with your working ?

[cerulean goblet]

i’m aiming for a 9 i got an 8 in my recent mocks

[amber loom]

if you got an 8 just now you'll get 9 for sure

[cerulean goblet]

hopefully, i’ve been making a tonne of silly mistakes lately

[amber loom]

alright so once you got your angle you're right in doing cosine rule

[cerulean goblet]

okay

[amber loom]

let me do it rq

that should be ~22.7

[cerulean goblet]

the length opposite the angle we worked out?

[amber loom]

yes the final side of the rightmost triangle

[cerulean goblet]

okay i got that correct too

[amber loom]

and then you're gonna have to do cosine rule again to calculate the angle opposite the side we just worked out in the second triangle

and I think this is where you made your mistake

[cerulean goblet]

yeah me too

i repeatedly got a maths error on the calculator

[amber loom]

so you know when labelling angles the capital A is always opposite to the side a

you have three sides here so it's doable for sure

so we are going to work out the angle labelled DAB

[cerulean goblet]

yeah i labelled the triangle myself, i ignored the labels given in the question

[amber loom]

yeah that's what you should do

okay so do you remember the equation for the angle from cosine rule or do you remember cosine rule and rearrange it yourself to get the angle

[cerulean goblet]

i rearrange it myself

[amber loom]

yeah okay so what's that formula from rearrangement

(that's what i do too)

[cerulean goblet]

wait let me quickly do it

i got this

oh my god i’m so stupid

[amber loom]

Yeah I don't think that's correctr

[cerulean goblet]

yeah that’s exactly where i went wrong then

[amber loom]

$(b^2+c^2-a^2)/(2bc) = cos(A)$

[ocean seal]

Karatesam100

[cerulean goblet]

oh

wait why

[amber loom]

Do you want me to go through the steps for that

[cerulean goblet]

yeah please

[amber loom]

$a^2 = b^2 + c^2 -(2bc)(cos(a))$

[ocean seal]

Karatesam100

[amber loom]

so that's where we start from right?

[cerulean goblet]

yup

[amber loom]

obviously a meant to be A at the cos A bit

[cerulean goblet]

yeah

[amber loom]

first I would move the b^2 and c^2 to the side with a^2

[cerulean goblet]

okay

[amber loom]

$a^2-b^2-c^2 = -2bc cos(A)$

[ocean seal]

Karatesam100

[cerulean goblet]

yeah

[amber loom]

at this point (this isn't necessary) I multiply by -1

this gives you

$b^2+c^2-a^2 = 2bccos(A)$

[ocean seal]

Karatesam100

[amber loom]

agreed?

[cerulean goblet]

i didn’t know you could do that

but okay that makes sense

[amber loom]

it makes it much easier to look at

and then you can do the rest

also what you did from this point was correct but it could be done more easily

you tried to find one angle and then did sine rule right?

[cerulean goblet]

yeah

[amber loom]

so you can, instead of doing that, just set your angle A as the angle you're trying to find, and then name your side accordingly

just skips a step

[cerulean goblet]

oh

[amber loom]

DW about it though, you're in year 11?

[cerulean goblet]

ohhh waitt

i think i done that way but i rearranged the formula wrong

and that’s why i got the maths error

[amber loom]

so x = A in your formula ?

[cerulean goblet]

yeah

[cerulean goblet]

yeah

[amber loom]

you'll 100% get the hang of it, you got like 95% of the way there before making a mistake

and you probs only did non-right angled trig only recently

[cerulean goblet]

tbh we haven’t even done it yet

my dad taught me all of this

it’s crazy

[amber loom]

yeah literally don't worry then you'll do fine lol

good that you asked for help though

[cerulean goblet]

yeah i’ve got 3 other questions lmao but i’ll use another channel

[amber loom]

if you want you can ask here

[cerulean goblet]

thanks a ton for your help dude

[amber loom]

I'm not busy

[cerulean goblet]

if you’re down?

oh alright

[amber loom]

did you get anywhere with this or were you just stuck completely

[cerulean goblet]

i got two marks for this question but it was clueless me in the exam trying to get any marks i could

[amber loom]

alright no worries

[cerulean goblet]

that was my working out

[amber loom]

edexcel ?

[cerulean goblet]

yup

[amber loom]

tough

give me a sec to work it out and then I'll tell you

[cerulean goblet]

alright

[amber loom]

I hate this question we don't do anything like this in A level

[cerulean goblet]

if you don’t know what to do it’s completely fine dw

[amber loom]

yeah go to the next one lol

I'll think it over

[cerulean goblet]

it’s cool

[amber loom]

if i get somewhere i'll dm you

[cerulean goblet]

alright

normally i’m decent with vectors but this one just hurts my head

[amber loom]

okay so ON has to be some multiple of OY

and MN has to be some multiple of MZ

[cerulean goblet]

yeah

[amber loom]

these questions are hard man what's with edexcel

i could do this with A level maths

but I'm trying to do it a way that you'd do in gcse

[cerulean goblet]

ah it’s cool then

i doubt they’ll give us questions like this anyways

[amber loom]

they are rough

[cerulean goblet]

they’re horrible ibsr

[amber loom]

how many marks is this one

[cerulean goblet]

5

it’s the very last question on the paper

i’m happy i even got one mark

[amber loom]

what equations and stuff do you know for vectors from edexcel

i did aqa so im not sure what edexcel people are meant to know

[cerulean goblet]

no equations whatsoever

[amber loom]

just that they have direction and stuff

[cerulean goblet]

yh

idk if we’re allowed to use a level maths methods to do working out

[amber loom]

you're not

[cerulean goblet]

oh damn

[amber loom]

i tried doing differentiation to work out one of my ones i dropped a mark

got the q right though

??

[cerulean goblet]

oh

i mean i would rather know how to do something and get marks for potential workng out and an answer tbh

so i’ll probably start teaching myself vector methods

[cerulean goblet]

yeah that’s just stupid ibsr

[amber loom]

i think the answer is 3/7

[cerulean goblet]

you’re probably right atp

[amber loom]

I'll keep working on it let's close the channel i'll dm you if i get it with gcse

[cerulean goblet]

alright

that’s fine, thanks a tonne for your help

icl idk if random people can message me but you can try

see ya in a bit, maybe

[amber loom]

if ur in the server

[cerulean goblet]

alright, anyways we’ll see

see ya

.close

[lone heart]

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

hello

[austere linden]

I need help, I have a reasoning problem

so i have to work with Q(X) = constant where Q is a polynom in Rm+1[X]

and Q(X) = a0 X1/1 + a1 X2/2 + ... + am Xm+1 / m+1

a0,a1... in Rm+1

in fact Q(X) = f(0) where f is a continuous function in [0,1]

the question is, determine the biggest m in N verifying this equality

and my brain tell me it's 0

(i'm fr so my mathematic english may be horrible)

so i don't know how to tackle with the pb

[mortal trellis]

$Q(x) = a_0 \frac{x^1}{1} + a_1 \frac{x^2}{2} + \ldots + a_m \frac{x^{m+1}}{m+1}$ or what?

[ocean seal]

Denascite

[austere linden]

yes it is

[mortal trellis]

and this is supposed to be constant?

[austere linden]

= f( x=0 ) where f is a continuous function on [0,1]

the right term is evaluated on x = 0, and the left one is for all x in [0,1]

[mortal trellis]

do you not know yet that only polynomials of degree 0 are constant?

[austere linden]

so the biggest m in N verifying this equality is 0 ?

but Q(x, m = 0) = a0 x1 / 1 which is not equal to a constant in [0,1]

that's my pb

[mortal trellis]

well then no m satisfies this equality

not sure why you start with a_0 x^1 instead of a_0x^0

[austere linden]

and what if all x = 1, like a0 / 1 + ... + am/m+1 = f(0), how can i determine m ?

[austere linden]

it's from an integration

[mortal trellis]

can you show the original problem statement

[austere linden]

it's in french...

and here are the terms definition 😅

[mortal trellis]

ok something something quadrature

[austere linden]

yea

[mortal trellis]

ok, so the point of quadrature is to approximate the integral int_0^1 f(x) dx by evaluating f(x) at certain points and then taking a linear combination of those, yes?

[austere linden]

i guess, it's the first part of the pb

[mortal trellis]

what exactly do you mean with pb

[austere linden]

it's a 39-questions pb

is problem not accurate in english ?

exam

hw

i dunno lol

[mortal trellis]

oh it's just short for problem? ok

[austere linden]

yes

In(f) is the quadrature formula

[mortal trellis]

yes

ok so in the problem they give you the quadrature formula I_n(f) = f(0)

[austere linden]

e(f) define its associated error

[mortal trellis]

and generally a measure of how good these formulas are is to check for how many polynomials they are accurate

[austere linden]

I0(f) = f(0)

exactly!

[mortal trellis]

good, so we want to know for which polynomials this particular formula is accurate

in other words, if $Q(x)$ is a polynomial, when is $Q(0) = \int_0^1 Q(x) dx$

[ocean seal]

Denascite

[austere linden]

and the first question is to determine the biggest m in N where the quad formula is accurate so the biggest m in N where e(P) = 0 where P is in Rm[X]

hmm no, f(0) = int(0,1) Pdx

P in Rm[X]

[mortal trellis]

but f is P

[austere linden]

i didn't understood this way

oh

so f is polynomial..

so the biggest m where e(f=P) = 0****

here f(0) = a0 for sure

[mortal trellis]

no. e is the error

the biggest m so that the integral is equal to f(0)

do you know that these quadrature formulas are linear

[austere linden]

what do you mean by linear

[mortal trellis]

if you want to approximate int f+g dx, you have I_n(f+g) = I_n(f) + I_n(g)

[austere linden]

oh sure but are we using it for the first question?

[mortal trellis]

so if you can approximate both f and g already, then you know that you can also approximate f+g

the point is that for polynomials this means we only need to approximate x^k

we don't have to worry about all polynomials, just about a basis

[austere linden]

i do not understand your idea

like

to sum up

we have f(0) = a0 = int(0,1) f

if the formula is accurate

and where f = a0 X0 + a1 X1 + a2X2 + ... + am Xm

so we have a0 = a0 1^1 / 1 + ... + am 1^m+1 / m+1 isn't it ?

when you integrate between 0 and 1

[mortal trellis]

the point I want to get to is that we don't have to worry about these big ass sums

they are annoying

if the quadrature formula is accurate for $1, x, x^2, \ldots, x^m$, then it is also accurate for any linear combination $a_0+a_1x+a_2x^2+\ldots+a_mx^m$

[ocean seal]

Denascite

[mortal trellis]

that means we only need to test for x^k

[austere linden]

ok i understand that point

[mortal trellis]

which is much easier

[austere linden]

ok so we have to check if e(x^m) = 0

so we can find the biggest m verifying this equation

[mortal trellis]

yes

so lets start slow

k=0

if $Q(x)=1$, is $\int_0^1 Q(x) dx = Q(0)$ ?

[ocean seal]

Denascite

[austere linden]

yes bc Q(0) = 1

[mortal trellis]

Q(x)=1 so yes Q(0)=1

ok good

so the formula is accurate for 1=x^0

next k=1

if $Q(x)=x$, is $\int_0^1 Q(x) dx = Q(0)$ ?

[ocean seal]

Denascite

[austere linden]

are we doing a recurrence ?

Q(0) = 0 while the integration wants Q(0) to be 1/2

[mortal trellis]

no

[austere linden]

so we get m=0

[mortal trellis]

yes

[austere linden]

ohh okkk

[mortal trellis]

the formula is only accurate for constant polynomials. which is pretty shit but then again if you just take f(0) what really do you expect

[austere linden]

so if i want to right it correctly

I_0 (f) = f(0) means int(0,1) {a0 + a1X + ... + amXm} = f(0) = a0

<=> f = a0 for all x

which is logical but i could not correctly establish my reasoning

[mortal trellis]

well written like this it's not true

the integration formula will still be true for some polynomials of arbitrary degree

namely if the sum of the a_i just happens to be equal to a0

the point is that you want it to be true for arbitrary polynomials

[austere linden]

yeah but int(0,1) {f} = int(0,1) {a0 + a1X + ... + amXm} = a0 + .... + am / m+1

and also equals to f(0) = a0

only if m=0

[mortal trellis]

or if the a_i just so happen to satisfy this

[austere linden]

a1/2 + a2/3 + ... + am/m+1 = 0

is it possible?

i mean of course

[mortal trellis]

a1=2, a2=-3

[austere linden]

but how can i dodge this issue

[mortal trellis]

give an explicit counterexample of a polynomial for which it doesn't work

which is what we did

it works for q(x)=1, but not for q(x)=x

[austere linden]

oh ok

and BECAUSE it works for q=1

it woks for all linear combinations

[mortal trellis]

yes

[austere linden]

ok tysm

one last thing

[mortal trellis]

this example is imo a tiny bit too simple to completely see the picture. but you'll probably do other examples later

[austere linden]

could you write properly your path of thought ?

like 1. understand the pb , 2. see that the formula is linear ... ?

[mortal trellis]

well I mean mostly I remember it from doing it before

so not exactly sure what you want to hear?

[austere linden]

the way you thought abt giving a counterexample

it's a brilliant idea i would never have had it

[mortal trellis]

which counterexample?

but again, I know this stuff already

it's something I've done before

in this way

[austere linden]

oh ok so i guess i will be better in maths w/ the experience

thank you for helping me

.close

[lone heart]

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[mortal trellis]

yes

experience helps a lot

[lone heart]

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[jagged forge]

I got this wrong… not sure why

[novel stag]

hi

[jagged forge]

Hello

[novel stag]

ur so smart

no sarcasticly

idk how to do that

[jagged forge]

Neither do I lol

[plain flame]

you need to write dx

$\int \frac{16}{x^2 \sqrt{16 x^2-100}}$

[ocean seal]

heavy

[plain flame]

this doesnt mean anything

$\int \frac{16}{x^2 \sqrt{16 x^2-100}} \dd x$

[ocean seal]

heavy

[plain flame]

we all know what you mean even without the dx

but your teacher will deduct points for it

[frigid stream]

Oh you just started caring about the notation finally Heavy 😏 Good !

[jagged forge]

Yea bad habit

[frigid stream]

On the second line, shouldn't you square 10/4?

[plain flame]

you applioed the right substitution

[frigid stream]

On the left

[frigid stream]

A* right substitution 🤓

By saying "the", you are assuming that only one substitution is working here

[jagged forge]

I don’t think so. That would make it 1000 and then I couldn’t pull it out of square root

[frigid stream]

x^2 is (10/4 sec(theta))^2

Also, you didn't compute the square correctly in the square root

[jagged forge]

Sqrt[100 tan^2] is not 10 tan

?

[frigid stream]

I am on the second line

There are two mistakes there

[jagged forge]

Oh

[frigid stream]

Look out at how you applied the squares, and look at at the brackets in the square root

(so 3 mistakes actually 👀)

But I am sure you can correct them easily

[jagged forge]

I see it

Thank u

[lone heart]

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[jagged forge]

.close

[lone heart]

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[high ether]

what's the integral of $\frac{t-i}{t^2+1}$

[ocean seal]

dabble

[high ether]

i would like to simplify by t-i and integrate to get ln(t+i)

but i believe i do not have the right to do that since the logarithm of a complex number is not defined

So what do i do?????????????????????????????????????

[limpid turret]

It is defined

But you don't need it here I don't believe

[high ether]

what do i do then

i am trying to solve a differential equation and i need to integrate this somehow

[limpid turret]

Split your fraction and you should get inverse trig integrals if my intuition is correct

,w derivative atan(x)

[ocean seal]

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[high ether]

idk what that means

[limpid spade]

lol i was going to do partial fractions

[high ether]

Ah

so it's 1/2ln(t^2+1)-iarctant?

[limpid turret]

Split your t and i into separate fractions

[high ether]

yeah ig so

[limpid turret]

,w derivative ln(t^2+1)/2 - i*arctan(t)

[ocean seal]

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[limpid turret]

Looks good

Btw log of complex is defined

[limpid spade]

,w integral of (t-i)/(t²+1)

[ocean seal]

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[high ether]

okay ill see if it gives me a good answer

[limpid turret]

,w log(a+b*i)

[high ether]

mmm... okay ill try both

[ocean seal]

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[limpid turret]

Okay Wolfie is being rude

But it exists

[high ether]

i am solving this system btw, just so you know

Ill see if the integral works

[limpid turret]

System of differential equations

Beautiful

[limpid spade]

lol

[limpid turret]

Just do matrix odes

[high ether]

yeah i am doing that

[limpid turret]

Replace current nightmares for new ones

Solodes are interesting but oh so much work

[high ether]

I fucking hate them

it's a time black hole

You spend hours staring at equations you're sure you can solve them BUT there's always this tiny fucking detail you missed

[limpid turret]

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[tardy flower]

is it possible for me to manipulate $1 + (1 + \lfloor log_2(n) \rfloor)$ into $1 + \lfloor log_2(n+1)\rfloor$?

[limpid turret]

[tardy flower]

my bad

[ocean seal]

Smish

[tardy flower]

there we go

[plain flame]

that seems false

[tardy flower]

sadge

thank you

.close

[lone heart]

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[tardy flower]

.reopen

[lone heart]

✅

[tardy flower]

wait

what if it was $\forall n \geq 1$

[ocean seal]

Smish

[plain flame]

still false

[tardy flower]

sadge

[outer lark]

its probably true for a very small region around 1

[plain flame]

try n = 5

its actually true for $\bigcup_{n=1}^{\infty} [2^n-1, 2^n]$

[ocean seal]

heavy

[limpid turret]

Real question : why?

[plain flame]

actually i think one of those ends of the interval is open

anyways

[tardy flower]

I'm trying to prove the correctness of an algorithm for all n greater than one

with induction

[plain flame]

just master method it

[tardy flower]

It's not divide and conquere

[plain flame]

what exactly is the alg

[plain flame]

i dont see how this cant be done with the master method

first you prove the recursive formula

[tardy flower]

I don't think I know master method

[limpid turret]

Have you mapped out what you get for several values?

[tardy flower]

yes

Also I have to do it with induction

[limpid turret]

What do you believe the function does?

[tardy flower]

It returns the number of digits it takes to convert a decimal number into binary

I did some searching around and found that the pattern was $\lfloor log_2(n)\rfloor$

[limpid turret]

n=0

[ocean seal]

Smish

[tardy flower]

the binary representation of the number 10 is 1010 which is four digits

[tardy flower]

sry my bad $\forall n \geq 1$

[ocean seal]

Smish

[limpid turret]

You're given n=0 tho

[tardy flower]

I asked my professor and they said I can return 0 for the base case even though it is undefined for formula

.close

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[high ether]

why if the sum of the coeficients is 0 then e^x is a solution

[high ether]

.close

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[versed loom]

i got 3/2 but i dont know what im doing wrong

[versed loom]

nvm i got it its 16/3

.close

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[honest umbra]

what variable am i supposed to derive with respect to?

[honest umbra]

...

[torn prairie]

do you guys know how to solve this?

[vale wigeon]

@torn prairie this channel is occupied, please move.

[lone heart]

@honest umbra Has your question been resolved?

[lone heart]

@honest umbra Has your question been resolved?

[limpid turret]

Looks like the subscripts define the index of which variable to choose

That is my guess at least

[lone heart]

@honest umbra Has your question been resolved?

[honest umbra]

the subscripts r 0 and 1 tho?

[lone heart]

@honest umbra Has your question been resolved?

[desert zephyr]

In the screenshot I see 1 and 2 for the derivatives and 0 for the point of which the neighbourhood is taken

[honest umbra]

.clsoe

.close

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[alpine sable]

Does this look right?

[keen mason]

Yep looks good to me

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@alpine sable Has your question been resolved?

[alpine sable]

Perfect thanks

[lone heart]

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[spare fern]

Hey guys, how would I know which rule to use for this differentiation question?

[proven crypt]

product rule and a little chain rule

[spare fern]

I rewrote it as 3x(4-2x)^1/2

[spare fern]

but how would I know this?

[median rain]

It is the product of two functions and there is a function to the power of something

[pliant cedar]

u can see theres a product involved

[spare fern]

I know it is a composite function since it is to the power of 1/2

hence I use the chain rule

although how do I know I'm supposed to use the product rule?

[median rain]

Use that rule when you see something like h(x) = g(x)f(x)

When a function is being multiplied by another function so : 2xlnx or xsinx or in your case that function

[spare fern]

so in my case I have f(x) = 3x(4-2x)^1/2

does this mean it is a function of a function?

[median rain]

Nah

You can see it as 3x = g(x) and (4-2x)^1/2 = h(x)

[outer lark]

the (4-2x)^(1/2) does need ⛓️

[median rain]

So f(x) = g(x)h(x) in your case

[spare fern]

oh ok is that because of the 3x or (4-2x)?

[median rain]

Both

They multiply each other

[spare fern]

and do I use the chain rule first or the product rule first?

[pliant cedar]

product rule coz its a product of two functions

[median rain]

Well you use the chain rule within the product rule

[pliant cedar]

and to differentiate one of these two function u use the chain rule

[spare fern]

ok so after differentiating I got this

-18x(4-2x)^1/2 * 1/2(4-2x)^-1/2

can anyone verify if this is correct?

@median rain

@pliant cedar

[median rain]

Holdup

[median rain]

Nah something wrong with your stuff

If you can show your work ill tell your whats up

[spare fern]

ok I'll send it through

[median rain]

I dont understand your last step

And your x variable being similar to your multiplication sign isn't particularly optimal

[spare fern]

I just did 3x * 3 * -2 = -18x

[median rain]

You cant do that since these arent multiplicators in both terms

[spare fern]

oh I see

but I can multiply 3 and -2 though right?

[median rain]

Id put your terms in fractions

[median rain]

-2 , 3x and 1/2

-2 and 1/2 cancel out to make -1

[spare fern]

so -3x

[median rain]

Yessir

Now since youre looking for a fraction put your terms in fractions

[spare fern]

alright

How is this?

[median rain]

Not quite

[spare fern]

What is wrong?

[median rain]

Here your first term isnt on the same denominator as your second

Therefore you have to put it on the same denominator
Do you know how to?

[spare fern]

yes

[median rain]

Great do that and try again you should be able to get the right answer

[spare fern]

why is it -3x?

shouldn't your drawing on the left be -3x

and the right should be positive 3x

[wary stream]

Also I hope you know this is not correct, you did uv and not u'v

[median rain]

His x variable and his multiplication symbol can be confused for each other he did that right

[median rain]

No

[spare fern]

I'm confused

[median rain]

As to how to put on common denominators or how i got to where i am?

[spare fern]

look at the line above the root fraction

did I make a mistake?

[median rain]

Yeah

[spare fern]

hmm

[spare fern]

ok so basically let's start from over here f'(x) = 3 * (4-2x)^1/2 + 3x * 1/2(4-2x)^-1/2 * -2

when I multiply 3 * -2 * 1/2 I get -3 correct?

[wary stream]

That's how that simplifies

[spare fern]

yes, so I should be left with -3(4-2x)^1/2 + 3x(4-2x)^-1/2

right?

[wary stream]

You put the negative with the wrong term

[spare fern]

.

[wary stream]

It's 3x * 1/2 * -2 = -3x

[median rain]

Yep youre putting your neg to the wrong term

[wary stream]

You were referring to the second term

Yet you put the negative sign on the first time

[spare fern]

oh I see

my bad

just curious though why can't I multiply all of them together 3, 3x, 1/2 and -2?

cause 3 is being multiplied, 1/2 is being multiplied, -2 is being multiplied

and isn't 3x being added?

[wary stream]

The 3x is part of 1/2 and -2

All getting multiplied

The 3 is on the first term

Separate from 3x, 1/2, and -2

[spare fern]

oh ok

How is this?

[lone heart]

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[reef karma]

Dy over dyx

[reef karma]

Derivative at a point gives us slope of tangent line

Is the point pt. Of tangency ie lying on curve

Or any point even outside curve?

[surreal meadow]

the point of tangency lies on the curve

[reef karma]

The actual question is can i find slope of tangent passing through a point not lying on curve using differentiation

?

[surreal meadow]

there may be multiple answers

[reef karma]

So differential method wont work?

[surreal meadow]

well i can't say for sure. i know it'll sometimes work, and sometimes there just won't be a line

i don't know if for your purposes that counts as "not working"

[reef karma]

Thanks

[gray isle]

consider the slope of your sought after line in two different ways

let P(p, f(p)) be the point of tangency and Q(q, f(q)) be the point not on the curve
equate f'(p) with slope of PQ from slope formula

@reef karma

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[serene junco]

Let b represent Bob's current age

and k represent Karen's current age

Can you write an equation or two that relates them based on the information?

consider this

if b is bob's current age, then (b+3) is Bob's age in 3 years

The first sentence could be re-written:
"(Bob's age in three years) is twice (Karen's age in three years)"

What do you fill that chart in with?

variables or like actual constant numbers?

🤔 okay... but Karen's current age is what we want to find, so surely that should be our variable, not her age in three years

This seems like an awfully convoluted way to do this lol

what would you usually do from here?

like in the two years ago column would you do 3y and y ?

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[serene junco]

Hm

what have you tried?

[cosmic crow]

I can't even start

[gray isle]

try expand and factorise the whole thing

[serene junco]

The quadratic they gave you is not in a usual form. Maybe if you expand the brackets and simplify it, you'll be able to factor it

[lone heart]

@cosmic crow Has your question been resolved?

[cosmic crow]

I'm not able to factorise it

[crimson totem]

what equation did you get?

[cosmic crow]

Oh wait

Wrote the thing wrong

Hold on

(x+10)(x-9)

[hazy apex]

simply (x-7)(x+8) - 34 .... then factorise it again
u'll get ur answer

[cosmic crow]

Oh yeah

Thanks

What about this one

[hazy apex]

use prime factorisation

u get 2^a x 3^b x 5^c x 7^d
so the no. of divisors will be (a+1)(b+1)(c+1)(d+1)

[cosmic crow]

I still don't get it

The options are 130 , 330 , 210 , 303

The answer dosent match

[hazy apex]

what are u getting after prime factorisation

[cosmic crow]

2^8x5^2x3^2x7

[hazy apex]

6^4 = 2^4 x 3^4

[cosmic crow]

Yeah

[hazy apex]

so

2^10 3^4 5^2 and 7^1

[cosmic crow]

Guess I'm just dumb

[cosmic crow]

Then what

[hazy apex]

use this
10+1 x 2+1 x 4+1 x 1+2

[cosmic crow]

Oh yeah

330

Sweet

What about this one

[hazy apex]

use m = sum of terms/ no. of terms

[cosmic crow]

Yeah

So a+b+c+d+e=30

[hazy apex]

hmm

[cosmic crow]

And after the subtraction 7 =x

Then the total number of divisors = 2??

Am i right?

[hazy apex]

x = 6 🥹

[cosmic crow]

Ok

The answer is 2 tho right

[cosmic crow]

Right ???

[hazy apex]

4*

1 , 2, 3, 6

[cosmic crow]

I don't get it

Oh wait

I'm just dumb

Alright what about this one

.close

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[paper sage]

How do I take scular product between u=(1,2,3) and n=(4,1,-2)

[paper sage]

I think I solved the problem I just did usual multiplying

And I got 0

[alpine sable]

\begin{gather*}
\mathbf{u} = \langle u_1 ,u_2 ,u_3 \rangle \quad \mathbf{n} = \langle n_1 ,n_2 ,n_3 \rangle \[3mm]
\to \mathbf{u} \cdot \mathbf{n} = u_1 n_1 + u_2 n_2 + u_3 n_3
\end{gather*} If i understood what you mean?

[paper sage]

Which means they are parallel

Yes

[ocean seal]

Lixera

[vale wigeon]

no, it means they are perpendicular.

[paper sage]

Oh yes ofc

[hazy apex]

cosA = 0

[vale wigeon]

and the dot product of these guys is indeed 0.

[paper sage]

But the question was if U was parallel with my pi

then u must be orthogonal to the normal vector! so the scalar product must equal zero

.close

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[naive valley]

it's the area of the two green regions above the x axis, minus the area of the middle green region below the x axis

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[alpine sable]

hello! very simple question, what would be the formula to solve this?

[echo socket]

I'd first evaluate S(10k) + S(10k + 1) + ... + S(10k + 8) + S(10k + 9) for a digit k

Should be k + (k + 1) + ... + (k + 8) + (k + 9)

(k + k + ... + k + k) + (1 + 2 + ... + 8 + 9)

So 10k + 45

(The first sum contains 10 k's and the other is just 9*(9+1)/2)

So we're adding 45 + (10 + 45) + (10 * 2 + 45) + ... + (10 * 9 + 45)

Which is 900

[alpine sable]

yeah thats what we thought before but i was just seeing if theres a shorter formula of some sort lol

thank you so much!

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[gusty coral]

hello:)

[gusty coral]

could anyone please help me find my mistake?

[daring zenith]

send along with question

[vale junco]

Send a question and we’ll try our best to help

[gusty coral]

A soft ball has a mass of 0.25 kg and is thrown horizontally with a speed of 110 km/h. At the moment the ball hits a plate, it has already lost 10% of its speed. Estimate the average force of air resistance during the throw, assuming the distance between the plate and the batter is approximately 15 m and ignoring gravity.

[gusty coral]

the bottom picture is the correction key, i understand their way of doing it but not sure why mine wasnt right

[surreal meadow]

did you change the velocity you were given into meters per second?

[gusty coral]

Yes i have

[surreal meadow]

i got a very different number for mv^2/2

well very is a strong word

off by a factor of 100

which i assume is the only issue

since 1.47 ~ 1.5

[gusty coral]

for mv^2/2 i had (25)(27.5)^2/2

[surreal meadow]

why 25?

the ball is .25kg

[gusty coral]

oh my

i am

so sorry

[surreal meadow]

don’t be

[gusty coral]

thank you sm:)

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[queen raven]

im confused about this question

[modern sedge]

Do you know what domain and range is?

[queen raven]

the domain is restricted to -1,1

and the range is -pi/2, pi/2 right?

so arcsin(theta) must be in that range

[modern sedge]

I am pretty sure domain is incorrect...

[queen raven]

if the domain of sin is -pi/2 to pi/2

then the domain of inverse sin would have to be -1 to 1, no?

so the inside function must comply with that domain, right?

as the overall function is arcsin(x) ?

[modern sedge]

Oh sorry, you are right

[queen raven]

how would i go about finding the domain of the inside function that complies with the outer function?

-1 < cosx < 1

right?

so arccos(-1) < x < arccos(1) ?

i dont know if im on the correct track?

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@queen raven Has your question been resolved?

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[oblique knot]

Hi how do I prove this?

[wet reef]

Write down the terms on either side of the equation

See what u get

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@oblique knot Has your question been resolved?

[mortal trellis]

why all over n

$\sum_{i=1}^n x_i^2 + 2\mu x_i + \mu^2$

[ocean seal]

Denascite

[mortal trellis]

now you can split this up into three sums

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[woven bloom]

hello