help-23

[opaque lintel]

but that twould put it in quadrant 4, so how would sin = 1/6 there if it's positive?

Sin could be 1/6 in Q1 and Q2 right?

So I should be looking for 4 solutions

I tried doing pi - arcsin(1/6) to get the other angle that sin could be 1/6 in Q2

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[tacit pendant]

can you help me to understand how can i apply chain rule here

$(x-2)^2$

[flat frigate]

odokawa

[icy lance]

let u=x-2

y=u^2

then what?

[tacit pendant]

xd

[lean otter]

Do u still need help

[tacit pendant]

i dont know man

people here are too verbose

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[trim jackal]

How do you achieve the odd permutations for the symmetry group of a tetrahedron? I have the 6 transpositions as single reflections through planes, but how do you achieve the remaining 6 that are the 4-cycles?

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[grizzled topaz]

I don't even know where to start

[grizzled topaz]

i made 4a into 2a and then used double angle + compound angle but im left with something really ugly 💀

[forest gust]

$\frac{1+\cos\left(4a\right)}{\sin\left(4a\right)}\to\frac{2\cos^{2}\left(2a\right)}{2\sin\left(2a\right)\cos\left(2a\right)}$

[flat frigate]

Combustion

[forest gust]

keep using double angle identities until you get to "a"

[forest gust]

also cancel the 2cos(2a)

also if you're wondering how i got it to a single fraction

it's just $\frac{1}{\sin\left(4a\right)}+\frac{\cos\left(4a\right)}{\sin\left(4a\right)}$

[flat frigate]

Combustion

[grizzled topaz]

howd u get the next part?

is that just an identity or smth??

[forest gust]

1+cos(4a)=2cos^2(2a)

it's the double angle identity

[grizzled topaz]

what

how

[forest gust]

cos(2x) = 2cos^2(x)-1

[grizzled topaz]

yeah

[forest gust]

add 1 to both sides

[grizzled topaz]

bro i cannt see it

[forest gust]

cos(2x)+1 =2cos^2(x)

[grizzled topaz]

💀 im special

[forest gust]

i mean like you can just sub in cos(4x)=2cos^2(2x)-1 and the ones will cancel

[grizzled topaz]

yeah

writing down working rn

[grizzled topaz]

wait dont u also ahve to mulyiplt the 1 by

2

[forest gust]

❓

[grizzled topaz]

oh wait mb its just multiplying the angle by 2

[grizzled topaz]

@forest gust do i just use double angle from now?

[forest gust]

yeah

[grizzled topaz]

or divide by 2 first

ig

[forest gust]

yeah and cos(2x)

[grizzled topaz]

so im left with cos (2x) / sin (2x)

[forest gust]

yep

[grizzled topaz]

bro again how'd u see that lmao

i fully expanded it 💀

[forest gust]

🤷‍♂️

[grizzled topaz]

thanks bro

ur so smart

[forest gust]

you're smart too

also

a good thing to do when there's sec/csc/tan/cot is turning everything into cos/sin, it works a good amount of time

sometimes it doesn't work sometimes it does

you'll probably know when it will and when it won't

[grizzled topaz]

bro not as cracked as u

u just saw it like that

even with the perfect squares thing

[forest gust]

you'll be even better lol

[grizzled topaz]

r u in harvard

lmao

[forest gust]

no lmao

[grizzled topaz]

is it hard to get into harvard

idk im not in us

[forest gust]

don't know i'm not american

[grizzled topaz]

o

did u do 4u maths

[forest gust]

don't know what that is

[grizzled topaz]

oh alright well the highest maths u can take for ur yr 12 ig

@forest gust idk how to approach this

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[desert pond]

hi, why is this not correct

[desert pond]

i cant wrap my head around it

why is it -1/pi * cos(u)

instead of -pi * cos(u)

[potent seal]

$\int \frac {\cos\frac{\pi}{x}}{x^2} dx$

[gritty glacier]

dx

[potent seal]

Ye

[gritty glacier]

[potent seal]

That's the q?

[gritty glacier]

yes ig

[desert pond]

yes

my bad

[gritty glacier]

take u = pi/x

[flat frigate]

Lorentz

[desert pond]

yeah i get du = pi * -1/x^2 dx right

then i just substitute inside the current equation

[gritty glacier]

so

,, \frac{-1}{\pi} \int \cos u \dd u

[desert pond]

where did -1/pi come from

[potent seal]

Why the -pi within the integral

[desert pond]

my du is - pi * 1/x^2

[gritty glacier]

right

[flat frigate]

Bettim

[potent seal]

Yeah

[gritty glacier]

now it makes sense ig

[desert pond]

so my current equation is cos u dx right i havent subbed du in

[potent seal]

There's dx/x^2

[desert pond]

ohh i got it confused

[gritty glacier]

before subbing it you wouldhave /x^2 in denom

[potent seal]

That becomes -du/pi

[desert pond]

dammit ok

i get it now

thx

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[gentle zodiac]

I don't get the solution for part a? I know wronkskian is just a 3x3 matrix here since it's 3rd order DE, so W = det(insert 3x3 matrix), but i'm confused with how part a equation was derived for each of these two cases

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[dapper venture]

chain rule

that's not e^x^2 that's e^(2x)

first try to differentiate e^(2x)

it's just off by a constant so you just need to divide

since d/dx e^(2x) = 2e^(2x), integral of e^(2x)= e^(2x)/2

$e^{2x}$

[flat frigate]

WhereWolf(ping if needed)

[dapper venture]

e^x times e^x is e^(2x)

no

say we have 2^3 times 2^3

do you get 2^6 or 2^9

you add the exponents

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[flint flicker]

Hello

[flint flicker]

I am unsure on how to do part C

Like how would I calculate that

[placid oak]

if you have your model parameters from part a, just extrapolate and find where it predicts the area is zero

[flint flicker]

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[fathom goblet]

So I have the double integral: $\int^{2}{0} \int^{\sqrt{16-x^2}}{x*\sqrt{3}} e^{x^2 + y^2} dy dx$
how do I rewrite this integral using polar coordinates?

[flat frigate]

alani

[fathom goblet]

I got $\int^{\pi/2}_{\pi/3} \int^4_0 e^{r^2} * r dr d\theta$ and solved it to get $\pi/6 ({(e^{16} - 1)}/{2})$ though this is my first time doing polar coordinates on my own, so I'm not sure of the answer. Can someone check?

[flat frigate]

alani

[plucky elk]

Did you draw a picture of the domain being integrated over

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[signal cove]

hello

[signal cove]

can someone tell me why we are dividing by 2 here

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[signal cove]

<@&286206848099549185>

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[vagrant ice]

(1, 0, 0, 1) dot (1, 0, 0, 1) is 2

similarly (2, 1, 0, 0) dot (0, 1, 0, 0) = 1

and (2, 1, 0, 0) dot (1, 0, 0, 1) = 2

[signal cove]

i mean why are they dividing it by 2

[vagrant ice]

cause you divide by v1 dot v1

I mean u1 by u1 in your example

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[uncut juniper]

yo

[uncut juniper]

a straight line intersects the x-axis at the point (2,0). The graduation on the x- and y-axis is equal and the angle between the x-axis and the line is 45 degrees. Determine the equation of another line, which passes through the point (4,0) and is parallel to the first line.

[vagrant ice]

from this info, what's the slope of the line?

[uncut juniper]

which line

[vagrant ice]

the line through (2, 0)

[uncut juniper]

yeah i don’t know the slope of the line

i need 2 points to figure it out

[vagrant ice]

The graduation on the x- and y-axis is equal
The angle between the x-axis and the line is 45 degrees.

[uncut juniper]

if it was 90 degrees it would have been a straight line

just saying

[vagrant ice]

like from this, if x changes by 1 unit, y changes by 1 unit

if x changes by 2 units, y changes by 2 units

and so on

[uncut juniper]

yes

[vagrant ice]

so what must be the slope of the line?

[uncut juniper]

1

[uncut juniper]

but we don’t know this

[vagrant ice]

we do

cause it's 45 degrees

45 degrees is a right-angled isoceles triangle

[uncut juniper]

that’s new

alright

[vagrant ice]

it's like this

but flip it so it goes diagonally up

45 is half of 90

so it's half of a square

[uncut juniper]

wait let me send a pic

[vagrant ice]

okok

[uncut juniper]

if there was a 90 degree distance between the x axis and rhe line

i have to take half of that?

[vagrant ice]

yeah

half the right angle

[uncut juniper]

who’s to say that 45 degrees means a slope of 1

[vagrant ice]

you can draw it out

with a compass and ruler you can bisect the right angle

into 2 equal parts

[uncut juniper]

i think I get it now

draw a right angle triangle

[vagrant ice]

yes

[uncut juniper]

alrighty

[vagrant ice]

that's easier actually

[uncut juniper]

tysm

y = x - 4

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[deft pivot]

general term of associated sequence is 1/3n

and it's limit is 0

so the series might be convergent

[hot remnant]

take out a factor of 1/3

and you will see why it is divergent

[lean otter]

Is there anyone here thats good with SPSS in quantitative math ? 🙂

[deft pivot]

aha because we will have inside brakets harmonic series ?

[hot remnant]

yup

if you want a proof u just use same proof as harmonic series diverges i guess

[deft pivot]

ok thanks

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[modest holly]

help?

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[lone arch]

Regarding Koch's curve

[lone arch]

I have following for the area

*as n -> infinity I meant to say

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[plucky elk]

https://en.wikipedia.org/wiki/Koch_snowflake
did you follow their proof

[lone arch]

No, I made my own, as above

But it seems it doesn't agree with wikipedia

[plucky elk]

where does sqrt(3) / 4 * (1/3)^2 come from

[plucky elk]

where does wikipedia say it's sqrt(3)/20

also which koch curve? the triangle?

[lone arch]

German Wikipedia has that

With a = 1, we have sqrt(3)/20

[lone arch]

Yes, the triangle

[plucky elk]

read the paragraphs after that

2 sqrt(3) / 5 agrees with the english for a=1=s

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[undone matrix]

I'm familiar with being able to split up the summations, however how do i get from that step to the next

[undone matrix]

the only thing thats coming to mind is the n(n+1)/2

wait i brain farted, theres a different for ^2 and ^3

my bad

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[turbid path]

given these 2 integrals

[turbid path]

how do i find the distance from 2 to 5

like i know what i need i just dont know how to do it

[rough storm]

you mean integral from 2 to 5?

[turbid path]

correcy

yeah

bad wording

[rough storm]

well, use fundamental theorem, rewrite integral in terms of antiderivative, that may help

i.e. F(2)-F(-1)=9

[hallow shuttle]

@turbid path a visual approach is pretty good

so if we only want that region from 2 to 5, what operation can we do with the areas

[turbid path]

i thought it was something like that so you just area of the lower integral from the area of the larger one

i mean that makes sense now that i think abt it

[hallow shuttle]

yes exactly

[turbid path]

its just the smaller area subtracted from the total area

[hallow shuttle]

that what the picture was hinting at

[hallow shuttle]

yup

[turbid path]

yeah so its -3

schweet

thanks bossman

[hallow shuttle]

np

[turbid path]

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[turbid path]

ok for this one

when applying linearity to b

you get integral from -1 to 5 (2x)dx and the same but (2)dx now i dont have anything that helps me solve for those values

so im not sure what to do here..

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[muted sapphire]

Why should the 2x + 2 matter in this case?

Can't you just solve the integral as normal?

Utilising this fact

@turbid path

[turbid path]

hey

yeah thats what i did

im just not sure how do i get a value from integral -1 to 5 of lets say bx dx

where bx is 2x

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[marsh walrus]

I'm curious if anyone has any thoughts for how to approach $$\text{Re} \qty( \frac{1}{i\pi} \int _0 ^\infty e^{i x \cosh (i+s)} \dd s ) = 0$$

[flat frigate]

jan Niku

[marsh walrus]

this is from an asymptotics book, so I think techniques like that are on the table, but I'm not really certain what to apply here

it does seem like the integral should become really oscillatory for big s, but is that enough to provide that we could just use a small neighborhood right of i

@rough storm its not super clean, i was trying that

$\cosh (i+s) = \cosh i \cosh s + \sinh i \sinh s$

[flat frigate]

jan Niku

[rough storm]

ye

[marsh walrus]

its not super clear

or is it

wait

yea im not sure here since now we have this product

its less clear what to do with the integrand

It doesn't look like the phase is ever stationary

and laplace is obviously out

does this mean you need steepest descent?

or is there some nicer way of finding the real/imag parts of an integral like this

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@marsh walrus Has your question been resolved?

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@marsh walrus Has your question been resolved?

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@marsh walrus Has your question been resolved?

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@marsh walrus Has your question been resolved?

[orchid dust]

Uh x = 0 or is that cheating?

I just turned it into a regular integral by using Eulers formula and looking only at the real part

(Imaginary part blows up tho)

But holy moly this integral is probably not possible to solve

Guess you're supposed to turn it into a residue integral somehow

This shid can not be turned into a residue integral cuz it got no poles

Man its been a while since I've dealt with complex integrals

Wait..

Nope sorry no clue

[marsh walrus]

yea thats kind of my issue

i have no idea what other path youd want to use

[marsh walrus]

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[meager crane]

i understood everything but where did the -1/9 go and how did it become -1

[quiet plume]

-1/9 is from completing the square

[hasty falcon]

9 * (-1/9)

[quiet plume]

Then it is multiplied by 9

[meager crane]

only 1/9 got multiplied by 9

and not the (x-1/3)^2

why?

[quiet plume]

It has been multiplied though.

[meager crane]

ohhhhh

[quiet plume]

There's a 9 in front of it

[meager crane]

yeah yeah

thanks alot

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[bleak harbor]

Let ABC be a triangle, M N and P distinct points of the vertices of the triangle, belonging respectively to (BC), (CA), (AB)
We assume in this question that the lines (AM) and (BN) and (CP) are parallel
show that: MB/MC × NC/NA × PA/PB = 1

[bleak harbor]

i need urgent help with this problem pls :<

[safe radish]

@bleak harbor Has your question been resolved?

[safe radish]

@bleak harbor Has your question been resolved?

[safe radish]

@bleak harbor Has your question been resolved?

[safe radish]

@bleak harbor Has your question been resolved?

[wheat condor]

Do you have a sketch of this? Difficult to picture.

Also I understand "distinct from" but not "distinct of".
Is that my problem?

[safe radish]

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[devout steeple]

this is what my math final exam will cover

Partial fractions

Spring problems

Binomial Series

Find foci and vertices of ellipse hyperbola or parabola

Arc length for rectangle, polar, parametric equations

Integration by parts

Series tests, limit test, p series

Find first and second derivatives for parametric equations and polar, or slope and line at a point

Find area between two curves, area with polar, and area with parametric

Integration by triangle and trig identities.

Volume with shells and washers.

Integrals with infinity as upper bound or an asymptote

Limit that uses power series

[devout steeple]

my teacher is letting us make a formula / notes sheet to bring to the exam, and my weak point is integrals

anyone know which formulas or integrals i should have written down

i am sure i need to memorize trig integrals

for

Integration by triangle and trig identities.

[wispy sentinel]

[devout steeple]

.close

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[lean otter]

I need help w math

[lean otter]

That’s the clearest pic I could get of it

I don’t know how to do any of it

[obtuse plover]

There’s no problem shown

[lean otter]

You don’t see the pic?

[obtuse plover]

I do, but what do u wanna do with those numbers and shapes

[lean otter]

Solve it

I don’t know any math

[obtuse plover]

Solve what is what I’m asking

[lean otter]

oh idk

[obtuse plover]

Just cuz there’s numbers and letters doesn’t mean there’s anything to solve

[lean otter]

Ohh okay

Thanks

[safe radish]

@sharp rain Has your question been resolved?

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[subtle robin]

Question: How many ways can I select 3 pencils from 6 different pencils?

[subtle robin]

Solution:

im confused to why 6x5x4=120 isnt the answer

[copper kernel]

if we choose the 1st pencil, then 2nd, then 3rd, that would be the same as choosing the 2nd, 3rd, then 1st

[subtle robin]

how did they get that equatino at the bottom though

equation

[untold sky]

i choose pencils 1, 2, and 3 in order
i choose pencils 3, 1, 2 in order
they’re the same 3 pencils and so you’re overcounting

this is the main distinction between permutations and combinations

[subtle robin]

yeah i understand the difference now

this doesnt matter the order and permutations does

but i dont understand how they solved it too

[copper kernel]

for each possibility you are overcounting by a factor of 6; for example, if I want to choose pencils 1,2,3 I can choose it in 6 ways:
1,2,3
1,3,2
2,1,3
2,3,1
3,1,2
3,2,1

each possibility is counted 6 times, so we must divide by 6

[subtle robin]

how do you know its a factor of 6

huh

is it always factor of 6

[copper kernel]

because when we choose 3, there are 3! ways to arrange it

[subtle robin]

oh

yeah

should i just memorise this formula n!/k!(n-k)!

it works for all the combination problems of this type right

[copper kernel]

you should 100% memorize that, it is super important

but make sure that you know why it works as well

[subtle robin]

and the permutation on is n!/(n-k)! right

and circle permutation one is (n-1)!

[copper kernel]

yep

those are less important to memorize

[subtle robin]

ok thanks

ill try find some videos that explain why they work

.close

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[plain hedge]

sorry to ping you my last thing timed out. I need to figure out the matrix which implements these but it seems impossible due to the fact there are contradictions

<@&286206848099549185>

[plain hedge]

this is the problem I ran into

0 = -2 is a contridictio n

r,s dont exist

[safe radish]

@plain hedge Has your question been resolved?

[plain hedge]

it too late for the mods

guess I will have to just accept that this will get a 50 or below

should have prepared better

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[slate galleon]

how do i tell if i should use 1 sample z test or 2 sample z test

[slate galleon]

(pls help i have exam tmrw)

[marsh walrus]

iirc it really just depends on how you are considering your samples/populations

like in a 2 sample z test scenario youre going to be comparing 2 populations that are both real i guess

and in a 1 sample one of the will be hypothetical

thats not exactly how the test is structured

but its one way to think of it

[slate galleon]

so 1 sample is checking if it is true

the hypothesiszed value

[marsh walrus]

well one sample would be like

"do cats that live around my house weigh less than 5kg"

and in two sample is like

"Do cats that live around my house weigh more than cats that live around my cousins house?"

[marsh walrus]

here you have a hypothetical, perfect second sample, do you see what i mean?

[marsh walrus]

here both populations have uncertainty

[slate galleon]

ohhhhhhh

alr bet

ty

[marsh walrus]

good luck

[slate galleon]

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[slate galleon]

ty

[safe radish]

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[fickle crane]

i got the system of equations
a+8b−3c=21
−8a−c+by=−23+5y
−3a+6b−6c=−18

but i dont know what to do from there

i used a matrix a;b;c and multiplied that by A

[plain hedge]

you could try moving the variables based terms on one side and leaving the constants on the otherside

then sovle using guassian elimination

but I would try to check that cuz I could be wrong

oh nvm

I didnt see that by

[fickle crane]

yeah i feel like the method im doing so far is working but im not sure where to go from here

[plain hedge]

actually it might be possible it you treat the y like an ordinary number

when ever you end up with a variable in the coefficients in a augment matrix you will at some point have to choose an arbitary variable and see if it works

[fickle crane]

okay thanks

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[fickle crane]

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[chilly atlas]

I'm trying to solve this simple limit but I don't know where to go from here

I know the answer is 1/2, but I don't know how you would get there from this

[icy lance]

try multiplying the numerator and denominator by 1/x

[hazy elbow]

or simply dividing num and denom by x

[sharp sandal]

le'hospital

[mellow rune]

or use l'hop if you're allowed to

though i'd assume otherwise since it's a very simple problem if you use it

[chilly atlas]

oh yeah, l'hopital's rule works very well

forgot about that

but im a bit confused on why you would multiply the numerator and denominator by 1/x

i've never heard of that strategy

[mellow rune]

when the limit goes to infinity doing that

then terms become constant/infinity

which is 0

except for x/x and 2x/x

which become 1 and 2 respectively

giving lim (1+1/x)/2 as x->inf

and 1/inf=0

so answer is 1/2

the strategy can work for certain problems where l'hop can't solve it (cases where the derivatives start looping)

divide numerator and denominator by the highest degree in the denominator (if the highest degree of the fraction is in the numerator then limit is infinity)

[chilly atlas]

wouldnt the highest degree of the denominator just be 1

2x^1

[quiet juniper]

lhopital kinda overkill for this 1

[chilly atlas]

online im seeing that 1 + (1/x) is the "highest denominator power" or "biggest power common" but i dont know what that means

ive never heard of such a concept

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[gritty pivot]

how to solve using product rule? i am completely lost

[lean otter]

either use implicit differentiation or apply chain rule to e^ln(f(x))

[gritty pivot]

what would t hat look like

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[empty gyro]

|x-4|<1

[slate thorn]

$\implies -1<x-4<1 \to 3<x<5$

[flat frigate]

Joshii

[empty gyro]

Still technically true since -3<3, but perhaps it was a typo. Hard to know with certainty without seeing whole problem

[slate thorn]

is there a reason why that could be useful?

[empty gyro]

Hard to say without seeing the whole problem

[slate thorn]

just proving the limit

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[manic cliff]

Hey guys, need someone's help completing this please.

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[meager crane]

i first used sin opp/hyp to get angle CDB

and this chapter is all about radians

so 90 degrees is pi/2 radians

i got angle z 0.92

and since 180 radian = pi

0.92-pi=2.214

so angle W=2.214

and by using the equation for arc length=angle times radius

i multiplied 2.214 by 5

and got 11.071

am i correct?

[meager crane]

but is the way i calculated the arc length correct?

[meager crane]

and then i calculate length AC by using pythagoras theorm

and add arc lenth with AC

[potent seal]

180⁰ is pi radians you mean?

[meager crane]

yes

[potent seal]

Yeah

Seems right

[potent seal]

The steps are right yeah

[meager crane]

ok thanks

[meager crane]

i have exam tmrw and i lost the answer sheet for this worksheet 😭

thanks alot for the help

.close

[potent seal]

Np

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[potent seal]

Good luck

[meager crane]

thxx

[safe radish]

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[prime solar]

im supposed to calculate this. brought it down to 48cos20/sqrt(3)+1 but idk what to do next

[prime solar]

is it impossible to simplify after that?

[potent seal]

Could you show your solution once?

[prime solar]

sin55*sin35 = 1/2(cos20 - cos90) = 1/2cos20 on top it beomes 48cos20. below sin50+sin110 = 2sin80cos30. the sin80s cancel out

maybe theres some other way of doing it im missing

[potent seal]

Right, so it's
$\frac{48\cos{20}}{1+2\cos{30}}$

[prime solar]

2cos30

[potent seal]

Mb

[flat frigate]

Lorentz

[potent seal]

Hmm

[prime solar]

maybe its impossible to go forward from there but the question is to "calculate" it

[potent seal]

Ig you could substitute the values now

[potent seal]

Coz it would be (48cos20)/ (1+sqrt3), like you said

If you want the exact value then I don't see a way apart from finding the values of cos 20 and sqrt 3

[prime solar]

makes sense, maybe the question was meant to be to just simplify

[potent seal]

Or that yeah

[prime solar]

ok thanks

.close

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[potent seal]

Np

[prime solar]

.close

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[lavish storm]

Exercice 2: Question 9.ii)
Prove (Un) is increasing and (Vn) is decreasing

[lavish storm]

@dry sierra

[dry sierra]

Do you have the table of variations of f ready?

(I don't know what the function is, but my guess is: f changes sign on both [0,1] and [1,+oo])

[lavish storm]

It does hold on

[dry sierra]

E.g. if it has a min at x=1 and has f(x)->+oo

[lavish storm]

We have f' and it's related to g(x)

[dry sierra]

Like: $$f(0)>0, f(1)<1,f(x)\tendsto{x\to+\oo}\ell>0$$

[flat frigate]

Matplotlib

[lavish storm]

what the heck is this

[dry sierra]

And f is strictly decreasing on [0,1] and strictly increasing on [1,+oo[

Is that the case?

[lavish storm]

It is

J'ai pris 2 restrictions

[dry sierra]

Then it's just a case of "use the intermediate value theorem" twice

[lavish storm]

Une sur ]0;1] et une sur [1;+oo[

[lavish storm]

Why?

TVI ne marche que pour les intervalles fermés

[dry sierra]

Wait, it's f(x)=n you're trying to solve, not f(x)=0. But that should work the same. What is the table of variations of f?

[lavish storm]

Or ici on a ]0;1] et [1;+oo[

[lavish storm]

It's what you said

I just proved it using bijections

[dry sierra]

Si $$f(x)\tendsto{x\to0^+}-\oo\text{ et }f(x)\tendsto{x\to1^-}+\oo,$$
alors tu peux aussi utiliser le TVI pour dire qu'il existe $x\in]0,1[$ tel que $f(x)=\text{[ce que tu veux]}$

[lavish storm]

2 restrictions and proved that they're bijective

[flat frigate]

Matplotlib

[lavish storm]

Il faut avoir f(0) et f(1)

[dry sierra]

Non ; regarde les limites que j'ai imposé à f

[dry sierra]

Tu peux essayer de démontrer ce fait en utilisant le TVI

[lavish storm]

On utilise pas de limites pour le T.V.I

[lavish storm]

Je ne comprends pas cette notation

[flat frigate]

Matplotlib

[lavish storm]

Le problème c'est pas de montrer ces solutions existent

C'est la monotonie

[dry sierra]

Montre-moi le tableau de variations de f que tu as obtenu

[lavish storm]

J'ai déjà montré ces deux solutions grâce à la bijection

[lavish storm]

Ok att

@dry sierra

[dry sierra]

Donc c'est exactement ce que je te disais

[lavish storm]

J'ai déjà montré l'existence des deux suites

[dry sierra]

Applique le TVI sur ]0,1] et sur [1,+oo[

Ah !

[lavish storm]

Mec le problème c'est la monotonie

DES SUITES

La question d'après

[dry sierra]

"Mec", sois plus clair la prochaine fois, et montre un peu plus d'efforts

[lavish storm]

I was clear

[lavish storm]

???

[dry sierra]

La monotonie de u_n et de v_n vient de la monotonie de f

[lavish storm]

On me dit de monter que Un est croissante mais je trouve qu'elle est décroissante

[dry sierra]

Mais je pense que l'énoncé est faux, je crois que u_n est __dé__croissante et v_n est décroissante

[lavish storm]

Exactement

[dry sierra]

Ou bien ton tableau de variations est faux et f(x)->-oo et pas +oo quand x->0 ?

[lavish storm]

Donc l'énoncé est faux 😭

[lavish storm]

C'est pas logique

[dry sierra]

,w plot f(x)=log(x)/(x-1)+log(x)/2

[flat frigate]

Results provided by WolframAlpha

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

Requested by matplotlib

[dry sierra]

Non, c'est bien l'énoncé qui est faux

[lavish storm]

c'est pas log c'est ln

[dry sierra]

Log c'est Ln pour un anglophone (mais comme tu es francophone tu ne le sais pas !)

[lavish storm]

Pourquoi est-ce comme ça

[raven heart]

et puis bon même si c'était un log avec une base différente ça changerait rien au résultat

[lavish storm]

J'ai toujours pas fait le cours de log_a

Ok c'est bon, je termine la série mtn

@dry sierra je te montre ma méthode d'abord

[dry sierra]

En fait, si tu avais réfléchi un peu plus et regardé la question iv), tu aurais vu que v_n ne peut pas être décroissante

[lavish storm]

De même pour Vn

[lavish storm]

Bah pourquoi? La suite Vn peut décroître vers e^n

[dry sierra]

Aaaaah, des belles feuilles Seyes comme on ne trouve qu'en France ! (bref, pardon)
Si v_n>e^n, alors la limite de v_n ne peut être que +oo, et il n'y a pas de suite décroissante de limite +oo

[dry sierra]

C'est correct sinon

[lavish storm]

CHUI PAS FRANÇAIS

[lavish storm]

Vrai

Ok c'est bon merci. La série était facile!!

[safe radish]

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[safe radish]

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[unreal jewel]

How can the uniqueness of the solution f(x) = 1 - x be proven? I observed that f(x) = f(x + 1) + 1 numerically during my experimentation, but I couldn’t prove it using the conditions in the question. I’m guessing that if I could prove this, that would in turn prove that f(x) = 1 - x is the only solution, but I’m not sure. I saw induction being suggested online for such a problem, but I can’t really understand how that would prove uniqueness. Thanks in advance! :)

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[safe radish]

@unreal jewel Has your question been resolved?

[safe radish]

@unreal jewel Has your question been resolved?

[limber lynx]

i think induction is the way to go

but we do induction on |n| instead of n

let me type out what i think should lead to the solution and you'll know what i mean

put x=0 in the second equation to get f(0)=1
apply f on both sides to get f(f(0))=f(1)
but f(f(0))=0
so, f(1)=0
this also gives f(-1)=-2
now, f(0)=1=1-0, f(1)=0=1-1, f(-1)=-2=-1-1
so the base case of |n|=1 and |n|=0 is done
assume that f(k)=1-k for all integer k satisfying 1<|k|<=n
we want to show that if |k|=n+1, then also, we have f(k)=1-k
let f(n+1)=y
then, f(y)=n+1 and f(-y)=1-n
apply f to get -y=f(1-n)
hence y=-f(1-n)
but, |1-n|<n and hence f(1-n)=n
so, we get y=-n
hence, f(n+1)=-n
from the second equation we get f(-n-1)=2+n, as expected

this kills it

the key idea is to use induction on the modulus of the input instead of the input itself

but that is not a bad thing to try given that integers are involved and not just natural numbers

[limber lynx]

what i have proven is that if such a function were to exist it must be f(n)=1-n

you still have to show that f(n)=1-n works

this is easy to do

@unreal jewel

[unreal jewel]

yeah i managed to show that f(n) = 1 - n works

the trouble i was having was proving that that was the ONLY possiblility

[unreal jewel]

i'll read through this now wait

[limber lynx]

yes

[unreal jewel]

maybe it's just because i'm tired but i don't understand why you brought the absolute value into this

[limber lynx]

becauuse induction is exclusive to N

a well ordered set

we cant directly use induction on integers

but the absolute value of an integer is always a natural number (with the exception of 0)

[unreal jewel]

my idea was to do induction on the positive integers and then do induction on the negative ones (by doing stuff with -n and -n -1. would that work too?

[limber lynx]

i am not sure because in my solution i needed the negative and positive vallues between -n and n to be true at the same time

that's also another reason i brought the modulus in

[limber lynx]

i needed the statement to be true for 1-n as well for induction to work

this is not a natural number if n is a natural number

thats why i brought in the modulus

the modulus is just a fancy way to say "let the statement be true for all integers between -n and n, where n is a natural number"

this is now in the domain of induction, because it is a statement concerned with natural numbers (not integers)

that's not to say your approach of doing positive and negative wont work tho

[limber lynx]

this approach is likely not to work because the second equation in the question deals with both positive and negative values at the same time

and we need the second equation because f(x)=1-x is not the only involution (function whose inverse is itself) on Z

[unreal jewel]

but how does this prove that such a function MUST be f(n) = 1 - n

this is just showing that f(n) = 1 - n satisfies the conditions, no?

[limber lynx]

nope; i have shown that f(n) must be 1-n for all n

so no other function can exist

forget about integers for a sec

i have basically shown that f(n) is something

then f(n+1) must also be that something but with n+1 subbed in

[unreal jewel]

but if a function is defined as f(n) = n - 1, then obviously when you plug in n + 1 it's gonna be the same thing

i'm sorry that i'm completely missing this lol

[limber lynx]

no it's alright; it's a subtle thing

im thinking about how best to convince you

you agree that f(1) and f(0) and f(-1) satisfy f(n)=1-n right

[unreal jewel]

f(1) as defined by the conditions in the question?

[limber lynx]

yes

[unreal jewel]

yeah

that's how i guessed f(x) = 1 - x

[limber lynx]

yes

now what i have shown is that if -1,0, and 1 satisfy f(n)=1-n then -2 and 2 must satisfy that as welll

then using -2,-1,0,1, and 2 i show that -3 and 3 satisfy that as well

and so on

so for every integer x, f(x)=1-x

in other words, the value of f(3) is uniquely determined by the values of f(2),f(1),f(0),f(-1), and f(-2)

and it is uniquely determined to be 1-3

if there were some other distinct function (say g) satisfying this condition then there must be some integer k such that g(k)=/=1-k

but i have shown that that cannot happen

what is the source of this question?

do you have access to the official solution

so you can get an idea for the kind of proof they are looking for

[unreal jewel]

unfortunately not

it's a qualfiication test for the irish maths olympiad

[limber lynx]

ah i see

[unreal jewel]

q2 is the only one i had no idea about

i got the rest eventualyl

[limber lynx]

think about other functional equations you have solved using induction

[unreal jewel]

ah okay, i think my soruce of confusion is that i wasn't sure when you were talking about f(x) = 1 - x and when you were talking about f(x) as defined in the question

[unreal jewel]

none 🗿

[limber lynx]

i see

[unreal jewel]

functional equations is my weakest "section"

[limber lynx]

that's alright you'll get better with time

there are good resources for them

[limber lynx]

the sketch of my solution is that i first showed -1,0, and 1 satisfy f(x)=1-x for f being the function as defined in the q
then i assumed the same happens for -n,-(n-1),...-1,0,1,...,n-1, and n
using this i showed the same happens for n+1 and -(n+1)
i concluded with f(x)=1-x for all integers x

[limber lynx]

how did you do q1?
my first instinct is to consider all the ways we can get x-y=4 with x and y being whole numbers less than or equal to 8

we can do 4-0, 5-1, etc

and then count the number of scoreboards for each

[unreal jewel]

so you showed that f(x) = 1 - x satisfies these conditions

no?

[limber lynx]

no it is subtle but we showed that if such a function were to exist then it must ONLY be f(x)=1-x

think about it for a sec

see the statement "i concluded with f(x)=1-x for all integers"

there is ONLY ONE function f such that f(0)=1, f(1)=0, f(-1)=-2, f(2)=-1, f(-2)=3, ...

im sorry but i dont know how to explain it properly

but im confident that my approach proves uniqueness as well

[unreal jewel]

nah it's okay

[limber lynx]

ive seen this exact approach used for many other functional equations

[unreal jewel]

i'm probably just too tired to think aboutg it properly lol

[limber lynx]

yeah prolly

[unreal jewel]

what are some of the resources?

AOPS is really the only one i've been recommended

[limber lynx]

im indian so ive got books written by indian authors
the chapter on functional equations in the book "challenge and thrill of pre college mathematics" is excellent

if you want a more comprehensive resource

[unreal jewel]

is there an ebook version of that?

[limber lynx]

try "functional equations" by venkatachala

[limber lynx]

there should be

[limber lynx]

one of the earliest questions in that chapter uses induction in a very similar way to what i did and it proves uniqueness as well

one sec let me get that book out and find that exact question

@unreal jewel on page 638 of my edition of the book we are given this question:
Q. Find all functions f:N->N such that f(2)=2, f(mn)=f(m)f(n) for all m,n and f(m)<f(n) whenever m<n

they then find the value of f(1) to be 1 using f(mn)=f(m)f(n)

then they assume f(1)=1, f(2)=2, f(3)=3, ... , f(2k)=2k for some natural number k

they then show that f(2k+2)=f(2)f(k+1)=2k+2 (from induction hypothesis)

[unreal jewel]

actually i had a question about this notation that might clear up my confusion

[limber lynx]

this forces f(2k+1) to be 2k+1 as the third condition gives 2k<f(2k+1)<2k+2

[unreal jewel]

if you say f: N -> N, does that mean EVERY natural number must be some output of an input?

[limber lynx]

no no

i just mean that the function is defined for all natural numbers

and that the outpiut is always natural

[unreal jewel]

yeah okay that's what i thought

but i wasn't sure

[limber lynx]

it can be the case that some natural number is not an output at all

for example, f:N->N given by f(n)=n+1

[unreal jewel]

that's why with my question i wasn't sure if there could be some other function that doesn't go through all the natural numbers as an output

[limber lynx]

1 is not an output

(im taking 0 to not be a natural number btw)

[unreal jewel]

right yes

just wasn't sure about that

[limber lynx]

yep

alr i gotta go and do some fuck all essay about colonialism or some shit

[unreal jewel]

okay one more question that i think might help em to understand this better

[limber lynx]

yes

[unreal jewel]

if the original question was from N to N, would the solution be the same just without the modulus?

because if i can understand that it'll be easier to understand the full soluition

[limber lynx]

it cannot be from N to N since the second equation talks abot -x

about

[unreal jewel]

oh yeah lol

[limber lynx]

but suppose there were some other question from N to N

then modulius would never come in

modulus*

[unreal jewel]

yes i get that

i'll read over your soltuon tomorrow probably when my head is clear. thanks for your help and patience, i really appreciate it :)

and good luck with your essay

[limber lynx]

no prob

fun question

[unreal jewel]

yeah functional equations seem cool

even tho i'm bad at them

[limber lynx]

i never thought to use induction on the modulus of the input before

[unreal jewel]

they're all very unique compared to other questions imo

[limber lynx]

it's a powerful idea tho

[limber lynx]

yes

i love functional equations

i have a relatively easy one that you may wanna try

you want it?

(hint: this one uses induction too)

[unreal jewel]

yeah sure

can i add you and ask about it then after i give it a go?

(which probably won't be today lol)

[limber lynx]

sure

add me

[unreal jewel]

epic

[safe radish]

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[lost marsh]

how do i simplify square roots along with how do i do it with the numbers in the front aswell

[granite idol]

example?

[safe radish]

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.

[lost marsh]

i dont have exacts but

[plucky elk]

Where do you see this then

[safe radish]

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[woeful kite]

Not sure what im missing

[safe radish]

@woeful kite Has your question been resolved?

[woeful kite]

<@&286206848099549185>

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[woeful kite]

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[latent eagle]

this is such a basic integral but i just get get a non zero answer

i understand the integration by parts set up

and i understand how the integral of cos(two pi x /a) ends up being zero in the bottom line

but i cant get the left part of the bottom line to not be zero

[empty gyro]

https://tenor.com/view/jake-peralta-squinting-read-trying-to-read-brooklyn-nine-nine-gif-13387080

[copper ginkgo]

is that the a coefficient of the fourier series?

[latent eagle]

yeah sorry professors handwriting lol

[latent eagle]

no it's not. here specifically, it's the length dimension between two rigid walls containing a particle

this is literally just a freshman level calc II integration by parts

and i keep fucking it up

[copper ginkgo]

what's the point of calculating the integral?

do you want it to be equal to zero, or do you want to find a value for a, for example?

[latent eagle]

i do not want it to be equal to zero

[copper ginkgo]

and that's true for every a that's different than 0, then

[latent eagle]

the context behind this particular integral is im trying to find the x uncertainty for a particle between to walls

[copper ginkgo]

I skipped some parts given that you are comfortable with the integration by parts

[latent eagle]

ok

first step

for cos argument

cos( two pi) = one

cos(0) =

one

one minus one is 0

thats the part im fucking up

OH

[copper ginkgo]

ahahaa

[latent eagle]

because there's a fucking zero in front of it

[copper ginkgo]

YEAH ahahah

[latent eagle]

jesus

thank you

[copper ginkgo]

np, you did all the work

[latent eagle]

thanks again pardner

.close

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[clever storm]

I would like help, I am not sure how to turn it into a power series

[rough storm]

,calc 9*15

[flat frigate]

Result:

135

[rough storm]

(9x+135-135)/(x+15)

[clever storm]

Hm, how did you get that?
My idea was to try to turn it into a geometric serieis by mutiplying both sides by 1/x

[rough storm]

do you see how writing it like this helps though?

[clever storm]

Let me try iit out rq

[rough storm]

$\frac{9(x+15)-135}{x+15}$

[flat frigate]

Moosey

[clever storm]

Okay, let me see

still stuck, i'll keep trying though

no dice

[rough storm]

$\frac{9(x+15)}{x+15}-\frac{135}{x+15}$

[flat frigate]

Moosey

[clever storm]

So then you have 9/x+15 - 135/x+15
which OHHH

OH

I forgot what fractions can do

[rough storm]

no the (x+15) cancels on the 9

[clever storm]

Now it's in the form ohh

[rough storm]

so its just 9- (135/x+15)

[clever storm]

So should we find the power series of that -135/x+15?

then subtract from 9

[drowsy karma]

i do not want to disturb but maybe this way:

$\frac{9x}{x+15}=9x\cdot \frac{1}{15+x}=\frac{9x}{15}\cdot \frac{1}{1-\left( -\frac{x}{15} \right)}=\\\frac{9x}{15}\cdot \sum_{n=0}^{\infty }\left( -\frac{x}{15} \right)^{n}=\sum_{n=0}^{\infty }9\left( -1 \right)^{n}\frac{x^{n+1}}{15^{n+1}}\\\text{ }\text{ }\text{ for }\text{ }\left| \frac{x}{15}\right|<1\Leftrightarrow x\in \left( -15,15 \right)$

[flat frigate]

Joanna Angel

[clever storm]

Hm, I never thought of that method

Question though, why wasn't the 1 in the numerator divided by 15

[drowsy karma]

that is based

on what you wanted

geoemtric series

i hope oyu can see it at once

rememebr every such simpel rational function

we always

transform

to form to sue geoemtric series

that is knwon trick

and now, you only need to compare with the right side of yoru task from your test

[clever storm]

Okay, I just want to know why the one in the numerator for the 3rd step wasn't also divided by 15
I thought what happened in the denominator had to happen in the numerator also

[drowsy karma]

i did that but

automatically

i moved it

can you see it ?

[clever storm]

Did you move it so that it would be 9x times 1/15?

[drowsy karma]

yes means

9x/15

[clever storm]

Oh yes I understand now

[drowsy karma]

such simple calculations are never written

[clever storm]

Okay that makes sense, thank you

[drowsy karma]

yvw 🙂

[clever storm]

So for the 1st 5 terms, I assume I would just start at 0 for the power series

[drowsy karma]

yes but you also need to observe waht your instruvtor wrote in test

the form of the right side

he wrote this:

$\sum_{n=0}^{\infty }c_{n}x^{n}$

[flat frigate]

Joanna Angel

[drowsy karma]

so now, you should compare

my result and his form

[clever storm]

that would be a maclauren series right?

[drowsy karma]

yes if center = 0, then itis Maclaurin

[clever storm]

Oh, I get it now

So then it should turn into, one sec i wrote it down

[drowsy karma]

its ok

[clever storm]

wait hold on

the thing on the bottom is what matters

[drowsy karma]

minus must be inside of parenthesis, you agree ?

[clever storm]

Yes

[drowsy karma]

(-1)^n

[clever storm]

so not -1/15^n?

[drowsy karma]

yes i told you

(-1/15)^n

always parenthesis if needed )

[clever storm]

Like this?

Ignore the arrow

[drowsy karma]

this form you have written, is equivalent to mine

it is the same expresion )

[clever storm]

Okay

[drowsy karma]

the most important is the method

i showed you

=

how to find maclaruin series

of the ratinal function

if assume, the denominator contains trinomilal sqaure

then you also need to use partial fractions too