help-49

[strong lava]

Yes

sinon, version courte,
t'as bien fini ta sup là ?

[pulsar sun]

I think I understand

so the condition is that the points are not "reversed"

[strong lava]

that's merely a very simple counterexample
It is absolutely not a necessary condition

[pulsar sun]

its not the unique condition ?

[strong lava]

va falloir bosser tes compétences en logique car après 1 an de prépa, je sais pas comment t'en est encore là

[pulsar sun]

?

[strong lava]

t'es bien en sup dans une petite prépa?

[pulsar sun]

oui je me débrouille bien en plus

oui

désolé mais je n'arrive pas à me l'imaginer quand je bute sur quelque chose j'ai l'impression juste quand je le code de toute façon je fais déjà une rotation

donc forcément elle existe

[strong lava]

SL2(R) est naturellement isomorphe au groupe (R / (2piZ), +), lui même isomorphe à R / Z aussi connu comme S^1, le cercle
L'isomorphisme en question se base sur le fait qu'une rotation est définie par l'image de (1, 0)
Pour tout autre point, il existe un changement de base orthogonale (plus précisément la composée d'une homothétie et d'une rotation) qui l'envoie sur (1, 0). Donc une rotation est aussi définie par l'image d'un point fixé, en regardant où elle l'envoie sur le cercle du bon rayon.

Bref, une rotation centrée en l'origine est donc déterminée par l'image de n'importe quel point différent de 0, définissant ainsi une fonction f : R* x R -> SL2(R) U {pas de rotation} surjective
Il est alors naturel que pour une paire quelconque donnée, il est fort peu probable que f(x, x') = f(y, y') soit dans SL2(R)
Encore moins 3 points

[pulsar sun]

le problème étant que quand je l'ai codé effectivement je connais tout les points avant la rotation forcément et j'applique une rotation d'angle theta aléatoire à ces points donc la rotation à déja bien lieu, le mieux serait alors plutot de prendre 3 points avant et après sachant que ceux après conservent la même distance par rapport à l'origine et les distances entre eux, il n'y aurait alors pas de rotation que quand il n'y a permutations de 2 points après

[strong lava]

donc tu construits tes points de tel sorte que la rotation existe toujours

[pulsar sun]

c'est le problème de mon programme je viens de me rendre compte

c'est pour cela que je ne comprenais pas

[strong lava]

après construire des points de manière aléatoire de telle sorte qu'il existe une isométrie les envoyant comme tu veux, ça parait compliqué

[pulsar sun]

faudrait juste que je dise ok je conserve les distance par rapport à l'origine et entre eux et voyons voir si une tel rotation existe

non si c'était vraiment aléatoire j'en aurais jamais des rotations

[strong lava]

jusqu'à ce que tu te rappelle de ce que sont les isométries de R²

[pulsar sun]

pas vu encore

[pulsar sun]

c'est le dernier chapitre à voir cette année

ok

on conserve la norme dans les isométries ?

[strong lava]

par définition tu préserve la distance

[pulsar sun]

une isométrie directe

[strong lava]

la notion de directe est arbitraire et inutile ici

[pulsar sun]

ça conserve l'orientation de la figure si j'ai bien compris

donc c'est plutot à partir de 2 points ou plus qu'on a pas forcément rotation

pourquoi m'a t'il parlé de 3 points ou plus ? je n'en ai aucune idée

[strong lava]

C'était ton idée ?

[pulsar sun]

oui enfin c'est mon prof qui m'a dit qu'a partir de 3 point ou plus ça va poser des problèmes

je lui demanderai ce qu'il voulait dire par là

[strong lava]

La question est bc plus intéressante si tu considères des rotations de centre quelque

[pulsar sun]

c'est ce que je voulais faire au début, pourquoi donc ?

[strong lava]

Car c'est soudainement plus trivial

[pulsar sun]

mon prof m'a juste dit que je devait rester avec l'origine car $f : x \mapsto xe^{i\theta}$ était une app linéaire

[strong lava]

Oui sinon 0 n'est plus fixe

[grand pond]

phoestaclies

[pulsar sun]

meme un isomorphisme ici

[strong lava]

Mais les espaces affines c'est pas le diable
C'est juste hors programme

[pulsar sun]

on en a parlé très vite fait oui

bon bah merci bien, j'éclaircirais tout ça demain avec mon prof. T'as fait prépa aussi alors ?

[strong lava]

Comme tu le sais

À moins qu'on n'ai pas parlé depuis 1 an car j'ai changé de pseudo ici

[pulsar sun]

tu as déjà une école, enfin tu es peut etre meme déja beaucoup plus loin que ça

nan je me rappelle pas de notre discussion

[strong lava]

Non je finis la L3 là

[pulsar sun]

juste de ton pseudo

[strong lava]

LLG -> Ulm

[pulsar sun]

c'était il y a deja longtemps je pense, après j'ai la mémoire courte pour certain trucs

tu était à llg et tu est allé à ulm c'est ça

bonne fin de soirée, merci encore

👍 je tacherais de ne pas oublier ce pseudo

.close

[midnight plank]

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[spice prairie]

could somebody explain how the 4/log2x-2 3 becomes log3 2x-2

i dont understand how

2x-2log3 becomes 4log3/ 4log2x-2 right

[vagrant marlin]

look

do you know that log base a of b

is log b over a

@spice prairie

[spice prairie]

sorry im back

[vagrant marlin]

yes

[spice prairie]

so

[vagrant marlin]

so do you know change of base formula

[spice prairie]

alogb = clogb/cloga

[vagrant marlin]

this thing

[spice prairie]

oh yea

[vagrant marlin]

yeah

so it becomes 1 / log3/log(2x-2)

do you follow

[spice prairie]

yeah

[vagrant marlin]

so it reciprocal so we flip

[spice prairie]

oh and thats log2x-2

[vagrant marlin]

so log(2x-2)/log3

[spice prairie]

/ log3

[vagrant marlin]

yeah

[spice prairie]

ohh

aightt

[vagrant marlin]

u get it now

[spice prairie]

and then 3log3 is just 1

[spice prairie]

yeahh

i have one more question tho

2sec

actually 2 more steps that i dont understand

first this

why does the above become that

and then just this

[vagrant marlin]

because

he exponentiated both sides with a base of 3

[spice prairie]

can you just do that

is it a rule or

[vagrant marlin]

yes

do you know the natural logarithm

[spice prairie]

ln

right

[vagrant marlin]

yes

so if you have lnx=2

for example

you exponentiate both sides with base e

to get x = e^2

[spice prairie]

doesnt it become e^lnx = e^2

[vagrant marlin]

yeah

do you know that

b^log base b of a = a

[spice prairie]

uhh

wdym

[vagrant marlin]

one sec

[spice prairie]

😭sorry

[vagrant marlin]

lemme write it out

[spice prairie]

aight

[vagrant marlin]

like this

oh sh

why is the left grey

[spice prairie]

😭

[vagrant marlin]

wait do you get it

[spice prairie]

the left a at the top

is the base right

or

no

wait

[vagrant marlin]

a is the base

[spice prairie]

yeah

[vagrant marlin]

and also the base of the logarithm

[spice prairie]

ohhh

omd

[vagrant marlin]

its a to the power of log base a of b

[spice prairie]

aighttt i understand

[vagrant marlin]

yeah

so thats what happened basically

[spice prairie]

isnt there like another way to do that

or do i have to apply that rule

[vagrant marlin]

its easiest to apply that rule

[spice prairie]

aight aight

and one more

the right side is making me trip

[vagrant marlin]

dp you know the property

log(a^b) = bloga

[spice prairie]

yeah

[vagrant marlin]

so sqrt x

is x^1/2

which they took out of the log

and then they squared it

[spice prairie]

why does it become 1/4 tho

[vagrant marlin]

because they square it

squared

on the right

[spice prairie]

for what

😭

ohh

cuz it goes

[vagrant marlin]

do you see that the expression is squared

[spice prairie]

out of the parentheses

[vagrant marlin]

yh

[spice prairie]

oh aight

so thats it basically

[vagrant marlin]

yh

[spice prairie]

i hate logs bro

its mad confusing

[midnight plank]

@spice prairie Has your question been resolved?

[midnight plank]

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[mighty pelican]

is the radius 8.9?

[tiny token]

CAN SOMEONE HELP ME WITH 1+1

[sweet wing]

do you have it in the form (x-a)^2+(y-b)^2=r^2?

[mighty pelican]

yes

(x-4)^2 + (y-8)^2 = 16 +64=8

[sweet wing]

from there you can directly read off the radius

since the RHS is radius^2

[deep vine]

let's back up for a second. are you sure it's 16+64?

[mighty pelican]

let me redo it

1

sec

[deep vine]

make sure you're accounting for the constant correctly

what i suggest is that your first step is to move the constant to the other side of the equation:
x^2+y^2-8x+16y = -16

[mighty pelican]

is the radius

just

8

[deep vine]

yes

[mighty pelican]

ok thx

the 16 and -16 cancelled out

[deep vine]

gj!

[mighty pelican]

and everything else was right

like the center and stuff

[deep vine]

yep

[mighty pelican]

is this correct

[sweet wing]

looks right

[mighty pelican]

thank u

correct?

[sweet wing]

submit and find out

[mighty pelican]

that doesnt tel me if its right or not

[sweet wing]

you dont ever get feedback?

[mighty pelican]

No

idk why she just doenst let us see our grade immediately

to stop cheatin gig

ig

[sweet wing]

but you see it at the end?

[mighty pelican]

No

[sweet wing]

never?

[mighty pelican]

it shows u after the due date

liek

if u submit

u can see after the thing is due

[sweet wing]

yes, so afterwards

[mighty pelican]

i mean like

so its due at midnight tonight

i wouldnt be able to see it til it was midnight

[sweet wing]

that makes sense

[mighty pelican]

should

i keep the

^2

on the 13

or no

[magic rampart]

yeh you have to

[sweet wing]

go to the form (x-a)^2+(y-b)^2=r^2

youve found a and b

now if you plug in the other point

you can find r

[mighty pelican]

i have everything

[runic yoke]

honestly same

[runic yoke]

so friends cant tell each other the answer

[mighty pelican]

yeha

but its a tedt

test

[runic yoke]

exactly

[mighty pelican]

where the questions r mainly diff for each person

[runic yoke]

i mean you can still ask friends for help

ah wait

nvm that wouldn't work

if the questions are diff then there is no point in hidiing the grade

[mighty pelican]

i agree

its also in a different order

[sweet wing]

if you know the answer is right, then you know (almost certainly) that the method is correct

[feral meadow]

hello, could anybody help me out with this question?: A store offers a one-day 10% discount on all cell phones. They also offer a warranty that costs either 5% of the discounted price or $35. Write an equation based on the original price that models each possibility for the warranty.

(sorry if this isn't how asking questions is supposed to work I can delete my message if so)

[sweet wing]

create your own channel under Math Help (Available)

[feral meadow]

ahh ok sorry i should have read that first

[sweet wing]

find an open channel and ask your question

[midnight plank]

@mighty pelican Has your question been resolved?

[midnight plank]

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[twilit field]

Consider the following subset of the XY plane $S={|z-iz|,|z|^2:z \text { is a complex number}}$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

then which statment is correct

1. S denotes an ellipse but not a circle

2)S is a parabola

3. s is a circle

4. S is a hyperbola

so $|z-iz|= |x+iy-i(x+iy)|$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

whic is

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

[grand pond]

ƒ(Why am. I here)=I don't know

[hard umbra]

can you give a geometric interpretation to |z - iz|

[twilit field]

it feels like it should be a hyperbola

[hard umbra]

geometrically, what is |z - iz|

[twilit field]

so z and iz are perpendiular to one another, right?

[hard umbra]

yes

[twilit field]

so it should be a stright line?

no an ellipse

[hard umbra]

no tell me what |z - iz| is first

[twilit field]

this

[hard umbra]

z and iz are perpendicular of the same length

[twilit field]

yes

[hard umbra]

not algebraically

you can do this purely from the geometry

what is |z - iz| then

[twilit field]

I don't really know the equations of conics in $\C$

[grand pond]

ƒ(Why am. I here)=I don't know

[hard umbra]

knowing this

you don't need to

this is all real stuff

z and iz are perpendicular and of the same length
what is |z - iz|

[twilit field]

hmm

a circle

[hard umbra]

no

it's just a real number

i want the value

the distance between z and iz

[twilit field]

$\sqrt{2|z|}$

[hard umbra]

z is outside with a | |

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

right

my bad

$\sqrt{2}|z|$

[grand pond]

ƒ(Why am. I here)=I don't know

[hard umbra]

so the x coordinate is just sqrt(2)|z|

and the y coordinate is |z|^2

i think at this point it's obvious what S is

[twilit field]

a parabola

[hard umbra]

evidently

[twilit field]

thanks

.close

[midnight plank]

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[twilit field]

$let f \Q \rightarrow \Q$ be a function such that $f(x+y)=f(x)+f(y)$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

where $x,y\in Q$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

and f(1)=10

prove the function is a bijection

so to start

[surreal moon]

[twilit field]

f(2)=f(1)+f(1)...

[misty gorge]

(this function happens to be Q-linear)

[twilit field]

no idea what that means

[misty gorge]

it's linear

since it's a map from Q -> Q

[twilit field]

this is supposed to be solved using HS level knowledge

[misty gorge]

yeah

[twilit field]

makes sense

If I redefined it to be a map from R to R

[misty gorge]

don't

[surreal moon]

Define $x$ as a ratio of two integers $x=\frac{p}{q}$

[twilit field]

it's derivative will be constant

[grand pond]

SWR

[misty gorge]

it's not linear if it's a map from R -> R

[surreal moon]

consider what you can do with that

[misty gorge]

it might not be continuous

[twilit field]

so this proves it's increasing on $[1, \infty)$( can be shown by induction)

[misty gorge]

what

[grand pond]

ƒ(Why am. I here)=I don't know

[surreal moon]

I believe you are overthinking it

Prove that $f(nx)=nf(x)$

[grand pond]

SWR

[twilit field]

can I present my method first please

[surreal moon]

Then prove $f(\frac{x}{n})=\frac{1}{n}f(x)$

[grand pond]

SWR

[surreal moon]

sure

[twilit field]

sorry

$f(1-0)=f(1)-f(0)$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

this proves f(0)=0

and similarly I can show It's odd about 0

so this can be extended to prove surjectivity

[misty gorge]

how

how do i use what you said to show that there's some x such that f(x) = 5

[twilit field]

f(3)=f(2+1)

f(4)=f(3+1)=f(3)+f(1)

f(5)=f(4)+f(1)

f(3)=f(2)+f(1)

[surreal moon]

[misty gorge]

none of these yield an x such that f(x) = 5

f(3) is 30 for example

(which you CAN get from what you said, i agree, but surjectivity won't come from that)

[twilit field]

for instance for $\Q^+$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

$f\left(n\right)=\sum_{i=1}^nnf\left(1\right)$

[grand pond]

ƒ(Why am. I here)=I don't know

[misty gorge]

yes i agree, but that doesn't get me 1/2 * f(1)

as an output

[twilit field]

I'm not done

$f\left(1\right)=2f\left(\frac{1}{2}\right)=4f\left(\frac{1}{4}\right).....$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

this isn't enought though

I agree

should probably just find a general formula for $f(n) , n\in Q$

[grand pond]

ƒ(Why am. I here)=I don't know

[surreal moon]

any thoughts?

[twilit field]

I'm thinking

any rational can be expressed as the sum of an integer and a rational number

I guess I use that?

[surreal moon]

Probably not the best route imo

[twilit field]

why not?

$f(\frac{p}{q}) = f(floor(\frac{p}{q}))+f(frac(\frac{p}{q}))$

[surreal moon]

Because then you're still left with a rational number between 0 and 1. It'll end up taking the same amount of effort to prove. It is wasteful.

[twilit field]

hmmm

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

could I have a hint

[surreal moon]

$f\left(\frac{p}{q}\right)=f\left(\floor{\frac{p}{q}}\right)+f\left(\floor{\frac{p}{q}}-\frac{p}{q}\right)$

[grand pond]

SWR

[surreal moon]

Sorry that's not a hint. I was just writing out your tex

[surreal moon]

Do this first

[twilit field]

I mean this is still enough

[surreal moon]

enough what?

[twilit field]

$f\left(\frac{p}{q}\right)=f\left(\left[\frac{p}{q}\right]\right)+f\left(\left[\frac{p}{q}\right]\right)-f\left(frac\left{\frac{p}{q}\right}\right)$

[grand pond]

ƒ(Why am. I here)=I don't know

[surreal moon]

This is my hint for you btw.

[twilit field]

I already did that for $x\in \Z$

no?

[grand pond]

ƒ(Why am. I here)=I don't know

[surreal moon]

No. You did it for x=1 only

[twilit field]

$f\left(nx\right)=f\left(\frac{n}{2}x\right)+f\left(\frac{n}{2}x\right)=2f\left(\frac{n}{4}x\right)+2f\left(\frac{n}{4}x\right).....$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

like this?

[surreal moon]

Not quite. No. You proved $f(nx)=2f\left(\frac{n}{2}x\right)$. I'm asking you to prove that $f(nx)=nf(x)$

[grand pond]

SWR

[twilit field]

then I'm not sure

[surreal moon]

Hint: You'll need induction

[twilit field]

okay

hmm

f(2)=2f(x)

I'm not sure honestly

[surreal moon]

Okay, for starters, what is your base case?

[twilit field]

f(1)

[surreal moon]

Our base case is to prove that $f(1x)=1f(x)$

[grand pond]

SWR

[twilit field]

yes

[surreal moon]

I'm huessing that doesn't take a lot of effort to prove

[twilit field]

isn't that trivially true

1x=x

[surreal moon]

Yes. Exactly.

[twilit field]

do f(1x)=f(x)=1f(x)

[surreal moon]

Yes. Perfect

Base case proven.

Now, for induction, what do we need to assume is true and what do we need to prove is true from that assumption?

[twilit field]

f(nx)=nf(x) is true

and then prove that f((n+1)(x))=(n+1)f(x) is true

[surreal moon]

exactly right

[twilit field]

that's trivially true too

but why does induction work though

[hard umbra]

it's certainly not "trivially" true

[surreal moon]

Trivial-ish. You still need to use your given property of f

[hard umbra]

even if it is "trivially" true, you should be even more so able to write a proof

[twilit field]

hmm

okay

so f((n+1)(x))= (n+1)f(x)

which is nf(x)+f(x)

[quaint aurora]

Does anyone know about groups? I'm just looking for more information, nothing more (if you know Spanish better, the truth is, English is difficult for me xd)

[surreal moon]

!occupied

[midnight plank]

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).

[surreal moon]

Alguien más ya está usando este canal de ayuda. Si necesita ayuda con una pregunta, abra su propio canal o hilo de ayuda (consulte #❓how-to-get-help para obtener instrucciones).

[surreal moon]

How did you get this?

[twilit field]

(n+1)=t

uh, I think i've spent too much time on this problem

[surreal moon]

Hint: $f((n+1)x)=f([nx]+[x])$

[grand pond]

SWR

[hard umbra]

brackets

this is pretty unfortunate notation, since a lot of people use [] for the floor function

[surreal moon]

Hint: $f((n+1)x)=f((nx)+(x))$

[twilit field]

might aswell solve this later

sorry

[grand pond]

SWR

[twilit field]

i'll do this later

sorry

.close

[midnight plank]

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[last slate]

I'm lost on part b. I tried plugging in 5 for x in the equation above.

[knotty forge]

Did you solve

.05 = (0.8409)^x

?

Because that will get you a different answer.

[strange zodiac]

https://www.desmos.com/calculator/n2klsc6ggh

[last slate]

yess it was equal to o.05. Thank you that makes sense!!

.close

[midnight plank]

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[midnight plank]

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[manic bison]

How to solve this?

[twilit field]

you sum up each element individually if I'm recalling the notation right

[manic bison]

How tho

Like there is an inequality at the bottom

[twilit field]

can you re-write the question more clearly please

can't read it

[manic bison]

Alright wait

[twilit field]

or type it out

[manic bison]

This one

[twilit field]

,rotate

[manic bison]

The notation here is Sigma btw

[twilit field]

oh

so i<j

I'm not sure then

sorry

[manic bison]

It's fine dw

[sinful river]

you do it step wise
as in first j takes a value (subsequent sum of all is)
then j steps up by 1

[manic bison]

I can't understand this fully

[last slate]

But what happens when j takes the value of 1 ?

[sinful river]

so your final sum is like this (all iterations are wriiten as (i+j))
(j=1 nothing)
(j=2) = i =1
j=3 i =1 and i =2
j=4 ; i=1 i=2 and i=3
j=5 not possible

[last slate]

Oh

[sinful river]

nothing i cant take values so no sum

[last slate]

Okay

[manic bison]

Oh I see

Wait lemme try this

[last slate]

Do you know any other method?

Something that helps when the range is large

I struggle with these kind of problems too

[sinful river]

mostly discrete summations use the nested loop method
the thing to do is just nest the summations
so you will get it as sums of sums

but if the value is large people generally arnt sadistic enough to make you do it by hand

[manic bison]

Will the answer be something like this then?

[sinful river]

no 0

i cant take 0

and its always 2 values

[manic bison]

What will i be there

[sinful river]

it will be like this
(1+2) + (1+3) + (2+3) + (1+4) + (2+4) + (3+4)

thats it

notice the ranges of i and j

minimum is 1 and i can never be equal to j as its strictly less than)

[manic bison]

Oh I see

[manic bison]

This will be the final answer right

[sinful river]

yes

the easier method is to expand it into 2 summations

[manic bison]

How do we do that tho like I know the identity but how do we do it in this particular question

[sinful river]

please excuse the handwriting

[manic bison]

Double Sigma?

[sinful river]

yep

[manic bison]

Won't it be multiplication here then

[sinful river]

the inner sigma is done first then the outer sigma

[sinful river]

youve done some basic coding?

[manic bison]

Ohhhh

[sinful river]

like c++ or java

its a nested loop

[manic bison]

Nope ;-;

This was done yesterday in my math olympiad preparation class

[sinful river]

ohk so basically it is always summation but the step up for the outer sigma will happen once when the entire case of the inner sigma is evaluated

so j will increase by 1 after summing all possible i values in the range
if you have i<j<k then its 3 sigmas deep
so k will increase once when all of J finishes which it turn steps once when all of i finishes

but the internal function always remains same

[manic bison]

Ohhhh

Wait lemme try it

[sinful river]

this is very important

[manic bison]

i +j will be the same in both the sigmas then?

[sinful river]

wait ill give you a sample question tell mw how you will solve it

[manic bison]

Alr

[sinful river]

same sigma
the condition is (2 <= i < j <= k < 10)

the function is (i+2j+k)

step 1 write down the ranges of each variable starting from the rightmost variable

[manic bison]

k will be [j,10)

[sinful river]

think again

it takes everything from 2 to 10

cuz j can even be 2

[manic bison]

Ohhhh

[sinful river]

the outermost variable has no issues

so you can blindly just put the exact range mentioned in the question

no comes the tricky part
try for j

[manic bison]

I see

[manic bison]

From i to k?

i not included

[sinful river]

again no lower variable its 2 to k

the k is correct though

[manic bison]

I see

[sinful river]

basically the same reason works i can take values from 2

[manic bison]

Wait I get it now it's cuz 2<= i

[sinful river]

yep

[manic bison]

Yea

[sinful river]

now i's range?

[manic bison]

2 to j

[sinful river]

equal to j?

[manic bison]

Nope

J not included

[sinful river]

Perfect

now step 2
expand the summation into n seperate summations and add the previous results

innermost variable is ?

[manic bison]

k?

[sinful river]

check again

[manic bison]

i + 2j +k?

[sinful river]

thats the function lmao

the innermost variable is i

outermost is k

[manic bison]

Mb I'm dumb af ;-;

[sinful river]

terminology only its fine

so write the summation with only

i

[manic bison]

Hmm ok wait

This maybe?

[sinful river]

Perfect

now add the j summation next to it

[manic bison]

Hmmm alr

[last slate]

[
\sum_{1\leqslant i < j < 5} (i + j)
]

[grand pond]

mid

[sinful river]

and then k

and you are done

[manic bison]

I see

[sinful river]

now simply just go summation by summation for each subsequent j and k value

again KEEPING THE FUNCTION THE SAME

the (i + 2j +k ) part

and you are set

[manic bison]

This the main issue

Like

I have to go one variable at a time?

[sinful river]

yea 😅

[manic bison]

Wait lemme try to solve it

@sinful river I'm clueless ;-;

Do I use n(n+1)/2 or something

[sinful river]

no expansion

start with k =2
no j or i possible

[manic bison]

Ohhhh

Wait brb

[sinful river]

then k =3 j = 3 and i will be 2

and so on

[old storm]

Isn't this just a nested for loop

[sinful river]

Basically

raged summations are nested for loops
the problem comes when you dont know what a for loop is XD

[manic bison]

Got it finallyyyyy

TYSM

[manic bison]

Yea idk what this is ;-;

[sinful river]

lessgo

[manic bison]

TYSMMMMM

.close

[midnight plank]

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[midnight plank]

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[twilit field]

so to start, I approximate sin(sin(kx)) as kx

so I get

$(kx+cos(x)+x)^{\frac{2}{x}}$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

which is in the form $1^{\infty}$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

so the limit is effectively

[subtle blaze]

well cos x is essentially 1

[twilit field]

$e^{\frac{2}{x}\left(kx+x^2+\cos\left(x\right)-1\right)}$

[subtle blaze]

[grand pond]

ƒ(Why am. I here)=I don't know

[subtle blaze]

nah nah

[twilit field]

the limit of this at 0

[subtle blaze]

i'd rather try find a form of $\left(1+\frac{a}{n}\right)^n$

[grand pond]

Frosst

[twilit field]

hmm, but is this approximation valid

[subtle blaze]

it's this + some o(x^2) terms or whatnot

which disapear in the limit

[twilit field]

yeah

[subtle blaze]

sort of

the top converges faster is the proper math

[twilit field]

$e^{\frac{2}{x}\left(kx+x^2+1-\frac{x^2}{2}-1\right)}$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

I just need the answer for now, have an exam in 2 weeks

I'll care about rigour later

[subtle blaze]

nah i'd do something like at this stage, let y = 2/x

[twilit field]

so k=3

[subtle blaze]

then go $(1 + (k+1)x)^y$

[grand pond]

Frosst

[subtle blaze]

then x = 2/y

[twilit field]

huh, okay

[subtle blaze]

$\left(1 + \frac{2(k+1)}{y}\right)^y$

[grand pond]

Frosst

[subtle blaze]

take the limit you get $e^{2(k+1)}= e^6$

[grand pond]

Frosst

[subtle blaze]

because x goes to 0, 2/x goes to inf

[twilit field]

so k=2

[subtle blaze]

yes

[twilit field]

wait

oh right

yeah, missed an x in my approximation

thanks

.close

[midnight plank]

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[subtle blaze]

👍

[twilit field]

sorry for not caring about the rigour atm

[subtle blaze]

the answer is here tho

because you have e^6 as what you want to get to

[twilit field]

yeah, got it

thanks

[midnight plank]

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[undone pawn]

Hi! Does anyone know what N is?

[junior flower]

beats me

i really can’t think of what that could be. do you have a statement of the intermediate value theorem somewhere?

[undone pawn]

Nothing with N in it 🥲

[junior flower]

loll

yea, i dunno then

[undone pawn]

funky

i'll ask my prof tmr

ty tho!

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[midnight plank]

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[jovial birch]

Could someone help me determine what approach I should use when solving this? ie. u substitution, integration by parts

[obsidian glen]

(1-cos^2(x))sin(x) is your numerator

then split it

[jovial birch]

I don't see that identity on my sheet, which one are you taking that from?

[obsidian glen]

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

[jovial birch]

ah ok

[jovial birch]

I split it like you said but I am not sure what to do about the denominator

[obsidian glen]

split as in distribute

also this is wrong

I meant something like\ $\int \frac{(1- \cos^2(x)) \sin(x)}{\sqrt{\cos(x)}} \dd x = \int \frac{\sin(x)}{\sqrt{\cos(x)}} \dd x - \frac{\cos^2(x)\sin(x)}{\sqrt{\cos(x)}} \dd x$

[grand pond]

ginny

[obsidian glen]

integral of (ab) is not equal to (integral of a)(integral of b) in general

[jovial birch]

May I ask why the 'dx' gets placed next to both terms?

oh I guess it's just how you've written it

usually it is factored out: (a-b)dx

but you wrote it a(dx)-b(dx)

regardless, I am hitting a brick wall with the square root of cosx in the denominator

[obsidian glen]

how about a u substitution to get rid of that then

[midnight plank]

@jovial birch Has your question been resolved?

[jovial birch]

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[twilit field]

suppose $log_a(b)+log_b(a)=c$the smallest integral value of c $\forall a,b >1$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

so to start

[last slate]

x = log_a(b)

do this substitution

[twilit field]

$\frac{\ln\left(b\right)}{\ln\left(a\right)}+\frac{\ln\left(a\right)}{\ln\left(b\right)}$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

AM>GM?

[last slate]

u are overcomplicating

[twilit field]

okay

oo

so I have to minmise $x+1/x$

[grand pond]

ƒ(Why am. I here)=I don't know

[last slate]

yes

find critical points

[twilit field]

cool, the min would be 2

thanks

[last slate]

you welcome

[twilit field]

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[midnight plank]

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[midnight plank]

Please don't occupy multiple help channels.

[craggy yew]

how is it a quadratic

[twilit field]

Try expanding the RHS

[craggy yew]

kk

got a cubic

(might be wrong)

[twilit field]

,w (n+2)(n+6)(n+20)

[grand pond]

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[twilit field]

so you have $n^3=n^3+28n^2+172n+240$

[grand pond]

ƒ(Why am. I here)=I don't know

[craggy yew]

yes

oh wait

the n^3s get cancelled

ohhhhhhh

i get it now

.close

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[craggy yew]

.reopen

[midnight plank]

✅

[craggy yew]

how did 151/7 come

[twilit field]

I don't know the context

[craggy yew]

this is the question

solution

[twilit field]

notice how the equations are in the from $a^3=2+abc$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

$b^3=6+abc$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

and $c^3=20+abc$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

do you follow so far?

[craggy yew]

...yes

[craggy yew]

continue

[twilit field]

now we have $abc=n$

[grand pond]

ƒ(Why am. I here)=I don't know

[craggy yew]

yes

[twilit field]

so we solve for n, yes?

[craggy yew]

just explain me this part

[twilit field]

now we have $a^3+b^3+c^3=28+3abc$

[grand pond]

ƒ(Why am. I here)=I don't know

[twilit field]

I think you should get the right answer from here

remember abc=n

[craggy yew]

im confused

[craggy yew]

i just wanna know why is the answer 151/7(i think i might have explained my question in a dumb way before so sorry)

[twilit field]

,w 28+3(-15/7)

[grand pond]

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[twilit field]

@craggy yew does this make any sense

[craggy yew]

ohhh

[craggy yew]

yes it does

thanks!

[twilit field]

cool, what exam is this?

some undergrad entrane ?

[craggy yew]

im prepping for ioqm

olympiad

[twilit field]

ah

as in the indian olympiad ?

[craggy yew]

yes

[twilit field]

cool, good luck

[craggy yew]

someone recommended me this book which has american problems as practice sums so im usin it

[twilit field]

also solve past putman problems IMO

[craggy yew]

?

[twilit field]

google past putman papers

[craggy yew]

if i recall correctly isnt putman an undergrad math exam?

[twilit field]

yes

but solve papers from the 60s

1960s that is

[craggy yew]

yes ik

are they easier?

[twilit field]

some of them have appeared in jee advanced papers

so yes

[craggy yew]

i think ill need to study wayy more to solve em cuz i just started oly prep and the only math knowledge i have is like 0.1% nt and 8-9th grade math

and school math is useless for olys anyways

[twilit field]

hmm, ok

[twilit field]

these were the kind of questions they asked in teh 50s

this in the 30s

[craggy yew]

still tough for me cuz i dunno calculus

[twilit field]

but yeah, do calculus first

[craggy yew]

this might be sort of solvable for me

[craggy yew]

what was the first question of the first putman paper?

[craggy yew]

lol i cant even understand the question

[midnight plank]

@craggy yew Has your question been resolved?

[midnight plank]

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[rare cloud]

A uniform sphere of weight W and radius 5 cm is being held by a string as shown in the figure. The tension in the string will be

[rare cloud]

Why is it at vertical equilibrium and not at equlibrium on the side

like not in the direction of string?

[sage helm]

cuz it's not moving?

if you mean translational equilibrium

i mean tbf there's equilibrium in all directions

work with what you please

[rare cloud]

yeah

how do you do it in the direction of string?

like if you take the component of W in the direction of string

W_s = Wcosx

T = Wcosx

[rare cloud]

but the answer is incorrect

[sage helm]

Ah okay

It's not "incorrect" per se

thing is if you're balancing forces along the string then you need to consider the normal due to the wall as well

[rare cloud]

oh

[rare cloud]

but you won't be able to calculate that

right?

T = Wcosx + Nsinx

[sage helm]

Yeah it's kind of hard to

Better to just balance the vertical

Might be able to do it with torque

[rare cloud]

alright

thanks

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[gaunt nimbus]

pls send help

[gaunt nimbus]

i dont know how to read this in order

For all epsilon > 0, there is some N > 0 such that for all n in R, if N < n, then 1/n is less than epsilon.

I just don't see why must 1/n be less than epsilon

[sweet wing]

you can choose N

and it can be in terms on epsilon

[gaunt nimbus]

wait

[sweet wing]

so, if oyu choose $N=\frac{1}{\epsilon} \leftrightarrow \epsilon = \frac{1}{N} < \frac{1}{N}$

[grand pond]

Flappie

[gaunt nimbus]

can i pick N to be 1/e

[sweet wing]

yes

[gaunt nimbus]

c

oh

thx

[sweet wing]

if epislon and N were swapped, then thats not allowed

[gaunt nimbus]

why tho

they are both > 0

[sweet wing]

but epsilon is stated earlier, the terms after are allowed to be dependent on it

[gaunt nimbus]

ic ic

so if I have for all in front of there exists, the variable for there exists can depend on the for all in front of it

[sweet wing]

if the statement was: $\exists N>0, \forall \epsilon, \forall n$ then $N$ is just some number

[grand pond]

Flappie

[sweet wing]

or if you have something like: $\exists x,\exists y,\exists z$ then z is allowed to depend on x and y, yis allowed to depend on x, and x isnt allowed to depend on anything (unless something else was stated earlier)

[grand pond]

Flappie

[gaunt nimbus]

ic ic

got it now

.close

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[pale ravine]

Hello! how to proof that this equation has only got solutions in real numbers when sin(omega) and sin(delta) are equal (didn't put /2 because sin is positive for all possible angles because they belong to first quarter)

[grand pond]

𝚃𝚛𝚒𝚜𝚝𝚎
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[midnight plank]

@pale ravine Has your question been resolved?

[midnight plank]

@pale ravine Has your question been resolved?

[pale ravine]

I've found only one solution which is close to the restriction which is pi/2. Are there 0 solutions if angles belong to (0; pi/2)?

[split jetty]

hey its you xdd

[pale ravine]

yes and again with weird questions XD

[split jetty]

btw $\omega$ and $\delta$ are supossed to /2?

[grand pond]

clonesolopros

[pale ravine]

both angles and both half-angles are in the first quarter and both must be half-angles

[split jetty]

let $a=2x, b=2y, c=2\omega, d=2\delta \implies$
$sin(a+b)cos(c)=cos(a-b)sin(d),$ where $a,b,c,d \in (0,\pi)$

[grand pond]

clonesolopros

[midnight plank]

@pale ravine Has your question been resolved?

[faint cairn]

This seems false. If you choose x = y then you get delta = 2 arcsin(LHS). Your choice of x and y are not unique, so you will get solutions where omega does not equal delta.

[pale ravine]

But what will happen if x can’t be equal to y?

[faint cairn]

One counterexample is enough. x = y is simpler because the cos term on the RHS goes away.

[pale ravine]

I understand it. but what counterexample will there be in case if x cant be equal to y?

[faint cairn]

The same argument would probably work except now it's uglier and you have to check for the constraints.

You just need to prove that for some omega there is more than one choice of delta.

[pale ravine]

You gave me an idea and I tried to redraw graph in Desmos. What will be a solution here if we look at graph? Everything in this crossing area? Or another function that should cross this red line?

Nvm, I think I understood what I was missing in this task and probably fixed it. At least it looks like I've fixed. After your message, I wondered what I was missing out on. So I decided to draw a full graph, made some constrains also took some things from the exercise and probably got what I needed. Thanks!

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[abstract aurora]

The table shows 3 values of x and their corresponding values of y, where y=f(x) +4 and f is a quadratic function. What is the y-coordinate of the y-intercept of the graph of y =f(x) in the xy-plane?

[sweet wing]

did you try to graph it?

[abstract aurora]

No

I was told I should find the inverse function

[sweet wing]

do you know the standard formula for a quadratic?

[abstract aurora]

And also told to put it in factored form then substitute from the values on the table

[abstract aurora]

Ofc

Lol

[sweet wing]

can you write it out please?

[abstract aurora]

AX^2+bx+c

[sweet wing]

great, now what happens if you plug in the numbers that you are given?

[abstract aurora]

I can't plug in the values

[sweet wing]

why not?

[abstract aurora]

I don't know a b and c

[sweet wing]

you have x and y values

doesnt matter if you dont know a b and c

thats what we are solving for

[abstract aurora]

Okay if I put the values

What should I do next

[sweet wing]

you should get a system of equations

you have 3 equations

and 3 unknowns

that means you can solve for the unknowns

do you know how to do that?

[abstract aurora]

I am not sure what u mean

Okay so I have 3 quad fn.

I think you meant to say

[sweet wing]

so, if you plug in the first point, you get $21^2a+21b+c=-8$

[grand pond]

Flappie

[abstract aurora]

Okay

[sweet wing]

and for the second one $23^2a+23b+c=8$

[grand pond]

Flappie

[abstract aurora]

Okay

[sweet wing]

and then also the third one

[abstract aurora]

Okay

[sweet wing]

you can solve this for a b and c

[abstract aurora]

Okay

[sweet wing]

if you do that, then you have a b and c for you quadratic equation

[abstract aurora]

Okay

Well I don't have a b and c 😅

I think you stopped at substituting the x and y values from the table

Sorry if there's a misunderstanding

😅😅

[sweet wing]

thats what you are solving

[abstract aurora]

Yea but i don't think I can find the a b and c

Because they are unknown 😅

[sweet wing]

you can also graph it

,w scatterplot (21,-8),(23,8),(25,-8)

[grand pond]

Results provided by WolframAlpha

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Requested by flappie.

[sweet wing]

like i asked you at the start

what do you notice about these points

[abstract aurora]

Okay, thank you.

😊🙏

I think I will close this ticket now 😅. Thank you sir 😊 @sweet wing

.close

[midnight plank]

Channel closed

Closed by @abstract aurora

Use .reopen if this was a mistake.

[analog saddle]

can i open a ticket?

[surreal moon]

#❓how-to-get-help

[abstract aurora]

Yes

[analog saddle]

so do i just send a question here?

[midnight plank]

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>.
• Type the command .close to free the channel when you're done.
• 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.

[analog saddle]

ah🙏thx

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[analog saddle]

i have finished 12i), i'm confused for the rest of the two parts. i have no idea at all on how to begin question 13, it looks horrific

oa is y = x(x+3)
ob is y = x - 1/4x^2
ab is y = x^2 - 6x + 9

[worthy kestrel]

coordinates would be at the point of intersection of the lines

[analog saddle]

oh ok

do i equate oa and ob?

[worthy kestrel]

no

doing that gets u the point of intersection which is 0,0 or point o

u want to find a and b

[analog saddle]

yhyh alrr

would ab and ob work?

[worthy kestrel]

yes

[midnight plank]

@analog saddle Has your question been resolved?

[analog saddle]

got x = 2 and x = 3.6

do i sub it back into any equation to find y?

2 x values means 2 points of intersection or am I confusing myself?

[worthy kestrel]

yes

they are the x values for a and b.

[analog saddle]

right ok

how would i be able to work out the area under graph?

i know its integration but not sure nhow

[midnight plank]

@analog saddle Has your question been resolved?

[worthy kestrel]

u finished part 2?

area = int oa+ab-ob

with appropriate bounds

[midnight plank]

@analog saddle Has your question been resolved?