help-39

[modest tartan]

so maybe misunderstanding

anyways @rough forge u a G

ima go do more questions

.close

[pearl pond]

Channel closed

Closed by @modest tartan

Use .reopen if this was a mistake.

[rough forge]

no i am dumb i

i was confusing something for a moment

[modest tartan]

smarted den meh

i got confused by the word "max"

lol

[rough forge]

0*10=0 🥀

[modest tartan]

remember the [x] when x belongs [1,2) = x

🥀

[pearl pond]

Available help channel!

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[rare holly]

why cant i do this

[thorn shale]

the ratio is number of coins, not denomination

when you set 8parts=100p you are saying each part has the same amount of value/money

but 1p and 5p have different values

[rare holly]

so when things are of diffrent value i cannot do this?

[thorn shale]

correct, you cannot do it since they have different values

but your idea is there: try to write an expression for the total value

[rare holly]

ok i know how do it now i got it right

thanks

.close

[pearl pond]

Channel closed

Closed by @rare holly

Use .reopen if this was a mistake.

[thorn shale]

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stone imp]

It’s telling me to prove that the triangles like

[rough forge]

what have you tried

[stone imp]

Okay so

I wrote

X^2+(x+2)^2=2x^2

[reef kestrel]

ledning: om du kan trigonometri använd det, annars Pythagoras sats

[stone imp]

Jag gjorde det men det blev helt konstigt

[reef kestrel]

här är sista termen fel

vad är (2x)^2

[stone imp]

4x^2🥲

Hoppsan

[reef kestrel]

aaaa det kanske hjälper? :)

[stone imp]

I parantesen gjorde jag kvadrerings regeln

Är det rätt

[reef kestrel]

precis! vad fick du då?

[stone imp]

Det blev typ x^2+4x+4

Tror jag

[reef kestrel]

ja precis så ska det vara

[stone imp]

Jag får fram det till 4x+4=2x^2

Kommer jag behöva använda pq formeln

[reef kestrel]

yes det är nog smidigast, du kan också använda kvadratkomplettering om du har koll på det

[stone imp]

Inte direkt är det enklare än pq

[reef kestrel]

ja ibland kan det vara enklare!

idén är att du utnyttjar kvadreringsregeln

[reef kestrel]

vill du först skriva om den här på ”vanlig” PQ form?

[stone imp]

Så typ

Oj nu blev den sne

[rough forge]

,rccw

[rough forge]

tuff

[reef kestrel]

merci

nu blev det ett litet teckenfel

titta noga på hur på formeln ser ut

aha jag skojar

såg inte minus tecknet! det ser rätt ut

[stone imp]

Sorri skriver lite knasigt ibland

[reef kestrel]

ingen fara! nu har du två lösningar för x eller hur

men här blir det viktigt att tänka efter: x representerar ju sidlängden av en triangel

[stone imp]

När jag genomför q delen det blir väl +2 då det är -(-

Nu är jag lost

[reef kestrel]

vad har du fastnat på?

[stone imp]

,rccw

[reef kestrel]

precis det ser rätt ut

[stone imp]

Hur tar jag roten ur 3 det blir decimal

[sturdy bane]

english please

[reef kestrel]

det behöver du inte beräkna

[smoky gull]

,w 4x+4=2x^2

[jolly parrot]

Results provided by WolframAlpha

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

Requested by u_seen_yaj

[smoky gull]

not neccesary

[sturdy bane]

its in the rules

[smoky gull]

idts

[stone imp]

She’s helping me in Swedish it’s easier esp with formulas

[reef kestrel]

rules say speaking in other languages is okay :)

[stone imp]

Jahaaaa

[sturdy bane]

yea im not swedish tho

[stone imp]

You’ll live

[sturdy bane]

https://tenor.com/view/cat-licking-popsicle-gif-3122588929262013745

[peak flint]

me neither

lol

[stone imp]

Så nu har vi ena x på 1-roten ur 3 och 1+roten ur 3

[reef kestrel]

om du läser uppgiften så vill man ju komma fram till en area som också är på liknande form

med roten ur 3 liksom

yes exakt men här måste du kolla på vilka x som är möjliga

[stone imp]

Hur gör jag detttt

[reef kestrel]

en liten följdfråga: kan en triangel eller kvadrat eller vilken form som helst ha en sidlängd som är negativ?

[stone imp]

Jag visste inte ens att de kan vara negativa

[reef kestrel]

det var frågan, tror du de kan vara negativa?

[stone imp]

Nej det skulle ju va konstigt

[reef kestrel]

precis, det kan man se enkelt i verkligheten att det är skumt om saker kan ha negative längder och så

[stone imp]

Jahaaa så det kna inte vara 1-rpten ur 3

[reef kestrel]

då måste vi kolla att de x vi har hittat är positiva eller hur

precis!

[stone imp]

Så nu har jag 1-roten ur 3

Och det är ett av kateterna

[reef kestrel]

plus ska det vara

[stone imp]

Som bara är c eller

Just det

Hoppsan

[reef kestrel]

ingen fara! lätt att blanda ihop

alright så nu har vi alla sidlängder på triangeln

[buoyant cypress]

We can talk other lang in help channels

[reef kestrel]

så nu är du i princip färdig. det som kvarstår är ju beräkning av arean

[stone imp]

Jaaaa tack sååå jätte mycket

[reef kestrel]

det var så lite :)

[stone imp]

Vänta jag ska försöka lösa arean ibland så gör jag fel i de enklaste områden

[reef kestrel]

ingen stress

[stone imp]

Har jag skrivit ut detta rätt

,rccw

[reef kestrel]

ja! nu gäller det att utveckla

[charred fern]

just multiply

[stone imp]

Hur multiplicerar jag in det där

[reef kestrel]

Tips: skriv parentesen som 3+roten ur3 så blir det lite enklare

[calm wing]

exakt det är fel utan parantes

[stone imp]

Jahaaaa

Så nu skrev jag det som 3

[reef kestrel]

ja och glöm inte parentesen runt första termen

[stone imp]

Vilken

[reef kestrel]

(1+sqrt(3))(3+sqrt(3))

så ska det stå

eftersom du multiplicerar hela basen med hela höjden

[stone imp]

Så här?

,rccw

[stone imp]

Delat på 2

[reef kestrel]

ja och delat på 2 också

[reef kestrel]

we typically domt show the answers right away but try to guide!

[stone imp]

Hur gör jag nu om jag ska multiplicera både parantesen

[reef kestrel]

det blir liknande som kvadreringsregeln fast lite mer komplicerad

[stone imp]

🥲

[charred fern]

oh yeah sorry i came back after a long time

[reef kestrel]

en minnesregel är FOIL: first outer inner last

jag ska hitta en bild brb

[stone imp]

Jahaaa jaa det kan jag men om jag multiplicerar med roten ur ändras siffran normalt som 3*roten ur 3 blir bara roten ur 9?

[reef kestrel]

inte riktigt utan det blir som du skrev 3*roten ur 3

om du vill skriva det som en rot gäller att

$3 \cdot \sqrt{3} = \sqrt{9} \cdot \sqrt{3}$

Hänger du med på vad jag gjorde?

[calm wing]

(det är sqrt(3) på den andra rotten)

[stone imp]

Tyyyyp inte

[reef kestrel]

aha oops

[jolly parrot]

plantsyavi

[reef kestrel]

så! tack

[stone imp]

Blir det så

,rccw

[reef kestrel]

de två sista termerna blir fel

eller snarare näst sista

sista ser ok ut men du kan också bara ta bort roten ur

[stone imp]

Hur menar du

Så det bara bli 3^2

[reef kestrel]

bara 3

$(\sqrt{4})^2$ vad är det

[jolly parrot]

plantsyavi

[reef kestrel]

sqrt4 =2, och sedan 2 i kvadrat blir 4

[stone imp]

Jag hänger inte med 🥲

[reef kestrel]

vad känns förvirrande?

[stone imp]

Vad som händer med de sista två termerna när de multipliceras

[reef kestrel]

börja med detta problem

vad blir dedär som jag skrev?

[stone imp]

Blir det roten ur 16

[reef kestrel]

ja precis, och vad är roten ur 16?

[stone imp]

Jahaaa

4

JAHA

Jag förstår

[reef kestrel]

så det jag ville förklara var att när du har roten ur och sen i kvadrat - det är som de kannelerar

[stone imp]

Men var roten ur 9 i mitt svar även fel

[reef kestrel]

ja den var fel,

[stone imp]

Hur skulle jag göra

[reef kestrel]

jsg skulle rekommendera att du först adderar lika termer

så du har ju nu 3 och 3 som du kan addera va?

[stone imp]

3+3?

[reef kestrel]

ja exakt

och sedan kan du också addera termer som har sqrt(3) bredvid

[stone imp]

Men roten ur 9 är kvar där vad ska jag ersätta den med

,rccw

[reef kestrel]

nja nästan men den termen ska inte vara sqrt9

Det ska stå 3*sqrt3

[stone imp]

Jahaaa

,rccw

[reef kestrel]

ja exakt. ser du något du kan förenkla nu

[stone imp]

Inte riktigt

[reef kestrel]

har du koll på hur man faktoriserar saker?

[stone imp]

Jag tror det

De va ju det med att bryta ut eller hur

[reef kestrel]

precis

bryt ut roten ur 3 från de två sista termerna

[stone imp]

Hur menar du

Hur bryter jag ur från en rot

[reef kestrel]

så du skriver

du kan bara behandla roten som ett tal

det är ju bara ett tal eller hur?

[stone imp]

Ja

[reef kestrel]

om du slår in det i miniräknaren så får du ett tal, runt 1.75 eller så tror jag

I matematik när du skriver sqrt(3) är det bara ett exakt sätt att skriva det talet

Som har många många många decimaler

men det blir inte magiskt på något sätt, det är fortfarande bara ett tall

[stone imp]

Jaaaa

Jag skulle betala dig att bli min matte lärare jag fattar lite mer nu

[reef kestrel]

hahaha tack 🥹🥹🥹

[stone imp]

Jobbar du som lärare kanske? Du förklarar ganska bra

[reef kestrel]

det gör jag faktiskt inte! jag pluggar på universitetet

[stone imp]

Jahaaaaa

Okej så hur gör jag med rötterna

Om jag bryter ur 3

[reef kestrel]

roten ur 3 ska du bryta ut

men bara från de 2 sista termerna eller hur? det är bara de som innehåller det

[stone imp]

Jaa

Eller Hur har ju två roten ur 3

[reef kestrel]

det här kanske hjälper: kan du förenkla $2a+6a$?

[jolly parrot]

plantsyavi

[stone imp]

2a+32*a

?

Ojsan

Nu gör discord det knasiga

[reef kestrel]

inte riktigt

[reef kestrel]

hur tänker du när du försöker förenkla det

[stone imp]

,rccw

[stone imp]

Så ?

Ojjj

Ni blev det fel igen

3 istället för 6

Tror jag

[reef kestrel]

ja exakt det ska vara 3

och det som står inne i parentesen blir 1+3

så totalt får du 8a eller hur

[stone imp]

Jaa

[reef kestrel]

så vi gör en till

$b+3b$

[jolly parrot]

plantsyavi

[reef kestrel]

du behöver inte bryta ut varje gång, du kan bara addera det som står framför b direkt

det är samma sak såklart

[stone imp]

Jahaaa

De var ju smidigare

[reef kestrel]

men förstår du varför det är samma sak?

[stone imp]

Ja jag tror det

[reef kestrel]

okej! vad får du här?

[stone imp]

B(1+3)

[reef kestrel]

precis och det blir?

[stone imp]

4b

[reef kestrel]

vi vill oftast skriva det så enkelt som möjligt, så då är det bäst att skriva såhär

[reef kestrel]

okej, nu om du bara ersätter b med $\sqrt{3}$ vad får du då?

[jolly parrot]

plantsyavi

[reef kestrel]

alltså vilket uttryck får du

[stone imp]

,rccw

[stone imp]

Så?

[reef kestrel]

känner du igen det?

[stone imp]

JAHA

jaaa

Så

Det blir bara

Roten ur 4

[reef kestrel]

näää

det enda vi gjorde var att ersätta b med roten ur 3

så om vi vet att b+3b=4b

och b är roten ur 3, vad får man?

[stone imp]

Roten ur 3 * 4?

[reef kestrel]

njaaa

eller vänta

menar du $\sqrt{4 \cdot 3}$?

[jolly parrot]

plantsyavi

[stone imp]

Nää..

[reef kestrel]

eller $4\sqrt{3}$

[jolly parrot]

plantsyavi

[stone imp]

Hur är det här bara 2 a poäng

[reef kestrel]

vi är snart färdiga

[stone imp]

Hur kom 4an

[reef kestrel]

b+3b=4b

det enda jag gjorde var att jag plockade ut b och så satte jag in roten ur 3

eller hur?

[stone imp]

Vi har ju 3* roten ur 3

[reef kestrel]

vi ville ju räkna ut

$3\sqrt{3} + \sqrt{3}$

[jolly parrot]

plantsyavi

[reef kestrel]

right?

[stone imp]

Vänta jag förstår hur det blir 4 gånger roten ur 4

3

[reef kestrel]

så då jämför jag med 3a+b=4b

där jag bytte b mot roten ur 3

[stone imp]

Okej ja då har vi 4*roten ur 3

[reef kestrel]

precis så vad har vi då för area uttryck?

[stone imp]

Ja duuu

[reef kestrel]

om du går upp lite hittar du resten

[stone imp]

Jahaaa

Då blir det

,rccw

[reef kestrel]

Njaa nu har du tagit med roten ur 3 igen

Men den finns redan i 4 roten ur 3

Vi adderade ju dem redan

[stone imp]

Så hur ska det bli

[reef kestrel]

kolla igen på detta uttryck

$6+\sqrt{3} + 3\sqrt{3}$

[jolly parrot]

plantsyavi

[stone imp]

Jaha

[reef kestrel]

sen sa vi de två sista blir $4\sqrt{3}$ eller hur?

[jolly parrot]

plantsyavi

[stone imp]

6roten ur34

Men va fan

[reef kestrel]

jag tror att du har rätt! om du menar såhär

[stone imp]

,rccw

[stone imp]

Den under

[reef kestrel]

precis

nu glömde vi att dela på 2

för att arean av en triangeln är ju halva

[stone imp]

Jaaa så

[reef kestrel]

så om vi bara delar på 2 är vi klara

[stone imp]

Omggg

HALLELUJA

gud det tog en timme

[reef kestrel]

bra jobbat! mitt tips skulle vars att du skriver upp en checklista på hur du löste uppgiften

det brukar hjälpa mig när jag lär mig nya saker

t.ex. steg 1: använd Pythagoras, steg 2: pq regeln

osv

[stone imp]

Kan jag lägga till dig om det finns små grejer som jag fastnar på ibland

[reef kestrel]

absolut! jag brukar dock inte vara online så ofta så du vet

[stone imp]

Det görrr inget de är inte jag heller men jag kollade lite på a uppgifter innan provet imorn och tänkte ändå att jag provar på mig några

I alla fall tack igen ha en trevlig kväll 🤗

.close

[pearl pond]

Channel closed

Closed by @stone imp

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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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• Be polite and have a nice day!

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[mortal crypt]

stuck a bit here, can i get a very vague hint which can direct my thinking and expand my mind towards the correct solution? I got first term and common difference by using the 2 terms provided. But im a bit confused how do i find the rational term

[plush bramble]

get the general term first and show your work here

[mortal crypt]

a moment, uploading it

A bit correction down, it's 31 - 64log10²

[plush bramble]

yea just plug your d

[plush bramble]

and corrected a

into a + (n-1)d and solve for when it must be rational

[pearl pond]

@mortal crypt Has your question been resolved?

[mortal crypt]

A moment I'll brb

[pearl pond]

@mortal crypt Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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• 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.

[pearl pond]

Please don't occupy multiple help channels.

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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[visual kernel]

So basically any way that I create any graph on my own?

[pearl pond]

Please don't occupy multiple help channels.

[honest vigil]

wait, which channel do you want to use? you have two channels active - pick one to proceed in.

[left steppe]

can you reword the question?

[visual kernel]

Oh so is there any way that I can create a graph in any kind

Like how I can create on my own

[honest vigil]

hi OP, if you have decided to proceed in this channel, you may want to close #help-36.

[left steppe]

are you talking about formulas?

@visual kernel come back here

"Ok yo so if I want to create a grafh lets say like a infinity symbol can I create it on my own?"

err

[visual kernel]

Yea

[left steppe]

like yeah

like drawing shapes?

[visual kernel]

Thats my qn

Yep

[left steppe]

yes you can create it on your own by using formulas

[visual kernel]

? Like curve and lines

[left steppe]

heres an example

[visual kernel]

Cant you create it by differentiation?

[visual kernel]

Nahhhh thats cool

[visual kernel]

?

[left steppe]

i think so i never tried it

[visual kernel]

Oh like the maxium point and minium point inflection points those stuff

[left steppe]

sure

[visual kernel]

Oh yea but it always bit tricky

[left steppe]

if you are done do .close

[visual kernel]

.close

[pearl pond]

Channel closed

Closed by @visual kernel

Use .reopen if this was a mistake.

[visual kernel]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[visual kernel]

.close

[pearl pond]

Channel closed

Closed by @visual kernel

Use .reopen if this was a mistake.

[sterile python]

!noclopen

[pearl pond]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[spare rivet]

im not sure where to even start to get this, is it some sort of am gm or cauchy schwarz, im very new to these type of inequalities

[pearl pond]

@spare rivet Has your question been resolved?

[austere tinsel]

Hm, I don't know if it's the intended solution, but I have a start on a geometric solution

If you take a 2x2 square and mark the sides according to the length of a,b and c,d, you could make right angle triangles with square side lengths (a,c), (a,c), (b,c), and (b,d)

[spare rivet]

interesting

[austere tinsel]

I honestly don't know if that's anything

But it's kinda neat

[lament pecan]

woah that's pretty dope

[austere tinsel]

Do you have further context for the question though? If it's for a class, what topic is it?

[spare rivet]

its about inequalities regarding qm-am-gm-hm, cauchy schwarz etc.

should be just clever manipulation for this

this was left as a particular exercise

cant even wrap my head around what to do tho

[fresh solar]

i tried applying ONLY am-gm but it didnt help

[lament dawn]

you can either use cauchy schwarz or QM-AM to bash this out

use cauchy on each pair such that you obtain the higher bound
(a^2 + c^2)(b^2 + d^2)
and
(a^2 + d^2)(b^2 + c^2)

[spare rivet]

wait its actually cauchy wtf

didnt think that

i actually tried with the pairs like you said but i couldnt think

[lament dawn]

you should get LHS is lesser or equal to (spoilers, try to work it out):
||LHS <= [(ab + cd)^2 + (ad + bc)^2][(ab + cd)^2 + (ac + bd)^2]||

[lament dawn]

err use the idea that (ad + bc) + (ac + bd) is actually just = (a+b)(c+d) = 2*2 = 4

[spare rivet]

i did it with something like ()() and ()() for cauchy so i was wrong

[spare rivet]

omg yeah idk why i didnt see this

[lament dawn]

no worries

after that you shoulllddd be able to maximise i think

[spare rivet]

i did see ad+bc+ac+bd=4 but didnt manage to link it to the equation

insane

thanks

[lament dawn]

QM-AM actually has a neater solution here i think

apply it to each bracket and you also get the product of the square of like each pairs

[lament dawn]

and then you just use this to bound it

[spare rivet]

LHS <= ((ab + cd)^2 + (ad + bc)^2)((ab + cd)^2 + (ac + bd)^2)

right

[lament dawn]

uh

a^2 + c^2 >= (a+c)^2 / 2

so on so on so on for the other pairs

then you get LHS >= (a+c)^2 (a+d)^2 (b+c)^2 (b+d)^2 all over 16

so now you just bound (a+c)(a+d)(b+c)(b+d)
to get 2(ab + cd) + 4

by amgm ab and cd are lesser or equal to 1

[spare rivet]

wait how do you bound (a+c)(a+d)(b+c)(b+d)

im not thinking very well today

[lament dawn]

(a+b)(b+d) = ab + ad + bc + cd

doing so for the other pair

adding the two results gets 2(ab + cd) + 4

[spare rivet]

oh

oh

sorry i need sometime to process all this

[lament dawn]

no worries 💛

wait this is like a gimmick i think using cauchy is the most intuitive method but QM-AM is a bit faster

yep i got 25 too

[spare rivet]

im so fried
after using cauchy to get LHS <= ((ab + cd)^2 + (ad + bc)^2)((ab + cd)^2 + (ac + bd)^2) what can i do to simplify

[spare rivet]

not sure where this would be applied

[lament dawn]

let x = ab+cd
y = ad + bc
z = ac + bd
y + z = 4
the expression thus is
(x^2 + y^2)(x^2 + z^2)
max this given y + z = 4, max occurs when y = z = 2
as such the equation is (x^2 + 4)^2 and ab and cd is lesser or equal than 1 such that x is lesser or equal to 2

[spare rivet]

wait shouldnt it be >= since its cauchy

i think lagrange would work here instead

qm am in this way also finds the lower bound instead no

idk anymore

lagrange yields lhs <= 64

very loose

[pearl pond]

@spare rivet Has your question been resolved?

[spare rivet]

maybe another day

.close

[pearl pond]

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[shadow elm]

i tried deriving frulani’s theorem from feynman’s method
but idk how to do partial derivatives

[lament dawn]

partial derivatives is kinda like a basic derivative but you treat the other variables as constants

suppose f(x, y) = x^2y + 3xy^2

so for example the partial derivative wrt to x is just 2xy + 3y^2

[shadow elm]

i got that
but when im integrating a partial derivative

its not behaving the way i want it to behave

[lament dawn]

where have you gotten til?

[lament dawn]

the integral is divergent so your definition of f(0) is wrong

[silk drift]

why what

[lament dawn]

the integral needs to be evaluated from -inf to 0 because its an improper integral at both bounds you must use limits for both lim b -> 0- ln|b| - lim a -> -inf ln|a|

you end up with -inf - inf which means the integral simply diverges

you cannot take the bounds of a divergent integral and then condense it into a single limit

[pearl pond]

@shadow elm Has your question been resolved?

[pearl pond]

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[royal galleon]

6.4 is $f(xy) = f(x) + f(y)$. So we hold y fixed and differentiate. Then he says it holds for all $y\neq 0$ but aren't we still holding y fixed?

[jolly parrot]

BigBen

[royal galleon]

or is it because you can hold the y fixed at any value

[summer imp]

Yeah. You can do this for any fixed value of y except 0

[royal galleon]

ok thx

.solved

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[tepid whale]

For this question here can i just multiply 96x32x48 and then just square root it to get the volume

[light helm]

yes

[tepid whale]

Ok thank you

[split void]

close the channel by typing .close if you're done

[tepid whale]

.close

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[rare holly]

how to do this

[sterile python]

What have you done?

[rare holly]

idk how to start

i only identified 111

where it is

[merry carbon]

And if you know the bearing of W from U, you also should easily be able to find the bearing of U from W (i.e. going the other way), right?

[honest vigil]

try bashing more angles out, as chartbit mentioned. there are quite a few you can get from the given information.

[sterile python]

Find UVW angle first might be a good start, OP!

Then you can find something useful in triangle UVW too 👀

[merry carbon]

(that might be a bit more challenging to start off with )

[sterile python]

https://tenor.com/view/shikimori-gif-25251680

Challenges are fun ☺

[pearl pond]

@rare holly Has your question been resolved?

[rare holly]

i got it

.close

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[woven river]

hi guys, I answered part b already and would like suggestions for a more refined solution

[woven river]

ill send my ans in a bit

I believe my answer is too long and ugly, so thats why I seek help here

ping me when answering as I will dive to other question first thanks

[warm patio]

if the infimum existed then there would be some lower bound on ncos(npi/2)

since you've shown that for n = 2k as n->infty we have that this T_2k goes to -infty no such bound exists

[woven river]

but not every even n goes to infinity

-inf

[warm patio]

we only need an infinite sequence that goes to -inf

[woven river]

thats why I created the "when n is even"

[woven river]

how to filter that

2pi, 6pi, 10pi,...

ah so do I make when n = 2 +4k with k = 0,1,2,..

?

😎

oh so by making a sequence out of the thing is enough to prove it?

out of T i mean

[warm patio]

yes

that gets arbirtarily small

so goes to -inf

[woven river]

I see, I understand it now

any more suggestion?

[warm patio]

dont write lim on the left, because this looks like you are evaluating the limit and you are not

[woven river]

I agree

should I name the sequence

like S, M, P

@summer imp I would gladly accept ur comment if u have any cuz I saw u typing

[summer imp]

Your first pic does have some issues in that you're defining the limit piecewise with respect to n, but n isn't fixed it's taken to infinity as k pointed out. You can prune a lot of that first pic by just writing the general value of n cos(npi/2) explicity depending on n, and then you can explain your choice of sequence accordingly.

[woven river]

I agree

[summer imp]

As for your sequence you could name it something like s_k and give the corresponding value of n*cos(npi/2) for n = 2 + 4k.

[woven river]

wait

how is it

hmm but I don't know if there should be something more before reaching to the conclusion that lim tends to -inf

I added extra brackets near cos for clarity ig

[warm patio]

I would evalute this cos first and then take the limit it is not immediately clear that this limit is -infty

If you're using this lim notation you have to keep writing lim until you evaluate it

so there should be a lim k->infty in the last line before (2+4k)cos((2+4k)pi)/2

[woven river]

oh I see

[warm patio]

alternatively you could skip writing lim and use the notation when you write an arrow and n->infty above it

[woven river]

the expression after the first "=" right?

[warm patio]

yes

[warm patio]

like this $w_n \xrightarrow{n \to \infty} -\infty$

[jolly parrot]

k

[woven river]

wha

isn

isn't the lim thing necessary for wn

[warm patio]

you could write the last line like this $w_{2+4k} = (2+4k) \frac{\cos (2+4k)\pi}{2} \xrightarrow{k \to \infty} -\infty$

[jolly parrot]

k

[warm patio]

its just other notation

[woven river]

man, ok but one last concern

do we really need to evaluate it

[warm patio]

wdym

[woven river]

the limit

before reaching -inf

[warm patio]

how else do you reach -inf

[woven river]

no like evaluating the whole cos thing

like u said

[woven river]

here

[warm patio]

well this cosine term could be zero

its not immediately clear it is not

[woven river]

no it cant

2+4k's sequence is 2, 6, 10, and plugging in to the cosine thing we always get -1

but do u think we should solve the thing before writing -inf

oh shoot

[warm patio]

right, but we should write this

[woven river]

yea I might fumble

wait ohh

I made a mistake writing the 1/2

not inside the cosing

[woven river]

noted

ok thanks bro

[warm patio]

youre welcome

[woven river]

I think i can hand;e this

by

e

.close

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[peak spire]

Hi guys could someone help me please? I'm doing a trig question and i dont like using CAST diagrams or using the graph

[peak spire]

and i use this to find the solutions

this is what the mark scheme says and idk why its +- 18.4 when i just put +18.4 , or why its +-180 when i just put +180, also idk where they got 161.6

[unborn abyss]

well so first of all you should have $\cos\theta = \green\pm\s\f9{10}$

[jolly parrot]

teacup kitten

[peak spire]

Oh wow how could i forget 🤣

Okayyyy i got positive 161.6

thank you for that

but how did they get +-

i mean -161.6

[humble tendon]

do you know cos(x) = cos(-x)

[peak spire]

no I don't

is that a trick or something

a hack

[humble tendon]

cos gives the same output for both positive and negative

you can see it in graph

[peak spire]

so whenever i find solutions it'll always be +-

?

[humble tendon]

for cos, yes

[peak spire]

only for cos

ok so thats just a rule thank u

so that explains +-18.4

what about sin?

whys it +-180

[humble tendon]

so for sin(x)=sin(180-x)

and sin(x)= -sin(180+x)

[peak spire]

so i should've done 0-180 as well?

[humble tendon]

180-0

[peak spire]

so its still the same as cos then?

still has positive and negative

[peak spire]

according to this line

[humble tendon]

but in this case, it doesn't make a difference as the value is 0

[humble tendon]

wdym

[peak spire]

okkkk thank you but isnt it the same as the cos rule of this then?

So basically

what im infering from this is that i should take both negative and positive values from sin and cos

but u said its only for cos

to take the negative value

[peak spire]

so why is this a rule

[humble tendon]

for cos(x)=cos(-x)
sin(x)=-sin(-x)

[humble tendon]

it is better to understand and also remember this by graph imo

[peak spire]

🤣 ok i probably should

Im tryna deep this

[humble tendon]

this might be of help

[peak spire]

Ohh the cast diagram

Yeah i should probably learn that then

[humble tendon]

yes

[peak spire]

What about b

https://media.discordapp.net/attachments/908078029778083872/1489292578984562828/image.png?ex=69cfe34c&is=69ce91cc&hm=cdac24ff097040642f82bc77ba5dcda2ce78e21780fd9fc18e0293a3665599dc&=&format=webp&quality=lossless

How would i go about doing part b?

[humble tendon]

notice 2x-50=2(x-25)

then use the formula for sin2x

[peak spire]

Oooo ur so smart

[humble tendon]

you'll get there eventually

it's just about practice

[peak spire]

🤣 Thank youuuu

Ur right

[humble tendon]

np

glad to help

[peak spire]

Ok i think im good to goo, thanks for the helppp

.close

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[hearty thicket]

Can someone verify this? I want to make sure d_omega is not exact

[sonic patrol]

This look correct

[hearty thicket]

Thanks!

.closed

[sonic patrol]

.close

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[sharp smelt]

I don't get how $L( \mu \mid x_1 \dots x_n)≥k(x_1 \dots x_n)$

[jolly parrot]

Wai

[sharp smelt]

and I'm quite confused by what the order relation even is ehre

like K(x_1,x_2...,x_n) makes no snese?

<@&268886789983436800>

[plush bramble]

k looks like a function of n variables

[sharp smelt]

I see

thanks

hmm, the choice of k* is intriguing

eh, make sense

thanks

.close

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[calm wing]

evans and rosenthal 🗣️🔥

goated book

[sharp smelt]

hell ya

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[shadow elm]

hey so
i tried to derive frulani's formula again using feynman's method
but when i integrate a PDE to basically get a function
the constant of integration should prolly be a function of the other variables

Can someone give me hints on how to derive this (i dont wanna look up a proof cuz doing it myself will be more rewarding i think)

[tropic mesa]

(Bear with me I only do partial integrations when it comes to field conditions in physics)

(So therefore I might not be fully accurate)

Essentially what you're trying to do is to differentiate under the integral sign (as is called feynman's trick) which you then use to simplify into a known integral ya

[shadow elm]

yes

[tropic mesa]

that works
and you are correct
if you integrate PDEs you do get a functional constant
But I don't think here it's necessary

you should be getting a f'(ax)/a term

check it

[shadow elm]

ok so the problem is

$$ \int_{-\infty}^0 \frac{e^{mx} - e^{nx}}{x} dx$$

[jolly parrot]

parthisjoking

[shadow elm]

i assumed the integral to be a function of m and n

so

$$ F(m,n) = \int_{-\infty}^0 \frac{e^{mx} - e^{nx}}{x} dx$$

[jolly parrot]

parthisjoking

[tropic mesa]

okay

continue

[shadow elm]

now i wanna do partial derivatives

since i know F(0,0) = 0

idk how to go forward from here

failed attempt number 7 ig

[tropic mesa]

give me a moment to process this lol

So you're partially integrating

and adding a constant function and also c

I think you're double counting somewhere

Are you sure you didn't double count anything?

wait a second

what's ln(0) doing there

@shadow elm

what is f(n)

did u find that?

[shadow elm]

idk how to do thay

this is my first time integrating a pde tbh

and maybe my 3rd time partially differentiating anything

[shadow elm]

i put 0 everywhere
F(0,0)

[tropic mesa]

just gonna send u this

[tropic mesa]

yeah red flag how is 0 in ln bro domain restrictions

[shadow elm]

not really tbh

if f(n) can be something

that cancels it out

we are fine ig

[tropic mesa]

yeah well ln(0) goes to infinity

infinity minus infinity is not zero

it's a undefined thing
it's a indeterminate form

Partial differentiation still obeys limit rules much like actual differentiation

so i think 0 being in ln..... would not work out

[shadow elm]

can you gimme a hint on how to approach this proof then

i dont wanna just look up the proof
since it is too much dopamine in a single day for me

[tropic mesa]

you got a funtion of m earlier

that's perfectly alright

If I were to just wing it I'd say m and n are symmetrical

and f(n) is just a -ln(n)

but you need a full fledged proof

[shadow elm]

no they arent 🥲
F(m,n) = - F(n,m)

[tropic mesa]

your integral is e^mx - e^nx no

do partial differentiation wrt n

i clearly see i will get a -1/n term

[shadow elm]

yea

but when i do it wrt m

i get a 1/m

[tropic mesa]

add them ig

f(m,n) anyway

split and add

one as a func of m

another as a func of n

ln(m/n)

[shadow elm]

that is definitely the answer , yes.

but

i dont know if the proof is very rigorous if i just add em
im missing something

[tropic mesa]

dude it's just a conservative field it obeys daum/daun=daun/daum

[shadow elm]

whats a conservative field

[tropic mesa]

remember electric field and magnetic field from twelfth grade?

Yeah well they follow this conservative fields idea

I could give u a physics explanation

but mathematically speaking

[tropic mesa]

if they obey this, then they are (mostly) conservative
(provided the field is defined on all points, of course)

(sometimes a loop may enclose the origin, so, well, if it does that, it's non-conservative)

I don't think you have to worry about that though since m and n

are both positive

Anyways I think I shall leave it to the more experienced people here to explain
Sorry but I can't exactly go beyond that

@shadow elm

[shadow elm]

thx 😇

i might need to ping <@&286206848099549185> sry

[tropic mesa]

No problem keep working chat

[shadow elm]

which jee

jee mains ?

so jee adv ?

IIT JAM you mean ?

oh
good for u

we might be redirecting tbh

!redir

[pearl pond]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[shadow elm]

its okay

its not that deep but

thanks

[grave mirage]

Use double integrals instead ig?

[shadow elm]

ok so

how

[grave mirage]

Write

f(ax)-f(bx) as

Integral of f'(t) dt under the limits ax and bx

Then switch the integral signs

[shadow elm]

ah

lemme try

[grave mirage]

Ye

[shadow elm]

did i uncook

oh wait

[grave mirage]

No no

Frullani is

f(ax)-f(bx)/x

You took f(x)/x as f(x) here

[shadow elm]

.

[shadow elm]

ah

[grave mirage]

Ye Ik

[shadow elm]

shi-

im new to this multivariate stuff

sry

lemme try again

[grave mirage]

Nah pretty good

[grave mirage]

Yes

Try making double integrals

[shadow elm]

,rcw

[shadow elm]

now the d/dx goes inside ?

nvmm

wrong again

wait wait

no comments

[grave mirage]

It would be better if you didn't let both integrals be in terms of x

[shadow elm]

,rccw

,rccw

[grave mirage]

Hmmm

[shadow elm]

but isnt this still wrong

[grave mirage]

yeah

[shadow elm]

cuz of the x

down below

[grave mirage]

I'll get pen paper

[shadow elm]

wait wait wait

i think i got it

i took the upper thingy as f(x) as u said

,rccw

[shadow elm]

idk if this is correct either

[grave mirage]

Yupppp

This

[grave mirage]

Correct

[shadow elm]

nvm

i didnt even know the full statement of frullani

proving the general form

would be hard imo

[grave mirage]

What are you referring to?

[shadow elm]

ok so

the problem i wanted to solve was the e^mx thingy

[grave mirage]

But ye y=tx and then change order of integration

[muted orbit]

<@&268886789983436800>

[grave mirage]

General f(x) is solved same way

[grave mirage]

Here you're solving the general formula

[shadow elm]

i took f(x) as e^mx

[grave mirage]

Yess

[shadow elm]

sry f(y)

[grave mirage]

f(y) is just e^y no?

[shadow elm]

no its e^my na then

[grave mirage]

It becomes that after you put limits

[shadow elm]

idk whats happening

🥲

[grave mirage]

Alr then

[shadow elm]

lemme restart and prove

[grave mirage]

Yes

[shadow elm]

the general form first

[grave mirage]

Yeah

Exactly

[shadow elm]

how do i "switch the integrals"

[grave mirage]

You first need to make a substitution

to get rid of the denominator x

And x in the limits

[shadow elm]

i cant think rn

there used to be a specific sub

to convert any limit to

1,0

[grave mirage]

Try y=tx

Then you switch integrals

Basically it means to integrate in a different order

You'll get smth like this

[shadow elm]

oh

x is a const in this int

i was applying prod rule

💔

[grave mirage]

f' btw

[grave mirage]

Rightt

[shadow elm]

that was pretty elementary tbh

just couldnt think cuz

[grave mirage]

True

[shadow elm]

i though of product rule

oh

now since

fxn is both in

x and t

i can switch

[grave mirage]

Yes

You need to do that or you'd have to integrate f(x)

[shadow elm]

ohhh

done

i int that

and back sub

y

so i get ln from that

[grave mirage]

Yeee

[shadow elm]

ugh

this was pretty much

basic stuff

💔💔

[grave mirage]

True

[shadow elm]

damn

also

can u tell me

when we convert a double int to polar

is dx dy and dy dx the same thing

[grave mirage]

Well ig the best hint you could get is the structure of that curve or smth

Well that surface area ig

[shadow elm]

cuz my teacher told me
$dxdy = rdrd{\theta}$

[jolly parrot]

parthisjoking

[grave mirage]

Yes true

[shadow elm]

$is dydx=rdrd{\theta} too ?$

[jolly parrot]

parthisjoking

[grave mirage]

Sub x = rcos(theta), y = rsin(theta)

[grave mirage]

They are commutative I believe if the integral comverges

[shadow elm]

well i have solved only

area under normal distribution

using this

[grave mirage]

Yeah that works there

It's related to jacobians and shit

Forgot exactly how

Search up fubini's theorem

[shadow elm]

,rcw

[shadow elm]

did i do smtn wrong

💔

[grave mirage]

You didn't integrate the first integral crazy

There should be a t in the denominator

f'(xt)dx

xt = k

[shadow elm]

no way

yea yea

reverse chain rule thingy with integrals

[grave mirage]

Yepp

Substitution is reverse chain rule

[shadow elm]

only linear sub is

[grave mirage]

No any sub I think

[shadow elm]

hmmmmm

yeah

[grave mirage]

int g'(f(u))*f'(u) = g(f(u))

D g(f(u)) = g'(f(u))*f'(u)

[shadow elm]

done fully

🙏🙏

finally

[grave mirage]

Nicee

[shadow elm]

i might go back to the basics and derive reduction formule + walli's formula now

[grave mirage]

Yeah walli's formula is fun

To derive

Without using beta fxn that is

[shadow elm]

i found a very interesting dounle integral today

its on daily integrals website

[shadow elm]

isnt it just IBP

[grave mirage]

Yeah IBP and then recursion

[grave mirage]

Lemme see

[shadow elm]

the toughest one

with floor and ceiling fxn

[grave mirage]

Can't see

[shadow elm]

nvm

it got reset

"daily" integral

[grave mirage]

🥀

[shadow elm]

damn

i cant seem to derive

wallis

i thought it was ez

reduction formule are ez

not wallis tho

[shadow elm]

HOW DO I USE RECURSION

[pearl pond]

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[dense storm]

How 1e wrong

[dense storm]

Answer had ln(2-x) + fraction instead of ln(x-2) - fraction

[flint basalt]

your work looks correct to me

[unborn abyss]

well first of all ln|2-x| is equal to ln|x-2|

is that why maybe?

[flint basalt]

^ this however idk why they would write it like that

[unborn abyss]

and also $\f{2}{x-2} = -\f2{2-x}$

[jolly parrot]

teacup kitten

[unborn abyss]

so it's probably equivalent

[dense storm]

Oh

I didnt see they had 2-x on bottom

Alr thanks

.close

[pearl pond]

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Closed by @dense storm

Use .reopen if this was a mistake.

[pearl pond]

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[swift spindle]

can someone verify thi

[unborn abyss]

have you tried evaluating it at that x value?