help-26

[olive charm]

it fin

[sharp shard]

16+36

[olive charm]

e

52

[sharp shard]

Do you know how to simplify squareroots?

[olive charm]

no

[sharp shard]

No worries

Here we have √52

[olive charm]

ok

[sharp shard]

√52 can be broken down

It can be written as √(2 × 26)

And we break the 26 again

[olive charm]

into 13

[sharp shard]

√(2 × (2 ×13))

[sharp shard]

Yess

[sharp shard]

So it's √(4×13)

√4 is 2

[olive charm]

yeah

[sharp shard]

So it's 2√13

[olive charm]

ohh

thank you

[sharp shard]

No problem!

[olive charm]

i tried something like this one and i got it wrong

[sharp shard]

find the other length of 7,10 triangle

[olive charm]

12.2

[sharp shard]

You don't need to go in decimals

[olive charm]

so just 12

[sharp shard]

If your questions don't explicitly mention it

[olive charm]

oh ok

[sharp shard]

The length should be √149

[olive charm]

oh yeah

[sharp shard]

Keep it in root

It's better

[olive charm]

ok

[sharp shard]

Now find the x

[olive charm]

√149+8^2?

[sharp shard]

Close it's (√149)^2 + 8^2

[olive charm]

oh ok

[sharp shard]

Is this step confusing to you?

Why we squared the value

[olive charm]

kinda

[sharp shard]

I may explain then

[olive charm]

ok ^^

[sharp shard]

In the 7,10 triangle

pythagoras theorem states
a² + b² = c²

a = 7 and b = 10

[olive charm]

oh ok i get why now

because we only found one side of it, so in order to find c we would have to square the answer

[sharp shard]

yess root value is the original value of the side to put it in pythagoras we need to square it

[olive charm]

from the first triangle

[sharp shard]

exactlyyy

[olive charm]

ok

so √213

[sharp shard]

Correct

That's the answer

[olive charm]

ok

next question

[sharp shard]

Yes

[sharp shard]

Just remember for these kinda questions

[olive charm]

ok

use the formula twice kind of

for the other question

[sharp shard]

a² + b² + c² = d²

[olive charm]

i meant

[olive charm]

i never learnt this

[sharp shard]

It's kinda like an upgraded pythagoras theorem

Where a b c are sides

of the cuboid

and d is the diagonal

[olive charm]

ah ok

[sharp shard]

You actually I think did when the gojo guy was explaining it to you

[olive charm]

gojo guy

funny

[sharp shard]

we need the value of d

[olive charm]

so im assuming

[sharp shard]

d = √(a² + b² +c²)

[olive charm]

do something similar to what we did for a^2 and so on

[olive charm]

that guy didnt explain too well

[sharp shard]

Oh I see it's alright

[sharp shard]

Is it understandable till this then?

[olive charm]

yes

[sharp shard]

noice

Then put the values of a,b nd c

[olive charm]

a is 8, b is 5, and c is 6?

[sharp shard]

Yes

[olive charm]

for these ones, does the number and letter matter?

[sharp shard]

Not necessarily it should just have to be sides of cuboid

[olive charm]

ok

8^2+5^2+6^2?

[sharp shard]

Yes

[olive charm]

so then 125 is what i got directly from this

[sharp shard]

d² = 125

d = √125

[olive charm]

11

[sharp shard]

No no

[olive charm]

oh

[sharp shard]

When I said don't go to decimals

[olive charm]

righttt

[sharp shard]

Dont just remove em, keep the roots

[olive charm]

we have to simplify the square root i guess???

[sharp shard]

Yes

[olive charm]

sorry im not used to keeping the root

[sharp shard]

I understand no issues, it just makes it clear and simpler if roots are present

[olive charm]

it does

[sharp shard]

How can you break 125

[olive charm]

i dont know if this would be any help but its some number in between 12 and 13

[sharp shard]

Well sorry but not really, whenever you're given something like this

[olive charm]

oh ok

[sharp shard]

check if it's divisible with numbers like 2,3,5,7

It's generally from these primes

[olive charm]

definitely by 5

[sharp shard]

certainly

[olive charm]

and thats it

[sharp shard]

√125 = √(5×25)

√25 is 5

so 5√5

[olive charm]

ok

this one looks confusing

[sharp shard]

Interesting

Let me think

[olive charm]

ok

[sharp shard]

Oh I was overcomplicating it

First look at the triangle

[olive charm]

thats fine

[olive charm]

ok

[sharp shard]

Triangle CDE

Can you figure out the data that is given to you for that triangle

[olive charm]

well we know d, b

and d, c

[sharp shard]

DC = 6 cm

[olive charm]

so d e is 4?

[sharp shard]

You're right

[olive charm]

ok

[sharp shard]

Now put the values in pythagoras theorem accordingly

And get that side

CE

[olive charm]

6^2-4^2=b^2?

[sharp shard]

Yup

[olive charm]

b^2 is 20

so b is 10

[sharp shard]

√20 ≠ 10

[olive charm]

huh?

[sharp shard]

b² = 20

[olive charm]

oh ok

[sharp shard]

b = √20

[olive charm]

ohh

[sharp shard]

Yes

We got the length of EC = √20 cm

[olive charm]

ok

[sharp shard]

Now look at the triangle AED

And find the value of AE

[olive charm]

oh we're trying to figure out a c

[sharp shard]

nope!

[olive charm]

the questions says that

[sharp shard]

DA is c

[olive charm]

that sounded rude i didnt mean to be rude my bad

[sharp shard]

It didn't don't worry

[olive charm]

wait does the 3,4,5 thing apply?

[sharp shard]

Certainly

[olive charm]

oh ok so then this one would be 3

[sharp shard]

You're correct

We have AE and EC

After adding them we'll get AC

[olive charm]

ok

so 3+√20?

[sharp shard]

Correct

However we can simplify

The root

[olive charm]

to 10?

[sharp shard]

Try to break it step by step

[olive charm]

2√10 i believe

oh wait

2√5?

[sharp shard]

YEAH

Good job

[olive charm]

yayyy

[sharp shard]

(3 + 2√5) cm

[olive charm]

that went to decimals, i got 7.47

[mystic ridge]

Yeah thats right

[olive charm]

wt

[sharp shard]

It's in cm

[olive charm]

oh ok

[sharp shard]

Oh no sorry

[olive charm]

but on the image it says cm

yeah

its fine

so 7.47 no rounding

no it would be 7.5

[sharp shard]

Yes

[olive charm]

2 more questions letffff

*left

[sharp shard]

It'll be done in no time

[olive charm]

its another kite one

[sharp shard]

Let's go

[olive charm]

i think i can try this on my own

[sharp shard]

Use the same principles

[olive charm]

ok

[sharp shard]

Absolutely

[olive charm]

ec is √5?

[sharp shard]

Yupp

[olive charm]

cb is 9?

[sharp shard]

Might be, but do we really need cb here?

[olive charm]

oh my bad

im supposed to do dae

[sharp shard]

ye

[olive charm]

4^2-√5^2?

[sharp shard]

Tell me the sides you're subtracting here

[olive charm]

da and de

[sharp shard]

check their lengths

[olive charm]

wait what am i doing

4^2-2^2?

[sharp shard]

Absolutely

[olive charm]

12

[sharp shard]

√12

is the length

[olive charm]

how?

[sharp shard]

a² + b² = c²
therefore, a² = c² - b² (that you did)

therefore, a = √(c² - b²)

[olive charm]

oh ok

[sharp shard]

Are you doubtful in something

[olive charm]

so √12+√5

[sharp shard]

You may ask I'll help

[olive charm]

yeah im kinda not understanding that

[sharp shard]

It's okay, let us work with your thought process then

[olive charm]

ok

[sharp shard]

How did you think and find the length

[olive charm]

which length?

[sharp shard]

In triangle DAE

[sharp shard]

AE

[olive charm]

so db was 4 cm, so half of that was 2cm

and then da is 4

so 4^2-2^2

thats what i got

[sharp shard]

absolutely correct

[olive charm]

ok

[sharp shard]

just one final step

when you do this

you're applying the pythagorean theorem

a² + b² = c² (general formula)

where a and b are sides of triangle and c is the hypotenuse(the longest side)

In triangle DAE

The longest side = DA = hypotenuse = 4cm

And you have one of the sides that is 2 cm

Put the values in pythagoras theorem

a² + (2)² = (4)²

a² = 4² - 2²

a² = 12

a = √12

Now tell me what part is troubling you

[olive charm]

why are we adding √?

[sharp shard]

a² = 12

but a is the length of the side

not a²

[olive charm]

ohh

[sharp shard]

so to remove the square we use √ both sides

[olive charm]

oh okk

i get why now

[sharp shard]

Got it completely?

[olive charm]

i think so

[sharp shard]

Then tell me what else you're unsure of

[olive charm]

so we just have to add ce and ae now? then the we remove the √?

[sharp shard]

nope

think of it like this

[olive charm]

ok

[sharp shard]

Why do we even put √

[olive charm]

uh...

[sharp shard]

It's because we want the actual length

Not squared length

In pythagoras theorem everything is squared

both the sides and hypotenuse

always use the formula for reference

[olive charm]

ok

[sharp shard]

Does it make sense till a² = 12?

[sharp shard]

Till here that is

[olive charm]

yes

[sharp shard]

yes

If the actual length is 10 then squared length is 100

It's a huge difference

[olive charm]

that makes a lot more sense ok i got that part

thanks

[sharp shard]

yes

that's why we need actual length and add them

√12 + √5

[olive charm]

im not sure but √17 or am i being dumb

[sharp shard]

You're not dumb don't say that

How did you approach till √17

[olive charm]

i added 12 and 5...

[sharp shard]

Well it works with numeric values but not in square roots lemme show how you why

√100 + √25 = 10 + 5 = 15
But if it is
√125 then it'll be = 5√5

5√5 = 11.1

[olive charm]

ah okk

[sharp shard]

So when we have two square root values

Either we leave them by simplifying them or we convert them into decimals

In this question it's better to convert I believe

[olive charm]

i think so too

[sharp shard]

As they want an approximate round off

[olive charm]

yeah

[sharp shard]

So √12 + √5

[olive charm]

how would we simplify again...sorry

[sharp shard]

No problem

Simplifying is only necessary when we want it to stay in root

We don't

Right now

Just convert them into decimals

And add them

[olive charm]

ok

5.7?

[sharp shard]

Yes

That's the answer

[olive charm]

oh ok!

[sharp shard]

don't forget the cm

It matters lol

[olive charm]

i was about to say

no i got that wronggg

[sharp shard]

?

[olive charm]

i got that problem wrong

[sharp shard]

Why may you think that

[olive charm]

im studying for a test so im on deltamath

and it tell you once you've got the problem wrong

[sharp shard]

Is there an explanation?

[olive charm]

i accidently skipped it...sorry

[sharp shard]

well no issues

[olive charm]

here let me show you what i see

[sharp shard]

Yes

[olive charm]

and i tried to understand the examples

that wasnt any help

[sharp shard]

Do you need to answer this question?

[olive charm]

yeah...

so basically i have 3 chances to answer 2 questions

[sharp shard]

Damn sorry

[olive charm]

nah its fine

[sharp shard]

I don't know how the previous question was wrong

It looked correct to me

[olive charm]

it may have been me...

i got excited to i may have typed it in wrong...sorryyy

[sharp shard]

nah it's okay

just check the question later

and the solution

[sharp shard]

What was the formula I told you

[olive charm]

uh i cant

a^2+b^2+c^2=d^2

[sharp shard]

a² + b² + c² = d²

Yess

d is what we need

put the values

[olive charm]

i was answer to that earlier

3^2+20^2+7^2

[sharp shard]

I think you made a typo

In middle it'll be 10²

[olive charm]

yeah i did, my bad

[sharp shard]

Np so what's the value

[olive charm]

153

[sharp shard]

You sure?

[olive charm]

wait do we need to add √?

[sharp shard]

that's a must too

but also that it's 9 + 100 + 49

[olive charm]

2^2 is 4

so 4+100+49

[sharp shard]

It's 3tho

In the image

[olive charm]

god dang it

you're right

i made a mistake my bad

[sharp shard]

It's alright

Part of life

So √158

[olive charm]

right

so 12.5 or no decimals yet?

[sharp shard]

It says simplest radical form

[olive charm]

so no?

[sharp shard]

Whenever you see that phrase you gotta keep roots and check if it's simplify-able or not

[olive charm]

so i can do 2

[sharp shard]

You can

Let's break it then

√(2 × 79)

Is there any further breaking point?

[olive charm]

i think thats all you can do

[sharp shard]

Exactly

We can't break it further

Means it's in its most simplest form as it is

[olive charm]

so 2√79

[sharp shard]

√158 that is

[sharp shard]

We only do that when

For example

√20

√(2×2×5)

see how the number repeats itself twice

that won't have the root but the number doesn't will

so 2√5

In this case

√(2×79)

There is no repetition of number

[olive charm]

ohhh

i get it

[sharp shard]

Hence no number will come out of root

[olive charm]

oh ok

[sharp shard]

Yes

[olive charm]

so the answer is √2x79

[sharp shard]

Yes that is √158

[olive charm]

ok next problem...

this again

[sharp shard]

same stuff

You can do it yourself

[olive charm]

a^2+b^2+c^2=d^2

[sharp shard]

If you feel something unsure I'll assist you

[sharp shard]

Yes

[olive charm]

ok thank you!

10^2+6^2+6^2=172

√172

86

43

[sharp shard]

don't forget the root

[olive charm]

i didnt i dont think

2√43?

[sharp shard]

YEAHHH

GOOD JOB

[olive charm]

YAYY

1 more problemmmm

[sharp shard]

Of course!

[olive charm]

its the same thing

so ill try myself

[sharp shard]

sure thing

for a little challenge give me your final answer directly

[olive charm]

ok!

d^2=√109

i cant do anything else

with 109

[sharp shard]

yes just a lil terminology mistake but your answer is done

It's √109

[olive charm]

ok

that was a simple one!

[sharp shard]

don't forget why we add root

d² = 109

we need d

so we √ both sides

√(d²) = d = √109

squareroot cancels the square

[olive charm]

yes i will remember thank you!

[sharp shard]

That is all that matters then

You did fantastic work

[olive charm]

thankss

is it ok if we freind request

because when you explain

i really can understand

i havent understood that much from others

[sharp shard]

I've no problem at all I just come really less online so like I might not be good of a help most of the time

[olive charm]

its ok

[sharp shard]

Thank you that really means alot

[olive charm]

ur welcome

so its ok if i request?

[sharp shard]

I did!

[olive charm]

thank you! you faster than me!

[sharp shard]

just sometimes

[olive charm]

alright well sorry for keeping you up late

do i have to do anything to like clsoe this chat or?

[sharp shard]

Not a bit of a problem and yes you need to

[olive charm]

uh how do i do that

this is my first time o nthe server

so

[sharp shard]

.close that is

Have a beautiful day ahead

[olive charm]

thank you you too!

.close

[topaz sinew]

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

I need help with this exercite, it wnats me to get it to a simple form ,idk it looks basic but I can t find an algoritm to get it down . I let what I have tried up ,if it helps...

[topaz sinew]

@digital sky Has your question been resolved?

[digital sky]

<@&286206848099549185>

[digital sky]

Please , if you guys could take a look, maybe you seen it în the past

[obsidian brook]

what is meant by simple form here ? is it talking about normal form ? in which diagonal elemnts are constants and other elements are zero ?

[magic rain]

i tried this

idk if it is good enough

[topaz sinew]

@digital sky Has your question been resolved?

[digital sky]

No , it means to be writed as a product of terms

[digital sky]

@obsidian brook do you have any idea?

[obsidian brook]

yes i think it would be good if we can simplify the determinant before doing main calculations in some way

[digital sky]

Yes

There is no other way

I have worked other exercices and that is the principle

But how?

[obsidian brook]

Subtract the first row from all other rows

[digital sky]

Ok

And after that?

[topaz sinew]

@digital sky Has your question been resolved?

[magic rain]

<@&286206848099549185>

[topaz sinew]

@digital sky Has your question been resolved?

[topaz sinew]

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• 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.

[lofty gate]

My question: What is a homogeneous equation?
Theme: exponential equations. That is, this is not trigonometry

[nimble hill]

is there somone who understand frensh

[acoustic pecan]

stop doing this

[nimble hill]

why

i need one

[acoustic pecan]

wait in your own channel, dont spam it in everyone else's

or open a new one if you wish to

[timber crystal]

If we have a function $f(x,y)$, then \
$f(tx,ty)=t^nf(x,y)$

[thorny flame]

AkitoLite

[timber crystal]

in other words, you are able to factor out some tⁿ from f(tx,ty) to obtain tⁿf(x,y)

an example is $f(x,y) = x^3 + y^3$ \
$f(tx,ty) = (tx)^3 + (ty)^3 = t^3(x^3+y^3)$

[thorny flame]

AkitoLite

[lofty gate]

I have task number 222 under number 2)
but I don’t get how to convert this equation into a homogeneous one. I thought that here I just need to equate the bases so that I can equate the degrees.
Do I still need to do something differently and not the way I think?

<@&286206848099549185>

[craggy haven]

um, 5^(x+4) = 5 * 5^(x+3)

[topaz sinew]

@lofty gate Has your question been resolved?

[lofty gate]

what does it mean? I meant that I understood that we need to solve this equation as 2^4x-6 = 4^x+1
2^4x-6 = (2^2)^x+1
2^4x-6 = 2^2x+2
4x-6 = 2x+2
4x-2x = 6+2
2x = 8
x = 4
like something like that, only the equation is what’s in the photo

[craggy haven]

you're going to have to use some more rules of exponents and try some things

[topaz sinew]

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[hard ice]

where did the summation come from

[mild hearth]

you're trying to find the sum of all the solutions

[hard ice]

shouldnt the upper limit be 6 then?

[mild hearth]

no, because 7^3 + 7 <= 500

[hard ice]

ok ok , i was doing 1 + 1 + 8 + 2 + 27 + 3 and so on untill the sum exceeded 500

Thanks

.close

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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.

[topaz sinew]

Please don't occupy multiple help channels.

[frank lagoon]

please help

[sweet shard]

close your other channel

[frank lagoon]

i cant find it on occupied

it does not appear

[sweet shard]

use your search history

[frank lagoon]

thx

[mortal thunder]

Where's this extra u coming from

[frank lagoon]

damn

ig another 2 hours incoming

[golden blade]

eisai ellinas?

[frank lagoon]

nai

[topaz sinew]

@frank lagoon Has your question been resolved?

[topaz sinew]

@frank lagoon Has your question been resolved?

[topaz sinew]

@frank lagoon Has your question been resolved?

[topaz sinew]

@frank lagoon Has your question been resolved?

[golden blade]

bro ti einai to prowlima edo?

[topaz sinew]

@frank lagoon Has your question been resolved?

[topaz sinew]

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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

[strange vigil]

Can anyone please explain calculus

[strange vigil]

Please explain calculus

[topaz sinew]

@strange vigil Has your question been resolved?

[deep crow]

Calculus is kinda broad do u have something specific

[hoary burrow]

@strange vigil for an intro ig

https://youtu.be/WUvTyaaNkzM?si=VmH3YbKk-CzpL6FS

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[bitter spindle]

Hi

[bitter spindle]

How to learn maths for beginners to take part in olympiad

[pearl fog]

practice questions

what level are you

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@bitter spindle Has your question been resolved?

[mortal thunder]

+_+ https://discord.gg/mods

Start with joining this. You'll get a good guidance as well as an assessment as to where you currently stand. ._> It'd be best to start first with an elementary Oly Math book

An Excursion in Mathematics

• is a really good beginner friendly book I believe

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[strange vigil]

What is stochastic Calculus

[strange vigil]

Please help

Please help <@&286206848099549185>

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[primal niche]

Difficult solution guys , I know how to do it just wanna see more try it

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[primal niche]

<@&286206848099549185>

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[timid halo]

Is it possible to figure out the function for the black line or does it look like im just supposed to use (base*height)/2 of the triangle ?

[sinful flint]

you can do both

they give you the same answer anyway

[timid halo]

so like how do I figure it out cause. I keep thinking its y(x) = 1-pi/2 * x

but it gives the wrong values

[odd pagoda]

y=mx+b, m=rise/run

have you seen that formula?

[timid halo]

yeah and its going downwards aat at pi/2 right?

crossing 1 at the y axis

[sinful flint]

i think itll just be easier to calculate the area of the triangle

itll make the integral easier aswell

<@&268886789983436800> ?

[odd pagoda]

how much does it go down

(and yes, viewing it as a triangle is certainly easier)

[timid halo]

well we only know it crosses the y axis at 1 and x axis at pi/2

[odd pagoda]

how much does it go down and how much to the right to get from the first to the second of these points

[timid halo]

I mean it goes down 1 and forwards n/2

if you want to be specific

[odd pagoda]

so rise=-1

run=pi/2

so m=(-1)/(pi/2)=?

[timid halo]

aint this just tan v = height/length to find the angle?

or where does the divison come from?

[odd pagoda]

y=mx+b and you plug the points (0,1) and (pi/2,0) in

you get a 2x2 system for m and b

solving for m gives you that formula

[timid halo]

wait so its delta y/delta x?

[odd pagoda]

the system is especially easy here cause of the 0s

[odd pagoda]

yes

[timid halo]

oh then I understand it

thanks

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[glass wraith]

How I solve this with u=vpg p=m/v

U=rw-aw

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[obtuse garden]

Hi! I need some help verifying whether my proof for a question is correct. Heads up It is a question for a summer programme but I don't need any help just need someone to confirm my reasoning

[sinful flint]

!da2a

[topaz sinew]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[sinful flint]

just send what you have

[obtuse garden]

yeah yeah ik but since it's a summer programme question I didn't want anybody to think im cheating

ill send it hold on

I got the answer which is all the numbers in S will be in the range [1/2.1] but I wanted to know if its possible to conclusively state which rationals in that range will be in S

if it is possible pls dont tlel me how to find it

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[errant hamlet]

so i tried to move one term in the other

sqrt(4x-y^2)=sqrt(4x^2+y)+sqrt(y+2)

square everything

and we got to 4x^2+2y+2sqrt(4x²y+y²+8x²+2y)-4x+y^2=0

[sweet shard]

solve for the square root term and square both sides again

[austere oyster]

if youre squaring remember to square in the form a^2 + b^2 + 2ab

[errant hamlet]

but its gonna be smth like (a+b+c+d)^2

i dont think THIS is the right way

[sweet shard]

if you solve it in a more efficient way ping me

[topaz sinew]

@errant hamlet Has your question been resolved?

[errant hamlet]

?

why

[mortal thunder]

just checking something -_-

[errant hamlet]

oh,ok

<@&286206848099549185>

[mortal thunder]

$\sqrt{2X + 1 + 2Y - Y^2} - \sqrt{Y + 1} = \sqrt{X^2 + 2X +Y}$

[thorny flame]

Arya

[mortal thunder]

$2X + 1 + 2Y - Y^2 = X^2 + 2X + Y + Y + 1 + 2\sqrt{Y + 1}\sqrt{X^2 + 2X + Y}$

[thorny flame]

Arya

[mortal thunder]

X² + Y² + 2√(Y + 1)(X² + 2X + Y) = 0

=> X = Y = 0

@errant hamlet

[errant hamlet]

yes

[mortal thunder]

oh btw I substituted X = 2x - 1, Y = y + 1

[errant hamlet]

wait a sec

[errant hamlet]

how did you get to this?

isnt there X^2+2Y+2√(Y + 1)(X² + 2X + Y)=0 ?

[mortal thunder]

You understood till here?

[errant hamlet]

yes

oh nvm

i get it

[mortal thunder]

$\cancel{2X} + \cancel{1} + \cancel{2Y} - Y^2 = X^2 + \cancel{2X} + \cancel{Y} + \cancel{Y} + \cancel{1} + 2\sqrt{Y + 1}\sqrt{X^2 + 2X + Y}$

[thorny flame]

Arya

[errant hamlet]

thank you so much!

[mortal thunder]

so on the RHS we only have +ve terms :)

on the LHS we require 0

So X = Y = 0 the only way tah go

[errant hamlet]

wow

thank you one more time!

have a nice day!

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[mortal thunder]

@sweet shard have a look once

squared but didn't solve !

[sweet shard]

crazy

that was genius arya

X = ax + b and Y = cy + d should always work.

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[manic iris]

Trying something to do with statistics:

• 50 slots
• 35 items (usable as much as possible)
• 11 items (all can only be used once)
  How many possibilities can there be + how do you reach the possibilites?
  I have done 35^50 but how would I add the other 11? 11!? C(22,11)?

[urban barn]

casework?

1 of the 11 items and all the remaining 49 slots with 35 items

and so on yeah

[manic iris]

yeah

[manic iris]

no not like that

I mean the 11 items can be used whenever but only one of each

and the 35 whenever

[urban barn]

no i get it

i’m saying consider casework where u include 1 of the 11 items

2 of the 11 items

3 of the 11 items

and so on

[manic iris]

oh right

yes

I get it now thanks

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[long stirrup]

there's 12 cases

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[acoustic wasp]

Heyo (I promise I've edited it so that there is little yapping now).

So, why can supremum of a set (the least upper bound) be infinity? That makes so little (intuitive, to me, at least) sense.

Also, how do you internalize proofs?

[strange whale]

That makes so little (intuitive, to me, at least) sense.
what's your intuition about sups then ?

it's not forbidden to yap here

[acoustic wasp]

Well supremum should be a real number, in my head that is

[strange whale]

you can leave your explanation about your 50 proofs in analysis class

cause it helps ppl who read you calibrate their answer

if anything that wasn't a yap at all

[acoustic wasp]

Well ty for being so kind first off, second, the sup - so, it should just be a real number. you know. Some number that no other number from the set can "overstep"

[opal vault]

do we agree that the supremum of a real number set should be the "smallest upper bound" of that set

[strange whale]

what should the sup of the set of all natural numbers be then

if it's supposed to be a real number

[acoustic wasp]

shouldnt exist

[opal vault]

when a set is not upper bounded

by definition

[acoustic wasp]

indeed

[opal vault]

we let the supremum = infinity

[strange whale]

+infty is just a way of saying DNE in this case yeah

[opal vault]

we define it like this for not upper bounded sets

if you want to look at it another way

consider the augmented real set $\bar \bR$

[thorny flame]

rafilou is not not born in 2003

[opal vault]

it's the set with real numbers

and we "add" infinity and -infinity as elements to it

in that case

the supremum of a real subset

[opal vault]

should coincide with its supremum "in the augmented real set"

[acoustic wasp]

Does then sup of R* exist? As in, infinity is "a part of the set" and the supremum, but not the max? I might be yapping, so you don't need to answer, now I am just curious, I got the answer already.

(R* is what we use to say like real with +-inf)

[opal vault]

R* is real with +-inf?

and not R without 0?

[opal vault]

I'll keep my notation just in case to avoid confusion

[acoustic wasp]

Nope, here it's real with +- inf, but our prof is weird as hell. Like we use log instead of ln and log10 as usual HS log.

[opal vault]

$\sup_{\overline{\bR}}\overline{\bR}= \infty$

[thorny flame]

rafilou is not not born in 2003

[opal vault]

no problem

it's even a maximum

since it's inside your set

[acoustic wasp]

You are absolutely correct

Is your major maths?

[opal vault]

kind of, yes

[acoustic wasp]

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[strange whale]

rooms are meant to be held lol, you're allowed to stay a bit more if you want, it's not like there's a shortage of rooms

.reopen

[topaz sinew]

✅

[acoustic wasp]

I just got no social understanding when it comes to online comunication yk

like me online and irl are two different entities

[strange whale]

I mean yeah it depends on the place sure

i'm telling you how it is here

I mean it's baby analysis, it's a bit boring yeah, you're studying things you already saw in high school essentially (calculus), now with added formalism on top

[acoustic wasp]

Alright, since we're open - second part of my question anyone? :D

[strange whale]

but like people have thought a long time about these things, in the 17th century newton and leibniz et al didn't give a damn about formalism, and then ppl in the 19th century found some really wacky examples that made them rethink what functions, limits, derivatives are

really bastardized history of early analysis here

[acoustic wasp]

Our prof. said something similar. He also pretty much ridiculed us that all this is 150+ years old, that only in the 3rd/4th semester do we start actually studying new stuff

[strange whale]

yeah this is just background really

[strange whale]

and HS is newton-leibniz'ing it essentially (depends where you are ig)

[strange whale]

I mean in general you don't try to remember a proof 100%

like most of the time, there's one or two important tricks that are essential to the proof that you might want to commit to memory

i.e. for the product rule the adding and substracting f(a)g(a+h)

the rest is routine, mechanical

[acoustic wasp]

Yeeah, I guess. I just feel the need to complain about having it hard when I chose a difficult uni and classes.

[strange whale]

and the big issue as a beginner is being familiar enough with the material to recognize what's routine and what's "trick"

[strange whale]

same thing for stuff like rolle of bolzano weierstrass as you mentioned

but these are more big boy theorems compared to the derivative rules so it might be worth studying these more

[acoustic wasp]

(undoxxing myself)

So his spirit might help me during finals

[strange whale]

bring a bolzano plushie or whatever

if it exists

[strange whale]

and that's why going to classes is important

if the profs are worth their salt, they'll try to emphasize what's important from what's very routine

& if the profs suck, well you do you

[acoustic wasp]

I go to every single one, even to extra classes, I chose the "hardest"/toughest TA for seminars, but like the lectures are a one man show. dude just tries to impress us by using 6 boards for a single proof, then saying that the rest is "up to us to do".

[strange whale]

so like a bad one man show

without describing what's "left to do" i suppose ?

[acoustic wasp]

To be fair to him, he usually suggests the main step, but says that it is just "primitive"

[strange whale]

maybe there's also other lecturers if there's enough ppl at your uni

is your TA ok in this regard at least ?

[acoustic wasp]

like with the arithmetics stuff, he shows one proof and says that the rest is too similar to spend time on, but then goes and proves a "sidenote" for half an hour.

[acoustic wasp]

it's the same guy :d His classes (seminars) are packed

[strange whale]

ah

[acoustic wasp]

because we all know that if you do not go to his seminars, you won't make it through. That is like the public secret.

He solves HIS problem sets there, which you can expect to see some variation of in the finals. And although he goes super fast, it it probably the best way to prepare for whats to come.

[strange whale]

well maybe you can check other people's shows in secret if it doesn't conflict

[acoustic wasp]

I've gone to a few, but most of the other ones are too slow :D

[strange whale]

yeah so you chose consciously it's not like you had no other choices got it

[strange whale]

aight

[acoustic wasp]

The "success rate" or wtvr when he holds the class for the year? 50%. The others have at least 67+ (and I mean, the students there are literal geniuses, I've never felt more stupid than when I talk with my mates in class)

[strange whale]

well he's not joking that's essentially the case for a first class, hopefully there's other ppl for other semesters

or maybe you'll get to like this guy who knows

[acoustic wasp]

Haha, that's the joke - there aren't. We are basically his "homeclass" for this series of classes. All continuations (unless you want to wait a year) are with him.

[strange whale]

ah that's how it works lol

well hopefully you don't get another prof then

[acoustic wasp]

That would suck, reaaaally suck. I'd have to prolong my Bc. to 4 years instead of 3 just because of that

At least we don't pay for our unis. Must suck for americans to get into debt. I love academic stuff, but idk if I'd get into debt just to study.

[strange whale]

yeah at least even an extra year doesn't cost you 2 legs and your soul if you ever have to do it

well anyway

you got other analysis questions or no ?

[acoustic wasp]

Yes, but I need to translate it for anyone to understand.

[strange whale]

yeah I gtg so I guess we can wrap this up

[strange whale]

it's in german I suppose ?

some german speaking ppl lurk around here you might get lucky

[acoustic wasp]

thanks, ybee

[strange whale]

but yeah it's better manners to translate

[acoustic wasp]

Got it already, imma open a new one so that I get to the top. Bye bye, ty

[strange whale]

aight see you around

[acoustic wasp]

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[brittle hawk]

S = ((1, 1, 1, 1, 0), (1, 1, −1, −1, −1), (2, 2, 0, 0, −1), (1, 1, 5, 5, 2), (1, −1, −1, 0, 0)) are vectors in R^5,

what's the best way to find basis of them? Like I know that I have to find which of them are linear independent, and I can put them in to a matrice and do gauss jordan elimination, but when I have a correct form, how do I determine which aren't linear independent and which are?

[sweet shard]

what was the result of your gauss jordan

[brittle hawk]

1 0 1 3 0
0 1 1 -2 0
0 0 0 0 1
0 0 0 0 0
0 0 0 0 0

and columns are vectors

My guess is that 1,2,5 are linear independent, but I don't know

[sweet shard]

rank 3

[brittle hawk]

yeah

[sweet shard]

oh the columns are vectors?

you can you use the matrix i think because it's clear columns 3 and 4 are linear combinations of columns 1 and 2. you can start by writing each vector as a linear combination of the canonical ones. then use the matrix to express those linear combinations of columns 3 and 4 in terms of the linear combinations of columns 1 and 2

absolute word salad but i'll try to think of a better way to express myself

or maybe it's the other way around. write the canonical bases in terms of the vectors in S.

[topaz sinew]

@brittle hawk Has your question been resolved?

[brittle hawk]

hmm, okay. But what if I would put vectors as rows?

wouldn't it be easier?

[sweet shard]

it's equivalent, but if it's easier for you go for it

[brittle hawk]

Okay, thank you!

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[wise quiver]

Help solve this integral

[pseudo horizon]

a u-substitution should help

so you can avoid expanding

[wise quiver]

By any chance could show me an example of using u-substitution

Bc

I was looking at examples

And I’m a little confused

About this part where du is introduced

[pseudo horizon]

i can walk you through it on this integral

[wise quiver]

Yes please 🙏

[pseudo horizon]

try u=1+x^2

then we differentiate to find du/dx

this is just like differentiating y=1+x^2

but we have u instead of y

it makes no difference

can you do this part?

[wise quiver]

Yes i can

[pseudo horizon]

ok let me know what you get

[wise quiver]

I got 2x

I think the main part is I get confused when actually substituting the u

[pseudo horizon]

before that, let's do this

du/dx = 2x

multiply both sides by dx

this is strange but this is the u-sub method

[wise quiver]

So I would get du= 2x dx

[pseudo horizon]

yes

we have a 5 in the integral instead of 2, so you can try to make it match

[pseudo horizon]

another way is just solve for dx here

i recommend just solve for dx

dx=du/2x

now you can substitute

[wise quiver]

Ohhh

I think I kinda get it

[pseudo horizon]

do you see where to substitute u?

[wise quiver]

Kind of

[pseudo horizon]

u=1+x^2

[pseudo horizon]

there's already 1+x^2 here

so remove it and replace it with u

so instead of 5x(1+x^2)^4 dx...

what will it look like?

[wise quiver]

5x(u)^4?

[pseudo horizon]

yes and the dx

5x(u)^4 dx

sorry i forgot to write it too

at first

now the dx is there and we know dx=du/(2x)

so you can remove dx and replace it with du/(2x)

what will that be?

[wise quiver]

5x(u)^4 . Du/2x

[pseudo horizon]

yeah 👍

$\int 5xu^4\cdot\frac{du}{2x}$

[thorny flame]

Axe

[pseudo horizon]

the x's will cancel

[wise quiver]

So then it’s du/2 and 5(u)^4

[pseudo horizon]

yes

we can move the 2 over

$\int \frac{5}{2}u^4 du$

[thorny flame]

Axe

[pseudo horizon]

can you integrate this?

[wise quiver]

Ahhhso it’s (1+x^2)^5/2 +c

[pseudo horizon]

looks right

[wise quiver]

Okok

Thank you

[pseudo horizon]

you're welcome

[tawdry lynx]

good job

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[bitter kraken]

Hi, I am new to this server. Do I ask questions for help on math problems in here?

[bitter kraken]

So I need some help with calculus, but my question is a little different. I have two problems that look kinda similar, but the methods to solve them are a little different. I was already able to solve them, but I want someone to walk me through why the process is a little different.

[rigid ivy]

Yes. Here. #❓how-to-get-help if you have questions about how to use the server

[bitter kraken]

Cool cool, can you help me on this? This is calculus

So my question is this:

Why don't we plug in the times for t in the formula for the velocity directly in problem 155 like we did in problem 157? Like how come in 157 we can just plug in 0.5 and 5.75 directly into the derivative of the given formula, but we can't plug in 0 and 14 directly into the derivatives of the velocity formulas in problem 155?

Lemme pull up my work on both questions too

[rigid ivy]

Why don't we plug in the times for t in the formula for the velocity directly in problem 155
You don't have the velocities yet. It's what you are trying to find.

[bitter kraken]

But like in 157 we can just plug it in directly?

As I showed here?

Essentially the processes look different, but why?

[rigid ivy]

oh you wanted to plug in the times

okay one sec lemme reread

[bitter kraken]

Yeah idk if my question makes sense

(These answers are all right btw, I checked with the answer key)

[rigid ivy]

Why don't we plug in the times for t in the formula for the velocity directly in problem 155 like we did in problem 157?
You can do it this way. It's how I would've done it. Let me see what you did

Did you not plug t directly into your velocity equation here?

[bitter kraken]

Wait

Oh wait, the difference is t was already given in 157

But in 155 we have to find t first?

[rigid ivy]

yup

[bitter kraken]

Ooooh I see

[rigid ivy]

[bitter kraken]

So we're sort of finding different things here?

Like in 155 we have to find the timepoint, while in 157 we have to find the position when it's at a certain time point?

[bitter kraken]

Wait, then what is s(t) in this case then?

[rigid ivy]

In 155, you want to find the velocity at specific positions (when $s=0$ or $s=14$). However, you don't have an equation relating velocity and position (you $\textit{could}$ make one, but it would be messy). But you $\textit{do}$ have an equation for position and time, and then for velocity and time. So, knowing your positing, you determine the time, and then you can use that time to find the velocity at that time, which corresponds to the position where you want to know the velocity.

[thorny flame]

SWR

[rigid ivy]

$s$ is your distance function.

[thorny flame]

SWR

[bitter kraken]

Oooh I see

So pretty much, in 155 you're trying to find t, while in 157 you're trying to find s(t)?

[rigid ivy]

in 155, you're trying to find t so you can use it in s'(t)

In 157, you're trying to find a lot of different things. It depends on which part a-f you are asking about specifically

[bitter kraken]

Part a and b in particular

In Part a we're trying to find s(t)?

[rigid ivy]

In part a, you simply want to find velocity at certain time, which is just calculating $\dot{s}(0.5)$ and $\dot{s}(5.75)$. In part b, you want the magnitude of the velocities.

[thorny flame]

SWR

[rigid ivy]

$s'$ (or $\dot{s}$, as is common notation for time derivatives)

[thorny flame]

SWR

[bitter kraken]

Sorry I mean comparing part a of 157 to 155

[rigid ivy]

In 157, you want velocity at a specific time, in 155, you want velocity at a specific position

[bitter kraken]

Since parts a and b of 155 are identical, just at different time points

[bitter kraken]

Ooooh I see. Is that why we have to calculate t in 155, but not in 157?

Because in 157, it's already given as 0.5 and 5.75?

[rigid ivy]

yes. $t$ is given in 157a, so there is no need to calculate for it.

[thorny flame]

SWR

[rigid ivy]

[bitter kraken]

I know this sounds kinda dumb, but is 155, parts a and b the same as 157, just with one extra step then?

[rigid ivy]

basically

just the extra step of finding t first

[bitter kraken]

Oooh I see, alright then. The concept and method should be identical though?

[rigid ivy]

yes

[bitter kraken]

Just derive if I want to find the velocity, plug it in, bada bing bada boom?

And velocity is just position', while acceleration is position'' ?

[rigid ivy]

In both problems, you are given position $s$ as a function of time $t$: $s(t)$. You use differential calculus to find the velocity, which is the derivative of position with respect to time: $\dot{s}(t)$. This derivative calculation gives you a function which represents velocity $\dot{s}$ as a function of time $t$. In 157, you are asked to find $\dot{s}$ at $t=0.5$, so you simply calculate $\dot{s}(0,5)$. But in 155, you are asked to find velocity at a specific position. You do not have velocity as a function of position. So, to use your velocity function, you must first find the time that satisfies the position you are looking at.

[thorny flame]

SWR

[bitter kraken]

And in order to find the velocity at a specific position in 155, you have to find t first, hence the extra step of finding t first?

[rigid ivy]

exactly

technically, you could find velocity as a function of time, but it's messy

[bitter kraken]

Aaaaaah alright that makes sense

[bitter kraken]

Ahh this is Calc 1, that's for another class later?

[rigid ivy]

In general, $x(t)=\frac12at^2+v_0t+x_0$, and $v(t)=at+v_0$. We solve for $t$ in terms of $v$, giving us $t=\frac{v-v_0}a$. Plug that into $x$ to get
$$x(v)=\frac12\left(\frac{v-v_0}a\right)^2+v_0\frac{v-v_0}a+x_0$$
You now have $x$ as a function of $v$. To get $v$ as a function of $x$, you need to solve the above quadratic, where you get
$$v(x)=\frac{v_0\pm\sqrt{v_0^2-2ax_0}}2$$

[thorny flame]

SWR

[rigid ivy]

It's just some algebra

[bitter kraken]

For sure.

And for 157 c, d, e, and f, it's just formulas I have to memorize?

-b/2a, quadratics, all that fun jazz?

[rigid ivy]

memorize, or learn to derive them yourself

[bitter kraken]

For sure

Man it's been 7 years since I did calc

I'm just rusty and want to make sure I know my stuff

[bitter kraken]

Finally, for b, literally the only difference between velocity and speed is speed is always positive, right?

So provided its the same times for both problems, the answer to b should always just be | a | in these types of problems?

[rigid ivy]

[rigid ivy]

[bitter kraken]

Perfect, those are all the questions I have. Thank you

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[topaz sinew]

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Use .reopen if this was a mistake.

[topaz sinew]

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