help-43

[quartz yoke]

ok

[open plaza]

we have angle EBD = FBC = 105

EB/FB = BD/BC = √2

find 2 similar triangles

[quartz yoke]

EBD

FBC

hmm

[open plaza]

have you made your final choice?

[quartz yoke]

uhh

ig

[open plaza]

alr correct

[quartz yoke]

YAY

that wasn't hard lol

[open plaza]

last part

[quartz yoke]

ok

[open plaza]

what is ED/FC?

[quartz yoke]

root 2

[open plaza]

now calculate the length of FC

[quartz yoke]

uh

(root(6)+root(2))/root(2)

[quartz yoke]

is it that?

[open plaza]

yea

simplify it

[quartz yoke]

root(3) + 1

[open plaza]

ok nice

[quartz yoke]

ok

[open plaza]

this problem is hell

[quartz yoke]

last

[open plaza]

[quartz yoke]

after this

ima play games

cuz my brain is clocked

my dad says 4th graders do this

[open plaza]

huhhhhh??

ok we have a rectangle

[quartz yoke]

yea

[open plaza]

and an equilateral inside

[quartz yoke]

ja

[open plaza]

this looks so deceivingly hard

[quartz yoke]

aww

need to do this in 10 mins

[open plaza]

10 minutes??

[quartz yoke]

its due at 11:05

its 10:55

im so cooked

[quartz yoke]

ja

ya

dk where to start

[open plaza]

?

it is a rectangle

[quartz yoke]

ye

[chilly basalt]

…..

[quartz yoke]

how can u prove square

if its true problem wud be ez

[nocturne pasture]

ohh

[open plaza]

there's 0 evidence that it's a square

[chilly basalt]

Sure, I admit that it’s hard

[chilly basalt]

Yee

[nocturne pasture]

i thought of pytha on right side too

[open plaza]

sorry I think I can't help you with such little time

[nocturne pasture]

where to start

[quartz yoke]

its okay

its a beast problem

it counts as xtra points

[open plaza]

oh wait

[nocturne pasture]

ohh

[open plaza]

you can set up equations

[quartz yoke]

so i'll get 1/2 as points

[quartz yoke]

ok

[open plaza]

call the long side at the left a

[quartz yoke]

ok

[open plaza]

and the short side b

[nocturne pasture]

is the left triangle scalene?

[open plaza]

the side at the right would be a+b

so we have
16 + a^2 = 25 + b^2 = 1 + (a+b)^2

[quartz yoke]

okay..

wait

cant we calculate a and b

[nocturne pasture]

the LR corner is not 90deg angle

[open plaza]

it is

[quartz yoke]

has to be

cuz quad their angles add to

360

[nocturne pasture]

tho yea

mbmb

[quartz yoke]

and 3 90's are given

[open plaza]

wth are you on about

[quartz yoke]

its kk

[nocturne pasture]

sry

[quartz yoke]

ok

ok now

$16 + a^2 = 25 + b^2 = 1 + (a+b)^2$

[boreal girder]

✪Royal✪

[quartz yoke]

we solve

[quartz yoke]

wait

i think a*b =4

[quartz yoke]

bros prob on geogebra

cookin

[nocturne pasture]

lol yea

[pallid rune]

explain why

[quartz yoke]

that means its wrng

but i will

16+a^2 = 25 + b^2 = 1+a^2+b^2+2ab

[nocturne pasture]

yea how'd you get a*b = 4

[quartz yoke]

can u subratact

OOH

if u do -b^2

u have to do it to all sides

so itz wrng

mb

[pallid rune]

mhm, can't play favorites sadly

but rn

you have a chain of equations

which means that you have 3 equations

[quartz yoke]

yea

[pallid rune]

which means you can solve and find a and b

[quartz yoke]

ok

past due at tis pint

point

[nocturne pasture]

aw

[quartz yoke]

a² + 2ab = 24

a² - b^2 = 9.

[quartz yoke]

i got 2 equation verify

[open plaza]

god damn it

[quartz yoke]

finally

ur back

[nocturne pasture]

lol

past due sadly

[open plaza]

no I haven't found away

[quartz yoke]

prob late for u

[quartz yoke]

does that mean itz impossible?

[open plaza]

other than bashing those equations

[nocturne pasture]

would be interested to know the answer tho

[nocturne pasture]

1208 for me

[quartz yoke]

hmm

[open plaza]

the solution is ugly as hell

[chilly basalt]

Hi hi hi, @quartz yoke can I teach you instead?

[quartz yoke]

11:09

[quartz yoke]

ok

[nocturne pasture]

am btw💀

[quartz yoke]

carbonite can listen

[quartz yoke]

same

but its midnight for u

[nocturne pasture]

eya

yea

[quartz yoke]

its noon fr me

[quartz yoke]

ok

[nocturne pasture]

I'll listen too

[chilly basalt]

So our goal is to set A+B = C

[quartz yoke]

ye,

ok

[chilly basalt]

But this time, we’ll use one variable instead

[quartz yoke]

ok

[chilly basalt]

Can you gimme the equation set with 1 variable?

[quartz yoke]

like

[chilly basalt]

Like this

[quartz yoke]

ok

x^2-16=y^2

[chilly basalt]

No, there is no y

Just x

[quartz yoke]

mb

[chilly basalt]

Purple + green = red

[quartz yoke]

yea

ok

[chilly basalt]

Done?

[quartz yoke]

no

wait how

[chilly basalt]

Use pyth

[quartz yoke]

ok

[chilly basalt]

Can you represent the purple side with x?

[quartz yoke]

like x^2-25

ands x^2-16

and re side

x^2-1

so

2x^2 - 41 = x^2-1

[chilly basalt]

No, where is your square?

[quartz yoke]

?

what

[chilly basalt]

It’s not correctly stated

The purple side should be sqrt(x^2 - 25)

[quartz yoke]

oh yea

[chilly basalt]

Do you agree?

[quartz yoke]

yea

mb

[chilly basalt]

Sure, now gimme the equation

[quartz yoke]

sqrt(x^2 - 25) + sqrt(x^2-16) = sqrt(x^2-1)

[chilly basalt]

Yeee

[quartz yoke]

ok

[chilly basalt]

Can you square both sides for me?

I’m serious.

[quartz yoke]

ok

[quartz yoke]

?

[quartz yoke]

?

[chilly basalt]

Square both sides

[nocturne pasture]

-# i think he means remove the sqrt

[quartz yoke]

k

[quartz yoke]

ye

[nocturne pasture]

-# right?

[chilly basalt]

$$[\sqrt(x^2 - 25) + \sqrt(x^2-16)]^2 = [\sqrt(x^2-1)]^2 $$

[quartz yoke]

(x^2-25)+(x^2-16)+2sqrt(((x^2-25)+(x^2-16))) = x^2-1

((x^{2}-25)+(x^{2}-16)+2\sqrt{(x^{2}-25)+(x^{2}-16)}=x^{2}-1)

[boreal girder]

✪Royal✪

[quartz yoke]

ok

[boreal girder]

SkyAndNight

[chilly basalt]

Lemme check my paper

[quartz yoke]

ok

[chilly basalt]

Yep correct

Simplify it

[quartz yoke]

ahh

ok

i started alrdy lol

[chilly basalt]

Done?

[quartz yoke]

no

[chilly basalt]

Alr

[quartz yoke]

4x^4-164x^2 = -80x^2 + x^4 prob wrng check

$4x^4-164x^2 = -80x^2 + x^4$

[boreal girder]

✪Royal✪

[chilly basalt]

Wait what? You’ve already removed the root

[quartz yoke]

too fast

prob wrng though

i was at 4(x^4 - 41x^2 + 400) = 1.6k -80x^2 +x^4

[chilly basalt]

Lemme verify, too fast

[quartz yoke]

$4(x^4 - 41x^2 + 400) = 1.6k -80x^2 +x^4$

[boreal girder]

✪Royal✪

[quartz yoke]

im used to problem like that

[chilly basalt]

Correct

[quartz yoke]

algebra is ez

geometry is hard

tbh

[quartz yoke]

what do we do now

[chilly basalt]

Suppose you can do the rest

[quartz yoke]

ugh

wait

once we get x we done right

[chilly basalt]

No, the problem you’re doing is quite advanced ngl 😂😂

[quartz yoke]

lol

[open plaza]

his advanced problems are disgusting

[chilly basalt]

Brother, you have this

[quartz yoke]

ja

so

[chilly basalt]

I’m not going to help you with this 😔
You have to think about it

[quartz yoke]

$3x^4 - 164x^2 + 80x^2 = 0$

[boreal girder]

✪Royal✪

[quartz yoke]

ima write steps down

jic

$ x^2 * (x^2 -28) = 0 $

x^2 = 0

or

x^2 -28 = 0

gtta use some formulas

[chilly basalt]

First of all, can x be either negative or zero?

[quartz yoke]

no

so 2root(7)?

[chilly basalt]

Sure, keep grinding

Oh wait, that’s fast

[quartz yoke]

did smth EXACTLY like that before

i had 8 algeb. classes so far

in 1 year

[quartz yoke]

is it wrng. or correct

[chilly basalt]

Correct

[quartz yoke]

my brain turned

on

[quartz yoke]

first time i've seen good algebra in geometry

well ty very much

[open plaza]

I fucking cant 😂

[chilly basalt]

I made a slight mistake in my calculation

[quartz yoke]

huh

why is solution so tiny

ima play some games ig

and then eat or take a nap

gn

[chilly basalt]

You deserve that

Cya

[quartz yoke]

ye

cya

.close

[compact pewter]

Channel closed

Closed by @languid condor

Use .reopen if this was a mistake.

Available help channel!

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Remember:
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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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[tough oak]

Can someone help me?

[tough oak]

I’m in 7th grade taking

[arctic portal]

!da2a

[compact pewter]

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/

[solemn matrix]

anyways

what's your question?

[tough oak]

8th grade standards

[solemn matrix]

alr

[tough oak]

And I just started

America

Btw

[arctic portal]

do you have a specific problem that you need help with?

[solemn matrix]

try giving us the table of contents

[tough oak]

No just want something to get started

[tough oak]

Huh

[nocturne pasture]

what do you need help with?

[solemn matrix]

like this one

[tough oak]

Yeah I know what it is

[solemn matrix]

yeah

[nocturne pasture]

yeah

[solemn matrix]

give us the table of contents

[open plaza]

[arctic portal]

if you need help with a certain problem, please post an image here. If you need help with a topic, i recommend you go to one of the topic channels

[solemn matrix]

so we can see what the 8th grade standards is

[tough oak]

But maybe some algebra

[nocturne pasture]

yes

[solemn matrix]

im not an american guy

[nocturne pasture]

same

[solemn matrix]

is there a system of equations?

[arctic portal]

can you post an example question 😭

[nocturne pasture]

or like give us the topics

[tough oak]

I’ll just leave cause I’m confused

[solemn matrix]

yeah

[solemn matrix]

dude

[arctic portal]

just type .close

[tough oak]

.close

[compact pewter]

Channel closed

Closed by @tough oak

Use .reopen if this was a mistake.

[arctic portal]

welcome to the mathcord btw!

[solemn matrix]

he left...

[nocturne pasture]

bruh

he actually left the server

[arctic portal]

oh

did we scare him away 😭 💀

[nocturne pasture]

lol

hope not

tho 3 people asking you questions is...

[arctic portal]

quite intimidating?

[nocturne pasture]

yeah i guess

[strange ermine]

I think he came here to learn something as a self-taught

[compact pewter]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[orchid lark]

i think this function has a removable discontinuity

[subtle helm]

why

might be jump

[kind viper]

is anything known about the function BESIDES this table of values?

e.g. a formula?

[orchid lark]

and this one has a jump discontinuity

[subtle helm]

this one looks more removable..

[orchid lark]

sorry my internet is bugging out one sec

[kind viper]

my question wasn't answered at all

[orchid lark]

nah this is it, just the table

[kind viper]

then there is not sufficient info even for a guess at what happens near 0.

!xy

[compact pewter]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[orchid lark]

my internet connection dropped for a minute right after i posted my question i'm sorry my message took a bit to send

[kind viper]

well i gave my verdict

gonna need you to post the question in full. yes, even if you're telling the truth about it not telling you anything.

[orchid lark]

this is it

sorry for the wait, it might not send though because I have one bar on and off

I think I might have to close this channel and ask again when my connection is more stable but just so you can see, really no other info or formulas

[sterile crag]

that's a funny question

[orchid lark]

yes it is, and those are supposed to be tables but they lack..... table

i'm gonna close the channel now but I'll try this again in a bit when i stop disconnecting

thanks

.close

[compact pewter]

Channel closed

Closed by @orchid lark

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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.

[cyan wren]

So I can ask any doubt here?

[subtle helm]

post ur question :3

[cyan wren]

Okey I'm gonna do some problems so I will let u know if I had some doubts k?

[subtle helm]

k

[humble mortar]

!da2a

[compact pewter]

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/

[cyan wren]

In real numbers,2 brands of chocolates are available in packs of 24 and 15 respectively.i need to buy equal number of chocolates then find the least number of boxes needed. Can u explain why is hcf is applied here? But not lcm since the word least Is given

[young raft]

show the solution you are looking at

[cyan wren]

I only have the question I don't have the answer but I know we have to hcf but I forgot why

So do u know?

Someone help me outttt

[cyan wren]

Yoo can you help me out here? It's kinda urgent

[subtle helm]

ye?

'highest common factor'

[subtle helm]

its not an exam, right?

[cyan wren]

Exam :((

In Monday!!

[subtle helm]

oh

as long as its not an ongoing exam rn

its fine

js post the questio

[cyan wren]

I don't have the question at the moment I gave the worksheet to my brother

I just remembered the question

[subtle helm]

okie

[cyan wren]

This is the question

[subtle helm]

are u sure

that we're not talking abt lcm

oh in real numbers

[cyan wren]

Yeah 100 percent

My teacher said why but I just couldn't understand and I don't remember

[subtle helm]

24 = 3 * 2 * 2 * 2 = 3 * 8
15 = 5 * 3 = 5 * 3

[cyan wren]

3 is hcf

[subtle helm]

,calc 1/8 * (24) = 1/5 * (15)

[boreal girder]

The following error occured while calculating:
Error: Invalid left hand side of assignment operator = (char 12)

[subtle helm]

so the least equal no. of chocolate u can get is 3

u can get 3 by buying 1/8 of the first box and 1/5 of the second box

[cyan wren]

Mhm yeah

But why r we doing hcf?

[subtle helm]

but its very weirdddd

buying 1/8 and 1/5 of boxes

are u sure its not lcm (for the last time)

[cyan wren]

Yes 1000000 percent

[subtle helm]

cuz i searched the problem online and they're using lcm

[cyan wren]

24/3 and 15/3 = u get the least number of boxes

[cyan wren]

Yeah I also checked but my teacher told this to me

[subtle helm]

this is correct if we use lcm as well

[cyan wren]

O

[subtle helm]

but using hcf is faster, ye

,calc lcm(15, 24)

[boreal girder]

Result:

120

[subtle helm]

cool

[cyan wren]

Oo

[subtle helm]

so dividing by the no. of each

,calc 120/24

[boreal girder]

Result:

5

[subtle helm]

,calc 120/15

[boreal girder]

Result:

8

[subtle helm]

lets see why this happens

lcm = hcf * non common factors

so

hcf = lcm/non-common factors

uhh

this will be very hard to write

lets say

${a = b \cdot c}$ and ${d = b \cdot e}$

[boreal girder]

k

[cyan wren]

I think it's not that complex yk..

My little brain cant take that much info

[subtle helm]

we can see that ${\text{hcf}(a,d) = b}$ and ${\text{lcm}(a,d) = b \cdot c \cdot e}$

[subtle helm]

nah its not much

[cyan wren]

K

[boreal girder]

k

[subtle helm]

so from this

the general procedure would be to divide lcm by no. of choc in each box

so

[cyan wren]

Mhm

[subtle helm]

For no. of choc in box ${a}$, $\frac{\text{lcm}(a,d)}{a} = \frac{\cancel{b \cdot c}\cdot e}{\cancel{b \cdot c}} = e$

[boreal girder]

k

[subtle helm]

doing similar thing to the other box

we will find

that u need to buy $c$ boxes of box ${d}$

[cyan wren]

Mhm..

[boreal girder]

k

[subtle helm]

lets retry our calculation if we divide each box by the hcf

[cyan wren]

It's 8 and 3

[subtle helm]

$\frac{b\cdot c}{\text{hcf}(a,d)} = \frac{b \cdot c}{b} = c$

[boreal girder]

k

[subtle helm]

similarly, u will find that by dividng box d with hcf, u get the no. for box a

so essentially both methods work

[cyan wren]

Oo

[subtle helm]

but u kinda have to switch the numbers for hcf

for example if we take the 24 from box a and do the divisin 24/3 = 8. we will get the no. for box b

the opposite thingy wont happen if we use the lcm method which is clearer

[cyan wren]

[subtle helm]

but this is js faster

[cyan wren]

Yepp

[subtle helm]

ok

u got it right? T^T

[cyan wren]

Mhm yeah

[subtle helm]

cool

anymore question?

[cyan wren]

I finished real numbers

I godda go to statistics

And then trigonometry..

Dayummm

I will let you know if I had doubts but tysm for ur time

[subtle helm]

aight

[cyan wren]

Really appreciate it

[subtle helm]

i suggest closing this channel and finding another helper cuz im really eepy

[cyan wren]

Samee here

What's the time there?

It's past midnight

[subtle helm]

2.50

.close

[compact pewter]

Channel closed

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Remember:
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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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[sudden spade]

Do you guys know about Brewster angle? I’m pretty sure my measured theta p at 33.5 is wrong somehow since it’s so far off from the 56.3 calculated value I think I did it wrong??

[sudden spade]

Do you think that the drawing is just not to scale?

[subtle helm]

is this the brewster angle from polarization from physics?

[sudden spade]

Yes

where the reflected and refracted ray make 90 degree angle

The lab manual gives this drawing so theta p is just the incident ray angle in my opinion

[subtle helm]

i believe that u got the wrong angle

@sudden spade the 56.3 is the brewster angle

however

${\varphi_p}$ is the angle of refraction

[boreal girder]

k

[sudden spade]

hold the phone mane.

but brewsters angle is the angle of incidence ?

[subtle helm]

my knowledge abt brewster angle is a little rusty

[sudden spade]

ok thanks anyways man big fan of ur work

[subtle helm]

shi

i really need to step up my game

@sudden spade are u sure u recorded the right angles

using snell's law (which im still familiar with tho lol)

$\frac{\eta_2}{\cancel{\eta_1}} = \frac{\sin(33.5^\circ)}{\sin(\varphi)}$. Therefore, ${\varphi = \arcsin\left(\frac{\sin 33.5^\circ}{\eta_2}\right)}$

[boreal girder]

k

[subtle helm]

,w arcsin(sin(33.5 deg)/ 1.53) in deg

[boreal girder]

Results provided by WolframAlpha

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

Requested by nuclearinteraction

[subtle helm]

180 should = 21.145 + 90 + brewster angle

,calc 21.145 + 90 + 33.5

[boreal girder]

Result:

144.645

[subtle helm]

there is smth missing

[boreal girder]

Result:

167.945

[subtle helm]

calculation involving the correct brewster angle:

,calc 33.155 + 90 + 56.8

[boreal girder]

Result:

179.955

[subtle helm]

gives 180

so u prolly recorded the wrong angle

heres the summary of my yappery:

• correct brewster angle: 56.8 deg approximately
• ur recorded angle of 33.5 is very likely wrong (u might perhaps recorded the angle of refraction)

[compact pewter]

@sudden spade Has your question been resolved?

[sudden spade]

I think ur wrong but i posted in the physcis channel ill see what they say

[subtle helm]

Okie

@sudden spade nvm I found out what I got wrong

It's leaving the prism to the air

Not going from air to prism

smh

[sudden spade]

yes i thoguth of that but im not sure. so i need to use arctan(air/rhombus) i believe but i wasnt sure

[subtle helm]

i was using rhombus/air this entire time

no wonder i got the wrong result

[sudden spade]

so ur saying that my measured value is right

[subtle helm]

smh. im so srry 🤦‍♂️

[sudden spade]

i think ur right now k

[subtle helm]

most likely

off by 0.4

[compact pewter]

Channel closed

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

Find the Area boundrr by the curve $x^2=4y$ and straight line $x=4y-2$

[boreal girder]

!

[quartz yoke]

I dont know how to start

I roughly plotted the graph

[inland nacelle]

hello

[quartz yoke]

Halo

[inland nacelle]

what you want to do is to express this area in terms of integrals

[quartz yoke]

Yes

[inland nacelle]

can you show your graph?

[inland nacelle]

Nice

[inland nacelle]

Can you do this?

[quartz yoke]

Yes but

The issue

Lemme consider the line as y2 and parabola as y1

[inland nacelle]

ok

[quartz yoke]

Is it y2-y1 or y1-y2?

[inland nacelle]

you want the area bounded right

[quartz yoke]

Ye

[inland nacelle]

from your diagram

wait i draw a diagram gimne a sec

[quartz yoke]

Kk

[inland nacelle]

you found your two roots (places where the values are equal) already

can you see that the integral of the parabola from root to root

is the green area only?

[quartz yoke]

How is the green area even considered?

[inland nacelle]

the green area is the area under the curve

it is the area under the parabola

can you see that?

[quartz yoke]

Ye

[inland nacelle]

thus it is the integral of the parabola from root to root

can you see that

[quartz yoke]

I dont know what root to root means

[inland nacelle]

Oh

you calculate the points where the curves are equal

[quartz yoke]

Ooh

Ye its -2 to +2

[inland nacelle]

the two red points essentially

[inland nacelle]

wait let me check this

[quartz yoke]

The left red point is -1 and right is +2

[inland nacelle]

yes -1

not -2

[quartz yoke]

Oh

[inland nacelle]

you are correct in this one

good job

[inland nacelle]

so you want to find the red area right?

[quartz yoke]

Ye

[inland nacelle]

and you know the green area already (by integrating the parabola from -1 to 2)

[quartz yoke]

Shud we even bother that?

[inland nacelle]

well i dont think there's a way to find the red area with a single integral

and note that the integral of our straight line is the red + green shaded area

can you see that?

[quartz yoke]

Ye but

Since parabola has extra area covered by the line

Wont it be y2-y1 as y2 is limiting

[inland nacelle]

yes

you found what i was trying to say

nice

[quartz yoke]

So the limits are

-1 to 2?

[inland nacelle]

the integral limits are -1 to 2 yes

[quartz yoke]

I have few more questions

Find the area of region enclosed by $y^2=4ax$ and line $y=mx$

[boreal girder]

!

[inland nacelle]

But you know how to do the previous one right

[inland nacelle]

this is similar, what have you tried?

[quartz yoke]

Ye solved and got ans

[inland nacelle]

Nice job!

[quartz yoke]

Ye i drew rough graph

My teacher taught like 2 to 3 questions based on area bw 2 curves and left

Ill open if i need any help

Thanks carbon

.close

[compact pewter]

Channel closed

Closed by @woven loom

Use .reopen if this was a mistake.

[inland nacelle]

np

good luck doing calc!

[quartz yoke]

.reopen

[compact pewter]

✅

[quartz yoke]

,rcw

[quartz yoke]

How is it y2-y1 bruh 😭

.close

[compact pewter]

Channel closed

Closed by @woven loom

Use .reopen if this was a mistake.

[compact pewter]

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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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[sleek burrow]

Define Pi as the set of values obtained by selecting each subset of {x1, x2, x3,....,xi} and summing its members.

Prove Pi = Pi-1 Union (Pi-1 + xi)

[timber cipher]

The + is weird. Is that supposed to be a different symbol?

Or is it just adding xi to every member of Pi-1

[sleek burrow]

So like if set is {6, 10, 14}

P0 = {0}
P1 = {0, 6}
P2 = P1 union P1 + 10
Which is
{0, 6, 10, 16}

[sleek burrow]

Yes

Sorry should have mentioned that

Also if possible guide me through the thought process

[timber cipher]

You seem to understand it. Was there anything you were unsure of?

Oh the actual proof

[sleek burrow]

Yeah

[timber cipher]

Do you have a method you're supposed to use?

[sleek burrow]

No, no such thing

Proving Pi-1 Union (Pi-1 + xi) to be subset of Pi,

I think this suffices?
"Pi-1 is subset of Pi because Pi-1 is a subset of Pi, so members of Pi-1 is a value included in by definition.

For each pair {Pi-1, xi} in Pi, will form a subset of Pi, so it's sum is also included in Pi

Actually, there's really only one pair {Pi-1, xi}

Now the other way around:
Proving Pi subset of Pi-1 Union (Pi-1 + xi)

Although I'd appreciate if there's a cleaner way to approach this proof

[compact pewter]

@sleek burrow Has your question been resolved?

[sleek burrow]

<@&286206848099549185>

[compact pewter]

@sleek burrow Has your question been resolved?

[ripe ether]

this is very vague

are x_1,...,x_i distinct here?

you are talking about subsets which means {1,1} is the same as {1}

if x_1=1, x_2=1, are you allowing 1+1, because the subset containing x_1, and x_2 is just {1}?

I am just assuming here you have a sequence $(x_i){i=1}^\infty$ and you defined $$P_i:=\left{\sum{j\in J} x_j: J\subseteq {1,2,\ldots, i}\right}$$

[boreal girder]

qwertytrewq

[ripe ether]

If my assumption is correct (that you are not really taking subsets but taking some elements with index less than i), then you should try case work, if $J\subseteq {1,2,\ldots, i}$ then either $J$ is a subset of ${1,2,\ldots, i-1}$ or $J$ contains $i$.

[boreal girder]

qwertytrewq

[compact pewter]

@sleek burrow Has your question been resolved?

[sleek burrow]

Hey

[sleek burrow]

Not necessarily distinct but monotonically increasing order

[ripe ether]

monotonically increasing means x_1<x_2<x_3 or x_1<=x_2<=x_3?

[ripe ether]

the main confusion comes from when you said subset. Taking subset is essentially throwing out the same element

[sleek burrow]

I think that it's supposed to be like a key value pair

Where all elements are distinct

But may have same value

But unique keys

[ripe ether]

so if 1=x_1=x_2=....., idk if, the way you defined it, P_2 is equal to {0, 1,2} or {0, 1}

[sleek burrow]

Yea it would be the former

[ripe ether]

ok so really your definition of P_i is this

you are taking a subset of the index, and summing over that index.

[sleek burrow]

Pi is listing all possible sums made up from its subsets

[ripe ether]

in that case, separate into two cases: the subset of index contains i, and the subset of index does not contain i

[ripe ether]

I don't think this is precise, again subset is throwing away equal elements.

[sleek burrow]

This was my logic

[sleek burrow]

The elements arent the values

But rather key value pairs

with each element having unique key

So like

{1,1,1} is really
{{1, UID1}, {1, UID2}, {1, UID3}}

And we represent the sums of the values of sums of subsets

[ripe ether]

its better to write it as sequences

this is the same as sequences

but i get what you mean

[sleek burrow]

Doesn't order matter in a sequence

But I guess for the purposes of this problem it's irrelevant

[ripe ether]

UID 1, UID2,... is an ordering

[sleek burrow]

No I am not saying there's an order

This was just an example

I'm just claiming UIDs are unique

They coudl be anything

[ripe ether]

to some extent you are using order to extract all the element x_1,...,x_i, but thats besides the point, it shouldn't matter as you've pointed out

[sleek burrow]

I'm not sure how were using order

ith element isn't ith in order within the data structure, but the ith iteration

[ripe ether]

we don't need order for the argument, but given your definition we can impose an ordering by listing the element as x_1,x_2,...

we are not ordering x_i by their size

just listing out the elements in order

its index is in order

[ripe ether]

and this is a definition for P_i

where you are taking a subset of the indicies (which is the same as picking out those x_j's), and then summing them

[sleek burrow]

Yeah

[ripe ether]

using this interpretation you can formally prove your statement (which is already somewhat outlined in your original idea): for an index subset $J\subseteq {1,\ldots, i}$ , either J contains $i$ or it is a subset of ${1,\ldots, i-1}$

[boreal girder]

qwertytrewq

[ripe ether]

this is to show that $P_i\subseteq P_{i-1}\cup {p+x_i:p\in P_{i-1}}$

[boreal girder]

qwertytrewq

[sleek burrow]

What is i here

[ripe ether]

same i in your definition of P_i

[sleek burrow]

Because we are talking of sums of subsets

J contains i isn't making sense to me

Because once you have put the ith element in to Pi, you must also put all elements corresponding to every element in Pi-1 added with the ith element

[ripe ether]

we are using the definition listed here

where J is used

its just a set of indices

either you have this index i (in which case you are including x_i in your sum) or you are not

[sleek burrow]

Ok true

Ah okay

[ripe ether]

Lemme be more formal: Suppose that an element is in $P_i$, then by definition it is equal to $\sum_{j\in J}x_i$ where $J\subseteq {1,\ldots, i}$

[boreal girder]

qwertytrewq

[ripe ether]

hopefully this clears things up

now you case work on whether J contains i or not

[sleek burrow]

Uh

Thisbis confusing me

Its sum over all xjs right

Not xis

[ripe ether]

yes

for each $j\in J$ you add $x_j$ to the sum

[boreal girder]

qwertytrewq

[ripe ether]

$J$ is merely selecting what things to sum over

[boreal girder]

qwertytrewq

[sleek burrow]

Okay alright

I got the one side of the proof

Now what about the other way

Pi-1 Union Pi-1 + xi being a subset of Pi

[ripe ether]

this is proving the other subset inclusion

[sleek burrow]

Wait what

[ripe ether]

picking any element in $P_i$, we show that its in $P_{i-1}\cup (P_{i-1}+x_i)$

[boreal girder]

qwertytrewq

[ripe ether]

which is essentially showing that $P_i\subseteq P_{i-1}\cup (P_{i-1}+x_i)$

[sleek burrow]

Yeah

Yeah

But we also need the other way around right

[boreal girder]

qwertytrewq

[ripe ether]

now pick any element in $P_{i-1}$ or $P_{i-1}+x_i$ and show that its in $P_i$

[boreal girder]

qwertytrewq

[sleek burrow]

Hmmm

True

Honest I'm just utterly unfamiliar with proofs

I need a way to formalise it properly

Thanks a lot tho

[ripe ether]

yeah but to do that you need to formally state the problem: thats why i gave the precise definition of $P_i$ so it is easier to come up with a formal proof

[boreal girder]

qwertytrewq

[sleek burrow]

Hmmm

[ripe ether]

Once your define
$$P_i:=\left{\sum_{j\in J} x_j: J\subseteq {1,2,\ldots, i}\right}$$
and define
$$(P_{i-1}+x_i):={p+x_i:p\in P_{i-1}}$$
then the proof that $P_i=P_{i-1}\cup (P_{i-1}+x_i)$ is more or less just definition verification

[boreal girder]

qwertytrewq

[compact pewter]

@sleek burrow Has your question been resolved?

[sleek burrow]

Thanks a lot @ripe ether

.close

[compact pewter]

Channel closed

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[lost jackal]

Given that a and b are odd coprime numbers such that a, b >= 1
and an integer L that is either
L = 2m if L is even
L = 2m + 1 if L is odd
for m >= 1
Can I prove that
$$ am^2 + bm^2 + am \not\equiv 0 \mod(L)$$

[boreal girder]

Sherif Player

[lost jackal]

For even L we can prove that by contradiction
let
$$am^2 + bm^2 + am \equiv 0 \mod(2m)$$
then
$$am + bm + a \equiv 0 \mod(2)$$
so it needs to be an even number
$$am + bm = (a + b)m$$
would always be even because both a and b are odd
which means (a + b) is always even
so
$$ a \equiv 0 \mod(2)$$
which contradicts the given information
thus proven by contradiction

[boreal girder]

Sherif Player

[lost jackal]

For odd L
Let
$$am^2 + bm^2 + am \equiv 0 \mod(2m + 1)$$
$$am+bm+a \equiv 0 \mod(2m+1)$$
$$ (a+b)m + a \equiv 0 \mod(2m+1)$$
Because (a+b) is even as both a and b are odd
Then it is possible for the statement to be true if and only if the 2 factors of (a+b) are 2 and a only.
But because b is a coprime of a then (a+b) is not divisible by a
thus the statement can't be true.

[boreal girder]

Sherif Player

[lost jackal]

Is my solution correct?

I am a bit not sure about the last conclusion of the odd L proving

[compact pewter]

@lost jackal Has your question been resolved?

Channel closed

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Remember:
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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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[strong zealot]

Hi. I'm stuck with this problem. I'm pretty bad at proving questions and also the sigma notation, so I don't know what to do period.

Part A and C I understand, but cannot/dunno how to prove algebraically.

[bleak dock]

part a) is just a "write down" question so you don't need to show work

how about part b? what have you written down for that?

[compact pewter]

@strong zealot Has your question been resolved?

[strong zealot]

I'm stumped for part b

[tall zephyr]

hello I need help to solve this math exercice : If it's possible to send me the answer on my privat channel ?

[strong zealot]

I tried to expand the LHS first to see if i could get the sum of a AP or GP series but could not find a difference or ratio

[bleak dock]

okay, so the hint $\theta \ge \frac{1}{2} \tan \theta$ applies to all $\theta$ in that domain

so $\theta_1 \ge \frac{1}{2} \tan \theta_1$

$\theta_2 \ge \frac{1}{2} \tan \theta_2$

[strong zealot]

But i can see that every angle of theta is less than pi/2

[boreal girder]

south

[bleak dock]

what happens if you add all of these inequalities from theta_1 to theta_n?

[bleak dock]

indeed, all thetas are angles in a right triangle and that's why

the maximum possible for any theta is (180 - right angle) degrees

[strong zealot]

Sorry, i don't get it yet

[bleak dock]

well and then it's the smaller angle in the right triangle so by symmetry that restricts theta to being 90/2 = 45 degrees max

anyways that's just me talking about the domain

[bleak dock]

this is the key step

[strong zealot]

How would you format your working out though if you were in an exam?

(If i may inquire)

[bleak dock]

you need to understand what you are writing down first

if you understood, then you should see how $\sum_{n = 1}^k \theta_n \ge \sum_{n = 1}^k \frac{1}{2} \tan \theta_n$

[boreal girder]

south

[strong zealot]

I need help understanding....

[bleak dock]

okay, so if you add the bottom two lines together, you get $\theta_1 + \theta_2 \ge \frac{1}{2} \tan \theta_1 + \frac{1}{2} \tan \theta_2$ right?

[boreal girder]

south

[strong zealot]

right

[bleak dock]

yeah, and what happens when you add this new inequality to $\theta_3 \ge \frac{1}{2} \tan \theta_3$?

[boreal girder]

south

[strong zealot]

it supports this?

oh

OH.

Wait, I starting to understand

[bleak dock]

yeah, so we've shown that when you add theta_n for n = 1 to 3

that's more or equal to the sum of 1/2 tan theta_n, again from n = 1 to 3

(n = 1 to 3 means n = 1, 2, 3)

[strong zealot]

Thank you

[bleak dock]

right, and the next property is actually the distributive law

so if I have $\frac{1}{2} x_1 + \frac{1}{2} x_2 + \frac{1}{2} x_3 + \frac{1}{2} x_4$

[boreal girder]

south

[bleak dock]

that's just equal to $\frac{1}{2} (x_1 + x_2 + x_3 + x_4)$

[boreal girder]

south

[bleak dock]

in summation notation, we'd write $\sum_{n = 1}^4 \frac{1}{2} x_n = \frac{1}{2} \sum_{n = 1}^4 x_n$

[boreal girder]

south

[bleak dock]

on the left-hand side, we have the sum of (1/2 x_1), so 1/2 is on the inside of the summation

on the right hand side, we take the 1/2 out and multiply

[strong zealot]

right, i understand up to here

[bleak dock]

right, so we have $\tan \theta_n = \frac{1}{\sqrt{n}}$ from part a

[boreal girder]

south

[bleak dock]

and we can take the 1/2 out

[boreal girder]

south

[strong zealot]

I got up to that, yeah.

How do we turn 1/sqrtn to 1/n

[bleak dock]

right, so the technique we used so far was to look at an inequality which would be true for every single n

so then we would just add the cases n_1, n_2, n_3 and so on all together to make the summation

[bleak dock]

can you guess what the inequality is for this step then?

[strong zealot]

no,

[bleak dock]

well if we can prove $\frac{1}{2} \sum_{n = 1}^k \frac{1}{\sqrt n} \ge \frac{1}{2} \sum_{n = 1}^k \frac{1}{n}$ we're done

[boreal girder]

south

[strong zealot]

right?

[bleak dock]

convince yourself first

[strong zealot]

hmm.

[bleak dock]

cause then you can just 'chain' the inequalities

[strong zealot]

Chain?

[bleak dock]

$\sum_{n = 1}^k \theta_n \ge \sum_{n = 1}^k \frac{1}{2} \tan \theta_n \ge \frac{1}{2} \sum_{n = 1}^k \frac{1}{\sqrt n} \ge \frac{1}{2} \sum_{n = 1}^k \frac{1}{n}$

[boreal girder]

south

[bleak dock]

you can combine the two inequalities like so

pick up from the third term

[strong zealot]

I haven't heard of combining inequalities like this before

INteresting

[bleak dock]

like okay let's just use numbers to illustrate the concept

if 10 > 9 > 8

and 8 > 7

you can combine these two to get 10 > 9 > 8 > 7

that's what we did

you combine the two inequalities where the 8 is

[strong zealot]

nvm i'm listeninh

listening*

[bleak dock]

btw that was different

[strong zealot]

got it.

oh wait.

so, possiblly

[bleak dock]

so from $\theta \ge \frac{1}{2} \tan \theta_n$ we went to $\sum_{n = 1}^k \theta_n \ge \sum_{n = 1}^k \frac{1}{2} \tan \theta_n$ right?

[strong zealot]

We have a>b, and we need to prove a>c, so we have to prove b>c to prove a>c?

[boreal girder]

south

[bleak dock]

yes! that's the reasoning

so what are b and c here?

[strong zealot]

from this;

b and c respectively?

[bleak dock]

wait not that one

this one instead

[bleak dock]

here

[strong zealot]

So we're done?

[bleak dock]

no

[strong zealot]

since sum of 1/sqrt n is >= sum of 1/n

by whatever proof

[bleak dock]

we still need to prove that step, like yes it's true, but why?

[strong zealot]

hmm.

[bleak dock]

do you agree we just need to show that 1/sqrt(n) >= 1/n?

[strong zealot]

yes, that was what i was thinking

[bleak dock]

yeah

well there's a condition and that condition is n > 0

but I mean this is a summation, so we're guaranteed that

[strong zealot]

I'm just stumped as to what is sufficient working out

I may ask chatgpt to save us time

[bleak dock]

well as I said, it starts with you understanding why first

[strong zealot]

since just by looking at it, it should be true for all n>0

[strong zealot]

good point...

[bleak dock]

the key reasoning is if I have 2 <= 3 for example

1/2 >= 1/3

the inequality sign flips

[strong zealot]

so you're saying we're supposed to state that n>=sqrtn

[bleak dock]

yes! that's why

so now we can look at how to write this properly

so $\sqrt{n} < n$ for all (real) $n > 0$

[boreal girder]

south

[bleak dock]

the for all part is really important

[strong zealot]

duely noted.

[bleak dock]

thus $\frac{1}{\sqrt{n}} > \frac{1}{n}$

[boreal girder]

south

[bleak dock]

and now we can just state this step as a consequence

move from the inequality to the summation

[strong zealot]

wow

[bleak dock]

also there's a minor point in that if $x > y$, that means $x \ge y$ automatically by definition

$x \ge y$ means $x = y$ or $x > y$

[boreal girder]

south

[bleak dock]

you don't have to point that out

just for your understanding

[strong zealot]

hm?

[bleak dock]

you don't have to mention this in your working

[strong zealot]

oh i get it

[bleak dock]

what do you mean by ''that"?

we covered a lot of things

oh

[strong zealot]

no sorry i was confused

I was confused in how a>b is automatically assumed to be a>=b. Could you elaborate a bit more?

[strong zealot]

idk why i changed the variables. mb

[bleak dock]

oh yeah it's because of (mathematical) logic

perhaps I'll draw something to illustrate

yeah so as you can see, x > y is a subset of x >= y

whatever is inside the blue circle, must also be inside the black circle

[strong zealot]

ah. Again, its something i haven't been taught before.

[bleak dock]

yeah so this relates to the concept of sets ofc

[strong zealot]

But, if i were asked to solve for ax^2+bx+c>0, for example

aren't my answers for x still >

[bleak dock]

yes

[strong zealot]

In interval notation, they would be )(, not [ ]

[bleak dock]

yeah so what I mean is that if I have x is in (2, 3)

x is automatically in the closed interval [2, 3]

[strong zealot]

alright. "by definition"

[bleak dock]

because this is just the same as:

1. x is in (2, 3), or
2. x = 2
3. x = 3

[strong zealot]

Anyways, moving onto part C, i think drawing a graph would be suffice? As the question says to show, and this is simply visual proof that trapezoidal rule is less accurate than integration

[bleak dock]

if we use a simpler example, if I say that x = 3 for example

that automatically means that x is one of 3, 4, 5

we're always going from a case with fewer numbers to a case with more numbers when I say 'automatically'

[strong zealot]

right

[bleak dock]

yeah I trust that you've drawn it correct

the summation is a left Riemann sum to be precise

[strong zealot]

Which?

[bleak dock]

you should have drawn something like this

[bleak dock]

for part c

[strong zealot]

Is the integral sign or the summation sign the reimann sum

[bleak dock]

the summation sign is the Riemann sum

[strong zealot]

oh. mb lol.

[bleak dock]

yeah then if you integrate 1/n dn from 1 to k

but then you take the limit as k to infinity

you get ln |n| + c which goes to +infinity

and then from part b, the sum of the angles is even larger than the integral = +infinity

[strong zealot]

I understand, but is that what the question was asking for?

[bleak dock]

if you reread question d, yes

it's subtle but if it's asking about the "total" angle, that's the sum of the angles from 1 to n

but then it's implied that you can take n to infinity for the limit

like "if you keep extending the pattern forever and ever, does the total of the angles ever reach a constant number?"

[strong zealot]

oh. Similar to a limiting sum?

[bleak dock]

yeah

like "is there a limiting sum?"

[strong zealot]

Here is a recap of what I understand for question D:

Since the integral (k,1) 1/n dn goes to infinity rather than a single constant, the sum of theta must be " greater than infinity" or at least, does not have a limiting sum because of the inequality "Chain"

so therefore, as the sum of theta -> infinity, it does not approach a "limit" i.e constant

?

[bleak dock]

yep! that's correct

so another way of thinking about of the chain is that the lower bound of the sum of theta is +infinity

that's the minimum already, +infinity

it has to be infinite

[strong zealot]

right.

I suppose thats it

thanks a lot. for real.

.close

[compact pewter]

Channel closed

Closed by @strong zealot

Use .reopen if this was a mistake.

[compact pewter]

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[rose goblet]

hey i need help with integral of sqrt(ax^2 +b) dx

[kind viper]

what's known about a and b?

[rose goblet]

i have seen different approaches on this depending on a,b

[rose goblet]

nothing

[kind viper]

not even that they are real numbers?

[rose goblet]

oh yea that

well not that

but lets do that first

[kind viper]

...

what

sorry, you've confused me.

do we know $a, b \in \bR$ or not?

[boreal girder]

Ann

[rose goblet]

lets start with this case

[kind viper]

what other cases even exist

do you want to allow a and b to be fucking complex or something

this is extremely suspicious!!!

!xy

[compact pewter]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[rose goblet]

why i cant?

[kind viper]

complex numbers and square roots are quite a big can of worms.

show the original problem please.

[rose goblet]

there isnt one

but lets discuss real numbers

i wanna see if there is a general approach

[kind viper]

no there isn't

[rose goblet]

because in some cases i saw this substitution

[kind viper]

you have to do case work depending on if the integral looks like sqrt(x^2 + 1) or sqrt(x^2 - 1) or sqrt(-x^2 + 1)

you're gonna get different-looking shit

up to some constant BS

sqrt(-x^2 - 1) is not real so probably not under your consideration

[kind viper]

no, there is no general approach. you have to do it differently in each specific case.

does this answer your question?

[solemn matrix]

please post your original problem

your problem looks confusing

[kind viper]

OP said there isn't one.

[solemn matrix]

it just looks like you made one

[rose goblet]

yes

[young raft]

there is a closed form tho, no?

[young raft]

if you want it to solve general case (if we take "nice" a and b and it's an indefinite intergal), then it's equivalent to solving $\int \sqrt{x^2 + c} dx$

[boreal girder]

artemetra

[young raft]

for some c and up to scaling yk

[rose goblet]

sub i was suggested by book was z-x root(a)= root(ax^2+b)

[young raft]

okay so there is a book and there is context