help-49

[midnight plank]

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[graceful ferry]

$$\iint_D3x^2dxdy$$
D is the domain bounded by:
$y=x^2$ $2y=x^2$ $y=\frac{1}{x}$ $y=\frac{3}{x}$

[grand pond]

prograce

[graceful ferry]

Change of variables
u=xy
$v=\frac{y}{x^2}$
I got $J=\frac{y}{x^2}(1+\frac{2}{x^2})$
How do I cancel terms or something

[grand pond]

prograce

[graceful ferry]

f(u,v) is hard to write

Nvm i calculated J wrong

J=1/3v

[graceful ferry]

Help

[midnight plank]

@graceful ferry Has your question been resolved?

[languid mica]

Polar ?

You dont need change of variables i think

$$\int_{\frac{3}{\sqrt[3]{6}}}^{1} \int_{\sqrt{2y}}^{\sqrt{y}} 3x^2 , dx,dy$$

[grand pond]

taebek

[languid mica]

I think its this

[graceful ferry]

But its way easier with cjange of variables

1/2<=v<=1, 1<=u<=3

[graceful ferry]

I have to do thid? Is this easier than finding f(u,v)?

[languid mica]

What did you set

[languid mica]

What was the change of variables?

[midnight plank]

@graceful ferry Has your question been resolved?

[shadow schooner]

compute v/u to isolate x

[languid mica]

Make sure det =/ 0

Cause then its irreversible so wont work

[graceful ferry]

u=xy
v=y/x^2

[graceful ferry]

Yea J=1/3v does not equal zero

[graceful ferry]

Did that and got f(u,v)=1

,,\int_{1/2}^{1}(\int_{1}^{3}\frac{1}{3v}du)dv

[grand pond]

prograce

[languid mica]

Solve this for x in terms of u,v

[graceful ferry]

x^3=u/v

[graceful ferry]

Im getting a wrong answer

[languid mica]

Then x= (u/v)^ 1/3

[graceful ferry]

Oh

[languid mica]

So f(v,u) = (u/v)^2/3

[graceful ferry]

mulitply by 3

[languid mica]

Yes

Sorry forgot 3

[graceful ferry]

,,\int_{1/2}^{1}(\int_{1}^{3}\frac{u^{\frac{2}{3}}}{v^{\frac{5}{3}}}du)dv

[grand pond]

prograce

[graceful ferry]

This is wrong too bruh

[last slate]

Hello

[raven zenith]

hello, this channel is occupied atm. if u need help with a math problem, see #❓how-to-get-help . otherise, go to #360643390594875392 for chat welcome to the mathcord

[graceful ferry]

Oh my god ithe function is 3x^3

It is right

.solved

[midnight plank]

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[graceful ferry]

Thanks

[midnight plank]

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[novel flax]

heyo i wanted to ask a ques abt physics, simple one conceptual, related to fluids
and its pretty urgent

[novel flax]

will a drop of water immersed in kerosene be the same as the mass of a normal drop of water?

i had an experiment on findind the surface tension and interfacial surface tension of water and water with kerosene

i got different values of 1 drop of water

[raven zenith]

idk if this is math, you should maybe check out some servers in #old-network

[novel flax]

.close

[midnight plank]

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[midnight plank]

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[vast loom]

...

[midnight plank]

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[vast loom]

<@&268886789983436800>

again bro

wtf

3x in a row

[midnight plank]

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[bronze mason]

how is the second blank not mean to be x? I've watched a bunch of videos on inverse functions and none of them seem to be helpful

[tired osprey]

look at the part behind the info text

[bronze mason]

this part?

[tired osprey]

ye

[bronze mason]

ummm

[tired osprey]

if sin^-1(x) = y, the sin(..) = ??

[bronze mason]

ummm

it would be

[steel crest]

maybe do something to both sides to isolate x

just a thought

[sharp coral]

look closely at the previous equation

[bronze mason]

wait but I dont have an x on both sides?

[steel crest]

that's good, you only want x on one side

[bronze mason]

oh wait

multiply by sin^-1? (on both sides?

[steel crest]

sin^-1 isnt a number, you dont multiply things by it

$$f^{-1}(x) = y$$

[grand pond]

gfauxpas

[steel crest]

if you have something like this

what happens if you apply f to both sides?

[bronze mason]

nothing right?

[steel crest]

:O

nothing?!!?

[bronze mason]

uhhh

I think so??

[steel crest]

why would applying a function do nothing?

[tired osprey]

do you know what $$f{f^{-1}(x)} $$ is?

[lavish venture]

()

not {}

[grand pond]

Hylke54

[steel crest]

if you apply f to both sides you get

[tired osprey]

omg

[lavish venture]

nope

[bronze mason]

umm

[steel crest]

$$f(f^{-1}(x))=f(y)$$

[grand pond]

gfauxpas

[steel crest]

do you see why that's true?

[bronze mason]

yeah

but

[steel crest]

but?

[bronze mason]

how do I do that here?

[steel crest]

well

what's the definition of $f^{-1}$?

[grand pond]

gfauxpas

[steel crest]

there are two common definitions, which one were you taught?

[bronze mason]

ummm

I don't believe I was really taught either I skipped precalc

[steel crest]

well I think it' sa good idea to understand what the notation means before solving a problem with it

[bronze mason]

just the inverse of f?

[lavish venture]

lmao

[lavish venture]

how is the inverse of f defined though

[steel crest]

I kind of assumed you knew what the problem meant if you were trying to understand the problem

[lavish venture]

how are f and f^-1 related

[steel crest]

$f(f^{-1}(x)) = x$ for all $x$ in the domain of $f^{-1}$, and

[grand pond]

gfauxpas

[steel crest]

$f^{-1}(f(x)) = x$ for all $x$ in the domain of $f$

[grand pond]

gfauxpas

[bronze mason]

I don't know what the fancy math term is called but 1 will be the same number but negative

[steel crest]

that's the definition of a function inverse

[lavish venture]

no

[tired osprey]

<@&268886789983436800>

[steel crest]

<@&268886789983436800>

[lavish venture]

that would be -f

[lavish venture]

they can see deleted chats

[little anvil]

<@&268886789983436800>

[steel crest]

oh, what power

look at these carefully madeye

[lavish venture]

thanks mate

[steel crest]

there are the definition of function inverse

do you understand the notation?

[bronze mason]

I think so?

[steel crest]

great, so

if you have

$$\sin^{-1}(x)=y$$

[grand pond]

gfauxpas

[steel crest]

what happens when you apply sin to both sides?

[bronze mason]

we basically just replace f for sin right?

[steel crest]

sin is an example of a function

f is a generic function symbol

[bronze mason]

ok wait

so

its x

[steel crest]

do both sides of the equation at the same time

here

if you understood this

then do the same for the equation:

$\sin^{-1}(x)=y$

[grand pond]

gfauxpas

[steel crest]

apply sin to both sides

[bronze mason]

sin(y)?

[steel crest]

where'd the equation go

you lost half of it

[bronze mason]

well that was right : /

not sure I fully understand it though

[steel crest]

maybe because you're not following my attempts to teach you and answering my questions...

:/

[bronze mason]

I'm trying : /

I have no idea how to follow them

[steel crest]

okay let's try this simply

[bronze mason]

ok

[steel crest]

$x=2y$

[grand pond]

gfauxpas

[steel crest]

apply the function f to both sides, what do you get

[bronze mason]

f(x)= f(2y)

[steel crest]

great!

now apply the function sin to both sides, what do you get

[bronze mason]

sin(x) = sin(2y)

[steel crest]

🎉

apply the function sin to both sides

what do you get?

[bronze mason]

sin(sin^-1(x)) = sin(y)

[steel crest]

yes!!!

now

we can simplify this

by using the definition of a function inverse

here, f = sin, f^-1 = sin^-1

[bronze mason]

ok now I'm confused though

[steel crest]

at which part?

[bronze mason]

on how to use this

to simplify

[steel crest]

$f(f^{-1}(x)) = x$ for all x in the domain of $f^{-1}$

[grand pond]

gfauxpas

[bronze mason]

this

[steel crest]

so if f = sin, then

$\sin(\sin^{-1}(x)) = x$ for all $x$ in the domain of $\sin^{-1}$

[grand pond]

gfauxpas

[bronze mason]

ok?

yeah so far so good

[steel crest]

so this simplifies

to what?

[bronze mason]

sin^-1(x) = y

: O

[steel crest]

NO

;__;

[bronze mason]

what

[steel crest]

just replace sin(sin^-1(x)) with "x"

because thye're equal

[bronze mason]

ok

[steel crest]

$$\sin(\sin^{-1}(x)) = \sin(y)$$

[grand pond]

gfauxpas

[steel crest]

becomes what?

replace sin(sin^{-1}(x)) with just x

what do you get as an equation?

[bronze mason]

ohhhh

x = sin(y)

I get it

so bassically

[steel crest]

https://tenor.com/view/confetti-purple-ubam-celebrate-gif-7733297067685171475

[bronze mason]

inverse function just deletes function

[steel crest]

undoes is a better way to think of it

but "for all ... in the domain of ..." is important

[bronze mason]

ok

thank you so much!!!

[steel crest]

that's why at the beginning of your question it says we're only considering the domain of sine limited to [-pi/2,pi/2]

because otherwise things break

np man

glad you fi nally got t

[bronze mason]

yeah I am too

so much math to do today ❤️

.close

[midnight plank]

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[steel crest]

.reopen

[midnight plank]

✅

[steel crest]

@bronze mason

[bronze mason]

yeah?

[steel crest]

just a concrete example of what goes wrong if you ignore the part about "for all blah in the domain of blah"

let h(x)=x^2 and let h^-1(x) = sqrt(x)

then h^{-1}(h(-1)) = 1

$$h^{-1}(h(-1)) = \sqrt{(-1)^2}=1$$

[grand pond]

gfauxpas

[steel crest]

so things can break

okay, that is all

[bronze mason]

uhhhh

[steel crest]

you dont follow?

[bronze mason]

no

[steel crest]

better example

$$\sin(2\pi)=?$$

[grand pond]

gfauxpas

[bronze mason]

atleast I just don't see why this doesnt work

[bronze mason]

ohhh

I can see why this 1 doesnt work

[steel crest]

similar reason

[bronze mason]

beause we know what sin(2pi) is

[steel crest]

becasue h(1)=h(-1)

and sin(2pi)=sin(0)

but sin^{-1}(sin(2pi)) = 0

because sine only has an inverse if you limit the domain to something like [-pi/2,pi/2]

anyway, if i confused you, just forget it, just wanted to give you a caution

you'll see more of stuff like that if you continue studying inverses

[bronze mason]

ok

[steel crest]

gl

[bronze mason]

I'll probably just forget about this until I actually need it

[bronze mason]

thank you! Ima need it 💀 this is problem 1 of 36 that I need to do today

.close

[midnight plank]

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[midnight plank]

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[quaint shoal]

How do I do these if the common unit circle angles can’t add up to the angle I’m solving for?

[quaint shoal]

Do use something like this and subtract the appropriate amounts??

[raven zenith]

you basically add and subtract them, yes

what specific problem is the one you're referring to that "can't add up?"

[quaint shoal]

so for 195, I would do 195-180 which is 15

question 2

[raven zenith]

actually

[zenith snow]

Try using 90 instead

[raven zenith]

psh psh victimizer, too many cooks in the kitchen

[zenith snow]

Ay fine fine
I m off

[raven zenith]

yes, because sin 195= sin 15

[quaint shoal]

why 90??

[raven zenith]

ty, explain why 90 tho

so @quaint shoal you actually only need to find sin 15

and you know sin 30, right?

(these are all in degrees btw)

[quaint shoal]

How would I use angle sum identity to find this

-45+30?

[raven zenith]

technically, sin 195=sin 15 already

bc on the unit circle

you're looping one more time

so you're actually landing back at the same place

[zenith snow]

So if u use 90 u can swap it off
For example
Sin(195)=sin(90+105)
And we know if there is an odd multiple of 90 in these cases
Sin becomes cos
And u could proceed with
Cos(105) like the first one u did

[raven zenith]

^^^^

[raven zenith]

-# that's kind of complicated 😓

[raven zenith]

@quaint shoal

[quaint shoal]

1/2

[raven zenith]

yes

and 15+15=30

[zenith snow]

I think they asked for exact values
Do the angles which give u that r 30,60,45,90...etc

[raven zenith]

so if u know 15+15=30, you can do sin(15+15)=sin 30

and u can use the addition formula, while plugging sin 30 in

so like sin(15+15)=1/2

wait a sec

sin 30

is not 1/2

[zenith snow]

It is

[raven zenith]

it is sorry

I was thinking in radians

[raven zenith]

does this make sense @quaint shoal ?

[quaint shoal]

Not rly

sry im rly drained today

[raven zenith]

that's ok

[quaint shoal]

trig identities got to me I think lol

[raven zenith]

do you agree that sin(15+15)=sin(30)?

[quaint shoal]

yea

[raven zenith]

alr

so the value u want is actually sin 15

so we don't simplify the LHS

instead, we plug in sin 30º, and we get sin(15+15)=1/2

and then, you can use the addition identity

sin(a+b)

[zenith snow]

May i intrude once again

I think by angles
They want 30,60,45,90

[raven zenith]

yeah ik he can use the half angle identity

[raven zenith]

?

[zenith snow]

Like exact values was mentioned in the question

[quaint shoal]

yea like the common angles

[zenith snow]

Yea i get u

[quaint shoal]

because this is meant to be no calc apparently

[raven zenith]

exact value of sin 195 right

[zenith snow]

Yeah i understood that

[raven zenith]

we're getting there

[quaint shoal]

would -30+45 work?

[raven zenith]

sure, but you also could have done sin(15+15)

[zenith snow]

I mean he cant use use calc

[raven zenith]

mmm yk u got a good point, 15+15 is much more complicated

tho we could use the pythagorean identity

[raven zenith]

so yeah, do -30+45

and expand that using the addition identity

[quaint shoal]

ok lemme try that rq

[quaint shoal]

Something went wrong tho

[raven zenith]

Mhm, that looks correct

[zenith snow]

That's the value of sin 15

[quaint shoal]

isn’t that what I’m solving for?

[raven zenith]

^^

[raven zenith]

sorry, I tripped earlier

sin 15 ≠ sin 30

[zenith snow]

No u see u did fin
Sin(180+15)

[quaint shoal]

wait wait so is doing 195-180 wrong?

[raven zenith]

on the unit circle, it's opposite

[zenith snow]

No u r on the ryt track

[raven zenith]

no, no, it was just not complete yet

[raven zenith]

you can do this ^^^

[zenith snow]

U see 180+ refers to 3rd quadrat where the value of sin is negative

So basically sin(180+15)=-sin15

[quaint shoal]

yeah

[zenith snow]

This was the way of finding rhe sum u needed to use

[zenith snow]

-sin15=-sin(45-30)

Do we agree?

[raven zenith]

@quaint shoal ?

[quaint shoal]

back

[quaint shoal]

yes

[zenith snow]

And u found the value of sin(45-30) already

[raven zenith]

(a.k.a. sin 15)

[zenith snow]

Here

[quaint shoal]

yep

[zenith snow]

All it's missing a minus dont u think?

[quaint shoal]

so all I do is flip the signs?

[zenith snow]

Yes since there is a minus b4 sin15

When u find sin195 on basis of 180

[quaint shoal]

would that apply for every instance like this?

let’s say for cos255

[zenith snow]

There r other alternatives but yes this will coincide with many

[zenith snow]

Yes try to take it in 180 form as well since we r doing 180

@raven zenith ,can u take over?
Igtg

[raven zenith]

sure no prob

[zenith snow]

Thnx

[raven zenith]

baii~

[zenith snow]

Bye

[quaint shoal]

thanks for the help

[quaint shoal]

ok so

[raven zenith]

mhm?

[quaint shoal]

I got this

and then I flip the signs

[raven zenith]

is this for sin 15 or sin 195?

(sin 195, aka -sin 15)

[quaint shoal]

This is a new problem cos255

[raven zenith]

ah

[raven zenith]

well, assuming that your calculations are correct, you can just flip the signs, yes

[quaint shoal]

alright my answer lines up with the answer key

and then of if I have something in the 4th quadrant or first

would I also flip the signs?

[raven zenith]

yeah

think of the unit circle

the cos is the x value

if you added 180º, think abt what would happen to the x value

[quaint shoal]

alright I see

so always flip the sign after calculating

[raven zenith]

yeah, if you're adding 180º

if you're adding other values, it's a different matter

[quaint shoal]

like what? 90, 270, etc?

[raven zenith]

adding 90º is different—sin(90º+theta)=cos theta

cos(90º+theta)=-sin theta

adding 270º is also different too

but you can figure out using the unit circle or addition identity

[quaint shoal]

Okay, I’ll do some practice and see how it goes

[raven zenith]

[quaint shoal]

I might be back soon with questions about tan but we’ll see

[raven zenith]

feel free to ask anytime, I might still be here, but if I'm not, others will

[quaint shoal]

@raven zenith what if it’s like sin105, where 45+ 60, in that case there’s no need to switch signs right?

[raven zenith]

yeah, since in the unit circle, the y values are still positive

[quaint shoal]

Got ittt

how about when doing difference of sum?

let’s say

Sin(105

so I get sin(-45-65)

oh wait nvm

[raven zenith]

it all just depends on whether or not the y value is positive

and when you're working on cos, it's whether the x value is positive

on the unit circle

[midnight plank]

@quaint shoal Has your question been resolved?

[midnight plank]

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[quaint shoal]

I’ve been doing some practice and came to realize that it just depends where each function is positive right? After you subtract let’s say 180-the angle it’s positive only the function is sim?

[quaint shoal]

might be explaining it wrong, but that’s how it works right?

[raven zenith]

wdym the function is positive?

oh ic what u mean

yeah, if, after adding an angle, it lands in the quadrant that's positive like u wrote there, then it's positive

[quaint shoal]

okkk I get it now

[raven zenith]

if ur not sure

u can always check with the addition formula

[quaint shoal]

yepp

thanks for all the help, I really do appreciate it

[turbid solstice]

yo am i correct

for m angle f

F

[blissful trench]

!occupied

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[blissful trench]

close the help channel if ur done

[midnight plank]

@quaint shoal Has your question been resolved?

[midnight plank]

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[gusty sky]

Can anyone help with question 2

[gusty sky]

I dont even know how to approach this problem

No matter what i try difference is not getting created for using telescoping series

[lyric charm]

do these [] mean floor or are they just brackets?

[gusty sky]

They are just normal brackets

[steel crest]

these arent series

[gusty sky]

Isnt this a Trignometric series ?

[steel crest]

these are sums

[lyric charm]

it's not an infinite series

they say clearly it's a sum that is only 9 terms long

[lilac current]

I think the answer is something between 6-7

[gusty sky]

Ohhkk i dont know the difference between sum and series i thought both meant adding together some values till a finite or infinite amount

[lilac current]

6-8

6-7

[gusty sky]

But how do I do it ?

[steel crest]

series implies infinitely many terms as defined by a limit

[lilac current]

Put on your calc

[gusty sky]

Hmm

[lyric charm]

!nosols

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[gusty sky]

So can someone pls tell how do I approach this problem

[lyric charm]

my only idea rn is to write the sum in sigma notation: $$S=\sum_{k=1}^9 \frac{\sin^4(12k\dg)}{2^{2k-2}}$$

[grand pond]

Ann

[lyric charm]

not that it seems to do much

then we will need to calculate this times 2^18/sin^2(12°)...

idk probably some trig bs with the sin^4 is gonna have to happen but i don't see it playing out nicely

[steel crest]

its not asking for you to evaluate the number

just determine what integers divide it

[drifting cloak]

its some typical "haha im gonna make the question harder by making the numbers scary" highschool bs

[gusty sky]

But how do I do that without evaluating this

[steel crest]

thinking

[drifting cloak]

try to factorize out all the options and see which one works

but its probably 4

[gusty sky]

Okk lemme try

[drifting cloak]

cus if its divisible by 21, then its divisible by 3 and 7

so

lol

so 21 is definitely not the answer

[gusty sky]

This is a multi correct question 😭

[drifting cloak]

oh

[cerulean halo]

that’s a really cool and interesting math question

[lyric charm]

do we even know P is itself an integer???

cause i don't know that it is

[steel crest]

by the implication of the question, itsnot obvious to me it is tho

i agree its not obviois

[gusty sky]

Im not able to create any difference in the terms in this due to it being bi quadratic

Idk what more to try

[steel crest]

de moivre?

[gusty sky]

No i dont think it uses it here as we had not been taught that prior to this test

[drifting cloak]

such an overkill

[steel crest]

[drifting cloak]

i mean

[raven zenith]

say ur suggestion then

it's not like it's a really hard thm either

[drifting cloak]

$2^{11} \times x \times y$

is always divisible by 4

[grand pond]

az·ez inlana vev inpeno

[drifting cloak]

if we assume it all becomes an integer

it will be even anyway

[lyric charm]

what are x and y

[drifting cloak]

whatever cos sin bs

its all a product anyway

so it doesnt matter

[lyric charm]

these are between 0 and 1 are they not

if they are not integers then they could be something that cancels out all the 2s

so i disagree with this conclusion as it stands rn

[drifting cloak]

$P = 2^{11} \times S \times \text{cosec(12)}^2$

if P is an integer

[lyric charm]

?? why S twice

[drifting cloak]

then P is divisible by 2

[lyric charm]

since when.

[drifting cloak]

oh shit just 1

my bad

[grand pond]

az·ez inlana vev inpeno

[drifting cloak]

typo

[steel crest]

suppose S=2^{-11}

[lyric charm]

your declaration implies S*csc^2(12) is an integer!

[drifting cloak]

point is

its even

its 2 x smth

if its even

then the only option is 4

[lyric charm]

that "smth" HAS TO BE AN INTEGER

we DON'T KNOW THIS YET

[drifting cloak]

u assume P is integer

[lyric charm]

we don't know even that yet

[drifting cloak]

look at the options

[lyric charm]

but we also cannot brazenly assume S/sin^2(12) is an integer

like you're doing

[drifting cloak]

yeah but out of all the options

it wouldnt make sense

if

3 or 7 or 21 is

the answer

so by elimination its just 4

[gusty sky]

Why are we just guessing now 😭 it's multi correct

[lyric charm]

wishy washy reasoning.

[drifting cloak]

unless u can choose "nothing works"

[lyric charm]

this WILL NOT DO.

[steel crest]

suppose S=7/2¹¹

[lyric charm]

we are NOT doing vibes based math here mate

[drifting cloak]

oh yeah true

[drifting cloak]

im just brainstorming

is brainstorming not allowed in this server?

[lyric charm]

you are presenting it as if it were actual reasoning

[drifting cloak]

^

from this point I was exploring ideas

[lyric charm]

you were sharing ideas that were half-baked if not wrong outright

[drifting cloak]

but i wasnt presenting them

as if its the correct solution

i expected argument

and thats what maths is about?

[drifting cloak]

see here?

u didnt have to mock me.

[lyric charm]

YALL P IS NOT EVEN A FUCKING INTEGER

[lavish venture]

[drifting cloak]

cooked

[gusty sky]

Wtf

[drifting cloak]

maybe its a trick question

[lyric charm]

this question is wrong

[drifting cloak]

and u just select none of the answer

[lyric charm]

🗑

[lavish venture]

you’re in degrees right?

[lyric charm]

yes ofc i am

[lavish venture]

lmao

is this a jee question

[gusty sky]

Ya

[lyric charm]

ah wait no

i put the summation wrong

the angles don't go 12, 24, 36, ...

actually this changes everything

nop still

even after putting the correct angles

P is still not an integer

(the angles go 12, 24, 48, 96, ... so they are 6*2^k)

[steel crest]

maybe [] means floor after all

but hten theyre just 0

[gusty sky]

Idk it is not explicitly mentioned here

[lyric charm]

there would be a giant disclaimer like "OH BY THE WAY [.] MEAN GIF FUNCTION"

[steel crest]

also it woul dbe very weird to make them all 0 like that

[civic lynx]

not to mention the first term is not rounded so you'd still get 2^18 * sin^2(12)

[lyric charm]

1/sin^2(12)

[civic lynx]

oh i thought it was sin^4 * csc^2?

regardless

[steel crest]

maybe it's a stupid typo and its supposed to be geometric

[gusty sky]

Nvm I will just ask my teacher

Thank you all

[fallow scarab]

!done

[midnight plank]

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[gusty sky]

.close

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[molten bay]

How can I start it?

[delicate sage]

Yk about resolution of forces ?

[molten bay]

Yaah a little bit

[delicate sage]

Move the forces, such that they act on a point. Then you can break them into components along x and y axis.

Finally you can get the resultant.

[molten bay]

I see

Thanks

Let me try

.close

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[molten bay]

x(1+.05)^4=500000

[molten bay]

so is there any chance to simplify (1+0.05)^4?

[lyric charm]

where do you see (1 + 0.5)^4? your original message said (1 + 1.05)^4

(and now i think you managed to make typos in both)

[molten bay]

Actually i was solving a question related to interest

They were asking that what will be the money approx in decimal which will be 500000 after 4 years

With rate of 5% per annum increased each year

[viral dagger]

1+0.5=1.5

[molten bay]

Yeah

[viral dagger]

huh is it .05 or 0.5

[molten bay]

0.05

[lyric charm]

no, there is no way to simplify 1.05^4

,calc 500000/1.05^4

[grand pond]

Result:

4.1135123739594e+5

[viral dagger]

,calc (500000)/(1.05^4)

[grand pond]

Result:

4.1135123739594e+5

[viral dagger]

well ill be damned

[lyric charm]

nah it's just that this calculator goes into scientific notation at 100k

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

I got something interesting at math exchange

.close

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[graceful ferry]

Given the series $S= 1+ \frac{1}{3} + \frac{1}{5} - \frac{1}{2} - \frac{1}{4} + \cdots$ (the changing harmonic series with different order of three positive then two negatives and repeat)
How do I know what this equals ?

[grand pond]

prograce

[graceful ferry]

The changing harmonic series is conditionally convergent so by theorem it could be equal to anything, how to calculate it?

[spare beacon]

Isn't it just (-1)^n+1/n

Or is it a typo?

It fits with a certain log function expansion ig

[dreamy lichen]

How do you even know this converges?

[lyric charm]

hold on

i don't get what the pattern here is

[gaunt flicker]

what does it continue with

[lyric charm]

you mean alternating harmonic series?

we need to make clear what the intended pattern of your series is before we can make a judgment

[gaunt flicker]

ye i think so

[nimble leaf]

there has to be a pattern. besides you can rearrange it so that it diverges too.

[spare beacon]

I'm seeing 1 - 1/2 + 1/3 - 1/4 +.....

But that's just log(2)

[lyric charm]

@graceful ferry?

[nimble leaf]

you don't know if you can calculate it for all patterns

[spare beacon]

Oh three positives

[nimble leaf]

I think his question is that, for conditionally convergent series, you can rearrange them to get other values

[lyric charm]

yall. let OP tell us something!

[spare beacon]

Yeah just realised

[nimble leaf]

you mean the reimann rearrangement theorem?

[graceful ferry]

Yes

[graceful ferry]

Leibniz

[graceful ferry]

Yes

[graceful ferry]

Just repeat the pattern

[lyric charm]

repeat the pattern?

so like, blocks of 5?

[graceful ferry]

Yes

[spare beacon]

Three odd - two even

[lyric charm]

1/(5n+1)+1/(5n+3)+1/(5n+5)-1/(5n+2)-1/(5n+4)?

or what

[graceful ferry]

Yes continue now three positives then another two negatives but like the harmonic series still

Like +1/7+1/9+1/11-1/6-1/8...

Repeating the pattern...

[lyric charm]

uhh

[nimble leaf]

,calc 1 + 1/3 + 1/5 - 1/2 - 1/4 + 1/7 + 1/9+1/11-1/6-1/8

[grand pond]

Result:

0.83654401154401

[lyric charm]

ok not the way i wrote it then

[nimble leaf]

,calc ln(3/2)

[grand pond]

The following error occured while calculating:
Error: Undefined function ln

[dreamy lichen]

https://larryriddle.agnesscott.org/series/rearrang.pdf

Might be relevant, this solves the general case of p positive terms followed by n negative

[graceful ferry]

Welp

[nimble leaf]

,calc log(3/2)/log(e)

[grand pond]

Result:

0.40546510810816

[lyric charm]

so then each block is

1/(6n+1) + 1/(6n+3) + 1/(6n+5) - 1/(4n+2) - 1/(4n+4)

for n going from 0 to infinity

,w simplify 1/(6n+1) + 1/(6n+3) + 1/(6n+5) - 1/(4n+2) - 1/(4n+4)

[graceful ferry]

Yes but also you forgt the first 5 terms

[lyric charm]

ok so that shit behaves like 1/n^2

[lyric charm]

what first five terms?

oh crap i typoed

[graceful ferry]

19 pages

[grand pond]

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[dreamy lichen]

roughly the idea of the paper i sent writes it out like this and then somehow uses some approximations or sth to get in the macaroni constant and some other harmonic stuff $\lim_{N\to\infty}\left(\sum_{n=0}^{3N}\frac{1}{2n+1}-\sum_{n=0}^{2N}\frac{1}{2n+2}\right)$

[grand pond]

MathIsAlwaysRight

[graceful ferry]

1+1/3+1/5-1/2-1/4

[lyric charm]

n=0 lol.

[graceful ferry]

Oh

[lyric charm]

anyway uhhh. hm.

[graceful ferry]

It looks similar to telescopic sum

[lyric charm]

,w sum[n=0, infty] (1/(6n+1) + 1/(6n+3) + 1/(6n+5) - 1/(4n+2) - 1/(4n+4))

[grand pond]

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[lyric charm]

apparently log(6)/2.

who would have thought.

[graceful ferry]

How would I know onpen and paper lol

I had options for the answer btw

[dreamy lichen]

if you know something about that harmonic stuff, you might be able to compute it

[dreamy lichen]

then just approximate

[graceful ferry]

Oh nice

[graceful ferry]

I need some time to understand this

[dreamy lichen]

i need some knowledge to understand that

[graceful ferry]

The image?😭

[dreamy lichen]

yeah, idk how those error terms work

https://math.stackexchange.com/questions/2967089/compute-1-1-3-1-2-1-5-1-7-1-4-1-9-cdots

slightly different question, but could be relevant. One of the answers used an approach i actually understand

[graceful ferry]

Is it not from the paper you sent

[dreamy lichen]

yeah, it is from the paper I sent

and I dont understand the paper I sent

[graceful ferry]

Aha

This question is from a past exam no way it is this complicated?

[dreamy lichen]

perhaps you were meant to approximate

[graceful ferry]

This way also looks plausible but how to find the function for the integral

[graceful ferry]

Idk how tho without complicated ways I was shown

[dreamy lichen]

literally just compute it

the options were ln3 / 2, ln6, ln6 / 2?

[lethal path]

$x + x^2 - 2 x^3 + x^4 + \cdots$ integrated from $0$ to $1$ gives that, hmm

[graceful ferry]

Yes or diverges

[grand pond]

south

[dreamy lichen]

well, we know it converges

[graceful ferry]

Manually?

It could have been divergent tho

[trim quest]

Rewrite the partial sum as H(6N)-(1/2)H(3N)-(1/2)H(2N), where H(N)=1+1/2+1/3+...+1/N is harmonic series. So, by its asymptotic H(N)=ln(N)+gamma+o(1) we have the original sum is ln(6)/2

[graceful ferry]

.close

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[unkempt skiff]

how do I do this im so confused

lim as x -> pi/2 (tan(2x)/(x-pi/2))

[unkempt skiff]

I tried substitution and set n = x-pi/2

But I just ended up getting back to the og expression..

[lethal path]

yep, so what did you get after that?

[gaunt nimbus]

[ \lim_{x \to \pi/2} \frac{\tan(2x)}{x-\pi/2}]

[grand pond]

k

[gaunt nimbus]

looks like a defn of derivative to me

[lethal path]

check what the denominator becomes now

[lethal path]

another nice way, yes

[gaunt nimbus]

!show

[midnight plank]

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

[gaunt nimbus]

@unkempt skiff

also u should try [\tan(x) \approx \sin(x) \approx x] for ${x \to 0}$

[grand pond]

k

[gaunt nimbus]

for lazy ppl

[spare beacon]

tan(t) = -tan(pi-t)

@unkempt skiff

[gaunt nimbus]

or that

[vivid yoke]

I think it's better to sub n=2x-pi rather than n=x-pi/2

[spare beacon]

Eh it's just factoring out a 2

Both go to 0 anyways

[gaunt nimbus]

anything works

u can even do

gulps

l'hopital rule

[spare beacon]

BURN THE WITCH

🥀

[gaunt nimbus]

[midnight plank]

@unkempt skiff Has your question been resolved?

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[inland patio]

Consider $$\sqrt{1-t} = \sum_{k=0}^{\infty} \left[\prod_{j=1}^k \left(\frac{j-1-\frac{1}{2}}{j}\right)\right]t^k=\sum_{k=0}^\infty c_kt^k,$$i.e. the Maclaurin series for $\sqrt{1-t}$. This is a power series, and by the ratio test I can show that its radius of convergence is $|t|<1$. This means it converges uniformly on $|t|\leq\rho$ for any $0\leq \rho<1$. Now, using the monotone convergence theorem, one can show the series also converges at $t=1$, and hence it converges absolutely for $|t|\leq 1$. My question is; does it converge uniformly on $[-1,1]$? Why?

[grand pond]

psie

[midnight plank]

@inland patio Has your question been resolved?

[humble torrent]

Doesn’t uniform convergence follow directly from Weierstras M-test or whatever it’s called

[inland patio]

Yes, you're right. 👍

[inland patio]

I have a more challenging question.

How can I see that the series converges to the function at t = -1?

[humble torrent]

Oh, I thought you claimed it did converge absolutely on [-1,1] ?

Oh nvm

[inland patio]

Yes, the series converges absolutely on [-1,1], but it's not certain the function value at -1 agrees with the series value, or?

[humble torrent]

?

[fresh sparrow]

ignore that lol mb

[humble torrent]

Right yeah didn’t see what u asked

I would say it would have to

Since a power series is in particular continuous if it’s uniformly convergent there, no?

And so the endpoint would be forced to be whatever the limit is

Which we know is whatever the function has

If it’s anything else it would be discontinuous at that point which contradicts uniform convergence iirc

Maybe there’s some more conditions on continuity if a power series is uniformly convergent, don’t recall exactly

[inland patio]

Hmm, what you're saying makes kind of sense. I think the power series is continuous within its radius of convergence, which is |t| < 1. I don't recall much what happens on the boundary |t|=1.

[crystal plover]

There's Abel's theorem for the behaviour at the boundary

[humble torrent]

Did it involve anything about uniform convergence?

,w Abels theorem

Oops

[inland patio]

Yeah, right. I'm familiar with this theorem as t approaches 1. Does it hold too for t approaching -1?

[humble torrent]

Oh hm

I think there’s issues there

Look into the proof and see if it works

[crystal plover]

I don't think it should change at all (?)

[humble torrent]

Let me check

Yeah okay it seems to still hold

[inland patio]

Thanks everyone. I think Abel's theorem it is. 🙂

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[rancid vigil]

the question is to find the values of n so that the expression is simplifiable , the solution is quite sus

[rancid vigil]

n shouldnt be a variable

it only have a bunch of values

and i got them all

8,20,9,33

,$\frac{26}{n-7}$ is simplifiable $\implies n-7 | 26$

[rancid vigil]

but they written it in reverse

idk if im missing something

[gaunt nimbus]

what does 'simplifiable' mean

[rancid vigil]

have a value that is uh

lol

its like a natural vaule

you dont get something like 3,5

[rose pilot]

i thought simplifiable is gcd is not 1

[rancid vigil]

thats what they said in the solution

ohhhh

you're right

[rose pilot]

so 14/16 also simplifiable in this terms right?

[rancid vigil]

exactly

thx

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[languid mica]

Whats the command for solving inte?

[fallow scarab]

???????????????

can you be more specific

or give an example

[languid mica]

,w int x dx

[grand pond]

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[languid mica]

Oh yes thus

,w int 1/(x^24 +1) dx

[grand pond]

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[fallow scarab]

going directly to the website is usually faster

[languid mica]

Oh i didnt know there is one

[languid mica]

Idk if i have same result

[fallow scarab]

https://www.wolframalpha.com/

[languid mica]

Cause they use csc notation etc and have split cos

[prime hornet]

that is... a lot of terms you computed

[languid mica]

This is basically my result

[languid mica]

Not really with a lot of help i did it without expanding the sum at all

With some tricks through complex numbers

This is what i did

[fallow scarab]

you can plug it into desmos to verify the graphs are the same

[languid mica]

How ?

[fallow scarab]

https://www.desmos.com/calculator

[languid mica]

@prime hornet it only took 5 pages i think it was good

And i leave a lot of gaps

[languid mica]

So what do i do

[fallow scarab]

https://help.desmos.com/hc/en-us/articles/4406040715149-Getting-Started-Desmos-Graphing-Calculator#h_01HSV06SHGA9ZQJG46QF8TJ4R2

[languid mica]

Never used ut

Ok so

You know any app to get the thing i wrote in copy form

Cause i cba fr writing that with parenthesis

Any picture option or some

[fallow scarab]

write one and learn copy paste

[languid mica]

Seems a bit off

Idk if i missed some parenthesis or some

[steel crest]

chatgpt o4mini

not regular chatgpt

and you will need to check it ofc

[languid mica]

How can i copy to see it cause jts so confusing on phone

[fallow scarab]

yea looks different

[languid mica]

But maybe i miss parenthesis

[fallow scarab]

but that doesn't mean the sums aren't equal

[languid mica]

Wdym

It is the sum

[steel crest]

what's n^-1?

[fallow scarab]

this looks like one term

[fallow scarab]

you have .... a lot of terms here

[steel crest]

why don't you start with a simpler problem with just the 6th roots of -1

[languid mica]

Unless that doesnt denote sum?

And i need to specific n and n is the same

Also c doenst seem to move it

Maybe i wrote something wrong notation wise

[fallow scarab]

it does denote sum yes

[fallow scarab]

unclear what green is here

[languid mica]

1/x^24 +1

[languid mica]

This is the graph i got

I might have missed some parenthesis or some

[fallow scarab]

yea that makes sense why c doesn't affect the graph. d/dx of any constant is zero

[languid mica]

I found

I put - instead of +

Yes its the same

Thanks yall

.close

[midnight plank]

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[midnight plank]

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[strange crest]

in this q, by obs i can say that x=1 is a valid root for the second eq bc subsituting x =1 satisfies our first proof condition now if i put x =1 in first eq i get 1+b+c =0 which when i sq both the sides i get b sq + c sq + 1 = -2(b+bc+c), this -2 extra sucks

[silent zodiac]

Just solve for common root

U should get the condition

[strange crest]

ikr but a thought arose in my mind like this, i don't even know if i am going right...

[silent zodiac]

Bhai practice more 🗣️

[strange crest]

??

[silent zodiac]

Oh u dont know hindi

‘Bro practice more’

[strange crest]

are yaar thought ko justify kaise karu>

[hallow hazel]

They have given you those conditions right

[silent zodiac]

Think abt it

[spare beacon]

Selection hoga yahi se? 🥀

[silent zodiac]

What other things can u apply

[strange crest]

the common roots condition, i know that

par socha raha hu agar aise hi ban jaye to faltu ka kya hi ghusna udhar

[silent zodiac]

Should just continue with it

Until u hit a complete blank

[strange crest]

.

[spare beacon]

If 1 is the solution of the first equation then c is the other root

c/a = product of roots

Nvm

Btw it's written "or" so these conditions will not be true simultaneously

[hallow hazel]

Use the condition

[silent zodiac]

Use that condition of common roots formula

[strange crest]

konsi class ke he bhaiye?

[hallow hazel]

If you square one equation you will just get that equation in other form you won’t get the other case that is possible

[strange crest]

looks like My thought is correct, bc x =1 might not, be a root of the first eq

am i right?

[hallow hazel]

You are correct 1 is a common root

But there is one more also

[hallow hazel]

And you cannot prove this as of now

So use the condition

[strange crest]

yeah but it almost looks like the other case, may be we can manipluate it such to remove the -2

if possible

[hallow hazel]

Not possible

[spare beacon]

Abhi hi JEE dia gng

[strange crest]

alr, i'm going with conditions👍 Thnaks

[strange crest]

Hogyi IIT

[spare beacon]

Typhoid ho gya advanced mai 🥀

Dia ni

[strange crest]

ch*d g guru

soja vros👍

[spare beacon]

ISI bhi ni de paya 🤣

Ch*d toh gae

[spare beacon]

Kis class mai ho?

[strange crest]

We are bound to dammation

11th naam ka hu😂, knowledge jh barabar hai

kuch samajh nh aata

[spare beacon]

Oh well

Bhai iss question mai

jo do statements hai vo mutually exclusive hai

[silent zodiac]

Bhai are yaar

[strange crest]

faltu hai IIT, pressurize hokar padh raha hu mai to..

maan nh krta maths aur physics chodkar

thoda bahut

[spare beacon]

Try for ISI and CMI if you like maths and physics

[spare beacon]

ye dono saath mai true nahi hai

So you can't use one statement to prove the other

Dono same case mai exist hi ni karti hai

[strange crest]

yeah.

[spare beacon]

b+c+bc=0 vale case mai x=1 is not the common root

[strange crest]

Based

Byy byy, loop me bhi fasna hai to sona padenga

[spare beacon]

Real

Bye

[strange crest]

padhai ka loop

[spare beacon]

Close chat

[strange crest]

.close

[midnight plank]

Channel closed

Closed by @strange crest

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[languid mica]

How is the end of a sum denoted

[humble torrent]

What is the end of a sum?

[blissful pier]

like (\sum_{k=0}^{\mathbf{n};\mathrm{(this?)}}f(k))?

[autumn canopy]

No mathbf needed

[languid mica]

How do you know what expression the sum is for

[grand pond]

PajamaMamaLlama

[blissful pier]

emphasis

[languid mica]

Like i wanna multiply a sum by number 5

[humble torrent]

In what context?

[autumn canopy]

Oh I thought you were the OP and asking about how to properly write it haha

[languid mica]

How do i put 5 out of the sum

[blissful pier]

no worries :)

[languid mica]

To what does the sum affect on

[blissful pier]

wrote it poorly to begin with anyways

[languid mica]

If that makes sense