help-17

[tidal salmon]

no?

[azure fox]

with 15 and 20

so

[tidal salmon]

you can use the same similar triangles

[azure fox]

huh

[tidal salmon]

remember you solved for b

read through this lol

i explain every step

[azure fox]

so wait

i need to find the ratio

?

[tidal salmon]

yup

[azure fox]

so it would be uhhh

wait

would it be like

the bottom and the c side thing

[tidal salmon]

remember with similar triangles, 16/20 = 12/d = 20/c

[azure fox]

ohhhhhhhhh

so now i have to do the other one

[tidal salmon]

since both the triangles have the same angles

[azure fox]

i feel like i know what i gotta do its just not there

[tidal salmon]

by other one you mean c right?

[azure fox]

yeah

[tidal salmon]

yup

d=15, like you got before

so do the same thing with c

[azure fox]

so 12/d = 20/c

wait

no i can do the

16 ratio

[tidal salmon]

yes

[tidal salmon]

yup

[azure fox]

hmm so

16/20 = 20/c

16c = 400

[tidal salmon]

yup

[azure fox]

25

[tidal salmon]

yup

[azure fox]

ohh i understand it now

[tidal salmon]

yea

[azure fox]

i had to make the ratios first

[tidal salmon]

mhm

for a, you can solve in two ways

the simple way is obvious

[azure fox]

subtract

[tidal salmon]

c = 16+a

or

you can use similar triangles again

since the other side is another similar triangle to the big one

they share the right angle and Z this time

[tidal salmon]

but that would be the most simple way lol

[azure fox]

so 9

[tidal salmon]

yup

[azure fox]

c = 16 + a

wait how do you get that eqaution tho

oh nvm

im stupid

dont answer that

so would be like

7 - x / 13 - x

x^2 + 5 / x^2 + 20

i gotta multiply

then factor

this one easy

[tidal salmon]

yup

lemme know if u get stuck

[azure fox]

5/2 and -5

i think both are reasonable

because

none ends up

negative

[tidal salmon]

you asking me to verify?

@azure fox if you dont have any more questions, can you say .close

[azure fox]

.close

[vocal sleet]

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[brittle zodiac]

WHATS MY R FOR 7 IM GONNA CRASH OUT

[hoary blaze]

isn't it e minus when ln(x) = 0? maybe i'm stupid idk

[tropic reef]

no how is it possible ln =0

maybe i am stupid too

[vague chasm]

When x is 1?

[tropic reef]

yaa

and minus e ??

[hoary blaze]

nah e - 1

that's the inner radius (idk what it's actually called)

[vocal sleet]

@brittle zodiac Has your question been resolved?

[brittle zodiac]

<@&286206848099549185>

i tried using every r i can think of and nothing worked

[vocal sleet]

@brittle zodiac Has your question been resolved?

[vocal sleet]

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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.

[dense jacinth]

can anyone explain me how to know, if i should define 2 variables by t and s or if i should do one variable by t

[dense jacinth]

im gonna be back in like 15 minutes

[dense jacinth]

like basically i dont get the difference between the 2 different systems in this case

back

[atomic pine]

it depends on the system of equations and whether they are degenerate (for example, in your second example, the two equations are practically the same)

[dense jacinth]

yes

but

the first one is degenerate too right

[atomic pine]

not really, because I and II are not equivalent

[dense jacinth]

by degenerate do you mean the variables or the 0=0

[atomic pine]

in some sense, I gives some restrictions, II gives MORE restrictions on top of I

[atomic pine]

it is a little bit more complicated than that, but let me try to explain

Firstly let us name our equations A_1, A_2,..., A_n (n equations) instead of I,II, for simplicity sake

[dense jacinth]

yeah

[atomic pine]

So you know, in the second example: A_1+4A_2 gives you the "degenerate equation" 0=0 right?

[dense jacinth]

yes sir

[atomic pine]

in general if, there is a combination c_1 A_1+c_2A_2+...+c_nA_n (well we have to make sure c_1,...,c_n are not all 0's), that gives you "0=0" we say A_1,...,A_n contains degeneracy

for your first example, we can say I, and II does not contain degeneracy, since aI+bII gives you the equation 0=0 if and only if a=b=0

this might be too much notation at once thrown at you

[dense jacinth]

yesh

i just dont understand how to know what to do

[atomic pine]

We say A_1,...,A_n is impossible to solve, if c_1A_1,...,c_nA_n gives you the equation 0=C for some constant C.

[dense jacinth]

yeah

[atomic pine]

for example, a+b=1, and a+b=2 is impossible to solve since A_1-A_2 gives you 0=1

[dense jacinth]

yeah

[atomic pine]

Here is a very neat fact: if $A_1,...,A_n$ are linear equation (not degenerate or impossible) with m variables, then we must have $n\leq m$ and the number of variables we need to introduce is precisely $m-n$

[twin meteor]

qwertytrewq

[atomic pine]

you can think of each non-degenerate equations as "removing one variable of freedom"

[dense jacinth]

yeah

[atomic pine]

so in your first example, since the two equations are not degenerate, you are removing two "variables of freedom", so you have one variable left to introduce

in the second example, I is the same as II, so you are just removing one "variables of freedom" so you will need to define 2 variables.

[dense jacinth]

yesss

i got that

i understood

[atomic pine]

studying whether equations are "degenerate" or "impossible" is a part of the study of linear algebra (they use very different notations, but I am defining things like this so that it is more formiliar to you)

[dense jacinth]

so in the first example is ent

sent

u basically try to express every variable through t?

[atomic pine]

here is an example for you: a+b+c+d=1, a+b+2c+2d=5, 2a+2b+c+d+=-2.

[atomic pine]

yeah!

because you have freedom over the value of z.

[atomic pine]

or freedom over the value of x (if you pick x as the value you have freedom over)

[dense jacinth]

man i fucked up

[atomic pine]

but you can only choose 1, since you can express everything else in terms of x (or z)

[dense jacinth]

i have an exam in 30 minutes

[atomic pine]

ok I can give you a tip on how to solve these equations

you do it by "reduction"

it is a method of getting rid of unwanted terms

[dense jacinth]

yeah just eliminate u mean

[atomic pine]

yeah

[atomic pine]

you can pick A_1, and try to eliminate the "a" term out of both A_2 and A_3

@dense jacinth can you try and do that?

[dense jacinth]

okay

[atomic pine]

Ill provide another example in the meanwhile

[dense jacinth]

yeah so multiply the first equation by 2?

[atomic pine]

can you elaborate?

maybe I should be more clear on the procedure

[dense jacinth]

i can get rid of c and d

if i do that

[atomic pine]

oh we are just removing the "leading term" (which is a) from the rest of the equations

to keep things organized

so we subtract A_2 with cA_1 to get a new equation B_2 where B_2 does not have "a" in it

what would c be?

[dense jacinth]

whats cA_1?

[atomic pine]

a constant multiple of A_1

like I-2II

[dense jacinth]

oh

[atomic pine]

the "2" in "2II"

[dense jacinth]

c+d=-4

?

-c-d*

[atomic pine]

shouldn't it be 4?

5-1

[atomic pine]

oh yeah that works

[dense jacinth]

oh

[atomic pine]

here is a neat fact: Solving A_1,A_2,A_3 is the same as solving A_1,B_2,A_3

so solving a+b+c+d=1, a+b+2c+2d=5, 2a+2b+c+d=-2 is the same as solving a+b+c+d=1,c+d=4,2a+2b+c+d=-2

now can you use A_1 to "reduce" A_3 (remove the occurence of a in A_3 by subtracting some copies of A_1)

[dense jacinth]

uhh

[atomic pine]

i think we can do this beforeyour exam

[atomic pine]

there are 2a's in A_3, and one a in A_1, so how do you remove the a's?

how many copies of A_1 do you need to subtract from A_3?

[dense jacinth]

oh

ok waot

[atomic pine]

is this confusing?

[dense jacinth]

yeah the A_2

stiff

[atomic pine]

pls lmk where you got confused on

[dense jacinth]

😭

i dont get what the A is about

[atomic pine]

A_2 is just referring to the equation a+b+2c+2d=5

Im numbering the equations 1,2,3 and denoting them as A_1, A_2, A_3

[dense jacinth]

oh

[atomic pine]

so A_1 is the equation a+b+c+d=1, A_2 is a+b+2c+2d=5, A_3 is 2a+2b+c+d=-2

[dense jacinth]

2

[atomic pine]

our first step is to get rid of the a's from A_2 (using only A_1) so we subtracted one copy of A_1

so we got a new equation B_2 which is c+d=4

[atomic pine]

good! what equation do you get then?

A_3-2A_1 is?

[dense jacinth]

wait

ill be right back

[atomic pine]

is it exam time?

[dense jacinth]

c+d =0

[dense jacinth]

no 😭

[atomic pine]

its 2a+2b+c+d=-2 I think you might have messed up some computation

2a+2b+c+d- 2(a+b+c+d)=-2-2

[dense jacinth]

oh

[dense jacinth]

-c-d=0

[atomic pine]

-2-2 is not 0

[dense jacinth]

ohhh

shit

-4

😭

[atomic pine]

yeah so you get -c-d=0

[dense jacinth]

yeah

we also have the other

thing

[atomic pine]

so the neat fact tell us that solving for A_1,A_2,A_3 is equivalent to solving A_1, c+d=4, and -c-d=-4

[dense jacinth]

OHHH

yes

i understand

like understand that it makes sense

[dense jacinth]

now we get 0=0 right?

[atomic pine]

no not yet

the next step is to organize our equations

and order it by leading terms

so in this case we don't need to order it

since a+b+c+d=1, c+d=4, -c-d=-4 is already ordered by the leading term

we now can take the second equation, and ask the same questions: How can I remover "c" (the leading term of the second equation) from both A_1, and the third eqution

So if we set B_1 to be a+b+c+d=1, B_2 to be c+d=4, B_3 to be -c-d=-4, we need to reduce B_1, and B_3 using B_2

I'll save you the trouble and tell you that you will get a+b=-3, c+d=4, and 0=0

[atomic pine]

And from here the solution is very easy to see: set t=b, s=d then a=-3-t, c=4-s

[dense jacinth]

my bad

i was gond

im bacj

[dense jacinth]

yes

ty

[atomic pine]

So the steps are:

1. Order your equations by the leading terms
2. take the first equation, and use that equation to "reduce" the other equations (so that the leading term of the first equation does not appear on the rest)
3. Order your equations again.
4. Take the second equations, and use that equation to "reduce" the other equations (so the the leading term of the second equation does not appear on the rest)
5. rinse and repeat

[dense jacinth]

but i think he wont do 4x3 system

only 3x2

[atomic pine]

if it is only 4*3 equations you will only need to do steps 1-4

if you have more equations, then you may need to do more steps

[dense jacinth]

also

[atomic pine]

but this strategy works for all system of linear equations (after the process, you will ALWAYS end up with a nice system of equations)

[dense jacinth]

this is an overdefined system

can u maybe summarize me

how to go forward on these

[atomic pine]

Let us try to solve this using the method I summarized above

order the equations first (in this case there is no need in ordering the equations since they all start at x)

[atomic pine]

Now the second step: can you reduce II and III using I?

you should get ||0=0 and 0=-1||

[dense jacinth]

wait ill be right back

[dense jacinth]

okay so

i mean the solution is on the screenshot😭

i would just do it like how my teacher did it

but 1 thing i dont understand,

[atomic pine]

yeah but our method is more systematic and always gives you a solution

(it might be more difficult to just "play around and find a way to reduce" compared to a systematic way of reducing our equations)

[atomic pine]

which part?

[dense jacinth]

wdym by using I sgain

[dense jacinth]

like the solution is 0=0, why dont we take that as solution

like usual

[atomic pine]

remove the x term from II, III, by subtracting some copies of I

[atomic pine]

because we get 0=2 for I-III*3

this is saying "if x,y satisfy I, III, then 0=2"

but this cannot be, so there cannot be any simultaneous solutions to I and II

I, II gives you degeneracy, but I,III give you an impossible solution

so I,II,III cannot have solution simultaneously

GL on your exam

[dense jacinth]

so

[dense jacinth]

yeah man ty

[atomic pine]

oh u r still here i thought u went to take the exam

[dense jacinth]

no like im in the bus 😭

[atomic pine]

Oh

[dense jacinth]

i understand it ngl

theres a chance the test is more focused on trigonomic functions

which im better at kinda

ok so generally

[atomic pine]

good stuff

[dense jacinth]

if we have a 3x4 system

i mean

4x3

[atomic pine]

you just do this process over and over again: ditch every 0=0 you see, and whenever you see 0=1 or 0=some non-zero value, you would know that there is no solution

[dense jacinth]

but we could get single values for x y and z right

[atomic pine]

right

for example the system of equations x=1, y=2, z=3

has single value solutions

[dense jacinth]

yeah

i doubt that 4x3 gonna come tho

we never did that

but maybe

😭

[atomic pine]

its all the same ngl, same process, just a little bit more steps

5*4, 5*5,...., you name it

[dense jacinth]

wait so

i can just memorize

a overdefined function cant have unlimited solutions?

[atomic pine]

no this is not true

x=1, x=1, x=1, x=1, x=1 is overdefined

but they still have the solution x=1

if a overdefined system of equations do not contain degeneracy, then it will never have solutions. (overdefined means there are strictly more equations than variables)

[dense jacinth]

yeah here for example

idk why hes calling the first eqution boss 😭

[atomic pine]

looks bossy ngl

[dense jacinth]

i think

we did it so

we like kind of neglect a function

[atomic pine]

again you just apply the same rules: reduce II and III using the equation I

and you realize II and III gets reduces to 0=0

so by the "neat fact" I stated a while back, it just suffices to solve equation I

[dense jacinth]

can you like write down what u mean by reduce II and III using the equation I

[atomic pine]

remove the "x" term from II and III, using I

so II+2/3 I

and III- 1/3 I

you get 0=0 and 0=0

[dense jacinth]

oh

[atomic pine]

refer to this process

the 2nd step says: take the first equation and remove "x" from II and III

[dense jacinth]

but here

why did he stop at 0=0

i dont understand 😭

[atomic pine]

he should have one more line saying something like 2I+3II gives 0=0

[dense jacinth]

ohh

so herr

every line gives 0=0

[atomic pine]

anyway what he is saying is I, II, III are essentially the same equations

so we just need to solve for I

[atomic pine]

its the same as what I did

we "reduce" II and III from I

we get 0=0, throw them out.

[atomic pine]

you should try doing examples using this process

and see what answer you get.

[vocal sleet]

@dense jacinth Has your question been resolved?

[dense jacinth]

last question

the differnece here is

that the one is infinite

as solution

omg

i understand now

the one is infinite and the other one with the z=t

is also infinite?

or what is ot

[vocal sleet]

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[paper depot]

omg is that the old notation for factorials

[ripe iron]

@worthy ridge

<@&286206848099549185>

[paper depot]

!noping

[vocal sleet]

Please do not ping individual helpers unprompted.

[ripe iron]

he is my freind lmao its cool

[worthy ridge]

Why?

[ripe iron]

lmao

[hybrid rampart]

yo can i grab a hand

[paper depot]

!occupied

[vocal sleet]

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).

@ripe iron Has your question been resolved?

[peak matrix]

I somehow managed to show that it equals this sum:
$\sum_{k=1}^n \binom{2k}{k} \cdot 4^{-k} = \sum_{k=1}^n \binom{-\frac{1}{2}}{k} \cdot (-1)^{k}$

[twin meteor]

MathIsAlwaysRight

[peak matrix]

idk if thats any helpful tho

[ripe iron]

what

[paper depot]

it would be much easier if it said to find the sum to infinity

[peak matrix]

it diverges, doesnt it?

I just tried putting it into wolfram alpha

it gives a closed form

it's pretty ugly tho

[ripe iron]

we should create a telescopic series

sheesh

ig no one knows how to solve

[vocal sleet]

@ripe iron Has your question been resolved?

[vocal sleet]

@ripe iron Has your question been resolved?

[neat rose]

@ripe iron whats the question

[neat rose]

what the gibberish?

make a valid question

[elder mountain]

My eyes

wtf

Ask chat gpt

Idfk

Js say bismlah and move on

😭

[ripe iron]

its not giberrish lma

tht is just factoril

factorial

its the old notation for factorial

[neat rose]

yeah, use latex to create a valid modern question

[ripe iron]

ugh

fine

$\frac{1}{1!}(\frac{1}{2}) +\frac{1.3}{2!}(\frac{1}{2})^2 +\frac{1.3.5}{3!}(\frac{1}{2})^3 + \dots \text{upto n terms is..}$

[twin meteor]

rj._.

[ripe iron]

@neat rose

[neat rose]

try harder make it something more readable

[ripe iron]

i got better things to do

[neat rose]

same here

good luck

[ripe iron]

:/

thanks ig

gl to you aswell

[vocal sleet]

@ripe iron Has your question been resolved?

[paper depot]

!nogpt

[vocal sleet]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[elder mountain]

Wdym why not

[elder mountain]

K so

[paper depot]

they often generate output which "sounds correct" but has numerous factual or logical errors.

[elder mountain]

So

Idc

[paper depot]

if you wanna argue this, go to #discussion or something.

somebody else's help channel is not the place.

[elder mountain]

Look no chat gpt go to deep seek

Dug

[paper depot]

can also call the mods if you really want to argue.

[elder mountain]

Yeh cool

[paper depot]

actually yeah <@&268886789983436800> we got someone here who knows more about our own rules than us

[elder mountain]

U did say no chat GPT but no deep seek

Anyway

Imma chill

Bye 😘

[versed bane]

if people wanted to use AI, they would just use it and not come here.

[elder mountain]

Ig

Anyway

U can also use daddy khan

He always comes to the rescue

Imma gamble my sons child support

🎀

[flat whale]

if you're not gonna help, can you not spam OP's help channel

go to #serious-discussion

[elder mountain]

K yall r sooo boring 🥱

Bye bye ig

[topaz eagle]

i think of wallis formulas

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

you may try to comprehend
the partial sums can be rewritten as a trigonometric polynomial (with degree 2n)

you'll transform the partial sum
1/2 + 1 ⋅ 3 / (2 ⋅ 4) + 1 ⋅ 3 ⋅ 5 / (2 ⋅ 4 ⋅ 6) + ⋯ + 1 ⋅ 3 ⋯ (2n − 1) / (2 ⋅ 4 ⋯ (2n))
to an integral with the integrand
cos² x + cos⁴ x + cos⁶ x + ⋯ + cos²ⁿ x

[topaz eagle]

You may try to use some math to capture the graphical observation that "the peak gets higher and higher"

[topaz eagle]

hint: ||for any δ > 0, consider the sequence of definite integrals whose integrand is cos² x + cos⁴ x + cos⁶ x + ⋯ + cos²ⁿ x over [δ, π / 2]. (think: why do we want to do so?) the integrand forms a sequence of functions. it's easy to establish uniform convergence to its pointwise limit. in fact, the integral of this pointwise limit over [δ, π / 2] is actually the limit value of this sequence of definite integrals. the rest shld b easy||

[vocal sleet]

@ripe iron Has your question been resolved?

[ripe iron]

this is jee question lmao they dont even allow a calculator

XD

[topaz eagle]

you don't need a calculator to observe this

the desmos graph is just to provide a graphical intuition
and you can sketch a similar graph without any calculator

[topaz eagle]

you just need to substitute x = 0 into the integrand to observe this phenomenon

as n gets "larger and larger", the peak gets "higher and higher"

[candid magnet]

whats the answer

[ripe iron]

'

[ripe iron]

well the answer is still wrong

[topaz eagle]

Oh I misread the question by overlooking what's on the right of ...

[ripe iron]

btw i got the answer from elsewhere

im just seeing if someone else cracks it

is that against the server rules or something

then ill take it down

[candid magnet]

no

its not

[ripe iron]

ok

[candid magnet]

can u send the link to the video

[ripe iron]

its in hindi

[candid magnet]

i can understand a bit

[ripe iron]

sure

https://www.youtube.com/watch?v=lyHn_BTvVGQ&t=5417s

question number 4

in the

time stamp

[candid magnet]

thanks

i think u shld be able to do it in the way

the perosn above mentioned

[topaz eagle]

I misread the question

I thought it's to find the limit value of this infinite series, by only looking at the symbols

I don't think there's an easy way to evaluate the integrand for any partial sum above

[candid magnet]

i think u can

get it

let me try

@ripe iron

,rccw

[candid magnet]

honestly it was basically that guys idea

[ripe iron]

yeah i understood his idea

what

how do you haave pi

in your answer

[candid magnet]

it gets cancelled out

when u open the integration with wallis u get this

multipled by pi /2

[ripe iron]

this is crazy stuff

man

your smart

[candid magnet]

honestly it was his idea

there was like a tougher version of this

in the adv sample paper they released for 2025

[ripe iron]

u in 12th>

?

[candid magnet]

ye

[ripe iron]

damn gl man

[candid magnet]

thanks

[ripe iron]

how was mains(if you dont mind me asking)

[candid magnet]

honestly it was bad

mains so annoying with those terrible

calculations

[ripe iron]

yeah its a pain

im in 11th

[candid magnet]

ic

[ripe iron]

2026 aspirant

[candid magnet]

which college if u dont mind

[ripe iron]

iit kgp

[candid magnet]

i meant

[ripe iron]

possibly

[candid magnet]

like rn

[ripe iron]

college?

[candid magnet]

school

[ripe iron]

oh

im in kerala

st vincent

[candid magnet]

ic

[ripe iron]

wby

[candid magnet]

fiitjee

[ripe iron]

oh

damn delhi>

?

[candid magnet]

no telengana

[ripe iron]

im in brilliant pala

hmm

[candid magnet]

fiitjee has many branches

[ripe iron]

st vincent is my school not coaching lma

[candid magnet]

ic

fiitjee

is like both here

[ripe iron]

yeah i know i have a friend studying in delhi

[candid magnet]

school plus coaching

[ripe iron]

mm

i dont go to school

[candid magnet]

they only do both

[ripe iron]

lmao

[candid magnet]

honestly doesnt matter

if ur coaching is good

[ripe iron]

yeah its alright

i mean its not the best

or anything

but its ok

once we had an air 3

[candid magnet]

understandable

[ripe iron]

once

[candid magnet]

mine is also somwhere along that

[ripe iron]

in 2020 or something

i dont know for sure

[candid magnet]

its not thta good compared to delhi

[ripe iron]

yeah

there is a fitjee near my place

[candid magnet]

thats still very impressive

[ripe iron]

also

i dont go there

[candid magnet]

oh

lmao

[ripe iron]

yeah but that guy used multiple coaching

[candid magnet]

ic

[ripe iron]

and lbh air under 10s have some special formulas

lol

[candid magnet]

real

u just got be born different

[ripe iron]

fr

[candid magnet]

no offence but like

[ripe iron]

theres one dude here bro just gets every single question

idk how

[candid magnet]

air under 10 is just ur born smart hard work only takes u so far

[ripe iron]

some guy here preparing from 6th

its crazy

no life

[candid magnet]

thats sad honestly

i started in 11th

[ripe iron]

same

[candid magnet]

and its already depressing enough

i couldnt imagine like 4 more yrs

[ripe iron]

yeh fr

[candid magnet]

than this

[ripe iron]

anyway man

gl

[candid magnet]

thanks bye

close the channel

[ripe iron]

go get iit

hmm

[candid magnet]

if ur done with it

i try

[ripe iron]

.close

[vocal sleet]

Channel closed

Closed by @ripe iron

Use .reopen if this was a mistake.

[vocal sleet]

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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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[bleak prawn]

u = 256 - x^2

[atomic jasper]

continue

[bleak prawn]

du = -2xdx

-(x*sqrt(u))/2x

-sqrt(u)/2

[wary hull]

4096/3

[atomic jasper]

continue

[atomic jasper]

!nosols

[vocal sleet]

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.

[bleak prawn]

-(u^1/2)/2

[atomic jasper]

that is the same thing, yes

[bleak prawn]

Oh wait

I think I see my mistake

1 moment

[atomic jasper]

did you forget the bounds?

[bleak prawn]

Ummm I did make a mistake but I still didnt get the right answer

So ill just continue

[atomic jasper]

can you show your derivation?

[bleak prawn]

Derivation of what sry?

[atomic jasper]

your working out

[bleak prawn]

Ah I just typed it haha so not really its deleted

[atomic jasper]

...

are you never writing these down?

[bleak prawn]

Sometimes I do

-(u^1/2)/2

[atomic jasper]

you should almost always write it donw

[bleak prawn]

Integrates to
-2(u^3/2)/2(3)

-(u^3/2)/3

[atomic jasper]

yes

[bleak prawn]

Which is negative...

[atomic jasper]

.

[bleak prawn]

Ok, I dont understand the bounds thing

Last time I did it and they told me not to do it

[atomic jasper]

the bounds are from x=0 to x=16

[bleak prawn]

1 moment ill find the chat

[atomic jasper]

but now we have u=256-x^2

if we fill in x=0 and x=16

we get the new bounds: from u=256 to u=0

u=256-0^2=256
u=256-16^2=256-256=0

[atomic jasper]

thats probably wrong

[bleak prawn]

I cant find it 😡

But yeah I did it and got the wrong answer, then they told me not to do the bounds and I got the right answer

[atomic jasper]

i cant think of a context where you dont change the bounds

[bleak prawn]

So now im lostttt

1 moment ill search again

[atomic jasper]

does what i wrote make sense?

this

[bleak prawn]

#help-15 message

Found it

[atomic jasper]

there you also plug in the bounds lol

[bleak prawn]

But I did?

[atomic jasper]

i dont see the issue

you first had 0 to pi/2

then you had 0 to 1

?

[bleak prawn]

Yes and it was wrong

[atomic jasper]

thats changing the bounds

[bleak prawn]

They wanted to just leave it as pi/2

Wait I will find that question again 1 moment

[atomic jasper]

oh i see now

i suppose you can do the whole calculation without changing the bounds

but you have to be precise with what youre doing

also, trig subs are quite a bit different and tricky sometimes

[bleak prawn]

How do I know when and when not to

[atomic jasper]

you got -u^(3/2)/3

if you dont change the bounds

then you need to fill back in u=256-x^2

to get -(256-x^2)^(3/2)/3 from 0 to 16

if you do change the bounds, then you can keep it at it was before

make sense?

[bleak prawn]

I thought this WAS what it was before?

[atomic jasper]

wdym?

[empty frigate]

yeah the issue isn't exactly that you did or didn't change the bounds, it's that you weren't consistent with what you were doing

[bleak prawn]

Like we need to convert it from u no matter what

[atomic jasper]

you dont necessarily need to

you can keep it in terms of u

[bleak prawn]

I dont understand

[empty frigate]

if you take bounds of x=0 to x=pi/2, and plug it into sin^3(x)/3 - sin^5(x)/5, that's fine

if you take bounds of u=0 to u=1, and plug it into u^3/3 - u^5/5, that's fine

[bleak prawn]

So what about the current question?

[atomic jasper]

if oyu wanna be super precise you should do $\int_{x=0}^{x=16} \frac{-\sqrt{u(x)}}{2}\dd u(x)$

[empty frigate]

if you take bounds of u=0 to u=1, and plug it into sin^3(x)/3 - sin^5(x)/5, that's complete nonsense, it's sort of the same kind of mistake as taking a formula that expects a length in inches and putting in a length in cm

[twin meteor]

Bonk

[empty frigate]

well it's the same kind of thing, you have to remember what the bounds you have actually mean and be consistent about it

[atomic jasper]

then you get $\left[\frac{-u(x)^{\frac32}}{3}\right]_{x=0}^{x=16}$

[twin meteor]

Bonk

[bleak prawn]

Thats what I did no?

[atomic jasper]

no, observe that u is a function of x

we are not plugging in u=0 and u=16

[bleak prawn]

So I DO take the bounds

For this one

[empty frigate]

...

[atomic jasper]

we are plugging in u(0) and u(16)

[empty frigate]

there are two approaches that work

[atomic jasper]

you always take the bounds

[empty frigate]

for every problem, both approaches work, it's not a matter of "in this case, what do you do"

[atomic jasper]

does this combo make sense?

[empty frigate]

it's just that you're instead using an approach that never works, by mixing the two approaches in a way that doesn't make sense

[atomic jasper]

-# maybe in special cases

[empty frigate]

-# yeah occasionally it will get you the right answer by numerical coincidence, or because the bounds wouldn't have been updated anyway, but still

if the bounds you have are x=0 to x=16, then you evaluate those bounds at a function of x

[atomic jasper]

i can show you the general form i foyu like odie

[empty frigate]

alternatively, you can figure out the bounds in terms of u, which are u=256 to u=0, and then evaluate those in your expression in terms of u

pick one of those two messages and follow through with it the whole way, don't try to do both at the same time

[bleak prawn]

The thing im confused about is I feel like I used the same method for both of them. With the trig example if I took the bounds 1 and 0 (changing the bounds) I got the wrong answer. If I DIDNT change them, taking pi/2 and 0, I got the right answer.
For this question, if I DIDNT change the bounds (0 and 16) I got the wrong answer. if I DO change them (0 and 256) I got the right answer. Im confuseddddd

[atomic jasper]

i think in the trig example you just did it wrong

or you misunderstood smth

we can go through the general form if youd like

[bleak prawn]

sin^2(x)cos^3(x)
sin^2(x)(1-sin^2(x))cos(x)
(sin^2(x)-sin^4(x))cos(x)
u = sin(x)
du = cos(x)dx
u^2 - u^4
Integrates to (u^3)/3 - (u^5)/5
The bounds also change due to the integration u = sin(x)
sin(pi/2) = 1
sin(0) = 0
so the final sum is:
sin^3(1)/3 - sin^5(0)/5

[bleak prawn]

Which step is wrong

It gave the wrong answer so I must have done something wrong

[empty frigate]

"1 to 0" is bounds in terms of u, which means you put them into u^3/3 - u^5/5, because that's the result in terms of u

[atomic jasper]

first you have u, then you use x

your u bounds are from 1 to 0

your x bounds are from 0 to pi/2

you changed the bounds correctly to the u bounds

but then you filled in the x equation

[bleak prawn]

So it should be sin^3(sin(x))/3 ?

[atomic jasper]

its either $\left.u^3/3-u^5/5\right|_1^0$ or $\left.\sin(x)^3/3-\sin(x)^5/5\right|_0^{\pi/2}$

[twin meteor]

Bonk

[atomic jasper]

you took the formula with x, and the bounds of u

[bleak prawn]

1 moment im just rereading everything 100 times 😵‍💫

[atomic jasper]

if you have a question, just ask

[bleak prawn]

Bro its nearly 1am my brain is so cooked, I think I will come back to this one in the morning and see if I can understand it then with a clearer head

[atomic jasper]

you sure?

i think we can bang it out in 10 minutes

for every integral

(general form)

[bleak prawn]

Tbh I think youve explained it in a way I can understand already, my brain just isnt functioning at this hour LOL

If I reread all this tomorrow morning I think ill be able to get it

[atomic jasper]

100%?

[bleak prawn]

Yeah I think so haha

But thank you both for your help until this point!!

[atomic jasper]

alrighty

[bleak prawn]

❤️

.close

[vocal sleet]

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[atomic jasper]

gn!

(in advance)

[bleak prawn]

Tot morgen hehe

[atomic jasper]

:p

[vocal sleet]

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[vast shale]

how solve this with integration by parts

[vast shale]

With usub it is easy

i dont get how this is possible with int by parts tho

[regal bane]

Do you know that it can be done?

[vast shale]

yes

yeah nvm

its stupid

.clode

.close

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[vast shale]

hey guys i have a questio

question

how do you solve this with integration by parts

[regal bane]

In this case, you take the derivative of 3-x, and the integral of e^2x

Much simpler integral after that

[vocal sleet]

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[pseudo rapids]

The grades of 300 students are normally distributed with a mean of 50 and a standard deviation of 5. 60 students were higher than Paul. What grade did Paul get?

[pseudo rapids]

i found the Z score

its 2

[livid lodge]

i learnt this last year but wasnt here so idk

can u look at mine tho

cause i tried urs so

#help-3

[pseudo rapids]

bro idk

just ask chatgpt

its easy

[quiet echo]

!nogpt

[vocal sleet]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[pseudo rapids]

@quiet echo can you help me?

[quiet echo]

Not right now.

[pseudo rapids]

@open sundial

<@&286206848099549185>

[coral chasm]

60 were higher than paul which means 60/300 or 20% of the population is higher than him which means paul is in the 80th percentile. Get the z score of the 80th percentile and just use the z score formula

[pseudo rapids]

om g thxcx

dadd

0.84

[coral chasm]

Then use this

[pseudo rapids]

0,84 = (300-p)/5

[coral chasm]

Yes

[pseudo rapids]

im getting -295,8

[coral chasm]

Wait, no

It should be 0.84 = (x - 50)/5

[pseudo rapids]

we dont know his mean grade

[coral chasm]

The mean that is referring to is the mean of the population

Which is given

[pseudo rapids]

fine i.g..

.close

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[pseudo rapids]

.close

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[sage wind]

isnt this question solved directly by using the theorem that i posted?

just apply theorem 3.50 to Σ|a_n|, Σ|b_n| and Σ|c_n| with c_n given as in 3.50

[regal bane]

No, since 3.50 assumes both converge, but the question only assumes one does

absolutely*

[sage wind]

but if the sum converges absolutely then it converges

[regal bane]

The question is basically asking "why does the cauchy product fail if we don't assume absolute convergence?"

Or, well, prove it fails

OH sorry, I have the pictures swapped

[sage wind]

np it is mb that i sent them like together

[sage wind]

isnt it just this statement ?

[regal bane]

My issue is that 3.50 doesn't say the cauchy product converges absolutely

Just that it "converges" which is annoyingly vague

[sage wind]

yes but you are given that Σ|a_n| and Σ|b_n| converge, you also know that Σ|a_n| converges absolutely since Σ|a_n|=Σ| |a_n| |

so replace Σa_n by Σ|a_n| and Σb_n by Σ|b_n| in 3.50

then c_n=Σ|a_kb_{n-k}|

so that Σ|c_n|=Σ| Σ|a_kb_{n-k}||=ΣΣ|a_kb_{n-k}|=Σc_n which is convergent

this means that Σc_n converges absolutely no ?

[vocal sleet]

@sage wind Has your question been resolved?

[vocal sleet]

@sage wind Has your question been resolved?

[vocal sleet]

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[tawny nacelle]

okay, quick question

[tawny nacelle]

I have started this proof as so

now, I cannot simply say that F is smooth because each component is smooth, right?

[silk rampart]

starry

hello

idk what ur studying so i cant help u with this

so can u tell me what u are studying

[tawny nacelle]

like, I need to find the coordinate rep of this map wrt to the charts on TS^1 and S^1 x R

and then show that that is smooth

is that correct?

[tawny nacelle]

idk, tangent spaces rn

[gusty cobalt]

write in differential geometry

[silk rampart]

that sounds advanced

good luck

[gusty cobalt]

these are for pre-uni

[tawny nacelle]

I am allowed to ask here

[gusty cobalt]

• These channels are for pre-university homework-type questions. Broad conceptual questions, as well as questions in university-level topics — are more appropriate in topic-specific channels (Pre-University, Early University, Advanced Mathematics) and in ⁠math-discussion.

[tawny nacelle]

I'm pretty sure this is the next thing I need to do, but I'm lacking confidence

[gusty cobalt]

but sure

[tawny nacelle]

I'm not unaware of this

but I choose to ask here anyways

[inner osprey]

we really need to do away with that section of text

[tawny nacelle]

meta suggestion ig

[jagged cargo]

it's not like anyone enforces them

[livid summit]

find charts of both manifolds and show the map is smooth in these coordinates

thats all

[tawny nacelle]

yeah so basically what I said then, yeah?

[livid summit]

it doesnt even need to be all combinations of charts

[tawny nacelle]

just one

one pair of charts

[livid summit]

one pair of charts, as long as you have enough charts to cover the domain

[tawny nacelle]

I'm 100% sure we do

thanks metal, and happy birthday

.solved

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[livid summit]

ty

[vocal sleet]

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[ember verge]

im supposed to evaluate this

[livid lodge]

bro

[ember verge]

i thought it would be 1/2 ln e

[livid lodge]

can someon ehelp me

[ember verge]

so 1/2

[livid lodge]

ive been waititng for years

help-25

[heavy yoke]

!noad

[ember verge]

the ans is 2

[gusty cobalt]

this is correct

[shut canyon]

its wrong you can check on calculator

[heavy yoke]

the answer key is wrong unfortunately

[shut canyon]

1/2 is correcy

correct

[ember verge]

oh ok

ty

[hallow meteor]

are you sure you didn't accidentally check the answer for the second question

[ember verge]

no these were the first under the 8.3 chapter and the in the answer key they were also the first

also the first two were right

idk why

[hallow meteor]

alright cool

just making sure

[ember verge]

ty anyway

.close

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[vague vessel]

Is this right

[hallow meteor]

nope wrong

x^7 doesn't have a dx

you can't just add it with 6dx

[vague vessel]

I thought I put all x with dx

[hallow meteor]

only if all x are with dx

its like 5x + 6

[vague vessel]

Then what do with 6

[hallow meteor]

its a different solution

wait I'm sorry

i got confused cuz you didn't multiply dx with x^7e^y

[vague vessel]

So mines right?

[hallow meteor]

you forgot a minus

final step

-e^-y = x+x^8/8 + c

you have to multiply by minus to apply ln on both sides

[vague vessel]

When I did ln I bring the -y forward and it become positive

[hallow meteor]

there's a ln on rhs

it would be -ln(-6x-x^8/8+c)

[vague vessel]

Oh so that’s correct answer then?

[hallow meteor]

yep

[vague vessel]

Ok so it asks for y(0)=5

5=-ln(c)?

How I find what c is

[hallow meteor]

multiply by minus and take e power on both sides

so c = e^-5

[vague vessel]

Wdym take e power both side

[hallow meteor]

to cancel out ln

e is the inverse function of ln

e^lnx = x

[vague vessel]

Ok so the answer is y= -ln(-6x -x^8/8 + e^-5?

[hallow meteor]

yep

[vocal sleet]

@vague vessel Has your question been resolved?

[vocal sleet]

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[finite hatch]

Hello

[finite hatch]

How would i put this into the last formula

Im confused on the 1/q

1/q = 5

[gentle sleet]

we want z right?
so we can raise both sides to the 1/5 power

[finite hatch]

oh

ok thanks!

[gentle sleet]

np!

[finite hatch]

.close

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[fallow sky]

can anyone help me with this, its in the 'proof by contradiction' chapter of my textbook and im not very confident in proof and dont know where to start or handle this

[fallow sky]

idk what to do for eitehr

[cedar bone]

What does it mean for a number to be odd?

@fallow sky

[fallow sky]

2n

2n + 1?

sorry i have really bad internet

[cedar bone]

Yep, so Even numbers look like 2n, and odd numbers look like 2n+1. Also, being prime means that its only factors are itself and 1. Keep these definitions in mind. To prove this by contradiction, you first assume the opposite is true: There is an even prime that is > 2

Does that much make sense?

[fallow sky]

yeah that all clear

[cedar bone]

Alright, so this number is by assumption even, so what does that say about it?

[fallow sky]

its divisible by 2

[cedar bone]

Yep, but what did we say the definition of prime was?

Its not divisible by anything but itself and 1 right?

[fallow sky]

yeah ok cool

so we cleared that prime numbers > 2 cant be even

[cedar bone]

Again, let's prove b) by contradiction. What's the opposite of "there are no two primes whose sum is 1001"?

[fallow sky]

there are two primes that add to 1001

[cedar bone]

Yep, and often with these kinds of problems its useful to specify that symbollicaly as well. So one way to do that would be to write that there are two primes p and q such that p+q=1001

What can you tell me about 1001 as a number?

[fallow sky]

its odd

[cedar bone]

Perfect

If p+q= an odd number, what can you say about p and q?

[fallow sky]

since they are primes they have to be odd since they have to > 2 so that would mean
2n + 1 + (2n + 1) = 2n + 1
4n + 2 = 2n + 1
2(2n + 1) = 2n + 1 which is a contradcition

[cedar bone]

You've got the main idea, but one thing's missing

there's an even prime that you need to account for

so assume one of p/q is that prime, and show that it still doesnt work

[fallow sky]

ok at this point im a little stuck

[cedar bone]

So you've made the assumption in what you wrote that the primes are >2

but ther's one prime that isn't >2, which is 2

so you have to account for that separately

So assume that one of p or q is in fact 2 (for contradiction)

Say p

then, q+2=1001

with q prime

what's wrong with that?

[fallow sky]

q would have to be 999 with isnt prime

[cedar bone]

Perfect

That completes it

[fallow sky]

alright so my conclusion statement would be that 2 primes cannot equal an odd intergar and with the exception of 2 it still cannot be

[cedar bone]

Oh, one more thing I should note on this sp[ecific step is that for strict correctness, instead of n in the writing of all three numbers, you should use a different letter. As written, it looks like all three are the same number, so the first step should instead read something like 2n+1+(2k+1)=2j+1 but you can reach the same kinda contradiction

[fallow sky]

alrigth so they all represent a different interger but its showing that the interger its representing is something different for each variable?

[cedar bone]

Yeah the last few sentences of the proof once you write it out formally will look something like "... which is a contradiction. Thus, 1001 is not the sum of two primes."

[cedar bone]

Well the definition of a number a being odd is that there's some n such that a=2n+1, bt for a different odd number b, there's still some number that that equation holds for, but it would be different. Thus, its inaccurate to write both a=2n+1 and b=2n+1

[fallow sky]

yeah that makes sense

alright cool

thanks for the help

[cedar bone]

no worries, best of luck

[fallow sky]

.close

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[vocal sleet]

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[autumn flax]

hello, i am currently studying for the Singapore and Asian Schools Math Olympiad, while looong at past paers, i came upun this question which i could not answer

[karmic crane]

try to look at corresponding spots in the first two columns of a given row

look what the result is in the third column when both the triangles are facing the same direction, and when they arent

or when there is a tringle in one column but not in the other

[autumn flax]

I belive its either A or D, dont know exactly

Im leaning on D

Tell me if its correct

[karmic crane]

D is what I got

[autumn flax]

i have some more questions

First, what is a crypathim

[vocal sleet]

@autumn flax Has your question been resolved?

[pallid temple]

Bassically a puzzle that replaces digits with letters

[autumn flax]

So, how would the above mentioned crypathim work

I have to go now, we will talk later

[vocal sleet]

@autumn flax Has your question been resolved?

[pallid hedge]

The first clue is from g becoming d which means o + d > 9 and g +1 = d

After that you substitute for o + d = 10 + g

[vocal sleet]

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[turbid snow]

If we have the expression -(x+3)(x-3), does the negative distribute to both terms or only just one?

[tidal umbra]

one

[languid grotto]

Think of it as (-1)(x+3)(x-3)

[turbid snow]

.close