help-28

[queen crater]

It does

[long helm]

Cause now i have $\sin(x)\cdot\cos(x) - \int\sin(x)\cdot(-\sin(x)) dx$

So i need to do IBP again

Right?

[vital citrus]

yea

[long helm]

Ohh..

Okai

And then again

[queen crater]

Not sure how you got that

[long helm]

Well i did IBP

Thats how i got that

[glossy valve]

dream

[queen crater]

That's better

[long helm]

But if i now integrate -sin^2(x)

Then i would need to integrate again

[queen crater]

Don't

[long helm]

I believe

[queen crater]

What is the most famous trig identity

[long helm]

And so i dont think IBP is a good method

[long helm]

sin^2(x)+cos^2(x)=1

Am i right?

[queen crater]

Right

[long helm]

Okai

[queen crater]

So replace that sin^2, then split the integral

[sick karma]

you can do another IBP for this to get integral trig repetition but you could solve this in 2 lines with the double angle identity

[thorn tartan]

Half angle formula prolly

[long helm]

But then i need to integrate sin^2(x)

[queen crater]

... no

[long helm]

Double angle identity?

[thorn tartan]

cos^2(pi/2 - x) = sin^2(x)

Could work?

[queen crater]

That's awful

[thorn tartan]

Just use double angle tho

[long helm]

But what do you mean?

[thorn tartan]

They are very weak at trig

[sick karma]

$\cos^2(x) = \frac{1}{2}+\frac{1}{2}\cos(2x)$

[glossy valve]

Triaxyz

[thorn tartan]

From what I've seen

[sick karma]

that's it

[thorn tartan]

They didn't know what sec was

[long helm]

Yes i dont know what sec is

[queen crater]

You have $\int \sin^2(x) dx$ do you not? So replace $\sin^2(x)$ using $\sin^2(x)+\cos^2(x)=1$

[glossy valve]

Nel

[long helm]

1/cos(x)=sec?

[thorn tartan]

Ye

[long helm]

This is the problem

[thorn tartan]

Are you familiar with this identity? @long helm

[long helm]

$\int\cos^2(x) dx$

[glossy valve]

dream

[sick karma]

@long helm

[queen crater]

I give up, there's too many distractions anyway

[long helm]

Noo

Wait

$\int\cos^2(x) dx$

[glossy valve]

dream

[long helm]

I need help with this one

[vital citrus]

okay dream. we integrated the square of cos(x). we got to the integral of the square of sin(x). we want to replace sin^2(x) in terms of cos^2(x) using pythagorean theorem

[long helm]

So i know that i can replace cos^2(x) with 1-sin^2(x)

@queen crater

[sick karma]

dream

the identity I showed you will turn this into a 3 line problem

[vital citrus]

okay wait no

god this is getting cofusing

please write your work down

[queen crater]

I'm not going to do this until the channel clears up

[vital citrus]

or we can go the simpler route to reducing the exponent @long helm uk the formula for cos(2x), right?

[long helm]

This is where i struggle…

And this is the original problem

3.107 (b)

@queen crater

Hope its ok i pinged youu

[vital citrus]

maybe its the handwriting maybe its the german, but does the question ask for the integral of the square of f(x) or the integral of f(x)

[long helm]

The square

Because they want me to find the volume

[sick karma]

yeah it's the disk method

[long helm]

Its not German 😭😭😭

[queen crater]

It's pi r^2 over [0,2pi]

[long helm]

Disk method?

[vital citrus]

alr sorry def looks eastern to me tho

[long helm]

Bahahah

Its ok

[sick karma]

it's exactly the process you wrote out already, don't worry about the name since that's what we call it in the US

[vital citrus]

please tell me

[long helm]

Ohh okai

[queen crater]

Do you understand the way to integrate using the double angle formula?

[long helm]

I am super confused now

What was the double angle formula again?

[vital citrus]

well either way. first tell me what cos(2x) is. that makes this so much more simpler

[long helm]

Wait let me search it up

[sick karma]

@long helm

[long helm]

Ohh

No i dont think i am familiar with that formula

[sick karma]

it's easy to derive if you know angle sum identities already

[queen crater]

So what do you want to do? Learn that formula or proceed with IBP?

[long helm]

Or waait

I think i know

Learn that formula

[long helm]

The formula

[queen crater]

Then you don't need me

[sick karma]

$\cos^2(x)
\ \cos^2(x) = \

frick

give me a second

[long helm]

Do you mean this formula: cos^2(x) = cos2x +sin^2(x)?

[sick karma]

$\cos^2(x)
\ \cos^2(x) = \cos(x) \cdot \cos(x)
\ \implies \cos^2(x) = \cos(x) \cdot \cos(x) - \sin(x) \cdot \sin(x) + \sin(x) \cdot \sin(x)$

[glossy valve]

Triaxyz

[sick karma]

yes, that's part of it

[sick karma]

do you agree with this process?

[long helm]

No

[sick karma]

explain

[long helm]

Where did - sin^2(x) + sin^2(x) come from?

Thats just zero

[sick karma]

what happens if you add them?

[long helm]

Ohh okai

[sick karma]

yeah

[long helm]

Yes

I agree

[sick karma]

if you add zero to the function then nothing changes

ok so

[long helm]

With the process

[long helm]

Yess

[sick karma]

now, what identity does $\cos(x) \cdot \cos(x) - \sin(x) \cdot \sin(x)$ remind you of?

[glossy valve]

Triaxyz

[long helm]

cos(x+x)

[sick karma]

yes

so now we have

$\cos^2(x) = \cos(2x) + \sin^2(x)$

[glossy valve]

Triaxyz

[long helm]

Wow

Yes

[sick karma]

with this line above you have proved the equation you stated

now

[long helm]

Ohh okai

[sick karma]

what can we change sin^2 to?

[long helm]

Cool

[long helm]

Well we want to remove the exponent?

[sick karma]

yes we want to relate the square cosine to something of power 1

so what can we do to the sine to clean the equation up a bit?

[long helm]

sin^2(x) = -(cos2x -1)/2

Right?

[sick karma]

it would be the negative of that

[long helm]

Ohh yes

[sick karma]

I was thinking to use 1-cos^2 instead, because I figured you were unfamiliar with the double angle identities

[long helm]

But how do you remember all the identities?

[sick karma]

part practice part good memory

[long helm]

Oh okai

[sick karma]

also if you forget one then deriving helps a lot

[long helm]

But can we use this one?

[sick karma]

you can

[long helm]

I am just super confused when it comes to the english names haha

[sick karma]

or you can also use sin^2 = 1 - cos^2 up to you

[long helm]

But then we have cos squared again

[sick karma]

I will show you why that does not matter

[long helm]

Oh okai

Then lets do this one

[long helm]

This one

The one you wanted

[sick karma]

$\cos^2(x) = \cos(2x) + 1 - \cos^2(x)$

[glossy valve]

Triaxyz

[sick karma]

if you combine like terms watch what happens

$2\cos^2(x) = \cos(2x) + 1$

[glossy valve]

Triaxyz

[long helm]

Omg

And then you divide by two on both sides

[sick karma]

yes

[long helm]

Wow

[sick karma]

now you have a relation for cos^2 and a cos term of power 1

[long helm]

Wow

Thanks a loooot

[long helm]

Yess

But its possible to do the same for sin^2(x)?

Right?

[sick karma]

yes

[long helm]

Okai

Thank you so so soooo much!

[sick karma]

you just use cos(2x) = cos^2 - sin^2 to move things around from here then you get to the double angle identity for sin^2

np

or rather start from cos^2 = 1 - sin^2 since you now know what cos^2 is

[long helm]

But sin(2pi)=sin(4pi)?

[brittle crystal]

yes, both equal zero

$\sin (n \pi) = 0$ for all integers $n$

[glossy valve]

Tillman

[long helm]

Yess thats true

Thanks

[long helm]

Look

I have done something super silly somewhere cause the answer is not correct

The volume cant be zero

Ohh i see something

I switched from 1/4 to 1/2

But that cant be the only mistake

[sick karma]

first of all, you are multiplying the integrals with bounds applied rather than subtracting them

[long helm]

Ohhhh

You are right

Silly meeee 😭 😭 😭

[long helm]

This is also incorrect

[sick karma]

right here

@long helm you didn't multiply the integral of 1 by 1/8

[long helm]

Ohhhh 😭😭😭

Thank youuu so so much

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[ruby fable]

Can someone explain this to me.

[ruby fable]

30

[gritty rose]

do you know what R^3 is in terms of x, y, z?

[ruby fable]

i m confused

how to draw the diagram of this on pape

[grave elm]

"Describe in words"

[ruby fable]

but my teacher added an extra req to draw it too

lol

[grave elm]

sketching in 3D is kinda painful..

[ruby fable]

can u help

he said rough sketch

[grave elm]

do you know what z=-2 represents?

is it a point / line / plane / sth else?

[ruby fable]

yeah no i m so bad at this

z is out of page

right

[grave elm]

Well, you could think about it like that, but if you wanna draw it on your 2D page, you need to draw the z-axis on the paper as well

the coordinate system can look like this

now consider all the points with z coordinates = 2, what shape do you think these points will make

[ruby fable]

like a point

arent we doing x < 4

[grave elm]

have you done z = -2 already?

oh you're supposed to do 30

okay yeah, lets do x < 4 then

[ruby fable]

yeah we need to solve 30

but w a graph

[grave elm]

i'd start by thinking about x = 4 first

[ruby fable]

okay

x = 4 is on the diagonal

[grave elm]

dont think of it like a diagonal, imagine it in 3D

and there are more points with x-coordinate 4

all these 3 points have x-coordinate 4

here is a side view

[ruby fable]

but i wanna learn how to draw on paper

[grave elm]

yeah ik

but you need to learn what shape will it make first

[grave elm]

now imagine a set of all such points with x coordinate 4

do you think it will be a single point? Line? Plane? Or something else?

[ruby fable]

plane

[grave elm]

okay great

now x < 4 means that the x coordinate is not equal to 4, but smaller than it

[ruby fable]

so how do we describe it

behind the plane

?

[grave elm]

it will basically be all the points behind the plane

[ruby fable]

okay one more question

if i m to sketch it on paper

[grave elm]

more precisely, you can say that it's the half of the space which contains the origin (because "behind" is dependent on where you look from)

[ruby fable]

the key is confusing : it says represents halfspace consisting of all points behind the plane x = 4. The half space is confusing me

[grave elm]

the plane divides the space into 2 "halves"

thoes halves are called halfspaces (one of them is behind, the other one is in front of the plane)

similarly to how lines divide planes into halfplanes

[ruby fable]

can i just avoid the term though its confusing

[grave elm]

as for how to sketch it, its kinda difficult, but I'd probably do sth like this

and then you can shade the "cube" i suppose

[grave elm]

you should probably get used to it if your teacher uses it

[ruby fable]

he doesnt but the key said it so wanted to clarify

[grave elm]

its not that difficult of a concept

just like lines divide plane into 2 half-planes, planes divide space into 2 half-spaces

[ruby fable]

Thanks got it

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[gusty bramble]

for the 1st one I got not atleast x for all y not(if Ox then Ey)

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[prisma heart]

Can someone help me how to do algebraic expression with multiplication, division, addition and subtraction

[placid oak]

!da2a

[full forum]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

@prisma heart Has your question been resolved?

[full forum]

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[gusty bramble]

im on part B

[old cave]

So theres a slight issue there

for example, the statement "x < y" is one predicate

if you negate it, what would be true?

How you do say x is not smaller than y? what must be true then?

[gusty bramble]

then x is larger than y?

[old cave]

yes, >= to be specific

[gusty bramble]

ohh greater or equal to

so every time you negate a > or < it flips it and turns it equal to?

[old cave]

yup, because thats the opposite statement!

[gusty bramble]

ahh okay so other than that its correct?

[old cave]

yeah if you do it for all the inequalities

[gusty bramble]

alright bet I did that

so on c now so far I have there is n (

ohh wait let me try something'

here i have this but im not sure how to show the predicate

[old cave]

you can use english

well depends

if they let you

[gusty bramble]

Oh im not sure the directions are vague

[old cave]

also where does the x come from? make sure to define it

[old cave]

actually it is probably better if you keep it like this

for understanding

There is a number n

you've got that part

[gusty bramble]

how would I define the x? put a for all x behidn the parentheses?

[old cave]

Yup after the exists n

because its "for no other number"

But you gotta think about it

other

[gusty bramble]

alright thank you

and then the last one like this?

[old cave]

hold on with c

its not quite right

the question taken literally was:
"there is a n such that for no other number x, x <= n"

*other * is a special word

what do you think that means for x?

[gusty bramble]

wait would I have to put another set of parentheses? like in b?

[old cave]

if x is another number than n, what condition do you need to put on x before you make your statement?

what has to be true about x in order for the given statement to be made?

[gusty bramble]

x is greater than n?

[old cave]

other as in different

[old cave]

thats equal to not (x <= n) which is fine

but which xs are we talking about?

other number x

[gusty bramble]

other numbers?

[old cave]

if x is another number than n, what is true?

what does that mean in simple terms

[gusty bramble]

that its greater than n?

[old cave]

hint: implication

[gusty bramble]

waittt

[old cave]

and just not equal

different

not the same as n

!=

$x \ne n$

[glossy valve]

flynger

[gusty bramble]

there exists n for all x if x >n then not(n>=x) ?

[old cave]

just if x is not n (aka different than n)

then not(n>=x)

which you can simplify the not(n>=x) to just x > n

[gusty bramble]

im sorry im super confused is it just the simplification? or is there more

it felt like we were talking about more

[old cave]

there is a number n

$\exists n$

[glossy valve]

flynger

[old cave]

such for that for all x

$\forall x$

[glossy valve]

flynger

[old cave]

if x is not n

$x \ne n \implies$

[glossy valve]

flynger

[old cave]

then $x > n$

[glossy valve]

flynger

[old cave]

The other than n part is reflected by if x is not n

and then you can make the statement like normal

[gusty bramble]

do you show the if x is not n part with a not equal sign?

[old cave]

sry i have to go for a bit

[old cave]

yup

[gusty bramble]

its okaywill you be bakc or no?

[old cave]

because thats the opposite of same (=)

not equal

[gusty bramble]

alright i didnt know we were allowed to put those

[old cave]

so yeah it should be $\exists n \forall x (x \ne n \implies x > n)$

[glossy valve]

flynger

[old cave]

In simple terms, there is a number n such that for any other number x, x is bigger

[gusty bramble]

oh wait if i have the (xnot = n) then x>n is that fine?

[old cave]

You shouldn't because we are just translating sentences

we dont know the context of what x and n can actually be

[gusty bramble]

okay ill fix that then

[old cave]

or wait im confused

what you asked but

yeah try to simplify if theres a not in front

[gusty bramble]

oh its okay it was probably wrong

[old cave]

if you can

its not wrong just cleaner if you rewrite equals into not equals

or <= into >

[gusty bramble]

okay did you get a chance to look at d?

[old cave]

yes

[gusty bramble]

i need to pass the not along

[old cave]

reposting

[gusty bramble]

here let me send the updated one

[old cave]

Can you translate the statement directly into math first?

send your result

[gusty bramble]

math like symbols or words?

[old cave]

just send d as is without simplifying the negate

just in math

[gusty bramble]

okay thats the one on the d line

the one udner that is the simplified version

[old cave]

there are two other numbers

is that a for all are you sure?

[gusty bramble]

ohh wait

i slipped up

x and z are there is

and it wouldnt be >= it would just be >

[old cave]

well hold on

first

x < n < z is equivalent to x < n and n < z

so negate it properly without any shortcuts

[gusty bramble]

wait do you mean give them the not? im confused

[old cave]

do you agree that x < n < z

means

x < n and n < z

[gusty bramble]

yea

[old cave]

whats the negation of x < n and n < z

theres an invisble "and" there basically

so its going to turn into an or

[gusty bramble]

x>n or n>z

[old cave]

=

since its the opposite of <

[gusty bramble]

would the = make sense though because its saying in between it cant be equal to then?

[old cave]

yeah but you negated the entire thing

[gusty bramble]

oh

[old cave]

and the lack of = from before comes from them being different to n right

but after you negate its different

its not between

so either one of them is the same as n (=) or n is bigger than both or smaller than both

$\neg \forall n \exists x \exists z (x < n < z)$

[glossy valve]

flynger

[old cave]

$\exists n \forall x \forall z \neg (x < n < z)$

[glossy valve]

flynger

[old cave]

$\exists n \forall x \forall z (x \ge n \vee n \ge z)$

[glossy valve]

flynger

[old cave]

Wait

that doesnt work

does it?

[gusty bramble]

.. wait why

[old cave]

there exists n

such that for all x, z

x or z is the same as n

or

x is greater

or z is smaller

why does that feel iffy

[gusty bramble]

wait you dont have to seperate the greater than and equal to right?

[old cave]

you dont

im trying to see if it lines up with intuition

[gusty bramble]

okay good

[old cave]

of what the opposite should be

so this is sorta a weird case imo

im not sure whats going on exactly

your original statement seems intuitively right but its simplifed

just saying each n has some x smaller than it and some z bigger than it

but once you reversed it seems to break down because

there can still be a z smaller than n and and an x bigger than n

that still puts n in between right

so thats not strong enough for a counterexample

[gusty bramble]

yeah wait

[old cave]

So you should start with

x < n < z or z < n < x

and i think thatll fix it

but its hella weird

[gusty bramble]

i think thats what i was saying earlier with the <=

[old cave]

n is in between x and z: x < n < z or z < n < x

that definition will work i think

[gusty bramble]

so still do the there is and for all at the start

okay like this you're saying?

on the d line

[old cave]

well

sorry

you'll have to negate x < n < z or z < n < x

from scratch

since we didnt have that before

(x < n and x < z) or (z < n and n < x)

not ((x < n and x < z) or (z < n and n < x))

[gusty bramble]

alright set them >=?

[old cave]

when you negate the ands and ors flip tho

[gusty bramble]

alright! and its good now?

[old cave]

yea it would be correct

...unless you want to simplify

[gusty bramble]

Awesome..

[old cave]

using propositional logic

[gusty bramble]

umm like what

[old cave]

i dont know if you need to do it

based on the question i gues you dont need to?

[gusty bramble]

yk what its fine thank you for helping me i appreciate it

[old cave]

honestly you probably didnt need to negate the insides

since the question didnt ask

sorry

you couldve probably left it outside hte parenthesis for these ones i guess

[gusty bramble]

nahh thats okay without you i still wouldve been at the start

im bombing my test on this either way ...

[old cave]

dont think like that

its not that bad

once you think of it as formalizing what you already understand

to be more precise

and yea its a little bit more confusing

[gusty bramble]

this is just one part thats the thing theres a statements part implications part rules of logic proofs and proofs on discrete structures

do you have any study recomendations?

[old cave]

hm

i have no idea sorry..

[gusty bramble]

dang alright its okay

[old cave]

I would just make sure that for each of the content topics you fully understand it

if something is amiss, figure out what is missing in your understanding

Practice helps but the goal of practice is just to understand smth intrinsically

[gusty bramble]

hmm alright you helped me with understanding some parts more, thanks

[old cave]

Yeah don't be scared, this is supposed to help you understand logic better

its a little dense at times but

how about starting with easier ideas and practice with those?

and then trying more complicated statements

[gusty bramble]

Ill try to keep that in mind..

[old cave]

everything complex is just a bunch of simple things combined after all

[gusty bramble]

im writing that down hold up

Alright have a good night thanks again

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

I hope this type of question belongs here, as it is not necessarily about a problem/certain subject
I'm currently failing maths pretty badly, with my average being 4 (5 is required). For americans, my grades are basically a B+ and 3 Fs
So my question is, how do I actually get good at maths. I also want to apply for a scholarship after I finish highschool on top of trying not to fail, So I need to study for the current lesson as well as taking highschool maths from the beginning. I have no idea how to study, we rarely get homework or practice exercises, the teacher speeds through lessons and I can't really understand anything. I don't have a study routine either and apart from just doing exercises and hoping for the best (which got me an F the last 3 times) I have no idea what to do

[oblique basin]

the main thing you can do is practise and practise

just blindly do problems, make mistakes

and then check where you went wrong

[brisk apex]

where can i get problems from

[oblique basin]

books

or tell chatgpt to give you some

tell the chapters. I can suggest some books

[brisk apex]

Would help a lot

and about taking highschool maths from zero, do you think Khan Academy is good enough? Because sure it's helpful but I feel like its not enough to make me above average

The scholarship I want to apply for is called MEXT and the maths exam is pretty hard

A bit below olympiad level id say

[oblique basin]

If you wanna pass, then yes

if you want scholarships, then no

[young sinew]

what topics are we talking about

[brisk apex]

highschool maths

[young sinew]

so like algebra, calculus

[brisk apex]

yep

trigonometry, geometry

although i dont think geometry is that important rn

[oblique basin]

trig, do S.L.Loney

[brisk apex]

I should probably also mention that I'm from romania and I don't know how many of these books I can find

online resources would be of great help

[oblique basin]

you can download pdf

Geometry is Euclidean or Coordinate?

[brisk apex]

for the scholarship i think it's both

[young sinew]

most books have online versions nowadays

[oblique basin]

euclidean, do evan chen

coordinate do S.L.Loney

[brisk apex]

I don't think we've started calculus in school yet so I can probably worry about that next year

right now we're on trigonometric equations and that typa stuff

exponential equations etc

and I also really really suck at logarithms

[oblique basin]

only thing hard about log is calculation

[brisk apex]

Is it?

[young sinew]

for calculus there are quite a lot of books with exercises
Spivak, Lang, Apostol

[brisk apex]

I usually remember the formulas but fail to make good use of them

As in I don't realize that I'm supposed to use it when solving problems

[young sinew]

that's what you learn by practise

[brisk apex]

Makes sense

[young sinew]

as for the books/problemsets often they are made for specific exams in mind (at least on high school level)

we have like hundreds pages long books with exercises in poland for high school exam

so you can try googling for the name of the exam you'll be taking (or some other exam as long as you know the language)

also if you have a scientific bookstore somewhere nearby you might check there

or at a page of scientific publishers (romania probably has one)

[full forum]

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[cinder thorn]

if $e \in L$, then we must have $e = r + q(\pi+i e)$ for some rationals $r, q$

[glossy valve]

bloubbloub

[cinder thorn]

actually nvm

me stupid

but here's an answer that uses something else https://math.stackexchange.com/questions/683084/field-not-closed-under-complex-conjugation

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@trim osprey Has your question been resolved?

[trim osprey]

Thanks. Discussion continuing in groups, rings and fields

.close

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[neon surge]

\textbf{Problem 7} (1 pt.). Let $(a_n)$ be a sequence of real numbers. We consider the following two conditions:
\begin{enumerate}
\item[(a)] The series $\sum_{n=1}^{\infty} a_n$ is convergent.
\item[(b)] For every non-decreasing sequence of numbers $b_n > 0$, such that $b_n \to \infty$, we have $\lim_{n \to \infty} \frac{1}{b_n} \sum_{k=1}^n a_k b_k = 0$.
\end{enumerate}

[glossy valve]

k

[neon surge]

I need some help with b->a

so I guess I suppose A_n is divergent

and then try to find some b_n sequence

but idk how to choose it

[narrow ermine]

What exactly is the question asking?

To show that they’re equivalent?

[neon surge]

yes

and I already got a->b

[grave elm]

and so you're trying to do it by contradiction?

[neon surge]

yes

[grave elm]

hmm have you tried looking at some specific example? Even something as simple as (an) = 1

[neon surge]

no, let me try this

b_n = n would suffice but not in the general case, now ill try a_n = 1/n

[grave elm]

try something oscillatory after that

like (-1)^n

btw do you know the abel summation thing?

its doable without it, but it can help

(summation by parts is what i mean)

[neon surge]

this?

[grave elm]

yeah, this

you can try applying it

If you needed another hint, once you rewrite it with abel, you can ping me

[neon surge]

ok I got $\frac{1}{b_n} \sum_{k=1}^n a_k b_k = A_n - \frac{1}{b_n} \sum_{k=1}^{n-1} A_k (b_{k+1} - b_k)$

[glossy valve]

k

[neon surge]

and imma need another hint yea

[grave elm]

okay great

the first term is just the partial sum of A, you cant really control that

but the second term can be controlled very well

if you keep bk constant most of the time, you can control exactly which terms Ak will appear in the sum

because constant bk means that bk+1 - bk = 0

and you can decide when to make jumps, which will add the Ak to the sum

[neon surge]

oooh

okay okay nice

[full forum]

@neon surge Has your question been resolved?

[grave elm]

want another hint?

[neon surge]

maybe I can use the reverse of the Cauchy theorem thing somehow to know which A_n to control

idk just a thought

[grave elm]

not sure which one you mean now

[grave elm]

oh hmm

havent thought about that, but it could work

What I did was considered some "nice" subsequences of (Ak) which converge to where I want and then worked with them

[neon surge]

so we would need Ak bounded for this right?

that would do that case\

[grave elm]

well, kinda

if you're relax the defns a bit, you can consider the subsequences which "converge" to infty

[grave elm]

by only keeping bn constant and only sometimes increasing it by +1, the 2nd sum becomes simply an average of the chosen Ak, at which bk increases

[grave elm]

and if the chosen Ak converge to some c (or infty), then then the averages will also convverge to c (or infty)

so you gain a complete control over the 2nd sum

[grave elm]

,texsp ||$\frac{1}{b_{n}}\sum_{k=1}^{n-1}a_{k}\left(b_{k+1}-b_{k}\right)=\frac{1}{p}\sum_{k=1}^{p}a_{n_{k}}$||

[glossy valve]

MathIsAlwaysRight

[grave elm]

||this is what i meant, once you pick the subsequence, you can pick bk such that the sum becomes a simple average, the bk increases by 1 at every an_k u wanna include||

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[dry arch]

How would you show
$$\lim_{n\to\infty}\frac{\ln(P_n)}{n^2}=0$$
where
$$P_n=\prod_{i=1}^{n}(1-(-1/\phi^2)^{i})$$

[glossy valve]

Axe

[brittle steeple]

My first instinct would be $\ln(P_n) = \sum \ln(1-(-1/\phi^2)^i)$

[glossy valve]

Dreyuk

[dry arch]

it should suffice to show P_infinity converges, right?

or that the sum you wrote converges

[brittle steeple]

Oh wait that's true isn't it

It would suffice to show it's bounded

[dry arch]

oh

[brittle steeple]

Well ig it's monotone so that's the same thing

If I were trying to convince myself I could prove it I'd just Taylor expand the ln

and I'd only care about the first term

but that could get annoying to justify

[dry arch]

$\sum\ln(1-(-1/\phi^2)^i)\leq\sum\ln(1+(1/\phi^2)^i)$

[glossy valve]

Axe

[brittle steeple]

Left is negative, right is positive

-x < ln(1-x) < 0 for x in (0,1)

if you use that to bound it should be sufficient

[dry arch]

thanks

i think i need a break ol

[brittle steeple]

no problem

[dry arch]

.close

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[mighty sleet]

can someone help me here? i figured out the second part being wrong, idk why the third part is. ill send a screenie of my wokr

[robust slate]

did you misread your 5 as a 9 near the end

[mighty sleet]

...

yes.

shi

.close

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[sacred solstice]

yo, im doing question 8a and I can get $arg(z-1)=\pi arg(z)$ but what am i supposed to do with the pi infront, i can do the graph for arg(z-1)=arg(z) but idk how to do it when I have the pi in front (if that changes anything of course)

[glossy valve]

KB

[mossy stratus]

arg(Z-1) - arg(z) = pi surely follow the rule arg(z-a) - arg(z-b) = alpha

which the position is (a,b) well i can see it starts at (1,0)

the alpha angle is pi (180 degree )so the the vectors oa oz directions are opposite, true, but z must on the line that connect those points

[sacred solstice]

so Z is on (0,1)

[mossy stratus]

yes why you have the two vectors start from different positions

[sacred solstice]

bc Z-1 is one unit to the right and Z "+ 0 " is starting from the origin

so Z would be something like this then ?

[mossy stratus]

hmm angle is 180 degree so Z must lay on between 0 and 1

wait i draw

I think it is like this

[sacred solstice]

yea so i got it correct then?

[mossy stratus]

yess

[sacred solstice]

ok ty

[mossy stratus]

no worries

[sacred solstice]

.close

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[sacred solstice]

ok i got another, for 8B how does it work now?

the lines are flipped:

.reopen

[full forum]

✅ Original question: #help-28 message

[sacred solstice]

i feel like i made it worse

[mossy stratus]

i think they form a semicircle

[sacred solstice]

a semicircle ?? why cant anything behave nicely

[mossy stratus]

well you can see the angle is $\pi/2$

[glossy valve]

Minhh

[mossy stratus]

which the angle formed by Z must look at the line AB as 90 degrees

this imagine of a circle, if an angle that is 90 degrees, so it must cover the diameter, which in this case is AB

let AB be two points on Re, which are (-2,0) and (2,0)

[sacred solstice]

so something like this ?

[mossy stratus]

no opposite

since pi/2 is a positive angle

[sacred solstice]

and we know its a semi circle becuase why ?

[mossy stratus]

.

[sacred solstice]

yea, i mean it half makes sense but its just strange

[full forum]

@sacred solstice Has your question been resolved?

[mossy stratus]

what else is strange?

[sacred solstice]

just odd that it comes out to be a semi cirlce, idk its just odd

and its hard to tell what shape it should be from looking at the stuff

[mossy stratus]

thats what i imagine of im not really sure but semicircle seems reasonable

[hardy charm]

Let me see see. 😁

[mossy stratus]

i dont think a triangle would would work lol ig

[hardy charm]

Which one?

[mossy stratus]

HELLO !! "See see"

8b

i am unsure if the semicircle should be drawn, but it seems reasonable enough

[sacred solstice]

bc. even for 8c i can see what it is but idk what shape fits Z properly

it would be a semi circle from past question, but its up by one so how does that change the graph of Z i just dont understand getting the set points of Z Which satisfy it

[hardy charm]

It is a semicircle.

I normally just convert them to Cartesian coordinates

[sacred solstice]

but how do i find Z bc i've got no clue

[hardy charm]

Wait you need to find z?

[full forum]

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[sacred solstice]

.reopen

[full forum]

✅ Original question: #help-28 message

[hardy charm]

What for?

[sacred solstice]

Z as in for when the expression is defined

[woeful girder]

Before slayla comes, good handwriting

[hardy charm]

Oh.

[sacred solstice]

when arg(z-i) - arg(z+3) = pi/2

for what Z make it true

[hardy charm]

Oh this one.

[sacred solstice]

also sucks how i got no way of checking what it should look like bc Desmos doesn't like it : (

does it have something to do with this?

[hardy charm]

Yes.

Normally it's a circle.

[sacred solstice]

so that's if they're on the same "level"

but what about when they are not

is it like a rotated semi circle?

[hardy charm]

I show you how I convert the stuffs.

arg(a) - arg(b) = arg(a/b)

So arg(z - i) - arg(z + 3) = arg[(z-i)/(z+3)].

Then let z = x+iy.

[sacred solstice]

yea that makes sense

[hardy charm]

Simplify such that the denominator is a real number.

[hardy charm]

Finally, since the argument is equal to π/2, the result is completely imaginary.

[sacred solstice]

is the circle from A-B the solution?

[hardy charm]

Yes.

But z cannot be the end point.

[sacred solstice]

is that just in general

[hardy charm]

So it should not be a solid dot.

[sacred solstice]

so when solving these, and they are equal to some angle say theta, i draw lines meeting up "in the middle of the points" and then draw a circle passing through all 3 and then thats my sol.?

[hardy charm]

Why three points?

[hardy charm]

$\tan(\arg(z-i) - \arg(z+3)) = \tan(\frac{\pi}{2})$\

[glossy valve]

Restarter

[hardy charm]

$\tan(A-B) = \frac{\tan(A) - \tan(B)}{1-\tan(A)\tan(B)}$

[glossy valve]

Restarter

[hardy charm]

Let $z = x+iy$, then $\arg(z) = \arctan(\frac{y}{x})$.

[glossy valve]

Restarter

[sacred solstice]

Gimme 1sec

[hardy charm]

Yeah.

[sacred solstice]

I understand how to calculate the arg, it’s just arctan, but when it comes to one minus the other so I just use arctan(a-b)?

[hardy charm]

No.

arctan(a) - arctan(b) ≠ arctan(a-b).

It's $\arctan(\frac{a - b}{1 + ab})$.

[glossy valve]

Restarter

[sacred solstice]

where a is arg(a) and b is arg(b)?

$arg(z-i)-arg(z+3) = \pi/2$

$tan(arg(z-i)-arg(z+3)) = tan(\pi/2)$

$(a-b)/(1+ab) = tan(\pi/2)$ where a = tan(arg(z-i)) and b = tan(arg(z+3))

[hardy charm]

No.

$a = \tan(\arg(z-i))$

[glossy valve]

Restarter

[sacred solstice]

so like... we get it in this form and how does it help us solve the problem?

[hardy charm]

Refer to this formula.

[glossy valve]

KB

[hardy charm]

Remember.
arg(z) = arctan(y/x) if z = x+iy.

So what is the value of tan(arg(z))?

[sacred solstice]

tan(arctan(x+yi)) = y/x

[hardy charm]

Correct.

[hardy charm]

So.

How do we simplify this using the fact I showed?

[sacred solstice]

tan(arg(a)) = -1/1 = -1, tan(arg(b)) = 3/1

[hardy charm]

Oh no.

You don't know what z is.

So z = x + iy.

[sacred solstice]

and we subst. z and then subst. x,y and then solve for z

[hardy charm]

To be specific, you're solving range of values of z.

Which is expressed in terms of x and y.

I'll just write my solution later on.

[sacred solstice]

but tan(pi/2) is undefined

[hardy charm]

Exactly.

At what scenario will a fraction be undefined?

[sacred solstice]

if it approaches infinity or 0

[hardy charm]

Sorry, fractions, not functions.

My bad.

[sacred solstice]

so 0 then

[hardy charm]

What 0?

[sacred solstice]

the denominator

[hardy charm]

Yes.

So let denominator = 0.

[sacred solstice]

which is our x

[hardy charm]

You got yourself a function in terms of x and y.

[hardy charm]

In which, it would be easier to graph compared to complex numbers one.

Am I right?

[sacred solstice]

yes

but itsnt the work to get here needlessly complex

oh so we get Z = yi bc x is 0

[hardy charm]

Wait what?

No.

[sacred solstice]

ok nvm, this makes no sense

so, we use tanget formulas and stuff to get $(a-b)/(1+ab) = tan(\pi/2)$

[glossy valve]

KB

[sacred solstice]

and tan(pi/2) is undefined so we conlcude y/x is undefined => x=0

and then from there we have a function in terms of x and y ?

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@sacred solstice Has your question been resolved?

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[frosty cipher]

does anyone know what assumption im making for method 1 bc it looks correct

[fleet relic]

include the question for context maybe idk

[frosty cipher]

oih ye

does anyone know what assumption im making for method 1 bc it looks correct

but method 2 is correct

[marsh trellis]

It’s wrong because the red angle in that small triangle is also 30

Clearly the hypotenuse is 24 and the opposite side is F11

[fleet relic]

can you draw the FBD for an object in an inclined plane rq

[fleet relic]

can you try writing some of the forces in terms of mg?

@frosty cipher in case you didnt see my question btw

[full forum]

@frosty cipher Has your question been resolved?

[foggy prairie]

Bro is still doing vectors

[foggy prairie]

Can you post the full question

[foggy prairie]

Oh here it is

Okay so

@frosty cipher

Let me know when you’re here hopefully if I didn’t fall asleep I could help you out

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[midnight stone]

Help

[midnight stone]

Pls tell me correct option

Google and chat gpt are telling wrong

[frigid trail]

which question?

[midnight stone]

Every

[frigid trail]

1. Rational
2. √2
3. 0
4. 5
5. Infinte
6. 5/3

[midnight stone]

5 in page 1

[frigid trail]

Sorry 1. Irrational

[midnight stone]

Hello

???

Bro 14 left

[void nova]

!noans

[full forum]

The purpose of this server is to help you learn; please don't ask for direct answers. Ask for guidance, explanations, or feedback instead.

[void nova]

And also, do not cheat during exam

[bright bronze]

I think exam got over and dude merely wants the answers checked

[midnight stone]

Yes

I am failing

Pls

If I my 12mcq are correct I will be passed

[sudden niche]

hello

i just landed

i am an +2 student

can i help with anything

[midnight stone]

Pls

[sudden niche]

ok

which question

[midnight stone]

7-20

All mcq

Pls

[sudden niche]

okay no problem

if u can give me 2m i ll download discord in my phone to see from laptop and type there

[midnight stone]

Ok

[sudden niche]

7)4
8)1 and 3
9)postulate
10)120

[bright bronze]

you got 3,7,8,14,15,16,17 all wrong. 9th is tough to tell since postulates and axioms are basically the same things. Maybe see what your textbook says about it. Cant talk about 19 and 20 since you kinda didnt mark

[sudden niche]

🥀

[sudden niche]

How is 7th wrong?

Yea it should be 4

[bright bronze]

abscissa is x coordinate

[sudden niche]

Ik thanks

[midnight stone]

So if I consider both assertion reason wrong it will be 11

[sudden niche]

How is 8 wrong

I think it is correct

[bright bronze]

depends on what you answered for those tbh

[bright bronze]

point line plane, all things are defined

[sudden niche]

Fr?

[midnight stone]

I wrote both are correct and reason is correct for assertion

[sudden niche]

They r undefined bro

[bright bronze]

then 19 is wrong

[sudden niche]

@bright bronze check on Google or smth ig

[bright bronze]

not according to euclid. He gave like 20+ definitions in his book

[midnight stone]

Then 12

Yay 12

[sudden niche]

In geometry they r undefined bro

[bright bronze]

verbatim from ncert which I believe OP is using based on their name

[sudden niche]

Yo bro can u please confirm answer for 8?

[midnight stone]

I will do rd sharma and rs agrawal in 10th if paased 9th

[bright bronze]

I can confirm that its not c

[sudden niche]

Which board u giving bro

[midnight stone]

Cbse

[sudden niche]

Google says otherwise

I checked rn

[bright bronze]

idt google is OPs textbook

[sudden niche]

I think google is universal textbook

[bright bronze]

feel free to contest that then. I got no opinion on what you have to say

[sudden niche]

I apologise if I am wrong but if u think it's 1 then I don't have anything to argue

[midnight stone]

I believe I can get 16.5 in written

[sudden niche]

What about assignment u have those?

[midnight stone]

Case based?

[sudden niche]

Assignments written one

[midnight stone]

Oh those I should get 20 on 20

[sudden niche]

Passing grade is.?

[midnight stone]

26.5 or 27.5

[sudden niche]

Then it shouldn't be hard passing

[midnight stone]

Let me calculate wait

Yeah I am getting 30

In 80

How shameful

[sudden niche]

Hm ig u ll have to perform better in future

[midnight stone]

Yes I need to do rd and rs

Like look at this question

[sudden niche]

These were pretty basics

[midnight stone]

Look Q34

[sudden niche]

Start grinding brother

[midnight stone]

Yeah

I have never been taught that question in school

How can I get perimeter from perpendicular

[sudden niche]

Wdym

U have to find area

[midnight stone]

But how from from perpendicular of triangle

[sudden niche]

U need to find side of triangle from perpendicular

Then use root 3/4 a sq

[midnight stone]

I have never solved such a question

[sudden niche]

U solved every question of ur book?

[midnight stone]

Only ncert

This question was not in ncert

Not even this type

[sudden niche]

Tbh for me this question is quite simple

[midnight stone]

I believe I will get 2 mark for herons formula and s formula

[sudden niche]

U might

[midnight stone]

Why might I should get

[sudden niche]

Depends whether u did them right or not

[midnight stone]

I wrote correct formula and solved the question by perpendicular

[sudden niche]

Then u would

[midnight stone]

In 20 marks of long questions I believe in 8 marks

5 for graph

Very easy

1 1 1 mark from other 3 5 mark question

[sudden niche]

Then I ll pass

Getting 6-7 from mcqs

[midnight stone]

12

[sudden niche]

18-19 from assignment

[sudden niche]

We should assume the worst

[midnight stone]

Assuming the worst already

I need 26.5 in this 80 mark

[sudden niche]

Hm

[midnight stone]

The assignment is different

[sudden niche]

Oh damn

[midnight stone]

12 in mcq 100%

8 in short answers

I mean 7

4.5 in 3 mark questions

8 in 20 mark

2-3 in 12 mark case study

[sudden niche]

Well u ll pass

[midnight stone]

I should

31.5 minimum

.close

[sudden niche]

Al the best

[full forum]

Channel closed

Closed by @midnight stone

Use .reopen if this was a mistake.

[full forum]

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[limpid monolith]

What do you call these shapes?
I'm not talking about exponential vs logarithmic, but rather the way the graphs open up

[limpid monolith]

Like they're both increasing, but one gets increasingly large, meanwhile the other one slows down the rate at which it's increasing.

If you understand me

[umbral dome]

concave up vs concave down (or convex vs concave)

[limpid monolith]

Thank you.

.close

[full forum]

Channel closed

Closed by @limpid monolith

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[gritty rose]

That's virtually the definition of e

[sacred valve]

i dont understand why is it tho

[signal arrow]

Don’t understand why that equals e?

[sacred valve]

yeah