help-49

[astral talon]

does this make sense

[tidal turret]

xi = 30m
xf = 40m
xf - xi = 10m

[astral talon]

yes, but it helps to be able to visualise the object starting from a 0 point

and then going
(here) origin

[tidal turret]

0-----------------30m-----------------40m
|---------deltax=10m--|

[astral talon]

and then ending up
. origin (here)

[astral talon]

yep

alright

so let's apply this to your question

can we find the separate displacements

note me using the word separate because when the velocity is positive it signifies a motion in a different direction than in negative

[tidal turret]

idk

[astral talon]

okay let's take it slower for a second

[astral talon]

do you know how to find the displacement from a velo time graph

[tidal turret]

well

this graph is y axis what and x axis what?

y axis is velocity
and x axis is time?

wdym?

[astral talon]

yep it says in the first sentence

"graph of "speed" against time" where speed is the incorrect translation of velocity as established

[tidal turret]

fr

so (x,y) = (time, velocity)

[astral talon]

yes

[tidal turret]

v(t)

what does the area under the curve denote

what is the integral of velocity over time graph

[astral talon]

exactly

[tidal turret]

?

[astral talon]

do you not know the answer?

[tidal turret]

I am not sure what the derivative and the integral do

[astral talon]

oh okay that's fine

[tidal turret]

like is related but idk

[astral talon]

so you can define acceleration as the rate of change of velocity, so that's the derivative

[tidal turret]

velocity is in m/s
time is in s

(m/s)*s = m so the integral is the displacement

[astral talon]

yep

[tidal turret]

how did you noticed?

[astral talon]

a = dv/dt, the gradient

[tidal turret]

in the case of the area under the curve, you need to multiply base and height of the triangle so the result is in meters

[astral talon]

exactly

[tidal turret]

?

[tidal turret]

??

what are u talking about

[astral talon]

ignore this for now it's unrelated to your current qs

it was in response to smth else you asked which i misunderstood

you can ignore it

[astral talon]

youre on the right track with this

so can we find the area?

[tidal turret]

10-0 = 10
10-0 = 10

base = 10s
height = 10m/s

area triangle = 1/2 . base. height = 1/2 . 10s . 10m/s = 50m

thats for the pink thingy

[astral talon]

perfect

[tidal turret]

now what?

@astral talon

[astral talon]

what about the blue area

[tidal turret]

15-10 = 5s = base
0--5 = 5m/s = height
area triangle = 1/2 . base . height = 1/2 . 5s. 5m/s = 12.5m

[astral talon]

total blue area

[tidal turret]

I made a mistake

now its fixed

[astral talon]

oh i was waiting for the total area 😭 thats why i didnt check, proud of you tho

[tidal turret]

for the light blue is

5m/s . 10s = 50m

[astral talon]

fixed it again, nice!

so the total comes out to

[tidal turret]

lmfao

total of displacement is

50m + 12.5m + 50m = 112.5m

[astral talon]

nono

notice that when the graph goes under the curve, it's a negative velocity, correct?

meaning now you're moving in the opposite direction

[tidal turret]

fuckkk

50m-12.5-50m=-12.5m

[astral talon]

perfect. now let's look at what the question says.

"If $x_o$ at t=0s = 10m"

[grand pond]

Sam | TBND

[astral talon]

there's a reason they specified that the INITIAL position of the object is 10m

let's take positive as right and negative as left, okay?

from 10m, you moved 50m to the right so coming to 60m,
and then you moved a total of 62.5 to the left, meaning you end up at -2.5m

now, applying THIS

knowing that xf = -2.5m and xi = 10m

can you find displacement and consequent average velo?

[tidal turret]

wdym?

can we go slower?

I got lost at x0(t=0) = 10m

that means initial position at 0 seconds is 10m wrt origin

[tidal turret]

ok perfect

-2.5m - 10m = -12.5m

[astral talon]

yes

[tidal turret]

we already knew this, btw

[tidal turret]

-12.5m/25s = ?

,w -12.5 / 25

[grand pond]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

average vel = -1/2 m/s

@astral talon

[astral talon]

yep

[astral talon]

it's not conceptually correct to look at it that way, i avoid it, although yeah you're right

[tidal turret]

my bad

[astral talon]

in that case you're assuming that x=10 is the origin

meaning you're saying x=10 is the new 0

[tidal turret]

thats exactly what x0(t=0)=10m

means

no?

[astral talon]

well if this question had further parts, it wouldnt be as simple

[tidal turret]

oh ok

[astral talon]

for instance, if I asked you instead of average velocity, i asked you for the final position of the object

in that case that's not what it means

the final position is not 50-62.5

it's 10 + 50 - 62.5

[tidal turret]

ok

physics is hard

ngl

or is it not? @astral talon

[astral talon]

it depends from person to person, i for one enjoy mechanics :D

[tidal turret]

I appreciate the help

will continue with homework

.solved

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[tranquil turtle]

if you change the <--> to -->, wouldn't the statement still be true (last line)?

[tranquil turtle]

it uses the biconditional because we're only interested in sets in R, not anything outside of R?

[surreal moon]

There is no set of all sets

[surreal moon]

I don't get what you are trying to ask

[tired osprey]

that's what it 's trying to prove no

it's not really mathematical tho

A ∉ A doesn't make any sense

[midnight plank]

@tranquil turtle Has your question been resolved?

[exotic stratus]

Yes but $\iff$ is a stronger predicate than $\implies$

[grand pond]

HitenTandon

[exotic stratus]

Hence it's preferred

[midnight plank]

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[broken rose]

I am really confused on how to figure out what transformation would convert the region into a square region

[midnight plank]

@broken rose Has your question been resolved?

[broken rose]

<@&286206848099549185>

sorry if I'm doing something wrong on discord side

(acquired help from a different resource)

.close

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[sour sierra]

Sorry I just need some help guiding how to do these questions

I am new to verifying or simplifying trigonometry and I just watched 2 video and I got this question

Absolutely stuck

,rotate

[livid python]

well what identity relates cot(x) and csc(x) ?

[sour sierra]

This is only what I managed to do

[sour sierra]

Is it like this

[livid python]

you do this when proving an identity yea
but in multiple choice questions i prefer using tricks (like multiplying by -1 etc.)

[sour sierra]

Oh

Let me see…

[livid python]

back to this question

[sour sierra]

Uhmmmm

let me check the video

[livid python]

you could memorize the sin and cos one then divide by either sin^2 or cos^2 to get the others

[sour sierra]

Sorry I don’t really get what u mean

[livid python]

you could get it by
sin^2+cos^2=1 divide it by sin^2
1+cot^2=csc^2
(not writing the argument cuz it saves time)

[sour sierra]

Ohhh

Thanks you so much that is so good to rmemeber

[livid python]

and the sec and tan one just divide by cos

so lets continue

[sour sierra]

Wait is there a sec and tan one

(Sorry)

[livid python]

1+tan^2=sec^2

[livid python]

nah it's ok

[sour sierra]

Thank you thank you

Ohhh

Okay!

[livid python]

rearranging the csc cot identity we get
1=csc^2-cot^2
it looks similiar to our expression here

but the problem is negative is not where we want it to be

[sour sierra]

Oh ye

So the answer is 1 - tanx?

Lol

[livid python]

we could derive another identity from this by multiplying by -1

[livid python]

no it's the same expression

[sour sierra]

Oh is it

[livid python]

when multiplying by -1 we get
-1= -csc^2 + cot^2

[sour sierra]

O

[livid python]

knowing addition is commutative
-1=cot^2 - csc^2

now this is the same expression in our original equation

[sour sierra]

Ohhh thanks for ur help

But

[livid python]

and then the answer should be ?

[sour sierra]

Tanx - 1

[livid python]

mmhm ?

[sour sierra]

But why not 1 - tanx

[livid python]

we subbed -1 not 1

[sour sierra]

Oh wait nvm lmao

okay thanks for ur help

Really appreciate

🙏

Let me try the next question

[livid python]

yeah in multiple choice question just know you could do these tricks instead of writing it in terms of sines and cosines

sometimes the latter is eaiser

[sour sierra]

Oh

[sour sierra]

Oh ye one question please

[livid python]

yeah ?

[sour sierra]

Actsully my exam will be like verifying Trigonometry

Is it okay if I learn this

(Like idk if it will help or what)

[livid python]

in verifying (with showing work) sadly you need to write most of the expressions in terms of sines and cosines

[sour sierra]

oh

So am I learning on the right track

[livid python]

yep writing it in terms of sines and cosines and getting used to it is pretty helpful on the long run

but these tricks mostly help in multiple choice question

[sour sierra]

Like the things I am doing rn

The exam will only be verifying thou

Ok… maybe I am stuck on this question

[livid python]

use the fact that
tan(90-t)=cot(t)
cot(90-t)=tan(t)
sin(90-t)=cos(t)
cos(90-t)=sin(t)

[sour sierra]

wow I didn’t know about that

Oh wait

[livid python]

you could pick which every you like to construct either
the sec tan identity or csc cot depening on your prefrence here

[sour sierra]

oic

[livid python]

sec(90-t)=csc(t) and same for csc

[sour sierra]

this is too comiplicated

[livid python]

i'll show why this works

[sour sierra]

okay

[livid python]

you don't need to memorize if your understand the idea behind it

[sour sierra]

Oh

Wait

I think I get it

[livid python]

You could go on and show the rest if you want

[sour sierra]

Ohhhh

Okay I get it now thank you for showing me

let me try the question again now

[livid python]

going good

[livid python]

ok

[sour sierra]

Is it -1

[livid python]

yep

[sour sierra]

Ohh

Thank you so much

Btw

I am still worrying

[livid python]

about ?

[sour sierra]

These types of question are my exam questions

am I even learning the right things rn

[livid python]

you are doing good

[sour sierra]

uhhh

But there are no sec or cosec or

[sour sierra]

(In the exam)

[livid python]

$\frac{a+b+c}{z}= \frac{a}{z}+ \frac{b}{z}+ \frac{c}{z}$

[grand pond]

yoboiqimmah

[sour sierra]

Wait uh

U think I can solve that?

I can try thou

lmao

[livid python]

give it a go

[sour sierra]

Ok….

Does this simplify to 4?

[livid python]

yep

[sour sierra]

forgot to write “s”

Thank you!

,rotate

[sour sierra]

Am I on the right track

[fallow scarab]

i have one of those lcd writing tablets

[sour sierra]

haha is it good

[fallow scarab]

it's fire

[sour sierra]

very reliable

WAIR

WAIT

[livid python]

-2 here looks wrong

[sour sierra]

I see what to do next

O

[sour sierra]

Wait is that sin^2(x) - 4

Or Wait is that sin^2(x - 4)

[livid python]

.

[sour sierra]

okay

,rotate

[sour sierra]

Is this currently right?

Yay

Let’s goooo

[livid python]

now that I look at it simplfiying the fraction might've been the shittiest way to go about it

[sour sierra]

I think it is ok

[livid python]

well while verifying I don't that is accepted

[sour sierra]

OPPs

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

a^2 + b^2 = c^2

[livid python]

no not the identity but rather adding/subtracting to both sides

[sour sierra]

Ohhhh

U mean the 3sinx + sinx = 4sinx?

[livid python]

yep i don't they give the marks doing it like this

[sour sierra]

Oh

Does it not work thou

Or is it like I am skipping steps

[livid python]

it works it proves it but it's like not accepted for some reason

[sour sierra]

Oh why not

[livid python]

maybe cuz like adding outside elements may ruin the verification process somehow

[sour sierra]

oic I don’t get it

[livid python]

don't ask me that's it in like every explanation in the topic of it

[sour sierra]

Ohhh okay!

What should I do now

Feel like I have mastered it with ur help

Thanks so much once again

Or else I did be struggling so hard

[livid python]

going back here i think recombining the fraction would help

[sour sierra]

Oh

Maybe yah

[sour sierra]

Is this true?

[livid python]

-4cos^(t)+4 specifically

[livid python]

notation wise no

cos^-1(x) refers to the inverse cosine of x

so cos^-2 is the square of the inverse

[sour sierra]

Oh…

[livid python]

(cosx)^-2 is the best

[sour sierra]

Ohh

[livid python]

but going back and recombining the fraction

cuz we getting side tracked

[sour sierra]

Do u think I should use SEC COT and Cosec when verifying

[sour sierra]

Yes

[livid python]

sometimes they accept it but you are better of with sines and cosines in the long run

[livid python]

You can continue from here

[sour sierra]

Okay

Oo

(Forgot to write -4)

[livid python]

Same concept can’t "randomly" divide and multiply

[sour sierra]

So what can I do

[livid python]

1-Keep the fraction
2-look at 4cos^2(x)-4

[livid python]

Yep

[sour sierra]

Do I do this

Oh

What could I do now

[livid python]

Simply the left side

[sour sierra]

Ayo

[livid python]

(The numerator only)

[sour sierra]

Wat the hellll

ayooooo

let’s goooo

[livid python]

Works

[sour sierra]

so when dealing when verification

[livid python]

https://tenor.com/view/horse-nodding-nodding-yes-horse-nodding-silly-gif-5152303604886252799

[sour sierra]

I can never multiply?

this is unrealllll

[livid python]

Can’t add multiply/divide/add/subtract both sides

[livid python]

Ikr

[sour sierra]

okay

Thank you so much for ur help once again

If the exam won’t contain cot sec cosec should I even learn them rn

[livid python]

Learn them even if it doesn’t contain them

[sour sierra]

Okay

Thank you so much for ur time and help!

Thanks!!!

[livid python]

It’s nothing really

[sour sierra]

thanks for ur patience fr

na took u so long

To teach me

Thanks so much

[livid python]

It’s ok man I was even doing another thing while I’m at it

[sour sierra]

https://tenor.com/view/pray-feet-pepe-pepe-feet-pray-gif-13367702538580662263

thank you so much

Have an amazing day or night

[livid python]

https://tenor.com/view/monkey-gif-22052184

[sour sierra]

Byeee!

[livid python]

Don’t forget to .close the room

[sour sierra]

okie

.close

[midnight plank]

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[untold bane]

Hello! I have no idea how to approach the second part of this question, question (b).

[untold bane]

omg i got it now i feel so stupid lol

.close

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[south delta]

Can someone tell me if my integral is correct? I need to do a revolution with x=8.5 but im not sure if this is the right formula for tubes

[tired osprey]

https://revisionmaths.com/advanced-level-maths-revision/pure-maths/calculus/volumes-revolution

[south delta]

Theres is with disk but where is with tubes

[tired osprey]

https://www.contemporarycalculus.com/cc/hoffman_calculus_5_5.pdf

Looks good

[south delta]

Then how come when i do the integral its not the right answer

And its very long

This is my final answer but when i go on integral calculator its not right

<@&286206848099549185>

[young mortar]

you didnt distribute the engatvie

[south delta]

Negative?

[young mortar]

yes

[south delta]

Which negative sign

[south delta]

But i still dont have an exposant 4

Only goes to 3

Oh nvm after the solving

But i have a cos

[south delta]

Here

[young mortar]

idk you shoudlve just made it into a fraction

its really confusing working with decimals

[south delta]

So i transform everything to a fraction?

[young mortar]

i woudlve from the start its hard to follow what you did wrong

[south delta]

Ok i will start from the beginning

[midnight plank]

@south delta Has your question been resolved?

[south delta]

The fractions didnt help

<@&286206848099549185>

[copper notch]

Use substitution of variables

@south delta

[south delta]

How so?

[copper notch]

set (x-7/2)=y

it will make things way easier

[south delta]

Ok i will try

Wait

Its not 7/2

Its 17/2

[copper notch]

You have to solve 2 integrals

ah, ok

[south delta]

With the u sub?

[copper notch]

It would be beter if I could see the original problem.

because there's a cosine

and you've erased some parts

[south delta]

This is the original

I inversed the order cuz it wasnt right

The order of the functions btw

[copper notch]

You have to calculate 2 integrals

[south delta]

But then how do i get rid of the x

If i do a u sub i have to isolate all of the xs and it’ll mak it harder

I dont think u sub is better

Ill have to transform them all to ys

[copper notch]

just a moment

[south delta]

Ok

[copper notch]

You have to do this kind of substitution

in order to have a simpler function to integrate

@south delta

[south delta]

But the x is 17/2-x

So i have to multiple everyone by it

[copper notch]

it's (x-3,5) in here

[copper notch]

where?

[south delta]

No i meant the rayon

The radius

For my revolution

On x=8.5 so i have to do 8.5-x

Or 17/2-x

Its the formula for the tube method

So it ends up being not 2 integrals but 9

[copper notch]

Is it rotating around an axis parallel to the y axis?

[south delta]

Yes

[copper notch]

ah

ok

[south delta]

I drew a rotating arrow to show it

[copper notch]

I see now

That's the problem, you're using the integral, as if it were a revolution around the x axis

[south delta]

So i have to switch everything to y?

[copper notch]

You will have to change coordinates, it will be easier if you shift the whole surface some units to the left, as if it were rotating around the y axis itself.

[south delta]

So i find the reciprocal functions?

So x=…

I dont have to do that

Its useless

It complicates things

Id need to do arccos

Over complicates everything

I just want to know why my integral isnt right

<@&286206848099549185>

<@&286206848099549185> ??

.close

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[molten bay]

How to solve it?

[thorny plinth]

what happens to f(g(x)) as u increase x

[molten bay]

u?

when i increase x g(x) will decrease

and when i increase x f(x) increase

But here it will decrease

[rocky copper]

When you say g(x) is decreasing, it specifically means that g decreases as x increases.

[thorny plinth]

correct

oh i see the issue now

[molten bay]

F(x) is increasing

But here x is decreasing

So f(g(x) will decreasing

[rocky copper]

Correct.

And it is 0 at x=0

[thorny plinth]

the options are incorrect

[molten bay]

So what next?

f(g(x)=0

[rocky copper]

What does it mean for the interval [0, inf)?

[molten bay]

it is our interval

[rocky copper]

These are all values > 0

[molten bay]

What rest it can mean

[rocky copper]

So for any value in the interval, what can you say about f(g(x))

Say for 5

At x=5

What will have happened to g(x)?

Compared to x=0

[molten bay]

Say about?

[rocky copper]

Going from 0 to 5

x has increased right

Going from 0 to anywhere in the interval [0, inf)

x will have increased

Consequently g(x) will ....?

[molten bay]

I can say nothing

F(g(x) is 0 and we are getting it is decreasing

f(g(x) will be <0

[rocky copper]

Wdym you can say nothing?

[rocky copper]

This is correct.

You can say something.

You can clearly see that it's decreasing, and it's maxima is 0, so it's always negative.

Whats causing the confusion ?

Also note that h(0) = f(g(0))

=0

[molten bay]

Yes

No option?@rocky copper

[rocky copper]

?

No option as in?

[rocky copper]

B option?

Always negative.

Whats wrong with it

0 is classified as both +ve and -ve

[molten bay]

Opps

Hmm right

[rocky copper]

If you have confusion tell me

[molten bay]

No no

Can i do it with derivative?

@rocky copper

F'(g(x)=f'(g(x)g'(x) and f'(x) is + and g'(x) is negative so hence we get multiply is <0 so option B

[rocky copper]

You could. The derivative of a decreasing function is negative over the domain on which it's decreasing

[rocky copper]

Yyup

[molten bay]

Tq

.close

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[molten bay]

.reopen

[midnight plank]

✅

[molten bay]

Small doubt@rocky copper

f(x) is defined only in [0,infinity) no?

But we will get g(x) negative values

[rocky copper]

why would g(x) have negative values?

It's simply said that g is decreasing, it's never given any value of g in particular, we only have a value of h

g can be ranging from say (6,5) as x ranges from [0, infinity)

[molten bay]

I am slightly confused at another thing

g is decreasing so when x increases

What f(g(x) will be?

What will be range of our g(x) it is [0, infinity) am I right

[sly burrow]

from the question yes

[sly burrow]

decreasing

[molten bay]

I am thinking it by drawing

[sly burrow]

g(x) is decreasing, right?

f(x) is increasing when x increasing

so when x decreases, it should be decreasing

makes sense?

[rocky copper]

Not necessarily

[sly burrow]

I think it's mentioned in the statement

[molten bay]

Why?

[rocky copper]

Oh true

It's mentioned in the problem mb

[molten bay]

So range is [0,infinity)

Here points are reversing

@rocky copper

[midnight plank]

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[untold sentinel]

\

[untold sentinel]

Here is my current working. I am unsure of how to evaluate the very last integral.

Please also check my working.

[turbid dove]

First one is correct

[pearl hull]

<@&268886789983436800>

[turbid dove]

What did u do in the second one last step?

[turbid dove]

Third is not correct

[turbid dove]

Hint: ||the first step is right||

[untold sentinel]

c(x)^(-2)

[untold sentinel]

wait how did I do it wrong

[turbid dove]

Wrong

[untold sentinel]

You're just saying it's wrong and I don't know what's wrong

I took quotient rule on s/c, ended up with 1/c^2

[turbid dove]

The wrong is simple, u applied the concept well

[untold sentinel]

That's definitely right

The question asks it in that form

[turbid dove]

It's so common between students

[untold sentinel]

It's not to me, that's why I'm asking for help

And you're just saying I'm wrong without clarifying whatever

So I don't trust you

[turbid dove]

Gimme time dude I'm reading

(C(x)³+s(x)³) / c(x)²

How to simplify this ?

[untold sentinel]

so you didn't even read the question

[turbid dove]

I'm reading ur answers

The other questions

[untold sentinel]

first three lines

those are properties relating c(x) to s'(x) and vice versa

[turbid dove]

Oh didn't see that

[untold sentinel]

Bruh

...

[turbid dove]

I was reading this

[turbid dove]

In (ii) the first part is correct

But the second not

[turbid dove]

.

[turbid dove]

Go up here

[untold sentinel]

c(x)³ + s(x)³ = 1.

can't you read the question 😭

so [c(x)³ + s(x)³]/c(x)² = 1/c(x)²

[turbid dove]

Well i had another way but u r good

[turbid dove]

I'm on the bus going back home from university

💀

[untold sentinel]

If you want to help, then do it properly please

[turbid dove]

Well I'm helping a little bit, u were waiting for 20mins

[untold sentinel]

Thanks at least

[turbid dove]

Could u explain how did u integrate this pls

[untold sentinel]

no figure it out yourself

I'm the one asking for help

[turbid dove]

iv ✅

[turbid dove]

Just making sure I'm not getting smth wrong

[turbid dove]

Where did c'(x) go
Did u integrate it?

[untold sentinel]

u = c(x)
du = c'(x) dx

.

[turbid dove]

Well it's all done here

V is correct

Great job

And thanks for ur warm thanks

Lol

[untold sentinel]

no it isn't

i havent finished it

I still need assistance with the last integral

[turbid dove]

Wtf why ur question has a lot of parts

Lemme see

[untold sentinel]

Thanks

[turbid dove]

Don't mind me for not reading the question 💀

[untold sentinel]

Its fine

[turbid dove]

I see

[turbid dove]

The problem Start here

U just make it more complicated

Do u want a hint or think about it again

The last two lines u wrote are wrong

[untold sentinel]

...

[turbid dove]

Hint: ||change sec'(x) not c(x)||

[turbid dove]

[untold sentinel]

Where tf does sec come from

I'm closing it

I solved it anyways

.close

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[empty ivy]

I don't really understand the question

[verbal pumice]

I think you're meant to say what sections are represented

Depending on lambda, with respect to a and b

for example, at lambda = 0, you have r = a, and at lambda = 1, you have r = b, so for lambda in between 0 and 1, it describes the region of the line between A and B

[midnight plank]

@empty ivy Has your question been resolved?

[empty ivy]

Ohh that makes sense

But how would you describe part b and c?

Like if i sub lambda = 2

r= 2b-a

but idk what thats meant to mean

[runic hamlet]

where on the line is that

in relation to A and B

[empty ivy]

Oh i get it thanks! and thanks @verbal pumice

.close

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[last slate]

Need to find CD

[nimble leaf]

Maybe you could coordinate bash

If you set B as the origin, AB as the y axis

C is at (6,0)

So find D

[last slate]

Bruh masochism

[nimble leaf]

AD is at an angle of 30 from the horizontal

Idk what AD is so

Nvm

[last slate]

It is one of tough ways to solve this problem

[slim eagle]

treat them as vectors

[stark tartan]

what if you drew a line from B to O (O being the point where AE and DC lines meet)?

[last slate]

Let me check it

[stark tartan]

you get BCO as an isosceles maybe an isosceles triangle can help somehow

[last slate]

I think we got even equilateral triangle

[stark tartan]

we are also given that ABC=AEC=ADC=90 degrees, right?

[last slate]

Absolutely

I think I am close to the answer

[stark tartan]

what troubles me is that if this figure is drawn to scale and the BC and CO lines are equal in length then they must look like they have equal length as well

[last slate]

Damn yes CD is 9

[stark tartan]

yes BCO is an equilateral triangle

[last slate]

.close thank you a lot for hints I appreciate it

[midnight plank]

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[stark tartan]

np man

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[pastel wharf]

Does this work?

[midnight plank]

@pastel wharf Has your question been resolved?

[pastel wharf]

.close

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[fervent burrow]

for this, am i supposed to write two values for x and two values for y?

[fervent burrow]

because in the end u get two x values

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[fervent burrow]

nvm

[paper inlet]

nice balls but this channel is about to close

so you should open another one

[paper inlet]

?

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[fervent burrow]

i dont get part b

[fervent burrow]

like what do u mean find an expression in terms of y

i completely don't understand this wording and what it is asking me

my brain can't process this question

[midnight plank]

@fervent burrow Has your question been resolved?

[crisp valley]

y is the length of CD

[fervent burrow]

so how do we get it

[crisp valley]

the answer would require someone who cares enough to solve it

which i am not

[paper inlet]

dude doesnt know how to solve it and is tryna act nonchalant

[crisp valley]

no i deadass have to do something

[paper inlet]

just admit it

you are even bothered to reply

[crisp valley]

got a date buddy

fuck off

[paper inlet]

alright dude we believe you chill out

[crisp valley]

it's glitched smartass

[crisp valley]

they're similar triangles by virtue of i have eyes
CE = 4 CD
= 4y

you can know it's 4 because 12 = 4(3), AE = 4 BD

[crisp valley]

in seriousness they're similar because bc is on ac, cd is on ce, and bd is parallel to ae

[midnight plank]

@fervent burrow Has your question been resolved?

[midnight plank]

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[twilit field]

I start by listing the first few terms

1,2,2.5

We use induction, the base case is triviakly true

we then assume $a_{n}-a_{n-1} >0$
\
$a_{n+1} -a_{n} = 3 -\frac{1}{a_n} - 3 + \frac{1}{a_{n-1}} = \frac{1}{a_{n-1}}- \frac{1}{a_n} >0$. Thus it's increasing

[grand pond]

What a wonderful world !

[twilit field]

Now to prove it's bounded

For this it's sufficent to show a_n>0

as a_1=1, and a_n is increasing, it follows that a_n>0, moreover, it thus follows a_{n+1} ≤3

It's thus monotone, and increasing

Thus it converges by the monotone convergence theorm

We find the limit now

$a_n=3-1/a_n \implies a_n^2-3a_n+1=0$, so $a= (3+\sqrt{5})/2$

[grand pond]

What a wonderful world !

[blissful pier]

well you found the limit of the sequence lol

[twilit field]

Are all my steps fine though

[blissful pier]

appear to be at a glance

[twilit field]

Thanks

Another question

The limit isn't too hard to find imo

$

$a_n = \sqrt{2a_n}$

[grand pond]

What a wonderful world !

[twilit field]

But proving it converges...

I feel like the monotone convergence theorm will help here

$a_2 =\

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[livid crystal]

Hi guys.. need help

[livid crystal]

I can show what I have so far

Now I don’t know

[tacit lodge]

Progressions?

[livid crystal]

Do I do log

[livid crystal]

What’s that

[tacit lodge]

Is that arithmetic progressions?

[livid crystal]

Geometric sequences

And series

[tacit lodge]

Oh geo progressions

Im not good at that sry

[livid crystal]

It’s ok

Anyone else out there that can help?

[blissful trench]

you can

[livid crystal]

Is there smt else I can do

[tacit lodge]

Isnt the formula for geo progressions sum

a ( (r^n-1)/(r-1))

[tacit lodge]

When |r| is 4

Wait is it |r| or just r

Oh wait nvm

[blissful trench]

no

log is fine

[livid crystal]

How do I do the log

Do I make it 5*n

Because the parenthesis are confusing me

How do I proceed

[tacit lodge]

This is what i got so far

They said ur allowed to use a calculator

[blissful trench]

the power rule

[tacit lodge]

Wait shit

[livid crystal]

Which one

[tacit lodge]

Theres a mistake

[livid crystal]

Which power rule

[blissful trench]

,tex .log rules

[grand pond]

KωΚιχ

[livid crystal]

So I do use 5^n?

And then do that

I got a weird number doing that

[blissful trench]

where are you getting 5^n from?

[livid crystal]

From the 1- -4

Two negatives

Make a five

[blissful trench]

no that's wrong

!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.

[tacit lodge]

Oh right sorry

[blissful trench]

you've said that as hint

better delete it

[blissful trench]

you know how to move things right?

[livid crystal]

I guess but not here

Idk what to do with the negatives and parenthesis

[tacit lodge]

Try simplifying the RHS on the 2nd step

[blissful trench]

leave the 1 to the other side

[livid crystal]

Done

16384 = -(-4)^n

Is what I have now

I see a way out

Yay

[tacit lodge]

So just 16,384 = 4^n

Use a log and ulk get the answer

[livid crystal]

I got it!

[blissful trench]

good job

[livid crystal]

Tysm KwKix

.close

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[viral dagger]

given 2 positive real sequences sequences $a_1\leq a_2\leq \dots \leq a_n$ and $b_1\leq b_2\leq \dots \leq b_n$, for any permutation $\sigma:{1,2,3,\dots,n}\to {1,2,3,\dots,n}$ prove that
$$(a_1+b_1)(a_2+b_2)\dots(a_n+b_n)\leq (a_1+b_{\sigma(1)})(a_2+b_{\sigma(2)})\dots(a_n+b_{\sigma(n)})$$

[grand pond]

skissue.in.a.teacup

[viral dagger]

not rly sure how to do this

[zealous schooner]

have you tried proving it by induction?

[keen rock]

How do you prove a base case

[zealous schooner]

well the base case is just n=2

[viral dagger]

uhh lemme try

[zealous schooner]

the rearrangement inequality is definitely gonna be used somewhere in there

[viral dagger]

ok base case works

[zealous schooner]

it's not that obvious, but yeah

[viral dagger]

assume for some k-1 and k it holds
then we want to prove for k+1 it holds
if sigma(k+1)=k+1, then it obviously holds
if sigma(k+1)≠k+1, say sigma(k+1)=m, then you can seperate all terms except those with a_m and a_k+1 with the fact that it holds for k-1 to get
nvm this doesent work

i thought you could prove that (a_m+b_m)(a_(k+1)+b_(k+1))<=(a_m+b_sigma(m))(a_(k+1)+b_m) but im not sure how

[zealous schooner]

this is just the n=2 case!

[viral dagger]

well no not really, for n=2 you know that sigma(m)=k+1 cause there arent anu other choices right? and also i dont think you can just cancel it all for k-1?

[zealous schooner]

well didn't you define m so that sigma(m) = k+1?

and wdym can't cancel it for k-1

[chrome vessel]

i think he just used n as k

[viral dagger]

i defined sigma(k+1)=m

it should be $$(a_1+b_1)\dots(a_k+b_k)(a_m+b_m)(a_{k+1} + b_{k+1})\leq (a_1+b_{\sigma(1)})\dots(a_k+b_{\sigma(k)})(a_m+b_{\sigma(m)})(a_{k+1}+b_{m})$$ which implies with $k-1$ case
$$(a_m+b_m)(a_{k+1}+b_{k+1})\leq (a_m+b_{sigma(m)})(a_{k+1}+b_m)$$ which i dont think is nescicarily true

[grand pond]

skissue.in.a.teacup

[chrome vessel]

maybe letting $k+1=\sigma(u)$ in the case where $u\neq k+1$ might help. Think about $(a_u+b_{\sigma(u)})(a_{k+1}+b_{\sigma(k+1)})$ vs $(a_u+b_{\sigma(k+1)})(a_{k+1}+b_{\sigma(u)})$.

[grand pond]

qwertytrewq

[viral dagger]

ohh fuckk uhh i forgot to specify 2 positive real sequences

[viral dagger]

well its >= since the whole thing is equivalent to a_u>=a_k+1 which is true

[chrome vessel]

you mean >=?

[viral dagger]

er whar

[chrome vessel]

$b_{\sigma(u)}=b_{k+1}\geq b_{\sigma(k+1)}$

[grand pond]

qwertytrewq

[viral dagger]

oh yea

[chrome vessel]

now induction will carry through

uve managed to get b_k+1 to the end

[viral dagger]

it should be $$(a_1+b_1)\dots(a_k+b_k)(a_u+b_u)(a_{k+1} + b_{k+1})\leq (a_1+b_{\sigma(1)})\dots(a_k+b_{\sigma(k)})(a_u+b_{\sigma(k+1)})(a_{k+1}+b_{\sigma(u)})$$ which implies with $k-1$ case
$$(a_u+b_u)(a_{k+1}+b_{k+1})\leq (a_u+b_{\sigma(u)})(a_{k+1}+b_{\sigma(k+1)})$$
$$(a_u+b_u)(a_{k+1}+b_{\sigma(u)})\leq (a_u+b_{k+1})(a_{k+1}+b_{\sigma(k+1)})$$

uhhhh

[chrome vessel]

i think you should write (a_1+b_{sigma(1)})...(a_{u-1}+b_{\sigma(u-1)})(a_{u+1}+b_{\sigma(u-1)})...(a_k+b_{sigma(k)}) * (a_u+b_{sigma(k+1)})(a_k+1+b_{sigma(u)}) instead

otherwise it would look as though u r multiplying by both (a_u+b_{sigma(u)}) and (a_u+b_{sigma(k+1)}) on the RHS

oh wait u r

wait no you arent

[grand pond]

skissue.in.a.teacup

[chrome vessel]

why is the inequality in the first line true

[viral dagger]

n=k+1 case?

from a_1 to a_n i just pulled out a_u in there

[chrome vessel]

I think you should write a more clear solution, its kind of hard to understand what you are doing

As well you multiplied a_u+b_u twice on the LHS

[viral dagger]

aight

[chrome vessel]

If I understand correctly, here is a way of writing what you wanna put:
$$(a_u+b_u)(a_{k+1}+b_{k+1})\left(\sum_{\substack{1\leq i\leq k\ i\neq u}} (a_i+b_i)\right)$$

[grand pond]

qwertytrewq

[midnight plank]

@viral dagger Has your question been resolved?

[viral dagger]

honestly im not too sure of the "thus" part and weather i can simply divide it or not

[chrome vessel]

this first inequality is by n=k-1?

[viral dagger]

n=k+1

[chrome vessel]

oh wait nvm i misread

[chrome vessel]

the n=k-1 case is wrong

because sigma restricted on ${1,2,\ldots, k+1}\setminus {u,k+1}$ is not be bijection between ${1,2,\ldots, k+1}\setminus {u,k+1}$ and itself

[grand pond]

qwertytrewq

[viral dagger]

i was kinda hoping since there are k-1 terms and a_1<=a_2<=...<=a_u-1<=a_u+1<=...<=a_k+1 it can hold

[chrome vessel]

well unfortunaly you need sigma to be a bijection too

[viral dagger]

ok

[chrome vessel]

hint: you need the $a_u+b_{\sigma(k+1)}$ term

[grand pond]

qwertytrewq

[chrome vessel]

$\sigma(1),\ldots, \sigma(u-1),\sigma(k+1),\sigma(u+1),\ldots, \sigma(k)$ is a permutation of ${1,\ldots, k}$ (why?)

[grand pond]

qwertytrewq

[viral dagger]

whats with the arrangement of sigmas

[chrome vessel]

to correspond with the a_i's

$a_1+b_{\sigma(1)},\ldots, a_{u-1}+b_{\sigma(u-1)},a_u+b_{\sigma(k+1)},\ldots$

[grand pond]

qwertytrewq

[chrome vessel]

ok this should be a pretty big hint for you, try and amend your proof

[viral dagger]

also i still dont get what you mean with this? isnt that just the definition of what the sigma function is

[chrome vessel]

sigma is a permutation of {1,...,k+1}, but I deliberately modified the function, to derive a permutation of {1,...,k}. The purpose is to apply the n=k case

[chrome vessel]

as you can see here, I threw out sigma(u)=k+1

and I reordered it a bit

like putting sigma(k+1) at position u

[viral dagger]

for this did you use

[viral dagger]

this?

[chrome vessel]

yeah

[viral dagger]

so a_u is a lil diffrent but a_k+1 is also a lil diffrent

[chrome vessel]

yeah but after the transformation, we get $(a_{k+1}+b_{\sigma(u)})$ which is nice since this is just $a_{k+1}+b_{k+1}$. However, $a_u+b_{\sigma(k+1)}$ might still be weird.

[grand pond]

qwertytrewq

[viral dagger]

ok i noticed this looks simmilar to n=k case?

btw with the sigma having a bijection problem how are you supposed to apply IH to it

[chrome vessel]

prove this hint

[chrome vessel]

we arranged it in a way to make it where b_k+1 is pair with a_{k+1}. If we dont chage anything else, you should be able to prove that the remaining terms forms a bijection

which is exactly the hint

[viral dagger]

i mean sigma(1),...,sigma(u),...,sigma(k+1) is a bijection to (1,2,...,k+1) and we know sigma(u)=k+1 so if we get rid of sigma(u) and k+1 then it becomes the hint no?

ohh so we proved sigma(1),...,sigma(u-1),sigma(k+1),sigma(u+1),...,sigma(k) is a bijection to (1,2,...,k) so by IH it holds?

[chrome vessel]

yeah, so, if you want to be a little bit more formal, if we have a function $\tau:{1,\ldots, k}\to {1,\ldots, k}$ defined by
$$\tau(x):=\begin{cases} \sigma(x) &x\neq u\ \sigma(k+1) &x=u\end{cases}$$
then $\tau$ would be a bijection

[grand pond]

qwertytrewq

[chrome vessel]

the proof of this just pretty much follows from the bijectiveness of sigma and sigma(u)=k+1, as you pointed out

[viral dagger]

ohhh ok i see

alright thank you

.solved

[midnight plank]

Channel closed

Closed by @viral dagger

Use .reopen if this was a mistake.

[chrome vessel]

np!

[midnight plank]

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[verbal pumice]

"This is like expanding a vector in terms of basis vectors"

[verbal pumice]

Could someone explain what that and the subsequent statement means exactly

[lyric charm]

the sinusoids are vectors in a particularly nice basis for the function space

[verbal pumice]

a function space is a vector space?

[quiet hinge]

Well, yes

You can easily verify with the definitions

[verbal pumice]

If I consider $\mathbb R^2$, I could write a vector $3\mathbf i + 4\mathbf j$, are the sine waves here the "unit vectors"?

[grand pond]

dyxn

[quiet hinge]

The Bn act as basis vectors' components

[lyric charm]

kinda

[lyric charm]

sure is!

[verbal pumice]

What's the analogy for the magnitude?

And I don't quite get how B_n becomes that integral

[quiet hinge]

You define a "dot product" and get the magnitude through that

[verbal pumice]

Is there a standard idea

Oh wait is the integral the dot product

Because it's getting us the value of the components

Ahh yeah I think it is it makes sense

[lyric charm]

sqrt of integral of the square

[lyric charm]

yes

[quiet hinge]

The dot product of f and g here (real valued functions) will be defined: $\langle f,g \rangle = \int_{0}^{L} f(x)g(x)\mathrm{d}x$

[grand pond]

Asteroid

[verbal pumice]

But the magnitude is not 1 right

Ah wait, doesn't matter

[quiet hinge]

And the magnitude is defined ||f|| = sqrt(<f,f>) as it is in the euclidean case

You can scale the inner product appropriately to make the vectors you want into magnitude 1

[verbal pumice]

in this case theres a sqrt(2/L) on th outside right

thats intrinsic to this inner product

because the basis hasn't been normalized I suppose

[quiet hinge]

Yeah thats the scaling referred to here

[verbal pumice]

yeah okay

so all the snazzy properties of vector spaces now hold

makes sense, thanks

.close

[midnight plank]

Channel closed

Closed by @verbal pumice

Use .reopen if this was a mistake.

[last slate]

can anyone help me in determinants ?

[sudden yacht]

Read #❓how-to-get-help

[midnight plank]

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[twilit field]

This makes no sense to me, could someone explain it?( I just need the formula, no proof)

[twilit field]

the s is confusing me

more context

is s the number of iterations

if so, I'd like someone to walk me through this

so I have $1+ \sum_{k=1}^{5} \binom{5}{k} \Delta^{k}$?

I'll drop it to the first two points because I'm doing this by hand

so like 1.22