help-17

[bitter pilot]

you cannot add a scalar with a vector

[sinful magnet]

Oh

but cant you just leave it as

A scalar and a vector

Or is that just impossible

[bitter pilot]

It's undefined

[sinful magnet]

If you add a scalar to a vector its undef?

[bitter pilot]

yes

[sinful magnet]

kk

What about this vector times vector = scalar

[bitter pilot]

scalar multiplication is defined

[sinful magnet]

scalar times vector

but its undefined

in this case

not sure why though

[sinful magnet]

photomath says its undefined because of rows and columns

but i dont get it

[bitter pilot]

because it doesnt treat it as a dot product

it treats it like a product of two matrices

and you cannot multiply a 3 by 1 matrix with itself

[sinful magnet]

so wait is the answer -1(1 2)

[bitter pilot]

ye

you have to have as many columns as the other one has rows

for matrix multiplication to be defined

So for a 3 x 1 matrix you need a 1 x n matrix

[sinful magnet]

what does l and m mean

[sinful magnet]

can u show me an example

[bitter pilot]

the 1st letter denotes the amount of rows, the second columns

l x m
l rows and m columns

m x n
m rows and n columns

[sinful magnet]

ah

so wait the lets say if it isnt in the same dimensions, the dot product wont work

[sinful magnet]

is that what u mean

[bitter pilot]

the dot product is only for vectors (aka matriced with only one column defined)

but that is different from actual matrix multiplication

[twin meteor]

bacc

[bitter pilot]

First one is defined by matrix multiplication because we have the 3 x 1 and 1 x 3

So the result is a 3 x 3 matrix

[sinful magnet]

ok so if the second is a dot product i know how to do it

nvm ima look up online how to solve matrix multipli

[bitter pilot]

ye

watch a video

[sinful magnet]

okay ty

[bitter pilot]

essentially, you multiply each row with each column, respectively

[vocal sleet]

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[bitter pilot]

can you express the numerator as a sum

[untold arrow]

Aren't you going to sleep🧐

[plain aurora]

What does the top look like?

You should get a nice simplification

[untold arrow]

I get it, thx

[vocal sleet]

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[fervent wasp]

Just type .close next time

[untold arrow]

Ooh

I thought they're the same

[fervent wasp]

Yeah, both works

[untold arrow]

Sometimes I'm just too lazy to type close 😂

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[ivory fulcrum]

I'm having trouble with an arithmetic sequence problem. The sequence a_n is defined by a_1 = 1/2 and
a_n = a_{n - 1}^2 + a_{n - 1} for all n greater than or equal to 2.
The goal is to prove that
1/(a_1 + 1) + 1/(a_2 + 1) + ... + 1(a_k + 1) < 2 for all k greater than or equal to 1.

So far, I've gotten to
1/(a_{n - 1}) - 1/(a_n) = 1/(a_{n - 1} + 1). Can somebody tell me how to substitute this into the original sum? This might be very straightforward but my brain isn't working very quickly right now...

[brisk moss]

this is hard to read

[ivory fulcrum]

im sorry i genuinely have no idea how to make it more readable

[brisk moss]

picture?

[ivory fulcrum]

I'll add what I've done so far:

This is where I am right now

I know it gives me a substitution I can put into the original sum, but i'm not sure how to do that

[inner osprey]

this is fine

the last equation you have is what's important as you've noticed

using it, you can consider what 1/(a1 + 1) + 1/(a2 + 1) + .... + 1/(ak + 1) would be

[ivory fulcrum]

I know its just really not coming to me right now.. like I can see that it has the same form as the terms in the sum so I can probably substitute it somewhere but I can't think of it for some reason,,,
like I know 1/(a_k + 1) = 1/a_(k-1) - 1/a_(n) and then 1/(a_(k-1) + 1) = 1/a_(k-2) - 1/a_(k-1) and so on but i dont know how to show that leading to an answer?
Could you maybe give me a hint if you know? It's probably obvious but I'm unable to see how i can use 1/(a_n + 1) = 1/a_(n-1) - 1/a_(n)

[flat whale]

You should learn latex or just write it and take a picture

[bitter pilot]

you deleted the question 😂

[inner osprey]

replace 1/(a1 + 1) with 1/a1 - 1/a2 and etc

[ivory fulcrum]

Everything I wrote was originally in latex, but I didn't know it worked for discord?

[flat whale]

$1/a_(n-1)$

[twin meteor]

riemann

[flat whale]

Yea no

[ivory fulcrum]

$\frac{1}{a_{n-1}}$

[twin meteor]

night night

[ivory fulcrum]

oh wait yay

[flat whale]

[ivory fulcrum]

🎉

[ivory fulcrum]

im sorry that was really obvious... i

will work on that and see if I finish

[inner osprey]

neat u get it

also no worries

[vocal sleet]

@ivory fulcrum Has your question been resolved?

[vocal sleet]

@ivory fulcrum Has your question been resolved?

[ivory fulcrum]

OK so I was able to find the terms and a telescoping series in the sum, but I still don't know if this shows that the sum will always be less than 2?

This is what I've done

[vocal sleet]

@ivory fulcrum Has your question been resolved?

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[untold arrow]

if f is continuous on [a, b], then S = {x in [a, b] : f(x) >= 0} can be seen as an at most countable reunion of intervals? How to prove that

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[halcyon compass]

Can someone help me with simpson rule (numerical method) that approximates integrals

[halcyon compass]

How is the formula derived from the parabola equation based on 3 points?

[vocal sleet]

@halcyon compass Has your question been resolved?

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[vast shale]

was learning integration by partial fraction. when i form partial fractions. just for simplicity sake lets say $\frac{2x+1}{(x+2)(x+3)} = \frac{A}{x+3} + \frac{B}{x+3}$ why can I put any values of x? to find A and B?

[twin meteor]

Matchstick

[vast shale]

my teacher said something about it becoming an identity

but i cannot fathom what difference that could possibly make

[edgy gulch]

because you're turning that original expression into another form, but you can go back to the original by algebraic manipulation

[vast shale]

how is an equation any different in that respect

[edgy gulch]

not much

its just an equation that holds true for all x values

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

okay

thank you so much

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[dapper olive]

hello is this available?

[dapper olive]

is there anyone here?

help me with this question

i have the solution for this question but i dont understand it

<@&286206848099549185>

[toxic river]

share the solution.

[dapper olive]

@toxic river

and the ans is option B

<@&286206848099549185>

where are the helpers :((

[dapper olive]

@vast shale

i have the solution but i dont understand it

and i believe you wont too

[pallid forge]

there was no other information before this?

[dapper olive]

no

you need to find the value of that expression

[pallid forge]

do you understand the first half of this?

[dapper olive]

no, i understand the second half

[pallid forge]

that's about where i'm at as well

[dapper olive]

after they took log

[pallid forge]

idk how you were supposed to know to do all that in the first half

[dapper olive]

if you got any other solution

to begin with

this is high school level problem

i am thinking of finding a different method

like either writing a general term then take summation and then use integral sum of limits that way

where n tends to infinity

[dapper olive]

@pallid forge can you help me with the general term

if the powers are not visible

[pallid forge]

well the numerator you can find in the third line of the given solution

[dapper olive]

but i need to understand how i got to the third line

[pallid forge]

well you could work backwards i guess

[dapper olive]

can you explain the first two lines

[pallid forge]

recognize that the numerator is the derivative of the denominator

my guess is that they worked backwards originally to find the solution

and then for some reason presented it in a forward manner like it comes naturally to multiply three polynomials together and know the equivalent polynomial

[dapper olive]

thats not how you are supposed to make questions

[pallid forge]

from here

maybe try starting here

[dapper olive]

can you think of any other method?

[pallid forge]

not really no

someone else might

[dapper olive]

do you understand the first two lines?

@pallid forge

can any other helper help me out

<@&286206848099549185>

😦

.close

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[thin root]

how did they get this

[hushed bobcat]

are you familiar with partial differentiation?

[thin root]

yeah

[hushed bobcat]

so the first equation is the partial differentiation of r(theta,phi) wrt theta. So you basically assume that phi is a constant while differetiating wrt theta

[thin root]

my calc 3 is very rusty i forgot what this is called

[hushed bobcat]

in the first term or r, 2sin(phi) would be constant and you then differentiate cos(theta)

[thin root]

yeah i can do that but idk why

ohhhhhh

nvm

it's just notation

was this called directional derivative?

[hushed bobcat]

nope

its partial derivative

[thin root]

what is this $\vec{r_{\theta}} \cross \vec{r_{\phi}}$ ?

[twin meteor]

ahtrader

[hushed bobcat]

are you familiar with cross product of vectors?

[thin root]

yeah

[hushed bobcat]

those two equations give you the vectors r(theta) and r(phi). now you apply cross product between those two vectors

[thin root]

yeah but why do i need it for the surface integral

srry i think im asking a dumb question

[hushed bobcat]

the surface element dS is given by magnitude of (r(theta)xr(phi))d(theta)d(phi)

so when you do zdS and substitute dS in terms of the parameterized form you would have to compute r(theta)xr(phi)

[hushed bobcat]

ofc not

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@thin root Has your question been resolved?

[vocal sleet]

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[hollow umbra]

For combination and permutation, can anyone define what does it mean by order matter

[warm vine]

if order matters then each arrangement is considered unique

for example you want to permute 2 numbers lets say 1 and 3

if order matters then (1,3) and (3,1) are both different arrangements

but if order doesnt matter then (1,3) and (3,1) are the same thing

[hollow umbra]

So what type of question would be like tho

For both

[hollow umbra]

What if you have like 6C5 and 6P5 what does that mean

[warm vine]

P is for permutations

and C is for combinations

[hollow umbra]

No I mean the orders

[warm vine]

Permutation when order matters
Combination when order doesnt matter

[hollow umbra]

Like so (12345) and (54321) are same thing for combination?

[warm vine]

you can remember as "permutation is an ordered combination"

[warm vine]

yes

[hollow umbra]

Oh ok

Ty for your help

[warm vine]

[hollow umbra]

.close

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[viral copper]

[ \sum_{k \geq 0} a_k < \infty \iff \sum_{k \geq 0} \arcsin(a_k) < \infty ]

[twin meteor]

ananas

[viral copper]

0 <= a_k < 1

I think it's true because sin(x) = x => x = arcsin(x)

for small x ofc

it's easy to prove it backwards since arcsin(x) >= x on [0, 1)

not sure about the forward direction

[woeful igloo]

Is this a part of the problem? Coz otherwise it seems like a wrong conclusion to make

coz if sum of a_k is finite, so is the sum of 1000000000*a_k. So, I wouldnt arrive at this as a conclusion

I mean I get that a_k gotta be smaller than 1 for arcsin to exist but still, I feel like that has to be part of the problem

[hard atlas]

well at least eventually you have a_k < 1 anyway

and then wlog you can just look at the end of the series where that is true

[dense eagle]

well sin x <= x so you can use x <= arcsin(x) for one of the directions

similarly, arcsin(x) <= 2x so u can use that for the other direction

[vocal sleet]

@viral copper Has your question been resolved?

[vocal sleet]

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[eternal glacier]

would you agree with the following? or see any issues

[silk osprey]

why not submit and find out?

[eternal glacier]

it doesn't tell me

[silk osprey]

it doesn’t tell you if you were correct

[eternal glacier]

yeah

[silk osprey]

seems like a quiz/test to me

[eternal glacier]

and it's timed, 15 minutes

online

[silk osprey]

online quiz

[eternal glacier]

yeah

[silk osprey]

<@&268886789983436800>

[eternal glacier]

i didnt know it's against TOS

.close

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[silk osprey]

cheating?

[eternal glacier]

i already answered, i'm just making sure

it's already submitted btw

[silk osprey]

that’s not cheating?

[eternal glacier]

how is it cheating if i already submitted?

i can't resubmit

[silk osprey]

https://discord.com/channels/268882317391429632/486128019652476939

[silk osprey]

well now you did

[eternal glacier]

i'm just looking for feedback

[eternal glacier]

wdym?

[silk osprey]

you submitted it after the fact yes?

after i pinged mods

[dull bear]

.reopen

[vocal sleet]

✅

[near smelt]

I was going to ban you for cheating, I'm not buying that it's necessarily submitted so you're not getting any feedback on it for now

[dull bear]

As in, was this in the test you did, and you're checking after you've done it that the answers you gave were right?

[silk osprey]

you believe that?

i’m with ghost

[near smelt]

Cheating is normally an instant ban, I'm timing you out for now so you can't post about that question while the test is still active and consider that a very close warning

[dull bear]

@eternal glacier has been around for long enough

[silk osprey]

if it’s already submitted they shouldn’t be able to have access to it or if they did they should know the answers..

it says "clear my choice"

[hard atlas]

yeah a bit surprising that someone with >40k messages hasnt heard about the no cheating rule.

[silk osprey]

💀

[silk osprey]

^

but they submitted already

😂😂😂

be fr

[near smelt]

I'm being pretty lenient considering cheating and slurs are like the two things we're pretty insta ban happy on

No point continuing this now though, just don't ever do something like that again or you will be banned

.close

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[near smelt]

If you do want feedback genuinely you can post about it when the timeout expires

[vocal sleet]

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[median thorn]

Pls help me with the exercise d) I can see in the graph that it is decreasing starting at the point f(0), but I don’t know how to write it in math terms

[median thorn]

Like {f(0);to +infinite}?

[spice perch]

you just write "in the interval (a, b)"

[median thorn]

I don’t get it

what a and b?

[spice perch]

the 2 endpoints of the interval

like x=0 and infinity

[median thorn]

it does not have endpoint

aa

[spice perch]

yep so thats just infinity instead

[median thorn]

Or it has?

[spice perch]

make sure to do ([

or whatever notation u do

since infinityt is not included

[median thorn]

I see that does not go to infinity it ends in y=-1

I don’t understand 😞

[spice perch]

Okay I thought the graph just ended there

just do whatever is true for the function

[median thorn]

I don’t know now

Is it ends there

Or not

[hardy thicket]

how would you define domain and range?

[median thorn]

you mean exercises e and f?

[hardy thicket]

oh nvm i thought you meant e

[median thorn]

now i am on d

[hardy thicket]

from what i see

(X, Y) are the 2 values i assume

[hardy thicket]

the point F is decreasing is 3 blocks up, so Y should be 3

and X should be 0

so interval is (0,3)

[median thorn]

[0;3] & [4;-1]

is that how the answer should look?

[hardy thicket]

do you mean this point by 4, -1

[median thorn]

No

It is little bit right

one square right

[hardy thicket]

nvm thats 3-1

this one?

[median thorn]

Yes

Or we should write [f(0);f(4)]?

[spice perch]

no

[hardy thicket]

The question is asking on what interval the F is decreasing, to me it can be all of these infinite intervals

[spice perch]

that is y coordinates

[median thorn]

oh yes sorry

[hardy thicket]

but it is saying "on what interval" rather than "on what intervals" so its talking about a single interval so what it could mean I think is when it is starting to decrease

and that only being the (0, 3) interval

[median thorn]

You mean it is all are intervals after point (0;3)?

but I think it is one big interval

[hardy thicket]

the 0,3?

[spice perch]

0 is not included since the function is not decreasing at x=0

[median thorn]

No 0;3 and 4;-1

[median thorn]

Yes right

[spice perch]

what do you mean 4;-1

[median thorn]

It is the point the interval ends

[median thorn]

This

[spice perch]

the interval is juyst for x

you just put the x coordinates

[hardy thicket]

1 second let me recheck the intervals defination

[spice perch]

oh

so its gonna be

[hardy thicket]

nah nvm

[spice perch]

sorry

i didnt see it extended further

thats ur answer

[median thorn]

Yes i get it but why I put only x coordinates

[spice perch]

the notation for intervals might be different for u

the interval is not open in the negative direction

that is to say 0 ius not included

[hardy thicket]

Interval can be only 1 point i believe, and since in question it is saying interval rather than intervals, so it could be either only 1 interval or the question is wrong

[median thorn]

(0;4] we write it like so

[spice perch]

give me 10m

ill explain

[median thorn]

ok

[spice perch]

the function assigns values of y and x to each other basically

like ur used to

f(x)

[median thorn]

so if I put x=0 then it gives me y=3?

that’s why I should look for x coordinates?

[spice perch]

erm

its askins u in which interval

that is between which 2 points

is f decreasing

that is, f(x)

so between which points x_0 and x_1 is f(x) decreasing

[median thorn]

ah I see

Thx you

.close

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[kind rivet]

.

Need help with some INTIGRALS

[kind rivet]

How is this solved (USEING identity too)

[grim lotus]

king's rule

$$ \int_a^b f(x) \dd{x} = \int_a^b f(a+b-x) \dd{x} $$

[twin meteor]

mayer-vietorUs

[kind rivet]

WOW I DIDNT KNOW ITS CALLED THAT IN SCHOOL THEY TEACH LIKE IDENTITY 5 ETC ETC ITS ASS ACCUALLY

@grim lotus how do I make it into like being able to put in identity

Anyone?

[hushed bobcat]

wdym?

[warm vine]

wdym? you just have to replace x with (a+b-x)

[kind rivet]

2 pi - x right

[warm vine]

yes

so sin(2pi-x) is?

[kind rivet]

Sorry I am shit in Intigrals school kinda rushed it I got this practice sheet from school so like I am trying to learn from this

[kind rivet]

-sin?

[warm vine]

yes

[kind rivet]

Now what do I do I am completely clueless

WAIT

DO I

[warm vine]

so now u have $\int_{0}^{2\pi}\frac{dx}{e^{-sinx}+1}$

[twin meteor]

Astar777

[kind rivet]

YES

NOW

[warm vine]

try to think what u can do here

[kind rivet]

I/ E^ -SIN X? LIKEIN THE BOTTOM

[warm vine]

1/e^sinx* but yes

[kind rivet]

SORRY YES

[warm vine]

$\int_{0}^{2\pi}\frac{dx}{\frac{1}{e^{sinx}}+1}$

[twin meteor]

Astar777

[kind rivet]

SO NOW WE FIND LCM AND AFTER LCM THE DENOM GOES UP WITH DX

[warm vine]

yes

$\int_{0}^{2\pi}\frac{e^{sinx}dx}{e^{sinx}+1}$

[twin meteor]

Astar777

[kind rivet]

😭 WHATEVER THIS IS ITS A SKILL DAAAAM

[warm vine]

its ez actually

[kind rivet]

what do we do now tho do we take SMT as t?

[warm vine]

nah

[full ferry]

sum the two integrals you have

[kind rivet]

it cam be simplified more?

[warm vine]

sum your initial integral and this

[kind rivet]

Huh

[full ferry]

if we let the initial integral be I, then 2I = [ the two values of I you have]

[warm vine]

I = $\int_{0}^{2\pi}\frac{dx}{e^{sinx}+1}$ and I = $\int_{0}^{2\pi}\frac{e^{sinx}dx}{e^{sinx}+1}$

[twin meteor]

Astar777

[warm vine]

sum these 2

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[fallen rain]

.reopen

[covert turtle]

Is there any way to check if you solved a constrained optimization correctly?

[lime gorge]

Plug ur solution back into ur objective and constraint

@covert turtle

[covert turtle]

Yeah I already did that and it checks out. Just wanted to make sure there wasn’t something I was missing. Thanks!

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[short kraken]

i dont understand what they did here

position function is just -4.9t^2 + vt + C

did they set v to 0

and our constant is 1900?

[twin ginkgo]

hey can i get help

[fallow kindle]

Ask in one of the available channels, this one's taken

[snow trout]

initial velocity is 0 cuz ur dropping it

height or distance is 1900 cuz thats how high ur dropping it from

[cunning basin]

guys help me

[snow trout]

!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).

[cunning basin]

2+2=5

right?????

[short kraken]

bro leave

[cunning basin]

is 2+2=5

[short kraken]

ah

[flat whale]

<@&268886789983436800>

[cunning basin]

mb sry

i was going to leave :/

[short kraken]

so because we're looking for t and we're only given that the height is 1900 we set v to 0

to find the initial time?

like thats whats going on here

[outer warren]

Don't troll

[cunning basin]

ok

[snow trout]

they used a kinematic eq

the third one here i believe

[short kraken]

that looks overly complicated

[snow trout]

its really not

[short kraken]

its not pretty to look at. its a lot easier for me to go "oh just set v to 0 and find t"

[snow trout]

i dunno how ur planning to solve this w/o the kinematics eq but ok

cuz thats literally what they used

[short kraken]

fine

what does the 0 subscript mean

[snow trout]

typically initial

x_i or x_0 denotes initial value

in this case distance/position

so here x_0 (or y_0 is more fitting ig) is 1900 as thats ur initial height

[short kraken]

canyon floor is position 0

my position 0 is s_0 right

[snow trout]

?

u r starting at height of 1900

[snow trout]

ya they put 0 at the left side to work out how long itd take for it to hit the ground

[short kraken]

right

but why did my initial velocity go away

[snow trout]

cuz ur just dropping it remember

ur not pushing the rock downwards w/ extra force

[short kraken]

i thought velocity was the change in y

[snow trout]

the instant ur dropping it, its velocity is 0

[snow trout]

yes the rate of change of displacement

for these kinds of qs its safe to assume when they say "drop" that v0 = 0

[short kraken]

so the velocity picks up as it goes further down? and the velocity is 0 at the exact moment my hand lets go of the rock

idk physics isnt my strongsuit lol

[snow trout]

yes thats right

[short kraken]

okay then i get it

[snow trout]

and ofc ur acceleration in the y-direction is plain old gravity

[short kraken]

so just to reiterate, t is how long it takes to reach the bottom, v is the rate at which its falling, and s is the initial height

[snow trout]

t is how long it takes to reach the bottom
not necessarily, its just a variable, here f(t) is a function of height wrt time

v is the rate at which its falling
rate of change of the height ya

s is the initial height
well s_0 is... s is something else

[short kraken]

oh

thanks for the help

.close

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[silver python]

i got 3x2 can anyone verify?

[flat whale]

explain your answer

because it's wrong

[vocal sleet]

@silver python Has your question been resolved?

[silver python]

actually i got undefined because in the final step multiplying by p1 is not possible

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[delicate edge]

im confused

not of the problem but how to input it

the system stated that for b, -4/3, -3/2, and -5/3 were all incorrect

[twin horizon]

Have you tried doing the two point form of the equation of a straight line

[delicate edge]

y2-y1/x2-x1?

(-2, -3), (3, 1)

4/3

1--3

[twin horizon]

$\frac{y_2 - y_1 }{x_2 - x_1}$ = $\frac{y - y_1 }{x - x_1}$

[twin meteor]

David

[delicate edge]

it states 4/5x+(4/3) is incorrect

as well as 4/5x+(-4/3)

[twin horizon]

Identify y2,y1,x2,x1

[delicate edge]

1,-3,3,-2

right?

[twin horizon]

Right

[delicate edge]

1--3/3--2

ik the slope though

i need help with the addition part

cause rn my input is 4/5x+(-4/3)

in slope intercept form

[twin horizon]

So m is 4/5

Now use the fact that
m = $\frac{y - y_1 }{x - x_1}$

[twin meteor]

David

[twin horizon]

Where x and y are just the usual variables

[delicate edge]

i dont seem to understand what you're asking of me

im sorry

[twin horizon]

Replace y1 and x1 in here

And replace m in here too

[delicate edge]

4-(-3)/5-(-2)

7/7

?

[twin horizon]

$\frac{4}{5}$ = $\frac{y - 1}{x - 3}$

[twin horizon]

No

[delicate edge]

OH

[twin meteor]

David

[twin horizon]

Then cross multiply

[delicate edge]

-3y/-x

[twin horizon]

No

[delicate edge]

confused!

[twin horizon]

$\frac{a}{b}$ = $\frac{c}{d}$ $\implies$ ad=bc

[twin meteor]

David

[delicate edge]

-3y=-x

[twin horizon]

From here

[delicate edge]

4/5=-3y=-x

[twin horizon]

No

Make the right side c/d

And the left a/b

[delicate edge]

OHHH

4x-12=5y-5

[twin horizon]

Yes now add 5 on both sides to make 5y on its own

[delicate edge]

(4x-7)/5=y

[twin horizon]

Distribute the ÷5

Or rather separate the left into two fractions

[twin horizon]

Good

[delicate edge]

4x/5-7/5=y

[twin horizon]

Yay

[delicate edge]

UGH

LETS FUCKING GOOOO

[twin horizon]

):

[twin horizon]

Looooool

[delicate edge]

sorry! two sided coin

cause now i am allowed to continue further into the course and NOT understand stuff even more 😭

RAHHHH

LEARNING

LETS GOOO

oh and question, do the teachers ever actually do their job of teaching in uni or is it usually just the students teaching themselves? @twin horizon

[twin horizon]

Well I think it depends on the uni

[delicate edge]

we're a research 1 university

so they pride themselves on self taught learning 🥲

ugh but the people are SO nice and the campus is gorj

anyways THANK YOUUU

i will 1000% be back

intro to calc here we come

[twin horizon]

You are welcome

Type . close if you are done

[delicate edge]

.close

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[lethal heart]

how do I calculate the sum of the series $\sum_{n=1}^\infty \frac{x^{2n}}{n+1}$?

[twin meteor]

card

[hybrid flicker]

do you know power series of ln?

[lethal heart]

No

[hybrid flicker]

power series of 1/(1-x)?

[lethal heart]

No

Ah...

But would it be Taylor's developments?

[vocal sleet]

@lethal heart Has your question been resolved?

[lethal heart]

<@&286206848099549185>

[flat whale]

the power series of natural logarithm is the same as the taylor series

,w taylor series log(1+x)

[twin meteor]

Results provided by WolframAlpha

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

Requested by rie.mann

[lethal heart]

I know It!

[flat whale]

what you have is close

[flat whale]

find something similar to this to make the expansion match yours

[lethal heart]

Mmm

$x^2=y$

[twin meteor]

card

[lethal heart]

$n=1-m$

[twin meteor]

card

[lethal heart]

$n+1=m$

[twin meteor]

card

[lethal heart]

$\frac{1}{y}\cdot\sum_{y=1}^\infty \frac{y^m}{m}$

[twin meteor]

card

[lethal heart]

$\frac{1}{y}\cdot\ln(1-y)$

[twin meteor]

card

[lethal heart]

$\frac{1}{x^2}\cdot\ln(1-x^2)$

[twin meteor]

card

[lethal heart]

Right ?

But if so, how do I use it if x=0?

[vocal sleet]

@lethal heart Has your question been resolved?

[vocal sleet]

@lethal heart Has your question been resolved?

[vocal sleet]

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[sage badger]

help, i dont know where to start

[flat whale]

do you know the definition of uncertainty

[sage badger]

kind of

like its a percentage of something not working

its the wording of his question that confuses me

[flat whale]

you need the definition

percentage of something specific

[sage badger]

ah

im like probably disabled so go easy on me

well

5% of 90km/h is 4.5km/h

so is the range like 90 plus, minus 4.5km/h

so 85.5km/h to 94.5km/h?

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[fair mesa]

the angle would be theta - pi/3 right??

not the other way

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[peak rain]

what would be the equation of this graph

[inner osprey]

first does it look like any type of graph you are familiar with?

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[unborn prism]

Does anyone know any textbook for garde 12 problems??

[unborn prism]

https://www.doubtnut.com/qna/103515
https://www.google.com/url?sa=i&url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DC--cqPgpSaU&psig=AOvVaw31Yw6amFWPi2Cj8wpA2uTq&ust=1726986855463000&source=images&cd=vfe&opi=89978449&ved=0CBcQjhxqFwoTCPDM_dS304gDFQAAAAAdAAAAABAl

Where can I find a book with this kind of problems?

[vast shale]

if you really want some pain in the ass problems

which will get you prepared for jee mains

and concept problems

for advanced

by cenagge

cengage

but

YOU HAVE TO DO ALL OF THEM

sometimes questions seem too easy and you are likely to skip the problems

but don't

every problem matters

[unborn prism]

OK
but which textbook can be helpful?

[vast shale]

you are preparing for jee right ?

[unborn prism]

Yes

[ruby flicker]

داش سلام خب چرا مشکلت رو به چت جی پی تی نمیدی

اوهوم

[woeful igloo]

I'd suggest you join some institutes' distance learning programs. They would provide you with lots of varied material for practice problems, which in my personal experience better curated and covers everything. And would be cheaper than getting multiple books for PCM.

[unborn prism]

I have learned everything well and I want to solve problems to practice more.
I would use PDF instead of buying paperbook so this won't make it expensive!

[woeful igloo]

Back when I was prepping, I was offered free online material as a promo by some online company called embibe. idk if you heard of it, they werent great for advanced, but they had a ton of mains level problems

No idea how they are doing rn, but you can check them out

Or you can just browse amazon for which publishers have ebooks for jee prep

[unborn prism]

Thanks

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[winged urchin]

Stuck.

[lavish river]

what is the area of one triangular surface?

[winged urchin]

It doesn't tell me

the only value i have is the surface area & the width

[lavish river]

what surfaces are there

[vocal sleet]

@winged urchin Has your question been resolved?

[winged urchin]

what?

there are 5 sides

[lavish river]

yes and what do we know about them

[winged urchin]

nothing

[warm vine]

4 are triangle

one is square

[lavish river]

yea that

what do we know about the square

what do we know about the triangles

[vocal sleet]

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[proven kettle]

where to start to factor this one?

looks v messy

[outer warren]

it is messy, you are not required to find the actual solutions

you are asked to apply IVT

[proven kettle]

hmmmm, i need to find an interval first though

can i just try and plug in numbers to see?

[outer warren]

yes

(you do have to be careful due to the discontinuity though)

[proven kettle]

at x = 2?

[outer warren]

yes

[proven kettle]

ahh okay, thanks!

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[mint lark]

for Q19, I am able to solve (a), but i dont know how to do (b)

[mint lark]

I am not really sure how to get started

[small quarry]

what you can think, is how the series related to what u have proved

[mint lark]

yeah it took out the odd terms

[small quarry]

You see in (b), all the numbers are even numbers

[mint lark]

thats where im stuck

[lavish river]

replace n with 2n

[small quarry]

ys

For = 2k

[lavish river]

u can rewrite the top in terms of summation (n^3)

the second can be thusly written as summ (2n)^3

[mint lark]

oh

[lavish river]

fill in the upper value of summation and change the formula appropriately that u found

[mint lark]

alr

got it

thanks

.close

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[mint lark]

.reopen

[vocal sleet]

✅

[mint lark]

hi maybe i am getting things wrong, but the answer is not correct

[split wind]

it should be (2k)³ instead of
2 sum k³

[mint lark]

oh

bruh

[loud walrus]

Yes

[mint lark]

thanks guys

.close

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[hearty sable]

Hi I have a question regarding surjective, injective functions etc.:

Define
f: A --> B
g: B --> C

If g o f is injective, then prove that f is injective and that g is not injective

I have a hard time proving this and making it intuitive

[mild flower]

are you missing a part?

imagine A, B, and C are all R (the real numbers)

and let f and g both be the identity function

then everything is injective

[hearty sable]

Huh

I don't think A, B and C are R

It would be stated in that case

[mild flower]

it doesn't even matter what A B and C are

!original

[vocal sleet]

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

[hearty sable]

Number 9a and b

[lavish river]

,rcw

[mild flower]

does als...dan mean if...then?

[lavish river]

why are there 4 parts

[hearty sable]

Yes

I will translate it

Let f: A --> B and g: B --> C be two functions. Prove or disprove:
(a) if g o f is injective then f is injective
(b) if g o f is injective then g is injective

The answer sheet tells me a is true, b is not true

But I don't understand why

[mild flower]

ok wow your original question was way off

okay, b will be easier

because all you need to do is to find a counterexample

might be useful to draw diagrams of A, B, C, f, and g

[hearty sable]

So for (a) this was my thought proces:

g o f: (A --> C) (actually A --> B --> C)
We know that A --> C is injective st every element a e A resonates with an element c e C, but since this has to go between B this means that A -- B is injective, hence f is injective

Not sure if this makes sense I'm very new to proofs and pure math etc.

Like this is my 1st week

But the same logic did not apply to (b) so I got stuck

[mild flower]

the way that I would do (a) is to show that if f(x) = f(y) then x = y

at least that's the way I would write it

as for b, the proof is very different: you just need a single counterexample

[mild flower]

hint on this: you know things about g(f(•))

[hearty sable]

(g o f)(x) = g(f(x)) and x e A, f(x) e B, g(f(x)) e C
g o f is injective s.t. g(f(x)) = g(f(y)) implies that f(x) = f(y)
If f was not injective, f(x) could not equal f(y) because one output cannot have two different inputs
This means x has to be equal to y and hence f is injective

Is that valid?

no wait this is circle logic I think

Lemme try again

g o f is injective, thus A --> C is injective. Hence:
for x e A there is only one g(f(x)) e C, and similarly for y e A there is only one g(f(y))
Let's say that g(f(x)) = g(f(y)). Since g o f is injective, this implies that the inputs have to be the same f(x) = f(y), and this also implies that x = y since they both give the same final output. Hence, f is injective

I think I got it now

Now (b)

[vocal sleet]

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[lavish river]

could someone explain the class equation with S3

[lavish river]

the class equation is

in this case |Z(G)| is 0

and x1 would represent all orbits

so we just get 6?

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[fresh wing]

I drew out the graph and I'm so confused

I got k as 30°

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[fresh wing]

<@&286206848099549185>

Hello

Okay, I'm done

[flat whale]

!done

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[crude wing]

anyone know how to solve this

[crude wing]

i got 0 idea

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<@&286206848099549185>

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[graceful dove]

if x = a / b, what is dx/da and dx/db?

[warm vine]

if you differentiate x = a/b with respect to a, 1/b is a constant

so what do you think it should be

[graceful dove]

oh, so dx/da = 1/b then?

I forgot a lot of this from calculus

maybe product rule if I remember correctly? (it's been like a year since I done this)

x = a * (b^-1)

dx/da = 1 * (1/b) + a * (-1/(b^-2))
dx/da = 1/b - a/(b^2)
?

feels very wrong

[cosmic cloud]

yea prod rule

a*1/b

1/b differentiation is zero

so the second term

becomes zero

[graceful dove]

ohhh

[cosmic cloud]

just 1/b

[warm vine]

yep

[graceful dove]

ok so dx/da = 1/b

[cosmic cloud]

yes

[graceful dove]

Cool, how about dx/db?

Thank you

[cosmic cloud]

same use product rule(if you want)

or just take a*1/b

keep a outside the differentiation

because its a constant

[warm vine]

differentiating wrt b so a is constant

[graceful dove]

x = a * 1/b
dx/db = 1 * 1/b + a * (-1 * / b^0)

[warm vine]

why is it 1*1/b

and also it should be b^2 in second term

[warm vine]

differentiation of a is zero

so it should be 0*1/b

we are differentiating with respect to b, so a is constant here

and differentiation of a constant is zero

[graceful dove]

if x=a/b then dx/db is zero?

sorry

[warm vine]

no

you are differentiating wrt b

so a is constant

keep it out of differentiation

and just differentiate 1/b

constant multiple rule

[graceful dove]

ohhhh

1/b = b^-1

-1 / b^0
-1

[warm vine]

why b^0

[warm vine]

see the power rule

[graceful dove]

Power rule? I forgot a lot 😦

[warm vine]

and try again

[graceful dove]

ohhhh

ok

thank you

1/b = b^-1

-1 * (b^-2)

-1/(b^2)

Ok, that seems more correct?

[warm vine]

yes

[graceful dove]

Then multply by a to get dx/db right, so dx/db = -a/(b^2)

?

[warm vine]

yes

[graceful dove]

ok, thank you!

[warm vine]

!done

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..close

.close

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[toxic girder]

how do i check for which angles this equation is defined for?

[toxic girder]

https://i.imgur.com/zsPQLDb.png

for example^

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<@&286206848099549185>

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[bitter pilot]

hiiiiiiiiiiiiiiiiiiii

😂

[tardy peak]

guten abend

[bitter pilot]

Was sind deine Zweifel?

[tardy peak]

b

[bitter pilot]

Kannst du mir die Ebene aus a) nennen?

[tardy peak]

(6|-3|2)X = 10 Normalform

X = (4|0|-7) + (-3|4|15)u + (-4|-2|9)v

[bitter pilot]

ok soweit passt die a)

jetzt wollen wir im prinzip zwei parallele ebenen bestimmen die einen abstand von 21 haben

[tardy peak]

ist andere ebene parallel?

[bitter pilot]

ja

weil sonst schneiden die sich sonst

dann gibts keinen abstand wirklich

Davon gibts 2 stück einmal von links und von rechts

[tardy peak]

bedeutet das, dass beide ebene gleiche normal vektor haben?

[bitter pilot]

ja

[tardy peak]

achso

[bitter pilot]

Du musst im Prinzip die neue Ebene parallel verschieben bis sie Abstand 21 haben

[tardy peak]

wie?

[bitter pilot]

Dein Kreuzprodukt ist ja (6,-3,2)?

[tardy peak]

nein

(66| -33| 22)

[bitter pilot]

ja

geteilt durch 11

(6,-3,2)

[tardy peak]

ja

[bitter pilot]

Also sieht die Gleichung so aus

6x-3y+2z+d=0

[tardy peak]

ist d 31

[bitter pilot]

nein

wir müssen normieren

Die eine Ebene lautet ja
6x-3y+2z=10

Die Norm ausrechnen

,calc sqrt(6^2+(-3)^2+2^2)

[twin meteor]

Result:

7

[bitter pilot]

Teilen durch 7

10/7 is der Abstand zum Ursprung von der Ebene a)

Ja genau

6x-3y+2z=10
und
6x-3y+2z=d

d muss 10+21*7 oder 10-21*7 haben

Wir multiplizieren mit den Norm weil wir die Ebenengleichung nicht normiert haben

Das wäre das gleiche wie wenn man alle normiert also durch 7 teilt und dann +-21 hinzufügt um den Abstand zum Ursprung um 21 Einheiten zu vergrößern

Hier eine Veranschaulichung

[tardy peak]

eigentlich habe ich nichts verstanden

jetzt verstehe ich

danke schön

[bitter pilot]

soryyyyyyyyyyyyyyyyyyyyy

ich war kurz wegi

[tardy peak]

na gut.

[bitter pilot]

was sind die zweifel

[tardy peak]

ich habe es verstanden.

[bitter pilot]

🥹

[tardy peak]

wie kann ich den Winkel finden?

[bitter pilot]

Markier mich bitte sonst lese ich das gar nicht ^^

[tardy peak]

ok

[bitter pilot]

Schnitt mit der z-Achse impliziert einen Punkt der Form (0,0,z)

[tardy peak]

z=5

[bitter pilot]

E : 6x-3y+2z=10

[bitter pilot]

haha ja

(0,0,5)

Lass mich das zeichnen

[tardy peak]

wie finde ich den winkel

[bitter pilot]

das sieht bissel kacke aus

Im Prinzip kann man das mit dem Skalarprodukt ausrechnen

[bitter pilot]

Also wir wissen aus der a) die Richtungsvektoren, die in der Ebene sind, wir können einen nehmen, sagen wir (-4,-2,9)

Und dann einen Vektor der parallel zur z-Achse ist, also (0,0,1) z.B

[tardy peak]

cosA = (6|-3|2)*(0|0|1)/7

[bitter pilot]

HÖR AUF SO SCHNELL ZU SEIn

😂

[tardy peak]

was

ist es falsch

[bitter pilot]

nein

du bist so schnell haha

war gerade am erklären

[tardy peak]

ok erklär

[bitter pilot]

Ich check mal ab was du da hast

,calc sqrt((-4)^2+(-2)^2+9^2)

[twin meteor]

Result:

10.049875621121

[bitter pilot]

Ich denke die 7 ist falsch, die falsche Norm

du willst die Norm von (-4,-2,9)

nicht von dem Normalvektor von vorhin

,w sqrt((-4)^2+(-2)^2+9^2)

[twin meteor]

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Requested by adoniszgks

[bitter pilot]

Die Norm von (0,0,1) ist trivial

[tardy peak]

aber warum (-4|-2|9)