help-28

[torn jolt]

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[paper lagoon]

can someone check my answer please

[golden shore]

What if you did not get the answer right

is there a punishment or sth

[paper lagoon]

yes 100 lashes

[golden shore]

really? Im being serious

[paper lagoon]

my grade drops

[golden shore]

I see

[golden shore]

does it affect you to get into a good college in future

[paper lagoon]

yes

[golden shore]

I suppose it only depends on the test result?

[paper lagoon]

can you just check my answer man

[golden shore]

curiosity has predominated my mind

[paper lagoon]

are you finna help me ?

[golden shore]

yes

[paper lagoon]

can you check my answer pls

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[mental rover]

assuming h is height and t is itme in seconds, what is the maximum height?

h=-5t^2 +15t +0.5

[full forum]

Please don't occupy multiple help channels.

[mental rover]

can anyone help?

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[jolly anvil]

Alrighty

[jolly anvil]

so we have:

y / (1-y) = c * e^t

How do we isolate y? Is this even possible?

Because they want teh equation in terms of y, but I don't see how to isolate y?

[rain wyvern]

u can multiply both sides by (1-y)

[jolly anvil]

but then you get y on both sides?

ohhh then we divide by y

thank you

[onyx glen]

not quite

y = (1-y) ce^t
=> y = ce^t - ce^t * y
=> (1 + ce^t)y = ce^t

[jolly anvil]

I was thinking of

y / (1-y) = c * e^t
y = c * e^t * (1-y)
1 = c * e^t * (1/y-1)
1 / (c * e^t) + 1 = 1/y
y = 1 / (1 / (c * e^t) + 1))

but my way seems more error prone & much uglier

[onyx glen]

you are right on both counts

[jolly anvil]

I think i'll take a stab at sleeping

and then hound the MATH_HELP chat

With Linear Algebra Problems tmmw

🤩

so excited wooooo

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[rocky saddle]

idk how to solve

[rocky saddle]

got this so far

bruh I swear no one is gonna help me

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@rocky saddle Has your question been resolved?

[near mirage]

Have you tried solving for alpha and beta?

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[fossil dagger]

Hello, does someone know how to solve the following question?

[granite torrent]

i do not believe it is reasonably possible to solve this equation algebraically

[hallow walrus]

$6\sin x\tan x-18\tan x+12\sin x=3$

[glossy valve]

kheerii

[hallow walrus]

is this the equation

?

[granite torrent]

it is what is written

[hallow walrus]

yeah this doesn't seem doable

[floral frost]

its trigo?

find x?

[granite torrent]

in problems like this we normally find either x or sin(x) or something

but for this problem in particular sin(x) ends up only be expressible in an irreducible quartic

which… although it technically has an algebraic solution, it is essentially the same as being unsolvable

[floral frost]

so whats the ques tho

okay

one sec

lemme try

[floral frost]

shall i do from here?

[hallow walrus]

i wouldn't suggest it

the question seems flawed

[floral frost]

whats the question tho

[granite torrent]

the question’s been posted twice what do you mean?

[granite torrent]

from here

[floral frost]

kk

got it

[granite torrent]

and restated here

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@fossil dagger Has your question been resolved?

[floral frost]

still doing

i think i got it

i got smth so damn weird

@granite torrent @hallow walrus

check it once

[granite torrent]

check what

[floral frost]

sending

i got it till there

pretty sure i over complicated it tho

[granite torrent]

that’s not solved

[floral frost]

yeah

i tried uptill there

im still doing

figured smth is better than nothing

[fossil dagger]

Ah, okay

[granite torrent]

our point is that this can’t be solved using normal algebraic methods

[floral frost]

maybe

[granite torrent]

what do you mean “maybe” that’s a fact

[floral frost]

probably yeah

but im still gon try it to waste my time

ive got an idiotic answer

sinx - 3 = cos x

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[tardy pine]

?

Is that true

[tardy pine]

W means true

F means false

[granite torrent]

everything in that table looks right

[tardy pine]

?

lol

[granite torrent]

why do people delete pings... at least edit message to say smth like "disregard this msg"

[tardy pine]

perfect

ty

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[dull mauve]

Could someone suggest me some research work that has a practical part? It is a very important topic that I have to have chosen tomorrow but at the moment I can't think of anything that interests me.

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[dull mauve]

<@&286206848099549185>

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[hazy sand]

https://tenor.com/view/tesla-albert-einstein-madwolfofōkami-gif-13692030474614117380

[torn jolt]

Do you have a question or ?

[hazy sand]

wtf

[torn jolt]

Close the channel if you don't need any help, or it will be considered as trolling and I'll have to ping the mods.

@hazy sand

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[worthy otter]

I’ve been researching the FFT, and during its derivation I came across the attached equation. How do those exponents add to get to that? Thanks!

[worthy otter]

https://brianmcfee.net/dstbook-site/content/ch08-fft/FFT.html

[stiff musk]

c(2k+1) = c2k + c
so
e^(c(2k+1)) = e^(c2k + c) = e^(c2k) e^c

apply this with c = -2pi j m/N

you're just using the property e^(a+b) = e^a e^b

[worthy otter]

Oh, I think my intuition thought you could combine the C terms, but I see now that you can’t

Thanks!

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[velvet tulip]

If v = 5/(4+s) and s(0)=5. How do I know the distance (s) at t=6?
I was wondering why stating ds = vdt doesn't work????

[velvet tulip]

I noticed you have to calculate dt = 1/v*ds. Seems logical if you look at it, but why is the other way not possible?

[devout valley]

Well seeing that ds/dt is v, how do you mean doesn't work? and how do you mean dt = 1/v*dt?

The latter doesn't seem right, the former does

[velvet tulip]

Well if you intergrate the first one you get s = vt =5t/(4+s). This doesn't give the correct answer.

Wait, there was a typo in the second one. Now fixed.

[devout valley]

Remember that v is not constant, and is a function of t, so you don't get vt from integrating

[velvet tulip]

V isn't a function of t, right??? I see no t in it? So why don't you get s =vt ?

[devout valley]

In any case, you're basically working with $\dv{s}{t} = \frac5{4 + s}$, which you can e.g. separate

[glossy valve]

@devout valley

[devout valley]

It is, assuming you're using v for velocity and s for position/displacement

[velvet tulip]

Yes, but why is that.
Like I know integrals, and use them a lot, but I'm not really sure if I fully understand them.

For example, if you just have the mathematical integral of xydz, it would give xyz right? As it's not depended on z?

Like that's how I know it?

[devout valley]

Well if you're assuming that x and y don't depend on z, that would be right, but if they did, it wouldn't be

[velvet tulip]

But in the original formula, yes v is depended on s. But how do you make up that v is depended on t?

[devout valley]

Here remember that s, the position, is a function of time (notice how they tell you that s(0) = 5, meaning that when t = 0, s = 5), and that v is the derivative of s with respect to t (so is a function of t)

[devout valley]

v depends on s, which depends on t (and as per before, v is the derivative of s with respect to t)

[velvet tulip]

Ahja I see it now.

So basically, within dynamics you always have to pay extra attention to what is depended on oneanother? And in this case with speed, velocity and distance, it always is depended on eachother, so you have to make sure to seperate the terms? Am I correct?

[devout valley]

Well you know how e.g. each of those terms are defined (velocity derivative of position wrt time, speed absolute value of velocity) at least

How do you mean separate?

[velvet tulip]

Yeah what you sent here. You get the function with s together with ds.

Like always group them together.

Think I got it. Cheers for your help.

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[crystal hatch]

I'm lost as to why my answer to this question was marked wrong

[onyx glen]

show exact question, input and error msg.

i think it may be some kind of input error, or maybe they're expecting some format you're not complying with.

[crystal hatch]

The exact question is the indefinite integral at the top of the image... there is no error message, other than my answer at the bottom was marked Incorrect

[onyx glen]

screenshot the whole thing.

we MUST rule this out before the math is called into question.

this is absolute and inviolable.

[crystal hatch]

Unfortunately the window had to be closed; it's a simple question provided for our online proctoring setup verification, and since it's a mock test, we're only told "Correct" or "Incorrect" with no explanation

[onyx glen]

well

[crystal hatch]

Yeah

[onyx glen]

your answer appeared correct

but now it seems it's impossible to diagnose what went wrong

[crystal hatch]

Ever used Knewton Alpha? I hate it.

[onyx glen]

knewton alpha?

what is this

[crystal hatch]

It's the online... "textbook" and "homework" my college requires us to use for all of our Calc II assignments

I'll sometimes use integral-calculator.com to check my work, and it's giving me weird results

So... I understand what it's trying to do, with u substitution... but the problem is that (5x^2 - 5)^2 at the bottom...

If we FOIL that out, it's 25x^4 - 50x^2 + 25

Also, it takes the derivative of (5x^2 - 5), 10x, but never seems to do anything with it...

Is this just a rogue answer?

bump

bump

<@&286206848099549185> Any other ideas?

Afternoon

[cloud ember]

Hello

[crystal hatch]

Hi there

Oh well

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[sweet jungle]

Can anybody help me with this problem?

I found it 1

Math community do not have any helpers at this difficulty I guess lol

I mean its easy tho

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[viscid epoch]

anyone here know hypothesis testing?

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[viscid epoch]

not yet

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[bronze forum]

Someone please help I don’t know how this is wrong I cannot see it

[bronze forum]

<@&286206848099549185> I gotta turn this in by 00:00 EST 😭

I’ve been at it since like 8

[inland moth]

are you sure the first quartile is 10?

[bronze forum]

If the median is 12 then yea(?)

3 on both sides, right?

I don’t know when to do the finding of the true value thing

Im guessing you’re implying I do that?

[inland moth]

the first quartile is the median of all the data before the median yes?

[bronze forum]

Yes

[inland moth]

how many data do you have before the median

[bronze forum]

6

[inland moth]

and since its 6 we find the average of the 3rd and 4th data

[bronze forum]

so when we have even numbers we find the true value ??

[inland moth]

idk what you mean by true value

[bronze forum]

Find average of 3rd n 4th

Add them then divide

[inland moth]

yes

so whats your 3rd and 4th data

[bronze forum]

8

Oh wait

[inland moth]

thats your q1

[bronze forum]

6 and 10

[inland moth]

yes

6 + 10 is 16

take their averages

8

[bronze forum]

Ah okay

So everything else is good?

[inland moth]

yeah

[bronze forum]

Okay 🙏🏽

Im gonna try this last one by myself

I’ll ping again if I get it wrong

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[winged hamlet]

Not even sure where to start, and can't find any answers online. We've never dealt with problems with multple variables like this.

[winged hamlet]

the answer is not 2 by the way, that is unrelated

[ancient sandal]

Ima read over this question one sec.

[winged hamlet]

thank you so much

[ancient sandal]

On the denominator is it saying a - ax

or d - ax

[winged hamlet]

d-ax.

[ancient sandal]

okay.

I reasoned that it should be 1/4

[winged hamlet]

that is the correct answer

[ancient sandal]

when dealing with limits at infinity we are dealing with horizontal asumptotes

Here is my work.

[winged hamlet]

let me look it over. thank you so much

[ancient sandal]

Np.

I'll break it down to you, so when dealing with horizontal asymptotes at infinity there are 3 cases we must consider

ax^n/bx^c, if c > n then y = 0

If ax^n/bx^c, if c = n then y = a/b

ax^n/bx^c, if n > c then there is no horizontal asymptote

Now when you observe what I did where I negative bx +c as well as + d, the reason we can do that is because at infinity their impact will be so minuet that we can negate it. It has virtually no impact.

[winged hamlet]

yes. that makes sense

I'm having trouble understanding where you go after getting sqrt(a)/a=2

[ancient sandal]

Here I'll show yah one sec.

[winged hamlet]

thank you so so much

[ancient sandal]

Since x is approaching negative infinity we have to keep that in mind, thus when we break the absolute value we have to multiply the x by a negative.

Does it make sense now?

[winged hamlet]

i'm sorry not really

so

[ancient sandal]

what confuses you?

[winged hamlet]

root a - 2a = 0

is where i am at

[ancient sandal]

sqrt(a) - 2a = 0, we can algebraically factor that.

are you confused as to how I factored it, how I obtained the root?

[winged hamlet]

nooo okay

so once you solve for a then it gives you the 1/4?

what about the other favtor?

0, right?

[ancient sandal]

Your confused on the zero product property ight ima explain it to yah rq.

I see.

[winged hamlet]

i'm sorry.

[ancient sandal]

Why are you sorry? Don't be sorry lol. Ur here to learn.

[winged hamlet]

what is m and n?

ohh yeah i see

[ancient sandal]

n is a real number and m just represents a variable.

[winged hamlet]

how does that relate though?

[ancient sandal]

Well it seems you were confused as to how we got the two solutions, so I was clarifying that.

Or r u confused on why we negate the a = 0

[winged hamlet]

no, i'm just confused why the only solution is 1/4, instead of both 1/4 and 0.

[ancient sandal]

because if a was 0 then ax^2 would be 0 and a would be -ax would be 0 so we would have lim x --> -infinity sqrt(bx + c)/d.

ima show

why that wouldn't work

[winged hamlet]

thank you

[ancient sandal]

lol.

it wouldn't equal two if we plugged it back in.

plus we would have a negative square root which is impossible lol, this is completely meaningless.

[winged hamlet]

thank you 🙂

[ancient sandal]

Yeah.

[winged hamlet]

that makes perfect sense

thank you so much!

[ancient sandal]

No worries. Any other questions u wanna clarify?

[winged hamlet]

nope 🙂

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[ancient sandal]

ight gl to u. pce.

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[thick hedge]

I was wondering if someone could explain the logic behind bournoulli's differential eqn?

[thick hedge]

The method to be precise

[gritty rose]

https://tutorial.math.lamar.edu/classes/de/bernoulli.aspx

[thick hedge]

ah, thanks!

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[earnest arch]

Can someone check if what I am doing is right?

[plush egret]

this seems like a weird way to do this problem

why not just use a standard geometric series?

[earnest arch]

Standard?

[plush egret]

yea

$\sum ar^n$

[glossy valve]

jan Niku

[plush egret]

something like this

wait is this gonna work

[earnest arch]

I

[plush egret]

yea i think this makes sense

[earnest arch]

I have never seen that symbol in my life 💀

[plush egret]

oh, sorry

it just means sum

[earnest arch]

OHHH

[plush egret]

a geometric series is like-

$ar^0 + ar^1 + \dots + ar^n$

[glossy valve]

jan Niku

[plush egret]

something like this

[earnest arch]

Mhm

[plush egret]

you just need to adjust the numbers to fit the problem

[earnest arch]

I see

[plush egret]

these are nice because theyre very standard and its just known that $$S_n = \sum _{k=0}^n ar^k = \frac{a(1-r^{n+1})}{1-r}$$

[glossy valve]

jan Niku

[plush egret]

am i going completely out of the material youve seen in class

[earnest arch]

No, it's okay

I have done this

[plush egret]

ah okay

so can you tell what a should be?

[earnest arch]

2400

This is the formula I used in the uhh pic I sent you

[plush egret]

oh

i see

well its almost right

but i really should have recognized that, sorry

i think you have an $n$ in the bottom, where you should have an $r$?

[glossy valve]

jan Niku

[plush egret]

so the bottom of the fraction should be 1 - 0.5

not 14-1

[earnest arch]

Ohhh

OH YEAH

😭 I am so lost, what am I doing...

Tysm, tysmmm

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[delicate ore]

Hi, stuck here not sure what I did wrong to get - n/3 instead of + n/6

[delicate ore]

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

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

for number 6

im

fucking braindead

and also 16 hrs into my day

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[violet verge]

unsure on where to start with this one

i know what associative and commutative mean but without knowing what the star operation does im stuck

[idle kelp]

You don't need to know what the operation does. Just start on the left handed side and apply those two properties to "move symbols around".

[onyx glen]

that's not a star, that's a diamond.

but also yeah the point is you don't need to know its inner workings

you just need to push symbols around and explain in detail how each step happens

[full forum]

@violet verge Has your question been resolved?

[violet verge]

not sure what you mean by push symbols around

[devout valley]

Do you remember the definitions of commutativity and associativity? Basically applying those

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[next basalt]

Find the inverse to: $y = x^2+4x+5$

[glossy valve]

Merineth

[next basalt]

How do i find the inverse when i have more than one x?

Because if i had only one x i'd just solve for it but

[spice orchid]

Are you sure that's the only information you were given

[next basalt]

oh sorry, no

where x >= -2

!occupied

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[flint crane]

Complete square?

[next basalt]

wdym complete square?

[flint crane]

Like

turn x^2+4x+5 into (x+2)^2 + 1

so y=(x+2)^2+1

[next basalt]

No way wrong

[flint crane]

?

That's not the final answer

[next basalt]

oh

[flint crane]

You still have to find the inverse by representing x in terms of y

[next basalt]

ooh so

$x = \sqrt{-1} -2$

[glossy valve]

Merineth

[flint crane]

What

typo?

[flint crane]

that'll be i-2 lol

[next basalt]

$y = x^2+4x+5 \
y = (x+2)^2 +1 \
y - 1 = (x+2)^2 \
\sqrt{(y-1)} = x+2 \
x = \sqrt{(y-1)} -2$

[glossy valve]

Merineth

[flint crane]

small mistake in line 3

negative squared can also give that

so you need +-

[next basalt]

Hmm

I'm not sure what you mean

[flint crane]

x^2 = a, x = +-sqrt(a)

[queen crater]

where x >= -2
you sure of the direction?

[queen crater]

Ok so you know x+2 >= 0

So you line 4, when you pick between +sqrt(y-1) and -sqrt(y-1), you can pick the positive one

[next basalt]

ooh i see what you mean

I though it'd be fine to keep it in sqrt form

$y-1 = (x+2)^2$ I take sqrt on both sides

[glossy valve]

Merineth

[lime grove]

Sorry im new to channel

[queen crater]

You can, and the equation would still be valid, but you would be throwing away half of the solution space

When you have a = b^2, you always need to consider both a = sqrt(b) and a = -sqrt(b)

If you then know that a >= 0, you can just pick sqrt(b)

If you know that a <= 0, you can pick the other one, -sqrt(b)

[next basalt]

aah okay that makes sense

and since i was told it should be bigger or equal than -2

[queen crater]

If you don't have that information, then you can't pick, you just keep both (can write it as +-sqrt(b))

[next basalt]

Can i even disregard in my case?

considering i have

x >= -2

i disregard the - right?

[queen crater]

Because x+2 >= 0

Yes

It gives you x >= -2 because there is no inverse otherwise

[next basalt]

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[mild iron]

Somehow I cant getting from the first thing to the second thing?

[hallow walrus]

!xy

[full forum]

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.

[hallow walrus]

what's the other side of the equality?

[mild iron]

0

The origional problem

[hallow walrus]

okay

[hallow walrus]

but what's the other side of this inequality?

....= sum....

it's cut off

[mild iron]

Sorry this

[hallow walrus]

oh I see

[hallow walrus]

combine the two summations

ah wait let me think about this

[mild iron]

Ofcourse

[hallow walrus]

well

take one of the terms to the other side for convenience

$(1+x+\frac{x^2}{2}+\frac{x^3}{6}...)\sum_{n=0}^{\infty} a_nx^n=\sum_{n=0}^{\infty} a_{n+2}(n+2)(n+1)x^n$

[glossy valve]

kheerii

[hallow walrus]

how can you create an x^0 (i.e constant term) on the LHS?

[limpid moat]

only by 1 times a0 x^0

[hallow walrus]

correct

and on the RHS?

[limpid moat]

ohh you are a big brain

[mild iron]

Sorry, I dont get it

[hallow walrus]

oh wait you weren't OP lmfao

[hallow walrus]

well, how can you create a constant term on the left?

[limpid moat]

i tricked you

[mild iron]

if x=0?

[hallow walrus]

no.

[mild iron]

n=0

[hallow walrus]

both sides of this equation are polynomials

we aren't inputting any value of x

[hallow walrus]

yes, but which term will it multiply with?

as in, we have 1+x+x^2/2...

[mild iron]

by 1

Only the first term

[hallow walrus]

yes

so what's the term there?

[mild iron]

a0?

[hallow walrus]

right

and on the right hand side?

[mild iron]

2a2

As n=0

[hallow walrus]

yes

so we get a_0=2a_2

does that match the first condition given there?

[mild iron]

It does

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[dark wigeon]

hi

[thick hedge]

[dark wigeon]

can someone helps

[thick hedge]

ask your question if anyone knows the answer, they will help

[dark wigeon]

ok thank

ive stuck in this problem for days

[thick hedge]

I'm not too sure, sorry

[dark wigeon]

of what?

[thick hedge]

of how to solve it I have some ideas but the steps will be very long

[dark wigeon]

ok np

[granite torrent]

there is not really an easy way to do this problem algebraically and completely

i would just plug in integers until finding a solution

[dark wigeon]

thats not professional

i have gave it to my teacher and he started laggin lol

[granite torrent]

i mean the true solution involves solving a quartic

and that's not very pleasant

[granite torrent]

if you want to be slightly more rigorous about it, you can notice that these two equations both describe half of a quadratic, therefore they have 1 real solution at maximum

then all you need to do is find that 1 solution and you are done (and this is best accomplished by assuming that x and y are both integers)

[dark wigeon]

the problem is how to find that solution

[granite torrent]

i mean

[dark wigeon]

@thick hedge where are u?

[granite torrent]

a more reasonable algebraic solution might look like this, but it's really pointless

rewrite x in terms of y in 1 equation, substitute into other equation, continually rearrange and square both sides until you are left with a polynomial
(this polynomial is a quartic)
depress the polynomial to make it easier to work with
apply rational root theorem to find any potential roots

[dark wigeon]

would that even work?

[granite torrent]

why would it not?

i'm telling you that this problem does not have a clean solve solution

it's just "wow math"... for people to put on youtube thumbnails and get views because it looks hard

[dark wigeon]

this is not a thumbnail

[granite torrent]

i was making an analogy/exaggerated statement

[dark wigeon]

aha sorry my english is not so good

[granite torrent]

oh no worries

if you still are looking for a satisfactory answer, you can probably search for this problem online and read through the results

(most solutions have to assume x and y are integers... or at the very least, rational... though)

because if you don't assume x and y are integers or rational, this problem becomes very very unpleasant

[dark wigeon]

so what is your advice to somone that want be good with equations?

[granite torrent]

at a school level or a math competition level?

asking because those two goals have very different pathways to achieve

[dark wigeon]

a math competiton level

[granite torrent]

hm i see

[dark wigeon]

we dont see so much in school level

[granite torrent]

that's reasonable

i'm not sure how good of an answer i can provide so i'll just say what ideas i have

i think the biggest way people start to improve at first is to simply go through a bunch of competitive mathematics problems
these can be accessed in many online archives... most competitions have problem archives, and AoPS has problem lists for AMC/AIME

for every problem someone get wrong, they go through the solution and try to understand two things:

1. how the solution actually works
2. how they could have come up with the solution on my own without any assistance/help

over time, i think doing this helps you see common tricks that are used in problem manipulation, and you slowly improve as you learn more

also, being surrounded by other math-talented individuals usually helps people learn due to motivational factors (if real life doesn't work, then finding a problem-solving discord might help)

there also exist many books that people use to learn (but personally i think some of these are annoying)

overall, i'm not really sure how to describe it as the process is usually really slow for most people, but it might help give you an idea

[dark wigeon]

oh thank you so much for your time

[granite torrent]

welcome

you could ask this question to 100 people and they would all give you a different answer but i think everyone's first experience has involved these factors before

[dark wigeon]

i appreciate your help you are good personne and again thank you

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[desert musk]

its correctly understood that to show that a matrix is surjective (in terms of linear transformations) i either need to show that the co-domain is equal to the rank of the RREF matrix or that the RREF matrix only has leading one's?

[desert musk]

<@&286206848099549185>

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[desert musk]

<@&286206848099549185>

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[dark wigeon]

$someone knows sicence naturell$?

[glossy valve]

brahim3579

[onyx glen]

this is not how you're supposed to use dollars for the bot.

[dark wigeon]

$sorry an=pi/{n+3}^n$

[glossy valve]

brahim3579

[onyx glen]

dollars mark the start and end of mathematical formulas

[dark wigeon]

how

[onyx glen]

and the formulas themselves are written in a markup language called LaTeX

[velvet sedge]

see pinned messages in #latex-help

[onyx glen]

and you can (and should) put them mid-message

like this: $a_n = \frac{\pi}{n} + 3^n$

[glossy valve]

Ann

[dark wigeon]

where can i learn it

[onyx glen]

#latex-help has a cheat sheet, otherwise google "LaTeX typesetting tutorial"

anyway,

is this sequence part of a problem/exercise/question that you're doing rn and want help with?

[dark wigeon]

nah i was just trying to write something

using that bot

like math

[velvet sedge]

go to #latex-help and click here to see pinned messages it has some resources to learn LaTeX. Overleaf has many articles on it, and there are books on it as well if you prefer

[dark wigeon]

aha thanks

[dark wigeon]

when you guys use it i think << damn how ,this guys are hackers>>

[onyx glen]

#bots and #latex-testing

for that

[onyx glen]

i mean i personally have several years of experience with it

but it is not... terribly hard to learn i dont think?

[dark wigeon]

ok thanks

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[balmy coral]

with this how am i supposed to express the solution?

[full forum]

@balmy coral Has your question been resolved?

[balmy coral]

<@&286206848099549185>

[devout valley]

,w rref [[1,0,1,2,1], [1,1,0,1,1],[2,0,1,1,1],[1,0,1,1,1]]

[glossy valve]

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[devout valley]

Interesting they say “multiple”

[devout valley]

Otherwise you could have what you have, or that the solution is (0,1,1,0) (note that your work shows x3 to be 1)

[balmy coral]

yeah i wrote on my paper x3 = 1

how do i express the multiple solutions then

with like t*(0,1,1,0) or osmething

[devout valley]

There’s only one solution to this, you don’t have a line of them from this equation (if I’ve copied it right) as you have no free variables left

[devout valley]

You’d e.g. want the rref to have zero rows (and no “0=1” ones while we’re at it)

Just (0,1,1,0) by itself is the solution you have here

[balmy coral]

yeah i would have wrote (0,1,1,0) just seems odd the exam paper stats "mutiple"

nvm it seems i dont speak my own languag properly

its means all solutions, which i suppose is just the one we've found

[balmy coral]

while we are at it, what do i do if lets say s=1

[minor crater]

you mean simultaneous?

[balmy coral]

no, it just means all i guess

i just didnt think that word meant all but i asked around and it does

[devout valley]

Yea “all” solutions sounds more reasonable I would think, probs what they intended i guess(?)

[minor crater]

nvm i misread it

they are indeed saying all

[devout valley]

Isn’t that basically what you’ve done? Not sure I get the question?

[balmy coral]

sorry

i meant if s = s

like just any varaible

find for every s in the real numbers, how many solutions

in the last row it has two s's in, so at some points it will say like s-2 = s-1 and if s=2 then it has no solutions, right?

so for s not equal to two, i should just do guass and find some solution set

[devout valley]

Yeah basically like that: choose different values of s, like if s=1 you get a solution but otherwise you won’t

[balmy coral]

which means i get a solutions one solutions for all numbers other than 2?

[devout valley]

Actually wait a moment, I’m being slow

But basically it’s kinda like that, choose different values of s, you may get e.g. no solutions or infinitely many etc etc based on where you get to

[balmy coral]

$$[
\begin{align*}
x_1 &= 1 \
x_2 &= \frac{s-1}{s-2} \
x_3 &= \frac{-2s+1}{s-2} \
x_4 &= \frac{s-1}{s-2} \
(1)
\end{align*}
]
$$

[glossy valve]

akiyama
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[balmy coral]

$
[
s \neq 2
]
$

[glossy valve]

akiyama
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[devout valley]

With $s\neq 2$

[glossy valve]

@devout valley

[devout valley]

The bot doesn’t render tex if there’s spaces

[balmy coral]

yes, what do i now conclude

from this solution set

[devout valley]

That would be a unique solution if you have that, as s is fixed in value

[balmy coral]

yeah, so since s = s we have either one solution with s =/ 2 and no solution with s = 2

[devout valley]

Yep

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[devout valley]

Didn’t even notice the capri sun

[limpid moat]

gg

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[split swift]

note that you have a zero row

that indicates that your transformation isn't full rank, so your vectors don't span

also, you can see that the second vector is a multiple of the first, so they can't be LI

so you effectively only have 2 vectors there, which can't span R3

the rank is the dimension of the column space of the matrix

(the space formed by using the columns as basis vectors)

also the augmented part isn't really relevant

you only need the rref of the matrix made from the vectors you're checking

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[wicked frost]

how do u know that the moment caused by f3x is counterclockwise and not clockwise

[hollow zenith]

it doesn't really matter how you see these forces and if its counterclockwise or clockwise you just have to choose one and then stick to that

[wicked frost]

it does tho cause the sign changes

but I don't see how f3x is counterclockwise

[hollow zenith]

is C the point of rotation for f3x?

[wicked frost]

idk tbh

[hollow zenith]

so for every point of rotation (I assume C,D and E), you calculate the negative and positive signs (by convention like this) around your point of rotation

[grim skiff]

Which value are you referring to?

The red one, right?

[wicked frost]

Yeah

[grim skiff]

That's not the moment of F3

[wicked frost]

its the moment of f3x

[grim skiff]

It's CCW because of the line of action

Line of action at point A is here

So the Moment of F3x would go CCW

Like this

@wicked frost

[wicked frost]

idk I would still think it would be cw

[grim skiff]

Why?

It goes CCW in reference of point A

That's where the reference is

[wicked frost]

idk how to explain it, it just looks like it would be cw

[grim skiff]

You can think it all you want, it goes CCW, because of the reference for point A

Do you see the line of reference for Point A, the green line I drew in? Do you see how it would go CCW around that line?

[wicked frost]

I still don't see it

is there like another way of doing it rather than just by looking at it and trying to tell the direction?

[grim skiff]

I just showed you how it's CCW

See the green line in my picture?

[wicked frost]

ye

[grim skiff]

See how F3, is under that line?

[wicked frost]

ye

[grim skiff]

F3x goes to the right, and that line is the reference point

Notice how the F3x would rotate CCW, around that line

If you still don't see it, I suggest asking your teacher, for help, but they are probably going to say something similar as that's how you determine direction

[split swift]

it might help if you draw a line from A over to the base of F3x, and notice that F3x is counterclockwise from that line

[grim skiff]

A similar way is shift that reference point, that you are focused on, down the beam, horizontally, to each force. If you shift point A along the line so it's in line with F3, you can see that F3x is CCW

[split swift]

if a force were pulling directly along the purple line, it would just be pulling away from the pivot and not contribute in either direction, but F3x is above that line; that is, it has some component that is CCW relative to the point it is pulling from

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@wicked frost Has your question been resolved?

[wicked frost]

Yeah this makes sense thanks

Does this method always work for any moment?

[split swift]

yes

[wicked frost]

Tysm

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[lethal moss]

Can someone tell me what did I do wrong? Why do I get a negative answer? I need to find the surface created by rotating the function along the OY axis

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@lethal moss Has your question been resolved?

[wise gate]

You should integrate from t=0 to t=π/2

[lethal moss]

why? isn't the integral before the change from 0 to a?

[wise gate]

But the integral is with respect to t, not x. You are not changing the variable
Also, when finding the area or the volume of something you must always integrate in positive order (that is, if t ∈ [a,b], then t varies from a to b, and not from b to a), otherwise you obtain the same number but negative

[wise gate]

And be careful, x goes from -a to a, not from 0 to a

[lethal moss]

Yeah this makes sense

Thnx

Could you help me with something else? Is another integral but it's kinda different

[wise gate]

Yes

[lethal moss]

I just don't know how to do this.
I need to find the volume of the body formed by rotating y^2=2px around y=-p, limited by x=p/2

Like if it was rotating around y=0 then I would know how to do this but I dont know how to do it around y=-p

[wise gate]

Well, let z = y+p

Then you are just rotating around z=0

And you just applied a translation, so the shape, the area and the volume did not change

[wise gate]

Rewrite y^2=2px in terms of z and do as usual

[lethal moss]

ok thnx

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[plush plaza]

Can someone please explain what happened here
How did the a become 0-1

[buoyant pewter]

on your photo i guess youve started to cope with Laurent series?

if so, then analyse this:

[plush plaza]

Im doing z transform

[buoyant pewter]

ok then )

[plush plaza]

I believe they used the formula of S infinity

But the first term in this case shouldve been aZ

[buoyant pewter]

on yoru phot, there has been sued geomtetric series,

[buoyant pewter]

and its sum

[buoyant pewter]

$\sum_{n=0}^{\infty}z^{n}=\frac{1}{1-z}\text{ }\text{ where }\text{ }\left| z \right|<1$

[plush plaza]

Yes
S = a/(1-r)

[glossy valve]

Joanna Angel

[plush plaza]

Thats true for the first part

[buoyant pewter]

the second series wud look liek this:

$\sum_{n=0}^{\infty}\frac{1}{z^{n}}=\frac{1}{1-\frac{1}{z}}\text{ }\text{ where }\text{ }\left| \frac{1}{z} \right|<1$

[glossy valve]

Joanna Angel

[plush plaza]

The limits were different tho

Heres the complete question

[idle kelp]

You can just set $m = -n$ so that $$\sum_{n=-\infty}^{-1} (a^{-1}z^{-1})^{n} = -1 + \sum_{m=0}^{\infty} (az)^{m}$$

[buoyant pewter]

yes that works like:

$\sum_{n=-\infty}^{-1}z^{n}=\sum_{n=1}^{\infty}z^{-n}=\sum_{n=1}^{\infty}\frac{1}{z^{n}}=\frac{1}{z-1}$

[glossy valve]

Joanna Angel

[plush plaza]

Ohh

Thanks but let me first understand this more

[buoyant pewter]

ofc

[glossy valve]

JessicaK

[plush plaza]

How did you separate out this -1?

and then how did you change the limit?

[wise gate]

$\sum_{n=-\infty}^{-1}\left(a^{-1}z^{-1}\right)^n=\left(a^{-1}z^{-1}\right)^{-1}+\left(a^{-1}z^{-1}\right)^{-2}+\dots$
Then apply the rule $\left(a^{-1}z^{-1}\right)^{-n}=(az)^n$ to obtain $(az)^1+(az)^2+\dots$, which is $\sum_{n=1}^\infty (az)^n$

[glossy valve]

d

[wise gate]

Then, the sum from n=1 to infinity, is the same as the sum from zero, but subtracting the term with n=0, which is (az)^0=1

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[plush plaza]

Oh thankss

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[muted crypt]

is

[glossy valve]

j💫

[muted crypt]

a exponential function

Or does the + 3 make it not one

Asking bc i wanna check my answer

Pls Lmk if i missed one

[tired sedge]

I thought exponential is $a^x$

[glossy valve]

Zemitrix

[muted crypt]

I learned exponential function is always in the form $a \cdot b^x$

[glossy valve]

j💫

[muted crypt]

they're essentially the same thing

[tired sedge]

Not necessary

Yeah

[tired sedge]

Why you didn't choose 5 one

First is indeed correct

[muted crypt]

So the first option is an exponential function?

[muted crypt]

Oh is that one? i got confused bc the exponent was inside the square root

[tired sedge]

Yrah

[tired sedge]

I guess yes lemme check graph

Ig it's exponential

[muted crypt]

Hm Okay

Thanks for the help

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[delicate ore]

can someone help with this

[delicate ore]

im not even sure wha it's asking me to prove

https://i.imgur.com/JKlc6AC.png

for (c)

if x = a then (x - a) = 0 and 0 * any function = 0

and if f(x) is a nth degree polynomial then h(x) is n-1th degree polynomial by definition (and also by what was proved in part (a))

and we don't actually care at all about what's in that h(x)

other than that it's 4th degree

so what do I prove?

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@delicate ore Has your question been resolved?

[tepid plover]

For (c), you just apply (a) to p(x). Then substitute x-a to x to get back to f(x), and use (b) to say that g(x+a) is a polynomial of degree n-1.

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[delicate ore]

ok yeah I see it

thanks

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[signal shell]

This question is related to Calculus 3 and more specifically Triple Integrals. I want to know how I can set up the integral when a sphere is shifted.

The question is:

An object is defined by eqn x^2 + y^2 + (z-1)^2 = 1 in the first octant. How do I find volume? I have tried spherical and cylindrical coordianates and both lead to nighmare integrals.

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@signal shell Has your question been resolved?

[signal shell]

<@&286206848099549185> If anyone wants to do some multivariable calculus...

[hollow cave]

nope

[sand sedge]

have you tried parameterizing the curves of the sphere?

[red blade]

Show ur work

Or im@not helping u

[signal shell]

1sec

[red blade]

I refuse to help you actually if you don’t show ur work

[signal shell]

Sorry it's quite a mess

I started off with cylindrical, moved to spherical and then jumped back and forth..

Please ignore page 92, I was trying to relate it to another problem

[unborn quarry]

Hello! Have you tried first drawing a sketch of the surface? 🤔

[signal shell]

Hi! Yes I tried. I know its a sphere thats shifted up by 1 on the z-axis. I also confirmed it by using an online tool.

Because its shifted up by one, the Phi is not a constant and that is what is stumping me.

[unborn quarry]

Ok I think cylindrical would be the easiest way 🤔, if you want to solve it with triple integrals and with the given surface

[signal shell]

I tried using cylindrical, cutting the sphere into half and making it like a bowl by adding an imaginary plane at z = 1. I was thinking of then doubling it to get total volume.

But it leads to a very weird integral

[unborn quarry]

Wait how are you getting those bounds for z

[unborn quarry]

From this eqn, you get 2 functions for z, roof and floor surfaces. Just put those as the bounds of z

[signal shell]

z > 0 and z < sqrt(1-x^2-y^2) + 1?

[unborn quarry]

No

Remember that when you have
a² = b
then
a = b, a=-b

[signal shell]

Yes

[unborn quarry]

Think of when you want yo find the functions that form a circle x²+y² = 1
y = √(1-x²)
y = -√(1-x²)

It's kind of the same algebra

Find those 2 functions for z, then use them as the bounds for z

[signal shell]

Do you mean this?

[unborn quarry]

z > 1-√(...)

[signal shell]

Sorry about the typo

[unborn quarry]

Ok dw

[signal shell]

I would then sub the x^2 + y^2 with r right?

[unborn quarry]

r²

[unborn quarry]

Ok

[boreal bluff]

dude it says first octant

since you cannot take negative

we dont have to actually go for 2. part

sorry for interrupting by the way

[unborn quarry]

Dw

The surface is completely above the xy plane 🤔

Since it's a sphere with radius 1, z-1 goes from -1 to 1, so z goes from 0 to 2

Is that what you mean (?

[unborn quarry]

Set the triple integral with this bounds for z and the ones you already calculated for r and theta. Then the integral is kind of nice

[signal shell]

You're right about kind of. There is a sqrt(1-x^2) integral and I despise those lol.

But yes, it gives me some semblance of an answer

Thank you so much!

I am getting the answer that was given.

[unborn quarry]

Ok great 👍🏼

[signal shell]

Appreciate it!

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[gilded tapir]

This is a right parallelepiped 80 20 and 5 how do i find the hight

[gilded tapir]

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[wicked frost]

Can’t find my mistake

answer should be 36root2 I got 12

[gritty rose]

plot one line at a time as a function of t and see where it goes wrong

or just use wolfram to simplify

,w simplify (3t^2-3)^2 + (3t^2+3)^2 + 36t^2

[glossy valve]

Results provided by WolframAlpha

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Requested by rie.mann

[wicked frost]

hmm its the same as what I got

then I divided everything by 18 or can I not do that since its in a root?

[gritty rose]

18 is under the square root

[wicked frost]

ah I see now

I should've factored out an 18

then taken the square root of 18

right?

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[solar karma]

I did the question but why is 15 not a solution?

[solar karma]

sin(x/3) = 1/roo2

you get x/3 = 45

x=15?

but if you sub 15 in you dont get 1

[deft tulip]

is this a case of mixing up radians and degrees maybe? the problem is using radians and you seem to be using degrees

[solar karma]

hm

ill try it radians again rq

nah radians doesnt work

I get 0.7853 smthing

which is def not

plus I got 135 which was one of the solutions so

[urban condor]

hello man

sin^2(x/3) = 1/2

[plain fox]

x/3 = 45
x = 45*3
no?

[urban condor]

sin(x/3)= +1/√2(or)-1/√2

so

x=45x3 (or)135x3

[solar karma]

💀

[deft tulip]

right you were multiplying both sides by 3 and you divided 45 by 3 instead of multiplying

[solar karma]

nvm I kept dividing 45 by 3 😭

[urban condor]

listen man you get even negative

don't forget that

[solar karma]

yea i kept that in mind

alr I got it ty

[urban condor]

bye

[deft tulip]

so 3pi/4 and -pi/4 should be solutions in radians

[urban condor]

nah

x/3 is there

you have to solve for x

not x/3

[solar karma]

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[celest hearth]

teacher's key

[thin quartz]

yeah there's a way

[little mica]

Replace x to (x-1)/2 both sides

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[cunning bane]

How do I do this?

[plain fox]

do u know how to find f(3+h) and f(3)?

[cunning bane]

F(3) is just 3^3 right

[plain fox]

yea

[cunning bane]

And f(3+h) = (3+h)^3?

[plain fox]

yeah and now use those values in the expression

[cunning bane]

Oh so u get like

27h+9h^2 + h^3 / h

And then u can factor that into

27 + 9h + h^2 right

[plain fox]

(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

[cunning bane]

Yeah but then we also have a^3 -a^3 right

Cause -f(3)

[plain fox]

oh u simplified it

[plain fox]

yeah and thats simplified

[cunning bane]

Alright thanks

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[empty trail]

find the largest rectangle that can be inscribed under the curve y = e^-x2 in the first and second quadrants

[empty trail]

???

ive got no clue how to do this

[hot knoll]

Imagine the height of the rectangle is the y coordinate and the width of the rectangle is the x coordinate

[empty trail]

umm

[hot knoll]

These types of problems are done usually with symmetric curves

So you can make a formula for half of the rectangle by doing

x*y

Then you can do 2*(xy)

[empty trail]

wait so

2 (e-x2)?

e^-x2

[hot knoll]

x * y

y is e^-x^2

[empty trail]

ohh

so

x & y

x(e^-x2)

[hot knoll]

Yea cause x is the width