help-19

[lavish jackal]

"undergraduate math"

[stiff hazel]

Ian are you trolling

[mystic saffron]

Yes I think so

[lavish jackal]

have you learned something f(x) and whats this?

[mystic saffron]

Ikr?

[odd edge]

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[lavish jackal]

lmao

[stiff hazel]

what was your answer?

[stiff hazel]

.

sir did you just use ai

a

ok

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[viscid crest]

I cant solve questions 2 n 3

I did more work on two of them but deleted them

This is whats left from my work

Doesn’t make any sense

[little tulip]

well

starting from 2nd

sinx/(cosx - cos^2x) = (cosx+1)/(sinxcosx)

is good, but the question is if it leads to something

I'd stay with:

$$\frac{1+ \cos x}{\sin x \cos x}=\frac{\tan x}{1- \cos x}$$

[clever fjord]

Modus

[little tulip]

now cross multiplication

@viscid crest try to proceed?

ah, it's about proving an identity 🥲

then, e.g. if you have a form (1+cosx)/(sinxcosx) you're almost done

divide by cosx and you'll end with:
1/sinx * (1 + 1/cosx) which is what you're looking for

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[rigid sonnet]

Hi. It needs to be calculated. The correct answer is 1. But I can't get the answer 1, I'm at a dead end in solving it.

[lavish jackal]

do you mind showing your work

[rigid sonnet]

Yes, now, wait a bit

[clever fjord]

faiyrose

[rigid sonnet]

but it turns out the same thing, what's the point?

I get a fraction of 10/6

[clever fjord]

faiyrose

[violet sundial]

2/3?

[rigid sonnet]

5/3 ?

[violet sundial]

no

[rigid sonnet]

what?

[violet sundial]

10 and 6 can simple for 2

[clever fjord]

faiyrose

[rigid sonnet]

Do we get 5/3^-1 ?

[clever fjord]

faiyrose

[rigid sonnet]

does the degree of minus 1 mean that I need to divide 5/3 by -1 ?

I don't understand, do I need to add 5/3 and 0.4?

Omg, I'll just get 3/5

[clever fjord]

faiyrose

[rigid sonnet]

yes, then we will get 1

Thank you so much for your help?

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[lone zinc]

Is the way i proved this valid?

[odd edge]

@lone zinc Has your question been resolved?

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[torpid vapor]

Am i wrong?

When i check online calculator like wolfram it looks like im wrong

[green sage]

you factored wrong at the top

you only took out the 5 from the x, not the 10

you should get (x-2)(x-5) as your factorization

[torpid vapor]

Omg ty

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[torpid vapor]

I dont know how i would approach this

I think i have to use partial fractions but it seems impossible when i look at it

[dawn tiger]

what does your integrla become after the substitution?

[torpid vapor]

This

But then it doesnt make sense bc the degree of the bottom is smaller than the top

[uncut loom]

can sum help me

[torpid vapor]

Bro…

[uncut loom]

??

[torpid vapor]

You’re in my help channel

[uncut loom]

OH

Oops

😭

[torpid vapor]

You need to find a different one

[uncut loom]

mbb

[dawn tiger]

polynomial long division and then partial fractions

[torpid vapor]

I get a remainder of 27

@dawn tiger

[dawn tiger]

lemme double check your substitution, if you let x = u^6 then dx = 6u^5du and your integral becomes 6u^5 du / (u^3 - 3u^2) which simplifies to 6u^3 / (u - 3)

great so you did that right

what is the full result of your long division?

u^3 divided by u - 3

[shy saddle]

do another substitution it gets so easy

[shy saddle]

bro do this it’s way easier

[torpid vapor]

The question requires me to express the integrand as a rational fraction

[shy saddle]

ahh

sorry

[torpid vapor]

I wouldve done it that way if i could

[mystic saffron]

hi

[torpid vapor]

@dawn tiger

My ipad is dead rip

[dawn tiger]

yeah that looks right to me

that means that u^3 / (u - 3) can be expressed as u^2 + 3u + 9 + 27/(u-3)

you don't even need partial fractions

[torpid vapor]

Hmm

Ok so simplified

u^2 + 3u + 9 + 27/(u-3) = u^2 + 3u + 36/(u-3)

And then just integrate this?

u^2 + 3u + 36/(u-3)

@dawn tiger

[dawn tiger]

no

9 + 27/(u-3) is not the same as 36 / (u-3)

but you would integrate 6 (u^2 + 3u + 9 + 27/(u-3)), yes

[torpid vapor]

And i would do that by splitting the integrals into 2

And adding

After

Then sub u for 6th√x ?

[dawn tiger]

yes so separately integrate term by term and then resubstitute for x

[torpid vapor]

Do you mind integrating it for me on the left side?

@dawn tiger

[dawn tiger]

you don't know how to integrate those? it's just power rule and then natural log

[torpid vapor]

9u+27u right

And then i can add them no?

36u

@dawn tiger

[dawn tiger]

the 9 becomes 9u, yes, but 27/(u-3) doesn't integrate to 27u

[torpid vapor]

Wait

Its 27/u-3

Ohhh

I thought it was

[torpid vapor]

(u^2 + 3u + 9 + 27)/(u-3)

Thats why i was getting confused

[dawn tiger]

so you can finish the rest of the problem off?

[torpid vapor]

So to clarify the integrand is

u^2 + 3u + 9 + (27/(u-3))

This

@dawn tiger

Brother

Please find an empty help channel

@cold sparrow

Im using this one

[cold sparrow]

please i am in a little hurrry am not here often

[torpid vapor]

I dont know

Looks like physics

[cold sparrow]

okay

sorry

[cold sparrow]

yeah

[torpid vapor]

Gl

[dawn tiger]

yes, all of that multiplied by 6, look up top where we pulled the 6 out of the integral

[torpid vapor]

This

[dawn tiger]

yes

[torpid vapor]

Ok ez

1 sec

@dawn tiger

Verify?

[dawn tiger]

looks good to me

[torpid vapor]

👍

Thank you

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[mystic saffron]

If the independent term of the following
polynomial:

is 23, determine the value of

[long tinsel]

independent term?

Oh constant

Okay

term independent of x

3 - 5 = -2

not -3

[mystic saffron]

oh yes

so n = 4

A

thanks

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[wise jasper]

how do you create the T symbol

[wise jasper]

in latex

this is a TeXnique problem btw

[ember oak]

$A^{T}$?

[clever fjord]

SWR

[ember oak]

Oh I see

[lavish jackal]

#latex-help

[ember oak]

$A^{\top}$

[clever fjord]

SWR

[wise jasper]

that renders differently

[ember oak]

$A^{\intercal}$

[clever fjord]

SWR

[ember oak]

Ew

Yeah go to #latex-help

[wise jasper]

intercal did it!

thanks!

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[paper cape]

30 children are supposed to line up in 3 rows. How many ways are there to line up?

[paper cape]

Shouldn't that just be 30! ?

[ember oak]

30 factorial?

[paper cape]

Yeah.

[ember oak]

If there were 3 students lining up in 3 rows, would the number of ways be 3 factorial?

[paper cape]

Yeah or am I stupid? 😂

[hardy panther]

(30C10)(20C10)(10!) 🤔

No

That's not right

It would be 30!

[merry finch]

But you need to permute the 10 people too

[paper cape]

My idea was (30C10)(20C10)(10!)(10!)(10!) which ultimately cancels to 30!

[hardy panther]

Every position is unique

[merry finch]

It ends up being 30! I think

[paper cape]

Yeah that's what I thought.

[hardy panther]

Oh!

I wondered this

Good job!

[merry finch]

Cos when you swap each line it should be different

Ah

But no there should be more

[paper cape]

And then I thought "Wouldn't it be the same as if the 30 children just lined up in a single row instead of three" which also leads to 30!

[merry finch]

Because the lines need not be the same size

[hardy panther]

Yeah, you have to arrange WITHIN each selection

But you know

What that also is?

(30P10)(20P10)10!

[merry finch]

It’s more of a stars and bars moment I think

[ember oak]

Oh I'm dumb. Ignore me

[hardy panther]

...pretty sure

[paper cape]

What exactly is the difference between (30C10) and (30P10)?

[hardy panther]

C -> order of selection doesn't matter, just what is in the group
P -> selection order does matter

[merry finch]

For every arrangement of 30 people, how many ways are there to put 2 bars between them such there there are at least 1 person in each “bin” (or line)

[hardy panther]

; ordered vs unordered selection

[paper cape]

Thank you. 🙂

[merry finch]

I doubt an empty line counts as a line

Otherwise putting 30 students in 2 lines would count as putting them in 3 lines

[merry finch]

This is only true when there are exactly 10 students in each line

Or exactly some number of students

Imagine 6 students 123456

[paper cape]

Can we not treat the three rows as one big row and look at the permutation, therefore 30! ? Or is that wrong?

[hardy panther]

Confirmed that 30! Same as (30P10)(20P10)10!

[merry finch]

12 34 56 and 123 4 56 is the same arrangement of 6! But not the same lines

There definitely should be more than 30! ways

[paper cape]

Ah.

[paper cape]

I thought I did something wrong because 30! looked way too much to me. 🥲

[merry finch]

It’s too little

30! Doesn’t care about where each line starts and ends

But where each line starts and ends very much matters

They count as distinct arrangements of the students

[merry finch]

As shown here

[merry finch]

Read from here if you haven’t already

[paper cape]

I did.

I think we can assume that there should be 10 children in each row.

I don't see how else to solve it tbh.

[merry finch]

We can’t

[merry finch]

.

[paper cape]

I meant in my homework. 😂

[merry finch]

Well it doesn’t say there needs to be 10 in each line

It is ridiculous to make such an assumption

[paper cape]

Never heard of that one.

[merry finch]

.

[paper cape]

Yeah just googled it.

But I'm thinking whether we need theorem 1 or 2.

[merry finch]

No idea what that means

[paper cape]

Nvm then.

[merry finch]

“Theorem 1” can refer to almost anything lol

It’s not a name of anything

[paper cape]

Looking at the wiki article "Stars and bars (combinatorics)" rn, that's why I said theorem one and two. ^^

[merry finch]

Oh

[hardy panther]

Oh...shit. yeah, doesn't say how long each line should be

Who said "stars and bars" earlier?

That would be part of it

[paper cape]

@merry finch

[merry finch]

We want theorem 1

Theorem 2 is for weak compositions that allow for 0’s

That’ll translate to allowing 0 people in a line to count as a line

We don’t want that

[paper cape]

Alright.

So we got n = 30 and k = 3 because we got 3 rows, right?

[merry finch]

Yes

[paper cape]

But 29C2 is not enough because the order matters.

[merry finch]

For each arrangement of 30 students in 1 line (30! ways), there are 29 choose 2 ways to cut them into 3 lines with at least 1 student in each line

[paper cape]

Wait.

30! * 29C2?

[merry finch]

Yes

[paper cape]

That is such a big number man. 😭

[merry finch]

Yes it is

[paper cape]

Is that even realistic?

Damn.

[merry finch]

Wdym realistic

30 is a big number

[paper cape]

Idk I can't imagine it lmao.

[merry finch]

Try it with 6

[paper cape]

Yeah you're right.

[merry finch]

With 6 you can see there’s already hundreds

Maybe trying with the smallest number of students 4

Will make more sense

[paper cape]

Let's say there are two students who are in love and want to stand next to each other, how many possibilities do we have then?

Can we just look at them as "one pair"? So that our n becomes 29 instead of 30?

[merry finch]

Btw you’re not actually wrong, 30! isn’t correct, although it does line up with the answer for 3 students

[merry finch]

Yeah that’s how I’d think about that

[hardy panther]

* 3 rows of 10

[merry finch]

Turn them into a box

You put the 2 students in a box and then now you have 28 students and a box

So 29 students

Oh

But you need to multiply by 2 at the end

[paper cape]

But the two can change places with each other, right?

[paper cape]

Yeah right.

[merry finch]

Because AB and BA is the same box but not the same arrangement for the lines

[paper cape]

So we'd get 29! * 28C2 * 2.

[merry finch]

Yes

[paper cape]

Thank you very much Sir, you are a genius. ❤️

[odd edge]

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[floral surge]

2/0,06 * 0,532/5,32 * 0,18/0,9

[floral surge]

Could someone help me with that?

I find 0,19152/0,28718 when multiplying all, which makes no sense. The answer is just 2/3

[agile crag]

I think they do this in Europe, but the , are decimal points, right?

[floral surge]

yes

[agile crag]

Cool

Let me make a step-by-step image of my solution

[floral surge]

i'm a total newbie in math and just dont know how to solve this :(

[static totem]

0.28728

[agile crag]

I'll try and make it simple

[floral surge]

tyvm

[static totem]

you lost a 1 somewhere

[floral surge]

i mean still the answer is just 2/3

i must have been doing it wrong

[static totem]

yes

wait ok

[agile crag]

Do you know how to convert decimals to fractions?

[floral surge]

yes

[agile crag]

Cool

I'll use it in my image

[agile crag]

What about reciprocal?

[static totem]

0.19152/0.28728 is exactly 2/3

[floral surge]

i dont know what it means

i might know but

im not learning math in english

[static totem]

you just gotta do divisions first to see it

[agile crag]

Like 1/(1/a) = a

[floral surge]

it isnt practical to divide 0.19152 to 0.28728

[floral surge]

yes

[agile crag]

Also that's impressive, what language are you natively learning it in?

[agile crag]

Cool, I'll use it

[floral surge]

turkish

thank you guys!

im grateful for all the assistance

[static totem]

1/0.03 ×
1/10 ×
0.2
= 0.2 / 0.3

[floral surge]

wow wait

it looks way simpler

when visualizing like this

all numbers are just multiplers

but i mean

when multiplying fractions

[agile crag]

I decided to not touch the third fraction in case you wanted to try it, but it follows the same style

[floral surge]

ty! let me check it

[agile crag]

I made a little mistake

I copied 0,9 as 0,19

But the rest of it should check out

[floral surge]

why dont we divide 5,32 to 100? it would give us 532/532 = 1 but i mean when multiplying we only operate between numerators and denominators, right?

so i have no idea why we're doing this here

wasnt that only valid for addition and substraction?

[agile crag]

That's valid, and also faster

[floral surge]

the only thing

i didnt understand is

we can do it like 532/532 = 1

but afaik

we're only allowed to

do the multiplication

between numerator and denominators

would it still be valid

if we simplificate idk

[agile crag]

From this step to the next?

[floral surge]

yes, i just dont understand why theres 532/1000 * 10/532

isnt it just 532/532 after the simplification

[agile crag]

Yeah, except we can't remove the 10/1000 immediately

[floral surge]

i was told that i should just slide comma by the amount of zeros

still confused but im trying to understand it

[agile crag]

The sliding the comma works better than what I did

[floral surge]

thank you!

[agile crag]

What I ended up doing was "proving" the slide-the-comma by expand the fractions and then canceling. So both works, but sliding the comma is faster

[floral surge]

i understand

so is this correct?

i was already doing that but

i thought i wasnt allowed to do that in multiplication

but rather only in addition and substraction

[agile crag]

Yup

[floral surge]

ok i'll try with this now

i still get huge numbers with 7-8 digits, feels like there's a trick for that.

[agile crag]

May I see an example?

[floral surge]

i mean

200 * 532 * 18 = 0,19152

[agile crag]

Yeah, there's a trick

It may actually overshadow the slide-the-comma, but it is usually easier to think with whole numbers instead of decimals

What I did to the bottom fraction was divide the numerator (top of the fraction) and denominator (bottom of the fraction) by 2 because 200/2 and 6/2 are whole numbers

And if you do that for the other two fractions, the numbers become small enough so you can multiply them easily

[floral surge]

so simplification

[agile crag]

Yeah

[floral surge]

i think the second one is 1/10

[agile crag]

Yup

[floral surge]

the last one is 1/5

[agile crag]

Yeah

[floral surge]

i get it now

i alrerady do simplification and expanding thingy but

for some reason

i've always thought

im not allowed to do these operations in multiplication

let me try again

thank you very much! @agile crag

i couldnt do that without you

im grateful for that

hope i can learn math extensively

[agile crag]

Yeah

Math is filled with neat little tricks like these, many of them are still probably undiscovered even today

[floral surge]

math is fascinating for me

a man made, flawless science

so now im closing this

goodbye

.close

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[agile crag]

Goodbye

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[cold tusk]

i cannot understand the proof

.close

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[heady plover]

since $f'(x)=e^x^2, f'(c)=e^c^2$, $xe^c^2=xf'(c)$, but then how to continue

[clever fjord]

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

[odd edge]

@heady plover Has your question been resolved?

[heady plover]

<@&286206848099549185> why noboday checked my q

[green elm]

so you want to show that $f(x) = xf'(x\theta(x))$ for some unique $\theta(x)$

[clever fjord]

Bungo

[green elm]

where $f(x) = xe^{c^2}$ and $xf'(x\theta(x)) = xe^{x^2 \theta(x)^2}$, so you will have equality if and only if $c^2 = x^2 \theta(x)^2$, or equivalently $\theta(x) = c/x$

[clever fjord]

Bungo

[green elm]

you have 0 < c < 1, so $0 < \theta(x) < 1$

[clever fjord]

Bungo

[green elm]

just need to check whether $\theta(x)$ is unique

[clever fjord]

Bungo

[heady plover]

so this assume that c=(x\theta(x))?

[heady plover]

wait doesn't it should be 0<c<x

[green elm]

well c is already defined as a function of x, so i'm seeking to find what theta(x) would satisfy that equation

[green elm]

oh yea typo, i meant 0 < c < x

[heady plover]

so we assumed that they're equal in order to find theta(x)?

[green elm]

yea that's what i'm doing, seeing if there is a theta(x) that makes it equal

[heady plover]

Good, i see it now

[green elm]

then the remaining thing is to check that it's unique

to do that it suffices to show that given x, c is unique

[heady plover]

hmmmm

how to show it?

maybe by some kinda theorem

[green elm]

it's not true in general for the mean value theorem, there could be multiple c that work

but in this case it's actually easy to see:

the mean value theorem says as follows:

$\frac{1}{x}\int_0^x e^{t^2},dt = e^{c^2}$ for some $0 < c < x$

[clever fjord]

Bungo

[green elm]

now the right hand side e^(c^2) is monotonically increasing with c (for positive c)

so in particular it's one to one (injective)

so different values of c give different values of the right hand side

so only one value of c can make the right hand side equal the left hand side

[heady plover]

fair enough

Thank you !!

[green elm]

yw, cheers

[heady plover]

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[heady plover]

I really need someone to help me understan this

Idk why they want to show that \bar a1 \bar a2>=a1a2? this seems completely unrelated to GM-AM equality

[mystic saffron]

if you repeat this exact process multiple times, you can prove that the geometric mean is smaller than the arithematic mean

and this could be done by a proof by induction

https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://4dspace.mtts.org.in/expository-article-download.php%3Fai%3D5&ved=2ahUKEwi6sI-Z-ZaHAxX7T2wGHTsMCiEQFnoECCAQAw&usg=AOvVaw1IcEGleiO1-68h_jy2lqFq

found this proof online

[heady plover]

Thank you ! but still confused, what's the point of making \bar a1 and \bar a2?

[mystic saffron]

bar a1 and bar a2 are the arithematic means A_1 and A_2

[heady plover]

\bar a2 means A2?

[mystic saffron]

isn't that so?

I must be confused

[mystic saffron]

oh jeez I misread

mb

[heady plover]

so idk why they assume \bar a1 \bar a2 like this

I don't see any meaning or regularities in such assumption

<@&286206848099549185>

[bright hinge]

ok so consider a case where all of a_1...a_n are the same

clearly, we have A_n = G_n = a_1 = ... = a_n

now what we will do, is we're gonna change a_1 and a_2 slightly while keeping the arithmetic mean the same. Clearly, this means we subtract a little from a_1, and add the remainder to a_2.

now we want to define two more numbers, we have \bar{a}_1 = A_n, \bar{a}_2 = a_1 + a_2 - A_n. We will note that \bar{a}_1 + \bar{a}_2 = a_1 + a_2.

in particular, this means that the arithmetic means of a_1, a_2 and \bar{a}_1, \bar{a}_2 are the same

[heady plover]

subtract a little from a_1, and add the remainder to a_2 ? Btw why do we keep AM the same instead of keeping GM the same?

[bright hinge]

it's easier to show that changing a sequence while keeping the AM the same must decrease the GM

than the other way around

that's really what we're trying to do here; we are trying to show that given a sequence with equal AM and GM, any change to the sequence which keeps the AM the same must make the GM decrease

other than if one of them is like negative but we assume this isn't the case

[heady plover]

So bar a2 means subtract bar a1 from the AM of a1 a2 ?

[bright hinge]

no, it means what they said it means

[heady plover]

I think i really need an example I'm too dumb to understand this just by words

why do we must keep AM the same

[bright hinge]

take a_1 = 2, a_2 = 4. Compute \bar{a}_1 and \bar{a}_2

[heady plover]

\bar{a}_1 = A_n = 3
\bar{a}_2 = a_1 + a_2 - A_n = 2 + 4 - 3 = 3
So, for a_1 = 2 and a_2 = 4, we have \bar{a}_1 = 3 and \bar{a}_2 = 3.

[bright hinge]

The basic course of this proof is that we have a given sequence, and we show that we can increase the Geometric mean until it is equal to the arithmetic mean, while keeping the AM the same

this means that the AM must have been greater than the geometric mean to begin with

does that make sense? It's a bit of a weird and subtle proof

[bright hinge]

to be noted is that we have kept the arithmetic mean mean the same, while the geometric mean has increased. Infact, they are now equal. This means that the geometric mean used to be lower and the arithmetic mean was not, which means that the arithmetic mean was higher than the geometric mean

we can keep doing this, for pairs a_i,a_j where a_i < A_n < a_j. We replace one element with A_n and another with the corresponding term needed to keep the AM the same. We note that all these replacements increase the GM while keeping the AM the same.

after n-1 such replacements, we will have replaced every single element with the arithmetic mean. At this point, A_n = G_n, but up to now A_n was constant and G_n was increasing, meaning that G_n was lower than A_n

the only sequences which we can't do this trick on to keep A_n the same and increase G_n are ones where all the terms are the same, whence A_n = G_n

[odd edge]

@heady plover Has your question been resolved?

[heady plover]

Thank you, Lemme diggest a bit I'm too dumb

[odd edge]

@heady plover Has your question been resolved?

[bright hinge]

It's ok, if you're not used to this sort of proof it is very strange

[heady plover]

Does this proof have a formal name?

[bright hinge]

I don't know of one

[heady plover]

Thank you....I just tried to find out if i can visuallize this proof

why any change to the sequence which keeps the AM the same must make the GM decrease ?

[bright hinge]

it doesn't, this was never claimed

[heady plover]

Can you help me check if i understand it correctly ?

[bright hinge]

all this is correct. If we increased the GM by changing the sequence without changing the AM, and now they're equal, what can we conclude about the AM wrt to GM for the sequence?

[heady plover]

the arithmetic mean is always greater than or equal to the geometric mean

[bright hinge]

well it couldn't have been equal, because the GM just got bigger and now they're equal

[heady plover]

Before the process, GM<AM

After the process, AM=GM

[bright hinge]

yeah

[heady plover]

Thank you very much !

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[mortal trench]

quick question

[mortal trench]

let's say you don't know the taylor series for the sine

how would you derive it given only the geometric definition

[weary pelican]

well you would start by finding its sequential derivatives (squeeze theorem for the first derivative and then it's easy)

[mortal trench]

ah i see

[weary pelican]

and then you would use taylor-lagrange formula with integral remainder

and show that the integral remainder goes to 0 when you add more terms

[mortal trench]

once we have the derivatives we can use the taylor formula. but the big problem is proving that the derivative of the sine is the cosine

[weary pelican]

sin(x)/x goes to 1

and (cos(x)-1)/x goes to 0

those are from the squeeze theorem geometrically

then focusing around a point "a"

(sin(a+h)-sin(a))/h

sin(a+b) formula

it's sin(a)*(cos(h)-1)/h + cos(a) sin(h)/h

boom limit is cos(a)

same thing for (cos(a+h)-cos(a))/h

limit is -sin(a)

[mortal trench]

ok got it thanks

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[glacial timber]

i'm confused, how do i know if those two curves intersect?

[worn sandal]

If there's a solution when you set them equal to each other

For example 2x^2+1 and 4x^2+3 don't intersect because no real value of x makes them equal

[twilit garnet]

That is, if $Ax^2+B=Cx^2+D$ has a solution

[clever fjord]

nameless individual

[glacial timber]

ah so they intersect if both equations are equal? okay thanks

u learn new stuff everyday

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[worn sandal]

Uh not quite but ok

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[wraith shuttle]

Hello guys
Just one question:
If X and Y are negative numbers, which of the following is negative:
a) xy
b) x+y
c)x/y

[slow sandal]

x+y

two negative numbers when multiplied by each other or divided by each other always result in a positive number

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@wraith shuttle Has your question been resolved?

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[wary forge]

q29

[wary forge]

q29

how to approacj

[wary forge]

<@&286206848099549185>

[earnest ether]

What is that

[wary forge]

hyperbola

my phone is on 8% 😰

[wary forge]

its q29 only

[earnest ether]

I ain't a helper

Sorry bud

<@&286206848099549185>

@wary forge needs help

[wary forge]

its okayy

bro is NOT gonna get helped 😰🤓

ping me whoever wants to help

😹😹😹 ima ask my sir

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[glass rose]

<@&286206848099549185> is this step I've circled crucial for solving this? I prefer it without since I already know the answer (I'm relearning Algebra 💀 )

[halcyon hazel]

umm

[nimble blaze]

!15m

[odd edge]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[nimble blaze]

crucial no

[glass rose]

My bad, I didn't know

[halcyon hazel]

x^8+9=x^8 * x^9

[nimble blaze]

if you don't need it, you don't have to write it

[halcyon hazel]

its fine

[halcyon hazel]

I think

[nimble blaze]

it's clear for most people reading it

include it if the teacher would otherwise dock marks

[glass rose]

x^8+x^9=x^17

[glass rose]

Ah alright

[halcyon hazel]

Wait a sec

Is that symbol minus or multiply

[halcyon hazel]

Nope

It's x^8 * x^9=x^17

[halcyon hazel]

Include it

Better than not

You won't lose anything

[glass rose]

Oh wait my bad I messed up the sign

I use x for multi

[halcyon hazel]

If express it with dot, make the dot clear.

(and preferablly bigger)

so the teacher can tell

[glass rose]

Alright, thank you

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[stuck hare]

Hello

[stuck hare]

Anyone here knows how to do double entry or make t account?

[pastel matrix]

can someone plz help

@everyone

[odd edge]

@stuck hare Has your question been resolved?

[cloud steppe]

all the terms have an x, factor it out, it forms a quadratic, use the quadratic formula or the factoring method to determine the real and complex roots

$126x-3x^2-3x^3 = -3x^3-3x^2+126x = 3x(-x^2-x+42)$

Set this equal to 0: $3x(-x^2-x+42) = 0$

Now use the quadratic formula, as recalled to be $ax^2 + bx + c = 0 \implies x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Now that we know what x is equal to, flip the sign to turn it into a factor, assume our x's here are some d and -e, our factors become $(x-d), (x+e)$

[clever fjord]

The Cat Collective

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[little tulip]

recall factored form of a polynomial

[lyric glade]

i had
a (x+2)^2(x-3)(x-1)-4

i know its a reflected 4th degree

so the leading cof will be negative

not sure if i made a mistake somewhere

[little tulip]

why -4?

[lyric glade]

is it -6?

i thought the constant term

is the y intercept

[little tulip]

yes but in expanded form

[lyric glade]

so what do i do with it when its not expanded? leave it out

[little tulip]

it's unnecessary adding (subtracting) a constant term in this case

[lyric glade]

okay so do i solve for a without the -6?

[little tulip]

but the y-intercept might be useful

[lyric glade]

did i get it right factored?

[little tulip]

yes (x+2)^2(x-3)(x-1) is valid

[zinc glacier]

watch your signs

[lyric glade]

is that the final factored form? do i not add the a to solve for the leading coefficient

[zinc glacier]

yes you need the a

[lyric glade]

let me try

[lyric glade]

im getting a negative half?

[zinc glacier]

seems right

[lyric glade]

i assume the negative is write because its reflected

right

[little tulip]

yup -1/2 seems correct

[lyric glade]

so factored its
-1/2(x+2)^2(x-3)(x-1)

thanks guys

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[sullen radish]

Hello! How can I find the rest from dividing a polynomial to another of the form (x-a)^n where n takes values such as 1,2 or even 3

[odd edge]

Please don't occupy multiple help channels.

[sullen radish]

I know how to do it for n=1

But for a higher grade it gets a bit harder

And so, I was wondering if there is a technique

I know Bezout's theorem

And I know how to divide two polynomials, but, let's say we have a trickier one

Let's say x^10 or something

[little tulip]

well, think about derivatives

if for example we have a polynomial P(x)

when dividing by (x-a) we can write down:
P(x) = (x-a)Q(x) + R(x)

where R is just a constant in this case

P(a) = R(a) = R

(remainder theorem)

when dividing by (x-a)^2 we can write down:

P(x) = (x-a)^2 * Q(x) + R(x)

where R(x) is a linear function let's say mx + b

then:
P(x) = (x-a)^2 * Q(x) + mx + b

P(a) = m * a + b
P'(x) = 2 * (x-a)Q(x) + (x-a)^2 * Q'(x) + m
P'(a) = m

and therefore we can obtain:
P(a) = P'(a) * a + b
B = P(a) - P'(a) * a

come back to R(x), finally:
R(x) = mx + b = P'(a) * x + P(a) - P'(a) * a

and same thing can be done when dividing by (x-a)^n for higher n's

[sullen radish]

I see, I see

Indeed, while it is a good method, I don't find it as efficient for very high powers of n, but in some cases maybe you can work it out with induction

Thank you, I didn't know of this method

[little tulip]

well, I don't think it would be useful for higher powers

[sullen radish]

Yes, I was thinking there may be some kind of formula for such types of divizions

But this method is great

Thank you!

Have a good night<3

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[hollow wasp]

I need help with mathlab, I have this problem to solve but I am not familiar with the program

[hollow wasp]

this is the example that I have but I dont know how to use it

[vital cobalt]

is this matlab?

[hollow wasp]

im supposed to use the program to do the exercise

but since I have no experince with it im finding difficult to do the operations

[hollow wasp]

yes

[vital cobalt]

word, matlab is a weird "language" ig. but its arrays are just x=[1 2; ....]

spaces move you down the row, ; for down the columns ig

[hollow wasp]

oh I see is a different syntax

can u explain me what is 'repmat(u,8,1) ?

[vital cobalt]

looks like its just creating an array full of those two mean values, but im not super familiar with it

https://www.mathworks.com/help/matlab/ref/repmat.html

[hollow wasp]

gotcha

another question if u know by any chance

any idea where this 7 comes from?

[vital cobalt]

well its n-1 but idk where that n comes from, maybe number of entries?

if thats attached to the original problem we have 8 vectors so ig that fits

[hollow wasp]

make sense

sorry if I just copy paste like that but the instructor is just bad...

ty for the help tho!

[vital cobalt]

no problem 🙂

[hollow wasp]

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[warm moat]

Find a basis for the subspace H when $H = Sp(\vec{a_1}, \vec{a_2}, \vec{a_3})$, with $\vec{a_1} = (1, -2, 1), \vec{a_2} = (2,-1,3), \vec{a_3} = (-1,-1,-2)$

[clever fjord]

Michael

[warm moat]

The solution is: "For example $\vec{a_1}$ and $\vec{a_2}$"

[clever fjord]

Michael

[warm moat]

But why isn't it all three? Don't you need all three vectors to make the subspace?

If I row reduce the matrix containing the vectors I get the identity matrix (if I did it correctly), so doesn't that mean they're linearly independent -> can form a basis?

[latent scaffold]

You rref'd it wrong

[raw quiver]

for it to be a basis, they all need to lineary independant

or find the lineray indepdant ones

[odd edge]

@warm moat Has your question been resolved?

[warm moat]

ah welp, probably should've checked that ._.

[warm moat]

yeah I know

Well thanks

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[high jay]

https://www.khanacademy.org/math/calculus-2/cs2-series/cs2-lagrange-error-bound/v/error-or-remainder-of-a-taylor-polynomial-approximation

[high jay]

can someone help me better understand the explanation he gives at 6:30?

why do the terms from f''(a) onwards just disappear?

[dawn quest]

P(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)/2! ...
so P(a) = f(a) + f'(a)(a-a) + f''(a) (a-a)/2!... = f(a)
thus, P(a)-f(a) = 0

[high jay]

that I understood

but how does f'(a) = p'(a), f''(a) = p''(a), etc. ?

this is what i cant wrap my head around

its just the product rule actually i think

makes more sense

thanks

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[dawn quest]

oh, wouldn't it just be the power rule?

(f'(a)(x-a)^2)/2! = f'(a)/2! (x-a)^2

for any power of (x-a) really

[high jay]

dk

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[hushed schooner]

What does this function mean?

[hushed schooner]

I've never seen something like this where there is a + in front of the function

it's always been a coefficient

can I just rearrange it and say y = tan theta - 1?

[latent scaffold]

yes

[hushed schooner]

okay so it's the same thing as that then

[broken schooner]

well

[latent scaffold]

It's just adding things

[broken schooner]

isnt it just a tangent graph but everything is shifted down one unit?

[latent scaffold]

Whether it's functions or constants it doesn't matter

[hushed schooner]

so y = -1 + tan theta is the same thing as y = tan theta - 1?

[latent scaffold]

Yes, just like -a + b = b - a

[hushed schooner]

I was confused because I've never seen any problems written this way and thought I missed something lol

[broken schooner]

bro is so advanced that he needs to work on his basics

[hushed schooner]

thanks!

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[quasi monolith]

So I integrated x/(x - 1) using u substitution

u = x - 1

du = dx

u + 1 = x

(u + 1)/u = 1 + 1/u

Integrate that gives us E^(u + ln|u|)

u * E^u = (x - 1) * E^(x - 1)

okay so they ended up just getting e^x instead of e^(x - 1)

im guessing that's because they did E^(u + ln|u| + C) --> u * E^(u + C) --> (x - 1) * E^(x - 1 + C) and they combined the -1 and C?

[mortal trench]

x/(x-1) = 1 + 1/(x-1)

[quasi monolith]

oh they just straight up polynomial divided?

ill try this

[latent scaffold]

That's it yes. The constant term manifests itself as an extra $e^{-1}$ factor in the exponential in your end result. Realistically, the most general answer would be $De^x (x-1)$ where $D > 0$.

[clever fjord]

Azyrashacorki

[quasi monolith]

Good to know. The Polynomial division works perfectly tbh though since I can avoid the extra steps. It was part of a Reduction of Order problem and I was using the formula for it

[latent scaffold]

Yeah it's a better approach to do that. x = x - 1 + 1

[quasi monolith]

ty

.close

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[lusty acorn]

help

[lusty acorn]

someone help me with how to do pascals triangle, like i know how to make the triangle but i dont get the variable part and the powers n stuff

<@&286206848099549185>

[azure igloo]

I think it's a trick to learn the power and trick doesn't have any meaning

[lusty acorn]

wdym''

[onyx turret]

pascals triangle can represent the coefficents of a binomial to the n-1th power

[lusty acorn]

so is it like n-1 each time?

like for example the 4th line

1
11
121
1331

what would 1331 be as the equation what power would it start with

1n^3+3n^2+3n^1+1^0 i know this isnt the correct equation but if it went down by 1 each time it would be something like this right

the other equations i saw which were the full one had like 2 variables together also

[onyx turret]

yes

I'm trying to figure out how the coefficients can be written into a general form

[odd edge]

@lusty acorn Has your question been resolved?

[onyx turret]

I don't think I am doing this right

[lusty acorn]

oh

right now im just reviewing polynomial division and rational roots ill get back to that later

[thin kelp]

3

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

remember that pascals triangle is indexed from 0

0: 1
1: 11
2: 121
3: 1331

[odd edge]

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[lusty acorn]

.reopen

[odd edge]

✅

[lusty acorn]

but how do u know what powers to put for a and b though

[onyx turret]

look at each term

a reduces by 1 degree as b increases by 1

[lusty acorn]

like i know the n-1 but like how do u know when to put a^2b or like ab^2

[onyx turret]

if b is a term then it is ab^2

[lusty acorn]

oh wait i think i got it

[lusty acorn]

so its when the degree for the a value goes down and b goes up and when they are the same value it becomes ab^2

let me try to make one for 1 4 6 4 1 and see if i get it

[onyx turret]

$(c_1x + c_2y)^n$

[clever fjord]

bamidbar_

[thin kelp]

this is wrong

[lusty acorn]

(a+b)^4 = a^4 + 4a^3b + 6ab^2 + 4ab^3 + b^4

is this correct

[onyx turret]

how so?

[thin kelp]

nvm ur right because of the symmetry of pascals triangle, but it is backwards tho

[lusty acorn]

a^4 + 4a^3b + 6ab^2 + 4ab^3 + b^4

is this correct for the 4th degree

[thin kelp]

middle term is a bit ambiguous do u mmean 6a * b^2 or 6(ab)^2

[lusty acorn]

yeah

[thin kelp]

yeah is not an answer?

[lusty acorn]

i put 6ab^2 but idk which one its supposed to be im pretty sure its 6(ab)^2

[thin kelp]

the latter is correcct

[lusty acorn]

which one is it supposed to be

[lusty acorn]

i just put 6ab^2 not sure which one it should be

[thin kelp]

the 2nd

[lusty acorn]

oh bet ima do the 5th row 1 5 10 10 5 1 and if i get that then i think im good

i already know how to do it when plugging in

a^5 + 5a^4(b) + 10a^3(b)^2 + 10a^2(b)^3 + 5(ab)^4 + b^5

this seem right?

[thin kelp]

almost

ur 2nd last term is incorrect

(ab)^4 = a^4 * b^4

[lusty acorn]

but the degree for a decreases by 1 each time though

thats for the 5th power btw

like the one after this one

10a^2(b)^3

would be

5a^1b^4

so would that just make it 5(ab)^4 ?

[onyx turret]

interesting

I don't really understand rational roots

[thin kelp]

no

(ab)^4 = a^4 * b^4

[lusty acorn]

yeah it takes a long time just to do 1 of those questions because u have to plug in so many different values for the gcf of the constant and the coefficient and the factoring

[thin kelp]

u distribute the power to both terms

[lusty acorn]

oh

[thin kelp]

however 5ab^4 is correct

be careful with parenthesis cuz grouping them together implies a^4*b^4

[lusty acorn]

wait actually i think its 5a^4b^1 i just put the perenthesis to put it on its own but i made it look more confusing but i got it now

apperciate the help from u guys

i checked it online and it was that

[odd edge]

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[thorn cloak]

Is it possible to use chebyshev polynomial to approximate cos(x) on 0 <= x < π and get better result than Taylor approximation? From wikipedia:

Polynomial expansions such as the Taylor series expansion are often convenient for theoretical work but less useful for practical applications. Truncated Chebyshev series, however, closely approximate the minimax polynomial.

[thorn cloak]

Better here means "less terms".
In particular, I want log_2 of maximum absolute error value to be less than -16.
For example Taylor approximation requires up to x^9 degrees but with 5 terms (because it skips even terms) to reach this error size.

[odd edge]

@thorn cloak Has your question been resolved?

[faint knot]

wikipedia seems to be indirectly referring to this: https://mathworld.wolfram.com/ChebyshevApproximationFormula.html

[thorn cloak]

Thanks. Is it possible to change the range?

The article is making polynomial on [-1, 1]

[faint knot]

modifying f(x) would work to do that

[thorn cloak]

Right right!

Seems to be solved. Let me close then?

[faint knot]

if its closed to you, you can .close

[thorn cloak]

.close

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[robust cape]

context, this is the paragraph leading to defining the alternating group

[Abstract Algebra by Dummit and Foote 3rd edition, sec 3.5, page 109]

[robust cape]

so what I don't understand is how $(i j) = \lambda (1 2) \lambda$

[clever fjord]

suckywuckyuwu

[robust cape]

like i understand that λ maps 1 -> i and 2 -> j, but if its a permutation, what does it map i and j to

[low locust]

it interchanges 1 and i

so 1->i and i->1

[robust cape]

so then wont λ(1 2)λ just map 1 -> i -> 1

[low locust]

well yes

[robust cape]

basically λ = (1 i)(2 j) right

[robust cape]

then how is λ(1 2)λ = (i j)

[low locust]

well (i j) also sends 1->1

[robust cape]

oh 😭 yes

shit okay yeah

i should be checking for i not 1 lmao

[low locust]

well you need to check all of them

[robust cape]

yeah I mean I was checking for 1 when I thought I was checking for i

😭 basically i was being stupid. tysmm

.close

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[unique fulcrum]

Definite integral of x^2 -2x from -1 to 2

[unique fulcrum]

,w Definite integral of x^2 -2x from -1 to 2

[clever fjord]

Results provided by WolframAlpha

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

Requested by ilovecocomelon_

[unique fulcrum]

How can area be zero???

The answer is 8/3 in my book

[zinc glacier]

you have to be careful with signed area

whenever the function goes below the x axis, the area becomes negative

im assuming the problem was something like "find the area between this function and the x axis on the interval provided"

!original please

[odd edge]

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.

[lone elbow]

,, \int x^2 dx + 2\int xdx = \frac{x^3}{3} + x^2

[clever fjord]

938c2cc0dcc05f2b68c4287040cfcf71

[unique fulcrum]

Here

[lone elbow]

,rotate

[zinc glacier]

what does the text say

[unique fulcrum]

So the positive and negative areas cancel out?

[zinc glacier]

ok yeah, you need to find integral of |f(x)| not integral of f(x)

[unique fulcrum]

I thought areas are considered scalar in integration

[unique fulcrum]

Wdym

[zinc glacier]

scalars can be negative

[unique fulcrum]

But here areas of opposite direction cancel out

That's a vector property

[zinc glacier]

they want the area, not signed area

[unique fulcrum]

So how am I gonna find

The other area

The one you're talking abt

, rotate

[unique fulcrum]

Check my work

[zinc glacier]

you need to find two separate integrals to find integral of |f(x)|

[unique fulcrum]

What if I put the absolute value symbol instead of these squared brackets

That would make a positive 4/3

And my ans this time would be 8/3

Cuz the main problem is that area over the abscissa and under it are cancelling out

So i can just take the modulus

(magnitude)

[zinc glacier]

integral of |f(x)| is not always equal to |integral of f(x)|

this is an example of it

[unique fulcrum]

Whats the example?

[zinc glacier]

why are you getting 0 area

[unique fulcrum]

Cuz of the negative sign

[zinc glacier]

|0| = 0 yeah?

thats |integral of f(x)|

if you consider integral of |f(x)|, i.e. the total area, thats different

[unique fulcrum]

I may not be understanding your way garlic, but I did the question my own way

So thanks

🧄

.close

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[zinc glacier]

ok if you say so

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