help-17

[long flame]

,w cos120 = -cos60

[twin meteor]

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[long flame]

😄

okay then it makes sense my calculations were wrong

thanks

.close

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[marsh bluff]

how is expansion of binomials related to combination?

[round plover]

if you look at the binomial (a+b)^n

when you're expanding it, you're basically picking between a or b, n times

if you pick a, k times, you have b n-k times, so you're on the a^kb^{n-k} term

[marsh bluff]

what do you mean by picking

[round plover]

but there are n choose k ways to pick k as from the n total brackets

(a+b)^n = (a+b)(a+b)...(a+b)

[marsh bluff]

ye

[round plover]

you can multiply this out by picking either a or b from each bracket

[civic otter]

Choosing, selecting, taking

[marsh bluff]

i dont understand

[round plover]

if you pick a every time, you get a * a * a * ... * a = a^n

[marsh bluff]

why are picking it out

[round plover]

look at the n=2 case

(a+b)(a+b)

to multiply it out, we do a*a + a*b + b*a + b*b

[marsh bluff]

yes

[round plover]

the a*a term is the same as picking a from both brackets

we have to multiply every term in every bracket by the terms in every other bracket

where do we get a*a + a*b + b*a + b*b from

[marsh bluff]

from a(a+b)+b(a+b)

[round plover]

are you familiar with FOIL

you don't need to expand the brackets like that

[marsh bluff]

ye

[round plover]

you multiply the first terms of each (same as "picking" the first term of each bracket, then multiplying them), then multiply the first term of the first bracket and last term of the second bracket, etc.

each multiplication requires picking a term from each bracket

in the longer set of brackets, (a+b)(a+b)(a+b)...(a+b), it's the same thing

you need to multiply a term from each bracket, which is the same as picking a or b for each bracket

[marsh bluff]

ok

[round plover]

so a term in the full expansion comes from a choice of a or b from each bracket

if you pick k a's, there must be n-k b brackets remaining

so you have an a^kb^{n-k} term

but there are n choose k ways to pick the k a's from the n brackets

[marsh bluff]

ok this is confusing

[round plover]

so this term has an additional coefficient of nCk

[marsh bluff]

im confused on this part

what is the a term?

the a^2 term?

[round plover]

a isn't referring to the variable

it's just a term as in, some term

[marsh bluff]

ok

[round plover]

I'll swap variables

(x+y)(x+y)...(x+y)

[marsh bluff]

ok

[round plover]

a term in the expansion of this comes from picking some x or y

like, with (x+y)(x+y), we can pick x from both

to get x^2

or x from first, y from second to get xy

etc.

so each term, like x^2, xy, y^2, comes from these choices

in the general case (x+y)(x+y)...(x+y) with n brackets

we can pick x from k of these brackets to get x^k

but if we didn't pick x, that means we picked y for that bracket

[marsh bluff]

i think u mean a term in the expansion of this comes from picking some an x with some other x or y

why is it x or y

[round plover]

I mean x or y from each bracket

[marsh bluff]

and it x and some other term

[round plover]

because each bracket is (x+y)

we can pick one from each bracket

[marsh bluff]

ok

so all terms in the expansion comes from two terms in the brackets

[round plover]

it'll come from one choice of x or y from each of the n brackets

we can look at the cubic case: (x+y)(x+y)(x+y)

if we pick x from all 3, we get x^3

if we pick x from the first, y from the second and third, we get xy^2

right?

[marsh bluff]

ye

we pick 1 term from each bracket

[round plover]

so with n brackets, (x+y)(x+y)(x+y)...(x+y)

if we pick k x's from among the n brackets

that means there are n-k brackets left over from which we pick y

so we have a x^k y^(n-k) term

[marsh bluff]

i cant understand the k x's part

what does k x's mean

[round plover]

we pick x from k of the brackets

where k is just some number less than n

in the cubic case, we had k = 3, which is picking x from all 3

and k = 1, which is picking x from 1 of the brackets

it just means how many x's you have

[marsh bluff]

so n=k

[round plover]

no

[marsh bluff]

since n is also the number of brackets

[round plover]

k is the number of x we pick, only

we don't have to pick x from every bracket

[marsh bluff]

ok

so k<=n

[round plover]

in the cubic case, we could pick x from the first, and y from the last two

to get xy^2

this is an example of k=1

[marsh bluff]

ok

[round plover]

if we pick x from exactly k brackets

there are n-k brackets remaining from which we pick y

in the cubic case of n=3 and k=1, we had xy^2; note that 2 = 3-1

[marsh bluff]

im having trouble to understand k brackets

if k is the number of x

[round plover]

we have n brackets total

we're picking x from k of them

[marsh bluff]

ok

[round plover]

in this example

we have n=3 (because we have 3 brackets)

and k=1, because we picked 1 x

[marsh bluff]

k can be 1, 2 or 3

[round plover]

yes

any number less than n

[marsh bluff]

ye

but it can be 3

[round plover]

yes

if k = 3, then we pick x from all 3 brackets

and we get x^3

[marsh bluff]

ok

i get this part

[round plover]

if we don't pick x from a bracket, we have to pick y from that bracket

because we need something from each bracket

[marsh bluff]

ok

thats when k = 0

[round plover]

no

even when we had k = 1

we picked x from one of the brackets

that means we need y from the other 2

[marsh bluff]

yes but

when k = 0 we pick all other y

[round plover]

yes

but I'm saying you must have y in there even if k is not 0

we had y in there when k = 1

the only time you don't have any y is when you pick x from every bracket, so k=n

[marsh bluff]

ok

[round plover]

the number of x and y must add up to the number of brackets

because you need exactly one x or y from each bracket

so if you have n brackets, and you pick x, k times

you have n-k brackets where you must pick y

because k + (n-k) = n [read this as number of x + number of y = number of brackets]

is this okay?

[marsh bluff]

ok

ye

[round plover]

cool

so this choice of x, k times, and y, n-k times

[marsh bluff]

ye

[round plover]

gives a $x^ky^{n-k}$ term

[twin meteor]

Desync

[marsh bluff]

yes

[round plover]

but, there's more than 1 way to pick x k times from n brackets

there are exactly n choose k ways

wait, I'm assuming you know the choose function

since you asked about combinatorics

is this correct?

[marsh bluff]

ye

im confused

[round plover]

alright cool

[marsh bluff]

theres 1 than 1 way to pick x k times from n brackets

[round plover]

sure, lets drop back to the cubic case

we have n = 3 here: (x+y)(x+y)(x+y)

[marsh bluff]

ok

[round plover]

and say we have k=1

so we want 1 x

well, we could pick the x from the first bracket

or from the second

or from the third

there are 3 choose 1 ways to pick that single x from the 3 brackets

[marsh bluff]

ok

ok

so this is where the coefficient appear

[round plover]

so more generally, there are n choose k ways to pick k x's from the n brackets

because we're choosing k brackets to pick x from out of n total brackets

[marsh bluff]

ye

[round plover]

that's exactly the binomial coefficients

[marsh bluff]

ye but why

why does it represent the coefficient of the terms

if its telling us the number of ways to pick x from the brackets

[round plover]

okay, so in the cubic case

we can pick x from the first bracket to get x * y * y = xy^2

[marsh bluff]

or

[round plover]

or we could pick x from the second to get y * x * y = yxy = xy^2

[marsh bluff]

yyx

[round plover]

yes

[marsh bluff]

3 choose 1

[round plover]

but all of these rearrange to xy^2

so you have (3 choose 1) * xy^2

[marsh bluff]

ye

then u have to sum them up

to simplify

xyy+yxy+yyx=3xyy

[round plover]

so that 3 is the coefficient

and it matches up with the number of ways to pick one x from three brackets

[marsh bluff]

im still confused

how does it represent the coefficient

if its the number of ways

[round plover]

each way of picking

gives us a copy of xy^2

okay?

[marsh bluff]

yes

[round plover]

so the number of ways of picking

gives the coefficient on xy^2 when you sum them upp

[marsh bluff]

ok

[round plover]

because if you have 3 ways of picking, you have 3 copies of xy^2

[marsh bluff]

if u phrase it like that

ye

[round plover]

the "ways of picking"

is given by 3 choose 1

because the choose function is defined to give the number of ways to pick

so if we have (3 choose 1) ways of picking

we have (3 choose 1) copies of xy^2

[marsh bluff]

ye

[round plover]

so the coefficient is (3 choose 1)

[marsh bluff]

interesting

so how is this related to the pascal triangle

[round plover]

pascal's triangle is just a way of writing out (n choose k) for various values of n and k in a systematic way

[marsh bluff]

how does a triangle made from adding the numbers above it gives us the coefficient of binomial expansions

why not give us the coefficient of trinomial expansion

oh u know

is there a trinomial theorem?

[round plover]

there is

[marsh bluff]

like

(a+b+c)^n

[round plover]

there's a general multinomial theorem

[civic otter]

It's a property of the binomial coefficient (which in turn arises from the factorials)

[marsh bluff]

oh

is there a trinomial pyramid

[round plover]

I guess it'd be more of a tetrahedron

[marsh bluff]

oh

whats that

[round plover]

"pyramid" with 3 sides

"pyramid" technically should have 4 sides

[marsh bluff]

oh ok

[round plover]

I don't know how to draw one, and I don't think they're really used in practice

because we have a formula for the coefficients

[marsh bluff]

ok

is there a tetrahedron for the coefficient of trinomial expansion?

[round plover]

yeah, there probably is

but we don't use it

[marsh bluff]

maybe it can tell us some number sequence

[round plover]

you can go explore that if you wish

[marsh bluff]

ok

jk

thanks for rthe help btw

i really appreciate your help

and it really makes me understandthe topic better

[round plover]

np

combinatorics is fun, but a lot of the proofs are really condensed and hard to get into

[marsh bluff]

oh ok

have a nice day

[round plover]

you too, take care

remember to .close the channel if you don't have any more questions

[marsh bluff]

ok

.close

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Channel closed

Closed by @marsh bluff

Use .reopen if this was a mistake.

[marsh bluff]

.open

.open

.open

[round plover]

it's .reopen

[marsh bluff]

oh ok

[civic otter]

.reopen

[vocal sleet]

✅

[round plover]

did you have a follow up?

[vocal sleet]

@marsh bluff Has your question been resolved?

[vocal sleet]

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[elfin moon]

Here given rank of matric is 2

So i tried to 3×3 and determinant 0

How to find T it then

[full hatch]

Is the determinante only 0 for t=0?

[golden night]

find the det in terms of t

[elfin moon]

I got t=0

I got t=3

By other matrix

[full hatch]

Wait are those two different matrizes?

[woeful cloak]

You should use the word submatrix here to fit the context. Saying "matrix" for a submatrix will confuse everyone

[woeful cloak]

anyway, this is not 0 for just t=0, and note that this is not the only submatrix of dimension 3 either

[elfin moon]

That's why I got

T=3

[woeful cloak]

yes, t=3 should be the only values that makes every submatrix of dimension 3 have a determinant of 0

[elfin moon]

How to solve this one

I am not good at row operations

[woeful cloak]

well I don't see any other way than doing row operations. You could find the determinant of E if you are eager to do it by hand

[elfin moon]

Can you show me how to do it by row operations?

I can write one step

[woeful cloak]

There is probably a way of making two of any columns/rows be dependent, but then that would involve solving a system, so that's not really any better

anyway, start out by choosing a pivoting column

[elfin moon]

Multiplying first row with-1 and adding gives many zeroes

I don't know if it's a good idea

[woeful cloak]

well the first thing to notice here is that when t=0, we will have all of the rows/columns be independent, meaning the rank of this matrix is 4 in this case. So, now we want to check for when t!=0. What this means is that you can now multiply/divide a row by t without worrying if it'll change the result

so now, can you find operations that makes all of the elements in the first column (except for the first one) to be 0?

[elfin moon]

I don't know what to do with it

If we minus 4 column with 1st

[woeful cloak]

ok, I need some confirmation first, are we doing row operations or column operations?

[elfin moon]

I was doing row operations but when you are suggesting column operations

[woeful cloak]

oh oops, I meant this to be a pivoting row, excuse me there

technically doing column or row operations is fine, I just need us to agree on one

[elfin moon]

Let's go with row

[woeful cloak]

alright then

anyway, let's continue with the case when t!=0.
As you can see, a row operation R3-R4->R4 will make the first and second element on row 4 to be 0. Continue this with row 2 and 3 makes the first element on row 3 be 0. Now how can we make the first element on row 2 be 0?

[elfin moon]

If we multiply first row 2 with a t then subtraction with first row

[woeful cloak]

sure

ok, tell me what you have after doing those operations

[woeful cloak]

alright looks good

now before moving on, let's see what happens when t=1

what would be the rank of E when t=1?

[elfin moon]

Rank will be <=3

[woeful cloak]

yes, but further note that when t=1, the bottom three rows will be all 0, what this means is that there is only one vector that spans this matrix

meaning the rank of E is 1 when t=1

so now we need to consider the case where t!=0,1

[elfin moon]

Yes rank will be 1

[woeful cloak]

you can see what the pattern should be, just keep working down the rows and consider the cases until you have reached the final row

shouldn't be too bad since our matrix is very nice

[elfin moon]

Can you elaborate more about cases?

What do you want me to do with rows?

If I put t=0

Then rank is 0

Null matrix

[woeful cloak]

again no, like I have said, when t=0, all of the row vectors are independent

this is trivial when you write it out, so in this case the rank is 4. Meaning you don't care about the value t=0 anymore

what this essential does is it allows you to multiply by the scalar t. You usually can't do this when t=0, because then that operation wouldn't make sense

so a similar thing happens for t=1, what this just does is to make sure that t=1 is not a value that we care about, so we can multiply by the scalar (t-1) or (1-t)

our goal is still trying to row reduce the matrix

(note that you only need to do one more row operation and you are good to go)

[elfin moon]

Now idea

Because each operation is disappearing leading zeroes

And this is taking so much time for me to solve just 5 minute thing

.close

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[unborn locust]

Can someone teach me the steps for this?

[boreal path]

@unborn locust

[unborn locust]

Yes

[boreal path]

How do you find the area of a square base 2 and height 3

[unborn locust]

2x3

[boreal path]

Exactly

Base times height

Now hwat is our base here

[unborn locust]

2y+3

[boreal path]

remember multiplication doesn’t matter the order

2x3 equals 3x2 so it doesn’t matter what you choose as the bad

now what’s our heigh

t

if you chose 2y+3 as base

[unborn locust]

y-1

[boreal path]

So we just multiply them

[unborn locust]

Yeah

[boreal path]

Do you know to multiply (2y+3)(y-1)

[unborn locust]

Wdym?

[boreal path]

Like how to expand those brackets

[unborn locust]

aren't I supposed to expand and simplify it

Oh ok

[boreal path]

What did you get

[unborn locust]

2y^2 - 2y + 3y-3

[boreal path]

Collecting like terms what are we left with ?

[unborn locust]

Could you expand on that?

[boreal path]

Collecting like terms is like

x+3x+3 simplifies to 4x+3

The terms ahve the same pronumeral which is x

so we just add them

[unborn locust]

Oh so like 2y2 - 2y + 3y

[boreal path]

That simplifies to what

[unborn locust]

2y^2 - 5y = 3y

[boreal path]

How’d you get that

[boreal path]

go back here

and just simplify like you normally do

i think i confused you

[unborn locust]

2y^2 - 5y - 3?

[boreal path]

Close

-2y+3y isn’t -5y

[unborn locust]

Ohhh right nvm

[boreal path]

Do you underdtand the problem?

[unborn locust]

Yeah

Wait

Is it supposed to be 1y

[boreal path]

1y is equal to y

[unborn locust]

Yeah

[boreal path]

Yes

So we just write it as y

but you are correct

[unborn locust]

How would I put that in the equation?

[boreal path]

Well why do you think there’s an equation

there’s no equal sign anywhere

[unborn locust]

Oh right

[boreal path]

It’s simply 2y^2+y-3

[unborn locust]

Oh ok

[boreal path]

as the expression for the area

[unborn locust]

Ohh

[boreal path]

hope this jelped

[unborn locust]

Yeah thanks

There is nothing else right?

[boreal path]

Nope

[unborn locust]

Is that the final answer?

[boreal path]

ur done

[unborn locust]

Oh ok

It was actually simple, thanks a lot

[boreal path]

You can also solve similar problems

for example if you are asked that for a ttiangle

right angled

we know the area formula for a triangle

so we do the same thing

[unborn locust]

Oh ok, thank you

.close

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[naive sable]

https://youtu.be/Ip3X9LOh2dk?t=504
I don't understand what he meant by this -- "If both b and c are non-zero then that b*c term tells you how much this parallelogram is stretched or squished in the diagonal direction."

[errant warren]

it is more of a rough intuition than an exact description, but as the terms b and c grow very large compared to a and d the parallelogram will sort of tend to be longer and thinner

if you consider how the matrix translates the vectors (0,1) and (1,0) which will end up being the sides of the parallelogram that is perhaps a way to visualise it

[naive sable]

hmm also ig the orientation of the area bc is opposite (to the area ad) so it's subtracted from b*d ?

[errant warren]

yeah that’s a good way to describe it

[naive sable]

Alright thanks

.close

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[remote vector]

Hey, I'm trying to understand a proof which I found online and I'm stuck at an apperently fact, namely that the red and the yellow angles are the same. Unfortunately the proof doesn't discuss it further and I can't figure out why the two angles of the same color are the same

[vocal sleet]

@remote vector Has your question been resolved?

[vocal sleet]

@remote vector Has your question been resolved?

[frozen cave]

cyclic quadrilaterals my frend

BCQP is cyclic

so opposite angles are supplementary

cyclic means it can be inscribed in a cicrcle

and the diagram shows it being inscribed in a circle

@remote vector

so BPQ + BCQ = 180 so BPQ = 180 - BCQ = BCA

so BPQ = BCA

the yellow angles are equal

same for the red

does this answer your question?

https://en.wikipedia.org/wiki/Cyclic_quadrilateral

[vocal sleet]

@remote vector Has your question been resolved?

[remote vector]

Yes, thanks a lot!

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[vast shale]

Tried integrating a semi circle got this weird thign

[vast shale]

$$\int_0^r (2\pi r \cos(x \frac{90}{r}) dx$$

$$\frac{2\pi r^2}{90} + C$$

[twin meteor]

᲼

᲼

[vast shale]

the divison shouldn't be there right?

[blissful sentinel]

why are there degrees

[vast shale]

$$ u = x \frac{90}{r}$$
$$ du = \frac{90}{r} dx$$
$$ dx = \frac{r}{90}du$$
$$\frac{2\pi r^2}{90} \int_0^r cos(u) du$$

[twin meteor]

᲼

[vast shale]

I did $$\frac{90}{r}$$ because that's how you umm

[twin meteor]

᲼

[vast shale]

you

find the radius of a cross section

with cos()

i think

cos(90) = 0 which would be the very end of the semi circle

[blissful sentinel]

Is your cosine in degrees??

If so sine is NOT an antiderivative

By the chain rule, the derivative of sine (in degrees) would be pi/180 cos (in degrees)

[vast shale]

wut?

[blissful sentinel]

The whole reason to use radians is that the derivative of cosine is -sine and the derivative of sine is cosine

If you scale the input differently, you need to use the chain rule to calculate what the result should be

[vast shale]

This was mentioned nowhere before

Ty

[blissful sentinel]

It works the same as any other type of scaling a function

👍

No problem

It can be trippy

[vast shale]

.close

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[strange tiger]

hello could i get help with this question? the proposition hint is included

[strange tiger]

used lim laws to get to here but now i am stumped 💀

[ebon rapids]

Ok, so this is a misguided idea

You can’t use limit laws in the original just yet because your denominator will approach 0

[strange tiger]

ohh

[ebon rapids]

The limits won’t exist if you split it like this

[strange tiger]

my original idea was to say that since f(x) is a polynomial u could simplify the expression in the numerator and then factorize

and pray to god that one of the factors are x-3

[river minnow]

What should the nominator approach so that the overall limit may exist given the denominator approaches 0?

[ebon rapids]

The key is to use the proposition hint

[strange tiger]

both the top and bottom has to be 0 for the limit to exist?

sometimes

[river minnow]

Yeah, so f(x) - x - 4 should approach 0 as x -> 3

[twin meteor]

CST

[ebon rapids]

You should be able to answer the question by now

[strange tiger]

OH

OHHH

THANK U

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[strange tiger]

wow

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[urban obsidian]

how would i do this?

Well intuitively i made the inscribed squares vertexes touch the midpoints of the bigger square

but idk how i could justify that

[rugged orchid]

Have you learned calculus

[urban obsidian]

sure but i have like 1 minute for the problem

surely i can't be lost in the weeds of optimization

[rugged orchid]

That’s different to having to justify it

You just need to find the area

This is obvious yes?

[vocal sleet]

@urban obsidian Has your question been resolved?

[urban obsidian]

yes

[rugged orchid]

Use Pythagoras

If it’s multi choice

You just need to find the answer asap

Justification is not needed

But if you did want to justify I would probably use calculus

[urban obsidian]

oh no

i know how to find the area

of the inscribed square

i mean how do i know

which one minimizes the area?

like what orientation

[urban obsidian]

i did this too and found the area

it was correct but it was just a hunch tbh

[rugged orchid]

You just need a hunch

For MCQ

If you want to justify it you probably need calculus

[urban obsidian]

by hunch you mean

i assumed the orientation was 90 degree

well "sleeping rectangle"

whatever you call it

exactly what you drew tbh

[rugged orchid]

Like this right?

[urban obsidian]

is there anything i can "memorize" though?

[rugged orchid]

Yeah

[urban obsidian]

yeah

i can do calc yeah but is there any facts i can memroize

[rugged orchid]

I mean that’s the obvious hunch

[urban obsidian]

about maximizing shit

[urban obsidian]

it also is the easiest one ig lol

[rugged orchid]

I guess you can try to visualise it

They won’t make it too hard in mcq

That’s not the point of mcq

MCQ pretty much (in my experience) are always the ones where they mostly have tricks you can use

Like there’s a fast and short way

And if you do the short way it gives you more time for other questions

[urban obsidian]

you're right

i just saw some people were spreading "concepts" relevant to this

https://cdn.discordapp.com/attachments/423244559682764800/1138013643137634335/image.png

something like this

ofc i guess i could prove via calculus

or could i

i might be rusty actually so maybe not

but where does this originate from

[rugged orchid]

Well all this can be proven via calculus

As I’m dumb I would do it with calculus

But maybe there’s another way geometrically

I’m too dumb to see one if they exist

[urban obsidian]

welp guess i'll do that as well then

[rugged orchid]

Mainly because I know I can do it with calculus and I won’t run into problems

[urban obsidian]

onc ei prove it once ig it gets easier to remember

[urban obsidian]

i could probably but not sure tbh, i'll try and see

was looking more for geometric proof ig

[rugged orchid]

Although I do remember that the biggest volume for surface area is a sphere

And likewise perimeter and area for circle

Because bubbles are spherical

It’s got the least potential energy or something

Something physics something idk

[urban obsidian]

hmmmmm

welp lemme try

thanks for the lead

i guess you're right

in that the orientation that you chose

[rugged orchid]

There’s also one about hexagons

[urban obsidian]

it's probably the easiest to do in terms of finding the area

[rugged orchid]

They tile the plane with the least perimeter

[urban obsidian]

ooo never heard of that tbh

[rugged orchid]

Something about honeycombs I think

[urban obsidian]

sure i'll try to check it out if necessary

anywa ythanks

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[rugged orchid]

But this is really just random reading and stuff so I don’t expect schools to expect this to be just a known fact unless you specifically covered this

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[craggy imp]

Don’t know what to do

[lilac plaza]

,rccw

[lilac plaza]

I'm not sure but does knowing 2sin(Φ)cos(Φ) = sin(2Φ) help..?

[sudden cloud]

Hi

Use this formula

Products to sum

Nvm I read the q wrong

[craggy imp]

so

what do i do?

[sudden cloud]

Theta = A

And the other = B

So its 1/2(a+b)

[craggy imp]

ahh

i see

[sudden cloud]

👍

[craggy imp]

i will try doing it now'

[sudden cloud]

👍

.close

yoo can I not close it

[craggy imp]

i got it it is D

[craggy imp]

thanks mate

[onyx gulch]

[sudden cloud]

Yup 👍

[craggy imp]

thanks

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[vast shale]

can anyone help me

get another channel

[open spade]

#❓how-to-get-help

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[twin citrus]

and then autor writes this :

so my issue here is that in the first formula x goes to 0 from negative direction as well

and therefore I can not understand how we get second equation from the first one

[flat whale]

do x = -1/n

[twin citrus]

so what next

[flat whale]

well you wanted x from 0^-

so x=-1/n will work for that direction

what does n approach?

writing n -> +- inf doesn't quite make sense

[twin citrus]

so this both are true

but i want to get second formula of e

not this

[peak matrix]

1st limit converges to e => 1st limit from positive direction converges to e => 2nd limit converges to e

Shouldn't this be sufficient actually?

[flat whale]

you already have 2?

[flat whale]

why

[peak matrix]

I think they just didnt understand how was the second formula derived from the first one

[flat whale]

oh

[twin citrus]

these two are not the same

[flat whale]

is that right? you don't understand how to get eqn 9?

right. so what's your question

[twin citrus]

i want to get secod eq from the first

[peak matrix]

The fact that first limit converges from both negative and positive direction is more than sufficient to conclude that the second limit converges

To make that conclusion, you need just the positive direction

[twin citrus]

aa that is good explanation

i understand now

tnx

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[dim silo]

Why am i not getting y as 3.16

[mild flower]

you didn't do the power rule right

,tex .log rules

[twin meteor]

hayley!

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@dim silo Has your question been resolved?

[dim silo]

what should i have put?

[mild flower]

$2\log y$ is equal to $\log(y^2)$

[twin meteor]

hayley!

[dim silo]

oh ok

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[lilac plaza]

,rccw

[knotty snow]

How do we approach this

I don't understand

and I have a quiz today

pls help

[lilac plaza]

I hope this helps

I'm a bit rusty on this topic

[terse forum]

your handwriting looks neat

[knotty snow]

this does, but why did you switch the function of the 3rd part to 3x/2 + 2a

[lilac plaza]

Thanks 🙂

[lilac plaza]

I don't think I switched it?

[obtuse sierra]

for the function to be continuous on all real x, it should also be continuous on those points where it is changing its definition.

[twin meteor]

Dyssrupt

[knotty snow]

yeah

but then how do u solve for a and b? do you set it equal to 4 as mentioned in the 3rd piece

[obtuse sierra]

what is the definition of function at x = -1?

[knotty snow]

ax-b if x <_ -1

[obtuse sierra]

and for x>-1?

[knotty snow]

2x^2 + 3ax + b if -1 < x <_ 1

4, if x > 1

[obtuse sierra]

so now just do this

[knotty snow]

but a and b are not defined

[obtuse sierra]

just substitute the value of x

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[naive sable]

$$\Gamma(\pi) = \int_{0}^{\infty} t^{\pi - 1} e^{-t}dt$$ $$Let u = e^{-t}$$ $$\implies du = -e^{-t}dt$$ $$ln(u) = ln(e^{-t}) = -t$$ $$\implies t^{\pi - 1} = [-ln(u)]^{\pi - 1}$$ $$\Gamma(\pi) = -\int_{0}^{\infty} [-ln(u)]^{\pi - 1}du$$

just a sec

[undone pond]

0 and infinity are limits correct? and not 0^infi

[naive sable]

yes

[twin meteor]

Climate

[naive sable]

What should I do now ? Any substitution?

[undone pond]

if I remember the formula correctly
lets assume
f(t) = t^PI
g(t) = e^(-t)

Integration by parts rule:

First as it is, Integration of Second, minus Integration of (Integration of second times derivative of first)

[naive sable]

ahh alright let me try ibp

yeah done

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[bleak crow]

Find all continuous equations $f : \mathbb{R} \to \mathbb{R}$ such as:
$f(\sqrt{2}x)=2f(x)$ and
$f(x+1)=f(x)+2x+1$
For all $x \in \mathbb{R}$

[twin meteor]

BeeReallyYum

[bleak crow]

I figured that f(x)=x^2 is a solution

but it was just a guess and I couldn't prove it

I got 4f(x)=f(2x)

[hybrid flicker]

this will be helpful

[hybrid flicker]

prove that this has to be true for integers first

[bleak crow]

right I think I can prove it easily with induction

but I don't think that's enough

because that's only one solution there could be more

but yeah i'm kinda stuck there

[hybrid flicker]

prove that f(n) = n^2, you're almost there

[distant eagle]

Since f(0) = 0 because of the first equation, the second equation gives you a unique solution f: Z -> R, then use 4f(x) = f(2x) to extend uniquely to the rationals of the form k/2^n and then use continuity

[bleak crow]

and would that be enough?

like that doesn't really account for the other possible solutions

[distant eagle]

So there is at most one solution

[bleak crow]

wait i'm sorry could you explain how that works

[hybrid flicker]

ok so are you aware of the diadic numbers?

it's the base 2 version of decimal numbers basically

[bleak crow]

so a number's binary representation?

[hybrid flicker]

nono

Decimal numbers can be written as (integer).abcd or integer + a/10 + b/100 + ... AS LONG AS the sum is finite

diadic numbers are basically the equivalent with powers of 2 :

it can be written as integer + a/2 + b/4 + c/8 + ... with a finite sum

Or even better, as k/(2^n) as explained above

[bleak crow]

oh alright got it

[distant eagle]

Actually those are not 2-adic numbers but rather dyadic rationals

[bleak crow]

that makes sense

[distant eagle]

2-adic numbers are a different, more complicated thing

[hybrid flicker]

sorry got confused with their french name

we get confused between diadic and diadic very easily

anyways

[hybrid flicker]

first prove this is true for integers

[bleak crow]

yeah so how does that help in this problem

[hybrid flicker]

then prove it is true for all diadic numbers

then, use that the set of diadic numbers is dense in R

[bleak crow]

alright yeah got it

but I guess I can't prove that it's true for all diadic numbers using induction right?

[hybrid flicker]

you can, because all diadic numbers can be written as k/2^n

so your induction will be on n

[bleak crow]

ohh alright that makes sense

but then how would I prove it's also the only solution

[hybrid flicker]

this is the analysis synthesis reasoning we're doing

you're showing that f solution => f(x) = x^2

so there is at most one solution which is the square function

reciprocately, you can show that this possible solution works

and so x^2 is the only solution

[bleak crow]

i'm sorry i'm kinda confused

how can we prove there is only one solution by showing it works

[hybrid flicker]

that's not what we're doing

we're showing "if f is a solution, then it HAS to be this one"

Let S be your set of solution functions. You want to show S = {x-> x^2}

so you show $S\subseteq {x\longmapsto x^2}$

[twin meteor]

rafilou2003

[hybrid flicker]

because you know the other inclusion is true, so you only have to prove this one

[bleak crow]

oh alright yeah that's clear

thank you for your help!

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[glacial meadow]

What do we have to do to the conversion factor if we are dealing with units that measure area?

[glacial meadow]

cube it?

or swuare it

[pale perch]

the factor tends to be (the conversion factor between the units)^2, so say you have 1m^2, thats 10000cm^2 since 1m=100cm

[thin vale]

1m=100cm, so 1m*1m=100cm*100cm => 1m^2=10,000cm^2

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[sweet minnow]

I'm really stumped with this

[sly sierra]

does $\theta^2 \tan \theta$ do anything naughty at $\theta = \pi/6$?

[twin meteor]

Bungo

[sly sierra]

any discontinuity, does it blow up to infinity, etc

[sweet minnow]

Honestly I'm not sure

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[vast shale]

Why is this definite integral not equal to 1 - pi/2 ? I can't come up with the solution wolfram alpha suggests

[paper depot]

show work?

[vast shale]

gimme a sec

[solid sage]

,w integrate x^2/(1+x^2)

[twin meteor]

Results provided by WolframAlpha

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

Requested by yooyooyoyooyooyooyoo

[solid sage]

Lmao yeah thought so, arcsin is the root a^2-x^2

[vast shale]

means?

[solid sage]

You used the wrong inverse trig identit

Derivative of arcsinx is not that

It’s arctan which you’re meant to use

[vast shale]

ahhh you're right haha

thank you!

[solid sage]

Yeah silly mistakes

Alg

[vast shale]

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[dusk jackal]

how do you solve part a and b

something to do with Normal forces and stuff

i get that the weight of the person is 50*9.8

meaning force down

but the other force which is the normal is not equal to WEIGHT on the person cuz the lift is going down?

not sure where to go from there

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@dusk jackal Has your question been resolved?

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@dusk jackal Has your question been resolved?

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[simple mason]

@dusk jackal
I recommend you to watch this video that will help you understand how to solve this kind of questions
https://youtu.be/sVVKpRvuNG0

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[woven gate]

(\phi:\bR[X,Y]\to \bR[t]:X\mapsto t^2,Y\mapsto t^3)
prove that Kernel of (\phi) is ((X^3-Y^2))

[hybrid flicker]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[woven gate]

I how do I show there isn't any other polynomial without X^3-Y^2 as a factor in the kernel

[hybrid flicker]

You want to show Ker(phi) = (X^3-Y^2)

so it's a double inclusion principle

you show Ker(phi) subset of (X^3-Y^2)

[woven gate]

yea that idk how to do

[worthy citrus]

the obvious hint here is a bit revealing so i'll try and be annoyingly vague: you're trying to show that something in the kernel, f, is divisible by (x^3-y^2), assume it's not, then how can you write f? (not actually a proof by contradiction)

[paper depot]

im not sure thats correct

[hybrid flicker]

I had a much more difficult proof so I'll delete

[woven gate]

i dont really see it

[hybrid flicker]

We're dealing with polynomials, so we have division

[woven gate]

do I quotient R[X,Y] by that ideal, then look at all the other elements except the 0 element

[worthy citrus]

another annoying vague hint: 7 = 3*2 + 1

[woven gate]

alr so div algorithm

[worthy citrus]

yeah

[woven gate]

I've not seen it for 2 variable polynomials

how should the degree of r[X,Y] look like?

[worthy citrus]

it wont matter

[hybrid flicker]

well it will if you want to apply division

[worthy citrus]

actually its a good point whether it does general to multivariate polynomials i dont know off the top of my head

[hybrid flicker]

the degree would be deg(X^nY^m) = n+m

[worthy citrus]

since R is a pid then probably does

i was just mirroring what you do in the single variable case but here you might need grobner basis shite

[woven gate]

[hybrid flicker]

nah bruh just deal with it as if deg(X^n Y^m) = n+m

[worthy citrus]

sure i think its fine, you deffo can write f = g*(x^3-y^2) + r(x,y)

the degree of r wont matter as you'll soon see

[worthy citrus]

no im still unsure this guarantees r is a polynomial and not a rational function

[woven gate]

alr so
p=q(X^3-Y^2)+r
where r has degree <3?
how would X^5= X^2(X^3-Y^2 )+X^2Y^2
deg of r is 4

[hybrid flicker]

if r is of degree < 3, then r = aX^2 + bXY + cY^2 + ...

[woven gate]

wot, yea but what went wrong in my example

[worthy citrus]

if this is fine, then you just apply the map, you dont need to worry about the degree of r beforehand

[woven gate]

is there some concrete reason why phi(r(x,y))!=0

[untold surge]

You can just show that
$\mathbb{R}[X,Y]/(X^{3}-Y^{2}) \to \mathbb{R}[t]$ is injective actually

[twin meteor]

Cogwheels of the mind

[untold surge]

Because if (X^3-Y^2) is strictly contained in the kernel, this homomorphism isn’t injective

[woven gate]

I think I should get it now

I'll do the rest, thanks a lot

[untold surge]

x=X+(X^3-Y^2), y=Y+(X^3-Y^2). The image of f(x)+g(x)y, which is f(t^2)+g(t^2)t^3, being zero must imply that f(x)=g(x)=0 otherwise there is some even degree polynomial =odd degree polynomial

[vocal sleet]

@woven gate Has your question been resolved?

[woven gate]

tyty, this was a really kool way

[untold surge]

Np

[woven gate]

thanks yall @worthy citrus @hybrid flicker

I'll close now

.close

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@quaint creek Has your question been resolved?

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[vagrant cypress]

pls help

[vagrant cypress]

function question btw

[karmic imp]

What's the goal?

Factor?

[vagrant cypress]

determine if the relation is a function

then find the domain and range

i did the first one

y= 2x^2 -3

?

[harsh canopy]

yeah

is that a function?

[vagrant cypress]

yep

[harsh canopy]

ok

[vagrant cypress]

question is to determine if the equation is a function or not

[harsh canopy]

so that question has been answered

anything else neeeded?

[vagrant cypress]

now i need to find the range

[harsh canopy]

ok

[vagrant cypress]

i already have the domain

(-inf , inf)?

[harsh canopy]

do you know where the minimum value is?

[harsh canopy]

yeah

[vagrant cypress]

yes

OHHH

I SEE

so

[-3, inf)

[harsh canopy]

yes

[vagrant cypress]

question

do we isolate y on one side

?

[harsh canopy]

with this one yes

[vagrant cypress]

would i do the same with c?

the 2nd equation in the picture

[harsh canopy]

with c it might be more helpful to isolate x

[vagrant cypress]

oh why could u explain

[harsh canopy]

because you have y^2, when you square root you will get +-

so it might be easier to get x = ....

[vagrant cypress]

how about e?

do i add the 5x to 2x

[harsh canopy]

yeah

with e, i would isolate y

[vagrant cypress]

so c

i got

-6x = -3y ^2 -9

then I divide all by -3?

[harsh canopy]

divide it by -6

to get just x =

[vagrant cypress]

so we divide to get the first variable by itself?

[harsh canopy]

yeah

[vagrant cypress]

x = .5y^2 +1.5 ???

is that right

after dividing by -6

[harsh canopy]

yeah

so think of what the equation y = 0.5x^2 + 1.5 would look like

[vagrant cypress]

not a function

[harsh canopy]

this would be a function

[harsh canopy]

the equation x = 0.5y^2 + 1.5 would be this curve on its side

[harsh canopy]

because of the variable switch

yeah like that

[harsh canopy]

are you saying that x = 0.5y^2 + 1.5 isnt a function?

[vagrant cypress]

my teacher said to use the vertical line test

[harsh canopy]

yeah

thats the easy way to spot if it is or not

[vagrant cypress]

and a function is supposed to have only 1 output

[harsh canopy]

true

so it isnt a function

[vagrant cypress]

could u help me with smth else

f(-x) = |x|-3

i got -1?

[harsh canopy]

wdym?

[vagrant cypress]

like the

f(x) but inside the x is a value

and u subsitute the x value in an equation with the x value from f(x)

[harsh canopy]

so what are you trying to figure out?

[vagrant cypress]

5 a

[harsh canopy]

evaluate it at the specified value?

is that what that says

[vagrant cypress]

ya

[harsh canopy]

but f(-x) is a graph

[vagrant cypress]

i’m rlly not sure

we did this in class but with a diff equation

and did nothing with graphs

E

[harsh canopy]

oh it means like that

well then the graph is exactly the same

if you sub -x into that equation of |x| - 3

[vagrant cypress]

would there be a domain and range for this

[harsh canopy]

yes there is always a domain and range

[vagrant cypress]

(-1.3, inf) ??

[harsh canopy]

is there any boundary on what x can be?

[vagrant cypress]

wdym by boundary

[harsh canopy]

as in can x be any number, and if not, is there a boundary (or an interval) on what x is

[vagrant cypress]

x is approximately -1.286

[harsh canopy]

that doesnt have any relevance to the domain and range of this graph though

[vagrant cypress]

is domain

(-inf, inf)?

[harsh canopy]

yeah

[vagrant cypress]

and the range would be um

after doing it

i got

6y = 7x + 9

and dividing that by 6

the range would be 7/6?

cuz y = mx + b

waiy

No