help-19

[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.

[acoustic nexus]

Dym DM and EN intersect at K lies on AB

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[zealous vale]

can some1 explain this

[zealous vale]

i mean yea they showed it but like

are u supposed to just guess that n is an exponent for the denominator?

[amber schooner]

bruh you do you get a_n = inf

[zealous vale]

thats what i thought

[amber schooner]

🤔

a_0 = 3

?

[zealous vale]

Yep

[amber schooner]

they literally gave a counterexample

3 ≠ inf

[zealous vale]

wait where was the counter example

[amber schooner]

you claim that the explicit formula for the nth term in the sequence is infinity..

they tell us that for n = 0 it’s 3, not infinity

anyways

how did you get it right the second time?

[zealous vale]

google

id rather not lose points

[amber schooner]

[amber schooner]

ok so is there a particular part of this argument you don’t follow?

this is also just straight up the form for a geometric sequence

[zealous vale]

I mean its saying a_n = an-1 / 8

So i was assuming what it was asking me to do is apply infinity to where n is

but then it said that was a nogo and i was lost

because then its a_0 = 3

so we get 3= a_n-1 / 8

which would be like 24/8

so the nth term was like 25

but then i googled it

and it said nuh uh

so idk how to do it

[amber schooner]

no. they gave a recursive definition for the nth term (one which depends on one or more of the previous terms)

they are asking for a formula strictly in terms of n

[amber schooner]

what?

the formula doesn't apply for n = 0

you can find a_1 using a_0

a_1 = a_0/8 = 3/8

[zealous vale]

wat

wait does recursive definition mean

u use the last term

as the an term

[amber schooner]

i literally defined it

did you see what i said in the parentheses

[zealous vale]

oh

no i didnt

https://tenor.com/view/cat-thousand-yard-stare-thousand-yard-stare-cat-ptsd-cat-soldier-cat-war-gif-9110270742088459827

kthxbye

.close

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[ruby prairie]

Is there a relationship between Narayana numbers and Bernoulli numbers?

[odd edge]

@ruby prairie Has your question been resolved?

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[icy dagger]

.close

.close

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[robust mirage]

how do i find the dx/dt of this problem?
i got this
i sort of remeber of the equation would look like this

[robust mirage]

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[vernal yacht]

Can you show the original question?

[buoyant hollow]

this is the question gng

[odd edge]

Show your work, and if possible, explain where you are stuck.

[buoyant hollow]

ohh

[vernal yacht]

Oops, not this one

[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.

[buoyant hollow]

okay

Prove that PA²+PC²=PB²+PD²

this is what we have to prove

i can prove that if only i probed that angle APC=angle BPD=90⁰

proved*

[buoyant hollow]

which im not able to

[tepid pelican]

not surprising, its false.

[buoyant hollow]

er

[gloomy tapir]

You named vertices wrong

[buoyant hollow]

the diagram isnt even bro

i js drew it js incase

[tepid pelican]

the labelling matters though

A and C are supposed to be opposite

[buoyant hollow]

o

okay ill keep that in mind

[tepid pelican]

Anyway, you're right that if you can prove that those are right angles, then the statemmetn follows

the only issue is that they are clearly not right angles

so you'll have to figure out a different approach

[buoyant hollow]

yeah that is what bothers me

[tepid pelican]

you can try using coordinates

if you can use that

B = (0, 0), C = (x, 0), A = (0, y), D = (x, y)

[buoyant hollow]

i also dont think ABC is a traingle

[buoyant hollow]

erm

[tepid pelican]

it indeed isnt

[buoyant hollow]

we're supposed to use pythagoras theorem

[tepid pelican]

i see, then there is the second approach

[buoyant hollow]

uhh

[gloomy tapir]

Yeah this way it is really easy

[buoyant hollow]

i dont see a right triangle tho

[buoyant hollow]

i have not been taught that

[tepid pelican]

now there are lots of right triangles

[buoyant hollow]

yea

[tepid pelican]

take use of those triangles

[buoyant hollow]

gotta name the constructions?

[tepid pelican]

try expressing PA^2 + PC^2 and PB^2 + PD^2

[tepid pelican]

sure, you can label the points on those sides

[buoyant hollow]

alright so

[tepid pelican]

and make sure to use the right labelling of vertices

[tepid pelican]

here its wrong again

[buoyant hollow]

yeah

got it

so

PA²=AG²+PG²

PB²=PG²+BG²

[tepid pelican]

can u write those without using P?

actually nevermind, its fine

[buoyant hollow]

okay

[tepid pelican]

it shoudl work both ways

[buoyant hollow]

PC²=CH²+PH²

and

PD²=DH²+PH²

there should be an easier way than this right?

[tepid pelican]

this is fine and its easy enough

now you just have to put the pieces together

u're like 1min away

[buoyant hollow]

o

really

damn okay

[tepid pelican]

yeah, try rewriting PA^2 + PC^2 = PB^2 + PD^2 using what u have now

[buoyant hollow]

how

thats not possible is it

[tepid pelican]

u know what PA is, you know what PC is, u know what PB is, you know what PD is

[buoyant hollow]

yeah

[lapis compass]

oh these kind of qs, are you a 10th grader by any chance

[tepid pelican]

so use these equalities and do the substitution

[buoyant hollow]

so

[buoyant hollow]

9th

[lapis compass]

i see

[buoyant hollow]

PA²+PB²=AG²+2GP²+GB²

and

PC²+PD²=CH²+2PH²+DH²

[tepid pelican]

yeah, but u need PA^2 + PC^2 and PB^2 + PD^2

[buoyant hollow]

oh

my bad

uh

PA²+PC²=AG²+GP²+CH²+PH²

PB²+PD²=GP²+GB²+PH²+DH²

[tepid pelican]

yeah, and u wanna prove that those 2 quantities are equal

[buoyant hollow]

yeahh

[tepid pelican]

can u see what it simplifies to if you put them equal?

some stuff will prolly cancel

[buoyant hollow]

will they.?

i dont think so..

[tepid pelican]

you know what, draw the diagram and actually highlight (with e.g. red color) every length present in PA^2 + PC^2

then do the same with PB^2 + PD^2 with a different color

so like highlight AG, GP, CH, PH red and GP, GB, PH, DH blue

[buoyant hollow]

yeah but

i dont think we had to draw perpendiculars..

ill still draw the diagram as u wish but i dont think its what we have to do

[tepid pelican]

its not entirely necessary, but it helps

[buoyant hollow]

o

okay

[tepid pelican]

the alternative would be expressing AP as AG^2 + AH^2

[buoyant hollow]

yeah

theres two ways of expressing all of them

[tepid pelican]

indeed

here is how it should look after the highlighting btw

[buoyant hollow]

o

i was drawing all the points

nv,

nvm

[tepid pelican]

its better to label them

my diagram is just illustrative

[buoyant hollow]

yeah

[tepid pelican]

anyway, can you see why the sum of blue lengths squared will be the same as the sum of red lengths squared?

[buoyant hollow]

no..

well

they do seem equal

visually atleast

well they are

[tepid pelican]

what lengths look equal?

[buoyant hollow]

AG=RH

BH*

[tepid pelican]

indeed

can you say why?

[buoyant hollow]

HC=GD

[buoyant hollow]

bc

AR IS PARALLEL TO GH

and

BH IS parallel to AG

and all the angles are 90 in them

so its a rectangle

[tepid pelican]

yep

[buoyant hollow]

and opp sides in a rect is same

same goes for

GD AND HC

as for

GH

[tepid pelican]

so AG^2 = BH^2, PG^2 = PG^2, PH^2 = PH^2, CH^2 = DG^2and so their sum is equal

[buoyant hollow]

theyre js same

[tepid pelican]

yeah

[buoyant hollow]

GP =HP

gp

[tepid pelican]

and summing equal things will result in equal sums

[buoyant hollow]

and

HP=HP

[tepid pelican]

this one is not necessarily true

oh you meant GP = GP, then its fine

[buoyant hollow]

no gp=gp

mybad

HP=HP

[tepid pelican]

yeah, then its fine

[buoyant hollow]

theyre both common for red nd nlue

but

we're wrighting everything individually

[tepid pelican]

now you can just sum those equalities

[buoyant hollow]

and

this is going yo be a long ass sum 😭

to

[tepid pelican]

yes, but remember what it was

[buoyant hollow]

yeah

got it

thanks 😭

ill closethis

.close

[odd edge]

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[tepid pelican]

we've shown that the right sides here are equal, so the left sides muse be equal too

so PA^2 + PC^2 = PB^2 + PD^2 and that concludes the proof

[buoyant hollow]

yes

thanks

wait wait

we dont require the

horizontal construction right

[odd edge]

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[wide drift]

This is a question and part of its solution

I dont really understand the proof here

I understand why they took the (X\Y) v (Y\X) ^ (Y\Z) v (Z\Y) bit

But I dont get the steps after that

the steps labeled 1 and 2

[willow zealot]

There are two cases here right

In plain English X and Y are related if they both contain 3 or neither contains 3

Do you get that from the definition of the relation?

[wide drift]

I thought it was if one of them doesnt contain 3

[willow zealot]

No

Assume only X has 3 and Y doesn't

Then X - Y will have three

So they will not be related

3 cannot be in the union of their differences

this is obviously gonna happen if neither has three

but they can also both have three and then the differences, X - Y and Y - X will not have three

[wooden python]

remember how i told you to rephrase the stmt of the relation itself "X R Y <=> 3 belongs to both X and Y or neither X nor Y"

thats what is being used in the proof

[wide drift]

OK I think this makes sense

So they're saying for the first bit that its in X and Y therefore it has to be in Y and Z

[willow zealot]

we only assume its in X

but X relates to Y so it follows that is in Y

[wide drift]

But then it has to be in Y?

Yeye

[willow zealot]

and Y relates to Z so on

Yep

[wide drift]

OK I get it

[willow zealot]

and then its the same reasoning in part 2

[wide drift]

Alright makes sense

Thank you!

❤️

.close

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[willow zealot]

no problem!

[odd edge]

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[bitter lion]

hey

[bitter lion]

can anyone help me idk how to draw this

like i get stuck during the ebaring of a from the mast is 160

also theres another queston that idk how to draw its

three towens A B C are positiojned relative to each other as follow:
Town B is 68km from A on a bearing of 225
Town C is on a bearing of 135 from A
Town C is on a bearing of 090 from B

how do i draw theese stuff help

[frank root]

what does your drawing look like at the moment?

[bitter lion]

a triangle

[bitter lion]

but when i want to draw the bearing of a from the mast 160 its like more than 180 cause like a is on the bottom left of the mast

[frank root]

Try put A on the right hand side :)

[bitter lion]

wait

this doesnt seem correct

[frank root]

It's not quite there yet

remember B is on the same side as A

[bitter lion]

oh

[frank root]

try draw your bearings for A and B then draw your points

[bitter lion]

i didnt see that

nb

make sense now

[frank root]

looks right!

[bitter lion]

yes thankyou

[bitter lion]

what about the other question pls

[bitter lion]

this

[frank root]

What does your diagram for that one look like atm?

[bitter lion]

like just a triangle

not right angle

just triangle

[frank root]

It should be a triangle- yes

[bitter lion]

is it like this?

[frank root]

yes exactly

good work

[bitter lion]

how do i find distance bc

[frank root]

you have a right angled triangle

try rotate it so angle A is on the bottom right

[bitter lion]

wait

how do i do that

in paint

[frank root]

oh i havent used paint in ages wait

okay disregard the rotation

[bitter lion]

ok

[frank root]

do you see the right angle in the triangle?

[bitter lion]

yes

[frank root]

try to use trig functions again

[bitter lion]

but i only have 1 side and 1 angle

[frank root]

one second...

see the angle i've drawn?

do u see any way you could find that one?

[bitter lion]

180-135

[frank root]

yeah exactly

[bitter lion]

oh

so angle c is 45

that mean its an isoceles

[frank root]

Yep

[bitter lion]

ohh ok

make sense now

thankyou

[frank root]

no worries

[alpine spindle]

.close

[odd edge]

@bitter lion Has your question been resolved?

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[mystic saffron]

who needs help w math i gotchu

[acoustic nexus]

[nimble blaze]

If you want to help, wait for people to post and claim their own personal channel.
And provide your input there.
For how this server works and guidelines, please read up on #rules and #❓how-to-get-help .

[mystic saffron]

i can go on the available ones and open my own one from there?

[nimble blaze]

no, if you're purely helping others, you shouldn't be opening channels

[mystic saffron]

oh im sorry im new here

[nimble blaze]

people who want help, get a personal channel dedicated to them for the duration

[mystic saffron]

ive joined like 5 mions ago

[acoustic nexus]

You need help or want to help...

[mystic saffron]

i want to help

[acoustic nexus]

wait why don't he have green leaf

[mystic saffron]

idk

bug or smth

[nimble blaze]

so like i mentioned, if you're new read up on the links
#❓how-to-get-help gives guidelines on how people should ask for help
which implies how you should be providing help (e.g. don't give answers / fully worked solutions directly)
and #rules for how to behave in general

if you don't have any other concerns, close the channel with
.close

[vernal yacht]

@mystic saffron

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

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[wide drift]

This is a question and its solution

[wide drift]

I dont understand its solution

How do we know that x => 1?

[wooden python]

x ∈ N

[wide drift]

Ahhhh ok

[wooden python]

and N, presumably, starts at 1

[wide drift]

Yeye

Makes sense

Thanks!

❤️

.close

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[merry birch]

i need to solve

[merry birch]

I have this

[lean ivy]

Bring the similar terms on one side

[lapis compass]

simplify x and constants first

and then square again

its prob a q from quadratic

[merry birch]

So 2x +4 =4rootx-2

[lapis compass]

wait

no

[lean ivy]

No

[lapis compass]

shouldnt it be 2x-4?

[merry birch]

ya

2x-4 = 4rootx-2

so do i square both sides

[lean ivy]

Yes

[merry birch]

actually

should i divide by 4

[lean ivy]

That will give you a half for x's coeffifcient

Try to keep it simple

[merry birch]

ok ill jsut square

[lapis compass]

yeah

[acoustic nexus]

$2x -4 =4\sqrt{x-2}$

[warped glacier]

easier is 2x - 4 = 4 sqrt(x - 2)

[acoustic nexus]

this what you got?

[warped glacier]

now you can square

[clever fjord]

Fionna The Unemployed

[acoustic nexus]

then just t=x-2

[warped glacier]

ooh that's nice

even better, $t = \sqrt{x - 2}$

[clever fjord]

south

[acoustic nexus]

Oh I meant that

[lapis compass]

oh thats smart

[acoustic nexus]

[lapis compass]

true

would save the expansion

[lean ivy]

Then what would you consider x as?

[acoustic nexus]

use this sub

then solve quadratic

[merry birch]

i have

4x^2-16x+16=idk

[lean ivy]

Show your work

[merry birch]

what

[lapis compass]

show the calculations

[merry birch]

i just squared 2x-4

[lapis compass]

the quadratic you obtained is

incorrect

[merry birch]

oh

[lean ivy]

Nah just show what you did

[merry birch]

i did in my head

[lapis compass]

💀

damn

[lean ivy]

Do it on a paper

[merry birch]

is that not right?

it is

right

[lean ivy]

It is

[merry birch]

then i can simplify

x^2-4x+4

[lean ivy]

Yup remember to do that on both sides

[merry birch]

how do i square 4-rootx-2

is it 16-x-2

[lean ivy]

Yep

Its just not 16-

16 would be in multiplication with the whole bracket

[merry birch]

what

[lean ivy]

Its not 16-square root of x-2,right?

[odd edge]

@merry birch Has your question been resolved?

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[valid lintel]

Calculate. Present the result in algebraic form. i did this problem on my own but i'm missing a minus in the result and i have no clue why it's supposed to be there...

[warped glacier]

how about if you write this as $(-3)^6 \cdot (1 + i)^6$?

[clever fjord]

south

[warped glacier]

like, what steps did you do exactly

[valid lintel]

i calculated Z and then used this formula

not forgetting about fi

[warped glacier]

cool, so $(1 + i)^6 = (\sqrt{2})^6 (\cos(6 \cdot \pi/4) + i \sin(6 \cdot \pi/4))$ right?

[clever fjord]

south

[valid lintel]

hope you don't mind my writing 🙏 , similar but yeah

there's 5/4pi

and as for the rest i can belive you can read just fine

[tacit wasp]

This is correct, yes

[valid lintel]

second parentheses is 1 so technically the answer is just (3sqrt{2})^6

but

in the answears there's a minus

[warped glacier]

double check what sin(6pi/4) is again

or sin(15pi/2), same thing

[valid lintel]

says it's -1 but why do i need to know the value of sin what about cos and i

[warped glacier]

ah, didn't you already observe that cos(15pi/2) = 0?

[valid lintel]

oh 😭

i see

okay

so cos is 0 sin is -1 and i just gonna disappear?

[tacit wasp]

No, it doesn't disappear 🤔

[warped glacier]

$(3 \sqrt 2)^6 (0 - 1i)$

[clever fjord]

south

[valid lintel]

yeahh but i can't multiply them because the first one is to the power of 6

what would i need to do next

[warped glacier]

the answer is $-(3 \sqrt2)^6 i$ and it's fine to leave it in that form

[clever fjord]

south

[valid lintel]

if you say so then sure. Thank you guys so much for help ❤️

[warped glacier]

no worries!

(you can check yourself on Wolfram Alpha, Symbolab, or Desmos complex mode)

or if you have a handheld calculator that supports complex numbers, that too

[valid lintel]

sure!

[warped glacier]

if you trust me

[valid lintel]

have a nice day

[warped glacier]

you too! if you're done type .close

[valid lintel]

.close

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[molten vigil]

hello, i need help to calculate an integral, which is the integral from 0 to pi, of e^x*cos(x)

[molten vigil]

idk how to use latex so i cant write it properly

[mystic saffron]

parts

all the way

[molten vigil]

yeah but i mean idk the primitive of either e^xcos(x) or
e^xsin(x)

or all of that

[mystic saffron]

wait

is it

[molten vigil]

sorry the * disapeared

[mystic saffron]

e^(x*cos(x)) or e^x times cos(x)

[molten vigil]

2nd one

[mystic saffron]

you dont need to know that then

u = cos(x) , v= e^x

[molten vigil]

but i'd get something i cant primitivate too

[mystic saffron]

false

[molten vigil]

if i get e^x * sin(x)

[mystic saffron]

yes

and you do by parts again

[molten vigil]

and i'd have e^x * -cos(x)

and idk how to do that too .-.

[stoic cloud]

$\int_0^\pi e^x\cos(x)\di x$

[mystic saffron]

and that's something familiar to you isn't it

[clever fjord]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[molten vigil]

it's - the integral i had in the first place??

[mystic saffron]

yes

[molten vigil]

ig i'll try like that but im not sure of what im about to do lmao

but thanks yeah i didnt see the method

[stoic cloud]

you should let 𝑢 = cos(𝑥) and 𝑑𝑣 = 𝑒ˣ.

do you know the integration by parts formula?

[mystic saffron]

I have already told him that.

[stoic cloud]

Sorry. Didn’t see the message.

[mystic saffron]

No worries:)

[stoic cloud]

you wrote 𝑣 though. I got confused about that a bit.

[mystic saffron]

(e^x)' = e^x so it doesnt matter whether dv = e^x dx or v = e^x, does it?

[stoic cloud]

Oh. Yeah you’re right

[molten vigil]

.close

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[odd edge]

@thin dagger Has your question been resolved?

[thorny spindle]

diagonal

its saying that all the elements lower than a11

can be divided by the diagonal

so we wrote d as in diagonal

[odd edge]

@thin dagger Has your question been resolved?

[lyric marlin]

I think d is supposed to be the a_ij of minimal length from the previous paragraph?

but then how are they getting the 0 there if they need a11 specifically

I would guess they're just using d = a11 but stating it that way for nicer further use

[odd edge]

@thin dagger Has your question been resolved?

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@thin dagger Has your question been resolved?

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[frank tide]

very random, question source is from me

since Lim ( f(x)^2 ) = ( Lim f(x) ) ^2 - as x approaches some value

shouldnt that mean that Lim ( 1/f(x) ) = ( Lim f(x) ) ^-1

[blazing shuttle]

its true given lim f exists and is nonzero

[frank tide]

does the ⚠️ mean im in trouble

did i break a rule

[blazing shuttle]

ur claim is true as long as u include my stipulations

[brittle beacon]

(no, you're fine, I was about to mention what @blazing shuttle just said now )

[frank tide]

oh i was talking about server rules 😭

[frank tide]

hmmm feels a bit like cheating but sure

[brittle beacon]

Nah you're all good, you aren't in trouble

[frank tide]

lovely!

ty people

.close

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[blazing shuttle]

np

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[marsh lichen]

Can I have some help please?

[jagged maple]

where are you stuck

[marsh lichen]

Well everythiing, I don't understand what's going on in the question.

you can ping me btw.

what the?

[jagged maple]

well

if you have some function

${f(x)\over P(x)Q(x)}$ where P and Q are polynomials and you’re trying to split the ratio, you should search for something like ${g(x)\over P(x)} + {r(x)\over Q(x)}$

[clever fjord]

Médicis

[jagged maple]

therefore you need to search some functions $g, r$ such that $P(x)r(x) + Q(x)g(x) = f(x)$

[clever fjord]

Médicis

[marsh lichen]

So you’re saying we just split the fraction into two pieces like magic?
What even is g(x) and r(x)? are those just random letters or?

[jagged maple]

random letters yes

[marsh lichen]

So they’re just placeholders? But, how do we know what numbers or stuff to put there?

[amber veldt]

specifically theyre placeholder polynomials

[nocturne brook]

$g(x)$ and $r(x)$ are names of two polynomials for which [\frac{f(x)}{P(x)Q(x)}=\frac{g(x)}{P(x)}+\frac{r(x)}{Q(x)}]

[clever fjord]

Flip

[nocturne brook]

in this case, f(x) = x^3, P(x) = x^2+1, and Q(x) = x^2+4

[marsh lichen]

So g(x) and r(x) aren’t just numbers, they’re whole polynomials? How many letters can a polynomial even have? I feel like I need a diagram or something lol.

also, Why does splitting it like that even help?

[nocturne brook]

one letter is enough

well, it tends to help because the polynomials g and r are actually of degree 1 or less

[jagged maple]

I think it’s a deadend actually

it’s not the way

[amber veldt]

typically one letter represents the polynomial and one letter represents the independent variable, if theres only one variable

[nocturne brook]

it's worth doing because goldtrain doesn't know about it

[amber veldt]

whats the original question

[jagged maple]

okay

[jagged maple]

.

[marsh lichen]

Just one letter?? seriously? So like g(x) = x or something? and that’s enough??

[nocturne brook]

for the name g? sure

[jagged maple]

I tried substituting $x^2 + 1$ by $u^2$, which makes it slightly easier

[clever fjord]

Médicis

[jagged maple]

and splittable

[twin inlet]

j’ai le même tableau

[nocturne brook]

that's an x^3 at the top

[jagged maple]

très bon goût

[jagged maple]

yes

[marsh lichen]

Dead end? so we’re going in circles here? What question are we even answering?

[jagged maple]

to split the integral in two

[marsh lichen]

Ah, vous parlez francais?

[jagged maple]

oui

[twin inlet]

oui

[amber veldt]

@marsh lichen have you studied number theory at all

[marsh lichen]

Interessant, Mr. Le chat.

[amber veldt]

I can draw a parallel if you have

[nocturne brook]

ah, désolé, je ce vois

[marsh lichen]

number theory? I don’t think i’ve really done that. Does it help with x^3 on top of fractions?

[amber veldt]

no but it helps understand why partial fraction decomposition works

[marsh lichen]

Draw a parallel? like a picture or like a math connection??

[amber veldt]

math connection

[marsh lichen]

Also, can someone remind me which integral we’re actually splitting? i think i got lost somewhere between partial fractions and u-substitution.

[marsh lichen]

So number theory just explains why the magic works?

[jagged maple]

okay I think it is the actual proper method

[marsh lichen]

Et aussi, les fractions partielles, c'est comme prendre x^3 et le couper en morceaux plus petits ?

[amber veldt]

im just gonna say it because i think explaining about how I will explain it will not be helpful

[twin inlet]

tu prends une fraction et tu la décomposes en deux autres fractions

[marsh lichen]

Maybe if someone writes out the actual split with numbers i can see it… right now it’s just letters and confusion everywhere..

[jagged maple]

${ln(x)x^3\over P(x)Q(x)} = ln(x) ( {A(x)\over P(x)} + {B(x)\over Q(x)})$

[amber veldt]

okay so watch this gold I think it might help

[clever fjord]

Médicis

[amber veldt]

a diophantine equation is one where all the variables are integers, and I'm going to discuss an important theorem about diophantine equations, and then show how it's related

a diophantine equation is linear when all the variables are to the first or zeroth power, like so:

4x+7y=-23

[marsh lichen]

Okay, so we just put the ln(x) outside and then split the x^3 fraction. I think i kinda see it, maybe.

[amber veldt]

here x and y must be integers

without asking what the solution is, I want to know, IS THERE a solution at all?

[marsh lichen]

Alors on coupe une fraction en deux ? Est-ce qu’elles s’additionnent comme par magie ou bien?

[amber veldt]

Bezout's Lemma says that these equations have solutions if and only if

gcd(4,7)|-23 in this case

in general

ax+by=c

has a solution iff gcd(a,b)|c

do you know this notation?

[marsh lichen]

Now we’re talking about diophantine equations? Ithought we were splitting x^3 fractions.

[amber veldt]

youre calling partial fractions magic

i'm trying to show you theyre not

[twin inlet]

par exemple $\frac{1}{x(x+1)}=\frac{a}{x}+\frac{b}{x+1}$, pour trouver a et b tu peux utiliser quelques techniques comme par exemple mettre tout au même dénominateur et identifier les coefficients. ici $\frac{1}{x(x+1)}=\frac{a}{x}+\frac{b}{x+1}=\frac{a(x+1)+bx}{x(x+1)}=\frac{(a+b)x+a}{x(x+1)}$ et en identifiant on trouve que $a=1$ et $a+b=0$ donc $b=-a=-1$

[marsh lichen]

Can someone tell me if the fraction version even has a “solution” too? i feel like i missed a memo.

[clever fjord]

tm

[twin inlet]

justement c’est pas sous le même dénominateur donc ça s’additionne pas si facilement

[amber veldt]

do you accept Bezout's lemma as I stated it, and do you know the notation I used?

I'm getrting there I promise

[marsh lichen]

So partial fractions aren’t magic? ok, i guess i’m starting to see that… but my brain is still doing backflips between fractions and integers.

[amber veldt]

but answer my yes/no questions please so I know if youre followinmg

[marsh lichen]

Donc on fait correspondre les coefficients après avoir tout mis sur le même dénominateur ? C’est plus logique. peut-être que je peux suivre cette partie.

[twin inlet]

oui on identifie

la partie constante vaut 1 d’un côté et de l’autre c’est a, et la partie correspondant au x ça vaut 0 à gauche et à droite ça vaut a+b

[amber veldt]

goldtrain

I'm showing you the m agicians tricks that make it all make sense but you have to work with me if you want my help

[marsh lichen]

I think i kinda follow Bezout’s lemma? like, gcd(a,b)|c means there might be integers that work?

[amber veldt]

gcd means greatest common divisor

the biggest number that is a factor of both numbers

[marsh lichen]

ahhh ok… donc la partie constante d’un côté est égale à 1, et de l’autre côté c’est a… donc a = 1 ? 😳

[amber veldt]

| means divides, meaning

[marsh lichen]

Oh ye I remember that.

[amber veldt]

x | d is another way of writing "d is divisible by x"

[twin inlet]

exactement

[amber veldt]

so Bezout tells us that something like

[twin inlet]

i let u with @amber veldt

[amber veldt]

3x-2y=101 has a solution

but notice this can be written as

x/2 -y/3 = 101/(6)

agree?

[marsh lichen]

ok.

[amber veldt]

I just divided both sides by 6

$$\frac x 2 + \frac {-y} {3} = \frac {101}{3 \cdot 2}$$

actually i'll write it this way so it's a plus sign, doesn't matter

[clever fjord]

gfauxpas

[amber veldt]

do you agree this diophantine equation has a solution by Bezout's lemma?

[marsh lichen]

So you just divided everything by 6 and now it’s a plus instead of a minus?
I kinda see that, but my brain is still half in fractions and half in integers .

[amber veldt]

whether I write -y or +(-y) doesnt matter, I never made a claim that y is positive or negative

I did turn it into fractions, but it came from a diophantine equation that's the important part

it came from this

3x-2y=101

$$3x-2y=101$$
$$\iff$$
$$\frac x 2 + \frac {-y} {3} = \frac {101}{3 \cdot 2}$$

[clever fjord]

gfauxpas

[marsh lichen]

Okat, so the signs don’t matter, it’s still a solution either way . I guess i’m starting to see why the diophantine part is important,but my brain is still juggling fractions and x^3 lol.

[amber veldt]

okay so

know what it means for a polynomial to be a divisor or another?

x divides y if x is a factor of y

here's the cool thing

Bezout's lemma also works for polynomials

let's say you have three known polynomials, f(x) and g(x) and h(x)

and two unknown polynomials, fall them P(x) and Q(x)

I want to know, does this equation have a solution?

$$f(x)P(x)+g(x)Q(x)=h(x)$$

[clever fjord]

gfauxpas

[amber veldt]

for example, I might say that

$$(x-7)P(x)+(x^2+1)Q(x) = 2x+5$$

[clever fjord]

gfauxpas

[amber veldt]

does this equation have a solution for polynomials P and Q?

yes, because gcd(x-7,x^2+1) = 1, and 1 divides 2x+5

here's the connection to PFD

$$(x-7)P(x)+(x^2+1)Q(x) = 2x+5$$
$$\iff$$
$$\frac{P(x)}{x^2+1} + \frac{Q(x)}{x-7} = \frac{2x+5}{(x-7)(x^2+1)}$$

[clever fjord]

gfauxpas

[marsh lichen]

so basically, if the gcd of the denominators divides the numerator, we can find P(x) and Q(x) that work? So that’s why partial fractions always has a solution. I think i’m starting to see the “why”.

[amber veldt]

yes!

[marsh lichen]

We done?

[amber veldt]

yeah

you can go back to calculus now

[marsh lichen]

https://tenor.com/view/wolf-of-wall-street-lets-goo-gif-11742735383203550293

[amber veldt]

so

for this integrand

forget log for the moment

$$\frac{x^3}{(x^2+1)(x^2+4)}$$

[clever fjord]

gfauxpas

[amber veldt]

let's just do PFD on this

because the numerator is a degree lower than the denominator, let's assume that all fractions are

[marsh lichen]

what does PFD mean?

[amber veldt]

partial fraction decomposition

[marsh lichen]

( \frac{(x^2 + 1)(x^2 + 4)}{x^3} = -\frac{3(x^2 + 1)}{x} + \frac{3(x^2 + 4)}{4x} )

we can integrate term by term:

$$
\int \frac{x^3}{(x^2 + 1)(x^2 + 4)} , dx = \int \left( -\frac{x}{x^2 + 1} + \frac{4x}{x^2 + 4} \right) dx.
$$
$$
\int \frac{(x^2 + 1)(x^2 + 4)}{x^3} , dx = \int \left( -\frac{3(x^2 + 1)}{x} + \frac{3(x^2 + 4)}{4x} \right) dx.
$$

$$
= -\frac{1}{3} \int \frac{x}{x^2 + 1} , dx + \frac{4}{3} \int \frac{x}{x^2 + 4} , dx = -\frac{1}{3} \int \frac{x^2 + 1}{x} , dx + \frac{3}{4} \int \frac{x^2 + 4}{x} , dx
$$

$$
\int \frac{x}{x^2 + a^2} , dx = \frac{1}{2} \ln \left| x^2 + a^2 \right| + C
$$
$$
\int \frac{x^2 + a^2}{x} , dx = \frac{1}{2} \ln|x^2 + a^2| + C
$$

$$
-\frac{1}{3} \int \frac{x}{x^2 + 1} , dx = -\frac{1}{3} \cdot \frac{1}{2} \ln \left( x^2 + 1 \right) = -\frac{1}{6} \ln \left( x^2 + 1 \right)
$$
$$
-\frac{1}{3} \int \frac{x^2 + 1}{x} , dx = -\frac{1}{3} \cdot \frac{1}{2} \ln(x^2 + 1) = -\frac{1}{6} \ln(x^2 + 1)
$$

$$
\frac{4}{3} \int \frac{x}{x^2 + 4} , dx = \frac{4}{3} \cdot \frac{1}{2} \ln \left( x^2 + 4 \right) = \frac{2}{3} \ln \left( x^2 + 4 \right)
$$
$$
\frac{3}{4} \int \frac{x^2 + 4}{x} , dx = \frac{3}{4} \cdot \frac{1}{2} \ln(x^2 + 4) = \frac{3}{2} \ln(x^2 + 4)
$$

$$
\int \frac{x^3}{(x^2 + 1)(x^2 + 4)} , dx = -\frac{1}{6} \ln \left( x^2 + 1 \right) + \frac{2}{3} \ln \left( x^2 + 4 \right) + C
$$
$$
\int \frac{(x^2 + 1)(x^2 + 4)}{x^3} , dx = -\frac{1}{6} \ln(x^2 + 1) + \frac{3}{2} \ln(x^2 + 4) + C
$$

[clever fjord]

Gold/Train

[marsh lichen]

Is this correct?

[amber veldt]

not sure wahts happening

what is all this

[marsh lichen]

Lemme try to polish it:

We start with the partial fraction decomposition for the integrand:
$$
\frac{x^3}{(x^2 + 1)(x^2 + 4)} = -\frac{x}{x^2 + 1} + \frac{4x}{x^2 + 4}.
$$
This allows us to integrate term by term:
$$
\int \frac{x^3}{(x^2 + 1)(x^2 + 4)} , dx = \int \left(-\frac{x}{x^2 + 1} + \frac{4x}{x^2 + 4}\right) dx.
$$
Step 1: Factor constants
$$
\int \frac{x^3}{(x^2 + 1)(x^2 + 4)} , dx = -\int \frac{x}{x^2 + 1} , dx + \int \frac{4x}{x^2 + 4} , dx.
$$
Step 2: Apply standard formulas

Recall the standard integrals:
$$
\int \frac{x}{x^2 + a^2} , dx = \frac{1}{2} \ln \left| x^2 + a^2 \right| + C.
$$
Using this:

First integral:
$$
-\int \frac{x}{x^2 + 1} , dx = -\frac{1}{2} \ln \left| x^2 + 1 \right|.
$$
Second integral:
$$
\int \frac{4x}{x^2 + 4} , dx = 4 \cdot \frac{1}{2} \ln \left| x^2 + 4 \right| = 2 \ln \left| x^2 + 4 \right|.
$$
Step 3: Combine results
$$
\int \frac{x^3}{(x^2 + 1)(x^2 + 4)} , dx = -\frac{1}{2} \ln \left| x^2 + 1 \right| + 2 \ln \left| x^2 + 4 \right| + C.
$$
And that’s the final integrated form.

[clever fjord]

Gold/Train

[amber veldt]

first of all, the PFD is wrong

second of all, why are you doing all that

that's not the integral youre trying to solve

[marsh lichen]

Oh ye, my bad.

[amber veldt]

so we don't know the degree of the numerators but we know they're proper fractions because the original expression is a proper fraction right

so let's write

$$=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+4}$$

[clever fjord]

gfauxpas

[amber veldt]

and actually it will be helpful to write it in Bezout lemma form

$$x^3=(x^2+4)(Ax+B)+(x^2+1)(Cx+D)$$

[clever fjord]

gfauxpas

[amber veldt]

now the key is to simplify this by using that it holds for all x

because this is an equality of polynomials

for examkple, it holds if x=0, which gives us:

$$0 = 4B+D$$

[clever fjord]

gfauxpas

[amber veldt]

do you see how I got that?

anyway im gonna go now

gl

[marsh lichen]

you plug in x=0 to simplify and get 0 = 4B + D. I see how that works now.

I guess I'm done here.

.close

[odd edge]

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[odd edge]

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[modern garden]

i need some help with Intermediate value theorem

[graceful viper]

what about it

[modern garden]

wait let me send u the question

here u go

prove that the equation have a unique solution

[late dust]

(x+1)^3 = 4+3x with x in R+?

[modern garden]

i think we wont use Intermediate value theorem here

[modern garden]

yeah

[late dust]

Well, you can

[modern garden]

R+ is not a segment

[late dust]

What does that mean

[modern garden]

The interval should be closed from both directions

[late dust]

I mean... R+ is closed, but even then, it doesn't matter, you don't need a closed interval, you can use limits

Or, if you want a closed interval, you can make one

[modern garden]

Thanks for your help, man. I think I'll get some sleep

I don't know why they're teaching us this in high school 😭

[storm void]

bro

nel is full of shit

[graceful viper]

wot

the advice was correct

[modern garden]

😭

[odd edge]

@modern garden Has your question been resolved?

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[ember wharf]

9

[woeful briar]

chain rule can help

||it lets you describe dy/dt in relation to dy/dx and dx/dt||

[ember wharf]

i dont know where to start

[ember wharf]

im stil confused how that helps

i also dont understand how to do 11

[woeful briar]

stick with 9

we can move on to 11 after

[ember wharf]

ok

[woeful briar]

what does chain rule tell you?

[ember wharf]

you do the derivitave of the outside function applied to the inside function times the derivitave of the inside function

[woeful briar]

yeah

if you consider y as a function of x which is a function of t, then
dy/dt (t)= dy/dx (y(x(t)) * dx/dt (t)

and you know the sign of dx/dt

and you know from the graph a lot about dy/dx

so you should be a able to make a conclusion about dy/dt

[ember wharf]

ngl im completely lost

[ember wharf]

why do you consider y as a function of x which is a function of t

[woeful briar]

the relationship between x and y (per the graph) is that of a function (y being a function of x)

i.e. for every x there is exactly one matching y

[ember wharf]

correct

wait so is x a function of y

or is y a function of x

y is a function of x cuz x is the input

right?

[woeful briar]

for this you would need for every y there is a unique x, which might not happen

[woeful briar]

(in this case it does happen, but for the question its not what we care about)

[woeful briar]

you consider x to be the input, and a function must have a unique output (a value for y) for every value for the input (x)

[ember wharf]

ok

wait

does that mean x^2 isnt a function

cuz its like mirrored

[woeful briar]

x^2 is a function

[ember wharf]

when y is 4 , x can be 2 or -2

thats 2 inputs

and the same output

[woeful briar]

you but you are flliping the quantifiers

I said for every x, there is y

not the opposite

and indeed you are right, x^2 is not injective in R

[ember wharf]

hm

ok

i think i see waht you are saying

[woeful briar]

now getting back to the question.

dx / dt is everywhere negative

what can you say (from the graph) about the sign of dy/dx

[ember wharf]

so we have the graph of y(x)

[woeful briar]

yes

[ember wharf]

we dont have the graph of dy/dx

so how do i know about the sign of dy/dx

[woeful briar]

what does it mean for dy/dx to be positive/negative?

its not just anything

its the derivative

it is much related to y(x)

[ember wharf]

its negative

dy/dx is negative

correct?

[woeful briar]

yes

[ember wharf]

i imagine tangent liners

[woeful briar]

why is that (I am testing you)

[ember wharf]

and the slope is negative

i imagined tangent lines and it looked to be negative

[woeful briar]

ok you have the correct idea

dy/dx is negative as y(x) is always strictly decreasing

now what can you say about dy/dt (using the chain rule)

[ember wharf]

so we know that dy/dx is neg and dx/dx is neg

do i set them equal to eachother

then differntiate

[woeful briar]

bumping this

take a look

[ember wharf]

ok

ima do it by myself gimme 3-5 min

do u use product rule

cuz its dy/dt * t

[woeful briar]

its not a product

I marked a product with *

its evaluation at t

[reef plank]

halp

[woeful briar]

!occupied

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[ember wharf]

oh ok

so whats the derivitave of dy/dt

with what respect of variable

[woeful briar]

well no need to differentiate this further

as we already have a relation between dy/dt, dy/dx and dx/dt

also this second derivative may not be defined

[ember wharf]

can we move on to 11 i dont think im going to get it ima js have my teacher explain it to me later

[woeful briar]

we are really close to finishing

I will just explain this a bit faster

and we will move on after

so you know dy/dx and dx/dt are negative

[ember wharf]

ok

[woeful briar]

their product, which by the chain rule is dy/dt, is positive

thats it

its fine if you don't fully understand this

but eventually it will click

[ember wharf]

ok

[woeful briar]

now for 11

first understand what it means for x and y to change at some constant rate

do you know what it means?

[ember wharf]

when the coefficents

are the same?

lino

no thatss wrong

when they are proportionally changing

at the same speed

[woeful briar]

ok I meant just for one of them

what does it mean for a function y to change at a constant rate of k

||I am looking for an answer in terms of derivatives||

[ember wharf]

when the derivitaves are the same>

?

[woeful briar]

same as what?

[ember wharf]

like the slopes of the tangent line

the derivivtaves are same value

[woeful briar]

when it is constant

yes

[ember wharf]

when a x value is plugged in

[woeful briar]

so a function f(t) changes at constant rate k if df/dt is always k for all t

[ember wharf]

yes

[woeful briar]

so now, according to the question, we assume this is true for x

so dx/dt = k

what can you say about dy/dt?

||again, chain rule is your friend||

[ember wharf]

dy/dt=k

[woeful briar]

how did you arrive at that?

[ember wharf]

im ngl i think im not understanding this whole part at all im going to go review the basic concepts more cuz i dont think im ready for these types of problems

[woeful briar]

sure you can review it later as much as you like

again, you are not supposed to understand everything immidietly

a first exposure can help you understand the second time

but really here it is the same trick from before

y is again a function of x so
dy/dt = dy/dx * dx/dt

and you know dx/dt = k, and dy/dx = a (by the equation) so dy/dt is k * a

[ember wharf]

ok, ty for the help you have gave so far im going to go ahead and close this channel

.close

[odd edge]

Channel closed

Closed by @ember wharf

Use .reopen if this was a mistake.

[odd edge]

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[jagged maple]

wow somebody had the exact same problem as yours

Are you trolling?

#help-19 message

this is the same pic

same framing

[stoic cloud]

This is a troll no doubt

[unkempt lichen]

troll or not, what about the explanation above do you not understand?

oh cool. so we're done?

.close

[odd edge]

Channel closed

Closed by @unkempt lichen

Use .reopen if this was a mistake.

[stoic cloud]

Was the troll banned?

[unkempt lichen]

dkdc

[amber schooner]

"stop yapping"

[stoic cloud]

hopefully

all their messages were deleted so.

good riddance

[odd edge]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[shrewd trellis]

Hi, is my answer for c correct?

[safe vapor]

not sufficient

[modest nacelle]

Why are you doing it upside-down? You should've integrated the f'(x), finding out that f(x)=... + C, and then you find the C

[amber schooner]

no

[safe vapor]

if you do not know integration, you can differentiate the ansatz and show that the y-int is correct and the derivative matches

[amber schooner]

you've shown that f as given has that y intercept rather than showing f as described in the previous parts with the added condition that its y intercept is 4.5 is precisely the function they gave

[modest nacelle]

This is not something you would ansatz with. It would suspiciously look like you looked at the answer key.

[safe vapor]

I meant that since the question asks you to show that f(x) = ... satisfies the equation, I wouldn't fault someone if they differentate the (given) f(x) (if it is truly given), and then showing the constant also matches. Integrating is obviously more straightforward, if op knows it.

[shrewd trellis]

whats ansatz?

[safe vapor]

Forget ansatz, do you know integration? If you do, forget my message completely.

[shrewd trellis]

yes

ik integration

[safe vapor]

do this then

[shrewd trellis]

but why integrate if they already gave us the original function, f(x)

[modest nacelle]

They didn't give you that

They gave you f'(x), not f(x)

[shrewd trellis]

oh ok I understand

my answer for D is correct but I want to check if I used the right notation/process

because Im unsure if this was the right thing to do

[modest nacelle]

Infinity is not just a large number.

Have you learned about Extended Real Numbers?

In Baby Rudin, Extended Real Numbers are the first thing you learn on the very first chapter.

[unkempt lichen]

let's not go there for this OP.

[modest nacelle]

Why not?

[unkempt lichen]

I don't think introducing extended reals to this particular OP is beneficial at this stage.

[odd edge]

@shrewd trellis Has your question been resolved?

[short terrace]

You can't just check the value at one point and say that's the limit

We need a proof

In particular, what does the graph of e^(-2x) look like

Does it have any asymptotes

[unkempt lichen]

btw if I may say this

your working for c) looks like you went the other way round

you took f(x) as gospel and showed that f(0) = 4.5
they want you to take f(0) = 4.5 (plus info from f'(x)) and show that f(x) is what they say it is

[shrewd trellis]

Give me a moment I’m reading this

[short terrace]

Yes, what you have done is proved that the given function works. But you haven't shown that it's the only function that does

You have to work forward

[modest nacelle]

Alright, here's some important theorems about limits.
If
$$
\lim_{x\rightarrow+\infty} f(x) = C
$$
and
$$
\lim_{x\rightarrow+\infty} g(x) = D
$$
Then
$$\lim_{x\rightarrow+\infty} f(x) - g(x) = C-D$$
$$\lim_{x\rightarrow+\infty} \frac{f(x)}{g(x)} = \frac{C}{D}$$

[clever fjord]

Xwtek

[shrewd trellis]

But the limit is correct, it’s 5

[faint knot]

theres a more rigorous way of doing the problem than plugging in bigger numbers

[shrewd trellis]

I know about

Factoring out

[faint knot]

but theres nothing to factor here

let me continue,

[shrewd trellis]

And plugging in infinity after simplified

alr

[faint knot]

thats not good either, thats a shortcut for something else

let me continue,

if you graph f(x) or plug in big values, youll see that the limit should be 5
but we still havent proven that as such

[modest nacelle]

That's almost like using Extended Real Numbers, but I've been advised not to continue on that route.

[faint knot]

one way is just to memorize limits you already know are true, then solve for problems using that

[modest nacelle]

Instead, use my theorems above

[faint knot]

for example, you can use as a given that $\lim_{x\to\infty}e^{-x}=0$

[clever fjord]

mtt

[faint knot]

proving this takes some effort, so you can take it as true for now

now based on this fact, heres how you show that $\lim_{x\to\infty}e^{-2x}=0$

[clever fjord]

mtt

[faint knot]

one way is: $\lim_{x\to\infty}e^{-2x}=\lim_{x\to\infty}(e^{-x})^2=\left(\lim_{x\to\infty}e^{-x}\right)^2=0^2=0$

[clever fjord]

mtt

[faint knot]

moving the limit inside the function was done through composition limit rule, which you can search up

another way is: $\lim_{x\to\infty}e^{-2x}=\lim_{x\to\infty}e^{-x}e^{-x}=(\lim_{x\to\infty}e^{-x})(\lim_{x\to\infty}e^{-x})=0\cdot0=0$

[clever fjord]

mtt