help-39

[hoary nacelle]

If I were to choose in maybe somewhere next year I don’t even think I’m ready

[rough forge]

Emma you are being a bit hard on yourself

[hoary nacelle]

I hope you’re telling the truth

It’s 50% dropout rate 😭

[rough forge]

and 50% pass rate

[hoary nacelle]

At even Harvard and I am not even from Harvard

[rough forge]

flip a coin

[hoary nacelle]

Yes math students pass

I just don’t know the more exercise I do the less I feel like I know

Crazy

[rough forge]

crazy part is you will never know enough

[hoary nacelle]

Like this book for first year course is literally illegal

[rough forge]

you stole it?

[hoary nacelle]

I doubt anyone can master it

[rough forge]

achso

[hoary nacelle]

What the fuck about factorial

[rough forge]

you can be proud that you are capable of even working with it

[hoary nacelle]

This is a textbook

For ra

[rough forge]

do you agree 1 = 0!

[hoary nacelle]

Yes

It’s defined

[rough forge]

damn you know

thought i could do this: 🤯

say whaat 0 = 1

[hoary nacelle]

But how to show uniform convergence for sum x^k/k!

Fuck thsi factorial sign

[rough forge]

this is the sum of e^x if k goes to inf

[hoary nacelle]

Yes true

I know x! Is approximately the same order as e^x

Anyway I should go to supermarket for now

Buy something and eat

I must master at least this uniform convergence part before end of Christmas

[rough forge]

,, f_n(x) = \sum_{k=0}^n \frac{x^k}{k!}

[jolly parrot]

𝔸dωn𝓲²s

[hoary nacelle]

Whenever I see it I found it horrifying

[rough forge]

If you consider the limit n -> oo it converges to e^x

and this taylor series is defined for all x on R

[hoary nacelle]

I will attempt this one after arctan one that one seems easier in term

Anyone thanks so much and thanks for helping me with checking you’re a hero

[rough forge]

checking my hero?

[hoary nacelle]

I am too frustrated with my exercise

[rough forge]

me too

[hoary nacelle]

On the books

But you’re a mathematician I think it should be easier right 😭😭

[rough forge]

applied mathematician but ermmm idk

life is silly

[hoary nacelle]

After cal(1,2,3) and LA I don’t have any math lessons anymore in mandatory fashion

Actually even cal3 and la aren’t mandatory

[rough forge]

and since life is silly and boring you choose RA

i like you

[hoary nacelle]

Thanks

[rough forge]

😂

[hoary nacelle]

And it’s true that Econ students take RA

For the sake of taking it

[rough forge]

prove it 🫵🏻

[hoary nacelle]

I think it’s quite well-known though

Google

[rough forge]

q.e.d.

[hoary nacelle]

And I feel like this is actual mathematics

Like Linear algebra calculus they are just calculation

But this is too different and too difficult

[rough forge]

ermm linear algebra i would argue

depends how you teach it

in school vector geometry I agree but in university hmm not really

vector geometry will be like a grain of salt in university lol

[hoary nacelle]

last question

How difficult is RA s final

In university (us standard)

[rough forge]

you are asking the wrong one lol

maybe @prime bramble knows

[hoary nacelle]

in any university

[rough forge]

it prob depends but dont let that discourage you

[hoary nacelle]

Like it will just like any other courses that assumes that you mastered the textbook right

Got it

😭😭😭

[plush bramble]

Just google example final exams

No one here knows what you consider hard

[prime bramble]

I do not

I’m not American either

riemann’s suggestion is a good one though

[pearl pond]

@hoary nacelle Has your question been resolved?

[pearl pond]

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[stoic imp]

Find the sum of all positive integers such that their expression in base $7$ digits is the reverse of their expression in base $16$ digits. Express your answer in base $10$.

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

<@&286206848099549185>

[distant narwhal]

is calculator or computer allowed?

[stoic imp]

I guess? should be doable without it but what is your idea?

[distant narwhal]

i'm just thinking about write a program to solve it🤣

[stoic imp]

should be doable, but idk about coding

[inner granite]

$7^n \cdot a_n + 7^{n - 1} \cdot a_{n - 1} + \ldots + a_0 = 16^n \cdot a_0 + 16^{n - 1} \cdot a_1 + \ldots + a_n$

[stoic imp]

what is your idea?

[inner granite]

Idk, I am expressing it in an equation so it will be clearer to solve

[stoic imp]

is okay, feel free to brainstorm, every idea is valuable

[jolly parrot]

Roman_Garland

[inner granite]

$7a_1 + a_0 = 16a_0 + a_1 \implies 6a_1 = 15a_0 \implies a_1 = \dfrac{5}{2} \cdot a_0$

[jolly parrot]

Roman_Garland

[stoic imp]

this is assuming it's a two digit number i suppose

[cunning comet]

if you assume that the expression in base16 digits has no leading zeros you can calculate a maximum value for the number of digits.
then you can do look at this case by case.

• one digit numbers should be trvial.
• for two digit numbers you can solve 16a+b = 7b+a
• and so on till the maximum number of digits

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

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[wild fable]

I’m not sure what I’m doing wrong with the negative sign in the sqrt

[steel dagger]

Have you tried letting x-2 = -2sec(theta) instead

Oh nvm that wouldnt help

[rigid mist]

x-2=2 sin(theta)

[steel dagger]

I dont see how you can avoid having i in denominator here, but an alternative sub is letting u= 4x-x² and manipulating the numerator

[wild fable]

Why sin?

[rigid mist]

just try x-2 = 2 sin(theta) and you will see why

[wild fable]

Ok I’ll try it

[rigid mist]

sorry im bad at explaining stuff

[steel dagger]

I think even x-2 = tan (theta) would work

A more trivial one than yajat's

Since you sub x = tan (theta) wherever you see 1+x² @wild fable

[wild fable]

ok thanks it worked

cuz of the negative in the square root a different sub was needed

[rigid mist]

right

[wild fable]

thankjs gng

.close

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[pearl pond]

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as this action is irreversible, and this channel may abruptly lock.

[lean orbit]

.close

[pearl pond]

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[lean orbit]

any help?

[lean orbit]

i keep messing up

[errant fable]

isnt 3rd step wrong

[lean orbit]

idek anymore

[errant fable]

derivated one side but simplified other side

[lean orbit]

which?

oh don't mind that

i derivated the next step

ill remove the y'

from the third step

[errant fable]

alternate solution would be

[lean orbit]

mhm

[errant fable]

using (f(x)/g(x))’=(f’(x)g(x)-f(x)g’(x))/g^2(x)

and working it out tirgonometrically

[lean orbit]

oh

lemme try that one

[errant fable]

i found 1/(s^2+c^2)

[lean orbit]

uh

ig mine's wrong

[errant fable]

i can check your work

[lean orbit]

please do

[errant fable]

this is correct

i was mistaken

[lean orbit]

hm?

[errant fable]

also sin^2(x)+cos^2(x)=1

[sinful nebula]

you can do from this method too

[lean orbit]

oh i forgo

righr

[sinful nebula]

in third step instead of braking ln (x/y)

[lean orbit]

so 1/(sinx+cosx)²

[lean orbit]

yeahh

[lean orbit]

mhm

[sinful nebula]

differentiate ln() and then use chain rule

[lean orbit]

wait so like 1/(sinx/sinx+cosx) * (sinx/sinx+cosx)'?

[sinful nebula]

ye

then multiply by y both sides

But here using quotient rule is better

[lean orbit]

ohh

[lean orbit]

yeah ig so

well ig i got the answer

thanks y'all

can i close?

imma close

.close

[pearl pond]

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[midnight haven]

what the heck is this

[sterile turtle]

how do i integrate this

[midnight haven]

what is integrate ( im in 7 class )

[viscid sierra]

add this to the original form of I

also since u is a dummy variable it can be replaced by x

[sterile turtle]

?

add them together for 2I?

[sharp ivy]

Yes

[viscid sierra]

yes

[sterile turtle]

could i get an explanation on this

ik its a dummy var, but i thought it had a set correlation to x so i couldnt say it can be x

[sharp ivy]

It can, just thought of it as integrating on x-axis

[sterile turtle]

hmmm

so is this true?

[sharp ivy]

Yes

You can just change u to x and du to dx

[sterile turtle]

ic

thx for that

.close

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[sly plover]

Yeah this is supposed to be a question on "series and progressions"

I won't even try to know why

[pearl pond]

@sly plover Has your question been resolved?

[sly plover]

.close

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[strong ether]

Can someone help me

[pearl pond]

Please don't occupy multiple help channels.

[strong ether]

New task

A circle C is defined by the equation x²-6x + y²+8y+20 = 0 A line I is defined by the equation y=x-4

(a) Determine the coordinate set for each of the intersection points between I and C

(b) Show that C has center at the point (3; -4) and radius r = √5

(c) Determine the values ​​of b for which the line with the equation y = 2x + b is a tangent to C

task a is done

How do I do b and C?

would it be nice to just write, yes that's fine blah blah... in the task b

[sly plover]

b can be done by the general form and comparison

[strong ether]

so just a descriptive text?

[sly plover]

Do you know what the general equation for a circle is?

[strong ether]

Im not sure.

but what do we do with the equation?

[sly plover]

x2 + y2 + 2gx + 2fy + c = 0

That's the general education

[strong ether]

what

[sly plover]

-g and -f are the cordinates for the centre

[strong ether]

but that has nothing to do with the question, right?

[sly plover]

I dunno man 😭

Maybe if the line is tangent you can find the centre

[strong ether]

😂😂

[sly plover]

I can't really think of anything else

[strong ether]

b) Show that C has center at the point (3; -4) and radius r = √5

[sly plover]

r will be √ g^2 +f^2 -c

[strong ether]

How do we make it?

[sly plover]

By expanding (x-g)^2 + (y-f)^2 = r^2

I don't really know another way except the general form

[strong ether]

but why, I don't understand at all

<@&286206848099549185> help

[sly plover]

Welp

[strong ether]

I only need the two questions.

<@&286206848099549185>

I need help with the last one.

<@&286206848099549185>

[strong ether]

BRO YOU RIGHT

wait a minut

@sly plover stay

you mean this

(x-a)^2 + (y-b)^2 = r^2

(x-3)^2 + (y-4)^2 = r^2

Is it in the center and radius of 5?

[sly plover]

Oh lmao

[sly plover]

Yeah

[strong ether]

what now

is it done correctly

[sly plover]

This thing?

The radius is √ (3)^2 +(4)^2 - 20

Which will come out to be √5

[sly plover]

You can solve the general equation of a line with the general eqn of a circle to obtain a general eqn for a tangent

[strong ether]

yeah

[sly plover]

Lot of general equations ik

[sly plover]

Should be

For y = 2x + b

[strong ether]

so I get 3 equations? what

[sly plover]

Nope

[strong ether]

okay, I'm just trying

[sly plover]

The slope of the tangent should be 2

And b = +- r√(1+2^2)

Which will be uhh

[strong ether]

huh you get this

[sly plover]

+- √5 . √5

[strong ether]

I not get this

[sly plover]

That's just +-5

[sly plover]

Basically

Assume line

[strong ether]

ahh

[sly plover]

y =mx + c

and circle

Then

Make quadratic in 1 variable

[strong ether]

i know but

[sly plover]

Use D = 0

[strong ether]

(x-3)^2 + (y-4)^2 = r^2

[sly plover]

So only 1 point where intersect

[strong ether]

(x-3)^2 + (y-4)^2 = sqrt5^2

[sly plover]

Yeah

[strong ether]

but i get these

I get two points

[sly plover]

Are these the points

Which lie on circle

[sly plover]

For c?

[strong ether]

yeah

mhm

Am I doing something wrong?

[sly plover]

What 😭

What exactly

Did you do

[strong ether]

I use

equation of the circle

[sly plover]

Mhm

And then?

Did you just

Make quadratic

And solved?

[strong ether]

then I put the values ​​in and get this

[strong ether]

yea

What do you do, exactly?

Show your example of solving the task.

[sly plover]

This

[sly plover]

Uhh

idk man

I'm not really good at explaining and stuff

😭

[strong ether]

I think you are good.

What is this formula called:

y =mx + c

[sly plover]

That's just

General form

Of a lime

Line

With slope m

[strong ether]

mmm

Can you show your example? such small cubes

<@&286206848099549185>

[sly plover]

Maybe try googling general equation of a tangent?

[strong ether]

Why is the tangent interesting here?

Shouldn't we prove whether it is in the center?

[sly plover]

Isn't c asking the values of b for which it is a tangent

[strong ether]

I don't think so.

b) Show that C has center at the point (3; -4) and radius r = √5

[sly plover]

Ohhh

You're talking about b)

Well i actually gtg now

But I think you probably understand what to do

Good luck

[strong ether]

ahh

What task did you think we were talking about?

Don't say task a...

<@&286206848099549185>

Helpp pleaassee

[strong relic]

what id the question? its really high up

[frail zephyr]

@strong relic

[strong ether]

^^

[strong relic]

which part?

[strong ether]

b and c

part a is done

(b) Show that C has center at the point (3; -4) and radius r = √5

c) Determine the values of b for which the line with the equation y = 2x + b is a tangent to C

[strong relic]

do you know how to find the center of a circle

and its radius

you need a specific form of the equation

[strong ether]

Isn't that the equation of a circle?

[strong relic]

yes

$(x-h)^2 + (y-k)^2 = r^2$

[jolly parrot]

ashy!

[strong relic]

how do you grt your equation into this form

[strong ether]

and we have a point given (3; -4)

(x-3)^2 + (x-4)^2 = sqrt(5)^2

so I get two points

[strong relic]

you shouldnt use the information they give you to prove it

[strong ether]

mhm

What will you do then?

[strong relic]

use the equation they give you

[strong ether]

x^2 - 6x+ y² +8y +20 = 0?

This?

[strong relic]

yes

[strong ether]

what

[strong relic]

put it into this form

[strong ether]

You want me to put that into the formula.

Give me 2 seconds.

[strong ether]

I don't know how...

[strong relic]

complete the square

[strong ether]

what you mean?

Can you show me how, because I don't know. @strong relic

[strong relic]

the method of factorizing quadratics

are you not aware of that?

[strong ether]

before we can plug it into the formula, do we need to factorize the equation?

[strong relic]

we need it in that form

and we need to factor it to get it into that form

[strong ether]

Can you show me a start?

so I know how

@strong relic

[strong relic]

look up completing the square

[strong ether]

I don't understand what you mean by "completing the square"

[half inlet]

how can i help you/

[strong ether]

@strong relic

[strong ether]

go up

[strong relic]

google it

[strong ether]

Can't you give me a clue or a start...?

"...."

[strong relic]

i really do not want to type out the whole process of completing the square

because it is well established across the internet

[strong ether]

Like this:

(x−3)^2 −9+(y+4)^2−16+20=0

@strong relic

[strong relic]

yes

now bring the constants to the right

[strong ether]

okay

and I get (x-3)^2 + (x-4)^2 = 5

So it's on point?

[strong relic]

yup

from that u can get the coords of the crnter

and the radius

[strong ether]

That's what I sent a long time ago, right?

Why are we doing double work?

Can you explain why we did what you want me to do?

@strong relic

[strong relic]

you used the information they gave you

(3, -4) and r = sqrt 5

and showed that it was correct

[strong ether]

yeah?

mhm

[strong relic]

but what if they didnt give you that information

and asked you to find the center and the radius

with just the equation

[strong ether]

ahh okay i understand

Now this

c) Determine the values of b for which the line with the equation y = 2x + b is a tangent to C

so task b is finished?

[strong ether]

yeah

[strong relic]

yes

do you know what tangency is based of the discriminant

[strong ether]

What is tangency?

I'm not sure.

you mean ax^2 + b + c = 0

@strong relic

[strong relic]

no

[pearl pond]

@strong ether Has your question been resolved?

[strong ether]

How so?

@strong relic

Can others help me? It can't be right that you're never given explanations on how...

<@&268886789983436800>

Can you take over, I've been waiting for 8 hours for an answer.

[toxic fractal]

moderators are not to be pinged just to request help, they deal with rules non-compliance

[strong ether]

ahh okay

but then you can't help me

<@&286206848099549185>

Combine the equation? What should I combine with? so y= 2x + b with what we got?

oh, so there will be 2 equations with 2 unknowns?

@strong relic

[strong relic]

set the discriminant to 0

one equation in x with a constant b

[strong ether]

can you show me?

[strong relic]

subsitute y into the first equation

[strong ether]

with what equation and why can't you show me how?

[radiant terrace]

If you end up getting pings from me at some point barda it was because discord lagged and I saw the mod ping but nothing following it.

[strong relic]

the second equation into the first one

[strong ether]

It's so nice, but you can't take over, can you?

[strong ether]

Ashy, I don't understand when you write, you can't show me what you actually mean.

[strong relic]

im not going to do it for you

y = 2x + b

substitute all instances of y with 2x + b in your circle equation

[radiant terrace]

In general you should not ping mods for math help. We generally use the mod ping to do things like respond to (i.e., mute/ban) trolls, spammers and other rule breakers. So if you ping us for math help it makes it harder for us to actually do that.

[strong ether]

I'm sorry. I didn't mean to ping you, I've just been sitting here for hours and still haven't solved the problem. Even though it's a question from.

[strong ether]

so I need to substitute y into the equation there?

Shouldn't it be combined?

and do we know our y-value?

[strong relic]

we subsitute it so we have one variable

do it first and show me

[strong ether]

x^2−6x+(2x+b)^2+8(2x+b)+20=0

Like this?

@strong relic

[strong relic]

okay

now expand it all

[strong ether]

x^2 −6x+4x^2+4bx+b^2+16x+8b+20= 0

[strong relic]

im assuming thats correct

now combine like terms

[strong ether]

5x² + (4b+ 10)x+ (b² + 8b+20) = 0

[strong relic]

alright

do you know the expression of the discriminant

[strong ether]

No

d = b² - 4ac?

@strong relic

[strong relic]

yes

when a quadratic is tangent

what does that tell you about the discriminant

[strong ether]

What are my b and a and c then?

[strong ether]

mhm

[toxic fractal]

"b" would be the coefficient of x. In your case, (4b+10)

"c" would be the independant term. In your case, (b^2+8b+20)

[strong ether]

ahh

So like this:

[toxic fractal]

you should not learn the letters for formulas, you should always learn their meaning

[strong ether]

d = (4b+10)^2−4⋅5⋅(b^2+8b+20) = 0

[toxic fractal]

why are you equaling it to 0?

[strong ether]

I know the formula well, just not in the mood right now...sorry.

[strong ether]

because of the square root rule

[strong relic]

thats right, as its tangent

[strong ether]

yea

[toxic fractal]

hm, i would not do that in the same line, but okay

[strong ether]

so what now

we have the line

what we do now

[toxic fractal]

now you get b from here

[strong ether]

(4b+10)² - 20(b² +8b+20) = 0

That's the equation for b?

[toxic fractal]

so now you expand and solve that

[strong ether]

b1 = - 15

b2 = -5

so we get 2 values

points

[strong relic]

i dont think there should be 2 valhes of b

wait no nvm thats right

[strong ether]

so im done?

[strong relic]

yup

[strong ether]

^

YESSS

[pearl pond]

@strong ether Has your question been resolved?

[pearl pond]

@strong ether Has your question been resolved?

[pearl pond]

@strong ether Has your question been resolved?

[pearl pond]

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[pearl pond]

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[devout yarrow]

can anyone slve this elaborately

[devout yarrow]

.close

[pearl pond]

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[pearl pond]

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[limber monolith]

hello

[limber monolith]

am Maltiti, anyone interested in free maths lessons of any degree should dm now

[tropic saddle]

thats not how this here works

[limber monolith]

i don't get you

[fringe raft]

what is your math question?

[tropic saddle]

these channels are for when you need help

[pearl pond]

@limber monolith Has your question been resolved?

[autumn fossil]

You can check the other channels here though and try to help the people who asked in them

[pearl pond]

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[midnight haven]

yo

[midnight haven]

why solution must be less than 2?

[light helm]

range of abs val

[compact ridge]

Otherwise LHS is nonnegative but RHS will be negative for x > 2

[vagrant relic]

because on LHS, there is a mod, meaning that 2-x > 0

[compact ridge]

Also check x = 2

[midnight haven]

why didnt we do that here?

shouldnt we do 0<x+3?

[compact ridge]

It's good practice to check that x = -1/2 in fact satisfies the original equation

[compact ridge]

Ah no either abs(x - 2) = -(x - 2)

Or abs(x - 2) = (x - 2)

Those are the only two possibilities

Ofc not all of the solutions will be valid for the original equation

[midnight haven]

oh

so when there is more than 2 solutions

we do the 0<equation on the right

[compact ridge]

If you have a quadratic somewhere yeah

So these are absolute value linear equations

A linear absolute value function is made up of 2 straight lines

Two lines can cross at most once

[midnight haven]

but isnt this a linear as well?

[compact ridge]

So line 1 and line 3, line 2 and line 3

[midnight haven]

ohhh

nvmnvmnvm

[compact ridge]

Two intersections at most yeah

[midnight haven]

first question has lines right?

second question only has 2

[light helm]

shouldnt we do 0<x+3?
<=
but yes, either that or check whether the value works

[midnight haven]

when we have more than 2 lines we do 0<equation on the right?

if we dont we just check it ourselves

thanks

.close

[pearl pond]

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[pearl pond]

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[midnight haven]

so when we divide by a quadratic the remainder is a linear

so (x^2 - 1) q(x) ax+b

so (x+1)(x-1) q(x) ax+b

we find the remainder of dividing by x+1

using the whatever theorm

4(1)^2023 + 2(1)^2024

which equals 6

then do the negative

which equals 2

now what

oh so

6 = (1-1)(1+1)q(1) + a(1)+b

so a+b =6

then do the other equation

-a+b = 2

b=4 and a =2

let me check mark scheme '

yup correct thanks guys

.close

[pearl pond]

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[sharp smelt]

Prove that if $(x_n)$ converges to x, and if $(y_n)$ diverges, then $x_n+y_n$ doesn't converge.
\
\
Proof : Lets suppose that it convegres , we then have $x + lim(y_n) = z_n$, or $lim(y_n) = z_n-x$, which contradicts the fact that it diverges

[jolly parrot]

ƒ( wai ina teacup)= I don't know

[fringe raft]

what is z_n

[sharp smelt]

oops

it should be z

[fringe raft]

what is z

[sharp smelt]

the limit of the sum of the sequence

[fringe raft]

right, but then you only have $\lim_{n \to \infty} (x_n + y_n) = z$

[jolly parrot]

rbit

[sharp smelt]

yes

and that x_n converges , to say x

[fringe raft]

but you cannot just move that x out

[sharp smelt]

right

so I have to use epsilonics

[fringe raft]

oh no no

[fringe raft]

from here, can you get just lim y_n on the left side somehow?

[pearl pond]

@sharp smelt Has your question been resolved?

[sharp smelt]

sorry, was and am eating, wull be back in a bit

[pearl pond]

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[midnight haven]

ff

[midnight haven]

i have no idea what is happening here

how did we get from line 2 to line 3

[strong relic]

factoring

if its confusing you can do a substitution

u = log2 x

[midnight haven]

ohh quadratic

didnt see that

thanks

.close

[pearl pond]

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[pearl pond]

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[pulsar stump]

im trying to prove a^n >= n^2 for a>1, can someone help?

[pulsar stump]

im trying to use induction from N=ceil(a^2)

but it doesnt work

[rough forge]

Is n natural

[pulsar stump]

yes

[cunning comet]

1.1^2 should be > 2^2?

[rough forge]

a = 2, n = 3 you get 8 >= 9

[strong relic]

does a have to be natural as well

[rough forge]

This inequality seems to hold rather for sufficiently large values of n

Or if you choose a special start value like a >3 instead, it seems to be strictly increasing

[cunning comet]

any further discussion is more or less useless until @pulsar stump clarifies the question.

[pulsar stump]

a > 1 real, i need to find N natural so for every n>=N the inequality above holds

my attempt was with induction and didnt work

[cunning comet]

i guess there is no such N. you could find an N depending on a, maybe.

[pearl pond]

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.

[rough forge]

ah ok

[cunning comet]

there is no such N (for all real a > 1), again the question doesnt make sense.

[rough forge]

There can because a^n is exponentially growing and will overpass n^2

[brave sluice]

for a particular a, yes

[cunning comet]

for every N you can find an a small enough (>1) such that the inequality is wrong.

[rough forge]

That's not the question

[cunning comet]

you can find an N for a given a, not for all a.

[rough forge]

He just wants to find an N where the inequality holds true

https://www.desmos.com/calculator/bve2qlggia

You can find that N

[brave sluice]

it will be a function of a

maybe that's ok though

it's ok if the question is like "for all a>1 show there exists N such that..."

[pearl pond]

@pulsar stump Has your question been resolved?

[pulsar stump]

it is

but what is that N

[pulsar stump]

it doesnt say it here

and in general i need to do real math to get it

[rough forge]

it's not like i was providing you the solution but to get the idea of your problem

[pulsar stump]

ik the idea of my problem

[pearl pond]

Channel closed

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[crimson hemlock]

can anyone please give me the answer for this question

[crimson hemlock]

I tried at many places but still every AI was somewhere confused

[iron stream]

I can give youuuuu

[crimson hemlock]

sure

[iron stream]

Simplest way is to

Expand along C2

,w -b[(b+c)(a+b)-b(a+b)]+ a[(a+b)²-(a-c)(a-b)] -c[b(a+b)-(a-c)(b+c)]

[jolly parrot]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by xor_11

[sharp smelt]

Try using the properties of dets

[crimson hemlock]

okay I am trying that

[crimson hemlock]

like?

[sharp smelt]

https://www.cuemath.com/algebra/properties-of-determinants/

[pearl pond]

@crimson hemlock Has your question been resolved?

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[torn crane]

Hello there.

[rustic garnet]

...hi?

[native coral]

send ur question 🙂

[torn crane]

Thanks for your care.

[iron stream]

Hiiii

Why didn't the lucky angel appear on Christmas 😭😭

[torn crane]

Sorry🫡

[torn crane]

This is mathsmatics channel. right?

[iron stream]

Yeahh

[iron stream]

But I'm up for any sort of conversation, as long as it's Ok with the moderators

[torn crane]

So, I think maths is the base of coding.

[iron stream]

Maths is the base of everything

[torn crane]

You are right.

[iron stream]

As Galileo said, "Mathematics is the language in which God has written the universe"

[torn crane]

Great. I love maths too.

[iron stream]

That takes you to this untravelled road friend.

[torn crane]

You are so kind. Excuse me where are you from. I'm from Japan.

[iron stream]

I'm from India, friend !

[torn crane]

I see. You are mathematician?

I'm Program developer

[iron stream]

Just a learner

[torn crane]

Good

[iron stream]

Great, I'm Electronics undergraduate too

[dusty heart]

yo wassup

[torn crane]

Great, I know a little about the Electronics. I've learned about some digital circuits and so on.

[dusty heart]

im trying to get to computer science uni actually rn

[surreal relic]

What is the question

[torn crane]

Good. How can I help you?

[surreal relic]

O-O

[dusty heart]

theres no question on this channel, i got a question tho in one of the channels

its something with integrals, im just here cuz im bored and waiting for someone to try to respond to my question

[surreal relic]

@torn crane if you have a question post it, if not please close the channel

[dusty heart]

i was just saying, i dont need any help

[iron stream]

@torn crane you can move to discussion page ...

[torn crane]

Got it.

[iron stream]

To close the channel

[acoustic belfry]

#discussion for this typa stuff

[iron stream]

Type ".close"

[acoustic belfry]

use this to ask for help

[surreal relic]

.close

[pearl pond]

Channel closed

Closed by @surreal relic

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[autumn fossil]

I'm struggling with part a)

[autumn fossil]

K is compact, closed and bounded

[solid pier]

what is theorem 3.3.8

[autumn fossil]

the exericse is only proving that (i) and (ii) imply (iii)

[solid pier]

yes, the other way is harder

[autumn fossil]

it seemed easier to me

[solid pier]

oh wait, hhhhhh sorry, there is one direction that's harder than the other one

idk my brain is off, sorry 😭

[autumn fossil]

You could be right, maybe im just having issues with the direction that's supposed to be easy

[limber oasis]

build In+1 such that if it's finitely coverable, so is In

[autumn fossil]

So it should "differ" from In by finitely many open sets i suppose

[limber oasis]

I let you figure out the details and see if you need more help

[autumn fossil]

oh wait, if the intervals are closed, aren't they all finitely coverable?

[limber oasis]

well, we are trying to get a contradiction

[autumn fossil]

oh

[limber oasis]

we said I0 isn't

because a finite cover of I0 covers K

but yes, a bounded interval is compact, ofc

but that's basically just "a simpler case"

which I think is what b and c do

[autumn fossil]

Oh so they actually mean that the In's can't be covered by any finite subcover of the open cover of K

[limber oasis]

and a) is about bringing K down to the simpler case of intervals

well, isn't that what's written (up to the intersection with K)?

#❓how-to-get-help

this channel is occupied

[autumn fossil]

they said that In intersection K cannot be finitely covered

[limber oasis]

with the given open sets

[autumn fossil]

oh, that makes a lot more sense now

[limber oasis]

context matters

[autumn fossil]

You can open your own help channel and get help there. E.g. #help-32 is currently available

If I take I0 = K, it will do no harm, right?

[limber oasis]

irrelevant. This isn't the place,
this is about keeping channels clean by not having parralel discussions

[faint spruce]

ok

[limber oasis]

(and also about not forcing helpers to work on problems they don't want to work on)

[limber oasis]

"interval"

[autumn fossil]

oh

so I could take something like
I0 = [inf K, sup K]

[limber oasis]

I0 exists because K is bounded yes

[autumn fossil]

supK is contained in K

because it's closed and bounded

therefore there is some open set covering it

say it's (a, b)

[limber oasis]

(one day you'll enjoy the privilege of saying "by definition")

[autumn fossil]

Wdym?

is there a definition of boundedness that says that the set is contained within an interval?

[limber oasis]

in a general metric space, a set is bounded iff there's a ball that contains it

[autumn fossil]

ah

[autumn fossil]

Anyway, I could now take I1 = [inf K, a]

[limber oasis]

what?

[autumn fossil]

wait open set

im confusing sets and intervals

so there is some open set (in the cover of K) that includes supK

[limber oasis]

my question still stands with "open set"

[limber oasis]

yes

[autumn fossil]

I could take the infimum of that set

call it a

oh it doesnt have to be bounded

[limber oasis]

you can always intersect it with I0

now every set is bounded

[autumn fossil]

would that help here?

my thought was basically successively removing intervals from the edges

[limber oasis]

it solves that problem: you may consider every set in this problem to be bounded

[limber oasis]

that is, in fact, how one builds decreasing nested intervals

[autumn fossil]

I could take an interval (supK - ε, supK + ε) which is a subset of the open set covering K and including supK

and then let I1 = [infK, supK - ε]

[limber oasis]

you need to ensure it's not finitely coverable

[autumn fossil]

the only problem is that if I build it like this, then i have no guarantee that in the limit i will get them to be length 0

[autumn fossil]

It's not, because if it was then by adding the open set of which (supK - ε, supK + ε) is a subset, i would get a finite cover of K

i should probably start naming stuff

[limber oasis]

yeah ok for some epsilon

[autumn fossil]

yep

but I have no idea how to make a sequence in which the length of the intervals will approach 0

Can I get a hint?

[limber oasis]

each time, take the largest fitting open set

then there is an argument

[limber oasis]

rather than any open set that contains the sup

[autumn fossil]

okay

what exactly is meant by largest?

the one in which epsilon can be made largest?

[limber oasis]

yes

[autumn fossil]

alright

[limber oasis]

you shrink the interval as much as you can by removing 1 open set

well, not the set

the connected component of that set that contains the sup

otherwise it's no longer an interval

[autumn fossil]

I now need to prove that the length of such I's will approach 0

and i have no idea where to start tbh

[limber oasis]

assume not

[autumn fossil]

oh i can kinda see where this is going

informally, there is some point where it gets stuck

and the epsilons will just start shrinking too fast

[autumn fossil]

specifically, this point is the infimum of upper bounds of In's

but there must be some open set in the cover which contains that infimum

[limber oasis]

yes

[limber oasis]

they can actually be as slow as they want

(well, a convergent series but still)

[autumn fossil]

ic

[limber oasis]

but yes that's the idea, now find the contradiction

[autumn fossil]

this set would presumably be "larger" (in the previously used meaning of larger) than one of the sets chosen to form In

[limber oasis]

yeah you'd end up choosing that set, and shrink more than you're supposed to (to respect the assumption -> contradiction)

[autumn fossil]

Yep, tysm

I think I can formalize it now

[limber oasis]

hence the intersection is not some [a,b] unless a = b
but the intersection is an interval
done
that's also b)

then c) is obvious

[autumn fossil]

Yeah, I've done b and c before a lol

[limber oasis]

but the intersection is an interval
This may or may not be obvious depending on your intuition

[autumn fossil]

is it just nested interval property?

wait no

nested compact sets

x must be in K as well

[limber oasis]

look I took RA 2 years ago, all I remember is the intuition
The names of the theorems I never saw? Forgor

[autumn fossil]

it actually must be [a, b] with a = b, doesnt it?

[autumn fossil]

lest it would be some other interval, which would have non-zero length (contradiction)

[limber oasis]

I also don't think so?

[autumn fossil]

Did you mean to reply to this?

My intuition (which is currently really off) suggests that the intersection will be just a single number

because the length of In goes to 0

and there must be some number in the intersection by this theorem

[limber oasis]

oh yes, I used that too I forgot
I just didn't notice

yeah then you need an argument for the limit fitting in a set, and that's the 0 length argument

[autumn fossil]

Okay, I think that's it. Tysm for your help, I really appreciate it

.close

[pearl pond]

Channel closed

Closed by @autumn fossil

Use .reopen if this was a mistake.

[pearl pond]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[dusty heart]

ive been trying this exercise for a while now

[dusty heart]

last time i asked of this exercise i remember someone telling me at some point in the exercise that i need to multiply by sec^2x

which ive never worked with so like, be patient with me, guys

[floral ember]

@dusty heart

multiply divide by

sec^4x

[iron stream]

sin⁴x + cos⁴x = 1- 2sin²xcos²x

[floral ember]

then you subsitute tan x as t

and then solve

that will help

[dusty heart]

right, i tried

[iron stream]

2sinxcosx. = Sin2x

[dusty heart]

mhm

[iron stream]

So reduce first

[dusty heart]

let me try

1 sec

[iron stream]

And then you can proceed

[floral ember]

@dusty heart try tis

[dusty heart]

i will afterwards, because i dont know the relation between sec and tan

[floral ember]

sec^2x = 1 + tan^2x

[iron stream]

sec²x = 1 + tan²x

[dusty heart]

oh, so sec^2x is the derivative of tan x

[floral ember]

yes

[dusty heart]

cuz i knew about this but written as tan'x = 1+tan^2x

imma try both of your solutions rn

[dusty heart]

ok

and now?

[iron stream]

Then you will get( 2 -sin²2x)*0.5

[dusty heart]

yea

[iron stream]

2 = 2sin²2x + 2cos²2x

[iron stream]

After that you need to multiply sec²2x to numerator and deno

[dusty heart]

so id get cos^2 of 2x in the end?

[iron stream]

Huh ?

[iron stream]

Then you will apply substitution

tan x = z
So sec² x = dz

tan 2x in this case *

[dusty heart]

ok so wait we had 1/ (1-sin2x)

[iron stream]

Why so ?

[floral ember]

bro mine is like the same ... but less steps...

[iron stream]

Nah Yoobie he will face difficulty later solving that integral

[dusty heart]

ill show you what i got in yours

[floral ember]

(1 + t^2)/(1 + t^4)

that you can do by partial fraction ig

[iron stream]

Huh ?

[floral ember]

consider t^2 as y

so t^4 will be y^2

then make partial fraction

[iron stream]

(1+t⁴) = ...

[floral ember]

then substitue y again as t^2

[iron stream]

That won't work

[iron stream]

You will get odd powers

I don't mean it's unsolvable but it's rather easy to resolve using trig identities

[dusty heart]

wait wait

[dusty heart]

do you get 1+t^2 in the end

[floral ember]

oh wait

yeah

you are right

[dusty heart]

nvm

[floral ember]

to solve 1 + t^2 / 1 + t^4

you have to multiply divide by 1/t^2

and then solve the integral

too much work

[dusty heart]

so

is this right?

[iron stream]

I wrote your name and it autocorrected y to b 😭 lmao

[dusty heart]

willddddddddddddddddddddddd

[floral ember]

😭

[dusty heart]

in the math chat toooo, bros freaky

[iron stream]

Bro expand 2 = 2sin²2x + 2cos²2x

[dusty heart]

i didddddddd

look at the pic i set

i forgot a 2 for cos^2x but dont mind that

[dusty heart]

yeye, i fixed it

[iron stream]

I've written sin²""" 2x """"

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so this is it right?

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This is what I mean

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oh wowwwww, thats clever

okk

i got it

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Did. It ?

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now what

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Yayy now divide numerator and deno by cos²2x

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do i substitute with tg(x/2)

[iron stream]

1st this

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so....by that sec thingy?

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1/cos²2x = sec²2x

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mhm, interesting

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lmao

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Imma solve you that integral anyway

This is the general way to solve integrals of the form

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ok, did it

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1/(a+bsinx + c cosx)

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So now substitution

t = tan 2x

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subtitute normally, with t=tgx/2?

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nvm

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Yeah send pic after you do

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kk

[iron stream]

Imagine playing a chess match, checking up on a gc and solving integrals. Multitasking at its peak !

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ight

this looks goooooood

[iron stream]

I believe in you

That you won't mess up with silly calculationz

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nvm i did smth wrong for sure

cuz my solution aint in the multiple choice

[iron stream]

It would take an hour by my sleepy eyes to decrypt that text

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right sooo

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Did you change the integration limits ? When you substitute?

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yah

dx= 1/2(t^2+1)dt

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No this can't be possible

t²+2 ?

We had 2cos²2x + sin²2x in the deno no ?

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tg2x=t => 2x=arctg t => x= arctg t/2 => dx= 1/ 2(t^2+1) dt

makes sense?

[iron stream]

I can't make sense 😭

Lemme grab pen and paper

Huhh !

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ok please

because also, tg 0 = tg 4pi = 0 so

thats a bit wierd

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Hey

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hi

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Is the answer 0 ?

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nope

i mean yes

thats what i got

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,w integrate from 0 to 2pi 1/(sin⁴x+cos⁴x)

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but its not in the multiple choice

[jolly parrot]

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Requested by xor_11

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Yeah wait one min

Don't change the limits we will convert to tan and apply limit again

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sure

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It's because we're using trig identities and they are periodic functions

So we have

Integration

1/(2+z²)

Which is of the form int (1/a²+x²)

That gives 1/a tan^-1 x/a

So 1/√2 tan^-1 z/√2