help-27

[stone stump]

I mean you dont need unique factorization

[fervent hornet]

True

[stone stump]

I suppose you probably need euclids lemma

not sure how to do it without that

and well if you have euclids lemma then you basically have unique factorization

hmm

[lost laurel]

if a|bc , with gcd(a,b)=1, then a|c

this?

[stone stump]

yes

[lost laurel]

I was going to argue that squaring the numerator and denominaitor doesn't add any common factors

is that enough?

[stone stump]

imprecise but thats the idea

[fervent hornet]

Like if there d|a and d|b
Means that d|a² and d|b² but here d=1 thus (a^2,b^2)=1

[stone stump]

that does not mean that the gcd is 1

that means that d is a common divisor of a^2 and b^2, not that it is the greatest common divisor

[fervent hornet]

Hmm true

[gritty terrace]

Yeah it seems like Euclid’s lemma is pretty essential

[lost laurel]

so essentially here I argue 1|a

1|b

1|a^2

1|b^2?

[fervent hornet]

No that doesn't work

[fervent hornet]

.

[lost laurel]

right

hmmm

I could use the hint perhaps

this

[fervent hornet]

That's possible

[lost laurel]

hmm

if I remember right , (b,a)=(kb,a)?

if a and b are co prime

[gritty terrace]

i mean if you prove that you’re basically done

but I think that’s false

gcd(a, b) = 1 would mean gcd(ab, a) = a

[arctic field]

not sure what hint to give without just outright giving away the answer

[lost laurel]

true

[gritty terrace]

so what you want to show is the following

suppose $d | a^2$ and $d | b^2$. Then $d | 1$

[woven radish]

Pseudonium

[fervent hornet]

Hmm I thought of a way but I think it's abusive

[gritty terrace]

that’s the universal property here

[lost laurel]

that would imply d=1, no?

[arctic field]

spoiler: ||as + bt = 1 => (as + bt)^3 = a^3 s^3 + 3a^2 s^2 bt + 3as b^2 t^2 + b^3 t^3 = a^2 (as^2 + 3s^2 bt) + b^2 (3as t^2 + b t^3) = 1||

[gritty terrace]

yes, it would!

what i would recommend is splitting into cases

[fervent hornet]

So this works

[gritty terrace]

either d = 1, in which case you’re done

[finite briar]

What?

[gritty terrace]

or d > 1, in which case, try to get a contradiction

[fervent hornet]

I thought it was abusive

[gritty terrace]

Wdym what

[finite briar]

What is this unholy algebra

[arctic field]

its just bezouts

[finite briar]

I didn't get what u did

[finite briar]

Why the

[topaz beacon]

i could see a contrapositive be useful here

[gritty terrace]

i stated something true?

[lost laurel]

why do you cube instead of squaring?

[arctic field]

cuz the other one doesnt work

[arctic field]

[finite briar]

Oh yes okay mb

[lost laurel]

yes?

[finite briar]

Isn't there a theorem saying that if GCD(a,b) = d then a=dx,b=dy. Can't we use that here?

[finite briar]

Lots of ways

[gritty terrace]

hint: ||if d > 1 then there’s a prime p which divides d||

[lost laurel]

one minute

why?

[topaz beacon]

fta

[gritty terrace]

This is just “||numbers have prime factors||”

[fervent hornet]

Snow does this work
ax+by=1
ax1+by1=ab,x1=abx and y1=aby
Thus 1|ab
And similar we prove ab|1 thus (a²,b²)=1

[lost laurel]

hmm, yeah, makes sense

[fervent hornet]

Or is that too abusive?

[arctic field]

similarly?

[lost laurel]

ah, so based on this I argue that the only common factor is 1

[gritty terrace]

yes

[lost laurel]

got it

thanks

[gritty terrace]

you show that if d > 1 is a common factor, you get a contradiction

[lost laurel]

ye

makes sense

[fervent hornet]

Yes like a²v+b²z=d and do it

[gritty terrace]

and my suggestion is to say “if d > 1, take a prime p dividing d, then…”

[lost laurel]

I think I got it

[gritty terrace]

We cannot have ab|1 unless ab=1

[fervent hornet]

That's why I asked

Is it abusive

[arctic field]

im not sure how you're gonna finish your proof though

if a = 2, b = 3, then ab isnt 1

[fervent hornet]

True

Wait i got it

Does this work @arctic field

[gritty terrace]

Don’t see how that works

How do you know that av + bz = 1

[fervent hornet]

From ax+by=1

[gritty terrace]

i don’t understand

you don’t know that v = x and z = y

[fervent hornet]

Hmm true

Ig I'll have to revise

[lost laurel]

let me try to write out a proof

so let d|aa

and d|b

so d|a^2

and d|b^2

but d=1 as that's the GCD

[gritty terrace]

I don’t understand your proof

[fervent hornet]

Hmm but v and z can be one of the many solution for

[lost laurel]

trying to base it off this

[gritty terrace]

You seem to be starting from $d | a^2$ and $d | b$

But why

[woven radish]

Pseudonium

[fervent hornet]

It doesn't work like that unfortunately

[lost laurel]

$d|a$ and d$|b$

[gritty terrace]

What you need to prove is

[woven radish]

ƒ(Why am. I here)=I don't Know

[gritty terrace]

If $d | a^2$ and $d | b^2$ then $d | 1$

[woven radish]

Pseudonium

[lost laurel]

hmm

a^2 and b^2 are co prime

[gritty terrace]

That’s the universal property

[gritty terrace]

That’s what you’re trying to prove

[lost laurel]

oh right

[fervent hornet]

.

[lost laurel]

hmm, yeah, that makes sense

[gritty terrace]

This doesn’t work, as someone pointed out

[fervent hornet]

But denascite said it doesn't mean that gcd will be 1

[gritty terrace]

Yeah so what you wrote doesn’t work

[fervent hornet]

So how is this different

[gritty terrace]

I’ve stated what we need to prove

I’ve not provided a proof of that statement

[fervent hornet]

Oh I see mb

[gritty terrace]

It’s what called the “universal property” of gcd

[lost laurel]

hmm

[fervent hornet]

Yeah snows proof works the best

[gritty terrace]

I have a different proof in mind

Using Euclid’s lemma

But if you want you can use snow’s proof

It also works

[lost laurel]

I would like to know how euclid.'s lemma works here

[gritty terrace]

So, what is Euclid’s lemma?

[lost laurel]

if a|bc , gcd(a,b)=1 a|c

if a|bc , gcd(a,b)=1 a|c

[gritty terrace]

Yes

There’s a special case of Euclid’s lemma I have in mind

What happens if a is prime?

[lost laurel]

it's only factors are 1 and itself

[gritty terrace]

Right

And then, for Euclid’s lemma?

[lost laurel]

c has to be 1

[gritty terrace]

No

[lost laurel]

as does b

[gritty terrace]

Also no

For example, 2 | 4 x 7

So what I’m asking is - suppose p is a prime, and p | bc. What can you use Euclid’s lemma to deduce?

[lost laurel]

the value of c?

[gritty terrace]

Try using Euclid’s lemma

[lost laurel]

p|c

and b is prime

[gritty terrace]

That won’t always be true

For example, 2 | 4 x 7

But 2 does not divide 7

Neither is 4 prime

[lost laurel]

true

in that case, I think I better understand euclid's lemma in greater depth before trying this

can I close this for now?

sorry

[gritty terrace]

Sure

[lost laurel]

.close

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[devout snow]

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[limpid sentinel]

does anyone have a clue as to how i could evaluate the sum of k!(n-k)! to infinity starting from 0 ?

[limpid sentinel]

idk what approach to take

[devout snow]

@limpid sentinel Has your question been resolved?

[limpid sentinel]

<@&286206848099549185> pls

[restive river]

could you show me the question?

nvm i have no clue

[lusty mantle]

what are we iterating

k or n

[limpid sentinel]

k

[limpid sentinel]

thats the queston

[lusty mantle]

and n is just some constant?

[limpid sentinel]

yes but it goes to inf

[lusty mantle]

yea k does but n?

[devout snow]

@limpid sentinel Has your question been resolved?

[limpid sentinel]

no n tends to

[restive river]

i'm just going to slam it into the prompts and see how it works out

[cunning viper]

hello

[devout snow]

@limpid sentinel Has your question been resolved?

[limpid sentinel]

wdym

[devout snow]

@limpid sentinel Has your question been resolved?

[devout snow]

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[subtle kindle]

I have to calculate this double integral

[subtle kindle]

Im not sure how to calculate the area of the integral

like that?

<@&286206848099549185>

<@&286206848099549185>

[copper sonnet]

Integrate it using integration by parts

[subtle kindle]

I tried but I don't know if I'm doing it right

when I differentiate the function, do I do it partially (with respect to x)?

[copper sonnet]

you are complicating things. Set x as you f(x)

It will be much easier

[subtle kindle]

Im not sure what you mean

You mean to integrate the function cos(x^2+y)?

[upper schooner]

(you don't want integration by parts at any point)

[copper sonnet]

Yeah, I am a dumbass. Just use u-sub

[subtle kindle]

u = x^2 + y or u = x^2?

[copper sonnet]

x^2 + y

[subtle kindle]

I'll try to do that

and when I calculate du, I differentiate with respect to x?

[copper sonnet]

Seeing x so pattern recognition suggest that I use integration by parts ☠️

[copper sonnet]

Yeah

just pretends y is a constant

[upper schooner]

it happens, it's kinda deceptive sometimes the insides being x^2 strongly hinted that it'd be some "specially crafted" question

[subtle kindle]

like that?

[copper sonnet]

Yes, that right

From this, it is pretty straight forward

[subtle kindle]

Is it correct so far?

[copper sonnet]

Looks correct

[subtle kindle]

I think I got it

[copper sonnet]

Yea, 2 is correct

[subtle kindle]

Last question: could I write the double integral like this?

I thought it was easier to calculate if it was the other way around

but are they interchangeable?

[copper sonnet]

In this problem, you can change order of integraion

but sometimes, when changing order of integration, you will need to change bound for it

[subtle kindle]

how do you mean?

[copper sonnet]

you can only change dxdy to dydx when it give you the same volume

[subtle kindle]

I understand

thank you!

.close

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[proven wing]

Best websites/books/resources to revise basic probability?

[faint zinc]

@proven wing Khan Academy https://www.khanacademy.org/math/statistics-probability

<@&268886789983436800> ^ sry, pls, thx

[dire forge]

I second khan academy ^^

[sacred hound]

I third khan academy.

But for like, at undergrad level, I might resort on slightly more advanced materials.

Penn state surprisingly has a nice note on statistics and probability you can look for.

For the book, some people would suggest Sheldon Ross' A first course in probability

Rick Durrett's Elementary probability for application is quite nice. But don't confuse with another book Probability:theory and example.

The latter is not introductory.

[proven wing]

Thank u

[devout snow]

@proven wing Has your question been resolved?

[devout snow]

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[solid osprey]

hrlp

like literally every question 🙏

for 25. x=21+1/23

is there a more elegant solution than plugging in x

[crisp zealot]

you could try multiplying the denominator by the conjugate

maybe you get lucky

[solid osprey]

wdym conjugate

rationalizing thing?

[crisp zealot]

yes

[solid osprey]

ok

[dense lynx]

you could plug in x = y + 1/y - 2 where y = 23 but regardless you have to simplify the expression

[solid osprey]

$$\frac{x+2+\sqrt{x^2+4x}}{x+2-\sqrt{x^2+4x}}$$
$$\frac{x+2+\sqrt{x^2+4x}}{x+2-\sqrt{x^2+4x}}\cdot \frac{x+2+\sqrt{x^2+4x}}{x+2+\sqrt{x^2+4x}}$$
$$\frac{2x^2+8x+4+2(x+2)(\sqrt{x^2+4x})}{x^2+4x+4-x^2-4x}$$

whar

[woven radish]

Skill_Issue

[solid osprey]

bot is 4

wait no

wait i think so you just () the x+2

[dense lynx]

yes that's a better idea than expanding

[solid osprey]

441+1/23^2+2×21/23+4×21+4/23

525+46/23+1/23^2
527+1/23^2

how do i simplify this?

[dense lynx]

it's equivalent to (23 - 1/23)^2

[solid osprey]

atleast, root this

[solid osprey]

oh thanks

[dense lynx]

yw

but ye reason like that is why it's convenient to use a diff variable like x = y + 1/y - 2

[solid osprey]

ok

[dense lynx]

but since ur alrd that far i think u can prob continue the problem however works now

[solid osprey]

x^2+4x+3 would be the end

530+1/23^2

ok this is ugly

(x+2)^2-1

did i go wrong somewhere?

axtualy this would be like damn near impossible to cross check lmao

[solid osprey]

how about other questions? i dont have any clue on how to do them

[devout snow]

@solid osprey Has your question been resolved?

[devout snow]

@solid osprey Has your question been resolved?

[tender cobalt]

@solid osprey found the intended way

youre not going to believe what you would have to do

[solid osprey]

what

[tender cobalt]

beforehand, did you simplify the fraction to this yet?

[solid osprey]

yep

[tender cobalt]

then this, right?

[solid osprey]

ye

[tender cobalt]

so this, right?

[solid osprey]

ye but only in my head ive thought of that

[tender cobalt]

look at that on the left

thats the quadratic formula, isnt it

inside the ()^2, I mean

[solid osprey]

it is?

no way

[tender cobalt]

this quadratic formula would be for what equation?

use s for the variable because youre going to square root it in the literal next step

[solid osprey]

n^2-(x)n+1=0

[tender cobalt]

what about s

[solid osprey]

what do you mean

[tender cobalt]

forget anything about s, nvm

now what would ((x + sqrt(x^2 - 4))/2)^2 be a solution to

[solid osprey]

this?

[tender cobalt]

theres a ()^2 there

did you see it?

[solid osprey]

o

hold on

i need to go rq for like 3m

[tender cobalt]

alr

[solid osprey]

@tender cobalt ok im back sorry took lobger than i thought

but uh

[solid osprey]

wont i need to expand this?

[tender cobalt]

no you dont

what would ((x + sqrt(x^2 - 4))/2)^2 be a solution to

you know (x + sqrt(x^2 - 4))/2 would be a solution to n^2 - xn + 1 =0

[solid osprey]

um

idk tbh

[tender cobalt]

how would you solve sqrt(n)^2 + x sqrt(n) + 1 = 0

[solid osprey]

idk

i set a be sqrt(n)

then solve normally

[tender cobalt]

so whats the solution to a^2 + xa + 1 = 0

[solid osprey]

imaginary

:p

[solid osprey]

does this equal to n^4-(x)n^2+1=0

im guessing you mean that?

[tender cobalt]

thats a mistake, also thats wrong

I mean a^2 - xa + 1 = 0

[solid osprey]

owh

[tender cobalt]

which has the solution a = (x + sqrt(x^2 - 4))/2

[solid osprey]

ye

[tender cobalt]

which is distinctly not imaginary

and also distinctly depends on x

[solid osprey]

i read that as a^2+a+1

[tender cobalt]

you assumed x < 2...

when x = 23 + 1/23

[solid osprey]

oh so this thing equals the quadratic thing swuared?

[tender cobalt]

I did inadvertently give you the equation by doing that, yes

[solid osprey]

cause a=sqrtn

oh

[tender cobalt]

n + x sqrt(n) + 1 = 0's solution is n = ((x + sqrt(x^2 - 4))/2)^2

whats x, again

[solid osprey]

23+1/23

[tender cobalt]

n + x sqrt(n) + 1 = 0

sub that in

rewrite n as sqrt(n)^2

rewrite this equation by expanding then factoring it

[solid osprey]

sqrt(n)^2+23sqrt(n)+sqrt(n)/23+1=0

how do i factor this lmao

[tender cobalt]

oh crap mb

[solid osprey]

uh

[tender cobalt]

sqrt(n)^2 - x sqrt(n) + 1 = 0

there we go

[solid osprey]

o

[tender cobalt]

you should also write 1 as 1/23 * 23

and write sqrt(n)/23 as 1/23 * sqrt(n)

to make it more obvious

[solid osprey]

(sqrt(n)-23)(sqrt(n)-1/23)

dis?

[tender cobalt]

yes

[solid osprey]

so n is either 23^2 or 1/23^2

[tender cobalt]

now look at ((x + sqrt(x^2 - 4))/2)^2 and consider which one it is

[solid osprey]

23^2?

[tender cobalt]

yea

[solid osprey]

cause 1/23^2 would mean that x<2

ok cool

thanks

[tender cobalt]

no it wont

it would mean you were looking at ((x - sqrt(x^2 - 4))/2)^2 instead

[solid osprey]

o

alr ty

[tender cobalt]

np

[solid osprey]

i need to go now, ill close this and when i can ill open a new channel

thans for your time, and your intresting sol, ive never seen that, looks kewl

.close

[devout snow]

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[tender cobalt]

np

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[zenith tartan]

Yo

[zenith tartan]

Q18

First question what does N(324, 3136) mean?

Is that the SD and Mean respectively?

[sacred hound]

N(324,3136) means your mean is 324 and your variance is 3136

N(a,b) means you have the mean a and variance b

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@zenith tartan Has your question been resolved?

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[devout snow]

@proper fractal Has your question been resolved?

[devout snow]

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[green crypt]

A chord of circle of radius 15 cm is parallel to the diameter of the same circle the length of the chord is 20 CM distance between chord and diameter is

[long sundial]

Draw it

(cosine rule destroys this)

[green crypt]

Oh ok

So i have it drawn

What do i do from here

[long sundial]

Your goal with problems like this is to form triangles using lines you know the length of

[green crypt]

Yep

[long sundial]

The construction in this case is:

Draw a circle
Draw the chord
Connect the centre to the two endpoints of the chord
The length of those two connections is 15
The length of the chord is 20

[green crypt]

Okay

Hm?

[long sundial]

Yeah then use Pythagoras

[green crypt]

Is it 5√5?

[solid osprey]

yea

[long sundial]

$\sqrt{15^2-10^2}$

[woven radish]

Max

[long sundial]

So yea

[green crypt]

Okay

Thx

.close

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[spring oak]

Option C is true?

[long sundial]

I don't think so, first note: $\cos^2x+\sin^2x=1$ and $\cos2x=1-2\sin^2x$, so we can create the following construction:
[\cos(2x)=(\cos^2x+\sin^2x)-2\sin^2x]

Actually I realized after typing this all we could've used: [\cos2x=\cos^2x-\sin^2x]

[woven radish]

Max

[spring oak]

Which means cos2x is made of other two functions

[long sundial]

Yea

Not linearly independent

[spring oak]

All others are wrong

[tall stirrup]

can you tell me what is the definition of linearly dependent?

[spring oak]

One vector is made of others

We can scale and add

[tall stirrup]

yeah so option c should be correct imo

[spring oak]

.close

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[smoky gyro]

not sure how to start

[long sundial]

What is formula for the sum of n terms?
What is the formula for the limiting sum?

Subtract them

[tall stirrup]

easy from here ig

[smoky gyro]

ohh

wait lol

yeah

true

.close

[devout snow]

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[wooden veldt]

Show your work

[long sundial]

What is the Taylor series of $\ln(1+y)$, then let $y=x^2$, that gives you the Taylor series for $\ln(1+x^2)$, then divide by x^2

[woven radish]

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

[devout snow]

@vocal sable Has your question been resolved?

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[modest lava]

Hi i need to show this :

[modest lava]

$\frac{n+1}{n+2}\leq\frac{k(2n-k+1)}{\sqrt{k(k+1)(2n-k+1)(2n-k+2)}}$

[woven radish]

UzuNGD

[modest lava]

for all integers k and n such that $1\leq k\leq 2n$

[woven radish]

UzuNGD

[modest lava]

Could i get some help

Squarring both sides I got

$\frac{(n+1)^2}{(n+2)^2}\leq \frac{k(2n-k+1)}{(k+1)(2n-k+2)}$

[woven radish]

UzuNGD

[modest lava]

And im stuck here

[long sundial]

What course is this from?

(to know which inequalities are likely to be used here)

[modest lava]

It's combinatorial but it's not the original problem

The problem is to show $C^{k-1}{2n+1} + C^{k+1}{2n+1} \geq 2 \cdot \frac{n+1}{n+2} \cdot C^k_{2n+1}.$

[woven radish]

UzuNGD

[modest lava]

with $1\leq k\leq 2n$

But i managed to get to $\frac{(n+1)^2}{(n+2)^2}\leq \frac{k(2n-k+1)}{(k+1)(2n-k+2)}$

[woven radish]

UzuNGD

[long sundial]

Yeah okay nice because I was about to recommend Cauchy schwarz or AM-GM after finding a manipulation but probably means u can do something even simpler

[modest lava]

Yeah i used AM-GM at the start

[long sundial]

Maybe try phrase this in a combinatorial argument?

[modest lava]

i used AM-GM on this to get $C^{k-1}{2n+1} + C^{k+1}{2n+1} \geq 2 \cdot C^k_{2n+1} \cdot \frac{k(2n-k+1)}{(k+1)(2n-k+2)}$

[woven radish]

UzuNGD

[modest lava]

So the only thing left to do is to prove this

it looks obvious so maybe im just missing something easy but i don't know how to prove it the right way

[long sundial]

You can rewrite this as [\frac{n+1}{n+2}\le \sqrt{\frac{k(2n-k+1)}{(k+1)(2n-k+2)}}]

[woven radish]

Max

[long sundial]

Hmm feels like Cauchy schwarz because of the square root but can't get it to work rn, (haven't got paper)

[modest lava]

Well i had this $\frac{n+1}{n+2}\leq\frac{k(2n-k+1)}{\sqrt{k(k+1)(2n-k+1)(2n-k+2)}}$

[woven radish]

UzuNGD

[modest lava]

i squarred both sides because i thought it would be easier

[long sundial]

Yea

Hmm I think Cauchy schwarz destroys it

Like do Cauchy schwarz on the RHS and you'll get very close or exactly to the RHS, I can't do it rn tho, but just look up CS and see if it works

[modest lava]

Ok i'll try i don't master very well this inequality but we'll see if it xorks

[modest lava]

you mean on this ?

[long sundial]

Yeah or your square version

[modest lava]

ok

[long sundial]

It won't work actually

[modest lava]

?

[long sundial]

I tried it and you don't quite get the correct numbers

[modest lava]

else i was thinking of using Tchebyshev

[long sundial]

Idk that inequality but if you're taught it in your course it's likely that's the trick

[modest lava]

well it's not in the course but i searched for it on the internet

$\frac{a_1b_1+\ldots+a_n b_n}{n} \geq \left( \frac{a_1+\ldots+a_n}{n}\right) \cdot \left( \frac{b_1+\ldots+b_n}{n} \right),$

[woven radish]

UzuNGD

[modest lava]

but i think it doesn't work either

it has nothing to do with it

We can also write : $\frac{(n+1)(n+1)}{(n+2)(n+2)}\leq \frac{k}{(k+1)}\cdot \frac{2n-k+1}{(2n-k+2)}$

[woven radish]

UzuNGD

[modest lava]

$\frac{k}{k+1}$ is capped at 1 and grow bigger when $k$ grows

[woven radish]

UzuNGD

[modest lava]

increase

but if this one increse then the other one decrease

if k increase 2n-k+1 deacrease

and when k=n we have equality

actually not exactly

RHS is still bigger

Maybe i have something

@Max do you think it's alright to say :

I take a function $f(x)=\frac{2x-k+1}{2x-k+2}$

[woven radish]

UzuNGD

[modest lava]

this function is just f(X)=X+1/X+2 but moved along the x axe

so that's the same that n+1/n+2

and then i say that k/k+1 is superior to n+1/n+2 because the function increase

@long sundial

it seems strange

[hidden adder]

try getting it as a quadratic inequality, cubic terms cancel out

[modest lava]

did you do it ?

ok 'i'll try

i developped it if that's what you meant

I get $3n^3k-n^2k^2+9n^2k-7n^3-6n^2-2nk^2+8nk-k^2+k+2n+2-2n^4$

[woven radish]

UzuNGD

[modest lava]

\leq 0

Are you sure it cancels out @hidden adder ?

wait i made a mistake

im redoing it

i get $2n^3+6n^2-4n^2k+2nk^2-8nk+6n+3k^2-3k+2 \leq 0$

[woven radish]

UzuNGD

[long sundial]

Don't quite think so

[long sundial]

Perhaps move everything to one side, might be easier to show something is positive

[modest lava]

doing that i get $4n^2k-2nk^2+8nk-3k^2+3k-2n^3-6n^2-6n-2$

[woven radish]

UzuNGD

[modest lava]

superior to 0

probably i need to factorize and use k<2n

wait actually it's wrong

for n=k=1

rip i probably have to redo the whole exercise

[devout snow]

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[steel fiber]

Hi

[steel fiber]

Can anyone help me with 15.12 question d and e

So far I’ve done the first ones a b and c

But i don’t understand d and e

Like d is basically

-3 = -3 + i0

That makes 0 = i0

And like what the heck is this how can i write this

And the solution is even more weird

[drifting sierra]

What do you find weird about it?

[zenith narwhal]

Complex number can have 0i

[drifting sierra]

sin(180º) = 0 and cos(180º) = -1 so it does give you -3

[zenith narwhal]

Complex number overshadows real numbers

So all real numbers are complex

[jaunty mantle]

That’s true

[devout snow]

@steel fiber Has your question been resolved?

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[steel fiber]

hi

[steel fiber]

again

the channel was closed

[steel fiber]

as i said

-3 = -3 + i0

and solution is

\theta = 180 and

r = v((-3)^2+0^2)

where did -3 came from ? according to the question its 0 + i0

if we move -3 to the other side

[drifting sierra]

It's not?

[steel fiber]

why do we think of it as -3 + 0i not 0 + 0i?

-3 = -3 + i0

-3 + i0 + 3 = 0

therefore, i0 + 0 = 0

no?

[drifting sierra]

It's 0i, not i0, but sure

I don't know what you don't understand about it

[steel fiber]

doesnt matter

[steel fiber]

it asks me to find polar version of i0 or 0i whatevery that is

how am i supposed to do that

and the solution involves -3

but we just made it gone like tf

[drifting sierra]

No, it's asking for the polar form of -3 + 0i

[steel fiber]

it counts as if the -3 on the other side doesnt exist

[steel fiber]

express each of the following complex numbers z in polar form

d ) -3 = -3 + 0i

[steel fiber]

.

[drifting sierra]

This is just telling you that -3 and -3 + 0i are the same number, since 0i is just 0

-3 is a complex number with an imaginary part of 0

[steel fiber]

so i just have to act like it doesnt appear there

like i just have to act like its not written there and the question is -3 + 0i?

[drifting sierra]

I mean yeah if it lets you understand the question better

[steel fiber]

yeah

okay then thanks

also i have got a different question

can u help me with it as well?

its de moivre's theorem

[drifting sierra]

Just open another channel

[steel fiber]

opening 2 channels is not allowe

allowed

[drifting sierra]

Close this one first

[steel fiber]

alright but why its a waste of time i was just wondering whether u have the time to help me or not

if not thats fine

[drifting sierra]

I don't

[steel fiber]

alr

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[zenith tulip]

how is 90/12 = 7,5? i keep finding 7,6

[polar chasm]

How do you keep finding 7.6?

[spring oasis]

!show

[devout snow]

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

[polar chasm]

Anyway, 90 / 12 = 30 / 4 = 15 / 2 = 7.5

[zenith tulip]

7x12 = 84

90 - 84 = 6

[polar chasm]

6/12 = 0.5

[zenith tulip]

why i do 6/12

confused atm

do you

i

just simplify

it

[polar chasm]

90 / 12 = (84 + 6) / 12 = 84 / 12 + 6 / 12 = 7 + 6/12

i thought that's what you did

[polar chasm]

you cant just take this 6 and say it's 7.6

[zenith tulip]

i mean it has remainder, right?

[polar chasm]

the remainder is 6, right

[zenith tulip]

so

how its 5

[eager nova]

It is not 5

[polar chasm]

he means 7.5

[eager nova]

You can also do like this

[polar chasm]

If you have 90 cakes and you are dividing them between 12 children, each child gets 7 cakes and you have 6 cakes left

then you need to divide those 6 cakes among those 12 children again, and add it to the 7 cakes they already have

6/12 = 0.5

[zenith tulip]

hmm

[polar chasm]

so each kid gets a bonus half of cake

[zenith tulip]

i mean

[polar chasm]

and they'll have 7.5 cakes each

[zenith tulip]

i get it but still makes no sense for me idk

oof

[eager nova]

120/12 = 10.
60/12 = 5.
30/12 = 2.5
(120-30)/12 = 10 - 2.5 = 7.5
Or
(60+30)/12 = 5 + 2.5 = 7.5

[zenith tulip]

thank you

maybe i should practice more

i mean

[polar chasm]

are you learning division without remainders?

[zenith tulip]

no i mean

im a math beginner currently at decimals

[polar chasm]

i see

[zenith tulip]

and this is the only question

i cant do division

it feels wrong

for some reason

[polar chasm]

try khan academy, their vids will explain the division

[zenith tulip]

i will check

thank you

.close

[devout snow]

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[native thistle]

can you guys tell me a simplified theory about this?

[acoustic totem]

need help?

[native thistle]

yes

[acoustic totem]

dang

i mean

who doesn't

we all need help at some point in life

everyone should seek help when they need it

[native thistle]

i dont understand why did it become (g(x))^2 either

[fervent cloud]

Cus

f(x)= x^2

f(g(x)) = g(x)^2

[sand dove]

yeah.
f(x) = x^2
so f(whatever you put in it) = (whatever you put in it)^2

take "whatever you put in it = g(x)"

[native thistle]

can you give me another example? cuz idk why the ^2 is outside

is the ^2 from the x^2?

[acoustic totem]

yes

[frozen aurora]

yes

if it helps, read it as (x)^2

[devout snow]

@native thistle Has your question been resolved?

[native thistle]

so of f(x) is 5x+2 it becomes f(g(x)) = (g(5x+2))

[devout snow]

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[frozen aurora]

no, the opposite: 5g(x)+2

[devout snow]

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[frozen aurora]

.close

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[whole hemlock]

I’ve honestly completely forgotten how to do this binomial distribution question

[rose rock]

The critical region is the set of values for which you reject $H_0$

[woven radish]

StrangeQuarkAL

[whole hemlock]

So 0.025 at least either side of 0.25?

Would make me reject it

[rose rock]

No, a probability no more than 0.025 on either side of the distribution

Do you remember how you typically find the test statistic?

[whole hemlock]

I’m good at the questions where it could say “rylan finds out 15 people have allergies, work out if his hypothesis is correct using the critical region” if that makes sense

Because you’d just put P(x<=15)

When you have to mess with the x value it messes me up

[rose rock]

That wouldn't involved the critical region necessarily but It is what I was referring to, so good.

Think about it like this

$$P(X <= c_1) = 0.025$$

$$P(X >= c_2) = 0.025$$
You have to find c. Normally, you would want to find a value of c that gives the greatest P(X <= c) less than 0.025 but that's not what this question askes for.
It says the probability in each tail should be as close to 0.025 as possible.

So all you have to do is test different values of c with P(X <= c) and pick the one that gives the probability closest to 0.025

[woven radish]

StrangeQuarkAL

[whole hemlock]

So like this?

[rose rock]

yeah

[whole hemlock]

Those are the closest values either side

[rose rock]

For the left tail, yes. you still have to do the right tail

[rose rock]

Also, you pick the one that's closer to 0.025 of the two

[rose rock]

That means testing P(X >= c) btw

[whole hemlock]

So like that?

Sorry about camera quality, my iPad isn’t designed for taking phoyos

Photos

And I’d pick the top value of either pair since they are closer

[rose rock]

that's fine. Yeah that's right

[whole hemlock]

It’s also kinda the question after that which throws me

[rose rock]

What is the critical region again?

[whole hemlock]

2.5%

[rose rock]

i mean the definition (plus that's significance level)

[whole hemlock]

If the answer was in the critical region doesnt that mean we reject the Ho

[rose rock]

yes it does. And if it wasn't, we wouldn't reject H_0.
So the probability of rejecting H_0 is the probability that some value lies in the critical region

Your critical region is X <= 6 and X >= 19 right?

[whole hemlock]

Correct, so I’d just right the probabilities for those numbers?

[rose rock]

So you're looking for P(...?)

[whole hemlock]

Between 6 and 19

[rose rock]

You would sum them

[whole hemlock]

Oh.. was the working out I did for the last question enough for 4 marks?

[rose rock]

P(X <= 6 and X >= 19)

[whole hemlock]

Or would I need to write something after

And 0.048 is the answer

[rose rock]

You should definitely do something like "Critical region: X <= 6 and X >= 19"

Other than that, your working seems alright

[whole hemlock]

Alright, thank you

I know I’ve taken up lots of your time already, but could I ask you about a kinematic question?

(And I’m very grateful)

[rose rock]

Sure

[rose rock]

no problem

[whole hemlock]

I believe my part a is correct

Or I hope

[rose rock]

Aside: are these edexcel questions?

[whole hemlock]

They are yes

[rose rock]

Ah they looked familiar

[whole hemlock]

I assume you took a level maths then

For part B- I have all of these values

And I’m working out t somehow, so not sure which to use

[rose rock]

yup

About your part a

[whole hemlock]

Is the displacement meant to be negative

Was a question I had

[rose rock]

It is, yes. Your answer for the displacement is correct

[whole hemlock]

Oh..

I thought I’d done that correctly

[rose rock]

You should probably write afterwards that the height is 98m

[whole hemlock]

Sorted

[celest finch]

Hi

[whole hemlock]

Hi

[rose rock]

Great

[celest finch]

I'd like to study some trigonometry from the beginning, specifically for game development, do you know some good courses face-to-face but online?
I'm pretty lazy to study something myself, I would like to have a tutor

[rose rock]

v = u + at
Remember that you're considering speed so v could be: v <= 24.5 or v >= -24.5

[whole hemlock]

Wouldn’t it just be negative?

Or would I do both

I think my U should be positive

And the V negative

I don’t know why I did that

[rose rock]

As the ball rises, its velocity will drop to and below 24.5, eventually becoming stationery. The negative case occurs on its way down

[whole hemlock]

Ohhh, I completely forgot about that way

[rose rock]

It's easy to miss out these things

[whole hemlock]

If I have a value for each suvat though

How would I know which equation to use?

[rose rock]

You could use anyone in that case but you don't in this case though

Yeah, you don't have the displacement. -98 is the displacement when it hits the ground as opposed to when its velocity is in the specified range

[whole hemlock]

If I work out t when it’s -24.5, it would be 10- that answer for that section right?

[whole hemlock]

Oh right

[rose rock]

That wouldn't be the right length of time. What you actually what to do is find the times for when it's -24.5 AND 24.5. The difference is the length of time you want

[whole hemlock]

I probably shouldn’t say T=x and just stick with t in the future

And the difference is 5, so that should be the anderr

Answer

[rose rock]

This works only because t is continuous(well doesn't break?) within -24.5 < v < 24.5

[rose rock]

Lemme work it out too, one sec

[whole hemlock]

Alright, for part C could you just say not using the exact value of gravity as an easy answer?

[rose rock]

Not sure if I've seen that in the mark scheme before

Might be allowed

[whole hemlock]

Only one way to find out

[rose rock]

The safe bet is air resistance anyway

[whole hemlock]

Oh yea

[rose rock]

Another one is the fact that it's modelled as the particle

[whole hemlock]

That’s usually my go to

I thought I’d always seen in markschemes rounding gravity to 9.8 affects the accuracy

Maybe that was when I did physics

[rose rock]

It's a bit a short while since I looked. I actually finished my alevels 2 weeks ago

[whole hemlock]

Oh you did? Very nice

[rose rock]

Mhmm

[whole hemlock]

Hope you found them okay

[rose rock]

edexcel Maths was pretty good so no complaints there

further was great as well

[whole hemlock]

Ohh the double maths

Hence the good help today

[rose rock]

Ahahah I guess

thanks

[whole hemlock]

That’s probably all the help I need today thanks unless you fancy having a look at a probability one but I’m almost 100% it’s fine

Extremely grateful

[rose rock]

you're very welcome

[rose rock]

Ehhh, never been a fan of stats but if you want to, Idm

[whole hemlock]

That’s fine then, you’ve helped enough

Have a great summer aswell lol

.close

[devout snow]

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[rose rock]

thanks! you too

[devout snow]

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[sinful gulch]

If lim f(x) > 0 can we say f(x)>0

[sinful gulch]

in particular I was trying to prove that if a function's first derivative is positive than that function is increasing

[trail eagle]

This doesn't really make sense. Where are you taking this limit at? And what do you mean by f(x) > 0?

In general, if the limit of a function is greater than 0, then we can't say that that function is greater than 0.

If you can use it, the mean value theorem can help proving what you want.

[sinful gulch]

okay let me elaborate

[sinful gulch]

yeah i know that proof

i had that problem in my exam today

and i wrote

lim h->0 (f(x+h)-f(x))/h > 0 implies f(x+h)-f(x) > 0

[devout snow]

@sinful gulch Has your question been resolved?

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[rough ingot]

Brand new to sequences and am kinda lost

[woven radish]

Ruby

[vale estuary]

a1 is equal to 1

so 3/1 plus 1, 3+1, 4

[rough ingot]

Alright that makes sense

So a2 is equal to 4 right

[vale estuary]

yes

then you do the same thing but a3=3/4 plus 1

or 1.75

[rough ingot]

So why does a3 have the denominator become 4

Because a2=4?

[vale estuary]

because $a_{n+1}=\frac{3}{a_n}+1$

[woven radish]

Ruby

[vale estuary]

when n=2, a3=3/a2 plus 1

[rough ingot]

Ok yea yea, so a3 will be 3/1.75

plus 1

[vale estuary]

a4 will be that

[rough ingot]

Yea that’s what I meant lol, ok I think I got it, thank you 🙏

[vale estuary]

yeah whenever you have an equation that relates the nth term to the n+1th term, you can plug in terms repeatedly to get the next one

[rough ingot]

Yea seems easy enough, just had to see it again to understand it, thanks dawg 🙌

.close

[devout snow]

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[devout snow]

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[jaunty abyss]

screams
How do I do this? We didn’t learn about this in class ughhhh

[jaunty abyss]

Help help help 🥲🥲🥲

I mean it’s a rectangle uh

All of the angles are 90?

Idk how that helps

[radiant anvil]

can u find congruent triangles?

[jaunty abyss]

Oh yes

But not here

Here I’m confused

[jaunty abyss]

Come backkkk

[radiant anvil]

u can find congruent triangles here

[jaunty abyss]

Ohh okay

[jaunty abyss]

Do I make an equation..?

Or smt

[radiant anvil]

find the congruent triangles first

then use those to form an equation

[jaunty abyss]

Are these the triangles??

The congruent ones

[jaunty abyss]

How bc I’ve tried to find one of the red ones using 45+45+3x+8 =180

Did not work

[radiant anvil]

im not sure how u got that..

[radiant anvil]

those are congruent yes

but how about NKM and LMK

[dire forge]

(the triangles they identified would also work, e.g. if we call the center P, just using the fact that KPN and LPM are congruent should work)

[jaunty abyss]

Is there an equation for that

[jaunty abyss]

But idk the other sides lengths

[radiant anvil]

yea i feel it might have been harder

ive to go but

[radiant anvil]

nvm im drunk here

NKM and NLM are cong

from length and angles being same parallel lines n stuff

from there KM=LN and solve

[devout snow]

@jaunty abyss Has your question been resolved?

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[devout snow]

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[hard ocean]

V is vector space , V= R3x3
U is subspace, how do I find an example M matrice?

[tender cobalt]

you only need to find one example of an M matrix?

[hard ocean]

Yea

[tender cobalt]

is that what the problem is asking you, or are you interpreting the problem that way?

[hard ocean]

No its asking me that

[tender cobalt]

think of a matrix that doesnt change anything when you multiply it

[tender cobalt]

its a simple one that youve memorized

[hard ocean]

It cant be scalar matrix

[tender cobalt]

they told you not to use scalar matrices?

[hard ocean]

Yeah

[tender cobalt]

@hard ocean have you considered representing B with a variable in each entry to turn "coming up with B" into "coming up with a solution"?

the equations you get will heavily reduce the amount of guesswork youd otherwise do with B

[hard ocean]

Oh i'll try it

[tender cobalt]

@hard ocean wait

have you tried just putting in this matrix

[hard ocean]

No?

[tender cobalt]

does it work?

[hard ocean]

Oh its right

How did u know

[tender cobalt]

look closer at what would happen if M was also the same matrix

[hard ocean]

Yes the same matrix

Any matrix like this works ?

[devout snow]

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[mental wagon]

Can someone please check this integral

And how In the world is this the correct ans

[faint hearth]

This is wrong

You cant just pull the x+1 out of the integral

[mental wagon]

I substituted the value

[mental wagon]

Oh

[faint hearth]

U changes with x, and x changes with u

[mental wagon]

So how do I do this

[faint hearth]

X+1 is not a constant with respect to u

[mental wagon]

Ok?

How do I intregate the second term then

[faint hearth]

Write $\int \frac{dx}{\sqrt{-x^2-2x+4}} = \int \frac{dx}{\sqrt{5-(x+1)^2}}$

[woven radish]

quickdoom

[mental wagon]

But....

[faint hearth]

I hope you can see where the sin^-1 comes from now.

[mental wagon]

How am I supposed to know that I gotta do this exact same step

[faint hearth]

Practice

[mental wagon]

How in the worl

Am I gon recognize

[faint hearth]

Have you heard of completing the square?

Its a fairly standard method

[mental wagon]

Do you know another name for it

[faint hearth]

And you should know it by now

[mental wagon]

Maybe ik abt it

[mental wagon]

What is it

[faint hearth]

Then you should recognize what i did there

[mental wagon]

Yea I understand the algebra

You did this

Do I also have to memorize the arcsin and all other inverse trig integrals

[faint hearth]

Firstly write :
$\newline -x^2-2x+4=-(x^2+2x-4)$. The expression in the bracket seems very close to $x^2+2x+1=(x+1)^2$, so lets create it, by adding and subtracting 1. So now we have $$-(x^2+2x-4)=-(x^2+2x+1-5)=-((x+1)^2-5)=5-(x+1)^2$$

[woven radish]

quickdoom

[faint hearth]

Yes

[mental wagon]

Ok thanks for that ig

[mental wagon]

.

What Abt this

[faint hearth]

Your choice

[mental wagon]

Someone told these formulas are for grad

Not for hs students

[faint hearth]

Definitely not true

Just read the textbook from which you got those problems. The formulas will be there

[mental wagon]

Yes

Btw Can I test you

[faint hearth]

Okay?

[mental wagon]

What's the integral of arccos

[faint hearth]

Ibp will work, but it will take a minute

Looks like xarccos(x)-sqrt(1-x²)

[mental wagon]

I always have to find them

[faint hearth]

,w differentiate xarccos(x)-sqrt(1-x^2)

[mental wagon]

These derivatives and integral

[woven radish]

Results provided by WolframAlpha

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

Requested by quickdoom

[mental wagon]

I dwrive them from the first step

It's so long

[mental wagon]

Wow

You're a smart undergrad lmao

[faint hearth]

🙂 thanks

[mental wagon]

How do I do this now

[faint hearth]

This is not a hard integral, though

[mental wagon]

I wanna be like you too

Calc pro

Yk

[mental wagon]

😢

It's the hardest one

I've ever seen

[faint hearth]

Practice your book

[mental wagon]

Ok

How much questions to become a pro??

[faint hearth]

When you have to intwgrate inverse trig functions, always do ibp. Let 1 be the function that is being intwgrated and differentiate the inverse function. After that you will have a fully algebraic functipn

[mental wagon]

I also wanna be like, looking at a hard integral, "umm yea this ones pretty ez"

[faint hearth]

Depends on the person. Just practice and do your part