help-10

[subtle whale]

can someone tell me what I did wrong in my steps in trying to find the covariance?

The answer is supposed to be 0.2 according to the steps provided by the book. I tried doing it using the table method as seen in IMAGE1 but I got cov=2 instead

[obtuse pebble]

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[dapper wagon]

How can I start it?

[dapper wagon]

since idk what's z

[obtuse pebble]

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[ebon orbit]

Is it fine to put brackets everywhere in math

[polar fossil]

there are definitely ways that you can write them and cause confusion or cause meaning to change

but using unnecessary brackets when the meaning doesn't change is fine

like if I said (5+3)+7 vs 5+3+7

those mean the same thing, that's fine

[ebon orbit]

Is it fine to do it with these topics

Actually its fine i got it, thanks for the help.

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[wild vigil]

So it says true or false: If The X of f(x)=2*1,3^x gets increased by 2 the f(x) will take 91% of the initial value off

[final thunder]

What do you think

[obtuse pebble]

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[timid silo]

x=-1/2 and y= 1 or x=1 and y=1

[obtuse pebble]

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[midnight orchid]

"solve diffequation" if y(0) = 1

idk how to even start here, very confused

[regal tundra]

dy/dx = x² * y²

You see what to do after this?

[midnight orchid]

not really

never done this type of diffequation

[regal tundra]

You can easily separate th variables here

[dense imp]

you can separate the variables so that you end up with a differential on each side that you can integrate

[midnight orchid]

yes bu

is ther anothe way

cuz ik u can like

get

1/y^2 dy = x^2dx

but i havent done it like that

in school

they havent taught me like this

so imn wondering if theres naother waey to this

[dense imp]

that's pretty much the standard way you would solve this

i mean it is very simple integrals, what's the problem with doing it that way? you just integrate both sides and get

$\int\frac{dy}{y^2}=\int x^2,dx$

[midnight orchid]

yes but

[warm shale]

Soosh

[midnight orchid]

ive never integrated "dx and dy"

ive been taught how to integrate like

uhh

[dense imp]

yes but you see that each one is a separate integral?

once you separate the variables, each one is like an individual problem of simple integration

if i just gave you: $\int x^2,dx$ you would know how to do that right?

[warm shale]

Soosh

[midnight orchid]

yes

x^3/3 ´+ c

right?

[dense imp]

yeah

[midnight orchid]

oh w

wait

so

whe uu

intergare

integrate that

dx gets rmoved aswell

obv

[dense imp]

yes

[midnight orchid]

hm

[dense imp]

and same with dy on the left

[midnight orchid]

how do we know its

dy/dx and not dx/dy

[dense imp]

well im assuming from the problem \ conventions that y is a function of x

you are right that the problem should be more precise in telling us that, but unless told differently, that's kind of the "regular" convention right?

y' = derivative of y right? i don't see any other variables in the problem except x so i guess it must be dy/dx

anyway i suggest you review notation of dy/dx, it is always the "what's being differentiated" at the top, divided by "differentiated in respect to what" on the bottom. also maybe review definitions of differentials (separate dx and dy) and how the notation dy/dx can be thought of as both the derivative and also a ratio of differentials (which allows you to separate them like we did for this problem)

[midnight orchid]

okay honestly

i understand it

evne if i havent learn it

thank you!

[obtuse pebble]

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[high python]

thank you sm

[obtuse pebble]

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[spiral timber]

for this integral would it not be dt at the end? because we are integrating wrt to t?

[spiral timber]

why is it written d(-st) at the end instead of dt?

[tardy epoch]

it's a measure theory thing

you can just think of it as change of variable in this case with

$u = (-st)$ so $du = d(-st) = -s dt$

[warm shale]

riemann

[spiral timber]

so when we write dt at the end we are not saying that it is being differentiated wrt but that dt symbolises something else

im a bit confused

[tardy epoch]

yes

[spiral timber]

if dt was written in place of d(-st) would it mean the same thing or no?

[tardy epoch]

no

[tardy epoch]

if you understand u-substitution for integrals, this is the same thing, just different notation / shortcut

[dense imp]

is it basically saying "differential of (-st)"?

[spiral timber]

so would this be correct use of the d at the end

basically shoving the exponential of the e in the bracket next to d?

[tardy epoch]

right

[dense imp]

yeah as long as A and B are constants in respect to t

[spiral timber]

sorted cheers guys

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[warm shale]

Clippy

[tardy epoch]

can you just show the original problem

[warm shale]

Clippy

[obtuse pebble]

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[honest flicker]

Idk how to do a.

[honest flicker]

Oh my God my connection 💀 😭

SENDDDD

[willow crater]

it can be written as (x+y)^2-2y^2

[naive heron]

is abcd system?

[honest flicker]

Idk what that us

[willow crater]

huh?

[honest flicker]

Is

[naive heron]

or just factor/simplify

[honest flicker]

Oki ty

[willow crater]

and then u can use the property a^2-b^2=....

[honest flicker]

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[grizzled shore]

Complete the square

[shell moth]

i did

thten i substitue whats in the brackets with u, no?

[grizzled shore]

Then use a substitution

[shell moth]

yah i do

but then

idk what to do after

which is where i need help

[grizzled shore]

What form you do want to get this in

What kind of structure allows you to solve this

[shell moth]

1/u^2 + 27/4

[grizzled shore]

$\int \frac{1}{x^2 + 1} , dx$

[shell moth]

i should probably get the 27/4 into a form that can be expressed as a squarE?

[warm shale]

Frosst

[grizzled shore]

Can you do this integral?

[shell moth]

yes

how do i get there though?

[grizzled shore]

So we want the integral to look like that

Do you agree?

[shell moth]

yes

[grizzled shore]

Ok that 27/4 is annoying

We want a 1 there not 27/4

Yeah?

[shell moth]

yes

this is also a formula i was taught to use

[grizzled shore]

Ok what if… we just let 27/4 (u)^2 = (x + 3/2)^2

Then when we do the substitution, we’ll be able to factorise the 27/4 out of the bottom

[shell moth]

do u mean 27/4 + (u)^2

?

[grizzled shore]

So really we want 3sqrt(3)/2 u = x + 3/2

[shell moth]

?

i dont understand

[shell moth]

why would you assume that first line?

[grizzled shore]

See now I’ve just pulled the entire fraction out and it’s an easy time

Because I wanted to pull the 27/4 out of the bottom

[grizzled shore]

Cos I want it to look like this

It’s a bit wack to do integrals especially with substitution

[shell moth]

why are u multiplying the fraction with u?

im so confused

[grizzled shore]

You need to know what you want to do the sub

[grizzled shore]

Because if I do

I can pull the fraction out

I could’ve written let $u = \sqrt{\frac{4}{27}}(x+3/2)$ as well

[warm shale]

Frosst

[grizzled shore]

But it’s a bit less obvious where that sqrt comes from

I mean at the end it is the same

[grizzled shore]

This generalises it’s for a ≠ 1

[shell moth]

i mean

solvinf using this is really easy

but

[grizzled shore]

Then you’re pretty much done if you’re happy using that formula

[shell moth]

i just cant wrap my head around thius moethod

[grizzled shore]

It’s the same thing really

If you can’t get it don’t worry about it unless you really care

It’s the same method in a different wrapper

[shell moth]

do you think you can find a video or website or seomthing that explains this in more detail?

first time i ever see something like this and ive reallly confused

how are you allowed to assume that equality?

and why would you do it in this specific way

[shell moth]

is there liek a formula or smoehting?

[grizzled shore]

Here I’ve proven that it’s the same thing

I’ve just shoved the a into the x term and did a substitution

[grizzled shore]

I get the exact same answer as this

[shell moth]

ok so youre basically factorising

the taking out the common factor

[grizzled shore]

Yes I factorise it into the x term

Then I substitute

But it’s the same thing

You can (and I did) prove they are the same

[shell moth]

@grizzled shore

would it be correct to think of it like this:

you do u = x + 3/2

which gives

integral 1 / [u^2 + (sqrt27/2)^2]

[grizzled shore]

dx

[shell moth]

then you factorise

[grizzled shore]

Yes

[shell moth]

and get

[grizzled shore]

You perhaps use v = sqrt(4/27)u

Yes

[shell moth]

yes

thats what i was getting it

thats also right?

[grizzled shore]

That’s exactly correct

[shell moth]

its lot more intuitive for me that way

[grizzled shore]

It’s like doing 3 + 2 + 5 = 10

Vs 3 + 2 + 5 = 5 + 5 = 10

And the other one you posted earlier is like “I know 5 + 5 is 10”

And 2 + 3 is 5

So the answer is 10

It’s just doing the steps but grouped differently

[shell moth]

i c

thanks alot

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[safe pier]

do your math

[obtuse pebble]

Please don't occupy multiple help channels.

[tardy epoch]

.close

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[ember kestrel]

I am trying to solve exercise 2.8 of Rudin's Principles of Mathematical Analysis, which asks the following: Is every point of every open set E in R^2 a limit point of E?

My proof is as follows: Let p in E and suppose p is not a limit point of E. Then, there exists a neighborhood of p, say N, such that N ∩ E = ∅. Thus, N is a subset of E^c. Therefore, p is in E^c. This is a contradiction since p can not be in both E and E^c at the same time. So we can conclude that p is a limit point of E.

What is the mistake in this approach?

[obtuse pebble]

@ember kestrel Has your question been resolved?

[ember kestrel]

<@&286206848099549185>

[obtuse pebble]

@ember kestrel Has your question been resolved?

[obtuse pebble]

@ember kestrel Has your question been resolved?

[ember kestrel]

<@&286206848099549185>

[obtuse pebble]

@ember kestrel Has your question been resolved?

[fathom flicker]

What is your definition of limit point

[obtuse pebble]

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[limber birch]

anyone knows how to integrate 1/(1+cosθ) to the form of asinθ/cosθ+b + c?

[limber birch]

i converted to 1/2 * sec^2(θ/2), then integrated to tan(θ/2) + c

then im hardstuck there

[tardy epoch]

,tex .half angle

[warm shale]

riemann

[chilly scroll]

you also could've rationalized multiplying 1-cosx both numerator denominator

in the original

[limber birch]

so i integrate 1-cosθ/1-cos^2θ?

[tardy epoch]

You already found the antiderivative here. What more are you trying to do?

[limber birch]

i need to get it in the form of asinθ/cosθ+b + c

[tardy epoch]

Try doing this

[limber birch]

so i integrate cosx-1/cos^2x -1

[chilly scroll]

Express 1 - cos²θ as sin²θ and integrate them serperately

i dont know how to use tex bot sorry

But you should get cosec²θ - cosecθcotθ now you can use the formulas

[obtuse pebble]

@limber birch Has your question been resolved?

[limber birch]

still trying to solve

oh shit

my answer is upside down now

im -cosx-1/sinx

[obtuse pebble]

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[hexed jungle]

what went wrong

[soft tiger]

you’re overcomplicating it

[warm shale]

mud ツ

[obtuse pebble]

@hexed jungle Has your question been resolved?

[hexed jungle]

@soft tiger yes

so i just take the derivative of the given function

[soft tiger]

yes

and you should end up with f(x)

[hexed jungle]

its two terms multiplying so i use product rule right?

[soft tiger]

yes

[hexed jungle]

thats what i did no?

[soft tiger]

how did you get 1/(x-1)

and (x+9)ln|x-1|?

[hexed jungle]

@soft tiger i wrote it on the right

product rule = u'v+uv'

[soft tiger]

yes

[hexed jungle]

derivative of ln(|x-1|) = 1/(x-1)

[soft tiger]

times (x+9)

[hexed jungle]

oh i messed up plugging it back in together

i see

[soft tiger]

yes

[hexed jungle]

i dont like how they use u's and v's, very similar looking characters

[soft tiger]

true

[hexed jungle]

thanks

.close

[obtuse pebble]

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[hexed jungle]

coulda j said messed up the order

was confused for a sec lol

[soft tiger]

my bad, i'll try better next time lol

[hexed jungle]

nah ppreciate u taking time out ur day to look over my work

🙆‍♂️

[obtuse pebble]

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[hearty basalt]

Compute the remainder when
3^333 is divided by 100

[hearty basalt]

whats the approach for this

like i would know how to do it if I calculated 3^333 but I shouldnt be doing that

[gilded drift]

CRT

[gilded drift]

Chinese remainder theorem

[hearty basalt]

chinese remainder theorem

oh ok

[gilded drift]

I keep doing crm because(chainsawman fan)

3^333=3 * (3^4)^83

[pine sail]

mid

[gilded drift]

wait, 3^3 * (3^5)^66

[gilded drift]

27 * (243)^66 has the same remainder when divided by 100 as 27 * (43)^66

[hearty basalt]

hmm ok I'll have to learn how it works

[gilded drift]

27 * (43^3)^22 do this and we get the remainder = 27 * 9^22 mod 100

[gilded drift]

and continue on from there, until you get the value, or you could use some tutorials online because I don't understand chinese remainder theorem

[hearty basalt]

I watched this video, this approach works right?

https://www.youtube.com/watch?v=EiZGCSsKtfM#:~:text=so what is 3 Square,leave the remainder as 9.

ya this approach is a little tough when the base number is big

[obtuse pebble]

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[gentle pumice]

hi i need help with this question, im not sure if i need to ignore cases like abu& abu being chosen.

[gentle pumice]

if i ignore those then it my probability for ai) will be 1/2 right?

but im not sure about the answers when multiplying it. cuz then i will be taking 1/4 x 3/3= 1/4

[obtuse pebble]

@gentle pumice Has your question been resolved?

[obtuse pebble]

@gentle pumice Has your question been resolved?

[obtuse pebble]

@gentle pumice Has your question been resolved?

[marble oar]

WELL IT SAYS TWO PARTICIPANTS REQUIRED, AND DUPLICATES WOULD JUST BE 1 PARTICIPANT, SO I SAY IGNORE THOSE CASES

[marble oar]

YE

[marble oar]

MULTIPLYING IT MATTERS WHEN YOU'RE CONCERNED WITH THE ORDER OF THE EVENTS.
EITHER WAY THE PROBABILITY WORKS OUT TO 1/2

-CHANCE OF ABU BEING CHOSEN FIRST AND BOY-GIRL PAIR = (1/4)(3/3) = 1/4
-CHANCE OF NUR BEING CHOSEN FIRST AND BOY-GIRL PAIR = (1/4)(1/3) = 1/12
-CHANCE OF LISA BEING CHOSEN FIRST AND BOY-GIRL PAIR = (1/4)(1/3) = 1/12
-CHANCE OF PING BEING CHOSEN FIRST AND BOY-GIRL PAIR = (1/4)(1/3) = 1/12

[obtuse pebble]

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[dense geyser]

<@&286206848099549185>

Guide me the way to solve this

[dire basin]

which step can you repeat@dense geyser

[dense geyser]

Can you tell how you did the 1 step last time you told the answer

I dont know that logarithmic rule

[dire basin]

well, let's recall the rule first.

[dense geyser]

Yes

[dire basin]

$$\log MN = \log M + \log N$$

[warm shale]

零下三度极寒

[clever pier]

!15m

[dense geyser]

Yes,

[dire basin]

$$\log \frac{M}{N} = \log M - \log N$$

[warm shale]

零下三度极寒

[lavish minnow]

assume log(x) = m then proceed it will solve easily

[clever pier]

!15minutes

[obtuse pebble]

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

[dire basin]

$$\log M^N= N * \log M$$

[dense geyser]

Oh

[dense geyser]

Yes

[warm shale]

零下三度极寒

[dense geyser]

Yes

[clever pier]

We can do this easily.

There's no need of these here.

[lavish minnow]

not forget to write x= e^m

[dire basin]

$$\log_{a}b = \log_{c}b / \log_{c}a$$

[dense geyser]

Wait, is this also a thing

[warm shale]

零下三度极寒

[dire basin]

yes

[dense geyser]

Oh this rule i didnt know

[dire basin]

in China, we call it bottom swap formula. I'm not sure what is it called in your country.

[dense geyser]

Ok ill too call it bottom swap formula

This is where i finally got , i cant tell what can i do here to simplify this equation

[dire basin]

you can see proof in wikipedia https://en.wikipedia.org/wiki/Logarithm#cite_note-6:~:text=Typical scientific calculators calculate the logarithms to bases 10 and e.[5] Logarithms with respect to any base b can be determined using either of these two logarithms by the previous formula%3A

[dense geyser]

Ok

[dire basin]

Now back to the qn, I'll show again.

[dense geyser]

Yes

[dire basin]

First, cause logx, $x > 0$, Let's try to move the unknown x as far to the left as possible and try to divide the right by the left, then we get $$x^{2\log x-2}=10$$, then $$2\log x-2=\log_{x} 10$$

[warm shale]

零下三度极寒

[dense geyser]

What you did here

[dire basin]

$$\frac{x^{2\log x}}{x^2}=10$$

[warm shale]

零下三度极寒

[dire basin]

understand

?

[dense geyser]

Yes

After this

The last line

[dire basin]

then use Bottom swap formula, we can get $$2\log x-2=\frac{\log 10}{\log x}$$

any question?

at this step

[dense geyser]

Wait let me process

[dire basin]

before x as bottom, now we use 10 as bottom

[warm shale]

零下三度极寒

[dire basin]

a small mistake on the left, I correct'

then our problem actually becomes a quadratic equation correspond $\log x$

[warm shale]

零下三度极寒

[dense geyser]

How you get the exponential 2logx-2 in the base

[dire basin]

cause $\frac{x^a}{x^b}=x^{a-b}$

[warm shale]

零下三度极寒

[dire basin]

this is how we get the exponential

[dense geyser]

I dont get it

I am very sorry if im being irritating

[dense geyser]

Can you show this using the question's example

[dire basin]

you mean you do not understand the the nature of math power exponential?

[dense geyser]

I do

I understand exponents

Im new with logarithms

[dire basin]

ok I will continue. you tell me you where you stuck.

this step has nothing to do with logarithms, donot worry

[dense geyser]

If i will get quadratic , then i can do it myself

[oak sun]

Opps

[dire basin]

yep

sry for my poor English. I am still not sure where you stuck.

[dense geyser]

Look ill show

[oak sun]

He is not able to solve quadratic equation

[dense geyser]

I dont get the relationship in these 2 equations

The 2nd one is bottom swap

[oak sun]

Log(a) /log(b) = log(a-b)

[dense geyser]

It will help me if you tell me the a b and c in the equation

[dire basin]

No,the step to get the exponential $2\log x-2$ did not use bottom swap

[warm shale]

零下三度极寒

[dense geyser]

Ok it used exponential law

[dire basin]

yes

I just use the exponential law on the left first

and we get $$x^{2\log x-2}=10$$

[warm shale]

零下三度极寒

[dense geyser]

Yes

And then you applied bottom swap

[dire basin]

No wait, now we just according to the Definition of logarithm

if $x^a=b$, then $a = \log_{x} b$

[warm shale]

零下三度极寒

[dire basin]

so we get socalled "a" here replace with $2\log x-2=\log_{x} 10$

[warm shale]

零下三度极寒

[dense geyser]

Yes

[dire basin]

any question

okok

[dense geyser]

No question till here

[dire basin]

we continue, now we use exactly the bottom swap just on the right

before the bottom is x, we use 10 as new bottom now!

[dense geyser]

Ok

[dire basin]

It should be noted that you can use any bottom. like 2 3 or even e

[dense geyser]

Ok we will use common bottom > 10

[dire basin]

but you noted that the left is $\log x$

[warm shale]

零下三度极寒

[dire basin]

so we use bottom 10

[dense geyser]

Yes

[dire basin]

no need to > 10, 2 also can use

[dense geyser]

Yes i mean arrow there but ok

10 is easy

[dire basin]

oh, 🤣 sry for my misconfused maybe you use → be better

[dense geyser]

I cant find that on my key board

Anyway

[dire basin]

okok, anyway now we can change the right to $$\frac{\log 10}{\log x}$$

[warm shale]

零下三度极寒

[dense geyser]

Yes

[dire basin]

now after this, I think you can make it.

just multiply $\log x$ to the left

[warm shale]

零下三度极寒

[dense geyser]

How does log x multiply with 2 log x

Is that 2 log ² x²

[dire basin]

logx can be seen as t=logx

[dense geyser]

Oh ok

[dire basin]

then $$(2t-2)t=\log 10$$

[warm shale]

零下三度极寒

[dense geyser]

Yes

[dire basin]

Think of logx as a variable like abc. here is t

so you can solve this equation to get t

[oak sun]

I was waiting to speak, but you solved the whole, great @dire basin

[dire basin]

then $x=10^t$

[warm shale]

零下三度极寒

[dense geyser]

Yes

I got 2t²-2t=1

[dire basin]

True Dudu

[dense geyser]

Should i replace log x here?

[oak sun]

No

Later

[dire basin]

No we have already replace logx before

[oak sun]

After getting the value of t

[dire basin]

yes @oak sun

Anyway you can try to replace logx here, you will find you move back to last step

[dense geyser]

Yes

[high python]

can you please help me with an exercise

[dire basin]

So just solve the equation correspond to t. I cannot wait it.

[high python]

if you r done

[oak sun]

Of course

[dense geyser]

Ok i have to solve for x in 2t²-2t-1=0

Ok thanks

🙂

.close

[obtuse pebble]

Channel closed

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[dense geyser]

Repost it

[obtuse pebble]

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[high python]

please help me

[oak sun]

What is the meaning of "montrer que"

@high python

[high python]

show that

it is in french

[oak sun]

Ok

[oak sun]

😅

[high python]

I tried using recurrence but it didn't work

[dire basin]

@dense geyser Ok, man, if you have any questions you can keep asking me, I'm an electrical engineer and just came to this community and would love to answer math questions because we deal with math a lot in our daily lives and math has caused us permanent head damage (PhD)🤣

[high python]

do you have an idea

[oak sun]

Me too

[high python]

?

[oak sun]

Wait a minute

[high python]

okay

thank youuu

[oak sun]

Okkk

There are 2 methods to solve it

[high python]

okay

[oak sun]

Use the concert of mathematical induction

For natural nums

Show that RHS will never be equal to 1

And the other method using progerssion

[high python]

is induction the same as recurrence ?

[oak sun]

Nope

[high python]

is there a simple way

[oak sun]

Nope

[high python]

something i would know

okay

[oak sun]

That's very complicated

[high python]

Thankyou so much

[dire basin]

what's the method using progerssion? Is this word "progerssion" spelled right. I cannot translate it right here myself.

[high python]

it is progression my friend 进展

[dire basin]

Oh, I see

Thx

[high python]

wlcm

you are chinese ?

[dire basin]

Yep

[high python]

which city r u from ?

[dire basin]

I was graduated from Wuhan. Now I come back to my Hometown.

[high python]

I am in Zhangzhou

[final thunder]

Ive not a contradiction proof for you.

[high python]

i love it here

[dire basin]

You mean Zhengzhou

?

[high python]

yes

[dire basin]

Oh so close.

[high python]

which is ?

[high python]

rly

cool

[dire basin]

So are you an international student?

[high python]

something like that

what about you

are u still a student

[dire basin]

No I join work in July this year

Electricity Engineering

[high python]

cool

I wich you luck

[dire basin]

No wonder I see the style of this math problem very similar to that of Chinese teachers.

[high python]

you seem nice

[dire basin]

Thx. I'm very glad to help others with math problems

[high python]

oh right

[high python]

do you have an english name ?

[dire basin]

Oh. This problem is actually really hard for most Chinese students.

[high python]

It is hard indeed

@final thunder will help me

[warm shale]

THEBATCH

[high python]

I hope

[dire basin]

haha

[final thunder]

To prove: every integer n can be expressed as $2^p(2q+1) \$
The cases where n is a power of two and where n is odd are trivial.
Consider the case where n is an even number and $n \neq 2^p$.
Assume to the contrary then there are numbers that cannot be expressed this way.
Then by well ordering principle, there must be a smallest such number $n$.
As n is even we can consider $2m = n$, which can be expressed in the form $2^p(2q+1)$ since $m<n$.
But then $n = 2^{p+1}(2q+1)$

[warm shale]

$Pure$

[high python]

how do you know that 2m can be expressed in the form 2^(p)(2q+1)

take 18 for example

[final thunder]

This is true by our assumption

since n is the smallest integer that cant be expressed in this way, and m < n we can write m in that way

[final thunder]

then m = 9 which is an odd number so clearly you can write it that way

2^0(2*4+1)

[high python]

yes yes I see

I still don't get the m<n part

why did we take n as the smallest

[final thunder]

Because if n is no the smallest element we can't be sure that m = n/2 can be expressed in that way

but since n is the smallest element that can't be expressed in that way m = n/2 definitely can be expressed in that way

[high python]

okay

thank you so much

[final thunder]

np

[high python]

I have another exercise if yo udon't mind

[final thunder]

you can post it

but ill have to go for a bit

[high python]

okay its fine

it says : let......with.....show that

[obtuse pebble]

@high python Has your question been resolved?

[dire basin]

I just take a shower and come back. And I have a little bit of an idea. Let's take a look together.@high python

First, we have $$xy+yz+xz=1$$

[warm shale]

零下三度极寒

[dire basin]

Square both sides, get $$(xy+yz+xz)^2=1$$

[warm shale]

零下三度极寒

[dire basin]

expand left sides, get $$x^2y^2+y^2z^2+x^2z^2+2y^2xz+2x^2yz+2z^2xy=1$$

[warm shale]

零下三度极寒

[obtuse pebble]

Channel closed

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[dire basin]

If you are still here, reopen the channel. @high python

[obtuse pebble]

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[waxen slate]

2x-\sqrt{2x^2}\cdot 4x+9=3 pls help

[waxen slate]

<@&286206848099549185>

[wise fossil]

what are you looking for ?

[waxen slate]

idk

[wise fossil]

haha are u solving for x ?

[waxen slate]

yes

[wise fossil]

can u send a ss of the qn cuz the text is kina confusing?

[ember frost]

$2x-\sqrt{2x^2}\cdot 4x+9=3$

[warm shale]

FungusDesu

[waxen slate]

yea

[wise fossil]

okie

[waxen slate]

complete solution

[ember frost]

take sqrt(x^2) out

$\sqrt(x^2)=|x|$

[warm shale]

FungusDesu

[waxen slate]

<@&286206848099549185>

[wise fossil]

wouldnt i tbe easier to expand the wholething first ?

[waxen slate]

pls help

to this

[wise fossil]

$4x^2-9x-(3/sqrt(2) - 2=0$

solve that

[warm shale]

literalyanything

[ember frost]

$4x^2-9x-(3\sqrt{2} - 2=0$

[warm shale]

FungusDesu

[wise fossil]

im still kina new to the server so trynna figure it out but thanks

[wise fossil]

sqrt 2 -2 is dividing 3

[ember frost]

im not sure where you get this from

[wise fossil]

break the sqrt 2x^2 down int xrt 2 then u move some stuff around

[obtuse pebble]

@waxen slate Has your question been resolved?

[wise fossil]

@waxen slate does it make sense now

[obtuse pebble]

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[obtuse pebble]

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[errant basin]

how do you do this without using triangle identities?

[errant basin]

ik you can just use the 30 60 90 triangle

but is it possible without it

[willow ravine]

What do you mean by “triangle identities”?

[errant basin]

the 30 60 90 triangle

[willow ravine]

I think all they did was divide by

sin(θ)

To get cos(θ) / sin(θ) = sqft(3)

Which is cot(θ) = start(3)

Then did the inverse of cotangemt on both sides

[errant basin]

um

so its only possible with a calculator in that case?

[obtuse pebble]

@errant basin Has your question been resolved?

[obtuse pebble]

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[bitter wraith]

Hi

[bitter wraith]

!closed

!solved

[sage lily]

@bitter wraith type .close

[bitter wraith]

.close

[obtuse pebble]

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[balmy meteor]

So I have a question about the limit comparison test. So you start out with a series and then you find a series that seems like it, such as a p series. If you do the limit comparison test and you get a finite non zero number, then they both do the same thing. What happens if you get infinity or 0 from the limit comparison test? How do you know what happens with the given series?

[balmy meteor]

I remember my professor said that if the limit comparison test gives you infinity or 0, and the p series diverges, the original series diverges as well, are there any other rules for this? Like what happens if you get infinity or 0 but the p series converges, does this mean that the original converges as well? Or does that just not tell us anything about it

[obtuse pebble]

@balmy meteor Has your question been resolved?

[balmy meteor]

<@&286206848099549185>

[obtuse pebble]

@balmy meteor Has your question been resolved?

[obtuse pebble]

@balmy meteor Has your question been resolved?

[obtuse pebble]

@balmy meteor Has your question been resolved?

[balmy meteor]

<@&286206848099549185>

[honest reef]

If the limit of the ratio of the terms of your series to the comparison series is zero, and the comparison series is convergent, then your series also converges

If the limit is infinity, it means your series’ terms are much larger than the comparison series’ terms. In this case, if your comparison series is divergent, your series will also diverge

[obtuse pebble]

@balmy meteor Has your question been resolved?

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[balmy meteor]

Thank you so much

[obtuse pebble]

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[placid smelt]

R we looking for the z value that corresponds to 0.81 or 0.19 ?

[native inlet]

0.81 because we are looking for what percent of the curve lies to the left of 81%

[placid smelt]

ok ty

would that value be 0.880?

[native inlet]

well I don't know z-score values off of the top of my head

but just look at your table :)

[placid smelt]

kk ty

.close

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[fathom sentinel]

can someone help me with this

[obtuse pebble]

Please don't occupy multiple help channels.

[fathom sentinel]

[fathom flicker]

.close

[obtuse pebble]

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[fathom flicker]

please stick to one channel #help-13

!occupied

[obtuse pebble]

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).

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[dull light]

for the last part

when u simplify u get:

sqrt(x^2)/(-10x)

which is just x/(-10x)

which is -1/10

correct answer is 1/10, where does the extra negative come from?

[violet sentinel]

nah it's definitely -1/10

oh wait

it's the limit to -inf

[dull light]

yeah i thin i now what i did wrong

1 sec

sqrt(-infinity)^2/(-10(-inf))

= sqrt(inf^2)/(10inf)
= inf/(10inf)

= 1/10

?

[violet sentinel]

yep 🙂

[dull light]

aight than u

.close

[obtuse pebble]

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[dense geyser]

Thanks man youre so nice , i have solved it already

[obtuse pebble]

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[tranquil dirge]

How do I solve part b?

[tranquil dirge]

I feel like I may be algebraically wrong lol

I did
257=-127.9+81.91Ln(x)
384.9=81.91Ln(x)
4.7=Ln(x)

[shut lagoon]

This looks alright. You just need to solve for x by exponentiating both sides now.

[tranquil dirge]

then I did Ln(4.7) which got me 1.5

my brain not really wrapping around how I put that back into the problem

[shut lagoon]

Oh but it's 257000 right

So you'll get a larger number

Mb I didn't read it properly

[tranquil dirge]

ur okai

[shut lagoon]

So yeah just do the same calculations as you did but with 257k instead of 257 and you should get a number of years that makes more sense.

[tranquil dirge]

.close

[obtuse pebble]

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[tiny tusk]

im kinda new to this whole trigonometry thing 🥲 need sm help

[sage dagger]

thats some neat handwriting you got there

[unreal musk]

You remember how sec and cosec are defined?

[tiny tusk]

Thanks g

[tiny tusk]

Well yeah ig

[unreal musk]

Uh-huh, how are they defined?

[tiny tusk]

Sec∅= 1/cos∅ and cosec∅=1/sin∅

Ryt

[unreal musk]

Cool cool, maybe try putting those in?

[tiny tusk]

Well yeah i kinda did

It left me with

Wait lemme show

Okay wait i think i myt hv got it

Aand done thank youu

[unreal musk]

Well done

[tiny tusk]

.close

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[weak light]

im pretty sure i am supposed to find the area in terms of y but idk how to set it up

[weak light]

ive also asked multiple people at my math library/tutoring program and they havent been able to get it either

[cursive creek]

do you know the formula for volume of a solid after being revolved around x/y-axis

[weak light]

yeah

[cursive creek]

you would have to calculate the definite integral from 0-1 for (e^-x^2)^2

and then multiply by pi at end

wait nvm since it's revolved around y-axis the formula actually needs to be f(y)^2 inside the integral

so you need to find f(y) and then same process

[weak light]

how do i deal with the x=1

because that is one of the curves that specifies the region

wait

nbope

[weak light]

.

[cursive creek]

do you get to see the answer?

[weak light]

got it

another volume question, how do i find the bounds

[obtuse pebble]

@weak light Has your question been resolved?

[obtuse pebble]

@weak light Has your question been resolved?

[obtuse pebble]

@weak light Has your question been resolved?

[old nymph]

Isnt it given

wait i think its improper

[obtuse pebble]

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[fathom flicker]

Could anyone check my proof for this?

the inequality should just be for n^2 btw, not n^2/2, small typo

[spice citrus]

Doesn't applying AM-GM on both factors also work?

[kind hawk]

maybe as a small note, those roots are real and positive. small technicality but needed for am gm

[fathom flicker]

hm but I wasn't using AM GM I was using cauchy schwarz, do they need to be positive for that?

[fathom flicker]

likely

[kind hawk]

wait I read amgm from jelle and mixed it up

sry

[fathom flicker]

So I'm good then?

[kind hawk]

yes

[fathom flicker]

Thanks guys

I have another one but I'll just open a new channel

.close

[obtuse pebble]

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[strange cape]

I have no idea on question 4b

[frozen thistle]

did you derive g?

[strange cape]

yes

[frozen thistle]

i think you made an error

the derivative of (n+1-x)^(n+1) should be negative

same with (n-x)^n

[fathom flicker]

Is this a test?

[frozen thistle]

looks like one

[strange cape]

nah just a homework

[frozen thistle]

@strange cape

you should try deriving g again

i just did, and the answer afterwards was pretty clear

[strange cape]

Alright

[frozen thistle]

in the second line, you factorised by (n-x)^n when the otherterm only has (n-x)^(n-1)

[strange cape]

and I put (n-x)^-1

[frozen thistle]

yeah my bad

thats correct, just develop the final expression

[strange cape]

I am thinking the relationship between the question

[frozen thistle]

once we find the monotonie of this function, we can use it to compare the 2 terms

since if you divide the second expression by the first, you'll get g(x) * [(n)^n]/[(n+1)^(n+1)]

[strange cape]

sorry , still thinking, I got no idea

[frozen thistle]

np, did u finish g'

[strange cape]

You mean the final ans?

[frozen thistle]

of this yes

[frozen thistle]

the n gets cancelled out too so only x remains but yeah this is correct

so what is the sign of g'?

[strange cape]

😓 no sense

[frozen thistle]

Is g' positive or negative?

[strange cape]

negative

[frozen thistle]

No, it's positive.
Look at the third line
There's a -n and +n

[strange cape]

yes but x<n

[frozen thistle]

Look at the third line, all the terms cancel out except X

$-n² + nx - n + x + n² + n - nx= x$

[warm shale]

lotus man

[strange cape]

oh nothing

the stupid mistake

[frozen thistle]

its ok

just make sure to double check your calculations

so g' > 0

[strange cape]

yes for now

[frozen thistle]

which means g is a strictly increasing function

so
For every x>0: g(x) > g(0)

and what's g(0)?

[strange cape]

I know that is zero slope

[frozen thistle]

i'm sorry i have to go

[strange cape]

((n+1)/n)^n

alright thanks a lot

[frozen thistle]

you can solve this question by dividing the second expression by the first one, it'll be written in terms of g(x). sorry i couldn't help you find the final solution

[strange cape]

sure thanks a lot!

[obtuse pebble]

@strange cape Has your question been resolved?

[strange cape]

.close

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[maiden bone]

i want to ask a dumb question
while factoring a number why only whole number matters?
why not decimal numbers?

google says its just because for precise calculations
but terminating decimal numbers are also precise?

[brazen gorge]

factoring a number means finding its prime factorizations. prime numbers are whole numbers

also, there is infinitely many ways to "factorise" any number if you include decimals

[maiden bone]

hmm this makes sense but
we can use fraction numbers to factorize , that would not create any problem?

[brazen gorge]

wdym?

show example

[chrome narwhal]

It wouldn’t cause problems per se, you could do it but it wouldn’t be very interesting because there’s like an infinite number of ways to do it for any number

It’s less that it’s wrong to do so, more that it’s just kinda pointless

So we only consider integer factorisations

[brazen gorge]

what you might be referring to is just division lol

[chrome narwhal]

We say that integers a and b are factors of x if ab=x

There doesn’t exist an integer a such that 10*a=24

It’s a definition

If you allow for fractions then there’s an infinite amount of choices a and b such that ab=x for any x so it’s not a particularly interesting thing to talk about

So yeah we define it to just be whole numbers

[maiden bone]

yeah just make things easy?

[chrome narwhal]

Not necessarily easy, more so that it makes it useful to talk about

[maiden bone]

And as now that a channel has opened, I have to ask one more dumb question . may i ?

[chrome narwhal]

Go ahead, no such thing as dumb questions

[warm canopy]

It's so we don't have to write down a million parentheses everytime we write down an expression

[maiden bone]

could you please explain?

[warm canopy]

Without any convention you would have to write 2+5*4 as 2+(5*4)

[maiden bone]

that what i am trying to say why do we even need parentheses ?
or BODMAS to obtain a desired answer?
in short i am just wondering that how can there be many different answers of a same equation and we need BODMAS to obtain a uniform answer every time ?
i am sorry for the senseless question , f* me

[warm canopy]

How would you know how to interpret 2+5*4 without parentheses or a convention like bodmas?

[chrome narwhal]

It’s just a convention, its like a non official agreement to make reading maths a little easier and more consistent without needing to use loads of brackets

[maiden bone]

hmmmmm
.............
thank you sir
@chrome narwhal @warm canopy

.close

[obtuse pebble]

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[thick oracle]

What kite angle

[obtuse pebble]

@timid silo Has your question been resolved?

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[steel urchin]

the last bit the h||e|^2

im just confused about how this is squared on the last bit i thought it was gonna be dropped because the l^2 is the space of square integrable functions right?

so you drop the square i thought? IDK

i mean it would make more sense to me if it was ||e^2| l^2

[obtuse pebble]

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[steel urchin]

.close

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[wild elm]

Hey. I found the questions while preparing for the Waterloo math test and I'm in 11th. Can someone help with the question because I do not even know where to start with such a question.

[obtuse pebble]

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[timid silo]

help

[timid silo]

pls

ok i have multiple questions for this one.

lol

i have abt 6

so this is one of my study questions

and this.

these are the first two i need help with

[pseudo swift]

right

what about the first one, any idea or no ?

[timid silo]

i think the first blank is "sum"

[pseudo swift]

I was talking about the first pic

sorry

[timid silo]

oh-

yeah no i got no clue.

[pseudo swift]

have you seen what polynomials are ?

[timid silo]

should i just put all the questions i have?

[timid silo]

ummmm i remember reviewing

[pseudo swift]

bit by bit is better, it's long to cycle between 6 images at a time

[timid silo]

ok then!

:))

so im thinking we replace x with one of the numbers right?

but solving it is very hard for me

I'm kinda slow.

[pseudo swift]

well a polynomial is some combination of 1, x, x^2, x^3, etc...

[timid silo]

yeah....

[pseudo swift]

the highest power that shows up in the polynomial is called the degree

[timid silo]

yup!

[pseudo swift]

so for example 5x^4 + x + 1, the degree is 4, the highest power of x is 4

[timid silo]

you lost me.

OH

nvm

lol

wait so it's 4?

[pseudo swift]

that's just an example, it's not the polynomial in the pic you sent

[timid silo]

oh

[pseudo swift]

so what about the thing in the pic now

-7x^2 + 2x^5 + 3 - 8x + 4x^3 - x^4

[timid silo]

um

.

got no clue

im a stupid junior 😭

wait so how would i even find the answer?

[pseudo swift]

well what are the powers of x that show up in that polynomial ?

[timid silo]

do i replace x with one of the numbers?

[pseudo swift]

no

[timid silo]

3, 4, 5, and 2

[pseudo swift]

(well technically there's also 1 (cause you have -8x^1) and 0 (from the +3x^0), but that's detail)

ok

[pseudo swift]

so what's the highest power here?

[timid silo]

5

[pseudo swift]

yep

and that's your answer

[timid silo]

oh!

[pseudo swift]

the degree of the polynomial -7x^2 + 2x^5 + 3 - 8x + 4x^3 - x^4 is 5

[timid silo]

OHHHHHHHH

OK!

:))

[timid silo]

so for this one, im pretty sure the first blank is "sum"

[pseudo swift]

is there a selection to choose from ?

or do you have to input yourself

[timid silo]

input myself

but i think it's sum for the first blank

[pseudo swift]

yeah that question is a bit crap but ok

[pseudo swift]

seems like a good idea

[timid silo]

ikr 😭

thank god it's only a study question!

[pseudo swift]

"the sum of multiplicities [of the roots]" vs "the degree [of the polynomial]"

can you relate those two ?

is one bigger than the other ?

are they equal ?

[timid silo]

um

???

i dont understand the question you are asking

[pseudo swift]

well you have to fill in the second blank right

[timid silo]

yes

[pseudo swift]

what I'm telling you is they're certainly expecting "equals" or "is less than" or "is more than" for that blank

[timid silo]

oh

ok i think i got it