help-17

[dry cedar]

Was bc

1/ (-1/2)^n-1 gives a whole number?

Or its bc they didnt do this in the first place

according to your explination

[outer warren]

missing () around the power

they essentially applied the negative exponent law

[dry cedar]

What do you mean

[outer warren]

and/or relation between multiplication and division

[dry cedar]

Ik a^m-n = a^m/ a^n

[outer warren]

missing ()

you reallly need to get in the habit of typing those in text

[dry cedar]

Ok, just to double check (using this and 'missing ()'

3/ (1/6)^n = 4

(18)^n = 4

and then (n)(ln(18)) = 4

Or no again

[outer warren]

no

how many times do i have to say that
3 * 6^n is NOT 18^n

[dry cedar]

Then could you please explain if you dont mind how they went from

First line, to second line

(dont worry about the third line), they just say RHS is positive therfore LHS must be positive

[its a Geometric Sequence question] ^

[outer warren]

$$\frac{1}{\br{-\frac12}^{n-1}} = 1024$$
$$\frac{1^{\gray{n-1}}}{\br{-2^{-1}}^{n-1}} = 1024$$
$$\br{\frac{1}{-2^{-1}}}^{n-1} = 1024$$
$$(-2)^{n-1} = 1024$$

[twin meteor]

ℝαμΩℕωⅤ

[vocal sleet]

@dry cedar Has your question been resolved?

[dry cedar]

THANK YOU VERY MUCH!!

Im a bit confused on how you got 2nd step to third step?

Hmmm

I dont think thats correct

1/ (-2)^-1 does make -2

But how did you get n-1 outside of it

[rapid swift]

the second step to the third step is the application of the fact that A^n / B^n = (A/B)^n

[dry cedar]

But there is no 1^ n-1

I’m also a bit lost on how there is one but it’s not black meaning you just need to imagine it but what

[rapid swift]

are you aware that 1 = 1^(n-1)?

[dry cedar]

How’d he get that

No

[rapid swift]

pull up a calculator

[dry cedar]

Oh ok

[rapid swift]

try multiplying 1 by itself (n-1) times

[dry cedar]

I can go on forever

=1

[rapid swift]

so there's where the n-1 comes from

[dry cedar]

If I may ask

This is another method but I’m a bit stuck

Can I do it through this way

Or impossible

I can rewrite it

To make it very clear

And more steps srry

[rapid swift]

I can't read what you wrote, sorry

[dry cedar]

,flip

,rotate2

[dry cedar]

@rapid swift I hope this is better

But basically I can’t divide the -ln 1024 with ln -1/2

Is there a way to make ln (-1/2) into an actual number

Is this impossible to solve through this method

due to ln (-1/2)

[rapid swift]

what, do you not have a calculator?

[dry cedar]

or can you change the ln (-1/2) into a number actually usable

[dry cedar]

doesnt work

Error

-ln (1024) / ln (-1/2) = error

[rapid swift]

oh that's because you have ln(-1/2)

[dry cedar]

Is there a way to change it?

Or something

[rapid swift]

I mean yeah if you recognize that the only integer values for (n-1) are positive even

[dry cedar]

mhm ^^ , its geometric sequence question, and n is basically nth term

[rapid swift]

then you replace -1/2 with 1/2

[dry cedar]

OH

[rapid swift]

huh???

[dry cedar]

Its a sequence question

and n is only positive

basicaly

[rapid swift]

they're even

you can't get a positive number from a negative number raised to an odd power

so if you want to solve (-1/2)^n = 1024, you have to realize that n is even anyway

in which case (-1/2)^n = (1/2)^n

[dry cedar]

Yeah but due to magic or whatever 'this server calls it logic'

Due to right hand side being 1024, left hand side has to be turned into positive

[rapid swift]

it doesn't have to

[rapid swift]

it's just something that you can do

[dry cedar]

ikik but for this at least it does

[rapid swift]

no it doesn't

[dry cedar]

and its screwing with my brain

[rapid swift]

you don't have to do anything

[dry cedar]

oh ok

[rapid swift]

it's just convenient

[dry cedar]

Oh

But apparently Its impossible to do otherwise no?

ohhhh maybe not

or something ok

[rapid swift]

wait also wtf are you doing at this point

why do you even need to solve (-1/2)^n = 1024

you started with 1/(-1/2)^n = 1024, no?

[dry cedar]

no?

I did this

so Q, is what I started with

[rapid swift]

oh you have (-1/2)^(n-1) = 1/1024

yeah I see

[dry cedar]

Yeah but then you stil have the issue

of -1/2

1/1024 / -1/2

^ thats what i did originally

And then i thought maybe if it do it the other way around

it might work

🤡

I dont know

[rapid swift]

well, you can't really take logarithms of negative numbers without picking the correct branch

so just change the -1/2 to 1/2 ffs

or do it the easy way and rearrange it to (-2)^(n-1) = 1024

[dry cedar]

If I may ask again lol

Let’s say you had

ln -3/4 = ln -4 + nln3

And you know n was positive

You just make the ln -3/4 into ln 3/4

And ln -4 into ln 4?

[rapid swift]

look

[dry cedar]

Even if it was like

Ln - . Ln - = ln - n

[rapid swift]

you can't take logs of negative numbers willy nilly

it just doesn't work like that

so stop writing them unless you want to go study stuff about complex numbers

[dry cedar]

👍 just asking , this is the first time i heard of ln (-)

Ty

I’ll stick to this

[dry cedar]

.

[rapid swift]

if you know that (-2)^(n-1) = 1024, you can't solve for n?

[vocal sleet]

@dry cedar Has your question been resolved?

[dry cedar]

Wym?

If you leave it as (n-1)ln-2 = ln 1024

You need to make it positive

[rapid swift]

stop writing negative logarithms please

just never do it

[vocal sleet]

@dry cedar Has your question been resolved?

[unique swan]

9/5C + 32 = 9/5(150) + 32 = 302 .... but going in reverse: 5/9(302) - 32 = ~135? What am I doing wrong here?

[vocal sleet]

Channel closed

Closed by @dry cedar

Use .reopen if this was a mistake.

[vocal sleet]

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.

[atomic lance]

How would I approach this question?

[crisp schooner]

You have to use the disk method

A = pi times the integral of the function squared

[atomic lance]

So would the shaded region be the region within the quadratic function and the lines?

So I'm guessing when it is rotated about the x-axis it will form some kind of spherical ellipsoid-like shape

[crisp schooner]

Yes the lower bound is -1/2 and the upper bound is x = 1/2

because you need to find the volume and average y only tells you the average area

[atomic lance]

Oh ok

So the radius would be -2x^2 + 1 right?

[crisp schooner]

yeah

i dont think so actually

[atomic lance]

Like this?

$$V\ =\ \int _{-\frac{1}{2}}^{\frac{1}{2}}\pi \left(-2x^2+1\right)\cdot dy$$

[twin meteor]

Lex1729

[atomic lance]

But I will need the -2x^2 + 1 in terms of y right?

[crisp schooner]

no

thats only if you revolve it around the y-axis

[crisp schooner]

this is right

except for the dy it should be dx

[atomic lance]

Oh I see

$$V\ =\ \int _{-\frac{1}{2}}^{\frac{1}{2}}\pi \left(-2x^2+1\right)^2\cdot dx$$

[twin meteor]

Lex1729

[atomic lance]

But shouldn't the radius be squared?

[crisp schooner]

Yeah sorry it should be

I forget that a lot

Basically it doesn't work because the average value theorem assigns equal weights to each y value but when trying to find the volume when revolving it around the x-axis they definitely dont have equal weights

because of the squared term in the cylinder formula

so its like by averaging it out you're pretty much decreasing it because the original higher ys would make the total volume higher because their squares would increase the volume more than the lower ys would decrease it

You could use it but I think it'd have to be weighted differently

Accounting for the fact that the higher y's will have a squared relationship kinda

it just means you use the original function

also the squared should be squaring the integrand not the entire integral

there are other problems where the y is like 1 and in that case you need to input the function minus the y value as your integrand

say it said y = 1 instead of y = 0

you would do

hold onj

$int\x$

[twin meteor]

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

[crisp schooner]

not sure how to do an integral

its definitely not

its definitely not

https://math.stackexchange.com/questions/1640995/why-cant-i-find-the-volume-of-a-rotated-graph-by-average-value-theorum

that and I'm taking a calculus class and that is definitely not the way you're supposed to do it

[atomic lance]

$$V\ =\ \int {-\frac{1}{2}}^{\frac{1}{2}}\pi \left(-2x^2+1\right)^2\cdot dx=\left[\frac{\left(-2x^2+1\right)^3}{-12x}\pi \right]{{{-\frac{1}{2}}}}^{^{^{^{\frac{1}{2}}}}}$$

[twin meteor]

Lex1729

[atomic lance]

Is this correct?

[crisp schooner]

uhh

you're not really supposed to do that with parenthesis...

[atomic lance]

Do what?

[crisp schooner]

integrate it as if its a single term

also put the pi on the outside of the integral

[atomic lance]

I just used reverse chain rule

[crisp schooner]

any constants that are multiplying the entire integrand you can just multiply it by the integral

[crisp schooner]

with parenthesis it doesn't work like that

because look when you try taking the derivative of that antiderivative

[atomic lance]

??

[crisp schooner]

you're gonna end up doing quotient rule and it's not gonna be the same

[atomic lance]

I don't get it

So how would you integrate this?: $$\int _{-\frac{1}{2}}^{\frac{1}{2}}\pi \left(-2x^2+1\right)^2\cdot dx$$

[twin meteor]

Lex1729

[crisp schooner]

first put the pi on the left of the integral

so $

sorry

[atomic lance]

$$\pi \int _{-\frac{1}{2}}^{\frac{1}{2}}\left(-2x^2+1\right)^2\cdot dx$$

[twin meteor]

Lex1729

[crisp schooner]

Yes

and then expand the polynomial

so you'll get -4x^4 - 4x^2 + 1

[atomic lance]

But why can't I just add one to the power and divide by the new power

then divide by the derivative of the inside function?

[crisp schooner]

uhhh

[atomic lance]

I don't see why that wouldn't work...

[atomic lance]

Okay, but my question is why wouldn't it work?

[crisp schooner]

Idk

[atomic lance]

I know it's wrong, but how is it wrong?

[crisp schooner]

You just have to have it in the correct form

[atomic lance]

How do I know whether it is in the correct form?

[crisp schooner]

Do you know about u-substitution

[atomic lance]

Yes

$$\int _{ }^{ }12\left(7+2x\right)^5\cdot dx$$

[twin meteor]

Lex1729

[atomic lance]

take this for example

[crisp schooner]

ok yeah so

[atomic lance]

Using the reverse chain rule I can add one to the power, then divide by the new power, then divide by the derivative of the inside function to get this:

[crisp schooner]

Ok I think I can explain it

[atomic lance]

$$\int _{ }^{ }12\left(7+2x\right)^5\cdot dx=\left(7+2x\right)^6+C$$

[crisp schooner]

Basically you're skipping a step

[twin meteor]

Lex1729

[crisp schooner]

So with that function you just showed, yes, you can do that

[atomic lance]

Yeah

[crisp schooner]

But that's because the x term becomes a constant when you take the derivative of it and do u sub

[atomic lance]

Oh I see

[crisp schooner]

With the other function it was -2x^2 + 1 so that means dx = du/(-4x)

[atomic lance]

So if the x term doesn't become a constant when I take the derivative of it, does that mean I can't use the reverse chain rule?

[crisp schooner]

yes

Or hold on

With parenthesis yes

But if it's like 2x^2 by itself that's what you're supposed to do

[atomic lance]

What do you mean "by itself"?

[crisp schooner]

If you have a single term that isn't being added to something else and those two things combined have an exponent

and the derivative of that expression ends up with an x

[atomic lance]

So something like this?

$$\int _{ }^{ }\left(2x^2\right)^5\cdot dx$$

[twin meteor]

Lex1729

[crisp schooner]

Yeah exactly

[atomic lance]

Oh ok, but what would be the use using reverse chain rule on that when we can just simplify it to: $$\int _{ }^{ }32x^{10}\cdot dx$$

[twin meteor]

Lex1729

[crisp schooner]

well you use reverse chain rule on that

now that its in the proper form

so that would equal (32/11)x^11 + C

but you can see if you had used antiderivative chain rule with your original (2x^2)^5 it wouldn't become that

it'd be like

((2x^2)^6)/24x

which definitely won't end up with 11 on the denominator

sorry if I knew how to use texit I could show it better but if you just do both ways you can see for yourself they're not equal

you should prob close the channel

[vocal sleet]

@atomic lance Has your question been resolved?

[atomic lance]

Oh ok, that makes sense.

Thanks for helping out 👍

.close

[vocal sleet]

Channel closed

Closed by @atomic lance

Use .reopen if this was a mistake.

[vocal sleet]

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.

[pearl current]

i confused

ohh nvm

i think i got an idea

.close

[vocal sleet]

Channel closed

Closed by @pearl current

Use .reopen if this was a mistake.

[vocal sleet]

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.

[slender ore]

Hey I’m having trouble understanding the process of solving this algebra question.
“Find the value(S) of p for which the graph of y = px^2 +2x-p-2 crosses the x axis twice.”
So I understand if finding p when y=0 but I’m just not sure of how to go about it

[solar needle]

if a quadratic function crosses the x axis twice then it must have 2 real solutions

[sullen thunder]

equate the equation with 0 - and solve how you normally solve a quadratic

[twin meteor]

Talent Unlimited

[slender ore]

Yeah I just thought I’d have to do something different given it’s asking for p

[sullen thunder]

hmm - wait a sec

the question actually asks to find the value(s) of p when the graph of the function crosses the x axis twice, this also means that the quadratic should have 2 distinct real solutions, and this is satisfied when the discriminant is greater than 0

[vocal sleet]

@slender ore Has your question been resolved?

[vocal sleet]

@slender ore Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[vocal sleet]

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.

[lean osprey]

When i plug in pi/2 for part a) i get undefined but i dont understand why that should make sense? What does it have to do with mapping a point near the North Pole?

[vocal sleet]

@lean osprey Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[verbal coyote]

How to find the volume of a rotated function? Using disk, washer, and shell methods?

[vast shale]

We can link you some videos about them if you want. These help channels are more applicable for specific questions rather

[verbal coyote]

Oh ok thank you

Could i get the videos?

[flat whale]

Just enter your search into YouTube

[flat whale]

Any of these words

[verbal coyote]

mm alright! I'll try doing that again!

[vocal sleet]

@verbal coyote Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[shrewd vortex]

Need help for this limit question

[shrewd vortex]

Number 22

My work so far

Am I doing it correct ?

[left lichen]

you dont need to use first principle

directly differentiate

then equate it to 0

[shrewd vortex]

That’s the only way we leanrt

So far

[left lichen]

alright

go ahead

[shrewd vortex]

It’s good so far right

[left lichen]

tell me the final equation

ill tell you

[shrewd vortex]

Okay

Doing it right now

Almost done

[left lichen]

ur taking too much time

honestly

[shrewd vortex]

It’s a big equation

ACC no I should’ve collected like terms first

And then cancel h out

[shrewd vortex]

What did you get

[left lichen]

x^3 - 5 + (4/x^2)

[vocal sleet]

@shrewd vortex Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[noble lynx]

Given that 𝑓(𝑥) = −2𝑥
2 + 6𝑥 − 1, find the expression for 𝑓
−1
(𝑥).

[noble lynx]

Given that 𝑓(𝑥) = −2𝑥^2 + 6𝑥 − 1, find the expression for 𝑓^−1(𝑥).

x = -2y^2 + 6y - 1

so far so good?

what do I do next?

x-1 = -2y^2 + 6y

I don't get what comes next then

do we take y common???

x = y (-2y + 6) - 1

that's not right

I need help figuring it out

[gaunt sparrow]

You would solve for y in terms of x here

But the function isn't injective

[noble lynx]

what's injective?

[gaunt sparrow]

Hum

Horizontal line test?

[noble lynx]

yes yes

[gaunt sparrow]

Yeah so if you think about it

This is a parabola, so the horizontal line test fails.

[noble lynx]

yeah ok so rn its failing the test

[gaunt sparrow]

Taking the inverse is essentially like flipping the axes, and now it means the inverse would fail the vertical line test, i.e. it's not a function because it has two outputs for a single value

You could still solve for y though, you'd just be stuck with two branches, probably when doing some square root operation.

[noble lynx]

oh okay

[noble lynx]

this I dont understand

but what do I do then?

i don't get where to begin'

[gaunt sparrow]

You have this equation

x = -2y^2 + 6y - 1

[noble lynx]

yeah

[gaunt sparrow]

Let's say x was just any constant

Could you solve for y?

[noble lynx]

yup

[gaunt sparrow]

Ok, so do whatever you would do to solve for y, but do it in terms of x

[noble lynx]

iiiii didn't catch that

do we use the quadratic equation or like what?

[gaunt sparrow]

For instance, you can start by moving everything to one side to get

2y^2 + 6y + 1 + x = 0

[noble lynx]

okay

but then it's still not a quadratic cus of the x

[gaunt sparrow]

You should think of x as a constant for now

So here, a = 2

b = 6

and c =

[noble lynx]

1+x?

[gaunt sparrow]

Yep

[noble lynx]

omg tough

-6 + root 6^2 - 4 x 2 x (1+x) / 2 x 2

-6 + root 28 +28x / 4

but like what next??

-6 / 4 * root 28 + 28x/4 is it?

<@&286206848099549185>

[grand spoke]

!orignal

[gaunt sparrow]

Mbmb hum I think I mixed up the sign of 6y in your equation

So the first term would be 6/4

Also this 6^2 - 4 x 2 x (1+x) is not equal to this 28 +28x

[noble lynx]

why won't it be -6/4?

[noble lynx]

oh

[gaunt sparrow]

I swapped the sign without realizing in the first bit above, that would just change the sign of this term

2y^2 + 6y + 1 + x = 0 should've been 2y^2 - 6y + 1 + x = 0

[noble lynx]

oh okok

-6/4 + root -8x + 28 / 4

[noble lynx]

what next tho?

[gaunt sparrow]

There would also be a plus/minus here since we don't know the sign of y (this is the branch thing I talked about earlier)

Apart from some minor simplifications, there's not much else to do

[noble lynx]

okok got it

-6/4 +- root -8x + 28 / 4

[noble lynx]

so am i done?

[gaunt sparrow]

Well you could, say, write -6/4 as -3/2

And take a 2 out of the square root

But that's about it

[noble lynx]

okok ty

[gaunt sparrow]

If you're not given any more info about the domain of x or y in the question, that is

[noble lynx]

uM

one thing

[noble lynx]

if -2x^2 + 6x - 1 was original eq

oh wait nvm

TY

[vocal sleet]

@noble lynx Has your question been resolved?

[noble lynx]

the function f(x) = 2x^3 - 5x^2+ax+b has )2x+3) as a factor, and when f(x) is divided by (x-2) the remainder is 7. Find a and b

<@&286206848099549185>

where do I even begin

[crystal surge]

you can first divide f(x) by 2x+3, since 2x+3 is a factor of f(x), th remainder should be 0

but there will be something else if you actually divide

so equate what comes as the remainder to 0

[noble lynx]

okok

lemme rty

we can use long division right

wait no how do we divide it if the function has ax+b in it???

[crystal surge]

then divide f(x) again with x-2, and set the remainder to 7

[noble lynx]

im so confused

[noble lynx]

this i understand, first step idk

[crystal surge]

you will get something in the remainder

[noble lynx]

remainder will have a+b tho

[crystal surge]

wait, i am giving you screenshot

[noble lynx]

okok

[noble lynx]

ok got it yeah

so thats theremainder

[crystal surge]

now do the same with x-2, except put it as =7 instead of =0

then solve the pair of equation by substitution or elimination method

[noble lynx]

HAHAHA

WHAT

so synthetic division?

2 |

[crystal surge]

?

[noble lynx]

I DONT GET IT

[crystal surge]

wait

[noble lynx]

wym do the same with x -2

put what as 7?

[crystal surge]

just wait

[noble lynx]

okok

[crystal surge]

after a-2 it is written x

[noble lynx]

i dont ge tit

like you divided both ok

what after that?

[noble lynx]

why did you do

(a-2) x + b = 7

[crystal surge]

then subtract both equations

[crystal surge]

like I showed here

got it?

@noble lynx ?

[noble lynx]

omg ok

TY

HONESTLY

I APPRECIATE IT SM

[crystal surge]

Welcome

and if you still have any doubts, you can ask me

[noble lynx]

Show that 3x^3+16x^2-22x can be written in the form (3x+1)(ax^2+bx+c)+d, where a,b,c are constants to be found.

see i get you could like divide f(x) with (3x+1)

but then what would I do about d?

[crystal surge]

ax^2+bx+c is the quotient and d is the remainder (if any)

[noble lynx]

OH

so i can just use long division and divide (3x+1) by 3x^3+16x^2-22x

and d will be the remainder

yes?

[crystal surge]

yes

[noble lynx]

wtf that was so simple compared to last q

i thought this was harder

[crystal surge]

in what grade are you?

[noble lynx]

11th

[crystal surge]

YO

i'm in 10th

[noble lynx]

bro

you finna make me cry im not kidding

where do they make you

[crystal surge]

your country?

[noble lynx]

India

[crystal surge]

hindi?

[noble lynx]

yupp

[crystal surge]

🤝

same

[noble lynx]

WHAT

HOW THE HECK ARE YOU THAT SMART THEN

WHAT DO YOU EAT

[crystal surge]

idk, that's just things are, ig

[noble lynx]

boutta kms

what textbook?

[crystal surge]

NCERT

but I.m studying 11th chemistry rn

organic stuff

and some maths as well

[noble lynx]

what the heck how the hell are you that far into the syllabus

[crystal surge]

curiosity

[noble lynx]

that's honestly so impressive

so you're studying beyond the syllabus rn?

[crystal surge]

yup

and i have my boards in like 2-3 days

[noble lynx]

you're smart as hell im so envious

wish i had your work ethic

[noble lynx]

you'll crush that shit jesus

[crystal surge]

don't get wrong idea

I.m lazy asf

[vocal sleet]

@noble lynx Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[opal flare]

Denote $0=3x^2y-y^3-3z$ and I'm trying to find a plane which is tangent to this surface and also parallel to $24x+7y-z=1$
I'm asked how many solutions are there, find one of the point of contact and find the plain.

[twin meteor]

mtr123

[vocal sleet]

@opal flare Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[fervent sun]

I tried searching it up but I don't know what to do with xi

It's a question about the definition of the definite integral

<@&286206848099549185> Hi!

[wraith python]

$x_i = a + \Delta x \cdot k$

[twin meteor]

Kookiemon

[fervent sun]

Is it possible you could explain why?

[wraith python]

Hmm, this would be easier to explain in a graph. You would have to give me time to make one.

[fervent sun]

If you're willing to I would be very grateful

[wraith python]

Ok, give me about 5 to 10 minutes.

[fervent sun]

👍

[wraith python]

And I wrote that equation wrong, it should be x_k.

https://www.geogebra.org/classic/uqgm5frv

[fervent sun]

Ooh this is cool

[wraith python]

So DeltaX is the length between each point on the x-axis.

[fervent sun]

Okay

That part I understand

Because it's (b-a)/n

[wraith python]

Correct.

The beginning of the interval begins at the left-most point of the interval which is x = 1 in your question.

[fervent sun]

Which is also a in your equation

[wraith python]

Correct.

[fervent sun]

Okay

How do I find the last part of it?

[wraith python]

?

[fervent sun]

x_k

That's the part I don't understand

[wraith python]

That part is DeltaX * k.

You know DeltaX which is 5/n and k is just k.

[fervent sun]

So the answer would be D?

[wraith python]

Correct.

[fervent sun]

Is it possible you could explain why?

Delta X is the change in x which makes sense

How does k come in?

[wraith python]

To calculate the next partition, you would multiple DeltaX by a sequence of numbers 1 through n.

1 + 1/5, 1 + 2/5, 1 + 3/5, 1 + 4/5, 1 + 5/5, etc.

At some point, the value will equal the right-most value of the interval.

[fervent sun]

$(1+\frac{5k}{n})^3

Oo

$(1+\frac{5k}{n})^3$

[twin meteor]

Meph

[fervent sun]

Is that y then?

Or rather

The "height"

[wraith python]

That will calculate the height at each interval point.

[fervent sun]

Okay!!! I got it!

That's really cool

Thank you!

[wraith python]

I'll update the graph later tonight. Check back on it tomorrow. It will have more information to hopefully clarify things.

[fervent sun]

Okay :D

.close

[vocal sleet]

Channel closed

Closed by @fervent sun

Use .reopen if this was a mistake.

[vocal sleet]

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.

[drowsy widget]

Consider strings over the alphabet {a, b, c}. An aa-free string is a string in which there are no substrings aa. Let Tn be the number of aa-free strings of length n. (a) Determine T1 and T2, and develop a recursive expression for Tn

[drowsy widget]

would T_1 be 3 and T_2 be 6?

[hybrid flicker]

T_2 is not 6

How many strings of length 2 are there ?

[drowsy widget]

we have a couple is there a way i can mathematically calculated it

we have 3 options

[hybrid flicker]

Yes so three options per letter

[drowsy widget]

would it be 3 choose 3?

thats just 1

[hybrid flicker]

No

Three options for each letter

[drowsy widget]

333

[hybrid flicker]

We have 2 letters

[drowsy widget]

?

[hybrid flicker]

No

Length 2, so 2 letters

And every time 3 options

[drowsy widget]

3 * 3

[hybrid flicker]

Yes

[drowsy widget]

so we should hv 9

[hybrid flicker]

Out of those, which ones have "aa" ?

[drowsy widget]

well would order matter here

is substring where order matters

[hybrid flicker]

Yes it's consecutive letters

Like in the word "hello", "ll" is a substring but "ho" is not

So, out of those 9 strings of length 2, how many have a substring "aa"?

[drowsy widget]

a is a substring of abc

but is (aa) a substring of (abc)?

its not right

[hybrid flicker]

No

[drowsy widget]

a, b, c, ab, ac, bc, ba, bc, cb?

[hybrid flicker]

You can't make another a magically appear

[drowsy widget]

those are all the substrinfs no

[hybrid flicker]

What are you listing?

[drowsy widget]

am i tripping

[hybrid flicker]

As I said

Order matters

Number of letters matters

[drowsy widget]

yeah thats whhy i listed both sides

[hybrid flicker]

Consecutive letters matter

[drowsy widget]

if order matter ab is dufferent to ba

right

[hybrid flicker]

Yes

[drowsy widget]

a, b, c, ab, ac, bc, ba, bc, cb so wouldnt these be substrings of abc with length 2

[hybrid flicker]

But I still have no idea what you're doing

[drowsy widget]

or am i off

[hybrid flicker]

Length 2

a, b and c are out of the deal

THEN

Consecutive letters

CONSECUTIVE

a and c dont touch each other in "abc"

So no "ac"

[drowsy widget]

lol

[hybrid flicker]

And again

[drowsy widget]

im twipping

[hybrid flicker]

Order matters

[drowsy widget]

i see now

[hybrid flicker]

a comes before b

Not after

So ba is out as well

Etc...

No the only substrings of "abc" of length 2 are "ab" and "bc"

[drowsy widget]

ab, bc,

so where did the 9 come from again

[hybrid flicker]

We're just counting possible strings of length 2

[drowsy widget]

so 9 is possible strings of length 2 in general?

[hybrid flicker]

yes

with 3 letters

aa, ab, ac, ba, bb, bc, ca, cb, cc

[drowsy widget]

but 2 is possible substrings of length 2?

[hybrid flicker]

You got it wrong

A substring is REFERRING to another string

It's a substring OF another string

I thought you were talking about substrings OF the string "abc"

[drowsy widget]

over the alphabet {a, b, c}

[hybrid flicker]

Like, you don’t talk about subsets without saying WHICH set it is a subset of

[hybrid flicker]

Then we talk about strings, not SUBstrings

[drowsy widget]

ok

[hybrid flicker]

There are 9 strings of length 2 made with the alphabet {a,b,c}

[drowsy widget]

aa, ab, ac, ba, bb, bc, ca, cb, cc

[hybrid flicker]

Yes

[drowsy widget]

and 1 is aa

so we actually have 8

[hybrid flicker]

Correct

Any string where you can see "aa" is out of question

[drowsy widget]

i see

so that means with length 3 we have 3 x 3 x 3

[hybrid flicker]

We start with the total amount of possible strings of length 3 yes

And then we remove the "aa"

[drowsy widget]

can i make the recursive expression with just T_1 and T_2

i feel like we need T_3, but T_3 can have at most 27 possibilities which is too much to write manually

[hybrid flicker]

How do you get a string of length n+1, already having a string of length n?

[drowsy widget]

fibonnaci

[hybrid flicker]

No

[drowsy widget]

im not sure then

[hybrid flicker]

Here's how I would do it:

Let's split the Tn strings into 2 groups:

One group has all the Tn strings that end with an a

Example : "cbacbbcba" etc... for n = 9

And the other group has all the Tn strings that Don't end with an a

So for n=9, an example is "cbacbcbac"

[hybrid flicker]

Name the number of strings in that group An

[hybrid flicker]

And the number of strings in that group Bn

Find An+1 and Bn+1 in terms of An and Bn

[drowsy widget]

would this be getting the reccurence expression

or just finding T_n+1?

cant i just do cases:

like case one:

we start with b. in that case we have n-1 length:

or we start with a, in that case we have n-2 length

[hybrid flicker]

Aha, very good question! What is the relation between Tn, An and Bn?

[drowsy widget]

T_n is all aa-free strings of length n

A_n is all T_n that end with a

and B_n is all T_n that dont end with a

[drowsy widget]

right?

[hybrid flicker]

That was an incentive to make you find the relationship between them

As in

Tn is the AMOUNT of aa-free strings of length n

A_n is the AMOUNT of...

So I'm expecting a math relation between those 3

[drowsy widget]

im not the brightest of the bunch so i cant rlly see a vivid relation here

[hybrid flicker]

Well

You split a group of T_n things into two

So when you join them back together, how many do you expect to get?

[drowsy widget]

T_n = a_n + b_n?

[hybrid flicker]

Exactly

So

If we find a recurrence relation for a_n and b_n

We just add them up

[drowsy widget]

how is that allowed

[hybrid flicker]

Why would it not be allowed?

[drowsy widget]

how can we just get the strings that end with a and strings that dont end with a and just say that its equal to t_n?

[hybrid flicker]

Wdym it's equal to T_n

The AMOUNT of strings of length n that dont have aa is T_n

Like

Say I have a class of students

28 dont have glasses

10 have glasses

How many students are there?

[drowsy widget]

oh so can i switch it and do:

T_n = a_n + b_n

where a_n is all t_n where strings end with c

and b_n is all t_n where strings dont end with c

[hybrid flicker]

Sure but there is no point in that

The reason why we look at if a is the final letter in the string is because it's relevant as to if there's gonna be an 'aa' problem when we add letters

[drowsy widget]

13

oh so when we want to add later on

[hybrid flicker]

Ok now you're just trolling

[drowsy widget]

i was jk btw

abt 13

[hybrid flicker]

Bru

[drowsy widget]

lool

srry

i just wanted to see ur reaction

bad time

[hybrid flicker]

So

Find An+1 in terms of An and Bn first

And just try to understand how we can build a string of length n+1, using strings of length n

[drowsy widget]

but isnt finding the ones that end with a redundant

as they just end with a and we know theyre part of t_n

so they cant end with for example aa

[hybrid flicker]

The point is that you might not be allowed to add just any letter to your string t_n

Like if your t_n string ends with a, you can't add another a to make it a string of t_n+1

You see the problem?

[drowsy widget]

ok

i see

i beliebe i got that

yea

[hybrid flicker]

Which is why it's crucial to distinguish between strings that and with a and the others

[drowsy widget]

so if we end with a

we cant add anoither a

we have 2 options

[hybrid flicker]

Yes

[drowsy widget]

mhhm

u make it way smoother

i like it

so now we have t_n = a_n + b_n

how would i go about finding a_n and b_n

[hybrid flicker]

That's the point

Find an+1 in terms of an and bn

This is the easiest out of the 2 in my opinion

How do you build a tn+1 string that ends with a using a tn string?

[drowsy widget]

so as in an+1 = a_n + a_n-1

[hybrid flicker]

Kinda, but

I expect the Right hand side to only make an and bn appear

[drowsy widget]

im ngl im kinda confused a lil bit srry

[hybrid flicker]

Ok

Let's take A3 for example

You have 8 strings in T2

ab, ac, ba, bb, bc, ca, cb, cc

Let me split them into A2 and B2

[drowsy widget]

2 end with 6 dont

[hybrid flicker]

A2 is ba, ca

[drowsy widget]

so 2 + 6?

[hybrid flicker]

Yes

B2 is ac, ab, bb, bc, cb, cc

So

Now we're gonna build a string of length 3 like this

We take a string of length 2, like bb

And we add a 3rd letter to the right

so we can create bba, bbb and bbc

Now, using only strings in A2 or B2 to make a string in A3

Do you think it's possible to make a string in A3 starting from a string in A2, using this method?

[drowsy widget]

A2 we have ba and ca, we can cant we or am i wrong

like cab or cac, and bab and bac

but we cant have caa or baa

[hybrid flicker]

So in terms of making a string in A3

Is it possible?

[drowsy widget]

yes??

[hybrid flicker]

Knowing that a string in A3 has to end with a

[drowsy widget]

oh no

ohh

its impossioble

because we cant end aa

[hybrid flicker]

Exactly

Because if you take a string in A2

It ends with a

And if you want to make it a string in A3

You have to add an a so it ends in a

And so it makes aa

Which doesn't work

So, the only way to make a string in A3 is if...

[drowsy widget]

is... uuhh. leaving out A2 and solely relying on B2?

[hybrid flicker]

Yes exactly

So to make a string in A3

You take a string in B2

And you add an a

So how many strings are in A3?

[drowsy widget]

A_3 = B_2 + 1?

[hybrid flicker]

Why +1?

[drowsy widget]

to add the a?

or not needed

[hybrid flicker]

No, you add the a and the end of each B2 string

[drowsy widget]

oh

so just B_2

in that case

same number of strings

just 1 bigger

[hybrid flicker]

Just 1 bigger in length

[drowsy widget]

yes

[hybrid flicker]

A3 = B2

[drowsy widget]

ye

[hybrid flicker]

Maybe then induce what An+1 is?

[drowsy widget]

A_n+1 = 0 x A_n + B_n

[hybrid flicker]

Exactly

Can you try to find a similar relation for B_n+1?

[drowsy widget]

B_n+1 = A_n + B_n?

or B_n+1 = A_n?

B_n + 1 needs to end with not an a

A_n ends with a but we can add b or c

so i think both is allowed

[hybrid flicker]

Both is allowed, but does each string in An only give us 1 string in Bn+1?

[drowsy widget]

i would say yes but i feel like ur going to say no becauseur asking that question xD

i feel like they would

why wouldnt they

[hybrid flicker]

for each string in A2

ba and ca

How many strings in B3 can we make from them?

[drowsy widget]

ab, ac ?

[hybrid flicker]

?

[drowsy widget]

ik we have 6

[hybrid flicker]

What are all the B3 strings made from A2?

[drowsy widget]

• 2 for A_n

oh B_3

[hybrid flicker]

Yes

[drowsy widget]

ab and ac

so abb, abc, acc and acb

right

[hybrid flicker]

Yes

So how has the size evolved?

[drowsy widget]

x 2

[hybrid flicker]

Yes

[drowsy widget]

or ^2?

[hybrid flicker]

×2

[drowsy widget]

ok

[hybrid flicker]

For each A2 string you either place b or c at the end

[drowsy widget]

so we multiply 2 * A_n?

[hybrid flicker]

So 2 possibilities

Yes Bn+1 = 2An + ...Bn

Try the same on Bn and see what it gets you

[drowsy widget]

ab, ac, bb, bc, cb, cc

abb, abc,
acc, acb
bbb, bbc
bcc, bcb
cbb, cbc
ccc, ccb

[drowsy widget]

also x 2?

[hybrid flicker]

Yes

[drowsy widget]

so b_n+1 = 2x a_n + 2 x b_n

so i have the two equations for an+1 and bn+1

wut shall i do with those

[hybrid flicker]

Ok, now the important part

Here's a little trick that we will use

[drowsy widget]

damn

[hybrid flicker]

Write a_n+2 + b_n+2 is terms of a_n and b_n

[drowsy widget]

a little trick

i like tricks

ok thats what i was thinking

so b_n is = T_n

[hybrid flicker]

B_n is not t_n

Recall that t_n is b_n+a_n

[drowsy widget]

oh

u wrote a_n+2

and b_n+2?

[hybrid flicker]

Yes

The trick is that T_n cannot be obtained with a 1st order recurrence

[drowsy widget]

woukld i do a_n+2 = a_n+1 + a_n?

and do the same thing as before

[hybrid flicker]

Well this is not true

But let me guide you

[drowsy widget]

kk

[hybrid flicker]

Let's first do something simpler

a_n+1 + b_n+1 in terms of an and bn

Easy right? We know the recurrence formula for each of those terms

[drowsy widget]

didnt we already solve that tho

[hybrid flicker]

Well no

We wrote a_n+1 in terms of ...

[drowsy widget]

we have a_n+1 and B_n+1

[hybrid flicker]

Yep

So to find an+1 + bn+1?

[drowsy widget]

ijm not sure my brain is kinda fried

[hybrid flicker]

Ok

x = 2

y = 3

x+y = ?

[drowsy widget]

5

[hybrid flicker]

Yes, how did you obtain that?

[drowsy widget]

getting the linear combination of them

[hybrid flicker]

So summing

[drowsy widget]

yes

[hybrid flicker]

?

[drowsy widget]

oh

we add them

then we will get T_n

i see

[hybrid flicker]

No that's not what I was going for

[drowsy widget]

darn

[hybrid flicker]

A_n+1 = 0 x A_n + B_n

[hybrid flicker]

b_n+1 = 2x a_n + 2 x b_n

a_n+1 + b_n+1 = ?

[drowsy widget]

0 x A_n + B_n + 2x a_n + 2 x b_n

[hybrid flicker]

And regrouping terms?

[drowsy widget]

2a_n + 3b_n

[hybrid flicker]

Yes

Now I want you to notice something very important

[hybrid flicker]

Can we write b_n+1 in terms of Tn directly?

[drowsy widget]

cant we isolate for B_n

[hybrid flicker]

No

[drowsy widget]

nvm

[hybrid flicker]

Bn+1 and bn are not the same

[drowsy widget]

then idk

[hybrid flicker]

Look at the right hand side

And remind me of the relation between an,bn and tn

[drowsy widget]

2x a_n + 2 x b_n this?

[hybrid flicker]

Yes look at this

And recall what tn is

In terms of an and bn

[drowsy widget]

oh so thats the answer no

t_n = 2 a_n + 2 B_n

[hybrid flicker]

No

Tn = An + Bn remember?

[drowsy widget]

yeah

[hybrid flicker]

So...

[hybrid flicker]

This is what?

[drowsy widget]

b_n+1

[hybrid flicker]

Yes, but also....

[hybrid flicker]

If you look at this...

[drowsy widget]

all t_n that doint end with a

im so srry i feel like im letting u down

lol

[hybrid flicker]

Yes

Xd

[drowsy widget]

:9

[hybrid flicker]

Tn = An + Bn

Bn+1 = 2An + 2Bn

See a link?

[drowsy widget]

yeah

we have a_n + b_n

[hybrid flicker]

And so....

[drowsy widget]

im ngl to you

[hybrid flicker]

What's the link between Tn and Bn+1?

[drowsy widget]

they are both aa free

lool

[hybrid flicker]

Bru

[drowsy widget]

that is true

u cant lie

[hybrid flicker]

Multiply it by 2

[drowsy widget]

oh

theyre multipled by

is the only difference

from n to n+1

we just have to multiply by 2

[hybrid flicker]

Listen

[hybrid flicker]

Rewrite this, multiply both sides by 2

[drowsy widget]

2Tn = 2An +2Bn

[hybrid flicker]

Yes

So Bn+1 = ?

[drowsy widget]

2?

[hybrid flicker]

Huh?

Bn+1 = 2An + 2Bn

[hybrid flicker]

Look at this

[hybrid flicker]

Then look at this

Back and forth how many times you wish

[drowsy widget]

im srry id rlly know what to say

2 x Bn+1 = 2 x (2An + 2Bn)

?

[hybrid flicker]

Plsss

[drowsy widget]

lool

[hybrid flicker]

Bn+1 = 2An+2Bn

[hybrid flicker]

And 2An + 2Bn is equal to...

[drowsy widget]

2An + 2Bn = b_n+1

[hybrid flicker]

I literally replied to the message you had to look at

Let's restart

Bn+1 = 2An+2Bn

[hybrid flicker]

And 2An + 2Bn is equal to...

[drowsy widget]

oh do we equate the equations

ohh

b_n+1 = 2T_n

right

[hybrid flicker]

Yes finally

[drowsy widget]

damn

im slow

srry

[hybrid flicker]

Ok now we go 1 up in speed bc i have to go to sleep soon

[drowsy widget]

ok

[hybrid flicker]

we use that :

An+1 = Bn

Bn+1 = 2Tn

[hybrid flicker]

But with the 2nd equation

We can say that Bn = 2Tn-1

This is just reindexing

So An+1 = 2T_n-1

Bn+1 = 2Tn

Finally, add em up

Tn+1 = An+1 + Bn+1 = 2Tn + 2T_n-1

And voilà

[drowsy widget]

oh me oh my