help-10

[ruby flame]

then

[latent walrus]

what?

what is (sqrt(2))^2

[ruby flame]

Oh that suppose to be a 3

[latent walrus]

there shouldnt be a sqrt2 under it anyway, the dot is multiplication

[latent walrus]

but this ^

[ruby flame]

So

i would this

208

[latent walrus]

no

why have you changed a minus into a multiplication

[latent walrus]

just riddle me this

[ruby flame]

?

You said the dot is multiplying

[latent walrus]

im referring to the one after 13 root2

[ruby flame]

2

[latent walrus]

and what is 2^3

[ruby flame]

8

[latent walrus]

so what is (sqrt(2)^2)^3 * sqrt(2)

[ruby flame]

8 root 2

[latent walrus]

yes

so what is the answer

[ruby flame]

5 root 2

[latent walrus]

theres your answer

finit

[ruby flame]

Thank you

[obtuse pebble]

@ruby flame Has your question been resolved?

[obtuse pebble]

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[chilly crater]

Need help pls

[obtuse pebble]

Please don't occupy multiple help channels.

[chilly crater]

?

[vast willow]

Ok so you know derivatives??

[chilly crater]

Nope

[vast willow]

Ok no problem

[chilly crater]

K

??

[native inlet]

do you know the formula for the vertex of a parabola?

[chilly crater]

Oo that

[vast willow]

So you are working examples on completing squares ?

[chilly crater]

Yh

Completing squares

Now??

@vast willow

[vast willow]

What did you get after completing squares

[chilly crater]

Not sure

[vast willow]

Do you know how to complete square

[chilly crater]

Complete the square
-5x^2 +80x=s

??

[vast willow]

No

[chilly crater]

Then for

S=5x^2+80x=5

??

@vast willow

[vast willow]

Wdym =5

[chilly crater]

Sry s

[vast willow]

What's s too

max distance ?

[chilly crater]

Yh

S=80t-5t^2

What equation

Then?

@vast willow

[vast willow]

Wait follow me

[chilly crater]

Alright

[vast willow]

80t-5t² you want to complete squares

So remember first that (a-b)²=a²-2ab+b²

[chilly crater]

K

Then

[vast willow]

Then look at 80t-5t²

Now 80t should be in the form of 2ab if you compare it with the one above

[chilly crater]

Okay

[vast willow]

Wait I made a typo

First take -5 as a common factor what do you get

[chilly crater]

16

[vast willow]

No

[chilly crater]

..

[vast willow]

You are taking -5 common factor from 80t-5t²

[chilly crater]

O

[vast willow]

So 80t-5t²=-5(...)

[chilly crater]

S= -5

Ohk

T^2-16t

[vast willow]

What should be in place of ...

[vast willow]

Correct

So -5(t²-16t)

Now compare the inside of the paranthesese with a²-2ab+b²

You will notice that a=??

[chilly crater]

Hn

16

[vast willow]

t²-16 and a²-2ab+b² a=??

@chilly crater

[chilly crater]

T

[vast willow]

Correct

[chilly crater]

Ohkk

[vast willow]

And 2ab=?

[chilly crater]

16

[vast willow]

No

We already no that a=t

So 2ab=2bt

So 2ab=?

[chilly crater]

So not 16t

8

[vast willow]

It is 16t in fact

[chilly crater]

Oohk

[vast willow]

Wait a sec

What is 8

You wrote 8 to refer to what

[chilly crater]

Not sure

[vast willow]

Then why did you say 8

[chilly crater]

I thought 2 x 8

Is 16

Something

Hn

[vast willow]

Look nvm

[chilly crater]

K

[vast willow]

So 2ab=16t did you understand why

[chilly crater]

Yh

[vast willow]

We have a²-2ab+b² we know that a² is t² we want to which of the other 2 refers to 16t

Since b² doesn't contain a in it then it doesn't have t

But 2ab on the other hand has a

So it has t

[chilly crater]

at

[vast willow]

So 2bt =16t

Thus b=??

[chilly crater]

8

[vast willow]

Great

[chilly crater]

Now

[vast willow]

Now what we want is to reach a form c(t-d)²+h

That's why we were talking about (a-b)²=a²-2ab+b²

[chilly crater]

-5(t-8)

[vast willow]

No

[chilly crater]

O nvm

[vast willow]

Wait dont be in a rush

[chilly crater]

Ok

[vast willow]

So we found out that a=t and b=8

So (a-b)²=(t-8)²=??

Expand

[chilly crater]

(T-8)^2 = t^2-16t+64

[vast willow]

Ok nice

We have t²-16t we are still missing +64 to get (t-8)²

To get this number we can simply add and subtract 64

So we are basically adding a version of 0 which doesn't change anything

[chilly crater]

Hm

[vast willow]

So we get -5(t²-16t+64-64)=-5((t-8)²-64)=-5(t-8)²+320

Is everything clear till now

[chilly crater]

Yh

[vast willow]

Ok nice

Now we got to the desired form

[chilly crater]

Hm

[vast willow]

Now look how does this form lead us to the solution

You have -5(t-8)²+320

[chilly crater]

Yh

[vast willow]

You want to maximize this

Now what's the sign of -5

[chilly crater]

A

[vast willow]

What's the sign of -5

[chilly crater]

Negative

[vast willow]

Ok what's the sign of (t-8)² for any t

[vast willow]

=0

It might be positive or 0 not just positive

[chilly crater]

Ok

[vast willow]

What's the sign of 320

[vast willow]

Ok

Now -5(t-8)² is either negative or 0 bc negative multiplied by positive gives negative and negative multiplied by 0 gives 0

[chilly crater]

Yh

[vast willow]

So there is no way to exceed 320

If -5(t-8)² is negative then -5(t-8)²+320<320

[chilly crater]

Yh so it's 0

[vast willow]

And if -5(t-8)²=0 then -5(t-8)²+320=320

[chilly crater]

Yh

[vast willow]

Now when is -5(t-8)²=0

For what value(s) of t

??

[chilly crater]

.

[vast willow]

?

[vast willow]

Do you know how to solve this equation

[chilly crater]

Yh

8

[vast willow]

Ok what is/are the solution(s)

[vast willow]

Correct

[chilly crater]

Hm

[vast willow]

Now for t=8 we reach the max distance

And we know that 0<8<16

We are done

[chilly crater]

That's it

Omg thank you so much

Bro

[obtuse pebble]

@chilly crater Has your question been resolved?

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[obtuse dust]

trying to find the cofunction in radians

i got sin 22.5pi/90 by

3(180)/8 getting 67.5, getting 90-67.5 to get 22.5 for the cofunction

so assuming thats all right trying to figure out a way to simplify sin 22.5pi/90

it twas not right that simplifies into pi/4 of course

so where did i mess up?

lol nvm 90 is supposed to be 180

its right

.close

[obtuse pebble]

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[light condor]

hi people can I get some help with this

I graphed the 2 functions, what do I do after that

[static marten]

you know what (f∘g)x means?

[light condor]

yes

f(g(x))

[static marten]

so we just need g(f(3))

no need for graph because the information's there

[light condor]

ohhh okay

lol

so how do I find f(3)? I don't have the function, I need to make one?

I'm guessing y2-y1/x2-x1 isn't gonna work here

[static marten]

huh

it gives different points for f(x)

[light condor]

yes

what do I do with those

[static marten]

like (-3,8), (3,1)

and we want f(3) first

[light condor]

so um f(3) = (3, 1)?

[static marten]

f(3)=1

[light condor]

OH

okay

and then you do g(1)?

[static marten]

yeah

[light condor]

hmmmm

-1?

[static marten]

yeah

[light condor]

it's that simple?

[static marten]

just realised the typo lol

yeah

[light condor]

okay I think I probably got the rest

you're the mvp chlamydia!

.close

[obtuse pebble]

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[chrome crypt]

Hi guys I am currently learning about application of derivatives and had a question on concavity, so I understand concavity, but to picturize it better I made up a scenario. Here is my idea. Lets say we have a wonky function with numerous turns where the Y axis represents Revenue relative to time. The goal is to sketch the function efficiently. If we had a humongous polynomial function it would be difficult as we would have to input numerous points to get an accurate representation of the graph which is not efficient. However if we take the first derivative and we will get information on the behavior of the function between certain intervals for when the derivative is equal to 0 or DNE. Now we come across a problem although we find the behavior of the function between certain intervals we could argue for each individual sector (interval) that the function can be drawn concave upwards or concave downwards and the behavior would remain the same. To understand whether or not its concave downward or concave upwards we need to take the second derivative which will give us the instantaneous rate of change of Revenue. Now throughout the function there exists a point where we have possible inflection points at f''0 or f'' DNE. Now after the possible inflection points we test points nearby the interval to observe the behavior of the slopes. If f'' > 0 I picturize as instanteous change in revenue being positive, so I think of that as our loss in revenue minimizing and becoming increasingly positive. If f'' < 0 the instataneous change in revenue is negative which means are losses will be increasing. After we conclude the concavity for all the intervals for f'', we can then determine the inflection points.

[chrome crypt]

The importance of inflection points is that if we think about it in terms of revenue we can determine the moment we start making positive revenue and the moment when revenue becomes negative.

I'd like some feedback. Perhaps I could have simplified what I wrote as it does seem a fair bit long, but in essence when we find the behavior of the function for intervals there are 2 arguments we can make, concave up or concave down and we cannot assume as it could be incorrect and we would be changing the data of the function which would lead to incorrect conclusions, this is where the concavity test is vital. I associate f'' > 0 in terms of revenue I picturize that as revenue becoming increasingly positive and f'' < 0 as our revenue dropping, instantaneous change in revenue negative, hence losing.

[grizzled shore]

First thing, polynomials are infinitely differentiable everywhere, so you never encounter a point where its derivative DNE

[chrome crypt]

I should have said functions my bad.

[grizzled shore]

Inflection points don’t determine the sign of the value

[chrome crypt]

I know that.

[grizzled shore]

It tells you when the sign of the rate of change change

[chrome crypt]

Its a transition point.

yeah I know that.

[grizzled shore]

So then it doesn’t tell you when you make positive or negative revenue

[chrome crypt]

I know it doesn't tell you, but I was asking myself what is the importance of it asides from the fact that we know its it the transition stage where the f'' sign changes

[grizzled shore]

Sometimes we aren’t too concerned with the places it transitions

[chrome crypt]

and because we know that an inflection point must be continous and have concave upwards and concave downwards before and after the point. So I was questioning how can I apply it.

[grizzled shore]

A big use of the 2nd derivative is telling us if a critical point is a local min or max

[chrome crypt]

Basically what I do is a bit strange at times but I like to apply real world scenarios a lot, cus my goal is to retain the info.

[grizzled shore]

If it’s concave down it’s a max

If it’s concave up it’s a min

[chrome crypt]

mhmmm.

[grizzled shore]

The sign of its concavity tells us the nature of critical points

[chrome crypt]

I have a question Frost I'd like to ask. I'd like you to tr and prod my analytical thinking to come up wit answer for the DNE.

Basically for the second derivative ik it says inflection point where f'' = 0 or DNE. I'm trying to think abt the DNE aprt and am thinking of a function where when we differentiate it from both left and right the derivatives don't match one another.

I'm just confused on pinpointing an exact function to show this, I can't think of one off the top of my head.

[grizzled shore]

,w graph y^3 = x from -5 to 5

[warm shale]

Results provided by WolframAlpha

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

Requested by frosst

[chrome crypt]

ahhh yes vertical tangent

[grizzled shore]

This function has an inflection point at x = 0

[chrome crypt]

Ooooooh thanks man.

[grizzled shore]

A more readily available example for 2nd derivatives is in physics

[chrome crypt]

i do love physics

[grizzled shore]

This is because the 2nd time derivative of position (given as a function of time) is acceleration

So you get a lot more problems regarding 2nd derivatives

[chrome crypt]

yeah i was thinking of using that scenario to explain concavity which helped.

One way that worked for me tho was thinking abt concavity in terms of revenue wit a wonky function.

[grizzled shore]

Well in SHM you have

$\ddot x = -k^2x$

[chrome crypt]

what is shm

[warm shale]

Frosst

[chrome crypt]

also r u using newtons notation for the derivative?

[grizzled shore]

Simple harmonic motion

No this is more common physics notation for time derivatives

Dots for time derivatives

[chrome crypt]

hmmmmm can I use higher order derivatives using Leibinz's notation?

[grizzled shore]

You can easily inspect that the acceleration is 0 when the object has position = 0

[chrome crypt]

Not familiar with harmonic motion.

[grizzled shore]

You can but physics uses a shorthand for time derivatives as they appear frequently

[chrome crypt]

Gotta learn that unit

[grizzled shore]

If I put something on a spring

And it moves or “bounces”

[grizzled shore]

It looks like this

It moves like this

[chrome crypt]

so ur analyzing the motion of the spring

hmmmm what comes to mind is sine waves and cos waves.

[grizzled shore]

Not the spring itself

But the actual particle on the spring

Or the end of the spring

[chrome crypt]

so if the particle is oscillating?

[grizzled shore]

Indeed we find that this is a 2nd order ODE with solutions that have sines and cosines

[grizzled shore]

Yes

[chrome crypt]

that is rlly god damn cool.

[grizzled shore]

Simple harmonic motion is things oscillating

[chrome crypt]

so what abt ocean waves or ferris wheels?

[grizzled shore]

Same thing

Well

Ocean waves are more dimensions

The differential equation solution is writing x(t) = ???

[grizzled shore]

This describes the motion

[chrome crypt]

Wow.

I am gonna defintely look into that.

[grizzled shore]

But this gives us explicitly how it moves

[chrome crypt]

I rlly like ur passion man. I too am passionate abt this, but a slow learner.

But I am very hardworking.

[grizzled shore]

Haha yeah this is all very fun and interesting stuff

[chrome crypt]

especially the proofs.

[grizzled shore]

The more you know the more it connect together

[chrome crypt]

The more you understand and struggle on a question is where the true learning occurs.

[grizzled shore]

This even goes into complex numbers in some cases

[chrome crypt]

i needa learn abt complex numbers n stuff.

Right now I am completing calculus 12 and am hoping to do Calc BC, although I think calc 12 comprises of calc ab and calc bc content.

kinda confused tho cus i keep hearing different stuff.

[grizzled shore]

Due to $e^{ix}=\cos(x) + i \sin(x)$

[warm shale]

Frosst

[chrome crypt]

i stands for imaginary number right?

never learnt complex numbers before but briefly remember from a video

sqrt-1 = i

[grizzled shore]

You might be able to see how you could get the cos and sin combinations in here

[chrome crypt]

or something like that.

[grizzled shore]

Yeah

[chrome crypt]

gotta learn this.

thanks mate for your time and assistance, just one last query. Yk for the local min and max points?

[grizzled shore]

Yeah what about them

[chrome crypt]

So basically when taking the second derivative I know its useful for pinpointing whether a function is concave up or down, but for maximum and minimum points I remember u said that its especially useful for determining max and min points.

All I have seen so far is that max points are just the critical points where the slope is 0 or DNE, but is there any application

of these local max and min points that I will later find in application of derivatives?

[grizzled shore]

Max points are not necessarily DNE

For continuous functions they are necessarily 0

Or on the boundary then it can have any slope

Well not really

On left side boundaries it’s non positive

On right side boundaries it’s non negative

The point of concavity is that when you find a place where your function has 0 slope

You know it’s a critical point

But you don’t know if it’s a local minimum or maximum

Concavity tells you this

[chrome crypt]

Right.

[grizzled shore]

It tells you via this

[grizzled shore]

And this

[chrome crypt]

Mhmmm.

that makes sense.

but is there any application

in word problems

for these points?

[grizzled shore]

Have you done the classic question about maximising ratio of volume of a cylinder to its surface area?

[chrome crypt]

not yet just a beginner yk what don't reveal spoilers for me ima figure it out.

[viscid gull]

don't spoil the manga for the anime only

[obtuse pebble]

@chrome crypt Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[quartz lantern]

I need to solve for x so that it makes the quadratic formula. I’m stuck on step 4. I need to complete the square but I think I’m doing it wrong

[native inlet]

nope that's right :) what would be that missing chunk?

[quartz lantern]

like that?

[native inlet]

make sure you add it to both sides of the equation

[obtuse pebble]

@quartz lantern Has your question been resolved?

[quartz lantern]

I know I need to square root both sides, but then what happens to the a/bx?

wait nvm

[obtuse pebble]

@quartz lantern Has your question been resolved?

[obtuse pebble]

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[jade summit]

im stuck, could someone help me finish this? (gaussian elimination)

[nocturne minnow]

You need to eliminate this 1 now

[jade summit]

yeah, but i just can't find a way to eliminate it

[nocturne minnow]

Use the third row

[jade summit]

if i subtract the third row, I'll get -1 on the other side, do I just revert it? idk if that would make sense though

[nocturne minnow]

Wdym other side?

[nocturne minnow]

Yeah this is fine to do

[jade summit]

okay

but if i add r3 again I'll just get the 1 back

[brave bramble]

Why isn't it fine to have a -1 there?

[jade summit]

isn't it supposed to all be ones?

what is it called again

i forgot the name

the first number after the zeros

[brave bramble]

You can multiply that line by -1 to fix it

[obtuse pebble]

@jade summit Has your question been resolved?

[jade summit]

wdym?

[brave bramble]

You're allowed to multiply a row by a constant

[jade summit]

could i also divide by a constant?

[brave bramble]

Yes!

[jade summit]

so I can just divide by -1?

R4

[obtuse pebble]

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[unreal whale]

how do i do part b

[tacit scarab]

Notice that g(x) and h(x) are constants

[unreal whale]

yea

[tacit scarab]

so substitute the values in the inequality

[unreal whale]

how about f(x)

[tacit scarab]

You should have $-8<f(x)\leq7$ now

[warm shale]

WhereWolf(ping if needed)

[unreal whale]

ye

[tacit scarab]

What's the x values that satisfy this?

You should be able to see it from the graph

[unreal whale]

7

[tacit scarab]

No

[unreal whale]

anything from -8 to 7

[tacit scarab]

Umm no

That's the range of y values

We want the x values

[unreal whale]

0?

[tacit scarab]

We want a range

[unreal whale]

mhm

[tacit scarab]

Ok let's forget the numbers

Do it graphically

g(x)<f(x)<=h(x) means that f(x) is between g(x) and h(x)

And when is that true

[unreal whale]

left side

or middle

??

@tacit scarab

[tacit scarab]

Yes the middle part

We want the range of x values of the middle part

[unreal whale]

so what would it be

[obtuse pebble]

@unreal whale Has your question been resolved?

[obtuse pebble]

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[timid silo]

need some help simplifying this

[timid silo]

unsure of how to handle the square on the last term

[errant lark]

Well, that should be taken care of, with the coefficient that you have in that ln term.

[timid silo]

so it would be -2ln for the last term?

[errant lark]

Why -2?

[timid silo]

cause i thought i was supposed to do something with the 1/2

[errant lark]

Yes, it does.

[errant lark]

You will multiply the 1/2 outside the bracket when you simplify.

That'll get rid of 2.

[timid silo]

overall cancelling leaving a - right

[errant lark]

Yes

[timid silo]

then just ln laws?

[errant lark]

Yes.

Show your work once you are done. Just in case.

[timid silo]

yeah im writing it out right now

i was just left on my last submision ill get back in a sec

Along the lines of this?

Can I cancel across the fraction like that?

[errant lark]

Show the full work. I dunno what you have done with 1/2 ln(x) term but it seems wrong.

[timid silo]

ill rewrite it so its more clear

?

do i have to distribute the 1/2 throughout the bracket?

[obtuse pebble]

@timid silo Has your question been resolved?

[errant lark]

Yes.

Your major error is here:

$$\frac 1 2\left[\ln{(x)} - 2\ln{(x^2 + 3x +2)}\right] \neq \frac 1 2\left[2 \ln{\left(\frac{x}{x^2 + 3x + 2}\right)}\right]$$

[warm shale]

Enemagneto

[errant lark]

Basically, identity $\ln{(a)} - \ln{(b)} = \ln{\left(\frac a b\right)}$ only works when there are no coefficients(more precisely coefficient is 1) with $\ln{(a)}$ and $\ln{(b)}$.

[warm shale]

Enemagneto

[timid silo]

ohhhhhhh

[errant lark]

Or they must have same coefficients.

[timid silo]

ahhhhhh

ok

i see the fault

[obtuse pebble]

@timid silo Has your question been resolved?

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[visual ridge]

e,f, and g. How do I find the domains if I don't even know what the functions are?

[visual ridge]

For e i got [-4,3] though i'm not sure if thats right

still not sure what to do for f

g is also ?

<@&286206848099549185>

[obtuse pebble]

@visual ridge Has your question been resolved?

[visual ridge]

.close

[obtuse pebble]

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[bold brook]

which one should I calculate first?

[static marten]

what do you mean by this

[bold brook]

inequality

[static marten]

there's one inequality there

[bold brook]

is This problem cannot be resolved?

[static marten]

no it can

[bold brook]

🥲

[static marten]

i just don't know 'which one' you're talking about

[bold brook]

8x-3 or x + 1

[stable barn]

,𝐴𝐷𝐶 = 𝐴𝐵𝐶. basically how do I prove b and d

[rich plume]

#❓how-to-get-help

This channel is already occupied

[rich plume]

That is a single inequality and there is nothing as such 'which one should I calculate first'. After simplifying both sides simutaneously, you get an interval for x
To start with, multiply BOTH sides by 5

[bold brook]

thank you my friend

is 8x-3 > x + 5 ?

[fossil crag]

if you're trying to multiply both sides by 5, write it first as $5\Big(\frac{1}{5}(8x-3)\Big) > 5(x+1)$

[warm shale]

rafilou2003

[fossil crag]

then, if you look at what the right hand side gives...

[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).

@bold brook Has your question been resolved?

[obtuse pebble]

@bold brook Has your question been resolved?

[obtuse pebble]

@bold brook Has your question been resolved?

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[grave harbor]

From this page is Russian: https://en.m.wikipedia.org/wiki/Binomial_theorem Why is the last step in the proof by induction possible?

[obtuse pebble]

@grave harbor Has your question been resolved?

[grave harbor]

<@&286206848099549185> Could you briefly explain why the summation can be expanded like in the part of my notes marked with ?

[obtuse pebble]

@grave harbor Has your question been resolved?

[obtuse pebble]

@grave harbor Has your question been resolved?

[grave harbor]

nvm so the first two in the sum are re-written a^(n+1-k) and b^k and also there's apparently the rule as above, and that's it

.close

[obtuse pebble]

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[cloud sparrow]

A car travels from C to E in 6 seconds, how long will it take to travel from E to B (refer to the diagram) note that $EC \neq 6$

[warm shale]

SirGareth

[vestal marsh]

What'd do is make |AB| = 1

And then see what ratio of |BE| is of |EC|

And then you should be able to figure something out

[cloud sparrow]

what do you mean by those "| |"

[vestal marsh]

Just means distance

[cloud sparrow]

ah

is the answer $\frac{2}{\sqrt3}$?

[warm shale]

SirGareth

[vestal marsh]

On my phone rn so can't verify

But it should be whatever the ratio is ×6

[cloud sparrow]

yeah that was wrong

[cloud sparrow]

just got this. thank you

[obtuse pebble]

@cloud sparrow Has your question been resolved?

[obtuse pebble]

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[quartz wave]

okay, firsly, very dumb question

[royal basin]

better a dumb question than one not asked at all

[quartz wave]

ok sooo

firsly

why can't we use the bernoulii

like for example for the P(X = 0), why can't we write it as 3C0 * 0.4^0 * 0.6^3

@royal basin any ideas>

[civic zealot]

probability isn't independent. Once you choose one tv it's taken out of the population changing the prob the next tv is defective or not

[royal basin]

was showering sorry

[rocky oak]

the sign on the blue oval, is that implies sign guys?

[obtuse pebble]

@quartz wave Has your question been resolved?

[obtuse pebble]

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[lethal surge]

why isn’t this the same?

I thought x/a/b was the same as x/ab

in the same way that 10/2/2 = 10/4

[lethal surge]

second equation is supposed to also say x= instead of just x

[nocturne minnow]

$\frac{x}{\frac{a}{b}} = x \cdot \frac{b}{a}$

[warm shale]

dldh06

[lethal surge]

but following that 10/2/2 would be 10*1

so what’s the difference between the 2?

[nocturne minnow]

Is that (10/2)/2 or 10/(2/2)?

Because one results in 5/2, the other 10/1

[lethal surge]

ahh so there have to be a set of parantheses

[nocturne minnow]

Yes

[lethal surge]

you can’t just equalize them like with multiplication

[nocturne minnow]

No

[lethal surge]

that explains everything, thanks

.close

[obtuse pebble]

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[amber axle]

so i got these right

[amber axle]

but i uh

dont understand why the sets of {A,B} are the same as the OPs of {A,B}

i also uh am not sure about this one

this is right

but im not sure how to count out the fact that the dupes are 2^n

<@&286206848099549185>

<@&286206848099549185>

[obtuse pebble]

@amber axle Has your question been resolved?

[amber axle]

<@&286206848099549185>

[olive kite]

sorry I can't help but also don't just spam the @

[amber axle]

i thought its fine to ping every 15 min

which is what ive been doing

if you check the timestamps :P

[olive kite]

pretty sure it means if no one responds within 15 minutes you can send out an @ only once

try asking about it in one the channels like proof-and-logic

[obtuse pebble]

@amber axle Has your question been resolved?

[obtuse pebble]

@amber axle Has your question been resolved?

[tall ore]

hi

@tall ore

[obtuse pebble]

@amber axle Has your question been resolved?

[obtuse pebble]

@amber axle Has your question been resolved?

[obtuse pebble]

@amber axle Has your question been resolved?

[candid sigil]

ordered pair of sets?

[blazing forge]

you want to know why there are 2^n ways to choose sets A and B, subsets of U, such that A = B? this is the same as just choosing a subset A of U

and there are 2^n subsets of U

if you didn't know that the cardinality of the powerset is 2^n, its because for each element in U, either its in A or its not in A. so for each element there are 2 choices, giving 2^n total possible subsets

[amber axle]

ah okay so i can say that:

because the set S of {a,b} where a=b & a,b subset U is just the set S={a} where cardinality of S = # of subsets of U (since a subset U?), and given that there are 2^n subsets of U, then the number of duplicates is 2^n (i get why there are 2^n subsets)

[amber axle]

im still confused about this one

im not sure why every OP of (a,b) where $a \in b \in U$ corresponds to every set {a,b} where $(a,b \in U) \land (a \in b \lor b \in a)$

[warm shale]

aidanlok

[amber axle]

??? why are you doing this in my help channel

[slim cove]

Please stop

!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).

[slim cove]

In the future, please use #bots to test the bots

[obtuse pebble]

@amber axle Has your question been resolved?

[blazing forge]

There is a bijection from the set of all ordered pairs (A,B) with A \subseteq B to the set of all sets {A,B} with A \subseteq B; it's defined by (A,B) |-> {A,B}.

Its definitely surjective. If (A, B) and (A', B') are two different ordered pairs that get mapped to the same set {A,B}, then {A, B} = {A', B'}. Thus, either A = A' and B = B' or A = B' and B = A'. We can't have the first case because then (A,B) = (A',B'), contradicting our assumption. But we also can't have the second case--if it were true, then B = A' \subseteq B' = A. This means that A = B = A' = B', so (A,B) = (A,A) = (A', B'), a contradiction as well. Thus, the map is injective, too.

Since there's a bijection, the cardinality of all such ordered pairs equals the cardinality of all such sets

[patent umbra]

Every ordered pair obviously gives you an unordered pair.

Conversely, every unordered pair has an unique ordering unless $A = B$. In which case $(A, B) = (B, A)$.

Equivalently, even though ${A, B}$ is unordered you can always order it by saying WLOG B is the larger set.

[warm shale]

chencking

[steep rock]

Hello

[amber axle]

okay I kind of get this

kind of

I got my friend to explain surjectives and stuff

I kind of get it

I’ll do some rereading and re read

[blazing forge]

if the bijection is too scary, chenckings explanation is simpler

the bijection just means for every ordered pair (A,B) there is exactly one unique set {A,B} and vice versa. so theres the same number of the pairs (A,B) as the sets {A,B}

[obtuse pebble]

@amber axle Has your question been resolved?

[limber pine]

.reopen

[obtuse pebble]

@amber axle Has your question been resolved?

[fiery prairie]

https://discord.com/channels/268882317391429632/1159940087237333012

[dusty canopy]

wait how do i use you

[latent walrus]

whats even going on down here

[gilded ferry]

Can I use you?

[balmy mortar]

#proofs-and-logic

[static patio]

can someone verify my proof from ToC?

[keen garnet]

!help

[obtuse pebble]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[static patio]

uff, i didnt noticed bad chat

thx

[modern cedar]

how can i test this integral for convetgence or divergence using lct or dct?

do i use lnx<=x<= ex?

[keen garnet]

!help

[obtuse pebble]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

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).

[modern cedar]

ah i see thx and sry

[trim grail]

.close

[amber axle]

[marble oar]

ONLY YOU CAN CLOSE IT BRUH

[brazen viper]

@amber axle is there still an open question? It's unclear to me.

[amber axle]

yeah I know

[amber axle]

yep still doing that assignment on at quantifiers

[brazen viper]

@amber axle the originally posted question?

[amber axle]

yes

the original OP -> S(Subsets) question

[brazen viper]

Q3.10? https://media.discordapp.net/attachments/903430332324405288/1156267593909420052/Screenshot_2023-09-26_at_12.34.34_PM.png?ex=6537f21c&is=65257d1c&hm=1a3f6a0f7df4d5135a71e1c3c272cdd724f819c1715461cf9670f4fc5be96ba5&

If we take the answer to Q3.8 as true, then we can consider the cases A <= B and B <= A, with each contributing 3^n by Q3.8.

But you notice that we have double counted the cases where A = B.

And those are 2^n (rather than 3^n) because we can only choose to include in A and B or neither A nor B.

I'm literally just restating the "explanation" section of the question, so I'm not quite sure if this is helpful.

@amber axle if this isn't sufficient, can you explain where this line of thinking loses you?

[amber axle]

this is where I don’t get it

I feel really dumb that’s a really simple tuple count

what are we including?

and why can it be in neither??

[brazen viper]

@amber axle members of set U (in sets A and B)

Sorry, I stepped away for a moment

[timid silo]

closed???

[brazen viper]

No, still open

[brazen viper]

@amber axle you still here?

[amber axle]

yep mb wasnt looking here

[brazen viper]

No biggie

[amber axle]

oh so its 2^n bc for every ele we have 2 choices? in both or missing from both? so, as a hypotheical, if we have ABC instead of AB, in the end, its still 2^n bc its in all or in none?

[brazen viper]

Yeah

As long as A = B = C

[amber axle]

hmmm ok

the guys above mentioned stuff about bijections

which we're just getting into

but maybe i just leave it as is with ur counting which is good

[brazen viper]

The bijection stuff isn't wrong or anything, I just thought the counting argument would be more understandable

[amber axle]

it is, for what ive learned so far

[brazen viper]

👍

Great, let me know if you have other questions

[amber axle]

yeah i do actually have some more counting questions lol

let me find it

[brazen viper]

Ping me when you find it please 🙂

[amber axle]

@brazen viper

(posting 4.2 for context, i understand that one fully)

when writing my explanation for 4.3

i was unable to explain precisely why we use the product rule to create the divisior we see in the formula

[brazen viper]

Sure, let's take the bookkeeper example

Because each e is indistinguishable, we need to divide by the number of ways we could distinguish it, which is 3!

And we also need to divide by the 2! ways to arrange the k, and the o

Because we are dividing by multiple things, this is the same as multiplying them, and dividing just once.

(a/b)/c = a/(bc)

@amber axle

[amber axle]

okay that makes sense

just to quote my rubric...

independence???

[brazen viper]

Ah,

Meaning the arrangement of the es is independent of the os and ks

[amber axle]

oh whaaaaat

okay

[brazen viper]

The internal arrangement that is

[amber axle]

so because the arrangement of each of the letters is independent we have to multiply their possibilities, but whats the other option? its not like we'd add...(?)

[brazen viper]

Like, if I labeled the es as e1 e2 e3, and the ks as k1 and k2 whatever order I put the es in, (e2 e1 e3 for example) I can still freely choose the order of the ks

If there was some rule where (for instance) maybe the k1 couldn't go before the e1 then in some configurations we couldn't use all possibilities

[amber axle]

OH

i see

okay

[brazen viper]

@amber axle were there any other lingering doubts?

[brazen viper]

Well, I think we've addressed everything, so I'm going to try to close this. Just open another help channel if you still have other questions. 🙂

.close

[obtuse pebble]

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[timid silo]

how exactly would I solve this

the y(0)=1 and y'(0)=0 is confusing me

what would I do

I know the homogenous equation is y''-4y'-12y = 0

[timid silo]

im confused on how I would solve it because of this

[alpine raven]

you mean homogeneous equation

[timid silo]

yes sorry

[alpine raven]

characteristic equation

r² - 4r -12 = 0

,w x² -4x -12=0

[warm shale]

Results provided by WolframAlpha

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Requested by herels

[timid silo]

right

[alpine raven]

then the homogeneous solutions are :
y(x) = A exp(-2x) + B exp(6x)
with A and B some constants

[timid silo]

right

[timid silo]

where do these come in?

[alpine raven]

they are just initial conditions bro...

[timid silo]

Ok

im supposed to solve it with the initial conditions but dont know how to

its says solve this IVP

[alpine raven]

they will be useful to find the constants we dont know

[timid silo]

initial value problem

[timid silo]

in what way?

[alpine raven]

.

[timid silo]

so those would be the constants

?

still confused sorry

[brazen viper]

@timid silo before you do that you need to find the particular solution

[timid silo]

Ok

[alpine raven]

the constants are A and B and we will need to find them with the initial conditions

[brazen viper]

But once you do, you can use the two initial conditions to find A and B explicitly

[alpine raven]

but we need the particular solution first

[timid silo]

Ok ill try to work on this now

and will be back with questions

[alpine raven]

take y = Cx exp(6x) and find C 🤔

[timid silo]

this would be the general solution

[alpine raven]

no, its the homogeneous solution

[timid silo]

homo

sorry got it confused again

[alpine raven]

already found it here

[timid silo]

Ok now im working on particular

[alpine raven]

.

[timid silo]

working on it

im stuck here @alpine raven

[alpine raven]

what is t tho

[timid silo]

t is x

[alpine raven]

then stick to one

dont work with both

[timid silo]

ok

[alpine raven]

also, what are you trying to achieve ?

🤔

[timid silo]

im trying to solve the particular solution

[alpine raven]

yea but why taking Ax exp(x)

🤔

[timid silo]

this is what im trying to do

[alpine raven]

yea but it doesnt mean you should always take A x exp(x)

nice try

you have exp(6x) in your equation

[timid silo]

right

[alpine raven]

6 is a root of the characteristic equation of our diff equation

[alpine raven]

.

[timid silo]

C= y/exp(6x)

right?

[alpine raven]

?

[timid silo]

I solved for C

[alpine raven]

if it was that simple :/

[timid silo]

im confused sorry

[alpine raven]

y'' - 4y' -12y
plug y in this

[timid silo]

so it would be the double derivative of exp(6x) -4(derivative of exp(6x) -12(exp(6x))?

double derivative of exp(6x) -4(derivative of exp(6x) -12(exp(6x))

[alpine raven]

woaw can you write in maths ?

[timid silo]

...

im trying to figure rhat out

would it be right?

[alpine raven]

write it in a better way, its hard to read how you wrote it

or take a paper and take a picture of what you did

[timid silo]

Ok

[alpine raven]

eh i dont download files

[timid silo]

resending

I understand

me either

[alpine raven]

where is the Cx in front of exp(6x) ?

y = Cx exp(6x)

$y = Cxe^{6x}$

[warm shale]

Herels

[timid silo]

better?

@alpine raven

[alpine raven]

why do you have three constants tho ?

cant it just stay C :x

[timid silo]

bc of this I thought

[alpine raven]

?

[timid silo]

I plugged in C everywhere

[alpine raven]

yea but what are those C1, C2 and C3

👁️

[timid silo]

Thats just C

ik its extra

[alpine raven]

remove the 1, 2 and 3

[timid silo]

Ok whats next

[alpine raven]

calculate it my dude

[timid silo]

and then that would be the answer to the particular

right?

[alpine raven]

keep your calm, the goal here is to find C

:x

[alpine raven]

calculate this first

[timid silo]

Ok

ill brb for a while but Ill work on it

[alpine raven]

i'll be asleep tho

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[steel goblet]

i have that velocity is <-6t + a, -4t + b, -4t + c >

and i know that r(0) is (7,8,5) but not really sure where to go from here

and i know that a^2 + b^2 + c^2 = 16, but that doesn't really help much

[obtuse pebble]

@steel goblet Has your question been resolved?

[steel goblet]

<@&286206848099549185>

[quaint wing]

doesnt the fact that the original position + acceleration = final position imply that like

a= -6w

b= -4w

c=-4w

where w is some number

basically the initial velocity vector must be proportional to the acceleration vector

[steel goblet]

i have no idea. But i did discover this, idk if its helpful.
So the velocity vector is <-6t +a, -4t + b, -4t + c>
since we know the intial condition for the position, we know that the position vector is
<-3t^2 + at +7, -2t^2 +bt + 8, -2t^2 +ct +5>, right?

[quaint wing]

yeah

[quaint wing]

^ i think i found a way to prove this algebraically

I got from line 3 to 4 by taking derivative with respect to time

[steel goblet]

wait im really confused. I have no idea how you got those equations, and i still don't know how to find a, b, and c

[quaint wing]

d=Vot+1/2at^2

from the big 5 equations

we start at (7,8,5), then go to (1,4,1)

so the displacement is (-6,-4,-4)

[steel goblet]

right

[quaint wing]

and its given to us that the acceleration vector is (-6,-4,-4)

[steel goblet]

right

[steel goblet]

but i still don't get how you get that equation

[quaint wing]

you mean d = Vot+1/2at^2?

[steel goblet]

yes

[quaint wing]

its one equation you can use when a is constant

lemme remember how to derive it

[steel goblet]

okay i understand that

wait no i dont never mind

but am i supposed to be using that formula?

i never learned that in class. And i looked that up and saw that its a formula used in physics but the class im taking is multivariable calc

[quaint wing]

hm

[quaint wing]

idk

there might be some other way to solve it

i was kinda approaching it the physics way

[steel goblet]

do you know if there's another way to solve it?

oh

[quaint wing]

i mean i havent taken multivariable calc so..

idk

[steel goblet]

so im trying to understand this. Is the initial velocity -6, -4, -4?

[quaint wing]

no

but its proportional to -6,-4,-4

[steel goblet]

oh now i see it's multiplited by t

but how does that help?

[quaint wing]

so you use the fact that at t = 0

|V| = 4

so switch -t to w just for simplicity

(-6w)^2+(-4w)^2+(-4w)^2=16

[steel goblet]

okay. so | <-6w, -4w, -4w> | is 4, right?

[quaint wing]

yeah

[steel goblet]

makes sense

wait, how do you know all those w's are the same?

wait do you calculus at all? like integrations?

[quaint wing]

because -t is a scalar not a vector

[quaint wing]

from there to there i took derivative with respect to time

[steel goblet]

but integrating the acceleration vector gives us <-6t +c1, -4t+c2, -4t+c3 > where c1, c2, c3 are all constants that may not be equal

[quaint wing]

true

but <c1, c2, c3> = Vo

[steel goblet]

right

but c1, c2, c3 may or may not be equal

[quaint wing]

so <c1,c2,c3> is proportional to <-6,-4,-4>

from this line

and c1^2+c2^2+c3^2=16

[steel goblet]

but then how do you find the values of c1,c2, and c3?

[quaint wing]

so we have this vector equation

but this is really three different equations we can split up

cuz yk if two 3d vectors are equal

all their components have to be equal

[steel goblet]

random question, but do you know at what time the particle is at point (1,4,1)?

[quaint wing]

i think its 4/sqrt(68)

because we get that 68w^2=16

so w = +/- 4/sqrt(68)

and we know -t=w

and t has to be positive so t is 4/sqrt(68)

[quaint wing]

^one way to think about this is it’s basically a taylor series

But because the third derivative is zero we can just write it like that

So it’s really just saying f(x)=f’(0)x+f’’(0) x^2/2!

[obtuse pebble]

@steel goblet Has your question been resolved?

[timid silo]

dang

[obtuse pebble]

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[timid silo]

is there an easy way to solve non homogenous ODEs?

[timid silo]

like a chart to follow?

or a specific method

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.close

[obtuse pebble]

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[dawn osprey]

how can I find x

[proven veldt]

how do i find domain and range and what points do i draw the upside down U

[crude coral]

Use trignometry

[dawn osprey]

how though

[crude coral]

We have
Base = x
Hypothenuse = x+6
Angle = 67

[proven veldt]

ngl i went to the wrong channel my bad

[crude coral]

Sintheta = perpendicular/hypotenuse
Costheta = base /hypotenuse
Tantheta = perpendicular/base

[dawn osprey]

just regular Cos?

[crude coral]

Yeah

[dawn osprey]

so

x/x+6

wait

OH

[crude coral]

Yeah, x/x+6 = cos67

[dawn osprey]

wait

nvm i got it thanku

.close

[obtuse pebble]

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[mighty pilot]

asking for me to convert to standard form, when it already is?

idk if im missing something

<@&286206848099549185>

[quaint wing]

try x^2+4x+0 maybe?

idk

[mighty pilot]

i did, came up wrong

im on my last attempt, used up 4/5 lol

[quaint wing]

hm

[mighty pilot]

some are saying its a()^2+c

idk

[quaint wing]

thats vertex form

[mighty pilot]

yea

[quaint wing]

might as well try it

[mighty pilot]

yea ig

came up wrong, oh well

.close

[obtuse pebble]

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[obtuse pebble]

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[late roost]

hi, im trying to find the readius and intervial for this problem, any idea where I may have messed up the limit?

[obtuse pebble]

@late roost Has your question been resolved?

[obtuse pebble]

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[long hedge]

im stuck on 312

[long hedge]

so if i use the chain rule

$sec^2(xy)(y+\frac{dy}{dx})=1\frac{dy}{dx}$

is this right

[warm shale]

putridplanet

[tardy epoch]

Right now use algebra to solve for dy/dx

[tardy epoch]

Should be x * dy/dx

[long hedge]

right

[long hedge]

idk what to do

divide sec^2(xy) ?

[tardy epoch]

It's just algebra

Set z = dy/dx and solve for z

[long hedge]

$\frac{y+xz}{z}=\frac{1}{sec^2(xy)}$

[warm shale]

putridplanet

[tardy epoch]

Distribute the sec first

[long hedge]

$sec^2xy sec^2y^2+sec^2x^2 sec^2yx \frac{dy}{dx}$

[tardy epoch]

Wot

[long hedge]

i distributed it into (y+x dy/dx)

[tardy epoch]

a^2(b+c) = a^2 * b + a^2 * c

[warm shale]

putridplanet