help-13

[tough dock]

In your table

[crimson sedge]

no they're not

[tough dock]

t

[crimson sedge]

the t is t, x and y are separate

[tough dock]

There is no 1

[tough dock]

You are trying to use t to find your equation

Since x1 and y1 given to be functions of t in your document

[crimson sedge]

i'm just putting the table into desmos how it was presented to me

and i have to make x longitude vs t

[tough dock]

Try just using t instead of t1

[crimson sedge]

same with y

[tough dock]

Oh

Oops, I think I got confused because you changed it from the first screenshot

Anyways, you have your equation for your longitudes now

[crimson sedge]

is the ~ a symbol i can use in an actual equation

[tough dock]

~ is a symbol special to Desmos

Which tells it to fit an equation

[crimson sedge]

so there's nothing before the at1^2 then? like no x= or smth?

[tough dock]

Outside of Desmos, I have seen ~ used to represent "is proportional to"

[tough dock]

That is what your x1 ~ is

~ replaces = in Desmos when you are trying to fit an equation

[crimson sedge]

i see

so the equation would be x1=at1^2+bt1+c then?

[tough dock]

To be consistent with your original table, you would just use x = at²+bt+c in your document

[crimson sedge]

oki

[tough dock]

And you would replace a, b, and c with their actual values given to you by Desmos

[crimson sedge]

i think it's just asking for the function, but i can do that when i go to solve the function

i think

[tough dock]

The screenshot you posted is just asking for the type of function

Which we found to be quadratic

So writing "quadratic" would probably work

[crimson sedge]

ah oki

would quadratic also be correct here too?

seems good to me but i'm not 100% bc i'm very confused on this lol

[tough dock]

Hmm

Let me check

[crimson sedge]

oki

[tough dock]

Because you might want an exponential here

I think a quadratic is fine there as well

[crimson sedge]

this is what exponential looks like

[tough dock]

You can add a + c at the the end of your exponential fitting, but the quadratic is already super accurate anyways

So using quadratic fitting for both your latitutde and longitude values should be fine

[crimson sedge]

oki

[tough dock]

Ah

You should probably write down the equation form somewhere just in case

[crimson sedge]

oki

question

[tough dock]

Here

[crimson sedge]

for this, would i put the numbers desmos gives me or just the equation

[tough dock]

The numbers

[crimson sedge]

oki

[tough dock]

You would use the numbers here

[crimson sedge]

so smth like this?

my bad forgot the 1 between 7 and 4 on the first number

[tough dock]

You are missing a t at the end of the -1.66286

[crimson sedge]

ty

[tough dock]

Also, I would recommend rounding to ~3 decimal places, since 5 is a bit much

[crimson sedge]

my teacher likes us using 5 decimal places for some reason

at least in our last assignment he had us do that

How would I use desmos to do this?

i think i understand what i need to do, i just dunno how to do it with desmos

[tough dock]

Hmm

This is slightly annoying, but it isn't too bad to do with desmos if you still have the tab open

[crimson sedge]

i do

[tough dock]

I would change the names of the coefficients to something like this first

The reason why we would do this is so that we can use them as variables in other equations

[crimson sedge]

i see

[tough dock]

because if you just use a (etc) for both of them, if you try to use them in outside equations, Desmos won't know what to do

So now, we can just rewrite the quadratics, but as actual equations (keep the fitting things, we still need those)

[crimson sedge]

i'm a bit confused xd

[tough dock]

It will end up looking something like this

x_p and y_p are going to be your "predicting" functions

Since they use the values that you got from fitting your equations

[crimson sedge]

should it look something like this?

[tough dock]

Since your argument to your predicting function is just t, use t instead of t1 for the equation

Since t1 is actual values

[crimson sedge]

oki

[tough dock]

You should end up with a graph that just draws over the quadratics you just fit

[crimson sedge]

now i'm assuming i just plug the values on each point into the table and then do the same for y?

[tough dock]

Yes, and you can do this quickly with the you have at the top

You can make the name of the column x_p(t1), and desmos will automatically fill in all the values for you

[crimson sedge]

oki

[tough dock]

Getting you something like this

[crimson sedge]

w

wait i can't highlight to copy from it

so sad

done :D

[tough dock]

Congratulations

[crimson sedge]

thank you :3

a few more questions on this thing

i meant i was done with that part

i'll try on my own and come here if i have questions

[tough dock]

Okay

Good luck

[crimson sedge]

thx

i have a question

what's rectangular form?

[tough dock]

Rectangular just means "normal" here

So can you plot the y values as a function of the x values

[crimson sedge]

which i don't think i can

seeing as they have to be directly related to t

[tough dock]

Not necessarily

You can check by plotting (x1,y1) in Desmos and seeing if they make a valid function

[crimson sedge]

oki

the top left is where they're appearing

but also this is how i'm doing it, idk if this is what you meant

[tough dock]

You can just write (x1,y1)

[crimson sedge]

oooh oki

the black one is (x1,y1)

[tough dock]

Okay, so now the question is

Do the black points make a curve that has one x value corresponding to 1 y value

[crimson sedge]

ye

so the answer would be yes then?

[tough dock]

Are you sure?

[crimson sedge]

ye

[tough dock]

?

[crimson sedge]

oh wait

i see what you mean

ye no

i thought you meant like one x value corresponding to the y value on the chart, not one y value overall

[tough dock]

Yeah

[crimson sedge]

does this look good? :3c

[tough dock]

Hmmm

It's a bit vague in my opinion

When you say "whenever an X value is graphed, ... associated in order to get the correct graph," I think this could be made more clear by specifying that there needs to be exactly one y-value for each x-value

[crimson sedge]

oki

better? :D

[tough dock]

Hmm

What do you mean by "there is a different Y value ... we're looking for"?

[crimson sedge]

i mean that there's a different y value that needs to be associated with every x value in order to get a graph that matches the chart that we're given

[tough dock]

Hmmm

Ah

[crimson sedge]

i added "from the given values from the chart" after "looking for" :3

[tough dock]

So you are saying that when we attempt to fit a graph to the points, we sometimes get multiple, different values than what we need for f(x) to be a valid function?

[crimson sedge]

ye

[tough dock]

I see

[crimson sedge]

lemme try to rewrite it so it's more clear

[tough dock]

Okay

I think using a CER format here could help you to be more clear

[crimson sedge]

i think this is better

[crimson sedge]

we were never taught that

imo if we weren't taught to answer in CER format then my teacher prob doesn't care that much as long as he gets that we understand why something does/doesn't work

[tough dock]

Okay

[tough dock]

Hmm

I don't think I understand the "every x value would need to have the same y value" line of reasoning

[crimson sedge]

that's the line of reasoning that you gave me xd

that for that type of function to work every x value would need one y value

[tough dock]

I think I might've phrased it a bit strangely

Sorry

[tough dock]

Every x-value would need a y-value assigned to it, but different x-value can have different y-values

Let me try to draw something

[crimson sedge]

oki

[tough dock]

This is a valid function, since it has a single y-value for each point

Hence, the red vertical lines only cross the valid graph once

[crimson sedge]

so it's the vertical line rule?

[tough dock]

Here, the red vertical lines cross the graph multiple times each

Which means that an x-value has multiple y-values assigned to it

[crimson sedge]

ah, we learned that as the vertical line rule

[tough dock]

Which is not allowed in a valid rectangular function

[tough dock]

yes

[crimson sedge]

so i could prob just write that it's because it doesn't pass the vertical line rule and he'd accept it most likely

[tough dock]

You can mention and use the vertical line rule here

Saying something like "A graph created using the longitude as an x-value and latitude as a y-value does not pass the vertical line test"

[tough dock]

Yes

Don't know why I typed "Some", oops

[crimson sedge]

it's oki

[tough dock]

Looks good I think

Short and to the point

[crimson sedge]

added a bit more

bc i really want that 100% lol

[tough dock]

probably want "invalid" instead of "non linear", since "non linear" excludes things that are not straight lines

[crimson sedge]

the graph isn't a straight line when we plot it out

which is why i said "non linear" :3

oh wait i see

i think i'll just stick with my short answer lol

[crimson sedge]

i see

thx for the visualization c:

[tough dock]

No problem

[crimson sedge]

also sorry i got mad earlier just kind of stressed bc last few days of school trying to catch up grades xd

[tough dock]

I always find visualizations helpful when trying to solve problems, so I tend to use them a lot

[tough dock]

It's fine

Happens to everyone

[crimson sedge]

c:

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[potent eagle]

need a little help on sequences

geometric sequences to be more specific

not sure where to begin here

besides that U6 = 64

[vague rapids]

the information is insufficient

[potent eagle]

really?

[vague rapids]

yes

you don't know the first term

[potent eagle]

but here's the answer

[vague rapids]

ahh

they used an example

you can come to infinite such cases

with first term being different in each case

[potent eagle]

what about this one?

not sure where to start here

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[coral jewel]

In a country, when moving from a province to another, there will be a guard to check at the gate. Sir Austin is sly alcohol distributor. Every time he passes a gate, he will bribe the guard by giving them a bottle of alcohol for each container. Given each container can hold a maximum of 15 bottles, what is the minimum amount of bottles Sir Austin needs to carry in order to go through 10 gates and supply 50 bottles for his customers?

[coral jewel]

id like to check my answer, which is 99

im not sure if its 99 or 60

[nova snow]

@royal loom sir?

[wicked stag]

x is number of containers
15x = 10x + 50 + k
or 5x - 50 = k
If x = 10 k = 0 (least non -ve value)
Therfore he carries 15*10 = 150 bottles

Got this ?

@coral jewel ^^^

k is the remainder of extra bottles if any

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[halcyon adder]

I have to find the laplace transform of this in matlab

what does the first equation mean

[subtle harbor]

dirac delta

[halcyon adder]

what does it stand for

what kind of function is it

is it a step or impulse function?

[subtle harbor]

impulse function

step function is usually denoted u(t) or H(t)

[halcyon adder]

so an impulse on t-1

[subtle harbor]

on t=1

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[hoary parrot]

Can someone help me solve part b

I have no idea how to approach it

[dusk goblet]

infinitely many solutions means they’re the same line

[hoary parrot]

Yeah, but how do I find what value of p makes them the same line?

[dusk goblet]

try making the two equations look the same

[hoary parrot]

I've tried setting both equations equal to 0, then putting both equations equal to each other, and then rearranging the equation but it got me nowhere

[dusk goblet]

do you have any work

[hoary parrot]

Yeah I'll send a picture althoughI crossed it out 😅

[dusk goblet]

what id do is

[dusk goblet]

divide the bottom equation by -2

and multiply by p

so that the left sides of both equations are equal

then i’d match the coefficients

[hoary parrot]

how does multiplying the bottom equation by (-2/p) make both LHS equal?

[dusk goblet]

because they’re both equal to p

[hoary parrot]

oh yeah that makes sense

thx

I'll try that

[dusk goblet]

it should work out nicely

[hoary parrot]

So would this be correct?

[late otter]

p = 2 is the right answer yes

[hoary parrot]

but was my method correct, of just looking at the coefficient of x towards the end?

[late otter]

i cant really tell what you have done here, sorry

[hoary parrot]

well once I got to -2px + 2y = (p²-4p)x + (p²-p)y
I took the coefficients of x and put them equal to each other so -2p = p² - 4p and then just solved for p

[late otter]

ohh youre on b) mb, give me a min

[hoary parrot]

Oh yeah sorry 😅
I already completed a

[late otter]

looks good, but you should test with the coefficients of y aswell, as p could also be equal to 0 in $p^2-2p = 0$

[wraith dagger]

caspar

[hoary parrot]

alright thanks

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[crimson sedge]

how do i do this

[granite knoll]

show the whole question

[crimson sedge]

"in the expansion of : "

[subtle harbor]

the coefficient of x^()?

[grim nimbus]

you know the equation of newton?

[crimson sedge]

i think 3 @twilit estuary

@subtle harbor

[subtle harbor]

you think????

[crimson sedge]

It’s 5

@subtle harbor

[subtle harbor]

there we go

now are you familiar with binomial theorem

[crimson sedge]

example?

[subtle harbor]

(x+y)^n=sum( ... )

[crimson sedge]

yes?

i've done binomial expansion before

[crimson sedge]

yeah ik that

$6C0 \cdot (-\frac{1}{2}x)^6 \cdot (3)^0$ +

$6C1 \cdot (-\frac{1}{2}x)^5 \cdot (3)^1$ +

$6C2 \cdot (-\frac{1}{2}x)^4 \cdot (3)^2$ +

$6C3 \cdot (-\frac{1}{2}x)^3 \cdot (3)^3$ +

$6C4 \cdot (-\frac{1}{2}x)^2 \cdot (3)^4$ +

$6C5 \cdot (-\frac{1}{2}x)^1 \cdot (3)^5$ +

$6C6 \cdot (-\frac{1}{2}x)^0 \cdot (3)^6$

[subtle harbor]

yes

what

[crimson sedge]

like this yeah?

[subtle harbor]

its all multiplication...

with addition between lines

[crimson sedge]

my ba

d

i realised lmao

[crimson sedge]

yes

[subtle harbor]

so yes, we only care about the x^5 coeff and the x^3 coeff

do you see why

[crimson sedge]

nope

[subtle harbor]

well look at this

[crimson sedge]

x^5

yes

x^3?? no

[subtle harbor]

think if it was just (5+8x^2)(x^5+x^3)

what would it look like when you expand that

[crimson sedge]

$8x^10$

?

[wraith dagger]

محمد احمد

[subtle harbor]

what

[crimson sedge]

idk but i get the point

[subtle harbor]

your 8x^2 power is going to add with the x^3 power to get you an x^5 power

so it'll add to the x^5 coeff

[wraith dagger]

محمد احمد

[crimson sedge]

yes it will

but thingy

mind re-explaining the

x^3

within context

$6C1 \cdot (-\frac{1}{2}x)^5 \cdot (3)^1$

$6C3 \cdot (-\frac{1}{2}x)^3 \cdot (3)^3$

[wraith dagger]

محمد احمد

[crimson sedge]

this is what we were after

[subtle harbor]

those are the coefficients yes

now you multiply them by each one to make them/keep them x^5 powers

[crimson sedge]

$-\frac{9}{16}x^5$ and $-\frac{135}{2}x^3$

?

[wraith dagger]

محمد احمد

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[crimson sedge]

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[short loom]

“There are two bowls filled with 10l of alcohol. The first bowl has an alcohol percentage of 20% and the second bowl x%. Two liters were taken from the second bowl and put into the first bowl and after mixing two liters were taken from the first bowl and put into the second one. If the alcohol percentage in the second bowl was then 70%, what is the sum of the digits of x?”

[flat prawn]

10 L each in one bowl right?

[short loom]

yes

[flat prawn]

The 10L of alcohol comprises of 20 percent of the whole mixture in the first bowl right?

Um sorry for asking so many questions

[short loom]

yes i think so

[short loom]

dw

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[short loom]

<@&286206848099549185>

[flat prawn]

60.606 percent?

x = 60.606

@short loom

[short loom]

just a sec

[short loom]

i just realised i translated the last part wrong

ill fix it lmao sorry

there

[flat prawn]

??💀

[short loom]

the question wasnt in english

[flat prawn]

Is it 6?

[short loom]

i read it wrong when translating

its supposed to be 9

[flat prawn]

Oh shit

[short loom]

oml

8*

damn

[flat prawn]

It's supposed to be a whole number then?

Like a two digit percentage and not a decimal

[short loom]

probably

if i knew i wouldnt be asking 🌝 idk where to start at the problem

[flat prawn]

I got the same equation again

I'm putting it in desmos agaij

78.95 percent

It's pretty close to 80

[short loom]

so

how did you get to that answer

[flat prawn]

Wait

[short loom]

nice

thank you

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[wise estuary]

When you're evaluating limits approaching infinite, and there's a sqrt in the numerator of the leading number that has a power of 6

[wise estuary]

that becomes power of 3 right

[humble pivot]

can you send the problem?

[wise estuary]

sure but trying to understand not just get answer haha

https://i.imgur.com/OJSxgO3.png

my intuition thought 1 and -1

OH

is it reverse

[humble pivot]

if im thinking correctly, it does become a x^3 term

[wise estuary]

yep ok

I misread the signs

it's -1 and 1

I struggle with ones that have an isolated sqrt

thanks

[subtle harbor]

when you pull it out of the square root it becomes |x^{3}|

[wise estuary]

Is khan a good resource to look at somewhere different for Continuity work? calc1 practice

got a lot of heavy math this fall

[subtle harbor]

i.e. $\sqrt{2+4x^{6}}=|x^{3}|\sqrt{\frac{2}{x^6}+4}$

[wraith dagger]

🫎MooseyMooseMooser 🫎

[wise estuary]

interesting

[subtle harbor]

https://tutorial.math.lamar.edu/classes/calci/limitsatinfinityi.aspx for these types of problems

[wise estuary]

lamar! that was the link I have been trying to think of

cheers

[subtle harbor]

glad to have helped :)

[wise estuary]

continuity is my biggest weakness right now I think

and various equations with sqrts

those fuck me up lol

done

[rough terrace]

Are you doing analysis type of continuity or just non-proof ones?

I could give you some advice on how to practice 🙂

[wise estuary]

Intro level, prior to this year I had math anxiety and irl stuff that made me withdraw from the class

So catching up on rudimentary

[rough terrace]

Ah! Then just keep practicing!

Just keep doing questions and you will ace it for sure.

Don't worry

You got this lol

[wise estuary]

Yep! Gotta practice practice

[rough terrace]

Yes yes yessss

Aight good luck!!!

Wish you all the best

Close the channel hahaha

That would be great

[wise estuary]

.close

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[meager flare]

need to find the perforation, this is what ive done but i cannot check the answers as i dont have them

let me send sec

so for x = 3log(16)/2 the numerator and denominator are 0

but when drawing it I cannot see the perforation, neither in desmos nor in my graphical calculator

[cedar kiln]

@meager flare Has your question been resolved?

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@meager flare Has your question been resolved?

[meager flare]

😢

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[rough terrace]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[tawny drum]

2

V=2/3 pir^3

then what

[thorn robin]

bro josen knows how to do that question

ask him

[tawny drum]

he just looked at the answers remember

and i cant understand it

[tough dock]

Then you have to find the maximum of that function

[thorn robin]

damn

[tough dock]

Oh wait

[tawny drum]

there cant be any r's

[tough dock]

You need to find the volume of the cylinder

[tawny drum]

need to get rid of the r

[tough dock]

as a function of radius

[tawny drum]

this is the solution

which i dont understand at all

[tough dock]

It first finds the volume of the cylinder as a function of radius

That is the pi r²(3-2r)

[tawny drum]

how is h=3-2r thugh

[tough dock]

It is determined in the graph in the top left

That is a 2d cross-section of the cone

Right?

So it uses that to determine the height of a cylidner that fits inside of it

[tawny drum]

yeah

i dont understand it

i still cant see where h=3-2r come frmo

[tough dock]

This is a cross section of your cone

Yes?

[tawny drum]

yes

[tough dock]

To fit a cylinder in the cone, we need the top to not extend outside the cone

[tawny drum]

yes

[tough dock]

That looks something like this

We can find out the maximum height for a given in radius of a cylinder using the top right point

Since you have that the value of the radius is just the x-coordinate of the point, if you find a function that describes the right side of the cone, then your height will be the value of the function at that x-coordinate

For example, if the radius is 1, then the maximum height is 1

If the radius is 0.5, then the maximum height is 2

As you can see from the graph, the right side is just a line with an y-intercept of 3 and a slope of -2

[tawny drum]

yeah

[tough dock]

So, the equation that describes this line is y = mx + b = -2x + 3

Since height can be found from the radius using this equation, you get that h = -2r + 3

Going back to this, the volume of a cylinder is just pi r²h

Now that we have an equation for h in terms of r, we can just plug that in

Getting our second equation

And then just expand

Now, to find the maximum value of this function V, you need to use some calculus

[tawny drum]

i got it

yeah

thanks

[tough dock]

No problem

Is that everything?

[cedar kiln]

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[cedar kiln]

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[shell grotto]

?help

[shell grotto]

:\int _0^{\pi }\frac{3}{2}cos\theta :d\theta

how to solve this

[torpid cipher]

integral of cos(a) is sin(a)

[shell grotto]

yeah how about the other?

[torpid cipher]

(inverse of derivative)

[forest dust]

3/2 is a constant it can be taken out of the integral

[shell grotto]

u gonna put the 3/2 in the outsiede right?

[torpid cipher]

yep

yess

[shell grotto]

then how about the pie

how will i solve it?

that's what i don't get

[torpid cipher]

evaluate it

oh

[forest dust]

that is a limit

[shell grotto]

i need to multiply it to the sin (a)

right?

but how

[torpid cipher]

2/3 (sin(pi) - sin(0))

[forest dust]

no

[torpid cipher]

nope

[shell grotto]

oh

[torpid cipher]

its the integral basics

[shell grotto]

OH OKAY

[torpid cipher]

F(b) - F(a)

[shell grotto]

i'll change the (a) of sin?

[forest dust]

okay so the limits tell you what points your theta is varying from

[torpid cipher]

basically

[shell grotto]

OHH OKAYY

THEN it will be canceled right?

[torpid cipher]

yeah

[shell grotto]

then 3/2 x 0

=0

AM I RIGHT?

[torpid cipher]

yeah

[shell grotto]

OKAY THANKS

yo @torpid cipher

can i pm u quickly?

[torpid cipher]

sure

[brittle oyster]

you can also use graphical approach @shell grotto

[cedar kiln]

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[crimson sedge]

Hi am having problem with this math question

[native ermine]

Find the area of all the different sites, and add them together

[crimson sedge]

how?

[native ermine]

Do you know how to find the area of a square and a rectangle?

[rustic raven]

Or just have formulas

Cube= 6a^2

[crimson sedge]

am not listening to my math teacher so idk💀

[rustic raven]

a is the length of the aide

Side

[rustic raven]

Yeah just try that 6a^2

[crimson sedge]

number 1?

[rustic raven]

For cuboids it's 2(lb+lh+bh)

[normal shore]

||start listening ||

[rustic raven]

6*81= 486 cu.units

[crimson sedge]

ohhh

i get it now

[cedar kiln]

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[proud mural]

That’s the Task the task I got but got not really a plan what I should do
Taylor polynomials of functions f (x) = x + 1 with m > 1 can be used to calculate roots numerically.
(a) Determine the value T8(a;a) and the value T8′(a;a) (derivative with respect to x) of the Taylor polynomial T8(x,a) of f.
(b) Determine all coefficients of the Taylor polynomial T3(x;a) of f.
(c) Give the remainder term of Lagrange R4(x;a) for T3(x;a) of f. For which ξ between x and a is |R4(x;a)| largest for x > a and for x < a?
(d) Approximate the value of √2 through T3(x; a). Choose a favorable (i.e. easy for the calculation) expansion point a > −1. According to subtask (b), what is a limit for the largest possible error that occurs with this approximation?

[proud mural]

And that’s what I got

[cedar kiln]

@proud mural Has your question been resolved?

[cedar kiln]

@proud mural Has your question been resolved?

[cedar kiln]

@proud mural Has your question been resolved?

[next palm]

for a), do you really need to calculate all those derivatives? from the general shape of taylor polynomials, you could notice something about what happens when you insert x = a

[proud mural]

sorry, but i dont know what you mean

[next palm]

If you have a Taylor polynomial, say, T200(x,a) of a function f, can you tell me what T200(a,a) is without performing any calculations?

[proud mural]

it´s f(a) or?

[next palm]

Yea

So to get T8(a,a) you don't need to calculate eight derivatives

You just plug into the original function

And what is the derivative T8'(a,a)?

[proud mural]

im not sure but is it not also f(a) ? (sorry but have to translate everything first)

[next palm]

That's ok; i speak german, dutch, and a bit of finnish too if that helps hahaha

[next palm]

No, not quite f(a). Consider an example, T2(x,a) = f(a) + f'(a) (x - a) + f''(a)/2 * (x-a)^2

If you took the derivative of this taylor polynomial and inserted x = a, what would that equal?

[proud mural]

T8(x,a) = f(a) + f´(a)(x-a)+(f´´(a)/2!(x-a)^2+........+f´´´´´´´´(a)/8! * (x-a)^8

oder liege ich da falsch?

[next palm]

jau stimmt schon

aber wenn du das am punkt x = a auswertest wird das sehr einfach

und das gilt auch für die ableitungen davon

[next palm]

nimm die formel mal und schreib auf was die ableitung davon wäre

[proud mural]

die formel T(x,a) soll ich ableiten oder?

[next palm]

jau

das möchte ja auch die aufgabe von dir, dass du T'8(a,a) ausrechnest

[proud mural]

T8´(x, a)= f(a) +f'(a) * (x-a) + f(a)/2! 3(x-a)^2+....+f´´´´´´´´(a)/7!8(x-a)^7

[next palm]

wo kommt das f(a) am anfang der summe her?

[proud mural]

gute frage, jedenfalls sollte nach erfahrung T8(a,a)=f(a) sein

[next palm]

ja das stimmt schon, aber die ableitung des taylorpolynomials ist nicht so

was wäre denn die ableitung des ersten taylorpolynoms T1(x,a) = f(a) + f'(a) (x - a)

[proud mural]

Okay, habe ich vergessen 😦 und wie bestimme ich jetzt den wert oder war das schon die bestimmung?

sorry bro, bin gerade ziemlich verwirrt also was soll ich als nächstes berechen?

[cedar kiln]

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[shy cedar]

hello

[shy cedar]

could anyone help with solving these?

ive tried once but noone that could help replied

[red stone]

German

[shy cedar]

its czech originally

<@&286206848099549185>

Yes

can you help ._.

yiipppeeee

you gonna read it or something?

yeah

idk how you did that but ok

oh

there are results but u need the whole calculation

[cedar kiln]

@shy cedar Has your question been resolved?

[shy cedar]

i got the 2nd and 3rd right but the rest i dont have a clue

[cedar kiln]

@shy cedar Has your question been resolved?

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[languid trail]

Hi i need some help.
I need to find f'(x) and calculate f'(2) when
A) f(x) = x^2
B) f(x) = 2x^3 - 3x
C) f(x) = 2
I thought i was doing good but it was all wrong, then realized some have f' and some just f

[languid trail]

I dont know what any of those mean

Heres a picture

[granite knoll]

f'(x) is the derivative

[languid trail]

What is a derivative

And how do i do it

[granite knoll]

you should review the concepts: https://www.khanacademy.org/math/differential-calculus/dc-diff-intro

[languid trail]

Alright ima try

Ty

.close

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[subtle grotto]

need help with c

not with a and b

[cedar kiln]

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[daring talon]

Determine the largest and smallest value, if any, of f(x, y) = 2x^2 + y^2 + 2y, on the part of the circular disc x^2 + y^2 ≤ 4 where x ≥ 1.

[daring talon]

When I want to optimize the circular arc

The answer say use

Why do you put the gradients vertical in the determinant?

and not horizontal? Maybe a stupid question.

[dawn junco]

it doesn't matter in this case

you just want to know where the two gradients are linearly dependent

(lagrange multipliers I suppose)

[daring talon]

Okey, so they just did vertical for no reason?

Im used to put them horizontal.

[dawn junco]

well the determinant doesn't change when you transpose a matrix..

[dawn junco]

ig

[daring talon]

Okey, think I get it. Thank you so much 🙂

.close

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[stoic void]

small question, during linear transformation, the dimentions do not have to be multiplyable like in the matrix multiplication case? This is a slide from one of my lectures

[granite knoll]

what do you mean "multipliable"

[stoic void]

i mean

when u multiply a matrix by a matri

x

the dimentions have to be

take mxn as first matrix dimentions

and second matrix has to be nxp

[granite knoll]

yes

[stoic void]

right

[crimson delta]

and here its 3x2 and 2x1

[stoic void]

oh my god im stupid

nvm me

[granite knoll]

ok

.close

[cedar kiln]

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[stoic void]

.close

[granite knoll]

i gotchu

[cedar kiln]

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[worthy basalt]

pls

[worthy basalt]

someone

explain...

WHERE DOES THE 2 COME FROM

[hidden mural]

there are many 2's

[trim jetty]

What is 2?

[muted bear]

https://youtu.be/aXbT37IlyZQ?si=LxlNe8G8c71ZSbiT try checking out this video

[worthy basalt]

wait

[hidden mural]

1+1

[worthy basalt]

ok here

how do u do this

ik

r^2+3r+2

Ik how to write their specific formulae

but the thing above

adds a 2 as well

why

[modern sparrow]

because

the series formula for r^2 and r start from r=1

but the sum in the question start from r=0

[worthy basalt]

what's that supposed to mean

[modern sparrow]

so you need to consider those as well in addition

[worthy basalt]

when r starts from 0 what does it mean

[modern sparrow]

same thing as starting from r=1 but add an extra term, i.e. if we let $f(r) = r^2+3r+2$, then the series is
$\sum_{r=0}^{n} f(r) = f(0) + \sum_{r=1}^{n} f(r)$

[wraith dagger]

lgkoo

[worthy basalt]

oh

that's it?

[modern sparrow]

yah

[worthy basalt]

oh okay thx man

cuz I was trying to solve it for hours

and I didn't get it

[cedar kiln]

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[fallen oyster]

I want to do a proper proof of this statement but im not sure if im supposed to start from the middle AGM then do two-by-two comparisons and then put them all together, or am i supposed to start with a statement with the 3 inequalities with basic facts and then move up to what i want to prove?

[cedar kiln]

@fallen oyster Has your question been resolved?

[fallen oyster]

<@&286206848099549185>

[flat prawn]

U can start from the first two and proceed rightways

[fallen oyster]

I see, another thing is i have to mention at the end when the equalities hold, do i do that seperately for each two or do i do one for all 4?

[flat prawn]

U can first prove (✓ab) ≥ 2/(1/a)+(1/b) , then a+b/2 ≥ ✓ab and soon

So ig seperately

I find it easier

[fallen oyster]

okay let me write that up rq

[flat prawn]

U can perform RHS - LHS and simplify it , in the end , it will show a result which can only be greater than or equal to zero
Therefore RHS - LHS ≥ 0
Which implies , RHS ≥ LHS
or , LHS ≤ RHS ( as in the question)

[fallen oyster]

wym? i was thinking of starting with the bf that for any real pos a b, (a-b)^2>=0 then derive all the two-by-two inequalities with that then connect them together

let me write the whole thing ill show

[flat prawn]

I see

I'll post mine too once u do it

Should I post mine?

[fallen oyster]

nope i was mad stuck on the first inequality

i just got it

rest should be more free

gimme like 5 🙏

okay done

@flat prawn

[flat prawn]

Oh damn nice

Wait

[fallen oyster]

this is a really clever way of constructing the inequality

never seen that before

thank you so much for your help 😇

[flat prawn]

Welcs

[fallen oyster]

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[astral relic]

[radiant fjord]

similar triangles!

[astral relic]

but tan b gives the ratio of ac to ba

[radiant fjord]

omg why am i stupid

you're so right 😭

one sec

[astral relic]

no u arent

🥰

[astral relic]

take ur time bud

[lone plaza]

3 -4 -5

9 - 12 - 15

[radiant fjord]

basically you have a pythagorean triple

the tan B = 3/4 tells you that both triangles BDE and BCA are similar to a triangle with (CCW from top) side lengths 4, 3, 5

[astral relic]

but how did u know

[radiant fjord]

pythagorean triple

[astral relic]

okay

[radiant fjord]

5^2 = 3^2 + 4^2

[astral relic]

ty!

[astral relic]

ty u too!

[radiant fjord]

hopefully you can solve for the rest now!

[lone plaza]

np

[astral relic]

Yes!!! Thank you a million

[astral relic]

^

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[civic orchid]

Hey, can anyone help me find the x intercepts in the equation "y=3x^3-14x^2-7x+10" (polynomial btw)

[void wasp]

Have you tried anything yourself?

[teal berry]

factor it

[civic orchid]

Yea I think one of them is (3x-2)

idk abt the other 2 tho

[undone halo]

Try dividing it by the factor you have

[void wasp]

Maybe I misunderstood the question but an x intercept is only a number not depending of x no?

[civic orchid]

Coordinate

like x,0

[void wasp]

Yeah but what's your x value

[undone halo]

thats what theyre trying to find

lo

they want to find the values of x such that y=0

[void wasp]

With these types of equations you have to guess the first value

[undone halo]

u can use rational root theorem for that

[void wasp]

I am talking about this. Idk what this is and how he got that @undone halo

[undone halo]

they say that one of the factors of the polynomial is (3x-2)

meaning when x = 2/3 the polynomial is equal to zero

I havent checked if thats true

Well it is actually

[void wasp]

explaining math in your second language is hard

Well I hope you figure it out with your way

[cedar kiln]

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[quasi scarab]

I know what to do I just can get the ans

[coral jewel]

what have you tried

[quasi scarab]

24/12=20/x-4

[coral jewel]

well? what do you think about it

[quasi scarab]

I'm getting 14

Oh

I just got it

Lol

Can I send another one?

[cedar kiln]

It is suggested that you limit yourself to one question per help channel, opening a new one once your original question is answered and your original channel has been closed. This is to make your channel easier to follow for potential helpers and can bring attention to the fact that your question has changed.

[quasi scarab]

.close

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[warm spruce]

Can somebody explain what exactly a characteristic polynomial is? How is it invariant under matrix similarity and how does it have the eigenvalues as its roots?

[cedar kiln]

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[cedar kiln]

@warm spruce Has your question been resolved?

[tawdry pagoda]

f(lambda) = det(A-lambda*I)

[cedar kiln]

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[hoary parrot]

I'm stuck on part b, please help

[languid bison]

Where did you get stuck? If you solved part a, I guess you had some progress with part b as well

[hoary parrot]

Ill send a picture of my working

I have no idea what to do

I tried doing m-2 = -2 but that gives m=0 which isn't a solution

[languid bison]

You want to get it so that there is no solution, right?

[hoary parrot]

no unique solutions yeah

so just find the value of m which gives a unique solution and say that m != that value

[languid bison]

Uh I don't think this is how it works

In most cases (for most m), there are unique solution

[hoary parrot]

That's how I got the previous answer for part A, which was correct 👀

@languid bison these are the answers provided by the texbook

[languid bison]

Huh, idk then.

I am half sure there is typo in the solution, but I don't want to do calculation right now. Maybe others could help with that

<@&286206848099549185>

[cedar kiln]

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[hoary parrot]

<@&286206848099549185>

[haughty turtle]

maybe like this?

[late otter]

the system will have no solutions for m = 2 and m = 3, so technically no unique solutions

[hoary parrot]

ok thank you

.close

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[errant eagle]

Find the value of x

[tepid quail]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[deft plank]

Shree dharacharya

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[errant eagle]

This is the question

Looks like my equation was wrong above

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[heavy crypt]

It is given that these matrices commute,
i.e. AB = BA. Is the following statement true or false?
(a) if x = 0 and y ≠ 0 then b = c

[heavy crypt]

I did the multiplication but I don’t see the connection between b and c

cause then itll only be 'b*y'

[flint burrow]

it's kinda weirdly stated but if they commute then AB=BA, you have done AB, do now BA, then let x=0, and given y ≠ 0 and b = c, is the statement true then? that is: is AB=BA?

[heavy crypt]

but if AB = BA then isnt the outcome the same

[flint burrow]

indeed, the question is basically asking: given x=0, y ≠ 0 and b = c, is AB=BA?

if yes then the statement is true, if not then it's false

[heavy crypt]

ahh

that order is confusing haha

ok ill try

ok ty

.close

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[bright flame]

Is it possible to go from the above general homogene characteristic equation to the general solution with the B?
Or did they skip the between steps? (I haven't solved a differential in a year but we need to solve one for electrical engineering) Because if they skipped steps I might as well learn this step by heart rather than try and solve it myself

And do you think the 2nd green arrow is a different way of writing the above (so also a solution) or is it an additional explanation for lambda?

[dawn junco]

No the part with the Bs is still just the homogeneous solution

If all the roots of the characteristic equation are distinct, each root r will incur an e^rt term in the solution

@bright flame

[bright flame]

Allrighty that does ring a bell (but like I can barely hear it ring 😉 ) from mathematics a year or so ago

I'm assuming that e^rt can be written as the 2nd arrow (because e^rt is another way of writing a imaginary number iirc right?)

dangit I should've tried to remember some of this from mathematics

[dawn junco]

r doesn't have to be imaginary

Wait wdym by second arrow

[bright flame]

it says r is either positively real or complex added imaginary so for some that would be the case

[dawn junco]

You mean the 3rd line you highlighted ?

[bright flame]

yellow

I have some extra notes on this on paper but I have zero clue what they mean, let me see if I can show them

I'm literally clueless what is what

[dawn junco]

Yeah the 2nd arrow is a way of extracting real solutions from the complex solutions e^(g+jd)t and e^(g-jd)t

[bright flame]

yes i think thats what I wrote also

but I have no clue what it means tho

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[dawn junco]

.reopen

[cedar kiln]

✅

[dawn junco]

Re my internet broke for a while

@bright flame

There's a small result that justifies that e^gt cos(dt) and e^gt sin(dt) are solutions

[bright flame]

I've gotten this from my math last year I think it's similar to this but still don't quite see how it's the same

[dawn junco]

i.e. that if U(t) + iV(t) is a complex solution to a linear homogeneous DE, then U(t) and V(t) individually are also solutions

[bright flame]

mhm

[dawn junco]

It's not hard to check at all

So if we focus on e^(g+jd)t

Its real part is e^gt cos(dt)

Its imaginary part is e^gt sin(dt)

You can do the same for e^(g-jd)t but it turns out its real and imaginary parts are just a scalar multiple of e^gt cos(dt) or e^gt sin(dt)

So it's redudant to count them also

[bright flame]

okay

I think I understand

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[analog sorrel]

would the acceleration-time graph look like this?

[modern sparrow]

yes

although, the acceleration-time curve shouldn't really intersect/touch the y-axis, since at t=0, the gradient is infinite

[analog sorrel]

so the acceleration line is just the gradient

[modern sparrow]

yh, acceleration is the rate of change of velocity (definition), so is just gradient

[analog sorrel]

igh i always struggle on graphs like idk when to find the area under the graph, when to just read off the graph and when to work out the gradient

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[languid latch]

help log(log(x-7))=0

[languid latch]

what is x?

or i know x is 3 but i dont know how to calculate

[clear umbra]

tip: log(1) = 0

[warm spruce]

what's the base of the logarithm?

[languid latch]

10

log(x-7)=0?

[clear umbra]

no

[warm spruce]

Lots of ways to approach it, but I like to go with the definition of log

[languid latch]

is that possible to calculate with no decimals?

[clear umbra]

yes

wait does this equation have a solution

[languid latch]

yes x is 3

3

[granite knoll]

plug that back in for x. what do you get?

[clear umbra]

oh nvm im dumb

[warm spruce]

x is not 3

[clear umbra]

its not 3

[languid latch]

must be, old national test

[grim yarrow]

u doing linear eqns?

show me question,they are so ez bro

[clear umbra]

if log(log(x-7)) = 0, what must log(x-7) be

[warm spruce]

log base a (x), is just what a needs to be raised to, to get x.

[clear umbra]

tip for this: log(1) = 0

[languid latch]

log(x-7)=1 ?

[clear umbra]

ok good

now can you solve this?

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[finite badger]

hey can anyone help me build the differential equation for this system? (entrance is u, x is the output)