help-0

[alpine sable]

I'll just draw it out

to make sure

we're thinking the same thing

[summer dirge]

i always called it ΔL, if that looks familiar?

alternatively, perhaps you've seen it as |S_1P_1 - S_2P_1|?

[summer dirge]

yes

[alpine sable]

yeah, so we want the hypotenuse

of the actual triangle

[summer dirge]

you want the difference in the length between the two lines

[alpine sable]

can anyone answer this for mr

[summer dirge]

!occupied

[lone heart]

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).

[alpine sable]

so hypotenuse minus opposite

or whatever the top one is

[summer dirge]

no

[alpine sable]

actually no it would be

adjacent

sorry

hypotenuse minus adjacent gives us difference

lets do that real quick

[summer dirge]

what is "adjacent"

[alpine sable]

bottom one is assumed to be lambda distance greater

just for description sake

it would be the top wavelength

above the hypotenuse

[summer dirge]

my issue is with your wording

the lines need not form a right triangle

this is your situation

it is not a right triangle, as you can see

theres no "hypotenuse" or "adjacent" side

[alpine sable]

yep

okay I think I need a bit of a push in the right direction

[summer dirge]

path length difference is just |S_1P_1 - S_2P_1|

[alpine sable]

I have no idea how to derive the path length

[alpine sable]

how would this

formula work

just asking from a point of wanting to understand

[summer dirge]

S_1P_1 is a distance

which you've been given

[summer dirge]

see here

you've also been given S_2P_1

[alpine sable]

oh completely looked over 10cm and 11 cm lol

let me update the labelling then

is the formula

[summer dirge]

yeah

so you have seen it before

[alpine sable]

okay so the absolute value part is essentially describing how the path length difference is equals to the set of variables

path length difference being the difference between the first distance indicated by PnS1 and the second distance PnS2

[summer dirge]

yes

well, the absolute difference

because we dont have negative length

[summer dirge]

you may use this to solve for lambda though

[alpine sable]

so in terms of visuals we're looking for the yellow line?

[summer dirge]

essentially yeah

[alpine sable]

why though

what significance does that serve

I would understand if we were looking for "L" being the distance between the points and the screen

idk im a total physics noob so

I'm just trying to understand why formulas are

[summer dirge]

hmm

[alpine sable]

also from what I'm seeing "n" is some form of constant

[summer dirge]

let's see if i can describe this to you

i havent done physics explaining for a bit

please bear with me

[alpine sable]

haha yeah no problem

[summer dirge]

n is the nodal line number

okay, here goes nothing

[alpine sable]

which is

"c" being the constant speed of light makes sense to me

[summer dirge]

when you're looking at a nodal line, you're looking at the line where the waves generated by S1 and S2 destructively interfere

[alpine sable]

okay I get destructive interference

where it makes a flat line because both amplitudes add up to that

[summer dirge]

since P1 and P2 generate waves at the same rate, this must occur at a half-integer multiple of the wave lengths of one of them

[alpine sable]

mhm because being 1/2 lambda more would cause destructive interference

[summer dirge]

yes

[alpine sable]

while 1 lambda more would cause constructive interference

[summer dirge]

so the path length difference between the two sources and the point on the nodal line must be equal to a half-integer multiple of the wavelength

[alpine sable]

but is this under the assumption that destructive interference is happening between the two points?

or does it always happen when you have two waves

[summer dirge]

this is what "nodal" line means!

destructive interference

all along the line

[alpine sable]

this is an image I like

[summer dirge]

mhm

[alpine sable]

is this what happens between

any two things

including in this instance

[summer dirge]

it turns out that you can take the number of the nodal line, cut 1/2 off of it, multiply it by lambda, and get the path length difference between the two sources and any point on the nodal line

i cannot exactly remember why, but it seems intuitively clear to me now

[alpine sable]

a lot of words here but I'll try clarifying:

so what your saying is the difference in between the two lines (path length difference) must be equal to a half integer multiple of the wave length

• my question is which wave length are we referring to

[summer dirge]

the wavelength of the waves generated by S1 and S2!

[alpine sable]

constructively I assume

right?

[summer dirge]

the path length difference is a bit more general than that

when you have constructive interference, the path length difference has to be equal to an integer multiple of the wavelength

does that make sense?

[alpine sable]

yep

agreed

[summer dirge]

so when you have destructive interference, the PLD has to be equal to a half integer multiple of the wavelength

[alpine sable]

correct

[summer dirge]

right, so what is the issue you're having rn?

[alpine sable]

so what I'm trying to understand is, what is the overall thing being formed from PS1-PS2

[alpine sable]

is it just a general wave length as shown here

probably shouldn't focus on that part too much then, in that instance lol

[summer dirge]

this picture makes it a bit hard to see what we mean by PLD

lemme see if i can find a more intuitive picture

i dont expect to though, since this idea isnt a very "clean" one to look at

okay nvm, this may work

[alpine sable]

now a nodal line would be as indicated by the image right here

where the red lines are the nodal line

aka where it's consistently destructive interference

[alpine sable]

what am i looking at

[summer dirge]

point C is on an anti-nodal line where constructive interference occurs

[alpine sable]

okay

[summer dirge]

you can physically count the number of crests to get there from both S1 and S2

[alpine sable]

oh 4 and 6

i see it labelled too

[summer dirge]

yeah

for a nodal line, you'd count half-integer multiples

the path length difference is just that

the difference in path length from S1/S2 to P

[alpine sable]

mm

so in the image you illustrated

it's 2 lambda?

[summer dirge]

thats right

[alpine sable]

okay perfect

[summer dirge]

in our question, they explicitly give you the distance from the sources to the point

[summer dirge]

i.e. they given you something like this

just on a nodal line instead

[alpine sable]

^^ just loaded it here for easier reference

[summer dirge]

and instead of 6 lambda and 4 lambda, they've given you 10cm and 11cm

[alpine sable]

mmm

okay

[summer dirge]

so now you can use this information to get lambda

[alpine sable]

so then our new formula is $1= (n-\frac{1}{2})\lambda$

[ocean seal]

mj

[summer dirge]

careful with the units

1 what?

[alpine sable]

cm

[summer dirge]

correct

but just be careful with units in general

n is unitless

[alpine sable]

okay

[summer dirge]

okay, so lambda = what?

[alpine sable]

so we have now $\frac{1 cm}{(n-\frac{1}{2})}=\lambda$

[ocean seal]

mj

[summer dirge]

yes

we know what n is though

[alpine sable]

$\frac{1 cm}{(1e-9-\frac{1}{2})}=\lambda$

[summer dirge]

it's the number of the nodal line we're interested in

[alpine sable]

wait so it's not

[ocean seal]

mj

[alpine sable]

1e-9?

wait a minute

[summer dirge]

where did that come from

[alpine sable]

so it isn't some constant

okay then

[summer dirge]

it's the nodal line number

which is given

[alpine sable]

hmmmm

I'm looking at what you sent

this here

there are multiple nodal lines though

[summer dirge]

thats right

but what does it say in your question?

[alpine sable]

okay we're given information

5 cm and 8 hz being

the frequency

and the distance between the two

[alpine sable]

which isn't important

redrew it

with out variables

our*

so why does the first nodal line matter, but not the second

[summer dirge]

P is stated to be on the first nodal line

which i've drawn here

[alpine sable]

mhm

so we don't actually know

the other one

ok kinda confused

[summer dirge]

yeah we have no info on the 2nd nodal line or 3rd, etc

[alpine sable]

oh nvm

ok

[summer dirge]

heres what we have

[alpine sable]

mmmmm okay

now

n is representative of nodal line

[summer dirge]

yes, the nodal line number

the first nodal line is always in the centreish

the other ones are further out

[alpine sable]

okay , and the nodal line in our instance is p

?

or the line that holds P

[summer dirge]

why? because the path length difference is minimal in the centre, so we call it the first

[summer dirge]

this

[alpine sable]

okay

so then would it be safe to assume

it's an average of the two

[summer dirge]

well, not quite

it's always about PLD, as i've said before

the nodal line is not necessarily dead centre

it could be in some cases, but it's very likely not in ours

[alpine sable]

hm okay

[summer dirge]

you can still use this btw

we know what "n" is

[alpine sable]

$\frac{1 cm}{(n-\frac{1}{2})}=\lambda$

[ocean seal]

mj

[summer dirge]

because we know which nodal line we're working with

[alpine sable]

first one?

[summer dirge]

thats right

which means n = 1

[alpine sable]

oh

I had thought of n being

the value

of the nodal line

not the nth nodal line

[summer dirge]

theres no definition of the "value" of a nodal line

[alpine sable]

yeah

[summer dirge]

nodal lines are nodal lines, we only given them numbers for convienence

[alpine sable]

so it's just 1 because it's the first nodal line

[summer dirge]

correct

if it were the 2nd, we would have n = 2, but the path length difference would also change in tandem

[alpine sable]

so if in the question it told us the second nodal line is located 10 cm from... we would use 2?

okok

[summer dirge]

and you'd get the same lambda in the end anyway

which makes sense, right?

[alpine sable]

$\frac{1 cm}{(1-\frac{1}{2})}=\lambda$

[ocean seal]

mj

[alpine sable]

it does

[summer dirge]

the wavelength doesnt change just because you sit on a diff nodal line

[alpine sable]

now hold on, we don't have seconds in denom

is that a problem?

[summer dirge]

wavelength has units of length

theres no need for time

[alpine sable]

ohh okay i see

so this is just

[summer dirge]

so you have 1cm/unitless

[alpine sable]

finding lambda

[summer dirge]

yeah

[alpine sable]

okay got you

so that would mean

lamda = 2 cm

[summer dirge]

correct

[alpine sable]

lambda*

alright, now we would sub that into our

$v=f\lambda$

[ocean seal]

mj

[alpine sable]

equation?

[summer dirge]

yes

and you're done!

f is given

[alpine sable]

woah

okay

this is some

neat stuff

[summer dirge]

here's my advice for high school physics problems

ALWAYS write down ALL of your givens first

[alpine sable]

yep, I even highlight them lol

[summer dirge]

it really helps you figure out which formulas you need

okay good

[alpine sable]

well thanks for your help 🙂

[summer dirge]

no problem

i did this problem 1 year ago now, so its been a while haha

[alpine sable]

haha atleast you won't have to relearn how to do it lol

always good to refresh

your memory

[summer dirge]

indeed

[alpine sable]

well thx again, gotta do the 30 other problems there are of this

.close

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[hot bison]

So for my math test I am allowed an 8.5 x 11 formula sheet. But no worked out examples or procedural steps, ect. Would the image below be considered a formula or something else?

[flint sonnet]

since there is no numbers ig its not an example

[hot bison]

so I could write that down?

[flint sonnet]

well the picture is not really necessary unless you really need it

[hot bison]

ok, thanks

.close

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[mystic holly]

im not sure where i went wrong

[fallen verge]

what is the antiderivative of 1/x

[echo socket]

Integrating x^{-1} should not yield 1

[alpine sable]

x^(-1) is not integrable via the common integration method for polynomials

[mystic holly]

well i dont have 1/x anythere

[alpine sable]

as i told you in your earlier help channels

[fallen verge]

you have (1/3)*(1/x)

[alpine sable]

,, \int x^n \dd x = \f{x^{n+1}}{n+1} + C \qq n \in \R\setminus\set{-1}

[alpine sable]

note the setminus of -1

[mystic holly]

so bascially we split it up into 2

[alpine sable]

elaborate.

[mystic holly]

you split up 1/3x into (1/3) and (1/x)

and then you took the intergral

true or false?

[fallen verge]

i guess

[mystic holly]

well then true or false

[fallen verge]

not how integration works

[mystic holly]

hum

could we do it befroe becasue it was a constant

[fallen verge]

\int f(x)g(x) dx \neq (\int f(x) dx)(\int g(x) dx)

$\int f(x)g(x) dx \neq (\int f(x) dx)(\int g(x) dx)$

[ocean seal]

ニンニク狼

[fallen verge]

ok sure

however, as you said, $\int c(f(x)) dx = c \int f(x) dx$

[ocean seal]

ニンニク狼

[mystic holly]

im not good at understanding letters notations

but basically, if its a constant you can split it and if it is not then you dont split it

yes no?

[fallen verge]

yes

[mystic holly]

alright, so goign back to the problem, uhh

(resending for convinence)

so i should split first and then integrate?

[fallen verge]

wdym by split

[mystic holly]

how does this look

[fallen verge]

thats good!

[mystic holly]

split like this

it does not look like any of these

closest seems like a

but we just imported an aboslute value sign

[fallen verge]

oh yeah

$\int \frac 1x dx = ln|x|+C$

[ocean seal]

ニンニク狼

[fallen verge]

the negative side has to have an antiderivative too

[mystic holly]

im confused

what

[flint sonnet]

the integral of 1/x is equal to ln|x|

[mystic holly]

.close

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[lethal spade]

hello I have to deal with the social surplus the problem is Prices and quantities demanded at dates t = 0, 1 are as follows:
P0 = 8, P1 = 10 and Q0 = 30, Q1 = 20.
What is the arc elasticity of demand?
I found 3.37

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Please don't occupy multiple help channels.

[alpine sable]

@lethal spade close either one of the channels

[lethal spade]

yes

[lone heart]

@lethal spade Has your question been resolved?

[lethal spade]

<@&286206848099549185>

[tacit arch]

Send the formulas for elasticity and work you did

[lone heart]

@lethal spade Has your question been resolved?

[lone heart]

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[quasi sonnet]

How do I continue

Mines the first picture

[tacit arch]

Stop closing and opening

No x is not a

[quasi sonnet]

Ok then x2=9 and x2=-4

Then what

[quasi sonnet]

?

.close

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[silent sundial]

Could someone help me with this problem: -5x^2-18xy+50x+8y^2-20y

[tacit arch]

What's the question

[silent sundial]

its just what i posted

you have to factorize

apparently

[tacit arch]

This is important to the question

[silent sundial]

Yeah

[lone heart]

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[silent sundial]

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[scenic wing]

where did the 2x and divided by 2 come from

[scenic wing]

i thought sin^2x +cos^2x = 1

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[austere compass]

Can someone help me go through this IVP exercise?\
\
\
Show that there's an interval $I \subseteq \mathbb{R}$ with $3 \in I$, so that the initial value problem
$$
\begin{cases}
x'(t) = (x(t) - t)e^{t^2+x(t)}\
x(3) = 7
\end{cases}
$$
has a unique solution.

[ocean seal]

Levens

[austere compass]

I tried having a go but I didn't get anywhere

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[austere compass]

<@&286206848099549185>

<@&286206848099549185>

[austere compass]

<@&286206848099549185>

man.. worth a try

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[alpine sable]

i was about to take a look at this 💀

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[vital wind]

I understand what the question is asking me to do, I just dont know how to do it.

Like I understand that I wont get an actual value for p so it will be in the answer I just dont get how im supposed to represent the area of the triangle.

[vital wind]

Poor dude was waiting for an hour 😭

[alpine sable]

i know 😭

[vital wind]

<@&286206848099549185>

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<@&286206848099549185>

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[alpine sable]

im working on it but got wrong answer

ln(x+8)/3 = ln x

.close

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[wooden topaz]

need help

[wooden topaz]

have no idea what to do

[tacit arch]

Calculate the derivative piecewise, then evaluate the one sided limits of the derivative

[wooden topaz]

?

find the derivative of each equation ?

i mean expression

[wooden topaz]

could you explain

[left isle]

yes

[wooden topaz]

4x-4 would be the answer ?

[left isle]

first find the derivatives of each, then plug in x=1 for each

if the answer is the same for both, then it exists and that's your answer

[wooden topaz]

-4x-4

okay well i got it wrong

how is this wrong

nvm

[tacit arch]

Pictures of screens are bad. Upload a screenshot instead

[wooden topaz]

oh okay

is this good enough

[wooden topaz]

lost on this one

<@&286206848099549185>

due at 11:59!

um

ok ig i fail

[left isle]

that isnt the derivative

[wooden topaz]

oh i already got it incorrect so nothing i can do now

im just on other problems

[charred jewel]

since you have a 8^h there, your guess should be an exponential function

you also have a 1 there

so what should your x be for your exponential function to be 1

or rather your a

[wooden topaz]

0 ?

[charred jewel]

ok good

now the first part of the limit definition of a derivative has a f(x+h)

since your x = a = 0, what should your f(x) be

remember you have a 8^h there

[wooden topaz]

0 ??

[charred jewel]

no...

[wooden topaz]

1?

[charred jewel]

no

f(x) is a function

not a value

[wooden topaz]

1 + 0 ?

[charred jewel]

no..

ok here we'll do a little guessing

since you have a 8^h there

lets guess that f(x) = 8^x

does that work?

[wooden topaz]

i think

well i tried that so no

[charred jewel]

really?

[wooden topaz]

yeah

for fx =8^h

[charred jewel]

its not h

[wooden topaz]

a= 1

[charred jewel]

its asking for f(x)

its input is x

so the argument for the function should be x

[wooden topaz]

oh

[charred jewel]

and a is also not 1

[wooden topaz]

-1

?

[charred jewel]

what did you tell me here before?

[wooden topaz]

a is 0 ??

[charred jewel]

yes

[wooden topaz]

do you know a good calc channel

calc 1 channel

i missed class last week

and also my professor isnt the best

[charred jewel]

organic chem tutor maybe

[wooden topaz]

one last thing unfortunately

i am on my 3 rd and last attempt

do i plug in 1

[charred jewel]

yeah

[wooden topaz]

thats it ?

2 plus 1

which is 3

then 3-1

[charred jewel]

its f(3)

[wooden topaz]

?

oh ok

sorry for pic but im in a rush

last attempt and last question

so this is correct

[charred jewel]

does the question want it in terms of f?

or a value

[wooden topaz]

uh

refer to this screen shot

i think just terms of f

[charred jewel]

then no

if h = 1

what is f(2+h)

[wooden topaz]

3

[charred jewel]

f(3)

remember you are still inside f

[wooden topaz]

f(3)- f(2)

divided by h

so 1

[charred jewel]

yeah

(but if you still get it wrong then im so sorry the question didnt really specify what it wants)

[wooden topaz]

its okay thanks for helping

[left isle]

is this a test?

[wooden topaz]

no homework

i got a 70

so its ok

just need more practice

.close

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[jovial flax]

6a)

[jovial flax]

i how do i calculate this im just not sure really

<@&286206848099549185>

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@jovial flax Has your question been resolved?

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[west brook]

I know this might be the wrong topic to ask, but is anyone here familiar with Digital Logic in computer science/engineering? This is a circuit and problem that I am working with, but I just need help understanding it

[median oar]

Are you trying to make a full adder?

[west brook]

yeah

[median oar]

Did you try the hint

[west brook]

yeah

i know it comes out to this

[alpine sable]

yeah

so now

k-map both the Sum and C-out outputs

[west brook]

but im just having trouble just understand how to get that into the whole circuit thing

[alpine sable]

do this

minimise using a k-map

[median oar]

Is that short for karnaugh

[alpine sable]

yes

[west brook]

uhh i dont think i learned k map or i mightve not reviewed it yet

[alpine sable]

well okay i guess we are doing the standard way then.

are you familiar with sum-of-products representation of boolean functions

[median oar]

(Karnaugh maps was invented to make this part of the problem easier)

[west brook]

yes

[alpine sable]

okay. So lets focus on the sum output for now. Write the 'sum' as a sum of products

[west brook]

you can just do this right:
A’ B’ C-IN + A’ B C-IN’ + A B’ C-IN’ + A B C-IN = sum

[alpine sable]

okay lets see

okay yes

thats right

lets replace C-IN with C for clarity sake

[west brook]

alright

[alpine sable]

you have [
A'B'C + A'BC'+AB'C'+ABC
]

[alpine sable]

now, its time to minimise this

do you know your boolean algebra

[west brook]

a little bit

[alpine sable]

here is a good starting hint

[west brook]

you can end up getting
C(A'B' + AB) + C'(A'B + AB')

[alpine sable]

yeah

[west brook]

and then C XOR (A XOR B).....?

[alpine sable]

yes!

that's the sum indeed

okay so we are going to do the same thing for C-out

list it out

[west brook]

so if you look back at this chart, you can derive
A'BC + AB'C + ABC' + ABC

and then you can simplify to AB + BC + AC

[alpine sable]

yeah

so the thing is though

what you wrote is correct

but there is a simpler way to view the circuit for the C output

We need to view the full adder as two half adders that are connected

which is true

do you know the logic diagram for a half adder?

[west brook]

um

i dont think i do

i just looked it up and it says this

[alpine sable]

yeah it is that

a full adder is basically two half adders constructed together

but anyways i guess this will be a bit hard to explain

[alpine sable]

this form is fine

[west brook]

okok

hello...?

@alpine sable hey are you there

[alpine sable]

what do you need

[west brook]

i was still stuck on the problem because i still dont really understand how making that truth table would go to the circuit

because im having trouble understanding the circuit

[alpine sable]

like just draw out the logic diagram now

you have the equations for both output minimised

[west brook]

.close

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[fathom ravine]

Hi, could you help me please? How can I resolve this exercise?

[fathom ravine]

Like I don't know what to do with logarithm

[exotic canopy]

use the right hand side of this rule on each of the fractions

[fathom ravine]

Hm, I still don't understand, could you write it out?

[exotic canopy]

81=ln(e^81)

[lone heart]

@fathom ravine Has your question been resolved?

[fathom ravine]

Could you write it out on paper? Like the whole resolved exercise, so I would be able to analyse it, because I still dont understand

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[cerulean junco]

What do i do next , ....

[mortal trellis]

multiply out the middle two brackets

and then immediately factor them again, but this time in the other order

(or in other words, just switch the two terms)

[cerulean junco]

ok ,but what is the logic behind taking out the "middle 2 brackets" ?

[mortal trellis]

well you want to switch them so that they will cancel with the other terms

either you know that you are allowed to do this if you only have combinations of A and I

or you do it the slow way by multiplying out and then factoring again, but in the other way around

[cerulean junco]

didn't get this point .....

[mortal trellis]

then ignore it

[cerulean junco]

u mean expanding !?

[mortal trellis]

yes

[cerulean junco]

,rotate

[cerulean junco]

.

[mortal trellis]

no

not sure what you did there

I mean only (I-A)(I+A)

show that it is equal to (I+A)(I-A)

[cerulean junco]

oh okay i get that ....

[cerulean junco]

got it ,
but can you tell what you said here ,i guess i missed some concepts ..

[mortal trellis]

notice that while you expanded, you only ever got terms of the form A^(i+j)

which is the same as A^(j+i)

so you know that you will be able to undo all the expanding, but with writing the terms the other way around

this will always happen if you multiply terms looking like this

like a polynomial in A, if that means something to you

some linear combination of I, A, A^2, A^3,...

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@cerulean junco Has your question been resolved?

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[warped topaz]

https://i.imgur.com/EGnG5jz.png

[warped topaz]

Im confused about the bit where you change from x to y

Like at which point do you do it

After you square f(x)?

After you integrate the squared f(x)?

CHARTBIT!!!!!!

😍

WELCOME BACK

[pseudo ice]

Why thank you

[pseudo ice]

Also how do you mean here? as in where you need to rearrange your function for $\pi \int x^2 \dd y$?

[ocean seal]

@pseudo ice

[warped topaz]

I think so

I just know that its x and we need to change it to y

Something like that

I dont really remember

[pseudo ice]

Yep, you’d need to change it to x = … before you put it in the integral, before you put it in the bit where you square it preferably

[warped topaz]

So (x^2)/9 = y?

3sqrt(y)?

[pseudo ice]

Yep like that then you put that as your x and square it (again )

As a “cheat” as well…

[warped topaz]

I like cheats...

[pseudo ice]

As the formula has an $x^2$ in it, you could just instantly say that $x^2 = 9y$ and then the integral $\pi \int x^2 \dd y = \pi \int 9y \dd y$ (with the limits of course)

[warped topaz]

😝

[pseudo ice]

Hahah it is I’m lacking today

[ocean seal]

@pseudo ice

[pseudo ice]

What i get for choosing to do tex halfway through a message

[warped topaz]

Wait sorry, I dont follow...

Why can you instantly say x^2=9y

[pseudo ice]

Multiply by 9 here

[warped topaz]

Omg 🤦‍♂️

Okkkkk

I understand the trick

Cause it needs to be squared anyways 😄

So saves time

Thank you sooooo much!!!

You are still the best helper ❤️

Really glad you are back

❤️

.close

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[loud monolith]

i have a question about area under curve

[loud monolith]

the formula for area under curve of two functions has a modulus if you dont want to graph em

is it okay to put the modulus outside the integral?

[livid sage]

i might be misunderstanding your question but i am pretty sure no

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[deft vigil]

Geometric Arithmetic
t3 = root2
t8 = 8
Find S8

[urban talon]

breh

[deft vigil]

I was absent during class and was given that for homework

mb

[urban talon]

i was mid typing out my question

anyway

its very clearly just doubling it each time

[deft vigil]

You can type it here I guess

[urban talon]

just do your typical steps

[deft vigil]

Yeah but I don't know my GP's

[urban talon]

aight so do you know how to find the progression

t3 is surd2, or 2^1/2, and t8 is 8, or 2^3

[deft vigil]

No

Oh yeah I get that

I know my indices

[urban talon]

or well

wait

wait what

[deft vigil]

?

[urban talon]

its not doubling mb

i

i feel like its obvious enough that i should just say the answer

[deft vigil]

Yeah but

Working out

Math teachers are hungry for that

[urban talon]

but at the same time idk what the process is like here

in your syllabus

cuz id just write the answer out personally

[deft vigil]

I'll work backwards from the answer

Something got to do with a(1-r^n)/(1-r) or without the denominator

[urban talon]

nah way simpler

[deft vigil]

WTf

What's your idea?

[urban talon]

just find any function f(n) where f(3)=surd2 and f(8)=8

hint is that both of those numbers are 2 to the power of something

[deft vigil]

Uh yeah we can't do that

We mathematically can

[urban talon]

why not

[deft vigil]

But the teacher won't allow it

He wants us to do it the GP way

[urban talon]

tell them to stick it where the sun dont shine

[deft vigil]

lol

He gives us basic examples so we can get use to it

Then we apply it to hardcore questions in the future

And I can't even solve a simple one bc I wasn't there 😔

[urban talon]

why hassle yourself with the way that is both more specific, and more of a hassle

o well

[urban talon]

anyway i havent touched that equation in like years so imma leave that to someone else to teach

i got my own questions n shit

[deft vigil]

Oh alright

Thanks for trying

Sorry for that

[urban talon]

all good lol

anyway

a is 1st term right?

r is ratio or whatever

[deft vigil]

I'm not sure if it's 1st term

oh

wait

nvm

t1 = a

yes

[urban talon]

you have n and term n for 2 equations

and a general formula

if you know how to sim eq that shit you can find both a and r

[deft vigil]

But I don't have t1

[urban talon]

you dont need t1

[deft vigil]

So d will be impossible to find

Are you sure?

[urban talon]

do you know simultaneous equations?

[deft vigil]

Yes

[urban talon]

if you have 1 unknown, you need 1 equation

if you have 2

you need 2

[deft vigil]

Yeah yesyes

[urban talon]

so here you have

a(1-r^n)/(1-r)

idk if thats right

but

[deft vigil]

Yeah that's right

[urban talon]

if a(1-r^3)/(1-r)=surd2 and a(1-r^8)/(1-r)=8

thats 2 equations and 2 unknowns

and you can sim eq that shit

[deft vigil]

mmk

I'll try that now

[urban talon]

or just tell your teacher to shove it

do what you love

later

[lone heart]

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[dusk wraith]

does anyone know how to find the range

[dusk wraith]

like the steps

[summer dirge]

If you can find the global minimum, you can use that to see that the range is all numbers greater than said global min

Alternatively, it’s probably possible to find the range of this graph by inspection

[dusk wraith]

thank you!

.close

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[rocky cloak]

Can a second order voronoi diagram be created using a first order voronoi diagram algorithm applied to a transformed version of the points?

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@rocky cloak Has your question been resolved?

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[lone heart]

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

[clear tendon]

can i get some help pls

<@&286206848099549185>

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[winter fable]

Guys how am i supposed to understand this sequence:

[winter fable]

2,2,4,6,10,16 ...

what is the nth term here

or ttt rule

and how am i even supposed to understand this

[lone heart]

@winter fable Has your question been resolved?

[brittle ember]

I think it is fibonacci sequence doubled

[winter fable]

Huh

[brittle ember]

do you know the rules of fibonacci sequence?

[winter fable]

nope

can you explain please

@brittle ember

oh

yeah i searched it up

yeah it does

tho how does that explain the sequence?

btwhow to get a helper role t

.close

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[winter fable]

oh

.reopen

[lone heart]

✅

[winter fable]

sorry yeah @brittle ember wat u sayin?

[swift grail]

$F_n = F_{n - 1} + F_{n - 2}$ where $F_0 = 0$ and $F_1 = 1$

[ocean seal]

casework

[winter fable]

what the fu-

[swift grail]

Fibonacci sequence

[winter fable]

oh

ok

[swift grail]

You know its like evolution of rabbits

First there is 2 then 3 then 5 .....

[brittle ember]

so 2+2=4, 2+4=6, 4+6=10...
on your sequence

[winter fable]

yeah

[winter fable]

OHHHHHHHHHHHHHHH

[brittle ember]

adding up before 2 terms

[winter fable]

OOOOOH I GET IT

Tysm man ur the best

btw how to get helper role

[brittle ember]

i forgor

[winter fable]

bru

[brittle ember]

idk in english

[swift grail]

You just get it from the roles button in the settings of the server

[winter fable]

oh

[brittle ember]

if you're done please close this thread

[winter fable]

yeah

.close

[lone heart]

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[robust light]

Hi, question coming up, gonna try LaTex to make it readable... 1 sec

[robust light]

Let
$$T : \text{P}_2(\C) \to \C^2$$ be defined by
$$[ T(p) = \bigl(p(0), p'(0)\bigr). ]$$
Further define the following ordered bases for $$\text{P}_2(\C) \C^2$$ respectivly.

[ocean seal]

binnet

Let
$$T : \text{P}_2(\C) \to \C^2$$ be defined by
$$\[ T(p) = \bigl(p(0), p'(0)\bigr). \]$$
Further define the following ordered bases for $$\text{P}_2(\C) \C^2$$ respectivly.
```Compilation error:```! Undefined control sequence.
l.50 $$T : \text{P}_2(\C
                        ) \to \C^2$$ be defined by
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.```

[golden canyon]

I think it's $\mathbb{C}$

[ocean seal]

Jelle

[robust light]

thanks,,,, writing the question is half the struggle... lol

Let
$$T : \text{P}_2(\mathbb{C}) \to \mathbb{C}^2$$ be defined by
$$[ T(p) = \bigl(p(0), p'(0)\bigr). ]$$
Further define the following ordered bases for $$\text{P}_2(\mathbb{C}) \C^2$$ respectivly.

\text{P}_2(\mathbb{C})&:\ B = (2+2x+x^2, 1-x, 1), &&E = (1,x,x^2), \
\mathbb{C}^2&: \mathbb{C} = \bigl((1,1), (0,-1)\bigr), &&F = \bigl((1,0), (0,1)\bigr).

[ocean seal]

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

[robust light]

i) SHow that T in a linear transformation
ii) Calculate the matrixrepresentation of T och T from the definition.
iii)Calculate the base change matrix id_(P_2(C)) and id_(C^2).
iv) Verify that the clause of base change for linear transformations is fulfilled for …

Let's see is this does it:

Let $$T : \text{P}_2(\mathbb{C}) \to \mathbb{C}^2$$ be defined by
$$[ T(p) = \bigl(p(0), p'(0)\bigr). ]$$.
Further define the following ordered bases for $$\text{P}_2(\mathbb{C}) \C^2$$ respectivly.

\text{P}_2(\mathbb{C})&:\ B = (2+2x+x^2, 1-x, 1), &&E = (1,x,x^2),
\mathbb{C}^2&: \mathbb{C} = \bigl((1,1), (0,-1)\bigr), &&F = \bigl((1,0), (0,1)\bigr).

i) Show that T is a linear transformation
ii) Calculate the matrixrepresentation of $$\matrixrep{T}{B}{C}$$ och $$\matrixrep{T}{E}{F}$$ from the definition.
iii) Calculate the base change matrix $$\matrixrep{\id_{\text{P}2(\C)}}{B}{E}$$ and $$\matrixrep{\id{\C^2}}{C}{F}$$.
iv) Verify that the clause of base change for linear transformations is fulfilled for $$\matrixrep{T}{B}{C}$$ och $$\matrixrep{T}{E}{F}$$ by comparing both sides of the equation in the clause.
(suggestion: i’ts easier to find the inverse of a 2x2 matrix than a 3x3 matrix).

[ocean seal]

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

[robust light]

nope...

[twin nimbus]

If you want inline equations use $ instead of $$

[ocean seal]

全能の存在
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[robust light]

I should have practiced this somewhere bforre.... oh, thanks! that helps!

How do you type a base change matrix on here?

I.e. [A]_B ^C if you get what ia mean...

[twin nimbus]

${[A]}_B^C$ doesn't work?

[ocean seal]

全能の存在

[twin nimbus]

I'm not certain what you're wanting the typesetting to look like, so

[robust light]

How ddid you write that?

[robust light]

THIS!

oh, i see!

[twin nimbus]

${[A]}_B^C$ doesn't work?

[robust light]

\matrixrep{T}{E}{F} is what i am used to, but it didnt work here... 1 sec cleanign up my question so it's readable...

[twin nimbus]

That's probably provided by some package that you have in your normal latex environment, but not in your texit preamble

You can add it to your texit preamble using the preamble command

[robust light]

yeah sfor sur

elemme close this until I got my text formating sorted and purmed myself some coffee...

.close

[lone heart]

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[robust light]

Ops, did I do this wrong too..? i meant to close my question not the room...

[twin nimbus]

You did it correct

[lone heart]

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[robust light]

Okay I am back, same proble, better formating!

[robust light]

Let $T : \text{P}_2(\mathbb{C}) \to \mathbb{C}^2$ be defined by
$$T(p) = \bigl(p(0), p'(0)\bigr)$$

Further define the following ordered bases for $\text{P}_2(\mathbb{C}) and \mathbb{C}^2$ respectively.

$$P_2(\mathbb{C}): B = (2+2x+x^2, 1-x, 1), E = (1,x,x^2) ,$$
$$\mathbb{C}^2 : \mathbb{C} = ((1,1), (0,-1)), F = ((1,0), (0,1)) .$$

i) Show that $T$ is a linear transformation.

ii) Calculate the matrix representation of ${[T]}_B^C$ och ${[T]}_E^F$ from the definition.

iii) Calculate the base change matrix
${[id_{P_2(\mathbb{C})}]}B^E$ and ${[id(\mathbb{C}^2)]}_C^F$

iv) Verify that the clause of base change for linear transformations is fulfilled for ${[T]}_B^C$ and ${[T]}_E^F$ by comparing both sides of the equation in the clause.

(suggestion: it’s easier to find the inverse of a 2x2 matrix than a 3x3 matrix)

[ocean seal]

binnet

[robust light]

I am stuggeling with these LA proofs 😦
Hopefulle having a readble question should make it a bit less of a stuggle though...

[summer dirge]

Have you done part i)?

[robust light]

Nope, I have done nothing on this problem yet, as I am stuck, sad and frustrated. I do have decent understanding but I am just not sure of how to approach this

[summer dirge]

okay, do you know the conditions you have to check to show that T is linear?

[robust light]

So the transformation T is a transformation of polynomials of degree 2 of less with complex coefficients? to what exactly though?

[summer dirge]

to C^2

[robust light]

TOocheck linearity do somehting like check that is associative and ...

[summer dirge]

not quite

[robust light]

(sorry, translation in progress...)

i'm not used to doign math in english...

[summer dirge]

I see

No problem

[robust light]

like, check the 8 conditions?

[summer dirge]

I think you’re getting confused with the axioms of a vector space

[robust light]

oh, shit, yeah i am, spot on!

isn't it somehting like check that T(a+b) = T(a)+ T(b)

[summer dirge]

We say a function that maps between vector spaces V and W is linear if for all x,y in V and c in F, we have T(x+y) = T(x) + T(y) and T(cx) = cT(x)

[robust light]

and check that cT(a) = T(ca)

[summer dirge]

yup

[robust light]

okay, this is all good, but I have no clue what the vector space C^2 is tbh...

[summer dirge]

think of it like R^2

But instead of having real numbers, you have complex numbers instead

[robust light]

okay, cool

impossible to visualise but sure

makes more sense now

[summer dirge]

it’s the set of all possible ordered pairs of complex numbers

[summer dirge]

Indeed, but that’s how it goes sometimes lol

[robust light]

okay, got it

[summer dirge]

alright, let’s check the first condition

Let’s pick some random x and y in P_2(C)

This means that x and y are both polynomials of degree 2, with complex coefficients

[robust light]

okay

[summer dirge]

let’s let x = ax^2 + bx + c and y = dx^2 + ex + f

[fiery halo]

can someone help me solve this im clueless.

A cone has its height x and radius y. if the value of x + y = s Find its height and radius when its volume is maximum for the different values of s

[summer dirge]

So now we get T(x+y) = T(ax^2 + bx+ c + dx^2 + ex + f)

[summer dirge]

!occupied

[lone heart]

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).

[summer dirge]

Do we get this point? @robust light

[robust light]

yeah, seems correct

so we need to also prove it is linear with multiplication with a scalar then?

[summer dirge]

okay, now the map says that we must send this to ((x+y)(0), (x+y)’(0))

[summer dirge]

Yes, in principle we can do both checks at once

[robust light]

sorry, hwere is this from?

[summer dirge]

This?

[robust light]

yeah, gimmi a sec

[summer dirge]

Comes from your question

[robust light]

using the E to F ?

[summer dirge]

oh nono

[robust light]

sorry, you've lost me a bit 😦

[summer dirge]

The very top of your question

Says that T(p) = (p(0), p’(0))

[robust light]

oh, that means polynomial and it's derivative at x= 0 right

[summer dirge]

that’s right

We have T(ax^2 + bx+ c + dx^2 + ex + f)

[robust light]

okay okay.. i'm back in it

[summer dirge]

And we must send it to (p(0), p’(0))

Where p is x + y

[robust light]

should be add coefficients first maybe?

[summer dirge]

you may if you’d like, but in my opinion, you can do this more easily than that

(x+y)(0) = c + f, if you notice that

and (x+y)’(0) = b + e

[robust light]

1 sec

yeah yeah!

got it1

[summer dirge]

okay, so we have (c+f, b+e)

[robust light]

and now I would just do the same, but separately? to show that they are equal?

T(x) + T(y)

[summer dirge]

you can, or you can keep going this way

[robust light]

okay, tell me 🙂

[summer dirge]

Either way is totally valid

[summer dirge]

okay, so imo the quicker way is to not compute T(x) + T(y) because that’s painful

we left off with (c+f, b+e)

let’s split this apart

[robust light]

(c, b) + (f, e)?

[summer dirge]

to (c,b) + (f,e)

Yes

I’m so slow at typing lol

Mobile sucks

[robust light]

oh, wow, on mobile! o_O that's fast for mobile!!!

[summer dirge]

notice that this is actually equal to T(ax^2 + bx+ c) + T(dx^2 + ex + f)

which is T(x) + T(y)!

[robust light]

is this not also equal to T(Qx^2 + bx+ c) + T(Sx^2 + ex + f) ?

can i really go in that direction for the proof?

[summer dirge]

yes, but that doesn’t matter

[robust light]

okay, as long as I show that linearity was preserved it follows I guess

[summer dirge]

the goal is just to show that T(x) = T(y), what you’re asking is whether T is injective

which is a different story

[robust light]

oh, okay cool

[summer dirge]

It’s okay if multiple elements get mapped to the same thing, because that doesn’t change the fact that those elements were mapped in a linear fashion

[robust light]

true true!

[summer dirge]

alright, condition 2 is very much the same

[robust light]

that uT(v) = T(uv)

I might be able to do that one on my own then.

[summer dirge]

try it out

[robust light]

Feels like part i) should be done with then. I think I need to take a pause, and write down this answer nice and tidy, will come back for part ii) - iv).
Thank you, HUGE HELP!

[summer dirge]

okay, I’ll be back later if you still need help, or perhaps others can help you with those parts

I’ve got a class in 5 min

[robust light]

thank you! should I maybe close this room for now then 🙂