help-13

[wraith dagger]

AustinU

[royal loom]

and you can then apply the ERF

for an easy particular solution

does that make sense Jbur or do you have question

[winged jewel]

what is the denominator of the particular solution fraction

[royal loom]

well, the sine and cosine have different values in the numerator so it depends on how you write it

[winged jewel]

the p()

[royal loom]

oh

it is the characteristic polynomial of the ODE

evaluated at phi

[winged jewel]

is that the same as auziliary equation?

auxiliary*

[wraith dagger]

AustinU

[winged jewel]

yes

[royal loom]

then yes

[winged jewel]

ahh

awesome

So how do you get an answer from that?

[wraith dagger]

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

[royal loom]

Except the important caveat is that sin(10t) is only the imaginary part of e^(10it) so at the end you must remember to take only the imaginary part of the y_p as your solution

[winged jewel]

Ahh ok

[wraith dagger]

AustinU

[royal loom]

but this obviously then will require some simplification

[winged jewel]

Ohh you sub in r as the power of the exponential

[royal loom]

as the coefficient of the power of the exponential yes

[winged jewel]

Yes

[royal loom]

and then you just simplify from here, until you get it into the form a+bi

[winged jewel]

Ahh that's neat

[royal loom]

taking just the "b" will give you the solution

because sin is the imaginary part of the complex exponential

[winged jewel]

Ahh ok

So how do you convert that into the other method

[royal loom]

What do you mean?

if you are asking if this will give the same solution as using the method of undetermined coefficients it will

[winged jewel]

But wouldnt one be in a complex number and the other isn't?

[royal loom]

so, after all of the simplifications

you will be left with this

[wraith dagger]

AustinU

[royal loom]

now here is the key point

when you take this left fraction

e^(10it) = cos(10t)+isin(10t)

right?

so only taking the imaginary part leaves you with from the left fraction 990sin(10t)/(99^2+100)

[winged jewel]

Yeah

[royal loom]

now for the right-hand fraction, the same thing applies except see that it is multiplied by i ? so when e^(10it) goes to cos(10t)+isin(10t)

and both are multiplied by "i"

it turns out that the cos(10t) becomes the imaginary part

because the i*i on the sine part will make it real

[winged jewel]

Ahh ok

Yeah that makes sense

[royal loom]

leaving only 100cos(10t)/(99^2+100)

from the right hand graction

so combining those is the completed answer

but if you wanted to do it yourself, from this step

1st expand out the denominator

2nd factor an 100 out of the denominator, then cancel it with a factor of 100 from the top

3rd multiply by the complex conjugate to make the denominator real

4th expand the complex exponentials in your new simplified fraction into their imaginary and real parts

5th take only the imaginary parts, this will be your solution

[winged jewel]

Ahh!

I can't lie it will take me a while to go through all this properly

I haven't seen this before

[royal loom]

and I'd reiterate again the reason we are taking the imaginary parts only is because at the very beginning we rewrote 1000sin(10t) as 1000e^(10it) which isn't necessarily true, the sine is only equal to the imaginary part of the exponential

so that is why we would only take the imaginary part at the end

[royal loom]

if you do happen to get used to it, this is so much faster than the undetermined coefficients

not tons of algebra and systems of equations

[winged jewel]

Thank you so much for your help man, I really appreciate it

When I fully go through it all in a bit, I need to head off now, do you mind if I dm you if I have any questions?

Again I really do appreciate all this help from you

[royal loom]

yeah feel free to DM me, no problem

https://www.desmos.com/calculator/icmf8oczrw

here is a graph of your answer in red, the book in blue

and wolfram / symbolab / my answer in purple

[winged jewel]

Oh wow I was miles off

And so was the book lol

[royal loom]

it is odd, the books answer differs only in period but not amplitude

I don't know why it is that way

[winged jewel]

Neither

The book always gets stuff wrong

It's really annoying

Does the erf method work for all differential equations?

Or only non homogenous ones

[wraith dagger]

AustinU

[royal loom]

so constant coefficients where the RHS is an exponential

or can be written as an exponential

so trig functions work too

[winged jewel]

So in this case you converted trig into the exponential

[royal loom]

yes

also in case you were wondering about

[wraith dagger]

AustinU

[royal loom]

the ERF would give you an answer of yaddayadda divided by 0

seems like it doesn't work?

but there is an extended formula

so even it that case it still works

[winged jewel]

Oh wow lol

How would that work

Because the solution would be undefined normally

[royal loom]

instead the solution would be given by

[wraith dagger]

AustinU

[royal loom]

for the case where p(phi) is a root of p(r)

and the pattern is fairly easy to memorize, because even if it so happened that p(phi) was a double root

you can still use ERF for a solution

it would then be

[wraith dagger]

AustinU

[royal loom]

for p(phi) being a double root of p(r)

and so on and so forth

[winged jewel]

Ahh

That's really neat actually

[royal loom]

https://tenor.com/view/cat-cool-lit-gif-13999500

[winged jewel]

In an exam, if they give a question like this, where I've only been taught the auxiliary equation method but I use thus

Would I still be marked correctly

This is in an official exam

Not an inschool test

[royal loom]

well, that would be up to the discretion of your professor, so I would ask in advance, but the answer will be correct and equal to the undetermined coefficients method

so it just depends on if they care what method you use to solve

[winged jewel]

It's a levels

So it's an external exam board which marks them

Not my teachers

So I'm not sure since it wouldn't be explicit on their markscheme

Since I technically wouldn't know this method

[royal loom]

hmmm[

I am not familiar with A-levels

sorry I wish I could let you know

it seems to me like this should be fine? it is just another method of solving ODEs

[winged jewel]

its an exam befor

before uni*

basically the last exam of secondary school/high school

in the uk

[royal loom]

if the question just asks to solve an ODE, then I would reason that it should be fine.

if the question says to solve using a particular method

and you use this isntead

then they might be pickky about it

[winged jewel]

yeah that wouldnt give me the marks

they likely will say something like

what is the complentary function

and then what is the particular integral

where I would need to use the other method i presume

[royal loom]

well that's not necessarily true, you can find the particular integral using this method (like we just did)

[winged jewel]

oh yeah thats true

im just working through it all now

I got this

As the imaginary part

[royal loom]

yes looks goo

good

that's correct

[winged jewel]

Oh nice

[royal loom]

just erase the "i"s and you are done

[winged jewel]

Can I just remove them?

[royal loom]

yup

[winged jewel]

Why can I just do that

Is it because we've already determined this is the imaginary part

So we are essentially just stating the whole value of the imaginary psrt

Ahh nice

[royal loom]

well, recall that we are just looking for the imaginary part of the solution, the imaginary part isn't isin(10t) or something it is just sin(10t)

that's the part, that is attached to the imaginary number

[winged jewel]

The I just signifies that part is imaginary

[royal loom]

yes

[winged jewel]

Ahh

[royal loom]

good work

[winged jewel]

Thanks!

How do I convert that into the other solution

From undetermined coefficients

As it looks fairly different

In terms of form

[royal loom]

I'm not sure what you mean

they are equal

what is the solution that you are getting from undetermined coefficients?

[winged jewel]

I mean this one, I'm not sure if undetermined coefficients is this method, but it's the one in my book

Even though the particular integral is incorrect

The rest of it I believe is correct as I got that too when I did it

[royal loom]

Okay so

the -.1cos(100t)

is incorrect

[winged jewel]

Yes

[royal loom]

if you are asking about the other part

[winged jewel]

I mean excluding that part

[royal loom]

that is the homogenous solution

[winged jewel]

Yeah

[royal loom]

that comes from solving the ODE set equal to 0

[winged jewel]

Yes

[royal loom]

what we just found is the particular solution

[winged jewel]

But we didn't have any restrictions

[royal loom]

you don't convert the particular solution into the homogenous solution ever, if that is what you are asking

[winged jewel]

Don't you need like limits per say to get the particular solution?

[royal loom]

no

[winged jewel]

Like y =0 when x =0

[wraith dagger]

AustinU

AustinU

AustinU

[royal loom]

and y_p is the particular solution

like the one we just found

so if you want the complete solution

you need to solve the ODE set equal to 0

and then add onto that, the part that we just found

[winged jewel]

Ohh so is this literally just the particular integral

What we found?

[royal loom]

yes

[winged jewel]

Ahhhh

[royal loom]

literally just the particular solution

[winged jewel]

I thought this was the whole solution

[royal loom]

no just particular

[winged jewel]

Sorry I thought by particular solution

You meant complete solution

I haven't heard that term before

[royal loom]

seems like the vocab for diffeqs is different in the UK

[winged jewel]

Yeah it is

This method does make sense though

I'm also trying to actually understand differential equations, but the methods behind them etc don't make much sense to me in terms of logic

I feel I just contradicted myself

What I mean is mathematically the methods are straightforward

But like logically I don't know why this works, or why we actually do these things

[royal loom]

well

the derivation is possible

but a bit more complicated than just solving it

[winged jewel]

Yeah definitely

I just find things alot easier when I understand why they occur

[royal loom]

this method works due to the linearity of the ODE

[winged jewel]

What does linearity mean

I've seen the term a few times

But I haven't been able to understand it fully

[wraith dagger]

AustinU

[royal loom]

it works out so that the ODE can be represented using a linear operator

if the ODE is linear

and a linear operator f(x) works like so: f(x+x1)=f(x)+f(x1) and f(5x)=5f(x)

here is a proof of the exponential response formula for your viewing pleasure

https://mitocw.ups.edu.ec/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/linear-operators-linear-time-invariance/MIT18_03SCF11_s17_6text.pdf

[winged jewel]

I'll take a look now thanks!

[royal loom]

I just looked it over, honestly not too complicated

if you understand linear operators and derivative operators it should make sense

[cedar kiln]

@winged jewel Has your question been resolved?

[winged jewel]

im looking over it now! thank you so much for your help man, i really appreciate it

[cedar kiln]

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[flat sparrow]

1

[cosmic steppe]

Factor the numerator and denominator

[flat sparrow]

alright

5 and 9

?

[obsidian coral]

That's not factoring

That's just numbers

[flat sparrow]

https://tenor.com/view/travis-scott-astroworld-apologize-gif-23759664

5(9)

[obsidian coral]

Those are still numbers

Take for example, x^2 - x - 6, factoring it would be (x + 2)(x - 3)

[flat sparrow]

ok

(y+4)(y-9)

[obsidian coral]

Where did 4 come from?

[flat sparrow]

the denominator and numerator

4y(5y)
9y(5y)

cancel the 5's

[obsidian coral]

So then why did you end up with (y+4)(y-9) and not a fraction?

[flat sparrow]

cus i alr knew how the answer needed to be formatted so i was just looking for the variables

/how to get them

pretty sure this is B

[obsidian coral]

Numerator is factorable

And denominator

[flat sparrow]

so would it actually be t+1/1

? didn't know if i could do that

[obsidian coral]

Nope

[obsidian coral]

Numerator of this

[obsidian coral]

Denominator of this

[flat sparrow]

ah ok

🫡

interesting

[obsidian coral]

Well, factor the numerator, what do you get?

[flat sparrow]

(t-1)(t-1)

[obsidian coral]

And the denominator?

[flat sparrow]

3

[obsidian coral]

Get a common denominator and add them

[flat sparrow]

ok

(Z+5)

16/7z

[cedar kiln]

@flat sparrow Has your question been resolved?

[flat sparrow]

how do you actually get 24 here?

<@&286206848099549185>

[paper frigate]

you can plug them in and move 1/c to the left

then you can multiply by b and you should have your answer

[flat sparrow]

alright

hmm

[paper frigate]

?

[flat sparrow]

1/12 = 1/8 + 1/b

[paper frigate]

you should have -1/12 +1/8= 1/b

[flat sparrow]

mmm

ah ic

[paper frigate]

did you get it?

[flat sparrow]

no

not sure what to multiply with -1/12 and +1/8

[paper frigate]

you stick b on the side of both of them wait gimme a second

1/8= 1/12 +1/b

1/8 -1/12= 1/b

then you combine like terms

then multiply by b

[flat sparrow]

so 20?

20(1/b)

1/20(1/b)

[paper frigate]

you would have .04167b= 1

then you divide by .04167 then you have it isolated

b=23.999

[flat sparrow]

where did you get .04167?

[paper frigate]

you get .4167 by subtracting 1/8 by -1/12

since we have to remove it from the right by subtracting the positive value we have to also subtract it from the left

do you have any doubts?

[flat sparrow]

oh i see

[flat sparrow]

no I think i get it n i can test rn

ima try this rq

[paper frigate]

alr dope lmk how it goes

[flat sparrow]

forgot to ask what you did with .4167 to get 23.999

but right now i got .2857

[paper frigate]

ohh since we mutiplied by b the .4167 is now .4167b=1 so now we have to divide the 1 by .4167 to isolate b and we get 23.999

[flat sparrow]

oh after you divided .4167 by 1 did you multiply 2.399 by 10?

so it would be 35

[flat sparrow]

for this?

[paper frigate]

it would be 30

[flat sparrow]

how so?

[paper frigate]

1/21=1/b 1/70

1/21 - 1/70= 1/b

1/30 =1/b ( multiply both sides by b)

1/30b = 1

then divide both sides by 1/30 to have b equal 30

do you understand?

[flat sparrow]

i mean i can see it i just don't know how you got 30

leme try again

1/21 = 0.04761

-1/70 = -0.01428

not seeing it

[paper frigate]

ohh okay is there an exact thing your stuck on or is it the whole thing?

[flat sparrow]

after these 2 parts

im confused

of course adding them together

but after that

[paper frigate]

ohh okay so 1/21 -1/70 = 1/b

so we first start by adding like terms which is the 1/21 -1/70. this becomes 1/30

so now we have 1/30=1/b

since b is in the denominator we can bring it out through multiplication

but we have to multiply both sides

so we have: 1/30b= 1

now 1/30b is really 1/30 times b

[flat sparrow]

how does it become 1/30

[paper frigate]

how do we get 1/30 or how does it become 1/30b?

[flat sparrow]

1/30

[paper frigate]

1/21 -1/70 equals 0.033333 which can be simplified as 1/30

[flat sparrow]

ah ok

[paper frigate]

do you have any doubts?

about everything mb

[flat sparrow]

nah its ok now i'm just gonna test it once more, I appreciate it

have a good evening

[paper frigate]

you too bro

[flat sparrow]

.close

[cedar kiln]

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[river lichen]

I don't know what the pattern is and Im stuck.

[violet rapids]

maybe?? idk lmao

unless theres something arithmetic

yea

121 - 110 = 11
110-99 = 11

[river lichen]

Oh okay thank you

.end

/end

how do i end

[upbeat dune]

Well its still 99

[river lichen]

yep got it

[restive robin]

Use .close

[river lichen]

oh thank you

.close

[cedar kiln]

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[light lintel]

Hi everyone! I need some help with this question if that's alg

[slender vapor]

Hi

[cedar kiln]

@light lintel Has your question been resolved?

[cedar kiln]

@light lintel Has your question been resolved?

[terse reef]

what counterexamples have you attempted to think about so far?

[light lintel]

none... i dont even know how to start attempting this question

[terse reef]

try thinking of some example sets for A and B

in proofs you really just need to try things without knowing if they’re right

it takes a bit of luck and intuition

[light lintel]

kashdkahskdjasd im so dumbbbbb

[light lintel]

can u give a random example of what an example could be?

[terse reef]

have you tried… anything?

like really any nontrivial example of sets you come up with will is likely to be a counterexample

loosely, sets are a collection of elements

so set A to be some collection of elements (start small)

[light lintel]

so A = Natural and B = integers

[terse reef]

how about even smaller

[light lintel]

is an example?

[terse reef]

sure that’s an example but it might be hard to prove whether P(A U B) = P(A) U P(B)

how about an example where A is only one element

[light lintel]

ok

let's say A is {1} and B is {1,2}

[terse reef]

try it with that, see if it works

if equality holds then you did not find a counterexample

[light lintel]

gotcha

[terse reef]

and you’ll need to keep trying new things

[light lintel]

i think it works/holds

so ill needa find another 2 sets

to prove it wrong

[terse reef]

yes

[light lintel]

I tried A = {1} and B = {2}

[terse reef]

and what happened?

[light lintel]

It seems like P(AUB) ≠ P(A)U P(B)

[terse reef]

indeed, they are not equal

[light lintel]

for P(AUB) I got {{1},{2},{1,2},{}}

for P(A)U P(B) I got {{1},{2},{}}

[terse reef]

yes so just put that into sentences and that would be your (dis)proof

[light lintel]

Thank you so very much fluff. Can I show you my solution to it afterwards and is it fine if I ask about some other questions?

[terse reef]

sure, you can ask in this channel again and i’ll answer if i’m around

[light lintel]

For a proof by contradiction, I should assume that notA is true and so if P(AUB) ≠ P(A)U P(B), then (A is a subset of B) or (B is a subset of A) is false?

[terse reef]

that’s not quite how you do proof by contradiction

for P -> Q, this means when you know that P is true, then Q is true

if you wanted to prove this by contradiction, you would say “assume P is true, but suppose for the sake of contradiction that Q was false…. we reach a contradiction so this we know that Q must be true” (completing the proof)

[light lintel]

Can we assume that there exists an element x in A but not in B. So since A is not a subset of B, there must be element y in A that is not in B.

A set {x,y} would be a subset of AUB right, since both x and y are in A. This means that AUB contains {x,y} and the power set P({x,y}), is a subset of P(AUB).

And so {x,y} is an element of P(AUB).

Looking at P(A)UP(B), we have x that is in A but not in B, so {x} is a subset of P(A) but not P(B)

And {y} would be a subset of P(A) but not P(B) because y is in A but not B.

and so there is {x,y} element of P(AUB) but {x,y} is not an element of P(A)UP(B)?

[terse reef]

i think you need a little more justification why {x, y} is not an element of PA U PB

[terse reef]

also minor typo with what your y is defined as i believe

[light lintel]

Can I say that {x,y} is not a subset of either P(A) or P(B), which implies {x,y} is not an element of P(A) U P(B)?

so {x,y} ∉ P(A) and {x,y} ∉ P(B)
{x,y} ∉ P(A) ∪ P(B)

[terse reef]

yes that would work

(and a very short reason for why {x,y} not a subset of P(A)then not a subset of P(B) would be nice)

[light lintel]

i wish i was as smart as u my sir

[light lintel]

ahh i missed it, should be y in B but not in A

[cedar kiln]

@light lintel Has your question been resolved?

[light lintel]

Not too sure about Q5

I was thinking that h is not injective because different inputs can give the same codomain

Then I asked chatgpt and they said h is injective....

for example 6-4 = 5-1

[light lintel]

.close

[cedar kiln]

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[lethal marlin]

How to determine the vertices of this one?

[gritty galleon]

!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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[lethal marlin]

3

[short blade]

what is your answer, and show your work if you have it

[lethal marlin]

Okay wait

[cedar kiln]

@lethal marlin Has your question been resolved?

[cedar kiln]

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[hollow kiln]

i tried that but got stuck midway

i think this is correct?

can someone approve please

<@&286206848099549185>

[cedar kiln]

@hollow kiln Has your question been resolved?

[cedar kiln]

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[coral valley]

Where is my mistake here?

the derivative will go on forever

[tropic oxide]

that's because you're a l'hop-basher

why did you l'hôpital (1/x)/(-x^-2) when it can and should be simplified lmao

[coral valley]

how would I know?

[tropic oxide]

it's $\frac{x^{-1}}{-x^{-2}}$...

[wraith dagger]

Ann

[tropic oxide]

it can be written as a single power of x

[coral valley]

how do I know I when I need to make it into another form vs just doing l'hop ?

[tropic oxide]

you never "need to" do anything

but you should treat l'hop as heavy machinery

whose use should be avoided if there are better things to do

[coral valley]

so it's all just blind trial and error

man I hate it

[south tundra]

Just simplify the fraction

[coral valley]

I don't know when to "just simplify the fraction" vs continuing the derivatives

you see the problem?

blind trials and errors

fkn calculus

maybe it's just the way it is, blind trials and errors lol

guess that's how math works

[coral valley]

I don't think anyone can answer this question either

.close

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[surreal moth]

not sure how to connect x and r to start differenciation

[drifting marlin]

Can you compute the areas of the circle and the triangle?

In terms of r and x respectively

[surreal moth]

yeah circle would be πr² and triangle √3/4(x²)

[drifting marlin]

Now recall that at the start, x = 6 and r = 8, and they grow at the same rate

So at time 0, what is the relation between x and r?

[surreal moth]

well

3r = 4x?

[drifting marlin]

I guess that's technically correct

However, you want a relationship that holds at any time

[surreal moth]

so just x = 3r/4? or

[drifting marlin]

After 10 seconds, r = 9 and x = 7

Does x = 3r/4 still hold?

[surreal moth]

nope

hmm

[drifting marlin]

The expression for dA/dr should give you a good idea about the intended relationship

[surreal moth]

uhhh, just looks to me like its area of circle - triangle, then differenciated

am i seeing something wrong?

[drifting marlin]

If you differentiated the area of the triangle wrt x, what would you get?

[surreal moth]

(√3/2)x

[drifting marlin]

And on the question that derivative is given by sqrt(3)/2(r-2)

So what do you think they're saying the relationship between x and r is?

[surreal moth]

oh im not smart

lmao

thanks mate

[drifting marlin]

All good now?

[surreal moth]

yep, thanks again

[drifting marlin]

[surreal moth]

.close

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[jagged yarrow]

Whats this supposed to mean

[jagged yarrow]

280 tons of sand corresponds to approximately 186.7 cubic meters in volume, and there would be approximately 3.35x10^17 sand particles in 279 metric tons of sand. The total volume of these particles would be approximately 176 million cubic meters.

[gritty galleon]

thats sounds worng

wait oh i get it

[gritty galleon]

it means that if you have 280 tons of sand, the amount of volume it will take is 186.7 cu. metres. so any container with V > 186.7 cu. m can contain all the sand

[jagged yarrow]

So

Is 280 tons of sand

176 million cubic meters or 186.7 cubic meters

[cedar kiln]

@jagged yarrow Has your question been resolved?

[jagged yarrow]

<@&286206848099549185>

My brain can’t comprehend it

[wraith burrow]

I guess it’s best to google the density of sand and calculate the volume oneself

[jagged yarrow]

I see

[gritty galleon]

b

since 176 million cu metres is the volume of 279 metric tons not 280 tons

[jagged yarrow]

I see tysm

I see everything now

[wraith burrow]

but aren’t tons almost the same as metric tons? I mean the problem here is mainly with million cu vs only cu

[jagged yarrow]

Oh

[wraith burrow]

I mean I speak only the metric SI system

[jagged yarrow]

Uhh so

[cedar kiln]

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@crimson sedge Has your question been resolved?

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[rocky wind]

Hello there

[rocky wind]

If in oblique triangle

a = 7 c = 5 and Angle A is 120 then side b = ?

With usual notation

Isnt this question incorrect? There shouldve been angle B as two sides given the angle should be included then one can calulate using law of cosines

[livid hound]

the question is fine

[rocky wind]

Then how to calculate length of side b?

[livid hound]

the cos rule just gives the relation of 3 sides and one angle
set that rule up depending on what's known and solve for your variable

[rocky wind]

Then we have two unknowns

Angle B and side b

Oh simultaneous solution

[livid hound]

yo don't need to use angle B

nor do you need simuls

consider what you can do with the rule using the angle A

[rocky wind]

You are talking about this one right

[livid hound]

yeh

[rocky wind]

Oh i can use angle A

Thank you!

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[lunar bobcat]

Hey There!

Right now I'm working on implementing ringsystems for my space simulator and I'm currently stuck on how the planet would cast a "shadow" on the ring system. Let me try to desribe the problem:

Right now I have a sun that casts light on a planet. This "light" is just taken as a direction and amplitude by the planet and adds lighting based on that on the planet. However, since the planet is in the way of the ring system in the back, and casts a shadow paralell to it's size, I figured I could just make it not render att all and just keep it black. The ringsystem is made up of a ton of rings, based on the distance to the camera, the set quality and the radius of the ring system. I use a system where I just draw every circle in a loop, where I set a variable to 0 and then repeat the function x amount of times (for example 360) and then update the variable at the end of every function. In this function I just set every point to be drawn with a cos and sin function for x and y. I wish I could illustrate the problem.

Basically, think of a circle with any radius, and then place a rectangle over one side. If the radius of the circle increases, the angles that the rectangle occupy go smaller and fewer. How would this be calculated?

Here's a visualization I did: https://scratch.mit.edu/projects/825950887

The left one is how you can think about the problem, where all angles occupide by the red block should not be drawn, and on the left is the wanted resualt (not the prettiest but you get me).

This doesn't take into considiration that the light might not always be paralell, since we only see it due to our distance from the sun, but I don't care for that right now and I just want to keep it parallel for now.

[cedar kiln]

@lunar bobcat Has your question been resolved?

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@lunar bobcat Has your question been resolved?

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[worthy ridge]

halpppp

[worthy ridge]

determinantal functions—

WHY does this need to be true for a determinantal function to exist?

Context:

'

[worthy ridge]

is it because a function must be "well-defined" (whatever that means) in order to "exist"?

[cedar kiln]

@worthy ridge Has your question been resolved?

[crimson delta]

yes it should be well-defined

which it wouldnt be if the value of it depended on which elementary matrices you used

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[stable portal]

How do i find the zero points of this graph and how can i know when the graph is growing or not

[south tundra]

Have you tried solving x^4 - 2x^2 = 0?

[stable portal]

yeah it ends with square root of 2

[south tundra]

Are you sure x = 0 and x = -sqrt(2) aren't also solutions?

[stable portal]

i really dont know if it can be cause i need to know when f(x) is larger than 0

and when its smaller

[south tundra]

Right, so keep in mind that the zeroes are -sqrt(2), 0 and sqrt(2)

Does it also ask you to solve f(x) > 0 and f(x) < 0?

[stable portal]

not solve for it asks me to decide

[south tundra]

"Decide"?

[stable portal]

yeah decide when f(x) >0 and when f(x) <0

[south tundra]

Ah, that's equivalent to solving though

Alright, let's do f(x) > 0 since the other one can be done using the same method

f(x) = x^4 - 2x^2 so this is the same as x^4 - 2x^2 > 0

We can factor x^2 out

Getting us x^2 * (x^2 - 2) > 0

Now, the thing is that x^2 is always gonna be positive for nonzero x

Meaning that we are gonna have to exclude 0 and divide both sides by x^2

Yielding x^2 - 2 > 0 or x^2 > 2

Solutions to which are x > sqrt(2) or x < -sqrt(2)

But, also considering that x has to be in [-2, 2],

We get that f(x) > 0 for x in [-2, -sqrt(2))U(sqrt(2), 2]

Meaning f(x) < 0 will be true for all other x values

I.e., x in (-sqrt(2), sqrt(2))

[stable portal]

and if f(x) <0 means that the graph is going down?

[south tundra]

f(x) < 0 means that the graph is below the x-axis

[stable portal]

wow i got it thanks for the help!

how can i solve if the graph is growing or not

is it correct to get the derivative of f(x)

[south tundra]

Yup

And see where the derivative is positive/negative

[stable portal]

maybe a stupid question but where do i go from f'(x)=4x^3-4x

[south tundra]

Solve 4x^3 - 4x > 0

[stable portal]

how do i solve x^2-1>0 and x^2-1<0

[south tundra]

x^2 < 1 when -1 < x < 1

[stable portal]

ok i got x is a member of (-1,0) and (1,infinite)

[south tundra]

Don't forget that x has to be in [-2, 2], so that has to be (1, 2]

[stable portal]

ok so its [-1,0] and [1,2]

[south tundra]

You didn't have to turn all () into []

[stable portal]

not really sure what the difference is

[south tundra]

E.g. (1, 2] includes 2 but (1, 2) does not include 2

[stable portal]

ohh

how can i write in a sentence where the graph is growing and when its not

based off what i got

[south tundra]

$f^{\prime}(x)>0 \iff x\in(-1,0)\cup(1, 2]$

[wraith dagger]

A Lonely Bean

[south tundra]

I would write it like this

[stable portal]

ok thank you alot

if it asks for a maximum point within the defined range how do i know if its global og local

i took the derivative og f(x) to solve for 0

that gave me x1=-1 x2=0 x3=1

how do i know if these points are global or local

.close

[cedar kiln]

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[unborn pond]

Honestly, i am messing up with the partial derivative of this eq.

[unborn pond]

can someone help

this is what my friend got but i am unsure if the partial derivative is correct

i even tried symbolab but tis super long

[cedar kiln]

@unborn pond Has your question been resolved?

[unborn pond]

Please help

[wraith dagger]

AℤØ

[crimson sedge]

it's the same as the second to last line

[digital cliff]

true

[crimson sedge]

I am not sure what the line above wants to tell us ...

[digital cliff]

yeah im not sure about that either

maybe they were writing what i did then decided to write it differently?

[crimson sedge]

But at least one can tell Tuyet that the partial derivative seems to be correct (as that was the thing thas was unclear) 🙂

@unborn pond is Vg the only quantity that can have a measurement error?

[unborn pond]

I will send you the question!

part a

[crimson sedge]

Alright, the problem text a answers my question.

[unborn pond]

so this is the answer?

[crimson sedge]

I think it is.

one can write it more neatly as

$$\Delta r = \frac{3}{2}\sqrt{\frac{\eta}{2pgV_g}}\cdot |\Delta V_g|$$

[wraith dagger]

Landau08

[crimson sedge]

But the solution as given by your friend seems correct.

[unborn pond]

Interesting

isnt vg suppoed to be in the square root?

rather than outside the square root?

oh wait because its originally vg^2? so sqrt of vg^2 is vg

hence this is the answer?

[crimson sedge]

in the last line it should be $\Delta V_g$

[wraith dagger]

Landau08

[unborn pond]

yes!

so this?

[crimson sedge]

and the square in the sqrt should be deleted

[unborn pond]

could you also

[crimson sedge]

it looks like the Delta is in the sqrt and V_g outside, is this by purpose?

[unborn pond]

oh my bad

[crimson sedge]

That looks good .

[unborn pond]

would you be able to look at my answer for part b

ill send it!

i think its correct! but idk

[crimson sedge]

There should be a + sign in the solution square root

wait the phone

[unborn pond]

?

[crimson sedge]

I see two problems:
$$\frac{\partial V_g}{\partial t} =-\frac{L}{t^2}$$

[wraith dagger]

Landau08

[crimson sedge]

the second is that in the error propagation formula you always have the square of the partial derivative which is always positive number => no minus sign inside the square root in the last line.

[crimson sedge]

That seems correct.

[unborn pond]

could you help me with the last part, part c. i dont know how to do the partial derivative of it

[crimson sedge]

I had to do that experiment once. You know the product rule?

$$(f(x)\cdot g(x))'=f'(x) \cdot g(x) + f(x)\cdot g'(x)$$

[wraith dagger]

Landau08

[unborn pond]

yes i do know the product rule

[crimson sedge]

In the example above $$ Q(r) = f(r) \cdot g(r)$$ where $$f(r)=\frac{d(6\pi \eta r)(V_g+V_E)}{V}$$ and $$g(r)=\frac{1}{\left(1+\frac{8.23\cdot 10^{-3}}{p r}\right)^{3/2}}$$

[unborn pond]

oh im not sure what this means, im sorry

[wraith dagger]

Landau08

[crimson sedge]

You can calculate the partial derivative of Q(r) using the product rule.

$$\frac{\partial Q}{\partial r} = \frac{\partial f}{\partial r} g(r) + f(r)\frac{\partial g}{\partial r}$$

[wraith dagger]

Landau08

[crimson sedge]

The other two partial derivatives are simpler because V_g and V_E appear in only one place.

$$\frac{\partial Q}{\partial V_g}=\frac{\partial Q}{\partial V_E} = \frac{d(6\pi \eta r)}{V}\frac{1}{\left(1+\frac{8.23\cdot 10^{-3}}{p r}\right)^{3/2}}$$

[wraith dagger]

Landau08

[unborn pond]

v

so this?

sorry it took me a minute to input it

[signal cave]

wow what an absolute beast

[crimson sedge]

That's correct but can be simplified. Because $$f(r)=\frac{d(6\pi \eta r)(V_g+V_E)}{V}$$, we have $$\frac{\partial f}{\partial r} = \frac{6\pi d\eta(V_g+V_E)}{V}$$

[wraith dagger]

Landau08

[crimson sedge]

and because $$g(r)=\frac{1}{\left(1+\frac{8.23\cdot 10^{-3}}{p r}\right)^{3/2}}$$, we have
$$\frac{\partial g}{\partial r} = \frac{3}{2}\frac{8.23\cdot 10^{-3}}{pr^2}\left(1+\frac{8.23\cdot 10^{-3}}{p r}\right)^{-5/2}$$

[wraith dagger]

Landau08

[crimson sedge]

(having used the chain rule and the like)

[unborn pond]

Okay give me one minute so I could type it ! Ty

[crimson sedge]

Sorry, I forgot to copy a factor of d in the derivative $\partial f/\partial r$.

[wraith dagger]

Landau08

[crimson sedge]

There is one little square missing.

on the r

[crimson sedge]

The square should be on the r in the denominator

[unborn pond]

is this good

[crimson sedge]

That seems to be correct (had a phone call again, which is why it took longer).

[unborn pond]

all good!

[cedar kiln]

@unborn pond Has your question been resolved?

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[winged pawn]

any idea on how i can solve this

[winged pawn]

im guessing the deformation principle and integrating over the circle

[cedar kiln]

@winged pawn Has your question been resolved?

[mental trail]

I'm going to make the assumption this is an integration over the complex plane.
The first part in doing path integration is finding a path over which to integrate

I'm letting you time to find an appropriate path for the ellipse

[cedar kiln]

@winged pawn Has your question been resolved?

[mental trail]

At least you need to find out if 0 (the only problematic point of the function exp(1/z)) is inside the ellipse

[cedar kiln]

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[outer vale]

how do i Find the partial derivative ∂𝑓/∂𝑥
of 𝑓(𝑥,𝑦)=𝑠𝑖𝑛(2𝑦)+4𝑥𝑐𝑜𝑠(𝑥𝑦)

[crimson sedge]

You can treat all other variables as constants and take the derivative with respect to x

[outer vale]

so would sin(2y) be 2cos2x ?

or is that wrong sorry

[crimson sedge]

You would treat y as a constant, which would make sin(2y) a constant, thus its derivative would be 0

[outer vale]

why would it be 0

or is that just a rule

[crimson sedge]

The derivative of any constant is 0

Think about the rate of change of a constant

[outer vale]

oh ye my bad

so to derive 4xcos(xy)

would it be -4xsin(xy) or is that wrong

[crimson sedge]

remember to use the product rule as its multipled by x

[outer vale]

okai em lemme try that

im a bit slow sorry

is the answer

4cosxy - 4xy sin xy?

[crimson sedge]

Seems correct

[outer vale]

ok gg i just have one more question

they gave me this reminder "Reminder: 𝑢𝑣
is written as u^v" do i need to put that in my answer or is it whatever

[violet rapids]

nice username

[outer vale]

ah cheers babe

[crimson sedge]

Im not sure

[outer vale]

ok gg

thank you sm

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[cedar kiln]

@modern flame Has your question been resolved?

[cedar kiln]

@modern flame Has your question been resolved?

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[tough plinth]

can someone explain to me how to find the radius of convergence and interval of convergence?

[tough plinth]

nvm i think i got it

.close

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[charred beacon]

hey can someone help me with part d?'

[charred beacon]

my friend sent me his owrking out but i don't understand it haha

i don't know how to equate the coeficcents to find the values of p, q and k

[cerulean sail]

Alright you can at least understand up to the derivative they got?

[charred beacon]

i don't understand how they got 8x^3

[cerulean sail]

Alright, we have $(4x^{3} + px^{2} + qx + 3) e^{-2x}$ as what we want to differentiate, you are happy with the product rule?

[wraith dagger]

@cerulean sail

[charred beacon]

ok yes got it

so do we put it all over e^2x

or use the u and v

thing

for the product rule?

[cerulean sail]

You can use the u and v thing for the product rule yea, e.g. set $u = 4x^{3} + px^{2} + qx + 3, v = e^{-2x}$ and do $u' v + v' u$

[wraith dagger]

@cerulean sail

[charred beacon]

ok!

i will try that now

[cerulean sail]

Cool cool, and try and simplify as much as you can!

[charred beacon]

ok! can i show u when im done lol

[cerulean sail]

Yes please just reply to this message or @ me!

[charred beacon]

@cerulean sail ok I think I got it!

however me and my friend have different answers for p because he got -6 and i got 6

so im not sure if i did something wrong?

thankyou!

[cerulean sail]

Let me take a look for you

[cerulean sail]

Seems like he actually got 6 as well, no? But I get p=q=6 overall too

[charred beacon]

but i got p = -6

[cerulean sail]

Ohhhh you got -6

[cerulean sail]

[charred beacon]

lol i was taking out a factor of -2x^2

[cerulean sail]

But you did it a bit naughtily that should be p-6 in those brackets!

But there you go

Also note that you seem to be missing the x terms too

[charred beacon]

wait i took out -2x^2 as a factor lol

oh

the x term cancelled out tho

NVM

UR RIGHT

it should be p-6

[cerulean sail]

Yep yep, there you go

[cerulean sail]

also, to here btw

[cerulean sail]

You should have the term 2(p - q)x [or equivalent] in here ideally

(well not "ideally", you have to lol)

But the benefit of that is that it provides you with a sanity check, because that should tell you that p=q, so you would have caught that a bit earlier

[charred beacon]

but -2qx + 2px = 0

so cant you just get rid of it

ohhhh ok

ohhhhhhh

OHHH THATS SMART

just for good measure

[cerulean sail]

I mean, it is, but you hadn't yet justified that it would have been by that point!

[charred beacon]

why do u write it as 2(p-q)x

oh ok nvm i think i get it

[cerulean sail]

You would have wanted to say something like
[
(-8x^{3} -2(p-6) x^{2} + 2(p-q)x + (q-6)) e^{-2x} = ke^{-2x}
]
so therefore $k=-8, -2(p-6)=0, 2(p-q)=0, (q-6)=0$ and all

[wraith dagger]

@cerulean sail

[cerulean sail]

And strictly speaking you didn't have to but I just chose to, you could have left it as (2p-2q)x as well

[cerulean sail]

Similarly like how you don't really need to have factored out that -2

[charred beacon]

oh ok yours looks much neater than mine

yeah thats smart

thats really smart thanku so much

[cerulean sail]

No worries, a pleasure Hope that helped for you!

[charred beacon]

THANKU OFC U DID

U HELPED SO MUCH THANKU!!!

[cerulean sail]

Brilliant have a wonderful rest of your day(/night!) catch you around hopefully!

[cedar kiln]

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[dreamy pendant]

i’ve got no idea if what i did was right

cant find an answer anywhere online

just looking for some clarification

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[lethal bison]

that's what I did I just want to know if what I'm integrating is correct. I'm pretty confident about the bounds but not the inside of the integral

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[mighty gorge]

Soft question about learning math

I have been reading about ungrading in college classes and the idea gave me some hope about learning math in school. I like learning math, but the environment where I have to perform well on tests and remember everything makes me panic.

So I thought about differential equations and I remember struggling with that in school. I pulled up Khan Academy, I am looking at the first video, and Sal encourages me to work out the derivative and second of e^-3x.

After the first derivative, (I think is -3e^(-3x) ) I realized that I don't think I can use chain rule and I don't know what to do. In that moment, I felt a huge amount of shame.

I think I know what to do now (multiplication rule), but has anyone else felt this way before? Where they feel shame for not knowing the answer?

[royal loom]

You should never feel ashamed for not knowing an answer! Although I am sure that myself and many others in here may have shared the sentiment at times. We all hold ourselves to much higher expectations than anyone else would, and what sometimes gets in the way of learning is when we feel like we are failing these expectations we hold for ourselves, or we compare ourselves to other people who we think know everything easily and we bash ourselves down for not being "good enough". But the truth is that it is such an important skill and foot in the door that you are trying to learn on your own and seeking knowledge and being willing to fail. You should never feel shame for getting stuck on something or not knowing something but feel encouraged that once you figure it out, this will be just another thing that you have added to your knowledge. You can't learn without failing, because if you get everything right then you must've already known how to do it all!

[mighty gorge]

...also I think I did the differentiation wrong

[wraith dagger]

AustinU

[royal loom]

is not a product of two functions, it is a constant (-3) multiplied by a function e^(-3x) what do you know about derivative rules for constants mutliplied by something?

[mighty gorge]

That would just be $e^(-3x)$

I was hesitant to do that because I know you need chain rule for the -3x...

oh wait do you need to do more than 1 chain rule?

[wraith dagger]

TheCollatzGriffin

[mighty gorge]

Amazing, I killed TeXit

[royal loom]

do you know that the derivative of e^u is e^u * du ?

if you have -3 * e^u (u being -3x) then the derivative is -3 * e^u * du

and what is du, if u=-3x?

[mighty gorge]

Oh, I didn't think to use substitution as a way to solve the derivative

Wait... I do that for integrals, but how does that work for derivatives?

Or are you doing that because the defEq has multiple derivatives, so that's why there's a dx?

[royal loom]

is there a reason you keep referring to this as a diffeq? is there a different original question?

and it was less of substitution, but more of a simpler way to right out the chain rule

it would not be right to say derivative of e^f(x) = e^f(x) right? it would be true if you applied chain rule and wrote the derivative of e^f(x) = e^f(x) * f'(x)

using the u instead of f(x) just makes it simpler to write that

[mighty gorge]

This is a diffEq:

$f(x) = e^{-3x}$

[wraith dagger]

TheCollatzGriffin

[royal loom]

ah , I hadn't looked at the image

so are you checking the solution you got then

[mighty gorge]

Yeah

[royal loom]

okay so where are you at with that now?

[mighty gorge]

I think when you used u instead of f(x), my brain melted and I thought oh crap u-sub

[royal loom]

XD

not real u-sub so no worries

just like referring to just saying like -3x is a function of x, might aswell call it u(x)=-3x and then just call u(x) u

for simplicity sake

[mighty gorge]

I need to re-read what you sent earlier understanding the u's won't hurt me

Hmm...

I'm going to go back and read the other stuff you sent 😅

Okay, I'm caught up

That's where I'm stuck

[royal loom]

wait hold on

again there is no product rule here

also

is -3 * -3 = -9?

[mighty gorge]

I think I got it

My math magic gets me again

[royal loom]

awesome

[mighty gorge]

I can't compute algebra to save my life

[royal loom]

happens to the best of us XD

are you self-learning diffeqs rn?

[mighty gorge]

I'm getting my feet a little wet. Mostly I want to get to the part where I couldn't figure it out and then I dropped the class

I had been going to office hours and tutoring every other day, at times every day, for the first two weeks of that semester

[royal loom]

ah dang, sorry to hear that you had to drop. Are you planning on taking it again soon?

[mighty gorge]

It makes me sad because difEq sounds cool

[mighty gorge]

No, that would be a terrible idea

I don't need to fail and then retake another math class

[royal loom]

I think diffeq is awesome

Hmm, well I see you have been looking at the khan academy resources

if you are interested in maybe some deeper learning I would strongly recommend this for self-learning diffeq

https://ocw.mit.edu/courses/18-03-differential-equations-spring-2010/video_galleries/video-lectures/

All lectures, HWs, PSETs, recitations, exams, included with the course syllabus and calendar totally free online. MIT OCW is a blessing

even the textbook is free online

also you can send me a DM if you ever have questions you are stuck on

[mighty gorge]

I appreciate it

I think I figured something out. While working on this, I'm getting really anxious and scared. I realize now that this is actually triggering a trauma response. Sob story short, failing in school caused a lot of problems.

On the bright side, I'm working in therapy and the trauma therapy is effective, just slow. I'm going to hold onto this link for future reference, I think it will be a lot easier for me to learn this once I have resolved the trauma stuff

[royal loom]

oh of course, learn at your own pace by all means! Sorry to hear that. Best of luck with whatever you decide to do in the future!

[mighty gorge]

Thank you, you too

.close

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[unborn pond]

i have a question

how did it go from the first step

to the second step

[cedar kiln]

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@unborn pond Has your question been resolved?

[hollow comet]

It's not tricky but whoever wrote this decided to leave out a lot of steps. When you apply the gradient to the exponential, it's the same as the 1-D derivative exponential but this time you have more components so you have to be careful.
Firstly remember r is given by xxhat + yyhat + z*zhat
So the derivative is just going to be 1 times the unit vector it is multiplying depending on which partial derivative you're doing. Simultaneously because you dot r with k, you are going to get in the derivative of the exponential a kx, Ky and kz factor out front each one multiplying an i because it's part of the exponential.

Remember that psi is the exponential solution, so after taking derivatives using the gradient, derivative of an exponential is an exponential times the derivative of whatever was in the exponential (because of the chain rule) so the person who wrote the notes put the psi BACK in place of the A*exponential

Does that help @unborn pond ?

Whoever wrote these notes got extremely sloppy and chose some bad notation. Based on the fact that the top right corner says "Optics", must be one of my sloppy physics brethren. 😉

[cedar kiln]

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[crimson sedge]

Where do I start with this?

[wraith dagger]

MathIsAlwaysRight

[dusk finch]

can it be in terms of t btw?

or does it have to be in terms of y and x?

[crimson sedge]

im not sure, it doesnt say

how do I find dy/dt and dx/dt?

[dusk finch]

just differentiate t/(4+t) for dx/dt

and sqrt(4+t) for dy/dt

[crimson sedge]

alright

i got 1/2sqrt(t+4) and 4/(t+4)^2

[crimson sedge]

is the answer just (1/2sqrt(t+4))/(4/(t+4)^2)

oh so it does have to be in terms of t

.close

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[opaque bane]

What exactly does “every circle has a unique circumcenter” mean? Like, if I just draw an equilateral triangle around that point, surely they have the same circumcenter?
Does it mean that every triangle has a unique relation between its centroid and its circumcenter?

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[lusty marlin]

I need to find the value of y in this diagram, in surd form

I found the value of x which is through pythagoras theorem and I end up with x=2^1/2