help-13

[austere hull]

what's your question anyway?

you can't remember the sum diff trig or what

,tex .sum diff trig

[wraith dagger]

Akira

[austere hull]

this is all you need to keep in mind, as far as I can tell

on link you provided

[teal magnet]

I see on the cos the +- is inverted while for sin it's the same, is the inversion on purpose?

so is the tan denominator

[austere hull]

yes

[teal magnet]

Can I get the script you used to make that image?

[austere hull]

it's secret helper knowledge https://github.com/riemann-discord/math-discord/tree/main

[teal magnet]

I'm guessing you don'T want t oshare the latex you used.

[austere hull]

you need preamble for this

have you read the introductions?

all you need to add the preamble.tex file and replace it

you could add your own as well

[teal magnet]

ah, it's a bit more involved than I hoped. well thanks for the image.

I have a latex software and libre office math.

[austere hull]

do you use overleaf

never heard of those you said

I'd say overleaf is best for latex

I can tell you how you can make this in your preamble in #latex-testing

make sure you got the file

[teal magnet]

No, I wanted it to use as plain latex.

but thanks for the offer.

[austere hull]

no worries

[teal magnet]

I use TeXstudio, I'll check out overleaf.

Anyway perhaps someone else know a page that I can learn about the subject by small steps.

[austere hull]

you can ask help in #latex-help

there's some people might help

[teal magnet]

no, small steps for what you made the image about.

[austere hull]

it's about factoid

[teal magnet]

I have a bad memory, I need to learn bit by bit.

[austere hull]

you understand this by reading all this

on README.md

[teal magnet]

no

[austere hull]

it literally says

[teal magnet]

what you made the image about

[austere hull]

which image?

[teal magnet]

look at the image.....

the one you made

[austere hull]

I did not make this

[teal magnet]

☝️

[austere hull]

yeah it was there

I added two things like

,tex .Shape

and

[wraith dagger]

Akira

[austere hull]

,tex .Psf

[wraith dagger]

Akira

[austere hull]

those what I did

[teal magnet]

That doesn't help with my question

[austere hull]

okay so you link you provided

does it make sense?

[teal magnet]

☝️ my question....

it does, but slower would be better

it's why I'm asking here

more steps

or more details to the steps

[austere hull]

okay

one last thing I can say

there's a khan academy explains this

[teal magnet]

also I noticed after posting it that the PDF is by the same authors, I'm assuming it's the same thing or a diferent version of it.

I was at khan academy, it ran through it, it made references to previous videos but there were no such videos in that course https://www.khanacademy.org/math/trigonometry/trig-equations-and-identities/xfefa5515:angle-addition-identities/v/trigonometry-identity-review-fun

I'm not studying to be a detective and have a super memory.

if you know of the videos he made reference to I'd be very happy to watch them.

or other videos that goes slower through it,.

[austere hull]

well the factoids for those who made this, it's pretty much khan academy video

and the pdf from here

it's a half angle

,tex .half angle

[wraith dagger]

Akira

[austere hull]

I'm not exactly sure what is your question

correct me if this is one example of your question

[teal magnet]

This is all too quick for me.

I guess if no one replies, I'll look at more websited tomorrow.

[austere hull]

I mean there's a slow speed you can do on youtube no?

[teal magnet]

more steps

The video has a transcript, I could go supuer duper slow.

slower through the subject.

I guess I'm dumb as a monkey, oh well!

[chrome bluff]

hello, I'm kira. I'm new here. can you guys help me with number 4?

<@&286206848099549185>

[cedar kiln]

Please read #❓how-to-get-help

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.

[crimson sedge]

okay , i can answer this question

do you want me to ?

[chrome bluff]

sure, thank you

[crimson sedge]

okay so , here we go

these both Luca and Ian answering right are independent events ,

got it ?

[chrome bluff]

owh, there are independent events

[crimson sedge]

yeah ,

so , probability that atleast one of them answers the question correctly will be = 1 - ( neither of them answers it correctly)

Luca getting it wrong = 1-5/7 = 2/7 ; Ian getting it wrong = 1 - 5/9 = 4/9 ; thus , both getting it wrong = (2/7)(4/9)= 20/63 , we multiplied directly because they are independent

[chrome bluff]

okay

so, the anwser to Qa is 43/63?

[chrome bluff]

(2/7)(4/9)=8/63

[slate lintel]

this looks cool
something's weird about the kerning on the Vertex line

@austere hull ^

[crimson sedge]

sorry , 8/63 , got electricity cut , so had to go

[crimson sedge]

thus , probability of getting at leat one correct answer = 1 - 8/63 = 55/63 , got it ?

this is the (i) part

[chrome bluff]

got it

[chrome bluff]

okay, thank you

[crimson sedge]

so , for the 2nd part

[crimson sedge]

probability of luca getting the answer correct is 5/7 , since the events are independent , the intersection of it with getting answer correct will also be the same , that is = 5/7

thus , if the question is answered correctly , the probability thatt Luca got the answer correct = (5/7)/(55/63)=9/11

got it ?\

similar goes for part c , for Ian , the answer will be = (5/9)/(55/63)=7/11

[chrome bluff]

got it

[crimson sedge]

and for the last part ,

intersection of being correct comes out to be = 25/63 , using the set theory-probability formula

thus , conditional probability comes out to be = (25/63)/(55/63) = 5/11

[chrome bluff]

got it, thank you @crimson sedge

[crimson sedge]

always 🙂

[cedar kiln]

@teal magnet Has your question been resolved?

[teal magnet]

Since some people decided I was done with my question and I'm not, I'll repeat it: Most of the pages on I found on trigonometric angle addition identities just list them and I can't remember that all straight through. I found the following page that seem to go through the material slower, do you know of something even slower through the subject? https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Precalculus__An_Investigation_of_Functions_(Lippman_and_Rasmussen)/07%3A_Trigonometric_Equations_and_Identities/7.02%3A_Addition_and_Subtraction_Identities

[wanton sail]

@chrome bluff This is someone else's help channel, please don't ask your question in it

See #❓how-to-get-help

[slim jasper]

Please avoid engaging when someone asks their question in someone else's help channel (other than redirecting them to open a channel of their own).

[wanton sail]

Personally I find it really helpful to start from the angle sum identities and derive the rest of the identities from there

Like for example if you know sin(a + b) = sin a cos b + cos a sin b

You can substitute in (-b) to get sin(a + (-b)) = sin a cos(-b) + cos a sin(-b)
or
sin(a - b) = sin a cos b - cos a sin b

I think KhanAcademy is a good resource for this sort of stuff, have you already tried it?

[teal magnet]

So if I manage to remember the 4 sin+- cos+- I'll manage to deduce the rest?

Khan Academy is my main source of information. It went too fast through it, you can look at the transcript on the tab for it; it's much too fast for me https://www.khanacademy.org/math/trigonometry/trig-equations-and-identities/xfefa5515:angle-addition-identities/v/trigonometry-identity-review-fun

[wanton sail]

Yeah I think the important ones to know are

Pythagorean
sin^2 x + cos^2 x = 1

Definition of tangent
tan x = sin x / cos x

Even/oddness
sin(-x) = -sin x, cos(-x) = cos x

Angle sum
sin(a+b) = sin a cos b + cos a sin b
cos(a+b) = cos a cos b - sin a sin b

Meaning of co-
sin x = cos(90 deg - x)

And then try to figure out the other ones (double angle identities, half angle identities, etc) based on these, and that'll help you remember them

[wanton sail]

Ah okay, I think Paul's Online Notes might also have something on it?

[teal magnet]

I don't think I've seen that. I've started with Math if Fun, which I often like, but too fast as well.

is*

[wanton sail]

If you look up a specific identity on youtube, you'll probably find in depth videos on it too

You might've already tried that, but just in case you haven't yet

Feel free to also ask any specific questions/confusions you have

[teal magnet]

I did not look at each ones, I looked at the assition subject as a whole. Are you referring to https://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/TrigFormulas.aspx ?

addition*

https://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/TrigIntro.aspx

[wanton sail]

Yeah

I recommend spending some time on justifying why each one is true

That will help you remember them much better

Like for example sin^2 x + cos^2 x = 1 is just because of the pythagorean theorem

[teal magnet]

It's what I'd like, how come therE's no page for each one individual?

I must be the only one that think all at once is too fast.

or at least a whole section of a page

[wanton sail]

Maybe this page? https://wumbo.net/examples/derive-trigonometric-identities/

I just looked up derivation of trigonometric identities

And got a bunch of results

This was one of the first ones

[teal magnet]

I haven'T looked at the page completly since I'm tired but it looked very promising!

very indeed

[wanton sail]

Awesome!

[teal magnet]

Well, provided it's indeed it, I don't need help anymore. If I do realized it's not enough tomorrow, mind if I ping you?

Also if tomorrow I realize it'S enough, I think thereS' a bot command to thank you, what is it and in which channel to do it?

Well... if you decide you want to, @ or DM me.

!close

.close

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[crude scroll]

If we let (A) be f(x), then (C) would be f'(x) = g(x), and then (B) would be g'(x) = h(x) right

[short blade]

yes

[crude scroll]

since for (C), the first third of the graph is positive, then negative, then positive

[crude scroll]

i see. ty

.close

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[winter sundial]

I’ve been reading “How to Prove It” and I’ve been reading up logic. I understand that an open sentence is a sentence that has a free variable and the sentence can either be true or false depending on the value of the free variable. I also understand that in order to make an open sentence a statement all the free variables will have to be quantified. Say I had this open sentence:
x + 0 = x
In addition say the universe of discourse is the real numbers.

I understand that this is an open sentence because x is a free variable due to there not being a quantifier for x. Since the universe of discourse is the real numbers, I can determine the truth set. The truth set is the real numbers. So the universe of discourse and the truth set turn out to be the same. My confusion arises in the fact that this is an open sentence but no matter what value I substitute for x in the universe of discourse the statement will be true. Can someone please explain? Thank you in advance!

[upper abyss]

I've not read "how to prove it" so take me with a grain of salt.

But if a sentence is always true, then it's definitely "either true or false"

Does Velleman require that there exists a value where the sentence is false?

[winter sundial]

In the book Velleman states: “When studying statements that do not contain variables, we can easily talk about their truth values, since each statement is either true or false. But if a statement contains variables, we can no longer describe the statement as being simply true or false. Its truth value might depend on the values of the variables involved. For example, if P(x) stands for the statement “x is a prime number,” then P(x) would be true if x = 23, but false if x = 22. To solve this problem, we will define truth sets for statements containing variables.”

[crystal raptor]

an open sentence is one that may be either true or false depending on the inputs, but that doesn't disallow it from being true for all inputs

[cedar kiln]

@winter sundial Has your question been resolved?

[winter sundial]

@wraith blaze So we need the quantifier to make it a quantified sentence but without it it still open.

[wraith dagger]

all lines are flat

all lines are flat

[winter sundial]

I understand it now! Thank you for that explanation

I know I asked you this question twice, I kept thinking about it and realized I didn’t quite understand it fully. Now I know we need to know the universe of discourse and how many “x’s” we are talking about

Right so if we choose all x in the universe of discourse or just one.

[wraith dagger]

all lines are flat

[winter sundial]

Could you explain the difference between the unique x and the for some x

[wraith dagger]

all lines are flat

[winter sundial]

I see that makes sense. Thank you!

[wraith dagger]

all lines are flat

[winter sundial]

That makes sense!

Thank you for that explanation

What book did you use to learn mathematical logic?

I’ll have to take a look at transition to advanced mathematics. Thank you for all the explanations! Thank you to everyone else who left a comment on my question!

.close

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[vital valley]

Why does e^lnx = x

[crimson delta]

definition

[vital valley]

Of what

[crimson delta]

either lnx or e^x

depends which one you defined first

[dire bronze]

let x > 0

e^lnx=x
ln(e^lnx)=lnx
ln(x) * ln(e) = ln(x)
ln(x) * 1 = ln(x)
ln(x) = ln(x)
x=x which is true
therefore e^lnx=x is true

[vital valley]

Interesting

Thanks

.close

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[naive zinc]

I do not know where to begin on this problem other than both a and b are true

[naive zinc]

The answer is d but idk why

[crystal raptor]

Did you send the right picture?

[naive zinc]

Yes, I moved answers but it is d

[rain drift]

so what does the integral mean? What does the integral represent?

[naive zinc]

The area under a curve in this case From 0 3

Why can’t it be greater than 7?

Or equal 7 I mean

[rain drift]

bingo! Now when it says that f(c) = 2 is a maximum, what does that mean?

[quaint forge]

it will be both incresing and decreasing in the interval [0,3] since the max value is at two the graph before it will be increasing and after it will be decrasing

[rain drift]

hint: What shape would give us a maximum area?

[naive zinc]

That it’s the highest y value on f from 0 3 and also the derivative is 0

Working backwards means it’s a 0 on the integral too I think ?

[rain drift]

yep! So is there a function you can think of that would give us the maximum area? Where every point in [0, 3] is the max?

[naive zinc]

Y=x?

[rain drift]

closeish. How about a constant?

[naive zinc]

Not quite sure what you mean “where every point is the max” that’s where im stumped

[rain drift]

how about the integral of y = 2 from [0,3]. What would that be?

[naive zinc]

2x

Is it 2x

Because integral of 2x is x^2

[rain drift]

that's the anti-derivative, but what about the area?

[rain drift]

This is what I mean.

[naive zinc]

Area of x^2 from 0 3 is

6

[rain drift]

right!

[naive zinc]

Thus it cannot be 7 but could be 5

[rain drift]

yeah exactly. Literally if from [0, 3] the max value for f(x) = 2 at point c, we can get the max area by making very point c = 2

hence the constant funciton

so the max area we can have is 6

[naive zinc]

I think I follow

[rain drift]

it can be 5, it can be 5.999999999...., 6.

But not things ike 6.00000000000000001, 6.00000000000000000000001. Those are invalid

6 is the max we could possibly do

[naive zinc]

You use the info given to make a conclusion then work backwards anti differentiating right?

[rain drift]

Not work backwards per se. I just saw that the max value y could've been was y = 2, so if we were to set every point f(c) = 2, that is, the constant function f(x) = 2, then we have a maximum area we could possibly have 😉

hopefully they make sense

[naive zinc]

Also correct me if I’m wrong but we essentially made a bounded region on the graph, or would it just be a hypothetical rectangle of our possible area?

[rain drift]

Yeah you can think of it as a "bound". f(x) = 2 would be our upper bound for y since f(c) = 2 is the maximum value we could have.

[naive zinc]

Thank you very much!

[rain drift]

yeah no problem! Nice work 🙂

did you have any other questions?

[naive zinc]

Yes actually if you’re willing

My conjecture was wrong I haven’t done any anti differentiation yet

And it’s the integral from -1 to 1 correct?

[rain drift]

one moment por favor

correct the bounds are -1 to 1

hmmm

it should be 0 me thinks

yeah it is 0!

[naive zinc]

I put 0 and it was incorrect 😭

[rain drift]

hmmm

okay let's work on it together then

[naive zinc]

The anti derivative is this yes?

[rain drift]

so if we were to solve this the old fashioned way, what method would we use to solve this integral?

[naive zinc]

U sub

Make 1-x^2 u

[rain drift]

yep exactly

[naive zinc]

Du= -2x

[rain drift]

yeah it's 0 lol

[naive zinc]

So we get a -1/2 constant

[rain drift]

maybe a parsing issue? Did you have a space after the 0 or something?

[naive zinc]

I did not

[rain drift]

yeah it's totally 0

[naive zinc]

Perhaps an error on the page?

This seemed like a look and know problem

Same negative area as positive and so it must be 0

[rain drift]

yeah the only way that would be 2/3 is if it were the absolute value of that function

maybe they mean total area?

the integral represents the net area

whatever the case either it's wrong or the problem was written in a confusing way

[rotund fox]

break the integral on 0 and then try.

[rain drift]

it's 1/3 unfortunately :/

[rotund fox]

For 1 side?

[rain drift]

yeah from 0 to 1

from -1 to 0 it'd be -1/3

[naive zinc]

Did I antidifferentiste properly btw?

[rotund fox]

I think we should add the area and also put absolute() on the broken integrals to get the area.

so that should give us 2/3

[naive zinc]

That’s one of the worst worded problems I’ve ever read lol

[rotund fox]

Its asking the area though. I suppose that if i were you i would not take that area as negative. might just be a me thing tho, i was taught to take abs when doing questions like these

[rain drift]

yeah indeed, lol.

I would put 2/3 for the credit, take a screenshot, and present it to your teacher that the problem might be incorrect or ask clarification for the problem.

[rain drift]

@naive zinc this is probably the most correct way to go about this

they more than likely mean "total area" rather than "net area". But I don't like that because the integral represents the net area lmao

[rotund fox]

oh thats a cool insight, they never asked to get the area by integral, which means that they 100% asked the total area

[naive zinc]

Unfortunately it’s for the clep exam which is all multiple choice which means come test day it won’t matter sadly

Thank you both very much

[rain drift]

yep np!

anything else? 🙂

[naive zinc]

For this problem I initially thought it was -1/12, in these types of questions is it always f(x)-g(x)?

I accidentally did x^4\4 - x^3/3 (integral from 0, 1 for both ofc)

[rain drift]

wait sorry which problem? Can you post a pic of the problem?

[naive zinc]

My bad

Does the rule of “greater minus lesser” apply in the area between two curved as well?

[rain drift]

yeah indeed. Whichever function is greater than is always first. If, for some reason the areas "switch", then you would have to do 2 separate integrals

so in this case, from [0, 1], x^2 is greater than x^3, so it would be

$\int_0^1 (x^2 - x^3) dx$

[wraith dagger]

mellowdramallama

[naive zinc]

Makes sense I remember now

This inverse function question is my last one, if you’re willing to help me one more time

[rain drift]

yep let's do it

[naive zinc]

H’ is 5x^4-2

And h’(0) is -2

So I tried doing where h’= -2

Whoops maybe I should do where it equals 0 instead?

[rain drift]

huh give me a second on this

yeah because h(x) doesn't have an inverse... I think

[naive zinc]

Why not?

Is there a formula for inverses?

[rain drift]

fails the horizontal line test, unless we limit the domain

it's not injective

[naive zinc]

Or do you just change y and x?

[rain drift]

nah. My guess is that we can play around with something about the chain rule or something

give me a moment

unless someone else viewing this thread has any ideas lol

[naive zinc]

Since we need h’ inverse at 0 we could make our domain the relative max and min?

[dire bronze]

Note that the function is increasing as x goes to infinity and decreases as x goes to negative infinity

and the derivative is increasing on both sides

[naive zinc]

Is it a slope of -1/2 or -1/3 right there when it crosses

At x=0

[dire bronze]

Also, the inverse is just the function flipped across the line x=y

[crystal raptor]

They probably want you to blindly use the inverse function derivative formula (ignoring woes about whether h inverse exists)

[rain drift]

oh say what? We completely skipped this in my calculus class. I had no idea this existed! Lmao

[naive zinc]

So did I lol

[crystal raptor]

Yeah its neat, you don't need the full inverse, just the preimage of the point you want

[naive zinc]

So I take the inverse at 0?

f(0)=1

So inverse is F^-1(1)=0

Correct?

So I get 1/f’(f^1(0) ?

[slate lintel]

yes

[wet haven]

You have to see if the function is a one to one function

[naive zinc]

We did by restricting the domain to (-.795, .795)

I’m trying to find where f =0 but I’m struggling atm

[slate lintel]

try rational roots first

you have $x^5 - 2x + 1 = 0$ so rational roots are either 1 or -1

[wraith dagger]

Hayley

[naive zinc]

f(-1)=2

And f(1)=0

[austere hull]

I'll fix it

[cedar kiln]

@naive zinc Has your question been resolved?

[cedar kiln]

@naive zinc Has your question been resolved?

[cedar kiln]

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[cinder crest]

Hi there! I need help with a question from diff eq. The prompt is 3. Solve the IVP. Final answers should be real-valued (no i involved).
2y′′′ + 3y′′ − 12y′ + 7y = 0, y(0) = 1, y′(0) = 0, y′′(0) = 2

[cinder crest]

I have the general solution I just need some help with the ivp part of the question

of solving for c1, c2, and c3

I'm stuck

[cinder crest]

I'm stuck on solving for A, B, and C

I've come up with numerous solutions that solve for the first 2 equations (y(0), and y'(0)) but I can never get the constants to fit all of the conditions (aka as well as the last equation of y''(0))

and I just don't know where to go from here

I thought that multiplying a 2 to all the constants in the 2nd equation and then subtracting that value from the last would let me solve for A, as I don't see any other way to solve for the other variables, as they aren't separable and I thought that -77/4 would work, as it fits all of the conditions, but it never ends up working, especially when I try and plug the values for A, B, and C back into the final equation

(basically I just need help with solving the constants of A, B, and C)

or just help to get a step into the right direction

[cedar kiln]

@cinder crest Has your question been resolved?

[cinder crest]

<@&286206848099549185> caN i get some help please

[scenic python]

what?

[fleet eagle]

Yeah ?

[cinder crest]

with the problem above please

[scenic python]

can you increase the size of the letter?, I'm without glasses

[cinder crest]

which letter

all of them?

[scenic python]

y

too small

[cinder crest]

oh

you mean the page?

[scenic python]

y

[cinder crest]

oh okay

yeah I gotchu

sorry about that

just gimme a second

is that any better?

[scenic python]

barely i can see

[cinder crest]

sorry I tried my best to resize the image

is there a way you can download the image on your pc/phone and just enlargen it from there?

[scenic python]

my sdd i almost full i i still can download

[cinder crest]

oh okay sweet

[scenic python]

is to see if are correct?

[cinder crest]

what?

[scenic python]

what you mean here

[cinder crest]

I need help solving for A, B, and C

and the prompt is there

like what I need to solve for etc.

I have the general equation and whatnot, I just need help solving for the constants of the IVP

[scenic python]

i will come later let me just find my glasses

[cinder crest]

alright ill just make a new thread then

.close

[cedar kiln]

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[placid tendon]

You have an system of 3 unknowns and 3 equations, just solve it

[cinder crest]

yeah I know that's what I'm supposd to do

but I need help solving it

thats what I was trying to get at

like what step do I take first

and is what I was doing valid?

[grand forge]

ODE's are a bit harder to solve than regular equations

[cinder crest]

^ facts

[placid tendon]

He’s referring to the coefficients

Not the ODE

[placid tendon]

Can you write the system of equations here ? It’s hard to see

[dawn junco]

we're already in #help-0 btw @placid tendon

you reopened the thing, but cart also opened another chan

I guess we can close that now

.close

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[olive meteor]

I'm confused

[slate lintel]

what i would do

is figure out that intersection point

[cedar kiln]

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[tulip harness]

Consider the following system of differential equations $x'(t) = Ax(t)$, where $A\in \mathbb{R}^{n\times n}$ and $A^3 = 0$. Show that the following second-order Runge-Kutta method is exact:\
$x_n = x_{n-1} + \frac{h}{2}(k_1 + k_2)$, where\
$k_1 = Ax_{n-1}$, $k_2 = A(x_{n-1} + hk_1)$

[wraith dagger]

casiel368

[tulip harness]

What in the name of math is this

I don't even know where to start

[cedar kiln]

@tulip harness Has your question been resolved?

[still barn]

The idea is that if you write down a Taylor expansion of x(t_(n+1)) you see that the O(h^3) and higher-order terms vanish because A is nilpotent. It'll help if you write down that expansion and see how it relates to the terms in your proposed runge-kutta method.

[tulip harness]

Okay in the mean time I got that idea, but what comes next?

If they gave me the Taylor 2 method it would be fine, but I only know they share order of convergence

I know how to prove that, but I don't see exactly how one translates into the other

[still barn]

I don't think I understand what you are asking.

You should know the solution to x'(t) = Ax(t) as a series. Write down that series and compare what it looks like to the RK method

[tulip harness]

Oh I really messed up in the middle before

It seems a bit clearer now

So now I'd like to prove that tau=0

[still barn]

It's been a little while since I've done anything like this so I don't know when it will all fall into place, but you can see that your hAx(t) is basically hk_1 and you can do a Taylor expansion on A^2 x(t) again to get something like A^2x(t) = Ax(t-1) + hA^2(x(t-1)) etc to get to your k_2. I think it's a bit early to just replace that with a tau.

An RK method really isn't that different from a Taylor method. You're just reapplying the Taylor method to your higher order derivatives like your A^2 x(t) and multiplying things all out. It's more of an annoyance than anything new.

[tulip harness]

Oh, I didn't know that

I knew that they share order because if you put the time into taking all multivariable derivatives it checks out, but that motivation seems cleaner

[still barn]

Yeah that's basically it. By rewriting it in this way, the Runge-Kutta method becomes a sort of weighted combination of the Ax(t) evaluated at a whole bunch of different points

[tulip harness]

You mean something like this?

oh it did really check out

thank you!

.close

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[reef sedge]

i dont understand for a,b

for a.)

why they do 2*4, the selection of number 4?, after choosing either a or i in the first letter of the word, you're left with 5 options, 4 constants ands 1 vowel

[eternal phoenix]

because you can start with A and I or I and A

[reef sedge]

i dont understand they start with 4 when there are 5 numbers to start with

oh wait

[eternal phoenix]

so double the cases

[reef sedge]

for b is it because 0 cant be the start of a number?

[reef sedge]

can you elaborate

[eternal phoenix]

so you can start with A and end with I

or start with I and end with A

we know that because that’s what the problem is saying

so that leaves us D, R, W, and N in between

which means that we have to figure out how many ways we can arrange those letters between A and I or I and A

which is just four factorial

but because we can start with A and end with I or start with I and end with A

we have to double the amount we can have

hence why 2 x four factorial

[reef sedge]

yes

uh

huh

i dont get it

wait

let me tr process this sorry

[eternal phoenix]

okay you’re good

just think about it

[reef sedge]

ohhhhhhhhhhhhhhh

i think i get it

since there are only two vowels

one must start with i or e

or start with e and end with i

if next letter is vowel, it doesnt meet the conditions

[eternal phoenix]

exactly

you got it

[reef sedge]

and for b

is it because

0 cannot be the start of a number

else if it was like 2,3,4,5,6

it would be

lets say it was digits 1,2,3,4,5,6

and we want odd 4 digit numbers

it woud be like

$6\cdot 5\cdot 4\cdot 3\cdot 3$

[wraith dagger]

fwgx

[reef sedge]

wait

is this it @eternal phoenix

[eternal phoenix]

ngl i have no idea for b😭

[reef sedge]

oO

[eternal phoenix]

0 can’t be the start of a number because it’s just a three digit number then

0123=123

[reef sedge]

yes

[eternal phoenix]

but idk how you get the answer for b

i’m not as good as most of the people in this server at math

[reef sedge]

oh

ok

its fine

[eternal phoenix]

the last number being 3 makes sense

because there’s only three odd numbers

but idk why the rest are the way they are

[reef sedge]

6 because at the start

there are 6 numbers to choose from

and so on till the last 3

maybe it should be two

[eternal phoenix]

i understand

so basically

at the end there are only three possibilities

because there are only three odd numbers

at the start

we cannot use 0

and we already used an odd number at the end because we know we have to

so there are four possibilities at the start

for the second digit

we have already used the first and last digit

[reef sedge]

nonon lets say

lets say it was digits 1,2,3,4,5,6

not 0-5

[eternal phoenix]

we need 0 because it can’t be used at the start

that’s why it’s so important

[reef sedge]

oh yes

[eternal phoenix]

so 3 at end

[reef sedge]

OHH

4 cant be used at the start

[eternal phoenix]

4 at start because we can’t use 0 or the one at the end

[reef sedge]

but it can be used in the middle 2

[eternal phoenix]

no 4 numbers can be used at the start

[reef sedge]

yes

[eternal phoenix]

then in second digit

[reef sedge]

4 can be used at the start

but at the start 0 is excluded

[eternal phoenix]

we can’t use ending digit or starting

but now we CAN use 0 because it’s not at the start

[reef sedge]

but then that means 0 can still be used 0 cos not start

[eternal phoenix]

so we have four numbers to use again

[reef sedge]

yeahhh

yes

[eternal phoenix]

and then third digit

we’ve already used three digits

so we can use any last three

and this can go in any order with the second and third digit because multiplication is commutative

so 4x4x3x3

[reef sedge]

whats commutative mean

[eternal phoenix]

of 4x3x4x3

that

4x3=3x4

[reef sedge]

?

oh

u mean

[eternal phoenix]

we can change the places of numbers and it’s the same

[reef sedge]

converse

?

[eternal phoenix]

i was always told commutative

just a different term

but yeah

[reef sedge]

ye

[eternal phoenix]

so if you choose last digit first, then first, you can go to the second or third digit and choose it next

doesn’t matter

[reef sedge]

hm?

oh

i wonder what it be

if it was

1,2,3,4,5,6

odd 4 digit

[eternal phoenix]

idkkk

[reef sedge]

e

ye

[eternal phoenix]

5x4x3x3 i think

three choices for last

five choices for first

four choices for second

three choices for third

[reef sedge]

isnt it 6 * 5 * 4 * 3

[eternal phoenix]

no

because the last has to be an odd digit

well

maybe

i’m not sure

i don’t think so because that would include the case where there’s three even numbers at the end

[reef sedge]

last one is *3 i think purely bc there are 3 odd numbers in 1,2,3,4,5,6

[cedar kiln]

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[verbal hazel]

Open.

[verbal hazel]

Evaluate the following.

It was too hard to type all of that so I sent the image.

[dire geode]

What's in the exponential

[verbal hazel]

Arc tan x

[dire geode]

Did you try u sub with u=arctan(x)

[verbal hazel]

Doesn't work.

[cedar kiln]

@verbal hazel Has your question been resolved?

[verbal hazel]

.close

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

Is there a formula for finding least common denomenator?

[stray otter]

LCM?

[crimson sedge]

How is that?

[stray otter]

This

[crimson sedge]

Don't understand this?

[stray otter]

Try this

https://youtu.be/BSLem5-aDpI

[cedar kiln]

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[obtuse heart]

hi can somone help me with this?

im not too sure about continous probability

[drifting marlin]

What must be true of a function for it to be a valid probability density function?

(They call it a distribution graph but that's stupid)

[obtuse heart]

im new to the topic so is it it cant be negative and the probabilty must be less than 1?

[drifting marlin]

The first is correct, the second is somewhat vague

What must the area under such a graph be?

[obtuse heart]

the probability right?

[drifting marlin]

The area under the graph is related to probabilities, yes

But I mean the area under the entire curve. It must be equal to a particular number for the curve to model a probability distribution. What number must it be?

[obtuse heart]

1?

[drifting marlin]

1

[obtuse heart]

ahh ok

[drifting marlin]

Find the value of n that makes the area under that graph 1

[obtuse heart]

oh so you find the value of 5

and then you minus the probaility?

[drifting marlin]

Find the value of 5?

[obtuse heart]

x = 5?

like find the probabilty of it

then take 1 - it?

[drifting marlin]

We aren't finding any probabilities

[obtuse heart]

oh

[drifting marlin]

Well I guess indirectly we are, since we're computing area

[obtuse heart]

so we find the area under 5

then we take 1- the area?

it will become n = 6 right?

[drifting marlin]

What do you mean "area under 5"

n=6 does sound correct, sure

[obtuse heart]

ok thanks for helping me to understand lol

[drifting marlin]

No problem, good luck

[cedar kiln]

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[dense cargo]

how is the (3,3) element not -2

[royal loom]

it very well should be if the operation being performed is what it says it is

so, a mistake was made

[dense cargo]

alright tysm

[cedar kiln]

@dense cargo Has your question been resolved?

[cedar kiln]

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[vivid forge]

Hi, I'm making an incremental game and I need some help with making a function that allows you to buy the max amount of the upgrade that you can afford with the current amount of currency you have. the series is demonstrated in the picture.

[crimson delta]

what is the exponent?

[vivid forge]

1.3

[crimson delta]

is performance crucial? otherwise you could just write a for-loop which increments n until it cant anymore

hmm 1.3 is not nice

[vivid forge]

I have considered that, but I was wondering if there is a slightly more optimal solution, if I can't manage that I'll just have to stick with a for loop

[crimson delta]

you could probably interpret the sum as a riemann sum to an integral which will give you an approximation of the correct value which will then give you better starting point for the for-loop

not completely sure tho

what do you think will be the average value of n for this function

like a+3 or a+500

in what ballpark

[vivid forge]

Currently that is a problem the fuction itself doesn't scale fast enough so I need to change the values, as of right now it's on the order of 100 to 1000

I was considering making increments that just caps it at n=100 and n=1000 to make it simpler

[gritty viper]

that's a strange price increase system

the more upgrades the user is buying at once, the more accurate an integral will be

so you could do the Riemann sum, then if the upgrade count turns out to be < 100, do a for loop for more accuracy

[vivid forge]

I see, also do you have a recommendation to change the price increase function perhaps?

[crimson delta]

exponential price increase system isnt really that strange

[gritty viper]

if it was ^2, there's a closed form formula for the sum of consecutive squares

[gritty viper]

Its not really exponential, it's level^1.3

Exponential I would think of as 1.3^level

[crimson delta]

ah whoops misread. but ok still not strange

[gritty viper]

maybe if you don't play incrementals it doesn't seem strange

[vivid forge]

Yeah I see, the idea was the make it more flat so it wouldn't go too fast but I ended up on the oppsite side

[crimson delta]

well it depends on the rest of the game

[gritty viper]

there's nothing wrong with it from a game design perspective

[crimson delta]

if the rest of the game has exponential increases, then yes this is strange

way too slow

[vivid forge]

Yeah, I had a friend play for a bit and he got to 10^1345

[gritty viper]

price increase is usually exponential, sometimes linear

[vivid forge]

which well, would take years

[gritty viper]

If the upgrade buff is linear, you could get the same effect by making it compounding instead

[vivid forge]

Yeah I'll rework it a bit that sounds easier than the other workaround

[gritty viper]

does it give a linear buff to production currently?

[vivid forge]

Yeah this upgrade itself gives +1 to base value with each upgrade

[gritty viper]

well one thing you could do is make the power ^1 or ^2 and use closed form

[vivid forge]

I was thinking of changing most of them to geometric series since that has a summation formula

[gritty viper]

The other thing you can do (this could preserve the exact power scaling you have now) is make the price increase, say, 2^x, and make the power increase (2^(1/2.3))^x

,w 2^(1/2.3)

[wraith dagger]

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[gritty viper]

so if the price doubles every upgrade you want a 35% compounding bonus

power = 1.35^level

[vivid forge]

Hmmm I see and then it won't ever exceed itself which I had a problem with as well

thanks a bunch for the help, I'll go think about it again and rework a few things

[gritty viper]

np, incremental game math is always interesting

[vivid forge]

.close

[cedar kiln]

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[cedar kiln]

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[half elk]

$\lim_{n→\infty}e^{a_n} = e → \lim_{n→\infty}a_n = 1$

[wraith dagger]

nigrumpapilionem

[half elk]

Can I do this for convergence?

[red pumice]

yes

it works because e^x is continuous

[slate lintel]

e^x is monotonic

[half elk]

Isn't deal of sequences is being defined only for natural numbers unlike functions?

[red pumice]

well yes

i mean to say that

suppose f is a continuous function

then $\lim_{n \to \infty} f(a_n) = f(\lim_{n \to \infty} a_n)$

[wraith dagger]

heavy0201

[astral bay]

...that's true if the limit on the right exists

but ``$\lim_{n\to\infty} f(a_n)$ exists" for $f$ continuous is not enough to conclude that $\lim_{n\to\infty} a_n$ exists

[wraith dagger]

bee [it/its]

[red pumice]

well yes that too

[astral bay]

i think it does work in the case of e^x but i'm not sure why exactly it's true

[slate lintel]

continuous and monotonic (derivative never 0)

so it has an inverse

[red pumice]

i would guess because inverse exists

ye

[slate lintel]

inverse maybe needs to not have a cusp as well? not sure

[astral bay]

ah yeah you can use ln as the continuous function

[astral bay]

in this

[slate lintel]

cusp is probably fine, log works and it's not uniformly continuous so

probably okay with a bunch of things

[astral bay]

$\lim_{n\to\infty} e^{a_n}=e \implies \lim_{n\to\infty} \ln(e^{a_n})=\ln(e) \implies \lim_{n\to\infty} a_n = 1$

[wraith dagger]

bee [it/its]

[half elk]

Does it work for every monotonic and continous function of a_n then?

But inverse of every monotonic function doesn't have to exist, right

[slate lintel]

,w define monotonic function

[wraith dagger]

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[slate lintel]

eh
i mean 'strictly increasing or strictly decreasing'

[cedar kiln]

@half elk Has your question been resolved?

[half elk]

Since monotonic function doesn't rule out f(a) = f(b) for a<b

I meant that

[slate lintel]

yeah i mean f(x) = 1 is monotonic and continuous

and f(g(x)) -> 1 for any g
but g definitely doesn't always converge

[half elk]

I see

Thanks for the help y'all

.close

[cedar kiln]

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[dreamy marsh]

Can someone help me with this

[dreamy marsh]

It's an I q question

I just need someone to explain this answer key

[livid hound]

that's just the aribtrary rule they came up with

which part of their explanation of their setup don't you understand

[muted bear]

I feel like this is far too few numbers for you to glom onto an alternating rule

[dreamy marsh]

Ah it's ok. I understand now. Ty btw

[livid hound]

||be psychic||

[dreamy marsh]

.close

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[cedar kiln]

Please don't occupy multiple help channels.

[thin hedge]

i don't understand how this is factored

how is this factored correctly

how does the negative go to cos x

its at sin x?

[crimson sedge]

cos(x) - cos(x)sin^2(x) = cos(x)(1 - sin^2(x))

Let a = cos(x) and b = sin(x)

then we have a - ab^2

factoring out a, we have a(1-b^2)

[thin hedge]

ohhh

i see it

so how did they use the variation

of the pythagorean identity

to get cos^2x

[crimson sedge]

because $\sin^2(x) = 1 - \cos^2(x)$

[wraith dagger]

kenfps

[thin hedge]

is it j subsituting

[crimson sedge]

and $\cos^2(x) = 1 - \sin^2(x)$

[wraith dagger]

kenfps

[crimson sedge]

yes

it's just subtituting

these are the two variations of the pythagorean identity

[cedar kiln]

@thin hedge Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[ebon grotto]

.

!status

[ebon grotto]

!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

[ebon grotto]

6

[rotund fox]

what is the question

[ebon grotto]

Light travels at a speed of 3 x 108 m/s. How far will light travel in the following times.
Give your answer in standard form
a) one hour b) one day c) one week

I want the

urm

working

for c part

scientific notation.

i need,

I wanna know how c part answer is 1.8144 * 10^14

ok

[astral bay]

(they said 6 "none of the above", not 5)

[rotund fox]

i think you could do 1 thing which is convert that speed into kmph or meter per hour and then solve it (easier method)

[rotund fox]

my bad

[ebon grotto]

I want working to know how c part is 1.8144 * 10^14 (scientific notation answers).

[ebon grotto]

Yeah no not allowed

[rotund fox]

wait what

[rotund fox]

in that case approximately convert week into seconds using scientific notations

[ebon grotto]

i need to convert day to week

but i frogot my working

[rotund fox]

1/7?

[ebon grotto]

i day = 2.592 x 10^13

i need answer for it

[rotund fox]

multiply that by 7

wait

no

[ebon grotto]

question : 3x10^8x60x60x7x24

[rotund fox]

1 day is 86000ish seconds

86400

seconds

[ebon grotto]

edited

I need working

question : 3x10^8x60x60x7x24

[rotund fox]

yep

[ebon grotto]

in scientific notation

[rotund fox]

60x60x7x24

solve this separately

[ebon grotto]

3600x168

604800

[rotund fox]

take the 100 out, multply that 100 with 10^8 to get 10^10

[ebon grotto]

no understand

[rotund fox]

604800 = 6048 x 100

= 3 x 10^8 x 10^2 x 6048

= 3 x 6048 x 10^10

[ebon grotto]

ok

18144 x 10^10

so 1.8144 x 10^10 + 10^4

so 1.8144 x 10^14

?

[rotund fox]

yeah

wait

[rotund fox]

10^4 isnt added

its multiplied

[ebon grotto]

no

[rotund fox]

yes

[ebon grotto]

oh shi

i did +

i meant x

t

ty

[rotund fox]

type .close if problem solved

[ebon grotto]

ye 1 sec

.close

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[tulip harness]

Consider the system of differential equations $x'(t)=Ax(t)$ with an initial value $x(0)=x_0 \neq 0$, where $A=\begin{pmatrix}
0 & 1\
-1 & 0
\end{pmatrix}$\
Analyse the behavior of the solutions obtained from the following approximation:\
$x_{n+1} = x_n + hA(x_n + x_{n+1})$, where $h> 0$

[wraith dagger]

casiel368

[tulip harness]

Okay so this exercise asked me to use Euler's method and implicit Euler before doing that and the norm of x goes to infinity or zero

I did prove that, and now I have to look at this new method

[wraith dagger]

casiel368

[tulip harness]

All of B eigenvalues have norm of 1

My intuition is that the DE describes the circular movement, because speed will always be a 90° rotation of the position vector, so using Euler or implicit Euler will over/undershoot the actual solution, but somehow the sum of both will be right on point, as shown by the fact that B is a rotation

The thing is, how do I finish this? What should I do next?

<@&286206848099549185>

[cedar kiln]

@tulip harness Has your question been resolved?

[cedar kiln]

@tulip harness Has your question been resolved?

[cedar kiln]

@tulip harness Has your question been resolved?

[tulip harness]

<@&286206848099549185>

[tulip harness]

<@&286206848099549185>

[cedar kiln]

@tulip harness Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[granite shuttle]

Hello I don't understand how to start this

am i adding -3 to 7 and ending with 4 as my first number

and -8 to 2i

am i just going to repeat this process on the -50

and then add the non square roots

[obsidian coral]

Yes

[obsidian coral]

Yes

[granite shuttle]

is there a formula i can follow for this that would make it easier to remember

[obsidian coral]

Not really, it's just adding/subtracting like terms

[granite shuttle]

wait oops

i didnt mean to put the negative on the 2i

is this right other than that

[obsidian coral]

Is the i under the root?

[granite shuttle]

oops

it is...

its not supposed to be right?

[obsidian coral]

No it's not

[granite shuttle]

so then its just (a+bi)+(c+di)=(a+c)+(b+di)i

[obsidian coral]

Yes

[granite shuttle]

is it ok if i use the plus minus symbol

as i do it

wait no thats wrong

to do

i wanted to do this but i realized its done

[obsidian coral]

Yes that plus/minus is wrong

[granite shuttle]

this feels really weird

but this is right?

[obsidian coral]

Yes

[granite shuttle]

thank you

i have a test im gonna take in an hour and im allowed to use like post it notes

do you think it would be a good idea to write these down

[obsidian coral]

Depends on what you know and don't know

[granite shuttle]

this is my first day with the material so i havent had much experience with it

im learning as i go

right now

[cedar kiln]

@granite shuttle Has your question been resolved?

[granite shuttle]

yes

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[heady mountain]

Find the area bounded by $r=7cos6\theta$
I have the integral $12 * 0.5 \int_{pi/12}^{3pi/12} (7cos6\theta)^{2} d\theta$ but I think this is wrong.

[heady mountain]

bounds are from when cos6theta =0 so 3pi/2 and pi/2 divided by 6

[wraith dagger]

kenny.xie

[cedar kiln]

@heady mountain Has your question been resolved?

[heady mountain]

<@&286206848099549185>

Ok so my “saved” answer in the software is the evaluation of this integral which got marked for credit

but the “solution” is 49pi/12

my integral evaluates to 49pi/2

.close

[cedar kiln]

Channel closed

Closed by @heady mountain

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[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[azure stirrup]

does anyone know how to do 7c and explain 11c

[azure stirrup]

i got this for 7a and b

[crimson sedge]

ok

[azure stirrup]

when it asks for to sketch the curve is it the original curve or the differentiated equation

<@&286206848099549185>

[cedar kiln]

@azure stirrup Has your question been resolved?

[azure stirrup]

<@&286206848099549185>

[tame heart]

prob g(x)

[azure stirrup]

what abt the tangents

[tame heart]

all the tangents at those points youve found in part b ig

[azure stirrup]

soo i just plot down those points..?

[tame heart]

nah i think you have to draw all the tangents at those specific points

[azure stirrup]

wdym all the tangents

[crimson sedge]

the tangents at those points, you have to draw them

[azure stirrup]

like a line…?

[crimson sedge]

yea

Do you know what a tangent is

[azure stirrup]

like that?

…

[crimson sedge]

yea but drawn to scale

[azure stirrup]

a straight line that touches the curve

[crimson sedge]

do you know how to write it's equation

[azure stirrup]

altho i’m not sure whether tje line is suppose to pass thru the curve like at 0,0 or 5,0

is that right?

[azure stirrup]

not too sure

[crimson sedge]

so at a point on the curve from where the tangent is draw, the slope of the curve at that point and the slope of the tangent will be the same

That's why it is a tangent

[azure stirrup]

icic

[crimson sedge]

so you know 2 things the slope at that point on the curve and you know the point where the tangent will touch it,

So you've got a point and the slope and you can write the equation of that line using that which is the tangent

[azure stirrup]

is the slope 2x-5

from q7a

[crimson sedge]

that's not the slope that is g'(x) but when you put a specific x-coordinate in it, it will tell you the slope at that point

[azure stirrup]

are the specific x-coordinates the coordinates from 7b?

[crimson sedge]

Yes

[azure stirrup]

so i put those x-coordinates into g’(x)

2(3)-5=1

?

[crimson sedge]

no you solved for those slopes which were already given to find the points

You already know them, they are given in the question

no need to solve for them again lol

[azure stirrup]

ohh the 1, -1 , 5 ,-5 and 0?

[crimson sedge]

yea

Using that you found the points in the first place

[azure stirrup]

and thats the gradient and i found the points so i put it into a formula