help-17

[severe dagger]

yeah this is where i was having trouble, wasnt sure how to approach it so i just gave it an attempt

[blissful sentinel]

Hmm

[blissful sentinel]

Have you gone over Rice's theorem in class?

If not, check out this StackExchange post about a very similar problem https://cs.stackexchange.com/a/85238 and let me know if you have any questions about what they did

I gotta go to bed, but feel free to ping me with any questions and I'll see it when I get up

[severe dagger]

yes, i take it my understanding is likely super limited lol 😭 ill revisit it and see how my answer changes

[severe dagger]

ty ill lyk (going to bed rn tho)

[viscid thicket]

Hi, not helping but I'm curious what subject is this ?🙂

[blissful sentinel]

theory of computation

[blissful sentinel]

np, I'm not sure how to use the hint tho

[severe dagger]

all of the problems we've done so far have been incredibly easy so putting it into practice for more complex ones makes me fall flat

[blissful sentinel]

Yes that's correct

[blissful sentinel]

That's exactly what this answer does

It doesn't use one of the sets you have on your bullet point list but you could modify it so that it does

[severe dagger]

honestly my problem with it is that hte structure of the answer differs from how we were taught to prove it so i dont fully understand how to reformat it

[blissful sentinel]

What is the overall format that you want to get it in

Right

so let's start the solution to this Q like that

Let's assume that CF_TM is decidable and M is a TM which decides it....

Then what's the next step

[severe dagger]

either construct a a TM "D" for its behavior or find something we can reduce it to?

[blissful sentinel]

Sure, so what's the TM that the answer constructs based on M?

[severe dagger]

like the whole definition or? sorry i don’t really get what you mean

[blissful sentinel]

You want to construct a TM D

What do you want to define D to be?

[severe dagger]

something that decides CFtm?

[blissful sentinel]

No, we already said M does that by assumption

You want to reduce CF_TM to some other problem

[severe dagger]

so like how i reduced it to A TM in my solution?

so i need to define a tm D that decides A TM using one that decides CF right

[blissful sentinel]

Yes that would be one way of solving the problem

I don't think the D in your solution decides A_TM, but if you could use M to define some sort of D that decides A_TM, that would be a successful reduction from CF_TM to A_TM, proving that CF_TM is undecidable.

[hidden turret]

golden

[severe dagger]

i understand the idea but the execution is kind of where im failing, how do i manage to make one that decides A through CF

like besides loosely connecting it

[blissful sentinel]

Have you read the stackexchange answer I linked? What does it do?

[severe dagger]

it simulates M on varying inputs and tests it behavior for non regular languages. depending on whether or not it halts on the input, it determines the regularity of the language

[blissful sentinel]

Okay so what is it reducing to what

[severe dagger]

the halting problem?

[blissful sentinel]

yeah, the halting problem on empty input, which is not one of your listed bullet-point examples

but you could think about adapting the same approach for your problem

[severe dagger]

ohh so you think i shoulder alter my proof to focus on setting up for the halting problem rather than A_TM?

[blissful sentinel]

(Sorry, I got the order of "reduces to" wrong, I should've said "reduction from A_TM to CF_TM, proving that CF_TM is undecidable" earlier)

I think it would be beneficial to first carefully read through how the answer there reduces {<M> | M is a TM and M halts on empty input} to {<M> | M is a TM and L(M) is regular} and then ask any questions about that that you don't understand, then try to adapt the same approach to reducing one of the problems on your list (such as HALT_TM, if that's what you choose) to CF_TM.

[severe dagger]

ill try that out, thanks

[severe dagger]

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[dull haven]

How do I interpret the results of this three factor ANOVA test?

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[hollow glen]

How would i find out the minimum value of b-a when a<b are in [0,1] and b^3-a^3=0.5?

[inland galleon]

I was about to answer this in discussion 2 lol

anyways do you know derivatives?

[hollow glen]

Yeah

[inland galleon]

Do you know how to find the minima and maxima of a differentiable function?

[hollow glen]

Yeah

[inland galleon]

Okay then this should be simple

You can factor b^3-a^3

into (b-a)(something)

[hollow glen]

I factored it to b2 +a2+ab

[inland galleon]

yeah

[hollow glen]

But... a<b?

[inland galleon]

we aren't using that yet

we have (b-a)(b2+a2+ab)=.5 right?

[hollow glen]

Yes

[inland galleon]

so what about moving (b2+a2+ab) to the right?

via division

[hollow glen]

Okay, in the sense f(x)=b-a?

[inland galleon]

not f(x)

f(a,b)

then we can find the minima of f(a,b) in [0,1]

[hollow glen]

Partial derivatives?

[inland galleon]

Yeah

I don't remember how to do that, so someone else can help you along with computing the minima and maxima

But it should be a straightforward computation

[hollow glen]

I learned pde and forgot everything after a year 😦

[vocal sleet]

@hollow glen Has your question been resolved?

[hollow glen]

I got b=2a, a=2b for fa and fb after i took the partial derivatives

Something's wrong

[inland galleon]

Hmmm I have time to spare so I'll try to relearn this

I am gonna follow this: https://sites.science.oregonstate.edu/math/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/min_max/min_max.html

[hollow glen]

Okay so im gonna post the whole question, this was the last part of it

[inland galleon]

https://www.symbolab.com/solver/partial-derivative-calculator/\frac{\partial}{\partial x}\left(\frac{.5}{\left(x^{2}%2B2xy%2By^{2}\right)}\right)?or=input

And it's symmetric, so the value should be similar for partial y

The partials are never zero

So we only need to check the boundary values

[hollow glen]

bit confused

[inland galleon]

So to find minima we need to check for critical points in the interior, and the value of the function in the boundary of its domain.

The domain is the square: [0,1] x [0,1]

actually no, it's smaller than that since a<b

[hollow glen]

Here the input is incorrect

It should be xy not 2xy

[inland galleon]

ah I see

Well then you do have some critical points to check. But you also need to check the boundary

https://www.symbolab.com/solver/partial-derivative-calculator/\frac{\partial}{\partial x}\left(\frac{.5}{\left(x^{2}%2Bxy%2By^{2}\right)}\right)?or=input

Here is the relevant part

[hollow glen]

After i obtain the partial derivatives, what happens if the system of equations dont have an unique solution?

Or in this case a=b=0

The trivial soln only?

[inland galleon]

Yeah I only get this solution

But if the system of equations have multiple solutions, then you have to check all of them.

[inland galleon]

It shows up as an example in the link I sent here

I have to go now good luck

[hollow glen]

Thanks!

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[leaden scroll]

hi how do i do this?

[silk osprey]

which

[leaden scroll]

oh whoops

a

[silk osprey]

i cant see the upper limit

[leaden scroll]

the top number is 2

[silk osprey]

ok

do you know what the definite integral represents

[leaden scroll]

no 😭

[silk osprey]

hmm

well

[wind geyser]

lmao

[silk osprey]

the definite integral between two points a and b represents the signed area the under curve between x = a and x = b

[leaden scroll]

is b 2 and a is 0

[silk osprey]

so in this case we are looking for the area bounded by g(x) and the x axis between x = 0 and x = 2

[silk osprey]

yes

can you tell me what that area is

it’s a common shape

you should know the area of a triangle

[leaden scroll]

1/2 b*h ?

[silk osprey]

yep

[leaden scroll]

do i use that in getting an answer

[silk osprey]

yes that’s the formula for the area of a triangle, now what’s the base length and height of this triangle

[leaden scroll]

cuz my prof showed us this and like idk what to do with any of it

[leaden scroll]

is the base 1 and the height 2?

[silk osprey]

yea this is useless in almost all problems you’ll encounter for computation

it’s just giving you a more formal definition of what the integral really represents

[leaden scroll]

ohh okay

[silk osprey]

just think of it as summing up infinitely thin rectangles

if you’d like an animation search up 3b1b

i’ll link a video

[wind geyser]

wonder what the * represents

[silk osprey]

https://youtu.be/rfG8ce4nNh0?si=97sLAOl8rCA8g2Z0

[leaden scroll]

oh i see

[leaden scroll]

?

[wind geyser]

ignore that

[leaden scroll]

okay

[silk osprey]

it’s just notation used for some point in the interval you’re summing usually used to distinguish between different summation methods

dw about it

[leaden scroll]

1/2 times 1*2 = 1

[silk osprey]

hmm how did you get a base of 1 and height of 2

[leaden scroll]

oh

1/2 timesw 2 times 4

[silk osprey]

yep

which is

[leaden scroll]

4

[silk osprey]

mhm

simple

can you do b now

what shape is it from 2 to 6

also note that we consider area below the graph to be negative

[leaden scroll]

would it be like a half circle?

[silk osprey]

yes exactly

do you know the formula for the area of a circle

let’s start there

[leaden scroll]

pi * r^2 /2

[silk osprey]

yea that’s for a semi circle

so what’s r here

[leaden scroll]

4?

[silk osprey]

hmm not quite

that’s the diameter

since it spreads from 2 to 6

the radius is the distance to the center of the circle

[leaden scroll]

2?

[silk osprey]

mhm

since the center would be halfway between

which is 4

the radius would be 2

[leaden scroll]

pi * 2^2 over 2

2pi

[silk osprey]

yep

not too bad

now let’s do the last part

but remember

[silk osprey]

this is -

for the integral

that’s the area, but since we’re evaluating the definite integral we consider portions below the x axis to be negative

[leaden scroll]

-2pi

[silk osprey]

yea

so now do part c

you can split it into parts btw

like

$\int_0^7 g(x) dx = \int_0^2 g(x) dx + \int_2^6 g(x) dx + \int_6^7 g(x) dx$

[twin meteor]

knief

[silk osprey]

we already determined the first two

[leaden scroll]

ohh i see

[silk osprey]

can you find $\int_6^7 g(x) dx$

[twin meteor]

knief

[silk osprey]

first of all what shape is it

[leaden scroll]

another tirangle?

[silk osprey]

yep

so we use 1/2 bh

where our base is what

and our height is what

[leaden scroll]

b is 1 and h is 2?

[silk osprey]

the base is indeed 1 but how’d you get the height to be 2

[leaden scroll]

do we not count under the graph?

[silk osprey]

we start at y = 0

that’s where we left off

from 6

[leaden scroll]

ohh

okay

[civic otter]

That's not a triangle...

[silk osprey]

yep

[leaden scroll]

oh

so it would be 1/2

[silk osprey]

mhm

[silk osprey]

so what does this become

[leaden scroll]

4 + -2pi + 1/2

[silk osprey]

yep

and we can combine the 1/2 and the 4

so

we get 9/2 - 2pi

[leaden scroll]

yuh

is that it?

[vocal sleet]

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[jagged cargo]

is W unique?

e.g. $U = {(x, y)\in\mathbb R^2: x + y = 0}\subseteq\mathbb R^2$

[sly sierra]

nope

for example say V is R^2 and U is the span of some nonzero vector v

then you can take any w that is not a scalar multiple of v, and then W = span{w} works

[jagged cargo]

ahh i see

thanks

.close

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[leaden scroll]

how do i do this

[golden herald]

Might be easier than you think

remember that
[
\frac{\sqrt{y} - y}{y^2} = \frac{\sqrt{y}}{y^2} - \frac{y}{y^2}
]

[twin meteor]

Aero

[leaden scroll]

is it * dy still

[golden herald]

yes

integration is distributive by the way

[leaden scroll]

wut

[golden herald]

you can do [
\int \left(f(x) + g(x)\right) dx = \int f(x) dx + \int g(x) dx
]

[twin meteor]

Aero

[leaden scroll]

ohhh okok

like that?

[golden herald]

yes

also,

although it doesnt matter that much

[leaden scroll]

what do i do after though 😭 like what am i solving for

[golden herald]

you just need to integrate

so uh

simplify y/y^2

what is that

[leaden scroll]

1/y

[golden herald]

ok

what is sqrt(y)/y^2 in the form of y^(something)

[leaden scroll]

y^-3/2

[golden herald]

ok great

so what is the integral of y^(-3/2)

[leaden scroll]

like the derrivative?

[golden herald]

no

the antiderivative

you need to apply the property that [
\int_a^b x^n dx = \frac{1}{n+1}\left[x^{n+1}\right]_a^b
]
for $n \neq -1$

[twin meteor]

Aero

[leaden scroll]

-2y^-1/2

?

[golden herald]

yeah you got it

thats the first antiderivative

the second integral, just use the other property that
[
\int_a^b x^{-1}dx = \left[\ln(|x|)\right]_a^b
]

[twin meteor]

Aero

[leaden scroll]

ln 2 ?

[loud walrus]

No abs value needed

In this one

[golden herald]

not in this example, but i am just giving the general antiderivative

[golden herald]

are you saying thats what the second integral evaluates to entirely?

[leaden scroll]

yes?

[golden herald]

Then no, thats incorrect

[loud walrus]

Show your work

You misscalculated something

[leaden scroll]

i just plugged in x to this

{(ln|2|)}

[golden herald]

you have 4 though, as the upper bound

not 2

[leaden scroll]

ohhhh

i think i got it?

1 - 2 ln(2)

[golden herald]

Bingo

[leaden scroll]

yayyy

tysm

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[cold obsidian]

If a function f(x-1) is reflected in the y-axis would the result be f(-x-1) or f(-(x-1))? Sorry im horrible with functions

[marsh charm]

if i had to take a educated guess

[tidal dock]

the latter

[marsh charm]

id say the 2nd one

[tidal dock]

f(-(x-1))

[cold obsidian]

But in desmos it gave me the first one

I thought its the second asw

[tidal dock]

show your desmos then

[marsh charm]

maybe u used desmos wrong

[cold obsidian]

The black and blue one reflected in y axis….

The green one reflected in y=1

[tidal dock]

oh well

i get it now

yes, it would be f(-x-1) = f(-(x+1))

[cold obsidian]

I dont wtf

[tidal dock]

f(x-1) is shifted one unit to the right

you want your thing to be reflected around y axis and shifted one unit to the left

so it will be f(-(x+1))

[cold obsidian]

Do u get this

[marsh charm]

i dont but i really dont need to

[cold obsidian]

So which one is right

[marsh charm]

dont ask me man

u seem to know more than me

[cold obsidian]

U,m

I think i kinda get it

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@cold obsidian Has your question been resolved?

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[torn tangle]

Can someone help me

I cant understand how to do it

I just get lost

[marsh charm]

do you know what a asymptote is

[torn tangle]

Ye, a "line" that a graph goes to but never will touch it

[marsh charm]

set the denominator

equal to 0

and solve for x

possibly a asymptote there

[torn tangle]

This

?

[marsh charm]

thats right

what do u think the next step is

[torn tangle]

Yyy, from what someone was trying to teach me, we do the lim of ±oo

[marsh charm]

yea

[torn tangle]

And calculate the domain

And then some lhopital rule and thats all i know

[marsh charm]

not infinity

from -2-

and -2+

basically just find out

if y

goes to positive or negative infinity

as x aproaches the e value

if they do

than its a asymptote

[torn tangle]

So calculate limits for this

Or what

[golden herald]

A few good rules of thumb to remember:

\medskip
For horizontal asymptote, you're always considering $\lim_{x\to \pm\infty}$

\medskip
For vertical asymptotes, you're considering $x = c$ where $c$ is where the denominator is 0. (You need to make sure that the numerator and denominator are coprime, though)

[twin meteor]

Aero

[torn tangle]

Coprime?

[golden herald]

like, there are no factors between the numerator and denominator because that wouldn't be a vertical asymptote. But anyways that's not your case

[golden herald]

For example, this value of x constitutes a vertical asymptote

[torn tangle]

What

[golden herald]

x = e^-2 is where your vertical asymptote is

[torn tangle]

Tell me if there is like one patern that i can follow for all calculations of asymptotes

[torn tangle]

Oh

Ye

[golden herald]

I mean I said the methodology here

[torn tangle]

Okay

So

So to calculate the vertical asymptotes i just take the denominator and make it =0

Yes?

[golden herald]

yes, but again make sure the fraction is coprime. What I mean by this is that make sure you don't have something like [
\frac{(x+2)(x+3)}{x+3}
]
because in this case $x = -3$ is not a vertical asymptote, it's a hole rather

[twin meteor]

Aero

[golden herald]

because the x+3's "cancel"

[torn tangle]

Okay

Sooo

Whats then i dont have a denominator or wtv

But I have Let's say y=xe^-2/x

[golden herald]

in that case x = 0 isn't an asymptote either, it's just a hole

wait

do you mean $x\frac{e^{-2}}x$ or $xe^{\frac{-2}{x}}$?

[twin meteor]

Aero

[torn tangle]

2nd one

^(-2/x)

I have smthing like this

[golden herald]

then yes x = 0 is an asymptote then

[torn tangle]

And have no clue what iv done here besides domain

[golden herald]

the actual procedure is that you need to calculate [
\lim_{x\to 0^+} xe^{-\frac{2}x}
] and
[
\lim_{x\to 0^-} xe^{-\frac{2}x}
]

[twin meteor]

Aero

[golden herald]

if they're infinity or -infinity

then there is an asymptote

[torn tangle]

I have this

Ik I have it solved but I'm trying to understand what to do, I need a scheme that I can follow

[golden herald]

I am sorry but I don't think I understand what you wrote in there? Anyways what I would do is utilise the fact that [
\lim_{x\to 0^+} xe^{-\frac{2}{x}} = \lim_{x\to 0^+}x \cdot \lim_{x \to 0^+} e^{-\frac{2}{x}}
]

[twin meteor]

Aero

[torn tangle]

I mean, in the [] you calculate the lim for this example 0‐, and it's 0- cuz it's -2/x and that's why there is this graph on the left, and cuz the lim was 0- * +oo it's undefinable so I did the lhopital or wtv it's called (the H above the =) and the do the lim of it

And somehow I got the asymptote

[golden herald]

I see. I think your notation might be your biggest enemy in what you wrote. Try not to substitute in stuff like "infinity" and "0^+" as it is both very improper and will confuse you. Try to keep everything in terms of x's when dealing with limits.

[golden herald]

Doing it like this might be more insightful for you, though. The limit of x will be 0 and to compute the limit of e^(-2/x), it's good to think of it inside out

as x becomes smaller and closer to 0 (say 0.00001), 2/x becomes huge (approaches infinity). so -2/x approaches -infinity, and the exponential function approaches 0 as its exponent approaches -inf

So your limit all in all is equal to 0 * 0

[vocal sleet]

@torn tangle Has your question been resolved?

[torn tangle]

Sorry for not responding

I'm walking to other campus

I'm gonna look into it later thanks

[vocal sleet]

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[twin meteor]

sotiris

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@stiff aspen Has your question been resolved?

[vocal sleet]

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[swift robin]

can sin^2y be written as (siny)^2?

[marsh charm]

no

siny ^ 2

is

siny * siny

yes

it can

[bronze osprey]

it can

[swift robin]

okok so as long as the whole thing is squared right

[bronze osprey]

$\sin^2 y = (\sin y)^2$

works for any exponent, except don't use -1 as the exponent

[twin meteor]

southlander!

[bronze osprey]

cause $\sin^{-1} y$ means $\arcsin y$

[swift robin]

why not -1?

[twin meteor]

southlander!

[swift robin]

oh

okay

[bronze osprey]

not 1/(sin y)

yeah

[swift robin]

arcsin

is

secant?

eh wait i meant cosecant

[bronze osprey]

no

I'm saying 1/(sin x) is cosecant

but 1/(sin x) is not arcsin(x)

[swift robin]

ohh

i see

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[cursive prairie]

Hello, I want to learn how to calculate the probability of reaching a step in a sequence of events, where if you fail one step, you return to the previous step.

Let's say we got 6 steps, everyone with a probability of success.

Step 1 60%
Step 2 40%
Step 3 40%
Step 4 20%
Step 5 20%
Step 6 9%

Every time you fail you return a step and lose a point, and every time you succeed you gain a point. I know you could multiply the chances and get a percentage chance of succeeding every step in a roll, but if you return one step, you get another chance, that would change your chances, right?

[cursive prairie]

Also, this is for a game, every step uses a resource. I want to learn the average use of resources to clear all steps.

[cyan talon]

right

[bronze osprey]

Markov chains

[cyan talon]

damn sniper w/ 2 word answer and nothing else 😉

[cyan talon]

but yeah markov chains represent exactly that kind of situation

I'm wondering if there's super intro resources/videos online about them tho, I only know of fat books abt them

[cursive prairie]

I didn't think this would be such a complicated problem

[cyan talon]

well it's not a complicated problem yes

like it would take me a few minutes by hand to do

but that's cause I know what markov chains are, and those questions you asked about the situation (average number of resources used, etc...) are pretty standard

[cursive prairie]

Got it, I can try to search a few videos about it. Is there anything specific I should search for, or just Markov chains?

[cyan talon]

just markov chains yeah

https://www.youtube.com/playlist?list=PLANMHOrJaFxPMQCMYcYqwOCYlreFswAKP
this series looks alright

& if you want something beefy but concise for more details, I found this pdf not too long ago http://www.statslab.cam.ac.uk/~rrw1/markov/M.pdf

@cursive prairie

[cursive prairie]

I don't know if I would understand a paper, I have only a high-school-level math knowledge. And i forgot most of the more complicated stuff already, haha

[cyan talon]

it's more like a small textbook, not a research paper

but yeah it's just an extra if you want to come back to it later w/ more math, I'm not suggesting you read it right now

[cyan talon]

is this for an actual game you're playing or just some example ?

[cursive prairie]

that example is from a game that exists, I'm trying to calculate resource expenditure

it's quite an old game, are you familiar with Ragnarok Online?

these are the values i'm working with

[cyan talon]

the shadow upgrades

[cursive prairie]

yeah

[cyan talon]

aight I mean I can give you the answer now for the average resources needed, if that helps you in the moment

but if you want to figure it out all by yourself it's fine too

[cursive prairie]

I know it's a dum reason, but I wanted to learn how to calculate it

[cyan talon]

even if I spit you out the answer, you can go read about markov chains afterwards lol

it's not either/or tbf

[cursive prairie]

yeah

[vocal sleet]

@cursive prairie Has your question been resolved?

[cyan talon]

@cursive prairie re had to go do something

resources from +3 +4 +5 +6 +7 +8 +9 to get to +10
[61.41, 60.41, 58.08, 52.08, 40.58, 36.46, 34.18]

here's what I got

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[nova crown]

We are supposed to proove that d/dx ln(x) is 1/x. We did case 1 when h > 0 and he said that case 2 when h < 0 might come on the test. But im having some struggles.

[nova crown]

We started by drawing a graph of 1/t

Then we compared the rectangles like such

So case one when h > 0 we can divide with h and get

Then let

$h \rightarrow 0^+$

[twin meteor]

vrsth

[nova crown]

Then by the squeeze theorem we get

So my question is when doing that h < 0. Do i need to make a seperate graph where the points are switched and how would i prove it? im stuck

[cursive turret]

for h < 0 you can do the same.

write k = -h, then you get pinot c and c-k instead of c and c+h, and as you said the points switch, c-k is left from c, but its the same sketch. (and 1/c is below 1/(c-k) and so on ...

[nova crown]

Right yeha

The points switch

Do i need to write k=-h or can i just directly write it as 1/c is below 1/(c+h)

[cursive turret]

you do not need to use a k, you can use h and argue that h< 0.

[nova crown]

Right

[cursive turret]

in the last step (division by h) you need to be careful since there are inequalities

[nova crown]

Yeah

they flip right

The graph is like this right now tho right

[cursive turret]

yes

[nova crown]

So just after switching the points to the picture above i just write up the comparisons of the areas

[cursive turret]

its the same way as in the case h>0

[nova crown]

The same comparison right?

so

and h < 0

Dont the ln() in the middle switch?

[cursive turret]

thats why i would prefer to use k = -h, you are comparing areas here so the most left and the most right side should be positive but ypu can do it this way, too.

[nova crown]

but this way its still correct?

The pic above

[cursive turret]

yes.

[nova crown]

Why do the ln no switch place?

like

ln(c)-ln(c+h)

[cursive turret]

in really they should switch and you should "-h" (interpreted as side of an rectangle should be positive) onm the most left and most right side.

but yes, you could to it your way. but still i would recommend the way k=-h a it would be much clearer. but again, you can do it with h as well.

[nova crown]

So with k = -h

its like this?

[cursive turret]

yes

[nova crown]

would it be like this?

because then we get the same thing that correct

[cursive turret]

are you posting from differnt accounts?

[nova crown]

yes

have 2 different acocunts so i can speak with ym friends on discord trough computer at the same time as i can screenshare the tablet

[nova crown]

But is it correct?

[cursive turret]

yas

[nova crown]

Okay!

so to finish it up

[nova crown]

Just ot be clear

[nova crown]

this pic

The inequalities witch when we divide by k since k is negative h

then i sub in -h in the spots where k is and the lns on top switch leaving us with the definition of derivative

and we can apply the squeeze theorem?

[cursive turret]

yes

[nova crown]

Thank you!!

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[rapid anvil]

Sanity check: If I have $\frac{f(x)g(x)}{g(x)h(x)}$, this is always equal to $\frac{f(x)}{h(x)}$ right?

[twin meteor]

imo2025

[lone linden]

As long as $g(x) \neq 0$, yes.

[twin meteor]

Civil Service Pigeon

[rapid anvil]

And if $g(x)$ can equal 0 at some x?

[twin meteor]

imo2025

[lone linden]

Then the domains aren’t necessarily the same

Unless all of the zeroes of g are also zeroes of h

[rapid anvil]

Prototypical example, $\frac{(x-2)(x-1)}{(x-2)(x-3)}$

[twin meteor]

imo2025

[lone linden]

Yeah something like that is what I’m getting at

[lone linden]

Because this doesn’t have 2 in the domain, but (x-1)/(x-3) does

[rapid anvil]

So there won't be an asymptote at x=2?

[lone linden]

This has a hole at x=2

(x-1)/(x-3) doesn’t

[rapid anvil]

i knew it

LOL

but apart from that no other information is "lost" when cancelling?

[lone linden]

Sure ig

[grizzled vault]

if you’re not canceling 0/0 then it’s okay i guess

lol

[rapid anvil]

lol i just used lhopital and showed there was a discontinuity but i was told i overkilled

oh well

thanks guys

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[hardy crater]

Hello, please could someone explain why dv/dt is equal to k/rooth ?

I understand everything else

[outer warren]

it says in the question

right above
a)

[hardy crater]

I see thanks

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[vale frigate]

how do i evaluate this (no lhopitals/series expansions)

[lethal heart]

Can you use notable limits?

[heavy yoke]

you will probably end up using these at some point

[somber portal]

i'd start by replacing cos(2x) with trig identities

[vale frigate]

yes

[vale frigate]

i used double angle identity and got (1-cosx(2cos^x -1))/x^2

[somber portal]

so you can now split that in two fractions

$\frac{1-cosx}{x}\cdot \frac{2cos^2x-1}{x}$

[twin meteor]

LordFelix

[heavy yoke]

not sure that works

[vale frigate]

yea

[crisp rapids]

hey

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@vale frigate Has your question been resolved?

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[frank bison]

3 idk how to do this

[frank bison]

This is what I’ve done

[flat whale]

This is separable

[frank bison]

This?

That’s a minus 1 mb

[flat whale]

That's not how separation of variables works

[frank bison]

it’s been a minute could you give me a hint plese 🙏

[flat whale]

Review
https://tutorial.math.lamar.edu/classes/de/SeparationofVariables.aspx

[frank bison]

Oh wait I remember now

Silly me

nevermind

[silk osprey]

or just do integrating factor but yes it’s also separable

[frank bison]

Thought that’s only for linears?

[silk osprey]

you still don’t know how to separate it?

oh wait

[frank bison]

I’m blanking

[silk osprey]

yea separate

divide

[flat whale]

My bad this was the PDE link. This one here is for ODEs
https://tutorial.math.lamar.edu/Classes/DE/Separable.aspx

[frank bison]

?

Do I just integrate or

[silk osprey]

bring the u^2 + 1 over

then divide by u^2 + 1

or divide by it now and subtract the 1 over after

[frank bison]

?

And integrate?

[silk osprey]

yep

id get rid of the negative tbh

1/(u^2+1) is nice

[frank bison]

Now wa

[silk osprey]

yep

well substitute back

u was y’ yea

[frank bison]

yea

?

Oh I’m forgetting constant

Integrate again?

[silk osprey]

does that say + 4 or + C1

[frank bison]

C1

yay?

[silk osprey]

well it should be -ln right

[frank bison]

negative sub

it’s a -x so let -x+c be u

[silk osprey]

oh wait

[frank bison]

negative comes out and it’s positive

[silk osprey]

no you’re right

[frank bison]

Oh ok

[silk osprey]

i forgot about the -x

[frank bison]

Let’s see if the book thinks so

Book says we good

Epic

Thank you

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[vast shale]

Help question 5

[vast shale]

I have no idea what we are trying to look for

[calm pecan]

!status

[vocal sleet]

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

[vast shale]

I dont understand both the question and answer

This is the answer

I dont know how to start

Nor what to find

I dont know im super confused

[calm pecan]

this looks pretty good

[calm pecan]

do you know differentiation

[vast shale]

Yes

[calm pecan]

do you know how to find extrema

[vast shale]

Find critical points then plug in second derivative to decide whether minimum or maximum then plug the criticals in original equation to get their y??

[calm pecan]

Oookay

you have $xy=147$

[twin meteor]

Percy

[vast shale]

Yes

[calm pecan]

you grab $x$ and write it as $\frac{147}{7}$

[twin meteor]

Percy

[calm pecan]

you need to minimise some function of x and y, $f(x,y)$

[twin meteor]

Percy

[calm pecan]

you replaced x with smth in terms of y so now it's just a function in terms of y, $g(y)$

[vast shale]

Where did the 7 come from?

[twin meteor]

Percy

[vast shale]

I dont understand

[calm pecan]

put in values, solve

[calm pecan]

i meant y, they're close on the keyboard

[vast shale]

Ok so after making x alone then plug it in x+3y=sum

Then what?

Finding derivative?

[calm pecan]

yes

[vast shale]

Okayy

Then criticals

Then what

[calm pecan]

then...

[calm pecan]

.

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@vast shale Has your question been resolved?

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[tight apex]

Can you help me with this question?

[tight apex]

0.4496751137945747 I found this number

Webwork says your numbers are very close

check your calculation

[flat whale]

show all your calculations

[tight apex]

can you give me the answer pls

[flat whale]

!noans

[vocal sleet]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[tight apex]

Z= (551−547.8)/25.3

Z score= 0.5503

[tight apex]

from the table

1 - 0.5503 = 0.449

[flat whale]

,w 1 - 1/2 * erfc(-32/253 * sqrt(2) / 2)

[twin meteor]

Results provided by WolframAlpha

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

Requested by rie.mann

[flat whale]

what does it say to round to?

[tight apex]

it says very close

[tight apex]

i didnt send this answer yet this is my last try

[flat whale]

do the instructions tell you to round or not?

[tight apex]

there is no info

what should i enter?

[flat whale]

no idea

only webwork knows

[tight apex]

can you look 4.28

i corrected it

but i didnt understand what should i do on question a

can i try 0.45

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[tender lotus]

For 16, what do I do after plugging in the g(x) function

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[tender lotus]

<@&286206848099549185>

[twin citrus]

you have to simplify

[tender lotus]

Simplify what

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@tender lotus Has your question been resolved?

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[open sage]

Let B be an orthonormal basis for a subspace V

B={v1,v2,v3}

given S = span{v1,v1+v2-v3} find an orthogonal basis for S which contains the vector v1

okay so a few things

given its orthonormal, the internal product of v1 v2 and v3 will be 0

(between themselves)

the dimension of S orthogonal will be 1

i need to find

<v1,u> = 0 and <v1+v2-v3,u> = 0

i just don't know how to force u to contain v1

[pale widget]

if the u's are your orthonormal basis you set u1 = v1 and then apply gram schmidt to get u2

[open sage]

why would i need u2

i've determined s orthogonal to be of dim 1

[pale widget]

How did you manage that when S is the span of 2 vectors of that form

[open sage]

V is a 3 dimensional subspace

since its basis is composed of 3 vectors

dim(V) = dim(S) + dim(S orth)

[pale widget]

I think I misread what you wrote, are you not looking for a basis of S

[open sage]

an orthogonal basis of S

i think i may be misunderstanding as well

am i being asked to find S orthogonal

[pale widget]

That's not asking for a basis of the orthogonal complement of S

[open sage]

or an orthogonal basis of S

i see

sorry about that

[pale widget]

thats asking for a set of basis vectors that spans S which happen to be orthonormal

[open sage]

okay

got it

okay so

i need gram schmidt

we've seen it in class

i've just never used it

i just say u1=v1

correct?

which i believe is the case when using gram schmidt anyways

you set one of the vectors to be the same

(the first)

[pale widget]

Honestly I would double check that something wasn't lost in translation but I don't see how that question would make sense as you initially interpreted it

[open sage]

no no i'm pretty sure you got it right

its just

2 am

and i've been cramming a lot

lol

"find an orthogonal basis for S which contains the vector v1"

[pale widget]

using that screenshot you would set q1 = v1 then calculate q2 where a2 = v1+v2-v3

[open sage]

im sure that's the correct translation

god that looks so tedious

alright

thanks!

[pale widget]

it can be tedious yes

but the power of it is that it always works

[open sage]

yeah you don't get these kinds of things named after you unless they work

or unless they're really stupid in some cases

i suppose gram and schmidt are two different people though

@pale widget

sorry to bother you

does that look fine

oh

i think i added an extra term

yeah that's right

from the solution sheet

thanks!

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[sturdy portal]

I am able to do the first two parts, but I struggle to continue with the third part, and I keep getting it wrong

[sturdy portal]

nvm i realized what I was missing

lol

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[sly basin]

I have to use the divergence theorem to find the flux

[tired gorge]

Where are you stuck?

[sly basin]

idk how to start this

i know to find flux

usually we paramterizie it

[heavy yoke]

can you state the divergence theorem?

[tired gorge]

Yeah that's a good place to start

[sly basin]

is it the triple integral and getting the div F

[tired gorge]

Yep

[sly basin]

so <d/dx, d/dy, d/dz> dot F

[tired gorge]

Correct

[sly basin]

ok

so do i calculate that first?

[tired gorge]

What do you think? Maybe write out the divergence theorem first

[sly basin]

i did

isnt it just tripple integral of <d/dx, d/dy, d/dz> dot F DV

[tired gorge]

Yep

But you should think about what volume you're integrating over

That's the hardest bit of this problem

[sly basin]

okok

so are u talking about like

the bounds?

[tired gorge]

Yes

[sly basin]

x is going from 0 to 4

so we just need to find y and z

so dzdydx

[tired gorge]

If you want to integrate in that order sure

[sly basin]

thats y i was wondering

isnt it wiser to

set up the integral (without bounds)

to see whats the easiest way to integrate

then based on that we can be picky with our bounds

[tired gorge]

Sounds like a good plan

[sly basin]

2 + 5 + 2z
7+2z

so i think its better to have dz inside right

get it out of the way

this way the and x will no powers

[tired gorge]

I think integrating over x and y may be smarter first, as if you integrate over z first you'll get the upper bound as a function of x and y and it'll make things harder later

But you can go about it either way

It's all polynomials so the integrals aren't bad

Remember your bounds aren't simple because your domain is a tetrahedron not a box

[sly basin]

jright

is there a mechanical way to solve this

or do i have to visualize it?

i was told in most of calc 3 I can get away without understanding

but just plugging and knowing formulas

[tired gorge]

It never hurts to draw a picture

[sly basin]

okok

heres the geogebra sketch 0, 4x 5y 7z

[tired gorge]

Do you know how to get your bounds?

[sly basin]

no idea man

[tired gorge]

My preferred method is to use projections, not as bad as it sounds

What is your integration order?

[sly basin]

what is projection method

I was thinking

we use the points

then view it at from xz plane

and use 2 points and the formua y2-y1=m(x-x2)

like that

[tired gorge]

Which bounds are you trying to calculate?

Your first step should be to decide your integration order

[sly basin]

but

how do u determine the best

integration order

im thinking

dxdydz

[tired gorge]

Like I said it doesn't really matter here but those look good enough you just have to make a decision

[sly basin]

i would rather have 1 fixed method

i dont really mind the integration

as long as its doable
by aprts, usub, trig sub, trig identity

idont really mind

[tired gorge]

Usually having your first integrals being over variables the integrand doesn't depend on is good but other times it depends on your volume, it's basically impossible to give you an order that works every time

That decision partially comes from experience unfortunately

Anyways since you've decided on your integration order it's time to work out your bounds

[sly basin]

right

so since its dxdydz
I know z is going from 0 to 7

i need x and y

[tired gorge]

My suggestion is to work inside out: start with the first integral (in this case it's over x) and write the bounds it can vary over in terms of the other variables

[sly basin]

Right

it can have xyz

but it cant have

x as is its bounds right?

after integrating it it should only be y and z

right

[tired gorge]

Yeah you can't have any xs lying around after the integral

[sly basin]

so since bound cant have x it has to have y and z

so we gotta look at it from like zy plane?

or somethig?

[tired gorge]

Sounds about right

[sly basin]

(0,0,0), (4,0,0), (0,5,0) (0,0,7)

So if w looking at yz plane. We ignore x

So (0,0), (0,0), (5,0) (0,7)

[tired gorge]

I'm not sure what you're doing there

[sly basin]

like

the points?

im not sure

how do i graph it

from the yz plane

given that (0,0,0), (4,0,0), (0,5,0) (0,0,7)

[tired gorge]

I'd suggest looking at it this way, for any point in the y-z plane there is a line in the x direction, that line is what you're integrating over first and the start and end points of that line will give you your bounds

[sly basin]

is there a mechanical approach?

[tired gorge]

Sort of

You see that line pierces two surfaces, you should solve for x given the other variables on those two surfaces and those are your bounds

In this case those two surfaces are the planes x=0 and the slanted plane

[sly basin]

wym

[tired gorge]

This is a little difficult to explain with words, but if you imagine a line in the x-direction that pierces your tetrahedron at two points

The x-coordinates of the intersection points (which will depend on y and z) are your start and end bounds for x

[sly basin]

im confused about something

(0,0,0), (4,0,0), (0,5,0) (0,0,7)

So given this can we not determine the plane?
there is like a x/4 + y/5 + z/7 = 1 plane?

[tired gorge]

Yes

That's where I'd start

[sly basin]

so can we always do that with tetrahedra?

[tired gorge]

You can determine equations for the sides of the tetrahedra always

And you need to

It's really nice when the tetrahedron has its points on the axes but in general it can be more difficult than that

[sly basin]

ok so using(0,0,0), (4,0,0), (0,5,0) (0,0,7)
x/4 + y/5 + z/7 = 1 plane?
We can solve for x
x/4 = 1-y/5-z/7
x = 4( 1-y/5-z/7)
x = 4-4y/5-4z/7

So x is going from 0 to x = 4-4y/5-4z/7

like that?

[tired gorge]

Looks right to me

[sly basin]

ok we just need. to focus on y

[tired gorge]

Yea

[sly basin]

can we just use the same plane equation

and substitute in values of y?

no huh

[tired gorge]

Not quite

Yeah

You look at the extent that y can vary depending on z

In this case we can just find y in terms of z for the line in the y-z plane

Or lines (we have y=0)

In essence we're projecting our shape down into the y-z plane

[sly basin]

since its dxdydz
when doing middle integral aka dy
it cant havex so only interms of y and z
thats y we doing yz plane which ignores all x"?

[tired gorge]

We already integrated out the x bit so the bound for y should only be in terms of z and the integrand in terms of y and z potentially

[sly basin]

okok

so how we find y

[tired gorge]

We "smoosh" our tetrahedron along the x-axis, in this case it'll leave us with a triangle

Then we repeat the process that we did for x, now in 2 dimensions, to solve for y in terms of z

[sly basin]

OHH

okok

I can kind of see it now

it basically becomes

triangle

going from up to down

right triangle

[tired gorge]

Yes

[sly basin]

so if we smoosh x axis that means we looking at yz plane
Which is a triangle with 3 points.
Now we gotta use those 3 points to find an equation for y correct?

[tired gorge]

Yep

[sly basin]

point A(0,0,0) point D(0,0,7) point B ( 0,5,0)

like that?

3 points

[tired gorge]

I think so

Looks right

[sly basin]

if its yz plane

dont we just have