help-10

[timid silo]

but i dont think i am right

can i have some advice

[timid silo]

<@&286206848099549185>

from the left use -(1+x) and from the right (1+x)

well either way it will diverge (if i can see it correctly)

Wdym by diverge

Also wouldn’t it just be undefined limit ?

[timid silo]

I’m getting undefined?

[hardy widget]

Should be that it doesn't exist

,w limit of (x^2 - 2)/(|1+x|) as x approaches-1

[timid silo]

Yeah ?

[warm shale]

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[timid silo]

yes divergent means that it doesnt exist (or goes to infinity)

Wait I’m getting positive infinity?

.close

[obtuse pebble]

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[strange yoke]

I put the question on the left, I don't understand what Im doing wrong

I divided every term by x^2 because its the highest power

[cedar lichen]

6 and 6/x² are not the same thing

[strange yoke]

I divided every term by x^2

so 6 turned into 6/x^2

[cedar lichen]

No you didn't

For the second fraction, you divided the top and bottom by x²

That's the same as dividing by 1

Besides, there's no need to divide at all

What's the lim x -> ∞ of 6?

[strange yoke]

6?

i thought I had to divide. that's how i was taught l'hopital rule and i thought i had to do that here

[cedar lichen]

There's no need for l'h here

It doesn't even apply, in fact

[strange yoke]

why not

x is approaching infinity

[cedar lichen]

L'h only applies in indeterminate forms

0/0, ∞/∞, etc

You don't have that

Besides, you wouldn't just be dividing by something. You'd divide the top and bottom by something, which doesn't change the fraction

Now, what's lim x->∞ 2/x²?

[strange yoke]

thats weird in my notes we used l'hopitals rule but the problems dont look like theyre indeterminate

thatd be 0

so the answer would be 6 cause 6+0

[cedar lichen]

They should be if you're applying l'h

And yes, the final answer's 6

[strange yoke]

the problem that we started practicing lh with is x/(x^2+5)

as x is apporaching infinity

is that indeterminate

[cedar lichen]

That's ∞/∞

[strange yoke]

howd you get that

[cedar lichen]

The numerator goes to ∞ and the denominator goes to ∞

[strange yoke]

why do they become infinity

oh because direct subsition

[cedar lichen]

You technically can't directly substitute infinity, but it's the same idea

[strange yoke]

ok so when i approach limit problems with infinity, i should try to "substitute" the infinity for x, and only use lh if its indeterminate, right?

[cedar lichen]

Yes

L'h gives wrong answers otherwise, since it only applies to indeterminate forms

It also only applies to single fractions. So if you had two fractions added together, you'd have to either combine them, or apply l'h twice

[strange yoke]

so like the problem that i was trying to solve from the picture, that would be like two fractions?

[cedar lichen]

Yep

You can't apply l'h to either, since neither are indeterminate

Also, and this is just my opinion, but I personally think you shouldn't resort to l'h unless you can't do it any other way. It gives the same answer, but I don't like l'h being used as a sort of crutch

[strange yoke]

if there was an x in the numerator and denominator then thatd be indeterminate right

[cedar lichen]

Yeah, but then you could just cancel them. Depends on how the x's are placed, I suppose

[strange yoke]

what other ways are there? ive just tried lh cause thats what we did last time in class so i thought it applied

[cedar lichen]

Depends on the situation

For rational functions, there's factoring, for instance

If you're going to infinity, you can divide the numerator and denominator by the highest power

If there's square roots, there's some possibility for simplification there

There's lots of different things

[strange yoke]

isnt that only lh rule?

[cedar lichen]

That's not l'h

L'h is when you take the derivative of the numerator and denominator

[strange yoke]

i dont think i was taught the right thing

i dont even know what a derivative is

[cedar lichen]

If you don't know what a derivative is, then you definitely were not taught l'h

[strange yoke]

this is overwhelming

i dont know ill try to figure it out

thanks for the help though

[cedar lichen]

The method you believed was l'h works for all rational functions when x-> ± ∞ btw

Not just indeterminate

[strange yoke]

like the thing where i can cross out 6/x^2 and make it 0?

[cedar lichen]

That's more just a general rule than anything

It's more like if you have (2x² + x)/(x² + 1), you could divide the top and bottom by x²

[strange yoke]

does it have to be indeterminate?

[cedar lichen]

No

[strange yoke]

so indeterminate only matters for lh rule?

[cedar lichen]

Yes

Which you'll presumably learn later

[strange yoke]

I hope so

its just overwhelming for me cause we were taught it was lhopital and it really isnt so now i dont even know what i know

[cedar lichen]

Yeah idk what your teacher is thinking

[strange yoke]

ive tried watching some videos online about lh and i dont understand it but ill try to spend some more time watching and see if i can understand

[cedar lichen]

Lh requires knowledge of derivatives first

So you probably won't be able to understand it now

[strange yoke]

i'll look into it

thanks

.close

[obtuse pebble]

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[rapid kindle]

does anyone know how to approach this problem? i know it has something to do with optimization but im not sure how to work through it

[rapid kindle]

the answer is 300 euros and 220 passengers

[obtuse pebble]

@rapid kindle Has your question been resolved?

[obtuse pebble]

@rapid kindle Has your question been resolved?

[hardy widget]

Let's let the number of price decreases be x

Then, the price is 500-10x

And the tickets sold is 180+2x

So the revenue is (500-10x)(180+2x)

You can expand and complete the square, and you should get this

,w maximize (500-10x)(180+2x)

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[hardy widget]

🤔

Hold on

The answer is wrong lol

300 Euros and 220 passengers means 66000

But 500 eurosand 180 passengers means 90000

[rapid kindle]

international baccalaureate math moment

¯_(ツ)_/¯

i graphed the (500+10x)(180-2x) and got (20, 98000) as the maximum as well

[hardy widget]

🤷

[obtuse pebble]

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[opal wagon]

hello

[opal wagon]

so lets say I bought a lightbulb and it will stop working on the 12500th day

what day of the week is 12500th day on?

how can I figure that out?

[high lily]

check remainder when divided by 7

[opal wagon]

ok its 1785.71429

so we can say 1785?

then what

[high lily]

find the remainder

[opal wagon]

oh

k

5

is the remainder

so its friday?

so how does it work

how does finding the remainder give us the answer

[high lily]

consider the number of days in a week

[opal wagon]

oh

[high lily]

so its friday?
depends on what day you start on

[opal wagon]

its on the 7th day

1785

and then p;us

plus 5 days is friday

right?

[opal wagon]

monday

[obtuse pebble]

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[barren junco]

m=1

4=3(1)+b

1=b

y=x+1

y=-x

-x=x+1

-2x=1

x=-1/2

y=1/2

(1/2,-1/2)(3,4)

thats what i got

i got it wron

g

idk how to do fraction distances

could someone help

[fierce lagoon]

I would just find the perpendicular line that crosses through (3,4).

This line and y = -x will intersect at (a,b); you can use distance formula there on

Which is what you seem to have done but uh

[fierce lagoon]

You need to use distance formula

[obtuse pebble]

@barren junco Has your question been resolved?

[barren junco]

yea i did

but i got 26.5

are the points correct?

[fierce lagoon]

Points are correct

You did distance formula incorrectly though

It's definitely not 26.5

[barren junco]

i need a unit answer tho

so im kinda confused

its 3-(-1/2)+4-1.2

[fierce lagoon]

There are no units

[barren junco]

3.5 and 3.5

[fierce lagoon]

No

[barren junco]

its asking for a unit answer

[fierce lagoon]

Well distance formula

[fierce lagoon]

What is $\sqrt{3.5^2 + 3.5^2}$

[warm shale]

Umbraleviathan

[fierce lagoon]

But also like

[barren junco]

24.5 but i already tried that

[fierce lagoon]

4-(1/2) is 2.5

Not 3.5

[barren junco]

h o w

[fierce lagoon]

Wdym how is 4-(1/2) = 2.5 lol

[barren junco]

thats the equivalent to 4-(.5)

[fierce lagoon]

Oh I'm blind

Ignore that lol

[barren junco]

ive been stuck on this for like 2 days id ont know how to do it if theres fractions involved

[fierce lagoon]

I saw the 1.2 and thought of 1.2 but mistaken the 2 is 1/2

[fierce lagoon]

Also no

Recall that $\sqrt{a^2 + a^2} = a\sqrt{2}$

[barren junco]

what the fuck

[warm shale]

Umbraleviathan

[barren junco]

idk what that is

ok

uh

idk how to do that

[fierce lagoon]

Somewhere you miscalculated the square root

Use your calculator

[barren junco]

ion got one

[fierce lagoon]

Online

Use desmos

[barren junco]

how do i turn 3.5^2+3.5^2 into a square root

[fierce lagoon]

sqrt

[barren junco]

oh fuck

[fierce lagoon]

sqrt(3.5^2 + 3.5^2)

[barren junco]

no ownder i was getting this bs wrong

i never squared it

[fierce lagoon]

You never took the square root

[barren junco]

god damn im dumb

jesus christ

[fierce lagoon]

It happens

[barren junco]

i was thinkin it was there for deco lmfao

[fierce lagoon]

I mean I legit did just say 4-(1/2) = 2.5

So like

Fuck ups happen lol

[barren junco]

people make mistakes ig

alr thats my problem why i was stuck on it for so long

ty

[fierce lagoon]

Np

[barren junco]

.close

[obtuse pebble]

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[obtuse pebble]

@gritty onyx Has your question been resolved?

[tardy epoch]

Try #advanced-analysis

[obtuse pebble]

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[timid silo]

im working on another integral hw. how would i go about integrating this?

[ruby path]

Use u sub

[timid silo]

below are my steps

-cos(2t) evaluated at pi/2 - -cos(2t) evaluated at 0

[ruby path]

No

[timid silo]

?

[ruby path]

Use u sub

Take 2t = u and then try

[ruby path]

If you haven't

Learnt u sub

I'll give you a shortcut

[timid silo]

we havent

[ruby path]

Ah

Alright

So listen

Whenever you have a linear function inside of another function, you need to integrate the major function and divide by the derivative of the linear function. That will give you the integral for the whole thing

In this case, sin 2t is the major function and 2t is the linear function

[timid silo]

ohhhh

[ruby path]

So what would the integral be?

[timid silo]

-cos(2t)/2

[ruby path]

Yup

Now you can evaluate the limits

And remember, this ONLY works for linear functions

[timid silo]

-cos(pi)/2 + 1/2

[ruby path]

Uh

[timid silo]

yep i got it

[ruby path]

$(-\frac{\cos \pi}{2}) - (-\frac{\cos 0}{2})$

Sorry for the slow LaTeX I'm not at my PC

[timid silo]

all good

i got 2pi/2 for the first part

• cos pi/2

[ruby path]

Yep my bad

[timid silo]

then i did leave change change for the signs

[warm shale]

NEONPerseus

[timid silo]

-cos pi/2 + cos0/2

cospi is just -1

cos0 is just 1

[ruby path]

Yup

[timid silo]

1/2 + 1/2 = 1

[ruby path]

Good going

[timid silo]

thanks

you are the best

[ruby path]

no u

[timid silo]

i didnt use simplified form originally

the system still accepted it

LOL

but its always a good idea to simplify

[ruby path]

xd lucky

Yes it is

[obtuse pebble]

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[bleak eagle]

I would really appreciate some guidance on some questions about generalizing the convergence or divergence in the iteration of linear functions

[obtuse pebble]

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[bleak eagle]

.close

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[void narwhal]

why this don't have some function infront of the $rdrd\theta$

[warm shale]

phngs

[void narwhal]

cuz i saw some double integrals in terms of polar cords, they wrote in this way

[paper socket]

Not sure I understand the question

You don’t need anything in front of rdrdtheta

You’re calculating area as well

Opposed to volume

[void narwhal]

then how do you explain this

[paper socket]

You can very easily write that as a triple integral

[void narwhal]

isn't this double integral

[paper socket]

And then there would just be rdrdtheta again

[void narwhal]

sorry but i dont understand well yet

[rich sparrow]

You can write anything you want in the function you're trying to integrate

[obtuse pebble]

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[trim jay]

I have no idea how should i approach this question

can anyone guide me through this?

[bleak eagle]

Try multiplying by the conjugate. So multiply by (7+sqrt10)/(7+sqrt10)

[trim jay]

ok ill try that

wait im confused.

I dont think it should work like this. Would you explain further so that i can understand more?

im not good in math, maybe you could explain it in a way that a kid would understand

[bleak eagle]

So you know how (x-1)*(x+1) = (x^2) - 1 if you foil it out?

[trim jay]

yepp

[bleak eagle]

It’s the same idea. You had the right idea in the pic you sent, except you did minus instead of plus

[trim jay]

oh

im still confused about why and how should i do that though.

would you explain more to me?

[bleak eagle]

So the question asks you to rationalize the denominator, which means you want to make sure that no square root signs appear in the denominator

[trim jay]

are we trying to cancel out the 7-(sqrt10) ?

[bleak eagle]

It’s okay if there is square roots in the numerator, but not the denominator

[trim jay]

oh, so thats what rationalize means

[bleak eagle]

Yes exactly

[trim jay]

ill try again, hold on

here, is it correct?

[bleak eagle]

Yep that’s right

[trim jay]

oh i see now

but how do i know when i should use the "(x-1)*(x+1) = (x^2) - 1" method you taught me?

[bleak eagle]

It depends on how high of a level you get into mathematics, but at this level, anytime you see the words “rationalize the denominator”, that should be the first thing you try

[trim jay]

thank you so much, i understand now : )

Would you mind me asking a new question?

It says i need to solve it by completing the square, but the answer shows a quadratic formula used. What should i do?

[small thicket]

i mean if u can only ansewr in quadratic formula

use quadratic formula

[bleak eagle]

Yep agreed

[trim jay]

oh alright then, ill do that

Thanks

.close

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[small thicket]

do we know what GCD means

[violet prism]

.close

[obtuse pebble]

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[violet prism]

greatest common devisor

i understand now but thanks anyway

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[royal basin]

well, how will you prove Q* isn't cyclic?

if you can prove that then you will have your nonisomorphism

...wait hold on

you confused me on the notation a little

is Q* meant to be Q \ {0}?

then (Q*, +) isn't a group at all!

do you have the problem exactly as written

to ensure this is a fuckup on the teacher's part and not on yours

post it here

okay yeah

the correct answer is "(Q*, +) is not a group at all"

[obtuse pebble]

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[onyx folio]

for the part in the general term, I don't understand how 11 was distributed to (n-1) and the final when simplified became 11n-5

[worn swan]

bc d=11

you substitute that and get 6+11(n-1)

using distributive property

you multiply 11 by both n and -1

and get 6*+11n-11*

[onyx folio]

ah I see it now

thanks

.close

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[coral junco]

This is the proof of the binominial coefficient (the thing on the very top left), there is this part of the proof which I didn't understand, I copied it off my professor's lesson, can someone explain it to me?

[coral junco]

It's a proof by induction

[obtuse pebble]

@coral junco Has your question been resolved?

[obtuse pebble]

@coral junco Has your question been resolved?

[coral junco]

Oh well I'm gonna close it, only my professor knows

.close

[obtuse pebble]

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[timid crow]

I originally psoted to this to stack overflow

[timid crow]

https://math.stackexchange.com/questions/4520261/proving-the-existence-of-a-certain-term-in-the-k-1th-derivative-to-find-riem

But no one could help

any help would be greatly appreciated 🙂

it deals with even values of riemann zeta

and an infinite expansion of cot(x)

[obtuse pebble]

@timid crow Has your question been resolved?

[timid crow]

<@&286206848099549185>

[obtuse pebble]

@timid crow Has your question been resolved?

[obtuse pebble]

@timid crow Has your question been resolved?

[obtuse pebble]

@timid crow Has your question been resolved?

[timid crow]

@timid silo the problem is that I’m not sure what type of math this is

Since this is beyond my own level

Alright, I’ll try

@timid silo thanks!

.close

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[upbeat silo]

I keep getting the wrong answer.

And I don't know what i am doing wrong.

[worn swan]

find the circumference

when you do, finding radius shouldnt be hard

[upbeat silo]

I did all of that and i get 3/32 for the radius

[worn swan]

is radius 6?

[upbeat silo]

ye

[worn swan]

360deg : 45 deg = x : 3/2 pi

solve for x

x=2r pi

[upbeat silo]

OHHH

SHOOT

IM SO DUM

I DID

45/360 = 8

LMAO

im an idiot

Tyyy

[worn swan]

np

[timid silo]

Hi what are these problems called? Radicals ? or what

[upbeat silo]

.close

[obtuse pebble]

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[undone mountain]

Hello

[undone mountain]

This is a question on superposition of two waves

Ive managed to work out the amplitude which is 5mm

but i have no clue where to start with the frequencies

<@&286206848099549185>

[obtuse pebble]

@undone mountain Has your question been resolved?

[obtuse pebble]

@undone mountain Has your question been resolved?

[obtuse pebble]

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[boreal cipher]

!help

[boreal cipher]

I don't know how to solve this question. Could someone please solve it with working?

[warm canopy]

replace x with -2

[boreal cipher]

-2f = 5x³

is it basically

-2.5x³ x (-2)

[warm canopy]

f(x) is not f times x

its a function f, evaluated at a number x

you're looking for f(-2)

replace all x with -2

[boreal cipher]

Ok for one second I thought i was smart

now i just feel dumb

so

F = 5 (-2)³

so its -40

idk

honestly

[warm shale]

ΣAC

[warm canopy]

which yes is -40

[boreal cipher]

How did u do that

command

[warm canopy]

#info message

[boreal cipher]

$ 4x-9 $

i literally dont understand

$ $ sdwa $ $

[warm canopy]

#latex-testing

if you're done with this channel then please close it

[obtuse pebble]

@boreal cipher Has your question been resolved?

[warm canopy]

.close

[obtuse pebble]

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[weary delta]

Any ideas?

[weary delta]

I figure that b_n is of the same order as 1/n^3, but I can’t seem to do much more

$\lim_{n \to +\infty} a_n = 0$ is I guess another thing that can be gathered from here

[warm shale]

Learath2

[gilded needle]

and from that, you know that in particular, a_n is a bounded sequence

try to use that to bound |a_n b_n| by something that is absolutely summable

[weary delta]

Hm, why is this the case?

[gilded needle]

any convergent sequence is bounded

the tail is bounded by, say L-epsilon and L+epsilon where L is the limit

and there are only finitely many elements not in the tail

[weary delta]

Okay, so in that case |a_n b_n| <= |\eps b_n| <= \eps |b_n| and since b_n ~ 1/n^3 this absolutely converges?

[gilded needle]

yeah, assuming eps is a bound for |a_n|

you might want to make the b_n bound a bit more explicit

but the reasoning is right

[timid silo]

so its about determining whether the b_n series converges?

|b_n| series

[weary delta]

Well we know as $n \to +\infty$ $a_n \to 0$, thus as you said $|a_n| \leq \epsilon$ maybe even $\forall \epsilon > 0$ ?

[gilded needle]

yes, this argument works as long as the |b_n| series is absolutely convergent

[warm shale]

Learath2

[weary delta]

Actually doesn't the e-d definition of limit guarantee that there exists an N > 0 forall epsilon s.t. n > N implies our bound works?

[timid silo]

well, is it perhaps possible to get an upperbound for |b_n| as K/n²?

[weary delta]

I think the 2nd given is enough to say that b_n is asymptotically of the same order as 1/n^3, so if |a_n b_n| is <= \epsilon |b_n| then the the convergence of this \epsilon |b_n| series is enough to show that the smaller |a_n b_n| also has to converge

[gilded needle]

that should certainly be possible

[weary delta]

if b_n is asymptotically the same order as 1/n^3 then |b_n| converges by the asymptotic comparison test

then \epsilon |b_n| converges, then |a_n b_n| converges, then our initial series absolutely converges

[gilded needle]

do you mean the limit comparison test? if so, i agree but you should first find an explicit bound for |b_n|, right now all you have is |b_n| ~ k/n^3, but you need something of the form |b_n| <= ...

it's not hard to convert from the ~ to an appropriate <=

[weary delta]

Why do I need to bound it? LCT already says that if one converges/diverges so does the other

[gilded needle]

right, but have you formally proved that sum(|b_n|) converges? I mean you're morally right but you should prove it

[weary delta]

Well, hm. $b_n \sim 1/n^3 \implies |b_n| \sim |1/n^3|$, no? and since as $n \to +\infty |1/n^3| = 1/n^3$ giving us $|b_n| \sim |1/n^3| = 1/n^3$

This is about what I was thinking

[warm shale]

Learath2

[timid silo]

|b_n|*n² goes to 0 right? so we can make |b_n|*n² arbitrarily small, therefore lets make it less then 1, so there is an N so that |b_n|*n²<1 for n≥N and therefore |b_n|<1/n². And then chose some K, so that |b_n|<K/n² for all n<N

maybe K=N²*sup{|b_n|: n<N}

[weary delta]

Is there something wrong with my line of reasoning?

[timid silo]

so you say that generally if the |a_n| series is convergent and |a_n| ~ |b_n|, then the |b_n| series is convergent too?

[weary delta]

Generally if a_n ~ b_n, if series a_n is convergent, so is series b_n. Don't need the absolute value

That's just the limit comparison theorem as other countries seem to call it

I guess the only question there is if you can abs both sides of an equivalence relation like that

[timid silo]

ohh i see

[weary delta]

$f \sim g \implies \lim_{n \to +\infty} f/g = 1 \implies |\lim_{n \to +\infty} f/g| = |1| \implies \lim_{n \to +\infty} |f/g| = 1 \implies \lim_{n \to +\infty} |f|/|g| = 1 \implies |f| \sim |g|$

[warm shale]

Learath2

[weary delta]

I think it does

[obtuse pebble]

@weary delta Has your question been resolved?

[obtuse pebble]

@weary delta Has your question been resolved?

[wild swallow]

what's the problem here?

[weary delta]

I'm a bit unsure if what I'm doing is sane

I started with the fact that $\lim_{n \to +\infty} a_n = 0$. Thus through the e-d definition of the limit I conclude that $\forall \epsilon > 0 \exists N > 0 : n > N \implies |a_n| < \epsilon$ thus $|a_n| < \epsilon_0$

[warm shale]

Learath2

[weary delta]

Then for the series at hand I decide to check for absolute convergence of the series $|(-1)^n a_n b_n| = |a_n b_n| <= |\epsilon_0 b_n|$

[warm shale]

Learath2

[weary delta]

Also given is $b_n \sim \frac{1}{n^3} \implies |b_n| \sim \frac{1}{n^3}$

[warm shale]

Learath2

[weary delta]

Thus I conclude $\epsilon_0 |b_n| \sim \frac{1}{n^3}$ which converges, thus our original series absolutely converges

[warm shale]

Learath2

[wild swallow]

so you're done

[weary delta]

Is my reasoning correct?

[wild swallow]

it is

[wild swallow]

except this should technically be 5/n^3

[weary delta]

Do constant factors really matter in this asymptotic notation?

[wild swallow]

well if you don't want them to then you gotta write O(n^-3)

[weary delta]

Hm, so only if the limit is one do we say $f \sim g$?

[warm shale]

Learath2

[wild swallow]

yes

[weary delta]

Shouldn't this also be $\frac{5\epsilon_0}{n^3}$? @wild swallow

[warm shale]

Learath2

[wild swallow]

yes

[weary delta]

One last question, do you happen to know if the limit comparison test applies to functions big-O of eachother aswell?

[wild swallow]

thats the same as asymptotically equivalent but with a constant

constants don't spoil convergence

[weary delta]

So I could aswell have said b_n = O(n^{-3})

[wild swallow]

yes

[weary delta]

Thank you

.close

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[hallow grotto]

hello i was wondering whether someone would be able to help me with smoothing functions like this

[hallow grotto]

or direct me to a resource i can use to learn more about them

i think this is a gaussian smoothing function but i cant find much helpful info on it

W is the smoothing function and h is the smoothing length

[obtuse pebble]

@hallow grotto Has your question been resolved?

[obtuse pebble]

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[spark plume]

hey guys, im not sure if im in the right place. and sort of a bit nervous to ask the questions as i have been out of school a long time and i have returned to school now but ive forgotten a lot of what i knew, so im trying to catch up using khan academy, and ive come across something im not really sure about, and thats finding the least common denominator. So they explain it pretty well, but they are mentioning i should use "multiples of" said number. Now wouldnt that mean for example 6 x 6 x 6 instead of 6x2 6x3 6x4 increments? sorry for the simplistic nature of my level of math and no this isnt a troll im genuinely interested in the answer here

[warm canopy]

Youre thinking of powers

6³=6x6x6

But 3*6= 6+6+6

[spark plume]

ahhhh ok, so when searching to match the common denominators do i just simply scale using the smallest one and go up till the least shared number is found?

[warm canopy]

Well you may have to scale both if one isn't a multiple of the other

You write out a list of multiples of both until you find the first number in both your lists

That's your lowest common denominator

[spark plume]

ahhh that makes total sense. awesome thank you. i was sort of looking at the video with the head tilt trying to figure out the meaning of what they were saying. Thank you for the solution 🙂 much appreciated

[hallow grotto]

no

pls help

.reopen

[obtuse pebble]

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[normal gate]

.close

[obtuse pebble]

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[tidal hawk]

yes?

oh nvm HAHA the timing

[normal gate]

sry lol

[tidal hawk]

issokii

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[novel bane]

Percentage formula.
I have this variable, whos value increases by 1 each second.
Id like the percentage to be 0.8 when the variable is at 1 and 0 when the variable reaches 3600 ( 1 hour).
Now is there a way to write this without using floating points?
Like instead of (x1.85)/100
~ (x185)/10000.
Im rly bad at math, so pls go easy on me with the explanations.

[shut marlin]

We usually use orders of magnitude to represent small decimal numbers:
10000% is 100, that is 10^2
1000% is 10, that is 10^1
100% is 1, that is 10^0
10% is 0.1, that is 10^(-1).

Things like 0.2%, which is equal to 0.002, would be represented as 2*(10^(-3)).

[obtuse pebble]

@novel bane Has your question been resolved?

[novel bane]

Hmm how would u write a formula for my problem tho?
If i want a 80% when the value is at 1 and 0% when at 3600?

We usually use orders of magnitude to represent small decimal numbers:
10000% is 100, that is 10^2
1000% is 10, that is 10^1
100% is 1, that is 10^0
10% is 0.1, that is 10^(-1).

Things like 0.2%, which is equal to 0.002, would be represented as 2*(10^(-3)).
@shut marlin
I will look this up ty

[shut marlin]

How much increment do you want each time the percentage increases?

*Decreases

Do you go from 80% to 79%, or from 80% to 79.5%, or maybe 80% to 79.9%?

[novel bane]

Hmm im trying to write this in code for a program that doesent accept floats

Whats the cheapest way?

[shut marlin]

Depending on how percise you want your code to be

The cheapest way is 80% at 1, and then directly drop to 0% at 1800

[novel bane]

80-79?

[shut marlin]

In that way you map half of the percentage to half of your values

[novel bane]

God i didnt even know abt this decrease 🤣

Yeah but then i could just use an if statement

[shut marlin]

Consider this case, and then generalize it to three, four, five

[novel bane]

I was wondering if a formula could do the work for me instead of just writing a ton of ifs

[shut marlin]

So in case of 3, you map 80% to a third of your values, some number in the middle to another third, and 0% to the last third

Eventually you'll get how to map 80%, 79%, 78%… on to 1, 2, 3…

[novel bane]

Idk much about compilers, im kinda self taught. So what ur saying is that using a formula would cost more energy than just 3 ifs?

Omg i dont understand

[shut marlin]

Kinda, I don't actually know what you're talking about when you say formula, I'm a CS student and I'll use a loop

[novel bane]

Have u heared about unix timestamp?

[shut marlin]

Yes

[novel bane]

I got 2 variables of them

To work with

So a prvious one and the current

So the current - the previous

Gives me a certain time

From that time i wanted to calculate a %

Based on how much had passed

[brave matrix]

Catching up on the conversation. You're saying you want the percentage to decrement by (1/3600) every second?

[novel bane]

Or well every time i call the function to calculate the percentage. The unix time will pass anyways. @brave matrix

[brave matrix]

So 3/3600ths if 3 seconds have passed, etc, right?

[novel bane]

Yes

[brave matrix]

How precise are you wanting it?

[novel bane]

I legit dont know, i didnt even know i had to factor that in. I thought there was a magic formula that calculated it perfectly. (80-79.999) but thats linear i guess and as i just learned that would cost more computing power? Which i kinda need to save a bit

80-70-60? How would that go?

[brave matrix]

Yeah, I mean you could just multiply by 10 (or 1000 or 1000000), then divide by that and round to int if you wanted

Just throwing out an easy solution, maybe not a perfect one :p

[novel bane]

How many digits can a int hold?

[brave matrix]

I'm a software developer and not a mathematician lol

32 bits

so about 2.something billion

[novel bane]

Oh ok thats enough

[brave matrix]

the value is 2.something billion, not digits

if you use uint, you'd have double that, up to 4ish billion, but you can't have negatives with uint, if that at all interests you :p

[novel bane]

Thx for the tip

[brave matrix]

Sure thing!

[novel bane]

Hmm u by any chance familiar with solidity?

[brave matrix]

Just had to look it up, that's the first I've heard of it. Looks very similar to Typescript, though

[novel bane]

Yeah rather typescript than js, its compiler is super strict

[brave matrix]

function calcPercentage(int elapsedTime){
    return round(8000 + (elapsedTime * (100/36))/10000);
}

something along those lines was what I was thinking

[novel bane]

If i already have 2 software eng's in here could u take a look at this and mb explain me a bit of whats happening here.

the << syntax is supposedly moving bits to the left?

[brave matrix]

any section in particular?

yep, that's a bit shift operator

[novel bane]

both encode functions

whats their purpose and whats like the purpose behind shifting bits?

i dont expect a whole course but if u could give me some points to go on about id appreciate it a lot. Im kinda learning by myself anyways but id like to know what to search or what this topic is exactly

[brave matrix]

I'm no expert on it, since I don't really ever use it in my work, but bit shifting can be very processor efficient if you have something that can be solved with 2^n, such as calculating Fibonacci sequences, etc.

I'm sort of taking a guess here, seems pretty specific to solidity, but if I had to take a guess without researching much, it looks like the first encode function is taking a uint112 (112 bit number? seems too big) and forcing it into a uint224 (twice as big I'd assume) then shifting it 112 times

[novel bane]

& all that for efficency purposes is ur guess

god

[brave matrix]

Yep, bit shifting would be more efficient then doing the squaring to get to that number

[novel bane]

damn thx a lot! I kinda understand the purpose of that code finally.

[brave matrix]

🥳

[obtuse pebble]

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[ruby skiff]

how did n-11 get divided out?

[royal basin]

they applied the rational root theorem and found that 11 was a root therefore they did the long division behind the scenes

[obtuse pebble]

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[lone blade]

find the orthogonal projection of the given vector on the given subspace W of the inner product space V .

[obtuse pebble]

@lone blade Has your question been resolved?

[charred plume]

I would start by finding an orthonormal basis for P_1(R). You know how to do that?

[obtuse pebble]

@lone blade Has your question been resolved?

[lone blade]

I got (1, sqrt 12 (x - 1/2)

[charred plume]

I think sqrt 12 should be 2/sqrt(3), but otherwise looks good.

(but I didn't write it down carefully so I could be wrong on the constant too...)

Hah, yeah I didn't do it carefully. Sqrt(12) it is.

Then take inner product of h(x) with each of those. Call the numbers you get A and B... Then the orthogonal projection is A + B sqrt(12)(x-1/2).

[lone blade]

isnt the projection the inner product (h,v) times v

[lone blade]

.close

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[dreamy current]

would i use a double angle formula or compound angle formula to simplify this expression?

[next sluice]

$\tan\left(\dfrac{\pi}{12}\right) = \tan\left(\dfrac{\pi}{4} - \dfrac{\pi}{6}\right)$

[dreamy current]

i just realised

[warm shale]

Alex88

[dreamy current]

thanks

im bad at noticing frations like these

i was tryna figure out what angles would work lol

.close

[obtuse pebble]

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[next sluice]

the only angles you can really try are 30, 45, 60, 90

[dreamy current]

yeah but i was working in radians

i know how to do it in degrees but i was tryna force myself to use radians

so id be more accommodated with them

[next sluice]

ahhh fair enough

[obtuse pebble]

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[timid silo]

$\ker S = {0} \implies \ker(T \circ S) = {0}$

[warm shale]

metnal

[timid silo]

let $S,T:V \to V$ be linear operators

[warm shale]

metnal

[timid silo]

how do i provide a counter example to this?

im not sure how to find the proper vector spaces and the proper linear transformations

[kind hawk]

what happens when you try to prove it. where does it go wrong?

try to understand that. then it's easy to construct a counterexample

[timid silo]

\begin{gather*}
S(\vec{v}) = \vec0 \implies \vec{v} = \vec0\
T(S(\vec{u})) = \vec0\
S(\vec{u}) = \vec0 \implies T(S(\vec{u})) = \vec0\
S(\vec{u}) \neq \vec0 \not\implies T(S(\vec{u})) \neq \vec0
\end{gather*}

this is what i arrived at

but im not even sure how to form a linear transformation

[kind hawk]

that last step is wrong

but why

[timid silo]

uh

[kind hawk]

you are using a property of T there

which it doesn't always have

[timid silo]

oh wait

its supposed to be \centernot

like

"not implies"

the bot didnt render it correctly

lemme fix it

[kind hawk]

ah ok

then that is correct. but why is it "not implies"

[warm shale]

metnal

[timid silo]

cuz u could be another vector and S(u) would be a nonzero vector and T could still output a zero vector for that

the kernel of T isnt limited to the 0 vector

[kind hawk]

yes

it's about the kernel of T

what's the easiest linear transformation with a non-trivial kernel

[timid silo]

i have no idea lol

[kind hawk]

what's the easiest linear transformation in general

[timid silo]

uhm

i dunno maybe something that just scales the vectors by a constant?

[kind hawk]

which is the easiest constant

[timid silo]

uh that would be 0 or 1 but im not sure these are useful here?

so maybe 2? idk

[kind hawk]

0

the easiest map is the zero map

[timid silo]

yes but doesnt it just shrink everything to 0

[kind hawk]

yes

that's the point

what happens if T=0 is the zero map

[timid silo]

ohhhhhhhhhh

so T(whatever) is gonna be 0

[kind hawk]

yes

[timid silo]

so the kernel of T is infinite?

i mean it contains infinite amount of vectors

[kind hawk]

well depends on which vector space

some vector spaces are finite

the point is that it's not trivial

[timid silo]

but it contains all the vectors in the vector space im guessing?

[kind hawk]

yes

[timid silo]

makes perfect sense

[kind hawk]

so for every x we have that T(S(x))=0

[timid silo]

yup

tysm

haveaa good day

.close

[obtuse pebble]

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[obtuse pebble]

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[unkempt jewel]

ABCD is a quadrilateral, EF makes a 90 degree angle with KF

it also says E, F, K, L is midpoints

the video solving this said to draw a line from E to K and that line would be equal to x

i dont understand how it is equal to x

[obtuse pebble]

@unkempt jewel Has your question been resolved?

[obtuse pebble]

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[signal trail]

Find the equation of a circle having a tangent line of y=3-5x which passes through the point (2,-7) and the center of a circle is contained on the line x=2y+19

[signal trail]

Good day, is there anyone who can help me with this problem? Anyone available

<@&286206848099549185>

[rain mountain]

First of all u have a tangent line slope so u have the inverse of it as well

Am i right?

so that vertical line to the tangent line we found has an equation for itself
its gonna be y = -x/5 + c , cause the slop of tangent line is 5
the vertical slope of the radius line will be -1/5

after finding c by putting the given point (2,-7) in the new equation.

the equation of radius line will have an intersection with the line of center of circle

so that we find the x,y of the circle

[gloomy valve]

Another way (without visual geometry) would be to write down the general equation for a circle centered at (x0, y0) and radius r:
(x - x0)^2 + (y - y0)^2 = r^2

Let y = y(x) and differentiate both sides with respect to x, that gives you tangent slope at every point of the circle. Plug in the given point (2, -7) with the condition that dy/dx = -5 and eliminate x0 by using x0 = 2y0 + 19. Then you can solve for (x0, y0), plug that back into the circle equation and solve for the radius r.

If you don't know calculus, go with XpertTiger's method

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

Thos was the working out done by my teacher. I'm not sure why they did divided it by 11.

What I would've done is 10% of 348.85 But I get the wrong answer

[royal basin]

the tax is 10% of the pre-tax total, not 10% of the post-tax total

[timid silo]

oh, so whwre did they get 11 from

[royal basin]

110% of pre-tax total = 348.85

[desert sinew]

The keyword is “including”

So let $$x$ be total amount.
We actually have $x + 0.1x = 348.85$, so $x = \dfrac{348.85}{1.1}$

Since we want the charged amount, we need to further multiply this quantity by 0.1 to get the final answer 🙂

[warm shale]

jimmy1234

[royal basin]

we actually just want the tax

[obtuse pebble]

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[agile robin]

Can someone explain how he went from indefinite integral cotx dx to ln sin x

[agile robin]

Got cut off buts it’s +c at the end

[fierce lagoon]

cot(x)dx = cos(x)dx/sin(x)

Set u = sin(x) so that du = cos(x)dx

Then you're integrating 1/u which is just ln|u| => ln|sin(x)|

[agile robin]

Ah I see how would I know I’d have to do the setting up u again?

[fierce lagoon]

You could

I mean it's one of the antiderivative identities

[obtuse pebble]

@agile robin Has your question been resolved?

[agile robin]

Ok

[obtuse pebble]

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[dull crypt]

how would I go about solving this?

i've worked the range out to be -4k ≤ y

oh also

$f(x) = k(x^2 + 4x)$

[warm shale]

manu

[obtuse pebble]

@dull crypt Has your question been resolved?

[dull crypt]

<@&286206848099549185>

(sorry)

[obtuse pebble]

@dull crypt Has your question been resolved?

[dull crypt]

<@&286206848099549185> pls ;(

[elfin cipher]

if u want to find k then plug any number in the function

like

$f(1) = k(1^2 + 4*1)$

[warm shale]

True

[dull crypt]

ahh thank you

.close

[obtuse pebble]

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[frail lake]

hey

[frail lake]

<@&286206848099549185>

[worldly fog]

!15m

[obtuse pebble]

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.

[frail lake]

ok but you avalaiable?

[worldly fog]

Well you didn't even ask a question

[frail lake]

well I wanted to start a conversation

but ok

here it it

I just dont know how to start

special right triangle

thats my teachers answers

[versed cave]

do you understand why (x)° is 45°?

[nocturne minnow]

If you read #❓how-to-get-help, you would know the way to start a channel is to post your question and wait, not start a conversation

[worldly fog]

Do you know the 90/45/45 triangle

[frail lake]

dam i just wanted to be friendly

yes how is it 45?

[worldly fog]

Because a square will have all 90 degree angles, correct?

And 90/2 is 45

So x is 45 degrees

[high lily]

pinging helpers before 15minutes and before asking a proper math question implying that you didn't read the rules or the auto generated message directly above may be considered hostile (quite the opposite of friendly)

[frail lake]

but it should all equal 180 yes? so it could be 30 and 60?

oh wait

no

dont say anything i get it

lol

[nocturne minnow]

Or because it's an isosceles triangle, you already know that one of the angles is 90, so you have two other angles that are equal to each other

[frail lake]

is me saying hey being unfriendly?

[high lily]

saying hey is friendly,
pinging helpers like that is not

[frail lake]

like dam guys i didnt mean to make yall upset

[worldly fog]

It's fine to start a conversation, #prealg-and-algebra or #geometry-and-trigonometry would be better suited for conversations.

[frail lake]

sorry I just needed help and I got it thanks

ok other question

if you jsut have that one leg

how would you tell if its supposed to be a 30 60?

[nocturne minnow]

Starting a conversation is better suited in #discussion or #serious-discussion
Getting help typically means posting your question first, what you tried and following the rules, like what that auto generated message states

[high lily]

it really depends on what info you're given and have to work with

if you're just given one leg of a triangle and literally nothing else, there isn't much to say about that triangle

[frail lake]

lets say this one

[worldly fog]

But because we know it's a square and we know the sides are congruent there is multiple ways to prove it's 45 degrees for both

[frail lake]

wait back up

sorry guys

another question on other problem

[high lily]

same principle as the previous question

[worldly fog]

Well, y will be 45 degrees which makes the other 45 degrees as well

[nocturne minnow]

IE this statement is useful

[frail lake]

so is there another way to show that x is 45?

[worldly fog]

So x is 17sqrt(2)

[high lily]

So x is 17sqrt(2)
no

[worldly fog]

Sorry, other way around, 17/sqrt(2)

[frail lake]

instaed of saying 90/2

is there another way of showing x is 45

[worldly fog]

Yes, as dldh06 said, isosceles triangle

[frail lake]

oh

yea

equals 180

[high lily]

90/2 is applied in one way or another

[frail lake]

but 30 and 60 equals 180 too

[nocturne minnow]

It's because it's an isosceles triangle

[worldly fog]

Well we know that the other two angles have to be the same

30 does not equal 60

[nocturne minnow]

Isosceles means that two sides are the same length

[nocturne minnow]

This is true but with the given image, the triangle has two equal sides hence an isosceles triangle

[frail lake]

ok im all good now thanks and sorry for upseting you guys

.close

[obtuse pebble]

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[obtuse pebble]

@royal rover Has your question been resolved?

[royal rover]

<@&286206848099549185> I do not know where to get started, I know the ending point and im pretty sure I am supposed to use SVD

[royal rover]

would differentiating this equation give x + lambda^T A?

<@&286206848099549185>

would this be correct?

.close

[obtuse pebble]

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[jaunty dew]

How can I find the coordinates for the point of contact, I have a line: y = mx and a curve: y= x^2+6x-4, the value of m is 2 or 10; how would I find the coordinates of the point of contact? Can someone please explain this to me, thanks.

[celest kettle]

point of contact

will be the solution of a simultaneous equation

ok here's an example

[jaunty dew]

ooohh but which equations should I take

[jaunty dew]

alright

[solar trellis]

the only two that you are given

[jaunty dew]

OOOOOOOOO

[celest kettle]

consider the point they intersect

[jaunty dew]

TY

[jaunty dew]

Tysm

[celest kettle]

cool

[jaunty dew]

.close

[obtuse pebble]

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[celest kettle]

glad I helped

[obtuse pebble]

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[sonic herald]

@celest kettle

[sonic herald]

i got a question

[celest kettle]

ok

[sonic herald]

i don't really understand what they're asking here

Find the measures of a complementary angle, a supplementary angle, and a vertical angle for an angle with the given measure: 47 degrees

@celest kettle

[celest kettle]

complementary is 43

supp is 133

[sonic herald]

wait so

their asking the measures of the 2nd angle

[celest kettle]

vertical...

[sonic herald]

i couldn't really understand what they were trying to ask

bc they just said find the measures of a complementary angle

like did they mean the angle that's not 47 degrees

[celest kettle]

if a is the complementary angle of b, a+b=90°

[sonic herald]

yea ik

but i didn't understand what they were trying to ask

ty

[celest kettle]

me neither natural language is hard

[sonic herald]

.close

[obtuse pebble]

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[celest kettle]

but I'm trying my best ig

[sonic herald]

wdym natural language

[celest kettle]

aka english

[sonic herald]

o

[obtuse pebble]

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[grave flicker]

So i know that if I plugged in 0 for x that the limits on some of these functions would not exist

but if i multiplied

0 by 0

it would return 0

[timid silo]

ok lets start with the first one, does the limit of floor(6x) exist?

[grave flicker]

It does exist

wait is it floor or abs value

?

[timid silo]

its floor

[grave flicker]

idek what floor is

wait what

We were never taught that

Could you try to explain what floor is?

[timid silo]

floor basically just rounds down your number,
floor(x) returns the maximum integer z with z≤x
floor(6.3) = 6
floor(4.9) = 4
floor(5) = 5
floor(-1) = -1
floor(-2.5) = -3

[grave flicker]

oh ok so here floor 6 = 6 got it

just like in programming

so back to your original question

[timid silo]

but now you got floor(6x)

what is the right side limit?

[grave flicker]

i would say floor 6x does not havea limit

DNE

[timid silo]

hmm, why would it be infinite?

[grave flicker]

wait

because

6 * x would

always produce an output

[timid silo]

we go x -> 0+

[grave flicker]

so as x approaches 0 from the right side?

[timid silo]

yeah

like floor(0.1), floor(0.01), floor(0.001)

[grave flicker]

so the limit

for that fn dosent exist right?

and same for g(x) on the first one

?

[timid silo]

the right side limit does exist

[grave flicker]

O_O

[timid silo]

what is floor(0.1)?