help-33

[past relic]

so now

wajt wha doesn't i2 go to the right

wait

ohhh

same direction

[trim stratus]

Yes, and the loop goes to the right too

[past relic]

cuzi f you draw counterclockwise

reverse

it goes same direction

so its -

wait dis might explain why my sub method didn't work

cuz the signs were off

wait chatgpt did say my signs were off

[trim stratus]

So what are your final two equations in terms of V1, V2, R1, R2, R3, I1, I2, and I3?

[trim stratus]

Lol

[past relic]

PERF

NOW I HAVE SIGNS CORRECG

AND CAN DO SUB METHOD

[trim stratus]

Yep!

[past relic]

thanks William! i hope you succeed in getting your PhD

[trim stratus]

I'm in high school 😭

[past relic]

oh damn

[trim stratus]

Lol

[past relic]

and i didn't even know how to do dis

[trim stratus]

Now you know

[past relic]

r u going to MIT

or Caltech

MIT campus is rlly nice

its so nice

[trim stratus]

I'm not American

[past relic]

museum is overrated

[past relic]

Cambridge?

[trim stratus]

Nope

[past relic]

hmmmmmmmm

your german

cuz germans invented physics

[trim stratus]

I'm not lol

[past relic]

ur norwegian

ok idk

all ik is you help w physics problem

i give you 5 stars

[trim stratus]

😭

[past relic]

@viscid flame #rate @trim stratus 5 stars

thanks have a gn!

[trim stratus]

Gn

[marsh citrus]

@past relic Has your question been resolved?

[marsh citrus]

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[flint onyx]

how do i factor polynomials with multiple variables

[flint onyx]

so currently my class is learning factoring stuff like
4u^2 + 4u +1

and i understand it (Although im kinda slow at it)

but she said there will be a extra credit problem on the quiz like
w^2 -x^2 -2x -1

and i wanna learn how

[queen spoke]

I can help

[flint onyx]

yay

so how do i even start something like that

[queen spoke]

I’ll send u a picture

Ok so start out by writing the eq

Then draw an x box

[flint onyx]

eq?

[queen spoke]

Like this

[flint onyx]

i know how to do the problem in the picture

[flint onyx]

its this thats hard

[flint onyx]

i learned that little tech heh 🥹

[queen spoke]

Draw that out then multiply the high pest variables coefficient with the lowest variable coefficient

And put that at top

Then put the number in the middle or b in the bottom

So put another 4 in the bottom

Now you want to find two numbers that multiply to the top which is four and add to the bottom which is also four

Can you give me two numbers?

[flint onyx]

i know how to do these 😭

[queen spoke]

Ok then what was the issue?

[flint onyx]

how to factor polynomials with multiple variables

w^2 -x^2 -2x -1

[queen spoke]

Ok give me example

So basically solve it like this

At the start don’t worry about the other variable

First factor

[flint onyx]

so factor one variable first

[queen spoke]

Yes factor the most common variable first

Then tell me what you get

[flint onyx]

okie

is it -(x^2)^2

wait

-(x+1)^2

[queen spoke]

Yes

[flint onyx]

okie cool

[queen spoke]

Don’t forget the w squared

[flint onyx]

yes

then what

do i distribute W?

[queen spoke]

Ok then

You need to recall a certain formula

Known as the difference of squares

[flint onyx]

isnt it just the conjugate

wait whatt

how is (x+1) a perfect square

[queen spoke]

It’s being squared which means it’s being multiplied by itself again

[flint onyx]

ohh okay

[queen spoke]

You use a squared minus b squared since a squared represents your w squared and b squared represents the x plus one squared

[flint onyx]

okay i think i understand

should i distribute the -1

[queen spoke]

Yes

[flint onyx]

so
(w-x-1)(w+x+1) is the answer

[queen spoke]

Correct

[flint onyx]

gosh thats weird

will it always be like that you think

what if its like

w^2 - 2w +2x +1

is that possible

if possible could you just send a question for me to factor that is similar because i dont know if it will be exactly the same

[queen spoke]

In that case you would do the same thing

[flint onyx]

but +2x is in middle

[queen spoke]

It’s would be a squared minus 2w plus 1

[flint onyx]

and how do i separate that

[queen spoke]

All in parenthesis

And plus 2x on the outside

[flint onyx]

could you type/ draw it

[queen spoke]

Sure thing

(W squared - 2w + 1) + 2x

Factor the inside of the parenthesis

[flint onyx]

ohh so you just move it out

okay that seems easyish

[queen spoke]

Yes when you two different variables, focus on the one that has majority and rearrange the equation if you must towards your favor

[flint onyx]

why does +2x go after and not before it

[queen spoke]

It can go before it or after it, it means the same thing

Whichever is easier for u

[flint onyx]

this would just factor to (w-1)^2 +2x since were not finding the difference

[queen spoke]

Correct!!!

[flint onyx]

okay im getting better at this, ima ask chat gpt to make questions like this so i can practice

do you have any tips for being able to get better at factoring with like mental math?

[queen spoke]

Asks ChatGPT difference of squares and addition of squares formulas and memorize them. Then you should be good for factoring with those

[flint onyx]

difference of squares are easy to do tho

[queen spoke]

So memorize the other one

[flint onyx]

what about if its like 3p^2 -10p -8

how would you just know how to mental math that

is it just practice?

[queen spoke]

Multiply a and c which makes negative 24 and put that on the top of an x in your head. Put b or negative 10 on the bottom and find two numbers which multiply to negative 24 and add to negative 10. However, I don’t recommend doing this in your head on an assessment if you’re just learning it as it’s not worth the consequences of missing points on the question.

Did u have any other questions?

[flint onyx]

no thats all, thank you for all the help also

have a good night

.close

[marsh citrus]

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[marsh citrus]

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[high needle]

would this be 0/0
and if that is correct is 0/0 nonexistent or 0

[fervent rampart]

0/0 is "indeterminate", which means that it could possibly be any limit and you don't know what it is yet

so you need to do some algebra to make it not 0/0 anymore

[timid rivet]

this is a disguised derivative

[stoic saddle]

not even that strongly disguised

[vernal mantle]

But you can solve this with addition formula too if you're like that frfr

[timid rivet]

yeah heh just doesn’t look like d/dx

[vernal mantle]

@high needle

[high needle]

how do i solve it

[high needle]

For the limit

How do we solve the derivative

Cuz i just plug 0 in for h

and i got 0/0

[vernal mantle]

Have you guys learnt the derivatives of trig functions in class

[high needle]

like sinx = cox

cosx

cosx= - sinx

that kind of stuff

?

[vernal mantle]

Well yea

And did you guys learn the first principle definition

[high needle]

Yeah

like slope of tangent line

equation

[vernal mantle]

So does this look familiar

[high needle]

BUt idk how this is a limits question can be derivative

[vernal mantle]

To the first principle definition

[high needle]

No

[vernal mantle]

So this is the first principle definition

If uh f(x)=sin(x)

And x=π/3

[vernal mantle]

What could you simplify this to

[high needle]

f(x)-f(x)/0

so 0/0

[vernal mantle]

Ok sorry I kinda need to go

You can ping another helper in a few minutes

[marsh citrus]

@high needle Has your question been resolved?

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[compact crescent]

How do I know (3)^(1/3) is bigger than 1

[leaden monolith]

Anything greater than 1 to any root is greater than 1

[harsh scroll]

$\sqrt[3]{3}>\sqrt[3]{1}$

[elfin berry]

Alexis_Fx

[elfin stone]

[ionic ferry]

Any number below 1 couldn't possibly get greater than one no matter how many times you multiply it to itself

[elfin stone]

You can cube both sides to verify

[leaden monolith]

That’s a better argument than what I had in mind

The cube function is strictly monotonic so it preserves < ordering

[ionic ferry]

Ahhh yeah or that

Also like logically, if the cube root of 3 was lesser than 1, then it would never reach 3 no matter how many times you multiply it to itself, it'd just keep on getting smaller

@compact crescent you still here mate?

[elfin stone]

There's a very high chance that he's not coming back lol

[marsh citrus]

@compact crescent Has your question been resolved?

[marsh citrus]

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[radiant plaza]

Could someone help me?

[sick walrus]

please send your question here.

[odd agate]

sure, question please.

[radiant plaza]

So, how do we solve this?

[stoic saddle]

do you know exponent laws / index laws

[radiant plaza]

Yeah a bit

I'm beginner

[stoic saddle]

ok can you name the laws that you know

just the laws, we'll come back to the question after this

[radiant plaza]

Yeah,

(a^m)^n = a^mn
a^-n =1/a^n
a^m . a^n = a^m+n

Actually I don't the name of that law

[marsh citrus]

@radiant plaza Has your question been resolved?

[stoic saddle]

ok good. sorry i was away.

you should also know a^0 = 1

simplify the stuff inside the brackets first. then show what you get.

[marsh citrus]

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[knotty trellis]

Given 2 cuboids of equal dimensions (say a <= b <= c), whats the minimal side length x of cube, in which they fit?

[knotty trellis]

I found out that if 2a <= c, x = c

And also that x >= c

And i also know that if 2a > c, then x <= 2a, conjecturing its exactly 2a

Imma return to this later

.close

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[rugged venture]

Hi, I want to understand how Ray tracing works with math , if anyone knows something about this feel free to tell me!

[elfin cairn]

If you have a light source S, we consider rays coming out of S, they relfect/bounce around and if they hit your camera then you can see them.
The color is of the rays is changed when they bounce of objects.

I don't know if this is 100% correct

But this is the idea

[harsh scroll]

-# I'm also curious, so i'll also wait here

[rugged venture]

Ahhh I get it

It is like this

[marsh citrus]

@rugged venture Has your question been resolved?

[iron marlin]

Usually you start from the camera, since what you are interested in is "what is the distribution of light coming in at this angle"

it's more efficient because it only considers rays that reach the camera

[elfin cairn]

And you only consider rays that hit the light source?

[iron marlin]

Pretty much. If you have diffuse surfaces, you want to prioritise rays which go to bright things to make your integration go faster.

Surfaces with diffuse scattering

Specular reflections can also be done

[marsh citrus]

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[humble osprey]

Hello, I have a weird question. Using diophantus' way of finding pythagorean triples, namely (k²+m², k²-m², 2km), is it true that if k and m have 1 as their only common divisor, then the triple created is a primitive pythagorean triple? If so, why? Also, if there is an article written about it, please send me a link or its name

[jagged relic]

,calc 5^2+3^2

[elfin berry]

Result:

34

[jagged relic]

,calc 5^2-3^2

[elfin berry]

Result:

16

[jagged relic]

,calc 253

[elfin berry]

Result:

30

[jagged relic]

No

[humble osprey]

huhhhh

our professor told us something like thaf

that*

i understood it wrong then

any idea what he might have meant?

[jagged relic]

https://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple

[humble osprey]

oh lol i thought it was diophantus' formula but its just the kind of equation

ill look it up and come back then

.solved

[marsh citrus]

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[wispy adder]

i keep getting 200/298

But i know its not the asnwer

[marsh citrus]

Show your work, and if possible, explain where you are stuck.

[wispy adder]

i got the answer now but i got it wrong on the first try because i didn't put parenthesis around 2x+2y dy/dx starting from line 2

.close

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[autumn stone]

(2x − 3)(x + 5)
_________________   = 0
x − 7

I would expect x to equal -5,7,3/2 however the answer is 5,3/2
My question is: Why isn't the x-7 on bottom when multiplied to the numerator create another solution for x?

[cunning fiber]

Because the original equation still has it in the denominator

meaning setting x=7 gives you division by zero

hence undefined

[autumn stone]

so when solving a question such as this each group of parentheses gets set on its own as equal to 0?

(2x-3) = 0;
(x+5) = 0;

actually bringing it back for a second, once you find a potential solution you have to plug it into the original question?

[cunning fiber]

In general, $\frac{a}{b}=0 \implies a=0$ and $b \neq 0$. Here, $a$ is a product of factors, so yes, you set each factor equal to zero (zero product property).

[elfin berry]

Civil Service Pigeon

[cunning fiber]

As long as you don't do steps that could introduce extraneous solutions, you don't have to worry about that.

Ex. If you noticed that $x-7$ being zero is not possible and didn't multiply by $x-7$, then you wouldn't have gotten an extra solution.

[elfin berry]

Civil Service Pigeon

[autumn stone]

I see ok so seeing as it was in the denominator I should of ignored it?

[cunning fiber]

In general, $\frac{a}{b}=0 \implies \boxed{a=0 \text{ and } \neq 0}$. Here, $a$ is a product of factors, so yes, you set each factor equal to zero (zero product property).

[elfin berry]

Civil Service Pigeon

[cunning fiber]

Note that whenever you set the numerator equal to zero, you need to make sure the denominator isn't zero

here, that's a non issue

but if you had something like $\frac{x^2}{x}=0$, then it would be

[elfin berry]

Civil Service Pigeon

[autumn stone]

Okay I get it, thanks

[cunning fiber]

[autumn stone]

.close

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[loud temple]

Can someone help me please

[marsh citrus]

@loud temple Has your question been resolved?

[regal coral]

i think you can look at 3 cases: cubes in the center (one face showing), cubes on an edge (two faces showing) and cubes in a corner

what is the minimum sum showing for each of these 3 cases?

[loud temple]

21?

[regal coral]

try one case at a time

[loud temple]

The middle is 3, the edges are 7, the corner is 11?

[regal coral]

the middle is 1

the 1 may be showing

[loud temple]

Oh

Ok then couldn’t the edges and centre be 2,3,4,5,6

Hm

Oh

[regal coral]

how could an edge show 2?

it can't

an edge may show a 1 and a 2, for a total of 3

[loud temple]

Mk

[regal coral]

an edge shows two numbers

a center shows 1 number

a corner shows 3 numbers

[loud temple]

Ohh

[loud temple]

.close

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[ornate leaf]

any restriction as to what a can be ?

[ornate leaf]

does it need to be an integer? Or is it in R or C

[static quarry]

probably it's assumed to be positive and real, since they say that e^at is "growing"

if it were negative then it would be decreasing

and if it were complex then "growing" doesn't even make sense

[ornate leaf]

if its complex

[ornate leaf]

is there any app that plots complex functions

like with a vector representation

[ornate leaf]

yea probably but is it integer

because its mentioned as order but how can we say its of the π order idk if it makes sense

[static quarry]

i don't see why it should have to be, the definition makes sense if a is any positive real number

[ornate leaf]

same as with like a derivative we cant use non integer

[ornate leaf]

because we call it order

[static quarry]

yea the name isn't the best

but the idea makes sense for any positive a

[ornate leaf]

alr

can you tell me any app that shows complex functions ?

i want to see what it would look like

[static quarry]

hmm i don't know offhand, can desmos do it?

[ornate leaf]

i dont think it can

i dont see a way to define a as complex

[marsh citrus]

@ornate leaf Has your question been resolved?

[marsh citrus]

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[pine ravine]

is it possible ot have a negitive degree polynomial? how would that even look like? Something in the set $\mathcal{P}_{-2}(\mathbb{F})$

[elfin berry]

BOSS

[vale iris]

I mean you can talk about expressions like $1+x^{-2}$ if you want but by definition they're not polynomials

[elfin berry]

depression

[marsh citrus]

@pine ravine Has your question been resolved?

[elfin stone]

polynomial cannot have negative degree

That's how they are defined

[unkempt locust]

You know by convention, we say that the polynomial 0 has degree -infinity

It’s the only polynomial I know with negative degree

[indigo nest]

Huh interesting

I understand not wanting it to be zero cuz euclid division by constants

[vale iris]

Yeah well spotted, I forgot about that

It's bc if you multiply two polynomials their degrees get added together

So the degree of 0 has to behave like -infinity

[indigo nest]

But negative infinity is interesting

[indigo nest]

Ah that use case makes sense

[marsh citrus]

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[pine ravine]

LADR puts "of non negitive degree" a lot so i was curious lol

[elfin stone]

What's LADR?

[indigo nest]

Linear algebra done right

[pine ravine]

Also, since lists are ordered pairs does this mean that only elements before v_k apply?

Like technically v_{k+1} is not on here but i imagine it sould sitll apply to that vector?

[vale iris]

Firstly lists aren't ordered pairs unless you're a set theorist

[vale iris]

And not necessarily

It just says there exists a k

The k+1th vector could be independent from everything else

[pine ravine]

I mean I was going off this definition

[vale iris]

Yeah ordered collection, not ordered pair

[pine ravine]

but ok it does kinda apply, like here i should think of it as a set?

[pine ravine]

whoops yeah

but for this set, I dont see why elements after v_k within the list cannot be considered for the lemme

[vale iris]

Okay let me give you an example

[pine ravine]

cool

[vale iris]

v1 = (1,0)

[pine ravine]

ok

[vale iris]

v2 = (1,0)

[pine ravine]

ok

[vale iris]

v3 = (0,1)

[pine ravine]

ok

[vale iris]

here, if you take k=2, then v2 is in the span of v1

But v3 is not in the span of v1 and v2

So you can't just use k+1

[pine ravine]

ok but whats stopping me from writing the list as

(0,1), (1,0), (1,0)

and taking the second coordinate as an example

[vale iris]

Oh absolutely nothing

But this implicitly comes with an ordering

If you reorder the vectors then you get a different k

[pine ravine]

yeah so the notation is implying, because lists are ordered, you can only use the span of elements before k

(0,1), (1,0), (1,0) is still a linearly dependent list in V

[vale iris]

Yes

But it's a different list

[pine ravine]

let k = 2 here

[vale iris]

In this different list, you have to take k=3

[pine ravine]

this is no longer true

[pine ravine]

no like lets erace everything before

[vale iris]

No no no

Think of this lemma as a machine

[pine ravine]

ok

[vale iris]

You feed it a list of vectors

And it spits out a k

If you feed it a different list of vectors, then it might spit out a different k

[pine ravine]

Ok

@vale iris im still lost

oh

we dont get to pick the k

[vale iris]

no the lemma provides that

[pine ravine]

so in this case k=3 makes it true

[vale iris]

Yes

[pine ravine]

yeah i see what ur saying

oh

because if certain elements are in a span there is always a furthest one

damn ok

makes sense

wasnt thinking hard enough there

tyty

[vale iris]

np

[pine ravine]

is this the proper logic

basically all elements of the list creating the span are before k

i guess ill do the proof to find out

.close

[marsh citrus]

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[vale iris]

Something like that yeah

The proof is pretty easy once you properly understand the lemma

[marsh citrus]

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[violet warren]

Can the 5 one be solved without substitution

[violet warren]

Like what i did was cube both sides

. close

[cunning fiber]

.close

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[wicked siren]

I need help

[ruby mulch]

!da2a

[marsh citrus]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[wicked siren]

I need help with physics

[elfin stone]

Hi, just post

[ruby mulch]

Just post the problem. Don't say "I need help"

[wicked siren]

for now on il do that.

[wicked siren]

Could you do it step by step

?

[indigo nest]

Well are you familiar with impulse

[wicked siren]

Not really

After all this is grade 12 physics

[indigo nest]

Yes grade 12 physics should have impulse

[wicked siren]

Well I’m not familiar with it, could you explain it

[indigo nest]

Impulse is the change in momentum

Essentially

It's also force integrated over time

[wicked siren]

Okay

What’s next?

Do I have to draw free body diagram for this?

Hello?

I think I would have to do kinematics for this

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[keen wyvern]

Am i setting H up correctly 🧍‍♀️

[wary lichen]

!original

[marsh citrus]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[marsh citrus]

@keen wyvern Has your question been resolved?

[keen wyvern]

Hello

[gentle siren]

did you figure it out or do you still need help?

[marsh citrus]

@keen wyvern Has your question been resolved?

[keen wyvern]

i did not figure it out

[gentle siren]

do you still want help on it

[keen wyvern]

yes

[gentle siren]

ok! so first find the intercepts of the two functions

if youre allowed it will be easy to do in desmos

[keen wyvern]

probably not a good idea because i wont be able to on the exam lol

y is 1 and -1

x positive 2

(2,1) (2,-1)

[gentle siren]

ok good!

do you know how to set up the integrals?

[keen wyvern]

idk i was thinking it was from x=0 to x=1

[marsh citrus]

@keen wyvern Has your question been resolved?

[marsh citrus]

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[karmic mulch]

Hey guys can anyone explain me how to find limits

[karmic mulch]

@everyone

[calm harbor]

Hi, I’m pinged. How can we help you today?

[karmic mulch]

Ah what

[calm harbor]

Do you have a specific question about finding certain limit since this one is too board

[karmic mulch]

Just how do you find a normal limit of a function

[karmic mulch]

,

[vernal forge]

send an example problem, it will be easier to help you this way

[calm harbor]

There are multiple approaches with respect to different circumstances. You’ll have to provide an example for us to evaluate which one is the most suitable.

[karmic mulch]

I don't know you can use any simple one

I just started learning limits and derivatives and i can't understand a word

@calm harbor @vernal forge

[vernal forge]

can you send an example

we aren't going to give you a full lecture

[vernal forge]

or an extract of the book that confuses you

[calm harbor]

It’ll take hours for us to give a full lecture.

The most effective way is that you provide us an example

[karmic mulch]

Please use any simple one

[vernal forge]

simple is subjective :^)

[karmic mulch]

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[calm harbor]

[marsh citrus]

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[fair remnant]

guys how to solve this ODE?

[marsh citrus]

@fair remnant Has your question been resolved?

[quick kindle]

maybe the sub v=y/x

[wispy stump]

Yes

Substitute v =y/x and then solve

Alsp dont forget to check the value of dy and substitute that aswell

[fair remnant]

i did but

the final integral is unsolvable that way

[marsh citrus]

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[grand tide]

hello, another question

[grand tide]

in the example given, why does (x + 9) need to be rewritten as (x - (-9))?

[late geode]

they're explicitly writing it in the form (x - c)

[sick walrus]

most likely to match the defn here.

[grand tide]

OH right

i overglossed that part

my bad 🙏

thanks guys

hehe

[late geode]

if you can see that -9 is a root seeing the factor of (x+9), there's no real need

[grand tide]

right

cause x = -9 any way

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[quaint arrow]

w0 = [1, 0, ..., 0].
So I am confused about this. Because the calculation i did was (1 + cosθ)^2 + (sinθ) ^ 2 + 2 < 3. This ultimately simplifies to cosθ < -1/2.
cosθ is only less than -1/2 for angles 120-240 and (120 degrees). So 120/360 = 1/3. Im not sure how the probability is being calculated in the problem

[quaint arrow]

also it looks like the solid black arc is the only part we can move to to decrease. by doing simple geometry, i have also calculated the angle of the arc is 120, and 120/360 = 1/3

[copper raven]

well they're not asking you to show that the probability is exactly sqrt(3)/4

you just need to show that it's less than sqrt(3)/4

you found that it's 1/3, well gg 1/3 < sqrt(3)/4

Im not sure how the probability is being calculated in the problem
but yeah I'm also wondering how they select the vector, what's that equation 2.31 they're talking about ?, is it just about the w0

[quaint arrow]

yes,

[copper raven]

do they describe their random search anywhere else ?

[quaint arrow]

yes, the random search algorithm is just picking a random direction and determining if the direction we moved in is less than the previous point

[copper raven]

ok but do they define how they select the random direction anywhere

[quaint arrow]

it's not defined explicitly, but the sense i got is that it's truly randomly selected

[marsh citrus]

@quaint arrow Has your question been resolved?

[marsh citrus]

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[elfin badger]

for this do i differentiate?

twice?

[jagged relic]

No, you check f(-x)

And -f(x)

[indigo nest]

And check if they're equal

[elfin badger]

im not sure how?

[indigo nest]

Okay firstly

What's an odd function

[jagged relic]

( if f(-x) = f(x) or if f(-x) = -f(x) )

[jagged relic]

By plugging -x into the definition of f(x)

[elfin badger]

thanks

so why is it -7x^2

and not positive?

[jagged relic]

Wdym? If you wrote it correctly, surely you know why

[elfin badger]

i dont understand it

properly

[jagged relic]

Why -7(-x)^2 = -7x^2 ?

[elfin badger]

yes

[jagged relic]

Because (-x)^2 = x^2

Because (-x)^2 = (-1x)^2 = (-1)^2 (x)^2 = 1 x^2 = x^2

[elfin badger]

yes i see now

i was adding them

mb

thank

thanks

.lose

.close

[marsh citrus]

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[native minnow]

Hello math fellows, I would like to have some help regarding (yet another) question I have in linear algebra (sorry I ask too many questions).
So let we have a field F, and we define a linear/vector space V on F.
Next we define a norm on V, and finally we can define an inner-product on it.
So first question, the inner product gives us the geometry?
Next, we have the following statement:
<a,b> = |a| |b| cos(t)
The <a,b> is define from the inner-product, but |a|, and |b| is define by the normed space(before inner-product joins the chat), so does it mean that the structure of the inner-product must be somehow compatible(or inherited) from the normed space?
And regarding the ratio cos(t) = <a,b> / |a| |b|. For any inner-product, and normed(metric) space (F,d), how is this proved that this ratio returns a value in [-1,1]?
Thank you in advance.

[leaden monolith]

It need not be compatible

But it doesn’t make much sense to prescribe conflicting information about the geometry of a space

[proud ice]

sorry I ask too many questions
No such thing

[glass silo]

(generally, if you have an inner product space and a norm, it's taken that the norm you're talking about is that induced by the inner product, rather than any general norm )

[leaden monolith]

If $\langle\cdot,\cdot\rangle$ is an inner product then $f(x) \coloneq \sqrt{\langle x, x\rangle}$ is a norm

[elfin berry]

frosst

[native minnow]

Yes, like defining the linear space, then defining an inner product on it which will define the norm as well. Pretty awesome.

[leaden monolith]

So if you prescribe an inner product on a vector space then it immediately also induces a norm

[native minnow]

The next question is, why specifically |a|=sqrt(<a,a>)? and not something else maybe?

[leaden monolith]

I mean you can pick something else but then it may have nothing to do with the inner product

[native minnow]

Like the statement cos(t)=<a,b>/(|a| |b|) may not essentially return [-1,1]? is it considered broken? or?

[native minnow]

Sorry, am a little confused. I'm wondered if this is possible that an inner-product satisfies all of required properties/axioms, but the statement/ratio <a,b>/(|a| |b|) does not essentially return [-1,1], then what? we have a new cosine?

[leaden monolith]

I’m not sure if this works in general

[native minnow]

Otherwise, if <a,b>/(|a| |b|) essentially returns [-1,1], then has it been proved already? or how to prove it?

[leaden monolith]

How can you define what cos(t) is if there’s no t on the right side of the equal signs

[native minnow]

Good question! now this is even more confusing now... wow!

[leaden monolith]

I’m pretty sure this only works in ℝⁿ if you take the dot product

There are some nice identities like Cauchy Schwartz

They won’t hold if the inner product has nothing to do with the norm

[native minnow]

Another question, so if we have a normed space (X,d), then defined inner-product must satisfies <a,a> = |a|^2 ? or this is a special case in R^n? since I havan't seen such conditions in lienar algebra books.

[leaden monolith]

Once again you can define the norm and the inner product separately

But then they will have nothing to do with each other

A lot of theorems rely on the fact that the inner product and norm and linked

[native minnow]

Yes, but we will have a geometry?

[leaden monolith]

Depends what you mean by geometry

Inner products give you an idea of what orthogonal means

[native minnow]

To find an angle between two vectors, and orthogonality for instance. The problem is that the distance is found by the norm space, but the angles are found by the inner-product, and they are not linked. that's my problem. what happens? is it actually possible? like we are going to have a chaotic geometry?

[leaden monolith]

What does an angle really mean

[native minnow]

That's another question.

[leaden monolith]

There are vector spaces where you can’t just take a protractor and measure the angle between vectors

[native minnow]

So they are not inner-product spaces?

[leaden monolith]

For example, consider the vector spaces of constant, linear and quadratic polynomials

[glass silo]

(also, as per the previous point, talking about a norm inside an inner product space almost certainly means you're referring to the norm that the inner product induces )

[leaden monolith]

Verify that this is indeed a vector space

And I can prescribe that $\langle f, g\rangle \coloneq \int_0^1 fg,dx$

[elfin berry]

frosst

[leaden monolith]

Then this is an inner product on the degree 2 and lower polynomials

Now I ask what does it mean if you ask me the angle between x² and 3x + 1

[native minnow]

Maybe finding the ratio cos(t) = <a,b>/(|a| |b|), and t=arccos(<a,b>/(|a| |b|)). Here comes my next question as, the cosine/arccos function must be defined already maybe? or we are defining it here? so if we assume cos() is already defined, then the ratio must return a value between [-1,1], and this means if the ratio returns something out this bound, then the geometry is broken, hence the inner-product still satisfies the axioms. So does it mean we have an inner-product, but we cannot have a proper geometry on it?

[leaden monolith]

You should notice as well that whenever we add structure to our sets we always require it be compatible with the existing structure, for example, a vector space is an abelian group under +, and then you add the scalar multiplication, and the distributivity axioms are the compatibility conditions

[leaden monolith]

It is not guaranteed that your right hand side is between -1 and 1

You can guarantee it if you use the induced norm, then Cauchy Schwartz can be shown to hold, then you can prove the right side is between -1 and 1

[native minnow]

Exactly, that's my question as well. We are working with standard inner-product and R^n, and everything is pretty cool and working. I'm thinking more abstarct(at-least trying)

[leaden monolith]

I mean as soon as you get a complex vector space as well, your inner product can return complex numbers

Now you’re trying to arccos a complex number let alone something not in -1 to 1

And this is even when you are using the induced norm

[native minnow]

Yes, then we will have another norm from complex to real! The next question is the norm to calculate the abs value of a complex must be induced from the norm we are using in inner product? or could be different then?

[native minnow]

Yes, R^n is pretty nice, very nice. That's the problem.

[leaden monolith]

You can certainly prescribe norms and inner products independently, but is it useful to do so? That’s really the question we ask

if you pick the induced norm then yeah on a real vector space you would get angles between vectors

[native minnow]

I tried to read many linear algebra books to find my answer, but they didn't go deep that way. For instance, Axle's stated that since norm in the R^n is not linear, so we needed inner-product to inject linearity. pretty cool pov.
Or Friedberg stated that to have a richer structure in order to have a geometry, we need that inner-product.

[leaden monolith]

What does he mean to inject linearity

[native minnow]

Well, that's the main question... what happends, and what if we go this way? IDK yet

[leaden monolith]

I’m not sure about that

[leaden monolith]

It turns out to not be very useful

[leaden monolith]

Oh into the discussion

Yeah if you have a vector space you have linearity

[native minnow]

Maybe it is, maybe not. IDK yet. I can't sleep, I need answers.

[leaden monolith]

You can’t do very much when Cauchy Schwartz doesn’t hold

When you introduce norms and inner products it’s because you want to talk about them and how they interact

But if you prescribed them separately and independently then the core theorems that relates them no longer hold

Then there isn’t much you can say because they haven’t got much to do with each other

You have no idea how they might interact because you’ve prescribed them independently from each other

[native minnow]

Yes, exactly. But it sounds very interesting.

[leaden monolith]

It is not..

Imagine a vector space defined where you didn’t need distributivity

Can you even prove that -1*v = -v?

Can you even prove that 0v = 0

[native minnow]

Great question actually, thanks for sharing your questions and thoughts.

[leaden monolith]

It’s good to give this a think no one really realises that our nice results happen because we forced them to appear

[native minnow]

Maybe I need keep studying more. Yes, it's never enough. Sorry for many questions.

[leaden monolith]

Don’t be sorry for questions!

[noble geyser]

studying doesnt help for math if you dont understand it

[native minnow]

Just too curious, and not that smart enough to find the answers easily from resources. But I love math, I have to success.

[noble geyser]

if you dont understand it continously then might give it up

you dont need to study math in uni

[native minnow]

Well, I'm a math student too.

[noble geyser]

not like im saying anything just im saying im in 10th grade and i dont study/learn anything i get A all the time

and i know uni year 2 things

[marble gazelle]

WHAT

[noble geyser]

studied everything until uni year 1

[marble gazelle]

you are in 10th?

wow

[noble geyser]

you didnt know?

[marble gazelle]

grade 11?

[marble gazelle]

no

[noble geyser]

i havent studied anything last year bruh

need to catch up until uni3 in 1 year

[native minnow]

I cannot get A, have level of infinity during exams. That's bad. But try my bst to understand stuff by studying and studying and asking and asking.

[marble gazelle]

write the topics you have learnt till date by that extra study

lemme see if it matches with me

[noble geyser]

asking is a very good habit

[noble geyser]

hmm lemme think i think it was integration by parts

[noble geyser]

you will learn more by asking thats a good think dont be ashamed of it

ok now

!done

[marsh citrus]

If you are done with this channel, please mark your problem as solved by typing .close

[noble geyser]

you come dm

[native minnow]

Am not ashamed at all, I ask my questions anyway. Thanks.

[noble geyser]

np im always here to help and im sure anyone that has the helper tag thinks the same!

[native minnow]

Thank you @leaden monolith , @glass silo , @proud ice , and @noble geyser . Really really appreciate your time and helps, awesome.
Math bless us all 🤘🤓🍻

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[marsh citrus]

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[pliant whale]

Is this right?

[stoic saddle]

,rccw

[stoic saddle]

y = log(cos(x^2))
y' = 1/(cos(x^2) * ln(10))

is that what it says?

not sure what the 90_ before it is meant to be

[pliant whale]

He gave y=log(cos(x^2)) and wants dy/dx the second step is from me idk if it’s right

[pliant whale]

That’s 20 question number my handwriting isn’t the best sorry

[stoic saddle]

long-winded way to say "yes that's what it says"

anyway, no, your derivative is incorrect.

[pliant whale]

alr

[stoic saddle]

yeah how did you make 2 look like 9

[noble geyser]

xDDDD

[pliant whale]

yeah thanks for the help guys really appreciated

[noble geyser]

you need double chain rule here

[pliant whale]

Yeah I don’t think so

[stoic saddle]

is this sarcastic or genuine

[noble geyser]

this is double chain rule

oh no sorry

its triple

[marsh citrus]

@pliant whale Has your question been resolved?

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[cedar field]

I am trying to solve this problem on a math/programming challenge website. There were many problems of this nature, but this one was the simplest. I know there is some kind of shortcut to finding the solution, since a brute-force approach would take billions of years to complete. I am looking for some information that would point me in the right direction. I looked at this wikipedia article about prime counting functions. Am I on the right track? https://en.wikipedia.org/wiki/Prime-counting_function

[elfin cairn]

you don't have to hide its Project Euler xD

[noble geyser]

lol

[elfin cairn]

anyway, there is no good algorithm to find pi(x) (I don't know any) but there are very good estimates

but they are only estimates

[rotund pike]

The shortcut is to count all the multiples of 2, then count all the multiples of 3 that aren't multiples of 2 * 3.

[cedar field]

up until 10 to the 12?

[rotund pike]

Yes, it will require some thought to take that and work it into a correct algorithm.

[cedar field]

Wouldnt all the multiples of 2 basically be 10 to the 12 divided by 2?

[rotund pike]

Yes.

But then you multiply that by 2 because the sum counts each one as 2.

[cedar field]

So the answer is approximately 10^12 / 2 + 10^12 / 3 - (10^12 / 6) ?

[rotund pike]

Well, that's only for the first two primes.

For example, 25 won't be handled there.

For the prime factor 5, you need multiples of 5 that aren't multiples of 2 or 3.

Oh, and it's more like 2 * (10^12 / 2) + 3 * (10^12 / 3 - (10^12 / 6)).

[cedar field]

to start off 2 * (10^12 / 2) + 3*(10^12 / 3) ... LPF * ( (10^12) / (Largest prime factor of 10^12) ) then that whole thing subtracted by all the shared factors?

[rotund pike]

No, you need to handle the shared factors inside each set of parentheses if you do it that way.

Think through the first few primes.

With 2, it's easy since it's the smallest prime.

You do 2 + 2 + 2 + ... + 2, one 2 for each even number.

With 3, it's still pretty easy.

[cedar field]

Largest prime factor of 10 to the 12 is 5 correct?

[rotund pike]

You do 3 + 3 + 3 + ... + 3, one 3 for each odd multiple of 3.

Yes, but why are you looking at the largest prime factor of 10^12?

To give an example of why we can't stop at 5 (if that's why you mentioned it), take 49, for example.

49 is within 2 through 10^12.

[cedar field]

right yeah

I know the mod 10 to the 9 means theres some sort of shortcut approximation I can take, because with modulo you lose information after the operation

[rotund pike]

No, that still requires an exact answer.

The last 9 digits have to be correct, but the only way you'll get them is to do the calculation correctly.

[cedar field]

is there some formula i should derive similar to how the sum from 1 to n is the same as n(n+1) / 2?

[rotund pike]

It also can introduce some complexity, like you might need to reduce mod 10^9 after each operation that can overflow your numeric type.

Well, the best way I can see to do it will have a short algorithm, but it won't quite be a closed form like that.

[cedar field]

This is my current code

void main()
{
    writeln(S(100)); // This is 1257
}

// Θ(n * sqrt(n)) = Θ(n^(3/2))
public ulong S(int n) {
    ulong sum = 0;
    for (ulong i = 2; i <= n; i++) {
        sum += smallestPrimeFactor(i);
    }
    return sum;
}

// Θ(sqrt(n))
public ulong smallestPrimeFactor(ulong n) {
    if (n <= 1) {
        return 1;
    }

    if (n % 2 == 0) {
        return 2;
    }

    for (ulong i = 3; i <= sqrt(cast(double)n); i += 2) {
        if (n % i == 0) {
            return i;
        }
    }

    return n;
}

Which completely slows down after about 100,000 or so

[rotund pike]

Yes, you want to avoid doing it that way because you have to partially factor each of 10^12 numbers that way.

For primes around 10^11, for example, your algorithm will take a long time just for one number.

This is why I recommend starting with the each smallest prime and getting all the numbers (no factoring required), rather than taking each number and getting the smallest prime.

[cedar field]

With the prime counting function?

[rotund pike]

I'm not sure the prime counting function can be used.

That'll tell you the number of primes below each number.

But they want the smallest prime of each number.

[cedar field]

Ohhh nvm

So the answer would be every y value on this graph up until x = 10 to the 12?

[rotund pike]

Yes.

[cedar field]

Is this the right track or am I just rephrasing what I already tried doing with my code

I'm just rephrasing

[rotund pike]

Well, it applies to both my method and yours.

It's essentially what the problem is asking rather than a method of figuring it out.

I'm assuming that's a graph of the least prime in each x.

[cedar field]

Yes

Is an integral involved?

[rotund pike]

I don't know a way to do it with an integral. I doubt it can be, but I'm not that advanced in math.

I mean, you can probably do the integral of smpf(floor(x)).

[cedar field]

Thats what I was thinking

Can this be solved on paper by hand or does it involve some kind of repetitive, heavy computing?

[rotund pike]

Well, it can be done on paper, but it'll take a while.

[cedar field]

Is it required to calculate fully S(10 to the 12) or is no number that big ever reached?

[rotund pike]

Well, that's what the problem wants you to find, but there are cheaper methods than finding the smallest prime of the numbers from 2 to 10^12.

[cedar field]

is this linear line something of use

[rotund pike]

Not really. The ones that stick up are just prime numbers, and their height is x.

So, the line is y = x.

[cedar field]

right

does the solution involve lots of exponents

[rotund pike]

No.

[cedar field]

I think its out of my reach, I will look at it tomorrow a little more but other than that I think i am biting off more than I can chew

I have to go to bed now

Thank you for the help!

[rotund pike]

You're welcome.

[cedar field]

.close

[marsh citrus]

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[marsh citrus]

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[loud temple]

number 3

[marsh citrus]

@loud temple Has your question been resolved?

[loud temple]

!helpers

[marsh citrus]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[loud temple]

!helper

[marsh citrus]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[loud temple]

<@&286206848099549185>

[marsh citrus]

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

[marsh citrus]

@loud temple Has your question been resolved?

[loud temple]

2

[strong loom]

Uh I think I know how to solve it

For the first part tbh u could just brute force it ish

U already know 10, 20, and 30 can’t be it cuz they already in other rows

And then u look for other multiples in row 7, 8, 9 that are also multiples of 10

I’ll just let u do the rest 😭

[marsh citrus]

@loud temple Has your question been resolved?

[woeful whale]

fr brute force looks like the only option left

[strong loom]

Fr all 3 can be solved using brute force

2nd one is like just prove smth greater than smth

[marsh citrus]

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[marsh citrus]

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Send your question here to claim the channel.

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

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

[ornate leaf]

hello in this ODE solving for y you can find a particular solution which then if you plug in the equation and take t=9 you get on the left side a number μ lets say and then δ(0) so we get δ(0) = μ in R. Is that not a contradiction ?

[ornate leaf]

and another question is if δ(t-9) = .. arethmitics with functions then doesnt that make δ a normal function but its not ?

(δ is the dirac delta "function" btw)

<@&286206848099549185>

[marsh citrus]

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[echo scaffold]

Could somebody check these?

[wispy stump]

On it

All correct, great job @echo scaffold

[elfin stone]

[ruby mulch]

But be careful on 49. By your writing, it can be pretty difficult to read whether you meant $\sqrt{x^2+2}$ or $\sqrt{x^2}+2$

[elfin berry]

Xwtek

[echo scaffold]

Ty!

[ruby mulch]

It just looks like $\sqrt{x^2+}2$, which looks weird.

[elfin berry]

Xwtek

[echo scaffold]

Kk

I’ll def be more careful on the test

[wispy stump]

Handwritten matters too

I remember losing couple marks for being misunderstood cuz of my handwriting🤣

[echo scaffold]

My handwriting is terrribllee luckily I have a typing accommodation on my iep.

It’s not because of my handwriting but it still helps that lol

[wispy stump]

Lol

@echo scaffold Pls close the channel by using .close if you’re satisfied with the answer

[broken hearth]

bro u can just see this

like

stare it

carefully

if there are even powers only it is even

if there is only cos function it is iven

if it is reducible to cos it is even

just stare it

like see in its soul

dont put

[marsh citrus]

@echo scaffold Has your question been resolved?

[marsh citrus]

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[rancid quest]

I'm entering a competition and I think knowing the quadratic formula is needed. I discovered it is used to find roots and I kinda see them in the old papers.

Does anyone have any tips on remembering the quadratic formula, or why it is like that, like the 4ac and 2a as the denominator

[sick walrus]

if you know how to complete the square from the general quadratic form $ax^2 + bx + c = 0$, that's exactly how you can derive the quadratic formula.

[elfin berry]

Lute

[fickle vessel]

https://www.mathsisfun.com/algebra/quadratic-equation-derivation.html

[rancid quest]

Oh alr thanks

[rancid quest]

Also thanks

Also what square?

Also what's an !xy

!xy

[marsh citrus]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[rancid quest]

oh

[jagged relic]

https://en.wikipedia.org/wiki/Completing_the_square#Overview
https://www.mathsisfun.com/algebra/completing-square.html

[rancid quest]

Ah ic thanks

.close

[marsh citrus]

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[marsh citrus]

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[ember garden]

Can someone help me with what to do next? I think im stuck on how to get y’ out of this

[ruby mulch]

Wait.
$$y = 2xe^y$$

[elfin berry]

Xwtek

[ruby mulch]

Then you take the derivative on both sides, right?

$$ y' = 2((x)'(e^y)+ (x)(e^y)')$$

[elfin berry]

Xwtek

[ruby mulch]

$$ y' = 2((1)(e^y)+ xe^yy')$$

[elfin berry]

Xwtek

[ember garden]

mhm

[ruby mulch]

$$ y' = 2e^y+ 2xe^yy'$$

[elfin berry]

Xwtek

[ruby mulch]

Then what are you aiming for?

Finding $y'$?

[elfin berry]

Xwtek

[ember garden]

oh

uh

i have this but

maybe iii

erm

...

divide by 2xe^y?

[ruby mulch]

I think you should move the $y'$-s to the same side of the equation.

[elfin berry]

Xwtek

[ember garden]

hm ok lemme try that rq

[ruby mulch]

No, not like that

[ember garden]

sorry

[ruby mulch]

Like this:
$y'-2xe^yy' = 2e^y$

[ember garden]

wait

oh then divide by y-2xe^y

but, how could we know that we can subtract the 2xey^y*y'

[elfin berry]

Xwtek

[ruby mulch]

Wait a typo

[ember garden]

oh

Then i can divide by parenthesis i think

[ruby mulch]

Yes

[ember garden]

wow ty

imma leave this open tho in case i have another quesrtion rq

Ok so i tried another one

Does this look good?

[ruby mulch]

Be careful of the bracketing.

[ember garden]

.close

[marsh citrus]

Channel closed

Closed by @ember garden

Use .reopen if this was a mistake.

[ember garden]

Ok actually another one so im finding the derivative of this one just a regular one

[marsh citrus]

✅

[cunning fiber]

what

the first looks like a typo

but what's going on here

[hasty flint]

I think you have made some errors

Mr etienne

The last step too, you cant cross out 2x and 2x, thats like saying (5 + 2)/ (1 + 2) = 5 (because you cross out the 2s). This is clearly not the case

[ember garden]

hm

ok ill try again thanks

[hasty flint]

The rule you used was good, but the derivative of x^2 + 1 is 2x

[fringe nacelle]

also , personally, i don't like the prime notation on parenthetical expressions, i think it's easy to miss

but that's not a mistake

[fringe nacelle]

i recommend D[1+x²], or Dₓ[1+x²], Euler's notation

you missed parentheses

[ember garden]

uh

hm where

[fringe nacelle]

oh well you put them back ay the end so

3rd step

[hasty flint]

Your derivative rule for qoutients was wrong

[ember garden]

time to redo lol

[fringe nacelle]

some students prefer never using the quotient rule btw

and writing instead

f/g=f g^-1

product and chain rule

[umbral oasis]

That is how you prove the quotient " rule "

There is no difference

[fringe nacelle]

ive seen proofs other ways

[umbral oasis]

Im saying those are not the correct ways

[fringe nacelle]

but anyway its one less rule to memorize, if a student is averse to memorization

[umbral oasis]

It should be seen as a corollary

[fringe nacelle]

it need not be

[ember garden]

I numbered my steps this time

[fringe nacelle]

i memorized it as

[ember garden]

If everything looks good now ill dp the algebra

[fringe nacelle]

D[f/g]=(f'g-fg')/g²
if yhat helps you

thats what my brain does to store the formula

[ember garden]

oh i thought thats what i did

[fringe nacelle]

it is

just different order of f'g and without (x) everywhere

this is a memorization technique, not a correction

[ember garden]

oki

[fringe nacelle]

you can also use partial fractions with complex numbers

(don't recommend that)

[fringe nacelle]

oof no not quite

g² is not differentiated

[ember garden]

;-;

[fringe nacelle]

so, easy to fix

[ember garden]

oops

can i put parenthesis on the bottom with '

so then diff it

[fringe nacelle]

theres no ' on the bottom

D[f/g]=(f'g-fg')/g²

[hasty flint]

Step 4 is wrong in the upper part too

[ember garden]

hm

ok ill go back to step 4 and start over

[fringe nacelle]

this gets a lot more routine abd easy with practice

[hasty flint]

I think you need to take a step back and think about the steps you've made. Doing extra steps isn't a bad thing if it prevents errors

[ember garden]

uh

im not sure how im getting the top part wrong