help-0

[terse meteor]

Do u realize z-2a is reflection of z respect to the imaginary axis?

[lone heart]

@noble vale Has your question been resolved?

[noble vale]

Yes so since z lies in the first quadrant would arg(z-2a) be -theta + pi?

[winter light]

Exactly

[terse meteor]

Yup

[noble vale]

tahnk u guys

.close

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[fast elbow]

this question is odd, three numbers make up a sequence, adding them will give 21 and multiplying them will give 231

[fast elbow]

so i solved this

assuming that the numbers are a, a+b, a+2b

and i got it wrong

yeah i forgot to mention

this is a special type of sequence where R_(n+1) = Rn+b

idk the name in english

[fast elbow]

my teahcer assumed that the three numbers were a-b , a , a+b

which made the whole thing a lot easier

so i find this question stupid

[sour mica]

Both ways would give you the answer

but yours and teachers would give different values for a

[fast elbow]

so i just calculated wrong

[sour mica]

you need to calculate the final numbers

like your a should match the teachers a-b

and so on

[fast elbow]

i mean- he got b = -4 , 4 and i got b = 10.42... and something

so i was wondering if the question is weird or i just calculated wrong

[fast elbow]

oh waait

[sour mica]

b = 10.42
Yea then you messed up your calculation

Yours and your teachers b would be same

[fast elbow]

and b= 1.527 approx

[sour mica]

just the value for a would be differnt

[fast elbow]

wow

[stiff ivy]

hey guys

[fast elbow]

okay ig ill just redo my calculations in english and see if i get the same result (ill send it here if i did)

[stiff ivy]

great job car and albarto

continue the hard work

❤️

[sour mica]

The name would be Arithmetic Progression jsyk

[fast elbow]

alright

[fast elbow]

sooo

just did the entire thing in english

and

💀 i got b= 4.945 , 23.055

should i show my calculations (TERRIBLE handwriting btw)

[sour mica]

You have a 7a^2 - 14a^2, which you kept as 7a^2
But that should be -7a^2

[fast elbow]

Oh…

Mhm i see

Its also a language barrier, signs in my language go on the right side of the number

So ty

[sour mica]

icic

[fast elbow]

yeah it works now

much appreciated

[sour mica]

[fast elbow]

.close

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[wind hill]

Hello! Do you guys also help validate statistical findings of a research? I just need to certify mine formally by a true statistician with the statistical tools I used (I have here a 2-page statistics of the result to be looked upon)

[lone heart]

@wind hill Has your question been resolved?

[lone heart]

@wind hill Has your question been resolved?

[upbeat hornet]

This isn’t really a good place for that, but if you send it here, maybe some basic errors can get caught

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@wind hill Has your question been resolved?

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[harsh tiger]

!help

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

[harsh tiger]

Left hand side 4 line to the bottom

Why is it 5/15

[tight pier]

1/3 = 5/15

[harsh tiger]

So 1/3 - 1/15 turned into 5/15

[tight pier]

no

they multipled the left by 5/5 to make the denominator common

the common denominator is 15

[harsh tiger]

Ok they made the denominator common so we have 1/3 turning into 5/15

[tight pier]

yes

[harsh tiger]

So we cancel one out after ?

Wait

[tight pier]

cancel?

[harsh tiger]

Nah

[tight pier]

we dont cancel anything

[harsh tiger]

So what happened with the 1/15

[tight pier]

it just stayed

[harsh tiger]

Aight

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[alpine sable]

im confused on what to do for this question

[alpine sable]

ping when reply

[fair dome]

The sequence converges to e^-2 @alpine sable

[pallid scarab]

you can try to find limit of ln(sequence)

then apply the exponential back

[alpine sable]

oh ok

also for this one

where did they get 1/3 from on the right end

[modest moon]

they moved 1/3 to the left here

[alpine sable]

huh?

[modest moon]

they moved 1/3 to the left in the same way they moved 3 to the left in the first equality

[alpine sable]

oh 3^-1

@modest moon how did they get 1/5 for the second part

[modest moon]

i dont know, sorry

i think it's a telescopic series where 1/3 and 1/5 in the summation are the only ones not being cancelled out, and the a_n of the sequence here goes to 0 so the series sum is 1/3+1/5+0=8/15

[modest moon]

oh i see why,

the series's pluses are:
1/[2n+1], 1/[2n+3], 1/[2n+5], ..., 1/[2n+1 +2k], ...
and the minuses are:
-1/[2n+5], -1/[2n+7], -1/[2n+9], ..., -1/[2n+5 +2k], ...

so the only ones not being cancelled out are
1/[2n+1], 1/[2n+3]

and since n starts from 1 to +inf, those are
1/[2*1+1], 1/[2*2+3]
which are
1/3, 1/5

[rocky musk]

we got regulus corneas doing math before gta 6

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[ocean fossil]

This might be a dumb question but here we go. I was randomly going through the MAT Oxford exam and saw some disputes between an answer.

If we have f'(x) and integrate it, Is it f(x) or f(x)+C? I thought there wouldn't be any +C but for some reason I saw a video that added it.

[ocean fossil]

And if it does, can you show an example where C≠0?

[crude parrot]

it is +C

f(x) = x^2 + 5

f'(x) = 2x

int f'(x) = x^2 + C

[ocean fossil]

we do not have any set value of fx

[crude parrot]

its just an example

[crude parrot]

^

[ocean fossil]

But the int of f'x is fx

and fx would be x²+5

[crude parrot]

yes

[ocean fossil]

So no +C

[crude parrot]

f'x is 2x

integral of f'x is x^2, and you add a constant C

[ocean fossil]

if the answer is f(x)+C

f(x) is x²+5

+C is just 0

[crude parrot]

5 is the C in this case

[ocean fossil]

but f(x) is x²+5 according to what you stated

[crude parrot]

since we don't know if there was a constant in the original function, we have to include the +C to account for the lost constants

imagine you werent given f(x) though

[ocean fossil]

but we already accounted for that by writing out the original eqn

[crude parrot]

you wouldnt know if there was a constant in f(x) unless you had the function given

when you take the derivative of f(x), any constant goes away because the derivative is zero

so when you integrate f'(x) back, you only get f(x) up to an unknown constant

[ocean fossil]

but fx already has the constant...

[crude parrot]

pretend i never gave you f(x) though

youre not getting the point

pretend i gave you only f'(x) and nothing else and told you to integrate

you wouldnt know if there was a constant or not in the original f'(x) equation

[ocean fossil]

by writing out f'x it is implied that f(x) exists and is 'given' although not exactly have an eqn

I would agree if it was the integral of fx

[crude parrot]

okay lets try an example

ill give you f'(x) = 6x^2

integrate that

[ocean fossil]

f(x)=2x³

[crude parrot]

okay good

however what if i told you that f(x) was actually 2x^3 + 500

f'(x) would still be 6x^2

if i said f(x) was actually 2x^3 + 12

f'(x) would still be 6x^2

that is why we include the +C

so the integral of that would actually not be just 2x^3, but rather 2x^3 + C

[ocean fossil]

Is it not implied that fx is "known" the same as f'x

[crude parrot]

no

the integral of f'(x) doesnt give you exactly f(x)

[ocean fossil]

so then why would it not just be integrate f(x)

[crude parrot]

why would you integrate f(x)?

that would give you the area under some graph, which has nothing to do with this

let me give you another example

[ocean fossil]

but the question itself doesn't have anything to do with the gradient etc... either so it would make more sense for it to be "integrate fx" instead of f'x

It seems like the f'x is there for something

[crude parrot]

f'(x) is the slope of the tangent line to the graph at some point

[ocean fossil]

yes I know but I'm using the same argument you used about how there would be no reason for it to be fx because it isn't exactly calculating area, by saying it doesn't ask for gradient either

It seems like a tricky question

Either that or it uses f'x instead of fx for no reason

[crude parrot]

your original question is "why do we add +C?" and the answer to that is because we don't know what f(x) would be when we integrate f'(x), because it loses the value of some constant C when we take the derivative of f(x)

try this: find the integral of f"(x) = 9x^2 for me (remember its second derivative)

i changed it to 9 so theres less fractions

[ocean fossil]

if I just want to find it and can call it in terms of fx (since the question doesn't have eqn for fx), I'd just call it f'(x) 🤷🏻‍♂️

[crude parrot]

im asking you to find f'(x) though, not just label it as f'(x)

obviously the integral of f"(x) is f'(x), but im trying to prove a point as to why we add C

[ocean fossil]

I think I know what differs. My POV is that because we call something f'(x), we must know f(x) in some sense

[crude parrot]

whenever we add +C is when we don't know what f(x) is. if we know f(x) and f'(x), then there is no need for the +C because you already would know the constant

because when deriving, there is a loss of some constant

which we wouldnt know if we werent given f(x)

[ocean fossil]

I know what you're saying. I'm just stating that imo it is more sensible to say that f'(x) implies fx. And I didn't say the answer is f(x)+C like you stated earlier, I said there has been disputes among it. (Seen both sides of the argument)

[ocean fossil]

And here I'm guessing you mean the integral of f''(x) is f'(x)+C and not just f'(x)

[crude parrot]

yes that is correct

[violet jetty]

I don't see how f'(x) implies f(x), or labelling soemthing "f'(x)" implies knowledge of what f(x) is. DEs is a whole field of problems where we have some information about f'(x) (or its higher order derivatives), yet oftentimes we can't analytically solve for f(x).

[barren mist]

$\int f'(x) dx = g(x) + C = f(x)$

[ocean seal]

aerial.ace

[ocean fossil]

But this is only in practice, a pure theoretical sided question shouldn't adhere to physics, is that wrong?

[barren mist]

you get g(x) using the integration rules. You can use any given conditions to find C which gives f(x)

[ocean fossil]

That pretty much is what I'm saying

[crude parrot]

Can you provide an example where you wouldnt put C and explain why?

[violet jetty]

DEs is math, not physics

[ocean fossil]

You know DE from deriving it from other things, not just.. knowing it

I know DE is math

But can you give an example of pure math where we for some reason have f'(x) and not f(x) without deriving it somewhere else

[ocean fossil]

if I have fx=x²+5, f'x=2x
the int of f'x is fx which is x²+5 QED

[tepid drum]

Wdym by for what reason , what reason makes you solve an equation for example what makes you solve 2x²+3x-5=0

[crude parrot]

thats what i explained earlier. you put +C when f(x) isnt given

[crude parrot]

almost all of my calculus exams after integrals have had questions like this

[ocean fossil]

Yes but you start somewhere for a reason. For example, f'x implies that fx is differentiable

[tepid drum]

You can just get a question to solve a differential equation for example

[frozen stump]

Here's an example that may change your mind; set f(x)=(sinx)^2
taking the derivative, we get f'(x)=2*sinx*cosx=sin(2x) (using the identity sin(2a)=2sin(a)cos(a))
now taking the integral without putting C, we obtain "f(x)=-cos(2x)/2"
applying the identity cos(2a)=1-2*(sina)^2 we find f(x)=-(1-2*(sinx)^2)/2=(sinx)^2-1/2
and just like that, we have f(x)=f(x)-1/2

[tepid drum]

Here you will start with derivatives

[crude parrot]

the pure definition of an indefinite integral is $\int f(x) dx = F(x) + C$

[ocean seal]

Krish

[ocean fossil]

ok this one actually makes sense but it just feels as though you're purposefully finding a way for it to make sense lol. I wouldn't have assumed you'd do all that. You could also just call them fx and f'x since no eqn is given (in my example), and then ATP you achieve the same result but without +C

[ocean fossil]

I stated earlier that I agree with that

just not for f'x

If I call f'x as gx

[crude parrot]

its the same for any order of derivatives, going from one the one below it

[ocean fossil]

Then the int of f'x is Gx+C

[tepid drum]

Take f(x)=x+2 and f(x)=x+5 you have f'(x)=1 for both of these. Now if you are given f'(x)=1 , how do you know if the original function f(x)=x or x+2 or x+5 or ....

All of these have the same derivative

They are different functions

[crude parrot]

which is why we add C to account for the lost constant

[ocean fossil]

Call them f(x) and g(x) instead to differentiate them and you should see the problem🤷🏻‍♂️

[crude parrot]

because we don't know what C could be, but f'(x) is the same for all of them

[tepid drum]

I don't see the problem

[crude parrot]

how does naming them differently change anything?

[ocean fossil]

f(x)=x+3, g(x)=x+5
f'(x)=1. If I assume the int of f'x=fx then we reach x+3, the same is done with gx
While if I assume +C, C=0

[crude parrot]

thats like saying f(x) = 2x, now solve for x = 2. now say its g(y) = 2y. solve for y = 2. same question

[tepid drum]

f(x)=x+2 and g(x)=x+5 , f'(x)=1 and g'(x)=1. Now suppose that you are only given f'(x)=1 and g'(x)=1.

[ocean fossil]

But I only use the functions themselves

[tepid drum]

How do you know that g(x)=x+5 and f(x)=x+2 and not g(x)=x+2 while f(x)=x+5 for example

You can't know

Because you are only given the derivative of f and g wrt x

You only need to satisfy the equations f'(x)=1 and g'(x)=1

g and f can be any functions that satisfy these equations

[crude parrot]

Think of taking a derivative. If you take the derivative of x, it’s 1. If you take the derivative of x+1, it’s also 1. If you take the derivative of x+60431, that is also 1.

Now think of it in reverse. What function gives you 1 when you take the derivative? We’ve just seen that x does. So does x+1. So does x+60431. In order to group this family of solutions together, we generalize the constant and call them “x+c.” Similar logic applies to other families of functions.

[ocean fossil]

that's not at all related to my example though, that's only in the case that f'(x) is defined

[barren mist]

@ocean fossil i think you are confusing anti derivative with integral

[tepid drum]

Unless you have more given like an initial condition (for example f(0)=2 and g(0)=5) you can't know the original functions

[ocean fossil]

But this image follows what I said though?lol

[tepid drum]

"if we have f'(x) and want to integrate it"

[barren mist]

if u said that then youre right

[tepid drum]

.

[ocean fossil]

The image shows the int of f'x is fx

[crude parrot]

its g(x) + C

its the green square

[ocean fossil]

And it says =fx

[violet jetty]

How do you define pure math? You can't abstract away generality, that's where the statement about WLOG come from. The fact that we use an "abstract" notation "f(x)" still necessitates that any operation involving it must hold for any f(x). If f(x) = x^2+5, f'(x) is 2x, yes. However the integral of f'(x) is not f(x), it is the antiderivative of f'(x) + C.

Here's a more "general" illustration
Let T be the differential operator s.t. T(f(x)) = f'(x)
Let S be the integral operator s.t. S(f(x)) = integral from 0 to x of f(t) dt

Consider the composition of operators ToS, meaning first you apply S and then you apply T. In this case, T perfectly inverts S. If you have a linear algebra background, you can see that this is true by determining the transformation matrices on some valid finite basis and show that their composition is the identity.

Now consider SoT, meaning first you differentiate, and then you integrate. Now, S does not perfectly invert T, the composition of the transform matrices do not yield the identity.

You can try by applying it to an arbitrary but defined polynomial, ie. P(x) = ax^3+bx^2+cx+d.
Integrate first and then differenitate, you end up with ax^3+bx^2+cx+d
Differentiate first and then integrate, you end up with ax^3+bx^2+cx+C, with the defined d being lost.

The point is that differentiation and integration are not opposite operations. Differentiation is the opposite of integration, and completely cancels out its effects, but integration is NOT the opposite of differentiation.

[tepid drum]

And my example shows that it is not

[ocean fossil]

only if we define them

[tepid drum]

Wdym by that

[barren mist]

it is though....

integral f'(x) gives you all possible solutions for f(x)

[tepid drum]

If you start with f(x) , differentiate it to become f'(x) now integrate to get f(x)+c

So it is not

Well you and me are saying the same thing

[barren mist]

no you dont get f(x) + c you get some g(x) + c

[tepid drum]

But different notation is making a confusion

[barren mist]

where g(x) + c = f(x)

[tepid drum]

Yes you do unless you use an initial condition

[barren mist]

yep thats what i was saying

[crude parrot]

this goes back to what we said about why naming them two different things doesnt really make a difference its just making it more confusing to understand

[tepid drum]

f(x)=2x implies f'(x)=2 but f'(x)=2 implies f(x)=2x+c

Unless you use f(0)=0 that you had satisfied in the initial f(x)=2x

You won't get 2x back after integrating

[crude parrot]

if you are just given an integral: integrate 2x. it is x^2 + C. if you are given f(x) = 2x+5 and told to integrate f'(x). it is 2x+5. if you are given a definite integral then there is no reason to have a +C because you will evaluate it to some number.

[barren mist]

ok nvm i understand now

my bad

[ocean fossil]

You explained it really well but if true, is there no reason as to why an examiner would write "integrate f'x" instead of "integrate fx"

[barren mist]

you were saying the same thing just differently

idk what akito means

[tepid drum]

no it is not your fault , you were using the initial condition implicitly so you werent wrong it is just that we were saying the same thing in a different way

[crude parrot]

again, naming them two different things doesnt matter. you could integrate f(x) and not have F(x). you could integrate f'(x) and not have f(x). you could integrate f"(x) and not have f'(x). naming them different functions doesnt make a difference in what the actual value is.

[ocean fossil]

The exact same image you gave. Int of f'x would give a function gx+C where gx is solely integrating f'x without fx in mind.

[tepid drum]

in general $\int f'(x)\dd x=f(x)+c$ but you also have something like $\int_0^xf'(x)\dd x=f(x)$

[violet jetty]

The naming is arbitrary. you can say "integrate banana(x)" and your task would still be the same. saying f'(x) really only implies that there exists some function that f'(x) is the derivative of. But how does this imply C = 0?

[tepid drum]

you should be convinced that the first is true from the series of examples that you were given

[barren mist]

if u take a function differentiate and then integrate it you wont get back f(x) you will get back f(x) + c

[crude parrot]

yes, because for definite integrals there is no constant of integration

[ocean fossil]

because the existence of fx implies that we can gain the original equation by just writing out fx as the answer to the int of f'x

[tepid drum]

i wrote the second because maybe he confused it with the indefinite integral idk

[crude parrot]

thats what ive been saying for like the past half hour but for some reason i think theyre confusing it with something else? not sure

[violet jetty]

Why? "we can gain the original equation by just writing out fx as the answer to the int of f'x"

[barren mist]

you cant gain back the original equation if you dont already know it or have information placing constraints on it

indefinite integration gives you a set of functions

[violet jetty]

You just know it exists, not that it can be obtained by integrating f'(x)

[barren mist]

one of which is your original function

[crude parrot]

^

[tepid drum]

as i told you , if you are given f(x) originally and then you differentiate wrt x to get f'(x), then if you want to integrate f'(x) wrt x then if you use the original f(x) that you were given to find a condition that will determine the constant of integration then you will get f(x) back after you integrate f' wrt x

[ocean fossil]

Isn't the whole purpose of integrating (indef, to be more precise) to get the original eqn? Isn't that why we write +C because the original function can be of any C constant

[tepid drum]

but if you dont do that then you will get a +C

[tepid drum]

but this wont ever happen with you

because the question will be meaningless

[crude parrot]

you will never get the original equation unless you are given some point for which you can plug in values for x and y in order to solve for C

[tepid drum]

you may have something like $f'(x)=2x+5,f(0)=0$ find $f(x)$

[crude parrot]

in that case you will get the original equation back

[tepid drum]

something like this would be meaningful

[crude parrot]

yes exactly this

[ocean fossil]

or I can just write f(x) which would be the original equation since I don't need to do all that since a function is given but not necessarily defined

[tepid drum]

but you wont encounter a question like $f(x)=x^2+5x$ find $f'(x)$ then find $\int f'(x)\dd x$

[crude parrot]

whats the point of integrating if youre gonna say something like the integral of f'(x) is f(x)? theres no point if you arent solving for the actual value.

[tepid drum]

this would be a pointless question

[crude parrot]

in this case you are right your answer would just be f(x), or x^2 + 5x

[tepid drum]

because of course you will get back f(x)

since you have an initial condition implicitly, namely f(0)=0 for example

[ocean fossil]

Uhh it's a question on the test? Like how am I supposed to answer that lol

[tepid drum]

!original

[lone heart]

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.

[crude parrot]

you will never have that. show an example for where you have that on an exam

[ocean fossil]

I'm on a car trip rn I can give it in about 10 mins

[crude parrot]

unless its to teach you a point, you shouldnt have to answer that kind of question

[ocean fossil]

But the question is about just that

And the dispute is around that

[tepid drum]

alright no problem when you are able send the picture

[crude parrot]

my guess is its either a pointless question where your teacher is trying to teach you this definition, or you are slightly mistaken in interpreting what the question is asking.

[violet jetty]

Nope. Integration is just a mathmetical operation, that is all. It can be used to recover the original eqn given additional info, but it is not able to do so by defult.

[ocean fossil]

(iii) s(x)² part

2018 Cambridge STEP 1 for those who wanna know

[crude parrot]

yes, you integrate that to get something + C, and then you solve for C using s(0) = 0

[crude parrot]

explained here

that will get you the original equation, but you still have to add +C and solve for C

[ocean fossil]

But the original purpose is to find fx no? Which is why we define it with +C

especially an indef int

[crude parrot]

integration is not to find an original equation

[ocean fossil]

indef

[crude parrot]

The indefinite integral is used to find the integral of the function, and the resultant expression represents the area enclosed by the function with reference to one of the axes. A definite integral has a defined value, so no +C.

[ocean fossil]

If its purpose is to find the area why put +C at all? Because it clearly is useless when finding area

[crude parrot]

it depends on what the question is

[ocean fossil]

By your definition, what purpose is +C

[ocean fossil]

I gave it 😭

[crude parrot]

im not sure why you arent understanding the purpose of +C still

we all explained it in several ways

[ocean fossil]

What is its purpose then?

[crude parrot]

scroll up and read the explanations

[ocean fossil]

You said that the resulting integral is to find area

There is no use of +C when finding area

[crude parrot]

you asked what the purpose of integrating is

not the purpose of +C

[ocean fossil]

here I asked for +C

[crude parrot]

yes, please scroll up higher and read all the explanations and examples that myself and a couple other people gave for this.

try to reread the entire thing and process what the purpose of it is

[ocean fossil]

can you link it/reply to it because there's... Too much

[crude parrot]

literally like anywhere in the entire conversation

[ocean fossil]

I mean surely if it's everywhere, you could link a specific message

[violet jetty]

Nope. That's not the "purpose" of an integral. The integral has various definitions, none of which follows from the definition of a derivative.

[crude parrot]

^

that's one of many times i explained it differently

since we have a loss of some constant when deriving. when we integrate we need to account for that.

that's the C

[violet jetty]

It just happens to be that the indefinite integral of a function can "cancel out" differentiation to an additive contstant, but it has no formal defintiion as "the opposite of the derivative"

[crude parrot]

We don't know what C is unless we are given some point which we can use to solve

[violet jetty]

Because the "opposite" (or "inverse") operation must be bidirectional, ie. integrate -> differentiate = differentiate -> integrate, but this is not the case

[ocean fossil]

From what I'm reading; the purpose is to show that there can be a constant C s.t the original equation could be the integral itself, plus a certain C

I'm paraphrasing

[crude parrot]

basically yes

[violet jetty]

purpose is a vague term, but it is able to do that, yes

[ocean fossil]

if you define it that way then you are indirectly showing that its purpose is to find the original eqn

[crude parrot]

integration doesn't have one set purpose

[violet jetty]

is the purpose of division to cancel out multiplication?

is that the only reason we have division for? just to reverse multiplication?

[ocean fossil]

Bassically what I'm asking is what other application/purpose would it have other than finding an original equation

[crude parrot]

finding area, volume, solving de, finding displacement, used a lot in phsyics, etc

[violet jetty]

finding the area under a curve

[ocean fossil]

are all of those needed of +C?

[crude parrot]

you are confusing yourself. are you asking the purpose of integration or of the constant of integration?

[ocean fossil]

it is about whether +C is needed. So to be even more precise, what is the use of +C other than finding the original eqn

[violet jetty]

yes? Results of des are highly dependent on ICs, and chaos theory shows that they can be VERY sensitive to c

[crude parrot]

it makes sure that all functions that can be a solution are included in a said set

[ocean fossil]

But are we really interested in what "can be a solution" when we already have the solution?

[tepid drum]

+C comes from the definition of an indefinite integral

[crude parrot]

im not sure why you still arent getting the point. its used when the solution isnt given

[ocean fossil]

This is starting to be a metaconvo🙏🏻

[violet jetty]

You should take an analysis course

[vast tapir]

the original purpose of integration was to find the area underneath a curve. it was later shown that integration and differentiation are related (FTC), one is sort of an inverse of the other. you can relate the rate of change of a function to the area under a different function. when you take an integral, what you get is a function which outputs the area under the curve. now if its a definite integral, then you know between which points you need to compute the area. but if its an indefinite integral, all you have is an area function, starting at an arbitrary point, and going forward a distance equal to the input. since you can start at any point, you can write +C to account for that. this is a convolute way of explaining what C is, the much more general way is, integration is antidifferentiation, but since the derivative of a constant is 0, you lose that much information

[violet jetty]

You would probably enjoy it

[tepid drum]

An indefinite integral of a function f is the set of all functions F that satisfy F'(x)=f(x)

It turns out that F(x)=f(x)+C

[ocean fossil]

But we do have the solution? here in the case of the question I posted, it just is fx

[vast tapir]

idts theres much use to thinking of the constant this way, but you can

[tepid drum]

C is any real number

You can check that for any constant C in R , F(x)=f(x)+C implies F'(x)=f'(x)

[crude parrot]

what if you arent given the solution though?? every integration question isnt the exact same as what you have in that one example.

[tepid drum]

If you want a specific value of C then you need more conditions

[ocean fossil]

integration is antidifferentiation
yes and an "antif'(x)" would make f'x become fx. (Yes I did read the other stuff lol)

[tepid drum]

You only need the value of f at a certain point

say f(0)=1 or f(1)=π or whatever

[crude parrot]

can you be clearer in what you arent understanding still?

[tepid drum]

Then after the integration you can find C by plugging the x and y of this point that you get which will leave you with one equation with one unknown , namely C

[ocean fossil]

It's about my specific question not in general lol

[crude parrot]

because saying the same definition in different words of why we use +C 1000 times isnt helping, so its clear that that's not what you're misunderstanding.

[crude parrot]

in your original question, that was general, which is why we got into this topic of why +C is added

we only got your specific question 20 minutes ago. the other 70 minutes before that you never mentioned once that it is about the specific question you are trying to answer.

[ocean fossil]

same thing as the initial pinned comment I said, but on a deeper level, the fact that the integral of a function looks for the original eqn, and as a by-product only obtains a set of possible solutions (that is, any C)

But if we have a solution that we can call fx, wouldn't adding +C be unnecessarily giving a "set of possible solutions" when we already have it at C=0

Bassically, +C adds the possible solutions. But if we already have it as in the Q I gave, isn't it unnecessary to have it add C

[crude parrot]

again, like we said earlier. if C is given or if some point is given, then there is no point of adding C at the end.

because after solving for C, you won't have another C after that, since you already have a constant

[ocean fossil]

I think it'd be better if I posted the question again so that you can look better into it.

(iii) s(x)²

[ocean fossil]

Yes but I later sent a picture of the main reason my question came to be since someone else wanted it

[ocean fossil]

I mean the question itself is pretty much what I'm asking.... Only difference is a substitution

[crude parrot]

yeah, so did we not answer it? what are you still confused about?

[ocean fossil]

ok like to this image I sent, should it have +C or not? So that I can know whether it follows the criteria you gave here.

[crude parrot]

in (i), it says that s(x)^3 + c(x)^3 = 1.

integral of s(x)^2 is s(x)^3/3

i believe it would be equal to 3 - 3c(x)^3

[ocean fossil]

what? s(x) is a function not a variable

[crude parrot]

wait disregard that my bad

so $\int s(x)^2 dx = -\int c'(x) dx$

[ocean seal]

Krish

[crude parrot]

therefore the integral of $-\int c'(x) dx$ would be $-c(x) + C$. So yes, this would have a constant of integration

[ocean seal]

Krish

[ocean fossil]

Is an integral supposed to ALWAYS give all possible sets of answers then?

[frozen stump]

see I think this is where you're mistaken; there is not such a thing as 'the' original equation
Here's an example; suppose two students are asked to integrate f'(x)=cos^3(x) (which is a rather hard integral)
one student has memorized his trig identities, and decides to use cos(3x)=4*cos^3(x)-3*cos(x). he rearranges this to get cos^3(x)=(cos(3x)+3cosx)/4, and integrates the RHS to get f(x)=sin(3x)/12+3sin(x)/4. here, f(0)=0.
another student has seen many trig integrals be solved with u-subs; he rewrites the function as cosx*(1-sin^2(x)), and substitutes u=sin(x). integrating the resulting polynomial 1-u^2 and substituting back, he finds f(x)=sinx-(sinx)^3/3. here, f(0)=1/3.
You'll find that those two expressions for f(x) are actually both valid; the derivatives of both are indeed (cosx)^3. Yet those expressions are different, and either could have been reasonably found by integrating f'(x).

[ocean fossil]

Even if we can easily know that the correct original eqn is just -cx

[crude parrot]

yes

if i am understanding your question correctly i believe so

for indefinite integrals

[ocean fossil]

that seems really dumb. Is there a specific use as to why we'd need to know the set of possible values instead of only the actual solution

[crude parrot]

it gives a general solution

[ocean fossil]

Is there any use for the general solution

[frozen stump]

because, as I wrote in that wall of text above, there is never an 'actual solution', only a set of solutions that are each equally valid

[crude parrot]

its used to get a particular solution

[ocean fossil]

But we already have the particular solution

[ocean fossil]

yes and what's the use of that

[vast tapir]

its useful in hmmm

[crude parrot]

you would still have +C because you dont know what c(x) or s(x) are

[vast tapir]

physics, engineering, maths, any kind of system where things are changing

it seems like youre asking what the use of calculus is

[frozen stump]

Anything! for one, you could use it to find definite integrals with FTC

[ocean fossil]

I am obviously asking where the "particular" solution cannot give the same advantage

[crude parrot]

yeah this is a very broad question

[crude parrot]

because its not accounting for every possible solution

a particular solution is just a solution at a certain point

keyword "particular"

[ocean fossil]

I know but is there a use of the other possible solutions that the "particular" solution specifically cannot do

[vast tapir]

when we are given initial values. lets say i know that f(0) = 2, and f'(x) = x^2, then f(x) = x^3/3 + C, plug in x=0, you get C=2, so the function i need is f(x) = x^3/3 + 2

[crude parrot]

it shows how some function can change or be manipulated under different conditions

[vast tapir]

now where i would get such conditions would be physics, eng, etc

[ocean fossil]

It is already assumed that we have the "particular" solution lol

[crude parrot]

and like stated before by someone, its very useful in linear difeq

[frozen stump]

It almost seems you're asking us what the point of algebra is when you can just use concrete numbers everywhere

just with general solutions to an integral vs particular ones

[ocean fossil]

except this one actually has a use though

[crude parrot]

you are confusing yourself by asking the same question again in a different way. i would say just know that +C is used for general solutions, and its used when you aren't given the original equation to begin with, or when you aren't given a set of points to plug in and find C.

i advise you to take that information and accept that its true

[vast tapir]

not really, we have an initial value which we know. lets say i know that a ball starts at velocity = 2m/s. and i know that the acceleration is given by a(t) = 2t - 3. we need to find the displacement at t=5s, how would you find it

[ocean fossil]

But in this case we know that the correct solution is -cx. So if you add +C, from this comment, you are adding it to find -cx which we've already found

[crude parrot]

c(x) is a function. we don't know if theres a constant to be added.

[vast tapir]

here you have to take into consideration the constant of integration
try and see what happens if you dont

ig this is an example of where the "particular solution" is not correct

[ocean fossil]

t=5m/s?

[vast tapir]

my bad, thats in terms of velocity. t = 5 seconds here

[ocean fossil]

I mean you just integrate it, put 5, add 2

[vast tapir]

displacement, not velocity

[crude parrot]

you are given a(t) = 2t-3, and v(0) = 2 (from what i understand)

not the best in physics

where v'(t) = a(t)

[vast tapir]

yeah

assume the ball is at the origin ofc (forgot to mention that haha)

[ocean fossil]

I don't see it

I'd just... Integrate twice

Then add 2•5

[crude parrot]

you have to plug in v(0) = 2

because $\int a(t) dt = v(t) + C$

[ocean seal]

Krish

[ocean fossil]

?

no

=2

[crude parrot]

yes whatever you get the point though

[vast tapir]

if you ignore v(0) = 2. then you just get v(t) = t^2 - 3t. then if you integrate again, you get s(t) = t^3/3 - 3t^2/2. however if you remember, v(0) = 2, so you need to plug in v(0) = 2, and the constant C = 2. so v(t) = t^2 - 3t + 2. now if you calculate the displacement, s(t) = t^3/2 - 3t^2/2 + 2t. see how different the solutions are

[ocean fossil]

I can do that

But you can also just... Multiply

And add

[crude parrot]

??

show how you would do the work for this

[ocean fossil]

speed times time is distance

😭

[ocean fossil]

integrate twice, plug in 5, add distance from 2(5)

[crude parrot]

you cant use that in this case

[vast tapir]

i mean you can

[tepid drum]

Btw this is only true when the speed is constant

[crude parrot]

its not uniform motion though

[ocean fossil]

there is acceleration

[tepid drum]

You will need to modify it a bit

[crude parrot]

the velocity changes over time

[ocean fossil]

That's why we integrate the eqn lol

[tepid drum]

So you can't use it (that's my point)

[crude parrot]

thats what we just said

[ocean fossil]

I asked where he, talking about the "particular eqn" being wrong

[vast tapir]

the particular equation is v = t^2 - 3t. (particular meaning ignore constants)

[ocean fossil]

you don't need to plug the v into the eqn

and also that wouldn't be pure integration, you're adding other conditions

[vast tapir]

exactly my point

[crude parrot]

you would end up with an incomplete solution that doesnt match the situation given in the problem

[ocean fossil]

in this case the integral isn't the solution, it's the displacement lol

[vast tapir]

in real life, we have initial value problems. i know that the function satisfies this, i know it is equal to this at this input. find me a function

[ocean fossil]

I'm not sure why you're writing it in terms of mechanics when you can write it in terms of graphs

[crude parrot]

if you skip. using the constant of integration you wont get the correct solution because youll miss the information that makes the solution unique to this particular problem

[vast tapir]

yeah its integrating twice

also wdym

i got a function which satisfies what i needed, i just call it displacement

its still a solution to the given criteria

[ocean fossil]

ie I'm asking why you said it in terms of physics and not just graphically

[vast tapir]

write it graphically, you can

[ocean fossil]

It's a solution to a different question

[vast tapir]

a = 2t - 3 would be a straight line. v would be a quadratic. the constant of integration would move the graph up and down, influencing how s(t) would look

[vast tapir]

how?

[crude parrot]

it would change the area underneath the graph.

which is what integration finds for us

[ocean fossil]

Since you defined the terms, and are adding conditions that aren't explicitly integration, there will be variations. That means even without adding a constant of integration, simply just adding V, would work

And in this case it isn't completely the same as the Q I sent but yeah

[crude parrot]

this isnt tying back to exactly the question in that problem set you sent. it ties back to why we use the constant of integration and why it makes a difference

[ocean fossil]

But logically you only need to add it.

If I already had a set integral that ALREADY had the 2 then I don't need a C

[crude parrot]

wdym, can you write out that integral

[ocean fossil]

I think this is just a case of "We mathematicians want to always have +C so we'll keep it that way"🤷🏻‍♂️

[vast tapir]

the C is 2...

[crude parrot]

no, we gave you several examples of why we use the constant and you refuse to accept it. this has nothing to do with what mathematicians "want" to do. thats how math works.

[ocean fossil]

Yes by deductive reasoning... You can use that same reasoning and not put it into the integral🤷🏻‍♂️

[safe reef]

Can someone help me

[crude parrot]

exactly, then youre still using C, just writing it a step earlier

[crude parrot]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[vast tapir]

the thing is... the deductive reasoning is also just integration

[ocean fossil]

Not when we already have the particular sols

I get the C aids in finding the particular sols. the question is why add +C when we can already define int of f'x to be fx

[vast tapir]

we have conditions the solution must meet. adding the constant makes it meet those conditions

[crude parrot]

we dont already have the particular solution in this case. you get the particular solution from plugging in 0 and 2

a particular solution isnt the points. after you plug in the points, you are able to derive the particular solution

[ocean fossil]

then that just proves it is tangent to my Q 🤷🏻‍♂️

[vast tapir]

i mean you did ask where it is useful to now just consider f(x) without the C

thats what we answered

[crude parrot]

yeah

[ocean fossil]

Yes I know I'm not sure why you're repeating since earlier

[crude parrot]

you are asking the same question over and over again in different words.

[ocean fossil]

I think you meant the opposite

[vast tapir]

well the integral of lets say x^2 is x^3/3, and also x^3/3 + 1, x^3/3 + 2... so unless 1=2, we include the +C as any arbitrary constant

[crude parrot]

plus we already answered this a WHILE ago.

[vast tapir]

i do mean the opposite, my bad i typed it wrong

[ocean fossil]

ok so the context behind the statements I gave was that in the Q, it should be implied that -cx is the particular sol to the int.
But since Krish said +C is required, I am asking where it is ever useful in the sense that a particular sols is not required.

In the case you provided, the "particular" sols is t^3/2-3t^2/2. Except you added a +2 in the first integral which changed the original particular sols. Instead of writing +C and then filling it in, why not just... Not write it, then add it due to other set conditions?

If I have 3a and someone gives me 5b, it doesn't change the fact that I originally had 3a. And neither did I write 3a+C just to prepare for the fact that I would obtain 5b. In this analogy, the 3a is the integral and the 5b is the added +2.

I get that you can add

[pearl kraken]

Are you just asking why +C was in the solution?

[ocean fossil]

yes

[crude parrot]

thats the same reason as the tens of reasons we gave earlier.

[pearl kraken]

Uh I mean usually I wouldn't consider those solutions to be the perfect solutions but in general there's just many reasons why they have + C and I wouldn't be too worked up over that

[ocean fossil]

Did you read the last paragraph? Can you tell me where I stopped being sensible

[crude parrot]

can you scroll through when you first asked for help and read the entire conversation and process it? if not, please watch a video on why the constant of integration is important.

[pearl kraken]

They just follow the convention and have + C unless you find the an actual direct value for constant

[crude parrot]

im not sure why you are refusing to believe that the constant of integration is a real mathematical concept

[ocean fossil]

ehhh you keep repeating the same thing and I keep agreeing. It's just that it's parallel to my question

[ocean fossil]

Never said that

[crude parrot]

you said this^

[ocean fossil]

I literally said earlier about integrating fx before and after you came lol

[pearl kraken]

Are you saying that the solution had + C redundantly

or do u mean something else

[ocean fossil]

The context

[ocean fossil]

This is the same thing the others have said but it still isn't COMPLETELY relevant to what I'm asking. If I already have a solution then how come we write +C? +C is added to find the particular solution. If I use the vel/acc the other person gave, he used v=2 as his C. Except this wasn't the particular solution. Rather, he just added +2 because... it makes sense. When I have 3 apples, I don't say "I have 3apples+C" just in the slight off-chance that someone could come and give me another one

[pearl kraken]

Its convention because mathematicians are precise

ik what you mean tho

[ocean fossil]

that was what I asked a little earlier due to needing to keep circling around. Is it another case of "I'm a mathematician and I want to write it that way"

[pearl kraken]

Yeah it pretty much is because they want to be precise with their solutions

even if you don’t have the + C

it’s not a big deal

its just a small detail for precision

if that makes sense

[ocean fossil]

It very much is in an admission test

[pearl kraken]

Yeah but don’t worry about it too much because nobody actually cares about that stuff that much to weigh it against you

you’re putting too much notice on something that is just convention

even if you don’t include it, as long it’s correct it’s not going to be a big deal

[ocean fossil]

I was asking the reason behind and they kept saying there was a use but never actually gave the one in my specific scenario🤷🏻‍♂️

[pearl kraken]

Well don’t worry about the reason because there might not actually be a good reason it’s a waste of time

[ocean fossil]

Learning the reasoning behind something is never a waste of time in my book

[pearl kraken]

Ok you can keep beating your head over + C if you want but im just letting you know its not worth the time

[crude parrot]

actually most professors will mark it incorrect if you don't include the constant

[pearl kraken]

yeah idk they don’t really care at my uni

[ocean fossil]

"fuck this guy, he forgot +0"

[pearl kraken]

you’re not failing an exam over a + C it’s just not worth the time to process and argue about why it’s there when you can just remember to put it

[ocean fossil]

"I'll take away all his 6 marks"

[crude parrot]

its not about the grade, its about just writing it down. whats the harm in drawing three more strokes with your pencil? theres no point arguing over that

[alpine sable]

can someone help me on algebra because i have trouble finding what X/ what ever my variebal may be in any case

plssss

help me

[ocean fossil]

.close

[lone heart]

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[ocean fossil]

Ask help in the open help channels

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[frosty gale]

How do i solve dis

[lone heart]

@frosty gale Has your question been resolved?

[crude parrot]

if you rotate that 90 degrees it looks like the x^3 function

this is the cube-root function, opposite of cubing

you can figure out if it is positive or negative, and then figure out how it is moved across the graph so that the point that is normally on the origin is now at the point (-2, -1)

[lone heart]

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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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[zenith flax]

okay so

do i get the area from 0 to -39/4 of 5y-y^2 (right) and y^2-8y (left), then get the area of 0 to 5 of y^2-8y?

@buoyant saddle you around to help brother?

[buoyant saddle]

hello

[buoyant saddle]

why?

just integrate dy

[zenith flax]

oh wait

i broke my own head

ignore me entirely

i just realized i started rotating about the y axis by accident

@buoyant saddle wait but

i think waht confused me is

is blue considered right?

wait

Is this a top-bottom problem or right-left

[buoyant saddle]

blue is 5y - y^2

[zenith flax]

t-b right?

[buoyant saddle]

right left

i mean you could do top bottom

but why

[zenith flax]

thats what i figured, but i cant tell which is right as red goes further out than blue

[buoyant saddle]

^

[buoyant saddle]

wdym?

[zenith flax]

like

[zenith flax]

this is a right - left problem

but aren’t both of them right at one point?

wait

is blue considered left and red right?

because of how far they extend?

or is it just for the bounds

[buoyant saddle]

🤔

i think you’re confusing dx with dy

[honest nacelle]

within the region, blue is always right and red is always left

[zenith flax]

i see

[median pier]

hi

can anyone pls explain what is parabola??

[lone heart]

@zenith flax Has your question been resolved?

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[fair crest]

I am getting an answer of -1/3i + lamda(1/3i + j)

[fair crest]

AM i doing it wrong here because i have no idea how the answer above comes

[strange schooner]

There's infinitely many vector equations for a single straight line

[pseudo ice]

Theirs is by e.g. noting that (x, y) = (x, 3x + 1) = ... = (0, 1) + x(1, 3)

[fair crest]

Oh so both mine and his is correct?

Does that also mean that here it could be r = -2i + 3j + lamda (4i - 5j)

[pseudo ice]

For which one, (i) or (ii)? Presumably the second one?

[fair crest]

Yes second forgot to mention

[pseudo ice]

(the direction vector is equivalent at least)

[fair crest]

Oh i see

[pseudo ice]

And yep, that would be fine too

[fair crest]

Its just i have a hard time figuring whats wrong and right when it comes to this

Any advice for that?

[strange schooner]

You define the line by a point and a vector

So, you just gotta check that
1): the point you chose is on the line
2): that for any given λ, that point + lambda times the vector is still on the line

As long as that happens, you've got a correct equation for the line

[fair crest]

So i have to be confidant in my method

Thanks guys

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[zenith flax]

🆘

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@zenith flax Has your question been resolved?

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[lone heart]

@white sparrow Has your question been resolved?

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[feral trellis]

Is there any shortcut to simplify
(x-y)^n where n≥3

[iron roost]

there are identities

the general form is x^n*y^0-x^(n-1)*y^1*...*x^1*y^(n-1)±x^0*y^n

so basically take x and y, their powers combined have to equal n, increase the power of x by 1, decrease the power of y by one each step, alternate the plus and minus

[gritty bramble]

binomial theorem

(x-y) -> (x+(-y))

[lone heart]

@feral trellis Has your question been resolved?

[vale wigeon]

misleading

[feral trellis]

I see

[feral trellis]

What do u mean

[lone heart]

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[vale wigeon]

.reopen

[lone heart]

✅

[vale wigeon]

hard to read + misses the binomial coefficients

also how come you closed the channel right after asking me that

[feral trellis]

I pressed the check emote from the bot before i noticed ur message

[vale wigeon]

k

[feral trellis]

May I ask how would you do it, cuz i do not understand it

[vale wigeon]

ok let's see

does the word "pascal's triangle" ring any bells to you

[feral trellis]

Nop

[vale wigeon]

ok hm actually

alright let's first consider the expansions of (x+y)^n for various n (we'll take care of the minus sign eventually)

do you know how (x+y)^2 expands?

[feral trellis]

Yes

x²+2xy+y²

[vale wigeon]

ok, and (x+y)^3 ?

[feral trellis]

Nop

[vale wigeon]

well, try multiplying out (x^2 + 2xy + y^2)(x + y) carefully

see what you get

(after collecting like terms obv)

[feral trellis]

x³+3x²y+3xy²+y³

[vale wigeon]

ah shit i forgot the 2 and you ran w/ it

my bad

[feral trellis]

Oh i didnt noticed

[vale wigeon]

redo that & collect like terms

uhh wait

how did those fours happen

i can tell you right away that's wrong

[feral trellis]

2x²y+2x²y

[vale wigeon]

how did you get 2x^2y in two ways?

[feral trellis]

that

let me redo mb

Oh

[iron roost]

I did forget the coefficients

[vale wigeon]

(x^2 + 2xy + y^2)(x + y) = x^3 + 2x^2 y + xy^2 + x^2 y + 2xy^2 + y^3
= x^3 + 3x^2 y + 3xy^2 + y^3

[feral trellis]

i forgot that there was a time when x²*y

[vale wigeon]

x^3 is not getting mulitplied by y here

[feral trellis]

i mean x²

[vale wigeon]

mm

anyway ok so like

$(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$

[ocean seal]

Ann

[vale wigeon]

this is the correct expansion

if you do it in a similar way (but properly --- and hopefully on paper and not direclty in discord) you can get $$(x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$$

[ocean seal]

Ann

[vale wigeon]

https://en.wikipedia.org/wiki/Pascal's_triangle

[feral trellis]

oh the amount of coefficients is n+1?

[vale wigeon]

well the number of terms is n+1 yes

as you can see even for n=2 which has three terms

[feral trellis]

Thats helpful

[vale wigeon]

in general, you can use something known as pascal's triangle -- an array of numbers constructed by the rule of "1's down the diagonals, and each number inside is the sum of the two numbers above it"

[feral trellis]

My wifi is trash

[feral trellis]

But i think i know that one now from that explanation

Ok the other concern i have is how is the coefficient relates to the (x+y)^n

like how is ^4 have 6 in the middle

and ^3 just have 3

[vale wigeon]

higher powers have more shit in the middle

[feral trellis]

Understandable

ight thx ima go back to class

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[rustic crest]

from the echelon form, I can that it's linearly independent. However, how do I find what is asked in part (d)?

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[paper cipher]

how e

[vale wigeon]

is that $\frac{\absv{x - |x|}}{x}$?

[ocean seal]

Ann

[paper cipher]

Yes

[vale wigeon]

ok, what answer do you think it should be?

[paper cipher]

d

[vale wigeon]

why?

[paper cipher]

put any number in there

[vale wigeon]

let's try to work it out together shall we

[paper cipher]

ok

[vale wigeon]

what happens if x = 6

|6 - |6||/6 = ?

[paper cipher]

0

[vale wigeon]

is that a positive integer?

[paper cipher]

no

[vale wigeon]

then how can D possibly be the right answer?

[paper cipher]

what if negative

[paper cipher]

mb

[vale wigeon]

ok, so what if x is negative

what'll |x| simplify to?

[paper cipher]

2

[vale wigeon]

just |x|.

[paper cipher]

-x

[vale wigeon]

is |x| = 2 or is |x| = -x?

[paper cipher]

6-(-6)/6

[vale wigeon]

...

[paper cipher]

-x

[vale wigeon]

i would prefer if you answered my question the first time around

yes, |x| = -x.

so what does |x - |x|| simplify to?

[paper cipher]

|2x|

[vale wigeon]

right, and that simplifies further to...?

[paper cipher]

2

[vale wigeon]

|2x| = 2 always?

[paper cipher]

2x

[vale wigeon]

no, |2x| ≠ 2x.

[paper cipher]

or -2x

[vale wigeon]

remember x is still negative

do not try to skip steps or run ahead of me.

right, |2x| = -2x.

and when you divide that by x, what do you get?

[paper cipher]

-2

[vale wigeon]

right

and is -2 a positive integer?

[paper cipher]

no

[vale wigeon]

so, does this answer your question as to why the right answer is E?

[paper cipher]

yes thanks

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[primal dagger]

i have to use the 10 axioms to see if its in the vector space or not. the answer key says that it is in vector space but just by doing axiom 1 wouldnt it be (1+1,y+y') = (2,y+y') which shows that axiom 1 fails?

[hushed locust]

no, because that's not how they defined addition

you have to base your answer on the definitions of vector addition and scalar multiplication provided

[primal dagger]

oh

ur so right

my bad 😂

thanks a ton

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[olive tundra]

Help

[olive tundra]

What am I doing wrong

I'm not seeing feedback

Someone help please

[vale wigeon]

what did you do for this problem thus far?

[olive tundra]

Ok so I've tried to do the similar sides thing

And tried to do Pythagoras theorem

But I think Im messing up

[olive tundra]

Could you let me know what to do

[vale wigeon]

hmm

ok, step 0 is to label all relevant points

on your diagram, please give each vertex of the square and the triangle a single-letter name

[exotic canopy]

spoiler: i have done this problem before and the final result is pretty disgusting, iirc it's some root of a quadratic or a quartic

so it's not super nice but definitely doable

[olive tundra]

Please show me

[exotic canopy]

do as Ann said first

[olive tundra]

Can I do 6/y = 6+x/6+y

Or is that wrong

[vale wigeon]

didn't do as i asked

[olive tundra]

?

You said label tho

[vale wigeon]

on your diagram, please give each vertex of the square and the triangle a single-letter name

[exotic canopy]

label points not lengths

[vale wigeon]

please read my message in its entirety and not only 10% of it

[olive tundra]

Ok give me a second

Forgive me but it's hard to do on a phone

Also I can't find a pen

[vale wigeon]

those should have been uppercase letters

but ok.

i think we can bring back your side labels of AZ = x and CY = y

[olive tundra]

Yes

[vale wigeon]

note that triangles ZAD and DCY are similar

this gets you that y/6 = 6/x

and applying Pythagoras to triangle BZY gets you (x+6)^2 + (y+6)^2 = 400

[olive tundra]

Quick question

Is it possible to use ZBY too?

[olive tundra]

Yes this is where I got stuck

Could I say y=36/x ?

[vale wigeon]

wdym by "use ZBY"

[olive tundra]

The bigger triangle

The one with a hypotenuse of 20

[vale wigeon]

i named it here

[olive tundra]

Ah my bad

[vale wigeon]

are you asking if it is OK to name it ZBY instead of BZY? yes it is

[olive tundra]

Alright

So back to the equation

We will have to substitute one of the values right?

[vale wigeon]

yes, and this is what i would suggest

[olive tundra]

But doing that will give u a very complex equation

[olive tundra]

But I'll listen to you

So what next

[vale wigeon]

yeah, 4th degree

no way around that unfortunately

[olive tundra]

Shoot

Ok so how will we simplify it

For us to get x

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@olive tundra Has your question been resolved?

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[idle gate]

hi, just wondering how to do this?

and verifying if my aswer is correct

i got 85096 ppl

.close

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[idle gate]

.reopen

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✅

[idle gate]

<@&286206848099549185>

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@idle gate Has your question been resolved?

[idle gate]

<@&286206848099549185>

[left isle]

can you show your integration

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@idle gate Has your question been resolved?

[idle gate]

Hi, I just put it into my calculator

Please ping me when u reply

[velvet flame]

ur integration is correct but in the question it is given that there were 1000 people when t=18 but u took 100
so make the calculation acc to that

[idle gate]

I checked n when I put it into the calc I put 1000

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[gleaming jacinth]

ive got a question regarding lstms if anyone can help me

[twin nimbus]

@gleaming jacinth hi, it's generally better to just ask your question. I might have been able to answer it with this message.

[gleaming jacinth]

alr

im not very good with latex so please excuse my incompetence

when is the weight matrix applied to the input, is it after the concatination or are there two separate matrices?

im very new to lstms and i dont fully understand how (and when) the weights and biases are applied

[twin nimbus]

Here's a block diagram of LSTM

[gleaming jacinth]

yep thats all im familiar with

well not all i suppose

but a lot of it

[twin nimbus]

So the four yellow boxes are four different perceptrons.

[gleaming jacinth]

alr

[twin nimbus]

or fully connected layers, I suppose.

They combine the short term memory output of the previous block with the input of this block.

[gleaming jacinth]

through concatenation?

[twin nimbus]

No, via matrix multiplication

with your weights

that's what it means to be a fully connected layer.

[gleaming jacinth]

ok

matrix multiplication between x_t and h_t-1?

[twin nimbus]

oh, I see what you mean now. No, those are concatenated

the matrix multiplication is your weights matrix.

[gleaming jacinth]

ok, and theyre matrix multiplied before or after the concatination?

yeah i figured

[twin nimbus]

It's w(x_t + h_{t-1}) + b

[gleaming jacinth]

oh just the previous hidden state?

ah nvm

how is the dimensionality of the weight matrix and bias vector determined

[twin nimbus]

if your hidden layer is hx1, and your input is gx1, then your bias vector is going to be hx1 and your matrix will be h x (h+g)

iirc

[gleaming jacinth]

?

[twin nimbus]

if I recall correctly

[gleaming jacinth]

ah

[twin nimbus]

those are the dimensions that make the matrix add and multiply work at least

[gleaming jacinth]

and this is of course assuming that all of your x_t+n's are all the same dimension?

@twin nimbus is it the same weight matrix for each gate?

[twin nimbus]

No, different ones

[twin nimbus]

It would be very difficult to handle if they weren't.

[gleaming jacinth]

during which parts of the cell are they used

[gleaming jacinth]

yeah... figured it'd be

like before each activation function, just sigmoid, just hyperbolic tangent?

[gleaming jacinth]

so x_t and h_t-1 are both the same length? considering h_t-1 is constructed using x_t-1, they must be the same length, right? and if so, what would h-t-1 and c_t-1 be equal to for the first cell, just [0,0,0...] for the length of x_t?

[twin nimbus]

no sorry plus in the middle there means concatenation

I just didn't have circle plus easily available on the keyboard, and I thought it would be understood from context.

[gleaming jacinth]

oh ok

[twin nimbus]

my bad.

[gleaming jacinth]

nah that's my fault

so the length of h_t-1 and x_t does not matter?

[twin nimbus]

your hidden state is however large you want to make it.

and your input is however large is appropriate for your input.

[gleaming jacinth]

how do i scale it (the hidden layer)?

add more cells, more gates?

[twin nimbus]

you scale it to be large enough to do the thing you want to do, but not too large that it makes it impossible to train.

[gleaming jacinth]

but how do i scale it

[twin nimbus]

I don't understand what you mean by scale it?

[gleaming jacinth]

unfortunately not

[twin nimbus]

???

[gleaming jacinth]

just add more cells?

[twin nimbus]

I don't understand what you mean by cells?

[twin nimbus]

let's do a toy example.

[gleaming jacinth]

each block i suppose

[twin nimbus]

oh

ok so let's say you're doing a language modeling task.

[gleaming jacinth]

im listening...

[twin nimbus]

your input is going to be tokens, and maybe your output will be a sentiment analysis.

So beyond your h_t you have, perhaps, a fully connected layer with a single output

your input is going to be a word embedding that you have to learn as well, so it takes a token and maps it to a, let's say, 384 length vector.

So each of your x_t is 384

and let's say your hidden state you've selected is 512 length vector

So we have the following input text:

"The quick brown fox jumped over the lazy dog."

This gets tokenized into:

"The | quick | brown | fox | jump | ed | over | the | lazy | dog | . |"

So the sequence of 11 tokens gets sent one at a time to your LSTM

"The" gets fed through your word embedding and becomes a 384 length vector

and this vector is X_0

h_{-1} is initialized to all 0s.

So your input to your first set of weight matricies is X_0 concatenated with h_{-1}

[gleaming jacinth]

what is the length of h_{-1}

384?

[twin nimbus]

512, the same length as all of your h vectors

[gleaming jacinth]

oh so it's a preset hyperparameter?

[twin nimbus]

yes

[gleaming jacinth]

kk

does this hyperparameter also effect c_t-1 or is that set separately?

[twin nimbus]

Because the operations in the cell combining h_t and c_t are pointwise, they need to be the same dimension

[gleaming jacinth]

by pointwise you mean elementwise right (not a correction im just clarifying)

[twin nimbus]

sorry, yes elementwise

[gleaming jacinth]

alr so after the concatination, then what

[twin nimbus]

then you multiply by the weight matrix w(h_{t-1} concat x_t)

which the weight matrix is 512 x 896 in this example.

so it takes your vector, which is 896 x 1, and turns it into a 512 x 1 vector

then we add the bias, which is also 512 x 1

[gleaming jacinth]

alr

[twin nimbus]

<@&268886789983436800> #help-0 message

❤️

Thanks again Hayley

[twin nimbus]

Did you have any other questions about this?

[gleaming jacinth]

then that's passed elementwise through sigmoid?

for the forget gate at least