Integrals

[lucid forge]

Hi, i have a question. So i think that indefinite integrals are not really integrals and i am apparently not the only one who thinks that way. So only real integrals are definite integrals based on that fact. I want to get some clarifications: Why are they denoted with the same symbol if they have completely different functions or operations?

[merry blaze]

Strictly speaking indefinite integrals represent a family of antiderivatives (family because of the constant you add at the end).

The Fundamental Theorem of Calculus relates the computation of a definite integral to the evaluation of an antiderivative at the bounds of integration. Since you (usually) need antiderivatives to compute definite integrals and indefinite integrals give antiderivatives, we use the same symbol without bounds to denote the process of generating such an antiderivative.

[lucid forge]

Ok, i think it makes sense.

[lucid forge]

But would it not make more sense if we called it differently? Instead of indefinite "integral", but somehow else? Can we make sense of this name?

[merry blaze]

I would argue they \emph{are} called differently. The sense in the name is exactly the sense given by the Fundamental Theorem of Calculus.

If you want more of a precise answer, the FTC also states that if $f$ is continuous the function $F(x) = \int_a^x f(t) \dd{t}$ is in fact an antiderivative of $f$ for a fixed $a$. This is sometimes not shown or avoided, but it is quite important.

The indefinite integral is the collection of all such functions for all choices of $a$, a.k.a. an \emph{indefinite} lower bound, so while they are technically different object, they're rooted in the same concept and it makes sense then to call them \emph{indefinite integrals}.

Here we're taking a to be a reasonable value that makes sense where f is continuous and all.

[night cave]

Azyrashacorki

[lucid forge]

Do you recommend first fundamentally understanding definite integrals before indefinite integrals or the other way around? Because what you are showing is definite integral if i am not wrong, but i do not quite understand how they work yet.

[merry blaze]

Usually you learn about definite integrals first because they're defined in terms of limits akin to how you learn about derivative first through their definition as limits.
In general, this limit isn't particularly easy or convenient to compute algebraically, just like the derivative isn't generally easy to compute from the definition, so you learn about the FTC which says that if you're able to get an antiderivative, then computation of the definite integral is very easy.
This shifts the problem of computing the limit to the problem of finding antiderivatives, but thankfully there's rules and tricks for that as well, which yield an indefinite integral - a family of antiderivatives - which you can then plug inside the FTC to compute your definite integral.

[lucid forge]

Alright, what are definite integrals? It is defined as limit for series? But again why we denote it ∫f(x)dx if we have lim(x to 0) big SIGMA (f(x)delta x) and delta x is a real number you shoud multiply with?

[night cave]

Azyrashacorki

Azyrashacorki

Azyrashacorki

[merry blaze]

I don't understand what you mean by "why" we denote it this way. Would you rather have to write the whole limit all the time?

[lucid forge]

No i am confused about why we have delta x in real series definition, but when we write dx at the right side of integral we never multiply anything with it

[lucid forge]

Why not just delta x = 1/n?

[merry blaze]

For all purposes at this level, dx is purely notation to explicitly state that this partitioning of the interval into smaller and smaller subdivisions is being done in the x-domain, i.e. that you're integrating with respect to x.

[lucid forge]

Ok, but again this notation for respect to x is misleading, is it not? Because we use also dx in other ways.

[merry blaze]

$\Delta x$ is the width of the rectangles. This is obtained by taking the interval $[a,b]$ and separating it into $n$ equally spaced subdivisions of length $\Delta x$, so each of those subdivisions has length $\frac{b-a}{n}$.

[night cave]

Azyrashacorki

[lucid forge]

Oh yes you are right

[merry blaze]

I would say it's helpful to state exactly what you're integrating with respect to. And in any case, you would never really have a dx floating around anyways so it doesn't strike me as ambiguous.

Moreover, it works well with Leibniz notation in the sense that things "cancel" the way you want them to if they were actual numbers (although this is justifiable without "cancelling").

[lucid forge]

But why we do not write as division this "respect to x" as for derivatives?

[merry blaze]

We do, in Leibniz notation.

The Newton notation is useful in 1-variable because it's often obvious what you're differentiating with respect to.

[lucid forge]

We do not for integrals, do we? We write ∫f(x)dx, not ∫f(x)/dx.

[merry blaze]

I thought you meant it was confusing because we have Leibniz notation for derivatives which we write with d/dx and such as well.
If you just have ∫f(x)dx what is misleading?

[lucid forge]

Well if we write d/dx for derivatives why not stay consistent and write ∫f(x)/dx also for integrals to say we integrate with respect to x? By misleading i meant that why we have dx that we never use in multiplication with anything?

[merry blaze]

The way the dx seems just multiplied there is because delta x is multiplied in the definition of the integral.

It wouldn't make sense to divide by dx if we've been multiplying by delta x in the definition.

And again writing it as something "multiplied" ends up making sense with Leibniz notation for derivatives in most cases.

[lucid forge]

Oh yes it actually makes sense. And speaking about multiplication, we never multiply, it is just a notation? But then in formal definition with limits, where does delta x disappear?

As for derivatives also, we have delta f(x)/ delta x where delta tends to 0, but then we get dy/dx, but we never use dx to divide anything when we differentiate?

[merry blaze]

You can think of it as the width of an infinitely small rectangle that we keep there just to signify what variable we were integrating against.

[lucid forge]

Maybe i will understand more if i understand how process of integration happens. So we take certain value a and put it into a function to get certain value and then multiply it by dx. Then next value which is infinitely close and so on until we reach b? So dx shows that we always multiply the result with dx? As for derivatives, we divide by dx every smallest function change?

[merry blaze]

It's a similar argument as to whether dy/dx is just notation or if it's a fraction.
It's clearly a limit of a fraction at least, and it's not different with integration, where you can think of dx as part of the limit of a product.

[lucid forge]

I am still confused with all that notation and fraction thing... How do all actions make even sense if we interpret is as a notation? And if we interpret it as a fraction, why do not we multiply or divide functions with these dx?

[merry blaze]

There's formal ways of defining what those are. In more advanced maths, you learn that f(x)dx is really something called a differential form, which comes with a meaningful notion of integration.
While you can't divide differential form literally, dy/dx can be made sense of in this context as well.

And again, regardless of whether or not you look at it as a fraction, the cancelling you do has a justification without cancelling. It just works out nicely.

[lucid forge]

So what you mean is that we can prove certain theorems and different things using notation and they look like cancellations?

[merry blaze]

Yes

Like how the chain rule looks like cancellation, but you can prove it holds without just relying on the fact it cancels.

[lucid forge]

Ah, yes it makes sense

[lucid forge]

But on the other hand about derivative notation: Can we treat dy/dx as fraction, because dy = f'(x)dx and it actually for some reason starts to make sense for me to denote differentation with this fraction, because when we write this fraction, it just means we differentiate with limit definition. So if we had dy = f'(x)du, then dy/du = f'(x) where u consists of x. However, how to understand dy/du = f'(x)?

But treating it purely as a fraction would justify proof of chain rule where you just multiply and divide by du and get the chain rule formula, would it not?

[merry blaze]

Yes, but to get to this point formally requires some amount of work which is inefficient compared to just accepting the fact that it's notation and that it, in most instances, works like a fraction.

Also dy = f'(x) dx makes sense mostly because in our notation dy/dx = f'(x).

[lucid forge]

No, but dy = f'(x)dx is a real formula and at least for me it makes sense and that is why i try to make sense of dy/dx from it.

[merry blaze]

It's a formula in the same sense that dy/dx = f'(x) is a formula.

[lucid forge]

Yes it makes sense now, but then why can not we just prove chain rule with multiplying and dividing with du if it is allowed to multiply by 1?

[merry blaze]

Because it's notation. Yes there is some meaning attached to $\dv{y}{x}$ that makes sense if we think of the derivative, and this is a good thing because it's very evocative of what we mean by a derivative, but you can't multiply by $\dv{u}{u}$ formally the same reason that you can't multiply by $\frac{(}{(}.$

At least not without arguing first that those symbols aren't just symbols but infinitesimals, and then you're in for a ride.

Or 1-forms

[night cave]

Azyrashacorki

[lucid forge]

I saw on reddit that people say dy/dx as a whole is just a symbol in reality and not a fraction... So we can really never use it as a fraction except for dy/dx = f'(x) definition?

[merry blaze]

You can use it like a fraction in most cases, but like I said those manipulations are purely convenient, they aren't formally valid unless you define what numbers dy, dx and du are.

It's just a nice notation after all because the rules we can prove often look like the manipulation of fractions.

[lucid forge]

If we multiply dy/dx with dx/dt will we get dy/dt or not? I think we should get it since they work like fractions, but why it is not the case for chain rule?

[merry blaze]

If x is also a function of t, then yes.

[lucid forge]

So why can not we do it in chain rule?

[merry blaze]

You can

$\dv{f}{t} = \dv{t} (f(x(t))) = f'(x(t)) \cdot x'(t) = \dv{f}{x}\cdot \dv{x}{t}.$

[night cave]

Azyrashacorki

[lucid forge]

What does f(x)*dx mean in math? Because apparently infinity these kinds of multiplications get summed to get an integral?

[merry blaze]

f(x)dx is what's called a 1-form. It's quite an abstract concept but it works out to what an integral does. The closed interval is what we call a "manifold with boundary" and in this particular case a theorem called Stoke's Theorem reduces down to what we know as the FTC.

Informally, you may think of it like the area of an infinitely slim rectangle and summing over those gives the area under the curve.

But at the end of the day it is just notation. $\int_a^b f(x) \dd{x}$ is shorthand notation for $\lim_{n\to \infty} \sum_{i=1}^n f(x_i^*) \Delta x$.
Just like $\left.\dv{f}{x}\right|{x=a}$ is shorthand for $\lim{x\to a}\frac{f(x) - f(a)}{x-a}$.

[night cave]

Azyrashacorki

[lucid forge]

But how can i intuitively understand f(x)dx without geometric interpretation(infinitely slim rectangle), that is, with numbers and logic?

[merry blaze]

See the issue is that this notation is so suggestive that it's hard to think of it as just notation, but it is at this point.
Your question is akin to asking what the l in lim means intuitively with numbers and logic. It's just there because we chose it.
f(x) dx doesn't "logically" mean anything on its own.

[lucid forge]

So it was basically created to solve physics problems? To multiply certain variables and sum them up? Because why would mathematicians need random meaningless function * dx sum?

[merry blaze]

It's notation that turns out to fit into a scheme in which the rules look like operations on fractions.

[merry blaze]

Yes. For instance, if you have a unit length wire that has a certain density at each point in [0,1] given by d(x). Then d(x)dx can be thought of as an "infinitesimal" mass at a point. Summing continuously over [0,1] you get the mass of the wire.

In its infancy calculus was based on infinitesimals, but it was clumsy and not very rigorous. The notation is vestigial in that sense.

[lucid forge]

So integrals were only invented for practical purposes. So how is it useful in math? Is it just math thing only meaningful for other sciences?

[merry blaze]

Integrals emerged from a need to compute areas and volumes. This definitely has physical applications.
It doesn't mean they are exclusive to physics and natural sciences.
There's tons of applications to probability, analysis, ODEs. They're used extensively in more abstract maths as well. For instance, in complex analysis integrals are very useful to compute some functions. In geometry you can tell things apart with how integrals behave on more abtract spaces.

[lucid forge]

I just realised that integration is just approxmation of area, but it does not calculate exactly? So why is it even useful if everything before was precise like limits and derivatives, but now intergrals are not precise at all?

[merry blaze]

The approximation is the finite sums.

The limit is not an approximation

[lucid forge]

But imagine small rectangles as areas below the curve. There is always space between 2 small rectangles at the top, no matter how slim they are?!

[merry blaze]

That's why we take the limit. It takes an approximation to an exact value.

[lucid forge]

But even with limit, if that triangle blank area gets infinitely small, there is still this area and it is not counted?

[merry blaze]

There is no blank area in the limit. The whole point is that this over/under-estimation has an error which vanishes in the limit.

Limits aren't approximations. They give a precise number which is fixed, regardless of the context.

It's the same reason why the derivative gives the exact slope of a function at a point even though we can't compute directly the slope between a point and itself.

Here we would like to compute a Riemann sum where the width of the rectangle is 0, but that's not possible.

The limit does it.

[lucid forge]

If at infinity small scale from 1 point to another it is just a straight line then why not use trapezoid area formula: (f(x1)+f(x0))/2*(x1-x0)? It would actually be exactly precise?

Wikipedia says that trapezoid rule is just an approximation.......

[merry blaze]

It is, because you're never at an infinitely small scale, so the area of the trapezoid you compute is never exact.

Just like the usual rectangle Riemann sums it's only exact in the limit.

[lucid forge]

But then how rectangles are different from trapezoid example?

[merry blaze]

At an infinitely small scale even, there isn't a trapezoid at all. It would not have a height.

The trapezoid rule just like the left and right endpoint rules for finite Riemann sums is an approximation because they are finite/

[merry blaze]

They aren't different. You're talking about different approximations of the area.
Taking the limit makes those approximations not approximations anymore.

[lucid forge]

Why taking limit is no longer approximation if even at infinity small scales we know there is empty area not counted in final area?

[merry blaze]

There is no empty area at the infinitely small scale, because it is infinitely small.

[lucid forge]

But no matter how small it is, it is there?

[merry blaze]

It is there if you stop at some point. The limit does not stop.

You can show mathematically that there is no empty area / error in the limit.

[lucid forge]

But here is a problem: If it never stops then there is always area that is empty that it should fill, but if it stops then there is also empty area. So in both cases there will always be empty area? Am i missing something?

[merry blaze]

It's not a process that takes time. It "never stops" in the sense that you're not fixing some n when you stop subdividing the rectangles. By taking the limit you're computing directly the result.

[lucid forge]

I read about definite integrals and finally understood them i think. So since they are defined as limits, it shows what value area under curve converges to! But it lefts me with final last confusion: If it converges to this value, how can we say it is exactly this value(the area)? Does contiunity explains this?

[sinful obsidian]

bro

[lucid forge]

What?

Thanks @merry blaze for helping me again!

.close

[obsidian tendon]

Solved

Post marked as solved by @lucid forge.

Use .unsolved if this was a mistake.

[merry blaze]

To reply to this quickly, it's exact in the same way that 0.999999999... is exactly 1. There is no real number between 0.99999999... and 1, so they are equal.
Similarly, there is no real number between 0 and the error in the area under a graph given by an integral, so the error is just 0.

[lucid forge]

I think 0,999999 is not exactly equal to 1, because then we would then be able to show 1 is equal to 1000, right? However i think i understand why there is no empty space between. So since they are so close, but not the same point(same point would mean distance 0 between them), there is no empty space.

[merry blaze]

You wouldn't be able to show that 1 is equal to 1000 because 500 is a real number strictly between 1 and 1000.
0.9999999... and 1 aren't "close" points. They are the same point. The distance between them is exactly 0.
The key point in analysis (in real numbers in general) is that "arbitrarily close" means equal. A limit is arbitrarily close to a value because it equals this value.
If you have some number x for which it is true that 0 <= x < epsilon for any given epsilon > 0, x is not "close" to 0. It is 0.

[lucid forge]

But logically if 0,99999... is equal to 1 then it means that 1,00000....1 is equal to 1 and 1,00000000...2 is equal to 1,00000...1, therefore also equal to 1. So we just found 3 numbers equal to each other?

Also if we looked at arithmetic series of number 1 and 2, but the one that has 2 as first number has difference 0, but the one that has number 1 has difference infinitely small number. So by definition of arithmetic series, if we plot these series on the same graph, the graph, which started at point 1 will cross point 2 at some point. So 1 = 2?

[merry blaze]

1.0000000....1 is not a thing. You're adding 1 at the end, but there isn't an end to the 0s. You're effectively adding 0.

[lucid forge]

Ok, then take example with 0,9999999... and it will be the same thing?

[merry blaze]

No, because here you're not adding anything at the end. The expression in its binary expansion is specifically $\sum_{n=1}^{\infty} \frac{9}{10^n}$.

[night cave]

Azyrashacorki

[lucid forge]

So you are saying that 0,99999... is the same as 1, but there is no such other number equal to 1? But since even 0,9999... is written differently then how can it be equal to 1?

Is 1,9999999... equal to 2?

[merry blaze]

Yes. It is a quirk of pretty much any expansion (in any base) that real numbers don't usually have unique representations.

[lucid forge]

I found on wikipedia proof and it convinced me by saying that in 0,999...9 9 never ends, therefore there is no distance between it and 1, so they must be equal... But it still sounds so unintuitive, because they look different, they should be different by intuition.

Are real numbers just not good for math?

[merry blaze]

This sits at the core of what it means in analysis to formalize notions of "arbitrarily close" things. Limits, and therefore derivatives and integrals are based on this idea.

The fact that 0.999999999... never ends and is equal to 1 doesn't mean real numbers are bad.
To the contrary, it specifically means there are no "holes" on the real number line, which is very important. It means the real number with its usual notion of distance is complete.

[lucid forge]

So i also saw an argument: If there is no number between 2 numbers then they must be equal. But in that case, how can we ever have a number and a next number without them being equal?

[merry blaze]

The whole idea is that there is no meaning in talking about a "next" number to a given value on the real line.

There is a continuous stream of numbers. You can't list them in order.

In fact this fails even in the rational numbers.

[lucid forge]

Ok, but even if there is no other number between 0,99999... and 1, they can be interpreted as next numbers, can not they?

[merry blaze]

No, they can't. They are two representations in decimals of the same number : 1.
Again the notion of "next" doesn't work in Q, let alone R.

[lucid forge]

So 1,00000...01 does not exist, because there is no position to put 1 at after infinity zeros? If so, then what is next number after 1? At least how can we describe it?

[merry blaze]

We can't. A "next real number" doesn't mean anything.

[lucid forge]

But is it not logical to at least describe next number after 1 to be something like 1,(0)n1, where n is arbitrary large number?

[merry blaze]

No because the number you're describing is still 1. Adding 10^{-n} for an arbitrarily large n is the same as adding 0.

You're essentially saying that $1 = 1 + \lim_{n\to \infty} \frac{1}{10^n}$.

[night cave]

Azyrashacorki

[merry blaze]

Which is true, the limit is 0.

[lucid forge]

So i think i understand this infinity 0 logic now: if we write infinity decimals after 1,... then there is no space for any other number. However, what if we talk about different size infinities now?

Well AI argues that 0's after decimal have no larger or smaller infinites, but there exists just 1 infinity, so there is no space.

[merry blaze]

Aside from the fact there is no meaning in doing this for larger infinities, this is already using the smallest infinity we have.

[lucid forge]

So 2 numbers are equal if and only if their difference is 0, that is a - b = 0, then a = b. So since 1 - 0,99999... = 0,0000... then also since after infinity 0's there can not be a number, then their difference is 0, therefore 1 = 0,99999..., right? So in philosophical sense they are like "different" numbers(no doubt why we write them differently), but their "effect" or value is the same?

[merry blaze]

They are two representations in decimals of the number 1.

[lucid forge]

So any number have 2 representations in decimals?

[merry blaze]

Yes

[lucid forge]

But all these facts are only true due to axioms we set?

[merry blaze]

They're true in any base.

In binary 1.11111111111 = 10 = 2

[lucid forge]

If 1 can be represented as 0,99999..., then there is another number that looks smaller than 0,9999... also like in 1 and 0,999... situation, but can be used to represent 0,99999...? But this means we can represent any number with any number?

[merry blaze]

There is a difference between a number, namely 1 and its decimal representations, namely 0.99999999... and 1.0000000...

The representations differ. The number they represent doesn't.

0.99999999... is already the number 1.
You can't represent 0.99999... because it's already a representation of some number.

[lucid forge]

If we can represent a as b and c and d if a=b=c=d, then why not with 0,9999...?

[merry blaze]

I don't understand what you're asking, sorry.
0.9999... is a decimal representation of the number 1.
If you're finding another representation for 0.999999..., then that's just another representation of 1.

[sinful obsidian]

hi

[lucid forge]

Like if we can represent 0,9999... as 1, then why can not we represent 0,9999... with number that by notation looks smaller, but is equal to it?

[merry blaze]

Because 0.99999... is 1.

Any other representation of it would be a representation of 1.

And hence either 0.99999... itself or 1.000000...

[lucid forge]

No, i mean if we had something like 0,9999..98, then it is equal to 0,9999... but their notation is different?

[merry blaze]

It's not because you're doing something "at the end"

[lucid forge]

But you said yes to this, therefore must it not mean that 0,999...98 also have a its own representation number like 0,999...?

Nvm i get it, if number really "ends" with 8 then it must have infinity 8's, but then it would be 0,999888..., which is definitely not equal to 0,999..., right?

[merry blaze]

Putting an 8 at the end is the same as adding 0

Just like putting a 1 at the end of 1.00000000..... is the same as adding 0

[lucid forge]

I think its rather that we can not possibly put that 1, because there are infinity 0, because if we were able to put that 1 then it would literary have 1 in it not 0?

[merry blaze]

Well those are two things that mean the same

You can't change the last 0 in 1.000000... to a 1 because there is no last 0.

Similarly you can't change the "last" 9 in 0.99999.... to an 8, because there is no last 9.

[lucid forge]

Yes, but about last thing i said: I meant if i really want a number to "end" with 8, then only possible solution is to make from certain decimal all next decimals equal to 8 or just infinity 8's.

[merry blaze]

Yes, if you want the decimal expansion to "end" in an 8, you need infinitely many 8's at some point.

[lucid forge]

If we talk about limits then since 0,999... = 1 then when limit converges to 1, it also never reach 0,999... at any point when converging?

[merry blaze]

$0.99999\ldots = \sum_{n=1}^\infty \frac{9}{10^n}$

[night cave]

Azyrashacorki

[merry blaze]

0.9999... is defined as a limit.

And the limit is 1.

[lucid forge]

But limit tells what it converges to, but never reaches. So if 0,999... = 1 then if we look this process of limit, then 0,999... is never seen in this process of convergence?

[merry blaze]

0.99999... is the result at the end. It's not "in the process" of happening

[lucid forge]

But that is what i am saying, am i not?

[merry blaze]

0.99... is never reached from a finite number of decimals, because it has infinitely many digits.

[lucid forge]

But limit is infintely large process, yet it is true that in this process we will not see 0,9999... since it is equal to 1, right?

[merry blaze]

The limit isn't an "ongoing" process. It's the result. An exact number.

[lucid forge]

Yes, but to know the result we go through a process

[merry blaze]

You don't have to that's what limits are for.

[lucid forge]

Ok, but if we intuitively look at it then we look at a process. For example, we say that x converges to 1 if x is 0,9 then 0,99 and so on

[merry blaze]

The limit computes the result at the end of the process. It itself is not a process.

Nor an approximation

The sequence given by ${a_n}$ where $a_n = \sum_{k=1}^n \frac{9}{10^n} = 0.999...9$ (with $n$ nines) is a sequence that converges.
The limit gives you the value at the end of this convergence "process." $\lim_{n\to \infty} a_n = 1$.

[night cave]

Azyrashacorki

[merry blaze]

In the language of analysis this means that $|a_n - 1|$ can be made arbitrarily small by taking sufficiently large $n$.

[night cave]

Azyrashacorki

[lucid forge]

So 0,999... and 1 difference convergres 0, but never is exactly 0?

[merry blaze]

The difference between 0.99...9 and 1 is nonzero for any finite n.

The difference between 0.99999... and 1 is 0.