Hi, i want to make sense of Leibniz notation. I find it really misleading and not intuitive at all. How to see Leibniz notation as logical? How to know when it can be used as a fraction?
#Leibniz notation
steady eagle
radiant hawkBOT
wide wasp
Just skimming this it seems to explain what it is and how it works quite well.
Essentially: it IS a fraction, BUT a fraction of infinitesimals.
steady eagle
Thanks, i will definitely read it. Can i ask you something about differentials?
wide wasp
Sure
steady eagle
why is dx = d(x+c)? I know the proof, but it makes no sense for me... Like why would infinitely small change of x be equal to x + c?
wide wasp
It's more that they are so close to being equal that you won't get a big error by saying they are equal 🙃
steady eagle
No, but equal sign says that they are exactly equal and says nothing about errors. Also math should be exactly precise?
wide wasp
Infinitesimals are values so small that we can't set them to exact values. 2 infinitesimals is still an infinitesimal value. It's weird, but you get used to it.
wide wasp
Estimates is a bigger part of math than any pure math enthusiasts like to admit, not everything can or should be made completely exact.
knotty bramble
its fine if its rigorous though
wide wasp
Infinities and infinitesimals are two sides of the same coin
This might be an easier way to explain indeed. Infinity +- constant is still infinity. Same thing applies to infinitesimals.
steady eagle
No, but it actually makes no sense, because lim(delta x go to 0)delta x = dx, but lim(delta x go to 0)delta x + 1000 = dx + 1000. So 1000,000...01 = 0,00000...01?
knotty bramble
No, but it actually makes no sense, because lim(delta x go to 0)delta x = dx, bu...
bro are you that guy who refuses to learn from a book but from youtube and ai?
the whole idea is that it is arbitary
steady eagle
bro are you that guy who refuses to learn from a book but from youtube and ai?
I am that guy who learns from sources that are valid, true and explains everything intuitively
What do you mean by arbitrary?
knotty bramble
I am that guy who learns from sources that are valid, true and explains everythi...
yeah but why not from a book
steady eagle
Who said i do not learn from a book? I love learning from books
knotty bramble
ok which books are you learning from rn?
steady eagle
It is not english book since i am not from english speaking country
But it is called just Calculus if you translate it
But lets not talk about it since it is not topic of this chat
steady eagle
Ok, but then in my example we got 1000 = 0,01, which is not true?
knotty bramble
I don't understand what you are trying to do here
steady eagle
I am using dx definition to show that 1000 = 0,01 which is not true. Since it is not the case i am misunderstanding something, but what is it?>
knotty bramble
No, but it actually makes no sense, because lim(delta x go to 0)delta x = dx, bu...
so are you saying that as x tends to a small number = dx, but dx + 1000 = dx + 1000. then you somehow conclud 1000.00000001=0.000001?
how does 1000 = .01?
what definition of dx?
steady eagle
so are you saying that as x tends to a small number = dx, but dx + 1000 = dx + 1...
Thats not what i said...
knotty bramble
Then I don't understand what you mean, dx is just representing a really small number
steady eagle
but dx + c can be very small number + 1000?
steady eagle
So since dx = d(x+c), we get that small number = small number + 1000?
knotty bramble
dx = d(x+c) is not a thing
steady eagle
it is
steady eagle
d(x+c) = (x + c)'*dx, therefore d(x+c) = dx, because derivative of x + c is just 1
knotty bramble
differential?
steady eagle
Yes
knotty bramble
you are differentiating twice
though
(x+c)'
thats is once
and then where did dx come from
Yeah
you are treating the differential like a fraction right?
steady eagle
Yes
knotty bramble
yeah but thats not what the differential means
steady eagle
Then what does it mean?
knotty bramble
it means when it changes by dx
i.e. a small value
what is your slope
This is what it means
steady eagle
I am talking about differential not derivatives?
knotty bramble
d(x+c) = (x + c)'*dx, therefore d(x+c) = dx, because derivative of x + c is just...
because you just said derivative
and now you are treating it as a differential
steady eagle
I think we keep misunderstanding each other
steady eagle
yes
knotty bramble
so thats a derivative
steady eagle
yes
steady eagle
yes
knotty bramble
so it makes sense
steady eagle
how
knotty bramble
f(x+h)-f(x) is small
given it is continuous
h * f' is small
its not actually a fraction
you cannot just treat the derivative like a fraction
because the d(x+c) is a function
in this case d(x)=f(x+h)-f(x)
steady eagle
dx = df(x)/f'(x)
steady eagle
The problem is that they do not explain, they just give proof
steady eagle
But i already said the proof before
knotty bramble
can i see the proof
knotty bramble
can you show me hte whole page
They shouldn't be going onto derivatives without talking about first limits and errors
steady eagle
Its not in english and it does not say anything else
knotty bramble
Its best get a better book then
steady eagle
Ok...
steady eagle
Ok, but back to the topic: How d(c+x) = dx? By proof it seems completely logical, but how it works if you put real numbers in?
knotty bramble
thats just this rearranged
d(x) is the function f(x+h)-f(x)
(c+x)' = 1
d(c+x)=(c+x)'dx => d(c+x)=1 * dx
d(c+x) = dx
Like i said, its not very helpful, even misleading
and introducing derivatives in limits is no book should ever do
steady eagle
Nvm, i just asked AI about differential of a constant and it says it is equal 0. So then it makes all true.
knotty bramble
whats the name of the book
steady eagle
Calculus
knotty bramble
get a new book
steady eagle
Ok
Well i got this book for "free" so i am using it
By "free" i mean i got it for free for limited time
knotty bramble
Check your dms
sullen patio
Ok, but back to the topic: How d(c+x) = dx? By proof it seems completely logical...
Informally, you can think of d(x+c) as df where f(x) = x+c
When you do a small change in x, the change in f is dx regardless of c (since it will cancel out)
steady eagle
Wait, can someone help me understand differentials? I thought about them and i got very confused. What do they even mean? What is their definition?
I always thought differential is just infinitely small number, but is that true?
It must be, because otherwise dy = y'*dx would not be true?
steady eagle
https://youtu.be/TaV9QlgXZV4?is=slEQHZwoTpMwgBHh
So if i understand correctly first minute of the video then it means that differential of x is infinitely small change in x? But what exactly does it mean? It means that they represent infinitesimal or a very small number?
wide wasp
They represent an infinitesimal, which is a very small number, so both.
steady eagle
But i get confused about example with x^2. When we find differential of it we get that d(x^2) = 2x*dx + dx^2, but we should get d(x^2) = 2xdx?
Like df(x) = f(x + dx) - f(x) = (x + dx)^2 - x^2 = 2x*dx + dx^2 and therefore we do not get df = 2xdx
wide wasp
Because they are infinitesimal values, they get the same issues as infinite values. Trying to do math directly with them will land you in trouble more often than not. That's why we mostly only work with dy/dx and not the infinitesimals themselves. We only acknowledge that that is what they are, so we can use the fraction relation in the chain rule and other useful situations.
steady eagle
Because they are infinitesimal values, they get the same issues as infinite valu...
Well dy = y'*dx seems logical and actually works, so why suddenly when we try this formula on real functions, it "breaks"?
wide wasp
It doesn't really break, but it depends on how rigid you need to be. If you can accept that dx^2 and higher powers are so small that we can discard them, then it should hold.
@steady eagle does that help?
steady eagle
It doesn't really break, but it depends on how rigid you need to be. If you can ...
Ok, i see, they are really small, but if we compare obtained expression to dy = y'dx formula, they are not equal?! So either formula is wrong, which would break all derivative notation thing, or something else is wrong... Lets look at x^3. We get that d(x^3) = 3x^2 * dx + 3x * dx^2 + dx^3, but we should get d(x^3) = 3x^2dx. So as we take larger and larger exponent, we get larger and larger errors if we remove those other terms?! So if n was really big it would not give precise result at all. So what is going on here?
wide wasp
Imagine dx=0,000000001. What is dx^2 or worse yet dx^3? It's extremely small values, and the coefficients would need to be very big for those to have any significant impact. Besides, the differential is defined as the linear part of the change. The full expression with dx^2 is the finite difference, not the differential.
So either explanation works. They are essentially zero, or: they're not part of how it's defined.
@steady eagle any clearer?
steady eagle
Imagine dx=0,000000001. What is dx^2 or worse yet dx^3? It's extremely small val...
Alright, they would be very small, but 2xdx itself will not be also very, very small? If we do not count other very small numbers, why even count this one(2xdx)? Of course others relative to this one will be much, much smaller, but since we are still talking about infinitesimals, it is also very, very small? Is dy = y'dx even correct formula?
wide wasp
Yes dy=y'dx is correct. It's the definition of the differential. Definition, not equation. What you're getting if you keep everything is the finite difference. Different things, even though they are approximately equal.
steady eagle
I asked AI and it said that dy = y'dx is just an approximation not completely precise? Is it true?
wide wasp
Yes and no. It can be taken as an approximation of the finite difference, but it IS the DEFINITION of the differential.
steady eagle
Yes and no. It can be taken as an approximation of the finite difference, but it...
Ok, so this approximation equation is just the definition of differential?
wide wasp
Yeah
steady eagle
Yet why we write dy/dx = y'?
wide wasp
Different form of same definition.
steady eagle
But derivative is exactly precise limit, but differential is not exactly precise, so how can we get from approximate equation something exact?
wide wasp
It's an approximation of the finite difference. It's exact for the differential. Different things. It is exact, because it is DEFINED that way.
You can't prove a definition. Just try to understand (which is what we have been trying to do here) or simply use.
hollow thunder
Hi, i want to make sense of Leibniz notation. I find it really misleading and no...
What is it
We meet again
wide wasp
What is it
df/dx
steady eagle
It's an approximation of the finite difference. It's exact for the differential....
Oh so equation is completely exact and true, but when we look at certain function differential it gets also errors? But do not all function behave exactly like straight lines or to be more precise linearly at infinitesimal sizes, so must it not obey this formula?
hollow thunder
df/dx
derivative
a notation is self defined
he defined d/dx as derivative not something he derived
wide wasp
Oh so equation is completely exact and true, but when we look at certain functio...
I'm too sleepy to continue this discussion. Best of luck.
steady eagle
I'm too sleepy to continue this discussion. Best of luck.
Oh ok, thank you for helping me
@hollow thunder maybe you know why dy = y'dx is true and not an approxmiation? If we look at x^2 we see it is not equal to that equation?!
hollow thunder
y' is dy/dx
steady eagle
Yes
hollow thunder
So i think i know what ur asking now
ur saying that
the derivative notation is a complete 'd/dx' not d upon dx
but when we write dy = y'dx we treat dy/dx as a fraction
right?
steady eagle
Yes
hollow thunder
see
when we're taking the derivative we write it as d/dx of y
d(y)
dx
like d / dx is a whole notation in itself
steady eagle
It means we differentiate with respect to x?
hollow thunder
yeah exactly
but it can also be used as an "exact differential" and when it is used in that manner , we can just write
Exact differential of y as
dy
exact differential of x as dx
in this manner it can be treated as a fraction
also a way of solving first order differential equations, treating them as exact differentials
steady eagle
Could you please explain what you mean by "exact" differentials?
hollow thunder
u can understand it by using geometrical interpretation
whats d(y)
a very small change in y
d(x) very small change in x
meanwhile
d/ dx of y is change in y with respect to change in x
hollow thunder
a very small change in y
thats exact differential
independent of changes in other terms
steady eagle
Oh ok, so we can treat it as a fraction not as a operation d/dx if they are independent differentials?
hollow thunder
yeah
steady eagle
But does not y always change only if x changes?
hollow thunder
it depends
dy can be 0
dx can be 0
dx ≠ 0 does not imply dy=0
if a particle is only moving along x axis then there is change in x but no change in y
another example take graph of
y = k
y is same but x can be anything
dx ≠0 dy = 0
when dealing with differential equations we dont talk about these stuff
generally
steady eagle
Hmm, yes, but when talking about differential of a function y, y definitely changes, does not it?
Nvm , it can also be 0, right?
steady eagle
Yes it is
steady eagle
Wait, but even if dy = 0, it is always still dependent on x, because we can get dy only if we get change in x?
hollow thunder
dy/dx is either
change in y / change in x
Or
change in y with respect to x
in first scenario its basically
0/any number
always going to be 0
steady eagle
Isn't dy = f(x + dx) - f(x)?
hollow thunder
are you referring to first principle
steady eagle
What is first principle?
steady eagle
yes
hollow thunder
first principle does not deal with exact differentials directly
it deals with rate of change with respect to something
It calculates the slope (rate of change)
steady eagle
Yes i understand derivatives. So if f'(x) is exact rate of change of function at point then if it moves by dx from that point, then we get the real change in function value, which is dy? Yet we do not say dy = y(x + dx) - y(x) for some reason?
hollow thunder
f'(x) is not dy
steady eagle
i know
hollow thunder
If f(x) is y then f'(x) is dy/dx
hollow thunder
Yes i understand derivatives. So if f'(x) is exact rate of change of function at...
youre mixing exact change with rate of change
theyre different
steady eagle
I am very confused, i do not what to say... I think i have to go sleep...
hollow thunder
yeah sleep
steady eagle
Ok, bye for now, thank you for helping
steady eagle
Oh so dy is an approximation
hollow thunder
in differentials it is exact in geomtrical sense it is approx
different things
i need to sleep now , also my battery is on 1 percent
steady eagle
oh ok, bye