Fundamental confusions of calculus

Discussion in 'Physics & Math' started by arfa brane, Feb 11, 2012.

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1. TachBannedBanned

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So, your contention has boiled down to the fact that you disagree with the notation, despite the fact that earlier you seemed to have no problem understanding it? Can you walk away for a few minutes from spending so much time and effort in proving me wrong and help Pete with his misunderstanding? It is a very simple issue but 100+ posts on this subject and he still fails to see it. THIS is what he really, really needs help with. Can you spare some time and help him?

3. PeteIt's not rocket surgeryRegistered Senior Member

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10,167
There's no problem with T being a function of z.
But don't you see that you are also specifying the z-coordinate of the bug's position as a function of t?
Like I said, z is constant over the bug's path. z=C. dz/dt = 0
If you don't know that, if you don't know how the bug moves (or not) in z, then how can you determine dT/dt?

Have you read temur's post, in response to my request for assistance?

Do you really want to understand this or do you want to be right no matter what? You might notice that I'm actually asking for help. Perhaps you should consider dong the same.

Thanks, I've been asking the same thing:

Is anyone other than Tach able to help me out with the questions in post 41 and post 37 (only after the second quote)?

I feel they've been lost in the noise.

5. TachBannedBanned

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No, I am not. Pete, it is very simple, assume that the temperature is a function of the altitude of the observer above the plane containing the bug, z is NOT the coordinate of the bug, actually it has NOTHING to do with the bug and its motion. For example, the temperature drops exponentially with the increase in altitude.

Try this (I know I am going to regret giving you an example):

$T(t,x,y,z)=e^{-z}(at+bx+cy)$
a,b,c are some constants
$x=f(t)$
$y=g(t)$

f,g are functions differentiable (at least) in the first order

Last edited: Feb 13, 2012

7. TachBannedBanned

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The (x,y) coordinate and its dependence of t is given. The temperature of the bug does not depend on z, the temperature T of the medium is what depends on z.

8. przyksquishyValued Senior Member

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No, you're revising history. I said that it wasn't clear just from the definition of your example that f in that example was a function of three variables called $\theta$, $u$, $v$. You are also completely evading that, even given that, just saying
$f \,=\, 3\theta \,+\, \sin(\theta)^{2} \,+\, \log(x)$​
alone doesn't tell anyone how much of that expression is supposed to be $f$ and how much of it is $u$ and $v$. This isn't just me being deliberately obtuse. My first reasonable "I know what you mean even if you're saying it wrong" guess was that $u(\theta) = \sin(\theta)^{2}$ and that you'd just forgotten about $v$ and were calling the third variable $x$ instead.

Instead of accepting this and moving on you responded to that with a couple of completely bogus and made up rules, that I responded to in the last two points [POST=2902983]here[/POST].

You also keep falsely insisting that the derivative of your example including the dependence on $\theta$ through $u$ is a total derivative. It's not, because you're still treating $x$ as an independent parameter.

How about you use standard notations and move on instead of generally flinging poo at everyone.

9. przyksquishyValued Senior Member

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Really? $x = 3$, $y = 2$, $t = 5$, $a = b = c = 1$ and $f(t) = g(t) = \sin(t/\pi)^{2}$. What's the bug's temperature?

10. PeteIt's not rocket surgeryRegistered Senior Member

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Hi przyk,
I wonder if you (or AlphaNumeric, or Guest254, or prometheus, or rpenner, or temur...) could run your eye over this and see if it makes sense?

When taking a partial derivative of an expression with both direct and indirect dependencies on the variable of interest and no other direct dependencies, my understanding is that it is implied that the indirect dependencies are ignored, eg:
\begin{align} y &= g(x) ; \frac{\partial}{\partial x} f(x, y) &= \frac{\partial}{\partial x} f(x, y)|_y \end{align}​
Is that right?

And in the case of an expression with only indirect dependencies on the variable of interest, my understanding is that it is implied that the chain rule should be applied, eg:
\begin{align} y &= g(x) \\ \frac{\partial}{\partial x} f(y, z) &= \frac{\partial}{\partial x} f(y, z)|_z \\ &= \frac{\partial}{\partial y} f(y, z) \, \frac{dy}{dx} \end{align}​
Is that right?

But in the case of an expression with both direct and indirect dependencies on the variable of interest, and other direct dependencies, then there is ambiguity, eg
$y = g(x) \\ \frac{\partial}{\partial x} f(x, y, z)$​
could potentially mean either
$\frac{\partial}{\partial x} f(x, y, z)|_z$ or $\frac{\partial}{\partial x} f(x, y, z)|_{y,z}$​

So it seems to me that for an expression like that you should be explicit, and that the two meanings are related like this:
$\frac{\partial}{\partial x} f(x, y, z)|_z = \frac{\partial}{\partial x} f(x, y, z)|_{y,z} \, + \, \frac{\partial}{\partial y} f(x, y, z) \, \frac{dy}{dx}$​
Is that right?
Or am I just hopelessly confused?

11. TachBannedBanned

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Err, why are you making x=3, y=2 , f=g when I you can clearly see that $x=f(t)$ and $y=g(t)$? Do you see the errors in your post?

12. temurman of no wordsRegistered Senior Member

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1,330
Yes. f is a function of 2 variables, and you are differentiating with respect to the first variable. This process is completely independent of the equation y=g(x), which defines a curve on the x-y plane.

Yes. What is really going on here is that you are considering the new function F(x,z) = f( g(x), z) and differentiating it with respect to x. Or you can say you performed a change of variable, replacing y with x.

I can assure you that you understand what is going on, so at this point it becomes just a notational issue. To me, $\frac{\partial}{\partial x} f(x, y, z)$ would always mean what you call $\frac{\partial}{\partial x} f(x, y, z)|_{y,z}$. What you denoted by $\frac{\partial}{\partial x} f(x, y, z)|_z$ is $\frac{\partial}{\partial x} F(x, z)$ with the new function $F(x, z) = f(x, g(x), z)$.

13. PeteIt's not rocket surgeryRegistered Senior Member

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10,167
Sweet, thanks temur.
It seems like notation can be one of the hardest things about approaching a new topic in an unstructured (ie self-taught) way.

In classes, you learn notation and concepts together at a nice, synchronized pace.
When you're on your own, you are more driven by concepts, and notation can get left behind.

What does $h = f \circ g$ mean?

14. temurman of no wordsRegistered Senior Member

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$h=f\circ g$ means that $h$ is a function defined by $h(y)=f(g(y))$ for each value of $y$.

15. arfa branecall me arfValued Senior Member

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It's notation for function composition.$f \circ g$ means "take g as an argument of f", conversely $g \circ f$ means "take f as an argument of g".

You "work" right to left in either case. We did this with what the lecturer called arrow diagrams, which map sets of points to sets of points. These are a good way to understand the concepts of "one to one" or "onto" functions between sets.

You just compose the diagrams in the order required.

16. AlphaNumericFully ionizedRegistered Senior Member

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6,702
Tach, I asked you a direct and relevant question. You replied multiple times to others over several hours. If you are unwilling to discuss things honestly and avoid addressing how your claims contradict the DEFINITION then you are trolling. If you post again and don't reply to this point then you're getting a trolling infraction. You have had plenty of notice and plenty of time. By throwing out your usual one line vitriol rather than replying to questions you are showing considerable dishonesty. Last chance.

17. PeteIt's not rocket surgeryRegistered Senior Member

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10,167
Thanks.
Next question...
Are $\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z$ and $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ the same as the orthogonal unit vectors $\mathbf{\hat{i}}, \mathbf{\hat{j}}, \mathbf{\hat{k}}$?

18. AlphaNumericFully ionizedRegistered Senior Member

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6,702
Yes, they are just different notation. People generally use the x,y,z or i,j,k versions if you're working in 3 dimensional systems. When you work in higher dimensional ones or arbitrary then you use the number index and use $\mathbf{e}_{i}$ as an unspecified term. For example, $\mathbf{e}_{i}\cdot \mathbf{e}_{j} = \delta_{ij}$ doesn't refer to the inner product between $\hat{i}$ and $\hat{j}$ but rather it tells you that if you put in say i=3 and j=25 (in a very high dimensional space) you'll get zero, since $\delta_{3,25} = 0$.

Once you're comfortable with index notation you might as well use it all the time. It helps you get into good habits when you get into things like relativity, where you deal with tensors a lot. Plus you can prove all sorts of otherwise very unpleasant expressions with it. For example. $\nabla \times (\mathbf{F} \times \mathbf{G})$ is a bugger to expand without suffice notation (it was an exam question in my undergrad days!)

19. PeteIt's not rocket surgeryRegistered Senior Member

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10,167
Thanks
What's $\delta_{ij}$?
It will always be zero, won't it? Or can you use $\mathbf{e}_i$ to refer to non-orthogonal basis elements?

What's suffice notation? (Google tells me nothing! This thread is the first hit after some random sites with "suffice. Notation")

Last edited: Feb 13, 2012
20. prometheusviva voce!Registered Senior Member

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2,045
$\delta_{ij}$ is the Kronecker delta symbol. It's defined to be 1 when $i=j$ and 0 otherwise.

21. przyksquishyValued Senior Member

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3,203
I've slept on this and here's how I'd put things.

First of all, if you look up how partial derivation is defined in standard math texts, then normally the thing you take partial derivatives of are functions. In physics we'd normally write something like $f \,=\, f(x,\, y)$ to say that $f$ is a function of two real variables, and we're letting $x$ and $y$ denote its first and second parameters, respectively. Then $f$'s partial derivatives are noted $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ (and preferably not $\frac{\partial}{\partial x} f(x,\, y)$). These are new functions, analogous to $f'$ in the single variable case. So you can write e.g.
$\frac{\partial f}{\partial x}(x_{0},\, y_{0})$​
to denote the partial derivative $\frac{\partial f}{\partial x}$ evaluated at the point $(x_{0},\, y_{0})$. This is analogous to writing something like $f'(x_{0})$ in the single variable case. For functions with longer names/definitions, you could write something like this:
$\Bigl( \frac{\partial}{\partial y^{i}} \, f \,\circ\, g \Bigr)(\bar{y}_{0}) \,.$​
This makes it clear that the function you're taking the partial derivative of is $f \,\circ\, g$, and then you're evaluating it at the point $\bar{y}_{0}$.

When you write something like this:
$\frac{\partial}{\partial x} \bigl( x^{2} \,+\, \sin(x + y) \,-\, yz \bigr) \,,$​
the thing between the brackets is an expression. In line with what temur told you, I'd read it as the definition of an unnamed function
$(x,\, y,\, z) \to x^{2} \,+\, \sin(x + y) \,-\, yz$​
that you're taking the partial derivative of, in this case with respect to $x$. So
$\frac{\partial}{\partial x} \bigl( x^{2} \,+\, \sin(x + y) \,-\, yz \bigr) \,=\, 2x \,+\, \cos(x + y) \,.$​
Or more correctly, the result is the function $(x,\, y,\, z) \to 2x \,+\, \cos(x + y)$.

With this convention, if $f = f(x,\, y,\, z)$ and $g = g(x)$, then:
$\frac{\partial}{\partial x} f(x,\, y,\, z)$ is synonymous with $\frac{\partial f}{\partial x}$,​
$\frac{\partial}{\partial u} f(u,\, v,\, w) \,=\, \frac{\partial f}{\partial x}$ (not $\frac{\partial f}{\partial u$ - why?),​
and
$\frac{\partial}{\partial x} f(x,\, g(x),\, z) \,=\, \frac{\partial f}{\partial x} \,+\, \frac{\partial f}{\partial y} \, \frac{\mathrm{d} g}{\mathrm{d} x} \,.$​

This is all pretty much in line with what temur has just told you, but hopefully it's still helpful.

Last edited: Feb 13, 2012
22. przyksquishyValued Senior Member

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3,203
He probably meant "suffix" notation.

23. TachBannedBanned

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Yes, you did work through but you worked incorrectly. In the book I cited, there are a lot of exercises, they all contradict your claims. Look, it is very simple:

$f=f(\theta, u,v)$

$u=sin^2(\theta)$
$v=ln(x)$
$f=3 \theta+u+v$

Therefore, contrary to your erroneous calculations, the partial derivative is:

$\frac{\partial f}{\partial \theta}=3$

It is the total derivative that has the sin function:

$\frac{df}{d \theta}=\frac{\partial f}{\partial \theta}+\frac{\partial f}{\partial u} \frac{du}{d \theta}=3+2 sin(\theta) cos (\theta)$

There are a lot of exercises in the chapter I linked in, they all contradict your erroneous calculations. Look, you are making a basic error, you have made others in the past, why don't you admit it and be done with it? You had no problem admitting to your errors when I pointed them out in the past, why are you so hard on denying this one? There is no shame in making an error, so admit it and be done with it.