# Fundamental confusions of calculus

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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:

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

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

Why can't I say $$f=f(\theta, u,v)$$

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

which is a perfectly valid function. According to your idea of the partial derivative $$\frac{\partial f}{\partial \theta}$$ is now zero which explicitly shows what you're saying is obviously rubbish.

I'm sorry, this isn't relevant or even very constructive but I can't help but be reminded by AN's post below of another one I made a while back (concerning the same poster, I might add...)
You are not actually engaging in a discussion here, you're just pointing at the book and saying "Look, look!". Yes, we've looked at it.
RJBeery said:
I have this comedic scene in my mind of you taking a debate class, where your turn at the podium consists of vigorous, silent pointing at textbooks.

Why can't I say $$f=f(\theta, u,v)$$

$$u=3 \theta + sin^2(\theta)$$

Because, if you do that,you no longer have $$f=f(\theta, u,v)$$ but $$f=f(u,v)$$
Because I already said (for Pete's benefit) that:

$$u= sin^2(\theta)$$

The example was constructed specifically to help him figure out the difference between partial and total derivatives.

According to your idea of the partial derivative $$\frac{\partial f}{\partial \theta}$$ is now zero which explicitly shows what you're saying is obviously rubbish.

Actually, any basic textbook would prove that , indeed, since you just made $$f=f(u,v)$$ =>$$\frac{\partial f}{\partial \theta}=0$$. Ain't basic calculus a bitch, eh?

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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:

$$\frac{\partial f}{\partial \theta}=3$$
You're only defining those details now, after the fact.

You are in no position to be telling anyone else they are in error. That's pathetic deflection. In post #18 you did not properly define the function $$f$$ and how it depended on $$u$$ and $$v$$. That is your own failure and nobody else's. The definition you have given here is not automatically implied by what you wrote in the example in post #18.

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.
In all of the examples you have cited, the total derivative is taken through all of a function's parameters. You are still leaving $$v$$ and/or $$x$$ independent of $$\theta$$ here. So your references do not support your case.

If there's one independent variable, you're taking a total derivative. If there are multiple independent variables and you're keeping all but one of them constant, you're taking a partial derivative. This really isn't complicated.

You're only defining those details now, after the fact.

Err, no, I did this for Pete three threads and hundreds of posts ago. If you do not understand something, just ask, don't jump to claim that you found an error.

You are in no position to be telling anyone else they are in error.

You mean, after being corrected multiple times, you still do not understand?

That's pathetic deflection. In post #18 you did not properly define the function $$f$$ and how it depended on $$u$$ and $$v$$.

So, you are still on that train, "I did not understand and I did not ask for clarification, therefore you are wrong"?

That is your own failure and nobody else's. The definition you have given here is not automatically implied by what you wrote in the example in post #18.

You had no trouble understanding it initially, why the trouble now?
After you have received repeated clarifications, do you still claim that there is an error?

In all of the examples you have cited, the total derivative is taken through all of a function's parameters. You are still leaving $$v$$ and/or $$x$$ independent of $$\theta$$ here.

Yes, the example is constructed to explain to Pete why there is no contribution in $$\frac{df}{d \theta}$$ from the functions (v in this case) that do not depend on $$\theta$$.

If there's one independent variable, you're taking a total derivative. If there are multiple independent variables and you're keeping all but one of them constant, you're taking a partial derivative.

The above would be true IF x were an independent variable. It ISN'T. There is only ONE independent variable in the example, and that is $$\theta$$.

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Actually, any basic textbook would prove that , indeed, since you just made $$f=f(u,v)$$ =>$$\frac{\partial f}{\partial \theta}=0$$. Ain't basic calculus a bitch, eh?

Except f is not independent of $$\theta$$. Are you telling me that you can't see that for $$f = f(u(x), v(x),y, \ldots)$$ the partial derivative wrt x is $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f }{\partial v} \frac{\partial v}{\partial x}$$?? That's the chain rule, and is about the first "compound" technique that is taught after simple differentiation.

lol!

Except f is not independent of $$\theta$$. Are you telling me that you can't see that for $$f = f(u(x), v(x),y, \ldots)$$ the partial derivative wrt x is $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f }{\partial v} \frac{\partial v}{\partial x}$$?? That's the chain rule, and is about the first "compound" technique that is taught after simple differentiation.

lol!

Err, you seem to be confused, the quantity $$\frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f }{\partial v} \frac{\partial v}{\partial x}$$ happens to be , contrary to your claims, not the partial derivative $$\frac{\partial f}{\partial x}$$ but the total derivative $$\frac{df}{dx}$$.
You are right on one thing, it is taught in basic calculus.

Err, no, I did this for Pete three threads and hundreds of posts ago.
If you did (I'll believe it when I see it), you failed to cite it.

So, you are still on that train, "I did not understand and I did not ask for clarification, therefore you are wrong"?
No, you are missing the point. It's not that you failed to properly define your example that I have a problem with here. It's that between when you posted the example and when you properly clarified it, you went around telling everyone else they were wrong when it was you who had posted an ambiguous example all along.

I also have a problem with you talking down to everyone, as if you were an authority on differential calculus (you're really not) when you failed to set up an example correctly, you cite examples that don't actually support your case, and even posted downright [POST=2902804]bullshit[/POST].

After you have received repeated clarifications, do you still claim that there is an error?
Again you're dodging the issue. Sure, if $$f \,=\, 3\theta \,+\, u^{2} \,+\, v$$ then $$\frac{\partial f}{\partial \theta} \,=\, 3$$. And if you'd said that in post #18 that would have been fine. But you didn't.

So, now that you have received repeated clarifications, do you understand that
$$\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, \sin(2\theta)$$ ?​
If you just say $$f \,=\, 3\theta \,+\, \sin(\theta)^{2} \,+\, \log(x)$$, do you understand why everyone who's ever passed university calculus is going to tell you $$\frac{\partial f}{\partial \theta} \,=\, 3 \,+\, \sin(2\theta)$$?

The above would be true IF x were an independent variable. It ISN'T. There is only ONE independent variable in the example, and that is $$\theta$$.
Bullshit. You've said yourself that $$x$$ was not a function of $$\theta$$. That makes it an independent variable.

What does $$h = f \circ g$$ mean?
That you need to turn your headlights on

Sure, if $$f \,=\, 3\theta \,+\, u^{2} \,+\, v$$ then $$\frac{\partial f}{\partial \theta} \,=\, 3$$.

Good, now that you have understood that you should stop trolling.

You've said yourself that $$x$$ was not a function of $$\theta$$. That makes it an independent variable.

You missed the point, x is NOT an independent variable in $$f=f(\theta,u,v)=3 \theta +u +v$$ .
$$\theta$$ is the only independent variable.
u is a function of $$\theta$$ and v is a function of x. We have been over this for about 50+ posts now.

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Good, now that you have understood that you should stop trolling.
Calling you out on your own immature behaviour is not trolling.

If you don't want me to point out where you're posting bullshit, don't post bullshit in the first place, and especially don't try to pretend it never happened while you insult everyone else.

You missed the point, x is NOT an independent variable in $$f=f(\theta,u,v)=3 \theta +u +v$$ .
$$\theta$$ is the only independent variable.
u is a function of $$\theta$$ and v is a function of x.
No, but $$v$$ is an independent variable of $$f$$. And $$x$$ is an independent variable in the expression $$f(\theta,\, u(\theta),\, v(x))$$, and in the expression $$3\theta \,+\, \sin(\theta)^{2} \,+\, \log(x)$$. That's why I said "You are still leaving $$v$$ and/or $$x$$ independent of $$\theta$$ here" in post #84. Either way, you have an independent variable, whether you want to make it $$v$$ or $$x$$.

We have been over this for about 50+ posts now.
Repeating yourself doesn't make you right.

By the way, you were very selective in responding to my last post. These questions were not rhetorical:
So, now that you have received repeated clarifications, do you understand that
$$\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, \sin(2\theta)$$ ?​
If you just say $$f \,=\, 3\theta \,+\, \sin(\theta)^{2} \,+\, \log(x)$$, do you understand why everyone who's ever passed university calculus is going to tell you $$\frac{\partial f}{\partial \theta} \,=\, 3 \,+\, \sin(2\theta)$$?

Yes, you did work through but you worked incorrectly.
Please point precisely where I made the mistake. Don't just say "The book! The book!". If I made a mistake where was it. Is the formula wrong? Did I accidentally say 1+1=3? Did I mistakenly use the wrong trig formula? Where, precisely, is the mistake.

All you're doing is saying "Wrong! The book!". And you skipped over how if you're right then all of Lagrangian/Hamiltonian mechanics is wrong.

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

$$\frac{\partial f}{\partial \theta}=3$$
Tach, I get the distinct impression you don't know any of this. You're doing a Reiku, you're just mindlessly quoting something someone else wrote, you're unable to engage in proper discussion and you're assuming your understanding is perfect. Seriously, you're arguing against the definition here!

Do you have any experience with Hamiltonian or Lagrangian mechanics? They completely invalidate your claims. Multiple people in this thread have papers published which use such mathematics. I personally spent the last fortnight doing Hamiltonian mechanics as part of my job. This isn't something on the periphery of our experiences, it's bread and butter stuff.

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.
Project much?

If my calculation was erroneous you should have no problem pointing to the specific line in the calculation which is mistaken. Don't just say "It's wrong" and move on, prove it wrong. If you can't then you're just pulling a Reiku.

The wikipedia article on partial derivatives explains all this quite well, and it isn't even confusing!

I like this part:
Basic definition

The function f can be reinterpreted as a family of functions of one variable indexed by the other variables:

$$f(x,y) = f_x(y) = \,\! x^2 + xy + y^2.\,$$

In other words, every value of x defines a function, denoted f[sub]x[/sub], which is a function of one variable.[2] That is,

$$f_x(y) = x^2 + xy + y^2.\,$$

Once a value of x is chosen, say a, then f(x,y) determines a function f[sub]a[/sub] which sends y to $$a^2 + ay + y^2$$:

$$f_a(y) = a^2 + ay + y^2. \,$$

In this expression, a is a constant, not a variable, so f[sub]a[/sub] is a function of only one real variable, that being y. Consequently, the definition of the derivative for a function of one variable applies:

$$f_a'(y) = a + 2y. \,$$

The above procedure can be performed for any choice of a. Assembling the derivatives together into a function gives a function which describes the variation of f in the y direction:

$$\frac{\part f}{\part y}(x,y) = x + 2y.\,$$

This is the partial derivative of f with respect to y.
--http://en.wikipedia.org/wiki/Partial_derivatives

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?),​
I'm missing some vocab and maybe concepts here, but I think...
It's because the function has been defined as f(x, y, z), so $$\partial x/\partial u$$ doesn't mean anything.
f(u,v,w) is a special case, shorthand for f(x=u, y=v, w=z)
(I guess this kind of parameter substitution could be useful if you want to consider the function over a limited range or something)
This is all pretty much in line with what temur has just told you, but hopefully it's still helpful.
Thanks przyk, it helped a lot to point me toward a more fundamental grasp of what a function is.

Calling you out on your own immature behaviour is not trolling.

You mean like this error-filled post?

No, but $$v$$ is an independent variable of $$f$$. And $$x$$ is an independent variable in the expression $$f(\theta,\, u(\theta),\, v(x))$$, and in the expression $$3\theta \,+\, \sin(\theta)^{2} \,+\, \log(x)$$. That's why I said "You are still leaving $$v$$ and/or $$x$$ independent of $$\theta$$ here" in post #84. Either way, you have an independent variable, whether you want to make it $$v$$ or $$x$$.

Actually, this video makes it clear that what you keep calling partial derivative is in reality the total derivative. So does eq (2) from this textbook. This is really simple stuff, I do not know why you are fighting so hard in admitting that you goofed.

Repeating yourself doesn't make you right.

This is because you repeat the same error. What is most puzzling is that you got this one right in the beginning.

By the way, you were very selective in responding to my last post. These questions were not rhetorical:

So, now that you have received repeated clarifications, do you understand that
$$\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, \sin(2\theta)$$ ?​
If you just say $$f \,=\, 3\theta \,+\, \sin(\theta)^{2} \,+\, \log(x)$$,

The point is that I don't, this has been explained to you repeatedly:

$$f=f(\theta, u(\theta), v(x))=3 \theta+u+v$$

where:

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

do you understand why everyone who's ever passed university calculus is going to tell you $$\frac{\partial f}{\partial \theta} \,=\, 3 \,+\, \sin(2\theta)$$?

Err, I missed these errors, it is just a repeat of the ones you did before. You need to remember that:

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

If you insist in ignoring the definition of f as a function of u and v and you keep insisting in making up your own definition, then, yes, your partial derivative is correct. But this wasn't the definition that Pete and I were working of.

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Please point precisely where I made the mistake. Don't just say "The book! The book!".

I didn't say "the book", I pointed out your mistakes very precisely in post 80. It is very simple, really, you clearly do not know the difference between partial and total derivative.

If I made a mistake where was it. Is the formula wrong? Did I accidentally say 1+1=3? Did I mistakenly use the wrong trig formula? Where, precisely, is the mistake.

You are calculating the total derivative but you claim it to be the partial derivative. Same mistake as przyk but he's cleverer about it, he's tried (repeatedly) to change the definition of the function f.

Tach, I get the distinct impression you don't know any of this. You're doing a Reiku, you're just mindlessly quoting something someone else wrote, you're unable to engage in proper discussion and you're assuming your understanding is perfect. Seriously, you're arguing against the definition here!

Resorting to personal attacks does not further your point, nor is it appropriate given your role as a moderator. You have made a simple math mistake, own it and be done with it.

Do you have any experience with Hamiltonian or Lagrangian mechanics?

Yes, quite a bit. I can post some of the work that I've done but what does it have to do with your basic confusion between the definition of partial vs. total derivatives?

If my calculation was erroneous you should have no problem pointing to the specific line in the calculation which is mistaken. Don't just say "It's wrong" and move on, prove it wrong. If you can't then you're just pulling a Reiku.

Personal attacks are unworthy for you given that you are a moderator. It is very simple, really:

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

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

Given the above, as per any textbook (or you may want to look at this video linked in by Pete, it does a very good job in describing the difference between partial and total derivatives):

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

where $$\frac{df}{d \theta}$$ is the total derivative and $$\frac{\partial f}{\partial \theta}$$ represents the partial derivative.
Your error is not really huge, you are mixing the definitions , the video will help you figuring out the difference. I track your error to the fact that you aren't considering the fact that f is a function of $$\theta$$ and $$u(\theta)$$ but rather you lump all the expressions in $$\theta$$ together. przyk is doing the same exact error.

I'm missing some vocab and maybe concepts here, but I think...
It's because the function has been defined as f(x, y, z), so $$\partial x/\partial u$$ doesn't mean anything.
If you meant $$\partial f / \partial u$$, then yes, something like that. When you write something like $$\frac{\partial f}{\partial x}$$, the x is really a sort of dummy variable. Its sole purpose is to indicate which parameter the partial derivative of $$f$$ is taken with respect to, according to some predefined convention. Beyond that the name is of no significance.

In fact as far as I know it's only in applied math that we really bother to name the parameters (e.g. to reflect their physical significance) at all. As far as I know, in pure analysis the mathematicians would just write $$\frac{\partial f}{\partial x^{k}}$$ or $$D_{k} f$$. Mathematically the parameter names $$x$$, $$y$$, and $$z$$ aren't part of the function. They're just extra baggage carried around with it that's only needed when we're taking partial derivatives in physics.

f(u,v,w) is a special case, shorthand for f(x=u, y=v, w=z)
(I guess this kind of parameter substitution could be useful if you want to consider the function over a limited range or something)
Not exactly. If you apply the convention I described, then $$\frac{\partial}{\partial u} f(u,\, v,\, w)$$ means the partial derivative of the function
$$(u,\, v,\, w) \to f(u,\, v,\, w) \,.$$​
This is a function that is identical to $$f$$, except that I'm using the convention of labelling its parameters $$u$$, $$v$$, and $$w$$ instead of $$x$$, $$y$$ and $$z$$. If you call this function $$\lambda$$, then $$\frac{\partial \lambda}{\partial u}$$ is the same thing as $$\frac{\partial f}{\partial x}$$.

As mentioned, some people have difficulty understanding what an independent variable is.

Penrose is quite good at setting examples that explain things like partial differentiation. Linear independence just means two contour lines (of constant potential, say) that intersect. You want to find a function that relates "variation" along one of the lines to a constant value along the other--you hold one variable constant. If Tach isn't confused about the dependence of parameters like time, or the variation of an angle, within some coordinate system, why is his debate with Pete getting nowhere?

The debate (in the Formal Debates thread, I'm not making this up) appears to be hung up on the definition of independence and what partial differentiation means, or how to do it properly.

And here we are. Tach appears to be unwilling to yield to any other authority. He more or less told me that Penrose didn't know what he was talking about. Now he's telling everyone who's responded to this thread the same thing.

How big a shovel do you need? I mean, really.

And here we are. Tach appears to be unwilling to yield to any other authority. He more or less told me that Penrose didn't know what he was talking about. Now he's telling everyone who's responded to this thread the same thing.

How big a shovel do you need? I mean, really.
Perhaps it would just be easier for Pete to arrange for a mirrored wheel to be relativistically accelerated in the lab?

How big a shovel do you need? I mean, really.
Hopefully, this video will help clarify your misunderstandings, it is really excellent. It is not that I disagree with Penrose, it is that you misinterpret what he's writing. It is very simple, really, if:

$$f=f(t, u(t))$$

then the total derivative of f wrt t is:

$$\frac{df}{dt}=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial u}\frac{d u}{d t}$$

where $$\frac{\partial f}{\partial t}$$ is the partial derivative of f wrt t.

Now, see if you can apply the above explanation to a simple exercise.
Let $$f=3t+u$$
$$u(t)=sin^2 (t)$$

Find the total and the partial derivative of f wrt t. What is the difference between the two?

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