Fundamental confusions of calculus

[RealityCheck]

Hang on, if you're substituting for y in the first equation, then a = b since y is constant. Otherwise are you saying y can have two values?

Hi arfa brane.

I think it is Trippy not Temur that you meant to quote there, mate?

Cheers!

.

 

[Tach]

You want to talk about bad faith?

First you say:



Which tells us nothing, other than what I quoted, because you neither define nor use u in this.

One really needs to be arguing in bad faith if one cannot identify $$u=sin^2(\theta)$$ when given the general form $$f=f(\theta,u(\theta),v(x))$$ IN THE SAME POST as
$$f(\theta,u(\theta),v(x))=3 \theta+sin^2(\theta)+ln(x)$$.

Especially AFTER I made clear that we are talking about the relationship between partial and total derivatives:

$$\frac{df}{du}=\frac{\partial f}{\partial \theta}+\frac{\partial f}{\partia u}\frac{\partial u}{\partial \theta}$$

 

[Tach]

So $$\partial_{\theta}\sin^{2}\theta$$ isn't zero, yet you said in post 18 it is. Sorry, you've just massively contradicted yourself.

BUT this is not what I said, you continue to grossly distort what has been posted.

Also, you continue to fail to address the potential term and by redefining the coordinate you're being dishonest because you're moving the goal post. In post 18 you take the $$\theta$$ derivative of a theta dependent function, namely $$\sin^{2}\theta$$ and you say it's zero.

One really needs to be arguing in bad faith if one cannot identify $$u=sin^2(\theta)$$ when given AT THE TOP OF THE POST the general form $$f=f(\theta,u(\theta),v(x))$$ IN THE SAME POST as
$$f(\theta,u(\theta),v(x))=3 \theta+sin^2(\theta)+ln(x)$$.

Especially AFTER I made clear that we are talking about the relationship between partial and total derivatives:

$$\frac{df}{du}=\frac{\partial f}{\partial \theta}+\frac{\partial f}{\partial u}\frac{\partial u}{\partial \theta}$$


Let me ask you (all) a simple question: identify $$v(x)$$ in post 18. Identify $$u(\theta)$$ in post 18.


You didn't show the "other 5 PhD's" the complete post 18, did you? Try again, by showing them the complete post 18, without clipping the top. At least one of them should be able to recognize that $$f=f(\theta,u(\theta),v(x))$$ PLUS $$f(\theta,u(\theta),v(x))=3 \theta+sin^2(\theta)+ln(x)$$ MEANS $$u=sin^2(\theta)$$ , thus explaining why $$\frac{\partial f}{\partial \theta}=3$$ without coming to the silly misinterpretation that I consider that $$\partial_{\theta}\sin^{2}\theta=0$$. Try it. I trust that you are honest enough to report back what the "other PhD's" said.

 

[rpenner]

One really needs to be arguing in bad faith if one cannot identify $$u=sin^2(\theta)$$ when given AT THE TOP OF THE POST the general form $$f=f(\theta,u(\theta),v(x))$$ IN THE SAME POST as
$$f(\theta,u(\theta),v(x))=3 \theta+sin^2(\theta)+ln(x)$$.

Then you are arguing in bad faith since in post #27 you wrote $$u = \sin \theta$$ and wanted us to read $$f(\theta, u, x) = 3 \theta + u^2 + \ln x$$ which is different than what you just said.

So quit shouting and read my (updated) post #176.

Noone, including yourself, knows what you meant when you wrote post #18 because you scrambled at least two (partial) definitions of f . Since then, you have presented contradictory definitions of f.
As a consequence you said $$\frac{\partial \quad}{\partial \theta} \left( 3 \theta + \sin^2 \theta + \ln x \right) = 3$$ which is not true in any context.

 

[Tach]

Then you are arguing in bad faith since in post #27 you wrote $$u = \sin \theta$$ and wanted us to read $$f(\theta, u, x) = 3 \theta + u^2 + \ln x$$ which is different than what you just said.

You surely realize that this produces the SAME exact result , $$\frac{\partial f}{\partial \theta}=3$$. It also produces the same exact $$\frac{df}{d \theta}$$

As a consequence you said $$\frac{\partial \quad}{\partial \theta} \left( 3 \theta + \sin^2 \theta + \ln x \right) = 3$$ which is not true in any context.

...which was sloppy. Yet, for anyone that understood that we are dealing with $$f=f(\theta, u(\theta), v(x))$$ and with the chain rule, would have had no problem understanding why $$\frac{\partial f}{\partial \theta}=3$$.
Paradoxically, przyk understood on the spot ,

I'll use one of Tach's examples to clarify things. This one's the most interesting because $$f$$ has both an explicit dependence on $$\theta$$ and an implicit dependence through $$u$$:

For convenience I'll shorten that to $$f(\theta,\,u(\theta),\, v)$$. Then strictly speaking, the function $$f$$ is a function of $$\theta$$, $$u$$ and $$v$$ and technically the partial derivative $$\frac{\partial f}{\partial \theta}$$ is just what Tach said it is: the derivative of $$f$$ only through its explicit dependence on $$\theta$$, with $$u$$ and $$v$$ held constant.

I am quite sure that you understand it just the same, especially after being explained so many times.

 

[rpenner]

Paradoxically, przyk understood on the spot ,

... only to go on the "Tach posted this, therefore it must be wrong" several posts later. I am quite sure that you understand it just the same, especially after being explained so many times.

Actually, he immediately started educating you in post #22 that if you fix u as a function of theta, then you have fundamentally started speaking about a different function than the original definition of f, which is the bulk of post #176's new material.

Strictly speaking if you substitute $$u = u(\theta)$$ you've defined a different function:

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

which you may also want to take partial derivatives of.

 

[Tach]

Actually, he immediately started educating you in post #22 that if you fix u as a function of theta, then you have fundamentally started speaking about a different function than the original definition of f, which is the bulk of post #176's new material.

The original function , as defined in post 18 is $$f=f(\theta,u(\theta),v,w...)$$:

The difference between total derivative and partial derivative is really simple, Pete, and it was spelled out quite clearly in the post you quoted: partial derivative is taken with respect to $$\theta$$ being an explicit argument, total derivative is the sum of the partial derivative (if it exists) plus the chain derivatives due to certain arguments being functions of $$\theta$$. You do not need to spread the same question over three threads in order to understand this.
In math terms: if $$f=f(\theta, u(\theta),v,w,...)$$

then:


The total derivative wrt $$\theta$$is:


$$\frac{df}{d \theta}=\frac{\partial f}{\partial \theta}+\frac{\partial f}{\partial u} \frac{du}{d \theta}$$

while the partial derivative wrt $$\theta$$ is:

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

If you think otherwise, then please explain. It takes an enormous amount of bad faith to argue that this is not the case. Why are you doing this? Especially after receiving repeated clarifications?

Yes, I could have written the example a little neater. No, I never imagined that so many people with math PhD's will not figure out that $$u(\theta)$$ is $$sin^2(\theta)$$ in the example. The most baffling thing is that I can count 11 times I explained this in subsequent posts, to no avail.

 

[Pete]

To everyone else I'm wondering whether people want this thread to continue. It's currently not a flame war but we are going round in circles with Tach. If people are content to go around and around for some more time, hoping Tach at some point actually tries to calculate something like the definition of the partial derivative applied to his function, then fine. If you think enough is enough then say so and when I get up in the morning I'll check the thread and see which way the wind is blowing.

I've got want I wanted. Thanks to all for your patience, and sorry for the shambles that developed (that wasn't my intention).

 

[przyk]

Paradoxically, przyk understood on the spot , only to go on the "Tach posted this, therefore it must be wrong" several posts later.

No, the first time in this thread I told you you got anything wrong was in [POST=2902775]post #26[/POST], in response to [POST=2902767]post #23[/POST], where you clearly and unambiguously did get something wrong. I elaborated in [POST=2902800]post #29[/POST].

The next time after that I told you you were wrong about something was in [POST=2902983]post #43[/POST]. That was in response to your [POST=2902804]post #30[/POST], where you wrote some complete and utter bullshit about partial differentiation.

Incidentally, my [POST=2902760]initial post[/POST] in this thread was a response to [POST=2902629]post #17[/POST] by Pete. I spent a while composing that reply and I don't think I'd even read your now infamous post #18 at that point.

 

[Trippy]

Hang on, if you're substituting for y in the first equation, then a = b since y is constant. Otherwise are you saying y can have two values?

No, in the first equation I'm saying that a=y and b=y[sup]2[/sup] (and implicitly that b=a[sup]2[/sup]). I thought about using a and a[sup]2[/sup] but I wanted to emphasize the point of treating them as constants.

 

[Tach]

No, the first time in this thread I told you you got anything wrong was in [POST=2902775]post #26[/POST],

Well, you understood it allright in post 22.

The next time after that I told you you were wrong about something was in [POST=2902983]post #43[/POST]. That was in response to your [POST=2902804]post #30[/POST], where you wrote some complete and utter bullshit about partial differentiation.

Yes, the post where you no longer admit that $$f=f(\theta,u(\theta),v)$$.

 

[Trippy]

The difference between total derivative and partial derivative is really simple, Pete, and it was spelled out quite clearly in the post you quoted: partial derivative is taken with respect to $$\theta$$ being an explicit argument, total derivative is the sum of the partial derivative (if it exists) plus the chain derivatives due to certain arguments being functions of $$\theta$$. You do not need to spread the same question over three threads in order to understand this.
In math terms: if $$f=f(\theta, u(\theta),v,w,...)$$

then:


The total derivative wrt $$\theta$$is:


$$\frac{df}{d \theta}=\frac{\partial f}{\partial \theta}+\frac{\partial f}{\partial u} \frac{du}{d \theta}$$

while the partial derivative wrt $$\theta$$ is:

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

Example:

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

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

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

One really needs to be arguing in bad faith if one cannot identify $$u=sin^2(\theta)$$ when given the general form $$f=f(\theta,u(\theta),v(x))$$ IN THE SAME POST as...

Someone really should confiscate your shovel.

This is not what you said:

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

This is:

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

This: $$f(\theta,u(\theta),v(x))$$ is a special case of this: $$f(\theta, u(\theta),v,w,...)$$

But that still does not change the example you presented.

 

[Tach]

This: $$f(\theta,u(\theta),v(x))=3 \theta+sin^2(\theta)+ln(x)$$ is a special case of this: $$f=f(\theta, u(\theta),v,w,...)$$

But that still does not change the example you presented.

So, can you identify $$v(x)=ln(x)$$? What is then the expression for $$u(\theta)$$?

 

[przyk]

Tach, your reply in post #23 and the bullshit in post #30 were unambiguously wrong regardless of how you defined $$f$$. If it's not clear, the part of post #30 I'm calling bullshit is all of this:

Either way , your claim that $$\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, \sin(2\theta)$$ is false since the partial derivative is always taken wrt the direct variable (in this case, $$\theta$$) and not through a function (in this case $$\sin(\theta)$$) . The moment you start using chain differentiation you cease calculating partial differentials and you are starting to calculate total differentials.

 

[Trippy]

One really needs to be arguing in bad faith if one cannot identify $$u=sin^2(\theta)$$

And yet:

Er, nope. As written, $$\frac{\partial}{\partial \theta} \Bigl( 3 \theta \,+\, \sin(\theta)^{2} \,+\, \ln(x) \Bigr) \,=\, 3 \,+\, \sin(2\theta)$$.

Err, wrong. Partial derivative means derivative wrt the explicit variable only. What you took is the total derivative wrt $$\theta$$. Think about:

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

where $$u=sin(\theta)$$.

Post #27

 

[Trippy]

So, can you identify $$v(x)=ln(x)$$? What is then the expression for $$u(\theta)$$?

Not what I stated, nor is it implied by what I stated - in fact there is a typo in that post, which I shall now rectify.

 

[Tach]

Tach, your reply in post #23 and the bullshit in post #30 were unambiguously wrong regardless of how you defined $$f$$. If it's not clear, the part of post #30 I'm calling bullshit is all of this:

...because you had just started refusing to accept that $$sin^2(\theta)$$ is $$u$$. It is a simple exercise in chain derivatives, post 18 is quite clear.

 

[Tach]

And yet:

Post #27

I already explained to you earlier in this thread that you can take $$u=sin^2(\theta)$$ in $$f=3 \theta +u+v$$ or $$u=sin(\theta)$$ in $$f=3 \theta +u^2+v$$ , it doesn't change ANYTHING. Remember the recent example we have gone over?

 

[arfa brane]

No, in the first equation I'm saying that a=y and b=y[sup]2[/sup] (and implicitly that b=a[sup]2[/sup]).

Ah yes. Sorry I should have had a[sup]2[/sup]=b.

But does introducing two "new" variables really say anything? Since a and b both depend on y, you might get confused about it and think you now have a function of three variables? I think, personally, that index notation shouldn't be unnecessarily cluttered that way, it's not something I've seen either.

Pedantically, if you have y constant then you have the "new" function where y = a:

f[sub]a[/sub] = x[sup]2[/sup] + xa + a[sup]2[/sup].

 

[przyk]

...because you had just started refusing to accept that $$sin^2(\theta)$$ is $$u$$.

The material I cited has nothing to do with how $$u$$ is defined.

Those are unambiguous errors you made that you can't dress up any other way. Live with it, Tach. This is only an issue at all because you keep trying to deflect them instead of owning up to them.