# Fundamental confusions of calculus

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

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1. ### RealityCheckBannedBanned

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Hi arfa brane.

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

Cheers!

.

Last edited: Feb 15, 2012

3. ### TachBannedBanned

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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:

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

5. ### TachBannedBanned

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

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.

Last edited: Feb 15, 2012

7. ### rpennerFully WiredValued Senior Member

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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.

8. ### TachBannedBanned

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

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

Last edited: Feb 15, 2012
9. ### rpennerFully WiredValued Senior Member

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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.

10. ### TachBannedBanned

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

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.

Last edited: Feb 15, 2012
11. ### PeteIt's not rocket surgeryRegistered Senior Member

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

12. ### przyksquishyValued Senior Member

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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.

13. ### TrippyALEA IACTA ESTStaff Member

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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.

14. ### TachBannedBanned

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Well, you understood it allright in post 22.

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

15. ### TrippyALEA IACTA ESTStaff Member

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Someone really should confiscate your shovel.

This is not what you said:

This is:
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.

Last edited: Feb 15, 2012
16. ### TachBannedBanned

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

17. ### przyksquishyValued Senior Member

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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:

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And yet:
Post #27

19. ### TrippyALEA IACTA ESTStaff Member

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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.

20. ### TachBannedBanned

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...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.

21. ### TachBannedBanned

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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?

22. ### arfa branecall me arfValued Senior Member

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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].

23. ### przyksquishyValued Senior Member

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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.