On Mon, 16 Nov 2009 14:32:00 +0100, Invisible <voi### [at] devnull> wrote:
>
> Damnit, I really suck at math. :'{

 From what I can tell, you basically suffer from a lack of education. That  
can be remedied, if you are willing to make the effort.



-- 
FE

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>> One formula appears to say the derivative is f(x) * g(x), and another 
>> appears to say it's f'(g(x)) * g'(x), which is different, but these 
>> are supposed to be different notations for the same formula...?
> 
> The first one is obviously wrong, unless you mixed up all the f's, g's 
> and little circles. It is difficult to tell without context.

Well, one formula says something like dy/dx = dy/du * du/dx, while the 
other says (f.g)'(x) = f(g(x)) * g(x).

>>> df(f(x))/dx = f'(f(x))*f'(x) = g(f(x))*g(x)
>>
>> Right. So basically, the function I'm looking for is f'(f(x))*f'(x)?
> 
> Yes.

Nice to know I got something right today...

>> And in general, the Nth composition will be ever more complex as N 
>> increases?
> 
> Yes.

Hmm.

>> ...what I *should* have done of course is
>>  http://www.wolframalpha.com/input/?i=derivative+f(f(x))
> 
> Except that it is the wrong answer, or rather the wrong question.

Yes. The query should be f[f[x]] instead of f(f(x)). If you ask for a 
higher iterate, it quickly becomes obvious that something is wrong.

> Just one more reason why it is better to learn the formulas yourself 
> instead of relying on a computer to help you all the time.

Oh, sure. But I find it useful to check against the computer. If the 
answers are the same, you probably did it right. If the answers are 
different, either you asked the computer the wrong thing or you stuffed 
up the math.

(For example, (Sqrt(3) a)^3 = 3 a^2. Or NOT, in fact - as Wolfram 
rightly pointed out...)

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Fredrik Eriksson wrote:

> From what I can tell, you basically suffer from a lack of education. 

Well, I _did_ go to a school for stupid people.

> That can be remedied, if you are willing to make the effort.

I'm more concerned about the time and cost aspect of it, but sure.

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On 16-11-2009 14:42, Invisible wrote:
> Fredrik Eriksson wrote:
> 
>> From what I can tell, you basically suffer from a lack of education. 
> 
> Well, I _did_ go to a school for stupid people.
> 
>> That can be remedied, if you are willing to make the effort.
> 
> I'm more concerned about the time and cost aspect of it, but sure.

It will cost time but not so much money. It is for your personal 
education, so you don't need to enroll in university. And even time is 
relative here, as long as you are trying to compute these things _not_ 
studying it will cost you more time.

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On Mon, 16 Nov 2009 14:41:12 +0100, Invisible <voi### [at] devnull> wrote:
>
> Well, one formula says something like dy/dx = dy/du * du/dx

Yes, but note the use of 'u', which in this case basically represents the  
inner function.



-- 
FE

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On 16-11-2009 14:15, Invisible wrote:
> Fredrik Eriksson wrote:
> 
>> Basic high-school level math. I am slightly surprised that you did not 
>> know it, though I am not in the least surprised that you did not 
>> bother to google for it.
> 
> ...what I *should* have done of course is
> 
> http://www.wolframalpha.com/input/?i=derivative+f(f(x))
> 
> *facepalm*

no, because that is a derivative with respect to f not x.
your answer might be 
http://www.wolframalpha.com/input/?i=derivative+f%20f%20x%20with%20respect%20to%20x
but then again it might not, depending on what you really want.

Many of my students are (hopefully: were) unable to perform a long 
division or even a long multiplication. They weren't though at school 
because 'everybody has a calculator at hand anyway'.

Someday in the near future all the rest of mathematics and calculus will 
not be though anymore because 'everybody has an internet connection at 
hand anyway'. That will be a sad day for our culture.

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>> ...what I *should* have done of course is
>>
>> http://www.wolframalpha.com/input/?i=derivative+f(f(x))
>>
>> *facepalm*
> 
> no, because that is a derivative with respect to f not x.

Yes, I figured that out already. (It's actually quite nice that 
Mathematica has an unambiguous notation for this - but you have to 
remember to use it!)

> Many of my students are (hopefully: were) unable to perform a long 
> division or even a long multiplication. They weren't though at school 
> because 'everybody has a calculator at hand anyway'.

At my school, we did nothing *but* long division! Pages and pages and 
pages of it, for years on end. From when I was 14 until I left school, 
_all_ we did in our maths lessons was long division. (Not to mention the 
homework, which was... long division.)

> Someday in the near future all the rest of mathematics and calculus will 
> not be though anymore because 'everybody has an internet connection at 
> hand anyway'. That will be a sad day for our culture.

I'm not sure they teach calculus *currently* in schools. Then again, I'm 
not sure because I didn't attend a normal school... I think in this 
country, if you want to learn about mathematics (not just arithmetic), 
you have to take a special course. I think it's outside the normal 
school curriculum.

Then again, how many people will ever need to know how to compute the 
integral of a polynomial? If you can't add, you're going to have a bit 
of a problem doing your weekly shopping, but there's not much need for 
higher math unless you're working in some specialist industry somewhere. 
(Or you're an academic, which nobody is.)

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>> Well, one formula says something like dy/dx = dy/du * du/dx
> 
> Yes, but note the use of 'u', which in this case basically represents 
> the inner function.

This is one of the more confusing aspects of calculus. The duality 
between "functions" and "variables that share a relationship". That and 
the multiple notations for the same concept...

I still don't entirely understand what you're saying, but presumably 
you're right.

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On Mon, 16 Nov 2009 15:11:13 +0100, Invisible <voi### [at] devnull> wrote:
>
> I still don't entirely understand what you're saying

(Excuse the mixing of Lagrange notation and ASCII-fied Leibniz notation)

First look at y = f(x). Then f'(x) = dy/dx.

Now try y = f( g(x) )

Let u = g(x). Then y = f( u ).

In f'(x) = dy/dx, we can trivially introduce 'u' thus:

  f'(x) = dy/dx * (du/du) = (dy/du) * (du/dx)

dy/du is obviously the derivative of 'y' with respect to 'u', i.e. f'(u).

Similarly, du/dx is the derivative of 'u' with respect to 'x', i.e. g'(x).

  f'(x) = (dy/du) * (du/dx) = f'(u) * g'(x)

Substitute u = g(x) to get:

  f'(x) = f'( g(x) ) * g'(x)



-- 
FE

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> Then again, how many people will ever need to know how to compute the 
> integral of a polynomial?

This is why it's not taught to everyone, and is reserved until when people 
*choose* they want to do maths (when you're 16 in the UK).

> If you can't add, you're going to have a bit of a problem doing your 
> weekly shopping, but there's not much need for higher math unless you're 
> working in some specialist industry somewhere.

You'd be surprised how many industries require specialist knowledge :-)  For 
example, I suspect the structural engineer that designed your building did 
plenty of calculations using calculus to estimate stresses based on the 
loading (simple cases can be looked up in tables, but anything unusual needs 
to be worked out manually).  Ditto for an electrical engineer who designed 
the power supply for your computer, the DCDC converter in your mobile phone 
and numerous other circuits - without an understanding of calculus you're 
going to be totally lost.

Of course computers can simulate and calculate stuff for you, but you're 
going to look a right idiot if you need to run ten 12 hour simulations to 
decide the correct structure or capacitor size when your colleague can work 
it out exactly in a few minutes with a piece of paper.

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