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June 27

Product Rule Proof

I'm learning about the product rule, and I am trying to understand a common proof of the rule. I understand that one way to prove the product rule is as follows:

$h'(x)=\lim _{a\to 0}{\frac {h(x+a)-h(x)}{a}}=\lim _{a\to 0}{\frac {f(x+a)g(x+a)-f(x)g(x)}{a}}$

$=\lim _{a\to 0}{\frac {f(x+a)g(x+a)-f(x)g(x+a)+f(x)g(x+a)-f(x)g(x)}{a}}$

$=\lim _{a\to 0}{\frac {[f(x+a)-f(x)]\cdot g(x+a)+f(x)\cdot [g(x+a)-g(x)]}{a}}$

$=\lim _{a\to 0}{\frac {f(x+a)-f(x)}{a}}\cdot \lim _{a\to 0}g(x+a)+\lim _{a\to 0}f(x)\cdot \lim _{a\to 0}{\frac {g(x+a)-g(x)}{a}}$

$=f'(x)g(x)+f(x)g'(x)$.

I'm wondering where $-f(x)g(x+a)+f(x)g(x+a)$ comes from in the numerator of the second step of the proof. I understand that both sets of terms cancel, but I am confused as to how one would know to multiply $f(x)$ and $g(x+a)$ and why this step is necessary.

Any help would be greatly appreciated.

Jtg920 (talk) 02:58, 27 June 2014 (UTC)

Well, it's needed to get to the next step. I'm not sure how the original concept behind the proof was discovered, but it most probably is due to Isaac Newton or Gottfried Leibniz, and it would be difficult to find their justification. — Arthur Rubin (talk) 04:03, 27 June 2014 (UTC)

Well, the concept is pretty obvious, isn't it? I mean, if you've seen it before. Just draw a square with base f(x) and height g(x), let x increase a tiny bit, and look at what happens to the area, ignoring the bit that's tiny in both dimensions.

Now, how to get from the concept to the algebra, that's just messing around until you get it right. --Trovatore (talk) 04:10, 27 June 2014 (UTC)

The article Product rule attributes it to Leibniz, and gives a reference, which appears no longer to be available online without a subscription or membership. — Arthur Rubin (talk) 04:12, 27 June 2014 (UTC)

The "trick" step is that $AB-ab=(A-a)B+a(B-b)$. To see this, consider a rectangle of sides A and B (we'll assume these are larger than a and b, respectively). Remove from that rectangle a smaller rectangle of sides a and b. The remaining L-shaped region can be cut into two rectangles of sides $(A-a),B$ and $a,(B-b)$. Sławomir Biały (talk) 20:53, 27 June 2014 (UTC)

Adding and subtracting the same terms is a part of the "standard arsenal" of tricks that mathematicians use. If you're asking how on Earth anyone knew to do that (which was my reaction when I was taking calculus), the answer is probably just trial and error, along with experience about what has worked in similar situations. OldTimeNESter (talk) 20:11, 27 June 2014 (UTC)

The two middle terms $-f(x)g(x+a)+f(x)g(x+a)$ are introduced for regrouping and factoring, a 'trick' which is somewhat similar to what you may recall from Algebra, in completing the square where you add a value so you must subtract the same value to keep the equation true. El duderino ^(abides) 07:31, 28 June 2014 (UTC)

June 29

Natural combinatoric numberings

1. The k-selections (from an ordered set of size n) in lexicographic order have a very convenient representation in base n because there are $n^{k}$ of them.
2. The k-permutations similarly use the factorial number system because there are $(n)_{k}$ of them.
3. The combinations are harder because ${\binom {n}{k}}$ does not(?) have a simple expression as a product of integers, but they have something almost as convenient (and still lexicographic) from the Pascal's triangle identity ${\binom {n}{k}}={\binom {n-1}{k}}+{\binom {n-1}{k-1}}$: you assign the first ${\binom {n-1}{k}}$ indices to the combinations that involve the first element, and so recursively among the sets with and without it.
4. The multinomial selections may be treated by combining the combination technique with a mixed radix system based on ${\binom {n}{k_{1},k_{2},\ldots ,k_{m}}}={\binom {n}{k_{1}}}{\binom {n-k_{1}}{k_{2},\ldots ,k_{m}}}$.
5. The multinomials may easily be extended to the case where some subset of s elements are taken to be indistinguishable by dividing by $s!$ — or even $\prod _{i}s_{i}!$. (Not so for the k-permutations or combinations with $k<n$, since not all of the indistinguishable elements are involved in each pattern.)

Questions, with a mind to writing code to interconvert between (subsets of) these arrangements and their indices:

1. Is my statement about factoring ${\binom {n}{k}}$ correct? (Of course it has a factorization! I mean a convenient one for this purpose.) Is there a better ordering (lexicographic or no) for combinations?
2. What natural ordering/numbering is natural for the multinomials with indistinguishable sets? --Tardis (talk) 03:47, 29 June 2014 (UTC)

July 1

What are these percents

I'm reading a message and it contains the following:

"Power to detect standardized effect is:

0.43 : 47% power to detect difference if 38 are in control and 38 have SE

0.86 : 96% power to detect difference if 38 are in control and 38 have SE

There is an 80% power to detect a standard difference of 0.64"

I understand how the values of 0.43 and 0.86 are calculated, using power calculations. It appears that there is a lookup table or something similar that converts 0.43 to 47%. However, I do not know of such a table. Does it make sense to anyone here?