Prove the Leibnitz rule of derivatives

[Karol]

Because i don't know what to do with that:
$$\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{(n+1)!}{(k+1)!(N-K)!}$$
So?
I am not guessing. n=5, k=3:
$$[(n+1) - (k+1)]! =(1\cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 - 1\cdot 2 \cdot 3 \cdot 4 )=(1\cdot 2 \cdot 3 \cdot 4 )(5 \cdot 6 -1 )=(k+1)![n(n+1)-1]$$

 

[Orodruin]

Because i don't know what to do with that:
$$\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{(n+1)!}{(k+1)!(N-K)!}$$

You have one more $n+1$ and one more $k+1$. Why did you not make the substitution there as well?

 

[fresh_42]

So?
I am not guessing. n=5, k=3:

$[(n+1)−(k+1)]!=(1⋅2⋅3⋅4⋅5⋅6−1⋅2⋅3⋅4)=(1⋅2⋅3⋅4)(5⋅6−1)=(k+1)![n(n+1)−1]$​

This is completely wrong, simple as that.
$(5-3)!=2!=2\neq 5!-3!=120-6=114$ and $(n-m)!\neq n!-m!$.

Repetition never makes something true, which has been wrong in the first place.

 

[Orodruin]

I am not guessing. n=5, k=3:
$$[(n+1) - (k+1)]! =(1\cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 - 1\cdot 2 \cdot 3 \cdot 4 )=(1\cdot 2 \cdot 3 \cdot 4 )(5 \cdot 6 -1 )=(k+1)![n(n+1)-1]$$

No, this is wrong. The factorial is outside of the parentheses! With $n = 5$ and $k = 3$ you have
$$
[(n+1)-(k+1)]! = [6-4]! = 2! = 2
$$
and
$$
(k+1)! (n-1) = 4! \cdot 4 = 96.
$$
Clearly not the same thing.

Edit: In addition, even if it did hold you cannot just take two arbitrary values for unknowns, show that it is true for that particular choice, and then expect that it holds for an arbitrary choice of the unknowns.

 

[SammyS]

$$\frac{n!(n+1)}{(k+1)!(n-k)!}=\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{(n+1)!}{(k+1)!k!}$$

$\displaystyle \frac{n!(n+1)}{(k+1)!(n-k)!}=\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}\ \ \ne\ \ \frac{(n+1)!}{(k+1)!k!}$

Now, take $\displaystyle \ \frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}\$ and make the substitution @Orodruin suggested:

...
What happens if you let $n+1 = N$ and $k+1 = K$ ?

 

[Karol]

$$\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{N!}{K!(N-K)!}=\frac{n!}{k!(n-k)!}=\frac{n!}{(k!)^2}$$

 

[Orodruin]

First step ok. The next two are clearly false. What is your argumentation? This is what I mean by "just guessing" because you can easily try a few values for $n$ and $k$ and see that it is non-sensical.

 

[Karol]

$$\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{N!}{K!(N-K)!}={n+1 \choose k+1}$$
$$\frac{d^{n+1}(uv)}{dx^{n+1}}=u^{n+1}v+\sum_{k=0}^{n-1} \left( \begin{array}{m} n+1\\k+1 \end{array} \right) u^{n-k}v^{k+1}+uv^{n+1}$$
But i have to prove:
$$\sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) u^{n-k}v^k\ \ \rightarrow\ \ \sum_{k=0}^{n+1} \left( \begin{array}{m} n+1\\k+1 \end{array} \right) u^{n+1-k}v^{k+1}$$
If i substitute k=0 the first term in the sum gives $~{n+1 \choose 1}~$ and doesn't equal 1, as needed

 

[Orodruin]

$$\frac{(n+1)!}{(k+1)![(n+1)-(k+1)]!}=\frac{N!}{K!(N-K)!}={n+1 \choose k+1}$$
$$\frac{d^{n+1}(uv)}{dx^{n+1}}=u^{n+1}v+\sum_{k=0}^{n-1} \left( \begin{array}{m} n+1\\k+1 \end{array} \right) u^{n-k}v^{k+1}+uv^{n+1}$$
But i have to prove:
$$\sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) u^{n-k}v^k\ \ \rightarrow\ \ \sum_{k=0}^{n+1} \left( \begin{array}{m} n+1\\k+1 \end{array} \right) u^{n+1-k}v^{k+1}$$
If i substitute k=0 the first term in the sum gives $~{n+1 \choose 1}~$ and doesn't equal 1, as needed

No, again you did not think it through thoroughly. Also, what you wrote down in the end is not what you have to prove (because it is false!). There should be a $k$ and not a $k+1$ in the binomial coefficient as well as in the derivative of $v$. You just want the implication of the theorem being true for $n$ meaning that it is true for $n+1$ - $k$ is a summation variable and does not play a role - nothing says that you should change the way $k$ appears.

 

[Karol]

I have to prove:
$$\sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) u^{n-k}v^k\ \ \rightarrow\ \ \sum_{k=0}^{n+1} \left( \begin{array}{m} n+1\\k \end{array} \right) u^{n+1-k}v^{k}$$
And i have:
$$u^{n+1}v+\sum_{k=0}^{n-1} \left( \begin{array}{m} n+1\\k+1 \end{array} \right) u^{n-k}v^{k+1}+uv^{n+1}=$$
$$=u^{n+1}v+\sum_{k=1}^{n-1} \left( \begin{array}{m} n+1\\k \end{array} \right) u^{n-k}v^{k}+uv^{n+1}=\sum_{k=0}^{n+1} \left( \begin{array}{m} n+1\\k \end{array} \right) u^{n-k}v^{k}$$

 

[Orodruin]

You forgot to make the shift in the term $u^{n-k}$ when you changed the summation variable.

You really should be able to find these errors yourself and not just assume that things are not the way they are presented in the problem. If you spent 10 minutes looking for this error, you might have found it.

 

[SammyS]

@Karol
You had this in post #10:

$$\frac{d^{n+1}(uv)}{dx^{n+1}}=\left[ \sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) u^{n-k}v^k \right]'=\sum_{k=0}^n \left( \begin{array}{m} n\\k \end{array} \right) (u^{n+1-k}v^k+u^{n-k}v^{k+1})$$
$$=u^{n+1}v+\sum_{k=0}^{n-1} \left[ \left( \begin{array}{m} n\\k \end{array} \right) + \left( \begin{array}{m} n\\k+1 \end{array} \right) \right] u^{n-k}v^{k+1}+uv^{n+1}$$

You got this (I suppose) by splitting the sum into two sums and taking out the k = 0 term from the first, and the k = n term from the second, then you shifted the index in the first sum so that you could combine them into one sum.

Instead of shifting the index so the sum starts at k = 0, shift it so that it starts at k = 1.

You get:

$\displaystyle \sum_{k=1}^{n} \left[ { n \choose {k-1} } + {n \choose {k}} \right] u^{n+1-k}v^{k}$​

I think you'll find that $\displaystyle \ { n \choose {k-1} } + {n \choose {k}} \$ is easier to simplify than the previous combination, and the index will easier to work with in your final result..

 

[Karol]

Note that i changed also the (n-1) above the ∑
$$=u^{n+1}v+\sum_{k=1}^{n} \left( \begin{array}{m} n+1\\k \end{array} \right) u^{n+1-k}v^{k}+uv^{n+1}=\sum_{k=0}^{n+1} \left( \begin{array}{m} n+1\\k \end{array} \right) u^{n+1-k}v^{k}$$

 

[epenguin]

I suggest that the simplest proof of Leibniz' theorem is by algebrisation. The student will in any case meet before too long in differential equations the very strong analogy if not more between differential operators and polynomials.

We start with
\begin{equation} D^{1}\left( uv\right) =uD^{1}v+vD^{1}u \end{equation}
where D¹ denotes the differential operator d/dx (so that also dⁱ/dxⁱ is $D^{i})$. These differential operators follow the same 'indices rule' as ordinary quantities raised to powers in arithmetic and algebra i.e.
\begin{equation} D^{n}D^{m}=D^{n+m} \end{equation}
All the calculus involved finishes there, and from there on it is algebraic.

I found it easier to reason with a slightly more complicated but explicit notation and rewrite eq (1) as
\begin{equation}D^{1}D^{0}(uv)=D^{0}uD^{1}v+D^{1}uD^0{v} \end{equation}
where $D^{0}$ represents differentiating zero times, so that $D^{0}u =u$ .

In the same way I represent the binomial usually written $(a+b)$ as
\begin{equation}(b^{0}a^{1}+b^{1}a^{0}) \end{equation}
Then if we accept as having been proved the usual binomial expansion formula (I don't know if the OP is there yet) which I will write in the form
\begin{equation}\left ( a^{1}b^{0}+a^{0}b^{1}\right) ^{n}=\sum ^{n}_{i=0}\binom {n} {i}a^{i}b^{n-i} \end{equation}
then Leibniz' theorem immediately follows because the formal algebra of the expansion of $(b^{0}a^{1}+b^{1}a^{0})^{n}$ and of $(D^{0}uD^{1}v+D^{1}uD^0{v})^{n}$ are exactly the same because they follow exactly the same index and other rules, so in equation (5) you can just replace $a^{i}b^{n-i}$ with $D^{i}uD^{n-i}v$ and the result is true.

I think at this argument is essential at some point in order to make clear that that the theorem depends on nothing but Equations (1) and (2) (maybe that is saying too much, but at any rate only on something minimal and familiar).

 

[Karol]

Why do i need to expand $~\left ( a^{1}b^{0}+a^{0}b^{1}\right) ^{n}~$ and not just (a+b)ⁿ?
Why do i need $~(D^{0}uD^{1}v+D^{1}uD^0{v})^{n}~$ and not just $~(D^{1}u+D^{1}v)^{n}~$?
$$(D^{1}u+D^{1}v)^{n}=\sum ^{n}_{i=0}\binom {n} {i}D^nn D^{n-i}v$$

 

[SammyS]

Why do i need to expand $~\left ( a^{1}b^{0}+a^{0}b^{1}\right) ^{n}~$ and not just (a+b)ⁿ?
Why do i need $~(D^{0}uD^{1}v+D^{1}uD^0{v})^{n}~$ and not just $~(D^{1}u+D^{1}v)^{n}~$?

... because
$\displaystyle D^{n+1}(uv)=D^n(D(uv))=... \$

 

[epenguin]

Why do i need to expand $~\left ( a^{1}b^{0}+a^{0}b^{1}\right) ^{n}~$ and not just (a+b)ⁿ?
Why do i need $~(D^{0}uD^{1}v+D^{1}uD^0{v})^{n}~$ and not just $~(D^{1}u+D^{1}v)^{n}~$?
$$(D^{1}u+D^{1}v)^{n}=\sum ^{n}_{i=0}\binom {n} {i}D^iu D^{n-i}v$$

That is just my idea actually which I haven't actually seen anywhere, not something you have to do. You can see, I hope, it's not wrong, at worst unnecessary.
The reason I thought of it is that I think it makes it easier - you can handle the two expressions blindly in exactly the same way., I thought that If you were treating
(uDv + vDu)ⁿ in the same way as (a + b)ⁿ a student might start writing things like (uⁿDⁿv + ...). Or at least wondering why he can't. Or to say it another way it makes the analogy between the two things more complete, and you have to think less when applying it. (Maybe you could do with that because actually the LHS of your last equation is not what we are looking for.).

I even thought one might make the analogy (equivalence? isomorphism?...? mathematicians say) more complete by making the ordinary binomial expansion in terms of operations rather than numbers, whatever they are, defining an operation E
E = (a + b)× where × is the multiplication symbol.

However this is open to mathematicians here to discuss what they think is the best way of formulating the argument, including didactically, and maybe do a better set out proof with the same idea - In fact that was I what hoping for.

 

[Karol]

Thank you all very much, Orodruin, Ray, Serena, Sammy, fresh 42 and epenguin