Bonus problem:
Let$$f(x)=x^p(1+x)^q$$find$$f^{(p+q)}(x)$$

Question

Bonus problem:
Let$$f(x)=x^p(1+x)^q$$find$$f^{(p+q)}(x)$$

turingtest · Answer

last time I posted this joemath just saw the answer, and Mr.Math just referred me to a link...
I wanna see some thought-process this time

kinggeorge · Answer

Looks like lots of chain rule is involved.

turingtest · Answer

...also directly from MIT OCW

kinggeorge · Answer

Let's see, since we're doing $p+q$ derivations, that means that we'll just have a constant term. since the degree of the polynomial would be $p+q$.

turingtest · Answer

^thinking along the right lines :)

kinggeorge · Answer

The minimum would be $$2\cdot p! \cdot q!$$ but I'm not sure if that's also the maximum value.

kinggeorge · Answer

If I had to take a wild guess, I'd say the answer would be $$p \cdot q \cdot (p! \cdot q!)$$

turingtest · Answer

not sure what you mean by minimum...
that's close to the answer I got the first time, but it's not quite right

anonymous · Answer

what math is this btw?

turingtest · Answer

Basic calculus =)
hint?
(small one, I promise)

kinggeorge · Answer

No hint yet. give me a couple more minutes for my next idea.

kinggeorge · Answer

Let's expand everything first and distribute, and see where we go.$$x^p(x+1)^q=x^p\left(\binom{q}{0}x^q + (\text{stuff})\right)=x^{p+q} + \text{stuff}$$That would then give us an answer of $$f^{(p+q)}=(p+q)!$$

turingtest · Answer

haha, I love that solution!
so different and simpler than the one I know

kinggeorge · Answer

I love using the word "stuff" in mathematical proofs.

turingtest · Answer

fyi, the technical way to write "stuff" is $O p(x)$ the big O symbolizing the junk terms
my solution is based on the Leibniz formula$$\sum\left(\begin{matrix}n\k\end{matrix}\right)u^{(n-k)}v^k$$which I'm sure you notice the similarities between that and the binomial theorem

turingtest · Answer

*oops
Leibniz formula:$$(uv)^{(n)}=\sum_{k=0}^{n}\left(\begin{matrix}n\k\end{matrix}\right)u^{(n-k)}v^k$$

turingtest · Answer

plug in n=p+q and you find that k=q is the only surviving (non-vanishing) term
$$\left(\begin{matrix}p+q\q\end{matrix}\right)u^{(p)}v^{(q)}=(p+q)!$$

kinggeorge · Answer

I still like my way better, but that's still really interesting.

turingtest · Answer

I agree, your's is simpler, which is more elegent