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

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

Mathematics
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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
Looks like lots of chain rule is involved.
...also directly from MIT OCW

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Other answers:

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 \(p+q\).
^thinking along the right lines :)
The minimum would be \[2\cdot p! \cdot q!\] but I'm not sure if that's also the maximum value.
If I had to take a wild guess, I'd say the answer would be \[p \cdot q \cdot (p! \cdot q!)\]
not sure what you mean by minimum... that's close to the answer I got the first time, but it's not quite right
what math is this btw?
Basic calculus =) hint? (small one, I promise)
No hint yet. give me a couple more minutes for my next idea.
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 \[f^{(p+q)}=(p+q)!\]
haha, I love that solution! so different and simpler than the one I know
I love using the word "stuff" in mathematical proofs.
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
*oops Leibniz formula:\[(uv)^{(n)}=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)u^{(n-k)}v^k\]
plug in n=p+q and you find that k=q is the only surviving (non-vanishing) term \[\left(\begin{matrix}p+q\\q\end{matrix}\right)u^{(p)}v^{(q)}=(p+q)!\]
I still like my way better, but that's still really interesting.
I agree, your's is simpler, which is more elegent

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