## TuringTest Group Title Bonus problem: Let$f(x)=x^p(1+x)^q$find$f^{(p+q)}(x)$ 2 years ago 2 years ago

1. TuringTest Group Title

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

2. KingGeorge Group Title

Looks like lots of chain rule is involved.

3. TuringTest Group Title

...also directly from MIT OCW

4. KingGeorge Group Title

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 $$p+q$$.

5. TuringTest Group Title

^thinking along the right lines :)

6. KingGeorge Group Title

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

7. KingGeorge Group Title

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

8. TuringTest Group Title

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

9. Bosniac32 Group Title

what math is this btw?

10. TuringTest Group Title

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

11. KingGeorge Group Title

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

12. KingGeorge Group Title

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 $f^{(p+q)}=(p+q)!$

13. TuringTest Group Title

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

14. KingGeorge Group Title

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

15. TuringTest Group Title

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

16. TuringTest Group Title

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

17. TuringTest Group Title

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

18. KingGeorge Group Title

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

19. TuringTest Group Title

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