## 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

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

Looks like lots of chain rule is involved.

3. TuringTest

...also directly from MIT OCW

4. KingGeorge

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

^thinking along the right lines :)

6. KingGeorge

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

7. KingGeorge

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

8. TuringTest

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

what math is this btw?

10. TuringTest

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

11. KingGeorge

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

12. KingGeorge

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

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

14. KingGeorge

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

15. TuringTest

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

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

17. TuringTest

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

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

19. TuringTest

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