Problem 43

ThePerfectHacker · #1 $Old$ 12-04-2007, 10:26 PM

1)Let $a_1,a_2,...,a_n$ be real (or complex) distinct numbers where $n\geq 2$ . Define $P_k = \prod_{j\not = k} (a_k - a_j)$ . Can it be that $P_1=P_2=...=P_n$ ?
(For example, let $a_1=1,a_2=2,a_3=3$ then $P_1 = (1-2)(1-3)=2$ , $P_2 = (2-1)(2-3)=-1$ , and $P_3=(3-1)(3-2)=2$ but all three are not the same in this case).

The next problem is for the younger kids so please give them a chance.

2)Let $f(x)$ and $g(x)$ be functions which you can differenciate. Note that $[f(x)g(x)]' = f'(x)g(x)+f(x)g'(x)$ . Can you find a formula for $[f(x)g(x)]^{(n)}$ where by $^{(n)}$ means to preform differenciation $n$ times repeatedly.

DivideBy0 · #2 $Old$ 12-04-2007, 11:28 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

2)Let $f(x)$ and $g(x)$ be functions which you can differenciate. Note that $[f(x)g(x)]' = f'(x)g(x)+f(x)g'(x)$ . Can you find a formula for $[f(x)g(x)]^{(n)}$ where by $^{(n)}$ means to preform differenciation $n$ times repeatedly.

After a little experimenting:

$[f(x)g(x)]'=f'(x)g(x) + f(x)g'(x)$

$[f(x)g(x)]''=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)$

$[f(x)g(x)]''' = f'''(x)g(x)+3f''(x)g'(x)+3f'(x)g''(x)+f(x)g'''(x)$

This reminds me of the binomial theorem:

$(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k} y^{k}$

So perhaps

$[f(x)g(x)]^{(n)} = \sum_{k=0}^n {n\choose k} f^{n-k}(x) g^{k}(x)$

Where $h^n(x)$ denotes the nth derivative of $h(x)$

Nice problem

P.S Is there any way to hide text?

topsquark · #3 $Old$ 12-05-2007, 08:11 AM

Quote:

Originally Posted by DivideBy0 $View Post$

P.S Is there any way to hide text?

Color the text white.

-Dan

ThePerfectHacker · #4 $Old$ 12-05-2007, 10:51 AM

Quote:

Originally Posted by DivideBy0 $View Post$

Nice problem

Good job. This result is called "Leibniz's Rule".

topsquark · #5 $Old$ 12-11-2007, 10:06 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

1)Let $a_1,a_2,...,a_n$ be real (or complex) distinct numbers where $n\geq 2$ . Define $P_k = \prod_{j\not = k} (a_k - a_j)$ . Can it be that $P_1=P_2=...=P_n$ ?
(For example, let $a_1=1,a_2=2,a_3=3$ then $P_1 = (1-2)(1-3)=2$ , $P_2 = (2-1)(2-3)=-1$ , and $P_3=(3-1)(3-2)=2$ but all three are not the same in this case).

Since I know the way your mind works, there is something wrong with what I am about to say...

Consider the case n = 3. Then the question is can
$a_1a_2 = a_1a_3 = a_2a_3$
where all three are distinct?

Obviously not since $a_1a_2 = a_1a_3 \implies a_2 = a_3$ .

Now, there's got to be something screwy here because I can generalize this argument to larger n and come up with similar results. But I can't believe it would be this easy...

-Dan

ThePerfectHacker · #6 $Old$ 12-11-2007, 10:37 AM

I came up with problem #1 accidently when I playing around with polynomials. Here is my original solution, however it seems to me that this problem is easy even if approached directly.

Proof:
Let $a_1,...,a_n$ be real (or complex numbers) and define $f(x) = (x-a_1)...(x-a_n)$ . The key step is to note that $f'(x_k) = P_k$ by using the general product rule for derivatives. Now, $\deg f(x) = n\geq 2$ thus, $\deg f'(x) = n-1\geq 1$ , this means that $f'(x)$ cannot attain the same values at $a_1,a_2,...,a_n$ (meaning $f(a_1)=f(a_2)=...=f(a_n)$ ) because otherwise the situation is that a degree $n-1$ polynomial attains the same value $n$ times for $n$ distinct numbers. Which is impossible. Thus, $P_1,P_2,...,P_n$ cannot all be the same.