# Am looking for minimum

1. Jul 12, 2004

### dedaNoe

I have
$$f(N)=\sqrt{Sum(\frac{A_i}{1-N}+\frac{B_i}{N})^2}$$
i=1 to 3

I need such N that gives minimum for f(N)?
Thanks!

Last edited: Jul 12, 2004
2. Jul 12, 2004

### AKG

$$f(N) = \sqrt{\sum _{i=1} ^3 \left (\frac{A_i}{1 - N} + \frac{B_i}{N} \right )^2 }$$

Now:

$$\frac{d}{dN} \left [ \sum_i g_i(N) \right ] = \sum g_i'(N)$$

Why is this true? Observe:

$$\frac{d}{dN} \left [ \sum _i g_i(N) \right ] = \frac{d}{dN}[g_1(N) + g_2(N) + g_3(N)]$$

$$\ = g_1'(N) + g_2'(N) + g_3'(N) = \sum g_i'(N)$$

Note, $g_i(N) = \left ( \frac{A_i}{1 - N} + \frac{B_i}{N} \right )^2$ and $\sum _i$ is just a shorthand way of saying $\sum _{i = 1} ^3$.

Now, f(N) reaches a minimum where f'(N) is zero or undefined. I've given you a way to easily find the derivative for f(N). You can tell where it will be zero or undefined. Take all those critical values for N where f'(N) is zero or undefined, plug those values of N into f(N), and choose the least value.

3. Aug 26, 2004

### Pi

It looks like AKG forgot to take the square root into account, but consider that when f is a minimum, f^2 is a minimum if f > 0, or a maximum if f < 0.