# Weighted Sum in Taylor Expansion (Partial Derivatives)

1. Jul 23, 2012

### unscientific

1. The problem statement, all variables and given/known data

From step 1 to step 2, what do they mean by "Taking the weighted sum of the two squares " ?

I tried and expanded everything in step 2 and it ends up as the same as step 1 (as expected),

3. The attempt at a solution

I tried looking up "weighted sum" and " weighted sum of squares", I think the more closely related one is here: http://en.wikipedia.org/wiki/Lack-of-fit_sum_of_squares

I read it and tried to understand it but it looks as if what they are describing has no relation to what's in the book..

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2. Jul 23, 2012

### clamtrox

You are overthinking it. It's just a convenient way of rewriting the first expression.

3. Jul 23, 2012

### unscientific

I'm sorry but I feel that if that was the case it would simply be written as " the expression can be rewritten to a form as: " instead of "weighted sum of the two squares".

4. Jul 23, 2012

### clamtrox

Do you at least agree that nothing magical has happened between the steps? The two expressions are equal.

The reason why you do this is you know always that the squared terms are positive. Then by requiring that R.H.S of the equation is always positive, you get useful constraints on the weights of the square terms. By requiring that
$$f_{xx}\left(\Delta x + \frac{f_{xy} \Delta y}{f_{xx}}\right)^2 + \left( f_{yy}-\frac{f_{xy}^2}{f_{xx}}\right) (\Delta y)^2 > 0$$
for all $\Delta x$ and $\Delta y$ you see that $f_{xx} > 0$ by setting $\Delta y = 0$ and $\left( f_{yy}-\frac{f_{xy}^2}{f_{xx}}\right)>0$ by setting $\Delta x = - \frac{f_{xy} \Delta y}{f_{xx}}$

5. Jul 23, 2012

### unscientific

I was confused by the term " sum of weights of the squares" as I didn't know what it meant. I tried using a different approach of aiming to "complete the square" first with (Δx) then i arrived at the same results.