# Find local maxima of this quadratic function

How can I find local maxima of this quadratic function?

$$f(x) = \sum _{i=1}^n -\frac{(z_i - x)_+^2}{2} - \left\{((\frac{(z_i - x)_+^2}{2})-(\frac{(y_i - x)_+^2}{2}) ) * c_i\right\}$$

which + means If $z_i$ is bigger than u its equal $z_i - x$ and else it's equal zero. Also $z_i$, $y_i$ and $c_i$ is a vector of values with same lengths.$z_i$ and $y_i$ is an interval.

$$z = (0.04124633, 0.05404163, 0.30939022, 0.57690859, 0.70211949, 0.73597262, 0.92130814,1.05226194, 1.68227033, 1.91068738)$$ $$y = (0.2112338, 0.4172289, 0.3991616, 0.8755037, 1.2448935, 1.3493179, 1.6522002, 2.4177490, 4.7770777, 2.3840472)$$ $$c = (5.337958, 2.275704, 10.64689, 2.873863, 1.293402, 1.03867, 0.6372883, 0.1032198, 0.000045619, 0.5670724)$$ I plot this quadratic function in terms of x which can be change between max finite value of $z_i$ and $y_i$ and see that theres is one local maxima but I do not know how can I find it matematically?

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Since the function is everywhere differentiable, it can have local extrema only at the points where $f'=0$. The formula for the derivative is $$\begin{split} f'(x) &= \sum _{i=1}^n (z_i-x)_+ + c_i \{(z_i - x)_+- (y_i - x)_+ \} \\ &= \sum_{i=1}^n (1+c_i)(z_i-x)_+ - \sum_{i=1}^n c_i (y_i - x)_+ \end{split} \tag 1$$ if I parsed your formula correctly (it appears to have a few unnecessary parentheses). This function is piecewise linear; its graph has corners at the points $z_i$ and $y_i$.

I don't think there is a very clever way to deal with such piecewise linear functions. I would list the numbers $z_i,y_i$ in increasing order, plug each into $f'$, and noted the signs. The maximum will be found within an interval where $f'$ is positive at the left endpoint and is negative at the right endpoint. The precise location is found by solving linear equation (since $f'$ is linear within the interval of interest).

For example, on the interval $(1.68227033, 1.91068738 )$ we have $$\sum_{i=1}^n (1+c_i)(z_i-x)_+ = (1+0.5670724) (1.91068738-x)$$ and $$\sum_{i=1}^n c_i (y_i - x)_+ = 0.1032198 (2.4177490-x)+0.000045619 (4.7770777-x) + 0.5670724(2.3840472-x)$$ So, you can simplify $f'(x)=\sum_{i=1}^n (1+c_i)(z_i-x)_+ - \sum_{i=1}^n c_i (y_i - x)_+$ by expanding the parentheses, and check if $f'$ turns to zero or not.

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Could you please give me an example to understand better your explanation? I dont know how to find and check each interval. Thanks. – rose May 14 '13 at 3:10
@rose Added an example. – 75064 May 14 '13 at 3:18
Thanks for your example. At the end for example I find that for this interval the value is equal -1.552835 and because its not in the interval (1.68227033,1.91068738) , so we do not have local maxima in this interval and I should continue for the rest 18 intervals. – rose May 14 '13 at 3:57
@rose Yes, and as I wrote, you can simplify the task by plugging in the endpoints before you try to solve the linear equation. Because if a linear function takes values of the same sign at both endpoints, then it does not turn into zero on that interval. – 75064 May 14 '13 at 4:05
yes. Thanks. Just I want to mention that your derivative for each parts neeed a minus. – rose May 14 '13 at 4:08