MHB What Is the Minimum Value of y in the Given Expression?

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The expression for y consists of the absolute values of x and several linear transformations of x, leading to a piecewise function. The number of terms, m, is questioned, with a suggestion that it should be 5152. The discussion highlights that the minimum value of y occurs at two adjacent integer points, indicating a global minimum. The slope of the function is noted to be -5152 for negative x, increasing by 2 at each integer transition. Understanding these properties is essential for determining the minimum value of y effectively.
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If y = \mid x \mid+\mid x-1 \mid+\mid x-3 \mid+\mid x-6 \mid+...+\mid x-5151 \midand m = no. of terms in the expression yand n = no. of integers for which y has min. valueThen \displaystyle\frac{m+n-18}{10} =
 
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jacks said:
If y = \mid x \mid+\mid x-1 \mid+\mid x-3 \mid+\mid x-6 \mid+...+\mid x-5151 \midand m = no. of terms in the expression yand n = no. of integers for which y has min. valueThen \displaystyle\frac{m+n-18}{10} =

Maybe it's me, but I find that incomprehensible.

For a start why is \(m\) not \(5152\)?

Do you mean \(n\) to be the number of integers corresponding to a local minima of \(y\)? You can show that there is a global minimum and it achived this at two adjacent integer points.

(the slope is -5152 for -ve \(x\), and increases by 2 when we pass a integer argument moving to the right ...)

CB
 
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Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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