Recall max$\{x_1,x_2\}=\frac{x_1+x_2}{2}+\frac{|x_1-x_2|}{2}$. Using this identity, derive a formula for max$\{x_1,x_2,...,x_n\}$

Question

I understand the identity when it's dealing with two numbers

Again, recall that max$\{x_1,x_2\}=\frac{x_1+x_2}{2}+\frac{|x_1-x_2|}{2}$

I get stuck when dealing with $n$ numbers, but I can get the first part down

Here's what I have: max$\{x_1,x_2,\ldots,x_n\}=\frac{x_1+x_2+\cdots+x_n}{2}+{}$ "stuff"

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I can't figure out how to use absolute value here.

Would the "stuff" be something like $\frac{|x_1-x_2-\cdots-x_n|}{2}$ or would it be

$$\frac{|x_1-x_2|+|x_2-x_3|+\cdots+|x_{n-1}-x_n|}{2} \text{ ?}$$

Or is it neither of those?

Answer 1

I suspect you may be expected to use things like $\max \{x_1,x_2, x_3\} = \max \{\max \{x_1,x_2\} , x_3\}$

That would then be $$\frac{\frac{x_1+x_2}{2}+\frac{|x_1-x_2|}{2}+x_3}{2}+\frac{\big|\frac{x_1+x_2}{2}+\frac{|x_1-x_2|}{2}-x_3\big|}{2} \\= \frac{x_1+x_2+|x_1-x_2|+2x_3+\big|x_1+x_2+|x_1-x_2|-2x_3\big|}{4} $$

etc.