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Friday, 12 August 2011
Math GRE - #11
For all real numbers $x$ and $y$, which of the answers below is the expression \[\frac{x+y+|x-y|}{2}\] equal to?
The maximum of $x$ and $y$
The minimum of $x$ and $y$
$|x+y|$
The average of $|x|$ and $|y|$
The average of $|x+y|$ and $x-y$
Solution :
The expression is equal to the maximum of $x$ and $y$.
Suppose $x>y$. Then $|x-y|=x-y$ and so \[\frac{x+y+|x-y|}{2}=\frac{x+y+x-y}{2}=x.\]
Now suppose $x<y$. Then $|x-y|=-(x-y)$ and so \[\frac{x+y+|x-y|}{2}=\frac{x+y-(x-y)}{2}=y.\]
\[\therefore\quad\frac{x+y+|x-y|}{2}=\max\{x,y\}.\]
Can you find a representation for $\min\{x,y\}$?
2 comments:
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This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.
This formula is also a nice quick way to see that the function defined to be the maximum of two (real-valued) continuous functions is also continuous (although it's not too bad to do this in a more general topological setting where one doesn't have the luxury of continuous binary operations, it's not nearly as quick). Another use for formula is in the proof for the Stone-Weierstrass theorem.
For the minimum of x and y, one would replace the expresion with $\frac{x+y-|x-y|}{2}$ (hopefully the tex works!)
@564687438437171172.0 Yup! You are correct! And the TeX works :)
2 comments:
Post a Comment
This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.
For the minimum of x and y, one would replace the expresion with $\frac{x+y-|x-y|}{2}$ (hopefully the tex works!)
Yup! You are correct! And the TeX works :)
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