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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\}?
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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 :)
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!)
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 :)
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!)
Yup! You are correct! And the TeX works :)
Post a Comment