# Writing out the minimum function

• Jun 19th 2010, 11:23 PM
Flemming
Writing out the minimum function
Hi you all

I have a quick and properly really easy question.
I saw around the internet that the minimum function can be written out:

$\displaystyle min\left\{ a,b\right\} =\frac{1}{2}\cdot(a+b-|a-b|)$

I haven't ever seen that before, so i am interested in knowing when it works and when it doesn't (and to some extent why it's true). I need to differentiate a minimum function in my exam in microeconomics so it would be a great help!

Best regards.
• Jun 19th 2010, 11:51 PM
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Quote:

Originally Posted by Flemming
Hi you all

I have a quick and properly really easy question.
I saw around the internet that the minimum function can be written out:

$\displaystyle min\left\{ a,b\right\} =\frac{1}{2}\cdot(a+b-|a-b|)$

I haven't ever seen that before, so i am interested in knowing when it works and when it doesn't (and to some extent why it's true). I need to differentiate a minimum function in my exam in microeconomics so it would be a great help!

Best regards.

Not sure about the differentiation part, but proving it's true is straightforward:

Case: a = b

Then $\displaystyle \frac{1}{2}\cdot(a+b-|a-b|) = \frac{1}{2}\cdot(a+b) = a = b$

Case: a < b

Then $\displaystyle \frac{1}{2}\cdot(a+b-|a-b|) = \frac{1}{2}\cdot(a+b-(b-a)) = \frac{1}{2}\cdot(2a)=a$

Case: a > b

Then $\displaystyle \frac{1}{2}\cdot(a+b-|a-b|) = \frac{1}{2}\cdot(a+b-(a-b)) = \frac{1}{2}\cdot(2b)=b$
• Jun 20th 2010, 01:21 AM
Flemming
Oh, you are right, I should have been able to figure that one out my self. But thank you.

Best regards,
• Jun 20th 2010, 04:00 AM
SpringFan25
if you have

a(x) = min(b(x),c(x))

Except in special cases, The function will not normally be differenciable at the point b(x) = c(x). it will be differenciable at other points provided b(x) and c(x) are differenciable.

What is the economics problem you are trying to solve? It may be easier to assume min(a,b) = a; and check that your solution is at a point $\displaystyle a \leq b$. If that doesn't work then assume min(a,b) = b and check that your solution is at a point $\displaystyle b \leq a$.
• Jun 23rd 2010, 03:30 AM
Flemming
I needed to differentiate a function with respect to two different variables and find the quotient of them in order to get a function which hopefully could tell me more about the function which i originally was assigned to find.
Even though it my problem seemed far fetched your hint of dividing into different situations actually worked like a charm. Thank you! :)