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Checking weather any positive number is zero or not 
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#1
Jul111, 12:00 PM

P: 2

I would like to have a function f such that f(0) = 0 and f(x) = 1 for any x>0. I can compute f in the following way:
f(x) = (2*x+1)/2  (2*x1)/2 . Here the division is integer division. But if x=0, here we divide (2*01)/2 or 1/2, which is a problem. Because we do not have any number 1 here. We can also compute f(x) as follows f(x) = ((2*x+3) % 2*(x1)+3)/2 . Here % is remainder. This is also a problem because we are doing something like x%y which is nonlinear. To be clear x%2 is allowed, but x%y is not allowed. So, can anyone help me constructing such a function which is linear and over positive integer and all operation will be integer operation? Or can anyone tell me that it is not possible to construct f with such restriction? I really appreciate any help. NB: f simply checks weather any positive number is zero or not and return 0 if 0 ;else return 1. Thanks a lot... 


#2
Jul111, 12:38 PM

P: 235

"So, can anyone help me constructing such a function which is linear"
The function you are describing is nonlinear by definition. So no, there is no linear function that can give the output: f(0) = 0 and f(x) = 1 for any x>0 There is a piecewise linear function: f(x) = { 0 if x=0 1 otherwise But someone who knows math better than me may demonstrate this wrong... 


#3
Jul111, 01:17 PM

HW Helper
P: 805

If you want it for all real numbers greater than 0, you could use the piecewise function the previous poster suggested. If you weren't looking for a piecewise function, a continuous function does not exist. By the IVT, it's impossible to have a continuous function with this property. With integers, it's possible. 


#4
Jul111, 05:28 PM

P: 2

Checking weather any positive number is zero or not
Thanks for your reply. "For all integers greater than 0 or for all real numbers greater than 0?" I am considering natural numbers including 0. "This example doesn't work. f(x) = (2*x+1)/2  (2*x1)/2 = x + (1/2)  x  (1/2) = 0, so really, this just says that f(x) = 0. " f(x) = (2*x+1)/2  (2*x1)/2 works because you have to consider (2*x1) as a whole number. you should not simplify it. look when x=2, f(2)= (2*2+1)/2  (2*21)/2 = 5/23/2= 21=1. "I'm not sure what your definition of remainder is here." c=a%b , d= a/b => a=d*b +c "Why not f(n) = [n/(n+1)], where [] is the ceiling function?" Since I am considering only positive integer number including 0, probably i cannot represent ceiling. Thanks... 


#5
Jul611, 12:52 AM

P: 3

f(x) = x^0 works but I don't know if that's quite what you're looking for.



#6
Jul611, 07:28 AM

P: 371




#7
Jul611, 07:33 AM

HW Helper
P: 805




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