| New Reply |
Checking weather any positive number is zero or not |
Share Thread | Thread Tools |
| Jul1-11, 12:00 PM | #1 |
|
|
Checking weather any positive number is zero or not
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*x-1)/2 . Here the division is integer division. But if x=0, here we divide (2*0-1)/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*(x-1)+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... |
| Jul1-11, 12:38 PM | #2 |
|
|
"So, can anyone help me constructing such a function which is linear"
The function you are describing is non-linear 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... |
| Jul1-11, 01:17 PM | #3 |
|
Recognitions:
|
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. |
| Jul1-11, 05:28 PM | #4 |
|
|
Checking weather any positive number is zero or notThanks 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*x-1)/2 = x + (1/2) - x - (1/2) = 0, so really, this just says that f(x) = 0. " f(x) = (2*x+1)/2 - (2*x-1)/2 works because you have to consider (2*x-1) as a whole number. you should not simplify it. look when x=2, f(2)= (2*2+1)/2 - (2*2-1)/2 = 5/2-3/2= 2-1=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... |
| Jul6-11, 12:52 AM | #5 |
|
|
f(x) = x^0 works but I don't know if that's quite what you're looking for.
|
| Jul6-11, 07:28 AM | #6 |
|
|
|
| Jul6-11, 07:33 AM | #7 |
|
Recognitions:
|
|
| New Reply |
| Thread Tools | |
Similar Threads for: Checking weather any positive number is zero or not
|
||||
| Thread | Forum | Replies | ||
| Why is (n^0=1)? where n is any positive number | General Math | 6 | ||
| Checking the answer to complex number question | Calculus & Beyond Homework | 1 | ||
| Checking the weather | General Discussion | 8 | ||
| Is 0 is an even and positive number? | General Math | 10 | ||