Say I have a set, denoted by A, which represent values(real numbers) close to 10. How can I represent it using the way of set representations.
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Say I have a set, denoted by A, which represent values(real numbers) close to 10. How can I represent it using the way of set representations.
After a brief search on "set representation" I didn't really find anything that explains this concept. Maybe you just mean set notation?
Even on set notation is fine, I'll work it on set representation later.
I'm not very confident that I understand the question, but... I'll give it a shot.
Let e and d be arbitrarily small numbers (e stands for epsilon, d for delta; like in epsilon-delta). Then the set S10 (the set of real values close to 10) is {x| abs(x-e) <= d} [x belongs to the reals]
Again I'm not sure that this is what you want but S10 is a set of real values that are close to 10....
If this wasn't what you wanted, could you provide an example? Perhaps then we could be on the same page.
Even what I know about sets, I believe this is correct. But I found one explanation like this,Quote:
Originally Posted by jemidiah
1 + (X - 102)(-1) where, X belongs to the reals.
I'm confusing with it, have you any comments.
That function you've listed confuses me too if it's supposed to connect to real values close to 10.
Depending on the X you choose, that function evaluates to anything (it's invertible except for X = 100). Here's some values I plugged in to get a feel for the function:
1 + (X - 100)^(-1) = f(X)
f(-1000) = ~0.999
f(-100) = 0.995
f(-10) = ~0.991
f(-1) = ~0.990
f(0) = 0.99
f(1) = ~0.990
f(10) = ~0.989
f(90) = 0.9
f(98) = 0.5
f(99) = 0
f(99.5) = -1
f(99.75) = -3
f(99.9999) = ~-10,000
f(100) = [undefined]
f(100.0001) = ~10,001
f(100.5) = 3
f(101) = 2
f(110) = 1.1
f(1000) = ~1.00
For most inputs of X it gives values very near 1; maybe that's the part that's supposed to help answer this question?
In any case I'm still confused by that function.
Actually I checked that function as you done. But still confusing. Anyway, you are same that what I thought. Thanks.