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#1 2007-11-20 09:28:13
- Kurre
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Kurres exercises
Okey so here are my random exercises:
Solve the equations:
1.
2.
3. Give all solutions for x
where k is any integer
4.Find a function that satisfies
enjoy!
Last edited by Kurre (2007-11-22 00:17:20)
#2 2007-11-22 03:58:49
- NullRoot
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Re: Kurres exercises
Interesting property of this one, Kurre...
Given
When X = -1, then:
When X = 1, then:
If you substitute for f(-1), using the right-most side from f(x), x=-1, then:
So:
I like it.
Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.
#3 2007-11-22 06:24:14
- Kurre
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Re: Kurres exercises
Actually i wasnt sure about the mathematics invloved in nr 4, it may need the condition |x|>1 (or |x|>p where p is somewhere between 1 and 2), but I decided to not post it at first because i wasnt sure 
Last edited by Kurre (2007-11-22 06:24:31)
#4 2007-11-22 10:47:24
- NullRoot
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Re: Kurres exercises
Actually, with a bit of afterthought, I'm perfectly happy with f(1) - 1 = f(1) + 1.
How about this? f(1) = √1
#4 doesn't hold up to that, unfortunately, unless you say:
f(x) = √1
For f(-x) < 0, x = √0.5
For f(-x) > 0, x = √z (such that 1/z = 0)
But that means making up crazy new Maths where we define the inputs for functions based on the output and 0z = 1.
And what kind of oddball would want to do that...? 
Last edited by NullRoot (2007-11-22 11:07:57)
Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.
#5 2007-11-25 21:05:40
- Kurre
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Re: Kurres exercises
5. Approximate
6. Which number is greater?
or
Last edited by Kurre (2007-11-28 02:19:09)