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Okey so here are my random exercises:
Solve the equations:
1.
2.
3. Give all solutions for x
4.Find a function that satisfies
enjoy!
Last edited by Kurre (2007-11-21 01:17:20)
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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 cant cope with is therefore your own problem.
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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 (2007-11-21 07:24:31)
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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 (2007-11-21 12:07:57)
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 cant cope with is therefore your own problem.
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5. Approximate
6. Which number is greater?
Last edited by Kurre (2007-11-27 03:19:09)
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