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**Kurre****Member**- Registered: 2006-07-18
- Posts: 280

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 (2007-11-21 01:17:20)*

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**NullRoot****Member**- Registered: 2007-11-19
- Posts: 162

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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**Kurre****Member**- Registered: 2006-07-18
- Posts: 280

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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**NullRoot****Member**- Registered: 2007-11-19
- Posts: 162

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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**Kurre****Member**- Registered: 2006-07-18
- Posts: 280

5. Approximate

6. Which number is greater?

or

*Last edited by Kurre (2007-11-27 03:19:09)*

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