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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 smile

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...? dizzy

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)

 

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