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For determination of the distance from a certain point, a function should be used. Namely, the input values 'x' are from N (1,2,3,4,5...), and the corresponding outputs 'y' from R, but such that the sequence 'y' (relative to 'x') is increasing;
[tex]
$y(1)<y(2)<y(3)<y(4)<...$ \\
but to satisfy \\
$y(1)>[y(2)-y(1)]>[y(3)-y(2)]>[y(4)-y(3)]>...$
[/tex]
The function that satisfies this is sqrt(x), but I'm interested in other possible functions that satisfy the above. The 'derivatives' of sqrt(x) should also work.
Thanks
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Hi onako;
I have cleaned up your latex.
The derivatives of the √ x do not work.
These work and so do many others.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thanks. I put 'derivatives' with '' emphasis, so it does not correspond to the real derivative of a function ('derivative of sqrt(x) might be sqrt(2x), for example, but this is a language obstacle').
In addition, I would like to extend the case for the following:
[tex]
$y(1)<y(2)<y(3)<y(4)<...<y(n-1)<y(n)$ \\
but satisfy
$y(1)=[y(2)-y(1)]=[y(3)-y(2)]=...=[y(k+1)-y(k)]>[y(k+2)-y(k+1)]>[y(k+3)-y(k+2)]>>...>[y(n)-y(n-1)]>$ \\
[/tex]
How could I extend the given functions to obtain this (for given k (and n)).
Thanks
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Hi;
This is how your latex should look. You are using the wrong tags for this forum. Over here we use the math tag instead of tex and $ are unnecessary.
but satisfy
There is no way to do what you ask. look closely at your statement is it not contradictory?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Note that
For the extension;
let y(1)=1; y(2)=2; y(3)=3; y(4)=4; I wonder what would be the function
y(x) for x>=5 that satisfy the above.
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Hi;
Yes, I see the error. I will recompute some other functions.
First, do you agree that it is impossible to do as you ask here?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Let
y(n)=n for n<=5;
So, now, if I specify the following values for y(5),...y(10)
y(6)=5.9
y(7)=6.7
y(8)=7.4
y(9)=8
y(10)=8.5
y(11)=8.9
....
so, the sequence is increasing in y(n)|n>=5, and the sequence of
differences is decreasing. Could we construct such a function for specific k and n (ending values).
I made a mistake above denoting all the functions with y().
Perhaps I'm missing something important here.
Thanks again
Last edited by onako (2010-11-01 01:57:59)
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Hi;
You are jumping ahead and there are many backed up questions. Let us do those first.
From post #1
Here are 2 new functions that I think work.
From post #6 do you agree that it is impossible to do as you ask here?
If you look at these and answer, then I am all caught up. Then we can get to the next questions.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Ok, I agree. I wrote that I made a mistake(typing). My last post should explain what I want.
Thanks for the functions. I should test their behaviour.
Perhaps a 'single' function might not work for the extension I'm asking for.
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Okay, test the new functions I have found to see if they work satisfactorily. Then you can explain exactly what you need for post #7. I am a little foggy there.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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In simple words:
we are trying to obtain a function whose behaviour should be like this (example):
y(1)=1
y(2)=2
y(3)=3
y(4)=4
y(5)=5
...............up to here the difference between two consecutive y(n) y(n+1) is equal (always equal to 1), and the y() sequence itself is increasing
y(6)=5.9 (difference is 5.9-5=0.9 and 0.9<1, where 1 corresponds to previous difference)
y(7)=6.7 (difference is 6.7-5.9=0.8 and 0.8<0.9, where 0.9 corresponds to previous difference (see above))
y(8)=7.4 (difference is 7.4- 6.7=0.7 and 0.7<0.8, where 0.8 corresponds to previous difference (see above))
y(9)=8 (same....)
Check that the sequence y() is increasing, and that, starting with n=6, the difference is decreasing.
This was an example. I hope I clarified what I look for.
I really appreciate your help.
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Hi;
Are you meaning that the first five must be exactly 1,2,3,4,5?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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That is for this specific example. I chose y(x)=x for x<=5
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The problem is is that for x = 1,2,3,4,5 you have a straight line and then suddenly the function goes off and does something else. Except for a spline or a piecewise function I have never seen that. It might not be possible.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Exactly. The point is to have approximately the same difference values for first k input values (say, for 1,2,3,4,5) satisfying y(k)<y(k+1), and then the difference should decrease gradually still maintaining y(n)<y(n+1).
Right now I'm trying to combine certain functions (sqrt, log, ...) with some linear to obtain approximately wanted behaviour. Any suggestions on what to try are welcome. Also, if there is a software for which I *might* draw a wanted line, and get the approximate function, please let me know.
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Hi onako;
I have not made much progress to finding a function like you want.
Also, if there is a software for which I *might* draw a wanted line, and get the approximate function, please let me know.
If you want one for your very own I am sure you can download a free grapher. It will probably only be able to get the equations for straight lines. To get the equation of a curve is a much more complicated task.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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