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Tetration extension for bases between 1 and eta
#3
I'm not sure how rigorous this is but here it is

Proof:

Assumption
(1)The function is a smooth, monotonic concave down function

Based on (1) we can establish
(2)Any line can only pass through f(x) a maximum of 2 times
(3)The intermediate value theorem holds for the entire domain of f(x)


Take the region R bounded on the x axis by x=-1 and x=0 and bounded on the y axis by y=f(-1) and y=f(0) ( y=0 and y=1).
Because of (3) every value, x, has a corresponding value, f(x), on the interval.
If we take a point (x,y) in R and assume that it is on the curve f(x) we can then use the relation and obtain the new point which equals . Applying this repeatedly we obtain the point .
We can now establish
(4)The point (x,y) in the region R is on the curve f(x) if is not on the secant line that touches the curve at 2 other known points of f(x) for any value of k.

Now we will find the equation of the secant line that touches f(x) at 2 consecutive known points. Using the point slope formula we find the equation to be . We must also note that if a point (x,y) is above the curve in the region R then is above the curve for any value of k.
we shall now extend (4) to say
(5)The point (x,y) in the region R is on or above the curve f(x) if for any value of k

Finally if we take the limit as k approaches infinity we will find that the slope of the secant line approaches zero and therefore follows f(x) exactly because f(x) has an asymptote as x goes towards infinity. Therefore for infinity large values of k

Now solving for y and taking the limit as k approaches infinity we obtain the desired result:

for

q.e.d
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Messages In This Thread
RE: Tetration extension for bases between 1 and eta - by dantheman163 - 11/05/2009, 11:53 PM

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