11/05/2009, 11:53 PM
(This post was last modified: 11/06/2009, 12:24 AM by dantheman163.)
I'm not sure how rigorous this is but here it is
Proof:
Assumption
(1)The function \( f(x)={}^x b \) 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 \( f(x+1)=b^{f(x)} \) and obtain the new point \( (x',y') \) which equals \( (x+1,b^y) \). Applying this repeatedly we obtain the point \( (x+k,{Exp}_b ^k (y)) \).
We can now establish
(4)The point (x,y) in the region R is on the curve f(x) if \( (x+k,{Exp}_b ^k (y)) \) 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 \( g(x)= ({}^k b-{}^{(k-1)} b)(x-k)+{}^k b \). We must also note that if a point (x,y) is above the curve in the region R then \( (x+k,{Exp}_b ^k (y)) \) 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 \( g(x+k) < {Exp}_b ^k (y) \)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 \( g(x+k) = {Exp}_b ^k (y) \) for infinity large values of k
Now solving for y and taking the limit as k approaches infinity we obtain the desired result:
\( {}^x b = \lim_{k\to \infty} (log_{b}^{ok}(x({}^k b- {}^{(k-1)} b)+{}^k b) ) \) for \( -1 \le x\le 0 \)
q.e.d
Proof:
Assumption
(1)The function \( f(x)={}^x b \) 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 \( f(x+1)=b^{f(x)} \) and obtain the new point \( (x',y') \) which equals \( (x+1,b^y) \). Applying this repeatedly we obtain the point \( (x+k,{Exp}_b ^k (y)) \).
We can now establish
(4)The point (x,y) in the region R is on the curve f(x) if \( (x+k,{Exp}_b ^k (y)) \) 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 \( g(x)= ({}^k b-{}^{(k-1)} b)(x-k)+{}^k b \). We must also note that if a point (x,y) is above the curve in the region R then \( (x+k,{Exp}_b ^k (y)) \) 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 \( g(x+k) < {Exp}_b ^k (y) \)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 \( g(x+k) = {Exp}_b ^k (y) \) for infinity large values of k
Now solving for y and taking the limit as k approaches infinity we obtain the desired result:
\( {}^x b = \lim_{k\to \infty} (log_{b}^{ok}(x({}^k b- {}^{(k-1)} b)+{}^k b) ) \) for \( -1 \le x\le 0 \)
q.e.d
