| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1732.1 | how about the old way | STAR::ABBASI | i am therfore i think | Wed Mar 17 1993 12:34 | 13 |
| >Prove that, for 0 <= x <= 1 and a positive integer k,
> (1+x)^k * [x + (1-x)^(k+1)] >= 1.
can one solve this by saying
y= (1+x)^k * [x + (1-x)^(k+1)]
find y' , make y'=0, solve for x, find y'', find which x from last
step that makes y'' >0, that is the min x, if the min x is not in
[1,0], then we are done, else plug x back in y to see that y>=1 at that
x.
\nasser
|
| 1732.3 | Inverse? | VMSDEV::HALLYB | Fish have no concept of fire. | Wed Mar 17 1993 17:06 | 11 |
| > Perhaps after differentiating, you get some interesting recursion
> relating f(k) to f(k-1). I can't think of any other way this apparently
> arbitrary formula could be interesting.
Implying that perhaps this is no more fun than proving |x| + 1 >= 1 ?
Gut hacking suggests for fixed k, (1+x)^k * [x + (1-x)^(k+1)] is monotonic.
If it has an inverse that is readily computed (and differentiated) it
may have applications as a "squashing function", used in neural networks.
John
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| 1732.4 | | STAR::ABBASI | i am therfore i think | Wed Mar 17 1993 18:07 | 17 |
| .3
>Gut hacking suggests for fixed k, (1+x)^k * [x + (1-x)^(k+1)] is monotonic.
yes, if it monotonic increasing, we know that x=0 the above is 1, so
we are done, but how to show that function is monotonic increasing or
decreasing over some range?
so for monotonic increasing , this means we have to find the points the
derivative is zero, and at each point check that y'' is always >0 (or
0?) at least ?
i have a feeling there is solution to this without using derivatives
and all, i think there is some trick or short cut to do it.
\bye
\nasser
|
| 1732.5 | Proof by Mathematical Induction | MARX::PADHY | | Fri Mar 19 1993 15:04 | 55 |
|
To prove that for 0 <= x <= 1 and a positive integer k,
(1+x)^k * [x + (1-x)^(k+1)] >= 1.
This could be easily proved by Mathematical induction.
Basis: k = 1.
------------
Then LHS becomes:
(1+x)^1 * [x + (1-x)^(1+1)]
= (1+x) * [x + (1-x)^2]
= (1+x) * [1 -x + x^2]
= (1 + x^3)
which is >=1 for 0 <= x <= 1. (Proved).
Next assume that the identity holds for any positive integer k:
--------------------------------------------------------------
i.e. assume for k > 0, and 0 <= x <= 1,
(1+x)^k * [x + (1-x)^(k+1)] >= 1 (1)
Now multiply both sides by (1-x^2) which is a positive value. Then (1) becomes,
(1+x)^(k+1) * [x(1-x) + (1-x)^(k+2)] >= (1-x^2)
=> (1+x)^(k+1) * [x - x^2 + (1-x)^(k+2)] >= (1-x^2)
=> (1+x)^(k+1) * [x + (1-x)^(k+2)] >= 1 - x^2 + (x^2) * (1+x)^(k+1)
=> (1+x)^(k+1) * [x + (1-x)^(k+2)] >= 1 + (x^2) * [(1+x)^(k+1) - 1]
Using Binomial theorm, it is easy to see that the RHS is always >= 1.
Hence, the identity holds for for k+1 also. (proved).
Conclusion:
-----------
The identity holds for all positive values of k, and x lying between
0 and 1.
Have fun.
Bud Padhy
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| 1732.6 | are you sure there is no typo here? | STAR::ABBASI | i am therfore i think | Fri Mar 19 1993 16:00 | 17 |
| > (1+x)^k * [x + (1-x)^(k+1)] >= 1 (1)
>Now multiply both sides by (1-x^2) which is a positive value. Then (1) becomes,
> (1+x)^(k+1) * [x(1-x) + (1-x)^(k+2)] >= (1-x^2)
i dont see how you did this step?
(1-x^2) * { (1+x)^k * [x + (1-x)^(k+1)] }
is the same as
(1+x)^(k+1) * [x(1-x) +(1-x)^(k+2)]
how, i dont see it?
thanks,
\nasser
|
| 1732.7 | Here is the step | MARX::PADHY | | Fri Mar 19 1993 16:15 | 20 |
|
re .6
Note that (1-x^2) = (1+x) * (1-x)
Hence,
(1-x^2) * { (1+x)^k * [x + (1-x)^(k+1)] }
= (1+x) * (1-x) * { (1+x)^k * [x + (1-x)^(k+1)] }
Now multiply (1+x) with (1+x)^k and (1-x) with [x + (1-x)^(k+1)]. The above
is then
= (1+x)^(k+1) * [x(1-x) +(1-x)^(k+2)]
Hope this helps!!
Bud
|