• Apr 12th 2008, 08:54 PM
uyanga
If $n\in N$, $n>1$ and $\alpha \in [0,1]$ prove that
$$(n^\alpha-(n-1)^\alpha)(2^\alpha+n-n^\alpha-(n-1)^\alpha-1)+(n-1)^\alpha \geq 1$$
• Apr 12th 2008, 09:16 PM
Mathstud28
?????????
Quote:

Originally Posted by uyanga
If $n\in N$, $n>1$ and $\alpha \in [0,1]$ prove that
$$(n^\alpha-(n-1)^\alpha)(2^\alpha+n-n^\alpha-(n-1)^\alpha-1)+(n-1)^\alpha \geq 1$$

With all these \$ I can't discern what the problem is
• Apr 13th 2008, 04:00 AM
angel.white
TRANSLATION:

If $n\in N$, $n>1$ and $\alpha \in [0,1]$ prove that
$(n^\alpha-(n-1)^\alpha)(2^\alpha+n-n^\alpha-(n-1)^\alpha-1)+(n-1)^\alpha \geq 1$
• Apr 13th 2008, 04:02 AM
Moo
• Apr 13th 2008, 04:08 AM
angel.white
Quote:

Originally Posted by angel.white
TRANSLATION:

If $n\in N$, $n>1$ and $\alpha \in [0,1]$ prove that
$(n^\alpha-(n-1)^\alpha)(2^\alpha+n-n^\alpha-(n-1)^\alpha-1)+(n-1)^\alpha \geq 1$

EDIT: This is wrong, I misread my own translation as $\alpha \in \{0,1\}$, this problem remains unsolved

Case 1. $\alpha = 0$
$(n^0-(n-1)^0)(2^0+n-n^0-(n-1)^0-1)+(n-1)^0 \geq 1$

$(1-1)(1+n-1-1-1)+1 \geq 1$

$1 \geq 1$

TRUE

Case 2. $\alpha = 1$
$(n^1-(n-1)^1)(2^1+n-n^1-(n-1)^1-1)+(n-1)^1 \geq 1$

$(n-(n-1))(2+n-n-(n-1)-1)+(n-1) \geq 1$

$(n-n+1)(2+n-n-n+1-1)+n-1 \geq 1$

$(1)(2-n)+n-1 \geq 1$

$2-n+n-1 \geq 1$

$1 \geq 1$

TRUE

Because Every value of alpha leads this to be true regardless of the value of n, this statement is always true.

Edit:
Quote:

Originally Posted by Moo

*sigh* stupid double posts >.<
• Apr 13th 2008, 04:11 AM
Moo
Hm, but nothing tells us that $\alpha$ is an integer ?

If you can prove that the function is increasing or decreasing, ok :p
• Apr 13th 2008, 04:17 AM
angel.white
Quote:

Originally Posted by Moo
Hm, but nothing tells us that $\alpha$ is an integer ?

If you can prove that the function is increasing or decreasing, ok :p

not the best morning for me, I read it as being an element of a set containing the integers 0 and 1.

Added a note stating it is not solved.
• Apr 13th 2008, 04:19 AM
Moo
Listen to the unicorn, she does know all :D

Now the problem is... understand what she says ^^
• Apr 13th 2008, 04:21 AM
Moo
Actually, your resolution can be used wisely...if the derivative (to $\alpha$) of the left member is more simple to study than the function itself... dunno, i didn't manage to solve it this morning :'(
• Apr 13th 2008, 04:46 AM
angel.white
I must be overthinking this, because the solution I was working on just now was much more complicated than a post in "Elementary and Middle School Math Help" should warrant :/
• Apr 13th 2008, 04:51 AM
Moo
Didn't notice the category :/
Perhaps the student went into the wrong one ?
• Apr 14th 2008, 06:02 PM
uyanga
I solved it
I solved it.Now it is't hard inequality.(Clapping)
• Apr 14th 2008, 06:19 PM
angel.white
Quote:

Originally Posted by uyanga
I solved it.Now it is't hard inequality.(Clapping)

Post it, please, I'm really curious now.