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Re: which values of m satisfies the inequality

  • To: mathgroup at smc.vnet.net
  • Subject: [mg104222] Re: [mg104202] which values of m satisfies the inequality
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sat, 24 Oct 2009 02:39:24 -0400 (EDT)
  • References: <200910230235.WAA27884@smc.vnet.net>

I am not sure what sort of general answer you are expecting since it  
is certainly impossible to "solve" this sort of inequality in any  
explicit way.

However, there are  (at least) two other things you can do. The first  
is to make a program that will determine the range of m's for any  
given x. For this purpose we define a function:

f[x_, m_] := (1 + x)^(m + 1) - (1 + x^m)*2^m

First suppose that m >0.
Let's show that the function has the property that if it is negative  
for some m then it will be negative for all m larger than that m. We  
do this by induction. So suppose we have

(1 + x)^(m + 1) < (1 + x^m)*2^m

Then

(1 + x)^(m + 2) == (1 + x)^(m + 1) (1 + x) < (1 + x^m)*2^m (1 + x)

so we need to show that (1 + x^m)*2^m (1 + x) < (1+x^(m+1))2^(m+1)
Simplifying

(1 + x^m)*2^m (1 + x) < (1 + x^(m + 1))*2^(m + 1) // Simplify

2^m (x-1) (x^m-1)>0

which is of course true, if x>1 and m>0.

So now for a given x, we only need to find the first m such that f 
[x,m] is negative.

FF[x_] := Module[{m = 0}, While[f[x, m] > 0, m++]; m]

For example:

FF[10]

5

This means that for x = 1 your inequality will first fail for m = 5,  
hence it only holds for m = 0,1,2,3,4.

For negative m you can similarly show that if the function f[x,m] is  
negative for some m, then it will also be negative for m-1. So in the  
same way you can write a program that, for a given x, will determine  
the smallest m for which the inequality holds. I will leave this to  
you. (Actually when I first tried solving this I forgot you were also  
interested in negative m so everything I have done assumes that m is  
positive (m=0 is trivial). But the above method certainly works for  
negative m also.)


As for the completely general case, the only thing one can do, as far  
as I can tell, is to get some necessary and some sufficient conditions  
for the inequality to hold. Assuming again that m is positive, I think  
the following condition is sufficient

m < (Log[x] - Log[2])/Log[2]

but not necessary. I also believe that

m < (-Log[x/2 + 1/2] - Log[2])/(Log[x/2 + 1/2] - Log[x])

is a sufficient condition (but not necessary). For example, for the  
case x=10 the first one gives

= m < (Log[x] - Log[2])/Log[2] // N

m<2.32193

which gives just m = 1 and 2  (remember I am assuming that m >0). The  
necessary condition gives:

  m < (-Log[x/2 + 1/2] - Log[2])/(Log[x/2 + 1/2] - Log[x]) // N

  m<4.01095

which is pretty good, since as we know the result is valid only for  m  
=1,2,3,4.

I obtained these conditions by hand so I am not giving the derivations  
here but they are easy (they might also be wrong as I have not really  
checked it carefully). Also, I have not tried to consider the case of  
negative m, but I sure one can obtain similar conditions.

Andrzej Kozlowski


On 23 Oct 2009, at 11:35, barefoot gigantor wrote:

>
> for what value or interval of m (-infinity < m < infinity) the  
> following is
> valid
>
> (1+x)^(m+1) > (1+x^m)* 2^m
>
> here x >=1
>
>



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