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From: Andrzej Kozlowski on 25 Oct 2009 01:17

On 24 Oct 2009, at 15:41, David W. Cantrell wrote:

> barefoot gigantor <barefoot1980(a)gmail.com> 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
>
> I shall assume that you really meant to have strictly x > 1. [That's
> because, if x = 1, then (1+x)^(m+1) = (1+x^m) * 2^m, regardless of m.]
>
> Answer:
>
> For x > 1 and m >= -1,
>
> (1+x)^(m+1) > (1+x^m) * 2^m
>
> David
>


Hmm...

(1 + x)^(m + 1) > (1 + x^m)*2^m /. {x -> 10, m -> 6}

False

Andrzej Kozlowski

From: David W. Cantrell on 25 Oct 2009 01:27
On 24 Oct 2009, at 07:24, Andrzej Kozlowski wrote:
> On 24 Oct 2009, at 15:41, David W. Cantrell wrote:
>> barefoot gigantor <barefoot1980(a)gmail.com> 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
>>
>> I shall assume that you really meant to have strictly x > 1.
>> [That's because, if x = 1, then (1+x)^(m+1) = (1+x^m) * 2^m,
>> regardless of m.]
>>
>> Answer:
>>
>> For x > 1 and m >= -1,
>>
>> (1+x)^(m+1) > (1+x^m) * 2^m
>
> Hmm...
>
> (1 + x)^(m + 1) > (1 + x^m)*2^m /. {x -> 10, m -> 6}
>
> False
>
Thanks, Andrzej! Yes, I mistakenly left off the upper bound for m.
I should have written:

For x > 1 and -1 <= m < c where c = 3.358...,

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

[barefoot: If it's important to you to have constant c determined with
greater accuracy, just ask.]

David


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