Prev: FromAdjacencyMatrix / FromAdjacencyLists - broken? Workaround?
Next: Symbolic matrix algebra - Sumopack
From: barefoot gigantor on 22 Oct 2009 22:44

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


From: Andrzej Kozlowski on 24 Oct 2009 02:44

On 23 Oct 2009, at 19:07, Andrzej Kozlowski wrote:

> 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).


It should have been "necessary but not sufficient".

Andrzej Kozlowski

From: David W. Cantrell on 24 Oct 2009 02:46
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

From: Andrzej Kozlowski on 24 Oct 2009 02:49
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
>
>


From: Andrzej Kozlowski on 25 Oct 2009 01:16

On 24 Oct 2009, at 23:18, David W. Cantrell wrote:

> 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
>

Yes, but this is not "if and only if".

(1 + x)^(m + 1) > (1 + x^m)*2^m /. {x -> 10, m -> 4}

True

but 4 >3.358...

See my post in this thread.

Andrzej




 |  Next  |  Last
Pages: 1 2
Prev: FromAdjacencyMatrix / FromAdjacencyLists - broken? Workaround?
Next: Symbolic matrix algebra - Sumopack