Any help with an logical/algebraic problem?

[Nihilist]

This isn’t exactly philosophy in the traditional sense, but it seems I know there are a lot of intelligent people here.

I am working on an translation/analysis of a Medieval treatise on music theory from the Migne Latin Patrology- and there is a particular proposition made by the author, about the relationship between musical intervals. The whole thing is easily converted to an algebraic expression, but I am having some trouble proving how it can be true (although it obviously is).

The problem will be solved if it can be shown that, for an any value of x equal or great than 2:
( x+2)/( x +1) > (x^2+2x+1 )/(x^2+2x)]^x

Is there a maths whizz out there who can suggest how the problem can best be approached?

I was thinking that, if it is shown that for x=2, the above holds true (which it does), then if there is no solution of the equation:
( x+2)/( x +1) = (x^2+2x+1 )/(x^2+2x)]^x,

for any value of x above 2, it would be proven, since the curves of the two functions would never intersect or cross. But how would this be done? How could it be shown that there is no solution to the above equation for values higher than x=2? Probably very easy for a maths expert, but way beyond my algebra.

I am sure there is someone one this forum who can tell me, and I would appreciate any advice (feel free to respond by private message- as this is problem not a scintillating forum topic). Many thanks in advance.

 

[TxGodfollower]

This isn’t exactly philosophy in the traditional sense, but it seems I know there are a lot of intelligent people here.

I am working on an translation/analysis of a Medieval treatise on music theory from the Migne Latin Patrology- and there is a particular proposition made by the author, about the relationship between musical intervals. The whole thing is easily converted to an algebraic expression, but I am having some trouble proving how it can be true (although it obviously is).

The problem will be solved if it can be shown that, for an any value of x equal or great than 2:
( x+2)/( x +1) > (x^2+2x+1 )/(x^2+2x)]^x

Is there a maths whizz out there who can suggest how the problem can best be approached?

I was thinking that, if it is shown that for x=2, the above holds true (which it does), then if there is no solution of the equation:
( x+2)/( x +1) = (x^2+2x+1 )/(x^2+2x)]^x,

for any value of x above 2, it would be proven, since the curves of the two functions would never intersect or cross. But how would this be done? How could it be shown that there is no solution to the above equation for values higher than x=2? Probably very easy for a maths expert, but way beyond my algebra.

I am sure there is someone one this forum who can tell me, and I would appreciate any advice (feel free to respond by private message- as this is problem not a scintillating forum topic). Many thanks in advance.

Actually need some calculus to show that; what your basically asking for is to solve the equation with X=infinity. Calculus has a function called a limit, which deals with this issue (and dividing by zero).
I was playing with the equation in excel, and I was able to solve it for numbers under 2 as well. For example:
X=1 you get (1+2)/(1+1) > ((1^2+2x1+1)/(1^2+2x1))^1 or 3/2 > 4/3. Changing to a common denominator, you get 9/6 > 8>6.

Also, if you make X really really really big, like x= 1x10^100, the two equations equal each other (according to excel). might just be a rounding error, not sure.

Hope that helps a bit.

 

[Tomdstone]

This isn’t exactly philosophy in the traditional sense, but it seems I know there are a lot of intelligent people here.

I am working on an translation/analysis of a Medieval treatise on music theory from the Migne Latin Patrology- and there is a particular proposition made by the author, about the relationship between musical intervals. The whole thing is easily converted to an algebraic expression, but I am having some trouble proving how it can be true (although it obviously is).

The problem will be solved if it can be shown that, for an any value of x equal or great than 2:
( x+2)/( x +1) > (x^2+2x+1 )/(x^2+2x)]^x

Is there a maths whizz out there who can suggest how the problem can best be approached?

I was thinking that, if it is shown that for x=2, the above holds true (which it does), then if there is no solution of the equation:
( x+2)/( x +1) = (x^2+2x+1 )/(x^2+2x)]^x,

for any value of x above 2, it would be proven, since the curves of the two functions would never intersect or cross. But how would this be done? How could it be shown that there is no solution to the above equation for values higher than x=2? Probably very easy for a maths expert, but way beyond my algebra.

I am sure there is someone one this forum who can tell me, and I would appreciate any advice (feel free to respond by private message- as this is problem not a scintillating forum topic). Many thanks in advance.

Let LHS= ( x+2)/( x +1).
Let RHS= (x^2+2x+1 )/(x^2+2x)]^x.
Let y=LHS/RHS. We must show y>1.
Verify y>1 for x=2.
Method: take ln and find d/dx of lny.
If d/dx of lny >0 for x>2, you are done, since y is increasing there.
If d/dx of lny <0 for x>2, find the minimum value and show for that minimum y>1.

 

[Just_Lurking]

Inequality: If a > 0, b > 0, and ab < 1, then (1+a)^b < 1/(1-ab).

The problem from the OP follows from this lemma by substituting a = 1/(x^2+2x) and b = x.

The inequality itself follows from (1+a) < e^a, so (1+a)^b < e^(ab), and e^(ab) < 1/(1-ab), where e = 2.718… These more basic facts follow most clearly from the infinite power series representations of e^x and 1/(1-x).

 

[Nihilist]

Actually need some calculus to show that; what your basically asking for is to solve the equation with X=infinity. Calculus has a function called a limit, which deals with this issue (and dividing by zero).
I was playing with the equation in excel, and I was able to solve it for numbers under 2 as well. For example:
X=1 you get (1+2)/(1+1) > ((1^2+2x1+1)/(1^2+2x1))^1 or 3/2 > 4/3. Changing to a common denominator, you get 9/6 > 8>6.

Also, if you make X really really really big, like x= 1x10^100, the two equations equal each other (according to excel). might just be a rounding error, not sure.

Hope that helps a bit.

Many thanks. Everything helps.

 

[Nihilist]

Let LHS= ( x+2)/( x +1).
Let RHS= (x^2+2x+1 )/(x^2+2x)]^x.
Let y=LHS/RHS. We must show y>1.
Verify y>1 for x=2.
Method: take ln and find d/dx of lny.
If d/dx of lny >0 for x>2, you are done, since y is increasing there.
If d/dx of lny <0 for x>2, find the minimum value and show for that minimum y>1.

Yes, I understand. Finding ln and its derivative which I will explore.

 

[Nihilist]

Inequality: If a > 0, b > 0, and ab < 1, then (1+a)^b < 1/(1-ab).

The problem from the OP follows from this lemma by substituting a = 1/(x^2+2x) and b = x.

The inequality itself follows from (1+a) < e^a, so (1+a)^b < e^(ab), and e^(ab) < 1/(1-ab), where e = 2.718… These more basic facts follow most clearly from the infinite power series representations of e^x and 1/(1-x).

Thanks. I don’t understand that immediately, but I will try to get my head around it.

 

[Tomdstone]

Inequality: If a > 0, b > 0, and ab < 1, then (1+a)^b < 1/(1-ab).

The problem from the OP follows from this lemma by substituting a = 1/(x^2+2x) and b = x.

The inequality itself follows from (1+a) < e^a, so (1+a)^b < e^(ab), and e^(ab) < 1/(1-ab), where e = 2.718… These more basic facts follow most clearly from the infinite power series representations of e^x and 1/(1-x).

This proof works nicely.

 

[lmelahn]

Thanks. I don’t understand that immediately, but I will try to get my head around it.

Just Lurking’s proof works quite nicely. Here is a graph of the two sides of your inequality. It holds for all x>0. (x=0 is a removable singularity.)

(Please Note: This uploaded content is no longer available.)

What Just Lurking was getting at is that you can expand any differentiable function to a power series.

For example, for all real numbers x:



Likewise, for |x|<1:



We can see immediately that e^x < 1/(1-x) when |x|<1. And we can also see immediately that, for all x>0, 1+x < e^x.