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Thread: geometric arithmetic means inequality

  1. #1 geometric arithmetic means inequality 
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    Okay, here goes nothin!
    Given: such that all are >0 for all , , and if and only if all are equal
    consider the case where , and allow such that then we have = for all m.

    now take the gradient of each

    for all and now let , and allow such that
    we observe that and so











    this does atleast imply that the two functions, when applied to the same sequence, have identical rates of change in every direction when every number in the sequence is the same. And, as shown earlier in the post, that, if a number is in , and must be a positive real number, then the only point where the geometric mean and the arithmetic mean of the sequence are equal is when all of the components of are equal. Please correct me if I'm wrong, but this should work for all positive real numbers in and it should work for any . All I'm doing right now is showing that for some sequence, the point where the two functions are equal is the point where all components of are the same and that the slope of the tangent hyper-whatever is the same in every direction for both equations. and please, tell me if I'm mistaken.


    (work in progress... close... will edit this post to finish my idea... this is a boat load of code, good practice though)


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  3. #2 Re: geometric arithmetic means inequality 
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    Quote Originally Posted by Arcane_Mathamatition
    Okay, here goes nothin!
    Given: such that , , and if and only if all are equal
    consider the case where , and allow such that then we have = for all m.

    now take the gradient of each

    for all and now let , and allow such that
    we observe that and so


    (work in progress... will edit this post to finish my idea... this is a boat load of code, good practice though)
    There is a problem with A. it should be



    Also there is no reason why equality of G and A at a point need require that all of the be equal. For instance, if any is 0 then G will be zero, but A can be zero in other ways. So in the case of three variables
    0=G(0,1,1)=A(1,-1,0).


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    it was part of the given that all numbers in the sequence are positive, and I probably didn't represent that right. This inequality was taken from the Real Analysis thread, I may have had some problems transcribing it. so that would break the given. and also, my product should be raised to the too, right?
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  5. #4  
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    Quote Originally Posted by Arcane_Mathamatition
    it was part of the given that all numbers in the sequence are positive, and I probably didn't represent that right. This inequality was taken from the Real Analysis thread, I may have had some problems transcribing it. so that would break the given. and also, my product should be raised to the too, right?
    You are correct, you did state that the numbers were positive. I was looking for a simple example to show that the equality of G and A does not require the equality of the point of evaluation. But my basic point still holds. Here is a better example

    1-G(1,1,1) = A(2,1/2.1/2)

    You don't need to raise your product to the 1/k power because you aready have each factor raised to that power, which is the same thing.

    I will be interested to see your proof, but I am a bit skeptical that it will work. The proofs that I know of eventually boil down to something involving the theory of convex functions and inequalities involving them (Jensen's inequality for instance). If you get interested this Wiki article has a few proofs, but don't look until you have worked our your own method or decided that it will not pan out.

    http://en.wikipedia.org/wiki/Inequal...eometric_means

    I don't know of a totally elementary proof from first principles. The Wiki proofs are not hard, but they do use some machinery (for which references are given).
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    it's the same set of numbers going into A as into G, the arithmetic mean of the sequence and the geometric mean of the same sequence
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  7. #6  
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    Quote Originally Posted by Arcane_Mathamatition
    it's the same set of numbers going into A as into G, the arithmetic mean of the sequence and the geometric mean of the same sequence
    Yes, but you cannot use your technique of taking the gradient without also defining your functions over the entire Euclidean space. I assume that you are going to come up with some inequality that applies there and then restrict it to the points of the sequence. So you have to start with premises that are provable for the entire Euclidean space.

    Perhaps I don't see where you are going. In fact I do not, and am interested in seeing that. But my point applies to the logic train and to your original premise.
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  8. #7  
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    I've edited my first popst a few times, please leave me some feedback on what I've done to it up to this point, thanks.
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  9. #8  
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    Quote Originally Posted by Arcane_Mathamatition
    I've edited my first popst a few times, please leave me some feedback on what I've done to it up to this point, thanks.
    I can't figure out what you are trying to do. I sppears to me to be invalid.

    There is no reason at all to believe that G=A if and only if the arguments are the same. In fact they are not in general equal, and that is the point of the geometric-arithmetic mean inequality. They are not the same function, and equally there is no reason to believe that their gradients are equal either.

    Try some examples to see this. but



    The rest of your computation, I have not looked at it in detail, is irrelevant unless you can get the initial premise squared away.
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    if I have 3 numbers, G(5,5,5)= and A(5,5,5)= in a statement, the mean of any set of identical numbers, no matter the size, will always be equal to every number in the set. G(C,C,C,C,C,C,...C)= A(C,C,C,C,C,C,...C)= where n is the number of terms in the set of numbers having their means, geometric and arithmetic respectively, taken. is that not true? I am using one set of numbers, and running it through 2 scalar value functions from in the event all of the terms in my number set being equal, the means, neccesarilly, will be equal to every other term in my set. and since the two scalar value functions have the same outbput in I am saying, I guess the better word is that they are equivalent at that point, meaning that they yeild the same value. Am I right? so I would say, for this case, Or am I still off on something. and again, the limiter of this equation is that all of the numbers in my set MUST be positive, or this no longer holds true, but otherwise this is the only case when the two means will be equivalent.
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  11. #10  
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    Quote Originally Posted by Arcane_Mathamatition
    if I have 3 numbers, G(5,5,5)= and A(5,5,5)= in a statement, the mean of any set of identical numbers, no matter the size, will always be equal to every number in the set. G(C,C,C,C,C,C,...C)= A(C,C,C,C,C,C,...C)= where n is the number of terms in the set of numbers having their means, geometric and arithmetic respectively, taken. is that not true? I am using one set of numbers, and running it through 2 scalar value functions from in the event all of the terms in my number set being equal, the means, neccesarilly, will be equal to every other term in my set. and since the two scalar value functions have the same outbput in I am saying, I guess the better word is that they are equivalent at that point, meaning that they yeild the same value. Am I right? so I would say, for this case, Or am I still off on something. and again, the limiter of this equation is that all of the numbers in my set MUST be positive, or this no longer holds true, but otherwise this is the only case when the two means will be equivalent.
    It is true that if G and A are applied to a sequence of numbers and if all of the numbers in the sequence are equal then G and A will also be equal. That is readily apparent from the definition of G and A. It is also true, but requires proof, that G and A are equal only on such sequences, that is a part of the theorem that you are trying to prove. You must be careful to not assume your conclusion within the hypothesis.

    It is not clear in your approach is what set it is that you consider to be the domain of G and A.

    For the purposes of the geometric-arithmetic mean inequality one would naturally consider them as defined on the set of all finite sequence of positive real numbers.

    But you seem to want to apply a gradient operator. That requires that the domain be an open subset of some for a fixed .


    Question: Are you experienced in and familiar with the development of rigorous mathematical proofs ? If so, what sort of classes in theoretical mathematics have you taken in which you were required to do proofs ?
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    I am not experienced in them, nor have I taken any classes that require proofs to be done by the student, outside of highschool geometry and the basic introduction to proofs we got there. I'm currently a student in a multivariable calculus course. after this semester I will be taking a course on linear algebra and differential equations. I only have an interest in this topic, I haven't got the honest to goodness education on it yet. And, if you don't mind my asking, I am curious about your backround in mathematics. From your posts, I would guess you are a college professor, and I'm just wondering, how much of yourself you have devoted to Math and how much education you have on the subject?

    no disrespect intended.
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  13. #12  
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    Quote Originally Posted by Arcane_Mathamatition
    I am not experienced in them, nor have I taken any classes that require proofs to be done by the student, outside of highschool geometry and the basic introduction to proofs we got there. I'm currently a student in a multivariable calculus course. after this semester I will be taking a course on linear algebra and differential equations. I only have an interest in this topic, I haven't got the honest to goodness education on it yet. And, if you don't mind my asking, I am curious about your backround in mathematics. From your posts, I would guess you are a college professor, and I'm just wondering, how much of yourself you have devoted to Math and how much education you have on the subject?

    no disrespect intended.
    To answer your last questoin first. I am retired from the aereospace industry. I did once upon a time teach university mathematics and conduct research in mathematics. I have a Ph.D. in that subject, as well as education and experience in engineering.

    The geometric-arithmetic mean inequality is not terribly difficult, but it is not the sort of thing that one would give to a typical undergraduate to prove. It requires some techniques and perspectives that would normally be found in students a bit more advanced than those usually found in the courses that you mentioned.

    That does not mean that the proof is beyond your intelligence or capabilities. It does mean that you might be better off starting with proofs of theorems that are bit simpler and more consistent with the mathematics that you have seen in the courses that you have taken so far. This one is more at the level of someone in a course in real analysis at an advanced undergraduate or beginning graduate level. The techjiques appropriate to this theorem show in books at the level of Rudin's Real and Complex Analysis or Real and Abstract Analysis by Hewitt and Stromberg. This is relatively advanced stuff, well beyond typical multivariable calculus, ODE and linear algebra.

    I can see from what you have done that you are interested in mathematics and have some aptitude. You are not afraid to tackle a hard problem and that is very important. But you also don't quite grasp the necessary ingredients in a rigorous proof, which not unusual for someone just starting to attempt proofs.

    I think that you would do well to take a class in which proofs are required of the students and get some experience with somewhat simpler theorems before taking on the more difficult ones. Such classes are usually at a junior or senior level and have a title like "Introductory Real Analysis". Such classes typicall start out by studying the point-set topology of which is interesting, not difficult and provides wonderful opportunity to learn to do proofs.
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    thank you
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