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God Has OCD Part Deux November 8, 2016

God Has OCD – Part Deux

Did you know that a shortage of remainders is why the hypotenuse can’t have prime factors in common with the legs of a right triangle?  Interesting eh?  Want to know more?  Then read on.

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In part 1 of God Has OCD we dealt with the Pythagorean Theorem    A2 + B2 = C2

and we proved that, for primitive Pythagorean Triples, A and B do not ever “share” prime factors with C. That is, A and B do not ever have a prime factor that is also a prime factor of C.

In this article we’ll look into the “sharing” of prime fctors for the more general Sum Of Powers formulas of which the Pythagorean Theorem is merely a special case.

We can express the general Sum Of Powers formulas like so:

T1n + T2n + T3n .. + Txn = Zn

In the above we’ll call T1n + T2n + T3n .. + Txn the “terms” of the equation.

We’ll demonstrate how, for “primitive” N-tuples, where the greatest common denominator of all the terms is 1,  that the terms may or may not, “share” prime factors with Z; except for, again, the special case of the Pythagorean Theorem where prime factors are never “shared” and where there are only 2 terms (A and B as we’re accustomed to seeing them called).

So where do we begin? As I often do, I experimentally test theories first, and then see if they can be proven. In this case, I could again use my SumOfPowers program (in C#).

With the SumOfPowers program we can specify:

        • How many terms to use
        • A range of values for the terms
        • The exponent (n) to use

 

 

The program would then find integer solutions if any exist. Follow the link below for an example screen print of the SumOfPowers program after an execution.

SumOfPowers Example

For this article I ran the SumOfPowers program and the results demonstrated that sometimes the terms T1..Tx will have a prime factor in common with Z and sometimes none of the terms will have a prime factor in common with Z.

That said, we now unequivocally know and can demonstrate the answer with a real example but the question is why? And then, why is this NOT true for the case of 2 terms (e.g. the Pythagorean Theorem)?

Anyway here are a few simple examples found by the SumOfPowers program:

T13 + T23 + T33 .. + T63 = Z3

13 + 23 + 33 + 63 + 163 + 173 = 213

13 + 23 + 63 + 63 + 63 + 73 = 103

33 + 43 + 53 = 63

23 + 53 + 123 + 123 + 153 + 173 = 243

T15 + T25 + T35 .. + T85 = Z5

85 + 145 + 165 + 205 + 205 + 225 + 245 + 305 = 345

 

 

So why is it possible for the Sum Of Powers for more than 2 terms to share prime factors with Z while it is NOT possible when there are just 2 terms?

T1n + T2n .. + Txn = Zn            << YES prime factors sometimes shared!

T1n + T2n = Zn                     <<  NO.  Less than  3  terms then Prime factors NEVER shared!

 

 

The reason is as follows… If we divide both sides of the equation by one of Z’s prime factors we get something like this (for example):

33 + 43 + 53 = 63                               Prime factors of 6 are 3 and 2. Let’s use 3…

33      +      43      +      53      =        63

—-             ——           ——-            ——–

3                 3                3                 3

 

 

3 divides 3^3 and it also divides 6^3

however (4^3)/3 and (5^3)/3 are not integers but the equation is still valid because

The sum of all of the remainders is an integer multiple of 3. For example:

(4^3)/3 = 21 + 1/3 and

(5^3)/3 = 41 + 2/3 then

1/3 + 2/3 = 3/3 = integer

So it all works out because we still get integers on both sides of the equation.

But note that this could never work out when there’s only 2 terms because then there would only be a single remainder after dividing the terms T1 and T2 by one of the prime factors of Z. And with only a single remainder, the left side of the equation can’t possibly be an integer!

For example, for A2 + B2 = C2

let’s assume that one of the prime factors of C was also a prime factor of A. Let’s call that common prime factor Pc. Then we would have something like the following if we divide both sides of the equation by Pc:

(Pc x Pa2 … x Pan)  x A       +         B2        =         (Pc x Pc2 … x Pcn) x C

——————————                    —————               ————————————-

        Pc                                           Pc                                    Pc

 

Integer + ( B2 / Pc) = Integer

But    B2 / Pc        yields some remainder after division; And

 

Unlike the case of more than 2 terms on the left side of the equation,

there are no additional remainders possible to add to this remainder

in order to obtain an integer that would make the equation valid!!!

 

Integer   +    ( B2 / Pc)        =     Integer

Integer   +   NON-Integer    =     Integer            <<=== a contradiction

 

 

So our original assumption of some prime factor of C also being a prime factor of A can NOT ever be true for the Pythagorean Theorem or for any Sum Of Powers formula with just 2 terms on the left of the equals sign.

 

 

And in some quirky and curious sense

we can say that this is because,

for just 2 terms (like the Pythagorean Theorem),

there will always be a “shortage of remainders”

that would prevent the validity of the equation!!!

All done.  We can go now.

Open The Gates!

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GOD Has OCD October 7, 2016

a2b2c2

God Has OCD

Pythagorean Triples and Their Prime Factors

Sum Of Powers and Their Prime Factors

pythagorasjpg

 

The Pythagorean Theorem is familiar to all of us who ever took Geometry in High School. But there’s more to it than we may have realized:

A2 + B2 = C2

This post is about yet another project where we investigate whether (or not) the terms of the equation “share” prime factors. In this case we consider Pythagorean Triples where A, B, and C are all integers (of course). It wasn’t the first time I had come across Pythagorean Triples but this time I decided it might make for a simple project to determine whether the terms of the Pythagorean Theorem “share” Prime Factors.

We’ll expand / extend the findings to apply to “Sum of Powers” equations in general. More on that later.

This question of “sharing” and “non-sharing” of Prime Factors among the terms of formulas of other “famous” theorems was the subject of other posts in this blog. Here are links to those other posts that investigate the sharing (or non-sharing) of Prime Factors for other notable formulas/theorems:

EVEN = HALF the difference of 2 squares

ODD = The difference of 2 squares

Prologue To Gerry’s Goldbach Theorems

Gerry’s Goldbach Theorem

Gerry’s Goldbach Weak Theorem

Chen’s Theorem and YAPFO

 

Pythagorean Triples

It is helpful to first read a little about Pythagorean Triples. Here is a link to an excellent and very interesting (yet short) treatment that’s well worth reading. It’s on Wolfram.com

http://mathworld.wolfram.com/PythagoreanTriple.html

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Primitive Pythagorean Triples

In this post/project we only deal with Primitive Pythagorean Triples (where A and B are relatively prime) because, as stated on the Wolfram page:

It is usual to consider only primitive Pythagorean Triples (also called “reduced”triples) in which A and B are relatively prime since other solutions can be generated trivially from the primitive ones.

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Step 1 – Generate Primitive Pythagorean Triples

The first step of this small project was to generate a bunch of Primitive Triples and to write them on a whiteboard; and then to manually determine whether they share any Prime Factors with the variable C (on the right side of the equation).

Keep in mind we are only considering Primitive Triples. Therefore A and B will NOT “share” any Primes because, by definition, they don’t share ANY factors (prime or otherwise); so we’ll be looking to see whether it’s possible for A and B to “share” a prime with C (on the right side of the equals-sign).

As it turns out, generating the Triples was very easy because I already had written a C# program that would do that. It was/is the “SumOfPowers” program that finds integer solutions for “Sum of Powers” equations like:

T1n + T2n + .. Txn = Zn

For this project I simply ran the SumOfPowers program specifying just 2 terms on the left side of the equation and an exponent value of 2. Like so…

A2 + B2 = C2

In this way the SumOfPowers program could instantly generate MANY combinations of A, B, and C (all of them integers) which I could then easily factor to see if they share any Prime Factors.

Here are a few very simple (and small) examples of the values for A, B, and C found by the SumOfPowers program.

3, 4, 5

5, 12, 13

11, 60, 61

119, 120, 169

48, 55, 73

36531, 36540, 51669

Of course, the SumOfPowers program could generate gigantic quantities of other examples where A, B, and C varied tremendously in size. Note that the SumOfPowers program is the subject of its own article (post) that you can read here:

http://itbegsthequestion.com/sumofpowers/

 

 

Step 2 – Manually Factor and Evaluate the Generated Examples

There’s not much to say here. We just manually reduce each term (A, B, C) down to their Prime Factors and see if any of those Prime Factors show up in more than 1 term. As it turned out, for all of the 20 or so sample Primitive Pythagorean Triples examined, there was never a case where a Prime Factor of either A or B was also a Prime Factor of C. I.e. for any particular Triple there was never a Prime Factor that showed up in more than 1 of the terms A, B, or C!

The fact that Prime Factors of A and B never appeared as a Prime Factor of C led to the next step which was to see if we could prove that this would always be the case.

Step 3 – Prove It

Here’s the proof (short and simple).

Let A, B, and C represent a Primitive Pythagorean Triple in the following equation (note that we simply put C on the left and put A and B on the right):

C2 = A2 + B2

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Show the above with C’s Prime Factors Pc1..Pcn exposed:

(Pc1 x Pc2 x .. Pcn) x C = A2 + B2

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Divide both sides by one of C’s factors (e.g. Pc1) and we get…

(Pc2 x .. Pcn) x C =    A2    +    B2

                                     —–            ——-

                                Pc1              Pc1

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If either A or B is to “share” a Prime Factor with C, then Pc1

1–  Must divide ONLY A or ONLY B, or it

2–  Must divide BOTH A and B.

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1–  If Pc1 divides just 1 term (ONLY A or ONLY B ) then we’d have the following (e.g. with Pc1 dividing ONLY A):

Integer = Integer +     B2 / Pc1

Integer = Integer + Non-Integer

Integer = Non-Integer  <<<========== contradiction

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2–  If Pc1 divides BOTH A and B then A and B are NOT relatively prime and therefore A, B, and C did NOT constitute a primitive triple after all which contradicts our original assumption/premise.

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Therefore, C can NOT “share” a Prime Factor

with either A or B.

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Pretty simple and straightforward, right? So what does it all mean?

Well, it should be obvious…

It means that God exists and has OCD.

ocdmeme

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It seems like we may be able to apply the above proof to Sums Of Powers in general. Instead of just having 2 terms on the left and one term on the right and an exponent of 2 like the Pythagorean Theorem:

A2 + B2 = C2

We can get more general with any number of terms and any exponent. Like so:

T1n + T2n + .. Txn = Zn

That will form the basis for the next project. I’ll let you know if it gets anywhere.

Open The Gates!

 

 

 

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Chen’s Theorem and YAPFO September 15, 2016

chen_jingrun

Chen

googlesearchfortheorems

Chen’s Theorem

This posting is about the Chen’s Theorem project.

Let’s start out by simply stating Chen’s Theorem.

 

In number theory, Chen’s theorem states that every sufficiently large even number can be written as the sum of either two primes, or a prime and a semiprime (the product of two primes). For example:

20 = 17 + 3      (2 primes)

20 = 5 + (3 x 5)       (a prime and a semiprime)

 

The part of the theorem about the sum of 2 primes we’ll ignore here because it’s the same as Goldbach’s Conjecture which was covered in an earlier posting

(see Gerry’s Goldbach Theorem)

 

The part of Chen’s Theorem we will cover here is the part about:

or a prime and a semiprime

 

The initial process of exploration used the usual “method.” I.e. “manually” play around with a few randomly chosen example numbers on a white board (EVEN numbers in this case) and try to spot patterns. If a meaningful/interesting pattern surfaces and it looks to be worth pursuing then the next step is to decide whether to write a computer program to test the perceived pattern over millions (or billions) of numbers to see if it still holds water. Writing the program(s) is never difficult but it’s always time-consuming; and I don’t want it to interfere with my watching of TMZ or AGT!!

Even if a pattern seems to hold true over billions of numbers, and even if some of those numbers are hundreds of digits long, it still doesn’t prove anything! It’s still just conjecture. We learned that from experience and it was touched on in a prior posting:

See the SumOfPowers article

Look for “Lander” or “Parkin” in the above posting/article.

 

Anyway… Even if the perceived pattern holds true across billions of tests, it’s still just conjecture. The next step is to actually Prove the truth (or falsity) of  our conjecture.

A perceived benefit of writing code to run thru billions of test cases is:

If our conjecture is false, we can greatly increase our chances of quickly realizing it by way of counter-example. We could save a lot of effort by not trying to prove something that can not be proved! Again, Look for “Lander” or “Parkin” in the posting/article SumOfPowers.

Of course, it may be that proving the conjecture could turn out to be easier than we think. If we attack the proof early on we could, perhaps, prove (or disprove) our conjecture with relative ease and thereby save ourselves the programming effort.

 

All that said…

The whole “Proof process” seems to have been getting easier lately so I’ve taken to trying to Prove the conjectures (pattern perception) in advance of writing any code to test billions of specific examples. And keep in mind that I would eventually have to go thru the whole Proof thing anyway! Why not at least give it a shot early on!? It might turn out to be relatively simple and I could save myself a lot of time.

So… for this project the following was the process…

 

Randomly choose an EVEN number that would be easy to work with via calculator (or in my head) and apply the following part of Chen’s Theorem:

the sum of a prime and a semiprime For example (and color coded):

The prime factors of 20 are 2, 5

20 = 5 + (3 x 5)            (sum of a prime and a semiprime)

20 = 11 + (3 x 3)

or

The prime factors of 68 are 2, 17

68 = 3 + (13 x 5)

68 = 11 + (3 x 19)

68 = 13+ (5 x 11)

68 = 17 + (3 x 17)

or

The prime factors of 12 are 2, 3

12 = 3 + (3 x 3)

12 = 2 + (5 x 2)

ETC.

 

The pattern noticed (i.e. the conjecture), is that, whenever one of the prime factors of the EVEN number on the left side of the equals sign shows up on the right side of the equals sign, then that prime factor will be in BOTH terms on the right of the equals sign. That is, that prime factor will constitute BOTH the prime, and, one or both of the primes that makes up the semiprime as shown in the examples above and color-coded in red.

Clearly it would be easy to write a program to spin thru bizillions of EVEN numbers to test the above conjecture. And if we were to encounter even a single counterexample it would be enough to PROVE the conjecture is false. But even if all of the examples tested conformed to the conjecture it would not be enough to actually PROVE it.   It would lend credence to the conjecture but it is not PROOF.

 

For this project I decided to try proving the conjecture in advance of any programming.    As it turned out, the proof came quickly and is very simple. Here it is; I call it:

 

Gerry’s Chen Based

YAPFO Theorem

Proof

YAPFO = Yet Another Prime Factor Oddity

 

 

First we restate Chen’s Theorem upon which this YAPFO Theorem is based:

In number theory, Chen’s theorem states that every sufficiently large even number can be written as the sum of either two primes, or a prime and a semiprime (the product of two primes). For example:

20 = 17 + 3 (2 primes)

20 = 5 + (3 x 5) (a prime and a semiprime)

The part of the theorem about the sum of 2 primes we’ll ignore here because it’s the same as Goldbach’s Conjecture which (again) is covered in an earlier posting (see Gerry’s Goldbach Theorem);

 

The part of Chen’s Theorem of interest here is the part about:

“or a prime and a semiprime “

Stated symbolically:

E = P + (Ps1 x Ps2)

where P is some prime, and, Ps1 and Ps2 are primes that make up the semiprime (Ps1 x Ps2). And E is, of course, some EVEN integer greater than 4.

The proof proceeds as follows:

E = Pf1 x Pf2 x Pf3… x Pfn

where Pf1 .. Pfn   are the prime factors of E.

Which leads to…

Pf1 x Pf2 x Pf3… x Pfn   =    +   (  Ps1 x Ps2  )

 

If we divide both sides by any one of E’s prime factors Pf1 .. Pfn   we get (using Pf1 for example)

 

                              P            ( Ps1 x Ps2 )

  Pf2 … x Pfn   =         ——    +     ———————–

                             Pf1                     Pf1

 

Note that the left side of the equal sign is an integer. But also note that:

If   P = Pf1  then for the right side of the equation to be an integer,  Pf1  must also equal Ps1 (or Ps2… it doesn’t matter which).

And likewise, if Pf1 equals Ps1 (or Ps2),   then in order for the right side of the equation to be an integer, Pf1 must also equal P.

In other words,    Pf1  can not evenly divide ( Ps1 x Ps2 )   unless Pf1 equals Ps1, or  Ps2, or both.  

 

So… for       E = P + (Ps1 x Ps2)

then if one of the prime factors of E equals any of the terms on the right side of the equals sign,  then it must equal at least 2 of the 3 terms,

one of which must be  (Pf1 must equal P).

 

By the way, here are some examples where one of the prime factors of E is equal to all 3 terms on the right side of the equal sign (they are easy to construct):

12  =  3 + (3 x 3)

6   =   2 + (2 x 2)

30 = 5 + (5 x 5)

 

 

 

*** end of proof ****** end of proof ****** end of proof ***

 

Ta Da!

heisrightyouknow

Now for an interesting (hopefully) note about the history of this short project…

It’s been awhile since working on a new project. I’ve been on the lookout for one (as always) but only a couple of days ago did I find Chen’s Theorem. Interestingly, while looking for a project subject, I thought that what might help is a central source for Number Theory formulas and theorems. I also thought “What took me so long to figure that out!?”

So I searched for that and voila. I immediately found the following:

https://en.wikipedia.org/wiki/Category:Theorems_in_number_theory

A valuable  find.

 

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What does that even mean!? September 14, 2016

I saw these cases of Mineral Water for sale at Whole Paycheck recently. I had to take a photo because I thought it was so unusual.

 

Lock The Gates!

wholepaycheckwaterwholepaycheckco2added

Note the ingredients… “Natural carbon dioxide”

wholepaycheckwaterco2

And note that the “natural carbonation” is “added.”

What does all that even mean?

 

But wait!  There’s More!

When I go out walking I will often stop at Whole Paycheck to buy a tasty Cherry-Vanilla soda.  Last week I took my soda up to the register to pay and the lady scans the can.  It doesn’t register.  She then looks up some code in a small book by the register and enters some numbers from the book….

Then she looks at me with a straight face and says

”  $4.25 please  “

 

And I responded “I don’t think so.  How about 74 cents?  They’re 74 cents total.”

What really struck me about  the situation was 2 things:

  1.  It never entered her mind that $4.25  was a bit high and was probably incorrect.
  2. She thought a customer would actually pay $4.25 for a can of soda (especially me!).

 

So… is it “… was 2 things”  or is it “… were 2 things?”    Let’s consult Google for “were vs was.”   The survey says…

 

Was is used in the first and third person singular past. It is used for statements of fact.

Were is used in the second person singular and plural and first and third person plural.

It is used in the subjunctive mood to indicate unreal or hypothetical statements.

I’m gonna stick with “was.”

 

Open The Gates!

 

 

 

 

 

 

 

 

 

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Divergence of Harmonic Series of Primes August 23, 2016

 

For Just the Proof of the divergence of the Harmonic Series of Primes

in just 5 easy steps  just scroll down to the section

Just The Proof

Continuing…

Story Background and History

About 6 weeks ago I ran across the Harmonic Series in one of my books. It wasn’t the first time I’d seen it but… anyway…

The Harmonic Series goes like this…

Divergence_HarmonicSeriesDef_01

Simple and straightforward, right? But what I found really interesting is a Proof that the series “diverges.” I.e. that the sum of the terms goes to infinity as n gets larger and larger (as n goes to infinity).  The most interesting, simple, and elegant Proof of the divergence of the Harmonic Series is the one by Nicole d’Oresme (ca. 1323-1382),

Oresme’s proof groups the harmonic terms by taking 2, 4, 8, 16, … terms (after the first two) and noting that each such block has a sum larger than 1/2,

Divergence_Oresme_1

 

and since an infinite sum of halves diverges, so does the harmonic series diverge. This proof is quintessentially simple and elegant. That said, here’s my story about how Oresme’s proof led me to search for another SIMPLE PROOF except that would be a SIMPLE PROOF of the divergence of The Harmonic Series of Primes.

 

Lock The Gates!

 

I presumed there would be a similar well-known series, except for Primes. So I googled “Harmonic Series Primes” and voila!! We find the Harmonic Series of Primes. Here are a couple of links:

http://mathworld.wolfram.com/HarmonicSeriesofPrimes.html

and

https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes.html

 

So here is what the Harmonic Series of Primes (HSOP) looks like…

Divergence_HSOP_01

All the way out to PN  (1/2  +  1/3  +  1/5  + …  1/PN)

Simple. Straightforward. And I quickly learned that this series also diverged (the sum of the terms of the HSOP goes to infinity). I could find various proofs of this but they were complicated and hard to follow… at least for me.  They weren’t nearly as simple as  Oresme’s proof for the Harmonic Series. Now that doesn’t mean a very simple proof of the divergence of the HSOP did not exist… Only that I did not find one.  If you find one the PLEASE let me know.

So that became my next project; a quest as it were to develop a SIMPLE proof.

The “quest” lasted about 6 weeks with ruminating about this while watching TMZ or while insomniating in bed at 3 in the morning listening to Art Bell or George Noory or the Tappet Brothers. I tried various approaches to the problem (multiple times and multiple variations for each approach)… a lot writing on white boards… but I didn’t get anywhere. It was frustrating.

In retrospect, as we’ll see shortly, I was hung up on the details of one particular part of one formula as opposed to what that part of that formula was, in general, trying to tell me!! Why  it took weeks to see it is beyond me.   However, there was one saving grace and it was that I intuitively felt that the Prime Number Theorem (PNT) would play a major (largest) role for a variety of reasons not the least of which is that it’s simple and elegant and packed full of information in a very small space!  That intuition would turn out to be very important!

 

Point 1.

All that said, here’s how it goes. Let’s start with the Prime Number Theorem (PNT).

The PNT tells us “approximately” how many Primes there are between 0 and some integer N. It’s a formula. And the formula is:

π(N) ~ N / ln(N),

where π(N) is the prime-counting function

and ln(N) is the natural logarithm of N.

 

Note that “ ~ means “approximates closely (asymptotically).” For example, how many Primes are there between 0 and 1 million?

π(N)   N / ln(N)

so for 1 million it’s…

π(1,000,000)  ~ 1,000,000 / ln(1,000,000)

 

So that is point 1.   That is,  The PNT tells us that the count of Primes between 0 and N is

 

N

___________

ln(N)

 

 

Point 2.

To compute the HSOP out to any Prime PN we can calculate an “average” value for the terms by dividing the sum of the terms by the number of the terms. For example, let’s take the HSOP out just 4 terms:

1/2 + 1/3 + 1/5 + 1/7 = (105 + 70 + 42 + 30) / 210 = 247/210 = 1.1762

and if we divide 1.1762 (the sum of the terms)   by 4 (the number of terms) ,    we get .2940476… which is the “average value” of the terms.

We can then multiply the average value of the terms by the count of the number of terms to obtain the sum of the terms (i.e.  to obtain the value of the series)!!

.2940476      x   4      =      1.1762

  Avg     x  Cnt   =    SumOfSeriesTerms

 

Think of it this way…

(TheSumOfTheSeriesTerms divided by TheCountOfTerms) = TheAverageValueOfTheTerms

or… Series / Cnt = Avg

which leads to

Series = Avg x Cnt      and earlier we said that  Avg x Cnt = Series     so…   Series = Series    

and that seemed trivial and meaningless so I would follow some bunny trail and look for a different approach.

 

 

 And that is what I was getting hung up on.

But it’s the wrong way to think of it!

Instead we need to think of it this way…

 

If we take the Harmonic Series of Primes out to the Nth prime then there is a number (a value) that is the “Average Value” of the terms of the series (call it AVG)

It turns out that the value of AVG is somewhere between ½ and PN/2 but I’ll leave that to the reader to ponder if they want to.

But we don’t really care what the formula for AVG is,  or what the value of AVG exactly is, just that it IS (Sounds like Bill Clinton doesn’t it?).

 

We also know, and saw demonstrated above,  that if we simply multiply AVG by the number of terms N (call it CNT) we get the value of the series out to PN (choose any N).

And we also know, and discussed above, that the number of terms (CNT) is obtained so very simply via the PNT (Prime Number Theorem) like so…

 

      PN

CNT  =   ______

        ln(PN)

 

Remember that I said earlier that I expected the Prime Number Theorem (PNT) to play a key role;  it just happened!  Anyway…

 

                                     PN

SeriesValue = AVG x CNT =   AVG  x  __________

                                     ln(PN)

                                  

As N goes to (and thus  PN  goes to infinity) so does the Series Value (the sum of the terms) thus proving the divergence of the Harmonic Series of Primes.

 

So that’s the story of the search for a SIMPLE proof of the divergence of the Harmonic Series of Primes.   Whew!!



Just The Proof

So considering all that background and explanation above,

here is the proof boiled down to just

5 Easy Pieces (the essentials):

 

But first we start with a reminder.   The Harmonic Series of Primes (HSOP) is

Divergence_HSOP_01

out to    1/PN

1. There is an “Average Value” for the terms out to PN (we’ll call it AVG) which is the sum of the terms divided by the number of terms.

2. The value of the series out to PN = AVG x (The number of terms out to PN)

3. From the Prime Number Theorem, the number of terms (call it CNT) out to Pis

PN

______

ln(PN)

4. Thus, The value of the series (the sum of the terms) out to PN is

SeriesValue AVG x CNT = AVG x    PN

                                  ______

                                  ln(PN)

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.

.

5. As N (and thus PN) goes to infinity so does CNT ( which per the PNT is       PN / ln(PN) )    which gives us

SeriesValue AVG x CNT =   AVG x infinity

 

Which, of course, means that the Harmonic Series of Primes diverges.

 

Open The Gates!

 

 

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Ads On The Internet August 17, 2016

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Sometimes the “Artificial Intelligence” used in placing ads can result in “strange” juxtapositions. I was researching something that Hillary Clinton said back in May 2008 about why she would not drop out of the race when it became obvious that Obama was going to win the Democratic Party’s nomination. She said…

Senator Hillary Rodham Clinton defended staying in the Democratic nominating contest  by pointing out that her husband had not wrapped up the nomination until June 1992, adding, “We all remember Bobby Kennedy was assassinated in June in California.” 

 The essence of what she said was that she would not drop out because Obama could be hit by a bus… or worse; and people say Trump puts his foot in his mouth; and some people just should not be able to get their hands on a gun; and I’ll bet she’s an exciting date; and… Stop! Just Stop!!

 

Anyway… then it hit me… The ad! The ad!!! Look at the ad that was inserted on the right side of the page as it was sent to my browser 8 years later!

ClintonRFK_MisplacedAd

I wonder if you would get the same ad if you pull up an article about Vince Foster, or Seth Rich,  or Bryan “The 5th” Pagliano, or some of the “Cosby Girls” like Juanita Broaddrick, or Monica “Vast Conspiracy” Lewinski,  or Paula “$850,000” Jones, or …

No… Seriously… Would you get a “She’s with us” ad?

The Internet… It Never Ceases To Entertain.

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Cleared For Takeoff August 11, 2016

This is not what you want to see

when you’re trying to take off.

 

I’ve seen this clip many times over the years and it always amazes me.

It’s from the series Victory At Sea – Suicide – Episode 25

Note the action at about the 30 second mark and again around the 49 second mark.

You can find the complete series on YouTube.

 

What must be going thru the American pilot’s head as he’s trying to take off!?

 

 

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Obama – Reporters – Talking Heads – and $400 million August 6, 2016

 

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Yep…  “The reason that cash was exchanged is because we don’t have a banking relationship with Iran.”

Yep… sure… uh huh.  That’s it.  Why didn’t I think of that?

So…    the $400 Million was for something agreed to last January (7 months ago!) and in the entire federal government   there wasn’t anybody who could find a way to transfer $400 Million to Iran except for flying a plane load of cash (in a variety of currencies).

 

Now it doesn’t take a Rocket Surgeon to figure out that was a blatant lie.  But we’ll put that aside.  What’s more telling, and which speaks humongous volumes, is that not one so-called “reporter” challenged him on the issue!!!

“But Mr. President; are you saying that in the last 7 months  there wasn’t anybody in your administration who could find a way to transfer $400 Million to Iran except for flying a plane load of cash?  And did that make sense to you either then or now?”

And in the days that followed, I have NOT seen/heard  ANY TalkingHead (so-called “journalist”) who has broached the subject.  Not one.  Not Anderson Cooper, not Megan Kelly,  not Charlie Rose…. not anyone!

The Blatant Lies

followed by

The Roaring Silence.

 

The Presidential shameless  lie is trivial when compared to silence that has followed since.  But then we’ve all been well aware of that for a long time now;  haven’t we?

And yet many are incredulous that others would/will give full support to Donald Trump in spite of whatever foibles he may be said to have (and they are said, 24 hours a day,  by the same silent TalkingHeads and so-called reporters).

 

So that said…

Who “gets it” and who doesn’t “get it?”

 

 

Open The Gates!

 

 

 

 

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Zika Conspiracy August 3, 2016

ZikaMosquito

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Conspiracy Nut!

Maybe yes and maybe no.  Anywhoo, here’s my theory. In a relatively short  time (within a year?) we could see the U.S. government and/or one or more state governments (as a federal surrogate) attempting to force the population to submit to blood testing for the Zika Virus. Not just a few folks in hospitals or in the military but rather squads of “bureaucrats” marching door-to-door to take people’s blood under the guise of public health necessity. Then, of course, the whole mess will make its way into the courts; and then…

Farfetched? Just another Conspiracy Theorist? Or… dot-connector extraordinaire.

We’ll see.

BTW, today is 8/2/2016. Set your alarm for 8/2/2017.

 

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Harmonic Series July 2, 2016

        

HA   =    1/2   +    1/3   +   1/4  +   1/5   +   1/6     …  +   1/n

HP   =    1/2   +    1/3   +   1/4  +   1/5   +   1/6     …  +   1/n

 

 

 

 

 

 

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