It Begs The Question
Hey…I'm just saying… And while we're at it, why are you defending them?

One More Thing February 9, 2019

Just one more thing sir. Would you happen to know how much
the Sum of the Reciprocals of Odd Composites is?

See the footnotes for links to the prior related posts that lead up to this one.


I don’t know how this happened but recently I was thinking about the whole “Math Mystery” thing which ended up being all about coming up with the “formula” (value) for the

Sum Of The Reciprocals Of Composites  

which I represent as Sc.  

 

That is, Sc, or the “Harmonic Series of Composites” (to coin a phrase)  
is simply the log of the Prime Counting Function!
That is,     Sc  =   ln(π(N))   or…   Sc  =   ln( N /   ln(N) )

Then it occurred to me that we now knew the following:

The Sum Of The Reciprocals Of All Numbers (aka the vanilla “Harmonic Series)
i
s  equal to ln(N)  which I will represent as Sn.  So…

Sn = ln(N)

and The Sum Of The Reciprocals Of The Odd Numbers, as represented by So,
is 1/2  of the The Sum Of The Reciprocals Of The All Numbers.  That is

So  =  Sn / 2
Or
So = ln(N) / 2

And the Sum Of The Reciprocals Of Just The Primes  is ln( ln(N) )
As I represent by  Sp.   

So,  the Sum Of The Reciprocals Of The Odd Composites  as represented by
Soc      should be “All of the Odd reciprocals  minus all of the Prime reciprocals

Soc  = So  – Sp

Soc  =  ln(N)/2   –  ln(ln(N) )

But using some “log arithmetic” we can rewrite the above equation as

Soc  =  ln(N1/2)  – ln( ln(N ) )                 

 

Using some more log arithmetic (See Footnote Log Arithmetic below) we get

Soc = ln(N1/2 / ln(N ) )                 

 

Or in English,  the Sum Of The Reciprocals of just the Odd Composite numbers (aka Soc),
is the log of (
the square root of N, divided by the log of N).
This is an interesting result when we consider that
the Sum Of The Reciprocals of ALL Composite numbers is

ln(  N / ln(N) ).


SOddComposites      = ln(  N1/2 /  ln(N) 

SAllComposites       =  ln(  N /  ln(N) )

                        

Who would have guessed.

Footnote on Log Rules:

1) logb(mn) = logb(m) + logb(n)

2) logb(m/n) = logb(m) – logb(n)

3) logb(mn) = n · logb(m)

Links to the first 4 episodes of the “Math Mystery”

which lead up to this post.

https://itbegsthequestion.com/ an-actual-math-mystery/

2 https://itbegsthequestion.com/ math-mystery-followup/

https://itbegsthequestion.com/ math-mystery-wrap-up/

4 https://itbegsthequestion.com/epilogue-to-the-composite-chronicle/

 

 

 

 

 

 

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