Adam Ledger

Mr. Adam Ledger

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7 years, 217 days
Perth, Australia

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@Carl Love  Ok this the general crux of the issue I have with understanding Asymptotics. In as much as I understand the basic principle of forming the relations based on the term which has the greatest exponent ie growth rate as x tends to infinity, when I write my number theory notes based on papers I read and wikipedia, I am unable to form a methodology for a "generic"  function and equation for determining the exact nature of the function to be placed inside the "O" and "o".

I am reading alot of papers written by Pierre Dusart, this week's focus is on the Chebyshev functions. I am having difficulty writing maple code for the Von Mangoldt function. 

Yes you are correct in seeing my intentions as using this function as my very first "worked" example in asymptotics, I wish to become proficient enough to be able to apply the principles of asymptotics to any such function I please, and so it's actual number theortic importance is not relevant.

@vv Ok thankyou Sir we will resume our discussion at a stage i have reached an adequate level of clarity to express my intentions in  natural language in an effective manner 

@vv  that's an exceptional point so should i then instead be replacing the expression on the lhs with it's least squares regression approximation to n? That is the real crux of the importance to PNT, how strong the least squares regression approximation is to n.

@Carl Love ok i did think that might be so but it's only day 2 of seriously studying asymptotics so i need to be told everything for some reason 

@Carl Love  yes these definitions are correct. Have i stated the asymptomic approximation correctly on my SE post?


ie which of these are correct:


$$( n\cdot\gcd(n ,\min(\mathcal D(n\cdot\bigl\lfloor \frac{p_n}{n} \bigr\rfloor) \backslash {\{1}\}))-\bigl\lfloor \sqrt {n} \bigr\rfloor\cdot\gcd(\bigl\lfloor \sqrt {n} \bigr\rfloor ,\min(\mathcal D(n\cdot\bigl\lfloor \frac{p_n}{n} \bigr\rfloor) \backslash {\{1}\}))\bigr)^{\frac{1}{2}}\sim n$$

$$( n\cdot\gcd(n ,\min(\mathcal D(n\cdot\bigl\lfloor \frac{p_n}{n} \bigr\rfloor) \backslash {\{1}\}))-\bigl\lfloor \sqrt {n} \bigr\rfloor\cdot\gcd(\bigl\lfloor \sqrt {n} \bigr\rfloor ,\min(\mathcal D(n\cdot\bigl\lfloor \frac{p_n}{n} \bigr\rfloor) \backslash {\{1}\}))\bigr)^{\frac{1}{2}}=O\Bigl(n\Bigr)$$



Was more of a broad request in asking Carl to give me clues because I know he is an expert in PNT.

But its find my friend gave a heap of papers to read so ill get by

Learning difficulties, late age cognitive development, and the sheer fact that I simply cannot seem to focus on anything other than what interests me passionately, I think maybe the nails in my coffin as far as ever being an expert in any particular field.

In heinsight, formal guidance in the way I learnt to use maple would have been a far better course of action what I choose. In my attempt to reverse the errors of my past ways, I will definitately be obtaining a copy of this publication to work from in the future.


Update: Just flicked my Dad an email asking if he would buy me a paperback copy and he ordered it! should be in the mail within the fortnight im so stoked!



you have shown that f(k) diverges to infinity as k is made infinitely large.


How does this prove that  f(k) is equal to 0 for a finite number of values on $\mathbb N$?

It is periodic at some k>177, how do we know that it does not oscillate from infinite to 0 an infinite number of times? you have assumed it to be strictly increasing, which it is not.

And if so, how does one calculate the  upper bound for S?

@vv ok i will discuss this in the morning with you thank you

@vv I put the corrected version up in a ME chat room, i will just put the latex here for you to copy and paste.


make not of the part below outline the part that still requires explicit proof.

We seek to prove:
$a | b {\rm \: \Rightarrow \:}\phi(a) | \phi(b)$       (1)
$\phi(a) | \phi(b) {\rm \: \not\Rightarrow \:}a | b$   (2) 
for (1) we have:
$a | b{\rm \: \Rightarrow \:} {rad(a)} |\, {rad(b)}$
${rad(a)} |\, {rad(b)}{\rm \: \Rightarrow \:}\prod_{p|a} p \, | \prod_{p|b} p$
$\prod_{p|a} p \, | \prod_{p|b} p{\rm \: \Rightarrow \:}\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b\cdot \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr)$
$\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b\cdot \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr){\rm \: \Rightarrow \:}{\Biggl\{\frac{b \cdot \prod_{p|b} (1-\frac{1}{p})}{\prod_{p|a} (1-\frac{1}{p})}}\Biggr\}=0$
${\Biggl\{\frac{b \cdot\prod_{p|b} (1-\frac{1}{p})}{\prod_{p|a} (1-\frac{1}{p})}}\Biggr\}=0 \land {\Bigl\{\frac{b}{a}}\Bigr\}=0{\rm \: \Rightarrow \:}{\Biggl\{\frac{b\prod_{p|b} (1-\frac{1}{p})}{a\prod_{p|a} (1-\frac{1}{p})}}\Biggr\}=0$
${\Biggl\{\frac{b\prod_{p|b} (1-\frac{1}{p})}{a\prod_{p|a} (1-\frac{1}{p})}}\Biggr\}=0{\rm \: \Rightarrow \:}{\Bigl\{\frac{\phi(b)}{\phi(a)}}\Bigr\}=0$
${\Bigl\{\frac{\phi(b)}{\phi(a)}}\Bigr\}=0{\rm \: \Rightarrow \:}\phi(a) | \phi(b)$
Thus establishing $a | b {\rm \: \Rightarrow \:}\phi(a) | \phi(b)$ is true.
We now look to establishing that (2) is true:
$\phi(a) | \phi(b) {\rm \: \Rightarrow \:}{\Bigl\{\frac{\phi(b)}{\phi(a)}}\Bigr\}=0$ 

Consider $z_1=x_1 y_1 \land z_2=x_2 y_2$ and $x_1,x_2,y_1,y_2 \in \mathbb Q$
And suppose we can assert ${\Biggl\{\frac{z_1}{z_2}}\Biggr\}=0$ knowing $z_2$ to divide $z_1$.
We then consider the individual divisibility criteria between the four integer variables in such a ratio with no remainder, as to allow us to see that the reason for the converse of (1) having a false truth value is elementary in nature, by finally substituting these generic variables for the four arithmetic functions appropriate in the division of the totients of $a$ and $b$, allowing us to see that either $a | b$ or it's negation can be true when only knowing $\phi(a) | \phi(b)$.

$\frac{x_2}{x_1} | \frac{ y_2}{ y_1}{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_2}{x_1}}\Biggr\}=0 \land {\Biggl\{\frac{ y_2}{ y_1}}\Biggr\}=0{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_1 y_1}{x_2 y_2}}\Biggr\}=0$
$\frac{x_2}{y_1} | \frac{ y_2}{ x_1}{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_2}{y_1}}\Biggr\}=0 \land {\Biggl\{\frac{ y_2}{ x_1}}\Biggr\}=0{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_1 y_1}{x_2 y_2}}\Biggr\}=0$
$\frac{x_2 y_2}{y_1} | {x_1}{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_2 y_2}{y_1}}\Biggr\}=0 \land {\Biggl\{{ x_1}}\Biggr\}=0{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_1 y_1}{x_2 y_2}}\Biggr\}=0$
$\frac{x_2 y_2}{x_1} | {y_1}{\rm \: \Rightarrow \:}{\Biggl\{\frac{x_2 y_2}{x_1}}\Biggr\}=0 \land {\Biggl\{{ y_1}}\Biggr\}=0 {\rm \: \Rightarrow \:}{\Biggl\{\frac{x_1 y_1}{x_2 y_2}}\Biggr\}=0$
${\Biggl\{\frac{z_1}{z_2}}\Biggr\}=0{\rm \: \Rightarrow \:}\frac{x_2}{x_1} | \frac{ y_2}{ y_1} \lor \frac{x_2}{y_1} | \frac{ y_2}{ x_1} \lor \frac{x_2 y_2}{y_1} | {x_1} \lor \frac{x_2 y_2}{x_1} | {y_1}$
${\Bigl\{\frac{z_1}{z_2}}\Bigr\}=0{\rm \: \Rightarrow \:}S$
$S:({\Bigl\{\frac{ y_2}{ y_1}}\Bigr\}=0 \land {\Bigl\{\frac{x_1}{x_2}}\Bigr\}=0)\lor({\Bigl\{\frac{x_2}{y_1}}\Bigr\}=0 \land {\Bigl\{\frac{ y_2}{ x_1}}\Bigr\}=0)\lor({\Bigl\{\frac{x_2 y_2}{y_1}}\Bigr\}=0)\lor({\Bigl\{\frac{x_2 y_2}{x_1}}\Bigr\}=0)$
Then finally we make the substitutions as follows:
$$y_1=\prod_{p|b} (1-\frac{1}{p})$$
$$y_2=\prod_{p|a} (1-\frac{1}{p})$$
$\phi(a) | \phi(b){\rm \: \Rightarrow \:}{\Bigl\{\frac{\phi(b)}{\phi(a)}}\Bigr\}=0{\rm \: \Rightarrow \:}S_{\phi(a),\phi(b)}$

$S_{\phi(a),\phi(b)}:S_{\phi(a),\phi(b),1}\lor S_{\phi(a),\phi(b),2}\lor S_{\phi(a),\phi(b),3} \lor S_{\phi(a),\phi(b),4}$
$S_{\phi(a),\phi(b),1}:{\Bigl\{\frac{ \prod_{p|a} (1-\frac{1}{p})}{\prod_{p|b} (1-\frac{1}{p})}}\Bigr\}=0 \land {\Bigl\{\frac{b}{a}}\Bigr\}=0$
$S_{\phi(a),\phi(b),2}:{\Bigl\{\frac{a}{\prod_{p|b} (1-\frac{1}{p})}}\Bigr\}=0 \land {\Bigl\{\frac{ \prod_{p|a} (1-\frac{1}{p})}{ b}}\Bigr\}=0$
$S_{\phi(a),\phi(b),3}:{\Bigl\{\frac{a \prod_{p|a} (1-\frac{1}{p})}{\prod_{p|b} (1-\frac{1}{p})}}\Bigr\}=0$

$S_{\phi(a),\phi(b),4}:{\Bigl\{\frac{a \prod_{p|a} (1-\frac{1}{p})}{b}}\Bigr\}=0$

And finally it is clear that:

$S_{\phi(a),\phi(b)}{\rm \: \Rightarrow \:} a | b \lor \lnot(a | b)$

$\phi(a) | \phi(b) {\rm \: \not\Rightarrow \:}a | b$

Show that:
$\Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b\cdot \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr) \Biggr)$
is true $\forall a,b \in \mathbb N$ such that $a |b$
Case 1:
$\lnot \Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b \Biggr)\land \lnot\Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) |  \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr) \Biggr)$
Case 2:
$\Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b \Biggr)\land \lnot\Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) |  \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr) \Biggr)$
Case 3:
$\lnot \Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) | \,b \Biggr)\land \Biggl(\prod_{p|a} \Bigl(1-\frac{1}{p}\Bigr) |  \prod_{p|b} \Bigl(1-\frac{1}{p}\Bigr) \Biggr)$

@nm  sol:= solve(x^2+1=0,x);


should work

@one man oh well it looks like you beat me to the mark here nops has been the most important thing about my code for a while now!

@tomleslie ok thankyou so much for this I have been trying to improve my propositional logic. I think I had in my head that if there are no brackets, because the logic operators are commutative and associative that order would be irrelevant but I do realise that what you are say must be the case, I just find it incredible I haven't run into this problem more frequently considering how often I do this sort of thing.

Well I'll be the first to admit alot of the things typed into my keyboard convey things implicitly that suggest that I am an amatuer enthusiast in number theory at this point, and this is absolutely correct. 

But I have spent alot of time think about the social issues involved with how my life went for me, and how it possibly could have been. I didn't make the right friends, all the way up to my time at university, I appreciated the wrong things about people on a personal level, essentially because I could handle their rejection, keeping in the back of my mind i secretly considered myself a "bit above them" in some sense. But the rejection of what I called "smart kids" I couldn't even fathom the thought of, and I don't always come across as too bright, so this generally lead to a belief I was stupid up until a year or so after i graduated highschool. 

Unfortunately at that stage in my social development, I still was not prepared to handle any direct criticism of my thoughts, and almost all of the nature of my social skills revolved around what is required to interact with people that many would classify as criminals, but it's really just tax evasion and a very questionable lack of product accountablility they were guilty of. 

Anyway, this separation worsed when i did eventually get into university, literally everything that came out of my mouth people looked either angry or disappointed by, for a number of factors now clear to me. But I do feel as if frightening people off via pretending they dont exist did become a tactic of most at some point, or there were subtle jokes made, I think maybe if both sides of the coin had seen each others situation not visible or readable in their interaction, a good chance improvements could have been made then.

But the problem reared it's head again when I decided I was contempt to study math at home, and interact with like minded people online. I would very often be extremely abusive in response to constructive criticism recieved, make vague remarks that implied some kind of social class conspiracy that is the real reason I am being criticised, I don't particularly want to recall exact exchanges of text that I am sure are some where in cyber space, but there was a clear attitude problem which is not going to be deleted with a single facebook account.

But I do feel as if I am making improvements in this regard, and it hasn't entirely  been only  because I was able to see that I had a problem, there has been some sort of "meeting half way" that has occurred, particularly in a few of the facebook groups i joined that were based in Boston? (i think) as well as this forum itself being far less authoritarian than the types of forums I originally chose to interact on.


I honestly do feel silly for alot of advice i recieve now that i can "feel" that i have been given and rejected many times over, so alot of the time when I  meet people younger than me that I can tell are quite deep thinkers, I will tend to try and lighten things be making a fool of myself in some respects, while trying to encourage their enthusiasm at the same time with the sincerely incredible things to learn.


Anyway I feel it to be an important subject, and the more people that are not afraid of looking stupid in deciding to take up even a hobbyist interest in mathematics, the more the  market value of everything that can be deemed an accessory or tool of such an interest, as well as that value of the on going customer support for such products,will increase.


So everyone is welcome to ignore the ramble prior to the last point if they like, more often than not it's internal dialogue because I tend to be absent minded, but i do believe the final point to be useful in terms of the maple forum's best interests.

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