But retractions are a good thing - they demonstrate that the scientific literature is self-correcting as objections are duly considered.
Getting an article past an editor and a couple of referees is only the first part of peer-review. The reviewing procedure proper begins when the article is published and has thousands of eyes on it.
The NASA paper is an example of the success of peer-review, not the failure: a flawed paper was published, these flaws were found and published to a receptive audience and the original paper was retracted. This is ideal.
The only possible improvement to this process is if no flawed papers were ever published. However, not only is this an impossibly high standard but it would be impossible to know if this standard was achieved without the second stage of peer-review.
The whole point of the Sokal affair (which is not an example of failure in scientific peer-review) is that nobody in that particular academic community caught the nonsense paper: in the end it was Sokal himself who came clean.
Nobody is claiming that scientific peer-review is equal to truth-detecting, but it is a great mechanism to find the errors in scientific papers, not least because an individual scientist can make a career out of finding these errors!
Referee's Report on the submission "Re:Peer Reviewed != True" by BMOC.
The article is to be commended for its brevity and clear layout.
However, it seems that the author makes the claim that peer-reviewed scientific papers do not contain evidence, a claim for which no reference is given and which we find to be unsubstantiated. We invite the author to consider that the "methods" and "results" section of a paper detail a set of observations. Short of performing every experiment and collecting observations personally, it is unclear to us what the author considers evidence to be.
Further, the author is clearly unfamiliar with the content of the referenced material. Indeed, with regard to the Sokal affair, the journal in question was neither a) scientific nor b) peer-reviewed. From the author's own reference:
Sokal submitted an article to Social Text, an academic journal of postmodern cultural studies,
and further,
At that time, the journal did not practice academic peer review and did not submit the article for outside expert review by a physicist.
In light of the above issues, which we feel are fundamental to the article and could not be addressed in a rewrite, we recommend against publication.
From the World Health Organisation Oral Health Database:
The metric used is the number of Decayed Missing or Filled Teeth in 12 Year Olds.
England has a mean DMFT of 0.7, and
USA has a mean DMFT of 1.19,
that is the average American 12 Year Old has worse teeth than the average English 12 Year Old.
£17.50 ($28) for an examination
£48 ($75) for simple procedure, such as root canal work, or removing teeth
£209 ($329) for anything else, such as crowns or dentures
I'm not sure what you mean by "not quite", as my objection to your proof is entirely valid.
Your omission doesn't fix things: the negation of commutativity is not "there exists b such that a+b != b+a for all a", which is trivially false, proved by taking a=0.
Commutativity is "a+b =b+a for all a,b" so the negation is "there exist a,b such that a+b != b+a", which you haven't shown.
I'd do something like the following inductive argument, if you define the integers as the cyclic group of infinite order. We first establish that every element commutes with 1: observe that
1= 1, so
1+1=1+1.
Next, assume that the element n commutes with 1, that is
1+n=n+1
and consider the next element, n+1. By associativity
1+(n+1) = (1+n)+1
and by the inductive hypothesis
(1+n)+1 = (n+1)+1.
We conclude that n+1 also commutes with 1 and so, by induction, every element commutes with 1.
We use this to prove commutativity between all elements:
Assume that every element commutes with m, that is
m+n=n+m for all n
Consider the element m+1. By associativity
(m+1)+n = m+(1+n)
but by the inductive hypothesis every element commutes with m, so
m+(1+n)=(1+n)+m.
We proved earlier that every element commutes with 1, so
(1+n)+m = (n+1)+m = n+ (1+m)
by associativity. Again using the fact that every element commutes with 1,
n + (1+m)= n+ (m+1)
and we have shown that every element commutes with m+1. Consequently, by induction, every element commutes with every other element.
In fact, this is not quite finished as we only induct in the positive direction so we've said nothing about negative integers. To fix this we simply repeat the argument replacing the 1's with -1's.
The UK survey was organised by a theological think tank called Theos. The questions found in Appendix B (pdf warning) are piss-poor:
They offer a false dichotomy between evolution being "part of God's plan" or "removes the need for God", which alienates any pantheistic beliefs or any belief in God that doesn't regard him as a creator.
The only decent question is number 6, which is the most relevant in terms of "Evolution Denialism" and I will repeat verbatim:
Darwinian evolution is the idea that life today, including human life, developed
over millions of years from earlier species, by a process of natural selection.
Which one of the following statements comes closest to your opinion of
Darwinian evolution?
1. It is a theory so well established that it’s beyond reasonable doubt
2. It is a theory that is still waiting to be proved or disproved
3. It is a theory with very little evidence to support it
4. It is a theory which has been disproved by the evidence
37% of the respondents answered 1.
36% of the respondents answered 2.
19% of the respondents answered 3.
No figure is given for those answering number 4, which I interpret as the "Evolution Denialism" position. Even if all the respondents answered this question this accounts for at most 8% of the survey sample.
So in summary, over a third of the population think that it is beyond reasonable doubt, and over a third would like to see more evidence (and bear in mind that these are people who may have not been looking for the evidence, nor presented with it.)
You've used the commutativity property without realising it: you need to justify `subtracting' a in this manner:
When we subtract a we are really adding the additive inverse (-a), and can do so on the left or the right of the equation. So either we either proceed by
a+b != b+a
a+b+(-a) != b + a + (-a)
a+b+(-a)!= b
or
a+b != b+a
(-a)+a+b != (-a)+b+a
b != (-a)+b+a.
At either stage we are stuck if we cannot use the commutativity property.
I was quite surprised that Google wouldn't be interested in getting the Ordnance Survey maps so I did a little digging: From the Ordnance Survey Blog:
The reality is that the problem has never been with Ordnance Survey, but with the terms and conditions of Google Maps. It has absolutely nothing to do with derived data or our licensing terms but everything to do with Google claiming the right to use any data you display in Google Maps in any way it sees fit, even if it doesn’t belong to them.
Frustratingly, this is only a problem that exists with Google Maps. No such clause appears in the terms of any other mapping API, including Bing Maps and our own OS OpenSpace.
Also by administering the sugar did you then start to notice his "nuts" behaviour when previously you weren't looking for it?
It's very important to blind (and double blind) studies to remove our biases. A better experiment would have been to prepare some sugary and sugar free candy, have an assistant randomly distribute them between the children (and keep track of which children get which candy). Only once you've rated the behaviour of the children do you find out what candy they were given.
I can't find any studies related specifically to red-light cameras so if you could provide some links that would be useful.
However, in 2005 the British Medical Journal published a systematic review "Effectiveness of speed cameras in preventing road traffic collisions and related casualties: a systematic review". This is a meta-analysis of all articles related to speed-camera sites, collisions, injuries and deaths. They conclude:
Existing research consistently shows that speed cameras are an effective intervention in reducing road traffic collisions and related casualties.
Whilst this doesn't directly address red-light cameras it suggests that introducing traffic cameras causes people to improve their driving habits.
It doesn't run Mathematica or Maple, but for those she waits until she gets home.
I run Maple 13 on my EEE 1000 (SSD). It struggles if you want to rotate 3D plots but apart from that it works fine.
I opted for the `linux' edition of the netbook but it shipped with some bastardised version of Xandros. Being new to linux I threw on the netbook remix of ubuntu 9.10 and couldn't be happier.
The Home Office received 95 extradition requests from the US between 1 January 2004 and 31 July 2009; 47 of these have taken place, with 36 ongoing, five withdrawn by the US and seven refused by UK authorities. The UK has made 42 extradition requests to the US during the same period; 27 of these have taken place, with 12 ongoing, three withdrawn by the UK and none refused.
Shamelessly lifted from the Guardian, one of the UK's better papers, in a column written by Sir Alan West, a minister in the Home Office.
So out of the resolved, non-withdrawn requests, the USA's requests are granted 47/54 times, whilst the UK's are granted 27/27 times.
Can anyone find a US source to verify these numbers?
(1/2)^3 = 1/8
(1/2)^4 = 1/16
A third of the way between 1/16 and 1/8 is 1/16 plus a third of their difference
1/16 + 1/3*[1/8-1/16] = 1/16 + (1/3)*(1/16)=1/12 which is rational
We can only make sense of a^p in the naive "multiply a by itself p times" for rational p. If p = r/q for integers r and q then a^p can be defined as "the q-th root of a^r".
However, this naive approach doesn't allow us to make sense of irrational powers (which clearly we need if we want to define (1/2)^t for t in some interval) Consequently we define a^p:= exp((p*log(a))) where exp is the exponential function (defined as the power series of e^x).
In fact, this is why e^x is called the exponential function as it allows us to define exponents (powers) for all numbers.
This definition makes sense for all p (rational, irrational or complex) and provides the same result as the naive method for rational p.
So, as the parent claims, we need a decent understanding of irrational numbers before we can understand exponential decay.
"Math" and "logic" aren't so much related as two different names for the same thing.
Not so, although it was firmly believed by most philosophers of mathematics in the period 1880-1930.
In a nutshell:
Frege thought that all mathematics could be reduced to logic but Russell's paradox completely undermined the naive set theory in Frege's work. Currently we use the Zermelo-Fraenkel set-theory axioms, in which Russell's paradox doesn't work.
The kicker: Whilst there are no known inconsistencies arising from the ZF axioms we cannot know that the axioms are consistent. This has been proved and is the content of Godel's second incompleteness theorem.
On the other hand, `logic' (say, predicate logic with equality which can be used to formulate propositions like "no slashdotter has had more than one girlfriend") has been proved consistent.
The storage size grows exponentially with its radius.
At a fixed data density (and a fixed number of platters), storage is proportional to the area of the platter, which is proportional to the square of the radius.
Storage size grows quadratically with its radius, not exponentially.
(+5) Interestingly enough, later on in his career Bobby Fischer thought that chess had become "just memorization and prearrangement" as people were playing well by simply memorising opening series of moves. He was a fan of variants of 'Shuffle Chess' where the back row is randomised so that a player couldn't just 'play by the book'
With this in mind, it could be interesting to study the teaching methods used by Polgar; my inner cynic suggests that as he owned thousands of books on chess, was regarded as an expert on chess theory but was only a moderate player his lessons may have consisted of "this is an effective but rarely used opening, memorise it!"
Logarithmic distribution of numbers is entirely independent from logarithmic distribution of leading digits. The former statement is the property that the whole sequence grows logarithmically when viewed over a large interval.
The latter statement is looking at small slices (the interval [10^(n-1),10^n) ) and saying that on each interval the distribution is approximately logarithmic.
The former is a macroscopic property telling us how the sequence grows. The latter is concerned with the fine detail on particular intervals.
Logarithmic distribution of leading digits does not imply logarithmic growth of the sequence.
Using words:
We can imagine a sequence with logarithmic distribution of leading digits with sufficient elements that the entire sequence grows rapidly (i.e. larger than logarithmic growth)
An example:
Let's choose a sequence so that we use all of the numbers with leading digit 1, plus some other elements. We won't specify the rest of the elements (as it would unnecessarily hide what's going on) as the important thing is the distribution of leading digits, and the total quantity of numbers.
Now for n digit numbers (i.e. those in the range [10^{n},10^{n+1}) ), we've already chosen all of the numbers with leading digit 1. It is clearly possible to select the remaining n digit numbers so that the distribution of leading digits is logarithmic (take about half as many numbers starting with 2 etc.)
Now, for n digit numbers, there are 10^(n-1) numbers with initial digit 1. With the logarithmic distribution of leading digits, these make up about 30% (http://en.wikipedia.org/wiki/Benford's_law) of the total selected n digit numbers, meaning that we've chosen an approximate total of 10^(n-1)/30*100 =1/3*10^n, n digit numbers.
This gives us the growth of our sequence as the number of digits increases, but we want the growth of the sequence as the number of numbers increases. But as there are x=10^n-10^(n-1) n digit numbers, the number of digits, n grows like n=log(x)-1 (engineer's log).
Consequently the number of numbers in our sequence grows like 1/3*10^((log x)-1)=(1/30)*x, which is linear growth.
Hence, we have logarithmic distribution of the leading digits, but linear growth of the sequence. So logarithmic distribution of leading digits doesn't imply logarithmic growth of the sequence.
This, of course, wasn't the parent's claim, but is interesting (and relevant nonetheless).
The parent's claim is negation of the following, (which has an easy proof):
Logarithmic growth of the sequence does not imply logarithmic distribution of the leading digits
Using words:
We can have a sequence that grows logarithmically but on the small intervals [10^n,10^n+1), has any distribution we like.
An example:
{9,99,999,9999,99999,...}
We take 1 element for each number of digits. The number of digits grows like log(x), therefore our sequence grows like log(x) - it has logarithmic growth. The distribution of the leading digits is left as an exercise for the reader (hint - it's not logarithmic).
Finally, let's not forget that the primes do not grow logarithmically, they grow sub-linearly O(n*log(n)), so the above doesn't apply directly to TFA.
Slashdot Mirror
Julio Frenk dean of the Harvard School of Public Health, is a co-author on the paper.
He's an academic, not a medic.
Oddly enough there are people in the world who are not motivated solely by profit.
But retractions are a good thing - they demonstrate that the scientific literature is self-correcting as objections are duly considered.
Getting an article past an editor and a couple of referees is only the first part of peer-review. The reviewing procedure proper begins when the article is published and has thousands of eyes on it.
The NASA paper is an example of the success of peer-review, not the failure: a flawed paper was published, these flaws were found and published to a receptive audience and the original paper was retracted. This is ideal.
The only possible improvement to this process is if no flawed papers were ever published. However, not only is this an impossibly high standard but it would be impossible to know if this standard was achieved without the second stage of peer-review.
The whole point of the Sokal affair (which is not an example of failure in scientific peer-review) is that nobody in that particular academic community caught the nonsense paper: in the end it was Sokal himself who came clean.
Nobody is claiming that scientific peer-review is equal to truth-detecting, but it is a great mechanism to find the errors in scientific papers, not least because an individual scientist can make a career out of finding these errors!
Referee's Report on the submission "Re:Peer Reviewed != True" by BMOC.
The article is to be commended for its brevity and clear layout.
However, it seems that the author makes the claim that peer-reviewed scientific papers do not contain evidence, a claim for which no reference is given and which we find to be unsubstantiated. We invite the author to consider that the "methods" and "results" section of a paper detail a set of observations. Short of performing every experiment and collecting observations personally, it is unclear to us what the author considers evidence to be.
Further, the author is clearly unfamiliar with the content of the referenced material. Indeed, with regard to the Sokal affair, the journal in question was neither a) scientific nor b) peer-reviewed. From the author's own reference:
Sokal submitted an article to Social Text, an academic journal of postmodern cultural studies,
and further,
At that time, the journal did not practice academic peer review and did not submit the article for outside expert review by a physicist.
In light of the above issues, which we feel are fundamental to the article and could not be addressed in a rewrite, we recommend against publication.
The system (basically) works.
From the World Health Organisation Oral Health Database:
The metric used is the number of Decayed Missing or Filled Teeth in 12 Year Olds.
England has a mean DMFT of 0.7, and
USA has a mean DMFT of 1.19,
that is the average American 12 Year Old has worse teeth than the average English 12 Year Old.
Further, NHS dentistry fees:
£17.50 ($28) for an examination
£48 ($75) for simple procedure, such as root canal work, or removing teeth
£209 ($329) for anything else, such as crowns or dentures
Consider yourself shown up.
Your (!P) is exactly what I wrote. Nice short-cut for the negative integers, though.
I'm not sure what you mean by "not quite", as my objection to your proof is entirely valid.
Your omission doesn't fix things: the negation of commutativity is not "there exists b such that a+b != b+a for all a", which is trivially false, proved by taking a=0.
Commutativity is "a+b =b+a for all a,b" so the negation is "there exist a,b such that a+b != b+a", which you haven't shown.
I'd do something like the following inductive argument, if you define the integers as the cyclic group of infinite order. We first establish that every element commutes with 1: observe that
1= 1, so
1+1=1+1.
Next, assume that the element n commutes with 1, that is
1+n=n+1
and consider the next element, n+1. By associativity
1+(n+1) = (1+n)+1
and by the inductive hypothesis
(1+n)+1 = (n+1)+1. We conclude that n+1 also commutes with 1 and so, by induction, every element commutes with 1.
We use this to prove commutativity between all elements:
Assume that every element commutes with m, that is
m+n=n+m for all n
Consider the element m+1. By associativity
(m+1)+n = m+(1+n)
but by the inductive hypothesis every element commutes with m, so
m+(1+n)=(1+n)+m.
We proved earlier that every element commutes with 1, so
(1+n)+m = (n+1)+m = n+ (1+m)
by associativity. Again using the fact that every element commutes with 1,
n + (1+m)= n+ (m+1)
and we have shown that every element commutes with m+1. Consequently, by induction, every element commutes with every other element.
In fact, this is not quite finished as we only induct in the positive direction so we've said nothing about negative integers. To fix this we simply repeat the argument replacing the 1's with -1's.
They offer a false dichotomy between evolution being "part of God's plan" or "removes the need for God", which alienates any pantheistic beliefs or any belief in God that doesn't regard him as a creator.
The only decent question is number 6, which is the most relevant in terms of "Evolution Denialism" and I will repeat verbatim:
Darwinian evolution is the idea that life today, including human life, developed over millions of years from earlier species, by a process of natural selection. Which one of the following statements comes closest to your opinion of Darwinian evolution?
1. It is a theory so well established that it’s beyond reasonable doubt
2. It is a theory that is still waiting to be proved or disproved
3. It is a theory with very little evidence to support it
4. It is a theory which has been disproved by the evidence
37% of the respondents answered 1.
36% of the respondents answered 2.
19% of the respondents answered 3. No figure is given for those answering number 4, which I interpret as the "Evolution Denialism" position. Even if all the respondents answered this question this accounts for at most 8% of the survey sample.
So in summary, over a third of the population think that it is beyond reasonable doubt, and over a third would like to see more evidence (and bear in mind that these are people who may have not been looking for the evidence, nor presented with it.)
You've used the commutativity property without realising it: you need to justify `subtracting' a in this manner:
When we subtract a we are really adding the additive inverse (-a), and can do so on the left or the right of the equation. So either we either proceed by a+b != b+a
a+b+(-a) != b + a + (-a)
a+b+(-a)!= b
or
a+b != b+a
(-a)+a+b != (-a)+b+a
b != (-a)+b+a.
At either stage we are stuck if we cannot use the commutativity property.
The reality is that the problem has never been with Ordnance Survey, but with the terms and conditions of Google Maps. It has absolutely nothing to do with derived data or our licensing terms but everything to do with Google claiming the right to use any data you display in Google Maps in any way it sees fit, even if it doesn’t belong to them.
Frustratingly, this is only a problem that exists with Google Maps. No such clause appears in the terms of any other mapping API, including Bing Maps and our own OS OpenSpace.
Also by administering the sugar did you then start to notice his "nuts" behaviour when previously you weren't looking for it?
It's very important to blind (and double blind) studies to remove our biases. A better experiment would have been to prepare some sugary and sugar free candy, have an assistant randomly distribute them between the children (and keep track of which children get which candy). Only once you've rated the behaviour of the children do you find out what candy they were given.
say to the person "It's a lot harder when it's your mom" and laugh out loud because you just said a joke do you?
The error is in laughing at your own jokes.
Mismoderated - posting to undo
However, in 2005 the British Medical Journal published a systematic review "Effectiveness of speed cameras in preventing road traffic collisions and related casualties: a systematic review". This is a meta-analysis of all articles related to speed-camera sites, collisions, injuries and deaths. They conclude:
Existing research consistently shows that speed cameras are an effective intervention in reducing road traffic collisions and related casualties.
Whilst this doesn't directly address red-light cameras it suggests that introducing traffic cameras causes people to improve their driving habits.
I run Maple 13 on my EEE 1000 (SSD). It struggles if you want to rotate 3D plots but apart from that it works fine.
I opted for the `linux' edition of the netbook but it shipped with some bastardised version of Xandros. Being new to linux I threw on the netbook remix of ubuntu 9.10 and couldn't be happier.
The Home Office received 95 extradition requests from the US between 1 January 2004 and 31 July 2009; 47 of these have taken place, with 36 ongoing, five withdrawn by the US and seven refused by UK authorities. The UK has made 42 extradition requests to the US during the same period; 27 of these have taken place, with 12 ongoing, three withdrawn by the UK and none refused.
Shamelessly lifted from the Guardian, one of the UK's better papers, in a column written by Sir Alan West, a minister in the Home Office.
So out of the resolved, non-withdrawn requests, the USA's requests are granted 47/54 times, whilst the UK's are granted 27/27 times.
Can anyone find a US source to verify these numbers?
(1/2)^3 = 1/8
(1/2)^4 = 1/16
A third of the way between 1/16 and 1/8 is 1/16 plus a third of their difference
1/16 + 1/3*[1/8-1/16] = 1/16 + (1/3)*(1/16)=1/12 which is rational
and further, (1/2)^t is defined by e:
:= exp((p*log(a))) where exp is the exponential function (defined as the power series of e^x).
In fact, this is why e^x is called the exponential function as it allows us to define exponents (powers) for all numbers.
We can only make sense of a^p in the naive "multiply a by itself p times" for rational p. If p = r/q for integers r and q then a^p can be defined as "the q-th root of a^r".
However, this naive approach doesn't allow us to make sense of irrational powers (which clearly we need if we want to define (1/2)^t for t in some interval) Consequently we define a^p
This definition makes sense for all p (rational, irrational or complex) and provides the same result as the naive method for rational p.
So, as the parent claims, we need a decent understanding of irrational numbers before we can understand exponential decay.
Potato, potahto.
"Math" and "logic" aren't so much related as two different names for the same thing.
Not so, although it was firmly believed by most philosophers of mathematics in the period 1880-1930.
In a nutshell:
Frege thought that all mathematics could be reduced to logic but Russell's paradox completely undermined the naive set theory in Frege's work. Currently we use the Zermelo-Fraenkel set-theory axioms, in which Russell's paradox doesn't work.
The kicker: Whilst there are no known inconsistencies arising from the ZF axioms we cannot know that the axioms are consistent. This has been proved and is the content of Godel's second incompleteness theorem.
On the other hand, `logic' (say, predicate logic with equality which can be used to formulate propositions like "no slashdotter has had more than one girlfriend") has been proved consistent.
The storage size grows exponentially with its radius.
At a fixed data density (and a fixed number of platters), storage is proportional to the area of the platter, which is proportional to the square of the radius.
Storage size grows quadratically with its radius, not exponentially.
(+5) Interestingly enough, later on in his career Bobby Fischer thought that chess had become "just memorization and prearrangement" as people were playing well by simply memorising opening series of moves. He was a fan of variants of 'Shuffle Chess' where the back row is randomised so that a player couldn't just 'play by the book'
With this in mind, it could be interesting to study the teaching methods used by Polgar; my inner cynic suggests that as he owned thousands of books on chess, was regarded as an expert on chess theory but was only a moderate player his lessons may have consisted of "this is an effective but rarely used opening, memorise it!"
Yes,
we call it "crossing the road"
j= - ((-1)^(0.5))
Math is the one exception, though.
and which other journals are worth reading? ;-)
The former is a macroscopic property telling us how the sequence grows. The latter is concerned with the fine detail on particular intervals.
Using words:
We can imagine a sequence with logarithmic distribution of leading digits with sufficient elements that the entire sequence grows rapidly (i.e. larger than logarithmic growth)
An example:
Let's choose a sequence so that we use all of the numbers with leading digit 1, plus some other elements. We won't specify the rest of the elements (as it would unnecessarily hide what's going on) as the important thing is the distribution of leading digits, and the total quantity of numbers.
Now for n digit numbers (i.e. those in the range [10^{n},10^{n+1}) ), we've already chosen all of the numbers with leading digit 1. It is clearly possible to select the remaining n digit numbers so that the distribution of leading digits is logarithmic (take about half as many numbers starting with 2 etc.) Now, for n digit numbers, there are 10^(n-1) numbers with initial digit 1. With the logarithmic distribution of leading digits, these make up about 30% (http://en.wikipedia.org/wiki/Benford's_law) of the total selected n digit numbers, meaning that we've chosen an approximate total of 10^(n-1)/30*100 =1/3*10^n, n digit numbers.
This gives us the growth of our sequence as the number of digits increases, but we want the growth of the sequence as the number of numbers increases. But as there are x=10^n-10^(n-1) n digit numbers, the number of digits, n grows like n=log(x)-1 (engineer's log).
Consequently the number of numbers in our sequence grows like 1/3*10^((log x)-1)=(1/30)*x, which is linear growth.
Hence, we have logarithmic distribution of the leading digits, but linear growth of the sequence. So logarithmic distribution of leading digits doesn't imply logarithmic growth of the sequence.
This, of course, wasn't the parent's claim, but is interesting (and relevant nonetheless).
The parent's claim is negation of the following, (which has an easy proof):
Using words:
We can have a sequence that grows logarithmically but on the small intervals [10^n,10^n+1), has any distribution we like.
An example:
{9,99,999,9999,99999,...}
We take 1 element for each number of digits. The number of digits grows like log(x), therefore our sequence grows like log(x) - it has logarithmic growth. The distribution of the leading digits is left as an exercise for the reader (hint - it's not logarithmic).
Finally, let's not forget that the primes do not grow logarithmically, they grow sub-linearly O(n*log(n)), so the above doesn't apply directly to TFA.