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# README for the SILVA v123 reference files

December 3, 2015 • PD Schloss • 9 min read

The good people at SILVA have released a new version of the SILVA database. A little bit of tweaking is needed to get their files to be compatible with mothur. This README document describes the process that I used to generate the mothur-compatible reference files.

## Curation of references

### Getting the data in and out of the ARB database

This README file explains how we generated the silva reference files for use with mothur’s classify.seqs and align.seqs commands. I’ll assume that you have a functioning copy of arb installed on your comptuer. For this README we are using version 6.0. First we need to download the database and decompress it. From the command line we do the following:

wget -N https://www.arb-silva.de/fileadmin/arb_web_db/release_123/ARB_files/SSURef_NR99_123_SILVA_12_07_15_opt.arb.tgz

tar xvzf *arb.tgz
arb SSURef_NR99_123_SILVA_12_07_15_opt.arb

This will launch us into the arb environment with the ‘‘Ref NR 99’’ database opened. This database has 597,607 sequences within it that are not more than 99% similar to each other. The release notes for this database as well as the idea behind the non-redundant database are available from the silva website. Within arb do the following:

1. Click the search button
2. Set the first search field to ‘ARB_color’ and set it to 1. Click on the equal sign until it indicates not equal (this removes low quality reads and chimeras)
3. Click ‘Search’. This yielded 526,361 hits
4. Click the “Mark Listed Unmark Rest” button
5. Close the “Search and Query” box
6. Now click on File->export->export to external format
7. In this box the Export option should be set to marked, Filter to none, and Compression should be set to no.
8. In the field for Choose an output file name enter make sure the path has you in the arb_ref_119 folder and enter silva.full_v123.fasta.
9. Select a format: fasta_mothur.eft. This is a custom formatting file that I have created that includes the sequences accession number and it’s taxonomy across the top line. To create one for you will need to create fasta_mothur.eft in the /opt/local/share/arb/lib/export/ folder with the following:

SUFFIX          fasta
BEGIN
>*(acc).*(name)\t*(align_ident_slv)\t*(tax_slv);
*(|export_sequence)

10. Save this as silva.full_v123.fasta
11. You can now quit arb.

### Screening the sequences

Now we need to screen the sequences for those that span the 27f and 1492r primer region, have 5 or fewer ambiguous base calls, and that are unique. We’ll also extract the taxonomic information from the header line. Run the following commands from a bash terminal:

mothur "#screen.seqs(fasta=silva.full_v123.fasta, start=1044, end=43116, maxambig=5, processors=8);
pcr.seqs(start=1044, end=43116, keepdots=T);
degap.seqs();
unique.seqs();"

grep ">" silva.full_v123.good.pcr.ng.unique.fasta | cut -f 1 | cut -c 2- > silva.full_v123.good.pcr.ng.unique.accnos

mothur "#get.seqs(fasta=silva.full_v123.good.pcr.fasta, accnos=silva.full_v123.good.pcr.ng.unique.accnos)"

#generate alignment file
mv silva.full_v123.good.pcr.pick.fasta silva.nr_v123.align

#generate taxonomy file
grep "^>" silva.full_v123.fasta | cut -f 1,3 | cut -c 2- > silva.full_v123.tax.temp

The mothur commands above do several things. First the screen.seqs command removes sequences that are not full length and have more than 5 ambiguous base calls. Note: this will remove a number of Archaea since the ARB RN reference database lets in shorter (>900 bp) archaeal 16S rRNA gene sequences. Second, pcr.seqs convert any base calls that occur before position 1044 and after 43116 to . to make them only span the region between the 27f and 1492r priming sites. Finally, it is possible that weird things happen in the alignments and so we unalign the sequences (degap.seqs) and identify the unique sequences (unique.seqs). We then convert the resulting fasta file into an accnos file so that we can go back into mothur and pull out the unique sequences from the aligned file (get.seqs).

### Formatting the taxonomy files

Now we want to make sure the taxonomy file is properly formatted for use with mothur. We’ll run the following code from within R:

tax <- read.table(file="silva.full_v123.tax.temp", sep="\t")
tax$V2 <- gsub(" ", "_", tax$V2)  #convert any spaces to underscores
tax$V2 <- gsub("uncultured;", "", tax$V2)   #remove any "uncultured" taxa names
#tax$V2 <- paste0("Root;", tax$V2)   #pre-empt all classifications with the Root level.

#we want to see whether everything has 7 (6) taxonomic levesl (Root to genus)
getDepth <- function(taxonString){
initial <- nchar(taxonString)
removed <- nchar(gsub(";", "", taxonString))
return(initial-removed)
}

depth <- getDepth(tax$V2) bacteria <- grepl("Bacteria;", tax$V2)
archaea <- grepl("Archaea;", tax$V2) eukarya <- grepl("Eukaryota;", tax$V2)

tax[depth > 6 & bacteria,] #good to go
tax[depth > 6 & archaea,]  #good to go
tax[depth > 6 & eukarya,]  #eh, there's a lot here - will truncate to the pseudo genus level
tax[depth > 6 & eukarya,2] <- gsub("([^;]*;[^;]*;[^;]*;[^;]*;[^;]*;[^;]*;).*", "\\1", tax[depth > 6 & eukarya,2])
depth <- getDepth(tax\$V2)
tax[depth > 6 & eukarya,]  #good to go

write.table(tax, file="silva.full_v123.tax", quote=F, row.names=F, col.names=F)

### Building the SEED references

The first thing to note is that SILVA does not release their SEED; it is private. By screening through the ARB databases we can attempt to recreate it. Our previous publications show that classify.seqs with the recreated SEED does an excellent job of realigning sequences to look like they would if you used SINA and the true SEED. Now we want to try to figure out which sequences are part of the seed. Earlier, when we exported the sequences from ARB, we included the align_ident_slv field from the database in our output. Let’s generate an accnos file that contains the names of the sequences with 100% to the SEED database and then use mothur to generate SEED fasta and taxonomy files. While we’re at it we’ll also generate the nr_123 taxonomy file as well. The following code will be run from within a bash terminal:

grep ">" silva.nr_v123.align | cut -f 1,2 | grep "\t100" | cut -f 1 | cut -c 2- > silva.seed_v123.accnos
mothur "#get.seqs(fasta=silva.nr_v123.align, taxonomy=silva.full_v123.tax, accnos=silva.seed_v123.accnos)"
mv silva.nr_v123.pick.align silva.seed_v123.align
mv silva.full_v123.pick.tax silva.seed_v123.tax

mothur "#get.seqs(taxonomy=silva.full_v123.tax, accnos=silva.full_v123.good.pcr.ng.unique.accnos)"
mv silva.full_v123.pick.tax silva.nr_v123.tax

### Taxonomic representation

Let’s look to see how many different taxa we have for each taxonomic level within the silva.full_v123.tax, silva.nr_v123.tax, silva.seed_v123.tax. To do this we’ll run the following in R:

getNumTaxaNames <- function(file, kingdom){
taxonomy <- read.table(file=file, row.names=1)
sub.tax <- as.character(taxonomy[grepl(kingdom, taxonomy[,1]),])

phyla <- as.vector(levels(as.factor(gsub("[^;]*;([^;]*;).*", "\\1", sub.tax))))
phyla <- sum(!grepl(kingdom, phyla))

class <- as.vector(levels(as.factor(gsub("[^;]*;[^;]*;([^;]*;).*", "\\1", sub.tax))))
class <- sum(!grepl(kingdom, class))

order <- as.vector(levels(as.factor(gsub("[^;]*;[^;]*;[^;]*;([^;]*;).*", "\\1", sub.tax))))
order <- sum(!grepl(kingdom, order))

family <- as.vector(levels(as.factor(gsub("[^;]*;[^;]*;[^;]*;[^;]*;([^;]*;).*", "\\1", sub.tax))))
family <- sum(!grepl(kingdom, family))

genus <- as.vector(levels(as.factor(gsub("[^;]*;[^;]*;[^;]*;[^;]*;[^;]*;([^;]*;).*", "\\1", sub.tax))))
genus <- sum(!grepl(kingdom, genus))

n.seqs <- length(sub.tax)
return(c(phyla=phyla, class=class, order=order, family=family, genus=genus, n.seqs=n.seqs))
}

kingdoms <- c("Bacteria", "Archaea", "Eukaryota")
tax.levels <- c("phyla", "class", "order", "family", "genus", "n.seqs")

full.file <- "silva.full_v123.tax"
full.matrix <- matrix(rep(0,18), nrow=3)
full.matrix[1,] <- getNumTaxaNames(full.file, kingdoms[1])
full.matrix[2,] <- getNumTaxaNames(full.file, kingdoms[2])
full.matrix[3,] <- getNumTaxaNames(full.file, kingdoms[3])
rownames(full.matrix) <- kingdoms
colnames(full.matrix) <- tax.levels
full.matrix
#           phyla class order family genus n.seqs
#Bacteria     66   169   321    654  2553 465754
#Archaea      19    47    37     68   138  21597
#Eukaryota    13    29    82    174   474  39010

nr.file <- "silva.nr_v123.tax"
nr.matrix <- matrix(rep(0,18), nrow=3)
nr.matrix[1,] <- getNumTaxaNames(nr.file, kingdoms[1])
nr.matrix[2,] <- getNumTaxaNames(nr.file, kingdoms[2])
nr.matrix[3,] <- getNumTaxaNames(nr.file, kingdoms[3])
rownames(nr.matrix) <- kingdoms
colnames(nr.matrix) <- tax.levels
nr.matrix
#          phyla class order family genus n.seqs
#Bacteria     64   159   310    622  2446 152308
#Archaea      16    31    30     58   134   3901
#Eukaryota    11    23    74    130   398  16209

seed.file <- "silva.seed_v123.tax"
seed.matrix <- matrix(rep(0,18), nrow=3)
seed.matrix[1,] <- getNumTaxaNames(seed.file, kingdoms[1])
seed.matrix[2,] <- getNumTaxaNames(seed.file, kingdoms[2])
seed.matrix[3,] <- getNumTaxaNames(seed.file, kingdoms[3])
rownames(seed.matrix) <- kingdoms
colnames(seed.matrix) <- tax.levels
seed.matrix
#          phyla class order family genus n.seqs
#Bacteria     52   101   193    387  1722  12083
#Archaea       7    13    18     30    81    294
#Eukaryota     7    14    25     47    86   2537

nr.matrix / full.matrix
#              phyla     class     order    family     genus    n.seqs
#Bacteria  0.9696970 0.9408284 0.9657321 0.9510703 0.9580885 0.3270138
#Archaea   0.8421053 0.6595745 0.8108108 0.8529412 0.9710145 0.1806269
#Eukaryota 0.8461538 0.7931034 0.9024390 0.7471264 0.8396624 0.4155088

seed.matrix / full.matrix
#              phyla     class     order    family     genus    n.seqs
#Bacteria  0.7878788 0.5976331 0.6012461 0.5917431 0.6745006 0.02594288
#Archaea   0.3684211 0.2765957 0.4864865 0.4411765 0.5869565 0.01361300
#Eukaryota 0.5384615 0.4827586 0.3048780 0.2701149 0.1814346 0.06503461

We see that our full-length database retains a significant majority of the taxa that were in the original NR database. The Archaea take a beating. If you are interested in analyzing the Eukaryota, I would suggest duplicating my efforts here but modify the screen.seqs and pcr.seqs steps to target your region of interest.

Finally, we want to compress the resulting alignment and this README file into the full length and SEED archives using commands in the bash terminal:

tar cvzf silva.nr_v123.tgz silva.nr_v123.tax silva.nr_v123.align README.*
tar cvzf silva.seed_v123.tgz silva.seed_v123.tax silva.seed_v123.align README.*

## Application

So… which to use for what application? If you have the RAM, I’d suggest using silva.nr_v123.align in align.seqs. It took about 10 minutes to read in the database file and a minute or so to align a 1000 full-length sequences. Here is an example workflow for use within mothur that will get you the V4 region of the 16S rRNA gene:

mothur "#pcr.seqs(fasta=silva.nr_v123.align, start=11894, end=25319, keepdots=F, processors=8);
unique.seqs()"

This will get you 104,711 unique sequences to then align against (meh.). Other tricks to consider would be to use get.lineage to pull out the reference sequences that are from the Bacteria, this will probably only reduce the size of the database by ~10%. You could also try using filter.seqs with vertical=T; however, that might be problematic if there are insertions in your sequences (can’t know a priori). It’s likely that you can just use the silva.seed_v123.align reference for aligning. For classifying sequences, I would strongly recommend using the silva.nr_v123.align and silva.nr_v123.tax references after running pcr.seqs on silva.nr_v123.align. I probably wouldn’t advise using unique.seqs on the output.

## Legalese

If you are going to use the files generated in this README, you should be aware of SILVA’s dual use license. We’ll leave it to you to work out the details.


text:
### Some(?) of the Parts

A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle

### Coke Machine

The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design...

### Pericut

Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?

# Garden Fence

##### Stage: 4 Short Challenge Level:
See all short problems arranged by curriculum topic in the short problems collection

A garden has the shape of a right-angled triangle with sides of length $30$, $40$ and $50$. A straight fence goes from the corner with the right-angle to a point on the opposite side, dividing the garden into two sections which have the same perimeter. How long is the fence?

If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.

This problem is taken from the UKMT Mathematical Challenges.

format: plain


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Препринт
Lazarev A. A., Lemtuzhnikova D., Werner F. Otto-von-Guericke Universitaet, 2019
We consider NP-hard multi{machine scheduling problems with the criterion of minimizing the maximum penalty, e.g. maximum lateness. For such problems, we introduce a metric which delivers an upper bound on the absolute error of the objective function value. Taking the given in- stance of some problem and using the introduced metric, we determine the nearest instance for which a polynomial or pseudo-polynomial algorithm is known. For this determined instance, we construct a schedule which is then applied to the original instance. It is shown how this approach is applied to di_erent scheduling problems.
Добавлено: 26 апреля 2020
Препринт
Lazarev A. A., Werner F. Otto-von-Guericke Universitaet, 2008. No. 15.
In this paper we consider a graphical realization of dynamic programming. The concept is discussed on the partition and knapsack problems. In contrast to dynamic programming, the new algorithm can also treat problems with non-integer data without necessary transformations of the corresponding problem. We compare the proposed method with existing algorithms for these problems on small-size instances of the partition problem with $n \le 10$ numbers. For almost all instances, the new algorithm considers on average substantially less "stages" than the dynamic programming algorithm.
Добавлено: 4 марта 2013
Препринт
Gafarov E., Lazarev A. A., Werner F. Otto-von-Guericke Universitaet, 2009. No. 38.
We consider single machine problems with opposite criteria, namely we consider the maximization of total tardiness, the maximization of the number of tardy jobs and the maximization of total completion time (in contrast to usual minimization problems)and a minimization version of the Knapsack problem.
Добавлено: 4 марта 2013
Препринт
Lazarev A. A., Werner F. Otto-von-Guericke Universitaet, 2008. No. 12.
The scheduling problem of minimizing total tardiness on a single machine is knownto be NP-hard in the ordinary sense. In this paper, we consider the special case of the problem when the processing times $p_j$ and the due dates $d_j$ of the jobs $j, \, j \in N = \{ 1, 2, \ldots, n \}$, are oppositely ordered: $p_1\ge p_2\ge\dots\ge p_n$ and $d_1\le d_2\le\dots\le d_n$. It is shown that already this special case is $NP$-hard in the ordinary sense, too. The set of jobs $N$ is partitioned into $\Bbbk, 1 \le \Bbbk \le n$, subsets$\mathcal{M}_1,\mathcal{M}_2,\dots,\mathcal{M}_\Bbbk$,$\mathcal{M}_\nu \bigcap \mathcal{M}_\mu=\emptyset$ for $\nu\ne \mu,$$N=\mathcal{M}_1\bigcup\mathcal{M}_2\bigcup\dots\bigcup\mathcal{M}_\Bbbk$,such that$\max_{i,j\in\mathcal{M}_\nu}|d_i-d_j|\le\min_{j\in\mathcal{M}_\nu}p_j$for each $\nu=1,2,\dots,\Bbbk$. We propose algorithms which solve the problem: in $O(\Bbbk n\sum p_j)$ time if $1\le \Bbbk< n$ in $O(n^2)$ time if $\Bbbk= n$ and in $O(n^2)$ time if $\max_{i,j\in N}|d_i-d_j|\le 1$. The polynomial algorithms do neitherrequire the conditions $p_1\ge p_2\ge\dots\ge p_n$ mentioned above nor integer processing times to construct an optimal schedule. Finally, we apply the idea of the presented algorithm for the case $\Bbbk = 1$ to the even-odd partition problem
Добавлено: 4 марта 2013
Препринт
Gafarov E., Lazarev A. A., Werner F. Otto-von-Guericke Universitaet, 2010. No. 20.
In this paper, we present a modification of dynamic programming algorithms (DPA), which we denote as graphical algorithms (GrA). For the knapsack problem and some single machine scheduling problems, it is shown that the time complexity of the GrA is less that the time complexity of the standard DPA. Moreover, the average running time of the GrA is often essentially smaller. A GrA can also solve largescale instances and instances, where the parameters are not integer. In addition, for some problems, GrA has a polynomial time complexity in contrast to a pseudo-polynomial complexity of DPA.
Добавлено: 4 марта 2013
Препринт
Gafarov E., Lazarev A. A., Werner F. Otto-von-Guericke Universitaet, 2010. No. 12.
In this paper, we consider the problem of maximizing total tardiness on a single machine, where the first job starts at time zero and idle times between the processing of jobs are not allowed. We present a modification of an exact pseudo-polynomial algorithm based on a graphical approach, which has a polynomial running time.
Добавлено: 4 марта 2013
Препринт
Gafarov E., Lazarev A. A., Werner F. Otto-von-Guericke Universitaet, 2010. No. 11.
In this paper, we consider some scheduling problems on a single machine, where a specific objective function has to be maximized in contrast to usual minimization problems. These problems are theoretically important and have also practical interpretations. We present some complexity results for such maximization problems with classical objective functions (e.g. total tardiness, number of tardy jobs and total completion time) and various additional constraints (e.g. deadlines, weights and/or release dates of jobs may be given). As a generalization, we consider a classical combinatorial problem with an opposite optimization criterion, namely a minimization version of the knapsack problem, for which we give an NP-hardness proof and an exact pseudo-polynomial algorithm.
Добавлено: 4 марта 2013
Препринт
Gafarov E., Lazarev A. A., Werner F. Otto-von-Guericke Universitaet, 2010. No. 8.
Problem RCPSP may be formulated as follows. Given a set $N=\{1,\dots,n\}$ of jobs. A constant amount (quantity) of $Q_k>0$ units of resource $k, k=1,\dots,K,$ is available at any time. Job $j\in N$has to be processed for $p_j\geq 0$ time units without preemption. During this period, a constant amount (quantity) of $q_{jk} \geq 0$ units of resource $k$ is occupied. Fur the rmore, finish-start precedence relations $i \rightarrow j$ are defined between the jobs according to an acyclic directed graph $G.$ The objective is to determine the starting times $S_j$ for each job $j=1,\dots,n,$ in such a way that: at each time $t$, the total resource demand is less than or equal to the resource availability for each resource type the given precedence constraints arefulfilled the makespan $C_{max} = \max_{j=1}^n C_j$, where $C_j=S_j+p_j,$ is minimized.
Добавлено: 4 марта 2013
Препринт
Gafarov E., Lazarev A. A., Werner F. Otto-von-Guericke Universitaet, 2010. No. 10.
In this note, we consider a single machine scheduling problem with generalized total tardiness objective function. An NP-hardness proof and a pseudo-polynomial time solution algorithm are proposed for a special case of this problem. Moreover, we present a new graphical algorithm for another special case, which corresponds to the classical problem of minimizing the weighted number of tardy jobs on a single machine. The latter algorithm improves the complexity of an existing pseudo-polynomial algorithm by Lawler.
Добавлено: 4 марта 2013
Препринт
Гафаров Е. Р., Лазарев А. А., Вернер Ф. Otto-von-Guericke Universitaet, 2010. № 7.
In this paper, we consider single machine scheduling problems with a non-renewable resource. This type of problems has not been intensively investigated in the literature so far. For several problems of this type with standard objective functions (namely the minimization of makespan, total tardiness, number of tardy jobs, total completion time and maximum lateness), wepresent some complexity results. Particular attention is given to the problem of minimizing total tardiness. In addition, for the so-called budget scheduling problem with minimizing the makespan, we present some properties of feasible schedules.
Добавлено: 4 марта 2013


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# Questions tagged [evolutionary-algorithms]

For questions about algorithms that exhibit characteristics of evolution in that the structure, form, or processes of that which is evolving incrementally improves, either by DNA mutation and selection, symbiosis, or other operations that can produce improvement.

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# In a study, researchers wanted to measure the effect of alcohol on the hippocampal region, the...

5 answers
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##### Find the equation of the hyperboloid shown in the figure below:Reciit[-/1 Points]DETAILSROGACALCET3 12.6.042.Find the equation of the quadric surface shown in the figure below where a = 5, b = 11, and â‚¬ = 7,.
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##### Write the Intermediate Value Theorem_ Now use IVT to show the following equations have at least one solution on [0, w]:3.coS t _ t3 0 =
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location:  Publications → journals
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Search: All articles in the CJM digital archive with keyword circle method

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1. CJM Online first

Ha, Junsoo
 Smooth Polynomial Solutions to a Ternary Additive Equation Let $\mathbf{F}_{q}[T]$ be the ring of polynomials over the finite field of $q$ elements, and $Y$ be a large integer. We say a polynomial in $\mathbf{F}_{q}[T]$ is $Y$-smooth if all of its irreducible factors are of degree at most $Y$. We show that a ternary additive equation $a+b=c$ over $Y$-smooth polynomials has many solutions. As an application, if $S$ is the set of first $s$ primes in $\mathbf{F}_{q}[T]$ and $s$ is large, we prove that the $S$-unit equation $u+v=1$ has at least $\exp(s^{1/6-\epsilon}\log q)$ solutions. Keywords:smooth number, polynomial over a finite field, circle methodCategories:11T55, 11D04, 11L07, 11T23

2. CJM 2013 (vol 66 pp. 844)

Kuo, Wentang; Liu, Yu-Ru; Zhao, Xiaomei
 Multidimensional Vinogradov-type Estimates in Function Fields Let $\mathbb{F}_q[t]$ denote the polynomial ring over the finite field $\mathbb{F}_q$. We employ Wooley's new efficient congruencing method to prove certain multidimensional Vinogradov-type estimates in $\mathbb{F}_q[t]$. These results allow us to apply a variant of the circle method to obtain asymptotic formulas for a system connected to the problem about linear spaces lying on hypersurfaces defined over $\mathbb{F}_q[t]$. Keywords:Vinogradov's mean value theorem, function fields, circle methodCategories:11D45, 11P55, 11T55
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# Asymptotic to a sequence of algebraic numbers.

Let $f(n)$ be the largest real solution of

$$x^n - x^{n-1} = 1$$

As $n$ grows to positive infinity we get the asymptotic :

$$f(n) = 1 + \frac{\ln(n)}{n} + \frac{\exp(2)}{n^2} + ...$$

Where the value $\exp(2)$ is optimal !

( and $...$ means smaller term(s) )

Notice $f(2)$ is the golden mean.

How to show this asymptotic ?

Edit

Corrected the formula.

• Wow, I suppose I take that back. The fact that $\exp(2)$ is optimal seems interesting... – Simply Beautiful Art Feb 17 '17 at 23:25
• I checked up to $n=10^{10}$, and for $n>11900$, $f(n)>1+\frac{\exp(2)}n$, and it seems to me that there is no constant $c$ such that $f(n)=1+\frac cn+\dots$, as after $n=2\times10^9$, or so, $f(n)=1+\frac{2.5\exp(2)}n+\dots$ looks more optimal. – Simply Beautiful Art Feb 17 '17 at 23:34
• This is incorrect: if $x=1+b/n$, then the LHS behaves like $e^bb/n$ for large $n$. I believe the correct asymptotics is $f(n)=1+\log n/n + o(\log n/n)$. – user138530 Feb 18 '17 at 0:19
• @ChristianRemling Pretty close to what I was thinking. Again checking, that holds much better, as far as I can tell. – Simply Beautiful Art Feb 18 '17 at 0:26
• making edits that totally change a possible answer to a questions are really crappy... – tired Feb 18 '17 at 11:07

A rough outline.

The first few terms in the asymptotic are

$$f(n) = 1 + \frac{W(n)}{n} + \frac{W(n)^2}{2n^2} + \cdots,$$

where $W$ is the Lambert-W function, so the stated asymptotic is incorrect.

First show that, with $x = 1 + \frac{W(n)}{n} + \frac{z}{n}$, where $z = O(1)$, we have

$$x^n - x^{n-1} - 1 \to e^z - 1$$

uniformly as $n \to \infty$. Conclude by Hurwitz's theorem that

$$f(n) = 1 + \frac{W(n)}{n} + \frac{\epsilon_n}{n}$$

with $\epsilon_n \to 0$ as $n \to \infty$. Substitute this into the equation

$$f(n)^n - f(n)^{n-1}-1 = 0$$

and apply asymptotic simplifications to conclude that $\epsilon_n \sim W(n)^2/(2n)$ as $n \to \infty$.

Let $f(n)=1+\epsilon(n)$. Then, $f^n(n)-f^{n-1}(n)=1$ becomes

$$(n-1)\log(1+\epsilon(n))+\log(\epsilon(n))=0$$

As $n\to \infty$, $\epsilon(n)\to 0$. Hence, we have

$$(n-1)\epsilon(n)+O(n\epsilon^2(n))+\log(\epsilon(n))=0 \tag 1$$

We can write $(1)$ equivalently as

$$(n-1)\epsilon(n)e^{(n-1)\epsilon(n)}=(n-1)e^{O(n\epsilon^2(n))}\tag 2$$

which using Lambert's W function is given by

$$\epsilon(n)=\frac{1}{n-1}W\left((n-1)e^{O(n\epsilon^2(n))}\right)\tag 3$$

Using the first term in the large argument asymptotic expansion of $W$ yields

\begin{align} \epsilon(n)&\sim \frac{1}{n-1}\log((n-1)e^{O(n\epsilon^2(n))})\\\\ &\sim\frac{\log(n-1)}{n-1}\\\\ &\sim\frac{\log(n)}{n} \end{align}

Hence, we find that the first two terms in the expansion of $f(n)$ for large $n$ is given by

$$\bbox[5px,border:2px solid #C0A000]{f(n)\sim 1+\frac{\log(n)}{n}}$$

And we are done!

• I am sorry to refer to this old post, but: How do you get from (1) to (2)? I do not see this immadiately. – mathfemi Jan 22 '18 at 11:05
• Exponentiate both sides of $(1)$ and multiply by $n-1$. – Mark Viola Jan 22 '18 at 16:38
• @MarkViola Another question: You are using the large Argument Expansion of W. Could you please explain why the Argument of W, which here is $(n-1)e^{O(n\varepsilon^2(n))}$, is "large" here? – mathfemi Jan 22 '18 at 19:47
• Note that we are analyzing the asymptotic behavior as $n\to \infty$. – Mark Viola Jan 22 '18 at 22:02
• @Rhjg I'll have a look. It's been quite a while since I posted this answer. – Mark Viola Sep 11 '18 at 14:40


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# Force and newtons law of motion

## Homework Statement

your car rolls on a level street at 50km/h. the coefficient of kinetic friction is 0.023. how far will the car go before stopping. (no breaking.)

## The Attempt at a Solution

F_fric=0.226m
F_a=(-1250/s)*m
i need to make F_a-F_fric equal to something to solve for s, the m's will cancel. I have no idea what it will equal since its not in equilibrium......

## Answers and Replies

tiny-tim
Science Advisor
Homework Helper
hi pat666!

use work done and the work-energy theorem

kk ill give that a whirl and tell you if i still cant do it... thanks

Hey, i still cant get this i tried finding the net work which i got -1250m J and then W=F/x with F=(-1250m)/x-0.226m. i got the acceleration (for F=ma) using kinematic equations.???????????????

tiny-tim
Science Advisor
Homework Helper
Hey, i still cant get this i tried finding the net work which i got -1250m J and then W=F/x with F=(-1250m)/x-0.226m. i got the acceleration (for F=ma) using kinematic equations.???????????????

Sorry, I don't understand any of that.

What is the energy? what is the force ?

If the distance is d, what is the work done ?

I just found the net work using 1/2mv^2-1/2mu^2. then i used W=F/x, to find the Force i just found the Force from friction and the "applied force". I might be doing something completely wrong, could you give me some more detailed instructions please.

ideasrule
Homework Helper
It's W=Fx, not W=F/x. Also, there's no applied force; the force on the car is just the force of friction.

ok thats where i screwed up. thanks.......


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### Homes

Six new homes are being built! They can be detached, semi-detached or terraced houses. How many different combinations of these can you find?

### Stairs

This challenge is to design different step arrangements, which must go along a distance of 6 on the steps and must end up at 6 high.

### Train Carriages

Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?

# Late Again

##### Age 5 to 7Challenge Level

David from Thomas Reade School sent in the following solution:

The shortest route is forward $1$, right $1$, forward $6$ and right $3$ which is $11$ spaces.
Another way is forward $1$, right $4$ and forward $6$ which is $11$ spaces.
Another way is forward $3$, right $4$ and forward $4$ which is $11$ spaces.

(This assumes that Moira is standing just outside the school gate and finished standing outside the school doors.) Thank you David.