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# A multi-sample approach increases the accuracy of transcript assembly

## Abstract

Transcript assembly from RNA-seq reads is a critical step in gene expression and subsequent functional analyses. Here we present PsiCLASS, an accurate and efficient transcript assembler based on an approach that simultaneously analyzes multiple RNA-seq samples. PsiCLASS combines mixture statistical models for exonic feature selection across multiple samples with splice graph based dynamic programming algorithms and a weighted voting scheme for transcript selection. PsiCLASS achieves significantly better sensitivity-precision tradeoff, and renders precision up to 2-3 fold higher than the StringTie system and Scallop plus TACO, the two best current approaches. PsiCLASS is efficient and scalable, assembling 667 GEUVADIS samples in 9 h, and has robust accuracy with large numbers of samples.

## Introduction

RNA sequencing (RNA-seq) has become the de facto standard in surveying the transcriptome of a cell, organism, or species, to determine the expressed genes and transcripts and their expression levels, and to enable differential and functional analyses1,2. A crucial step in virtually all RNA-seq data analyses is assembling the reads into full-length transcripts. The accuracy of transcript reconstruction is critical for quantification, detection, and characterization of alternative splice variants, and the identification of differences in gene expression and splicing patterns between tissues, developmental stages, and physiological or disease states.

Virtually all transcriptomic studies involve multiple samples. The current paradigm is to assemble the reads in each sample, then merge the partial transcripts (transfrags) across all samples to create a unified set of meta-annotations3, which is used as reference for downstream quantification and differential analyses. Most single-sample assemblers including Cufflinks, isoCEM, Scripture, Traph, CLASS, iReckon, CIDANE, FlipFlop, CLASS2, StringTie, Scallop, and TransComb4,5,6,7,8,9,10,11,12,13,14,15 build a graph structure from read alignments on the genome, then traverse the graph to select an optimized set of transcripts, represented as paths. Recent second-generation transcript assembly methods including StringTie, CLASS2, and Scallop have taken great strides toward increasing the accuracy and efficiency of assembly at single-sample level, and meta-assemblers such as StringTie(ST)-merge and TACO3 have led to more robust collections of meta-annotations. Despite these efforts precision remains low, with <40% of the predicted transcripts in a single-sample and <30% of transcripts in meta-annotations representing complete and accurate reconstructions3,16. Concomitantly, a handful of efforts have focused on designing methods that simultaneously assemble transcripts across multiple RNA-seq samples, notably CLIIQ, ISP, MiTie, and FlipFlop11,17,18,19. However, none of these programs has comparable performance or is equipped to handle more than a limited number of samples. (See Supplementary Methods for a more in-depth review of related work.)

We present PsiCLASS, based on an approach that simultaneously analyzes multiple RNA-seq samples, which achieves significantly higher precision at sensitivity comparable to the best current approaches, and significantly higher overall accuracy in its default setting. PsiCLASS differs from traditional assemblers in two ways. First, it is a combined assembler, reporting a set of transcripts for each sample, and meta-assembler3, producing a set of meta-annotations obtained by combining the individual samples’ outputs. Second, unlike traditional single-sample assemblers, it uses information from all samples to produce transcripts for individual samples and for the unified annotation set. PsiCLASS starts by selecting a set of high-confidence introns and subexons at each locus, using statistical models of introns and intronic read levels. In doing so, it first generates a set of features (subexons, introns) for each sample, and then corroborates information across all samples to select a high-confidence subset. Next, PsiCLASS builds a unified subexon splice graph that is used within a dynamic programming optimization procedure to select a representative set of transcripts in each sample (see “Methods” and Supplementary Fig. 1 for details). Lastly, PsiCLASS extracts a subset of meta-annotations from the aggregated transcript sets by voting. Salient features of PsiCLASS include: (i) an algorithm that assembles reads from multiple samples simultaneously, taking advantage of their redundancy and complementarity; (ii) significantly higher precision at meta-assembly level, up to 2–3 fold, over StringTie with ST-merge and Scallop with TACO, and high precision (>50%) overall; (iii) 22–140% increase in precision at meta-assembly level over StringTie with ST-merge, and 35–89% over Scallop with TACO, when matched to their sensitivity levels; (iv) higher overall performance in combined sensitivity and precision at the individual sample level; (v) improved consistency among the individual samples’ annotations and meta-annotations, and robustness with different meta-assemblers; and (vi) high efficiency and scalability, taking only 9 h to assemble 667 GEUVADIS samples, and robust accuracy as the number of samples increases. Overall, PsiCLASS is highly efficient and overcomes limitations in the existing methods, showing significantly higher accuracy, in particular precision, on data sets with a handful to hundreds of samples, thus providing an efficient method for RNA-seq data analysis.

## Results

### Performance evaluation on simulated data

We compared PsiCLASS with the best current approaches, namely StringTie and Scallop at the individual sample level, and the combinations of StringTie with ST-merge and Scallop with TACO, at the meta-assembly level. We also included FlipFlop, which was the only other competitive multi-assembler; because of excessive run times, however, FlipFlop was not feasible for our tests on real data sets. (Other combinations are shown in Supplementary Fig. 2). We first applied the methods to 25 RNA-seq samples simulated with Polyester20, where reads were aligned with two methods, Hisat221 and STAR22. Performance was slightly better for all programs when reads were aligned with Hisat2 (Supplementary Fig. 3), therefore we chose this alignment method for the rest of the analyses.

On the simulated data, PsiCLASS with default voting achieved 71.4% precision, which is 15.0% higher than the StringTie system, and 28.3% and 21.6% higher than Scallop and FlipFlop, respectively, combined with TACO, whereas sensitivity for all programs was roughly 50% (Fig. 1a). Even at the individual sample level, PsiCLASS had both the highest precision and the highest sensitivity: 73.8% precision on average, compared with 70.8% for StringTie, 62.9% for Scallop, and 50.7% for FlipFlop, and 53.4% sensitivity compared with 41.7% for StringTie, 46.2% for Scallop, and 36.1% for FlipFlop. Precision values for StringTie, Scallop, and FlipFlop, but to a lesser extent for PsiCLASS, dropped significantly after aggregation, hence PsiCLASS produces more consistent sets of transcripts between individual samples and the set of meta-annotations.

We further investigated the performance of methods based on the transcripts’ expression levels (Fig. 1b). Simulated transcripts were divided into low (463 transcripts; Fragments Per Kilobase (FPK) < 30), medium (658 transcripts; 30 ≤ FPK < 500) and high (322 transcripts; FPK ≥ 500) according to the predefined expression levels. PsiCLASS with voting reconstructs the largest fraction of high-expressed transcripts, 82.4%, and all three programs recover ~60% of the medium-expressed ones. FlipFlop has by far the highest sensitivity in detecting low expression features, 41.0%, compared with 10–20% for the other programs. Note that, because a reconstructed transcript’s expression level may fall in another class than the predefined one, precision cannot be rigorously evaluated.

Similarly, to assess the programs’ performance based on the gene’s splicing complexity, we divided genes into three categories by the number of their alternatively spliced transcripts: low (1 transcript per gene; 680 genes, 680 transcripts), medium (2 transcripts per gene; 166 genes, 332 transcripts), and high (3 or more transcripts per gene; 106 genes, 431 transcripts). While, StringTie showed the best overall performance on the single-transcript genes, PsiCLASS had the highest sensitivity and precision on the medium and high complexity classes, in particular a 16.8–24.5% gain in sensitivity and 21.0–31.8% in precision on highly expressed transcripts (Supplementary Fig. 4).

### Performance evaluation on real RNA-seq data collections

We next assessed the performance on two representative RNA-seq data sets, generated with two different library preparation protocols: 25 randomly selected sets from polyA-selected lymphoblastoid samples from the GEUVADIS population variation project, and 73 rRNA-depleted total RNA libraries from postmortem human liver samples with funding from the Stanley Medical Research Institute. At the meta-assembly level, PsiCLASS’s precision for the GEUVADIS set is 69.3% and 89.2% higher than StringTie’s and Scallop plus TACO’s, respectively, whereas sensitivity is higher or comparable, by 4.5% and −0.9%, respectively (Fig. 1c). Similarly, advantages of the multi-sample approach are seen for the liver total RNA data set, with more than twofold and threefold increases in precision, albeit at 18.6% and 35.3% lower sensitivity. Even when matched to the other programs’ sensitivity settings, PsiCLASS maintains 140% and 89% gains in precision over the StringTie system and Scallop plus TACO, respectively, albeit for this data set Scallop with TACO retains superior sensitivity over the entire parameter range (Fig. 1d). In such cases, the default voting setting may not present the best tradeoff, and as PsiCLASS’s precision remains significantly higher than that of its counterparts the user may choose a different cutoff. Even at the individual sample level, PsiCLASS’s performance exceeds that of its competitors’, with higher or comparable per sample average measurements in both components (Fig. 1c, d, boxplots).

### Performance evaluation on heterogeneous data collections

Most RNA-seq analyses are aimed at determining gene expression or splicing differences between two biological conditions. To explore the robustness of PsiCLASS when combining two-condition samples, we applied it and the other methods to RNA-seq samples from hippocampi of normal mice (24 samples) and mice with induced epileptic seizures (20 samples) (ref.23 and Source Data). The diagrams in Fig. 1e indicate that at the level of individual samples PsiCLASS has higher sensitivity than both StringTie and Scallop, by 26.2% and 8.4% on average, whereas precision exceeds Scallop’s by 42.3% and is comparable to StringTie’s at 66%. Moreover, after voting PsiCLASS’s precision at the level of meta-annotations is 45.7% and 55.0% higher than the other programs’, respectively, along with an increase in sensitivity (15.1% and 7.8%), therefore recommending it as the overall best performer. Thus, PsiCLASS can be effectively and more reliably used on the aggregate set of samples in a two-condition comparison.

We further explored the applicability of PsiCLASS to more divergent and heterogeneous collections of data. We applied PsiCLASS to single- and multi-tissue collections extracted from the GTEx24 repository (ten RNA-seq samples for each of six tissues: cortex, frontal cortex, cerebellum, heart, liver, and lung). We then aggregated the results in each collection in three ways, using PsiCLASS voting, TACO, and TACO applied to the samples assembled separately by tissue. In all cases, PsiCLASS with its default voting algorithm had better performance, with both higher sensitivity, and higher precision (Supplementary Fig. 5). Notably, our approach resulted in significantly more accurate transcript sets for all collections, with 11.2–16.8% increase in precision over TACO, and a significant 21.4–47.1% increase over TACO applied to the per tissue assembled collections. As the heterogeneity in the data increases, gains in sensitivity reach diminishing returns, while the increase in precision becomes more pronounced. These simple experiments show promise for extending PsiCLASS’s capabilities to more heterogeneous data sets.

### Deconvoluting the effects of algorithmic components

While the different components of the PsiCLASS algorithm are intercalibrated to work in harmony with each other, to guide future program development we attempted to deconvolute the effects on performance of the individual algorithmic components. In a first test, we comprehensively evaluated all combinations of individual sample (PsiCLASS, StringTie, and Scallop) and meta-assemblers (voting, ST-merge, and TACO) at multiple levels. Notably, voting significantly increased transcript level precision for all programs, albeit PsiCLASS maintained a clear advantage in all but one case (GEUVADIS-25), where Scallop’s performance was comparable (Supplementary Figs. 6 and 7). Conversely, unlike the other two programs, PsiCLASS rendered comparable results when its sample-level transcript sets were merged with the voting, ST-merge and TACO aggregation methods, demonstrating the robustness of its reconstructions (Supplementary Fig. 6). As conventional measurements based on the number of fully reconstructed transcripts only partially describe the accuracy of a data set, we also assessed the accuracy at the feature (exon, intron) and gene level. Indeed, even when the best meta-assembly method (voting) is used for all programs, PsiCLASS captures significantly more gene and transcript content, with 10–40% more introns and 10–50% more internal exons than StringTie and Scallop, at comparable 95–98% precision (Supplementary Fig. 8). Further, PsiCLASS reported significantly more genes that had at least one fully reconstructed transcript, 7.0–24.7% more than StringTie and 2.9–31.3% more than Scallop. Lastly but notably, PsiCLASS also detected many more genes: 3.9–9.6% more than StringTie and up to a very significant 46.3% more than Scallop, with the largest of these gains observed for the liver data (Supplementary Figs. 9 and 10). These improvements are the direct consequence of the completeness of PsiCLASS gene structure model as captured in the shared subexon graph, an algorithmic feature that particularly favors low expression genes.

### Performance evaluation with small numbers of samples

We next investigated the utility of the multi-sample approach for small data collections, including single samples. Table 1 shows the results for all methods on sets of 2, 3, 5, and 10 samples from our simulated set, averaged over five independent trials, and for single-sample sets, averaged over all possible 25 trials. The multi-sample approach shows benefits in all cases, most notably increased sensitivity (range 47–53% for PsiCLASS, compared with 42–48% for StringTie, and 43–48% for Scallop with TACO) at comparable or higher precision (range 67–70% for PsiCLASS, compared with 64–71% for StringTie and 57–62% for Scallop).

PsiCLASS is designed to take advantage of cross-sample gene information in a multi-sample experiment. Nevertheless, we also evaluated it and the other programs on isolated single-samples from three previously reported real data sets (Table 2). PolyA-selected RNA-seq paired-end reads from CD20+ (SRR534319, 25 million reads), CD14+ (SRR545695, 39 million reads), and IMR90 (SRR534291, 114 million reads) cell lines were previously used to demonstrate performance improvements in both the StringTie and Scallop publications. When PsiCLASS used solely information from that sample, performance across the three samples was similar between StringTie and PsiCLASS, whereas Scallop had significantly higher sensitivity albeit at slightly lower precision. However, since each data set was part of a multi-sample experiment (SRR534291-SRR534292, SRR545695-SRR545700, and SRR534319-SRR534324), we sought to assess the advantage of the cross-sample information. Even with the small (2–6) number of replicates, when all samples in an experiment were considered PsiCLASS’s performance for the three samples improved (up to 28.3% sensitivity increase over PsiCLASS single-sample, and up to 35.0% over StringTie), especially for the low coverage sample (SRR534319).

Overall, PsiCLASS showed competitive or higher performance on experiments with a small (2–10) number of samples, with benefits most evident for data sets of three or more samples. It also ranked comparable with StringTie and Scallop on single-sample data sets in our control experiments, and similarly to StringTie on the three samples for which the two reference programs were optimized. Therefore, our program can be effectively used for small RNA-seq collections and within the conventional assemble-and-merge approach.

### Performance evaluation on very large data collections

As the emerging landscape of RNA-seq foresees increasingly larger data sets from large patient cohorts and population variation studies, we aimed to assess the suitability of the multi-sample approach as the data set grows. We evaluated program performance on increasingly larger subsets of RNA-seq samples from the GEUVADIS population variation project, up to the full set of 667 samples (Fig. 1f). All methods show improvements in sensitivity as the number of samples increases, but while PsiCLASS and Scallop show further slight gains after 20–50 samples, the sensitivity of StringTie drops. Precision drops markedly for both Scallop and StringTie, to less than 35% for 50 samples and below 20% for the full set of samples. In sharp contrast, PsiCLASS’s sensitivity and precision remain almost constant with more than ten samples, demonstrating the robustness of this approach. Also, with sensitivity 9–40% higher than StringTie’s and precision (>50%) twice as high as that of the other two systems when the data set exceeds 20–50 samples, PsiCLASS is unequivocally the best suited for handling large RNA-seq collections. Lastly, PsiCLASS took only 9 h with 24 threads to process the 667 samples at a peak memory of 14 GB RAM on an 3.0 GHz Intel “Ivy Bridge” Xeon server, amounting to <1 min per sample. By comparison, StringTie took 34 h to process the samples sequentially, with 24 threads, at a peak memory of 524 MB, and Scallop, which is single-threaded, required 75 h with a peak memory of 5 GB.

## Discussion

Determining the set of expressed genes and transcripts in an RNA-seq experiment is critical for subsequent quantification and differential expression and splicing analyses. The conventional approach to process each sample separately and then merge the sets of transcripts to create a unified set of annotations has limitations, in particular low precision. We present PsiCLASS, a transcript assembler and meta-assembler that simultaneously analyzes all samples in an RNA-seq experiment. Algorithmic underpinnings of PsiCLASS include a global subexon graph, statistical methods for cross-sample feature selection, and dynamic programming optimization and voting algorithms for transcript selection at the level of individual samples and over the entire collection of data, which collectively lead to more complete, more accurate, and more consistent sets of annotations. In particular, our study suggests that voting-based approaches may have advantages over the current assembly-based aggregation methods in some cases, albeit their broader applicability needs to be further investigated.

PsiCLASS is designed to leverage the gene information across samples in a one or two condition experiments, to improve the accuracy of gene and transcript reconstruction for downstream differential gene expression and differential splicing studies, and may not be ideally suited to multi-condition experiments, such as annotation of genes and alternative splicing in large data sets such as GTEx24 or TCGA25. Nevertheless, our limited tests on collections of samples from 2 to 6 GTEx tissues reflect that PsiCLASS has better performance than TACO, and therefore show promise for exploring capabilities for handling more heterogeneous data sets in the future.

PsiCLASS had significantly higher precision at similar sensitivity when compared with current best methods, and showed consistently high precision (>50%) when tested on data from a variety of experimental conditions. PsiCLASS is scalable, efficient, and robust with large numbers of RNA-seq samples, thus providing a highly effective paradigm for large-scale analyses of collections of hundreds and thousands of samples. PsiCLASS is available free of charge from https://github.com/splicebox/PsiCLASS.

## Methods

### Algorithm overview

PsiCLASS builds a global subexon graph of a gene and its splice variants from genome-aligned RNA-seq reads in all input samples. It then traverses the graph to select a subset of the encoded transcripts in each sample. Lastly, it combines the predicted transcript sets across all samples, using a voting procedure to select a final set of meta-annotations.

Building per sample subexon graphs. PsiCLASS builds a subexon graph for each sample, then combines graphs across all samples to create a global subexon graph. In each sample, PsiCLASS uses candidate introns extracted from spliced read alignments to divide the genome into regions and subexons. A region denotes a maximal contiguous portion of the genome covered by reads. A subexon is a portion of a region delimited by two consecutive splice junctions and/or the end(s) of the region. A subexon graph has subexons as vertices, and two subexons are connected by an edge if they are adjacent in the same region or connected by an intron. Candidate splice variants are encoded as maximal paths in the subexon graph.

To build the sample-level subexon graph, PsiCLASS clusters read alignments along the genome that are colocated and on the same strand. Introns are extracted from spliced alignments and used to divide the region into subexons. A major confounding factor in determining subexons from RNA-seq data is the presence of intronic unprocessed RNA (‘noise’). To differentiate between intronic noise and signal, such as retained introns or alternative 5′ and 3′ gene ends, PsiCLASS assigns each subexon a score that reflects the probability that it is noise. In contrast to current single-sample methods, which simply discard a subexon if it fails sample-wide cutoffs, PsiCLASS then combines sample-level scores across all samples to determine a final label for the subexon and its inclusion in the global subexon graph.

More specifically, PsiCLASS computes the probability that a subexon is due to intronic noise using two models: (i) the exon–intron coverage ratio, and (ii) the intronic read coverage. Let ci be the average read coverage of (intronic) subexon i. In the coverage ratio model, PsiCLASS calculates a score that is equal to the coverage ratio of this subexon versus its flanking subexons: $$r_i = {\mathrm{min}}\left( {\frac{{c_i}}{{c_{i - 1}}},\frac{{c_i}}{{c_{i + 1}}}} \right)$$. The score is fitted to a mixture of two Gamma distributions, one representing signal and one noise: $$p\left( {r_i} \right) = \pi \Gamma _{\theta _0,k_0}\left( {r_i} \right) + \left( {1 - \pi } \right)\Gamma _{\theta _1,k_1}\left( {r_i} \right)$$, where π, (1 − π) are the prior probabilities that an intronic subexon is noise or signal, respectively, and $$\theta _0,k_0,\theta _1,k_1$$ are the parameters for the Gamma distributions, calculated with an expectation maximization algorithm. With these parameters, PsiCLASS can infer the probability that subexon i is noise according to the Bayes formula: $$P_R\left( {r_i} \right) = \frac{{\pi \Gamma _{\theta _0,k_0}\left( {r_i} \right)}}{{\pi {\mathrm{\Gamma }}_{\theta _0,k_0}\left( {r_i} \right) + \left( {1 {\,}-{\,} \pi } \right)\Gamma _{\theta _1,k_1}\left( {r_i} \right)}}$$.

The coverage ratio model above is insufficient when the overall gene coverage is low. Hence, the second model establishes a similar formula for coverage levels, $$P_C\left( {c_i} \right)$$, with $$\theta^\prime_0,k^\prime _0,\theta ^\prime _1,k^\prime _1$$ the parameters inferred using coverage. The final per sample subexon score then is $$P\left( i \right) = {\mathrm{max}}(P_R\left( {r_i} \right),\;P_C(c_i))$$.

Building the global subexon graph. PsiCLASS removes likely artifactual introns and intronic noise subexons by evaluating evidence across all sample, and builds a global subexon graph by combining individual samples’ graphs that share at least one intron. Multi-sample intron selection: To select a highly accurate set of introns, PsiCLASS assesses each candidate intron’s read support across all samples. Assume the experiment contains M samples, and denote each intron by its coordinates in the genome, e.g., (a, b). Let S (a, b) denote the total number of read alignments supporting (a, b) over all samples. Then the total number of alignments supporting its splice sites: $$S\left( a \right) = \mathop {\sum }\nolimits_{(a,y)} S\left( {a,y} \right),\;S\left( b \right) = \mathop {\sum }\nolimits_{\left( {x,b} \right)} S(x,b)$$. PsiCLASS keeps intron (a, b) iff: (i) $$\frac{{S\left( {a,b} \right)}}{M} \ge 0.5$$, indicating strong read support in one or a few samples, or consistent read support across multiple samples; and (ii) (a, b) appears in at least M0 samples, where $$M0 = \min \left( \lceil{\frac{M}{{50}}\rceil\left( \lfloor{\frac{{b - a + 1}}{{100,000}}\rfloor + 1} \right),\;M} \right)$$, if |ba| ≥ 100,000 (long intron). Condition (ii) is intended to filter out long intron-type alignment artifacts due to gene families and repeats, or from sequencing errors, which can lead to merged genes and transcripts. Multi-sample subexon selection: To determine a global set of subexons, PsiCLASS combines the subexon sets of individual samples with some modifications. Where multiple 3′ or 5′-end (i.e., subexons not delimited by a splice site) candidate subexons occur with the same endpoint and potentially different lengths among the samples, PsiCLASS creates a unique subexon with the median length. Further, to determine intronic subexons, PsiCLASS calculates a final score by combining all sample scores with a Bayesian formula. More specifically, let $$\bar \pi$$ denote the prior probability of intronic noise in the global model, calculated as the average of the mixture coefficients of the samples. Then the subexon score: $$P_n\left( {{\mathrm{noise}}|{\mathrm{data}}} \right) = \frac{{\bar \pi {P({\mathrm{data}}|{\mathrm{noise}}})}}{{\bar \pi P\left( {{\mathrm{data}}|{\mathrm{noise}}} \right) + \left( {1 - \bar \pi } \right)P\left( {{\mathrm{data|real}}} \right)}}$$ reflects the probability that the subexon is noise, where data is the observed information such as the coverage in each sample. We assume the samples are independent, hence $${{P}}\left( {{\mathrm{data}}|{\mathrm{noise}}} \right) = \mathop {\prod }\nolimits_{s = 1}^M G_0^{\left( s \right)}\left( i \right)$$, where s = 1,…, M denotes the sample. Here, $$G_0^{\left( s \right)}\left( i \right)$$ is $$\Gamma _{\theta _0^{\left( s \right)},\:k_0^{\left( s \right)}}(r_i^{(s)})$$ if the ratio model is used for subexon i in sample s, and $$\Gamma _{\theta \prime _0^{\left( s \right)},\;k\prime _0^{\left( s \right)}}( {c_i^{(s)}} )$$ if the coverage model is employed. Similarly, $$P\left( {{\mathrm{data}}|{\mathrm{real}}} \right) = \mathop {\prod }\nolimits_{s = 1}^M G_1^{\left( s \right)}\left( i \right)$$. In the end, the subexon is retained if it passes a predefined threshold.

Transcript selection. Candidate transcript models are represented as maximal paths in the global subexon graph, from a node with no incoming edges (source) to a node with no outgoing edges (sink). Since the graph generally encodes a much larger number of transcripts than is biologically possible, PsiCLASS identifies and selects a subset of transcripts that can explain all contiguity constraints from spliced reads and paired reads. PsiCLASS first predicts a set of transcripts for each sample, using a graph-based dynamic programming algorithm with the global subexon graph and the sample specific alignment data, then combines the individual samples’ transcript sets and selects a subset of meta-annotations by voting.

To predict a set of transcripts for each sample, PsiCLASS employs a SET_COVER framework and dynamic programming algorithms12, adapted for subexon graphs. Namely, we define a constraint as a cluster of read alignments with the same subexon pattern. Like CLASS2, PsiCLASS uses constraints to decrease the memory usage while preserving the structural and contiguity information contained in the full set of reads. For a given graph G, let C = {c1, …, cm} denote the set of constraints and T = {t1, …, tn} the set of candidate transcripts, encoded in the graph. Given a constraint ci, its abundance ai = a(ci) defined as the number of supporting reads (or read pairs) normalized by the number of possible start positions of the reads within the constraint’s subexons. To reduce the transcript selection problem to SET_COVER, we view each candidate transcript tj as the set of constraints that are compatible with its exon-intron structure: C(tj) = {c1, …,cnj}, where ci ~ tj. Constraint ci is compatible with the transcript tj’s exon-intron structure, denoted ci ~ tj, if its list of subexon intervals is contained in that of the transcript’s and all the exon–exon junctions correspond to introns in the transcript.

In the simplest formulation, the goal then is to select a minimal (parsimonious) subset of transcripts that satisfies all constraints. More realistically, to account for the different abundance of constraints, we define a transcript’s abundance as the minimum abundance among its set of constraints: Aj = min { ai| ci ~ tj }. The goal then becomes to determine a subset of transcripts that most closely explain the constraints and their abundance levels. PsiCLASS uses a greedy approximation framework to address this problem, iteratively selecting the transcript that covers the largest number of constraints weighted by the constraints’ abundance, then adjusting the constraints’ abundance levels before the next iteration:

While ({nondepleted constraints} $$\ne \, \emptyset$$):

Choose transcript $$t \in T$$ that maximizes $$|C\left( t \right)|\left( {1 + \frac{{A_t}}{A}} \right)$$

Update the constraints’ abundance:

$$x = \mathop {{{\mathrm{min}}}}\limits_{c \in C(t)} \{ a\left( c \right)\}$$
$${\mathrm{For}}\;{\mathrm{each}}\;c \in C\left( t \right)\!\! :$$
$$a\left( c \right) = a\left( c \right) - x$$

if $$a\left( c \right) \le 0$$, mark constraint c as depleted.

PsiCLASS implements the procedure above in two steps. First, it determines the candidate set of transcripts T, using either enumeration (for graphs with <200,000 transcripts) or a variation of the splice-graph dynamic programming algorithm in Song et al.12 that considers all reads single-end, for fast processing. Once the candidate transcript set T is determined, PsiCLASS applies the greedy SET_COVER approximation algorithm above.

For completeness, we include a brief description of the dynamic programming optimization procedure. The algorithm considers all subpaths L, and recursively calculates the maximum number of constraints f(L) for substranscripts starting with subpath L: f(L) = maxL { f(L′) + c(L, L′), if L′ exists; c(L), if L′does not exist }, where: (i) L′ is a subpath immediately following L so that all constraints compatible with L end before or within L′; (ii) c(L, L′) is the number of constraints starting in and (partially) compatible with L and L′, and compatible with the concatenated subpath L, L′; and (iii) c(L) is the number of constraints covered by subtranscript L. To take into account the abundance levels in the optimization process, at each sweep of the graph the algorithm excludes subpaths that cover constraints with abundance below a fixed value x; hence, the dynamic programming algorithm will return the best transcript with abundance greater than x (x-abundance transcript). With this modification, at each graph sweep the selection process selects an x-abundance transcript, starting with x0 = 0 (thus guaranteeing that such a transcript exists), and each selected transcript’s abundance value used as lower bound for the selection process at the following step: 0 =x0<x1<x2<xm, until no transcript can be found. The optimal transcript then is among those selected by the sweeps. More details, along with proof of correctness for the algorithm, can be found in Song et al.12.

Selecting a global set of transcripts. PsiCLASS selects a set of meta-annotations from the individual samples’ sets of transcripts by voting. Each transcript is assigned a score equal to the transcript’s average estimated abundance across the samples (i.e., every sample has one vote, weighted by the transcript’s abundance level; default cutoff: 1.0). As different cutoff parameters values might work best for data with specific characteristics, the user can readily adjust or recalibrate the voting parameters postassembly, starting from the already computed sets of transcripts for the individual samples.

### Sequence data

We generated 25 RNA-seq samples, with ~85 million 100 bp paired-end reads, using the software Polyester20 with the default gene and transcript distribution models and randomly sampling 10% of the transcripts (at 13,912 genes) from the human GENCODE v.2726 gene annotations. Reads were aligned to the reference genome hg38 separately with Hisat2 v.2.0.521 and STAR v.2.5.3a22. Chromosome 2 alignments were extracted and used in the assembly and evaluations. Human liver samples were obtained from the Stanley Medical Research Institute (SMRI) and previously sequenced by Dr. Sabunciyan’s lab. All ethical standards set forth by the Johns Hopkins University and local laws were complied with. The liver tissue was collected postmortem by the SMRI after obtaining consent from the family of the deceased. A detailed description of the consent procedure and the collection was published27. The project was reviewed by the Johns Hopkins University Institutional Review Board (IRB) and was given exempt status since only postmortem tissue was involved in the research. For sequencing, total RNA was isolated using the Qiagen RNeasy kit and libraries were constructed using the Illumina TruSeq Stranded Total RNA kit for Human/Mouse/Rat following the manufacturers recommended protocol. The resulting stranded, rRNA depleted liver libraries were sequenced on an Illumina HiSeq 2000 instrument. 667 RNA-seq samples from human lymphoblastoid cell lines part of the GEUVADIS population variation project were publicly available from ArrayExpress (accession: E-GEUV-6), and mouse hippocampus RNA-seq data were those reported in ref. 23 and available from GenBank (ProjectID: PRJEB18790). Lastly, RNA-seq samples from six human tissues (frontal cortex, cortex, cerebellum, heart, liver, and lung) used in the heterogeneity study were downloaded from the dbGAP GTEx24 repository. The lists of samples for each evaluation are included in the Source Data file.

### Performance evaluation

Once the reads were mapped to the genome, we used StringTie v.1.3.3.b and Scallop v.0.10.2 to assemble them into transcripts, for each individual sample. Transcript sets for all samples in an experiment were then merged with StringTie(ST)-merge and TACO v.0.7.3. For PsiCLASS v.1.0.1, reads were assembled simultaneously across all samples. To evaluate the accuracy of transcript assembly, we employed standard sensitivity (Sn) and precision (Pr) measures and evaluation criteria to assess the accuracy of transcript reconstructions by comparison to a gold reference, namely the set of simulated transcripts and the human GENCODE v.27 and mouse RefSeq gene annotations. A predicted transcript is deemed a true positive (TP) iff its intron chain fully matches that of a gold reference transcript. If N is the number of predicted transcripts, M be the number of ground truth transcripts, then Sn = TP/M and Pr = TP/N 5. Performance metrics at the transcript, exon, and intron levels were calculated with the tool grader included in the PsiCLASS package, and at the gene level with the tool Cuffcompare (v.2.2.1). For gene content analyses, only transcripts that were a full match, contained or splice variant transcript (codes ‘=’, ‘c’, and ‘j’) of a reference gene were considered.

### Reporting summary

Further information on research design is available in the Nature Research Reporting Summary linked to this article.

## Data availability

Raw sequence data for the GEUVADIS project can be obtained from ArrayExpress (Accession:E-GEUV-6), mouse hippocampus data from GenBank (ProjectID:PRJEB18790), and alignments of simulated data from Zenodo (https://doi.org/10.5281/zenodo.1407759). The liver data has been deposited in GenBank under accession PRJNA575230. In addition, the full set of evaluation results represented in Fig. 1, Tables 1 and 2, and Supplementary Figs. 2, 49 are provided as a Source Data file, and scripts for performing the evaluation and links to the assembled transcripts can be obtained from the project site in GitHub. All other relevant data are available upon request.

## Code availability

PsiCLASS is available under a GNU GPL license from https://github.com/splicebox/PsiCLASS.

## References

1. Wang, Z., Gerstein, M. & Snyder, M. RNA-Seq: a revolutionary tool for transcriptomics. Nat. Rev. Genet. 10, 57–63 (2009).

2. Pan, Q., Shai, O., Lee, L. J., Frey, B. J. & Blencowe, B. J. Deep surveying of alternative splicing complexity in the human transcriptome by high-throughput sequencing. Nat. Genet. 40, 1413–1415 (2008).

3. Niknafs, Y. S., Pandian, B., Iyer, H. K., Chinnaiyan, A. M. & Iyer, M. K. TACO produces robust multisample transcriptome assemblies from RNA-seq. Nat. Methods 14, 68–70 (2017).

4. Trapnell, C. et al. Transcript assembly and quantification by RNA-Seq reveals unannotated transcripts and isoform switching during cell differentiation. Nat. Biotechnol. 28, 511–515 (2009).

5. Li, W., Feng, J. & Jiang, T. IsoLasso: a LASSO regression approach to RNA-Seq based transcriptome assembly. J. Comput. Biol. 18, 1693–1707 (2011).

6. Guttman, M. et al. Ab initio reconstruction of cell type-specific transcriptomes in mouse reveals the conserved multi-exonic structure of lincRNAs. Nat. Biotechnol. 28, 503–510 (2010).

7. Tomescu, A. I., Kuosmanen, A., Rizzi, R. & Makinen, V. A novel min-cost flow method for estimating transcript expression with RNA-Seq. BMC Bioinform. 14 (Suppl 5), S15 (2013).

8. Song, L. & Florea, L. CLASS: constrained transcript assembly of RNA-seq reads. BMC Bioinforma. 14 (Suppl 5), S14 (2013).

9. Mezlini, A. M. et al. iReckon: simultaneous isoform discovery and abundance estimation from RNA-seq data. Genome Res. 23, 519–529 (2013).

10. Canzar, S., Andreotti, S., Weese, D., Reinert, K. & Klau, G. W. CIDANE: comprehensive isoform discovery and abundance estimation. Genome Biol. 17, 16 (2016).

11. Bernard, E., Jacob, L., Mairal, J. & Vert, J. P. Efficient RNA isoform identification and quantification from RNA-Seq data with network flows. Bioinformatics 30, 2447–2455 (2014).

12. Song, L., Sabunciyan, S. & Florea, L. CLASS2: accurate and efficient splice variant annotation from RNA-seq reads. Nucleic Acids Res. 44, e98 (2016).

13. Pertea, M. et al. StringTie enables improved reconstruction of a transcriptome from RNA-seq reads. Nat. Biotechnol. 33, 290–295 (2015).

14. Liu, J., Yu, T., Jiang, T. & Li, G. TransComb: genome-guided transcriptome assembly via combing junctions in splicing graphs. Genome Biol. 17, 213 (2016).

15. Shao, M. & Kingsford, C. Accurate assembly of transcripts through phase-preserving graph decomposition. Nat. Biotechnol. 35, 1167–1169 (2017).

16. Steijger, T. et al. Assessment of transcript reconstruction methods for RNA-seq. Nat. Methods 10, 1177–1184 (2013).

17. Lin, Y.-Y. et al. CLIIQ: Accurate comparative detection and quantification of expressed isoforms in a population. Algorithms Bioinform. LNCS 7534, 178–189 (2012).

18. Tasnim, M., Ma, S., Yang, E. W., Jiang, T. & Li, W. Accurate inference of isoforms from multiple sample RNA-Seq data. BMC Genomics 16 (Suppl 2), S15 (2015).

19. Behr, J. et al. MITIE: Simultaneous RNA-Seq-based transcript identification and quantification in multiple samples. Bioinformatics 29, 2529–2538 (2013).

20. Frazee, A. C., Jaffe, A. E., Langmead, B. & Leek, J. T. Polyester: simulating RNA-seq datasets with differential transcript expression. Bioinformatics 31, 2778–2784 (2015).

21. Kim, D., Langmead, B. & Salzberg, S. L. HISAT: a fast spliced aligner with low memory requirements. Nat. Methods 12, 357–360 (2015).

22. Dobin, A. et al. STAR: ultrafast universal RNA-seq aligner. Bioinformatics 29, 15–21 (2012).

23. Srivastava, P. K. et al. Genome-wide analysis of differential RNA editing in epilepsy. Genome Res. 27, 440–450 (2017).

24. Consortium, G. T. Human genomics. The genotype-tissue expression (GTEx) pilot analysis: multitissue gene regulation in humans. Science 348, 648–660 (2015).

25. Tomczak, K., Czerwinska, P. & Wiznerowicz, M. The Cancer Genome Atlas (TCGA): an immeasurable source of knowledge. Contemp. Oncol. 19, A68–A77 (2015).

26. Harrow, J. et al. GENCODE: the reference human genome annotation for The ENCODE Project. Genome Res. 22, 1760–1774 (2012).

27. Torrey, E. F., Webster, M., Knable, M., Johnston, N. & Yolken, R. H. The Stanley Foundation brain collection and neuropathology consortium. Schizophr. Res. 44, 151–155 (2000).

## Acknowledgements

Development and evaluations were performed on the Maryland Advanced Research Computing Center (MARCC). Work was supported in part by NSF grants ABI-1356078 and IOS-1339134 to L.F., and by NIH grants R01GM129085 and R01GM124531 to L.F. and Kathleen Burns. S.S. was supported by a grant from the Stanley Medical Research Institute.

## Author information

Authors

### Contributions

L.F. and L.S. conceived the project, L.S. developed the method, S.S. provided sequence data, and L.S., L.F., G.Y., and S.S. contributed to the evaluation. All authors wrote and approved the paper.

### Corresponding author

Correspondence to Liliana Florea.

## Ethics declarations

### Competing interests

The authors declare that they have no competing interests.

Peer review information Nature Communications thanks the anonymous reviewer(s) for their contribution to the peer review of this work. Peer reviewer reports are available.

Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Rights and permissions

Reprints and Permissions

Song, L., Sabunciyan, S., Yang, G. et al. A multi-sample approach increases the accuracy of transcript assembly. Nat Commun 10, 5000 (2019). https://doi.org/10.1038/s41467-019-12990-0

• Accepted:

• Published:

• DOI: https://doi.org/10.1038/s41467-019-12990-0

• ### FINDER: an automated software package to annotate eukaryotic genes from RNA-Seq data and associated protein sequences

• Sagnik Banerjee
• Priyanka Bhandary
• Carson M. Andorf

BMC Bioinformatics (2021)


text:
# zbMATH — the first resource for mathematics

On the conjugation of some class of one-dimensional endomorphisms with a class of piecewise expanding mappings. (English. Russian original) Zbl 0605.28016
Russ. Math. Surv. 40, No. 1, 211-212 (1985); translation from Usp. Mat. Nauk 40, No. 1(241), 187-188 (1985).
Let T be a piecewise continuously differentiable mapping of an interval $$[a,b]$$ into itself. The author gives conditions under which there is a non-singular conjugation of T with a piecewise expanding mapping. The results are then applied to the family of mappings $$T_ ax=ax(1-x)$$ of the unit interval into itself. It is shown that there is a continuum of values of the parameter a for which $$T_ a$$ is non-singularly conjugate to a piecewise expanding mapping.
Reviewer: D.Newton

##### MSC:
 28D99 Measure-theoretic ergodic theory 37A99 Ergodic theory
##### Keywords:
non-singular conjugation; piecewise expanding mapping
Full Text:

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 Presentation

Allostery

2017-10-27

You measure MM curves for a variety of enzymes (solid curves). You add molecule $A$ to each solution and remeasure its MM curve (dashed line). For each curve: What changed ($K_{M}$ or $V_{max}$)? Is this a competitive or noncompetitive inhibitor?
$K_{M}$, competitive
$V_{max}$, noncompetitive
$V_{max}$, activator

Competitive inhibitor

• Inhibitor competes with substrate for same site
• This raises $K_{M}$: lower apparent affinity for substrate because it has to compete
• No effect on $V_{max}$. If you add enough substrate, you swamp out competitor
• Inhibitor is chemically similar to the substrate
• Example: Ethanol competes with methanol for alcohol dehydrogenase, lowering the rate of formaldehyde production.

Noncompetitive inhibitor

• Binds distant from the active site, altering active site to turn off activity
• This leads to a drop in $V_{max}$. (lower $[E]_{T}$: Less of the enzyme is in the active form.)
• No effect on $K_{M}$. What active enzyme is around has exact same affinity for substrate
• The inhibitor need not have any chemical similarity to the substrate

Example noncompetitive inhibitor

Conceptual goals

• Understand that binding at one site in a protein can alter activity at another site
• This "allostery" arises because the "allosteric effector" interacts with one conformation, but not the other.
• Understand how this applies to hemoglobin (BPG).

Skill goals

• Determine how binding at on site affects activity at the other
• Predict the effects of adding allosteric effectors to a system of equilibria.

Hemoglobin transports $O_{2}$ from the lungs to tissues

stephaniefuturedoc

Hemoglobin is a tetramer four proteins

Credit: Janet Iwasa (Utah)
Credit: Janet Iwasa (Utah)
Credit: Janet Iwasa (Utah)

$E_{active} + I \rightleftarrows E_{inactive} + I \rightleftarrows E_{inactive} \cdot I$

$[E]_{active} = [E]_{T}\theta_{active}$

$\theta_{active} = \frac{[E_{active}]}{[E_{active}] + [E_{inactive}] + [E_{inactive}\cdot I]}$

What are the ingredients you would need for allostery?

Ingredients:

• Two different binding sites recognizing different things
• Equilibrium between two (or more) protein shapes
• Different "activities" (functions, properties, etc.) of each shape
• Binding to one shape, but not the others

Summary

Allostery is when binding at one site in a protein alters activity at another site

• Two different binding sites recognizing different things
• Equilibrium between two (or more) protein shapes
• Different "activities" (functions, properties, etc.) of each shape
• Binding to one shape, but not the others

Noncompetitive inhibitors

BPG allosterically regulates $O_{2}$ binding in hemoglobin

You can’t perform that action at this time.


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# Rotational Dynamics of a plank of mass

#### jgens

Gold Member
1. Homework Statement

A thin plank of mass M and length l is pivoted at one end. The plank is released at 60 degrees from the vertical. What is the magnitiude and direction of the force on the pivot when the plank is horizontal.

2. Homework Equations

N/A

3. The Attempt at a Solution

So, here's what I worked out ...

Reasoning:
I figured that when the plank is horizontal there are two forces acting on it: The graviational force and the force of the pivot on the plank (which is a radial force). Since the gravitational force has no component in the radial direction when the plank is horizontal, the force required to keep the plank in uniform circular motion must supplied by the force of the pivot on the plank. Using Newton's Third Law, we see that the force of the plank on the pivot is equal in magnitude and opposite in direction to the force of the pivot on the plank. Therefore, we just need to calculate the force of the pivot on the plank (the centripetal force needed to keep the plank in uniform circular motion) in order to find the desired quantity. Denote this force by F.

Calculations:
Let dF be the force exerted by the pivot on a small segment of mass dM at a radial distance r from the pivot. This means that ...

dF = w2r dM

where w is angular velocity of the rod when it's horizontal. Assuming that the mass density of the rod is uniform, we see that k = dM/dr = M/l. Using this information in the above equation, we get ...

dF = kw2r dr

Integrating this gives ...

$$F = \int_0^l k\omega^2r \, dr = k\omega^2\int_0^lr \, dr = \frac{k\omega^2l^2}{2} = \frac{M\omega^2l}{2}$$

Is this right so far?

Related Introductory Physics Homework Help News on Phys.org

#### jgens

Gold Member
Calculations cont:
We can find the net torque acting on the plank if we treat it as if all of the mass were concentrated at the center of mass. Therefore, this torque is given by ...

$$\tau = I\alpha = \frac{ML^2}{3}\alpha = \frac{Mgl}{2}\sin{(\theta)}$$

This can be written ...

$$\alpha = \ddot{\theta} = \frac{3g}{2l}\sin{(\theta)}$$

Therefore,

$$\dot{\theta}\ddot{\theta} = \frac{3g}{2l}\sin{(\theta)}\dot{\theta}$$

Integrating this, we see that ...

$$\frac{\dot{\theta}^2}{2} \bigg{|}_0^{\omega} = \frac{3g}{2l}\int_{\pi/3}^{\pi/2}\cos{\theta}\,d\theta$$

Solving this equation, we get ...

$$\frac{\omega^2}{2} = -\frac{3g}{2l}\cos{(\theta)}\bigg{|}_{\pi/3}^{\pi/2} = \frac{3g}{4l}$$

Is this right in the context of the problem?

#### tiny-tim

Homework Helper
hi jgens!
A thin plank of mass M and length l is pivoted at one end. The plank is released at 60 degrees from the vertical. What is the magnitiude and direction of the force on the pivot when the plank is horizontal.

Reasoning:
I figured that when the plank is horizontal there are two forces acting on it: The graviational force and the force of the pivot on the plank (which is a radial force). Since the gravitational force has no component in the radial direction when the plank is horizontal, the force required to keep the plank in uniform circular motion must supplied by the force of the pivot on the plank. Using Newton's Third Law, we see that the force of the plank on the pivot is equal in magnitude and opposite in direction to the force of the pivot on the plank. Therefore, we just need to calculate the force of the pivot on the plank (the centripetal force needed to keep the plank in uniform circular motion) …
that's really good!

unfortunately you then left the path of reason and went berserk with equations about torque

go back to your reasoned argument … horizontal force = mass times centripetal acceleration of the centre of mass …

so get the acceleration from the angular velocity, and get the angular velocity from conservation of energy!

(btw, you didn't need that integration in your first post … if you look at ∫ r dr, you can see it's bound to always give you the position of the centre of mass, in this case at L/2! )

#### jgens

Gold Member
Okay, so for conservation of energy, we can express everything in terms of translation and rotation of the center of mass. Initially, all of the energy is potential energy, and the potential energy at the CM is ...

$$Mgh = \frac{Mgl\cos{\theta}}{2} = \frac{Mgl\cos{\pi/6}}{2} = \frac{Mgl\sqrt{3}}{4}$$

By the Law of Conservation of Mechanical Energy, this is equal to the rotational and translational kinetic energy of the CM. Since the CM isn't rotating, this gives us the equation

$$\frac{Mgl\sqrt{3}}{4} = \frac{1}{2}MV^2 = \frac{1}{8}M\omega^2l^2$$

Solving for $\omega^2$ yields,

$$2\sqrt{3}\frac{g}{l}$$

This doesn't look right though, so I think that I've done something wrong. By the way, what doesn't work about my argument with torque?

#### tiny-tim

Homework Helper
Since the CM isn't rotating …
uhh? that doesn't mean anything!

the rod is rotating, so you do need to add the rotational KE

(and btw, either you used the wrong angle, or you used cos instead of sin)
By the way, what doesn't work about my argument with torque?
it's probably ok, but i couldn't be bothered to check it!

try again!

(and now i'm off to bed :zzz:)

#### jgens

Gold Member
Okay! Now I understand you. And going through the calculations it looks like it should give me the same result as my argument with torque. That's a relief.

Thanks!

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# Introduction to Training Predictions

#### 2020-02-21

Training predictions are the out-of-fold predictions on train data made by a model. That is, DataRobot can do 5-fold cross validation, where it trains on 80% of the train data and predicts for 20% of the train data. After doing this for each segment of the data, the five different 20% holdout sets can be recombined into a single file with a prediction for each row of the training data that was not made by a model that had trained on that row. This is important because predictions for rows that the model has trained on (in-fold predictions) will almost always overfit the data and not generalize well to new data. These training predictions are useful for further model validation and for blending the model with other models. Generating and retrieving these training predictions is now possible via the DataRobot API.

## Retrieving Training Predictions

Before you can retrieve training predictions, you must first request their creation. This is done on the model object you want training predictions for.

dataSubset specifies the subset of training data you want training predictions for, such as DataSubset$All for all training data (note this will retrain your model at 100%), DataSubset$ValidationAndHoldout will return predictions for solely data in validation and holdout sets, and DataSubset$Holdout will return predictions solely for the holdout set. models <- ListModels(projectId) model <- models[[1]] trainingPredictions <- GetTrainingPredictionsForModel(model, dataSubset = DataSubset$All)
kable(head(trainingPredictions), longtable = TRUE, booktabs = TRUE, row.names = TRUE)
partitionId prediction rowId
1 Holdout No 0
2 3.0 No 1
3 2.0 Yes 2
4 3.0 No 3
5 4.0 No 4
6 3.0 No 5

You may also find it valuable to split a call to request and get like this:

models <- ListModels(projectId)
model <- models[[1]]
jobId <- RequestTrainingPredictions(model, dataSubset = DataSubset$All) # can run computations here while training predictions compute in the background trainingPredictions <- GetTrainingPredictionsFromJobId(projectId, jobId) # blocks until job complete kable(head(trainingPredictions), longtable = TRUE, booktabs = TRUE, row.names = TRUE) partitionId prediction rowId 1 Holdout No 0 2 3.0 No 1 3 2.0 Yes 2 4 3.0 No 3 5 4.0 No 4 6 3.0 No 5 Or you can retrieve training predictions from a specific ID. trainingPredictions <- ListTrainingPredictions(projectId) trainingPredictionId <- trainingPredictions[[1]]$id
trainingPrediction <- GetTrainingPredictions(projectId, trainingPredictionId)
kable(head(trainingPrediction), longtable = TRUE, booktabs = TRUE, row.names = TRUE)
partitionId prediction rowId
1 Holdout No 0
2 3.0 No 1
3 2.0 Yes 2
4 3.0 No 3
5 4.0 No 4
6 3.0 No 5

DownloadTrainingPredictions(projectId, trainingPredictionId, "trainingPredictions.csv")


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# Performing skeleton tracking with ros-skeltrack

In response to the buyout of OpenNI/NiTE, I tried to explore an open source alternative Skeltrack. I came across a helpful ROS wrapper for Skeltrack here ( https://github.com/jmerdich/ros-skelt... ).

So as to provide the /tf topic that this package depends on, I ran openni_launch in one shell of my Ubuntu 14.04 system. In another shell I ran rosrun ros_skeltrack ros_skeltrack_node. I'm using an ASUS Xtion and not a Kinect, which is why I'm not using GFreenect (as recommended by the author of Skeltracker).

It seemed to compile, however I couldn't figure out how to visually detect or track a skeleton. My first instinct was to use the same settings on RViz -- as this worked with openni_tracker. However, the ros_skeltrack package did not seem to work the same way.

In fact, when I ran rosnode info skeletor_processor on the relevant node that was published, I got the following output:

--------------------------------------------------------------------------------
Node [/skeletor_processor] Publications:   * /rosout [rosgraph_msgs/Log]

Subscriptions:   * /camera/depth/image_raw [sensor_msgs/Image]

Services:   * /skeletor_processor/get_loggers  * /skeletor_processor/set_logger_level

contacting node http://[omitted]/ ... Pid: 10724 Connections:  * topic: /rosout
* to: /rosout
* direction: outbound
* transport: TCPROS  * topic: /camera/depth/image_raw
* to: /camera/camera_nodelet_manager (http://[omitted]/)
* direction: inbound
* transport: TCPROS

Unlike with openni_tracker, nothing was being published by the wrapper.

Any thoughts, hints in the right direction, or ideas would be very helpful and appreciated.

edit retag close merge delete


text:
Rs Aggarwal 2017 Solutions for Class 9 Math Chapter 16 Summative Assessment I are provided here with simple step-by-step explanations. These solutions for Summative Assessment I are extremely popular among Class 9 students for Math Summative Assessment I Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 9 Math Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Question 1:

Mark (âœ“) against the correct answer in each of the following:
Which of the following is a rational number?
(a) $\frac{2}{\sqrt{3}}$
(b) $\frac{\sqrt{2}}{3}$
(c) $3\sqrt{5}$
(d) $\frac{-3}{5}$

(d) $\frac{-3}{5}$

The number of the form $\frac{p}{q}$, where p and q are integers and $q\ne 0$, are known as rational numbers.
Here $-3$ and 5 are integers and $5\ne 0$.
Therefore, $\frac{-3}{5}$ is a rational number.

#### Question 2:

Mark (âœ“) against the correct answer in each of the following:
The value of k, for which the polynomial x3− 4x2 + 2x + k has 3 as its zero, is
(a) 3
(b) −3
(c) 6
(d) −6

(a) 3

Let f(x) = x3− 4x2 + 2x + k.
3 is the zero of f(x).
Therefore, f(3) = 0

#### Question 3:

Mark (âœ“) against the correct answer in each of the following:
Which of the following are zeros of the polynomial x3 + 2x2− 5x − 6?
(a) −2
(b) 2
(c) −3
(d) 3

(b) 2 and (c) −3
Let f(x) =
x3 + 2x2− 5x − 6.
We have to find the zeros of the polynomial.

⇒  f(x) = 0
x3 + 2x2− 5x − 6 = 0
x3 + 3x2x2 −3x −2x −6 = 0
x2(x + 3) −x(x + 3) −2(x + 3) = 0
(x + 3)(x2x − 2)
(x + 3)(x2 − 2x + x − 2)
(x + 3){x(x − 2) +1(x − 2)} = 0
(x + 3)(x − 2)(x + 1) = 0
Either x + 3 = 0 or x − 2 = 0 or x + 1 = 0

x = − 3 or 2 or −1

Disclaimer: Here both 2 and − 3 are correct answers.

#### Question 4:

Mark (âœ“) against the correct answer in each of the following:
The factorisation of x2 + 7x − 12 yields
(a) (x − 3) (x − 4)
(b) (3 + x) (4 − x)
(c) (x − 4) (3 − x)
(d) (4 − x) (3 − x)

(c) (x − 4) (3 − x)

x2 + 7x − 12 = − (x2 − 7x + 12)
= − (x2 − 4x − 3x + 12)
= − { x(x − 4) −3 (x − 4)}
=  − (x − 4)(x − 3)
= (x − 4)(3 − x)

#### Question 5:

Mark (âœ“) against the correct answer in each of the following:
In the given figure, BOC = ?
(a) 45°
(b) 60°
(c) 75°
(d) 56°
Figure

(b) 60°

Since, OAB is a straight line, the sum of all the angles
on the same side of AOB, at point O on it, is 180o.

#### Question 6:

Mark (âœ“) against the correct answer in each of the following:
In the given figure, âˆ†ABC is an equilateral triangle and âˆ†BDC is an isosceles right triangle, right-angled at D. Then ∠ACD = ?
(a) 60°
(b) 90°
(c) 120°
(d) 105°
Figure

(d) 105°

#### Question 7:

Mark (âœ“) against the correct answer in each of the following:
Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is
(a) 30 cm2
(b) 45 cm2
(c) 60 cm2
(d) 78 cm2
Figure

(c) 60 cm2

#### Question 8:

Mark (âœ“) against the correct answer in each of the following:
In an isosceles right triangle, the length of the hypotenuse is .
(a)
(b) 6 cm
(c) 5 cm
(d) 4 cm
Figure

(d) 4 cm

In a right-angled triangle,

#### Question 9:

If $x=7+4\sqrt{3}$, find the value of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$.

#### Question 10:

Factorise: $\left(7{a}^{3}+56{b}^{3}\right)$.

#### Question 11:

Find the value of a for which (x 1) is a factor of the polynomial (a2x3− 4ax + 4a − 1).

Let f(x) = a2x3− 4ax + 4a − 1.
(x 1) is a factor of f(x)
Therefore, f(1) = 0

#### Question 12:

In the given figure, if AC = BD, show that AB = CD. State Euclid's axiom used for it.
Figure

From the given figure,
AC  = AB + BC
BD  =  BC + CD
It is given that AC = BD.
Putting the values of AC and BD, we get:
AB + BC =  BC + CD..................(1)
According to Euclid's axiom, when equals are subtracted from equals, the remainders are also equal.
Subtracting BC from both sides in equation (1), we get:
AB + BC $-$ BC = BC + CD $-$ BC
AB = CD

Hence, proved.

#### Question 13:

In âˆ†ABC, if 2∠A = 3∠B = 6∠C, calculate the measure of ∠B.

In a âˆ†ABC, 2$\angle$A = 3$\angle$B = 6$\angle$C = xo (say)

#### Question 14:

In the given figure, BAC = 30°, ∠ABC = 50° and ∠CDE = 40°. Find ∠AED.
Figure

Side BC of $∆$ABC is produced to D.

Side CE of $∆$CED is produced to A.

#### Question 15:

If $x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ and $y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$, find the value of (x2 + y2).
or
Simplify: $\frac{\left(7+3\sqrt{5}\right)}{\left(3+\sqrt{5}\right)}-\frac{\left(7-3\sqrt{5}\right)}{\left(3-\sqrt{5}\right)}$

$\text{Given},\phantom{\rule{0ex}{0ex}}x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}},y=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\phantom{\rule{0ex}{0ex}}{x}^{2}={\left(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\right)}^{2}=\frac{5+3+2\sqrt{15}}{5+3-2\sqrt{15}}=\frac{8+2\sqrt{15}}{8-2\sqrt{15}}\phantom{\rule{0ex}{0ex}}\mathrm{And}\phantom{\rule{0ex}{0ex}}{y}^{2}={\left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)}^{2}=\frac{5+3-2\sqrt{15}}{5+3+2\sqrt{15}}=\frac{8-2\sqrt{15}}{8+2\sqrt{15}}$

OR

#### Question 16:

If 2 and $\frac{-1}{3}$ are the zeros of the polynomial 3x3− 2x2 − 7x − 2, find the third zero of the polynomial.

#### Question 17:

Find the remainder when the polynomial f(x) = 4x3− 12x2 + 14x − 3 is divided by (2x − 1).

By the remainder theorem, we know that when p(x) is divided by (2x - 1), the remainder is $p\left(\frac{1}{2}\right)$.
Now,

#### Question 18:

Factorise: (p q)3 + (q r)3 + (r p)3

(p q)3 + (q r)3 + (r p)
Let p q = x
q r = y
r p = z
Therefore, x + y+ z =
p q + q r + r p = 0
So,
(p q)3 + (q r)3 + (r p)3 = x3 + y3 + z3
= 3xyz   (when x + y + z = 0, then x3 + y3 + z3 = 3xyz )
= 3(p q)(q r)(r p)

#### Question 19:

In the given figure, in âˆ†ABC, it is given that ∠B = 40° and ∠C = 50°. DE || BC and EF || AB. Find: (i) ∠ADE + ∠MEN (ii) ∠BDE and (iii) ∠BFE.
Figure

#### Question 20:

In the given figure, âˆ†ABC and âˆ†ABD are such that AD = BC, ∠1 = ∠2 and ∠3 = ∠4. Prove that BD = AC.
Figure

#### Question 21:

In the given figure, C is the mid-point of AB. If DCA = ∠ECB and ∠DBC = ∠EAC, prove that DC = EC.
Figure

#### Question 22:

In âˆ†ABC, ALBC and AM is the bisector of ∠A. Show that $\angle LAM=\frac{1}{2}\left(\angle B-\angle C\right)$.
Figure

#### Question 23:

In the given figure, AB || CD, BAE = 100° and ∠AEC = 30°. Find ∠DCE.
Figure

#### Question 24:

Factorise: a3b3 + 1 + 3ab

#### Question 25:

If $x=\frac{1}{\left(2-\sqrt{3}\right)}$, show that the value of (x3− 2x2 − 7x + 5) is 3.
Or
Simplify: $\frac{1}{\left(1+\sqrt{2}\right)}+\frac{1}{\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{\left(\sqrt{3}+\sqrt{4}\right)}+......+\frac{1}{\left(\sqrt{8}+\sqrt{9}\right)}$

OR

#### Question 26:

If $x=\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}$ , then show that $b{x}^{2}-ax+b=0$.

#### Question 27:

If (x3 + mx2x + 6) has (x − 2) as a factor and leaves a remainder r when divided by (x − 3), find the values of m and r.

#### Question 28:

If r and s be the remainders when the polynomials (x3 + 2x2− 5ax − 7) and (x3 + ax2− 12ax + 6) are divided by (x + 1) and (x − 2), respectively and 2r + s = 6, find the value of a.

#### Question 29:

Prove that:
$\left(a+b{\right)}^{3}\left(b+c{\right)}^{3}+\left(c+a{\right)}^{3}-3\left(a+b\right)\left(b+c\right)\left(c+a\right)=2\left({a}^{3}+{b}^{3}+{c}^{3}-3abc\right)$.

#### Question 30:

On a graph paper plot the following points:
A(3,3), B(2, 4), C(5, 5), D(0, 2), E(3, −3) and F(−5,−5).
Which of these points are the mirror images on (i) x-axis (ii) y-axis?

Points A(3,3), B(2,4), C(5,5), D(0,2), E(3,-3) and F(-5,-5)
are shown on the graph paper below.

(3,-3) is the mirror image of (3,3) on the x-axis .

#### Question 31:

In the given figure, in a âˆ†ABC, BEAC, ∠EBC = 40° and ∠DAC = 30°. Find the values of x, y and z.
Figure

#### Question 32:

In the given figure, ABC is a triangle in which AB = AC. D is a point in the interior of âˆ†ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC.
Figure

From (1), (2) and (3), we have:

#### Question 33:

In the given figure, ABCD is a square and EF is parallel to diagonal DB and EM = FM. Prove that: (i) BF = DE (ii) AM bisects BAD.
Figure

#### Question 34:

In the given figure, AB || CD. If BAE = 100° and ∠ECD = 120° then x = ?
Figure

Draw $EF\parallel AB\parallel CD$.
Then, $\angle AEF+\angle CEF=x°$
Now, $EF\parallel AB$ and AE is the transversal.

Again, $EF\parallel CD$ and CE is the transversal.

Therefore,

#### Question 1:

Mark (âœ“) against the correct answer in each of the following:
An irrational number between 2 and 2.5 is
(a) $\sqrt{3}$
(b) 2.3
(c) $\sqrt{5}$
(d) $2.\overline{)34}$

(c) $\sqrt{5}$

An irrational number between a and b is $\sqrt{ab}$.

#### Question 2:

Mark (âœ“) against the correct answer in each of the following:
Which of the following is a polynomial in one variable?
(a) x2 + x2
(b) $\sqrt{3x}+9$
(c) ${x}^{2}+2x-\sqrt{x}+3$
(d) $\sqrt{3}+2x-{x}^{2}$

(d) $\sqrt{3}+2x-{x}^{2}$

Here, the powers of x in all the terms are non-negative integers.
Therefore, it is a polynomial in one variable.

#### Question 3:

Mark (âœ“) against the correct answer in each of the following:
$\frac{1}{\left(\sqrt{18}-\sqrt{32}\right)}=?$
(a) $\sqrt{2}$
(b) $\frac{1}{\sqrt{2}}$
(c) $-\sqrt{2}$
(d) $\frac{-1}{\sqrt{2}}$

(d) $\frac{-1}{\sqrt{2}}$

#### Question 4:

Mark (âœ“) against the correct answer in each of the following:
If $p\left(x\right)=\left({x}^{4}-{x}^{2}+x\right)$, then $p\left(\frac{1}{2}\right)=?$
(a) $\frac{1}{16}$
(b) $\frac{3}{16}$
(c) $\frac{5}{16}$
(d) $\frac{7}{16}$

(c) $\frac{5}{16}$

#### Question 5:

Mark (âœ“) against the correct answer in each of the following:
If p(x) = x3 + x2 + ax + 115 is exactly divisible by (x + 5) then a = ?
(a) 8
(b) 6
(c) 5
(d) 3

(d) 3

p(x) = x3 + x2 + ax + 115 is exactly divisible by (x + 5).
Therefore, p($-$5) = 0

#### Question 6:

Mark (âœ“) against the correct answer in each of the following:
The equation of y-axis is
(a) y = 0
(b) x = 0
(c) y = 0
(d) y = constant

(b) x = 0

The equation of y-axis is x = 0.

#### Question 7:

Mark (âœ“) against the correct answer in each of the following:
In the given figure, the value of x is
(a) 10
(b) 12
(c) 15
(d) 20
Figure

(d) 20

#### Question 8:

Mark (âœ“) against the correct answer.
In the given figure, CE || BA and EF || CD. If BAC = 40°, ∠ACB = 65° and ∠CEF = x° then the value of x is
(a) 40
(b) 65
(c) 75
(d) 105
Figure

(d) 105

#### Question 9:

Factorise: $\sqrt{2}{x}^{2}+3x+\sqrt{2}$.

#### Question 10:

Prove that $\sqrt{5}$ is an irrational number.

We will show that $\sqrt{5}$ is irrational by contradiction.
Let us assume that $\sqrt{5}$ is rational.

Therefore we can represent it in the form of $\frac{a}{b}$.
$\sqrt{5}$ = $\frac{a}{b}$
Let us assume that $\frac{a}{b}$ is in its lowest form, i.e. a and b are co-prime.

Therefore, 5 is the factor of ${a}^{2}$ . If 5 is a factor of ${a}^{2}$, 5 will also be the factor of a.

So ,we can write a = 5c.

Therefore b2 is divisible by 5.
Since, it is divisible by 5, b will also be divisible by 5.

Therefore, a and b have at least 5 as a common factor.

But this contradicts that a and b are co-prime.

This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational.

Therefore, $\sqrt{5}$ is irrational.

#### Question 11:

Draw the graph of the equation y = 2x + 3.

To draw the graph of y = 2x + 3, make a table of values of x and y satisfying the given equation.
Now, we have

 x y 0 3 $-\frac{3}{2}$ 0

#### Question 12:

If $x=\left(3+\sqrt{8}\right)$, find the value of $\left({x}^{2}+\frac{1}{{x}^{2}}\right)$.

#### Question 13:

Find the area of the triangle whose sides measure 52 cm, 56 cm and 60 cm.

#### Question 14:

In the given figure, AB || CD. Find the value of x.
Figure

Draw $EF\parallel AB\parallel CD$.
Now,  is the transversal.

Also,
$EF\parallel CD$ and FD is the transversal.

Therefore,

#### Question 15:

Find the values of a and b so that the polynomial (x4 + ax3− 7x2 + 8x + b) is exactly divisible by (x + 2) as well as (x + 3).

#### Question 16:

Using remainder theorem, find the remainder when p(x) = x3− 3x2 + 4x + 50 is divided by (x + 3).

#### Question 17:

Factorise: (2x3 + 54).

#### Question 18:

Find the product (a bc) (a2 + b2 + c2 + ab + ac bc).

#### Question 19:

In âˆ†ABC, if ∠A − ∠B = 33° and ∠B − ∠C = 18°, find the measure of each angle of the triangle.

#### Question 20:

In the figure, in âˆ†ABC, the angle bisectors of ∠B and ∠C meet at a point O. Find the measure of ∠BOC.
Figure

#### Question 21:

In the given figure, AB || CD. If BAO = 110°, ∠AOC = 20° and ∠OCD = x°, find the value of x.
Figure

Draw $OE\parallel AB\parallel CD$.
Now,  is the transversal.

Also,
is the transversal.

#### Question 22:

In a right-angled triangle, prove that the hypotenuse is the longest side.

#### Question 23:

In the given figure, prove that:
x = α + β + γ.
Figure

Join BD to produce it to E.
Side BD of triangle ABD is produced to E.

Side BD of triangle BCD is produced to E.

#### Question 24:

Find six rational numbers between 3 and 4.

#### Question 25:

If $\frac{\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)}=a+\sqrt{15}b$, find the values of a and b.
OR
Factorise: $\left(5a-7b{\right)}^{3}+\left(9c-5a{\right)}^{3}+\left(7b-9c{\right)}^{3}$.

OR

#### Question 26:

Factorise: $12\left({x}^{2}+7x{\right)}^{2}-8\left({x}^{2}+7x\right)\left(2x-1\right)-15\left(2x-1{\right)}^{2}$.

#### Question 27:

If (x3 + ax2 + bx + 6) has (x − 2) as a factor and leaves a remainder 3 when divided by (x − 3), find the values of a and b.

#### Question 28:

Without actual division, show that (x3− 3x2 − 13x + 15) is exactly divisible by (x2 + 2x − 3).

#### Question 29:

Factorise: a3b3 + 1 + 3ab.

#### Question 30:

In the given figure, AB || CD, ECD = 100°, ∠EAB = 50° and ∠AEC = x°. Find the value of x.
Figure

Draw $EF\parallel AB\parallel CD$.
Now,  is the transversal.

Also,
is the transversal.

Now,

#### Question 31:

Prove that the bisectors of the angles of a linear pair are at right angles.

#### Question 32:

In the given figure, AD bisects BAC in the ratio 1 : 3 and AD = DB. Determine the value of x.
Figure

In the figure,

Now, divide $72°$ in the ratio 1 : 3.

Hence, the angles are 18o and 54o.

Given,
$AD=DB\phantom{\rule{0ex}{0ex}}⇒\angle DAB=\angle DBA=18°$

In $∆ABC$, we have:

#### Question 33:

In the given figure, AM BC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, find ∠MAN.
Figure

#### Question 34:

If the bisector of the vertical angle of a triangle bisects the base, prove that the triangle is isosceles.


text:
# The quantum plane

Update of May 27, 2017: I dug out some older unpublished slides authored in 2015 and early 2016. I added something about the quantum gap and something on the quantum plane at the very end. Here is the presentation, just spoken now.

## The quantum line

In one dimension, there is a natural compact metric space D on which one has a translation group which features a smallest unit. The group D of dyadic integers is the Pontyagin dual of the Pruefer group and is a natural “one-dimensional quantum space”. The real analogue is the circle, a compact topological group which is the dual of the integers and a quotient of the real numbers. The translation on D is called the adding machine. It is known in ergodic theory as the von Neumann-Kakutani system. Part of my thesis dealt with this system as it appears naturally in the context of renormalization of Jacobi matrices $L \mapsto L^2 + c$, a complex dynamical system in a non-commutative space. The non-commutative Julia sets constructed there produced interesting examples of almost periodic operators. What I did in my thesis is to get from any random Jacobi operator L on (X,T) a new Jacobi operator D on (X1,T1) such that D^2+c = L holds. The matrices one gets in the limit have the spectrum of the Julia set of the map $z -> z^2+c$. The matrices had already been introduced by G.A.Baker, M.L. Mehta, D.Bessis, P.Moussa in the 80ies. A new point in the thesis was to see the renormalisation on a non-commutative Banach algebra of operators, crossed products of a commutative Banach algebra defined by an automorphism. Ergodic theory so naturally introduces the group of dyadic integers just by renormalization: the renormalization map is the 2:1 integral extension. Take two copies X1,X2 of the probability space, then map using T from X1 to X2, and then use the identity to get back from X2 to X1. What happens is that the renormalization picture directly gives the nature of the self-similar operators. They are almost periodic matrices. After having worked on Barycentric refinements of complexes, a new point of view appeared: the dyadic group of integers is the Barycentric limit of a one dimensional complex. It is the unique attractor of the Barycentric refinement operation. Here is a simple illustration of the non-commutative Julia renormalization: we see the Kirchhoff matrix of a circular graph $C_8$:

$\left[ \begin{array}{cccccccc}2&-1&0&0&0&0&0&-1\\-1&2&-1&0&0&0&0&0\\ 0&-1&2&-1&0&0&0&0\\ 0&0&-1&2&-1&0&0&0\\0&0&0&-1&2&-1&0&0\\0&0&0&0&-1&2&-1&0\\0&0&0&0&0&-1&2&-1\\-1&0&0&0&0&0&-1&2\\ \end{array} \right]$

The matrix 4 B – B.B decomposes into two copies of the Kirchhoff matrix of C4:

$\left[\begin{array}{cccccccc} 2&0&-1&0&0&0&-1&0\\0&2&0&-1&0&0&0&-1\\-1&0&2&0&-1&0&0&0\\0&-1&0&2&0&-1&0&0\\0&0&-1&0&2&0&-1&0\\ 0&0&0&-1&0&2&0&-1\\-1&0&0&0&-1&0&2&0\\0&-1&0&0&0&-1&0&2\\\end{array}\right]$

Indeed, the Julia set of the quadratic map $T(z) = z(4-z)$ is the interval [0,4]. This picture gives the density of states in the Barycentric limit of a one dimensional simplicial complex. Apropos: here are two unusual photos of Julia and Fatou. They are from the book of Michele Audin “Fatou, Julia, Montel”, Springer Lecture Notes in Mathematics 2014. Click on the picture to see it large:

## Two dimensional quantum space

We tried to get a grip on higher dimensional versions here [PDF]. The following construction was done in thesis (page 250) We proved that the renormalization map for Z^d systems has a unique fixed point. Lets look at the construction in two dimensions:

In October 2015, we started to investigate the Barycentric limit in the 2 dimensional case. It is a renormalizaation story on pairs (T,S) of commuting measure preserving transformations. This generalizes the von Neumann-Kakutani case. First make integral extensions T_1^2=T, S_1^2=S, U^2 = ST = TS.

Definition. Let T,S be a pair of commuting automorphisms (measure preserving invertible maps) of a Lebesgue space, (a standard probability space) X. Define a new system by taking a union of 4 copies of X1,X2,X3,X4.

On X1, define T1: X1 -> X2     x -> Tx
On X1, define S1: X1 -> X3     x -> Sx
On X2, define T1: X2 -> X1     x -> x
On X2, define S1: X2 -> X4     x -> Sx
On X3, define T1: X3 -> X1     x -> x
On X3, define S1: X3 -> X4     x -> Tx
On X4, define T1: X4 -> X3     x -> Sx
On X4, define T2: X4 -> X2     x -> Tx

Now, $T1^2=T1$, $S1^2=S1$ and $T1 S1 = T2 S1$. We have a renormalization map $(X,T,S) \to (X1,T1,S1)$ of commuting maps. It is a contraction on Z2 actions if we take the metric $d( (T,S),(U,V)) = m( \{ x \; |\; T(x) \neq U(x) or S(x) \neq V(x) \}$. The limiting system is a compact topological group with a $Z^2$ action having a unique invariant measure. There is an additional symmetry B, the Barycentric refinement. Still we struggle to get from that space pictures of Barycentrically refined geometric spaces. It seems that we have to introduce on each level more and more transformations (defined by the two basic transformations T,S). Here is one attempt to get a grip on the refinement step:

## A martingale idea

But that produces a mess of transformations. It would be nicer to have only two transformations and look at their orbits. Depending on the sigma-alrebra used, we would get the individual Barycentric refinements. This is kind of a Martingale idea: play with the $\sigma$– algebras rather than with the transformations. Here is the picture: we always work only with two transformations S,T and a helper transformation U=ST=TS. What happens after the first step is that we have now 6 triangles. The central face has degree 6. On the edge, say Tx, the outgoing transformation T gets us back to x because the sigma-algebra tells us so. On the next level each of the triangles will again be refined:

Here is an other doodle. It might be better to renormalize the Z2 action with tripling the probability space each time:


text:
# Weakly reversible and deficiency zero realizations: Structural characterization and uniqueness

Időpont:
2022. 11. 29. 17:15
Hely:
Weakly reversible and deficiency zero reaction networks form the main cast of the Deficiency Zero Theorem. At the same time, it has been recognized that to have the dynamics of a complex-balanced system, $(G,k)$ may not itself be complex-balanced, but is instead dynamically equivalent to a complex-balanced system $(G*,k*)$. In general, there may be multiple complex-balanced realizations for a given ODE. We proved that there can be at most one weakly reversible and deficiency zero realization, and uniqueness no longer holds when either assumption is dropped. From our result, we also obtained a simple algorithm for whether such a realization exists.


text:
# Vector grad as a normal vector

i have done a little on multivariable calculus in school -partial derivates and tangent planes to f(x,y), but now we have moved on to functions of more than 2 variables

my teacher, however, doesnt really teach us - she gives us equations and tells us to do questions and i usually find myself having to read the textbook to convince myself that i understand the equations. shes only part time so she is never about to help me at lunch time, so if you guys could help me that would be great

my textbook wasnt able to convince me why the vector grad is $$\nabla _{g}$$ is normal to the surface - surely if it has the same direction as the surface in x, y and z then it cant be going away from the surface at an angle of pi/2 rad? :S

i know that:

$$\nabla _{g}$$ = < $$\frac{\partial{g}}{\partial{x}}$$ , $$\frac{\partial{g}}{\partial{y}}$$ , $$\frac{\partial{g}}{\partial{z}}$$ >

but its not clear to me why this should be perpendicular to the surface. would you be able to help justify why it is? i dont know anything about vector spaces and stuff like that, because a level further maths is pretty limited.

Last edited:

arildno
Homework Helper
Gold Member
Dearly Missed
g(x,y,z) is no surface, but g(x,y,z)=CONSTANT is a surface!

And, the gradient of g at a point on this surface (called a level surface), indicates the direction along which g increases most swiftly.
Clearly, the direction that g increases most swiftly must be orthogonal to the direction(s) along which g retains its value (i.e, along the level surface g=CONSTANT).

Thus, the gradient of g is orthogonal to the surface g=constant.

g(x,y,z) is no surface, but g(x,y,z)=CONSTANT is a surface!

And, the gradient of g at a point on this surface (called a level surface), indicates the direction along which g increases most swiftly.
Clearly, the direction that g increases most swiftly must be orthogonal to the direction(s) along which g retains its value (i.e, along the level surface g=CONSTANT).

Thus, the gradient of g is orthogonal to the surface g=constant.

oh yeah sorry i meant that it was = constant.

im not sure i understand why it represents the direction along which the gradient increases most swiftly.. surely it just represents a vector which is travelling in the same direction as the surface in the x y z directions, if it has the same slope in all 3 directions?

arildno
Homework Helper
Gold Member
Dearly Missed
surely it just represents a vector which is travelling in the same direction as the surface in the x y z directions, if it has the same slope in all 3 directions?
Nope.
At every point (x,y,z) (incluing, therefore, every point ON a particular surface), the gradient tells us the direction in which g(x,y,z) increases most swiftly.

This is easily seen as follows.
Let h(t) be the function:
$$h(t)=g(\vec{x}_{0}+t*\vec{n}), \vec{x}_{0}=(x_{0},y_{0},z_{0}), \vec{n}=(n_{x},n_{y},n_{z}),||\vec{n}||=1$$
where $\vec{x}_{0}$ is some arbitrary fixed point, and $\vec{n}$ is some unit vector in some fixed direction.
h(t) is therefore an evaluation of g along a straight line going through $\vec{x}_{0}$.

Now, we have, by the chain rule (evaluated at t=0):
$$\frac{dh}{dt}_{t=0}=\nabla{g}_{\vec{x}_{0}}\cdot\vec{n}=||\nabla{g}_{\vec{x}_{0}}||\cos\theta_{n}$$
where the angle is between the line we are following and the direction of the gradient at our point.
This expression is maximized when $\vec{n}$ is parallell to the gradient of g.

If you, at some point walks along a level surface, the value of g won't change at all.
Thus, the directions along the level surface cannot coincide with the direction along which g increases most strongly; they have to be orthogonal to the gradient direction.

lavinia
Gold Member
df is zero on tangent vectors to the surface because f is constant on the surface. The directional derivative of f along the surface is zero. If you want to express this with an inner product with a vector then this vector is by definition perpendicular to the surface. If you use the standard metric on Euclidean space, this vector is the gradient.

LCKurtz
Homework Helper
Gold Member
I'll throw in 2 cents worth too. Your surface is g(x,y,z) = c. Suppose you have a smooth curve on the surface parameterized as

$$\vec R(t) = \langle x(t),y(t),z(t)\rangle$$

Then g(x(t),y(t),z(t)) ≡ 0 as a function of t. Therefore its derivative with respect to t must be identically 0. We have, by the chain rule,

$$g'(t)=g_x\frac{dx}{dt}+ g_y\frac{dy}{dt}+g_z\frac{dz}{dt}$$
$$= \langle g_x,g_y,g_z\rangle \cdot \langle \frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\rangle=\nabla g \cdot \frac{d\vec R}{dt} = 0$$

This tells you that the gradient of g is perpendicular to R'(t). But R'(t) is tangent to the curve R(t), which lies on the surface. Think of all the curves on the surface passing through a particular point P on the surface. The gradient at P is perpendicular to all these curves. The only way that could happen is if it is perpendicular to the surface itself.


text:
# Group order and left cosets

Let $G=S_3 \times \mathbb{Z}_4$ be a group and $H= \langle(1 2 3)\rangle \times \langle2\rangle$ and $K=\langle(1)\rangle \times \langle2\rangle$ be subgroups.
Find the order of $G$, $H$ and $K$.
Find the left cosets of $H$ in $G$ AND $K$ in $H$.

• Where are you stuck? – lhf Dec 5 '13 at 2:13
• i don't know how to start – mariert Dec 5 '13 at 2:17
• I know that S_3= (1)(12)(13)(23)(123)(321) and Z_4= {0, 1,2,3 } – mariert Dec 5 '13 at 2:19

The direct product of two groups $H$ and $K$ has the underlying set $$\{(h,k):h \in H \text{ and } k \in K\}$$ which has order $|H \times K|=|H|\,|K|$.

In this case, $S_3$ is the symmetric group on $3$ elements, so has order $3!=6$, and $\mathbb{Z}_4$ is the cyclic group of order $4$.

The notation $\langle h \rangle$ around a group element $h$ describes the subgroup generated by $h$. In this case, we have $\langle (123) \rangle=\{\mathrm{id},(123),(123)^2\}$ and $\langle \mathrm{id} \rangle=\{\mathrm{id}\}$, subgroups of $S_3$, and $\langle 2 \rangle=\{0,2\}$, a subgroup of $\mathbb{Z}_4$.

Cosets of subgroups are their "translates". Formally, (and using multiplicative notation) if $H$ is a subgroup of $G$, then the cosets are $gH=\{gh:h \in H\}$ for $g \in G$.

So, for example, the subgroup $\langle 2 \rangle$ of $\mathbb{Z}_4$ has the cosets $0+\langle 2 \rangle=\{0,2\}$ and $1+\langle 2 \rangle=\{1,3\}$ (we use additive notation for this group).

Your task is to apply these to the specific groups in the question. It's probably a difficult task if you haven't played around with these properties beforehand to get a feel for how they work.


text:
## Video and Examples

Writing ratios in simplest form. $\frac{12}{15}\,\rightarrow~\,\frac{4}{5}$

 In this figure, it is easy to see that the ratio of yellows to greens is $\frac{6}{8}$. If we put circles around each column of tiles we can see that the ratio of yellow columns to green columns is $\frac{3}{4}$. Since the number of tiles never changed - only how we looked at them - this shows that the ratios $\frac{6}{8}$ and $\frac{3}{4}$ are equivalent to each other. In other words, $\frac{6}{8}\,=\,\frac{3}{4}$ Ratios can be reduced just like fractions!

 Example 1 There are 12 boys and 16 girls in a class. What is the ratio of boys to girls in simplest form? What is the ratio of boys to girls to total in simplest form?

 Example 2 What is the ratio of spoons to glasses in simplest form? [show answer]   $\large\frac{\mbox{spoons}}{\mbox{glasses}}=\frac{2}{6}=\frac{1}{3}$

Ratios can be reduced or scaled up just like fractions!

Directions:

1. The ratio is randomly created and plotted on the graph.
2. Reduce the blue ratio.
3. Use the sliders to create the ratio in simplest form. Point B moves as you create the ratio.
4. The ratio in simplest form will ALWAYS lie somewhere on the dotted line.

# Self-Check

Q1: What is the ratio of yellows to the total number of tiles in simplest form? [show answer]

Q2: In the word "PROPORTION" what is the ratio of consonants to vowels in simplest form? [show answer]

Q3: The chess club at school has 18 boys and 15 girls in it. What is the ratio of girls to boys in simplest form? [show answer]


text:
## Chemistry: The Central Science (13th Edition)

$mass(Fe) = 244.62g$ $mass(CO_2) = 289.08g$
Balanced equation: (3.64a) $Fe_2O_3 + 3CO --> 2Fe + 3CO_2$ 1. Convert the mass of ($Fe_2O_3$) to nº of moles: $mm(Fe_2O_3) = 55.85*2 + 16*3 = 159.7g/mol$ $n(moles) = \frac{mass(g)}{mm}$ $n(moles) = \frac{350}{159.7}$ $n(moles) = 2.19$ 2. Use the proportions of the balanced equation to find the number of moles of $Fe$ and $CO_2$ formed: $Fe$: $\frac{1}{2} = \frac{2.19}{x}$ $x = 2*2.19 = 4.38 moles$ $CO_2$: $\frac{1}{3} = \frac{2.19}{y}$ $y = 3*2.19 = 6.57moles$ 3. Convert these numbers to grams: $Fe:$ $mm(Fe) = 55.85g/mol$ $mass(g) = mm * n(moles)$ $mass(g) = 55.85 * 4.38$ $mass(g) = 244.62g$ $CO_2:$ $mm(CO_2) = 12*1 + 16*2 = 44g/mol$ $mass(g) = mm * n(moles)$ $mass(g) = 44 * 6.57$ $mass(g) = 289.08g$


text:
# format numbers in reference so they sort correctly

This is a followup to question list of occurences of references. In the MWE below the index lists the Goals in alphabetic order where numeric is needed for the second token:

The solution I found (on SE - sorry I lost the link) for the chapter and section numbers for the subsidiary indexing doesn't seem to work for the main headings. I tried a few others, to no avail.

Edit: the solution to convert outcome of reference to regular number looks as if it ought to work but I tried something like that and it failed.

\documentclass[12pt]{book}

\usepackage{amsthm}
\usepackage{environ}
\usepackage{imakeidx}
\usepackage{refcount}
\usepackage{hyperref}

\makeindex[intoc]
\makeindex[name=goals,title=Goal References,columns=1,intoc]

\newcommand{\gref}[1]{%
[Goal~\ref{goal:#1}]
\index[goals]{Goal~\getrefnumber{goal:#1}!Section
\ifnum\value{chapter}<10 0\fi\arabic{chapter}.%
\ifnum\value{section}<10 0\fi\arabic{section}}
}

\makeatletter
\newcommand\l@goal[1]{%
\vskip \z@ \@plus \p@
\begingroup
\parindent\z@
\rightskip\@pnumwidth
\parfillskip-\@pnumwidth
\leavevmode #1\nobreak\hfil\nobreak\null\par
\penalty\@highpenalty
\endgroup}

% arrange goal numbering by chapter
\newtheorem{Goal}{Goal}[chapter]

% define a shorthand to be able to get at \BODY with \expandafter
\newcommand\indexgoal[1]{\goalindex[goals]{Goal~\theGoal! #1}}

% save a copy of \index
\let\goalindex\index
\NewEnviron{goal}[1]{%
\begin{Goal}
\label{goal:#1}%
% here we neutralize \index so that it won't do damages
\begingroup
\renewcommand\index[2][]{}%
% but the main command uses \goalindex, so it's safe
\expandafter\indexgoal\expandafter{\BODY}%
\endgroup
%{\em label:} #1 %comment in/out to restore/suppress printing label
\normalfont{}\noindent
\BODY
}
[\end{Goal}]
\makeatother

\newenvironment{goals}
{%
{\textbf Chapter goals:}
}
{}

\begin{document}

\tableofcontents{}

\setcounter{chapter}{1}
\chapter{TWO}

\tracingmacros=1

\begin{goals}

\begin{goal}{twofirst}
\end{goal}

\end{goals}

\tracingmacros=0

\begin{goal}{twosecond}
spend time on content, not \LaTeX
\end{goal}

\section{two-one}
stuff\index{stuff}
Reference to a chapter 11 goal: \gref{elevenfirst}

\setcounter{chapter}{10}
\chapter{ELEVEN}

\begin{goal}{elevenfirst}
have fun in any case\index{fun}
\end{goal}

Chapter contents here, referring to some goals:

\setcounter{section}{1}

\section{whatever}

Reference to a chapter 2 goal: \gref{twosecond}

Reference to a chapter 11 goal: \gref{elevenfirst}

\setcounter{section}{10}
\section{another}

Second reference to a chapter 2 goal: \gref{twofirst}

\printindex[goals]
\printindex

\end{document}

-
Would you please explain better the problem? –  egreg Jul 17 '12 at 20:33
@egreg The goals from chapter 11 (11.1, 11.2 etc.) come before those from chapter 2 (2.1, 2.2 etc.) since 11 precedes 2 in alphabetic (but not numerical) order. If there were more than 9 goals in any single chapter I would predict similar misbehavior (but haven't tested in the MWE). –  Ethan Bolker Jul 17 '12 at 23:33

You have to format the entries in such a way that the alphabetization is correct. So I use a trick:

\documentclass[12pt]{book}

\usepackage{amsthm}
\usepackage{environ}
\usepackage{imakeidx}
\usepackage{refcount}
\usepackage{hyperref}

\makeindex[intoc]
\makeindex[name=goals,title=Goal References,columns=1,intoc]

\makeatletter
\newcommand{\gref}[1]{%
[Goal~\ref{goal:#1}]
\edef\@goal@number{\getrefnumber{goal:#1}}%
\edef\@the@section{\thesection}%
\index[goals]{Goal~\expandafter\sortgref\@goal@number.0\sortgref!%
\expandafter\sortgref\@the@section.0\sortgref @%
Section~\thesection}%
}
\def\sortgref#1.#2#3\sortgref{%
\ignoresort{\ifnum#1<10 0\fi#1.\ifnum#2<10 0\fi#2}#1.#2%
}
\protected\def\ignoresort#1{}

\newcommand\l@goal[1]{%
\vskip \z@ \@plus \p@
\begingroup
\parindent\z@
\rightskip\@pnumwidth
\parfillskip-\@pnumwidth
\leavevmode #1\nobreak\hfil\nobreak\null\par
\penalty\@highpenalty
\endgroup}

% arrange goal numbering by chapter
\newtheorem{Goal}{Goal}[chapter]

% define a shorthand to be able to get at \BODY with \expandafter
\newcommand\indexgoal[1]{\edef\@goal@number{\theGoal}%
\goalindex[goals]{Goal~\expandafter\sortgref\@goal@number.0\sortgref! #1}}

% save a copy of \index
\let\goalindex\index
\NewEnviron{goal}[1]{%
\begin{Goal}
\label{goal:#1}%
% here we neutralize \index so that it won't do damages
\begingroup
\renewcommand\index[2][]{}%
% but the main command uses \goalindex, so it's safe
\expandafter\indexgoal\expandafter{\BODY}%
\endgroup
%{\em label:} #1 %comment in/out to restore/suppress printing label
\normalfont{}\noindent
\BODY
}
[\end{Goal}]
\makeatother

\newenvironment{goals}
{%
{\textbf Chapter goals:}
}
{}

\begin{document}

\tableofcontents{}

\setcounter{chapter}{1}
\chapter{TWO}

\tracingmacros=1

\begin{goals}

\begin{goal}{twofirst}
\end{goal}

\end{goals}

\tracingmacros=0

\begin{goal}{twosecond}
spend time on content, not \LaTeX
\end{goal}

\section{two-one}
stuff\index{stuff}
Reference to a chapter 11 goal: \gref{elevenfirst}

\setcounter{chapter}{10}
\chapter{ELEVEN}

\begin{goal}{elevenfirst}
have fun in any case\index{fun}
\end{goal}

Chapter contents here, referring to some goals:

\setcounter{section}{1}

\section{whatever}

Reference to a chapter 2 goal: \gref{twosecond}

Reference to a chapter 11 goal: \gref{elevenfirst}

\setcounter{section}{10}
\section{another}

Second reference to a chapter 2 goal: \gref{twofirst}

\printindex[goals]
\printindex

\end{document}

The in goals.idx you'll find

\indexentry{Goal\nobreakspace  {}\ignoresort {02.01}2.1! learn more \LaTeX  |hyperpage}{3}
\indexentry{Goal\nobreakspace  {}\ignoresort {02.02}2.2! spend time on content, not \LaTeX  |hyperpage}{3}
\indexentry{Goal\nobreakspace  {}\ignoresort {11.01}11.1!\ignoresort {02.01}2.1@Section\nobreakspace  {}2.1|hyperpage}{3}
\indexentry{Goal\nobreakspace  {}\ignoresort {11.01}11.1! have fun in any case\index {fun}|hyperpage}{5}
\indexentry{Goal\nobreakspace  {}\ignoresort {02.02}2.2!\ignoresort {11.02}11.2@Section\nobreakspace  {}11.2|hyperpage}{5}
\indexentry{Goal\nobreakspace  {}\ignoresort {11.01}11.1!\ignoresort {11.02}11.2@Section\nobreakspace  {}11.2|hyperpage}{5}
\indexentry{Goal\nobreakspace  {}\ignoresort {02.01}2.1!\ignoresort {11.01}11.1@Section\nobreakspace  {}11.11|hyperpage}{5}

so that the sorting will be done correctly: here is goals.ind

\begin{theindex}

\item Goal\nobreakspace  {}\ignoresort {02.01}2.1
\subitem Section\nobreakspace  {}11.11, \hyperpage{5}
\item Goal\nobreakspace  {}\ignoresort {02.02}2.2
\subitem Section\nobreakspace  {}11.2, \hyperpage{5}
\subitem  spend time on content, not \LaTeX  , \hyperpage{3}
\item Goal\nobreakspace  {}\ignoresort {11.01}11.1
\subitem Section\nobreakspace  {}2.1, \hyperpage{3}
\subitem Section\nobreakspace  {}11.2, \hyperpage{5}
\subitem  have fun in any case\index {fun}, \hyperpage{5}

\end{theindex}

The \ignoresort macro does nothing with its argument, so the entries will be printed without padding zeros.

-


text:
# Higher Green functions and their CM values

Posted in
Speaker:
Jan Bruinier
Affiliation:
Date:
Wed, 2018-09-05 10:15 - 11:00
Location:
MPIM Lecture Hall

The automorphic Green function for a modular curve $X$ is a function on
$X\times X$ with a logarithmic singularity along the diagonal which is a
resolvent kernel of the hyperbolic Laplacian. It plays an important role
in the analytic theory of automorphic forms and in the Arakelov geometry
of modular curves. Gross and Zagier conjectured that for positive
integral spectral parameter $s$ the values at CM points of certain linear
combinations of Hecke translates of this Green function are given by
logarithms of algebraic numbers in suitable class fields.
In certain cases this conjecture was proved by Mellit and Viazovska. We
report on joint work with S. Ehlen and T. Yang in which we establish new
cases and a generalization of the conjecture to orthogonal groups of
signature $(n,2)$.

 © MPI f. Mathematik, Bonn Impressum & Datenschutz


text:
Thread: I need help with this word problem in calculus. pretty please

1. I need help with this word problem in calculus. pretty please

The equation for free fall at the surface of the moon is s=31.8t^2in with t in sec. Assume a rock is dropped from the top of a 12,000 in cliff. Find the speed of the rock at t=4sec. Iḿ not sure if Iḿ over complicating things, but I do not know if this is plug or chug or if it is more complicated than that, because we have been working on limits and this has nothing to do with this.

2. Re: I need help with this word problem in calculus. pretty please

What do you get if you differentiate s with respect to t? What does this represent?

3. Re: I need help with this word problem in calculus. pretty please

Originally Posted by wbarner
The equation for free fall at the surface of the moon is s=31.8t^2in with t in sec. Assume a rock is dropped from the top of a 12,000 in cliff. Find the speed of the rock at t=4sec. Iḿ not sure if Iḿ over complicating things, but I do not know if this is plug or chug or if it is more complicated than that, because we have been working on limits and this has nothing to do with this.
The equation $s=31.8t^2\text{ in}$ means that at time $t$, the rock is at position $s$ inches. Let's determine how far the rock travels in 4 seconds (if it has hit the bottom of the cliff, that will certainly affect its speed).

$$s = 31.8(4)^2 = 508.8\text{ in} < 12,000\text{ in}$$

So, the rock is still falling after four seconds. Next, we have the point $(4,508.8)$. If we consider another point, $(t,s) = (t,31.8t^2)$, the average speed of the rock between those two points is the slope of the line between those two points. It is the change in distance over the change in time. This is written as $\dfrac{\Delta s}{\Delta t}$. The closer we get the second point $(t,s)$ to the point $(4,508.8)$, the closer this average becomes to the instantaneous rate of change at $t=4$. In other words, we are looking for the following limit:

$$\lim_{t \to 4} \dfrac{s-508.8}{t-4} = \lim_{t \to 4} \dfrac{31.8t^2-508.8}{t-4}$$

Can you solve it from here?