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IMR Press / JIN / Volume 21 / Issue 1 / DOI: 10.31083/j.jin2101005
Open Access Original Research
Development and validation of anti-human Alpha synuclein DNA aptamer using computer modelling techniques—an in silico study
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1 Human Sciences Research Centre, University of Derby, DE22 1GB Derby, UK
2 Department of Science & Engineering, Novel Global Community Educational Foundation, 2770 Hebersham, Australia
3 AFNP Med Austria, 1010 Wien, Austria
4 Department of Human Anatomy, College of Medicine, Taif University, P.O. BOX 11099, 21944 Taif, Saudi Arabia
5 Department of Pharmacology & Therapeutics, Faculty of Veterinary Medicine, Damanhour University, 22511 Damanhour, AlBeheira, Egypt
*Correspondence: <EMAIL> (<NAME>)
J. Integr. Neurosci. 2022, 21(1), 5; https://doi.org/10.31083/j.jin2101005
Submitted: 1 August 2021 | Revised: 26 August 2021 | Accepted: 18 September 2021 | Published: 21 January 2022
This is an open access article under the CC BY 4.0 license.
Abstract

Background: Biomarker detection strategies have, in recent years, been moving towards nucleic acid-based detection systems in the form of aptamers, short oligonucleotide sequences which have shown promise in pre-clinical and research settings. One such aptamer is M5-15, a DNA aptamer raised against human alpha synuclein ($\alpha$-syn) the causative agent in Lewy body and Parkinson’s disease (PD) associated dementia. While this aptamer has shown promise, in silico methodologies have demonstrated a capacity to produce aptamers that have higher affinities for their targets than in vitro generated sequences. Methods: A Python script random generated library of DNA sequences were screened based on their thermodynamic stability with the use of DINAMelt server-QuickFold web server. The selected sequences were examined with MFold in order to generate secondary structure data that were used to produce 3D data with the use of RNA composer software. Further on, the structure was corrected and RNA was replaced with DNA and the virtual screening for $\alpha{}$-syn aptamer took place with a series of molecular docking experiments with the use of CSD-Discovery-GOLD software. Results: Herein we propose an alternative in silico generated aptamer we call TMG-79 which demonstrates greater affinity for the target compared to M5-15 (M5-15 = –15.9 kcal/mol, TMG-79 = –17.77 kcal/mol) as well as better ChemPLP fitness scoring between the top poses (M5-15 = 32.33, TMG-79 = 53.32). Structural analysis suggests that while there are similarities, the greater potential flexibility of TMG-79 could be promoting greater affinity for the $\alpha$-syn compared to M5-15. Conclusions: In silico methods of aptamer generation has the potential to revolutionise the field of aptamer design. We feel that further development of TMG-79 and validation in vitro will make it a viable candidate for diagnostic and research use in the future.

Keywords
DNA aptamers
In silico design
Computer aided design
Alpha synuclein
Biomarker detection
1. Introduction

Biomarker detection has become a critical component in the diagnosis of disease and has emerged as a critical factor in deciding courses of treatment of patients. Over 30 years ago, the first monoclonal antibody was developed for biomarker detection with the first licensed products being deployed in 1986 [1]. Global market value for monoclonal antibodies surpassed $\$95 billion in 2017, with an expectation that it will reach $\$131 billion by 2023 [2]. High specificity and utility of monoclonal antibodies in both research and clinical settings fuels the demand for these detection strategies [3]. While monoclonal antibodies are still considered the gold standard in both clinical and research settings, being involved in a number of key molecular biology techniques like western blotting and immunofluorescence microscopy [4, 5], there are a number of drawbacks in relation to the technology. Monoclonal antibodies are exceptionally expensive to produce and purchase depending on the target protein, ranging between €53/g and €84/g depending on the production method chosen [6]. Reliance on animal models to produce these antibodies, which has been reduced to an extent due to the widespread adoption of hybridoma production strategies, also introduces problems. Species specific glycosylation of the antibodies can lead to batch-to-batch variations in antibody affinity [7] and the variable fragment crystallisable (fc) domain derived from the host species can cause immunogenicity problems in clinical settings [8]. Despite these issues, monoclonal antibodies have seen widespread adoption, but in recent years industry and individual researchers have been looking at potential alternative strategies for biomarker detection. Aptamers have emerged as a potential candidate to replace monoclonal antibodies.

1.1 Aptamers

Aptamers are defined as synthetic ligands composed of single stranded RNA or DNA which have unique structures that can be designed to target specific biomolecules with high affinity [9]. The production of aptamers is achieved through a process of systematic evolution of ligands by exponential enrichment (SELEX) and was discovered by three groups independently at the beginning of the 1990s, beginning with RNA aptamers then expanding into DNA and peptide-based aptamers (referred to as affimers) developed today [10, 11, 12]. Nucleic acid-based aptamers are single stranded nucleic acid structures which can form secondary structures by folding in on themselves into thermodynamically stable shapes [13]. These structures consist of double helices, and hair-pin structures of varying sizes governed by the number of mis-match pairing of bases which produce internal loops and three-way helical regions [14]. While SELEX has been used previously for the development of aptamers, there have been movements towards the use of in silico computer-based modelling to streamline the production and design of aptamers [15]. Techniques like M.A.W.S (making aptamers without SELEX) are the most notable recent examples of in silico aptamer development and has been seen as a gold standard for the development of aptamers in silico [16]. The focus of aptamer generation in silico looks to remove some of the initial high costs associated with aptamer selection, as well as looking to produce high affinity aptamers prior to the purchase of reagents needed for in vitro validation of aptamer binding [17].

One of the main advantages of aptamers over their monoclonal antibody counterparts is tissue penetration, as well as a capability to be raised against targets that do not elicit an immune response, such as charged ions, intracellular proteins and vitamins [18]. The use of aptamers has emerged as a potential detection strategy for a wide number of biomarkers, with particular interest in the detection of markers for neurological disease [19]. In particular, alpha synuclein ($\alpha$-syn) has emerged as a candidate in the diagnosis of neurological diseases such as Parkinson’s (PD) and Alzheimer’s (AD).

1.2 Alpha synuclein ($\alpha$-syn)

$\alpha$-syn is a 140 amino acid protein encoded by the SNCA gene located on chromosome 4 and has been implicated in the progression of Lewy body, Alzheimer’s and Parkinson’s disease associated dementia [20]. Under physiological conditions, $\alpha$-syn is maintained in an unfolded state but in disease states it is capable of forming toxic oligomers and aggregates with local proteins and has been linked to toxicity via mitochondrial damage, inflammation and axonal transport dysfunction (Fig. 1, Ref. [21]) [22].

Fig. 1.

Summary of the interactions in AD. Notably $\alpha$-syn is associated directly with synapse loss and neurodegeneration in AD, however this is separate from the formation of amyloid plaques which are typically associated with the condition as described in [21].

The link between AD and $\alpha$-syn has been established, however the link with PD has not been fully confirmed but has the potential to be used to establish differences in diagnosis with lower cerebrospinal fluid (CSF) levels being detected in PD compared to AD [23]. Efforts to prevent the aggregation of $\alpha$-syn, and by extension alleviate symptoms of AD, have focused on the use of aptamers due to their inherent ability to penetrate the blood brain barrier. Several aptamers have been identified which have successfully targeted $\alpha$-syn and have demonstrated mitochondrial protective effects as well promoting $\alpha$-syn degradation in vitro [22, 24, 25]. The approaches taken when producing anti $\alpha$-syn aptamers has been the traditional SELEX process, with minor modifications that have been incorporated over time to improve the methodology. With the advances in computer simulations and molecular docking experimentation, interest in improving the experimentally derived aptamers has increased exponentially.

1.3 In silico design

The concept of computer aided drug design, and by extension in silico aptamer design, has been gaining traction in recent years, specifically due to its ability to streamline the design of innovative treatment and detection strategies for a wide variety of applications. Advances in computer algorithms capable of predicting the folding characteristics of single stranded DNA, as well as increasing availability of these software, have been critical to the development of in silico methods [26]. The standardised approach to in silico aptamer design is summarised in Fig. 2 (Ref. [27]).

Fig. 2.

Summary of steps within in silico aptamer design. RNA folding algorithms have proven to be effective at predicting DNA aptamer structure despite minor chemical differences. It is common practice to correct these structures in post. Image based on information contained within [27].

Sequence generation can be upscaled to be comparable with in vitro SELEX, allowing for the screening of an exceptionally large number of sequences. The difficulties come when establishing selection criteria for the computer-generated sequences, and which ones to proceed with in a docking experiment as the large number of sequences make it impractical to dock every aptamer produced. A common approach is to look at the free energy potential of the aptamers secondary structure [26] and secondary structural motifs [28] to screen potential aptamer candidates. The second option is particularly useful when looking at aptamer optimization studies, as the original sequence can be mutated at specific points to see what impact this will have on overall binding.

Following sequence generation, aptamers are then docked to the target molecule in question. There are a number of programmes which are available for use in this field, some which are freely available and some requiring a subscription [28]. Irrespective of the programme used, the objective is to create a pose (two molecules which have formed a complex) with the lowest Gibbs free energy ($\Delta{}$G) binding energy [27]. Experiments themselves can take two main approaches, fixed or flexible docking which can be further sub-divided into 5 categories. The approaches are summarised in Table 1 [29].

Table 1.Summary of docking common docking approaches in silico design. Information in the table is based on [29].
 Docking approach Key characteristic Rigid docking - fixed receptor and ligand Both receptor and ligand bond angles and atomic positioning are fixed. This mimics the traditional “lock and key” theory for ligand/receptor interactions. Is the least resource intensive out of the approaches, but does not take into consideration structure flexibility. It also limits understanding of the solvents impact on docking “Soft docking” - rigid receptor, semi-flexible ligand Small degree of flexibility within a small area for the ligand, the receptor remains fixed “Side chain flexibility” - rigid receptor, side chain flexibility only Majority of the structures (both ligand and receptor) are fixed in position, with only the amino acid side chains able to flex Composite receptor modelling - flexible receptor Combined structural information breaking up larger receptors into smaller subunits. These subunits are then contained within a single docking “box”. Allows for degree of flexibility in the receptor while minimising computer load Ensemble docking - Flexible receptor Generation of a composite receptor structure from multiple components, with each subgroup capable of rotating freely through its structure. Ligand can be either fixed or flexible in these scenarios. Critically in this instance they are not confined to a single “box”, but are made up of multiple components that work independently. This is a processor intensive position and requires time and resources to run effectively

In silico docking is not without its limitations however, despite the advances in computer modelling and DNA/RNA folding algorithms. Issues with computer processing load [27], particularly with more flexible approaches to docking, as well as the difficulties in accurately mimicking the electrochemical environment these molecules would experience in vivo [30] do mean that at this stage, predictions need to be validated in vitro before definitive conclusions can be made. However, this does not detract from the usefulness of in silico methods for they have the potential to both streamline and improve the aptamer generation process.

1.4 Aims

In this study, an experimentally produced aptamer was docked alongside an in silico developed aptamer against human $\alpha$-syn. The intention of this study was to demonstrate a novel method by which aptamers can be developed in silico and validated against previously observed interactions, with the hope of refining that aptamer to improve the binding efficiency to its target. To our knowledge, this is the first investigation looking into the refinement of an anti-human $\alpha$-syn DNA aptamer to produce a structure with a predicted higher binding affinity to the experimentally derived sequence.

2. Methods
2.1 Generation and selection of sequences and structures

Aptamer sequences were generated using Python script to create a library of 1000 sequences with an N30 wobble region flanked by constant primer binding regions (ATAGTCCCATCATTCATT-N30-AGATATTAGCAAGTGTCA). Sequences were retained following prediction of thermodynamic stability using the DINAMelt server-QuickFold web server [31]. Retained sequences were required to have negative $\Delta{}$G values, using the parameters listed in Table 2.

Table 2.DINAMelt server-QuickFold parameters for 2D sequence generation.
 Folding temperature 25 ${}^{\circ}$C Ionic conditions [Na${}^{+}$] = 150 mM [Mg${}^{++}$] = 0 mM Distance between paired bases No limit on distance Sequence type Linear Maximum possible foldings 50

Secondary structural information for 50 structures, selected based on their $\Delta{}$G, was obtained using MFold [32]. Structure prediction in MFold used the same parameters as were used within the DINAMelt server-QuickFold.

2.2 Creation of 3D structures using RNAcomposer

Vienna sequence outputs from MFold were used to assist in the production of 3 dimensional models using RNAcomposer [33]. To account for this being a programme which folds RNA, all thymine residues within the sequences were replaced with uracil groups for the purpose of folding. Vienna sequences were input along with the base sequence data, with the top 50 sequences being run in batch format within RNAcomposer. Structure prediction used default parameters for the purposes of folding.

Protein data bank (PDB) output files of the predicted sequences were converted into mol2 format using CSD-Discovery-Mercury [34]. Structures were then modified to reintroduce the deoxyribose groups, replacing the ribose sugar groups within their structures using PyMol [35] via command line editing within the software (Schrödinger, New York, USA). Commands were set to remove O2’ and HO2’ atoms from all residues and to add a hydrogen to the C2’ atom of each residue. Uracil groups were replaced with thymine groups using the mutagenesis wizard also contained within the PyMol software. Files were then exported back into CSD-Discovery-Mercury to add any hydrogens to the structure. In addition, bond angles and types were corrected as some had been incorrectly modelled during the file conversion process. Of the 50 structures that were originally generated, the top 10 were selected to dock with the target molecule. Primary structural similarity to M5-15 was used as the primary criteria when selecting the aptamer, followed by thermodynamic stability. This is due in part to the nature and composition of free hydrogen linkage points within the short structure determining how the single stranded nucleotide will fold in on itself [36].

2.3 Virtual screening of anti-human $\alpha$-syn aptamer using CSD-Discovery-GOLD

Docking experiments were analysed determining the predicted interactions between the components of the final pose. Bond distance, nature of interactions and type of interaction were all documented and analysed. Overall fitness of poses was determined using ChemPLP fitness scoring within CSD-Discovery. Structures were virtually screened in blocks of 10 using LYS 96, O as a central atom and a radius of 81Å, encompassing the target protein completely. All rotatable bonds were fixed to create a rigid dock and search efficiency was set to low/virtual screening. Fitness scoring function was set to ChemScore with no re-scoring function selected.

Binding energies were determined using follow up confirmation docking experiments using AutoDock Vina. Docking parameters between the programmes were kept constant. The docking positions generated by CSD-Discovery were used to inform the choice of grid location for docking in AutoDock Vina. The top 3 aptamers selected from CSD-Discovery were docked against $\alpha{}$-syn using the following settings: grid size: 110, 110, 110; coordinates: center_x = 23.673, center_y = 56.090, center z = –15.985; energy range = 4; exhaustiveness = 8; ligand torsions set to zero during setup. A summary of data files used within the work can be found in Table 3 [37, 38]. A spinach protein aptamer, 4ts2, was chosen as a control as it has no specificity to the target protein. Comparisons with a sequence which binds to a protein which is in an unrelated organism gives an indication of the baseline interaction between the target and a nucleotide sequence.

Table 3.Record of all PDB files used in study and any modifications made.
 File name Modifications to file Reference 1xq8 (Human $\alpha$-syn) No modifications made [37] 4ts2 (Spinach RNA aptamer in complex with DFHBI ((5Z)-5-(3,5- Metal ions (Magnesium and Potassium) removed [38] difluoro-4-hydroxybenzylidene)-2,3-dimethyl-3,5-dihydro-4H-imidazol-4-one), magnesium ions) Ligands (beta-D-fructofuranose-(2-1)-alpha-D-glucopyranose and Cytidine-5’-Phosphate-2’,3’-cyclic phosphate, (5Z)-5-(3,5-difluoro-4-hydroxybenzylidene)-2,3-dimethyl-3,5-dihydro-4H-imidazol-4-one) removed Hydrogens added
2.4 Data analysis/structure analysis

Top poses from CSD GOLD were used to determine the best aptamers to move forward to docking in Autodock Vina based on a combination of Fitness score, number of hydrogens and average bond length of interactions. Analysis of variance was conducted on the data produced by Autodock Vina to determine any significant changes in binding energies in comparison to M5-15.

3. Results
3.1 In silico 2D aptamer structure prediction

The in silico selection of ssDNA aptamers against $\alpha$-syn using simulated SELEX produced numerous potential candidates which were shown to be comparable to the reference aptamer made by [37] within the limitations of the model. Virtual screening of the top 50 aptamers based on $\Delta{}$G produced a pool of 10 aptamers which are shown to bind to $\alpha$-syn (see Table 4). The pool of candidates displays a large variation in $\Delta{}$G due to differences in GC content and the presences of Watson-Crick base pairing.

Table 4.Data displaying the Gibbs free energy of the reference aptamer and the top 10 aptamer sequences based on $\Delta{}$G in ascending order.
 Aptamer Sequence ΔG kj/mol M5-15 Forward -GTATGGTACGGCGCGGTGGCGGGTGCGTGG- Reverse –6.39 120 Forward -CGGGTTAGGTGGAAGATCCCACCTATTCCG- Reverse –8.52 123 Forward -GTGGGGGTCCAGGCCTAGTACTGCTAAGGC- Reverse –7.70 36 Forward -CATAAGCCCATGCTAGACTTGCGAAAGGGT- Reverse –7.16 TMG-79 Forward -ACGTTGGTGCATGCCGAAGCCGGCGAATGC- Reverse –6.96 60 Forward -GGGGGATACTAACCCGTCCCCGGCGAGCGG- Reverse –6.27 54 Forward -ATAAGGGCCCGTGCCGTTCTACGGTTCCCG- Reverse –6.22 195 Forward -CCTTTCGGGGGCGGCCCCCCCTTCCGGTAT- Reverse –5.96 116 Forward -TAACTTTTAAAATCTGTAGAGCAGGTGCGA- Reverse –5.79 125 Forward -ACGTTTGGGGAAATAGTATCCCCCCAACAT- Reverse –5.28 30 Forward -CTGAAAGGGGCTGTGCCGCCGGTAATAATA- Reverse –4.83

The known sequence which binds to $\alpha$-syn was shown to form three hairpin/stem loop structures when folded at 25 ${}^{\circ}{}$C under the ionic conditions [37]. Many of the in silico sequences selected based on $\Delta{}$G have shown the formation of similar binding motifs which may be effective at targeting $\alpha$-syn due to structural homology (see Fig. 2). Lower $\Delta{}$G values were also shown to be heavily correlated to the formation of Watson-Crick base pairing as opposed to higher $\Delta{}$G values which are the result of linear regions within secondary structures.

Pairwise alignment (Fig. 3) was conducted to determine the sequence homology between each candidate and M5-15. When aligned, TMG-79 was shown to have the largest percentage similarity (51.4%), suggesting a degree of similarity between the structural motifs of TMG-79 and the M5-15 (see Figs. 2,4). Aptamers 123 and 125 both show lower similarities than TMG-79 which would explain structural differences within aptamer 123 and 125 such as the appearance of structural motifs not present in M5-15 and the change in relative position of stem loops.

Fig. 3.

Pairwise alignment of N30 wobble region for top three selected aptamers based on PLP Fitness score against experimentally created aptamer. Similarity: TMG-79 = 51.4%; Aptamer 123 = 20.8%; Aptamer 125 = 9.3%.

Fig. 4.

Predicted secondary structures for top 10 aptamers based on $\Delta{}$G predicted by mfold for sequences in digitally created library. M5-15, referred to here as the reference aptamer [37] raised against $\alpha$-syn, is included as a control within the process. (Energy rules: DNA, temp = 24 ${}^{\circ}{}$C, 125 mM [Mg], 0 mM [Na]).

3.2 Docking data

GOLD docking results show a large improvement in fitness scores when comparing positive and negative controls to in silico aptamers. TMG-79 PLP fitness (Fig. 5) indicates it could be a candidate for greater binding potential. Further analysis of docking data shows that aptamer complexes for candidates 123 and 125 have a larger number of predicted H bonds with a comparable average bond length (Table 5).

Fig. 5.

Data showing the ChemPLP fitness score for the top poses of the previously selected aptamer sequences. The reference aptamer and spinach aptamer serve as positive and negative controls, respectively. Each selected aptamer shows a higher docking score for their top pose when compared to the reference aptamer.

Table 5.Breakdown of GOLD docking data including H-bond contribution, number of H-bonds and the average bond length of these polar interactions.
 Aptamer PLP. Chemscore. Hbond (3.dp) Number of H bonds Average bond length (3.dp) Top pose binding energy (kcal/mol) Mean Binding energy (kcal/mol) (2.dp) Mean dist from rmsd l.b. Reference 0.008 2 2.948 –16.8 –15.90 27.312 4ts2 0.000 0 - –17.6 –16.22 21.583 79 2.000 2 2.884 –18.7 –17.77 13.801 125 1.089 3 2.944 –17.0 –16.40 16.830 123 4.976 5 2.734 –18.4 –16.48 16.314 rmsd, root mean squared distance.

Despite having a lower fitness score than that of TMG-79, aptamer 123 was predicted to form a larger number of H-bonds making it a promising candidate. This aptamer also appears to show some minor structural similarities to the reference aptamer, particularly when examining its hairpin loop structures. Large variations in ChemPLP score were observed across the poses for each aptamer due to the low search efficiency used for virtual screening.

In addition, follow up experiments using AutoDock Vina demonstrating top pose and average binding affinity for the reference, spinach and the top 3 in silico generated aptamer sequences. Binding free energy was lower for all generated sequences compared to the M5-15, these changes were not statistically significant (TMG-79, p = 0.18; aptamer 123, p = 0.17; aptamer 125, p = 0.42; Spinach, p = 0.18).

Further analysis of the reference aptamer and TMG-79 showed that interactions occurred mainly in or around the stem loop structural motifs of each aptamer. The reference aptamer was predicted to form H-bonds at residues LYS97 and LYS21. In contrast, TMG-79 was predicted to form H-bonds at protein residues LYS80 and GLN99 suggesting that it may interact in a similar region to the reference aptamer. Differences in the overall structure of TMG-79 compared to the reference aptamer support the observation that the interactions are similar, but at different points within the protein. Aptamer 123, whilst having a lower overall ChemPLP fitness score, had a greater number of interaction points in these simulations, importantly with polar side chains of amino acids such as THR81, suggesting stability of these bonds within the complex. However, given that ChemPLP score is a summary of critical factors, including binding energy, this suggests that while there are greater interaction points, these interactions are less likely to form spontaneously and be stable. Bond distances observed from the reference aptamer are also greater than in 2 of the 3 aptamers generated, and mean binding energy predicted greater affinity across all aptamers screened. This indicates a tighter binding potential with the generated sequences compared to M5-15. Structural comparisons of the poses suggests that all aptamers bind to a similar region within $\alpha$-syn, primarily to residues contained within the 2 alpha helix motif containing LYS21 LYS43 LYS45 THR59 LYS80 THR81 SER87 LYS96 LYS97 GLN99 ASN103. These residues appear to be critical to the stabilisation of aptamer binding within the protein; the poses containing the in silico aptamers (Fig. 6) share remarkable similarity to the reference aptamer. Prediction of the final location of the aptamer within the central cleft within the target molecule also suggests that this may be a preferential location for a DNA molecule within $\alpha$-syn.

Fig. 6.

Showing the binding position and orientation of each aptamer in their top pose when docked against alpha synuclein. Each aptamer is predicted to bind within the alpha helices of $\alpha$-syn similarly to the reference aptamer.

4. Discussion

The purpose of this investigation was to determine if it was possible, based on a template sequence derived experimentally, to improve upon the binding affinity of an aptamer to a given target. While there has been significant advancement in aptamer generation in recent years, any improvements that can be made will naturally be of benefit in the future. As part of this work, the team has managed to generate several potential aptamer candidates which demonstrated higher affinity for the target molecule than the experimentally derived aptamer within our model. Data presented here suggests that it is possible to produce an aptamer which has the potential to be comparable or better than the experimentally derived sequence. While the data presented here does not suggest a significant difference in predicted binding affinity, it does demonstrate that there is a possibility with a greater data set and use of the aptamer in real world settings that the aptamer could either perform as well or outperform the reference aptamer. Analysis of the predicted interactions could shed light on the reasons for the observed increases in affinity, despite this not being statistically significant in this case.

Nature of the observed interactions

When comparing the reference aptamer interaction with the generated sequences, there are similarities between the way they interact with $\alpha$-syn, specifically around the preference to bind within the alpha helix flanking structure within the protein. However, there are differences in the predicted binding between the reference and generated aptamers. Molecular docking relies on predicting the presence of intermolecular forces between two molecules, including van der Waals forces, hydrogen bonds, electrostatic interactions, side chain interactions, hydrophobic interactions and interaction of the solvent on docking (usually water) [39]. Scoring of a docking pose is a combination of all these factors, based on the assumptions and conditions that were imposed during the docking experiment [40]. The likelihood of a docking pose being true can be argued based on known interactions between amino acid residues, the involvement of any side chains and how they interact with DNA molecules. Within the docking poses proposed here, there are a number of interaction points that are observed particularly within amine groups and aromatic side chains between the DNA and the amino acids. These observations have been documented previously. Within all of the structures proposed, there were 5 amino acid residues which were consistently involved in some capacity. These amino acids were, threonine (Thr), lysine (Lys), glutamine (Gln), asparagine (Asn) and serine (Ser).

Thr and Lys interactions with DNA are well documented, with Thr interacting predominantly with phosphate backbone groups [41], and Lys predominantly interacting with guanine groups [42]. Specifically, when looking at Lys, it has been reported to interact with oxidized guanine residues (8-Oxo-G). Binding induces conformational change in the 8-oxoguanine DNA glycosylase (Ogg) required for the removal of 8-Oxo-G, an important DNA repair mechanism [42]. While there are no 8-Oxo-G residues present within the structure of the aptamer proposed here, it appears that there is a natural interaction between Lys and guanine. Thr based hydrogen bonding has the highest propensity in highly hydrophilic environments [43], and since the target protein is water-soluble and primarily found in neuronal tissue, there is a suggestion that this interaction may occur in vivo. Following on from this, we should consider the possibility of the predicted Gln, Asn and Ser interactions between the generated aptamer and the target. Gln has been reported previously to have strong association with polar side chains and phosphates contained within the backbone, as well as interacting with the nitrogenous base component of adenosine [41]. Asn is generally associated with base recognition binding, predominantly to adenosine, and has been demonstrated in synthetic zinc finger constructs like 1meyC [44]. These interactions have also been observed in B-DNA binding of transcription factor-like elements [45]. Ser rich proteins are essential pre mRNA splicing factors. They regulate constitutive splicing and influence alternative splicing processes through an extensive and controlled phosphorylation of the Ser residues [46]. Ser residues are highly conserved within the DNA binding domains of bZip transcription factors of Arabidopsis thaliana, and were reported to be in physical contact with DNA [47] specifically providing a binding point through the hydroxymethyl side chain, something that was observed here in our model.

Interestingly, the non-species aptamer 4ts2 demonstrated a greater binding affinity for the target M5-15 which was an unexpected outcome of the work. This may be in part due to the very nature of $\alpha$-syn and its ability to associate with other “sticky” proteins. The unfolded nature of $\alpha$-syn allows it to function as a molecular chaperone assisting in the transport of vesicles and facilitating SNARE formation [48]. This function could extend to the adhesion of nucleotides to $\alpha$-syn, and could also suggest that RNA has a naturally greater affinity for $\alpha$-syn and might be a viable alternative to the DNA aptamers raised here and experimentally. This has been observed experimentally, with $\alpha$-syn being capable of binding its own mRNA to prevent translation of further protein [49]. Using catRAPID, an algorithm used to predict protein/RNA interactions, demonstrated $\alpha$-syn easily binds to its own mRNA without the need for an intrinsic riboswitch sequence within the mRNA [50].

When considering the structures of the aptamers, it is important to consider the presence and size of key structural motifs within the aptamer. Loops, both external and internal, tend to be much more flexible in nature when compared to standard Watson and Crick paired sections or other loops such as stems and bulges [51]. This inherent flexibility, combined with the relatively small size of aptamers in three-dimensional space, allows for the aptamer to integrate with regions of proteins that larger more complicated structures, like antibodies, are simply incapable of achieving [14]. This has been demonstrated previously by Autiero et al. [52] when looking into RNA aptamer binding to S8 ribosomal proteins. Here the ability of the aptamer to adjust its structure to bind to the target not only improves the binding efficiency of the aptamer but is also a critical property of aptamers that warrants further investigation.

When looking at the reference aptamer and the top aptamer generated as part of this study, displayed in Fig. 7, the most notable differences between them are the extension of the stem loop between C7 and G25 and the incorporation of a greater number of Watson and Crick pairing between G27 and C48 in TMG-79. This increase in bulge size, and the incorporation of unpaired G44 and A45, gives greater flexibility to this extended region in comparison to the reference aptamer’s similar structure between C26 and G39. An additional consideration is the increase in the number of potential binding points which have been predicted within TMG-79. This may be due to the increased flexibility of the aptamer compared to the reference aptamer which, while not being a rigid structure, has a greater number of structural motifs limiting its flexibility [52]. It is also important to remember that, while the increase in flexibility is important, the structural motifs themselves provide unique binding properties which allow for aptamer/protein interactions. However, there is evidence to suggest that both stable and unstable secondary motifs are important in aptamer binding. Adachi et al. [53] demonstrated that their RNA aptamer’s inherent instability actually promoted tighter binding with higher specificity compared to aptamers which had more stable structures. This conflicts with Troisi et al. [54], where the structural motifs interact in a specific fashion to promote greater affinity for thrombin of NU172. The difference between these studies and their approach to the development of the aptamer may be down to either differences in complexity of the proteins or their specific locations. In both examples presented, they are looking at relatively complex proteins, either thrombin [54] or interleukin 17 receptor A/F (IL17A/F) inhibition [53]. Within the second example, the hope is to inhibit formation of IL17A/F, requiring only to be able to bind to one component to competitively inhibit this function. In contrast, Troisi et al. [54] identified sequence specific binding to the thrombin protein which could be significantly reduced based on single amino acid mutations. The conclusion that can be drawn from this is that location, structure and the complexity of the proteins involved dictates the requirement for either rigidity or flexibility within the aptamer’s structure which influences its ability to bind to its target. Considering the inherent properties of $\alpha$-syn being a mainly unstructured protein with the exception of the two alpha helix motifs, it is interesting to see that both the reference aptamer and TMG-79 both bind within this alpha helix “sandwich”. Incorporation into this region and the increased affinity for the target observed with TMG-79 suggests that, while the protein may be mainly in an unstructured state, this is a critical recognition element for biomarker detection. While this data is promising, the requirement to get a solved crystal structure for the interaction is necessary to confirm the interactions predicted as part of this work.

Fig. 7.

Binding interaction points highlighted in red. (A) M5-15. (B) TMG-79. (C) While there are similarities between the two aptamer structures (51.4% sequence homology), the notable difference is the lack of stem loop between G43 and C58 within the reference aptamer. This change results in a larger bulge region in TMG-79, allowing for increased flexibility of the overall aptamer structure. Top pose predictions of 4ts2 (left) and M5-15 (right) against $\alpha$-syn are shown.

5. Conclusions

While these models are based on artificial, computer-generated structures it does present an interesting case for moving towards a hybrid in silico/in vitro development strategy for aptamer selection. Within this work we have demonstrated using in silico methodologies that you can refine the structures of experimentally derived aptamers to provide a structure which has the potential to have greater affinity for the target. Predicted interactions, as well as calculated free energy, suggest that TMG-79 has the potential to be a more effective means of detecting $\alpha$-syn. Moving forward, this work opens up opportunities to develop aptamers exclusively in silico with all the potential time and cost saving benefits this could bring.

Author contributions

All authors contributed to the study conception and design. Model design, python scripting for library generation, docking simulations and data collection was performed by MR, TAI. MR and GDZ assisted in data interpretation and writing of the 1st draft of the manuscript. AFBDA, SD, AAlbrakati, GESB and AAlexiou assisted with final manuscript writing. All authors approve the final version of this manuscript.

Ethics approval and consent to participate

Not applicable.

Acknowledgment

The authors would like to thank the University of Derby for supporting the project. We would like to thank the Taif University Researchers Supporting Program (project number: TURSP -2020/151), Taif University, Saudi Arabia.

Funding

This work was supported by Taif University Researchers Supporting Program (project number: TURSP -2020/151), Taif University, Saudi Arabia.

Conflict of interest

The authors declare no conflict of interest.

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# Why is the definition of cardinal number as the set of all sets equivalent to a given set “problematical”?

In Fundamentals of Mathematics, Volume 1 Foundations of Mathematics: The Real Number System and Algebra, after defining set equivalence as the ability to put the elements of the related sets in one-to-one correspondence, the following statement appears:

The cardinal number $$\tilde{x}$$ of a set $$x$$ is then regarded as representing "that which is common" to all sets that are equivalent to $$x$$. Thus, we might say that the cardinal number of $$x$$ is simply the set of all sets that are equivalent to $$x$$, although such a definition is problematical on account of its relationship to the universal set.

The term problematical can have a slightly different connotation than the term problematic. The former implying requires expert handling. In other words, this may not be grounds for completely rejecting the definition. Unfortunately I do not have access to the German Language original to know what "problematical" was translated from.

Regardless of that nuance, the authors are certainly indicating that their proposed definition leads to difficulty in "relationship to the universal set". Is this difficulty simply Russell's antinomy?

• It should be noted that there is a way of getting around this issue via Scott's Trick. – Hayden Feb 17 at 7:50

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# Is there a formal definition of convergence of series?

One is often asked to check if a given series converges or not in one's first year of uni. Is there a formal definition that allows us to check this? We are only given a bunch of tests that are troublesome to remember (I've never liked cramming in math), and a Google-search only yields results regarding sequences.

-
The formal definition is that the sequence of partial sums converges. This is usually stated in calculus but often the students don't really understand the concept of sequential convergence anyway and just end up memorizing a bunch of tests which they don't understand. This is why I have a big issue with the way series are taught in calculus. In many ways it's the most interesting part of the course but it ends up being the most painful part. By the way you are not alone, I had no idea what it really meant for a series to converge until I took real analysis. –  Seth Apr 14 '14 at 19:12
A series is a sequence: the sequence of partial sums of another sequence. So the definition of convergence for series is the same as for sequences. –  lhf Apr 14 '14 at 19:12
@lhf A series is a sequence? Not quite! And neither is their theory of convergence –  imranfat Apr 14 '14 at 20:37

The formal definition is that $$\sum_{k=1}^\infty a_k$$ converges exactly if the sequence of it's partial sums converges, i.e. if the sequence $(s_n)_{n\in\mathbb{N}}$ defined by $$s_n = \sum_{k=1}^n a_k$$ converges. In other words, you per definition have that $$\sum_{k=1}^\infty a_k = \lim_{n\to\infty} s_n = \lim_{n\to\infty} \sum_{k=1}^n a_k$$

-
Oh, of course, that was simple, yet something I was completely unaware of consciously. I have to agree with @Seth, it is indeed weird that we haven't formalized it. –  <NAME> Apr 14 '14 at 19:16
Isn't that tautological? As far as I know, the definition of $\sum_{k=1}^\infty a_k$ is $\lim_{n\to\infty} \sum_{k=1}^n a_k$. –  <NAME> Apr 15 '14 at 0:33
@EuroMicelli Uh, yes, sure. But the OP asked for the definition of $\sum_{k=1}^\infty a_k$. I even emphasized that by writing specifically that "... you per definition have ...". –  fgp Apr 15 '14 at 0:37
Ok. I read question as him looking for the definition of convergence of a series (in line with the title), not the definition of infinite series. I see your interpretation. –  <NAME> Apr 15 '14 at 0:45
@EuroMicelli What I gave is the definition of convergence of a series. The definition of just an infinite series is that it's an element of $\mathbb{R}^\mathbb{N}$, i.e. a function from $\mathbb{N}$ into $\mathbb{R}$. –  fgp Apr 15 '14 at 0:48

A series $a_n$converges if the limit $$\lim_{n\to \infty}\sum_{i=0}^n{a_i}=L$$ exists and is finite. In other words, for all $\epsilon>0$ there exists an $N$ such that for all $n>N$, $$\left|L-\sum_{i=0}^n{a_i}\right|<\epsilon$$

-

Let $\sum_{n=m}^\infty a_n$ be a formal infinite series. For any integer $N\geqslant m$, we define the $N^{\text{th}}$ partial sum $S_N$ of this series to be $S_N:=\sum_{n=m}^N a_n$; of course, $S_N$ is a real number. If the sequence $(S_N)_{n=m}^\infty$ converges to some limit $L$ as $N\to\infty$, then we say that the infinite series $\sum_{n=m}^\infty a_n$ is convergent, and converges to $L$; we also write $L=\sum_{n=m}^\infty a_n$, and say that $L$ is the sum of the infinite series $\sum_{n=m}^\infty a_n$. If the partial sums $S_N$ diverge, then we say that the infinite series $\sum_{n=m}^\infty a_n$ is divergent.

from <NAME> - Analysis I.

-


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### Use MSE instead of RMSE

Hello,

I don't understand why we can use the MSE for update instead of the RMSE. Technically, it seems like they would lead to the same solution, but the expressions for the gradients are different, we have a leading constand when differentiating the RMSE.


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# Difficult Alligations or Mixtures Solved QuestionAptitude Discussion

 Q. A merchant mixes three varieties of rice costing Rs.20/kg, Rs.24/kg and Rs.30/kg and sells the mixture at a profit of 20% at Rs.30 / kg. How many kgs of the second variety will be in the mixture if 2 kgs of the third variety is there in the mixture?
 ✖ A. 1 kg ✖ B. 3 kgs ✔ C. 5 kgs ✖ D. 6 kgs

Solution:
Option(C) is correct

If the selling price of mixture is Rs.30/kg and the merchant makes a profit of 20%, then the cost price of the mixture $=\dfrac{30}{1.2}$ = Rs.25/kg.

We need to find out the ratio in which the three varieties are mixed to obtain a mixture costing Rs.25 /kg.

Let variety $A$ cost Rs.20/kg, variety $B$ cost Rs.24 / kg and variety $C$ cost Rs.30/kg.

The mean desired price falls between $B$ and $C$.

Step 1:
Find out the ratio $Q_a : Q_c$ using alligation rule.

$\dfrac{Q_a}{Q_c}=\dfrac{30-25}{25-20}=\dfrac{1}{1}$

Be warned if we do not simplify the ratio and let if remain $\dfrac{Q_a}{Q_c}=\dfrac{5}{5}$, it changes the final answer, which will be the COREECT answer as well. See at the end of this answer for more details.

Step 2:
Find out the ratio $Q_b:Q_c$ using alligation rule.

$\dfrac{Q_b}{Q_c}=\left(\dfrac{30-25}{25-24}\right)=\dfrac{5}{1}$

Step 3:
$Q_c$ is found by adding the value of $Q_c$ in step 1 and step 2 $= 1 + 1 = 2$

Therefore, the required ratio $= 1 : 5 : 2$

If there are 2 kgs of the third variety in the mixture, then there will be 5 kgsof the second variety in the mixture.

Note:
When appearing in actual exam problems of this kind are to be skipped, at least in the first go.

If you were able to solve decent number of other problems, then you should look at this problem. Moreover such tye of questions may have MORE THAN ONE SOLUTION which may not match the given answer choices with the correct alternate answer of yours. For an alternate (and CORRECT) answer check SundararajC's comment.

Edit1: Thank you VINAY and bimal for mentioning the anomaly, changed the option from choice $B$ to choice $C$

Edit2: As pointed by SundararajC, 1.6666 kg of 2nd variety is ASLO a correct answer (check out how) but since it is not given in the answer choices we have settled with 5 Kgs.

## (17) Comment(s)

<NAME>
()

easy way:A=20/kg,B=24/kg ,C =30/kg

let x kg be rice A ,y be rice B and 2 is rice C

also final CP = 25

therefore, 20x+24y+60=25(x+y+2)

solving we get, y=10-5x

since , y cannot be zero or negative

hence, x can only be 1 giving y = 5kg

Pragya
()

cannot we do it like

b:c=5:1

so c=2kg we have to find 2nd variety let x so

5:1=x:2

x=10?

Kiran
()

Can someone explain me why resultant of type 3(Q_C) is coming into picture in this problem and not in

Question no 3:

In what ratio must a person mix three kinds of wheat costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg?

Prince
()

the answer would be 5 (option c) if we assume first quantity is 1 kg.

Kiran- In what ratio must a person mix three kinds of wheat costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg?

Correct answer-Infinitely many solution and the solution given is wrong. 12:7:7 is one possibilty

SundararajC
()

sundar:

I got the answer 1.6666 kg as follows.

Cost price of mixture is 25 R/kg

20 5

24 5

25

30 1

30 5

totally 16 parts in which third variety contains 6 parts,2 nd variety contains 5 parts and 1st variety contains 5 parts.

6 parts is 2 kg.therefore 5 parts is 1.6666kg.

so 1 st and 2 nd varieties both having 1.6666kg each.

Deepak
()

SundararajC
()

profit 20 percent. So (selling price-cost price)/cost price=0.2

So cost price becomes 25r/kg.

now

rs/kg | ave |difference

20 | | 5

|

24 | | 5

| 25 |

30 | | 1

30 | | 5

so totally 16 parts.

totally 16 parts in which third variety contains 6 parts,2 nd variety contains 5 parts and 1st variety contains 5 parts.

6 parts is 2 kg.therefore 5 parts is 1.6666kg.

so 1 st and 2 nd varieties both having 1.6666kg each.

SundararajC
()

Depak i explain simply as follows.

3rd variety (30/kg variety) is 2 kg

Selling price of mixture is 30/kg

My answer is both 1 st and 2nd variety is 1.6666kg.

Let me take this as correct answer.

Now Cost price,

$=(1.66666 \times 20)+(1.66666 \times 24)+(2 \times 30)$

$= \text{ Rs.}133.29$

Total selling amount $=(2 +1.66666+1.66666)\times 30=\text{ Rs.}160$

Now let us calculate the profit percentage,

$=\left(\dfrac{160-133.329}{133.329} \right)\times 100=20\%$

Which complies with the question.

Deepak
()

Thank you SundararajC, for your input. There is no error with your solution, multiple answers to the given question is possible. Modified the solution to incorporate your input.

This might really help others, thank you for your inputs.

<NAME>
()

Answer is neither C nor B. The answer should be 10 Kg for the second variety of rice.

When Qa: Qc = 1:1 and Qb: Qc = 5:1, then it is clear that Qa: Qb: Qc = 1 : 5 : 1 in the final mixture

Now it is given that Qc = 2kg in the final mixture.

hence, Qb = 5 x 2 = 10 Kg in the final mixture.

Rakesh
()

I believe the solution makes it clear that the 3rd ratio is to be calculated by ADDING the values received of $Q_c$ from step 1 and 2.

That is $Q_c=1+1=2$, which makes final ratio to be, $1:5:2$ and not $1:5:1$.

So, 5 kgs appear to be right answer to me.

Maksud
()

why are few people suggesting addition of q(c)??

my ansns is 10.

ratio is 1:5:1

why are some of u guys adding 1+1 and making the ratio 1:5:2??

<NAME>
()

ans should option c ie 5

the required proportion would be 1:5:2 . it clearly indicates that if 3 quantity is 2 kg the 2 quantity must be 5 kg.

Bimal
()

is there any short-cut sir.......another thing the answer will be 'c' not 'b'....

Dheeraj
()

If we calculate the allegations of the 1st and 2nd mixtures(the ratio obtained is 1:5) followed by the 2nd and 3rd mixtures (the ratio is 5:1).

Thus if we proceed our calculations as per the original ratios concept we get 1:5:1 and the answer obtained is eqaul to 10.

How should we analyze beforehand that the question is of such type?

Naresh
()

$a:c = 2:2$

$c:b= 2:5$

So,

$a:c:b = 2:2:5$

or

$a:b:c = 2:5:2$

Naresh
()

Sorry, my mistake... since the mixture contains all the parts... it's 1:5:(1+1)
so 1:5:2 is correct ..


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# How to get the correlation between two large (sparse) matrices?

I have two large (sparse) matrices (500000*500000) and would like to perform the mantel test to get a similarity measure.

• Is there a way to break-up these matrices into smaller ones and perform the test?
• Or can you suggest a way to do it efficiently without running into memory problems?
-
 If you just want to estimate a "similarity measure," you don't need the Mantel test: just compute (say) the Pearson correlation between the two matrices. – whuber♦ Oct 17 '11 at 16:18

## migrated from stackoverflow.comOct 7 '11 at 0:37

However, as these are sparse matrices, it's better to go after a method for accessing the data via coordinate object lists (COO representation), i.e. (row, column, value), and implement a simple calculation for tabulating the test statistic across the two matrices. You can merge the two matrices' COO versions or perform more complex tabulations using the data.table package.


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User <NAME> - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:51:41Z http://mathoverflow.net/feeds/user/1583 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68218/is-there-an-algebraic-construction-of-the-quillen-determinant-line-bundle/68266#68266 Answer by Tam<NAME>l for Is there an algebraic construction of the Quillen (determinant) Line Bundle? Tamas Hausel 2011-06-20T04:48:33Z 2011-06-21T21:12:31Z
For $G={\rm PGL}_n({\mathbb C})$ there is no complex algebraic Quillen line bundle on the complex character variety $Y$ as the cohomology class $[\omega]=\alpha_2\in H^2(Y;{\mathbb Q})$ is not pure by Proposition 4.1.8 in http://arxiv.org/abs/math.AG/0612668.

EDIT: It is instructive to think about the case $G={ GL}_1({\mathbb C})$. Then the character variety is $M_B=GL_1^{2g}$ the complex torus while $M_{DR}$ is an affine bundle over $Jac(C)$. Analytically $M_B\cong M_{DR}$ but the universal bundle $L$ on $M_{DR}\times C$ which is just the pull back of the Poincare bundle from $Jac(C)\times C$ is not algebraic on $M_B$, for example because the cohomology class $[\omega]$ of the symplectic form (which shows up in $c_1(L)^2$) has weight $2$ in $H^2(M_{DR};{\mathbb Q})$ but has weight $4$ (homogeneous weight $2$) in $H^2(M_B;{\mathbb Q})$. Thus in particular there is no complex algebraic Quillen bundle on $M_B$ with first Chern class $[\omega]$.

EDIT 2: To take David's example let $g=1$, then $M_B={\mathbb C}^* \times {\mathbb C}^*$ which does not have non-trivial pure cohomology in $H^2(M_B;{\mathbb Q})$ and even the Picard group is trivial (see an argument here); thus $M_B$ does not support any non-trivial algebraic line bundle.
 http://mathoverflow.net/questions/62544/homotopy-type-of-hilbert-schemes-of-points-of-mathbb-c2/62578#62578 Answer by <NAME> for Homotopy type of Hilbert schemes of points of $\mathbb C^2$ Tamas Hausel 2011-04-21T21:53:04Z 2011-04-21T21:53:04Z
It is a general folklore result, that if ${\mathbb C}^\times$ acts on a smooth complex variety $X$ so that the fixed point set $X^{{\mathbb C}^\times}$ is proper and the limit ${{\rm lim}_{\lambda \to 0} \lambda z }$ exists for every $z\in X$, then the downward flow $$D:=\{ z\in X | {\rm lim}_{\lambda \to \infty} \lambda z \mbox{ exists}{\}}$$ is a retract of $X$. This can be proved by Morse theory arguments as in Kirwan's book or you can prove it by first showing that the imbedding induces

$$H_*(D;{\mathbb Z})\to H_{*}(X;{\mathbb Z})$$ an isomorphism (by induction with respect to an appropriate ordering of the set of components of $X^{{\mathbb C}^\times}$). Similarly you can show that the fundamental groups are also isomorphic. Then the relative Hurewitz theorem will tell you that they are weakly homotopy equivalent, and as they are varieties so CW-complexes, Whitehead's theorem imply that they are homotopy equivalent.

For the Hilbert scheme you can use the ${\mathbb C}^\times$ action induced from the dilation on ${\mathbb C}^2$ mentioned above, which will have the required property, due to the fact that the Hilbert-Chow morphism is proper.
 http://mathoverflow.net/questions/12377/decomposing-tensor-products-of-irreducible-representations-of-reductive-groups-ov/12391#12391 Answer by Tamas Hausel for Decomposing tensor products of irreducible representations of reductive groups over a finite field Tamas Hausel 2010-01-20T07:15:52Z 2011-03-16T08:50:11Z
Theorem 1.4.1 in arxiv:0810.2076 answers some of your questions for generic semisimple irreducible representations. <NAME> has hitherto unpublished results where he does answer your question for all generic irreducible representations in terms of intersection cohomology of certain quiver varieties. We did not know about other results on the representation ring of $GL_n({\mathbb F}_q)$. EDIT: but see Victor's answer for related results of Lusztig. EDIT 2 (added 16/03/11) Letellier's paper is now available at: http://arxiv.org/abs/1103.2759
 http://mathoverflow.net/questions/42577/homology-of-a-symplectic-leaf-in-gln/43033#43033 Answer by <NAME> for homology of a symplectic leaf in GL(n) Tamas Hausel 2010-10-21T10:18:33Z 2010-11-12T19:57:19Z
Below now is a proof that for sufficiently generic $z$, $b_1(Y(z))=n-1$.

First we show $b_1(Y(z))\geq n-1$.

Let $d=(d_1,d_2..,d_{n-1}): Y(z)\to({\mathbb C}^*)^{n-1}$. One can explicitly lift the loops in $H_1(({\mathbb C}^*)^{n-1})$ as follows. Fix $i$ between $1$ and $n-1$. Let $e(t)=exp(2\pi \sqrt{-1} t)$ and $$B(t):=\left(\begin{array}{cc} e(t) & 1\\ e(t)(z_i+z_{i+1}-e(t))-z_iz_{i+1} & z_i+z_{i+1}-e(t) \end{array}\right).$$ Then the characteristic polynomial of $B(t)$ is $(\lambda-z_i)(\lambda-z_{i+1})$. Form an $n\times n$ matrix $A(t)$ as having $z_1,\dots,z_{i-1}$ on the main diagonal, then $B(t)$, then $z_{i+2},\dots,z_{n}$ and zero everywhere else. This matrix has the property that $A(t)\in Y(z)$ (as we assumed generic z) and $d_j(A(t))=z_1\cdot\dots\cdot z_j$ unless $j=i$ when $d_i(A(t))=e(t)z_1\cdot\dots\cdot z_{i-1}$. Thus $A(t)$ is a loop in $Y(z)$ covering the $i$th generator of $H_1(({\mathbb C}^*)^{n-1})$.

Thus we find that $d_*:H_1(Y(z))\to H_1(({\mathbb C}^*)^{n-1})$ is surjective so $b_1(Y(z))\geq n-1$.

To argue that $b_1\leq n-1$ we first show that the divisors $D_i:=d_i^{-1}(0)$ are irreducible on $X(z)$ for sufficiently generic $z$. We note that $D_i$ is irreducible in $M_{nxn}$ which can be easily seen over a finite field by counting points on $D_i$. This implies that for a sufficiently generic $z$ the divisor $D_i$ is irreducible on $X(z)$. When $z_i\neq z_j$ for $i\neq j$ we see that $X(z)$ fibers over the full flag variety with contractible fibers, so it is simply connected. From the homology Gysin sequence we get that removing an irreducible divisor from a smooth variety can increase $b_1$ by at most one. By induction we can conclude that $b_1(Y(z))\leq n-1$.

As an example take $n=2$. Here we are removing a copy of ${\mathbb C}^*$ from $X(z)$ an affine line bundle over ${\mathbb P}^1$. This will have Betti numbers $b_1=1$ and $b_2=2$ with weights $q,q,q^2$ respectively, to give Serre-polynomial $q^2+1$. We will have the map $d=d_1:Y(z) \to {\mathbb C}^{*}$. The fibers of this map will generically be a ${\mathbb C}^{*}$ and will have two singular fibers, which will give the non-trivial fundamental group. The point is that one can lift the loop generating $\pi_1({\mathbb C}^*)$ as generically $d_1$ is a fibration with connected fibers.
 http://mathoverflow.net/questions/25313/finitely-many-arithmetic-progressions/25340#25340 Answer by <NAME> for Finitely many arithmetic progressions Tamas Hausel 2010-05-20T10:21:47Z 2010-05-20T10:21:47Z
This was one of my favourite problems in high school. My proof went like this: if you look at the problem modulo n where n is the least common multiple of the differences of the arithmetic progressions then you can rephrase the problem as follows: if the vertices of a regular n-gon are partitioned into regular k-gons centered at the origin then two of them will have the same size. To prove this arrange the regular n-gon to have vertices at the nth roots of unity in the complex plane and assign the monic polynomial to every individual k-gon whose roots are exactly the vertices of the polygon. This way you will get the expression $x^n-1=(x^{k_1}-\zeta_1)(x^{k_2}-\zeta_2)\cdots$. Multiplying out the RHS we see that if $k_1$ is the least of the $k_i$'s then the only way to cancel the term of $x^{k_1}$ from the RHS is to have another $k_i=k_1$ proving the claim.
 http://mathoverflow.net/questions/4331/is-the-wedge-product-of-two-harmonic-forms-harmonic/4533#4533 Answer by <NAME> for Is the wedge product of two harmonic forms harmonic? Tamas Hausel 2009-11-07T15:26:55Z 2009-11-07T15:26:55Z
Interestingly in (24) of hep-th/9603176 it is mistakenly claimed that the wedge product of harmonic forms is automatically harmonic. Because it is false we still do not know the predicted existence of those middle dimensional $L^2$ harmonic forms on these non-compact complete hyperkahler manifolds.
 http://mathoverflow.net/questions/68218/is-there-an-algebraic-construction-of-the-quillen-determinant-line-bundle/68238#68238 Comment by <NAME> <NAME> 2011-06-20T13:59:24Z 2011-06-20T13:59:24Z David, the universal bundle is only algebraic in the $M_{DR}$ algebraic structure - it is not algebraic in the character variety algebraic strucure, as its Chern classes are not pure. http://mathoverflow.net/questions/40337/ingenuity-in-mathematics/40373#40373 Comment by <NAME> <NAME> 2010-09-29T16:24:46Z 2010-09-29T16:24:46Z The chessboard problem can be nicely continued by proving that the chessboard can be tiled with dominos if two differently coloured squares are removed. Check out the solution of puzzle B at gurmeetsingh.wordpress.com/2008/09/12/… http://mathoverflow.net/questions/12377/decomposing-tensor-products-of-irreducible-representations-of-reductive-groups-ov/12422#12422 Comment by <NAME> <NAME> 2010-01-20T16:31:12Z 2010-01-20T16:31:12Z Interesting. Any chance to find the result for the character ring of $PGL_2({\mathbb F}_q)$ somewhere explicitly? http://mathoverflow.net/questions/4331/is-the-wedge-product-of-two-harmonic-forms-harmonic/4533#4533 Comment by <NAME> <NAME> 2009-11-07T18:39:42Z 2009-11-07T18:39:42Z No, the individual 2-forms are harmonic, and on Taub-Nut is is in $L^2$ as well. One can show that there are no other ones as in arxiv.org/abs/math/9909002 and arxiv.org/abs/math/0207169


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1. ## laplace help

f (t) = {1 0 < t < a
{0 a < t < 2a

If f(t + 2a) = f(t), t>0
f(t) is called: _______________

and its graph for 0 < t < 4a looks like:

and

L{f(t)} = ?

2. If You have a function defined as...

$\sigma(t)=\left\{\begin{array}{cc}1,&\mbox{ if }
0a\end{array}\right.$
(1)

... then is...

$\mathcal{L}\{\sigma(t)\} = \frac{1-e^{-as}}{s}$ (2)

Now if $f(t)$ is $\sigma(t)$ 'periodically extended' with period $2a$ is...

$\mathcal{L}\{f(t)\} = \frac{\mathcal{L}\{\sigma(t)\}}{1-e^{-2as}}= \frac{1-e^{-as}}{s\cdot (1-e^{-2as})}$ (3)

Kind regards

$\chi$ $\sigma$


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# 2707. Ground Works

The Hilbert Mole is a small and very rare mole. The first and only specimen was found by <NAME> at his backyard. This mole lives in a huge burrow under the ground, and the border of this burrow forms a Hilbert curve of n-th order (Hn).

Hilbert curves can be defined as follows. H1 is a unit square with open top side (fig. 1a), Hn consists of four copies of Hn-1: bottom left and bottom right are copied without changes, top left is rotated 90 counter-clockwise and top right is rotated 90 clockwise. These small copies are connected by three segments of unit length (fig. 1b,c,d).

Trying to exterminate the mole, Mr. Hilbert fills the burrow with water (fig. 2). But air inside the burrow prevents water from filling it entirely. In this problem we suppose that air and water are incompressible and cannot leak throw the borders of the burrow. Your task is to find the total area of the burrow, filled with water.

Note that water can flow over the obstacle only when its level is strictly higher. See examples on fig. 3 for further clarification.

### 输入格式

The first line of the input file contains two integer numbers: n and α ― order of Hilbert curve and slope angle of surface in degrees (1 ≤ n ≤ 12, 0 ≤ α < 90).

### 输出格式

The first line of the output file must contain a single real number ― the total area of the burrow, filled with water. The relative error of the answer must not exceed 1e-7.

### 样例

Input
5 30
3 45
4 10
3 0

Output
190.803847577293
15.5
91.573591766702
26.0

2 人解决，2 人已尝试。

2 份提交通过，共有 61 份提交。

8.6 EMB 奖励。


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# Math Help - Chance of success.

1. ## Chance of success.

In Left 4 Dead there is cool move when 4 hunters land on 4 different targets resulting in instant win.
I did some statistic research against good teams to find out how much hunters fail.

• There is 0.8 chance that hunter will get shot during the jump.
• There is 0.6 chance that hunter will get stopped after the jump.
• There is 0.6 chance that hunter will jump on wrong target(which makes whole attack fail).

So to make successful attack all 4 hunters must not get shot during the jump, must not get stopped if didn't got shot during the jump, must jump on right target.

How do I eliminate all failures and find out success probability?

2. P(not shot and not stopped and not miss) = P(not shot) * P(not stopped) *P(not missed) = 0.8 *0.6*0.6 =.288

I've assumed that the events are all independant

3. But when its 4 hunters it goes like this?
$(0.6*0.6*0.8)^4 = 0.00687970714$
So chance is 0.6%?

4. if the outcome for each hunter is independant, yes


text:
# Open and Equitable Scholarly Communications: Creating a More Inclusive Future – ACRL Insider

“ACRL is pleased to announce the release of “Open and Equitable Scholarly Communications: Creating a More Inclusive Future,” prepared for ACRL by <NAME> and <NAME> with <NAME>, <NAME>, <NAME>, <NAME>, <NAME>, and <NAME>. Developed over the course of a year with leadership from the Research and Scholarly Environment Committee (ReSEC) and with a high degree of community involvement, this powerful new action-oriented research agenda encourages the community to make the scholarly communications system more open, inclusive, and equitable by outlining trends, encouraging practical actions, and clearly identifying the most strategic research questions to pursue.

This report is an important contribution to ACRL’s core commitment to equity, diversity and inclusion which includes valuing different ways of knowing and identifying and working to eliminate barriers to equitable services, spaces, resources, and scholarship. The full research agenda is freely available on the ACRL website and will be available for purchase in print in the ALA store….”

# “Open and Equitable Scholarly Communications: Creating a More Inclusive” by <NAME>, <NAME> et al.

Abstract:  For many years, the academic and research library workforce has worked to accelerate the transition to more open and equitable systems of scholarship. While significant progress has been made, barriers remain. The Association of College and Research Libraries (ACRL) seeks to stimulate further advances through this action- oriented research agenda, which is designed to provide practical, actionable information for academic librarians; include the perspectives of historically underrepresented communities in order to expand the profession’s understanding of research environments and scholarly communication systems; and point librarians and other scholars toward important research questions to investigate. This report represents a yearlong process of reviewing the scholarly and practice-based literature to take into account established investigation coupled with extensive public consultation to identify the major problems facing the academic library community. Through interviews, focus groups, workshops, and an online survey, over 1,000 members of the ACRL community offered their thoughts and expertise to shape this research agenda. Incorporating guidance and input from ACRL’s Research and Scholarly Environment Committee and an advisory panel, this document recommends ways to make the scholarly communications and research environment more open, inclusive, and equitable.

# [Open letter to <NAME>]

“On behalf of the undersigned national and regional library, research, publishing, and advocacy organizations, we are writing to express our commitment to ensuring that American taxpayers are guaranteed immediate, free, and unfettered access to the results of scientific research that their tax dollars support, and to encourage the Administration to support continued progress towards this shared goal. We strongly endorse updating existing U.S. policy to eliminate the current 12-month embargo period on articles reporting on publicly funded research, and to ensure that they are made immediately available under terms and conditions that allow their full reuse. To unlock the full value of our collective investment in science, the underlying data needed to validate an article’s conclusions, along with any corresponding software or code, should also be made immediately available….”

# ACRL/SPARC Forum: Collective Reinvestment in Open Infrastructure (ACRL)

Libraries are increasingly considering scaling back their subscriptions or cancelling big deals altogether. Yet, the question of how and where to reinvest the resources that become available is both far from settled and increasingly pressing. As we start to move away from the subscription model, we should be intentional about crafting the vision for open research communication we strive to build and how we intend to build it.

This forum, “If I Had A Million Dollars: Collective Reinvestment in Open Infrastructure,” will invite active participation throughout the session in a facilitated discussion with experts representing both libraries and research funders. …”

# Open and Equitable Scholarly Communications: Creating a More Inclusive Future

“For many years, the academic and research library workforce has worked to accelerate the transition to more open and equitable systems of scholarship. While significant progress has been made, barriers remain. The Association of College and Research Libraries (ACRL) seeks to stimulate further advances through this action-oriented research agenda, which is designed to provide practical, actionable information for academic librarians; include the perspectives of historically underrepresented communities in order to expand the profession’s understanding of research environments and scholarly communication systems; and point librarians and other scholars toward important research questions to investigate.

This report represents a yearlong process of reviewing the scholarly and practice-based literature to take into account established investigation coupled with extensive public consultation to identify the major problems facing the academic library community. Through interviews, focus groups, workshops, and an online survey, over 1,000 members of the ACRL community offered their thoughts and expertise to shape this research agenda. Incorporating guidance and input from ACRL’s Research and Scholarly Environment Committee and an advisory panel, this document recommends ways to make the scholarly communications and research environment more open, inclusive, and equitable….”

# Feedback on Draft ACRL Research Agenda – Deadline Extended to January 11 – ACRL Insider

“ACRL’s Research and Scholarly Environment Committee together with <NAME> and <NAME> – the authors selected by ACRL to design, develop, and deliver a new ACRL research agenda for scholarly communications and the research environment – are grateful for the feedback received thus far on the draft document for public comment. To ensure the robust community engagement continues, the deadline to comment has been extended to COB Friday, January 11, 2019….

Your input is crucial for ensuring the final document is as helpful as possible both in guiding academic librarians on actions that can be taken now to promote a more open system of scholarship and identifying essential areas that merit further investigation….”


text:
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# If x and y are positive integers and x^2 – y^2 = 101, what is the valu

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21 Feb 2020, 04:47
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If x and y are positive integers and x^2 – y^2 = 101, what is the value of x^2 + y^2 ?

A. 5050
B. 5101
C. 5150
D. 5201
E. 5301

Are You Up For the Challenge: 700 Level Questions

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Re: If x and y are positive integers and x^2 – y^2 = 101, what is the valu  [#permalink]

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21 Feb 2020, 04:48
Bunuel wrote:
If x and y are positive integers and x^2 – y^2 = 101, what is the value of x^2 + y^2 ?

A. 5050
B. 5101
C. 5150
D. 5201
E. 5301

$$x^2 - y^2 = (x-y)(x+y)=101$$. 101 is a prime number, so it can be broken into the product of two positive multiples only in one way 101=1*101. Now, since x and y are positive integers, then x - y < x + y, thus x - y = 1 and x + y = 101. Solving for x and y gives x = 51 and y = 50. Thus $$x^2 + y^2=51^2+50^2=5,101$$.

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Re: If x and y are positive integers and x^2 – y^2 = 101, what is the valu  [#permalink]

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21 Feb 2020, 09:36

Solution

Given
In this question, we are given that
• The numbers x and y are positive integers
• Also, $$x^2 – y^2 = 101$$

To find
We need to determine
• The value of $$x^2 + y^2$$

Approach and Working out
The given equation can be written as $$x^2 – y^2 = 101$$
• Or, (x + y) (x – y) = 101

Now, 101 is a prime number and can be written as a product of two integers only as 101 * 1
Hence, we can say x + y = 101 and x – y = 1

Solving, we get x = 51 and y = 50
• Therefore, $$x^2 + y^2 = 51^2 + 50^2 = 2601 + 2500 = 5101$$

Thus, option B is the correct answer.

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Re: If x and y are positive integers and x^2 – y^2 = 101, what is the valu  [#permalink]

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21 Feb 2020, 14:27
Bunuel wrote:
If x and y are positive integers and x^2 – y^2 = 101, what is the value of x^2 + y^2 ?

A. 5050
B. 5101
C. 5150
D. 5201
E. 5301

We can PLUG IN THE ANSWERS, which represent the value of $$x^2 + y^2$$.
Since $$(x^2 + y^2) + (x^2 - y^2) = 2x^2$$, the correct answer must yield an EVEN INTEGER when added to $$x^2 – y^2 = 101$$.
Implication:
The correct answer must be ODD.
Eliminate A and C.

B: $$x^2 + y^2 = 5101$$
Adding together $$x^2 + y^2 = 5101$$ and $$x^2 – y^2 = 101$$, we get:
$$2x^2 = 5202$$
$$x^2 = 2601$$

Since $$50^2 = 2500$$ and $$60^2 = 3600$$, $$x$$ in this case must be between 50 and 60.
For its square to yield a units digit of 1, $$x$$ must have a units digit of 1 or 9.
Test $$x=51$$:
$$51^2 = (50+1)^2 = 50^2 + 1^2 + 2*50*1 = 2500 + 1 + 100 = 2601$$
Success!

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Re: If x and y are positive integers and x^2 – y^2 = 101, what is the valu   [#permalink] 21 Feb 2020, 14:27
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text:
# Mike is playing his speaker at an unknown frequency. His friend, Emma, gets on her bike and rides away from Mike at a uniform speed, holding an open-air column horizontally over her head. What was the frequency of the sound coming from Mike's speaker?

## Mike is playing his speaker at an unknown frequency on a hot ${31.0}^{\circ} C$ day. His friend, Emma, gets on her electric bike and rides away from Mike with a uniform speed of $16.0 \frac{m}{s}$, holding a $0.450 m$ open-air column horizontally over her head. The air rushing through the tube produces a third harmonic tone in the tube. Mike hears a beat frequency of $5.00 H z$. What was the frequency of the sound coming out of Mike's speaker?

Jun 27, 2016

$1109.33 H z$ or $1119.33 H z$
(significant figures excluded)

#### Explanation:

Order of Solving

$1$. Determine the speed of the sound waves coming from Mike's speaker according to an air temperature of ${31.0}^{\circ} C$.

$2$. Find the frequency produced from the sound waves going through the tube above Emma's head according to a third harmonic tone.

$3$. Find the frequency Mike detects when the sound waves travel through to him as Emma rides away.

$4$. Determine the possible frequencies that could have come out of Mike's speaker.

Step 1
The speed of sound waves in air depends on the air temperature. The higher the temperature, the greater the speed. At ${0}^{\circ} C$, the speed of sound is $331 \frac{m}{s}$. Each increase in one degree Celsius results in a speed increment of $0.6 \frac{m}{s}$. This can be represented by the formula:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{v}_{s} = 331 \frac{m}{s} + \left(\frac{0.6 \frac{m}{s}}{{\textcolor{w h i t e}{i}}^{\circ} C}\right) \times \text{temperature}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Using the formula, the speed of the sound waves coming from Mike's speaker was

$\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{{v}_{s}} = 331 \frac{m}{s} + \left(\frac{0.6 \frac{m}{s}}{{\textcolor{w h i t e}{i}}^{\circ} C}\right) \times {31.0}^{\circ} C$

$= \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{349.6 \frac{m}{s}}$

Step 2
The air travelling from Mike's speaker through the tube above Emma's head produces a third harmonic tone. Since the information regarding the tube's length and harmonic is given, we can use that knowledge to combine it with the speed of the sound waves to determine the frequency of the sound coming from the tube. This can be found by using the formula for the frequency of standing waves in an air column open at both ends:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{f = \frac{n v}{2 L}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\underline{\text{where:}}$
$f =$frequency
$n =$harmonic number
$v =$speed of wave
$L =$length of air column

In our case, we will let $f$ be ${f}_{t}$, the frequency of the tube, and $v$ be ${v}_{s}$, the speed of the sound waves.

$\textcolor{p u r p \le}{{f}_{t}} = \frac{n \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{{v}_{s}}}{2 L}$

$= \frac{\left(3\right) \left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{349.6 \frac{m}{s}}\right)}{2 \left(0.450 m\right)}$

$= \textcolor{p u r p \le}{1165.33 H z}$

Step 3
The Doppler effect causes a change in the frequency heard by Mike as Emma rides away from him. The frequency that he hears is lower than the actual frequency of the sound waves in the tube. To derive the frequency detected by Mike, we can use the Doppler effect formula for sound:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{f}_{d} = \left({v}_{s} / \left({v}_{s} \pm {v}_{o}\right)\right) f} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\underline{\text{where:}}$
${f}_{d} =$detected frequency
${v}_{s} =$speed of wave
${v}_{o} =$speed of object
$f =$actual frequency

In our case, we will let $f$ be ${f}_{t}$. The sign between ${v}_{s}$ and ${v}_{o}$ will be $+$ and not $-$ since the tube is moving away from Mike.

$\textcolor{t e a l}{{f}_{d}} = \left({v}_{s} / \left({v}_{s} + {v}_{o}\right)\right) \textcolor{p u r p \le}{{f}_{t}}$

$= \left(\frac{349.6 \frac{m}{s}}{349.6 \frac{m}{s} + 16 \frac{m}{s}}\right) \textcolor{p u r p \le}{1165.33 H z}$

$= \textcolor{t e a l}{1114.33 H z}$

Step 4
The frequency detected by Mike is different from the actual frequency of the tube, but it's also different from the frequency coming out of his speaker. Mike reports that he hears a beat frequency of $5.00 H z$, so we can use the beat frequency formula to determine the two possible frequencies coming out of his speaker:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{f}_{b} = | {f}_{1} - {f}_{2} |} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\underline{\text{where:}}$
${f}_{b} =$beat frequency
${f}_{1} =$first frequency
${f}_{2} =$second frequency

In our case, we will let ${f}_{2}$ be ${f}_{d}$, the frequency detected by Mike. The unknown variable, ${f}_{1}$, is the frequency of the sound coming from the speaker.

Since the equation has an absolute value sign, there are two possible scenarios.

$\underline{\text{Case 1:}}$
${f}_{b} = {f}_{1} - \textcolor{t e a l}{{f}_{d}}$

$5.00 H z = {f}_{1} - \textcolor{t e a l}{1114.33 H z}$

${f}_{1} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{1119.33 H z} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\underline{\text{Case 2:}}$
${f}_{b} = - \left({f}_{1} - \textcolor{t e a l}{{f}_{d}}\right)$

$5.00 H z = - \left({f}_{1} - \textcolor{t e a l}{1114.33 H z}\right)$

${f}_{1} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{1109.33 H z} \textcolor{w h i t e}{\frac{a}{a}} |}}}$


text:
# If a ray of light enters from alcohol to air. The refractive index of alcohol is 1.36. Calculate the speed of light in alcohol with respect to air.

19.7k views
in Physics
closed

If a ray of light enters from alcohol to air. The refractive index of alcohol is 1.36. Calculate the speed of light in alcohol with respect to air.

by (54.0k points)
selected by

Given;

Refractive index of alcohol (μ) = 1.36.

Speed of light in vacuum = 3×108m/s.

Refractive index (μ) = $\frac{c\,(speed\,of\,light\,in\,vaccum)}{v\,(speed\,of\,light\,in\,alcohol)}$

⇒ 1.36 = $\frac{3\,\times\,10^8}{speed\,of\,light\,in\,alcohol}$

⇒ speed of light in alcohol = $\frac{3\,\times\,10^8}{1.36}$

∴ Speed of light in alcohol = 2.21 × 108m/s

Speed of light in alcohol with respect to air is 2.21 x 108ms-1


text:
# France '98

Time Limit: 1000 ms

Memory Limit: 65535 ms

## Description

Today the first round of the Soccer World Championship in France is coming to an end. 16 countries are remaining now, among which the winner is determined by the following tournament:

 1 Brazil -----+
+-- ? --+
2 Chile ------+       |
+-- ? --+
3 Nigeria ----+       |       |
+-- ? --+       |
4 Denmark ----+	       |
+-- ? --+
5 Holland ----+	       |       |
+-- ? --+       |       |
6 Yugoslavia -+       |       |       |
+-- ? --+       |
7 Argentina --+       |	       |
+-- ? --+	       |
8 England ----+		       |
+-- World Champion
9 Italy ------+		       |
+-- ? --+	       |
10 Norway -----+       |               |
+-- ? --+       |
11 France -----+       |       |       |
+-- ? --+       |       |
12 Paraguay ---+	       |       |
+-- ? --+
13 Germany ----+	       |
+-- ? --+       |
14 Mexico -----+       |       |
+-- ? --+
15 Romania ----+       |
+-- ? --+
16 Croatia ----+

For each possible match A vs. B between these 16 nations, you are given the probability that team A wins against B. This (together with the tournament mode displayed above) is sufficient to compute the probability that a given nation wins the World Cup. For example, if Germany wins against Mexico with 80%, Romania against Croatia with 60%, Germany against Romania with 70% and Germany against Croatia with 90%, then the probability that Germany reaches the semi-finals is 80% * (70% * 60% + 90% * 40%) = 62.4%.

Your task is to write a program that computes the chances of the 16 nations to become the World Champion '98.

## Input

The input file will contain just one test case. The first 16 lines of the input file give the names of the 16 countries, from top to bottom according to the picture given above. Next, there will follow a 16 x 16 integer matrix P where element pijgives the probability in percent that country #i defeats country #j in a direct match. Country #i means the i-th country from top to bottom given in the list of countries. In the picture above Brazil is #1 and Germany is #13, so p1,13=55 would mean that in a match between Brazil and Germany, Brazil wins with a probability of 55%. Note that matches may not end with a draw, i.e. pij + pji = 100 for all i,j.

## Output

Output 16 lines of the form "XXXXXXXXXX p=Y.YY%", where XXXXXXXXXX is the country's name, left-justified in a field of 10 characters, and Y.YY is their chance in percent to win the cup, written to two decimal places. Use the same order of countries like in the input file.

## Sample Input

Brazil
Chile
Nigeria
Denmark
Holland
Yugoslavia
Argentina
England
Italy
Norway
France
Paraguay
Germany
Mexico
Romania
Croatia
50 65 50 60 55 50 50 65 45 55 40 55 40 55 50 50
35 50 35 45 40 35 35 50 30 40 25 40 25 40 35 35
50 65 50 60 55 50 50 65 45 55 40 55 40 55 50 50
40 55 40 50 45 40 40 55 35 45 30 45 30 45 40 40
45 60 45 55 50 45 45 60 40 50 35 50 35 50 45 45
50 65 50 60 55 50 50 65 45 55 40 55 40 55 50 50
50 65 50 60 55 50 50 65 45 55 40 55 40 55 50 50
35 50 35 45 40 35 35 50 30 40 25 40 25 40 35 35
55 70 55 65 60 55 55 70 50 60 45 60 45 60 55 55
45 60 45 55 50 45 45 60 40 50 35 50 35 50 45 45
60 75 60 70 65 60 60 75 55 65 50 65 50 65 60 60
45 60 45 55 50 45 45 60 40 50 35 50 35 50 45 45
60 75 60 70 65 60 60 75 55 65 50 65 50 65 60 60
45 60 45 55 50 45 45 60 40 50 35 50 35 50 45 45
50 65 50 60 55 50 50 65 45 55 40 55 40 55 50 50
50 65 50 60 55 50 50 65 45 55 40 55 40 55 50 50

## Sample Output

Brazil     p=8.54%
Chile      p=1.60%
Nigeria    p=8.06%
Denmark    p=2.79%
Holland    p=4.51%
Yugoslavia p=7.50%
Argentina  p=8.38%
England    p=1.56%
Italy      p=9.05%
Norway     p=3.23%
France     p=13.72%
Paraguay   p=3.09%
Germany    p=13.79%
Mexico     p=3.11%
Romania    p=5.53%
Croatia    p=5.53%

Ulm Local


text:
# fundamental_ion_collision_freq¶

plasmapy.formulary.collisions.fundamental_ion_collision_freq(T_i: Unit(‘K’), n_i: Unit(‘1 / m3’), ion, coulomb_log=None, V=None, coulomb_log_method='classical')

Average momentum relaxation rate for a slowly flowing Maxwellian distribution of ions.

3 provides a derivation of this as an average collision frequency between ions and ions for a Maxwellian distribution. It is thus a special case of the collision frequency with an averaging factor.

Parameters
• T_i (Quantity) – The electron temperature of the Maxwellian test ions

• n_i (Quantity) – The number density of the Maxwellian test ions

• ion (str) – String signifying a particle type of the test and field ions, including charge state information. This function assumes the test and field ions are the same species.

• V (Quantity, optional) – The relative velocity between particles. If not provided, thermal velocity is assumed: $$μ V^2 \sim 2 k_B T$$ where $$μ$$ is the reduced mass.

• coulomb_log (float or dimensionless Quantity, optional) – Option to specify a Coulomb logarithm of the electrons on the ions. If not specified, the Coulomb log will is calculated using the ~plasmapy.formulary.Coulomb_logarithm function.

• coulomb_log_method (str, optional) – The method by which to compute the Coulomb logarithm. The default method is the classical straight-line Landau-Spitzer method ("classical" or "ls"). The other 6 supported methods are "ls_min_interp", "ls_full_interp", "ls_clamp_mininterp", "hls_min_interp", "hls_max_interp", and "hls_full_interp". Please refer to the docstring of Coulomb_logarithm for more information about these methods.

Returns

nu_i

Return type

Quantity

Notes

Equations (2.36) and (2.122) in 3 provide the original source used to implement this formula, however, in our implementation we use the very same process that leads to the fundamental electron collison rate (2.17), gaining simply a different coefficient:

$ν_i = \frac{8}{3 * 4 * \sqrt{π}} ν(v_{Ti})$

Where $$ν$$ is the general collision frequency and $$v_{Ti}$$ is the ion thermal velocity (the average, for a Maxwellian distribution).

Note that in the derivation, it is assumed that electrons are present in such numbers as to establish quasineutrality, but the effects of the test ions colliding with them are not considered here. This is a very typical approximation in transport theory.

This result is an ion momentum relaxation rate, and is used in many classical transport expressions. It is equivalent to: * $$1/τ_i$$ from ref 1, equation (2.5i) pp. 215, * $$ν_i$$ from ref 2 pp. 33,

References

1

<NAME>. “Transport processes in a plasma.” Reviews of plasma physics 1 (1965): 205.

2

<NAME>. “NRL (Naval Research Laboratory) Plasma Formulary, revised.” Naval Research Lab. Report NRL/PU/6790-16-614 (2016). https://www.nrl.navy.mil/ppd/content/nrl-plasma-formulary

3(1,2)

<NAME>, Fundamentals of Plasma Physics draft material, Chapter 2, http://homepages.cae.wisc.edu/~callen/chap2.pdf

Examples

>>> from astropy import units as u
>>> fundamental_ion_collision_freq(0.1 * u.eV, 1e6 / u.m ** 3, 'p')
<Quantity 2.868...e-05 1 / s>
>>> fundamental_ion_collision_freq(1e6 * u.K, 1e6 / u.m ** 3, 'p')
<Quantity 1.741...e-09 1 / s>
>>> fundamental_ion_collision_freq(100 * u.eV, 1e20 / u.m ** 3, 'p')
<Quantity 63087.5... 1 / s>
>>> fundamental_ion_collision_freq(100 * u.eV, 1e20 / u.m ** 3, 'p', coulomb_log_method='GMS-1')
<Quantity 63085.1... 1 / s>
>>> fundamental_ion_collision_freq(100 * u.eV, 1e20 / u.m ** 3, 'p', V = c / 100)
<Quantity 9.111... 1 / s>
>>> fundamental_ion_collision_freq(100 * u.eV, 1e20 / u.m ** 3, 'p', coulomb_log=20)
<Quantity 95918.7... 1 / s>


text:
Imported by Threaded, since a TH splice can't be used in the module where it is defined.
forkOS requires the threaded RTS, so this TH fails if haddock was built without -threaded.


text:
# Optimization of sieve of Erathosthenes

#include <iostream>
#include <conio.h>
#include <windows.h>
#include <math.h>
using namespace std;
#define RUNS 1000
char z[100000];
int i,j,k,c;

void main(void)
{
DWORD starttime,endtime;
float totaltime;

starttime = GetTickCount();//get start time
for(k=0;k<RUNS;k++)
{
c=0;

//for (i=0;i<10000;i++) z[i]=0; //clear array
memset(z,0,100000);
z[0]=1; // 0 is not a prime
z[1]=1;  // 1 is not a prime
//now remove all evens from 4 up
for(i=4;i<100000;i=i+2) z[i]=1; //remove evens
//now loop through remaing up to square root of max number
for(i=3;i<316;i=i+2)
{
if(z[i]==0) for(j=2*i;j<100000;j=j+i) z[j]=1;
}
}

endtime=GetTickCount();//get finish time
//calc time
for(i=0;i<100000;i++)
{
if(z[i]==0) {cout<<i<<" ";c++;}
}
cout<<"primes found="<<c<<endl;
totaltime=((float)endtime-(float)starttime)/(1000.0*RUNS);//calculate total time in secs
cout<<"Totaltime="<<totaltime<<" sec\n";

printf("Press any key to end");

getch();
}

1. I'm trying to find any optimization for my sieve of Eratosthenes code for counting first 100000 prime numbers.
2. The program first mark all even numbers, than square root of max number.
3. The program already does take fraction of the seconds to count these prime numbers, but Im looking for any optimization to make it even quicker.
• Man this comes up a lot. Think its an article I will have to write about :-) – Martin York Feb 12 '15 at 22:34
• Small optimization to your increment. Rather than 'i=i+2 You can use i=i+2 one iteration followed by an i=i+4 the next iteration (because every third increment of 2 is also divisible by 3). So: 5 -> 7 -> 11 -> 13 -> 17 -> 19 ... Basically this is equivalent to cutting out all 2 and 3. – Martin York Feb 12 '15 at 22:37

if (z[i] == 0)
for (j = 2 * i; j < 100000; j = j + i)
z[j] = 1;

can be changed to

if (z[i] == 0)
for (j = i * i; j < 100000; j = j + 2 * i)
z[j] = 1;

as

• k * i (with k < i) is already set to 1 by sieve k.
• when j is odd, j + i is even and so already set by even sieve.

For readability, you may split into function and do some renaming, to have something like:

const int N = 100000;
const int SQRT_N = 316;

void shieve(char (&isNotPrime)[N])
{
memset(isNotPrime, 0, N);
isNotPrime[0] = 1; // 0 is not a prime
isNotPrime[1] = 1; // 1 is not a prime
// now remove all evens from 4 up
for (int i = 4; i < N; i = i + 2) {
isNotPrime[i] = 1; // remove evens
}
// now loop through remaing up to square root of max number
for (int i = 3; i < SQRT_N; i = i + 2) {
if (!isNotPrime[i]) {
for(int j = i * i; j < N; j = j + 2 * i) {
isNotPrime[j] = 1;
}
}
}
}

• Great stuff, it worked – skwiot86 Feb 17 '15 at 7:24

In C++ you should use constants instead of #define

#define RUNS 1000

should become

const int RUNS = 1000

Loop variables should be declared in the smallest scope possible, not at the top of the file.

Always use braces, the following line is not easy to modify:

if(z[i]==0) for(j=2*i;j<100000;j=j+i) z[j]=1;

using namespace std;

## Lack of modularization

All your code is top level , you should use more functions to facilitate reuse.

• the question was for optimisation of the code to find primes faster, not optimisation of whole c++ structure. – skwiot86 Feb 17 '15 at 10:21
• @skwiot86 here reviewers are free to comment on every aspect of the code. – Caridorc Feb 18 '15 at 16:54

## 1

At first, do not use using namespace std.

## 2

You do not use any function from "math.h" header so you do not need to include it.

## 3

Next, using #include <windows.h> is also bad. You use it only for time measure. Standard C++ library has a dedicated header for this purpouse. It will make code more portable. Example:

#include <chrono>

// ... some code

auto startTime = std::chrono::high_resolution_clock::now();

// do calculations

auto endTime = std::chrono::high_resolution_clock::now();

// output

auto duration = std::chrono::duration_cast<std::chrono::milliseconds>(endTime - startTime).count();
std::cout << "Total time: " << duration << " milliseconds.";

## 4

main function should always return int, and accepted arguments are either none or

(int, char* [])

## 5

Instead of #define macro, C++ uses const keyword:

const int runs = 1000;

• Your first and fifth point duplicate one of the already existing answers... Additionally I find your formatting somewhat over the top. Still: nice observations – Vogel612 Feb 15 '15 at 13:25
• @Vogel612 Was doing this on mobile, thus it was somehow tedious to make proper formatting. Also, I wanted to make it more complex - that's why my observations are duplicating. – enedil Feb 15 '15 at 14:14


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Posted on Categories:Numerical analysis, 数值分析, 数学代写

数学代写|数值分析代写Numerical analysis代考|MATH408 ERRORS: DEFINITIONS, SOURCES, AND EXAMPLES

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数学代写|数值分析代写Numerical analysis代考|ERRORS: DEFINITIONS, SOURCES, AND EXAMPLES

The error in a computed quantity is defined as
Error $=$ true value $-$ approximate value
The relative error is a measure of the error in relation to the size of the true value being sought:
$$\text { Relative error }=\frac{\text { error }}{\text { true value }}$$
To simplify the notation when working with these quantities, we will usually denote the true and approximate values of a quantity $x$ by $x_T$ and $x_A$, respectively. Then we write
\begin{aligned} \text { Error }\left(x_A\right) &=x_T-x_A \ \operatorname{Rel}\left(x_A\right) &=\frac{x_T-x_A}{x_T} \end{aligned}
As an illustration, consider the well-known approximation
$$\pi \doteq \frac{22}{7}$$
Here $x_T=\pi=3.14159265 \ldots$ and $x_A=22 / 7=3.1428571, \ldots$,
Error $\left(\frac{22}{7}\right)=\pi-\frac{22}{7} \doteq-0.00126$
$\operatorname{Rel}\left(\frac{22}{7}\right)=\frac{\pi-(22 / 7)}{\pi} \doteq-0.000402$
Another example of error measurement is given by the Taylor remainder (1.9).
An idea related to relative error is that of significant digits. For a number $x_A$, the number of its leading digits that are correct relative to the corresponding digits in the true value $x_T$ is called the number of significant digits in $x_A$. For a more precise definition, assuming the numbers are written in decimal, calculate the magnitude of the error. $\left|x_T-x_A\right|$. If this error is less than or equal to five units in the $(m+1) s t$ digit of $x_T$, counting rightward from the first nonzero digit, then we say $x_A$ has, at least, $m$ significant digits of accuracy relative to $x_T$.

数学代写|数值分析代写Numerical analysis代考|Sources of Error

Imagine solving a scientific-mathematical problem, and suppose this involves a computational procedure. Errors will usually be involved in this process, often of several different kinds. We will give a rough classification of the kinds of error that might occur.
(E1) Mathematical equations are used to represent physical reality, a process that is called mathematical modeling. This modeling introduces error into the realworld problem that you are trying to solve. For example, the simplest model for population growth is given by
$$N(t)=N_0 e^{k t}$$
where $N(t)$ equals the population at time $t$, and $N_0$ and $k$ are positive constants. For some stages of growth of a population, when it has unlimited resources, this can be an accurate model. But more often, it will overestimate the actual population for large $t$.

The error in a mathematical model falls outside the scope of numerical analysis, but it is still an error with respect to the solution of the overall scientific problem of interest.
(E2) Blunders and mistakes are a second source of error, a familiar one to almost everyone. In the precomputer era, blunders generally consisted of isolated arithmetic errors, and elaborate check schemes were used to detect them. Today the mistakes are more likely to be programming errors. To detect these errors, it is important to have some way of checking the accuracy of the program output. When first running the program, use cases for which you know the correct answer. With a complex program, break it into smaller subprograms, each of which can be tested separately. And when you believe the entire program is correct and are running it for cases of interest, maintain a watchful eye as to whether the output is reasonable.
(E3) Many problems involve physical data, and these data contain observational error. For example, the speed of light in a vacuum is
$$c=(2.997925+\epsilon) \cdot 10^{10} \mathrm{~cm} / \mathrm{sec}, \quad|\epsilon| \leq 0.000003$$

数学代写|数值分析代写数值分析代考|ERRORS: DEFINITIONS, SOURCES, AND EXAMPLES

error $=$ 真值 $-$ 相对误差是相对于正在寻找的真实值大小的误差的度量:
$$\text { Relative error }=\frac{\text { error }}{\text { true value }}$$在处理这些量时，为了简化符号，我们通常表示一个量的真值和近似值 $x$ by $x_T$ 和 $x_A$，分别。然后我们写
\begin{aligned} \text { Error }\left(x_A\right) &=x_T-x_A \ \operatorname{Rel}\left(x_A\right) &=\frac{x_T-x_A}{x_T} \end{aligned}作为一个例子，考虑众所周知的近似$$\pi \doteq \frac{22}{7}$$

$\operatorname{Rel}\left(\frac{22}{7}\right)=\frac{\pi-(22 / 7)}{\pi} \doteq-0.000402$另一个误差测量的例子是泰勒余数(1.9)。

数学代写|数值分析代写Numerical analysis代考|Sources of Error

$$N(t)=N_0 e^{k t}$$

$$c=(2.997925+\epsilon) \cdot 10^{10} \mathrm{~cm} / \mathrm{sec}, \quad|\epsilon| \leq 0.000003$$

MATLAB代写

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