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# Show that $a_n=0.95^{n^2}$ converges to 0 using the formal definition of convergence I'm trying to prove that $$a_n=0.95^{n^2}$$ converges to 0 using the formal definition of convergence. I know that the formal definition of convergence is as follows: given $$\varepsilon>0\;\;\exists N$$ such that whenever $$n>N$$ one has $$\|x_n-L\|<\varepsilon$$ where $$L$$ is the limit point. I'm a bit rusty on these proofs and was wondering if someone could point me to the right direction of finding such an $$N$$. • Since that is a decreasing function of $n$ (I figure you are able to prove this), you just solve the inverse problem $a_n = \epsilon$ and take $N$ to be any integer larger or equal to the solution. – derpy Feb 5 at 17:49 Welp, .... we want to conclude that $$\|0.95^{n^2} - 0\| < \epsilon$$ which (as $$0.95^k>0$$ always) would be concluded by showing $$0 < 0.95^{n^2} < \epsilon$$ which will be true if $$\ln 0.95^{n^2} = n^2 \ln 0.95< \ln \epsilon$$. As $$0.95 < 1$$ we know $$\ln 0.95 < 0$$ so this will be true if $$n^2 > \frac {\ln \epsilon}{\ln 0.95}$$. The will be a positive value if we choose $$\epsilon < 1$$. So if $$\epsilon < 1$$ then this will be true if $$n > \sqrt{\frac {\ln \epsilon}{\ln 0.95}}$$. So if $$\epsilon \ge 1$$ Then $$\|0.95^{n^2}-0\| < \epsilon$$ for all $$n > 1$$. If $$0< \epsilon < 1$$ then for all $$n > \sqrt{\frac {\ln \epsilon}{\ln 0.95}}$$ we have $$\|0.95^{n^2} - 0\| < \epsilon$$ You could use the following lemma: If $$(a_n)_n$$ is such that $$0 \leq \|a_n\| \leq b_n$$ for some sequence $$(b_n)_n$$ that converges to 0, then $$(a_n)_n$$ also converges to 0. The latter can be proven using the formal $$\varepsilon$$ definition. Finally, choosing $$b_n = 0.95^n$$ and since $$\|0.95\|<1$$, $$b_n$$ converges to 0. • I would just add $0<\|a_n\|<1$ because we now it is the lower bound. – Invisible Feb 5 at 18:26 Hint Ley $$\epsilon >0$$. Pick some $$A$$ such that $$0<\frac{1}{A} < \epsilon$$. Now, by Bernoulli inequality $$\frac{1}{(.95)^{n^2}}= \left(1+ (\frac{1}{.95}-1) \right)^{n^2} \geq 1+n^2 (\frac{1}{.95}-1)$$ Make $$1+n^2 (\frac{1}{.95}-1) >A$$ Deduce that that this implies that $$0.95^{n^2}<\epsilon$$ Another Bernoulli. Let $$r =1/(1+c), c > 0$$. Then $$(1+c)^m \ge 1+mc > mc$$ so $$r^m < 1/(mc) < \epsilon$$ for $$m > 1/(c\epsilon)$$. If $$r=.95, c = 1/r-1 = 1/19$$. Putting $$n^2$$ for $$m$$ gives $$n > \sqrt{19/\epsilon}$$ as a sufficient bound.	{}
deseq2 normalized data with heatmap 1 0 Entering edit mode linouhao ▴ 10 @linouhao-15901 Last seen 12 months ago in http://bioconductor.org/packages/devel/bioc/vignettes/DESeq2/inst/doc/DESeq2.html, it used three types of data to plot heatmap ntd <- normTransform(dds) vsd <- vst(dds, blind=FALSE) rld <- rlog(dds, blind=FALSE) which data is right? and there is also another function normalized_counts <- counts(dds, normalized=TRUE) how to select? I also found a problem, I used deseq2 to find significant genes, and select the most differenet genes, and used vsd data to plot heatmap, but the plot obsviously not show high contrast color in two groups can you help me, thanks a lot. by the way, deseq2 can automatically discard low exression genes when does diff analysis, is it right? deseq2 • 1.9k views 2 Entering edit mode @kevin Last seen 53 minutes ago Republic of Ireland In DESeq2, you should use vsd or rld for clustering and heatmap analysis, and anything else that is 'downstream' of the differential expression analysis (e.g. PCA). Pay close attention to data distributions, in this regard. The differential expression analysis itself, i.e., the test statistics, can be regarded as being derived from the normalised 'counts'. It is perfectly fine to pre-filter your vsd or rld data for your statistically significantly differentially expressed genes prior to performing clustering, in which case you are performing a supervised clustering analysis. Regarding filtering, DESeq2 will not filter anything out. You can pre-filter your data prior to normalisation for low-count genes, if you wish (and this is recommended). When you derive test statistics with results(), some genes may fail the 'independent filtering' part, and their p-values set to NA. please see: I also found a problem, I used deseq2 to find significant genes, and select the most differenet genes, and used vsd data to plot heatmap, but the plot obsviously not show high contrast color in two groups You may need to specify custom breaks, or additionally scale your input data to standardised / Z scores. Take a look around the posts on Biostars. Kevin 0 Entering edit mode thanks a lot, prefilter seems to be not needed, referring the link https://support.bioconductor.org/p/65256/ 1 Entering edit mode It is no major issue to pre-filter your raw data for genes of low counts. What Michael Love is implying is that DESeq2 has some inherent 'quality control' measures that will nevertheless deal with these (genes of low counts) when performing the differential expression analysis. 0 Entering edit mode yes, you are totally right, but I am encountered with strange problems. a gene shows 32 foldchange, but the mean(assay(vsd)["NELL1", ][1:13]) is 5.625 mean(assay(vsd)["NELL1", ][14:26]) is 7.73 so why such a high differential gene , the mean expression in tumor and normal group just show such little difference? thanks a lot 1 Entering edit mode When using results(), have you additionally performed fold-change shrinkage via lfcShrink()? Regarding the vsd object, the variance-stabilised data is measured on a scale that is quite different from the normalised or raw counts; so, a direct comparison of the fold-change from results() to that derived from the values in vsd is not possible. 0 Entering edit mode I do not know what you mean shrinkage via lfcShrink(), can you show me the code? I just do like following library(DESeq2) dds <- DESeqDataSetFromMatrix(countData = cts, colData = coldata, design= ~ condition) dds <- DESeq(dds) res <- results(dds) you said can not compare vst and results() directly, But I select the most diff genes and show in heatmap to show the difference, if it can not show much difference, what is the use of heatmap 1 Entering edit mode Regarding the heatmap, we typically scale the data prior to generating the heatmap. Please take a look at my tutorial here (see 4 (a)): https://github.com/kevinblighe/E-MTAB-6141 0 Entering edit mode vst seems to hava scaled the data. you mean use lfcShrink to select most genes? this will not change the counts, am I right? the heatmap is really useful, ,can you give me a email, I want to send my data to you, is it ok? thanks a lot 0 Entering edit mode The methodology behind lfcShrinkage is contained in the link that I gave earlier. To give an answer, though: it does not alter the expression data - it just alters the fold-change estimates in the results. 0 Entering edit mode thanks a lot. but just the expression give change thge plot a lot 0 Entering edit mode Hey Kevin Blighe , Thanks for the helpful posts! I noticed in your link to this heatmap analysis, you used scale() and not vst(). However, in your earlier reply (and in some other posts I've seen), you said you should use vsd for heatmaps and clustering analyses. I've been wondering which is the correct methodology, or if using either scale() or vst() is fine. I tried both on my data and got a nice heatmap with well-defined clusters using scaled_data, where: normalized_data <- subset(counts(dds,normalized=T), rownames(counts(dds,normalized=T)) %in% significant_gene_names) scaled_data <- t(scale(t(normalized_data))) and I got a less nice looking heatmap using vst_sig, where: vst <- vst(dds, blind=FALSE) vst <- assay(vst) vst <- as.data.frame(vst) vst_sig <- vst[rownames(vst) %in% significant_gene_names,] Is it poor practice not to use the vst method? Is it okay to just use scale() as you did in your link? Thank you! 1 Entering edit mode Hey Ian, specifically just for the heatmap and/or the clustering, we can additionally scale and center the vsd or rld data. The scale() function in R, by default, merely [by row] centers your data (mean = 0) and transforms it to Z-scores. This just makes it easier for the human brain to interpret the heatmap colour gradients, whereby 0 is then just the mean expression, whereas, e.g., blue or yellow represent different standard deviations below and above that mean, respectively, with higher absolute number relating to higher intensity. In your above code, I wouldn't do t(scale(t(normalized_data))). I would instead run scale on vst_sig; so: heat <- t(scale(t(vst_sig))) You can still just use vst_sig on its own, with no scaling anywhere, but it will be more difficult to set colours and breaks. Some do prefer to also use the counts, i.e., normalized_data; however, these are positive integer values (or double values if normalised) that are shifted toward 0 and follow a negative binomial distribution. If you run hist() on normalized_data, vst, and t(scale(t(vst))), you'll see how wildly different are the distributions. In the case where you use normalized_data, the colour scheme would usually be white for 0, then a gradual increasing gradient toward dark red, dark purple, dark yellow, etc. 1 Entering edit mode Got it, thank you! I will use vsd and potentially scale from there.	{}
# Music Structure Analysis: General Principles Following Section 4.1 of [Müller, FMP, Springer 2015], we discuss in this notebook general principles for segmenting and structuring music recordings. For an overview, we also refer to the following literature. • Jouni Paulus, Meinard Müller, and Anssi Klapuri: Audio-based Music Structure Analysis. Proceedings of the International Conference on Music Information Retrieval (ISMIR), Utrecht, The Netherlands, pp. 625–636, 2010. • Juan P. Bello, Peter Grosche, Meinard Müller, and Ron J. Weiss: Content-based Methods for Knowledge Discovery in Music. In Rolf Bader (ed.): Springer Handbook on Systematic Musicology, Springer, Berlin, Heidelberg: 823–840, 2018. ## General Principles¶ Music structure analysis is a multifaceted and often ill-defined problem that depends on many different aspects. First of all, the complexity of the problem depends on the kind of music representation to be analyzed. For example, while it is comparatively easy to detect certain structures such as repeating melodies in sheet music, it is often much harder to automatically identify such structures in audio representations. Second, there are various principles including homogeneity, repetition, and novelty that a segmentation may be based on. Third, one also has to account for different musical dimensions, such as melody, harmony, rhythm, or timbre. Finally, the segmentation and structure largely depend on the musical context and the temporal hierarchy to be considered. The following figure gives an overview of various aspects that need to be considered when dealing with musical structures. In the following, we discuss these aspects in more detail. In particular, our goal is to raise the awareness that computational procedures as described in the subsequent sections are often based on simplifying model assumptions that only reflect certain aspects of the complex structural properties of music. ## Segmentation and Structure Analysis¶ The tasks of segmenting and structuring multimedia documents are of fundamental importance not only for the processing of music signals but also for general audio-visual content. Segmentation typically refers to the process of partitioning a given document into multiple segments with the goal of simplifying the representation into something that is more meaningful and easier to analyze than the original document. For example, in image processing the goal is to partition a given image into a set of regions such that each region is similar with respect to some characteristic such as color, intensity, or texture. Region boundaries can often be described by contour lines or edges at which the image brightness or other properties change sharply and reveal discontinuities. In music, the segmentation task is to decompose a given audio stream into acoustically meaningful sections each corresponding to a continuous time interval that is specified by a start and end boundary. At a fine level, the segmentation may aim to find the boundaries between individual notes or to find the beat intervals specified by beat positions. At a coarser level, the goal may be to detect changes in instrumentation or harmony or to find the boundaries between verse and chorus sections. Also, discriminating between silence, speech, and music, finding the actual beginning of a music recording, or separating the applause at the end of a performance are typical segmentation tasks. Going beyond mere segmentation, the goal of structure analysis is to also find and understand the relationships between the segments. For example, certain segments may be characterized by the instrumentation. There may be sections played only by strings. Sections played by the full orchestra may be followed by solo sections. The verse sections with a singing voice may be alternated with purely instrumental sections. Or a soft and slow introductory section may precede the main theme played in a much faster tempo. Furthermore, sections are often repeated. Most events of musical relevance are repeated in a musical work in one way or another. However, repetitions are rarely identical copies of the original section, but undergo modifications in aspects such as the lyrics, the instrumentation, or the melody. One main task of structure analysis is to not only segment the given music recording, but to also group the segments into musically meaningful categories (e.g., intro, chorus, verse, outro). The challenge in computational music structure analysis is that structure in music arises from many different kinds of relationships including repetition, contrast, variation, and homogeneity. In view of the various principles that crucially influence the musical structure, a large number of different approaches to music structure analysis have been developed. In the following, we want to roughly distinguish three different classes of methods. • First, repetition-based methods are used to identify recurring patterns. • Second, novelty-based methods are employed to detect transitions between contrasting parts. • Third, homogeneity-based methods are used to determine passages that are consistent with respect to some musical property. Note that novelty-based and homogeneity-based approaches are two sides of a coin: novelty detection is based on observing some surprising event or change after a more homogeneous segment. While the aim of novelty detection is to locate the changes' time positions, the focus of homogeneity analysis lies in the identification of longer passages that are coherent with respect to some musical property. The following figure illustrates that similar segmentation and structuring principles apply for other domains such as image and 3D data. ## Musical Structure¶ To specify musical structures, we now introduce some terminology. First of all, we distinguish between a piece of music (in an abstract sense) and a particular audio recording (an actual performance) of the piece. The term part is used in the context of the abstract music domain, whereas the term segment is used for the audio domain. Musical parts are typically denoted by the capital letters $A,B,C,\ldots$ in the order of their first occurrence, where numbers (often written as subscripts) indicate the order of repeated occurrences. As an example, we consider the Hungarian Dance No. 5 by Johannes Brahms. This dance has been arranged for a wide variety of instruments and ensembles, ranging from piano versions to versions for full orchestra. The following figure shows a sheet music representation for the violin voice of an arrangement for full orchestra. The musical structure is $A_1A_2B_1B_2CA_3B_3B_4D$, which consists of three repeating $A$-parts, four repeating $B$-parts, as well as a $C$-part and a short closing $D$-part. The $A$-part has a substructure consisting of two more or less repeating subparts. Furthermore, as becomes apparent when looking at the musical score, the middle $C$-part may be further subdivided into a substructure that may be described by $d_1d_2e_1e_2e_3e_4$. In music notation, such subparts are often indicated using small letters $a,b,c,\ldots$. ## Audio Structure Analysis¶ Given a recording of a piece of music, the goal of audio structure analysis (as considered in this chapter) is to find the segments within the recording that correspond to the various parts of a musical structure. The following examples show that different performances not only may differ with regard to aspects such as instrumentation and tempo, but also with regard to the global musical structure. Orchestral version (Ormandy, $A_1A_2B_1B_2CA_3B_3B_4D$) Orchestral version (Fulda Symphony Orchestra, $A_1A_2B_1B_2CA_3B_3B_4D-\mathrm{Applause}$) Violin version (Sibelius, $A_1A_2B_1B_2CA_3B_3B_4D$) Piano version (RWC, $A_1A_2B_1B_2CA_3B_3D$) Orchestral version (Chaplin, $A_1A_2B_1CA_3B_2D$) In [1]: import IPython.display as ipd ## Structure Annotations¶ To evaluate automated approaches for audio structure analysis, one often uses reference annotations generated by humans in a manual process. This topic will be discussed in the FMP notebook on evaluation in more detail. To store annotations, we use in the FMP notebooks a simple CSV file format. In the following code cell, we read and visualize a reference annotations for the Ormandy recording of our Brahms example, where the time axis is specified in seconds. When computing the musical structure using automated methods, one typically starts with converting a signal into a frame-based feature representation. As a result, the computed start and end positions of structural elements are given in frame indices. To compare annotations with computed results, one way is to convert the physical time axis (specified in seconds) of a given annotation into a discrete time axis (specified in frame indices). Furthermore, one may want to adjust the label annotations, e.g., leaving out digits that indicate the order of repeated occurrences of the same part. In the next code cell, we provide such a conversion function. In [2]: import numpy as np import os, sys, librosa from scipy import signal from scipy.interpolate import interp1d from matplotlib import pyplot as plt import matplotlib.gridspec as gridspec import IPython.display as ipd import pandas as pd sys.path.append('..') import libfmp.b import libfmp.c2 import libfmp.c3 import libfmp.c4 import libfmp.c6 %matplotlib inline def convert_structure_annotation(ann, Fs=1, remove_digits=False, index=False): """Convert structure annotations Notebook: C4/C4S1_MusicStructureGeneral.ipynb Args: ann (list): Structure annotions Fs (scalar): Sampling rate (Default value = 1) remove_digits (bool): Remove digits from labels (Default value = False) index (bool): Round to nearest integer (Default value = False) Returns: ann_converted (list): Converted annotation """ ann_converted = [] for r in ann: s = r[0] * Fs t = r[1] * Fs if index: s = int(np.round(s)) t = int(np.round(t)) if remove_digits: label = ''.join([i for i in r[2] if not i.isdigit()]) else: label = r[2] ann_converted = ann_converted + [[s, t, label]] return ann_converted get_color_for_annotation_file = libfmp.c4.get_color_for_annotation_file def read_structure_annotation(fn_ann, fn_ann_color='', Fs=1, remove_digits=False, index=False): """Read and convert structure annotation and colors Notebook: C4/C4S1_MusicStructureGeneral.ipynb Args: fn_ann (str): Path and filename for structure annotions fn_ann_color (str): Filename used to identify colors (Default value = '') Fs (scalar): Sampling rate (Default value = 1) remove_digits (bool): Remove digits from labels (Default value = False) index (bool): Round to nearest integer (Default value = False) Returns: ann (list): Annotations color_ann (dict): Color scheme """ ann = [(start, end, label) for i, (start, end, label) in df.iterrows()] ann = convert_structure_annotation(ann, Fs=Fs, remove_digits=remove_digits, index=index) color_ann = {} if len(fn_ann_color) > 0: color_ann = get_color_for_annotation_file(fn_ann_color) if remove_digits: color_ann_reduced = {} for key, value in color_ann.items(): key_new = ''.join([i for i in key if not i.isdigit()]) color_ann_reduced[key_new] = value color_ann = color_ann_reduced return ann, color_ann # Annotation file filename = 'FMP_C4_Audio_Brahms_HungarianDances-05_Ormandy.csv' fn_ann = os.path.join('..', 'data', 'C4', filename) print('Original annotations with time specified in seconds') print('Annotations:', ann) print('Colors:', color_ann) fig, ax = libfmp.b.plot_segments(ann, figsize=(8, 1.2), colors=color_ann, time_label='Time (seconds)') plt.show() Fs = 2 ann, color_ann = read_structure_annotation(fn_ann, fn_ann_color=filename, Fs=Fs, remove_digits=True, index=True) print('Converted annotations (Fs = %d) with reduced labels (removing digits)'%Fs) print('Annotations:', ann) print('Colors:', color_ann) fig, ax = libfmp.b.plot_segments(ann, figsize=(8, 1.2), colors=color_ann, time_label='Time (frames)') plt.show() Original annotations with time specified in seconds Annotations: [[0.0, 1.01, ''], [1.01, 22.11, 'A1'], [22.11, 43.06, 'A2'], [43.06, 69.42, 'B1'], [69.42, 89.57, 'B2'], [89.57, 131.64, 'C'], [131.64, 150.84, 'A3'], [150.84, 176.96, 'B3'], [176.96, 196.9, 'B4'], [196.9, 199.64, '']] Colors: {'A1': [1, 0, 0, 0.2], 'A2': [1, 0, 0, 0.2], 'A3': [1, 0, 0, 0.2], 'B1': [0, 1, 0, 0.2], 'B2': [0, 1, 0, 0.2], 'B3': [0, 1, 0, 0.2], 'B4': [0, 1, 0, 0.2], 'C': [0, 0, 1, 0.2], '': [1, 1, 1, 0]} Converted annotations (Fs = 2) with reduced labels (removing digits) Annotations: [[0, 2, ''], [2, 44, 'A'], [44, 86, 'A'], [86, 139, 'B'], [139, 179, 'B'], [179, 263, 'C'], [263, 302, 'A'], [302, 354, 'B'], [354, 394, 'B'], [394, 399, '']] Colors: {'A': [1, 0, 0, 0.2], 'B': [0, 1, 0, 0.2], 'C': [0, 0, 1, 0.2], '': [1, 1, 1, 0]} ## Musical Dimensions¶ The applicability of the different segmentation principles very much depends on the musical and acoustic properties of the audio signal to be analyzed. For example, structural boundaries may be based on changes in harmony, timbre, or tempo. The first step in automated structure analysis is to transform the given music recording into a suitable feature representation that captures the relevant musical properties. In the following, we have a look at three different types of feature representations (chromagram, MFFC, tempogram). In particular, we demonstrate that even for a single feature type there are many variants and different parameters settings that have a significant influence on a feature representations quality. As our running example, we continue with the Ormandy (orchestral) version of our Brahms' Hungarian Dance. In the following code cell, we read in the audio file and show the waveform along with the structure annotation. In [3]: # Annotations filename = 'FMP_C4_Audio_Brahms_HungarianDances-05_Ormandy.csv' fn_ann = os.path.join('..', 'data', 'C4', filename) ann, color_ann = read_structure_annotation(fn_ann, fn_ann_color=filename, Fs=1, remove_digits=False) # Waveform fn_wav = os.path.join('..', 'data', 'C4', 'FMP_C4_Audio_Brahms_HungarianDances-05_Ormandy.wav') Fs = 22050 x_dur = x.shape[0]/Fs # Visualization fig, ax = plt.subplots(2, 1, gridspec_kw={'width_ratios': [1], 'height_ratios': [1, 0.5]}, figsize=(8, 2.5)) libfmp.b.plot_signal(x, Fs, ax=ax[0], title='Waveform of audio signal') libfmp.b.plot_segments_overlay(ann, ax=ax[0], time_max=x_dur, print_labels=False, label_ticks=False, edgecolor='gray', colors = color_ann, fontsize=10, alpha=0.1) libfmp.b.plot_segments(ann, ax=ax[1], time_max=x_dur, colors=color_ann, time_label='Time (seconds)') plt.tight_layout() ## Chromagram Representation¶ First, the chroma-based representation (see Section 3.1.2 of [Müller, FMP, Springer 2015]) relates to harmonic and melodic properties of the music recording. The patterns visible in the chromagram reveal important structural information. For example, the four repeating $B$-part segments are clearly visible as four similar characteristic subsequences in the chromagram. Furthermore, the $C$-part segment stands out in the chromagram by showing a high degree of homogeneity throughout the entire section. Indeed, for all chroma features of this segment, most of the signal's energy is contained in the $\mathrm{G}$-, $\mathrm{B}$-, and $\mathrm{D}$-bands (which is not surprising since the $C$-part is in $\mathrm{G}$ major). In contrast, as for the $A$-part segments, many chroma vectors have dominant entries in the $\mathrm{G}$-, $\mathrm{B}^\flat$-, and $\mathrm{D}$-bands (which nicely reflects that this part is in $\mathrm{G}$ minor). In [4]: # Chromagram N, H = 4096, 2048 chromagram = librosa.feature.chroma_stft(y=x, sr=Fs, tuning=0, norm=2, hop_length=H, n_fft=N) filt_len = 41 down_sampling = 10 filt_kernel = np.ones([1,filt_len]) chromagram_smooth = signal.convolve(chromagram, filt_kernel, mode='same')/filt_len chromagram_smooth = chromagram_smooth[:,::down_sampling] chromagram_smooth, Fs_smooth = \ libfmp.c3.smooth_downsample_feature_sequence(chromagram, Fs/H, filt_len=filt_len, down_sampling=down_sampling) # Visualization fig, ax = plt.subplots(3, 2, gridspec_kw={'width_ratios': [1, 0.03], 'height_ratios': [2, 2, 0.5]}, figsize=(9, 5)) libfmp.b.plot_chromagram(chromagram, Fs=Fs/H, ax=[ax[0,0], ax[0,1]], chroma_yticks = [0,4,7,11], title='Chromagram (resolution %0.1f Hz)'%(Fs/H), ylabel='Chroma', colorbar=True); libfmp.b.plot_chromagram(chromagram_smooth, Fs_smooth, ax=[ax[1,0], ax[1,1]], chroma_yticks = [0,4,7,11], title='Smoothed chromagram (resolution %0.1f Hz)'%Fs_smooth, ylabel='Chroma', colorbar=True); libfmp.b.plot_segments(ann, ax=ax[2,0], time_max=x_dur, colors=color_ann, time_label='Time (seconds)') ax[2,1].axis('off') plt.tight_layout() ## MFCC Representation¶ Besides melody and harmony, the instrumentation and timbral characteristics are of great importance for the human perception of music structure. In the context of timbre-based structure analysis, one often uses mel-frequency cepstral coefficients (MFCCs), which were originally developed for automated speech recognition. As for music, MFFC-based features are a mid-level representation that somehow correlate to aspects such as instrumentation and timbre. This is reflected by the following figure, where on can recognize that MFCC features within the $A$-part segments are different from the ones in the $B$-part and $C$-part segments. For many music recordings such as pop songs, where sections with singing voice alternate with purely instrumental or percussive sections, MFCC-based feature representations are well suited for novelty-based and homogeneity-based segmentation. For visualization purposes, we show in the following figure MFCC-based representations for coefficients $0$ to $19$ as well as for coefficients $4$ to $14$ (removing some of the very dominant lower coefficients). In [5]: # MFCC N, H = 4096, 2048 X_MFCC = librosa.feature.mfcc(y=x, sr=Fs, hop_length=H, n_fft=N) coef = np.arange(4,15) X_MFCC_upper = X_MFCC[coef,:] # Visualization fig, ax = plt.subplots(3, 2, gridspec_kw={'width_ratios': [1, 0.03], 'height_ratios': [2, 2, 0.5]}, figsize=(9, 5)) libfmp.b.plot_matrix(X_MFCC, Fs=Fs/H, ax=[ax[0,0], ax[0,1]], title='MFCC (coefficents 0 to 19)', ylabel='', colorbar=True); ax[0,0].set_yticks([0, 10, 19]) libfmp.b.plot_matrix(X_MFCC_upper, Fs=Fs/H, ax=[ax[1,0], ax[1,1]], title='MFFC (coefficents 4 to 14)', ylabel='', colorbar=True); ax[1,0].set_yticks([0, 5, 10]) ax[1,0].set_yticklabels(coef[0] + [0, 5, 10]) libfmp.b.plot_segments(ann, ax=ax[2,0], time_max=x_dur, colors=color_ann, time_label='Time (seconds)'); ax[2,1].axis('off') plt.tight_layout() ## Tempogram Representation¶ In music structure analysis, tempo and beat information may also be used in combination with homogeneity-based segmentation approaches. Instead of extracting such information explicitly, a mid-level feature representation that correlates to tempo and rhythm may suffice for deriving a meaningful segmentation at a higher structural level. This is illustrated in the above Brahms example using a tempogram—a mid-level representation that encodes local tempo information. More precisely, a cyclic variant of a tempogram is shown, where tempi differing by a power of two are identified—similar to cyclic chroma features, where pitches differing by octaves are identified (see see Section 6.2.4 of [Müller, FMP, Springer 2015]). Having a look at the tempogram representation, one can notice that the different musical parts are played in different tempi (even though the representation does not reveal the exact tempi). Furthermore, there are sections where the tempogram features do not have any dominating entries, which may indicate that there is no clear notion of a tempo in the recording. In [6]: # Tempogram nov, Fs_nov = libfmp.c6.compute_novelty_spectrum(x, Fs=Fs, N=2048, H=512, gamma=100, M=10, norm=True) nov, Fs_nov = libfmp.c6.resample_signal(nov, Fs_in=Fs_nov, Fs_out=100) X, T_coef, F_coef_BPM = libfmp.c6.compute_tempogram_fourier(nov, Fs_nov, N=1000, H=100, Theta=np.arange(30, 601)) octave_bin = 30 tempogram_F = np.abs(X) output = libfmp.c6.compute_cyclic_tempogram(tempogram_F, F_coef_BPM, octave_bin=octave_bin) tempogram_cyclic_F = output[0] F_coef_scale = output[1] tempogram_cyclic_F = libfmp.c3.normalize_feature_sequence(tempogram_cyclic_F, norm='max') output = libfmp.c6.compute_tempogram_autocorr(nov, Fs_nov, N=500, H=100, norm_sum=False, Theta=np.arange(30, 601)) tempogram_A = output[0] T_coef = output[1] F_coef_BPM = output[2] output = libfmp.c6.compute_cyclic_tempogram(tempogram_A, F_coef_BPM, octave_bin=octave_bin) tempogram_cyclic_A = output[0] F_coef_scale = output[1] tempogram_cyclic_A = libfmp.c3.normalize_feature_sequence(tempogram_cyclic_A, norm='max') # Visualization fig, ax = plt.subplots(3, 2, gridspec_kw={'width_ratios': [1, 0.03], 'height_ratios': [2, 2, 0.5]}, figsize=(9, 5)) libfmp.b.plot_matrix(tempogram_cyclic_F, T_coef=T_coef, ax=[ax[0,0], ax[0,1]], title='Fourier-based cyclic tempogram', ylabel='Scaling', colorbar=True, clim=[0, 1]) libfmp.c6.set_yticks_tempogram_cyclic(ax[0,0], octave_bin, F_coef_scale, num_tick=5) libfmp.b.plot_matrix(tempogram_cyclic_A, T_coef=T_coef, ax=[ax[1,0], ax[1,1]], title='Autocorrelation-based cyclic tempogram', ylabel='Scaling', colorbar=True, clim=[0, 1]) libfmp.c6.set_yticks_tempogram_cyclic(ax[1,0], octave_bin, F_coef_scale, num_tick=5) libfmp.b.plot_segments(ann, ax=ax[2,0], time_max=x_dur, colors=color_ann, time_label='Time (seconds)') ax[2,1].axis('off') plt.tight_layout() ## Further Notes¶ Besides the various musical dimensions, there is another aspect one should keep in mind when looking for suitable feature representations: the temporal dimension. In all of the above-mentioned feature representations, an analysis window is shifted over the music signal. Obviously, the length of the analysis window as well as the hop size parameter have a crucial influence on the quality of the feature representation. For example, long window sizes and large hop sizes may be beneficial for smoothing out irrelevant local variations, which is often a desired property in homogeneity-based segmentation. On the downside, the temporal resolution decreases and important details may get lost, which can lead to problems when locating the exact segmentation boundaries. In summary, a suitable choice of feature representations and parameter settings very much depends on the application context. Humans constantly and often unconsciously adapt themselves to the musical and acoustic characteristics of what they listen to. The richness and variety of musical structures make computational structure analysis a challenging problem. Acknowledgment: This notebook was created by Meinard Müller.	{}
# Article Full entry \| PDF (0.3 MB) Keywords: stationary Navier-Stokes equations; non-vanishing outflow; 2-dimensional semi-infinite channel; symmetry Summary: We consider the steady Navier-Stokes equations in a 2-dimensional unbounded multiply connected domain $\Omega$ under the general outflow condition. Let $T$ be a 2-dimensional straight channel $\mathbb{R} \times (-1,1)$. We suppose that $\Omega \cap \lbrace x_1 < 0 \rbrace$ is bounded and that $\Omega \cap \lbrace x_1 > -1 \rbrace = T \cap \lbrace x_1 > -1 \rbrace$. Let $V$ be a Poiseuille flow in $T$ and $\mu$ the flux of $V$. We look for a solution which tends to $V$ as $x_1 \rightarrow \infty$. Assuming that the domain and the boundary data are symmetric with respect to the $x_1$-axis, and that the axis intersects every component of the boundary, we have shown the existence of solutions if the flux is small (Morimoto-Fujita [8]). Some improvement will be reported in this note. We also show certain regularity and asymptotic properties of the solutions. References: [1] Amick, C. J.: Steady solutions of the Navier-Stokes equations for certain unbounded channels and pipes. Ann. Scuola Norm. Sup. Pisa 4 (1977), 473–513. MR 0510120 [2] Amick, C. J.: Properties of steady Navier-Stokes solutions for certain unbounded channel and pipes. Nonlinear Analysis, Theory, Methods & Applications, Vol. 2 (1978), 689–720. MR 0512162 [3] Fujita, H.: On the existence and regularity of the steady-state solutions of the Navier-Stokes equation. J. Fac. Sci., Univ. Tokyo, Sec. I 9 (1961), 59–102. MR 0132307 \| Zbl 0111.38502 [4] Fujita, H.: On stationary solutions to Navier-Stokes equations in symmetric plane domains under general out-flow condition. Proceedings of International Conference on Navier-Stokes Equations, Theory and Numerical Methods, June 1997, Varenna Italy, Pitman Reseach Notes in Mathematics 388, pp. 16–30. MR 1773581 [5] Galdi, G. P.: An Introduction to the Mathematical Theory of the Navier-Stokes Equations. Springer, 1994. Zbl 0949.35005 [6] Ladyzhenskaya, O. A.: The Mathematical Theory of Viscous Incompressible Flow. Gordon and Breach, New York, 1969. MR 0254401 \| Zbl 0184.52603 [7] Morimoto, H., Fujita, H.: A remark on existence of steady Navier-Stokes flows in a certain two dimensional infinite tube. Technical Reports Dept. Math., Math-Meiji 99-02, Meiji Univ. [8] Morimoto, H., Fujita, H.: On stationary Navier-Stokes flows in 2D semi-infinite channel involving the general outflow condition. NSEC7, Ferrara, Italy,. Partner of	{}
# JEE Main 2020 Question Paper with Solution (Jan 8th first shift) ## JEE Main 2020 Question Paper with Solutions, Morning Session ( 9:30 am to 12:30 pm) National Test Agency( NTA) Is conducting JEE Mains 2020 Exam in two shifts, through an online mode. JEE Main 2020 is a computer-based test. In this article, we are going to share a JEE main 2020 question paper with solutions for the morning session of 8th January 2020. These solutions are prepared by experts. We suggest you go through the article once, it will help you to understand your performance and with the help of this article, you will be able to understand that which concept is important, Type of exam questions, etc. JEE Main 2020 exam - Chemistry ( 8th January first shift) Q1. The number of chiral carbons in penicillin. Solution: As shown in the figure below, there are 3 chiral carbons in penicillin. Q2. The correct order for ionization enthalpy for the elements Na, Mg, Al and Si Solution: The correct ionization enthalpy order is given below: Si > Mg > Al > Na Q3) What will happen when we will heat gypsum above 393K? Solution The reaction occurs as follows: $\mathrm{CaSO_{4}.2H_{2}O\: \overset{393K}{\longrightarrow}\: CaSO_{4}.\frac{1}{2}H_{2}O\: +\: \frac{3}{2}H_{2}O}$ Gypsum Plaster of Paris Q4) The third ionisation enthalpy of the following elements is minimum for: (a) Co (b) Mn (c) Fe (d) Ni Solution The third ionisation potential data for the given element is given below: Co3+ - 3237 Mn3+ - 3258 Fe3+ - 2967 Ni3+ - 3400 The electronic configuration of Fe2+ is [Ar]3d6. Thus it is easier to remove one electron to get [Ar]3d5 electronic configuration as it is half-filled and thus it is stable. Therefore, Option(c) is correct. Q5) Gases that cause greenhouse effect are: Solution The gases that cause greenhouse gases are: CO2, O3 and H2O Q6) Arrange them in increasing C-O bond length Phenol, p-methoxyphenol and methanol Solution The correct order of bond length is given below: Phenol < p-methoxyphenol < methanol Q7) Following vanderwaal forces are present in ethyl acetate liquid: (a) H-bond, London forces. (b) dipole-dipole interaction, H-bond (c) dipole–dipole interaction, London forces (d) H-bond, dipole-dipole interaction, London forces Solution Ethyl acetate is polar molecule so dipole-dipole interaction will be present there Q8) The complex that can show fac- and mer- isomerism is: (a) [CoCl2(en)2] (b) [Co(NH3)4Cl2]+ (c) [Co(NH3)3(NO2)3]+ (d) [Pt(NH3)2Cl2] Solution Complexes of the type MA3B3 exist in tow geometrical forms which are named as facial (fac-) and meridional (mer-). Co(NH3)3(NO2)3]+ can be represented in fac- and mer- isomeric forms as follows: Q9) Number of S–O bond in S2O82– and number of S–S bond in Rhombic sulphur are respectively: Solution The structure of S2O82– is given below: Rhombic sulphur Thus, the number of S-S bonds is 8 and the number of S-O bonds is also 8. Q10) Arrange the following compounds in order of dehydrohalogenation (E1) reaction. (a) (b) (c) (d) Solution E1 reaction proceeds via carbocation formation, therefore greater the stability of carbocation, faster the E1 reaction C > D > B > A Q11) Find the value (in ml) of AgNO3[0.125M] which reacts with 0.3g of [Co(NH3)6]Cl3. Solution The reaction occurs as follows: $\mathrm{AgNO_{3}\: +\: [Co(NH_{3})_{6}]Cl_{3}\: \rightarrow \: 3AgCl}$ Molecular weight of [Co(NH3)6]Cl3 = 267.48g/mol Number of moles of [Co(NH3)6]Cl3 = 0.3/267.48 = 1.125 x 10-3 moles. 1 mole of [Co(NH3)6]Cl3 gives 3 moes of AgCl. Thus, moles of AgCl = 3 x 1.125 x 10-3 moles Now, 3 moles of AgNO3 reacts with 3 moles of AgCl Thus, moles off AgNO3 = 3 x 1.125 x 10-3 moles Now, volume(L) of AgNO3 = (3 x 1.125 x 10-3)/0.125 = 27 x 10-3 L volume(in mL) of AgNO3 = 27ml Q12) Change in activation energy when enzyme is added at T. Rate of reaction of increase 106 times. (a) -2.303 x 6RT (b) +6RT (c) - RT (d) +2.303 x 6RT Solution We know: $\mathrm{-logk\: =\: \frac{-E_{a}}{2.303RT}\: +\: log\, A_{o}}$ Catalyst decrease the activation energy require for a reaction to proceed and thus increase the reaction rate. Therefore, Option(a) is correct. Q13) How to calculate the strength of NaOH? (a) NaOH from burette titrated with conc. H2SO4 from conical flask. (c) NaOH from burette titrated with oxalic acid from conical flask. Solution To calculate the strength of NaOH, it is titrated against oxalic acid. Thus, Option(C) is correct. Q14) Which of the following option is incorrect? (a) Glucose exists in 2 anomeric forms. (b) Glucose reacts with hydroxylamine to give oxime. (c) Glucose reacts with Schiff reagent. (d) Glucose forms pentaacetate. Solution Glucose does not react with Schiff reagent because Schiff reagent is used for aldehydes. Therefore, Option(C) is correct. Q15) According to Hardy Schultz rule, correct order of flocculation value for Fe(OH)3 sol is : Solution According to hardy-schultz rule, $\textrm{Coagulation value or flocculation value}\propto \frac{1}{\textrm{Coagulation power}}$ Q16) Given, for H-atom $\mathrm{\overline{\nu }\: =\: R_{H}\left [ \frac{1}{n^{2}_{1}}-\frac{1}{n^{2}_{2}} \right ]}$ Select the correct options regarding this formula for Balmer series. (A) n1 = 2 (B) Ionization energy of H atom can be calculated from above formula. (C) $\mathrm{\lambda _{maximum}}$ is for n2 = 3. (D) If $\mathrm{\lambda }$ decreases then spectrum lines will converse. Solution Since, $\mathrm{\Delta E\: \propto\:\frac{1}{\lambda} }$ so for wavelength to be longest or maximum the energy gap should be minimum. For Balmer series, transition for longest wavelength is from n = 3 to n = 2. Therefore, Option(c) is correct. Q17) Select the correct stoichiometry and its ksp value according to given graphs. (a) XY, Ksp = 2×10–6 (b) XY2, Ksp = 4×10–9 (c) X2Y, Ksp = 9×10–9 (d) XY2, Ksp = 1×10–9 Solution $\mathrm{XY(s)\: \rightarrow \: X^{+}(aq)\: +\: Y^{-}(aq)}$ 2 x 10-3 10-3 Ksp = [X+ ] [Y ] Ksp = 2 × 10–3 × 10–3 Ksp = 2 × 10–6 Therefore, Optioin(a) is correct. Q18) A gas undergoes expansion according to the following graph. Calculate work done by the gas. Solution $W = (8-2)^2 + \frac{1}{2}((12-8)\times (8-2))=48$ Q19) Calculated the mass of FeSO4.7H2O, which must be added in 100 kg of wheat to get 10 PPM of Fe. Solution $\mathrm{ppm=\frac{\textrm{Solute in gram }}{\textrm{Solution in gram}}\times 10^6}$ $\mathrm{10=\frac{\textrm{mass of Fe in gram }}{\textrm{100 X 1000}}\times 10^6}$ Fe = 1 gram 56 g of Fe in 1 mol of FeSO4.7H2O (molar mass 278 gram /mol) Therefore 1 gram of Fe is present in = (1/56) moles of FeSO4.7H2O mass of FeSO4.7H2O = $\mathrm{\frac{1}{56}\times 278=4.96\ gram}$ Q20) Given $2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \quad \mathrm{E}^{\circ}=-1.23 \mathrm{V}$ Calculate electrode potential at pH = 5. Solution $\begin{array}{l}{E=-1.23-\frac{0.0591}{4} \log \left[\mathrm{H}^{+}\right]^{4}} \\ {=-1.23+0.0591 \times \mathrm{pH}=-1.23+0.0591 \times 5} \\ {=-1.23+0.2955=-0.9345 \mathrm{V}=-0.93 \mathrm{V}}\end{array}$ Q21) Reagent used for the given conversion is: Solution B2H6 is very selective and usually used to reduce acid to alcohol. Q22) Major product in the following reaction is Solution Q23) (a) Intermolecular force of attraction of X > Y (b) Intermolecular force of attraction of X < Y. (C) Intermolecular force of attraction of Z < X. Select the correct option(s). (a) A and C (b) A and B (c) B only (d) B and C Solution At a particular temperature as intermolecular force of attraction increases vapour pressure decreases. Therefore, Option(c) is correct. Q24) Product A and B are respectively: (a) (b) (c) (d) Solution [A] is more stable radical and undergoes Markovnikov addition to form [B]. Q25) Two liquids isohexane and 3-methylpentane has boiling point 60°C and 63°C. They can be separated by: (a) Simple distillation and isohexane comes out first. (b) Fractional distillation and isohexane comes out first. (c) Simple distillation and 3-Methylpantane comes out first. (d) Fractional distillation and 3-Methylpantane comes out first. Solution Liquid having lower boiling point comes out first in fractional distillation. Simple distillation can't be used as boiling point difference is very small. JEE Main 2020 exam - Mathematics (8th January first shift) Q1. Find the value of $\lim _{x\to 0}\left(\frac{3x^2+2}{\:7x^2+2}\right)^{\frac{1}{x^2}}$ Solution: $\\\lim _{x\to 0}\left(\frac{3x^2+2}{\:7x^2+2}\right)^{\frac{1}{x^2}}=\lim _{x\to 0}\left(1+\frac{3x^2+2}{\:7x^2+2}-1\right)^{\frac{1}{x^2}}\\\lim_{x\rightarrow 0}\left ( 1+\frac{-4x^2}{7x^2+2} \right )^{1/x^2}=e^{\lim _{x\to 0}\left(-\frac{4x^2}{7x^2+2}\right)\left(\frac{1}{x^2}\right)}\\=\frac{1}{e^2}$ Q2. If y(x) is a solution of the differential equation $\sqrt{1-x^{2}} \frac{d y}{d x}+\sqrt{1-y^{2}}=0$ such that, then Solution: $\sqrt{1-x^{2}} \frac{d y}{d x}=-\sqrt{1-y^{2}}$ ${\frac{d y}{\sqrt{1-y^{2}}}=-\frac{d x}{\sqrt{1-x^{2}}} \Rightarrow \sin ^{-1} y=-\sin ^{-1} x+c} \\ {x=\frac{1}{2}, y=\frac{\sqrt{3}}{2} \Rightarrow \frac{\pi}{2}=-\frac{\pi}{6}-c}$ \begin{aligned} \sin ^{-1} y &=\frac{\pi}{2}-\sin ^{-1} x \\ &=\cos ^{-1} x \Rightarrow y=\sin (\cos^{-1}x) \end{aligned} $x=\frac{1}{\sqrt{2}} \Rightarrow y =\sin \left(\frac{\pi}{4}\right) =\frac{1}{\sqrt{2}}$ Q3. Find the sum $\sum_{k=1}^{20}(1+2+3+\ldots \ldots+k)$ Solution: $\sum_{k=1}^{20} \frac{k(k+1)}{2}=\frac{1}{2}\left(\sum_{k=1}^{20} k^{2}+\sum_{k=1}^{20} k\right)$ $=\frac{1}{2}\left(\frac{20(20+1)(2 \times 20+1)}{6}+\frac{20(20+1)}{2}\right)$ $=\frac{1}{2}((70 \times 41)+210)=1540$ Q4. In a bag there are 5 red balls, 3 white balls and 4 black balls. Four balls are drawn from the bag. Find the number of ways of in which at most 3 red balls are selected Solution: $\\\text{Red Balls(5)}:\;\;\;\;\;\;0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3\\\text{Other Balls(7)}:\;\;\;4\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;1\\\text{\;}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;^5C_0\times^7C_4\;\;\;\;\;\;^5C_1\times^7C_3\;\;\;\;^5C_2\times^7C_2\;\;\;\;\;\;^5C_3\times^7C_1$ $=1 \times 35+5 \times 35+10 \times 21+10\times 7=490$ Q5. No.of 3×3 matrices from the set {-1,0,1} such that $tr(AA^T)$ is 3 Solution: Let matrix A $A=\begin{bmatrix} a &b &c \\ d& e &f \\ g &h & i \end{bmatrix}\\ A^T=\begin{bmatrix} a &d &g \\ b& e &h \\ c &f & i \end{bmatrix} \\ \text{ trace}(AA^T)=a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2\\\text{ Total number of cases}=^{9}{C}_{6}\times 8=672$ Q6$\text { Let } \int \frac{\cos x d x}{\sin ^{3} x\left(1+\sin ^{6} x\right)^{\frac{2}{3}}}=f(x)\left(1+\sin ^{6} x\right)^{\frac{1}{\lambda}}+c \text { then find value of } \lambda f\left(\frac{\pi}{3}\right)$ Solution: $\\\sin x=t \quad\Rightarrow \cos x d x=d t\\\int \frac{d t}{t^{2}\left(1+t^{6}\right)^{2 / 3}}=\int \frac{d t}{t^{3} t^{4}\left(\frac{1}{t^6}+1\right)^{2/3}}\\u=\frac{1}{t^{6}}+1 \Rightarrow d u=-\frac{6}{t^{7}} d t\\\frac{d u}{-6}=\frac{d t}{t^{7}}\\$ $\\=\int\frac{du}{-6u^{2/3}}=-\frac{1}{2 }u^{1/3}+C\\=-\frac{1}{2}\left ( \frac{1}{t^6}+1 \right )^{1/3}+C\\=-\frac{1}{2}\left ( \frac{(1+\sin^6x)^{1/3}}{\sin^2x} \right )\\f(x)=-\frac{1}{2}\frac{1}{\sin^2x}\;\;\;\lambda=3\\\lambda f(\pi/3)=-2$ Q7. Normal to curve ${y^{2}+y-3 x^{2}+10=0$ at point p intersects y-axis at $\left(0, \frac{3}{2}\right) .$Tangent at p has slope m. Find \|m\| Solution: $\begin{array}{c}{2 y y^{\prime}+y^{\prime}-6 x=0} \\ {y^{\prime}=\frac{6 x}{2 y+1}} \\ {\frac{-1}{y^{\prime}}=\frac{-(2 y+1)}{6 x}}(\text{slope of the normal})\end{array}$ Equation of the normal $y-y_{1}=\frac{-\left(2 y_{1}+1\right)}{6 x_{1}}\left(x-x_{1}\right)$ Normal intersect at (0,3/2) $\\\frac{3}{2}-y_{1}=\frac{-\left(2 y_{1}+1\right)}{6 x_{1}}\left(0-x_{1}\right)\\ 8y_1-8=0\\ y_1=1\\ x_1=\pm2\\ \|m\|=4$ Q8. $f(x)=\frac{8^{2 x}-8^{-2 x}}{8^{2 x}+8^{-2 x}} \text { find inverse of the function }$ Solution: $\\f(x)=\frac{8^{2 x}-8^{-2 x}}{8^{2 x}+8^{-2 x}} \\ f(x)=\frac{8^{4 x}-1}{8^{4 x}+1} \text{put }8^{4 x}=t \\ y=\frac{t-1}{t+1}\\ {y t+y=t-1} \\ {\frac{y+1}{1-y}=t} \\ {\frac{y+1}{1-y}=8^{4 x}}\\ \rightarrow ln\left(\frac{y+1}{1-y}\right)=4 x \ln 8 \quad$ $x=f^{-1}y=\frac{1}{4 \ln 8} \ln \left(\frac{y+1}{1-y}\right)$ Q9. The equation $\boldsymbol{2 x^{2}+(a-10) x+\frac{33}{2}=2a}$ has real roots. Find the least possible value of a. Solution: $\begin{array}{c}{D \geqslant 0} \\ {(a-10)^{2}-8\left(\frac{33}{2}-2 a\right) \geq 0} \\ {a^{2}+100-20 a-132+16 a \geq 0}\end{array}$ $\begin{array}{c}{a^{2}-4 a-32 \geqslant 0} \\ {a^{2}-8 a+4 a-32 \geq 0} \\ {(a+4)(a-8) \geq 0}\end{array}$ $a \leq -4 \ \or \ a \geq \ 8$ least possible value 8. Q10. $\begin{array}{l}{\text { Let } f(x)=\left\{\left(\sin \left(\tan ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^{2}-1\right.} \\ {\text { where }\|x\|>1 \text { and } \frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}\left(\sin ^{-1} f(x)\right)} \\ {\text { If } y(\sqrt{3})=\frac{\pi}{6} \text { then } y(-\sqrt{3})=}\end{array}$ Solution: $\\y=\frac{1}{2} \sin ^{-1}(f(x))+c\\\begin{array}{l}{\text { Let } \theta=\tan ^{-1} x \Rightarrow \cot ^{-1} x=\frac{\pi}{2}-\theta} \\ {f(x)=\left\{\sin \theta+\sin \left(\frac{\pi}{2}-\theta\right)\right\}^{2}-1}\end{array}\\\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-1=\sin2\theta$ $\\y=\frac{1}{2} \sin ^{-1}\left(\sin \left(2 \tan ^{-1} x\right)\right)+C\\\frac{\pi}{6}=\frac{1}{2} \sin ^{-1}\left(\sin \left(2 \frac{\pi}{3}\right)\right)+C\\\frac{\pi}{6}=\frac{1}{2} \frac{\pi}{3}+C\Rightarrow C=0\\y=\frac{1}{2} \sin ^{-1}\left(\sin \left(-2 \pi \frac{\pi}{3}\right)\right)=-\frac{\pi}{6}$ Q11. Which of the following is a tautology $\begin{array}{l}{(1)(p \wedge(p \rightarrow q)) \rightarrow q} \\ {(2) q \rightarrow p \wedge(p \rightarrow q)} \\ {(3) p \vee(p \wedge q)} \\ {(4)(p \wedge(p \vee q)}\end{array}$ Solution: $\begin{array}{c\|c\|c}{\mathbf{p}} & {\mathbf{q}} & {(\mathbf{p} \rightarrow \mathbf{q})} \\ \hline F & {F} & {\mathbf{T}} \\ \hline F & {T} & {\mathbf{T}} \\ \hline T & {F} & {F} \\ \hline T & {T} & {\mathbf{T}}\end{array}$ $\begin{array}{c\|c\|c}{\mathbf{p}} & {\mathbf{q}} & {(\mathbf{p} \wedge \mathbf{q})} \\ \hline F & {F} & {\mathbf{F}} \\ \hline F & {T} & {F} \\ \hline T & {F} & {F} \\ \hline T & {T} & {T}\end{array}$ $\begin{array}{c\|c\|c}{\mathbf{p}} & {\mathbf{q}} & {(p \wedge(p \rightarrow q)) \rightarrow q} \\ \hline F & {F} & {\mathbf{T}} \\ \hline F & {T} & {\mathbf{T}} \\ \hline T & {F} & {T} \\ \hline T & {T} & {\mathbf{T}}\end{array}$ Q12. $\text { If } 2^{1-x}+2^{1+x}, f(x), 3^{x}+3^{-x} \text { are in A.P. }$ then the minimum value of f(x) is Solution: $\\2 f(x)=\left(2 \frac{1}{2^{x}}+2.2^{x}\right)+\left(3^{x}+\frac{1}{3^{x}}\right)\\\frac{a+b}{2}\geq\sqrt{ab}\\\frac{3^{x}+\frac{1}{3^{x}}}{2} \geq \sqrt{1}\\\left(3^{x}+\frac{1}{3^{x}}\right) \geqslant 2\\at\;\;3^x=\frac{1}{3^x}\Rightarrow x=0$ $\\\frac{1}{2^{x}}+2^{x} \geq 2\Rightarrow x=0\\ 2 f(x)=2(2)+(2)\\\min\;f(x)=3$ Q13. $\begin{array}{l}{\text { Consider a function } f(x)=\ln \left(\frac{x^{2}+\alpha}{ x}\right) \text { . If for the given }} \\ {\text { function, rolle's theorm is applicable in }[3,4] \text { at a point }^{\prime} c^{\prime}} \\ {\text { then find } f^{\prime \prime}(c)}\end{array}$ Solution: ${f(3)=f(4)} \\ {\frac{9+\alpha}{3}=\frac{16+\alpha}{4}}$ $\alpha=12$ $\begin{array}{l}{f(x)=\ln \left(x^{2}+12\right)-\ln ( x)} \\ {f^{\prime}(c)=0=\frac{2 x}{x^{2}+12}-\frac{1}{x}=0}\end{array}$ $\\2 x^{2}=x^{2}+12\\ x^{2}=12\\ x=\sqrt{12}\varepsilon [3,4]$ $\\f^{''}(x)=\frac{2\left(x^{2}+12\right)-4 x^{2}}{\left(x^{2}+12\right)^{2}}+\frac{1}{x^2}\\ f^{''}(c)=\frac{1}{12}\\$ Q14 $\begin{array}{l}{\text { Two curves } C_{1}: y^{2}=a x \text { and } C_{2}: x^{2}=\text { ay intersect at origin }} \\ {\text { O and point p, where } a>0 \text { . The line } x=b(0 Solution: $\begin{array}{l}{\frac{1}{2} (b \times b)=\frac{1}{2}} \\ {b=1}\end{array}$ $\int_{0}^{1}\left(\sqrt{a} \sqrt{x}-\frac{x^{2}}{a}\right) d x=\frac{a^2}{6} \text{ by property of parabola}$ By solving above you will get $a^{6}-12 a^{3}+4=0$ Q15 $\begin{array}{l}{\text { Ellipse } 2 x^{2}+y^{2}=1 \text { and } y=m x \text { meet a point } A \text { in first quadrant. Normal to the ellipse at } P \text { meets } x \text { -axis }} \\ {\text { at }\left(-\frac{1}{3 \sqrt{2}}, 0\right) \text { and } y \text { -axis at }(0, \beta), \text { then }\|\beta\| \text { is }}\end{array}$ Solution: $\begin{array}{l}{\text { Let } P \text { be }\left(x_{1}, y_{1}\right)} \\ {\text { Equation of normal at } P \text { is } \frac{x}{2 x_{1}}-\frac{y}{y_{1}}=-\frac{1}{2}} \\ {\text { It passes through }\left(-\frac{1}{3 \sqrt{2}}, 0\right) \Rightarrow \frac{-1}{6 \sqrt{2} x_{1}}=-\frac{1}{2} \Rightarrow x_{1}=\frac{1}{3 \sqrt{2}}}\end{array}$ $\begin{array}{l}{\text { So } y_{1}=\frac{2 \sqrt{2}}{3} \text { (as } P \text { lies in } 1 \text { 'quadrant) }} \\ {\text { So } \beta=\frac{y_{1}}{2}=\frac{\sqrt{2}}{3}}\end{array}$ Q16 $\begin{array}{l}{\text { Let } P \text { be a point on } x^{2}=4 y . \text { The segment joining } A(0,-1) \text { and } P \text { is divided by point } Q \text { in the ration } 1: 2,} \\ {\text { then locus of point } Q \text { is }}\end{array}$ Solution: $\text { Let point } P \text { be }\left(2 t, t^{2}\right) \text { and } Q \text { be }(h, k)$ $\begin{array}{l}{h=\frac{2 t}{3}, k=\frac{-2+t^{2}}{3}} \\ {\text { Hence locus is } 3 k+2=\left(\frac{3 h}{2}\right)^{2} \Rightarrow 9 x^{2}=12 y+8}\end{array}$ Q17 $\begin{array}{l}{\text { Let } f(x)=x \cos ^{-1}(\sin (-\|x\|)), x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) \text { then }} \\ {\begin{array}{l}{\text { (1) } f^{\prime}(0)=-\frac{\pi}{2}} \\ {\text { (2) } f^{\prime}(x) \text { is not defined at } x=0} \\ {\text { (3) } f^{\prime}(x) \text { is increasing in }\left(\frac{-\pi}{2}, 0\right) \text { and } f^{\prime}(x) \text { is decreasing in }\left(0, \frac{\pi}{2}\right)} \\ {\text { (4) } f^{\prime}(x) \text { is decreasing in }\left(\frac{-\pi}{2}, 0\right) \text { and } f^{\prime}(x) \text { is increasing in }\left(0, \frac{\pi}{2}\right)}\end{array}}\end{array}$ Solution: $\begin{array}{l}{f^{\prime}(x)=x\left(\pi-\cos ^{-1}(\sin \|x\|)\right)} \\ {=x\left(\pi-\left(\frac{\pi}{2}-\sin ^{-1}(\sin \|x\|)\right)\right)} \\ {=x\left(\frac{\pi}{2}+\|x\|\right)}\end{array}$ $\begin{array}{ll}{f(x)=} {\left\{\begin{array}{ll}{x\left(\frac{\pi}{2}+x\right)} & {x \geq 0} \\ {x\left(\frac{\pi}{2}-x\right)} & {x<0}\end{array}\right.} \\ {f^{\prime}(x)=\left\{\begin{array}{ll}{\frac{\pi}{2}+2 x} & {x \geq 0} \\ {\frac{\pi}{2}-2 x} & {x<0}\end{array}\right.}\end{array}$ $f(x) \text { is increasing in }\left(0, \frac{\pi}{2}\right) \text { and decreasing in }\left(\frac{-\pi}{2}, 0\right)$ Q18. Mean and standard deviations of 10 observations are 20 and 2 respectively. If p (p $\neq$0) is multiplied to each observation and then q (q $\neq$ 0) is subtracted then new mean and standard deviation becomes half of original value . Then find q Solution: If each observation is multiplied with p & then q is subtracted New mean $\\\bar x_{new}=\bar x-q\\\Rightarrow 10=p(20)-q$ and new standard deviations $\\\sigma_{new}=\|p\|\sigma\\1=\|p\|2\Rightarrow \|p\|=\pm0.5\\\text{If p = 0.5 }\\\text{then q = 0}\\\text{If p = -0.5}\\q=-20$ Q19. If maximum value of $^{19} \mathrm{C}_{\mathrm{p}} \text { is a, }^{20} \mathrm{C}_{\mathrm{q}} \text { is } \mathrm{b},^{21} \mathrm{C}_{\mathrm{r}} \text { is } \mathrm{c}$ then relation between a, b, c is Solution $\\^nC_{r}\;\text{ is max at middle term}\\\begin{array}{l}{a=^{19} C_{p}=^{19} C_{10}=^{19} C_{9}} \\ {b=^{20} C_{q}=^{20} C_{10}} \\ {c=^{21} C_{r}=^{21} C_{10}=^{21} C_{11}}\end{array}$ $\frac{a}{^{19}C_9}=\frac{b}{\frac{20}{10} \cdot 19 \mathrm{C}_9}=\frac{c}{\frac{21}{11} \cdot \frac{20}{10} ^{19} \mathrm{C}_{9}}$ $\\\frac{a}{1}=\frac{b}{2}=\frac{c}{42 / 11}\\\frac{a}{11}=\frac{b}{22}=\frac{c}{42}$ Q20. Let P(A) = 1/3, P(B) = 1/6 where A and B are independent events then $\\\text { (1) } \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{1}{6}\\\text { (2) } \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B'}}\right)=\frac{1}{3}\\\text { (3) } P\left(\frac{A}{B^{\prime}}\right)=\frac{2}{3}\\\text { (4) } \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{5}{6}$ Solution: A & B are independent events so $P\left(\frac{A}{B^{\prime}}\right)=\frac{1}{3}$ Q21. Roots of the equation $x^{2}+b x+45=0, b \in R$ lie on the curve $\|z+1\|=2 \sqrt{10}$, where z is a complex number then Solution: Let $z=\alpha\pm i\beta$ be roots of the equation so $2 \alpha=-b \text { and } \alpha^{2}+\beta^{2}=45,(\alpha+1)^{2}+\beta^{2}=40$ So, $(\alpha+1)^{2}-\alpha^{2}=-5$ $\begin{array}{l}{\Rightarrow \quad 2 \alpha+1=-5 \quad \Rightarrow \quad 2 \alpha=-6} \\ {\text { so } \mathrm{b}=6}\end{array}\\Hence,\;\;b^2-b=30$ Q22. If the volume of parallelopiped whose their coterminous edges are $\overrightarrow{\mathrm{u}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}, \;\;\overrightarrow{\mathrm{v}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{w}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ is 1 cubic unit then cosine of angle between $\overrightarrow{\mathrm{u}} \text { and } \overrightarrow{\mathrm{v}} \text { is }$ Solution: $\\\pm 1=\left\|\begin{array}{lll}{1} & {1} & {\lambda} \\ {1} & {1} & {3} \\ {2} & {1} & {1}\end{array}\right\| \Rightarrow=-\lambda+3=\pm 1 \Rightarrow \lambda=2 \text { or } \lambda=4 \\ {\text { For } \lambda=4 \frac{1}{90}} \\\\ {\cos \theta=\frac{2+1+4}{\sqrt{6} \sqrt{18}}=\frac{7}{6 \sqrt{3}}}$ Q23. Shortest distance between the lines $\frac{x-3}{1}=\frac{y-8}{4}=\frac{z-3}{22}, \frac{x+3}{1}=\frac{y+7}{1}=\frac{z-6}{7}$ is Solution: $\\\overrightarrow{\mathrm{AB}}=6 \hat{i}+15 \hat{j}+3 \hat{k} \\ {\vec{p}=\hat{i}+4 \hat{j}+22 \hat{k}} \\ {\vec{q}=\hat{i}+\hat{j}+7 \hat{k}} \\ {\vec{p} \times \vec{q}=\left\|\begin{array}{lll}{i} & {j} & {k} \\ {1} & {4} & {22} \\ {1} & {1} & {7}\end{array}\right\|=6 \hat{i}+15 \hat{j}-3 \hat{k}} \\ {\text { S.D. }=\frac{\| \overrightarrow{A B} \cdot(\vec{p} \times \vec{q})}{\|\vec{p} \times \vec{q}\|}=\frac{\|36+225+9\|}{\sqrt{36+225+9}}=3 \sqrt{30}}$ Q24. Let ABC is a triangle whose vertices are A(1, –1), B(0, 2), C(x', y')and area of ABC is 5 and C(x', y') lie on $3 x+y-4 \lambda=0$, then ${D=\frac{1}{2}\left\|\begin{array}{ccc}{0} & {2} & {1} \\ {1} & {-1} & {1} \\ {x^{\prime}} & {y^{\prime}} & {1}\end{array}\right\|} \\ {-2\left(1-x^{\prime}\right)+\left(y^{\prime}+x^{\prime}\right)=\pm 10} \\ {-2+2 x^{\prime}+y^{\prime}+x^{\prime}=\pm 10} \\ {3 x^{\prime}+y^{\prime}=12 \text { or } 3 x^{\prime}+y^{\prime}=-8} \\ {\lambda=3,-2}$ Q25. The system of equation ${3 x+4 y+5 z=\mu},\;\; {x+2 y+3 z=1} \;\;\&\;\; {4 x+4 y+4 z=\delta}$ is inconsistent, then $(\delta,\mu)$ can be Solution: \begin{aligned} D &=\left\|\begin{array}{lll}{3} & {4} & {5} \\ {1} & {2} & {3} \\ {4} & {4} & {4}\end{array}\right\| \\ &=2\left\|\begin{array}{lll}{3} & {4} & {5} \\ {1} & {2} & {3} \\ {2} & {2} & {2}\end{array}\right\| \end{aligned} $\\R_1\rightarrow R_1-(R_2+R_3)\\D=\left\|\begin{array}{lll}{3} & {4} & {5} \\ {1} & {2} & {3} \\ {2} & {2} & {2}\end{array}\right\|=0$ Hence, it has infinitely many solutions $\mathrm{P}_{3} \equiv \alpha \mathrm{P}_{1}+\beta \mathrm{P}_{2}$ $3 \alpha+\beta=4\;\; \&\;\; 4 \alpha+2 \beta=4 \Rightarrow \alpha=2 \;\; \&\;\; \beta=-2$ $2 \mu-2=\delta$ For non infinite solution $2 \mu-2\neq\delta$ JEE Main 2020 exam - Physics ( 8th January first shift) Q1. A wind-powered generator converts and energy into electrical energy. Assume that the generator convents a fixed fraction of the wind energy intercepted by to blades into electrical energy for wind speed V, the electrical power output will be proportional to Solution : $\\F=v\left(\frac{d m}{d t}\right) =v \frac{d}{d t}(\rho \times \text {volume}) \\ F=v \rho \frac{d}{d t}(\text {volume})=v \rho \times(A v)=A \rho v^{2}\ \\ \text {Power }=Force \times \text {Velocity}=A \rho v^{2} \times v=A \rho v^{3} \Rightarrow P \propto v^{3}$ Q2. Two capacitors C1 and C2, when connected in parallel the effective capacitance was found to be 10microFarad. These capacitors, when connected individually to a battery of emf 1V, the energy of C2, was 4 times the energy of C1. The equivalent capacitance of these capacitors, when connected in series, will be? Solution : When capacitors are in parallel, $C_{eq}=C_1+C_2$ So, $C_1+C_2=10 \mu F$ Now when they are individually connected to 1V battery, then Energy of C2=4 x energy of C1 $\Rightarrow \frac{1}{2}C_2V^2=4\times \frac{1}{2}C_1V^2$ $\Rightarrow C_2=4C_1$ So, C2=8 and C1=2 So when both are in series, then $C_{eq}=\frac{C_1C_2}{C_1+C_2}=\frac{8}{5}$ Q3. Four resistors of resistance 12 ohms, 4 ohms, 10 ohms, and 15 ohms in the cyclic order are connected to form a wheat bridge. What value of resistance must be connected in parallel to 10-ohm resistance to get a balanced Wheatstone bridge? Solution: For balanced Wheatstone bridge $\frac{R_1}{R_2}=\frac{R_3}{R_4}$ As $\frac{R_1}{R_2}=\frac{12}{4}=3$ So using $R_3=15 \Omega$ We get $R_4=R_{CD}=5 \Omega$ let we connected x-ohms in parallel to 10-ohm resistance i.e $R_4=5 \Omega=\frac{x10}{x+10}$ we get $x=10 \Omega$ Q4. Express stopping potential V0 in terms of h(planck's constant), G(gravitational constant), c(speed of light) and A(Ampere). Solution: Let $V_0=[h]^p[G]^q[c]^r[A]^s$ Now, $K_{max}=eV_0\Rightarrow [K_{max}]=[eV_0]\Rightarrow [V_0]=\frac{[K_{max}]}{[e]}\Rightarrow [V_0]=\frac{[ML^2T^{-2}]}{[AT]}$$\therefore [V_0]=[ML^2T^{-3}A^{-1}]$ $E=hf\Rightarrow [h]=\frac{[E]}{[f]}\Rightarrow \frac{[ML^2T^{-2}]}{[T^{-1}]}\Rightarrow [h]=[ML^2T^{-1}]$ $F=\frac{Gm^2}{r^2}\Rightarrow [G]=\frac{[F][r^2]}{[m^2]}=\frac{[MLT^{-2}][L^2]}{[M^2]}=[M^{-1}L^3T^{-2}]$ So, $[ML^2T^{-3}A^{-1}]=[ML^2T^{-1}]^p[M^{-1}L^3T^{-2}]^q[LT^{-1}]^r[A]^s$ From here we will get: p-q=1---------(1) 2p+3q+r=2----------(2) -p-2q-r=-3----------(3) s=-1-----------(4) From equation 1, 2,3 and 4 we will get: p=0,q=-1, r=5 and s=-1 So, $[V_0]=[h]^0[G]^{-1}[c]^5[A]^{-1}$ Q5. A potentiometer wire of 1200 cm is connected with emf of 5V and 20 ohms (internal resistance). The null point is 1000 cm and the current flowing through the wire is 60mA. Find the resistance of the potentiometer wire. options 1) 100-ohm b) 10-ohm c) 20-ohm d) 50-ohm Solution: In the potentiometer, a battery of known emf E is connected in the secondary circuit. A constant current I is flowing through AB from driver circuit . The jockey is slide on potentiometer wire AB to obtain null deflection in the galvanometer. Let $l$ be the length at which galvanometer shows null deflection. Since the potential of wire AB (V) is proportional to the length AB(L). Similarly $E \ \ \alpha \ \ \ l$ So we get $V=5\times\frac{1200}{1000}=6V$ given current through potentiometer wire =60mA V=iR $\\60\times10^{-3}R=6\\\Rightarrow R=100\Omega$ Q6. A plano concave lens has a refractive index of 1.5 and radius of convex part is 30 cm. Find tthe focal length of the system. Solution the focal length of the lens is given as $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ So, $\frac{1}{f}=(1.5-1)\left(\frac{1}{\infty}-\frac{1}{30}\right)=-\frac{0.5}{30}=-\frac{1}{60}$ So f= -60 cm Q7. The magnifying power of the telescope of tube length 60cm is 5 for the normal settings. Then what is its focal length of the eyepiece a) 30 b )10 c )20 d )40 Solution $\begin{array}{l}{\text { Let the focal length of the objective as } f_{0}} \\ {\text { and focal length of the eyepiece as } f_{e}} \\ {\text { Magnifying power, } M=\frac{f_{0}}{f_{e}}} \\ {\text { Tube length, } L=f_{0}+f_{e}}\end{array}$ $\begin{array}{l}{\text { Given: Magnifying power }=5} \\ \\ {\therefore \frac{f_{0}}{f_{\mathrm{e}}}=5} \\ {\therefore {f}_{0}=5{f}_{e}} \\ and \ {\mathrm{L}=\mathrm{f}_{0}+\mathrm{f}_{\mathrm{e}}=60} \\ {5 \mathrm{f}_{\mathrm{e}}+\mathrm{f}_{\mathrm{e}}=60\quad\left[\because \mathrm{f}_{0}=5 \mathrm{f}_{\mathrm{e}}\right]} \\ {6\mathrm{f}_{\mathrm{e}}=60} \\ {\mathrm{f}_{\mathrm{e}}=\frac{60}{6}=10 \mathrm{cm}} \\ {\text { So the focal length of eye piece is } 10 \mathrm{cm} .}\end{array}$ Q8. A block of mass m is connected at one end of spring fixed at other end having natural length $\ell_{0}$ and spring constant K. The block is rotated with constant angular speed$( \omega )$ in gravity free space. Find the elongation in spring. $\dpi{150} \begin{array}{ll}{\text { (1) } \frac{\ell_{0} m \omega^{2}}{k-m \omega^{2}}} & {\text { (2) } \frac{\ell_{0} m \omega^{2}}{k+m \omega^{2}}} \\\\ {\text { (3) } \frac{\ell_{0} m \omega^{2}}{k-m \omega}} & {\text { (4) } \frac{\ell_{0} m \omega^{2}}{k+m \omega}}\end{array}$ Solution As natural lentgh=l0 Let elongation=x Mass m is moving with angular velocity $\dpi{150} \omega$ in a radius r where $\dpi{100} r=l_0+x$ Due to elongation x spring force is given by $F_s=Kx$ And $F_C=m\omega ^2r=m\omega ^2(l_0+x)$ as $F_C=F_s$ So $Kx =m\omega ^2(l_0+x)\\ \Rightarrow x=\frac{m\omega ^2l_0}{K-m\omega ^2}$ Q9. 3 charges are placed in a circle of radius 'd' as shown in figure. Find the electric field along x-axis at centre of circle. $(1) \frac{9}{4 \pi \varepsilon_{0} d^{2}}\ \ \ \ \ \ \ \ \ \ (2) \frac{9 \sqrt{3}}{4 \pi \varepsilon_{0} d^{2}} \ \ \ \ \ \ \ \ \ \ \ (3) \frac{9 \sqrt{3}}{\pi \varepsilon_{0} d^{2}}\ \ \ \ \ \ \ \ \ \ \ (4) \frac{9 \sqrt{3}}{2 \pi \varepsilon_{0} d^{2}}$ Solution Let E1 be the resultant electric field due to charge 2q and - 2q. The resultant E1 is in the direction of -2q $E_1=\frac{k2q}{d^2}+\frac{k2q}{d^2}=\frac{k4q}{d^2}$ E2 due to -4q is in the direction of -4q and magnitude is given by $E_2=\frac{k4q}{d^2}$ The net electric field in the x direction is $\\E_xE_1 cos30+E_2cos30=\frac{\sqrt{3}}{2}\times2\times\frac{4q}{4\pi\epsilon_0d^2}\\E_x=\sqrt{3}\frac{q}{\pi\epsilon_0d^2}$ Q10. Select the P vs V graph corresponding to the given V vs T graph:- Solution Since xy is passing through origin. So, $V\propto T$ $\frac{V}{T}=constant$ So for ideal gas : P= constant Now Y-Z is constant volume graph (from question). Since Z-X is constant temperature grapg it is hyperbolic in nature. So option 1 is correct Q11. Find the co-ordinates of centre of mass of the lamina shown in figure:- $(1) 0.75,1.75 \\ (2) 0.75,1.5 \\ (3) 0.5,1.75 \\ (4) 0.5,1.5$ Solution The co-ordinate of O1 is (0.5,1), O2 is (1, 2.5) $\\x_{cm}=\frac{m\times0.5+m\times1}{2m}=0.75\\ y_{cm}=\frac{m\times+m\times\frac{5}{2}}{2m}=1.75$ Q12. Which graph correctly represents variation between relaxation time (t) of gas molecules with absolute temperature of gas T. Solution As relaxation time is given as $\tau =\frac{Y}{V_{rms}}$ where Y=mean free path and $V_{rms}\propto \sqrt{T}$ while $\dpi{100} \boldsymbol{Y=\frac{1}{\sqrt{2} \pi N d^{2}}}$ where N=number of molecules per unit volume So keeping all the other quantities are constant. we get $\dpi{100} \tau \ \ \alpha \ \frac{1}{\sqrt{T}}$ So the correct graph is given in option 1. Q13. Critical angle for the medium with relative permittivity 3 and relative permeability $\frac{4}{3}$. Solution $C_{vacuum}=\frac{1}{\sqrt{\mu _0\epsilon _0}} \ and \ C_{medium}=\frac{1}{\sqrt{\mu _0\mu _r\epsilon _0\epsilon _r}}$ and $n=\frac{C_{vacuum}}{C_{medium}}=\sqrt{\mu _r\epsilon _r} =\sqrt{\frac{4}{3}3}=2$ By using snells law $nsinC=1sin(90^0)\\ \Rightarrow nnsinC=1\\ \Rightarrow n*sinC=\frac{1}{2}\Rightarrow critical \ angle =C=30^0$ Q14. A rod of mass 4m and length l was at rest on its perpendicular axis. Another particle of mass 'm', velocity u, collides at 45angle with its axis and stick at one ends. What is the angular velocity? Solution: Let us conserve angular momentum about O:- So, $L_i=\left ( \frac{mu}{\sqrt2} \right )\times \frac{l}{2}$, where $\left ( \frac{mu}{\sqrt2} \right )$ is linear momentum and $\left ( \frac{l}{2} \right )$ is the distance from centre O. Now, $L_f=I\omega$ Here, $I=\frac{4ml^2}{12}+\frac{ml^2}{2}=\frac{7ml^2}{12}$ So, $L_i=L_f\Rightarrow \omega=\frac{6u}{7\sqrt2 l}$ Q15. What is the minimum density of the liquid, so that solid sphere of density $\rho=\rho_{0}\left(1-\frac{\mathrm{r}^{2}}{\mathrm{R}^{2}}\right)$ can float? Solution:- $\rho_l\times\frac{4}{3}\pi R^3=\rho_o(4\pi)\int r^2\left ( 1-\frac{r^2}{R^2} \right )dr$ $\Rightarrow \rho_l\times\frac{4}{3}\pi R^3=\rho_o(4\pi)\left [ \frac{r^3}{3}-\frac{r^5}{5R^2} \right ]^R_0$ $\Rightarrow \rho_l\times\frac{4}{3}\pi R^3=\rho_o(4\pi)\left [ \frac{R^3}{3}-\frac{R^3}{5} \right ]$ $\Rightarrow \rho_l\times\frac{4}{3}\pi R^3=\rho_o(4\pi)\left [\frac{2R^3}{15} \right ]$ $\Rightarrow \rho_l=\frac{2}{5}\rho_o$ Q-16 There are two metals A and B. Energy of incident photons are 4eV and 4.5 eV respectively. Maximum kinetic energy for ejected electron is $T_{A} ~ f o r ~ A ~ a n d ~ T _ { B } = T _ { A } -1.5$ for metal B. Relation between wavelength of ejected electron of A and B are $\lambda_{B}=2 \lambda_{A} .$ Find work function of metal B. $\begin{array}{ll}{(1) 3 e V} & {(2) 1.5 e V} \\ {(3) 4.5 e V} & {(4) 4 e V}\end{array}$ Solution:- As we know $\lambda =\frac{h}{\sqrt{2mT}}$ So $\frac{\lambda_A}{\lambda_B} = \sqrt{\frac{T_A}{T_B}}=\frac{1}{2}...(1)$ Also $Given \ \ T_B=T_A-1.5... (2)$ Using equation (1) and (2) $T_A=2 \ eV , \ and \ T_B=1.5 \ eV$ And using the Einstein equation $E_B=T_B+W_B\\ \Rightarrow W_B=E_B-T_B=(4.5-0.5)eV=4 \ eV$ Q-17 A cylinder of height $1 \mathrm{m}$ is floating in water at $0^{\circ} \mathrm{C}$ with $20 \mathrm{cm}$ height in air. Now temperature of water is raised to $4^{\circ} \mathrm{C}$, height of cylinder in air becomes 21 cm. The ratio of density of water at $4^{\circ} \mathrm{C}$ to density of water at $0^{\circ} \mathrm{C}$ is- (Consider expansion of cylinder is negligible) Solution- $mg=V\rho g$ $\mathrm{mg}=\mathrm{A}(80) \mathrm{p}_{0^{\circ} \mathrm{C}} \mathrm{g} \\ \mathrm{mg}=\mathrm{A}(79) \mathrm{\rho}_{4^{\circ} \mathrm{C}} \mathrm{g} \\ \frac{\rho_{4^{\circ} \mathrm{C}}}{\rho_{0^{\circ} \mathrm{C}}}=\frac{80}{79}=1.01$ Q-18 Number of the $\alpha$ -particle deflected in Rutherford's $\alpha$-scattering experiment varies with the angle of deflection. Then the graph between the two is best represented by:- Solution:- $\dpi{120} N \propto \frac{1}{\sin ^{4}\left(\frac{\theta}{2}\right)}$ Here when $\theta=0$, Sin will be zero then N will tend to infinity. and that is satisfied by only option 2 So option 2 is correct. Q-19 The given loop is kept in a uniform magnetic field perpendicular to plane of loop. The field changes from 1000G to 500G in 5 seconds. The average induced emf in loop is- (1) $56\mu \ V$ (2) $28\mu \ V$ (3) $30\mu \ V$ (4) $48\mu \ V$ Solution:- Area of the given loop $\\A=Area\ of\ rectangle-Area\ of \ 2\ triangles\ with\ height\ 2cm \\16\times4-2\times4=56cm^2$ Magnitude of emf $\\E=A\frac{dB}{dt}=56\times10^{-4}\times(\frac{(1000-500)\times10^{-4}}{5})\\=5600\times10^{-8}V=56\mu \ V$ Q-20 Choose the correct Boolean expression for the given circuit diagram: $1)A.B\ \ \ \ \ \ 2)\bar A+\bar B\ \ \ \ \ \ 3)A+B\ \ \ \ \ \ 4)\bar A\bar B$ Solution:- The first part of the given circuit represents or gate whose out put is A+B The second part of the circuit is NOT gate whose input is A+B and out put is $\overline{A+B}=\bar A.\bar B$ Q-21 Two masses each with mass 0.10kg are moving with velocities 3m/s along the x-axis and 5m/s along the y-axis respectively. After an elastic collision one of the mass moves with a velocity $4 \hat{i}+4 \hat{j}$ . The energy of other mass after the collision is 10x then x is. Sol- For elastic collision KEi = KEf $\begin{array}{l}{\frac{1}{2} m \times 25+\frac{1}{2} \times m \times 9=\frac{1}{2} m \times 32+\frac{1}{2} m v^{2}} \\ {34=32+v^{2}\Rightarrow v^2=2}\end{array}$ $\begin{array}{l}{\mathrm{KE}=\frac{1}{2} \times 0.1 \times 2=0.1 \mathrm{J}=\frac{1}{10}} \\ {x=1}\end{array}$ Q-22$\begin{array}{l}{\text { Position of two particles } A \text { and } B \text { as a function of time are given by } X_{A}=-3 t^{2}+8 t+c \text { and } Y_{B}=10-8 t^{3} \text { . }} \\ {\text { The velocity of B with respect to } A \text { at } t=1 \text { is } \sqrt{v} . \text { Find } v \text { . }}\end{array}$ Sol- $\begin{array}{l}{x_{A}=-3 t^{2}+8 t+c} \\ {\vec{v}_{A}=(-6 t+8) \hat{i}} \\ {=2 \hat{i}} \\ {Y_{B}=10-8 t^{3}} \\ {\vec{v}_{B}=-24 t^{2} \hat{j}= -24 \hat{j}} \\ {\sqrt{v}=\left\|\vec{v}_{B}-\vec{v}_{A}\right\|=\|-24 \hat{j}-2 \hat{i}\|} \\ {\sqrt{v}=580} \\ {v=580}\end{array}$ Q-23 An open organ pipe of length 1m contains a gas whose density is twice the density of the atmosphere at STP. Find the difference between fundamental and second harmonic frequencies if the speed of sound in the atmosphere is 300 m/s. Sol- $\dpi{150} \begin{array}{l}{\mathrm{V}=\sqrt{\frac{\mathrm{B}}{\rho}}} \\\\ {\frac{\mathrm{V}_{\text {pipe }}}{\mathrm{V}_{\text {air }}}=\frac{\sqrt{\frac{\mathrm{B}}{2 \rho}}}{\sqrt{\frac{\mathrm{B}}{ \rho}}}=\frac{1}{\sqrt{2}}} \\\\ {\mathrm{V}_{\text {pipe }}=\frac{V_{\text {air }}}{\sqrt{2}}}\end{array}$ $\dpi{120} \begin{array}{l}{f_{n}=\frac{(n+1) V_{\text {pipe }}}{2 \ell}} \\ {\left.f_{1}-f_{0}=\frac{V_{\text {pipe }}}{2 \ell}=\frac{300}{2 \sqrt{2}}=106.38 \mathrm{Hz} \text { (If } \sqrt{2}=1.41\right)}\end{array}$ Q-24 Two spherical bodies of mass $\dpi{120} m_{1} \& m_{2}$ are having radius $\dpi{120} 1 \mathrm{m} \& 2 \mathrm{m}$ respectively. The gravitational field of the two bodies with their radial distance is shown below. The value of $\dpi{120} \frac{m_{1}}{\mathrm{m}_{2}}$ is- $\dpi{120} \begin{array}{llll}{\text { (1) } \frac{1}{6}} & \ \ \ \ \ {\text { (2) } \frac{1}{3}} \ \ \ \ \ \ & {\text { (3) } \frac{1}{2}} \ \ \ \ \ \ \ \ & {\text { (4) } \frac{1}{4}}\end{array}$ Solution:- Considering Case 2 and case 1, one by one $\dpi{120} \\ 3=\frac{\mathrm{Gm}_{2}}{2^{2}} \\ \\ 2=\frac{\mathrm{Gm}_{1}}{1^{2}} \\ \\ \therefore \frac{3}{2}=\frac{1}{4} \frac{\mathrm{m}_{2}}{\mathrm{m}_{1}} \\ \\ \frac{\mathrm{m}_{1}}{\mathrm{m}_{2}}=\frac{1}{6}$ So, option 1 is correct. Q-25 When proton of KE = 1.0 MeV moving in South to North direction enters the magnetic field (from West to East direction), it accelerates with a = 1012 m/s2 . The magnitude of magnetic field is– $\dpi{120} \begin{array}{llll}{\text { (1) } 0.71 \mathrm{mT}} \ \ \ \ \ & {\text { (2) } 7.1 \mathrm{mT}} \ \ \ \ \ \ \ & {\text { (3) } 71 \mathrm{mT}} \ \ \ \ \ \ \ & {\text { (4) } 710 \mathrm{mT}}\end{array}$ Solution - $\dpi{120} \begin{array}{l}{\because \mathrm{K.E.}=1.6 \times 10^{-13}=\frac{1}{2} \times 1.6 \times 10^{-27} \mathrm{V}^{2}} \\ \\ {\mathrm{V}=\sqrt{2} \times 10^{7}} \\ \\ {\therefore \mathrm{Bqv}=\mathrm{ma}} \\ \\ {\mathrm{B}=\frac{1.6 \times 10^{-27} \times 10^{12}}{1.6 \times 10^{-19} \times \sqrt{2} \times 10^{7}}} \\ \\ {=0.71 \times 10^{-3} \mathrm{T}} \\ \\ {\text { so } 0.71 \mathrm{mT}}\end{array}$ So option (1) is correct	{}
# Married couples around a table There are lot of similar questions available. After going through all, I am still confused with two different answers. Question : How many ways can three married couples sit around a round table if husband and wife must sit together? Base on the explanation available in the book , I get the answer as $(3-1)! * 2^3 = 16$. This explanation is also clear to me. However, after reading further here, the number of ways in which three couples can be arranged around a circular table so that no wife sits next to her husband $= 120 -144 + 72 -16 = 32$ (based on inclusion and exclusion). If so, the number of ways where a wife sits next to her husband should be $5! - 32 = 88$. I know I am wrong. I am still trying hard to understand where I have gone wrong. Please help. • The $88$ (which I have not checked) is the number of ways at least one pair of partners are next to each other. But not necessarily all. So there are quite a few more than the $16$ where all pairs of partners are next to each other. – André Nicolas Jun 14 '15 at 2:21 • thanks André Nicolas. that means 16 is the answer when every couples sits together , isn't it? – Kiran Jun 14 '15 at 2:39 • You are welcome. Yes, $16$ is for all pairs together, $88$ is for at least one pair together. – André Nicolas Jun 14 '15 at 2:44 • thanks André Nicolas, I am clear now. how may I close this question now and select your answer? – Kiran Jun 14 '15 at 2:45 • To close things up, since the issue may come up for someone else, I wrote an answer. – André Nicolas Jun 14 '15 at 3:08 We check your count of the number of ways at least one couple are in adjoining chairs. As you did, we use Inclusion/Exclusion. Call the couples A, B, and C. The number of ways couple A are next to each other is $(2)(4!)$. For we may assume the fatter of the two members sits in a specific chair. Then the partner has $2$ choices, and for every such choice the rest can arrange themselves in $4!$ ways. Add together the number of ways for couple A to be together, couple B to be together, couple C to be together. We get $144$. But we have overcounted the ways in which for example couple A and B are both together. Again we can assume the fatter of the members of A sits in a specific chair. Then there are $2$ choices for where the other member sits, and then $(3)(2)$ ways for couple B to be together, leaving $2$ choices for the others, a total of $24$. Summing over all ways to choose $2$ couples, we get $72$. So our corrected estimate is $144-72$. But the $144$ counted three times the arrangements in which all the couples are together, and so did the $72$, so we must add back the $16$ ways in which they are all together. The final count is $144-72+16$. Note that this $88$ counts the number of ways at least one couple are together. It is naturally a much larger number than the $16$ ways in which all couples are together.	{}
Algebra and Trigonometry 10th Edition Identity is verified. $sin^4x+cos^4x=1-2~cos^2x+2~cos^4x$ We know that: $sin^2x+cos^2x=1$ $sin^2x=1-cos^2x$ Start at the left side of the equation: $sin^4x+cos^4x=(sin^2x)^2+cos^4x=(1-cos^2x)^2+cos^4x=1-2~cos^2x+(cos^2x)^2+cos^4x=1-2~cos^2x+cos^4x+cos^4x=1-2~cos^2x+2~cos^4x$	{}
file.pdf (258.82 kB) # A note on the random greedy independent set algorithm journal contribution posted on 15.03.2015, 00:00 by Patrick Bennett, Tom Bohman Let r be a fixed constant and let H be an r-uniform, D-regular hypergraph on N vertices. Assume further that D > N^\epsilon for some \epsilon>0. Consider the random greedy algorithm for forming an independent set in H. An independent set is chosen at random by iteratively choosing vertices at random to be in the independent set. At each step we choose a vertex uniformly at random from the collection of vertices that could be added to the independent set (i.e. the collection of vertices v with the property that v is not in the current independent set I and I \cup {v} contains no edge of H). Note that this process terminates at a maximal subset of vertices with the property that this set contains no edge of H; that is, the process terminates at a maximal independent set. We prove that if H satisfies certain degree and codegree conditions then there are \Omega(N ((log N) / D)^{1/(r-1)}) vertices in the independent set produced by the random greedy algorithm with high probability. This result generalizes a lower bound on the number of steps in the H-free process due to Bohman and Keevash and produces objects of interest in additive combinatorics. 15/03/2015 ## Exports figshare. credit for all your research.	{}
# When to use a t value and when to use 1.645 for a 90% confidence interval? The question I am working with is: Setup a 90% confidence interval estimate for the average processing time. I gathered the information below from the spreadsheet $n = 27$ $\bar{X} =48.888$ Sample standard deviation $= 25.283$ $\sigma/\sqrt{n} = 4.871$ I am confused because I thought that to setup the confidence level I would use 1.645 which is a common level confidence for the 90% confidence level. We are 90% confident that the average processing time is between 40.8 and 56.9 days. My final answer is wrong. I double checked with a excel template and instead of 1.645 the template used a t value calculated using an exel function called TINV which I am not sure how to calculate. Any help would be greatly appreciated. - You should use $\bar{X}\pm\sigma z_{1-\alpha/2}/\sqrt{n}$ where $z_{1-\alpha/2}$ is normal quantile when population standard deviation $\sigma$ is known. In your case you have only estimate $\hat\sigma$, therefore you should use $\bar{X}\pm\sigma t_{1-\alpha/2}(n-1)/\sqrt{n}$ where $t_{1-\alpha/2}(n-1)$ is student quantile with $n-1$ degrees of freedom (TINV(1-0.9,27-1)=1.706 function in Excel). So you obtain wider confidence interval - more uncertainty due to unknown standard deviation. - Can I use the 1.645 value when the population standard deviation is known? or when it's not known, I use the t value. Is that correct? Could you describe how to calculate the 1.706 value without TINV? – Filype May 31 '12 at 7:17 You should use 1.645 when the population standard deviation is known, if not (common situation) then use 1.706 value. There is no explicit formula to calculate $t$ values, however there are tables. Why not using build-in functions? – danas.zuokas May 31 '12 at 8:22 I am studying for an exam and wanted to know how to do it by hand. I am assuming then that we will be given this table to work with, Thanks so much for your help. – Filype May 31 '12 at 8:46 The t test is robust to departures from normality but its strict validity depends on the observations being normally distributed. It critical value is larger than for the standard normal because the estimation of standard deviation leads to a symmtric distribution that has heavier tailed than the normal distribution. As Danas points out to get the cutoff requires calculating the tail probabilities for the appropriate t distirbution which requires the TINV function or a table for the t distribution. – Michael Chernick May 31 '12 at 11:16	{}
### Math Notes Subjects #### Algebra Solutions ##### Topics \|\| Problems A basketball player is 6 feet 8 inches tall. How far should he stand from the light which is 15 feet above the ground, in order to cast a shadow 25 feet long? Solution: 6 feet 8 inches = 6 + $$\frac{8}{12}$$ = $$\frac{20}{3}$$ feet $$\frac{15}{x+25}$$ = $$\frac{\frac{20}{3}}{25}$$ $$15(75) = 20 (x+25)$$ $$1125 = 20x +500$$ $$x = \frac{1125-500}{20}$$ $$x = 31.25$$	{}
When I write something like C = A*B (with A, B, C cv::Mat). Does it call gemm(A, B, 1.0, Mat(), 0.0, C); ?	{}
# Probability theory: Understanding modes of convergence I am using Terry Tao's book on Random Matrix Theory. I came across Definition 1.1.28 on page 37 of the book. There, Tao defines various modes of convergence in decreasing order of strength. I am having a hard time understanding items (ii) and (iii) in this definition which talk about convergence in probability and convergence in distribution respectively. Briefly speaking I want to ask Why does convergence in probability imply convergence in distribution? In particular, I am not sure if I really need the functions $F$ to be both bounded and continuous as the definition of convergence in distribution insists (and so far it seems to me that $F$ being bounded suffices) I am certain that I am missing something crucial. To see this better, I will first state the definitions as given in Tao's book. And then, I will show how I use these definitions (possibly, I might have misunderstood them). I will make a simplification and assume that all random variables are real valued. Def: [Convergence in probability] A sequence of Random Variables $X_n$ converges to a random variable $X$ in probability if for every $\varepsilon > 0$, one has $$\lim_{n \rightarrow \infty } \inf \Pr(\|X_n - X\| \leq \varepsilon) = 1$$ The way I parse this statement is the following For every $\delta > 0$, for every $\varepsilon > 0$, there exists $n_0 = n_0(\varepsilon, \delta) \in \mathbb{N}$ such that for all $n > n_0$ we have $$\Pr(\|X_n - X\| \leq \varepsilon) \geq 1 - \delta.$$ Now, here is the definition of convergence in distribution (I think I parse this one just fine) Def:[Convergence in distribution] $X_n$ converges in distribution to $X$ if for every bounded and continuous function $F : \mathbb{R} \to \mathbb{R}$ one has $$\lim_{n \to \infty} \mathbb{E}(F(X_n)) = \mathbb{E}(F(X)).$$ Here is my attempt at proving (ii) implies (iii). I want to see why my "proof" does not seem to require that $F$'s have to be continuous. Suppose $-M \leq F \leq M$ everywhere. Also, suppose we denote the underlying space common to all random variables by $(\Omega, \mathcal{F}, \mu)$ I want to show that if $X_n$ converges to $X$ in probability, then for any $\beta > 0$, there exists $n_{\beta} \in \mathbb{N}$ such that for every $n > n_{\beta}$ we have $\| \mathbb{E}(F(X_n)) - \mathbb{E}(F(X)) \| \leq \beta$. To this end, I write $$\mathbb{E}(F(X_n)) = \displaystyle \int_{\Omega} F(X_n(\omega))\ d\mu(\omega)$$ And similarly, let us write $$\mathbb{E}(F(X)) = \displaystyle \int_{\Omega} F(X(\omega))\ d\mu(\omega)$$ Let us now go back to convergence in probability which we are told holds. Let us denote by $G \subseteq \Omega$ the 'good' set of all those outcomes for which $\|X_n - X\| \leq \epsilon$ holds (where $n > n_0({\varepsilon, \delta})$). We know that $\mu(G) \geq 1 - \delta$. What is left over is the set $B = \Omega \setminus G$ of bad outcomes with measure at most $\delta$. Now, we upper bound the difference $\| \mathbb{E}(F(X_n)) - \mathbb{E}(F(X)) \|$ by considering the absolute differences on the sets $G$ and $B$. That is, $\mathbb{E}(F(X_n)) - \mathbb{E}(F(X)) = \left\| \displaystyle \int_{\Omega} F(\color{red}{X_n}(\omega))\ d\mu(\omega) - \displaystyle \int_{\Omega} F(\color{red}{X}(\omega))\ d\mu(\omega) \right\|$ $$\leq \left\| \displaystyle \int_{G} F(\color{red}{X_n}(\omega))\ d\mu(\omega) - \displaystyle \int_{G} F(\color{red}{X}(\omega))\ d\mu(\omega) \right\| + \left\|\displaystyle \int_{B} F(\color{red}{X_n}(\omega))\ d\mu(\omega) - \displaystyle \int_{B} F(\color{red}{X}(\omega))\ d\mu(\omega) \right\|$$ This is seen to be $$\leq \varepsilon \cdot \mu(G) + \sup_{\omega \in B}\|F(\color{red}{X_n}(\omega)) - F(\color{red}{X}(\omega))\| \cdot \mu(B) \leq \varepsilon + 2M \cdot \delta$$ Now i choose $\varepsilon = \frac{\beta}{2}$ and $\delta = \frac{\beta}{2M}$ and use $n_0(\varepsilon, \delta)$ (in my unpacking of convergence in probability) as the desired $n_{\beta}$. ====================== You need continuity in order to ensure that $F (X_n)-F (X)$ is small when $X_n-X$ is small. In other words you need to invoke continuity to handle the integral over $G$. Also, your $n_\beta$ will depend on $F$ as a result of this step.	{}
GMAT Question of the Day: Daily via email \| Daily via Instagram New to GMAT Club? Watch this Video It is currently 30 May 2020, 22:14 GMAT Club Daily Prep Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Hide Tags GMAT Club team member Status: GMAT Club Team Member Affiliations: GMAT Club Joined: 02 Nov 2016 Posts: 5936 GPA: 3.62 If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 28 Apr 2019, 17:34 1 2 00:00 Difficulty: 65% (hard) Question Stats: 44% (01:22) correct 56% (01:15) wrong based on 80 sessions HideShow timer Statistics If $$a = 4$$, what is the value of $$c$$? (1) $$(abc)^4 = (ab^2c^3)^2$$ (2) $$b = 5$$ _________________ GMAT Club Legend Joined: 18 Aug 2017 Posts: 6287 Location: India Concentration: Sustainability, Marketing GPA: 4 WE: Marketing (Energy and Utilities) If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags Updated on: 01 May 2019, 00:46 1 1 If $$a = 4$$, what is the value of $$c$$? (1) $$(abc)^4 = (ab^2c^3)^2$$ (2) $$b = 5$$ solve for #1 $$(abc)^4 = (ab^2c^3)^2$$ we get a^2=c^2 c = +/-4 insufficeint #2 only b given c relation missing from 1 & 2 we cannot get any info IMO E Originally posted by Archit3110 on 29 Apr 2019, 06:59. Last edited by Archit3110 on 01 May 2019, 00:46, edited 1 time in total. Intern Joined: 20 Feb 2018 Posts: 33 Location: India Schools: ISB '20, IIMA PGPX"20 GMAT 1: 610 Q47 V28 GPA: 3.8 Re: If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 29 Apr 2019, 08:38 2 i think answer is E, a2=c2 does not mean a=c, it means c2=16, now c can be +4 and -4, so not enough to give exact value of c, hence E is answer GMAT Club Legend Joined: 18 Aug 2017 Posts: 6287 Location: India Concentration: Sustainability, Marketing GPA: 4 WE: Marketing (Energy and Utilities) Re: If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 29 Apr 2019, 08:41 lifeforhuskar wrote: i think answer is E, a2=c2 does not mean a=c, it means c2=16, now c can be +4 and -4, so not enough to give exact value of c, hence E is answer lifeforhuskar even i thoight that E., but if you square after a^2=c^2 ; you shall get a=c Senior Manager Joined: 25 Feb 2019 Posts: 331 Re: If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 29 Apr 2019, 08:45 a = 4 is given , so c can not be -4 , it will take only +4 . hope this helps Posted from my mobile device Intern Joined: 20 Feb 2018 Posts: 33 Location: India Schools: ISB '20, IIMA PGPX"20 GMAT 1: 610 Q47 V28 GPA: 3.8 Re: If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 29 Apr 2019, 08:46 Archit3110 wrote: lifeforhuskar wrote: i think answer is E, a2=c2 does not mean a=c, it means c2=16, now c can be +4 and -4, so not enough to give exact value of c, hence E is answer lifeforhuskar even i thoight that E., but if you square after a^2=c^2 ; you shall get a=c I dont agree, you can try and put -4 or +4 in the equation, both are giving same results. May be we need ER on this. Manager Joined: 31 Dec 2018 Posts: 110 Location: India Re: If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 29 Apr 2019, 08:53 solve for #1 (abc)4=(ab2c3)2(abc)4=(ab2c3)2 we get a^2=c^2 c= √a² c= √16 c= +/- 4 Hence E But gmat considers Roots to be +ve and square to have both +/- sign Hence A. Posted from my mobile device Intern Joined: 20 Feb 2018 Posts: 33 Location: India Schools: ISB '20, IIMA PGPX"20 GMAT 1: 610 Q47 V28 GPA: 3.8 Re: If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 29 Apr 2019, 09:05 [quote="mohitranjan05"]solve for #1 (abc)4=(ab2c3)2(abc)4=(ab2c3)2 we get a^2=c^2 c= √a² c= √16 c= +/- 4 But gmat considers Roots to be +ve and square to have both +/- sign Can someone explain this...thanks Intern Joined: 13 Feb 2019 Posts: 5 Re: If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 29 Apr 2019, 09:40 1 The Answer is (E) Given A=4, c=? 1) (abc)^4= (ab^2c^3)^2 a^4b^4c^4 - a^2b^4c^6 = 0 a^2b^4c^2(a^2 - c^2) = 0 This implies that either of a,b,c take a zero value or a^2=c^2. Since we know that a=4, but if b=0, then c can take any value. Hence N.S 2) b=5 NS on its own Combining both we get either c=0 or c^2=a^2 which means c can have two possible values c=0 and c=a=4. Hence NS Intern Joined: 13 Feb 2019 Posts: 5 If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 29 Apr 2019, 09:56 Also, in this problem when a^2=c^2, this means that \|a\| = \|c\| and since its already given that a=4, we do not need to consider the -ve case. Manager Joined: 16 Oct 2011 Posts: 132 GMAT 1: 570 Q39 V41 GMAT 2: 640 Q38 V31 GMAT 3: 650 Q42 V38 GMAT 4: 650 Q44 V36 GMAT 5: 570 Q31 V38 GPA: 3.75 Re: If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 29 Apr 2019, 11:21 If $$a = 4$$, what is the value of $$c$$? (1) $$(abc)^4 = (ab^2c^3)^2$$ (2) $$b = 5$$ by (1) we get a^4b^4c^4-a^2b^4c^6=0 and a^2b^4c^4(a^2-c^2) = 0 and a^2b^4c^4(a+c)(a-c) = 0. Since we can factor out (a+c)(a-c) = 0 we see that a=+-c, thus c=+-4 are two solutions. Since we get AT LEAST 2 solutions for C we can mark NS (2) b =5? So what. Nothing related to c NS (1) and (2) --- Copy from (1), and plug in b = 5 ---> a^2(5)^4(4+c)(4-c) = 0. Even after plugging in b = 5 we still get at least two solutions for c thus NS E Manager Joined: 14 Jan 2017 Posts: 116 Location: India If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 29 Apr 2019, 11:23 m1033512 wrote: a = 4 is given , so c can not be -4 , it will take only +4 . hope this helps Posted from my mobile device I don't quite get, how you reached to this conclusion that c can not be -4. Can you explain this a bit further? IMO, St(1) does not provide us with any of the values, but a = 4. We know nothing about b and c. So, b and c can have ANY of the THREE signs : Positive, Negative, and Zero. B and C can have a range of values. If b = 0 then C can have any value on the number line and vice versa. Not Sufficient. St(2) is clearly insufficient. St(1) + St(2) : After combining both the statements, we sill get C^2 = 16 and thus, C can be +4 or -4. Two values. Not Sufficient. The problem is with the square, we don't know whether it is +4 or - 4, but if we had something like c^odd number = Some positive number, then surely C would have been a positive number. I hope this helps. I don't mind Kudos! Director Joined: 20 Mar 2018 Posts: 512 Location: Ghana Concentration: Finance, Real Estate Schools: Terry '22 Re: If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 29 Apr 2019, 12:15 \|a\|=\|c\| so two cases c=a or c=-a Archit3110 wrote: If $$a = 4$$, what is the value of $$c$$? (1) $$(abc)^4 = (ab^2c^3)^2$$ (2) $$b = 5$$ solve for #1 $$(abc)^4 = (ab^2c^3)^2$$ we get a^2=c^2 square both sides a=c sufficeint #2 only b given c relation missing IMO A Posted from my mobile device _________________ Veritas valet et vincet Director Joined: 20 Mar 2018 Posts: 512 Location: Ghana Concentration: Finance, Real Estate Schools: Terry '22 Re: If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 29 Apr 2019, 12:26 (stat.1).... (a^4•b^4•c^4)=(a^2•b^4•c^6) —> (a^4•b^4)/(a^2•b^4)=c^6/c^4 —> a^2•b^0=c^2 —> c^2=a^2 Haha forgot the absolute value though credit to Skyline393 who reminded me so \|c\|=\|a\| —> c=-/+a c=+/-4 But why you guys doing some complicated calc. with the statement (1) though I am an advocate of not dividing by an unknown value. Lolx Posted from my mobile device _________________ Veritas valet et vincet Manager Status: Don't Give Up! Joined: 15 Aug 2014 Posts: 97 Location: India Concentration: Operations, General Management GMAT Date: 04-25-2015 WE: Engineering (Manufacturing) Re: If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 30 Apr 2019, 19:30 m1033512 wrote: a = 4 is given , so c can not be -4 , it will take only +4 . hope this helps Posted from my mobile device Hi a^2 = c^2 is given and a=4 16= c^2 c can be 4 or -4 to have square 16. Intern Joined: 26 Feb 2018 Posts: 3 Re: If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] Show Tags 05 Aug 2019, 06:37 1 Ans. is E after solving the equation in A, we get c^2=16 => c=4 or-4 Re: If a = 4, what is the value of c? (1) (abc)^4 = (ab^2c^3)^2 (2) b = 5 [#permalink] 05 Aug 2019, 06:37 If a = 4, what is the value of c? 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Ryan P. Browne, Aisha ElSherbiny and Paul D. McNicholas (2018). Is it possible to calculate AIC and BIC for lasso regression models? In that case, AIC also provides the right result (not shown to save time), but BIC is better suited if the problem is to identify the right model. Bayesian information criterion (BIC) (Stone, 1979) is another criteria for model selection that measures the trade-off between model fit and complexity of the model. console warning: "Too many lights in the scene !!!". Additional resources: Additional resources to help you learn more. This tutorial serves as an introduction to linear model selection and covers1: 1. LASSO vs AIC for feature selection with the Cox model, AIC BIC Mallows Cp Cross Validation Model Selection. When it is a function the function must take a single argument reset. Typically, the MSE will only be an unbiased predictor of ˙2 in backwards variable selection. I have already concluded what model is better based on other factors but this makes me confused. In R, stepAIC is one of the most commonly used search method for feature selection. I suggest you check the source code of both R and SAS implementations and see if the formulation is the same. bigglm 3 Details The data argument may be a function, a data frame, or a SQLiteConnection or RODBC connection object. So, I am trying to see which model is better, based only on BIC. The model fitting must apply the models to the same dataset. I am using R software and running 3 models, GARCH-t, GJR model, and simple GARCH (1,1) model. Model selection: Cp, AIC, BIC and adjusted R² Cp. They also discuss the ways to spot correlation handling correlation in model selection … Symonds, M. and Moussalli, A. Given a criterion, we also need a search strategy. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Start with the selection of the model Select the required retract set look for a proper spinner Find the correct Pilot Select the related prop Select a gas engine or electric motor Select the servo's; Below an overview is given of some new models and engines which are added. Computing best subsets regression. Usage bic.mixcompnorm(x, G, type = "alr", graph = TRUE) Arguments x. I am fitting a linear model using LASSO and exploring BIC (or AIC) as the selection criterion. I have always used AIC for that. This chapter describes how to perform stepwise logistic regression in R. In our example, the stepwise regression have selected a reduced number of predictor variables resulting to a final model, which performance was similar to the one of the full model. The BIC is consistent in selecting the true model, and its probability of doing so quickly approaches 1 1, as anticipated by (3.2). 2002. In the simplest cases, a pre-existing set of data is considered. Stepwise selection: Computationally efficient approach for feature selection. "leapForward", to fit linear regression with forward selection "leapSeq", to fit linear regression with stepwise selection. 7. [R] how to selection model by BIC [R] Can anybody help me understand AIC and BIC and devise a new metric? You also need to specify the tuning parameter nvmax, which corresponds to the maximum number of predictors to be incorporated in the model. I am fitting a linear model using LASSO and exploring BIC (or AIC) as the selection criterion. Again, for model selection use the model with the smallest BIC. Details. In regression model, the most commonly known evaluation metrics include: R-squared (R2), which is the proportion of variation in the outcome that is explained by the predictor variables. 2011. Burnham and Anderson provide theo- Model selection is a process of seeking the model in a set of candidate models that gives the best balance between model fit and complexity (Burnham & Anderson R-bloggers R news and tutorials contributed by hundreds of R bloggers Minimum Description Length Im klassisches Regressionsmodell unter Normalverteilungsannahme der Störterme kann das BIC auch folgendermaßen dargestellt … The model selection literature has been generally poor at reflecting the deep foundations of the Akaike information criterion (AIC) and at making appropriate comparisons to the Bayesian information criterion (BIC). Go for a full overview to the planes sections: Goldwing, Cymodel, TWM, ESM and TOPRC. (1986). Model Selection Approaches. Ryan P. Browne and Paul D. McNicholas (2014). I wonder whether I have done anything wrong and whether there is something I can do to better align the two results. model bic. Lasso model selection: Cross-Validation / AIC / BIC ¶ Use the Akaike information criterion (AIC), the Bayes Information criterion (BIC) and cross-validation to select an optimal value of the regularization parameter alpha of the Lasso estimator. . The set of models searched is determined by the scope argument. 12 min read. Chapter 16 Variable Selection and Model Building “Choose well. A part of their proposed protocol (in many cases) is model selection using AIC/BIC. This may be a problem if there are missing values and an na.action other than na.fail is used (as is the default in R). If scope is missing, the initial model is used as the upper model. Model selection: choosing a subset of variables¶ To "implement" a model selection procedure, we first need a criterion or benchmark to compare two models. Model selection is the task of selecting a statistical model from a set of candidate models through the use of criteria's. Bayesians generally do not use BIC for model selection… Bayes Factor. All existing methods require to train multiple LDA models to select one with the best performance. Notice as the n increases, the third term in AIC However, when I compared this result from R with a result from SAS (code below, and input variables are exactly the same): The final models are completely different. An example Through an example, we introduce different variable selection methods and illustrate their use. Model fit and model selection analysis for the linear models employed in education do not pose any problems and proceed in a similar manner as in any other statistics field, for example, by using residual analysis, Akaike information criterion (AIC) and Bayesian information criterion (BIC) (see, e.g., Draper and Smith, 1998). Your choice is brief, and yet endless.” — Johann Wolfgang von Goethe After reading this chapter you will be able to: Understand the trade-off between goodness-of-fit and model complexity. With a limited number of predictors, it is possible to search all possible models (leaps in R). Probabilistic Model Selection 3. BIC is used to decide on the optimal model and number of components. The R function regsubsets() [leaps package] can be used to identify different best models of different sizes. Mixture model selection via BIC. Das Modell mit dem kleinsten BIC wird bevorzugt. AIC and BIC criterion for Model selection, how is it used in this paper? Model Selection Criterion: AIC and BIC 403 information criterion, is another model selection criterion based on infor-mation theory but set within a Bayesian context. das Modell. But you can also do that by crossvalidation. Despite the sample size n n doubling at each step, their probability of recovering the true model gets stuck at about 0.60 0.60. Select a single best model from among $$M_0$$, . Formula can be re-expressed using the model $$R^2$$, which is easier to calculate $$$\text{BIC} = n\ln(1-R^2)+(p+1)\ln(n)+\text{constant}, \tag{7.3}$$$ where the last term constant only depends on the sample size $$n$$, and the observed data $$y_1,\cdots, y_n$$. Share. Cite. I have already concluded what model is better based on other factors but this makes me confused. (in a design with two boards). information criteria: AIC, BIC, DIC, WAIC Simultaneous selection and estimation (LASSO, NNET, CART) Model selection — How? Mixture model selection via BIC. Model selection and multimodel inference, 2nd edn. What does it mean if they disagree? 3. You don’t have to absorb all the theory, although it is there for your perusal if you are interested. I often use fit criteria like AIC and BIC to choose between models. How to add aditional actions to argument into environement. Cross-validation, a non-Bayesian model selection technique, also picks 5. It is often the case that some or many of the variables used in a multiple regression model are in fact not associated with the response variable. You need to specify the option nvmax, which represents the maximum number of predictors to incorporate in the model.For example, if nvmax = 5, the function will return up to the best 5-variables model, that is, it returns the best 1-variable model … The most useful resource I have stumbled upon is this earlier question here on CrossValidated: Is it possible to calculate AIC and BIC for lasso regression models? Just think of it as an example of literate programming in R using the Sweave function. Model Selection Criterion: AIC and BIC 401 For small sample sizes, the second-order Akaike information criterion (AIC c) should be used in lieu of the AIC described earlier.The AIC c is AIC 2log (=− θ+ + + − −Lkk nkˆ) 2 (2 1) / ( 1) c where n is the number of observations.5 A small sample size is when n/k is less than 40. You can customize the criterion used (i.e. This method seemed most efficient. It only takes a minute to sign up. Model selection is a process of seeking the model in a set of candidate models that gives the best balance between model fit and complexity (Burnham & Anderson 2002). The Challenge of Model Selection 2. Thanks for contributing an answer to Cross Validated! Replication requirements: What you’ll need to reproduce the analysis in this tutorial. It is common to choose a model that performs the best on a hold-out test dataset or to estimate model performance using a resampling technique, such as k-fold cross-validation. The AIC can be used to select between the additive and multiplicative Holt-Winters models. A list including: A message informing the user about the best model. The vertical axis probably means "Drop in BIC" compared to the intercept-only model, not the model BIC. For instance, if your ideal model has a BIC of 1451.665, corresponding to a drop of 220. It is a bit overly theoretical for this R course. It is possible to build multiple models from a given set of X variables. AIC/BIC for a segmented regression model? Advances in Data Analysis and Classification, 8(2), 217-226. G. A numeric vector with the number of components, clusters, to be considered. Model Selection in R Charles J. Geyer October 28, 2003 This used to be a section of my master’s level theory notes. The right-hand-side of its lower component is always included in the model, and right-hand-side of the model is included in the upper component. When fitting models, it is possible to increase the likelihood by adding parameters, … The statistical analysis of compositional data. Springer, New York. An alternative approach to model selection involves using probabilistic statistical measures that attempt to quantify both the model A boolean variable, TRUE or FALSE specifying whether a graph should be drawn or not. Chapman \& Hall. [R] automatic model selection based on BIC in MLE [R] Stepwise logistic model selection using Cp and BIC criteria [R] problem with BIC model selection [R] Model selection with BIC [R] regsubsets (Leaps) [R] Generating a model fitness when score using svyglm? In some cases, the second peak may exceed the first. What are they really doing? I implemented @johnnyheineken's answer as follows: I basically tried a few lambdas, fine-tuned the ranged of lambda, and found the "best model" when lambda = 0.0001. A matrix with compositional data. Which is better? Keywords model selection, mixtures of normal distributions . R package version 1.5. Estimating Common Principal Components in High Dimensions. Specifically, Stone (1977) showed that the AIC and leave-one out crossvalidation are asymptotically equivalent. Note that since all of the models are fit to the same data set, you will get the same model selection results regardless of whether you use BIC or the normalized BIC … For the least square model AIC and Cp are directly proportional to each other. Signed, Adrift on the ICs MathJax reference. Venables, W. N. and Ripley, B. D. (2002) Modern Applied Statistics with S. Fourth edition. I ended up running forwards, backwards, and stepwise procedures on data to select models and then comparing them based on AIC, BIC, and adj. Why did Churchill become the PM of Britain during WWII instead of Lord Halifax? [R] Question about model selection for glm -- how to select features based on BIC? How do you say “Me slapping him.” in French? Using di erent selection criteria may lead to di erent models (there is no one best model). AIC, BIC, etc. Bayesian Information Criterion 5. 2. Asking for help, clarification, or responding to other answers. In general, it might be best to use AIC and BIC together in model selection. This tutorial is divided into five parts; they are: 1. p and F-tests use a \full" model MSE. Auch für das BIC gilt, dass das Modell mit dem kleinsten Wert des Informationskriteriums eine bessere Anpassung aufweist als die Alternativmodelle. Model selection or model comparison is a very common problem in ecology- that is, we often have multiple competing hypotheses about how our data were generated and we want to see which model is best supported by the available evidence. More technically, AIC and BIC are based on different motivations, with AIC an index based on what is called Information Theory, which has a focus on predictive accuracy, and BIC an index derived as an approximation of the Bayes Factor, which is used to find the true model if it ever exists. In a context of a binary classification, here are the main metrics that are important to track in order to assess the performance of the model. The BIC values for every possible model and number of components. You shouldn’t compare too many models with the AIC. R : Robust nonlinear least squares fitting of three-phase linear model with confidence & prediction intervals Hot Network Questions What does children mean in “Familiarity breeds contempt - … Akaike Information Criterion 4. Would having only 3 fingers/toes on their hands/feet effect a humanoid species negatively? (but not the type of clustering you're thinking about), Why are two 555 timers in separate sub-circuits cross-talking? , $$M_p$$ using cross-validated prediction error, $$C_p (AIC), BIC$$, or adjusted $$R^2$$. We suggest you remove the missing values first. [R] Which model to keep (negative BIC) [R] SEM model testing with identical goodness of fits Model selection is the task of selecting a statistical model from a set of candidate models, given data. For example, you can vary nvmax from 1 to 5. All standard LDA methods and parameters from topimodels package can be set with method and control. Since this constant does not depend on the choice of model, … I know that they try to balance good fit with parsimony, but beyond that Im not sure what exactly they mean. SBC usually results in fewer parameters in the model than AIC. LASSO Regression with AIC or BIC as Model Selection Criterion. Model selection: goals Model selection: general Model selection: strategies Possible criteria Mallow’s Cp AIC & BIC Maximum likelihood estimation AIC for a linear model Search strategies Implementations in R Caveats - p. 3/16 Crude outlier detection test If the studentized residuals are … R-sq. The fifth step is to compute the BIC for each model and then select the model which has the smallest BIC. An information criterion tries to identify the model with the smallest AIC and BIC that balance the model fit and model complexity. I am using R software and running 3 models, GARCH-t, GJR model, and simple GARCH (1,1) model. Model selection conducted with the AIC will choose the same model as leave-one-out cross validation (where we leave out one data point and fit the model, then evaluate its fit to that point) for large sample sizes. Model Selection in R Charles J. Geyer October 28, 2003 This used to be a section of my master’s level theory notes. A good model is the one that has minimum AIC among all the other models. [R] automatic model selection based on BIC in MLE [R] Stepwise logistic model selection using Cp and BIC criteria [R] problem with BIC model selection In multiple regression models, R2 corresponds to the squared correlation between the observed outcome values and the predicted values by the model. 3.2 Model selection. Murtaugh ( 2009 ) argued that ‘Simulation, in which the ‘true’ model is known, would seem the only definitive way to compare model‐building techniques’. The difference between the BIC and the AIC is the greater penalty imposed for the number of param-eters by the former than the latter. Here, we explore various approaches to build and evaluate regression models. Comparing models: Determining which model is best. Linear Model Selection. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. If scope is a single formula, it specifies the upper component, and the lower model is empty. Model selection: choosing a subset of variables¶ To "implement" a model selection procedure, we first need a criterion or benchmark to compare two models. Improve this question. How can we compare models? Results obtained with LassoLarsIC are based on AIC/BIC criteria. Model selection: goals Model selection: general Model selection: strategies Possible criteria Mallow’s Cp AIC & BIC Maximum likelihood estimation AIC for a linear model Search strategies Implementations in R Caveats - p. 10/16 Mallow’s Cp Cp(M) = SSE(M) b˙2 n+2 p(M): a plot with the BIC of the best model for each number of components versus the number of components. Eine sehr popul are Strategie in der Praxis ist es, Werte von R2 adj, AIC, AICc und BIC zu berechnen und die Modelle zu vergleichen, die AIC, AICc und BIC minimieren, mit jenem das R2 adj maximiert. Data Prep. It is defined as follows: Main metrics― The following metrics are commonly used to assess the performance of classification models: ROC― The receiver operating curve, also noted ROC, is the plot of TPR ve… The alr or the ilr-transformation is applied to the compositional data first and then mixtures of multivariate Gaussian distributions are fitted. Can a half-elf taking Elf Atavism select a versatile heritage? Just think of it as an example of literate programming in R using the Sweave function. Including such irrelevant variables leads to unnecessary complexity in the resulting model. Dennoch kann der Gesamterklärungsgehalt des Modells gering sein. How to accomplish? The different criteria quantify different aspects of the regression model, and therefore often yield different choices for the best set of predictors. Dimension reduction procedures generates and returns a sequence of possible models $M_0$ ... We want Cp, BIC to be as small as possible and adjusted R squared as large as possible. Then the model with just waist.girth and weight should have a BIC of about 1551. The term which is added (2dsigma²) is the... AIC ( Akaike Information Criterion). Using the all possible subsets method, one would select a model with a larger adjusted R-square, smaller Cp, smaller rsq, and smaller BIC. How to add ssh keys to a specific user in linux? In Chapter 2 we briefly saw that the inclusion of more predictors is not for free: there is a price to pay in terms of more variability in the coefficients estimates, harder interpretation, and possible inclusion of highly-dependent predictors. Unfortunately, manually filtering through and comparing regression models can be tedious. To learn more, see our tips on writing great answers. For those wishing to follow along with the R-based demo in class, click here for the companion R script for this lecture. Can someone identify this school of thought? Springer. The type of trasformation to be used, either additive log-ratio ("alr") or the isometric log-ratio ("ilr"). References. Unlike Bayesian procedures, such inferences are prior-free. There is a clear philosophy, a sound criterion based in information theory, and a rigorous statistical foundation for AIC. A numeric vector with the number of components, clusters, to be considered. glmulti: An R Package for Easy Automated Model Selection with (Generalized) Linear Models; ... BIC, and adj. Lets prepare the data upon which the various model selection approaches will be applied. Model selection is the problem of choosing one from among a set of candidate models. Best subset selection: Finding the best combination of the ppredictors. We ended up bashing out some R code to demonstrate how to calculate the AIC for a simple GLM (general linear model). Use MathJax to format equations. However it is far more costly to compute. But building a good quality model can make all the difference. This is a tutorial all about model selection, which plays a large role when you head into the realm of regression analyses. Recall that our I-T metrics, as well as likelihood ratio tests, used the value of the likelihood surface at the MLE. Somit zieht BIC eher einfache Modelle vor. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Note that BIC (Schwarz Information Criterion) is no more Bayesian than AIC. rev 2021.1.21.38376, The best answers are voted up and rise to the top, Cross Validated works best with JavaScript enabled, By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. . However, when I received the actual data to be used (the program I was writing was for business purposes), I was told to only model each explanatory variable against the response, so I was able to just call BIC ist sehr ahnlich zu AIC, nur ist der Faktor 2 im Strafterm jetzt durch logn ersetzt. Confusion matrix― The confusion matrix is used to have a more complete picture when assessing the performance of a model. Model Selection in R We will work again with the data from Problem 6.9, “Grocery Retailer.” Recall that we formed a data table named Grocery consisting of the variables Hours, Cases, Costs, and Holiday. With a limited number of predictors, it is possible to search all possible models (leaps in R). It is based, in part, on the likelihood function and it is closely related to the Akaike information criterion. My student asked today how to interpret the AIC (Akaike’s Information Criteria) statistic for model selection. For example, in selecting the number of latent classes in a model, if BIC points to a three-class model and AIC points to a five-class model, it makes sense to select from models with 3, 4 and 5 latent classes. See Also 4. It is a bit overly theoretical for this R course. The most useful resource I have stumbled upon is this earlier question here on CrossValidated: Is it possible to calculate AIC and BIC for lasso regression models? mixture: Mixture Models for Clustering and Classification. What are some "clustering" algorithms? It is computation intensive procedure and ldatuning uses parallelism, so do not forget to point correct number of CPU cores in mc.core parameter to archive the best performance. Since this is a very introductory look at model selection we assume the data you’ve acquired has already been cleaned, scrubbed and ready to go. (Poltergeist in the Breadboard). The evidence approximations can be computed directly from the eigenvalue spectrum and they are very fast. The above formula is for Cp, RSS is the same Residual sum of squares. Later, Burnham & Anderson suggested that the ‘proper way to compare AIC‐ and BIC‐based model selection is in terms of achieved performance, especially prediction …’. What's the ideal positioning for analog MUX in microcontroller circuit? However, the task can also involve the design of experiments such that the data collected is well-suited to the problem of model selection. R-sq. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. In statistics, the Bayesian information criterion or Schwarz information criterion is a criterion for model selection among a finite set of models; the model with the lowest BIC is preferred. R topics documented: ... fying a join or nested select k penalty per parameter for AIC quiet When FALSE, warn if the ﬁt did not converge... Additional arguments. I need 30 amps in a single room to run vegetable grow lighting. Is cycling on this 35mph road too dangerous? [R] automatic model selection based on BIC in MLE [R] Stepwise logistic model selection using Cp and BIC criteria [R] problem with BIC model selection [R] regsubsets (Leaps) [R] Generating a model fitness when score using svyglm? I always think if you can understand the derivation of a statistic, it is much easier to remember how to use it. We try to keep on minimizing the stepAIC value to come up with the final set of features. Making statements based on opinion; back them up with references or personal experience. What‘re we doing? Model Selection. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. This method seemed most efficient. 5. Given a criterion, we also need a search strategy. Aitchison J. Model selection concerns both the covariance type and the number of components in the model. Mobile friendly way for explanation why button is disabled. Model performance metrics. So, I am trying to see which model is better, based only on BIC. Who decides how a historic piece is adjusted (if at all) for modern instruments? Difference between chess puzzle and chess problem? Both methods peak at dimensionality 5, however BIC is increasingly unreliable for large dimensionalities. “stepAIC” does not necessarily means to improve the model performance, however it is used to simplify the model without impacting much on the performance. One quick check is to code up the BIC using both R and SAS, then compare the AIC/BIC values. ), how the output is reported, what’s included in the output (e ... Burnham K.P., Anderson D.R. There are several different formulations for information criteria. I implemented @johnnyheineken's answer as follows: The AIC and LOOCV are inconsistent in selecting the true model. Versus the number of components in the model, and adj balance the than... Aic/Bic values of recovering the TRUE model a model as likelihood ratio tests, used the value the... More Bayesian than AIC proposed protocol ( in many cases ) is model selection manually filtering Through and comparing models. Model than AIC one quick check is to compute the BIC using both R and SAS then... Best models of different sizes former than the latter that balance the model fit model... Aic/Bic criteria when it is a clear philosophy, a non-Bayesian model selection we try to balance fit. Model is empty on other factors but this makes me confused based only on BIC bigglm 3 Details data! Student asked today how to use AIC and Cp are directly proportional to other... And Cp are directly proportional to each other each step, their probability of recovering the TRUE model and regression... The difference!!! user contributions licensed under cc by-sa, which to. The number of predictors g. a numeric vector with the BIC of the.! 2014 ) you check the source code of both R and SAS implementations see... Kleinsten Wert des Informationskriteriums eine bessere Anpassung aufweist als die Alternativmodelle Finding the best performance the term... Need 30 amps in a single best model from among a set of candidate.. The PM of Britain during WWII instead of Lord Halifax vs AIC for a GLM... 5, however BIC is increasingly bic model selection in r for large dimensionalities learn more think if you understand! Values and the AIC is the same Residual sum of squares the R regsubsets. ’ s Information criteria ) statistic for model selection is the same proposed protocol ( in cases... Our I-T metrics, as well as likelihood ratio tests, used the value of most..., 8 ( 2 ), 217-226 on writing great answers choosing one from among (. To build multiple models from a given set of candidate models instance, if ideal. And exploring BIC ( or AIC ) as the selection criterion selection: Finding the best from. Compare Too many models with the best model from among \ ( bic model selection in r ), is... Say “ me slapping him. ” in French other models unfortunately, manually filtering Through and regression. Clear philosophy, a sound criterion based in Information bic model selection in r, although it is possible calculate! Of Britain during WWII instead of Lord Halifax the additive and multiplicative Holt-Winters models of,... Possible to search all possible models ( leaps in R using the Sweave function cc! Code to demonstrate how to add aditional actions to argument into environement ist der Faktor 2 Strafterm..., stepAIC is one of the likelihood surface at the MLE to balance good fit with,. Outcome values and the AIC how the output ( e... burnham K.P., Anderson D.R and the values. ) model Mallows Cp Cross Validation model selection am trying to see which bic model selection in r... 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## Section8.6Example: The Cipher Class Hierarchy Suppose we wish to design a collection of cipher classes, including a Caesar cipher and a transposition cipher. Because the basic operations used in all forms of encryption are the same, both the Caesar class and the Transpose class will have methods to encrypt and decrypt messages, where each message is assumed to be a string of words separated by spaces. These methods will take a String of words and translate each word using the encoding method that is appropriate for that cipher. encrypt: Take a sentence and encode each word. decrypt: take a sentence and decode each word. In addition to encrypt() and decrypt(), each cipher class will need polymorphic encode and decode methods, which take a single word and encode or decode it according to the rules of that particular cipher. encode: Take a word and encode it. decode: take a word and decode it. ### Subsection8.6.1Class Hierarchy From a design perspective the encrypt() and decrypt() methods will be the same for every class: They simply break the message into words and encode or decode each word. However, the encode() and decode() methods will be different for each different cipher. The Caesar.encode() method should replace each letter of a word with its substitute, whereas the Transpose.encode() method should rearrange the letters of the word. Given these considerations, how should we design this set of classes? Because all of the various ciphers will have the same methods, it will be helpful to define a common Cipher superclass (Figure 8.6.1), which will encapsulate those features that the individual cipher classes have in common—the encrypt(), decrypt(), encode(), and decode() methods. Some of these methods can be implemented in the Cipher class itself. For example, the encrypt() method should take a message in a String parameter, encode each word in the message, and return a String result. The following method definition will work for any cipher: public String encrypt(String s) { StringBuffer result = new StringBuffer(""); StringTokenizer words = new StringTokenizer(s);// Tokenize while (words.hasMoreTokens()) { // Encode each word result.append(encode(words.nextToken()) + " "); } return result.toString(); // Return result } // encrypt() This method creates a local StringBuffer variable, result, and uses StringTokenizer to break the original String into its component words. It uses the encode() method to encode the word, appending the result into result. The result is converted back into a String and returned as the encrypted translation of s, the original message. If we define encrypt() in the superclass, it will be inherited by all of Cipher's subclasses. On the other hand, the polymorphic encode() method cannot be implemented within Cipher, because unlike the encrypt() method, which is the same for every Cipher subclass, the encode() method will be different for every subclass. So, by declaring the encode() method as abstract, we will leave its implementation up to the Cipher subclasses. Thus, within the Cipher class, we would define encode() and decode() as follows: // Abstract methods public abstract String encode(String word); public abstract String decode(String word); ### Subsection8.6.2Class Design: Cipher Listing 8.6.2 provides the full definition of the Cipher class. The abstract encode() and decode() methods will be implemented by Cipher's subclasses. Note again that encrypt() and decrypt() call encode() and decode(), respectively. Java's dynamic binding mechanism will take care of invoking the appropriate implementation of encode() or decode(), depending on what type of cipher is involved. ### Subsection8.6.3Algorithm Design: Shifting Characters The Caesar class (Listing 8.6.3) extends Cipher and implements its own version of encode() and decode(). The encode() method takes a word and returns the result of shifting each of the word's letters. How do we do that? Fortunately, because char data in Java are represented as 16-bit integers. we can use some arithmetic to perform the shift. For example, the character 'h' has an ASCII code of 104. Adding 3 gives 107, the ASCII code for 'k', which is shifted 3 characters to the right of 'h'. The problem is this doesn't always work so simply. For example, the ASCII code for 'y' is 121. Adding 3 gives 124, the ASCII code for '\|', which is not our desired result. To fix this, what we need to do is “wrap around” to the beginning of the alphabet, so that 'y' gets shifted into 'b'. In order to accomplish this wrap around we need to do some modular arithmetic. The alphabet has 26 letters and for any positive integer N, N % 26 will always give a number between 0 and 25. If we number the letters 'a' to 'z' as 0 to 25, then 'y' would be 24. Adding 3 gives 27 and 27 % 26 would give 1, which is the letter 'b'. Thus, if we can convert the letters to numbers between 0 and 25, we can use simple arithmetic to perform the shift. This leads to the following algorithm. For step one of this algorithm we can convert any letter 'a' to 'z' into a number 0 to 25 simply by subtracting the letter 'a' from it: 'a' - 'a' = 0 'b' - 'a' = 1 'c' - 'a' = 2 ... 'z' - 'a' = 25 Adding the shift and dividing % 26 gives us a number between 0 and 25. To convert that back into a letter, we can just add the letter 'a' to it and use the (char) cast operator to convert it to a character. If we combine these steps, we get the following expression for shifting ch by 3: (char)('a' + (ch -'a'+ 3) % 26) which breaks down as follows: ch -'a' // Convert to 0 to 25 (ch -'a'+ 3) // Add shift (ch -'a'+ 3) % 26 // Divide % 26 ('a' + (ch -'a'+ 3) % 26) // Add 'a' (char)('a' + (ch -'a'+ 3) % 26) // Cast back into a letter To summarize, we can shift any character by 3 if we map it into the range 0 to 25, then add 3 to it mod 26, then map that result back into the range 'a' to 'z'. The algorithm for the reverse shift in the decode() method is similar. Accept in this case the reverse Caesar shift is done by shifting by 23, which is $$26-3\text{.}$$ If the original shift is 3, we can reverse that by shifting an additional 23. Together this gives a shift of 26, which will give us back our original letter. See this code in action below. #### Activity8.6.1. Run the code below. Change the text in main to another lowercase string and try again. This code will only work with lowercase letters. ### Subsection8.6.4Class Design: Transpose The Transpose class (Listing 8.6.5) is structured the same as the Caesar class. It implements both the encode() and decode() methods. The key element here is the transpose operation, which in this case is a simple reversal of the letters in the word. Thus, “hello” becomes “olleh”. This is very easy to do when using the StringBuffer.reverse() method. The decode() method is even simpler, because all you need to do in this case is call encode(). Reversing the reverse of a string gives you back the original string. ### Subsection8.6.5Testing and Debugging Listing 8.6.6 provides a simple test program for testing Cipher and its subclasses. It creates a Caesar cipher and a Transpose cipher and then encrypts and decrypts the same sentence using each cipher. If you run this program, it will produce the following output: ******* Caesar Cipher Encryption ***** PlainText: this is the secret message Encrypted: wklv lv wkh vhfuhw phvvdjh Decrypted: this is the secret message ***** Transpose Cipher Encryption ******* PlainText: this is the secret message Encrypted: siht si eht terces egassem Decrypted: this is the secret message See this code in action below. #### Activity8.6.2. Run the code below. Change the plain variable in main to another lowercase string and try again. This code will only work with lowercase letters. #### ExercisesSelf-Study Exercises ##### 1.Caesar Shift. Modify the Caesar class so that it will allow various sized shifts to be used, instead of just a shift of size 3. (Hint: Use an instance variable in the Caesar class to represent the shift, add a constructor to set it, and change the encode method to use it.) ##### 2.Transpose Rotate. Modify Transpose.encode() so that it uses a rotation instead of a reversal. That is, a word like “hello” should be encoded as “ohell” with a rotation of one character. (Hint: use a loop to append the letters into a new string)	{}
Corpus ID: 73662780 # Model theory of the field of \$p\$-adic numbers expanded by a multiplicative subgroup ```@article{Mariaule2018ModelTO, title={Model theory of the field of \\$p\\$-adic numbers expanded by a multiplicative subgroup}, author={Nathanael Mariaule}, journal={arXiv: Logic}, year={2018} }``` Let \$G\$ be a multiplicative subgroup of \$\mathbb{Q}_p\$. In this paper, we describe the theory of the pair \$(\mathbb{Q}_p, G)\$ under the condition that \$G\$ satisfies Mann property and is small as subset of a first-order structure. First, we give an axiomatisation of the first-order theory of this structure. This includes an axiomatisation of the theory of the group \$G\$ as valued group (with the valuation induced on \$G\$ by the \$p\$-adic valuation). If the subgroups \$G^{[n]}\$ of \$G\$ have finite… Expand #### References SHOWING 1-10 OF 24 REFERENCES THE FIELD OF p-ADIC NUMBERS WITH A PREDICATE FOR THE POWERS OF AN INTEGER A Version of o-Minimality for the p-adics • Mathematics, Computer Science • J. Symb. Log. • 1997 Presburger sets and p-minimal fields • R. Cluckers • Mathematics, Computer Science • Journal of Symbolic Logic • 2003 Solution of a Problem of Tarski • J. Myhill • Mathematics, Computer Science • J. Symb. Log. • 1956	{}
# How do you solve x - 2y + z = - 2, 2x - 3y + 2z = 2, and 4x - 8y + 5z = - 5 using matrices? Sep 8, 2017 The solution is: $x = 7 , y = 6 , z = 3$ #### Explanation: We have: $x - 2 y + z = - 2$ $2 x - 3 y + 2 z = 2$ $4 x - 8 y + 5 z = - 5$ Which we can write in vector matrix form: $\left(\begin{matrix}1 & - 2 & 1 \\ 2 & - 3 & 2 \\ 4 & - 8 & 5\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}- 2 \\ 2 \\ - 5\end{matrix}\right)$ Or: $\boldsymbol{A} \boldsymbol{\underline{x}} = \boldsymbol{\underline{b}} \implies \boldsymbol{\underline{x}} = {\boldsymbol{A}}^{- 1} \boldsymbol{\underline{b}}$ Where $\boldsymbol{A} = \left(\begin{matrix}1 & - 2 & 1 \\ 2 & - 3 & 2 \\ 4 & - 8 & 5\end{matrix}\right)$; $\boldsymbol{\underline{x}} = \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)$; $\boldsymbol{\underline{b}} = \left(\begin{matrix}- 2 \\ 2 \\ - 5\end{matrix}\right)$ We can find ${\boldsymbol{A}}^{- 1}$, or solve this linear system via any of several methods: Method 1 - Gaussian Elimination Here we form an augmented matrix of the equation coefficients. ( (1, -2, 1, \|, -2), (2, -3, 2, \|, 2), (4, -8, 5, \|, -5) ) We can now perform elementary row operations with the aim of getting leading zeros in progresisve rows: ( (1, -2, 1, \|, -2), (2, -3, 2, \|, 2), (4, -8, 5, \|, -5) ) stackrel(R_2-2R_1 rarr R_2)(rarr) ( (1, -2, 1, \|, -2), (0, 1, 0, \|, 6), (4, -8, 5, \|, -5) ) ( (1, -2, 1, \|, -2), (0, 1, 0, \|, 6), (4, -8, 5, \|, -5) ) stackrel(R_3-4R_1 rarr R_3)(rarr) ( (1, -2, 1, \|, -2), (0, 1, 0, \|, 6), (0, 0, 1, \|, 3) ) We can now use back substitution to get the values of $x$, $y$, and $z$: Row $3 \implies z = 3$ Row $2 \implies y = 6$ Row $1 \implies x - 2 y + z = - 2$ $\therefore x - 12 + 3 = - 2 \implies x = 7$ Thus we have a unique solution: $x = 7 , y = 6 , z = 3$ Method 2 - Matrix Inversion A matrix, $\boldsymbol{A}$, is invertible if and only if its determinant $\left\mid \boldsymbol{A} \right\mid \ne 0$. There are several ways to invert a matrix inclining finding the adjoint, row reduction or even a calculator. I will use the adjoint method, which has several steps; • Calculating the Matrix of Minors, • Form the Matrix of Cofactors, $c o f \left(\boldsymbol{A}\right)$ • Form the adjoint matrix, $a \mathrm{dj} \left(\boldsymbol{A}\right)$ • Multiply $a \mathrm{dj} \left(\boldsymbol{A}\right)$ by $\frac{1}{\left\mid \boldsymbol{A} \right\mid}$ to form the inverse ${\boldsymbol{A}}^{-} 1$ At some point we need to calculate $\left\mid \boldsymbol{A} \right\mid$, or $\det \left(\boldsymbol{A}\right)$, and this can also be used to test if the matrix is actually invertible so I prefer to do this first; $\boldsymbol{A} = \left(\begin{matrix}1 & - 2 & 1 \\ 2 & - 3 & 2 \\ 4 & - 8 & 5\end{matrix}\right)$ If we expand about the first row and "strike out" the row and column to form a smaller determinant and alternate signs we get; $\left\mid \boldsymbol{A} \right\mid = + \left(1\right) \| \left(- 3 , 2\right) , \left(- 8 , 5\right) \| - \left(- 2\right) \| \left(2 , 2\right) , \left(4 , 5\right) \| + \left(1\right) \| \left(2 , - 3\right) , \left(4 , - 8\right) \|$ $\setminus \setminus \setminus \setminus \setminus = \left\{\left(- 3\right) \left(5\right) - \left(- 8\right) \left(2\right)\right\} + 2 \left\{\left(2\right) \left(5\right) - \left(4\right) \left(2\right)\right\} + \left\{\left(2\right) \left(- 8\right) - \left(4\right) \left(- 3\right)\right\}$ $\setminus \setminus \setminus \setminus \setminus = \left\{- 15 + 16\right\} + \left\{10 - 8\right\} + \left\{- 16 + 12\right\}$ $\setminus \setminus \setminus \setminus \setminus = 1 + 4 - 4$ $\setminus \setminus \setminus \setminus \setminus = 1$ As $\left\mid \boldsymbol{A} \right\mid \ne 0 \implies \boldsymbol{A}$ is invertible, so we now calculate the matrix of minors by systematically working through each element in the matrix and "strike out" that row and columns and form the determinant of the remaining elements, as follows: "minors"(bb(A))=( ( \|(-3, 2), (-8, 5)\|, \|(2, 2), (4, 5)\|, \|(2, -3), (4, -8)\| ), ( \|(-2, 1), (-8, 5)\|, \|(1, 1), (4, 5)\|, \|(1, -2), (4, -8)\| ), ( \|(-2, 1), (-3, 2)\|, \|(1, 1), (2, 2)\|, \|(1, -2), (2, -3)\| ) ) $\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}- 15 + 16 & 10 - 8 & - 16 + 12 \\ - 10 + 8 & 5 - 4 & - 8 + 8 \\ - 4 + 3 & 2 - 2 & - 3 + 4\end{matrix}\right)$ $\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}1 & 2 & - 4 \\ - 2 & 1 & 0 \\ - 1 & 0 & 1\end{matrix}\right)$ We now form the matrix of cofactors, $c o f \left(A\right)$, by taking the above matrix of minors and applying the alternating sign matrix as in $\left(\begin{matrix}+ & - & + \\ - & + & - \\ + & - & +\end{matrix}\right)$ Where we change the sign of those elements with the minus sign to get; $c o f \left(\boldsymbol{A}\right) = \left(\begin{matrix}1 & - 2 & - 4 \\ 2 & 1 & 0 \\ - 1 & 0 & 1\end{matrix}\right)$ Then we form the adjoint matrix by transposing the matrix of cofactors, $c o f \left(A\right)$, so; $a \mathrm{dj} \left(A\right) = c o f {\left(A\right)}^{T}$ $\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\left(\begin{matrix}1 & - 2 & - 4 \\ 2 & 1 & 0 \\ - 1 & 0 & 1\end{matrix}\right)}^{T}$ $\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}1 & 2 & - 1 \\ - 2 & 1 & 0 \\ - 4 & 0 & 1\end{matrix}\right)$ And then finally we multiply by the reciprocal of the determinant to get: ${\boldsymbol{A}}^{-} 1 = \frac{1}{\left\mid \boldsymbol{A} \right\mid} a \mathrm{dj} \left(\boldsymbol{A}\right)$ $\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(1\right) \left(\begin{matrix}1 & 2 & - 1 \\ - 2 & 1 & 0 \\ - 4 & 0 & 1\end{matrix}\right)$ $\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}1 & 2 & - 1 \\ - 2 & 1 & 0 \\ - 4 & 0 & 1\end{matrix}\right)$ So then we get the solution the linear equations as: $\boldsymbol{\underline{x}} = {\boldsymbol{A}}^{- 1} \boldsymbol{\underline{b}}$ ..... $\left[\star\right]$ $\therefore \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}1 & 2 & - 1 \\ - 2 & 1 & 0 \\ - 4 & 0 & 1\end{matrix}\right) \left(\begin{matrix}- 2 \\ 2 \\ - 5\end{matrix}\right)$ $\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}\left(1\right) \left(- 2\right) + \left(2\right) \left(2\right) + \left(- 1\right) \left(- 5\right) \\ \left(- 2\right) \left(- 2\right) + \left(1\right) \left(2\right) + \left(0\right) \left(2\right) \\ \left(- 4\right) \left(- 2\right) + \left(0\right) \left(2\right) + \left(1\right) \left(- 5\right)\end{matrix}\right)$ $\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}- 2 + 4 + 5 \\ 4 + 2 \\ 8 - 5\end{matrix}\right)$ $\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}7 \\ 6 \\ 3\end{matrix}\right)$ Thus we have a unique solution: $x = 7 , y = 6 , z = 3$ Method 3 - Cayley Hamilton Theorem The CH-Theorem states that any square matrix satisfies its own characteristic equation. We can exploit this property to find a the inverse. We have: $\boldsymbol{A} = \left(\begin{matrix}1 & - 2 & 1 \\ 2 & - 3 & 2 \\ 4 & - 8 & 5\end{matrix}\right)$ So, the characteristic equation is given by: $\left\mid \boldsymbol{A} - l a m \mathrm{da} \boldsymbol{I} \right\mid = 0 \implies \left\mid \begin{matrix}1 - l a m \mathrm{da} & - 2 & 1 \\ 2 & - 3 - l a m \mathrm{da} & 2 \\ 4 & - 8 & 5 - l a m \mathrm{da}\end{matrix} \right\mid = 0$ If we expand about row 1 then we get: $\left\mid \begin{matrix}1 - l a m \mathrm{da} & - 2 & 1 \\ 2 & - 3 - l a m \mathrm{da} & 2 \\ 4 & - 8 & 5 - l a m \mathrm{da}\end{matrix} \right\mid = 0$ $\left(1 - l a m \mathrm{da}\right) \| \left(- 3 - l a m \mathrm{da} , 2\right) , \left(- 8 , 5 - l a m \mathrm{da}\right) \| - \left(- 2\right) \| \left(2 , 2\right) , \left(4 , 5 - l a m \mathrm{da}\right) \| + \left(1\right) \| \left(2 , - 3 - l a m \mathrm{da}\right) , \left(4 , - 8\right) \| = 0$ $\therefore \left(1 - l a m \mathrm{da}\right) \left\{\left(- 3 - l a m \mathrm{da}\right) \left(5 - l a m \mathrm{da}\right) - \left(- 8\right) \left(2\right)\right\} + 2 \left\{\left(2\right) \left(5 - l a m \mathrm{da}\right) - \left(4\right) \left(2\right)\right\} + \left\{\left(2\right) \left(- 8\right) - \left(4\right) \left(- 3 - l a m \mathrm{da}\right)\right\} = 0$ $\therefore \left(1 - l a m \mathrm{da}\right) \left\{- 15 - 2 l a m \mathrm{da} + l a m {\mathrm{da}}^{2} + 16\right\} + 2 \left\{10 - 2 l a m \mathrm{da} - 8\right\} + \left\{- 16 + 12 + 4 l a m \mathrm{da}\right\} = 0$ $\therefore \left(1 - l a m \mathrm{da}\right) \left\{l a m {\mathrm{da}}^{2} - 2 l a m \mathrm{da} + 1\right\} + 2 \left\{2 - 2 l a m \mathrm{da}\right\} + \left\{4 l a m \mathrm{da} - 4\right\} = 0$ $\therefore l a m {\mathrm{da}}^{2} - 2 l a m \mathrm{da} + 1 - l a m {\mathrm{da}}^{3} + 2 l a m {\mathrm{da}}^{2} - l a m \mathrm{da} + 4 - 4 l a m \mathrm{da} + 4 l a m \mathrm{da} - 4 = 0$ $\therefore - l a m {\mathrm{da}}^{3} + 3 l a m {\mathrm{da}}^{2} - 3 l a m \mathrm{da} + 1 = 0$ $\therefore l a m {\mathrm{da}}^{3} - 3 l a m {\mathrm{da}}^{2} + 3 l a m \mathrm{da} - 1 = 0$ (Side Note - If we solve this characteristic equation we get ${\left(l a m \mathrm{da} - 1\right)}^{3} = 0$ yielding three identical solutions $l a m \mathrm{da} = 1$, which are the eigenvalues of the given matrix). Then By CH-Theorem we have: ${\boldsymbol{A}}^{3} - 3 {\boldsymbol{A}}^{2} + 3 \boldsymbol{A} - \boldsymbol{I} = 0 \implies \boldsymbol{A} \left({\boldsymbol{A}}^{2} - 3 \boldsymbol{A} + 3 \boldsymbol{I}\right) = \boldsymbol{I}$ $\therefore {\boldsymbol{A}}^{- 1} \boldsymbol{A} \left({\boldsymbol{A}}^{2} - 3 \boldsymbol{A} + 3 \boldsymbol{I}\right) = {\boldsymbol{A}}^{- 1} \boldsymbol{I}$ $\therefore {\boldsymbol{A}}^{- 1} = {\boldsymbol{A}}^{2} - 3 \boldsymbol{A} + 3 \boldsymbol{I}$ And we can readily calculate this matrix product ${\boldsymbol{A}}^{2}$: ${\boldsymbol{A}}^{2} = \left(\begin{matrix}1 & - 2 & 1 \\ 2 & - 3 & 2 \\ 4 & - 8 & 5\end{matrix}\right) \left(\begin{matrix}1 & - 2 & 1 \\ 2 & - 3 & 2 \\ 4 & - 8 & 5\end{matrix}\right)$ $\setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}1 - 4 + 4 & - 2 + 6 - 8 & 1 - 4 + 5 \\ 2 - 6 + 8 & - 4 - 9 - 16 & 2 - 6 + 10 \\ 4 - 16 + 20 & - 8 + 24 - 40 & 4 - 16 + 25\end{matrix}\right)$ $\setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}1 & - 4 & 2 \\ 4 & - 11 & 6 \\ 8 & - 24 & 13\end{matrix}\right)$ And so we have: ${\boldsymbol{A}}^{- 1} = {\boldsymbol{A}}^{2} - 3 \boldsymbol{A} + 3 \boldsymbol{I}$ $\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}1 & - 4 & 2 \\ 4 & - 11 & 6 \\ 8 & - 24 & 13\end{matrix}\right) - 3 \left(\begin{matrix}1 & - 2 & 1 \\ 2 & - 3 & 2 \\ 4 & - 8 & 5\end{matrix}\right) + 3 \left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)$ $\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}1 & - 4 & 2 \\ 4 & - 11 & 6 \\ 8 & - 24 & 13\end{matrix}\right) - \left(\begin{matrix}3 & - 6 & 3 \\ 6 & - 9 & 6 \\ 12 & - 24 & 15\end{matrix}\right) + \left(\begin{matrix}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{matrix}\right)$ $\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}1 - 3 + 3 & - 4 + 6 & 2 - 3 \\ 4 - 6 & - 11 + 9 + 3 & 6 - 6 \\ 8 - 12 & - 24 + 24 & 13 - 15 + 3\end{matrix}\right)$ $\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\begin{matrix}1 & - 2 & - 1 \\ - 2 & 1 & 0 \\ - 4 & 0 & 1\end{matrix}\right)$, as above and we proceed with $\left[\star\right]$ in Method 2. Sep 8, 2017 I got $x = 7 , y = 6 , z = 3$ using Cramer's Method:	{}
시간 제한메모리 제한제출정답맞힌 사람정답 비율 2 초 256 MB116654.545% ## 문제 Since the Phage F-X174 was sequenced in 1977, the DNA sequences of thousands of organisms have been decoded and stored in databases. This sequence information is analyzed to determine genes that encode polypeptides (proteins), RNA genes, regulatory sequences, structural motifs, and repetitive sequences. A comparison of genes within a species or between different species can show similarities between protein functions, or relations between species (the use of molecular systematics to construct phylogenetic trees). With the growing amount of data, it long ago became impractical to analyze DNA sequences manually. Today, computer programs such as BLAST are used daily to search sequences from more than 260 000 organisms, containing over 190 billion nucleotides. These programs can compensate for mutations (exchanged, deleted or inserted bases) in the DNA sequence, to identify sequences that are related, but not identical. A variant of this sequence alignment is used in the sequencing process itself. The so-called shotgun sequencing technique (which was used, for example, by The Institute for Genomic Research to sequence the first bacterial genome, Haemophilus influenzae) does not produce entire chromosomes, but instead generates the sequences of many thousands of small DNA fragments (ranging from 35 to 900 nucleotides long, depending on the sequencing technology). The ends of these fragments overlap and, when aligned properly by a genome assembly program, can be used to reconstruct the complete genome. Shotgun sequencing yields sequence data quickly, but the task of assembling the fragments can be quite complicated for larger genomes. For a genome as large as the human genome, it may take many days of CPU time on large-memory, multiprocessor computers to assemble the fragments, and the resulting assembly will usually contain numerous gaps that have to be filled in later. Shotgun sequencing is the method of choice for virtually all genomes sequenced today, and genome assembly algorithms are a critical area of bioinformatics research. ## 입력 N lines of input file contain N strings of the length L each. Given strings consists of only English alphabet letters [A − Za − z]. 2 ≤ N ≤ 100 000, 2 ≤ L ≤ 100 000. Note, letters are case sensitive. In the test cases N × L ≤ 5 × 1 024 × 1 024 to reduce input/output. ## 출력 Output the string of the length L + N − 1 that all strings in the input file are its substrings, starting from different positions. It is guaranteed that the correct answer exists. In case of several possible answers output any of them. ## 예제 입력 1 abb bba ## 예제 출력 1 abba	{}
# Proof by induction that $3\mid n^3-n$ - confused about inductive step I have these notes, but I'm confused on what is happening at the inductive step. Theorem: $\forall n \in \mathbb{N} 3 \| (n^3-n)$ Inductive Step: For $n \geq 0, show P(n) \Rightarrow P(n+1) is True$ Assume P(n) true i.e. $3 \| (n^3-n)$ Examine: $(n+1)^3 - (n+1)$ = $n^3 + 3n^2 + 3n + 1 - (n+1)$ = $n^3 + 3n^2 + 2n$ = $n^3 - n + 3n^2 + 3n$ I understand what we're trying to prove but 1. where does $(n+1)^3 - (n+1)$ come from? 2. What is happening here $n^3 - n + 3n^2 + 3n$, how did we get there from the previous point? I assume $n^3 + 3n^2 + 2n$ is not obvious if it was divisible by 3, thus we assumed $3 \| (n^3-n)$ is true... But how does step 2 become 3? $P(n+1)$ is the statement that $(n+1)^3 - (n+1)$ is divisible by 3. Just fill in $n+1$ in $P(n)$. So you have to simplify this expression first. The equality between 2 and 3 is just algebra (you write $2n$ from the second term as $3n - n$, which is true for all $n$). Why would you do this? Because you want to use your assumption that $P(n)$ is true, which says that $n^3 - n$ is divisible by 3. The second term has no term $n^3 - n$ but it has an $n^3$ and then substracts $n$ to get the number $n^3 - n$, about which we know something (!), and add $n$ to another term to compensate and keep the equality. Now note that the last line is a sum of multiples of 3, the first one, $n^3 - n$, by the inductive assumption. 1. $(n+1)^3-(n+1)$ comes from the expression $n^3-n$ when you replace $n$ with $n+1$. This is the induction step - to show that $(n+1)^3-(n+1)$ is divisable by $3$ if we assume that $n^3-n$ is. 2. Here the following happens: $n^3-n$ is divisable by $3$ (by the inductive assumption), $3n^2$ is divisable, $3n$ is divisable, therefore, their sum $n^3-n+3n^2+3n$ is divisable too. (Note that $+2n=-n+3n$ here from 2 to 3.)	{}
## [astro-ph/0610007] Separating the Weak Lensing and Kinetic SZ Effects from CMB Temperature Maps Authors: Mario A. Riquelme, David N. Spergel Abstract: A new generation of CMB experiments will soon make sensitive high resolution maps of the microwave sky. At angular scales less than $\sim$10 arcminutes, most CMB anisotropies are generated at z $< 1000$, rather than at the surface of last scattering. Therefore, these maps potentially contain an enormous amount of information about the evolution of structure. Whereas spectral information can distinguish the thermal Sunyaev-Zeldovich (tSZ) effect from other anisotropies, the spectral form of anisotropies generated by the gravitational lensing and the kinetic Sunyaev-Zeldovich (kSZ) effects are identical. While spectrally identical, the statistical properties of these effects are different. We introduce a new real-space statistic, $<\theta (\hat{n})^3 \theta (\hat{m})>_c$, and show that it is identically zero for weakly lensed primary anisotropies and, therefore, allows a direct measurement of the kSZ effect. Measuring this statistic can offer a new tool for studing the reionization epoch. Models with the same optical depth, but different reionization histories, can differ by more than a factor of 3 in the amplitude of the kSZ-generated non-Gaussian signal. [PDF] [PS] [BibTex] [Bookmark] Discussion related to specific recent arXiv papers Antony Lewis Posts: 1610 Joined: September 23 2004 Affiliation: University of Sussex Contact: ### [astro-ph/0610007] Separating the Weak Lensing and Kinetic S This paper defines a simple component of the CMB 4-point function which they claim vanishes for lensing but provides a signature of kinetic SZ, hence allowing one to probe the latter independently. My first comment is that they are using a series expansion in the deflection angle. As they mention this is an approximation, and in fact is not really very good (e.g. see Fig 8 of astro-ph/0601594). However doesn't the lensed < T_lens(x)^3 T_lens(y) > vanish more generally (for uncorrelated lensing potential) because T_lens is linear in T, we assume T is Gaussian and isotropic, and <T(x)^2> is unchanged by lensing? My second comment is that the SZ signal will be correlated with the lensing signal to some extent (see e.g. astro-ph/0208325), which presumably complicates the picture. Christopher M. Hirata Posts: 3 Joined: January 18 2005 Affiliation: Princeton University Contact: ### [astro-ph/0610007] Separating the Weak Lensing and Kinetic S However doesn't the lensed < Tlens(x)3 Tlens(y) > vanish more generally (for uncorrelated lensing potential) Yes, I also get that it vanishes.	{}
## Intermediate Algebra (6th Edition) $20$ Cancelling the common factors between the numerator and the denominator, the given expression, $\dfrac{5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{5\cdot4\cdot\cancel{3\cdot2\cdot1}}{\cancel{3\cdot2\cdot1}} \\\\= 5\cdot4 \\\\= 20 .\end{array}	{}
# Property of norm Let $X$ be a compact Hausdorff space and let $C(X)$ denote the set of continuous complex valued functions on $X$. Define $$\\|f\\|:=\sup\{\|f(x)\|:x\in X\},$$ then prove that $\\|fg\\|\leq \\|f\\|\\|g\\|$. - What are you struggling with? Do you understand the definition of the norm? – Michael Albanese Jul 7 '14 at 8:04 yes i know the definition but i do not know why it is inequality,it should be equality.But now i understand that how this is inequality.Thanks to everyone. – Nannes Jul 7 '14 at 9:06 In future, you should include such concerns in the post so people can address them. – Michael Albanese Jul 7 '14 at 9:10 For every $x \in X$ we have $\|g(x)\| \leq \sup_{z \in X} \|g(z)\|$ by definition of the supremum, so for every $x \in X$ we may observe that $$\|f(x)g(x)\| = \|f(x)\|\|g(x)\|\leq \|f(x)\|\left(\sup_{z \in X}\|g(z)\|\right) =\|f(x)\|\\|g\\|,$$ Since this is true for every $x\in X$ we may take the supremum on both sides of the equation to get $$\\|fg\\| = \sup_{x \in X}\|f(x)g(x)\| \leq \sup_{x \in X} \|f(x)\|\\|g\\| = \\|f\\|\\|g\\|$$ - yes perfect.Thanx – Nannes Jul 7 '14 at 9:08 Alternatively, as $\{(x,x)\| x\in X\} \subset \{(x,y)\| x,y\in X\}$: $$\sup_{x\in X} \|f(x)g(x)\| \le \sup_{x,y\in X} \|f(x)g(y)\| = \sup_{x\in X}\|f(x)\| \sup_{y\in X}\|g(y)\|$$ - How do you justify $\sup_{x \in X} \|f(x)g(x)\| \leq \sup_{x,y \in X} \|f(x)g(y)\|$ ? – Surb Jul 7 '14 at 8:27 this is because $\{(x,x) \| x\in X\}\subset X\times X$. – mookid Jul 7 '14 at 8:49 ok, it is really alternative then :). – Surb Jul 7 '14 at 8:52 For all $x \in X$ we have $$\|f(x)\| \leq \\|f\\| ,$$ as well as $$\|g(x)\| \leq \\|g\\|.$$ Multiplying the two inequalities gives $$\|f(x)g(x)\| \leq \\|f\\|\\|g\\| \; \forall x \in X .$$ Thus $\\|f\\| \\|g\\|$ is an upper bound of the set $\{\|f(x)g(x)\|:x \in X\}$, hence the least upper bound can't be larger. -	{}
2020 NA Regionals Practice Contest 1 #### Start 2020-12-12 09:00 AKST ## 2020 NA Regionals Practice Contest 1 #### End 2020-12-12 14:00 AKST The end is near! Contest is over. Not yet started. Contest is starting in -289 days 14:23:32 5:00:00 0:00:00 # Problem HSaving For Retirement Alice is saving for her retirement. She hasn’t really decided how much she wants to save, but when she retires, she wants to have strictly more money than Bob will have when he retires. Bob is $B$ years old. He plans to retire when he becomes $B_ r$ years old. He saves $B_ s$ every year from now until then. Alice is $A$ years old. She wants to save $A_ s$ every year. When is the earliest time she can retire? ## Input The input is a single line consisting of $5$ space separated integers; $B$, $B_ r$, $B_ s$, $A$, $A_ s$. ## Output Output the age at which Alice can retire so that she has more money than Bob will have at age $B_ r$. ## Limits • $20 \leq B \leq B_ r \leq 100$ • $20 \leq A \leq 100$ • $1 \leq A_ s, B_ s \le 100$ ## Explanation of first example At the age of $25$ Bob has saved $5$ every year for $5$ years. This means he has $25$ saved up. At the age of $23$ Alice has saved $10$ every year for $3$ years. This means she has $30$ saved up, which is strictly more than $25$. Sample Input 1 Sample Output 1 20 25 5 20 10 23 Sample Input 2 Sample Output 2 20 28 5 30 9 35	{}
# Linux Package to Simulate Mechanical Systems 1. Dec 19, 2011 ### syberraith I want to start modeling mechanical systems, something say on the order of a motor, input torque only needed, driving a crankshaft that drives a piston. I'm looking to examine the forces generated in the crank, connecting rod and piston. I was wondering what open source Linux packages are available to do this. Would z88 do? I have years of experience with SPICE modeling, so I guess I'm looking for a mechanical equivalent. 2. Dec 19, 2011 ### Simon Bridge Military grade solid-modelling package. I don't know how complete it is.... In repos I see mostly physics engines for computer programmers. 3. Dec 19, 2011 ### syberraith That looks awesome. I've been looking for a CAD program for a long time. I actually stopped looking for a free one because they were so difficult to learn and different from each other. BRL-CAD looks worthy of the time and effort to climb up it's learning curve. If I may ask some clueless questions: In general how does a constructive solid geometry modeling system differ from a finite element analysis system? Do they serve different purposes? Can they both do dynamic force analysis? 4. Dec 19, 2011 ### Simon Bridge You'll see there is an offshoot specialized to simulate munitions. Whenever someone says "there are no good free/libre-software CAD systems" this is my goto model. The objections then tend to center around the lack of a GUI because what they really mean is that their personal favorite CAD system is not free software. CGS modelling is a computational convenience from it's original purpose in modelling munitions. The military application is central to the development - lots of the libs will be irrelevant to designing and analyzing an engine. But if you want details, I'm afraid you'll have to ask the devs. There is a lot of info on the site and tutorials around the place. It is a very long time since I've used anything like it, so I'm sure it is quite different now. If you have Ubuntu or derivatives, there is always: sudo apt-get install varkon* From the package description: VARKON is a high level development tool for applications in the area of Engineering, Computer Aided Design and Product Modelling. VARKON can be used as a traditional CAD-system with drafting, modelling and visualisation if you want to but the real power of VARKON is in parametric modelling and CAD applications development. VARKON includes interactive parametric modelling in 2D or 3D but also the unique MBS programming language integrated in the graphical environment. There's actually quite a few - search for CAD in synaptic. Most are specialized - like architecture or sailmaking. Nature of Open Source dev. I've never had to model more than the shape of something so I have no experience using CAD packages for analysis. Last edited: Dec 19, 2011 5. Dec 19, 2011 ### syberraith I've been reading up on this a little. I'm still looking for how to do dynamic analysis. The testing I've seen so far has to do with static forces and material strength rather then dynamic loading during operation. Anyway I can do the dynamic modeling crudely with math and make an animation with Blender, which I have used before, until I figure something else out. Thanks, fred 6. Dec 19, 2011 ### Simon Bridge That's largely what I do yeah. Why not ask them? I see BRL-CAD will do movement, so presumably you can apply a series of instantaneous models to get a dynamic result a la numerical methods. You can't be the first person to want to do what you want to do so ask. Similarly with Varkon, one of the tutorials is supposed to have an example of an engine model. I doubt you'll find any CAD system which does general dynamical analysis ... so you'll need to be specific. Good luck and happy hacking. 7. Dec 19, 2011 ### AlephZero There are a whole range of different "levels" of dynamic analysis that you could do on your type of structure. Probably a good place to start would be with software aimed at modeling robot arms and such like, which will treat your structure as a set of rigid parts with mass and inertia properties, jointed together. That will be able to calculate the accelerations of the parts through the complete rotation cycle, and the forces between them. If you then want to do a detailed analysis of an individual part (for example finding the stresses in the connecting rod bearings) you could take the forces and accelerations from your dynamics model and apply them to a detailed finite element model of just that part. The point of doing that is to reduce the dynamics analysis to equivalent static stress analyses on the various parts (each one at the worst loading conditions for that particular part, which will probably occur at different times in the rotation of the crank system), whcih is much cheaper than trying to run a dynamics analysis on a complete detailed model. There may be some manual iterations required here, because if you decide to change the shape of a part when you see the stresses, the change in mass will affect the loads in ALL the parts, not just the part you changed. To do the analysis "all in one go" you would need the sort of software that is used to do car crash simulations etc. But that is almost certainly overkill since (presumably) your crank system is not going to have any permanent plastic deformation, and running large models is time consuming. At work, we run those type of models on seriously large multi-CPU systems (e.g. 64 processors), and some of the models run for literally weeks. That's probably not the place you want to start from! 8. Dec 20, 2011 ### syberraith That sounds like just what I'm wanting. That would be overkill at this point. So you do car crash testing, huh? A few decades ago I almost go a job at Breed Corp in NJ as a fortran programmer back twhen they were using UNIVACs 6000s I think. The girl who was doing the hiring said she didn't want to hire anyone smarter then herself. Looking back, it was rather dumb of me to play the smart one... :) 9. Dec 20, 2011 ### syberraith I've downloaded the four BRL-CAD tutorials. I'll have to dig into that a bit. I'll look at the VARKON stuff too. 10. Dec 20, 2011 ### syberraith I believe all the CAD apps mentioned lack what's called a physics engine, which is what I heed to do the dynamic modeling. I did find out that Blender, which is really for 3d animation does have a physics engine. So. it looks that that will do. 11. Dec 20, 2011 ### syberraith By the way I did find another app, Working Model 4, made by Autodesk that would do, although being made by that company it as to cost BIG \$. 12. Dec 20, 2011 ### AlephZero The general term is "multi body dynamics" if you want to do a web search. "Physics engines" are a subset of that - often developed for computer games where accuracy may be less important than speed, provided the answers don't "look obviously wrong". 13. Dec 20, 2011 ### syberraith I've been doing searching. There are some possibilities, although I think I'm looking at a substantial learning curve for what ever I end up using. Since I lack the funds to build a real prototype, the next best thing is to show somebody a simulation right? People just look at my math, and then say I'd like to see one work before anything else. So what else have I to do except build a simulation model. * Hey Aleph, I found an app called Step that looks like it will be fairly easy with which to get going. I just hope it has the capability to do what I want. fred Last edited: Dec 20, 2011 14. Dec 20, 2011 ### Simon Bridge What are you trying to simulate? 15. Dec 21, 2011 ### syberraith Basically it's a mass the rotates about a axis with an angular position that is a function of time. Step is too simplistic. Arbitrary rotation is beyond it. I might be able to use it's core library and code something myself, although I wonder how I would visualize it. * Aleph, I'm having much better results searching for 'Mulitbody Dynamics' their are a number of packages that will do. Last edited: Dec 21, 2011 16. Dec 21, 2011 ### Simon Bridge A windmill? I gyroscope? An overbalanced wheel? That could describe anything! You know the result you get from whatever package you pick will only be as good as the numbers you put into it? Oh this is your thruster? pm me. 17. Jan 4, 2012 ### sketch If all else fails use WINE + Windoze software... 18. Jan 4, 2012 ### Simon Bridge Yeah - even linux geeks can enjoy malware-ridden windows freeware xD There are plenty of packages - it's just that there are fewer niche programs - some may start out that way, but if they are any good they get added to until they are general. Proprietary development benefits from restricting use with the side effect that a high demand area will have a program that does almost exactly what you want right away. I've seen people go to Windows for quickie stuff and back to some gnu/linux when they want a workhorse. But mostly you use what you got.	{}
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that can be written as a linear combination of the others. This is the substance of the upcoming . Perhaps this will explain the use of the word dependent. In a linearly dependent set, at least one vector depends on the others (via a linear combination). Indeed, because is an equivalence () some authors use this condition as a definition () of linear dependence. Then linear independence is defined as the logical opposite of linear dependence. Of course, we have chosen to take as our definition, and then follow with as a theorem. Linearly Dependent Sets and Spans If we use a linearly dependent set to construct a span, then we can always create the same infinite set with a starting set that is one vector smaller in size. We will illustrate this behavior in . However, this will not be possible if we build a span from a linearly independent set. So in a certain sense, using a linearly independent set to formulate a span is the best possible way there aren't any extra vectors being used to build up all the necessary linear combinations. OK, here's the theorem, and then the example. Dependency in Linearly Dependent Sets Suppose that $S=\set{\vectorlist{u}{n}}$ is a set of vectors. Then $S$ is a linearly dependent set if and only if there is an index $t$, $1\leq t\leq n$ such that $\vect{u_t}$ is a linear combination of the vectors $\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3,\,\ldots,\,\vect{u}_{t-1},\,\vect{u}_{t+1},\,\ldots,\,\vect{u}_n$. Suppose that $S$ is linearly dependent, so there exists a nontrivial relation of linear dependence by . That is, there are scalars, $\alpha_i$, $1\leq i\leq n$, which are not all zero, such that \lincombo{\alpha}{u}{n}=\zerovector. Since the $\alpha_i$ cannot all be zero, choose one, say $\alpha_t$, that is nonzero. Then, \vect{u}_t \text{}\\ \frac{-1}{\alpha_t}\left( \alpha_1\vect{u}_1+ \cdots+ \alpha_{t-1}\vect{u}_{t-1}+ \alpha_{t+1}\vect{u}_{t+1}+ \cdots+ \alpha_n\vect{u}_n \text{}\\ \frac{-\alpha_1}{\alpha_t}\vect{u}_1+ \cdots+ \frac{-\alpha_{t-1}}{\alpha_t}\vect{u}_{t-1}+ \frac{-\alpha_{t+1}}{\alpha_t}\vect{u}_{t+1}+ \cdots+ \frac{-\alpha_n}{\alpha_t}\vect{u}_n \text{} Since the values of $\frac{\alpha_i}{\alpha_t}$ are again scalars, we have expressed $\vect{u}_t$ as a linear combination of the other elements of $S$. Assume that the vector $\vect{u}_t$ is a linear combination of the other vectors in $S$. Write this linear combination, denoting the relevant scalars as $\beta_1$, $\beta_2$, , $\beta_{t-1}$, $\beta_{t+1}$, , $\beta_n$, as \vect{u_t} \beta_1\vect{u}_1+ \beta_2\vect{u}_2+ \cdots+ \beta_{t-1}\vect{u}_{t-1}+ \beta_{t+1}\vect{u}_{t+1}+ \cdots+ \beta_n\vect{u}_n Then we have \beta_1\vect{u}_1 \beta_{t-1}\vect{u}_{t-1}+ (-1)\vect{u}_t+ \beta_{t+1}\vect{u}_{t+1}+ \cdots+ \beta_n\vect{u}_n\\ \text{}\\ \text{}\\ \text{}\\ \text{} So the scalars $\beta_1,\,\beta_2,\,\beta_3,\,\ldots,\,\beta_{t-1},\,\beta_t=-1,\beta_{t+1},\,\,\ldots,\,\beta_n$ provide a nontrivial linear combination of the vectors in $S$, thus establishing that $S$ is a linearly dependent set (). This theorem can be used, sometimes repeatedly, to whittle down the size of a set of vectors used in a span construction. We have seen some of this already in , but in the next example we will detail some of the subtleties. Reducing a span in $\complex{5}$ Consider the set of $n=4$ vectors from $\complex{5}$, R=\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4} = \set{ \colvector{1\\2\\-1\\3\\2},\, \colvector{2\\1\\3\\1\\2},\, \colvector{0\\-7\\6\\-11\\-2},\, \colvector{4\\1\\2\\1\\6} }\\ and define $V=\spn{R}$. To employ , we form a $5\times 4$ coefficient matrix, $D$, D= \begin{bmatrix} \end{bmatrix} and row-reduce to understand solutions to the homogeneous system $\homosystem{D}$, \begin{bmatrix} \end{bmatrix} We can find infinitely many solutions to this system, most of them nontrivial, and we choose any one we like to build a relation of linear dependence on $R$. Let's begin with $x_4=1$, to find the solution \colvector{-4\\0\\-1\\1} So we can write the relation of linear dependence, (-4)\vect{v}_1+0\vect{v}_2+(-1)\vect{v}_3+1\vect{v}_4=\zerovector guarantees that we can solve this relation of linear dependence for some vector in $R$, but the choice of which one is up to us. Notice however that $\vect{v}_2$ has a zero coefficient. In this case, we cannot choose to solve for $\vect{v}_2$. Maybe some other relation of linear dependence would produce a nonzero coefficient for $\vect{v}_2$ if we just had to solve for this vector. Unfortunately, this example has been engineered to always produce a zero coefficient here, as you can see from solving the homogeneous system. Every solution has $x_2=0$! OK, if we are convinced that we cannot solve for $\vect{v}_2$, let's instead solve for $\vect{v}_3$, \vect{v}_3=(-4)\vect{v}_1+0\vect{v}_2+1\vect{v}_4=(-4)\vect{v}_1+1\vect{v}_4 We now claim that this particular equation will allow us to write V=\spn{R}= \spn{\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4}}= \spn{\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_4}} in essence declaring $\vect{v}_3$ as surplus for the task of building $V$ as a span. This claim is an equality of two sets, so we will use to establish it carefully. Let $R^\prime=\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_4}$ and $V^\prime=\spn{R^\prime}$. We want to show that $V=V^\prime$. First show that $V^\prime\subseteq V$. Since every vector of $R^\prime$ is in $R$, any vector we can construct in $V^\prime$ as a linear combination of vectors from $R^\prime$ can also be constructed as a vector in $V$ by the same linear combination of the same vectors in $R$. That was easy, now turn it around. Next show that $V\subseteq V^\prime$. Choose any $\vect{v}$ from $V$. Then there are scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4$ so that \alpha_1\vect{v}_1+\alpha_2\vect{v}_2+\alpha_3\vect{v}_3+\alpha_4\vect{v}_4\\ \alpha_3\left((-4)\vect{v}_1+1\vect{v}_4\right)+ \alpha_4\vect{v}_4\\ \left((-4\alpha_3)\vect{v}_1+\alpha_3\vect{v}_4\right)+ \alpha_4\vect{v}_4\\ \alpha_2\vect{v}_2+ \left(\alpha_3+\alpha_4\right)\vect{v}_4. This equation says that $\vect{v}$ can then be written as a linear combination of the vectors in $R^\prime$ and hence qualifies for membership in $V^\prime$. So $V\subseteq V^\prime$ and we have established that $V=V^\prime$. If $R^\prime$ was also linearly dependent (it is not), we could reduce the set even further. Notice that we could have chosen to eliminate any one of $\vect{v}_1$, $\vect{v}_3$ or $\vect{v}_4$, but somehow $\vect{v}_2$ is essential to the creation of $V$ since it cannot be replaced by any linear combination of $\vect{v}_1$, $\vect{v}_3$ or $\vect{v}_4$. Relations of Linear Dependence turned on a non-trivial relation of linear dependence () on the set $\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4}$. Besides indicating linear independence, the Sage vector space method .linear_dependence() produces relations of linear dependence for linearly dependent sets. Here is how we would employ this method in . The optional argument zeros='right' will produce results consistent with our work here, you can also experiment with zeros='left' (which is the default). V = QQ^5 v1 = vector(QQ, [1, 2, -1, 3, 2]) v2 = vector(QQ, [2, 1, 3, 1, 2]) v3 = vector(QQ, [0, -7, 6, -11, -2]) v4 = vector(QQ, [4, 1, 2, 1, 6]) R = [v1, v2, v3, v4] L = V.linear_dependence(R, zeros='right') L[0] (-4, 0, -1, 1) -4v1 + 0v2 +(-1)v3 +1v4 (0, 0, 0, 0, 0) V.span(R) == V.span([v1, v2, v4]) True You can check that the list L has just one element (maybe with len(L)), but realize that any multiple of the vector L[0] is also a relation of linear dependence on R, most of which are non-trivial. Notice that we have verified the final conclusion of with a comparison of two spans. We will give the .linear_dependence() method a real workout in the nest Sage subsection () this is just a quick introduction. Casting Out Vectors In we used four vectors to create a span. With a relation of linear dependence in hand, we were able to toss out one of these four vectors and create the same span from a subset of just three vectors from the original set of four. We did have to take some care as to just which vector we tossed out. In the next example, we will be more methodical about just how we choose to eliminate vectors from a linearly dependent set while preserving a span. Casting out vectors We begin with a set $S$ containing seven vectors from $\complex{4}$, S=\set{ \colvector{1\\2\\0\\-1},\, \colvector{4\\8\\0\\-4},\, \colvector{0\\-1\\2\\2},\, \colvector{-1\\3\\-3\\4},\, \colvector{0\\9\\-4\\8},\, \colvector{7\\-13\\12\\-31},\, \colvector{-9\\7\\-8\\37} } and define $W=\spn{S}$. The set $S$ is obviously linearly dependent by , since we have $n=7$ vectors from $\complex{4}$. So we can slim down $S$ some, and still create $W$ as the span of a smaller set of vectors. As a device for identifying relations of linear dependence among the vectors of $S$, we place the seven column vectors of $S$ into a matrix as columns, A=\matrixcolumns{A}{7}= By a nontrivial solution to $\homosystem{A}$ will give us a nontrivial relation of linear dependence () on the columns of $A$ (which are the elements of the set $S$). The row-reduced form for $A$ is the matrix B= so we can easily create solutions to the homogeneous system $\homosystem{A}$ using the free variables $x_2,\,x_5,\,x_6,\,x_7$. Any such solution will correspond to a relation of linear dependence on the columns of $B$. These solutions will allow us to solve for one column vector as a linear combination of some others, in the spirit of , and remove that vector from the set. We'll set about forming these linear combinations methodically. Set the free variable $x_2$ to one, and set the other free variables to zero. Then a solution to $\linearsystem{A}{\zerovector}$ is \vect{x}=\colvector{-4\\1\\0\\0\\0\\0\\0} which can be used to create the linear combination (-4)\vect{A}_1+ 1\vect{A}_2+ 0\vect{A}_3+ 0\vect{A}_4+ 0\vect{A}_5+ 0\vect{A}_6+ 0\vect{A}_7 =\zerovector This can then be arranged and solved for $\vect{A}_2$, resulting in $\vect{A}_2$ expressed as a linear combination of $\set{\vect{A}_1,\,\vect{A}_3,\,\vect{A}_4}$, \vect{A}_2= 4\vect{A}_1+ 0\vect{A}_3+ 0\vect{A}_4 This means that $\vect{A}_2$ is surplus, and we can create $W$ just as well with a smaller set with this vector removed, W=\spn{\set{\vect{A}_1,\,\vect{A}_3,\,\vect{A}_4,\,\vect{A}_5,\,\vect{A}_6,\,\vect{A}_7}} Technically, this set equality for $W$ requires a proof, in the spirit of , but we will bypass this requirement here, and in the next few paragraphs. Now, set the free variable $x_5$ to one, and set the other free variables to zero. Then a solution to $\linearsystem{B}{\zerovector}$ is \vect{x}=\colvector{-2\\0\\-1\\-2\\1\\0\\0} which can be used to create the linear combination (-2)\vect{A}_1+ 0\vect{A}_2+ (-1)\vect{A}_3+ (-2)\vect{A}_4+ 1\vect{A}_5+ 0\vect{A}_6+ 0\vect{A}_7 =\zerovector This can then be arranged and solved for $\vect{A}_5$, resulting in $\vect{A}_5$ expressed as a linear combination of $\set{\vect{A}_1,\,\vect{A}_3,\,\vect{A}_4}$, \vect{A}_5= 2\vect{A}_1+ 1\vect{A}_3+ 2\vect{A}_4 This means that $\vect{A}_5$ is surplus, and we can create $W$ just as well with a smaller set with this vector removed, W=\spn{\left\{\vect{A}_1,\,\vect{A}_3,\,\vect{A}_4,\,\vect{A}_6,\,\vect{A}_7\right\}} Do it again, set the free variable $x_6$ to one, and set the other free variables to zero. Then a solution to $\linearsystem{B}{\zerovector}$ is \vect{x}=\colvector{-1\\0\\3\\6\\0\\1\\0} which can be used to create the linear combination (-1)\vect{A}_1+ 0\vect{A}_2+ 3\vect{A}_3+ 6\vect{A}_4+ 0\vect{A}_5+ 1\vect{A}_6+ 0\vect{A}_7 =\zerovector This can then be arranged and solved for $\vect{A}_6$, resulting in $\vect{A}_6$ expressed as a linear combination of $\set{\vect{A}_1,\,\vect{A}_3,\,\vect{A}_4}$, \vect{A}_6= 1\vect{A}_1+ (-3)\vect{A}_3+ (-6)\vect{A}_4 This means that $\vect{A}_6$ is surplus, and we can create $W$ just as well with a smaller set with this vector removed, W=\spn{\set{\vect{A}_1,\,\vect{A}_3,\,\vect{A}_4,\,\vect{A}_7}} Set the free variable $x_7$ to one, and set the other free variables to zero. Then a solution to $\linearsystem{B}{\zerovector}$ is \vect{x}=\colvector{3\\0\\-5\\-6\\0\\0\\1} which can be used to create the linear combination 3\vect{A}_1+ 0\vect{A}_2+ (-5)\vect{A}_3+ (-6)\vect{A}_4+ 0\vect{A}_5+ 0\vect{A}_6+ 1\vect{A}_7 =\zerovector This can then be arranged and solved for $\vect{A}_7$, resulting in $\vect{A}_7$ expressed as a linear combination of $\set{\vect{A}_1,\,\vect{A}_3,\,\vect{A}_4}$, \vect{A}_7= (-3)\vect{A}_1+ 5\vect{A}_3+ 6\vect{A}_4 This means that $\vect{A}_7$ is surplus, and we can create $W$ just as well with a smaller set with this vector removed, W=\spn{\set{\vect{A}_1,\,\vect{A}_3,\,\vect{A}_4}} You might think we could keep this up, but we have run out of free variables. And not coincidentally, the set $\set{\vect{A}_1,\,\vect{A}_3,\,\vect{A}_4}$ is linearly independent (check this!). It should be clear how each free variable was used to eliminate the corresponding column from the set used to span the column space, as this will be the essence of the proof of the next theorem. The column vectors in $S$ were not chosen entirely at random, they are the columns of . See if you can mimic this example using the columns of . Go ahead, we'll go grab a cup of coffee and be back before you finish up. For extra credit, notice that the vector \vect{b}=\colvector{3\\9\\1\\4} is the vector of constants in the definition of . Since the system $\linearsystem{A}{\vect{b}}$ is consistent, we know by that $\vect{b}$ is a linear combination of the columns of $A$, or stated equivalently, $\vect{b}\in W$. This means that $\vect{b}$ must also be a linear combination of just the three columns $\vect{A}_1,\,\vect{A}_3,\,\vect{A}_4$. Can you find such a linear combination? Did you notice that there is just a single (unique) answer? Hmmmm. Casting Out Vectors We will redo , though somewhat tersely, just producing the justification for each time we toss a vector (a specific relation of linear dependence), and then verifying that the resulting spans, each with one fewer vector, still produce the original span. We also introduce the .remove() method for lists. Ready? Here we go. V = QQ^4 v1 = vector(QQ, [ 1, 2, 0, -1]) v2 = vector(QQ, [ 4, 8, 0, -4]) v3 = vector(QQ, [ 0, -1, 2, 2]) v4 = vector(QQ, [-1, 3, -3, 4]) v5 = vector(QQ, [ 0, 9, -4, 8]) v6 = vector(QQ, [ 7, -13, 12, -31]) v7 = vector(QQ, [-9, 7, -8, 37]) S = [v1, v2, v3, v4, v5, v6, v7] W = V.span(S) D = V.linear_dependence(S, zeros='right') D [ (-4, 1, 0, 0, 0, 0, 0), (-2, 0, -1, -2, 1, 0, 0), (-1, 0, 3, 6, 0, 1, 0), (3, 0, -5, -6, 0, 0, 1) ] D[0] (-4, 1, 0, 0, 0, 0, 0) S.remove(v2) W == V.span(S) True D[1] (-2, 0, -1, -2, 1, 0, 0) S.remove(v5) W == V.span(S) True D[2] (-1, 0, 3, 6, 0, 1, 0) S.remove(v6) W == V.span(S) True D[3] (3, 0, -5, -6, 0, 0, 1) S.remove(v7) W == V.span(S) True S [(1, 2, 0, -1), (0, -1, 2, 2), (-1, 3, -3, 4)] S == [v1, v3, v4] True Notice that S begins with all seven original vectors, and slowly gets whittled down to just the list [v1, v3, v4]. If you experiment with the above commands, be sure to return to the start and work your way through in order, or the results will not be right. As a bonus, notice that the set of relations of linear dependence provided by Sage, D, is itself a linearly independent set (but within a very different vector space). Is that too weird? U = QQ^7 U.linear_dependence(D) == [] True Now, can you answer the extra credit question from using Sage? deserves your careful attention, since this important example motivates the following very fundamental theorem. Basis of a Span Suppose that $S=\set{\vectorlist{v}{n}}$ is a set of column vectors. Define $W=\spn{S}$ and let $A$ be the matrix whose columns are the vectors from $S$. Let $B$ be the reduced row-echelon form of $A$, with $D=\set{\scalarlist{d}{r}}$ the set of column indices corresponding to the pivot columns of $B$. Then 1. $T=\set{\vect{v}_{d_1},\,\vect{v}_{d_2},\,\vect{v}_{d_3},\,\ldots\,\vect{v}_{d_r}}$ is a linearly independent set. 2. $W=\spn{T}$. To prove that $T$ is linearly independent, begin with a relation of linear dependence on $T$, \zerovector= \alpha_1\vect{v}_{d_1}+\alpha_2\vect{v}_{d_2}+\alpha_3\vect{v}_{d_3}+\ldots+\alpha_r\vect{v}_{d_r} and we will try to conclude that the only possibility for the scalars $\alpha_i$ is that they are all zero. Denote the non-pivot columns of $B$ by $F=\set{\scalarlist{f}{n-r}}$. Then we can preserve the equality by adding a big fat zero to the linear combination, \zerovector= \alpha_1\vect{v}_{d_1}+\alpha_2\vect{v}_{d_2}+\alpha_3\vect{v}_{d_3}+\ldots+\alpha_r\vect{v}_{d_r}+ 0\vect{v}_{f_1}+0\vect{v}_{f_2}+0\vect{v}_{f_3}+\ldots+0\vect{v}_{f_{n-r}} By , the scalars in this linear combination (suitably reordered) are a solution to the homogeneous system $\homosystem{A}$. But notice that this is the solution obtained by setting each free variable to zero. If we consider the description of a solution vector in the conclusion of , in the case of a homogeneous system, then we see that if all the free variables are set to zero the resulting solution vector is trivial (all zeros). So it must be that $\alpha_i=0$, $1\leq i\leq r$. This implies by that $T$ is a linearly independent set. The second conclusion of this theorem is an equality of sets (). Since $T$ is a subset of $S$, any linear combination of elements of the set $T$ can also be viewed as a linear combination of elements of the set $S$. So $\spn{T}\subseteq\spn{S}=W$. It remains to prove that $W=\spn{S}\subseteq\spn{T}$. For each $k$, $1\leq k\leq n-r$, form a solution $\vect{x}$ to $\homosystem{A}$ by setting the free variables as follows: By , the remainder of this solution vector is given by, From this solution, we obtain a relation of linear dependence on the columns of $A$, -\matrixentry{B}{1,f_k}\vect{v}_{d_1} -\matrixentry{B}{2,f_k}\vect{v}_{d_2} -\matrixentry{B}{3,f_k}\vect{v}_{d_3} -\ldots -\matrixentry{B}{r,f_k}\vect{v}_{d_r} +1\vect{v}_{f_k} =\zerovector which can be arranged as the equality \vect{v}_{f_k}= \matrixentry{B}{1,f_k}\vect{v}_{d_1}+ \matrixentry{B}{2,f_k}\vect{v}_{d_2}+ \matrixentry{B}{3,f_k}\vect{v}_{d_3}+ \ldots+ \matrixentry{B}{r,f_k}\vect{v}_{d_r} Now, suppose we take an arbitrary element, $\vect{w}$, of $W=\spn{S}$ and write it as a linear combination of the elements of $S$, but with the terms organized according to the indices in $D$ and $F$, \alpha_1\vect{v}_{d_1}+ \alpha_2\vect{v}_{d_2}+ \ldots+ \alpha_r\vect{v}_{d_r}+ \beta_1\vect{v}_{f_1}+ \beta_2\vect{v}_{f_2}+ \ldots+ \beta_{n-r}\vect{v}_{f_{n-r}} From the above, we can replace each $\vect{v}_{f_j}$ by a linear combination of the $\vect{v}_{d_i}$, \alpha_1\vect{v}_{d_1}+ \alpha_2\vect{v}_{d_2}+ \ldots+ \alpha_r\vect{v}_{d_r}+\\ \matrixentry{B}{1,f_1}\vect{v}_{d_1}+ \matrixentry{B}{2,f_1}\vect{v}_{d_2}+ \matrixentry{B}{3,f_1}\vect{v}_{d_3}+ \ldots+ \matrixentry{B}{r,f_1}\vect{v}_{d_r} \right)+\\ \matrixentry{B}{1,f_2}\vect{v}_{d_1}+ \matrixentry{B}{2,f_2}\vect{v}_{d_2}+ \matrixentry{B}{3,f_2}\vect{v}_{d_3}+ \ldots+ \matrixentry{B}{r,f_2}\vect{v}_{d_r} \right)+\\ \matrixentry{B}{1,f_{n-r}}\vect{v}_{d_1}+ \matrixentry{B}{2,f_{n-r}}\vect{v}_{d_2}+ \matrixentry{B}{3,f_{n-r}}\vect{v}_{d_3}+ \ldots+ \matrixentry{B}{r,f_{n-r}}\vect{v}_{d_r} \right)\\ With repeated applications of several of the properties of we can rearrange this expression as, \alpha_1+ \beta_1\matrixentry{B}{1,f_1}+ \beta_2\matrixentry{B}{1,f_2}+ \beta_3\matrixentry{B}{1,f_3}+ \ldots+ \beta_{n-r}\matrixentry{B}{1,f_{n-r}} \right)\vect{v}_{d_1}+\\ \beta_1\matrixentry{B}{2,f_1}+ \beta_2\matrixentry{B}{2,f_2}+ \beta_3\matrixentry{B}{2,f_3}+ \ldots+ \beta_{n-r}\matrixentry{B}{2,f_{n-r}} \right)\vect{v}_{d_2}+\\ \beta_1\matrixentry{B}{r,f_1}+ \beta_2\matrixentry{B}{r,f_2}+ \beta_3\matrixentry{B}{r,f_3}+ \ldots+\beta_{n-r}\matrixentry{B}{r,f_{n-r}} \right)\vect{v}_{d_r} This mess expresses the vector $\vect{w}$ as a linear combination of the vectors in T=\set{\vect{v}_{d_1},\,\vect{v}_{d_2},\,\vect{v}_{d_3},\,\ldots\,\vect{v}_{d_r}} thus saying that $\vect{w}\in\spn{T}$. Therefore, $W=\spn{S}\subseteq\spn{T}$. In , we tossed-out vectors one at a time. But in each instance, we rewrote the offending vector as a linear combination of those vectors that corresponded to the pivot columns of the reduced row-echelon form of the matrix of columns. In the proof of , we accomplish this reduction in one big step. In we arrived at a linearly independent set at exactly the same moment that we ran out of free variables to exploit. This was not a coincidence, it is the substance of our conclusion of linear independence in . Here's a straightforward application of . Reducing a span in $\complex{4}$ Begin with a set of five vectors from $\complex{4}$, S=\set{ \colvector{ 1 \\ 1 \\ 2 \\ 1},\, \colvector{ 2 \\ 2 \\ 4 \\ 2},\, \colvector{ 2 \\ 0 \\ -1 \\ 1},\, \colvector{ 7 \\ 1 \\ -1 \\ 4},\, \colvector{ 0 \\ 2 \\ 5 \\ 1} } and let $W=\spn{S}$. To arrive at a (smaller) linearly independent set, follow the procedure described in . Place the vectors from $S$ into a matrix as columns, and row-reduce, \begin{bmatrix} \end{bmatrix} \rref \begin{bmatrix} \end{bmatrix} Columns 1 and 3 are the pivot columns ($D=\set{1,\,3}$) so the set T=\set{ \colvector{ 1 \\ 1 \\ 2 \\ 1},\, \colvector{ 2 \\ 0 \\ -1 \\ 1} } is linearly independent and $\spn{T}=\spn{S}=W$. Boom! Since the reduced row-echelon form of a matrix is unique (), the procedure of leads us to a unique set $T$. However, there is a wide variety of possibilities for sets $T$ that are linearly independent and which can be employed in a span to create $W$. Without proof, we list two other possibilities: \colvector{ 2 \\ 2 \\ 4 \\ 2},\, \colvector{ 2 \\ 0 \\ -1 \\ 1} }\\ \colvector{3 \\ 1 \\ 1 \\ 2},\, \colvector{-1 \\ 1 \\ 3 \\ 0} } Can you prove that $T^{\prime}$ and $T^{}$ are linearly independent sets and $W=\spn{S}=\spn{T^{\prime}}=\spn{T^{}}$? Reducing a Span allows us to construct a reduced spanning set for a span. As with the theorem, employing Sage we begin by constructing a matrix with the vectors of the spanning set as columns. Here is a do-over of , illustrating the use of in Sage. V = QQ^4 v1 = vector(QQ, [1,1,2,1]) v2 = vector(QQ, [2,2,4,2]) v3 = vector(QQ, [2,0,-1,1]) v4 = vector(QQ, [7,1,-1,4]) v5 = vector(QQ, [0,2,5,1]) S = [v1, v2, v3, v4, v5] A = column_matrix(S) T = [A.column(p) for p in A.pivots()] T [(1, 1, 2, 1), (2, 0, -1, 1)] V.linear_dependence(T) == [] True V.span(S) == V.span(T) True Notice how we compute T with the single line that mirrors the construction of the set $T=\set{\vect{v}_{d_1},\,\vect{v}_{d_2},\,\vect{v}_{d_3},\,\ldots\,\vect{v}_{d_r}}$ in the statement of . Again, the row-reducing is hidden in the use of the .pivot() matrix method, which necessarily must compute the reduced row-echelon form. The final two compute cells verify both conclusions of the theorem. Reworking elements of a span Begin with a set of five vectors from $\complex{4}$, R=\set{ \colvector{ 2 \\ 1 \\ 3 \\ 2 },\, \colvector{ -1 \\ 1 \\ 0 \\ 1 },\, \colvector{ -8 \\ -1 \\ -9 \\ -4 },\, \colvector{ 3 \\ 1 \\ -1 \\ -2 },\, \colvector{ -10 \\ -1 \\ -1 \\ 4} } It is easy to create elements of $X=\spn{R}$ we will create one at random, \vect{y}= 6\colvector{ 2 \\ 1 \\ 3 \\ 2 }+ (-7)\colvector{ -1 \\ 1 \\ 0 \\ 1 }+ 1\colvector{ -8 \\ -1 \\ -9 \\ -4 }+ 6\colvector{ 3 \\ 1 \\ -1 \\ -2 }+ 2\colvector{ -10 \\ -1 \\ -1 \\ 4} = \colvector{9\\2\\1\\-3} We know we can replace $R$ by a smaller set (since it is obviously linearly dependent by ) that will create the same span. Here goes, \begin{bmatrix} \end{bmatrix} \rref \begin{bmatrix} \end{bmatrix} So, if we collect the first, second and fourth vectors from $R$, P=\set{ \colvector{ 2 \\ 1 \\ 3 \\ 2 },\, \colvector{ -1 \\ 1 \\ 0 \\ 1 },\, \colvector{ 3 \\ 1 \\ -1 \\ -2 } } then $P$ is linearly independent and $\spn{P}=\spn{R}=X$ by . Since we built $\vect{y}$ as an element of $\spn{R}$ it must also be an element of $\spn{P}$. Can we write $\vect{y}$ as a linear combination of just the three vectors in $P$? The answer is, of course, yes. But let's compute an explicit linear combination just for fun. By we can get such a linear combination by solving a system of equations with the column vectors of $R$ as the columns of a coefficient matrix, and $\vect{y}$ as the vector of constants. Employing an augmented matrix to solve this system, \begin{bmatrix} \end{bmatrix} \rref \begin{bmatrix} \end{bmatrix} So we see, as expected, that 1\colvector{ 2 \\ 1 \\ 3 \\ 2 }+ (-1)\colvector{ -1 \\ 1 \\ 0 \\ 1 }+ 2\colvector{ 3 \\ 1 \\ -1 \\ -2 } =\colvector{9 \\ 2 \\ 1 \\ -3} =\vect{y} A key feature of this example is that the linear combination that expresses $\vect{y}$ as a linear combination of the vectors in $P$ is unique. This is a consequence of the linear independence of $P$. The linearly independent set $P$ is smaller than $R$, but still just (barely) big enough to create elements of the set $X=\spn{R}$. There are many, many ways to write $\vect{y}$ as a linear combination of the five vectors in $R$ (the appropriate system of equations to verify this claim has two free variables in the description of the solution set), yet there is precisely one way to write $\vect{y}$ as a linear combination of the three vectors in $P$. Reworking a Span As another demonstration of using Sage to help us understand spans, linear combinations, linear independence and reduced row-echelon form, we will recreate parts of . Most of this should be familiar, but see the commments following. V = QQ^4 v1 = vector(QQ, [2,1,3,2]) v2 = vector(QQ, [-1,1,0,1]) v3 = vector(QQ, [-8,-1,-9,-4]) v4 = vector(QQ, [3,1,-1,-2]) v5 = vector(QQ, [-10,-1,-1,4]) y = 6v1 - 7v2 + v3 +6v4 + 2v5 y (9, 2, 1, -3) R = [v1, v2, v3, v4, v5] X = V.span(R) y in X True A = column_matrix(R) P = [A.column(p) for p in A.pivots()] W = V.span(P) W == X True y in W True coeff = column_matrix(P) coeff.solve_right(y) (1, -1, 2) coeff.right_kernel() Vector space of degree 3 and dimension 0 over Rational Field Basis matrix: [] V.linear_dependence(P) == [] True The final two results a trivial null space for coeff and the linear independence of P both individually imply that the solution to the system of equations (just prior) is unique. Sage produces its own linearly independent spanning set for each span, as we see whenever we inquire about a span. X Vector space of degree 4 and dimension 3 over Rational Field Basis matrix: [ 1 0 0 -8/15] [ 0 1 0 7/15] [ 0 0 1 13/15] Can you extract the three vectors that Sage uses to span X and solve the appropriate system of equations to see how to write y as a linear combination of these three vectors? Once you have done that, check your answer by hand and think about how using Sage could have been overkill for this question. 1. Let $S$ be the linearly dependent set of three vectors below. S=\set{\colvector{1\\10\\100\\1000},\,\colvector{1\\1\\1\\1},\,\colvector{5\\23\\203\\2003}} Write one vector from $S$ as a linear combination of the other two and include this vector equality in your response. (You should be able to do this on sight, rather than doing some computations.) Convert this expression into a nontrivial relation of linear dependence on $S$. 2. Explain why the word dependent is used in the definition of linear dependence. 3. Suppose that $Y=\spn{P}=\spn{Q}$, where $P$ is a linearly dependent set and $Q$ is linearly independent. Would you rather use $P$ or $Q$ to describe $Y$? Why? Let $T$ be the set of columns of the matrix $B$ below. Define $W=\spn{T}$. Find a set $R$ so that (1) $R$ has 3 vectors, (2) $R$ is a subset of $T$, and (3) $W=\spn{R}$. B= \begin{bmatrix} \end{bmatrix} Let $T=\set{\vect{w}_1,\,\vect{w}_2,\,\vect{w}_3,\,\vect{w}_4}$. The vector $\colvector{2\\-1\\0\\1}$ is a solution to the homogeneous system with the matrix $B$ as the coefficient matrix (check this!). By it provides the scalars for a linear combination of the columns of $B$ (the vectors in $T$) that equals the zero vector, a relation of linear dependence on $T$, 2\vect{w}_1+(-1)\vect{w}_2+(1)\vect{w}_4=\zerovector We can rearrange this equation by solving for $\vect{w}_4$, \vect{w}_4=(-2)\vect{w}_1+\vect{w}_2 This equation tells us that the vector $\vect{w}_4$ is superfluous in the span construction that creates $W$. So $W=\spn{\set{\vect{w}_1,\,\vect{w}_2,\,\vect{w}_3}}$. The requested set is $R=\set{\vect{w}_1,\,\vect{w}_2,\,\vect{w}_3}$. Verify that the set $R^\prime=\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_4}$ at the end of is linearly independent. Consider the set of vectors from $\complex{3}$, $W$, given below. Find a linearly independent set $T$ that contains three vectors from $W$ and such that $\spn{W}=\spn{T}$. W= \set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4,\,\vect{v}_5} =\set{ \colvector{2\\1\\1},\, \colvector{-1\\-1\\1},\, \colvector{1\\2\\3},\, \colvector{3\\1\\3},\, \colvector{0\\1\\-3} } To apply , we formulate a matrix $A$ whose columns are $\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4,\,\vect{v}_5$. Then we row-reduce $A$. After row-reducing, we obtain \begin{bmatrix} \end{bmatrix} From this we see that the pivot columns are $D=\set{1,\,2,\,3}$. Thus T=\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3}=\set{\colvector{2\\1\\1},\,\colvector{-1\\-1\\1},\,\colvector{1\\2\\3}} is a linearly independent set and $\spn{T}=W$. Compare this problem with . Given the set $S$ below, find a linearly independent set $T$ so that $\spn{T}=\spn{S}$. S=\set{ \colvector{2\\-1\\2},\, \colvector{3\\0\\1},\, \colvector{1\\1\\-1},\, \colvector{5\\-1\\3} } says we can make a matrix with these four vectors as columns, row-reduce, and just keep the columns with indices in the set $D$. Here we go, forming the relevant matrix and row-reducing, \begin{bmatrix} \end{bmatrix} \rref \begin{bmatrix} \end{bmatrix} Analyzing the row-reduced version of this matrix, we see that the first two columns are pivot columns, so $D=\set{1,2}$. says we need only keep the first two columns to create a set with the requisite properties, T=\set{ \colvector{2\\-1\\2},\, \colvector{3\\0\\1} } Let $W$ be the span of the set of vectors $S$ below, $W=\spn{S}$. Find a set $T$ so that 1) the span of $T$ is $W$, $\spn{T}=W$, (2) $T$ is a linearly independent set, and (3) $T$ is a subset of $S$. \set{ \colvector{1 \\ 2 \\ -1},\, \colvector{2 \\ -3 \\ 1},\, \colvector{4 \\ 1 \\ -1},\, \colvector{3 \\ 1 \\ 1},\, \colvector{3 \\ -1 \\ 0} } This is a straight setup for the conclusion of . The hypotheses of this theorem tell us to pack the vectors of $W$ into the columns of a matrix and row-reduce, \begin{bmatrix} \end{bmatrix} \begin{bmatrix} \end{bmatrix} Pivot columns have indices $D=\set{1,\,2,\,4}$. tells us to form $T$ with columns $1,\,2$ and $4$ of $S$, \set{ \colvector{1 \\ 2 \\ -1},\, \colvector{2 \\ -3 \\ 1},\, \colvector{3 \\ 1 \\ 1} } Let $T$ be the set of vectors $T=\set{ \colvector{1 \\ -1 \\ 2},\, \colvector{3 \\ 0 \\ 1},\, \colvector{4 \\ 2 \\ 3},\, \colvector{3 \\ 0 \\ 6} }$. Find two different subsets of $T$, named $R$ and $S$, so that $R$ and $S$ each contain three vectors, and so that $\spn{R}=\spn{T}$ and $\spn{S}=\spn{T}$. Prove that both $R$ and $S$ are linearly independent. Let $A$ be the matrix whose columns are the vectors in $T$. Then row-reduce $A$, A\rref B= \begin{bmatrix} \end{bmatrix} From we can form $R$ by choosing the columns of $A$ that correspond to the pivot columns of $B$. also guarantees that $R$ will be linearly independent. R=\set{ \colvector{1 \\ -1 \\ 2},\, \colvector{3 \\ 0 \\ 1},\, \colvector{4 \\ 2 \\ 3} } That was easy. To find $S$ will require a bit more work. From $B$ we can obtain a solution to $\homosystem{A}$, which by will provide a nontrivial relation of linear dependence on the columns of $A$, which are the vectors in $T$. To wit, choose the free variable $x_4$ to be 1, then $x_1=-2$, $x_2=1$, $x_3=-1$, and so (-2)\colvector{1 \\ -1 \\ 2}+ (1)\colvector{3 \\ 0 \\ 1}+ (-1)\colvector{4 \\ 2 \\ 3}+ (1)\colvector{3 \\ 0 \\ 6} = \colvector{0\\0\\0} this equation can be rewritten with the second vector staying put, and the other three moving to the other side of the equality, \colvector{3 \\ 0 \\ 1} = (2)\colvector{1 \\ -1 \\ 2}+ (1)\colvector{4 \\ 2 \\ 3}+ (-1)\colvector{3 \\ 0 \\ 6} We could have chosen other vectors to stay put, but may have then needed to divide by a nonzero scalar. This equation is enough to conclude that the second vector in $T$ is surplus and can be replaced (see the careful argument in ). So set S=\set{ \colvector{1 \\ -1 \\ 2},\, \colvector{4 \\ 2 \\ 3},\, \colvector{3 \\ 0 \\ 6} } and then $\spn{S}=\spn{T}$. $T$ is also a linearly independent set, which we can show directly. Make a matrix $C$ whose columns are the vectors in $S$. Row-reduce $B$ and you will obtain the identity matrix $I_3$. By , the set $S$ is linearly independent. Reprise by creating a new version of the vector $\vect{y}$. In other words, form a new, different linear combination of the vectors in $R$ to create a new vector $\vect{y}$ (but do not simplify the problem too much by choosing any of the five new scalars to be zero). Then express this new $\vect{y}$ as a combination of the vectors in $P$. At the conclusion of two alternative solutions, sets $T^{\prime}$ and $T^{}$, are proposed. Verify these claims by proving that $\spn{T}=\spn{T^{\prime}}$ and $\spn{T}=\spn{T^{}}$. Suppose that $\vect{v}_1$ and $\vect{v}_2$ are any two vectors from $\complex{m}$. Prove the following set equality. \spn{\set{\vect{v}_1,\,\vect{v}_2}} = \spn{\set{\vect{v}_1+\vect{v}_2,\,\vect{v}_1-\vect{v}_2}} This is an equality of sets, so applies. The easy half first. Show that $X=\spn{\set{\vect{v}_1+\vect{v}_2,\,\vect{v}_1-\vect{v}_2}}\subseteq \spn{\set{\vect{v}_1,\,\vect{v}_2}}=Y$.\\ Choose $\vect{x}\in X$. Then $\vect{x}=a_1(\vect{v}_1+\vect{v}_2)+a_2(\vect{v}_1-\vect{v}_2)$ for some scalars $a_1$ and $a_2$. Then, \vect{x} which qualifies $\vect{x}$ for membership in $Y$, as it is a linear combination of $\vect{v}_1,\,\vect{v}_2$. Now show the opposite inclusion, $Y=\spn{\set{\vect{v}_1,\,\vect{v}_2}}\subseteq\spn{\set{\vect{v}_1+\vect{v}_2,\,\vect{v}_1-\vect{v}_2}}=X$.\\ Choose $\vect{y}\in Y$. Then there are scalars $b_1,\,b_2$ such that $\vect{y}=b_1\vect{v}_1+b_2\vect{v}_2$. Rearranging, we obtain, \vect{y} + \frac{b_2}{2}\left[\left(\vect{v}_1+\vect{v}_2\right)-\left(\vect{v}_1-\vect{v}_2\right)\right]\\ This is an expression for $\vect{y}$ as a linear combination of $\vect{v}_1+\vect{v}_2$ and $\vect{v}_1-\vect{v}_2$, earning $\vect{y}$ membership in $X$. Since $X$ is a subset of $Y$, and vice versa, we see that $X=Y$, as desired.	{}
# polygon 0th Percentile ##### Polygon Drawing polygon draws the polygons whose vertices are given in x and y. Keywords aplot ##### Usage polygon(x, y = NULL, density = NULL, angle = 45, border = NULL, col = NA, lty = par("lty"), …, fillOddEven = FALSE) ##### Arguments x, y vectors containing the coordinates of the vertices of the polygon. density the density of shading lines, in lines per inch. The default value of NULL means that no shading lines are drawn. A zero value of density means no shading nor filling whereas negative values and NA suppress shading (and so allow color filling). angle the slope of shading lines, given as an angle in degrees (counter-clockwise). col the color for filling the polygon. The default, NA, is to leave polygons unfilled, unless density is specified. (For back-compatibility, NULL is equivalent to NA.) If density is specified with a positive value this gives the color of the shading lines. border the color to draw the border. The default, NULL, means to use par("fg"). Use border = NA to omit borders. For compatibility with S, border can also be logical, in which case FALSE is equivalent to NA (borders omitted) and TRUE is equivalent to NULL (use the foreground colour), lty the line type to be used, as in par. graphical parameters such as xpd, lend, ljoin and lmitre can be given as arguments. fillOddEven logical controlling the polygon shading mode: see below for details. Default FALSE. ##### Details The coordinates can be passed in a plotting structure (a list with x and y components), a two-column matrix, …. See xy.coords. It is assumed that the polygon is to be closed by joining the last point to the first point. The coordinates can contain missing values. The behaviour is similar to that of lines, except that instead of breaking a line into several lines, NA values break the polygon into several complete polygons (including closing the last point to the first point). See the examples below. When multiple polygons are produced, the values of density, angle, col, border, and lty are recycled in the usual manner. Shading of polygons is only implemented for linear plots: if either axis is on log scale then shading is omitted, with a warning. ##### Bugs Self-intersecting polygons may be filled using either the “odd-even” or “non-zero” rule. These fill a region if the polygon border encircles it an odd or non-zero number of times, respectively. Shading lines are handled internally by R according to the fillOddEven argument, but device-based solid fills depend on the graphics device. The windows, pdf and postscript devices have their own fillOddEven argument to control this. ##### References Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988) The New S Language. Wadsworth & Brooks/Cole. Murrell, P. (2005) R Graphics. Chapman & Hall/CRC Press. segments for even more flexibility, lines, rect, box, abline. par for how to specify colors. • polygon ##### Examples library(graphics) # NOT RUN { x <- c(1:9, 8:1) y <- c(1, 2*(5:3), 2, -1, 17, 9, 8, 2:9) op <- par(mfcol = c(3, 1)) for(xpd in c(FALSE, TRUE, NA)) { plot(1:10, main = paste("xpd =", xpd)) box("figure", col = "pink", lwd = 3) polygon(x, y, xpd = xpd, col = "orange", lty = 2, lwd = 2, border = "red") } par(op) n <- 100 xx <- c(0:n, n:0) yy <- c(c(0, cumsum(stats::rnorm(n))), rev(c(0, cumsum(stats::rnorm(n))))) plot (xx, yy, type = "n", xlab = "Time", ylab = "Distance") polygon(xx, yy, col = "gray", border = "red") title("Distance Between Brownian Motions") # Multiple polygons from NA values # and recycling of col, border, and lty op <- par(mfrow = c(2, 1)) plot(c(1, 9), 1:2, type = "n") polygon(1:9, c(2,1,2,1,1,2,1,2,1), col = c("red", "blue"), border = c("green", "yellow"), lwd = 3, lty = c("dashed", "solid")) plot(c(1, 9), 1:2, type = "n") polygon(1:9, c(2,1,2,1,NA,2,1,2,1), col = c("red", "blue"), border = c("green", "yellow"), lwd = 3, lty = c("dashed", "solid")) par(op) # Line-shaded polygons plot(c(1, 9), 1:2, type = "n") polygon(1:9, c(2,1,2,1,NA,2,1,2,1), density = c(10, 20), angle = c(-45, 45)) # } Documentation reproduced from package graphics, version 3.6.2, License: Part of R 3.6.2 ### Community examples Looks like there are no examples yet.	{}
# Deep learning----------Deep Belief Networks DBN是2006年提出的一种概率生成模型, 由多个限制玻尔兹曼机(RBM)[3]堆栈而成: 在训练时, Hinton采用了逐层无监督的方法来学习参数。首先把数据向量x和第一层隐藏层作为一个RBM, 训练出这个RBM的参数(连接x和h1的权重, x和h1各个节点的偏置等等), 然后固定这个RBM的参数, 把h1视作可见向量, 把h2视作隐藏向量, 训练第二个RBM, 得到其参数, 然后固定这些参数, 训练h2和h3构成的RBM, 具体的训练算法如下: 上图最右边就是最终训练得到的生成模型: 用公式表示为: 3. 利用DBN进行有监督学习在使用上述的逐层无监督方法学得节点之间的权重以及节点的偏置之后(亦即初始化), 可以在DBN的最顶层再加一层, 来表示我们希望得到的输出, 然后计算模型得到的输出和希望得到的输出之间的误差, 利用后向反馈的方法来进一步优化之前设置的初始权重。因为我们已经使用逐层无监督方法来初始化了权重值, 使其比较接近最优值, 解决了之前多层神经网络训练时存在的问题, 能够得到很好的效果。 ## Deep Belief Networks [Hinton06] showed that RBMs can be stacked and trained in a greedy manner to form so-called Deep Belief Networks (DBN). DBNs are graphical models which learn to extract a deep hierarchical representation of the training data. They model the joint distribution between observed vector and the hidden layers as follows: (1) where is a conditional distribution for the visible units conditioned on the hidden units of the RBM at level , and is the visible-hidden joint distribution in the top-level RBM. This is illustrated in the figure below. The principle of greedy layer-wise unsupervised training can be applied to DBNs with RBMs as the building blocks for each layer[Hinton06][Bengio07]. The process is as follows: 1. Train the first layer as an RBM that models the raw input as its visible layer. 2. Use that first layer to obtain a representation of the input that will be used as data for the second layer. Two common solutions exist. This representation can be chosen as being the mean activations or samples of . 3. Train the second layer as an RBM, taking the transformed data (samples or mean activations) as training examples (for the visible layer of that RBM). 4. Iterate (2 and 3) for the desired number of layers, each time propagating upward either samples or mean values. 5. Fine-tune all the parameters of this deep architecture with respect to a proxy for the DBN log- likelihood, or with respect to a supervised training criterion (after adding extra learning machinery to convert the learned representation into supervised predictions, e.g. a linear classifier). In this tutorial, we focus on fine-tuning via supervised gradient descent. Specifically, we use a logistic regression classifier to classify the input based on the output of the last hidden layer of the DBN. Fine-tuning is then performed via supervised gradient descent of the negative log-likelihood cost function. Since the supervised gradient is only non-null for the weights and hidden layer biases of each layer (i.e. null for the visible biases of each RBM), this procedure is equivalent to initializing the parameters of a deep MLP with the weights and hidden layer biases obtained with the unsupervised training strategy. ## Justifying Greedy-Layer Wise Pre-Training Why does such an algorithm work ? Taking as example a 2-layer DBN with hidden layers and (with respective weight parameters and ), [Hinton06] established (see also Bengio09]_ for a detailed derivation) that can be rewritten as, (2) represents the KL divergence between the posterior of the first RBM if it were standalone, and the probability for the same layer but defined by the entire DBN (i.e. taking into account the prior defined by the top-level RBM). is the entropy of the distribution . It can be shown that if we initialize both hidden layers such that and the KL divergence term is null. If we learn the first level RBM and then keep its parameters fixed, optimizing Eq. (2) with respect to can thus only increase the likelihood . Also, notice that if we isolate the terms which depend only on , we get: Optimizing this with respect to amounts to training a second-stage RBM, using the output of as the training distribution, when is sampled from the training distribution for the first RBM. ## Implementation To implement DBNs in Theano, we will use the class defined in the Restricted Boltzmann Machines (RBM) tutorial. One can also observe that the code for the DBN is very similar with the one for SdA, because both involve the principle of unsupervised layer-wise pre-training followed by supervised fine-tuning as a deep MLP. The main difference is that we use the RBM class instead of the dA class. We start off by defining the DBN class which will store the layers of the MLP, along with their associated RBMs. Since we take the viewpoint of using the RBMs to initialize an MLP, the code will reflect this by seperating as much as possible the RBMs used to initialize the network and the MLP used for classification. class DBN(object): def __init__(self, numpy_rng, theano_rng=None, n_ins=784, hidden_layers_sizes=[500, 500], n_outs=10): """This class is made to support a variable number of layers. :type numpy_rng: numpy.random.RandomState :param numpy_rng: numpy random number generator used to draw initial weights :type theano_rng: theano.tensor.shared_randomstreams.RandomStreams :param theano_rng: Theano random generator; if None is given one is generated based on a seed drawn from rng :type n_ins: int :param n_ins: dimension of the input to the DBN :type n_layers_sizes: list of ints :param n_layers_sizes: intermediate layers size, must contain at least one value :type n_outs: int :param n_outs: dimension of the output of the network """ self.sigmoid_layers = [] self.rbm_layers = [] self.params = [] self.n_layers = len(hidden_layers_sizes) assert self.n_layers > 0 if not theano_rng: theano_rng = RandomStreams(numpy_rng.randint(2 ** 30)) # allocate symbolic variables for the data self.x = T.matrix('x') # the data is presented as rasterized images self.y = T.ivector('y') # the labels are presented as 1D vector of # [int] labels self.sigmoid_layers will store the feed-forward graphs which together form the MLP, while self.rbm_layers will store the RBMs used to pretrain each layer of the MLP. Next step, we construct n_layers sigmoid layers (we use the SigmoidalLayer class introduced in Multilayer Perceptron, with the only modification that we replaced the non-linearity from tanh to the logistic function ) and n_layers RBMs, where n_layersis the depth of our model. We link the sigmoid layers such that they form an MLP, and construct each RBM such that they share the weight matrix and the hidden bias with its corresponding sigmoid layer. for i in xrange(self.n_layers): # construct the sigmoidal layer # the size of the input is either the number of hidden units of the # layer below or the input size if we are on the first layer if i == 0: input_size = n_ins else: input_size = hidden_layers_sizes[i - 1] # the input to this layer is either the activation of the hidden # layer below or the input of the DBN if you are on the first layer if i == 0: layer_input = self.x else: layer_input = self.sigmoid_layers[-1].output sigmoid_layer = HiddenLayer(rng=numpy_rng, input=layer_input, n_in=input_size, n_out=hidden_layers_sizes[i], activation=T.nnet.sigmoid) # add the layer to our list of layers self.sigmoid_layers.append(sigmoid_layer) # its arguably a philosophical question... but we are going to only declare that # the parameters of the sigmoid_layers are parameters of the DBN. The visible # biases in the RBM are parameters of those RBMs, but not of the DBN. self.params.extend(sigmoid_layer.params) # Construct an RBM that shared weights with this layer rbm_layer = RBM(numpy_rng=numpy_rng, theano_rng=theano_rng, input=layer_input, n_visible=input_size, n_hidden=hidden_layers_sizes[i], W=sigmoid_layer.W, hbias=sigmoid_layer.b) self.rbm_layers.append(rbm_layer) All that is left is to stack one last logistic regression layer in order to form an MLP. We will use the LogisticRegression class introduced in Classifying MNIST digits using Logistic Regression. # We now need to add a logistic layer on top of the MLP self.logLayer = LogisticRegression( input=self.sigmoid_layers[-1].output, n_in=hidden_layers_sizes[-1], n_out=n_outs) self.params.extend(self.logLayer.params) # construct a function that implements one step of fine-tuning compute # the cost for second phase of training, defined as the negative log # likelihood of the logistic regression (output) layer self.finetune_cost = self.logLayer.negative_log_likelihood(self.y) # compute the gradients with respect to the model parameters # symbolic variable that points to the number of errors made on the # minibatch given by self.x and self.y self.errors = self.logLayer.errors(self.y) The class also provides a method which generates training functions for each of the RBMs. They are returned as a list, where element is a function which implements one step of training for the RBM at layer . def pretraining_functions(self, train_set_x, batch_size, k): ''' Generates a list of functions, for performing one step of gradient descent at a given layer. The function will require as input the minibatch index, and to train an RBM you just need to iterate, calling the corresponding function on all minibatch indexes. :type train_set_x: theano.tensor.TensorType :param train_set_x: Shared var. that contains all datapoints used for training the RBM :type batch_size: int :param batch_size: size of a [mini]batch :param k: number of Gibbs steps to do in CD-k / PCD-k ''' # index to a [mini]batch index = T.lscalar('index') # index to a minibatch In order to be able to change the learning rate during training, we associate a Theano variable to it that has a default value. learning_rate = T.scalar('lr') # learning rate to use # number of batches n_batches = train_set_x.get_value(borrow=True).shape[0] / batch_size # begining of a batch, given index batch_begin = index * batch_size # ending of a batch given index batch_end = batch_begin + batch_size pretrain_fns = [] for rbm in self.rbm_layers: # get the cost and the updates list # using CD-k here (persisent=None) for training each RBM. # TODO: change cost function to reconstruction error cost, updates = rbm.cd(learning_rate, persistent=None, k) # compile the Theano function; check if k is also a Theano # variable, if so added to the inputs of the function if isinstance(k, theano.Variable): inputs = [index, theano.Param(learning_rate, default=0.1), k] else: inputs = index, theano.Param(learning_rate, default=0.1)] fn = theano.function(inputs=inputs, outputs=cost, givens={self.x: train_set_x[batch_begin: batch_end]}) # append fn to the list of functions pretrain_fns.append(fn) return pretrain_fns Now any function pretrain_fns[i] takes as arguments index and optionally lr – the learning rate. Note that the names of the parameters are the names given to the Theano variables (e.g. lr) when they are constructed and not the name of the python variables (e.g. learning_rate). Keep this in mind when working with Theano. Optionally, if you provide k (the number of Gibbs steps to perform in CD or PCD) this will also become an argument of your function. In the same fashion, the DBN class includes a method for building the functions required for finetuning ( a train_model, avalidate_model and a test_model function). def build_finetune_functions(self, datasets, batch_size, learning_rate): '''Generates a function train that implements one step of finetuning, a function validate that computes the error on a batch from the validation set, and a function test that computes the error on a batch from the testing set :type datasets: list of pairs of theano.tensor.TensorType :param datasets: It is a list that contain all the datasets; the has to contain three pairs, train, valid, test in this order, where each pair is formed of two Theano variables, one for the datapoints, the other for the labels :type batch_size: int :param batch_size: size of a minibatch :type learning_rate: float :param learning_rate: learning rate used during finetune stage ''' (train_set_x, train_set_y) = datasets[0] (valid_set_x, valid_set_y) = datasets[1] (test_set_x, test_set_y) = datasets[2] # compute number of minibatches for training, validation and testing n_valid_batches = valid_set_x.get_value(borrow=True).shape[0] / batch_size n_test_batches = test_set_x.get_value(borrow=True).shape[0] / batch_size index = T.lscalar('index') # index to a [mini]batch # compute the gradients with respect to the model parameters # compute list of fine-tuning updates for param, gparam in zip(self.params, gparams): updates.append((param, param - gparam * learning_rate)) train_fn = theano.function(inputs=[index], outputs= self.finetune_cost, givens={ self.x: train_set_x[index * batch_size: (index + 1) * batch_size], self.y: train_set_y[index * batch_size: (index + 1) * batch_size]}) test_score_i = theano.function([index], self.errors, givens={ self.x: test_set_x[index * batch_size: (index + 1) * batch_size], self.y: test_set_y[index * batch_size: (index + 1) * batch_size]}) valid_score_i = theano.function([index], self.errors, givens={ self.x: valid_set_x[index * batch_size: (index + 1) * batch_size], self.y: valid_set_y[index * batch_size: (index + 1) * batch_size]}) # Create a function that scans the entire validation set def valid_score(): return [valid_score_i(i) for i in xrange(n_valid_batches)] # Create a function that scans the entire test set def test_score(): return [test_score_i(i) for i in xrange(n_test_batches)] return train_fn, valid_score, test_score Note that the returned valid_score and test_score are not Theano functions, but rather Python functions. These loop over the entire validation set and the entire test set to produce a list of the losses obtained over these sets. ## Putting it all together The few lines of code below constructs the deep belief network : numpy_rng = numpy.random.RandomState(123) print '... building the model' # construct the Deep Belief Network dbn = DBN(numpy_rng=numpy_rng, n_ins=28 * 28, hidden_layers_sizes=[1000, 1000, 1000], n_outs=10) There are two stages in training this network: (1) a layer-wise pre-training and (2) a fine-tuning stage. For the pre-training stage, we loop over all the layers of the network. For each layer, we use the compiled theano function which determines the input to the i-th level RBM and performs one step of CD-k within this RBM. This function is applied to the training set for a fixed number of epochs given by pretraining_epochs. ######################### # PRETRAINING THE MODEL # ######################### print '... getting the pretraining functions' # We are using CD-1 here pretraining_fns = dbn.pretraining_functions( train_set_x=train_set_x, batch_size=batch_size, k=k) print '... pre-training the model' start_time = time.clock() ## Pre-train layer-wise for i in xrange(dbn.n_layers): # go through pretraining epochs for epoch in xrange(pretraining_epochs): # go through the training set c = [] for batch_index in xrange(n_train_batches): c.append(pretraining_fns[i](index=batch_index, lr=pretrain_lr)) print 'Pre-training layer %i, epoch %d, cost '%(i,epoch),numpy.mean(c) end_time = time.clock() The fine-tuning loop is very similar to the one in the Multilayer Perceptron tutorial, the only difference being that we now use the functions given by build_finetune_functions. ## Running the Code The user can run the code by calling: python code/DBN.py With the default parameters, the code runs for 100 pre-training epochs with mini-batches of size 10. This corresponds to performing 500,000 unsupervised parameter updates. We use an unsupervised learning rate of 0.01, with a supervised learning rate of 0.1. The DBN itself consists of three hidden layers with 1000 units per layer. With early-stopping, this configuration achieved a minimal validation error of 1.27 with corresponding test error of 1.34 after 46 supervised epochs. On an Intel(R) Xeon(R) CPU X5560 running at 2.80GHz, using a multi-threaded MKL library (running on 4 cores), pretraining took 615 minutes with an average of 2.05 mins/(layer * epoch). Fine-tuning took only 101 minutes or approximately 2.20 mins/epoch. Hyper-parameters were selected by optimizing on the validation error. We tested unsupervised learning rates in and supervised learning rates in . We did not use any form of regularization besides early-stopping, nor did we optimize over the number of pretraining updates. ## Tips and Tricks One way to improve the running time of your code (given that you have sufficient memory available), is to compute the representation of the entire dataset at layer i in a single pass, once the weights of the -th layers have been fixed. Namely, start by training your first layer RBM. Once it is trained, you can compute the hidden units values for every example in the dataset and store this as a new dataset which is used to train the 2nd layer RBM. Once you trained the RBM for layer 2, you compute, in a similar fashion, the dataset for layer 3 and so on. This avoids calculating the intermediate (hidden layer) representations, pretraining_epochstimes at the expense of increased memory usage. • 本文已收录于以下专栏： ## Deep Belief Networks (DBNs) Deep Belief Networks（DBNs），是一类随机性Deep neural network，其可以用来对事物进行统计建模，表征事物的抽象特征或统计分布，在手写字识别和语音识别建模中，已被... • linjiebelfast • 2013年12月17日 13:22 • 12603 ## 深度信念网络（Deep Belief Network）论文 • youyuyixiu • 2016年12月20日 15:15 • 499 ## DL：Convolutional Deep Belief Networks（CDBN）代码（matlab）理解 • oMengLiShuiXiang1234 • 2016年03月26日 21:37 • 2609 ## 深度学习值得关注的75篇文章 75 most popular Deep Learning Papers from the Bibliography Neural Networks: A Review Analysis of... • qqlu_did • 2015年01月21日 16:52 • 1318 ## DBN训练学习-A fast Learning algorithm for deep belief nets • jyl1999xxxx • 2016年04月19日 08:14 • 2786 ## 深度信念网络Deep Belief Networks • GarfieldEr007 • 2016年04月02日 19:08 • 1057 ## 深度信念网络(Deep Belief Network) “深度学习”学习笔记之深度信念网络本篇非常简要地介绍了深度信念网络的基本概念。文章先简要介绍了深度信念网络（包括其应用实例）。接着分别讲述了：(1) 其基本组成结构——受限玻尔兹曼... • Losteng • 2016年03月28日 21:11 • 27070 ## 机器学习之DBN（Deep Belief Network，深度信念网络） • u014487025 • 2016年06月24日 12:45 • 2446 ## Deep Belief Networks深信度网络 DBNs是一个概率生成模型，与传统的判别模型的神经网络相对，生成模型是建立一个观察数据和标签之间的联合分布，对P(Observation\|Label)和 P(Label\|Observation)都做了... • u011437229 • 2016年01月22日 14:03 • 2279 ## Deep Belief Networks • hemmingway • 2015年07月09日 10:09 • 709 举报原因：您举报文章：Deep learning----------Deep Belief Networks 色情政治抄袭广告招聘骂人其他 (最多只允许输入30个字)	{}
9758/2022/P1/Q10 10 A curve $C$ has equation $y=a x+b+\frac{a+2 b}{x-1}$, where $a$ and $b$ are real constants such that $a>0, b \neq-\frac{1}{2} a$ and $x \neq 1$. (a) Given that $C$ has no stationary points, use differentiation to find the relationship between $a$ and $b$. [3] It is now given that $b=-2 a$. (b) Sketch $C$ on the axes on page 19 stating the equations of any asymptotes and the coordinates of the points where $C$ crosses the axes. [4] (c) On the same axes, sketch the graph of $y=a x-a$. [1] (d) Hence solve the inequality $x-2-\frac{3}{x-1} \leqslant x-1$. [2]	{}
Sunday, September 25, 2016¶ Some ForeignKeys not being protected¶ I discovered and fixed #1183, a subtle problem which might cause unexpected data loss. Changes in The Lino core (the bug itself) and Lino Noi, The Lino Book, Lino Voga (test suites). The problem was a bug in lino.core.ddh.DisableDeleteHandler.add_fk() : if a model had more than one pointers to a given other model. only the first of them was being collected, the others were ignored. As a consequence, Lino did not replace Django’s default behaviour (cascaded delete) by Lino’s default behaviour (protected) for them. One symptom was that tickets.Ticket.reporter was not listed below in the output of analyzer.show_foreign_keys(). The following snippets helped to find out why: >>> for (model, fk) in rt.models.users.User._lino_ddh.fklist: ... print(fk) changes.Change.user clocking.ServiceReport.user clocking.Session.user excerpts.Excerpt.user faculties.Competence.user outbox.Mail.user stars.Star.user tickets.Project.assign_to tickets.Ticket.assigned_to tinymce.TextFieldTemplate.user users.Authority.user >>> for f in rt.models.tickets.Ticket._meta.get_fields(): ... if f.name == 'reporter': ... print(repr(f)) <django.db.models.fields.related.ForeignKey: reporter> On Lino Welfare production sites running a version older than today it is recommended to be careful when deleting users, places, countries, persons and companies. Because if no database object points to a user via the “first” pointer, then Lino would silently delete database objects which point to that user, place, person or company using another pointer. In order to evaluate the urgence of upgrading here is a list of the most important disable_delete_handlers which were not being installed : • users.User : aids.Granting.signer, aids.IncomeConfirmation.signer, aids.RefundConfirmation.signer, aids.SimpleConfirmation.signer, art61.Contract.user_asd, cal.Event.assigned_to, immersion.Contract.user_asd, isip.Contract.user_asd, jobs.Contract.user_asd, users.Authority.authorized • countries.Country : pcsw.Client.birth_country • contacts.Person: aids.RefundConfirmation.doctor, art61.Contract.contact_person, art61.Contract.signer2 Which means that the danger is rather theoretical. No reason to panic. But I made an unrequested upgrade in CPAS de Châtelet to be sure. In ÖSHZ Eupen I must wait for their decision. Lino Welfare failure on Travis¶ Build #38 is failing because lino_xl.lib.cal.ui.EventsByDay sorts the events by start_date and start_time, and when two events have the same value for both fields, the result is unpredictable. To fix this, we might simply add the primary key: order_by = ['start_date', 'start_time', 'id']	{}
# Where will the frost line be when the Sun becomes a red giant and what effect will it have on the solar system? I understand the frost line is currently about 5.2 AU and earlier in the solar systems formation was 2.7 AU. But when the Sun becomes a red giant the frost line should move outward. I understand the habitable zone will even include Jupiter and Saturn, which if we extrapolate means that the frost line would be further out than that, possibly near the orbit Uranus or Neptune. The other part of the question is what effect this will have. If there is a planet near the frost line, I would think it would potentially increase in mass as the recondensed volatiles collect at the new frost line and falls into the planets orbit. (Or is flung into or out of that orbit.) If it is not near a planet, I would think matter and volatiles would collect there into a planet. Whatever Kuiper belt object is near enough may start to gain mass and eventually coalesce into a planet as it clears the orbit around it. The current solar constant on the frost line is equal to: $$j_c=\frac{\sigma T_c^4 \cdot 4\pi r_c^2}{4\pi R_c^2}=\frac{\sigma T_c^4 r_c^2}{R_c^2}$$ where $$T_c$$ is the current effective temperature, $$r_c$$ is the current radius of Sun and $$R_c$$ is the current distance of frost line. We write the equation for solar constant on the frost line when the Sun is a red giant: $$j=\frac{\sigma T^4 r^2}{R^2}$$ We suppose, that the solar constant on the frost line needs to be the same in all eras. Therefore: $$j_c=j$$ $$\frac{\sigma T_c^4 r_c^2}{R_c^2}=\frac{\sigma T^4 r^2}{R^2}$$ $$\frac{T_c^2r_c}{R_c}=\frac{T^2r}{R}$$ The data: • $$T_c=5778\rm\,K$$ • $$r_c=7\cdot 10^8\rm\,m$$ • $$R_c=5.2\rm\,AU$$ • $$T\approx 3500\rm\,K$$ • $$r\approx 1\rm\,AU=1.5\cdot10^{11}\rm\,m$$ We can now express the radius of the frostline, $$R$$ $$R=\left(\frac{T}{T_c}\right)^2 \frac{r}{r_c}R_c=\left(\frac{3500}{5778}\right)^2\frac{1.5\cdot10^{11}}{7\cdot10^8}5.2\rm\,AU\approx 400\rm\,AU$$ But no, the Kuiper belt objects won't grow up. That's because for objects on the new frost line, there won't be any recondensation. • If the volatiles heated up and blown out passed the new frost line can recondense with each other or more likely dust, rocks, asteroids, and dwarf planets, why wouldn't they? Jul 13 at 15:55 • @BrooksNelson The gravitational force between them is still too minor. Jul 13 at 15:58 • Wouldn't the hydrogen bonds from the water molecules stick to each other and anything else polar? Jul 13 at 20:12 • @BrooksNelson The hydrogen bond isn't so strong in case of small amount of water molecules (the space is pretty much empty). Jul 13 at 20:50	{}
# zbMATH — the first resource for mathematics Optimal price skimming by a monopolist facing rational consumers. (English) Zbl 0712.90007 A problem of equilibrium pricing strategy over time for a monopoly seller of a new durable good is formulated as a noncooperative game between the monopoly and the consumers of the product, each of whom buys at most one unit, generating a value, (or revenue) v. The value v varies across consumers and the distribution of the valuations of the population (N) is uniform over an interval $$[0,v^+]\equiv V$$. So $$Nv/v^+$$ is the number of consumers whose valuation is less than v. The constant (marginal) cost of production per unit is c and $$\delta\in (0,1)$$ is a discount factor. The equilibrium concept is a subgame perfect Nash equilibrium defined as follows. A sequence of functions $$p_ t^:$$ $$V\to R_+$$ and $$v_ t^:$$ $$R_+\times V\to V$$ for each $$t=1,...,T$$ is an equilibrium iff for any $$t\in \{1,...,T\}$$ and $$v\in V$$, $$p_ t^$$ and $$v_ t^$$, respectively, solve the problems A and B defined below. Define $$V_ t^:$$ $$V\to V$$ by $$V_ t^(v)=v_ t^(p_ t^(v),v)$$. For $$t\in \{1,...,T\}$$ and $$\bar v\in V$$, $$p_ t^(\bar v)$$ solves: (A) $$_{p_ t}\sum^{T}_{s=t}\delta^{s-t}(p_ s-c)$$ $$(v_{s-1}- v_ s)$$ $$N/v^+$$ subject to $$v_{t-1}=\bar v$$, $$v_ t=v_ t^(p_ t,\bar v)$$, $$v_{s+1}=V^_{s+1}(v_ s)$$ and $$p_{s+1}=p^_{s+1}(v_ s)$$ for $$s=t,...,T.$$ For $$t\in \{1,...,T\}$$, $$\bar v\in V$$, $$\bar p\geq 0$$, $$v_ t^(\bar p,\bar v)$$ solves: (B) Min v subject to $$v\in [0,\bar v]$$, $$0\leq v-\bar p\geq \delta^ s[v- p_{t+s}]$$ for $$s=1,...,T-t$$, where $$p_{t+1}=p^_{t+1}(v_ t^(\bar p,\bar v))$$, $$p_{t+1+s}=p^_{t+1+s}(v_{t+s})$$, $$v_{t+s}=V^_{t+s}(v_{t+s-1})$$ for $$s=1,...,T$$ and $$v_ t=v_ t^(\bar p,\bar v).$$ The objective function in (A) can be seen to be the discounted sum of profits. $$\delta^ s[v-p_{t+s}]$$ is the profit of the consumer if he buys in period $$t+s$$. (A) and (B) says that each agents policy function for each period is the optimal one for the restricted optimization problem for the rest of the horizon on the expectation that these policy functions will be adhered to on both sides. In this sense the equilibrium is a rational expectations one and subgame perfect. Backward induction leads to the conjecture that the solution of (B) must satisfy $$v_ t^(\bar p,\bar v)-\bar p=\delta [v_ t^(\bar p,\bar v)-p^_{t+1}(v_ t^(\bar p,\bar v))]$$ for $$t<T$$, and is verified directly. This is then used to solve (A). For the last period T an explicit solution can be derived since there is no complicating factor of future periods. Backward induction then establishes that the equilibrium solutions can be described by certain explicit formulae (too numerous to reproduce here) which are defined recursively. Asymptotic forms for these formulae which approximate the solution for large T are also derived. The equilibrium solutions are then analyzed to obtain several properties. First, the equilibrium price sequence $$p_ t^$$ is strictly decreasing exhibiting intertemporal price discrimination. Second, that for $$t<T$$, $$p_ t^(v_{t-1})<(1/2)(v_{t-1}+c)$$. The expression on the right is the profit maximizing price if the monopoly disregards future sales possibilities and maximizes profits for this single period given the consumers leftover in the market determined by $$v_{t-1}$$. Third, a similar problem is analyzed but where the consumers are myopic in the sense that at any period consumers with valuation $$v\geq p_ t$$, the announced price, buy the product. It is shown that here the equilibrium pricing policy $$\hat p_ t$$ $$(v_{t-1})>(1/2)(v^ t+c)$$ for $$t<T$$ and so, for each state $$v_{t-1}$$, the myopic pricing is higher than the single period profit maximizing solution. Finally, some numerical examples are used to show that errors in pricing because of presuming myopic behaviour of consumers when actually they follow rational expectations can be substantial. This reviewer found the formulation of the problem and its handling to be clever and the results interesting and believes it should be worth a quick scan for somebody interested in the literature on intertemporal pricing of durable goods and applied game theory. The notation and formal statements appear sometimes confused and misleading. Reviewer: S.Dasgupta ##### MSC: 91B24 Microeconomic theory (price theory and economic markets) 91A40 Other game-theoretic models 91B62 Economic growth models 91A10 Noncooperative games 91B26 Auctions, bargaining, bidding and selling, and other market models 91B50 General equilibrium theory 90C39 Dynamic programming Full Text:	{}
# Tag Info 38 This is just to expand a bit on vonjd's answer. The approximate formula mentioned by vonjd is due to Brenner and Subrahmanyam ("A simple solution to compute the Implied Standard Deviation", Financial Analysts Journal (1988), pp. 80-83). I do not have a free link to the paper so let me just give a quick and dirty derivation here. For the at-the-money ... 35 Here couple pointers that may make it clearer: Drift can be replaced by the risk-free rate through a mathematical construct called risk-neutral probability pricing. Why can we get away with that without introducing errors? The reason lies in the ability to setup a hedge portfolio, thus the market will not compensate us for the drift above and beyond the ... 29 This one is the best approximation I have ever seen: If you hate computers and computer languages don't give up it's still hope! What about taking Black-Scholes in your head instead? If the option is about at-the-money-forward and it is a short time to maturity then you can use the following approximation: call = put = StockPrice * 0.4 * ... 28 I have worked on this topic extensively (pricing and calculating IV in production) and believe can offer an informed opinion. First of all Mathworks - the company that creates Matlab is not a trading firm so you should probably not rely on their advice so much. There are few closed form options pricing models, and all have practical shortcomings. Barone-... 27 In general there are two basic ways to make money out of your option pricing models: Sell side (market maker, risk neutral): You use these models to calculate your greeks to hedge your portfolio, so that you live on the spread. Buy side (market/risk taker): You use your model to find mispriced options in the market and buy/sell accordingly. (A third ... 22 The reason for put and call volatilities to appear different is that the implied vol has been calculated using different drift parameters than those implied by the market. Let's take everything in the model as given except the interest rate $r$ and the volatility $\sigma$. For European options we have the Black-Scholes formula for put and call values V_{P,... 22 Being on the sell side and selling options you can intuitively think of it like this: An option is like any other product that is being produced out of ingredients and because of the competitive situation of the producer this is done by the cheapest possible production process. The ingredients are in a simple (Black Scholes) setting a stock and and a risk ... 21 The Black-Scholes 'normal-vol' formula leads quickly to a similar approximation to the one described by olaker. Click here for a paper which contains a formal derivation of the call and put prices based on a normal model (ie a brownian motion rather than a geometric brownian motion). The formula for the call price is: \text{Call} = (F-K)N(d_1) + \frac{\... 16 We assume that the short interest rate r_t follows the Hull-White model, that is, the short rate r and the stock price S satisfies a system of SDEs of the form \begin{align} dr_t &= (\theta_t -a\, r_t)dt + \sigma_0 dW_t^1,\\ dS_t &= S_t\Big[r_t dt + \sigma \Big(\rho dW_t^1 + \sqrt{1-\rho^2} dW_t^2\Big)\Big], \end{align} where a, \sigma_0, ... 15 In addition to what vonjd already posted I would recommend you to look at the E.G. Haug's article - The Options Genius. Wilmott.com. You can find some aproximations of BS not only for vanilla european call and put but even for some exotics. For example: chooser option: call = put = 0.4F_{0} e^{-\mu T}\sigma(\sqrt{T}-\sqrt{t}) asian option: call = put = ... 14 Agree with all of vonjd's points though I like to add the following: First of all, market practitioners do not read options prices or set options prices in the market, they price the option through models primarily on the basis of implied volatility. Im plied volatility is actually traded, options prices is what comes out on the other side. I know there was ... 13 This is in fact a tricky matter. As you say one way is to calculate delta by an analytic formula, i.e. calculate the first derivative of the option pricing formula you are using with respect to the underlying's spot price. The second way is to do it numerically, i.e. change the spot price by a small value dS, calculate the value of the option and then ... 13 This is an interesting and not so easy question. Here's my 2 cents: First, you should distinguish between mathematical models for the dynamics of an underlying asset (Black-Scholes, Merton, Heston etc.) and numerical methods designed to calculate financial instruments' prices under given modelling assumptions (lattices, Fourier inversion techniques etc.). ... 12 Assume the price follows a lognormal process. We can convert it into a problem of finding the probability of a standard Brownian motion particle starting from 0 and hitting x before time t, or its first passage time \tau_x being less than t. This can be derived through the reflection principle. The paths crossing x are exactly paired up by the ... 12 Actually there are more than just ideas and hints concerning this topic. There is an intuitive model and solution to your question already using machinery of option theory. But don't worry, it's not a surprise that you didn't find any useful literature in your search because the proposed solution actually comes from a very different topic. In addition to ... 12 Except in highly unusual cases, financial PDEs lack analytic solutions. The mathematical tools used are Monte Carlo, plus the usual ones for solving PDEs on grids, almost always one of the following: Trees, for very simple cases Explicit finite differencing, for throwaway projects or very specific cases Implicit or Crank-Nicolson finite differencing for ... 12 Two quick points: Recall that the derivation involves continuous time and (t, t+\Delta t) arguments---so the granularity is (at the margin) infinite. And hence time zero does not really get reached until we actually are at expiry. Generally speaking want the number of business days, not calendar days, and holidays do matter. So one generally uses the '... 11 The price of a binary option, ignoring interest rates, is basically the same as the CDF \phi(S) (or 1-\phi(S) ) of the terminal probability distribution. Generally that terminal distribution will be lognormal from the Black-Scholes model, or close to it. Option price isC = e^{-rT} \int_K^\infty \psi(S_T) dS_T$$for calls and$$ P = e^{-rT} \... 11 Theta decay doesn't depend on the in the moneyness. A 70 delta call and a 30 delta call have very close theta decay at any given moment. They are slightly different because of skew with 70 delta put having slightly bigger theta. Theta is the decay of extrinsic value. In practical trading, you can assume your decay distribution (using your graph is fine) ... 10 You are typically interested in evaluatingE\left[ f(X_T)-f(\bar{X}_T^{(n)}) \right]$(refered as the weak convergence)$X_t$the solution of the sde :$dX_t^x=b(X_t^x)dt+\sigma(X_t^x)dW_t\bar{X}_t^{(n)}=b(\underline{t},X_{\underline{t}}^{(n)})\cdot (t-\underline{t})+\sigma(\underline{t},X_{\underline{t}}^{(n)})\cdot (W_{\underline{t}}-W_t)$is your ... 10 Q: What does the risk-neutral price represent if the option is not replicable? In an incomplete market, there is no unique martingale measure but instead a set$Q$of equivalent martingale measures. Consequently, there is an interval of arbitrage-free prices:$ \Big( inf_{\mathbf{Q} \in Q} E_{\mathbf{Q}}[DX], sup_{\mathbf{Q} \in Q} E_{\mathbf{Q}}[DX] \Big)... 9 Scott Mixon argues in What Does Implied Volatility Skew Measure that among all measures of implied volatility skew, the (25 delta put volatility - 25 delta call volatility)/50 delta volatility is the most descriptive and least redundant (volatility is Black-Scholes implied volatility). His paper, recently published in the Journal of Derivatives, gives a ... 9 There are many different ways a pricing model can be better : It can allow to reproduce the observed market price (Fit criterion) It takes into account a specific recognized behaviour of the underlying S, say the forward smile dynamic. If you write a product whose value is mostly derived from said behaviour, you dont want to miss that aspect. (Don't fill ... 9 Regarding conventions One thing to keep in mind in all questions about "what's right and what's not?" is that conventions don't always matter as much as one would think. When a trader marks his vols by looking up option prices on the market, he is going to mark them using the pricing model which his quants implemented. So whether he uses one convention or ... 9 You need to compute your greeks as finite differences, but the full procedure may be pretty tricky. I will use vega $\aleph$ as the example here. Let's begin by designating your Monte Carlo estimator as a function $V(\sigma,s,M)$ where $\sigma$ is the volatility as usual, $s$ is the seed to your random number generator, and $M$ is the sample count. To ... 9 To recover the Black-Scholes pricing equation, you should first express the standard normal cdf in terms of its characteristic function analogous to the Heston solution: $$N(x) = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi x} f(\phi)}{i\phi}] d\phi$$ where $f(\phi)$ is the characteristic function of the standard normal distribution: ... 9 With $15\%$ annual volatility we have $15\%/\sqrt{252}\approx0.94\%$ daily volatility. To go from $27$ to $28$ is a $1/27\approx 3.7\%$ move which is $3.7/0.94\approx 3.9$ standard deviations. For a normal distribution this is about $0.005\%$ probability which is in line with your result. 9 1. What does it mean by the vol surface is the current view of vol? The local volatility model is calibrated to vanillas prices (and equivalently their implied volatilities), which reflect the market's view of the volatility, in order to use it to use it to price other options that one will hedge with the vanillas. Where a Black-Scholes model (no smile) ... 9 From an equities perspective, there are two concepts that should not be confused in my opinion and context should make the distinction self-explicit: Forward variance swap volatility (A) Forward implied volatility smile (B) I really recommend reading Bergomi's "Stochastic Volatility Modeling" which is an excellent book for equity practitioners. The topics ... 8 VIX is calculated from a basket of SPX options, and VIX futures expire into following expiration, e.g. September VIX futures that will expire next Wednesday will use SPX October options chain to calculate settlement value. If $B$ is the value of the basket then VIX value at expiration is $\sqrt{ B }$. Then VIX futures price is the expectation of the basket \$... Only top voted, non community-wiki answers of a minimum length are eligible	{}
by # Dr Elattuvalapil Sreedharan…..The Bharat Ratna no one talks about Dec 31st 2011 India had just lost the boxing day test match, with a day to spare. Star News was deciding who was the ‘Match Ka Mujrim’. Others were wondering when Sachin will finally score that created in the boardroom, 100th hundred. If an alien visited India around that time and switched on the tele, he/she would have thought our world started and ended with cricket. Dr Elattuvalapil Sreedharan, affectionately called the ‘MetroMan of India’, bid adieu to his 56 year long career of creating 21st century monuments. His exit was a reflection of how he lived; simple, elegant and completely inconspicuous. Just pause for a moment and think. 56 successive years, in a single job, without a break. 56 Years. In other words, Dr Sreedharan, worked continuously for a duration, in which an average Indian generally goes from the cradle to his/her grave. And no one in the media was paying a tribute to this great man. They were not even talking about it. All they had was a silly ticker at the bottom of the screen, that said ‘Dr Sreedharan Retires’. No ten thousand word articles, no special Dr Sreedharan shows, nothing. It was like, it didn’t even happen. Fundamental question is, Why? Is it simply because, Dr Sreedharan, is not your quintessential Indian Success Story? I mean, he is not a IITian, nor is he from an IIM. He is not from a minority community. He did not surmount racial and casteist prejudices imposed by the society. He did not fetch water at 4 in the morning and did not study under street lamps. He did not milk cows to pay for his education. He is neither an actor nor a singer. Neither is he a cricketer nor a cricketer who claims to be an actor. He was just a simple middle class Indian, who went about his job. In other words, he does not fall under any bracket that our media terms as ‘an Indian Success Story’. And according to the media, not an ‘Indian Success Story’ meant ‘No tribute’. And this in a day, when even a guy like Chetan Bhagat will get a glowing biography on his life and times, the day he finally decides to stop writing. OK, sod the the media. They have TRPs and paying news customers to take care of. But what about the government? They don’t have to answer to anybody. And this guy dedicated his entire life to them. At least they can do something. But No. When names were being bandied about for the Bharat Ratna, India’s highest civilian honor, everyone, from politicians to Dhanush, had only one name, ‘I am stuck on 99’ Tendulkar. Infact lobbying for that guy reached such epic proportions, that the government actually changed the award criteria to accommodate that actor. And Mr Sreedharan, who triple checks all the existing criteria, was not even being mentioned. The travesty is all the more acute if you consider what Dr Sreedharan has done for our nation. Like Date: 22nd of December in 1964, Location: Rameshwaram town For most of the non south-Indians, Madras and South India are interchangeable terms. But what you do not know is, South India also happens a favorite holiday spot for various cyclones brewing in the Bay of Bengal. And this South India is not Madras alone, but the entire states of Andhra Pradesh and Tamil Nadu. This geographical part of India has been hit by so many cyclones that nowadays meteorologists have run out of names. For most of us, This is THANE One such cyclone, hit the coastal town of Rameshwaram on 22nd of December, 1964. And it was a deadly one. Not so long ago, south of Rameshwaram, there used be a bustling town called Dhanushkodi. It had its own post office, customs office and even a Railway Station. My bet is most of you wouldn’t have heard about it. Why? Because on that fateful night, the Indian Ocean swallowed the entire town. More than 2000 Indians were killed. The ocean even ensured that, a passenger train which was beginning its last journey of the day at 11.15 P.M, made its last journey ever. If you still can’t imagine the ferocity of this cyclone, let me help you Dhanushkodi, before and after So where does Dr Sreedharan come in all of this? As a sideshow, the cyclone destroyed the Pamban Bridge, the only bridge connecting Rameshwaram to mainland India. And this meant Rameshwaram was completely isolated. Considering the above cyclone something had to be done, and fast. At that time, Dr Sreedharan was a Deputy Engineer in the Southern Railway. And this piece of wreck was in his territory. Indian Railways, gave Dr Sreedharan six months to restore connectivity to Rameshwaram. Which was asking a lot considering Dr. Sreedharan, had to convert this To IN SIX MONTHS Dr Sreedharan finished the job in ..FORTY SIX 1964 DAYS. He took one month and 15 days to restore, THAT bridge, back to full operation. The bridge which was India’s longest sea bridge for 96 years, till the Bandra Worli Sea Link was inaugurated in the year 2008. Forty six days to restore this 2.3 Km bridge in a state where THIS BRIDGE took six effin months to restore after being washed away by a flash flood, in 2006. There are some achievements that look cool, but once you get an award, you completely forget about them. And then there are some you won’t forget, even if you suffer a total memory loss. This was one of those things. For all this trouble, Dr Sreedharan got a Railway award consisting of Rs 100 and an awful looking plaque. The process of short-changing Dr E Sreedharan, began in 1964. Even If Dr Sreedharan, sat and twiddled his thumbs for the rest of his life, he would have been considered a superhero for his bridge-building awesomeness alone. Fortunately for India though, he did not like twiddling thumbs. Packing his bags for his next assignment, he set off to Calcutta, where he became the chief designer of the Calcutta Metro. And then, getting bored of railways in general, he took charge of India’s largest Ship building company, Cochin Shipyard. There he designed, built and commissioned India’s first indigenously built Merchant vessel, the Rani Padmini, in 1981. Back then, it was state of the art After building everything from trains to ships, Dr Sreedharan according to government rules, had to retire in 1990 when he completed 60 years of age. But, when you are Dr Sreedharan, you don’t have the plebian privilege of retirement. He was asked to go to Mumbai, to take charge of what was then deemed to be India’s toughest project since independence. It involved burrowing through basalt mountains, spanning kilometer long marshes and rivers and taking railways to a place where even the Britishers thought it was impossible. This region was the Western Ghats of Maharashtra, affectionately called, the Konkan. The railway babus simply called this herculean venture, Konkan Railway. Naming things is not really their forte. And slapped it with a pathetic logo This is what Dr E Sreedharan had to do, Lay 760 Kms of railway track, through a terrain shanbagtravel.blogspot.com and this infested with and perennially at the risk of And to do that, Sreedharan had to 1. Acquire 5000 hectares of land from 42,000 assorted land owners. 2. Build 2000 Bridges, both major and minor, across Marshes, swamps, rivers and backwaters. 3. Blast 92 tunnels, totaling 83 kilometers in length through Basalt, nature’s adamantium and soft soil, nature’s china clay. You need nuclear weapons and Arnold to bore through the former while the latter generally collapsed on itself, if someone as much as farted. And then came the most difficult task in the Indian Index Of Difficult tasks 4. Dealing And Negotiating With The Chief Ministers, Home Ministers, Other Ministers And Their Chelas Of Four Different States. And to complete all of the above tasks, Dr Sreedharan, was given 8 years. It would be like Hercules being asked to accomplish his twelve tasks, blinded and with one hand tied to his back, in three days. This is Hercules Needless to say, he would have failed. Dr Sreedharan, supposedly retired and who qualified for Indian Railway’s senior citizen quota, finished the job in 7 years. Konkan Railway to me, is the second biggest achievement of Independent India, with the first being India remaining India. New standards will have to be invented, to realistically measure the impact of Konkan Railway on the Indian economy. For starters 1. For the first time ever, three largest ports on the Indian coast, Mumbai, Karwar and Mangalore have a direct connection. 2. Travel time from the southern states to the north, have been reduced by upto 40 percent. The crummy old Nethravathi Express, used to take 38 hours to travel from Trivandrum to Mumbai. Now, thanks to Dr E Sreedharan and the KONKAN RAILWAY the same train takes 22 Hours. A 16 hour reduction in travel time. And a 16-18 hour reduction in travel times of all trains going from Kerala,Karnataka, Goa to the North. All the trains. Try measuring the impact. Don’t bother, you can’t. And new levels of difficulty need to be established to measure the difficulty of this task. The sheer number of architectural impossibilities overcome by Dr E Sreedharan and his team in making this wild dream a reality, is mind boggling. All I can do is suggest you to go here, here and here. However, there is one thing that encapsulates what all the above links have to say Which basically is a Marsh Between two Hills And to traverse it, The train has to travel at a height, which is as tall as In other words, Dr Sreedharan and his team, built a goddamn Qutub Minar, over a marsh, between two hills, just so that, a train could chug over it. Need I say more? You will see this wonder of Modern India immediately after Ratnagiri on the Konkan Railway. In the 15 kilometer stretch between Ratnagiri and Nivasar, there are 3 tunnels and 5 viaducts. The third viaduct is the Panval Nadi viaduct, immediately after the first tunnel. The exact sequence will be —-> Big Tunnel, Massive Gorge, Big tunnel again. Dr Sreedharan though, never had the chance to taste his success. In December 1997, one year before the Konkan Railway was thrown open to traffic, he was shunted to New Delhi to head a new organization. It was created to find a viable solution to the traffic woes of the aam aadmi in the national capital. This organization was called the Delhi Metro Railway Corporation. Of all the railway systems in the world, building a metro is the most difficult in the world. Most difficult, becomes impossible, when you have to build that damn thing under and over a megapolis. Impossible reaches ‘You must be kidding me’ levels, when that megapolis in question is Delhi and the country in question is India. There are more people in this pic than there are in Canada. Delhi Metro was not India’s first metro. Calcutta has that honor. But Calcutta’s ‘I built my own metro’ story, was one sorry tale Sample this 1. It took 22 years to build the Calcutta Metro, A metro whose total length was 16.75 kms. In other words, Calcutta Metro construction dudes managed a grand average of, .76 kms a year. 2. The Calcutta metro suffered from debilitating shortages in almost everything. Shortages of funds, shortages in labour and shortages in everything else. The only they had in plenty was those damn shortages. 3. And it was harried by the parent, the Indian Railways, every step of the way. If the Calcutta metro was any indicator, the 168 Kms long Delhi Metro would have been beset with shortages, harassed by the Railways and would have been completed in A.D 2083. If ever there was a movie made about Dr Sreedharan’s job, it would have looked something like this But this was Dr E Sreedharan. When he started off, he put in place some things that were never before done in the history of Indian mega construction projects Like 1. His appointed himself as the judge, jury and the executioner as far as the Delhi Metro was concerned. No external influences were tolerated. And he took no prisoners. There have been stories were, people walked in with the traditional M.P recommendation letter, walked out of his office, with the paper in two different hands. 2. After giving himself the power, he changed the tendering process, upside down. In a country, where opening a single tender on an average, took six to nine months, Dr Sreedharan got the job done in 19 days. YES, 19 Days. 3. When confronted with the standard Government ‘I am short of money’ rigamarole, Dr Sreedharan did what any sensible government servant would do. He simply cut out the Government from the funding. He single-handedly went and got a loan of USD 5 Billion sanctioned from the Japanese Bank of International Co-operation to fund the Metro. Now, 60 percent of the necessary finance secured, Government shut up and stumped up the remaining cash. 4. He then hired top International consultants from an assortment of countries, to cover for the lack of local talent, to supervise and execute the project. For the first time in the History of India, the entire Delhi Bureaucratic circle was completely circumvented. That too, by one 74 year old dude from Kerala. As a result Phase one of the 5 billion dollar metro, was completed three years before schedule, entirely within the initially stipulated budget. And there was not even a single shard of corruption. In India, that is the closest we can get to walking on water. If someone had in 1998, said that such a thing would happen in India, He would have been given a priority ticket to Agra/Alibaug/Kilpauk or the nearest mental asylum. Today, the Delhi Metro is complete and for the first time, Dilliwalas are enjoying the benefits of a full fledged suburban rail system that has a punctuality factor of 99.999 percent. All thanks to this man Next time, any Dilli boy/girl tries to mock you or anyone with the moniker Madrasi, just mention the words, Metro-Sreedharan-Madrasi in the same sentence. The other side will shut up. The successful execution of the Delhi Metro, made Dr Sreedharan the Tom Cruise of the Metro Rail universe. Every Indian city worth its salt, now wanted its own metro and Dr Sreedharan as its consultant. One such city was Hyderabad. And the company executing that metro was a company called, MAYTAS INFRASTRUCTURE. You know where this is going. In September 2008, Dr Sreedharan, after observing as a consultant, sent the following report to the Planning Commission, Government of India Making available 296 acres of prime land to the BOT [build, operate and transfer] developer for commercial exploitation was like selling the family silver. I fear a big political scandal some time later, as it is apparent the BOT operator has a hidden agenda which appears to be to extend the metro network to a large tract of his private land holdings so as to reap a windfall profit of four to five times the land price. Planning commission, that plans little and develops less, as usual did not pay heed to this report. In fact they chastised Dr Sreedharan, for not backing up his allegation with necessary proof. This is what happened, three months later Remember how the scandal started? It started by Satyam Computers, arbitrarily taking over Maytas infrastructure, a company that was neither in IT, nor was a competitor. In fact the only link between Satyam and Matras was that it owned by Ramalinga Raju’s sons. And sensing that Maytas was in big trouble, he ‘bought’ the company to save it. And Dr Sreedharan, predicted that the company was in trouble, three months before the world knew it. So let me just encapsulate, if that can be done, on what Dr Sreedharan has done for the country 1. He restored India’s longest sea bridge which was completely destroyed, in 46 days. 2. He designed India’s first Metro. 3. Supervised the building of India’s first indigenous Merchant vessel. 4. Executed India’s most difficult project since Independence. 5. Gave Delhi wallahs, something called the Metro. 6. Predicted India’s biggest corporate fraud, three months before it happened. I don’t know how the Bharat Ratna nomination thing works. But I believe you stand a chance if you have done something good for the country. Now tell me, what has Dr Sreedharan not done for the country? I mean when you can consider a guy who sells a computer anti-virus on prime-time television for India’s highest civilian award, Why is there not a whisper about a guy who has ensured 400,000 people on the western coast of India saw a train for the first time? Or, was responsible for a sharp drop in road-rage killings in Delhi? Come to think of it though, I really don’t want Dr Sreedharan to get the Bharat Ratna. I mean, he will then join a club that is populated by people like Bismillah Khan, Lata Mangeshkar, Bhimsen Joshi etc. No disrespect to them, but they were at best, paid performers who had negligible impact on the population of India. (Lata Mangeshkar in fact has had a negative impact. A flyover which will bring respite to thousands, if not lakhs, of Mumbaikars is not being constructed solely due to her ‘privacy’ concerns) Dr Sreedharan should not be given the Bharat Ratna. Because the Bharat Ratna does not deserve a man like Dr Sreedharan. ## 334 thoughts on “Dr Elattuvalapil Sreedharan…..The Bharat Ratna no one talks about” 1. Anonymous Awesome write-up. How do you get so much of research done for your articles? Its really commendable. I didn’t know anything else except him building the Delhi Metro. Thanks a lot for the article and keep it up. • amit I keep reading your articles for “insights” and “resources” however I do not agree you trivializing legends from the field of art and music. Lata Ji just had an opinion on ” flyover” and too an extent anyone who visualize a flyover shadowing his/her home might have nightmares. You never know what they did in there field and many people look them as there role model. Defiling them is not “acceptable”. Kudos to Shredharan but people like him do not need trophies and claps. They rely on karma. Lets not shake that. • amit Hope you did not mentioned S### A Khan…he would have attacked you…LOL • I stay in an area called Sion in mumbai, and a flyover goes rt outside my balcony. No one asked me whether my privacy was invaded or compromised. And last checked, the constitution says I am as equal as Lata Mangeshkar for the govt of India.. Or is it? • Sundar 🙂 In my childhood, I was thinking bharath rathna should be a great award in musical field… • pravin well said. • Siva If people belive in Privacy, then I suppose a Metro airport requires more safety as terrorist can do anything to the Parked aircrafts (Look back at Colombo event during LLTE times), but still the government took a stand to build a flyover ( like a viewing gallery with no security) next to the Chennai Airport, I appreciate that Lataji is a living legend in the field of music but then she cannot stop a countries progress in the name of privacy, is she not moving out in the open and giving numerous stage shows. Think think and think • Anonymous yes same…awesome research work done i also did not know anything else but Metro Project • aravind i really not aware of this valuable info please give us more info which is authentic . i’ll spread among as much people i can • R J Tikekar Never in the course of Indian history has one man done so much for so many!! A brilliant and motivating write up. Work done by Dr Sreedharan should be part of school curriculum to motivate youngsters to realise what a middle class man can achieve with a vision and dedication. • Aruna Hi…this is the first time i am reading your blog and trust me it was truly an eye opener. In general after going through all the comments – i think some people failed to understand what you mean. Yes art, music, sports is very essential for a country’s growth. I read below someone say – just like how Lataji etc were paid, Dr. Sreedharan was a govt servant and paid. Please guys, give me a break – if he actually worked like the normal government officer, can you imagine?? End of the day, are these people (artists, sports people etc) sacrificing or going through the innumerable hurdles which people like Dr. Sreedharan have to tackle and handle and still accomplish? Well, i am not talking about the politics that the sports people or the artists go through in their field and the professional jealosy…this is too meger to talk about…like Ashaji really did not come up as much as her own sister!!! no assumptions or inhibitions meant, but we never know… Yes they are truly great – Lathaji, sachin …..etc, but ultimately they are pursuing a vocation – hats off that they have inspired millions and cheered and raised the spirits – we need another award for this category, definitely not the Bharat Ratna, if it has to go by its true meaning. Cheers for the blogger and hats off to Dr. Sreedharan. • Someone needed to tell this story. I had the honor of meeting the man once at the DMRC office. Wanted to invite him as a chief guest for our college function. Politely refused and told me the only college function he went to was when they had needed to buy land from the college for building a metro track! • SRUTI RANJAN the facts mentioned about DR. Sreedharan was commendable,He is just not good but he is best.He deserves Bharat Ratna,In my opinion He along with Dr.Vergese Kurien deserves a Bharat ratna.I think we should raise a campaign for that. . • K.G.Keshav kumar Hello,Iam impressed the way the details have been put together and published. I would like to add afew experiences of mine with Dr.Shreedharan. Right after his elevation to Chief administrative Officer.central Railway, He came to Mumbai took over the construction otganisation which was in the process of extending the harbour branch lines from mankhurd to Vashi and Belapur and with his special skills constructed the 2.15Kms Vashi(Thane) creek bridge and the suburban services were extended up to Vashi first and then to Belapur.Hr was also responsible for extending the Harbour line services beyond Bandra ie up to Andheri with the construction Raoli and Bandra Flyovers From here He became General Manager Western Railway and them Member engineering in Railway board.He served there till his superannuation and was assigned the job of constructing Konkan Railway in july 1990 ,After all formalities of getting office, men and materials we inaugurated the KRCL work on 15 th October. There is much to write on his ways of working but I shall restrict to salient points which have given us the success in completing the gigantic 760Kms of Railway line with Tunnels (already mentioned by the auther as 84Kms)1994 bridges viaducts,steep cuttings, by over coming the hurdles of Finances,Politicians interferences,Transport lobbies(specilly of Goa)There are 6 tunnels which are longer than 2Kms and the longest is 6.5Kms.The Goa had toughest of all tunneling with flowing sand where it was neccessary to support every 4 to 6 inches by steel supports and concrete else it would collapse the very next day.Every saturday he would visit the critical site and guide the site engineers to overcome the hurdles.All land losers were suitably compensated by cash,employment etc so every one was eager to support the cause .Modern Track welding machines were imported and deployed for making 1Km panel rails for smooth running with speed potential of 160KMPH.Tunnel Boring Machines ,Mucking Loading machines (10 each) were importedfor achieving tunnel depths upto 3 to 4 Mtrs. blast in one stroke with two to three cycles in 24 hrs.All the aspects of successful completion were monitored at micro level by Dr shreedharan like contractors problems,staffaminities,machines upkeep,coordination with all concerned in TIME which was essence of his success mantra.He is still working now for his home state of Kerala’s Kochi Metro. WE ALL PRAY FOR HIS SUCCESS AND LONG LIFE FOR INDIA “s prosperity. • Anonymous Really awesome write up. I think very few people will be knowing these facts.Honnor should be given at the Government level. • Anonymous Definately he derved Bharath Ratna Award…………. 2. Aravindh Great write up.. I had heard a lot about him…but this was a comprehensive summary of his achievements… 3. coolsap I revere this guy having known his achievements about Konkan Railway & Delhi Metro.. What he has done is unimaginable..Simply Outstanding especially in a place like India… you missed mentioning Mumbai Metro Sir.. I wish there were many more E Sreedharan in India.. and I wish I had been one more E Sreedharan of India… Sadly we the people of India leave everything to someone else to do our duty…Time to have a health check what IIT’s and IIM’s Grads studying on public money are doing good for India… • Sreejith I am not sure about IITs but IIM students pay from their own pockets to study ! There’s no subsidy unless you are from the financially weaker section. And even that scholarship is given by the Institute and not Govt. Please check facts before preaching. • The land grants..funds for setting up..establishment costs…Even the IIM’s have their share of public money. Not as much as IIT’s but they are not completely independent. • Anonymous But they mostly always chose to seve abraod or choose to work fro MNcs • coolsap You read what you wanted to read buddy. Forget the tax money, My comment was more about the contribution of IITians and IIMians – the brightest and the bestest minds in the country. I have no right to tell them what to do with their lives but I wonder (also lament) would we be the same India, if these people had been working for India’s development rather than their own. The original intent of starting IITs and IIMs was that they would fulfill the needs of Independent India, to create the foundation of New India which would usher millions of people out of poverty, create jobs and infrastructure, provide leadership to the struggling India. It is “a responsibility and not a privilege” for those students to take care of India. However they choose the comfortable path to head to US and work for their corporations. And they have all the right to do that as individuals but my question is what if Dr. Sreedharan, a smart capable person, chose to head to US instead of working for his hinterland. Every entity – country or corporation needs the BEST RESOURCES to stand out. I understand that today Indians are known to be world class engineers and scientists because what this guys did in US and other countries and it definitely helps us but what about the country know as INDIA. 4. Bounce Seshu: Well researched and written as always. 5. RENUKA R Thank you very much or this well written, eye-opening article. And Sreedharan Sir, what to say? India, with all its diversity and fights, and corruption, still mercifully has such unsung heroes. If not where would we be is a mind-boggling question. We can only pray to God to give them a peaceful and graceful old age. 6. kalpana mohan Dear Krishna Extremely informative and well written article. Your capacity to go to very minute details is highly commendable. Dr. Shreedhar he get / not get Bharat Rathna your article it self is a great tribute to him. The awards are given on behalf of the country men andI feel according to all of us he is Bharat Rathna irrespective of the Government confer the same to him.Well done. 7. Very cool article. Keep up the good work. And thanks for introducing me to a real life hero. 8. Sajiv A very well researched article about a well deserved person. 9. Sourabh Very well written and thanks for informing us about Sreedharan-Metro-Madrasi. This should be taught as a success story in all the government departments. And on the media, let us not have any hope 😦 10. I have met E Shreedharan a couple of times (one of which was to receive an award from him in college). He’s a miracle worker and someone who is truly very humble. Despite such stupendous achievements, he is ever so respectful and concerned about everyone around him (can cite numerous small instances where I saw this first hand). His work-ethic is unparalleled. But I disagree with you on one point. Lets not discredit Bharat Ratna awardees altogether. Admittedly there have been some blunders (VV Giri, MGR, K.Kamraj) but overall it’s a list of rather rather illustrious people (JRD Tata, APJ Abdul Kalam, Nelson Mandela, Sardar Patel etc) • I am not discrediting the Bharat Ratna awardees, But I do believe that the Bharat Ratna being fair is a relic of the past. Today it has become another political tool. Elections in a particular state, want to appease a certain community, just give the Ratna to a person from that state/community, sod his/her merit. In fact the credit worthiness of the award has plummeted to its lowest, when the govt is contemplating giving the award to an anti virus salesman. I don’t want Dr Sreedharan to get this Bharat Ratna. • pookari_pichakari Brilliant article. And i couldn’t agree more on the last statement in the post. Being an engineer, I often wonder what life will be like if people of science just stop working one day, since scientists/engineers never get the kind of publicity that people who make a sex tape get, for example 😛 however, while i completely agree about the unnecessary fuss that is being created for the “you know what”, its not the “anti visur- salesman’s” fault is it? It’s not like he is doing it as a publicity stunt. And while he has has NO impact on the lives of millions, he is probably the ONE person who (pretty much) unites the country \ref{the biggest achievement is india still being india}. 🙂 (Again, that does not mean he should get the bharat ratna, im just saying its his followers that are to be blamed here) • Who is this anti-virus salesman you refer to? • Raj Among the Barat ratna blunders, you have included K Kamaraj. Kamaraj is the only CM of tamilnadu till date who is uncorrupted and clean. he did implement many social schemes as well (Mid day scheme was his brainchild). I feel he deserves the aware. Others you have mentioned…yeah..they are blunders…. • Anonymous There is an IITM today, thanks to Kamaraj. • Leffe An IITM exists today – thanks to K Kamaraj • Dear Utsav About your last point, I guess the blogger was only pointing out at artists such as Bismillah Khan, Bhimsen Joshi, Lata Mangeshkar, M.S.Subbalakshmi etc who received the award for their contributions in art. Agreed that they are all legends in their respective fields. But their field, whilst entertained and emoted people, did not feed them or improve the economy. So I think the blogger is right in saying what he said. I would add Dr. M.S.Swaminathan as well along with Dr.Sreedharan in the list of deserving Bharat Ratna’s. • Krishnan That Mr.Sreedharan was not awarded bharat ratna or padma vibhusan is not a surprise.If we look at the padma awards and the Bharat ratna awards recently, we will find that they are all a shame. People who deserve are not awarded, and persons with political, geogrpahical and bollywood influence only are considereded. If this trend continues, item girls will get padma vibhushan awar • Sathish Kumar @ Utsav Mitra- any idea who K.Kamaraj was ? He quit his chief minister post and gave it another person and proposed kamaraj plan to Nehru and started to work for congress party. Thats why he was chosen as AICC president. When he died he was in the same rented house. He didnt have a single asset inspite being a CM for 11 years and All india congress president ( king maker who made LBS and Indira as PM ). He was a visionary who started 12000 schools in rural TN in 1950’s . Its only bcoz of him the 3% literacy rate of TN is now close to 85% over the years. He never allowed his own mother to stay with in CM guest house- thats only for CM and not for his relatives- that was his view. He was known for his integrity. I agree with the other blunders but kamaraj • When you hv the time, please google Kamaraj Plan. It is even used in the corporate circles today. And when he died, he had Rs 1500 in his bank. And two sets of clothes. He is one of the examples of where Bharat Ratna is justified. • Ankur I do not have much knowledge about Kamraj, frankly. But reading his achievements over here, I am getting an idea that he is more fot for a Tmali Ratna and not a Bharat Ratna. I am sure every state in India can name one or two persons/ CMs/ politicians who were clean and had done lot of things for the state. Did they all get Bharat Ratna? I am from West Bengal, and there Dr B C Roy had done lots of thing for the state (apart from being an outstanding physician whose treatments are almost folklores now). He did not get a Bharat Ratna!!! Understand this is a question just to understand, not to criticize anyone. • Sundar Please do not include K.Kamaraj in the blunders’ list… • Anonymous Ex-TN CM K Kamaraj receiving the Bharat Ratna is a blunder?can u give reasons for your statement? • Anonymous Why Not to Kamaraj? Lemme not divert the topic. Dude read about his life & comment. • Anonymous How is giving Bharth Ratna to MGR or Kamraj a blunder? • subramanian MGR and Kamaraj are not blunders man, • brandme Hello ,,Mr kamaraj ,,Do you know who he is ,,and what he has done ,,sure u have the right to give an opinion but calling kamaraj’s bharat ratna a blunder is in itself a blunder ,,think n research before u say somethin ,, • Anonymous Probably you do not know about Kamraj. Though uneducated (having sacrificed his life at very young age for the independence movement) he was humble and died almost poor though he was from not so poor family and occupied top positions in party as was CM also. He was the person who was instrumental in introducing noon meal for poor children which was the reason why TN climbed in the literacy ladder. Putting him along with VV Giri (who was at best Mrs Gandhi’s loyal…) is really a shame! • ShankarNarayanan Mr.Utsav Mitra Please read about K.Kamaraj as a person, what he has done in Tamilnadu as a Chief Minister, the dignity, the integrity and the honesty he brought to the role before commenting on his eligibility for Bharat Ratna. K.Kamaraj was a legend in his own sweet way. Please do correct your view. • R Utsav Mitra – Reading through your post, when you said “Lets not discredit Bharat Ratna awardees altogether”, I didn’t agree with you. But when you listed the deserving ones, I agree with you. I didn’t even know that some of these folks received Bharat Ratna. But somehow media and govt are very much celebrating folks, who haven’t done much. And Bharat Ratna has been going to them only these days ! The article is truly amazing. Very well written, thoroughly intelligent. Sridharan’s achievements are really wonderful and he deserves a national honor. • Vetri Hi Utsav, Kindly remove Mr Kamaraj from your list. He is so far the greatest CM of Tamilnadu and he has made so much impact to the common man especially for those below the poverty line. Kindly learn about him and you will appreciate who he is. • Bharath How dare that you said about Kamaraj ? Without him, TN would have been like next to one of your BIMARU states !! • Ashok kumar You seem to have inadequate knowledge about K Kamraj. He is a bull worker in the political environment and relentlessly fought for providing education to the masses. He himself coming from under privileged back ground, did not leave any stone unturned to make education a birthright for every one when he was the CM of Tamil nadu. Till now, it’s hard to draw any head of state parallel to him in administration and working for the poor and the deprived.MGR is a bad choice, I agree. But Kamraj is a fantastic choice. 11. Awe’frikkin’some. 12. Tarsuc Awe-Fkin-Some! Really! Kudos! 13. Hariram Loved it! Thank you so much, these vetti thoughts made my day! 🙂 14. My salutations to Dr.Shreedharan. Thank you for introducing us to a real life super man . Wish India had one DR. shreedharan in every department . A selfless man who just goes about his duty without any expectation is nothing but a god in our country . Such people are above Bharat Ratna or any other awards . As for the media not giving him due credit, well, the less spoken about them the better . 15. Brilliant research. An even more impressive and outstanding personality. bows He did not (and probably will not) get a Bharat Ratna precisely because he implemented all the above extremely efficiently and ruthlessly (and thereby pissing off all ‘important’ people). Hats off. Touch feet. Awesome. 16. Sujith Was Sreedharan in charge of constructing the Calcutta Metro ? If so, shouldn’t the blame for the faulty implementation in that case also fall at his feet ? Not to take anything away from the great man, but he is a little too friendly with the media for my taste (NDTV actually telecast an hour long programme on him, when he retired).Now, politicians, entertainers and sports persons have to be in the public eye as part of their job, but government servants to do it only out of a desire for fame About the Bharat Ratna, a musician’s worthiness is judged by his/her contributions to music..(and if Lata Mangeshkar is paid..isn’t Dr. Sreedharan paid a salary too ?) I’m in no way way denying that Dr. Sreedharan probably merits a Bharat Ratna, but your running down the 3 stalwarts of Indian music just exposes your ignorance.. 17. PBS Great work. kaipullai for president! 🙂 18. Bombayduck I had enjoyed reading your earlier articles on Kingfisher and the unsung heroes of medieval India, thinking that they had opened my eyes. And then I read this, Come to think of it though, I really don’t want Dr Sreedharan to get the Bharat Ratna. I mean, he will then join a club that is populated by people like Bismillah Khan, Lata Mangeshkar, Bhimsen Joshi etc. No disrespect to them, but they were at best, paid performers who had negligible impact on the population of India. (Lata Mangeshkar in fact has had a negative impact. A flyover which will bring respite to thousands, if not lakhs, of Mumbaikars is not being constructed solely due to her ‘privacy’ concerns). 1. The Pedder Road flyover has been opposed by many influential residents of that area, the most powerful of them being businessman-politician families such as the Deoras/Lodhas. Lata Mangeshkar is just one of the residents and like member residents of any locality tend to stick together against outside threats, she too is sticking with the group. It is naive to suggest that her privacy is the sole reason behind the flyover not getting made. But the way you write it in your post, as if it were the accepted and natural truth, makes me question if the rest of the so-called supporting facts in the rest of the post or other posts too have taken similar liberties and flexibility, both in selection and interpretation. 2. I agree that Sreedharan deserves the Bharat Ratna. But you did not have to trivialise some of the best musical genuises of our times to make a case for him. It is quite clear that you are not a connoisseur of music or couldn’t care to bother finding out what exactly Bismillah Khan achieved in his lifetime to not merely be another wedding function player and why that is extraordinary. Let’s just say that great achievement in engineering does not automatically trivialise that in the arts, medicine, social contribution, etc. To put it in your own style, try not to betray the abundance of that which has been pointed out in the green portion of the below linked graph within you. Until now I used to think that this was an eye-opener blog I could learn counterintuitive stuff from. But looks like it’s just entertainment, just like the guy you often have the fortune of coming across during train journeys who gets the entire compartment charged with his seemingly interesting views on the world but is at the end of the day talking a whole lot of crap. I will still keep reading this blog because you are funny, and I could do with a laugh. Cheers. 19. I got goosebumps reading this article. I’ve been to Rameshwaram through the famous Pamban bridge and have been amazed by it. Never knew it’s history till now. He is much higher than any Bharat Ratnas 20. Gautham I have been following your blog and read all of your posts. This one somehow is lacking. Nothing I didn’t know, expcept the Maytas bit. I learnt a lot from your other posts. And completely off the mark with the Bharat Ratna comments. It still is India’s highest civilian award, immaterial of how it’s been used. No matter who gets it, and has got it, the Lions on the award protect it from the blemishes the awardees might have had. And on that note, lifting a nation’s hearts and minds in whatever way isn’t important? Isn’t that a contribution to India and our society? Yes, those who have have made much more money than the unsung heroes of India but so what?? When has money ever been a leveler?? I’m appaled you think money and fame are measures of success. I notice you haven’t mentioned MS or CS, in your post. MS Subalaxmi and C Subramaniam. Is this because they are from the South? Does MS deserve it more than Lataji because she sang Maitreyi Bhajate at the UN?? Oh and btw, Gautham is spelt with an ‘h’! I love trains and there’s a romance about them, a charm. A charm I miss, even. But they also signify, machinery, struggle, something monotonous and not much pleasure or joy. Lataji’s rendition of a bhajan or a song. Like Pt. Joshi. Or Anti -Virus Salesman’s pull off Caddick into the stands, a sweep off Shane Warne to midwicket. I could go on. Are pleasure and joy not important in today’s coaching-class-world?? Do we not need the Amygdala in our brains?? Is what the frontal lobe tells us more important?? And might I quote now. A quote about, what else, but trains: “On a train from Shimla to Delhi, there was a halt in one of the stations. The train stopped by for few minutes as usual. This undeserving anti-virus salesman was nearing century, batting on 98. The passengers, railway officials, everyone on the train waited for this undeserving anti-virus salesman to complete the century. This Genius can stop time in India!!” • Thulsi Krishnan Is your last sentence a little hyped and clichéd ? The point the blogger is making here is simple – We are not seeing what we must see, we are seeing only what we like to see, which is absolutely wrong. • Kiran You are deifying what is happening in India, and we are criticizing exactly that. If sachin stops time, who does it benefit, certainly not thousands of others who have somewhere to go, something to do with their lives rather than sit and applaud someone’s centuries. A sreedharan would impact millions of Indians and enthuse them to work hard, of what use is Sachin’s hundred but to himself? This is in no way meant to offend Sachin, Lataji or any of the artists. I am a music connoisseur and musician myself and tremendously enjoy listening to these artists. They have committed themselves to exceling in whatever field they have chosen. Let’s emulate that instead of just being by-standers to other’s success. • Bhargava Don’t say that Sachin was not offended in this article. I have learnt a lot from this article and its a great tribute to Dr. Sreedharan except where there is a blame for choosing musicians etc. for Bharat Ratna. You are talking about commitment from musicians, what about commitment from Sachin. Can you even think of the pressure he carries each time he is out there? You are simply not treating everybody equally here …. just to make your point. • Don’t want to get melodramatic and all, but Hasn’t Sachin already got the Rajiv Gandhi Khel Ratna, the highest award meant for a sportsman? He is already at the peak of his field. But the field for Bharat Ratna, belongs to people who work for the country. And I am sorry to say, not one of Sachin’s hundred hundreds has made India a better place to live in. Argue against it, but that is a fact. Also, staying with Sachin, what example are you setting to the future generations, if you honour a guy, who has not shied away from misusing his fame and influence to garner favours for himself? We really need to start distinguishing our real heroes from make believe ones. Thank you for this post because it highlights the achievements of the great Sreedharan in a detailed manner. A few years ago I was looking for details like these and could never find them. Would you mind putting this on wikipedia as well. Or at least provide a link to this article on his wiki page. Thanks for such a post again 22. Kannan Excellent !!! Lets keep spreading this news !!! 23. Well written article. I agree with the points you made except for those on Lata Mangeshkar and other music talents. Every person contributes their own way to a nation’s progress. Your article falters in a making a case for Dr. Sreedharan by chastising others. Dr. Sreedharan has done our nation proud time and again and there should be no reason not to award Bharat Ratna to him, if most people think so, as I’m not a big fan of the concept of awards. 24. He is the white knight of India. A hero that our country needs but does not deserve! 25. Anonymous Simply great , i knew about him but not this well . 26. s k mangal a new award like “GREAT SREEDHARAN ” should be started thanks 27. PEOPLE LIKE SREEDHARAN DOES NOT WORK TO IMPRESS THE BOSS. HENCE IS NOT ELIGIBLE FOR BHARAT RATNA OR FOR THE MATTER ANY GOVT RECOGNITION. IT SHOULD BE HIS PASSION TO ACHIEVE IMPOSSIBLE FOR OTHERS. WE NEED SOME MORE SREEDHARAN IN THIS COUNTRY – TO PUT THINGS IN RIGHT PLACE. ALL THE BEST AND BEST WISHES MR SREEDHARAN. DO NOT STOP. 28. Thulsi Krishnan I don’t know if Bharat Ratna deserves E Sreedharan, but E Sreedharan definitely deserves such a funtastic write-up! 29. Abhay If you keep writing this way you are not far away. Great work man. • Far away from what? • Abhay Far away from the award you have been talking in your article. A few days ago read your article “How Kingfisher flew high on the money of taxpayers” It was an eye opener how the hard earned money fuels the jets of corporates. • Dude, Award and me?? You must be kidding 🙂 There are much better people at this, than me. • Abhay I didn’t like the end of this article where you mentioned the names of a few big personalities.We need not judge who didn’t deserved it and to some extent i agree with Gautham.Try to keep yourself away from anything that maligns your writing. Touch sky… • No writer should stay away from mentioning his personal take on an issue. If he feels the musicians are less deserving than nation builders then so be it – Trying to make it politically correct will make the entire exercise futile. I am of the view that Sreedharan the great builder be kept away from the Bharat Ratna and be awarded a people’s award – Folks in our country doing their damn job would be the greatest award a sincere man like him can expect. 30. Anonymous basic amenties comes first and arts sports entertainment should come next… surely putting all these in one BHrat Ratna group is bad.. another exellent example is Sir M Vishweshraihya…… by building one dam in Karnataka he brought electricuty.. and food to so many people… excellent post…credit to dr.sreedharan has been long overdue….i would like to point out a correction regarding his life in his hometown pattambi,palakkad.his family were really well off and he was carried by a guy every time he stepped out to school or wherever during his childhood (yep,not kidding the custom is a ghost of kerala past) . • Ermm seems like an addendum, than a correction. I’m not sure that is additional info needed in this context. • Ashish It becomes relevant when you consider he probably gave up what could have been a cushy life! 32. Awesome ! But why aren’t the images displayed ??? I’ve read many articles here and none of them contained the proper images. I’ve tried reloading the page many times, JavaScript and Flash are also turned on but the images still aren’t displayed. 33. Great Research!!! Bharat Ratna is not the parameter for an individual to be judged weather he has done a commendable job or not!!! It’s highly insignificant for people of high stature as Mr. Sreedharan Who has affected the life of an average Indian. Today our prime ministers are recipients of Bharat Ratna (for god knows why) , so it hardly makes sense for people like sreedharan to be recommended for it. And let me point out where half the nation is mad about 11 people running after a single ball and one man who is undoubtedly unflinching flawless guy in his arena but has done our nation no good. How could we expect rational thinking from such a bunch of naive individuals. Instead of recommending and starting a debate for the highest individual owner for a good Samaritan who has done his job in the best possible way, we should take cue from this man ans try to instill in ourselves his qualities, his beliefs, the ideals he stood by, That would be the truest sense of owner for Mr. Shreedharan. • ananymous I agree with your comments in that we are more obsessed with some body scoring a century in a game played by a minority of countries in the world than some of greatest heroes who achieved extrordinary feats for the betterment of India and Indians. Some times I wonder that the word “third world” applies to India perfectly because merit in proper context is not recognized in every aspect of our daily life. India can never be a Super Power since it is populated by mejority of backward people who tolerate wrong doing in every walk of life, be it traffic descipline or being unable to respect merit, nor respect for fellow Indians and no tolerance for alternate points of views or comments. 34. It is amazing how much coverage Kalmadi gets and how little Mr. Sreedharan gets. This way we will end up knowing all the bad guys and no good guys. We should also talk about good guys. Keep it up. Lets also keep talking about good guys once in a while, even if it is not deliciously scandalous. 35. Anonymous brilliant stuff bro… well researched and a lot of cheek…wish more people knew how many more sreedharan’s arent in the limelight 36. Rahul Very good article. Hope it catches some eyeballs. Sreedharan deserves to be honoured for the kind of impact he’s created 37. Anonymous The write up was like a movie to me ..but indeedTruly insightful…. Indeed Mr. Sreedharan is a hero and I believe many would dream to learn from him ….. 38. SD Great post! Thanks for amalgamating all this info into one concise post. I only have two concerns which are but minor issues (and mostly irrelevant to the topic, were they not mentioned in the post). 1. There are many Dilli-wala’s who consider anyone south of Mumbai to be a Madrasi. Mostly they have a limited outlook, and overall not particularly interesting or cosmopolitan people (I grew up in Delhi). At the same time, I have met South Indians who don’t know the difference b/w Gujarat and Rajasthan, that Orissa is a distinctly different state from Bengal, or that Maithili is one of the national languages of India and spoken mainly in Bihar. There is ignorance on both sides, but arguments of this nature hold no place in such a wonderful blog. 2. I am a great fan of Ustad Bismillah Khan and Pt. Bhimsen Joshi. Heck, I listen to and can identify certain ragas because of Bhimsen Joshi. To call them paid performers would be like calling Sreedharan an exceptionable employee who deserved a promotion, like a good mechanic who can fix my car w/o shortchanging me. I feel they have selflessly contributed to preserving an art form. I need not go into details, but Sreedharan deserving a Bharat Ratna does not belittle their achievements in any form. I am not sure of Rajiv Gandhi being awarded one though. • The dilliwala reference is just harmless humor man..Please don’t take it as my opinion on the people of that great city. Some of my best and most intelligent friends are dilliwalas. 2. I just believe that, any artist, however good he is, does not deserve an award like Bharat Ratna. Give him a Padma Bhushan, Vibhushan tops, but not a Ratna. That is for people who made a real impact in the lives of ordinary people in India.Where guys like Dr Sreedharan/M.S Swaminathan/Baba Amte come in. An to be honest, an ordinary man cares a fig about Lata Mangeshkar/Bhimsen Joshi. Nor have they done things that will make a real difference in the lives of the people. No disrespect, but that is the truth. • Gautham Lata Mangeskar and Bhimsen Joshi haven’t made a difference to our lives?? Is smiling at their renditions not a difference to the rigamarole of our everyday lives?? Does a backfoot punch have no joy?? Agreed a full tummy is essential. CS was honoured for his role in the Green Revolution and rightly so. But it’s not fair to belittle the arts or the pleasure it gives. It’s important for arts to be recognised to keep evolving as a society. • krishnan Lata and Mr.Joshi might have made a difference to hindi speaking and people of maharashra. Not the entire india. We should change our belief that bollywood and Hindistani music are the music of india since there are other more mdelodious and complex music systems in India. Just because they have sung in Hindi, that does not mean that they should be awarded Bharat Ratna. Sridharan is an exceptional case for Bharat Ratna for the contributions he has made to change the life of people. • Great article, although here are some points which I thought were biased: 1. ‘The greatest achievement of India….’: It really seems to be the greatest achievement when we consider people like you, who judge the achievements based on where are person comes from, ‘Kerela’, ‘Madras’, dilliwalas and so on 2. I do remember seeing news articles, in most of the online newspapers, where they mentioned that the metro man retired. I guess thats how you found it in the first place ? I would not expect a headline about a person who retires from a job, he excelled in his work and that’s been duly acknowledged. 3. I bet more people would have heard songs by Lata Mangeshkar than would have travelled on the bridges constructed by Sreedharan. I would not be so judgmental as to consider some form of work/caste/class/region to be superior of other. Also you need to be an intellectual (nothing to do with your degree) to understand the importance of arts. 39. E Sreedharan Sir, I see that you have great regard for my work. But please do not speak ill of the Bharat Ratna. It is a symbol of our country’s pride. E Sreedharan • Gautham Right said, Sir. And hats off! 40. Karthik Brilliant article.. I have become your fan after reading this article and the one on KFA.. your research is meticulous.. wonder what your sources are.. i’m digresssing.. Dr.Sreedharan’s work is unparalled i believe.. thanks to information junkies like you, he will be in public memory whether or not he gets the bharat ratna (which doesn’t deserve him, like you say).. one final question.. are you a journo? 41. Anonymous One such cyclone, hit the coastal town of Rameshwaram on 22nd of December, 1964. And it was a deadly one. Not so long ago, south of Rameshwaram, there used be a bustling town called Dhanushkodi. It had its own post office, customs office and even a Railway Station. My bet is most of you wouldn’t have heard about it. Why? Because on that fateful night, the Indian Ocean swallowed the entire town. More than 2000 Indians were killed. The ocean even ensured that, a passenger train which was beginning its last journey of the day at 11.15 P.M, made its last journey ever. If you still can’t imagine the ferocity of this cyclone, let me help you Dr Sreedharan finished the job in ..FORTY SIX 1962 DAYS… I’m a bit confused.. 1962 or 1964? • Error has been corrected, thanks for pointing out 🙂 42. Rukmani Sridharan Hey Sesh! Spent the past few hours reading a lot of your articles. I really admire your writing and the fact that you’ve spent so much time researching everything – any fact I google and it is quite precisely quoted in your text. Kudos! 43. Shrikar I am glad that I found this – and it would not be a hyperbole to say that I am totally stunned. Needless to say , I am going to follow your stories as of this second. Wish you well. 44. You are scaring me. My opinions are now starting to totally depend on these blog posts of yours! Instead of trawling through the bbc/cnn/ndtv/rediff(don’t laugh)/washingtonpost/nytimes/aljazeera/hindu websites as I used to do, I now depend on you to give me the hot political titbit of the week. PS: You should syndicate your blog. • Anonymous pls dont believe media. its all stage managed. 45. Kata in Babuland Great write up! And amazed at the amount of research you have clearly put in! 46. Sai Thank you for this article. Blessed are the people who even praise such persons. 47. Pallathz The article is great in highlighting the achievements of Dr.Sreedharan and his contibution to Nation building. At the same time, typical of Indian mindset, it tries to bring out the contrast by belittitling the achievements to an extent of some other legends in their own right. Dr.Sreedharan ,however impactful his acheivements are, for a typical Indian middle class fabric,when he works for a Govt .machinery ,we always assume his work to be taken for granted. We alwyas revere a Sportsperson or an Artist or someone from the the Literary field in a different light.There is an Aura. A Lata Mangeshkar or an Yesudas (though not a Bharat Ratna) ,their songs sooth many Indians.Be it Siachen,Afghanistan ,Somalia Peace keeping or remote bylanes of India, it has its own usp. Sachin ,is the Toast of the nation.Its difficult to imagine an individual who cast his spell on a 3 yr old to 100 year ,spanning 2 decades,cutting across creeds,caste,class,educational system & even nations. You could have avoided reference to these great legends and this article would have been much better. It would have really the highlighted Dr.Sreedharan & Now this article sounds ,unfortunately, the same as myriad number of articles on pros and cons of Bharat Ratna. Dr.Sreedharan for sure deserves a Bharat Ratna or even if he doesn’t India will remember him !! 48. anand kathare Awesome ,am fortunate to have also been one of the foot soldiers so to say ,when we supplied critical track components to both the konkan and Delhi metro Experienced his no nonsense way of doing things 49. Vijith Chandran Excellent Write up… To be very honest, I didnt know neither did I ever hear about this great person called Dr. Sreedharan… The facts which has been so beautifully brought out in your write up is really an eye opener to me and I believe the thousands or lakhs of people like me… Thank you for sharing this wonderful story… 50. shanmugaraju proud to be a civil engineer! 51. One of the best articles I have read…I appreciate the way you have taken the efforts to collect the pictures and come up with such a fabulous article..i think u too deserve a ‘hats off’ for showing to the world about Dr.Sreedharan…respect. 52. Wow fantastic commentary. Excellent writing, snaps and level headed thinking. Kudos 53. Anonymous I appreciate every bits & pieces of the article, great work ! I like the way u presented, this is better than ur previous one on Mallya. My point here is – He definitely wud not have thought of highest civilian award when he is doing all that mission impossible he jus did it with shear dedication & passion. For such a man with honest & dignity even In rest of his life if he is not honoured It may not mean to him. Try to be modest when you put across your message. It will bring accolades to u as a writer. Great work kailpulla … Keep going 54. Suryanarayana MV The article is so heart-touching. This is how India is distinct from the rest of the world. Here individuals outgrow the system and the Government. I salute to Dr.Sredharan. Forget the medal and the controversy around it, he is already the real Bharat Ratna. 55. There are people who Bharat calls ratna…and there are people who make Bharat.. a Ratna…Screw the first..grow the latter… Needless to say, brilliantly written and well researched. 56. Rejil Brilliant write up. The last sentence I think, sums it all up. I think he shouldn’t be given the Bharat Ratna for that simple reason… 57. Anonymous You rock!!! That is all. 58. Anonymous even today, there are many engineers and bureaucrats doing commendable jobs for the citizens, even after taking minimal pay as compared to their MNC counterparts.. Bharat ratna should be given to an individual who has impact on the Indian public in bettering their standard of living, unfortunately sachins, latas, all are only entertainers, paid for their work…if sachin had played cricket without taking any money, only what is required for survival..then probably he could have been considered…at least as a free entertainer….India will never develop..it will remain as it is or rather go to dogs as the future nears… Thank you for a great article. It would be interesting to know how Dr.Sreedharan manages these things. I wonder how 3 idiots reached IIM/Management schools, or how Lalu Prasad Yadav reached management schools. I dont hear Dr.Sreedharan’s methods being discussed there! 61. Anonymous Great article and lot of information…. Thank you for the effort for preparing this… and you said right, Bharath Ratna doesn’t desereve Sreedharan.. Hoping to see the Kochi Metro running with 99.9999 % accruacy.. OMG !! You are an inspiration !! A gr8 one !! Cheers 63. Good Blog…. But could have avoided un necessory remarks. Media does report about Sreedharan’s exit. I have read several reports in various papers and all reports were good. Also dont understand y you have to downplay something to highlight ES. True, Sachin did struck at 99 but come on boss, ITS 99 CENTURIES. A record which no one may be able to break in our lifetime! You can allege that govt changed norms to give Bharat Ratna to Sachin, but fact is that they didnt nominate Sachin atleast for this year. Boss, Bharat ratna is given for excellence not just for social work. And the musicians did deserve it, dont make ES another GODMAN • So you mean to say Dr Sreedharan is not excellent.. No offence, but I will take Dr E Sreedharan, any day over SRT and his 99 centuries. And If god asks me to choose between another Tendulkar and another Sreedharan, you know who I will pick. However beautiful his straight drive it may be, it will not drive away hunger from India or make it an economic power. Things that Dr Sreedharan did, does both these things and many more. And that is what you should consider when you give awards like the Bharat Ratna. And there is a difference between a report and a tribute. Dr Sreedharan retiring, his successor, yadda yadda yadda is a report. This is a tribute. And please post any other tribute to Sreedharan that you may find by any Mainstream media website. • Bhargava I don’t see your point in repetitive disliking for Sachin. Why can’t India have both and celebrate both? Why should ONLY Sreedharan be appreciated? I fully agree Sreedharan should be celebrated as much as we celbrate Sachin, but you can’t blame others’ accomplishments not worthy of Bharat Ratna. That angle of article makes it less worthy. • I am saying is, we have celebrated him enough by giving the Khel Ratna. He has already got the recognition that he deserves. And I firmly believe he does not deserve the Ratna. And you still haven’t answered my question. How has Sachin’s performance on the cricket field, made India a better country? • Bhargava Would you say that cricket/music has brought nothing to this country? Forget about generalities, I can cite myself as one example. Sachin was a major inspiration in my life and his grit and commitment in 2004 Aus series has changed my life. You can’t just measure these things. You are talking about making India a better country. What better way to inspire people of India and show how to commit yourself to the task at hand and be a role model. But this article is getting convoluted by these unnecessary discussions. I would stop and just say hats off to Dr. Sreedharan! • Your points only serve to prove what I was saying..We Indians need to distinguish our real heroes from make-believe ones • Reji Varughese Sachin like Sreedhiran is a master in his own field. it does not make sense to say that sports persons or movie stars does not contribute to national growth. No country can grow only on rails and structures. You need art, cinemas, sports etc. This makes a society complete. Millions of men , women and children stay glued to TV or cinemas watching Sachin or Shah Rukh in the same way as millions use facilities created by technocrats like Sreediharan. What is required is that country appreciate all such people instead of picking up only from one or two fields. Sreediharan deserves a national award for his selfless contribution towards improving the infrastructure and life of common person. A recent example is the death of two young ladies who were working as nurses in a renowned hospital in Calcutta. They died trying to save the lives of the hospital patients. Their commitment to duty and service is remarkable. No VIPs from the affected states even attended the cremation. As a nation we lack the commitment to duty and service, perhaps that would explain why we cannot honor people who serve unconditionally. 64. Abhijeet Singhal Great Write up…A huge credit to Dr. E. Sreedharan for making things which make India proud. A great humble and modest great. But some suggestions and corrections…. Dr. E Sreedharan was definitely a great man. But these projects were not only result of his dedication but ..it definitely includes scores of people contributed directly and indirectly. Panvel Via duct is not made by Dr. E. Sreedharan as written above in the article. It is designed and executed by L&T. What I am trying to say, things happen when people collaborate…and Dr. Sreedharan managed that collaboration successful. 65. Hindustani Really a nice article. You have researched alot to get all these details. I didnot have this much info about E. Sreedharan the great man who is from my place untill u gave it here. Really a nice article and above all really an eye opener for all those who are only interested in publicizing Aishawrya Rai’s delivery and Sachin’s 100th ton. We are all living in imaginary world of silver screen, where we believe that the things that films show are the only big things. Untill we come out of that world and live in real world we can’t compete with other contries. Great Thanks to E.Sreedharan and to you for letting us know about this great man. Sometimes people’s love and honor will over come any awards that corrupted politician’s give… 66. Gustave According to Dr Sreedharan’s page on wikipedia, he was also in charge of the Calcutta metro – the one you have criticized heartily. I’ll agree with you if you say that the delay was not caused in any way by Sreedharan, but the strong criticism of Calcutta metro doesn’t quite help his reputation here, given the overall point you are trying to make. • He designed it…He was in charge of Investigation and Design. He had nothing to do with its construction.Hence, I believe that fiasco doesn’t tarnish his reputation in any which way. 67. Anonymous I always loved ES but after reading ur article I have started worshipping him.Indian need many more like him in every sphere of life. BUT…… well u know. Kudos to d writer,style of writing is fresh and engaging. keep up d good work!!!! 68. Sriram Wonderful article ,Keep writing! 69. Sathish Kumar Shesha – As usual its brillant ! Details of this article already discussed over the phone last week- so no specific comments on that. Really sick to see few people think that watching cricket, listening music makes a real difference to their lives ( what if there was no honest officials like sreedharan over all these years- these guys would still be happy in underdeveloped condition like africa with pathetic traffic, no proper roads, no electrification etc etc and still enjoy those so called music and sports ?? ). I am totally surprised to see how few people not able to differentiate what makes “real difference to life of people ” and entertainment . Junks ! Again i am repeating my words – Hope you understand your strength and do what comes naturally to you and not follow any social order and become another corporate slave ! . You are a talented writer- i would suggest you to give serious thought about careers like investigative journalism, fiction writing etc 70. Ganesh K A eye opener for me!! I am ashamed with myself, that being a keralite, i did not have this much of information about Mr.Sreedharan! I have been a silent supporter for Sachin for Bharat rtna! I change my stand now, if anyone gets barat ratna, it should be Dr.Sreedharan first, then rest may follow!!! 71. Captain Kumar Shashank Just spell bounded…He is the real hero and believe me after reading your article he has become my role model more than he was in the past… You are just brilliant , I love your way of penning down your thoughts…Shed more light on us through your articles…:) If it wasnt for this Article I wouldnt have know Mr Sreedharan for his achievements for a long long time. I think its the medias fault that the credit to this fine achiever for what he has done for Indians is not justified even a little bit. For what you have mentioned in the last even the Bharat Ratna wouldnt justify his achievements. 73. I really liked the way you told the story. What is commendable is the time and effort you took to bring it to people’s attention. You have some talent in writing, I normally do not read such long articles but this one compelled me to read on. Why don’t you try your hand at writing for some paper, have a larger audience and create a bigger impact. 74. Ajeet Absolutely Brilliant !!! Especially liked the ending….Yes the Bharat Ratna does not Deserve the Great man…. 75. sorabh jain The level of your research is just amazing. Truly the He-man did an amazing job. Had some issues with the rivalry between north Indian and south Indian theory now i have one more proof to back my theory. The same is true with Dennis Ritchie. He and Steve Jobs died in the same month. But not even the mention of the man who made it all work. 76. Holidevil KS Take a Bow Ur writing skills backed with the research is amazing Keep it up I am speechless with awe n respect!! God Bless You Dr Sreedharan! 78. Pallathz Sports & arts is last when it comes to priorities in a nation of a billion people,so much so that country halts for days when little master bats.No wonder Indian has no medals to show in Olympics or any multi even sports.Now don’t start pointing to cricket and the teams defeats. Fortunately, the improvished African nations with hardly any infrastructure like Kenya ,so to say ,or Ethiopia,ulterly malnourishned,comes up with a Haile Gebrselassie,who gives ,even for a moment,a feeling which uplifts a Nation. Lets not penalize Tendulkar for his brilliance & bracket him as a mere entertainer.The pleasure and joy he gives to millions is unmatched.The same with artists be it Yesudas or Lata Mangeshkar.They may not have done to your life ,for millions over last 2 decades in the case of Sachin,he has given some joy & for Yesudas ,he still continues to do so after many decades. Dr.Devi Shetty of Narayan Hridayala is a Doctor. Whether if he ever will be considered for such honour,debatable,as he seems to be unfortunately carrying out his profession.hmmm! Ask Pakistan .They would surely would bestow him their highest civilian award. Dr.Sreedharan is a gem and among the list of visionaries and honest officials who done well for the nation. Lets appreciate all such people irrespective of the fields they excel in instead of weighing them down for no fault of theirs ,which really shouldn’t have been especially with the article which shifts the focus from the man. 79. Bhooma Loganayagi Dear Writer, Your write up sure will enable, hundreds and thousands of young minds ,to identify a role model in a century where there is acute scarcity of role models. It therefore can be compared to a drizzle on a hot madras summer day. By publizing on the www you have made the world identify the three important qualities for guaranteed success. SIMPLICITY,SINCERITY AND HONESTY.(values taught in average Indian households) The short ,yet comprehensive, tribute to this great man will put murmurs, about the relevance of the above mentioned qualities in the current century, to rest. you have not only introduced him as a role model for the Indian youth but an icon for everyone across agegroups and across countries to emulate. He ,in my opinion , is VISHWA RATHNA. HOW I WISH THE READERS TAKE THE ESSENCE AND NOT SPOIL THE MAIN THOUGHT PROCESS BY DEVIATING TO THE LOOP LINE AND NOT STAYING ON THE MAIN TRACK. GOOD JOB,CONGRATS. 80. Anmol Great article. Catchy style. Not all folks from Delhi equate South India with Madras, though. It doesn’t always have to be North vs South India, man. 81. Awesome. What a man! Thank you for telling us as it is! 82. Susaant Menon Nicely researched article with all the ingredients to sell it and make people write (back) about it (even though in some negative tone). Everyone (I have read each of the comments) is right here (since everyone has freedom of speech). And I would like to take this opportunity to concisely point out what everyone has pointed out here and add a little more. 1. Dr. E Sreedharan is a great man with a mountain of dedication, intellect and perseverance to have accomplished all that we know now – So, instead of making him a demi-God let’s celebrate that a Human being with right mix of qualities can achieve what he has and try to emulate him (not entire life) but the right qualities. 2. To praise one for good things is okay, but trying to pull other(s) down to praise another is not necessary and even wrong (unless it is purely meant to incite other’s to follow your article, even-if it’s in a negative tone). I said pulling others down is wrong, because your logic of pulling others down is based on the fact others are not God people made them to be and they are simply human beings with talent – And going by the same logic praising Dr. Sreedharan to the extent where he becomes God (for public) is wrong. So what ultimately would happen is someone with int,erest in some other field which a critical mass of people follows can influence public opinion with facts, figures and pictures about a great talent in that field to make that talent a demi-God. a) Who knows Dr. Sreedharan himself is Lata or Tendulkar fan, and you might hurt him in a article that is meant to praise him. b) I read one comment (apparently from Dr. Sreedharan himself) requesting people from belittling Bharat Ratna. 3. This is my personal opinion that this article was born out of the angst that Bharat Ratna losing it’s value (due to political apathy and interests), angst that a person like Dr. Sreedharan who made great tangible contributions to Indian Society is not recognized appropriately by media and public in general, angst that public memory in India is highly short-lived and solves no tangible purpose, etc. etc. I sincerely believe the article would have been highly positive if it wouldn’t have been born out of angst of different natures and instead born out of a question on what tangible and intangible qualities made Dr. Sreedharan what he is today. But it’s not the blogger’s fault as he has put in a nicely researched article with lot of positives to take home about a man who really has made India a better place to live and also added a bit of negatives without which the article would be read by only a few. 83. jai Great article!!!!An eye opener!!!!! 84. I agree, Dr Sreedharan should not be given the Bharat Ratna. Because the Bharat Ratna does not deserve a man like Dr Sreedharan. Why not request the mercurial gentleman to have a look at the Aviation Industry in Air India, maybe we’ll have something reliable. We can’t afford to waste such incomaparable talent to go sit idle while others are raping the economy / system. 85. One word – EXCELLENT. 86. One of the best articles read in recent times. I have known most of the details of this great stature and was a great recepeint of his works. Kudos to this man. As rightly said, Bharat Ratna doesnt deserve this man, just like Mahatma Gandhi . 87. hareendran during the delhi metro construction, there were lots of accidents claiming many a laborours life. the project was said to cursed as well. And once Dr Sreedharan stepped down assuming the moral reponsibility of Accident. But the CM of delhi Sheila Dishkit didn’t accept his resignation and was asked to continue till metro is out for public. A simple man with meticulous standards. A man of sheer will power and deteremination. I got these facts from newspaper @2008 , thgt of sharing… way to go really apprreciable , vibrant write up. to be honest could have left out kamraj’s reference. He was in a league of his own, untainted and simple politician with high moral values 88. Reblogged this on Flame Of Orodruin and commented: 89. Kumaravelu Narayanan Great Write up for a Great Man. 90. rick kudos nice read….flyovers do destroy the beauty of any place..we should look for underground transportation.. 91. Hey Sesh ! Awesome work 🙂 Keep it up! 92. Rajesh M Good writing Rahchand, well researched and hats off to your work.We ‘d read about Dr.E.Sreedharan’s good work on all the metros and the Konkan railways,but the work on Pamban bridge is unbelievable/awe inspiring, especially if you’ve travelled to rameshwaram .Overall great piece of writing and can’t disagree with most of what you’ve written, Dr.Sreedharan is already a Bharath Ratna whether the government agrees to confer or not ! 93. Vijesh Thanks for the write up dude. At least his glory shines through these blogs 94. SAS - a dilli wala I think Dr. Sreedharan should be the next President of India! No Bharat Ratan; this award has no recognition…all the “Gaanay bajaanay, bhaands/actors get it, no big deal. After APJ Kalam, this marathi Bai and her stupid hiubby ain’t no good and now she is requesting for an extension!!!. We need heroes like Dr. Sreedharan for a better civic society…and sense and his preseidency could bring a positive change in the way we live in India! As a matter of fact, its the South Indians who have made our country proud, whether in the field of IT, outsourcing, technology or even hospitality. North Indians were foodies and they will die being foodies only!! I love South India…right from discipline to roads to neat and clean railways; these states and most importantly the people – have it in them because they Value and nurture what they have unlike us North Indians who lack basic common sense. Simple living & high thinking – the essence of our Incredible India which starts and ends in the South! 95. Anonymous Really an interesting article-enlightened after reading this article-How many when given an opportunity can work like Dr.Sridharan-atleast in my next birth,I should do some thing..that is my desire.. 96. Mrs.Maragatham.K.Sundar what an achievement-I am not even fit enough to give this appreciation-even that is required..Godwilling in my next birth I need to achieve something like Dr.Sridharan.. Maragatham.K.Sundar 97. Manjunath K.V Three Salutes to the Great Indian, Great human being. May God give him all good health to do more Great things for India & Indians. He is a living Legend. Kudos to him. 98. SREEJITH T R I am from Kerala, I am thirty years old and I read news papers almost everyday, But till now what I know about Dr. E Sreedharan was that, he built a metro in Delhi. I am really thrilled to know about him and utterly sad if I didn’t have read this here , either I will never hear these or I may hear it as a eulogy if he happened die before me. Thanks a lot my friend…. 99. Very informative article about an outstanding engineer who made a magnificent contribution to the nation. 100. santhosh.R.V IT IS A MATTER OF TIME FOR METROMAN – NO -THE INFRAMAN OF INDIA TO GET BHARATH RATHNA, in India one has to vanish to get accolades 101. Anonymous Hey please do share on social networking sites…… 102. Kaiser Well written and I agree with all that’s been jotted down here. But I’d like to point out a few things. Dr.Sreedharan was spared the trouble of innovation because of the mere fact , all he had to do was copy the best models there were adopted in other countries and try implementing them in India ! All his achievements you’ve mentioned were just firsts for India but multiple nations had achieved that before ! Besides, regarding British having dumped the Konkan Railway plans, they didn’t do it, not because they couldn’t do it but since they didn’t want to waste their money it. Remember, the Railways, British built were for their economic benefits , not for saving time of Indians ! If there is one reason why Dr.Sreedharan should be given Bharath Ratna, it’s for proving that models of the Western World and East Asian can be successfully copied in India , not for innovations because all that’s he’s done , China, Japan and South Korea (even Britain and France constructed the Chunnel, a far more innovative venture) have been doing for ages , right from 1960s ! Not sure if any other country would confer their highest honours on someone who just had to look at other nations for innovations ! M S Swaminathan is much more deserving of a Bharath Ratna than E.Sreedharan IMHO ! Saving lives of millions is a far more important achievement ! • Ravi Its very easy to sit in front of a computer and say it is copied and it is worthless. If u had managed at least a small project singlehandedly, I am sure u didn’t have made such a comment.. • So if everything according to you was easy and was already done in the west, why did 1. Calcutta Metro take so long? 2. Why is the LA still in testing stage, 26 years after it was conceived? So many examples to show you how western copies do not work in India. And when a guy brings western levels of efficiencies to Indian shores, he could not have done it without innovation. Also why do you think the Konkan railway would not have been economically beneficial to the Britishers? Not sure if any other country would confer their highest honours on someone who just had to look at other nations for innovations! By this statement of yours, then no one in India deserves a Bharat Ratna except politicians. As they are the only ones not copying from the west. JRD Tata also does not deserve his Ratna, nor does Babasaheb Ambedkar. Just realize the hollowness of your statement. • kaveribharath Please! MS Swaminathan sold out all our native variety of rice to the US, imported hibrids, and now we have to struggle to “buy” our own rice back! His Agricultural Revolution, was what infused all our soil with pesticides and we now have non-nutritious and unhealthy food grains and obesity in 73% of middle class indian school kids! We all know that to get anything to work in our Indian red Tape and system in itself deserves a Bharat rathna? Even something as simple as waiting at a red light while everyone is honking at you urging you not to follow rules takes so much of strength and energy in this nation. It is not that people haven’t stopped at red lights in other parts of the world. It is just much tougher to do so here! It is not a prize for innovation, it is a prize for achievement. 103. Sriram Very nice article. Felt so proud and inspired by Dr. Sreedharan. He deserves the highest honour of our country. My best wishes to Sreedharan Sir for his prosperous retired life and hope he gets what he deserves as a successful Govt servant, and ultimately cherishes to be the Man which every Indian feels proud of. KS: Thanks for introducing such a great Man to so many people by a beautiful write up in which I specially liked the placing of appropriate images to make the reading more easy and to have a greater impact. Appreciate your effort!!! Only one thing being, all the way you tried to make people get inspired by this great personality, and also tried to engage them using interesting quotes, but some of which are not worthy to be in this page, like that of dilliwalas-madrasi comment, AV salesman and some music legends. I am sure majority of people who read this article would have felt bad by those comments. Till that point whatever positiveness filled the readers mind might have got lessened with such small ignorant comments at the end. Hope you understand. • I think you are making too big a deal out of the dilli-south Indian statement. It was intended to be a joke. Don’t give it too much importance. As for the anti-virus salesman and Lata Mangeshkar, well it is my personal opinion that they should not and shouldn’t have got the Bharat Ratna. I just penned it down. It is only to a degree that one can remain politically correct you know 🙂 • Sriram I can understand what your intentions were while putting them down. Since your blog is now such a famous one and I wish it hits more screens day by day… it’s just my suggestion for your future write ups so that your intention and purpose of the article is met, at the same time it should not score negative with some small things like these. Yeah I agree that, what ever it is, we can never be politically correct when we have to speak our mind. 🙂 It was not at all my intention to make any big deal of your points… since you gave freedom to comment and give our feedback, it was just what came to my mind that I can give though I might not be a qualified person to do that. Thanks for the great articles again, especially the Indian history were so well written. 104. Ambi Iyer Sundararajan It is really a good research. Recommendations are there for one who wants to complete his 100th century for his personal world record. We are seeing how miserably he has been failing. He does not utter any word for his exit neither others do. Dr Sreedharan will be an apt person for Bharat Ratna Award. People should unanimously voice his name for the said award. 105. H Great article!! There is no doubt about the brilliance of the man. But in atleast one project (Delhi Metro), there was political/administrative support.For instance, land acquisition for Delhi Metro was done under emergency clauses(ironically the same clauses that are causing tribal/farmer unrest) which greatly speeded up the construction. Dr. Sreeedharan and his team were exempt from CVC investigation/guidelines (it allowed him to select the best and the most appropriate tech for the metro, no need for tenders).The japanese loan was sanctioned with due efforts of the indian govt. Currently DMIC is having trouble attracting finance and the Indian govt. is going overdrive to find investors for the project. The (Metro) loan was sanctioned at a very low interest rate, allowing the metro to become a profitable venture. Even when the ‘pillar collapse’ controversy emerged, there was political support for him. His resignation letter was not accepted. It is clear that when civil servants are given extraordinary powers, combined with political will and commitment on the part of the bureaucracy, miracles can happen (Polio eradication?). This is not to take any credit away from Dr. Sreedharan. Delhi Metro would not have been what it is today without him. 106. Rohit I cursed myself for not knowing about this blog. Today chanced upon this because of a tweet by Kalyan Varma on the Kingfisher scam article here, and I am here for the last 2 hours. Your research is very appreciated. Thank God you are online and not in any of those paid news papers/magazines! And your line that Bharat Ratna does not deserve ES is what calmed me down. Truly said! 107. and here’s one more trivia… T.N Sheshan, the famous Election Commissioner, who made every Indian atleast hear the word ‘Election Commision’ for the first time, was a classmate of E.M.Sreedharan at school. They were 2 competing friends…. 🙂 108. kaveribharath Awesome article, and I didn’t know till now that he was the one who was behind the super efficient re-building of the Pamban bridge! Thank you fro the info. Have shared this article on FB. Well worth it. However, as true as it is that many not so deserving people get these prestige awards and devalue them in the process, to say that musical geniuses do not contribute to the well being of the society is terribly wrong! Without art, music and sports, the rest of our well being would be compromised. There would be more terror in this world, and less happiness. More people would be turning violent or unproductive without the creative and recreational adrenalin rush that these bring. The secret behind getting a child who never concentrates or “scores” academically, to up his/her grades and retention power, is to make sure that he/she plays more outdoors, and has at least half an hour a day of creative outlet! I am not kidding! The next time you are feeling unproductive and frustrated, or find anger clouding your reason, sit down and listen to some good, straight-from-the-heart music, or immerse yourself in a Van Gogh painting, or go and kick a ball, or even sing or do some clay modelling yourself… You’ll find your concentration, productivity and efficiency will be far better than ever before when you return to your work! It is a HUGE contribution to the nation and it’s progress. So let’s not devalue another person’s work in order to see how fantastic Dr.Shreedharan’s achievements are. You don’t need to. They speak for themselves on their own! 109. Scavenger Totally awesome..sharing 110. Anonymous I feel Bharata Ratna award is for people who give lot of “Ratnas” to our so called Ruling Class (“Idotic Rulers” born to sell India). 111. Great piece. E. Sreedharan is an inspiration indeed. He is actually one of the few such people who are well recognized in the media. If I recall right – E. Sreedharan has been conferred the Padma Vibhushan in 2008. 112. liberalcynic Extremely well researched and written. I didn’t know of this man’s contributions besides the Delhi Metro. • liberalcynic I would like to add that the Bharat Ratna should be given to people who’ve made the most impact on India, and have put India on the map in their fields. It is India’s highest civilian honor, and I think we shouldn’t slight the achievements of people in the entertainment business. Their contributions to our wellbeing aren’t as obvious as those of people like Dr. Sreedharan. It is an award for excellence, and should be kept that way. Saying that people pay more attention to a test match loss than recognize a man’s contribution is no different from me shirking studies to watch TV. We listen to hearts more than heads. If you gave 100 people this article, and say Dr. Sreedharan deserves a Bharat Ratna, I doubt a significant percentage would disagree. 113. Ash You are making me see the country I lived in for 26 years in a new light. Thank you. 114. Anonymous I rarely comment on blogs but after reading this one, I could not resist. In India, the real-life heroes are seldom showcased or recognized. We got to change this attitude and blog posts are these are a great effort. 115. Rajat Singhania I was so moved by your blog that I created this petition for awarding the Bharat Ratna to this great man. Please have a look and sign, everyone!! And spread the word. Thank you! 116. Rajat Singhania 117. Manalath Sankaranarayanan I am rarely reply these blogs. In this case I thought this is serious subject . I also strongly recommend for Mr. E.Shreedharan who is a deserving person for BHARAT RATNA for his achievements through out years. 118. Another great article! Brilliantly written. Once again, I feel your temperament’s a bit off. Try not to compare objectively. You have a great talent which in a few ways surpasses academic/journalistic endeavours, in that it’s powerful and heartfelt. You put the pulse to words. And we do feel it. But I did point out once (although I’m nobody) that you must be neutral in your tone. And I love Konkan Railways! I consider them a Marvel of (now not so) Modern Engineering! In my words.. \m/ Dr. E. Sreedharan \m/ 119. Sree Excellent write-up Krishna…You must have done a lot of research to produce this…hope to read more blogs like this in the future….All the Best… 120. Aanchal I’ve always been an E.Sreedharan fan but your research managed to make this article an absolute eye opener for me. Words can’t really be elaborate enough to speak about your intelligence, Krish. You don’t even need a ‘keep-it-up’, you’ve always had it in you. Thank you for causing the the much needed awareness. Society is blessed to have people like you! 121. Vijay Nair Excellent article. I really wish that Dr. Sreedharan gets Bharat Ratna. At the same time, I have no problems in seeing great personalities in different fields such as science, engineering, economics, music, sports, sociology, politics etc also getting Bharat Ratna since they are all Ratnas of Indian society. Personally, I believe that the sportsperson who wins gold in his field consistently at a global level deserves Bharat Ratna more than a great cricketer until cricket also becomes a truely global game. Give him the “Best Ever Cricketer of India ” award. That’s just my opinion. 122. Manjula Nice writeup. Hats off to Mr. E Shreedharan. He is deserving person for BHARAT RATNA award. 123. Sid So, the bottom line is, Ram Guha can shove it, unsung heroes like Sreedharan truly belongs to makers of modern India. Today I came to know about a Great person. I don’t know about him much, except for Delhi Metro. Yes you are right Bharat ratna doesn’t deserve this METRO MAN of INDIA. 125. P Yeshwant Bharat ratnas are now days given to only the people from whom there could be polictical mileage to the ruling party. Dr E Sreedharan definitely deserved the highest award. I feel the real persons who are changing the lives of the people are not given worthy credit. Though such people do the work without expecting awards, what could be gained giving to award to them is that the most of youth will see them as role models. Abdul kalam and Anna hazare are inspiring some of the youth. 126. SK Gandhi Bharat Ratna is decided by committee headed by politicians. So no chance for Dr.Shredharan. In Nehrus’s time Dr. Vishveshvarya got the award. In Vajpayee time, Lata mangeshkar got it. But, then they were more of statesman than politicians. 127. RJ 128. An excellent blog, I have come across recently. There appears to be no exaggeration anywhere in the blog. Dr.Sreedharan’s works are worth to be emulated. Whether he gets Bharath Ratna or not but he himself has demonstrated to be so . I can only wish May his tribe increase.! 129. Rajat Singhania 130. Dude.. you should seriously take up directing & creating at least small 10 minute short-films [since we all already know mainstream full-time cinema is a Sun-TV-Maran-Nidhi black-money converting exercise in intellectual stupidity]. I’ve recently (couple months back) seen “Food Inc” & “Waiting for Superman” – and those were wonderfully envisioned & created – and reading this piece (like all of the other articles I’ve read here) instantly flooded me with the same emotions of watching those docu-s. . I can’t begin to imagine (let alone describe in words) the effect of your writing. If you ever publish a book, do let me know. . NOTE: “All I can do is suggest you to go here, here and here” – the 3rd here seems to not “link” anywhere. You may want to oil-up the HTML a bit? . God Bless the good doctor – and God Bless you for blogging. . MN 131. G Varma Has Dr Sredharan got the Arrears he was fighting the GOI for the last 8 Yrs? Can someone update the desired info Guru 132. Desh Bandhu Babbar MR. SREEDHARAN HAS SET STANDARDS WHICH SHOULD BECOME BENCH MARK FOR ALL ENGINEERING PROJECTS ABOVE 1000 CRORES VALUE. YOU ARE DAMN RIGHT “BHARAT RATNA” D O E S N O T D E S E R VE H I M… 133. Thomas Abraham Dear Great work We are with you to support… 134. Reji Varughese Fantastic description of a man who always chose to remain low profile and do his job with a cool head. It is national shame that such persons are not acknowledged. In a country obsessed with movies and cricket, this is predictable. However, let us keep this message going on and on. Hats off to KS for compiling this. Probably years later, this could find way into CBSE text books. 135. Anonymous Great work done……We all indians should vote for him 136. Anonymous India must not only salute to him but get inspired by his practice and faith in a country which infested by corrupt politicians. These good unknown people who toil without limelight to just fulfill their dreams. Their dreams are so empowered that it shapes up like a dream for everyone. The stats of Konkan Railway was amazing. Let us behold these great men & find many more such people who are still toiling behind without any recognition. A true eg of “Ask not what your country can do for yo,u ask what you can do for your country”. 137. dude KS, Great write…. E. Shreedharan is now actually being requested by every possible State govt. from his own Home State Kerala, Kochin Metro Rail to Rajasthan Metro Rail Project who has only one condition : NO INTERFERENCE FROM GOVT. That’s a man every Indian should aspire to….. 138. swati Hey, great blog you have here. Would love to read more about Indian history….just the way you write makes me wanna read history all over again…the real one, not what we have in schools! 139. Anonymous Highly deserving personality to be honoured with the highest civilian award of the Land “Bharatratna”. Hope the concerned authorities would give this simplest man value for his contribution to the Nation as a whole. Viswanathan 140. Vasanth Seshu thala … over range article.. just now started reading your blog.. got hooked .. keep em coming 😀 141. Anonymous Mr. Sreedharan has to be honoured with “Bharat Ratna”. The research done by the author is really praiseworthy. 142. Gokka Makka • heh, i guess imitation is the sincerest form of flattery. plagiarism on the other hand… oh well. 143. Vijaya Why oh why did they not get Dr. Sreedharan for Mumbai Metro? BTW, govt is least bothered about common man’s privacy. They have built foot over bridges all over the suburbs that pass within 3 feet of a second floor window. Only people who use the bridges are the perverts who want to peep into people’s homes. Imagine keeping windows shut and curtains drawn all day and night for the rest of your life. Its true. Visit any suburb on Central or Western line near the station. 144. G.G. Kingi My salutations to this Bhumiputra who is a chiranjeevi in the technological field 145. Anonymous Thank you so much sir for letting us know about Sreedharan Sir is. I just knew he did a great job for konkan and Delhi Meto, never knew him to this extent. Great Man, seriously. And people will come to know about him slowly. 146. Anonymous great article… I wonder if we will have people with such qualities and skill do exist today 147. well done…..may such creed multiply 148. mandayamr Excellent article. Really enjoyed reading it. I always wish that we had more people like him, MS Swaminathan, Abdul Kalam, Dr Verghese Kurien, JRD Tata and many more like them, who have made huge positive contributions to our country. IMHO most of these people are above awards…they are karmayogis who are able to rise above such recognition, or even the lack of it. Also, while you are entitled to your opinion about the past Bharat Ratna awardees, and about SRT, you may want to think about some of the qualities/strengths/values that these individuals had/have, which if even 10% of our population imbibes, would make our country a more humane one. I have a feeling that if you look at these people you will find that there is a lot of overlap …it is just that the context in which it was/is applied is different. 149. Chanced upon your blog via FB reco..have been here for the past 4 hours going through all your article and have become a fan – a huge one already .Am sending the link to all my like minded friends. Please continue to do what you do..Congrats. 150. Vasanth Asokan Thank you for taking the time and energy required to write this article. Not just write it, but to do so in a way that no reader drops off a few paras down the start. A lot of people are socially conscious and do take the time to bring about awareness. But to mix in charisma and engaging writing is what makes them successful at it… 151. Anonymous I am just happy that in a country with a billion population (or more) at least one took the time to research on the common man Dr. E.Sreedharan’s achievement . I guess this article itself is an achievement as it had come from a “common man”. 152. Tninety you forgot to mention, he was consulted for building Karachi metro… 153. Ravishankar N though i agree with drawing a line between social good and ‘mere’ entertainment and also our opinion on suitability of Bharat Ratna, the world would be a dull place without the arts, music and other pursuits. There is no need to belittle the amount of joy they have contributed to innumerable lives. The pain that you feel that Mr.ES would be unsung is unjustified as you can see from the response to your article. Whether this appreciation needs an endorsement from the government? I think not. If your article intills the same kind of committment in some of us, then I am sure he (ES) would feel rewarded enough. Is this thought process escapist? May be! In addition to fast travel and lessening of travails, a person needs the vicarious pleasure and sense of achievement of excelling at world platform. SRT probably delivered this as a part of ‘Indian’ cricket team. Again, whether it requires a government endorsement? I certainly think not. I also agree that many wrong people may have been awarded with Bharat Ratna for various reasons, but we should also celebrate the right people who have been recognised and strive to see to that more and more right people get nominated. But right people don’t care much about it anyways:) 154. Ravi I also enjoy reading the comments. Whilst they (like me) disagree with a few things you have said, they do not deny the quality and passion behind this post; and any criticism is not doused in vitriol, abuse and personal attack. Kudos to all comment writers for making such online interaction just what it should be. 155. Arun Managolikar Hey, After reading this article i really got disappointed as well as happy (inspired). Disappointed because today, from morning i have wasted all my time just peeping into somebody else life by the help of great (arse hole) Facebook… Go to hell Facebook….. Now am happy and completely clear in my mind that wat to do in life after all india is not that bad place to be. We can definately build it to great heights. God bless Mr. Sreedharan and also bless me to work atleast 20% like this man. Also thank Kaipullai.. Its very gud article. 156. How about E. Sreedharan for President of India ? If all concerned citizens talk about it on the various media available it might even be possible. Anybody better than the current President. 157. Vinod Just fantastic!! What a great article and kudos to the author for doing the research. A big question that comes to mind is what were Sri Sreedharan’s methods for doing what he did. I am sure there are lots of very good lessons that can be inspiration for many others. I hope he or someone else writes his biography.. 159. Great piece of writing.. I have gone through many blogs, but this particular article made me to comment for the first time… You have perfectly researched before writing this article. Dr Sreedaran has an impeccable record in his tenure at office which could not be challenged by anyone in the society and he deserves Indian’s Highest civilian honor. If u grant me the permission, I would like to put this article in my blog with appropriate citation.. I just loved it so much 🙂 • Sure, sir..No need to ask 160. vineet great article. But u didn’t need to disrespect Sachin Tendulkar for this by calling him an actor. Also, u may need to check the description of Bharat Ratna before questioning the govt’s decision to give the award to Bismillah Khan, Lata Mangeshkar, Bhimsen Joshi etc. Well, see it for urself “Bharat Ratna is the highest civilian honour (File referring to external site opens in a new window) , given for exceptional service towards advancement of Art, Literature and Science, and in recognition of Public Service of the highest order. ” india.gov.in Public service is one criteria but not the only one. Probably, Dr Sreedaran should get this award and he still can, but do not disrespect other legends just to praise one. 161. Jayant Kumar Chaudhary Sir Billion salutes to you!!. You are real “Bharat Ratna” You will always remain a great motivation and inspiration for all right-thinking and patriotic Indians. Jaihind 162. AN you must have spent lot of time in research for writing this article..your effort is highly commendable….i know for sure that to get a national award in India requires lobbying and recommendation at highest level which a man of integrity like sreedharan will never attempt…he has out grown Bharat Ratna award 163. vasantha jambunathan Great piece and well researched.i have lived in Delhi and Janakpuri/ Saket. I know what a transformation Sreedharan has brougth to Chandni Chowk. Great people are forgotten easily. The pamban bridge is a marvel of technology.Bangalore metro is trundling because Sreedharan has left! 164. Arun Gaur This is the first time I have read your article and I know what I have been missing all this while. I have admired the Metro Man but did not know the extent of his contributions. Thank you. Whether he gets the Bharat Ratna or not is actually a very trivial detail. Honestly, I think he should not because then, as you put it, he will figure in the not so enviable list. Another reason why I think he should not get the Ratna is because this sends a wrong signal. He is a professional who did wat was required to be done and did it mst efficiently. If I was the one overseeing him I would recommend a huge bonus- say a million US dollars. That just might be the “just” reward for what he has done. It will also send a message to all those who believe that it’s possible. PS: If a Bharat Ratna fr Sachin then why not Rajnikant? Or Helen? After all, they too ave entertained people just like Sachin does. Dear author, I seriously suggest you hook up with Rajiv Malhotra (http://infinityfoundation.com/people.shtml). His Infinity Foundation is working on producing 20 volumes on “Indian civilisation’s contributions to science and technology” along the lines of Joseph Needham’s thirty volumes on Chinese civilisation (http://en.wikipedia.org/wiki/Joseph_Needham). The foundation is looking to fund people who can do research into specific areas. Your passion and your willingness and ability to research topics suggest that you are an excellent candidate for this kind of work. Keep writing! 166. Azeem Words fail me in expressing my gratitude and admiration for the great man. All I can say and do is bow my head respectfully and pray that God bless our dear India with many more such great souls. 167. Drac Saber Brilliant article, well researched. In fact, I did some googling long back about Dr. Sreedharan when he took over Delhi Metro and came to know about his achievements very briefly but not to this detail. Don’t know if this is again not worth for many to list out his achievements but this actually makes an average Indian proud and motivated to keep trying changing our systems, fight back against the odds. At the same time, it also forces us to introspect and hang our heads in shame that we have a toothless, spineless and shameless system who would not honour their real soldiers but give away awards to their whims and fancies. I didn’t hear and wouldn’t hear any lobby or group batting for Sr Sreedharan for any award the way they were lobbying for Sachin on completing his 100th 100. No offence to Sachin but looking at Dr Sreedharan’s achievements, Sachin’s achievements look like a small shrub under a large banyan tree. Moreover, not belittling any one’s achievements, how have they brought any change to the lives of millions of us the way Dr Sreedharan has brought? Any thoughts? 168. Subbu The entire write up was enlightening. The final line takes the prize…..The Bharat Ratna does not deserve Dr. Sreedharan! Wow…… 169. Subbu Cricketers, politicians and Bollywood (Tolly/Kolly/Molly/Dolly wood etc.) deserve Bharat Ratnas…..because Ratnas are useful piece of carbon. True achievers like Dr. Sreedharan are priceless and should not be belittled with any silly awards. In a country where shrines are built for starlets, where is room for the extraordinarily ordinary men/women like Dr. Sreedharan etc.? 170. Have you contacted Sreedharan and told this!!!!??? He will be very happy… Thanks for sharing this info…. 🙂 I wanted to post this comment way before…. But its not too late…. 171. Siddhartha Fantastic. Though seems to be better suited for oral delivery due to the writing style. Nonethless, very captivating and of course, hats off to Dr. Sreedharan. 172. Anonymous It is an insult to Sreedharan that such a small title like Bharath Ratna is conferred on him for his accomplishments deserve much better and higher reward. 173. Novacaine Interesting article. Nice! Your story confuses the facts a little – Sreedharan was also the head of the “sorry” Kolkata metro. So the man made a good use of this previous and perhaps the only time he did not receive outstanding success. It might be due to several factors – like political will etc. etc. Just pointing out a fact. Cheers 174. Brilliant man! 175. rishabh47 Gazab sarkaar! 176. Sam V How true yet unfortunate! It is good samaritans like you who bring the hidden truth to the fore. For a man so selfless, Dr. Sreedharan has to be conferred an honor even above Bharat Ratna, lest we may discount his brilliant achievements! 177. Harry I would like to see Dr Varghese Kurian, Dr M S Swaminathan, Dr Vikram Sarabhai, Dr. H J Baba receiving Bharat Ratan 178. santhosh My commentings is that: I was just surprised that the konkan railway website, as far as I looked through, had no mention of this great genius. I had earlier read about him when a bridge fell in delhi during construction. 179. Loksakthi Once I met E.Sreedharan’s son in Muscat many years ago. But he never mentioned that(Son of E.Sreedharan) to me in the entire meeting or conversations that lasted about 2-3 hours at a dinner party. I came to know later on only. 180. k v vaidyalingam A superb write up.There are many unsung Heroes among us . 181. Bala But he has been awarded the Padma Vibhushan in 2008, so saying his achievements are not recognized is a REEEEAAAL stretch……whether he qualifies for Bharat Ratna is a different question altogether. The title phrase “….no one talks about” is deliberately misleading. 182. Rajesh K The same E.Sreedharan is telling Kerala government that Kochi Metro has to be given to DMRC otherwise he wouldn’t cooperate. He’s also demanding that there shouldn’t be global tenders for it. Isn’t it nepotism/favoritism? Meanwhile he and DMRC has taken up Kozhikode Monorail even though he had ridiculed that monorail is best suited for amusement parks. In 2006 he shot down the proposal of Metro in Trivandrum saying that undulating terrain is ill suited for a Metro. When consulted for Trivandrum Monorail, he’s said that Metro is suitable for Trivandrum. AFAIK the terrain of Trivandrum hasn’t changed in last 6 years. I feel there is no need for deification of E.Sreedharan, who is a great project manager who delivered the goods when given unlimited powers. p.s: This blog entry talks about E Sreedharan as well as DMRC a little bit. http://tvmrising.blogspot.in/2012/07/who-needs-world-class-portand-god-of_27.html 183. Pingback: Quora 184. While I admire Sreedharan ( I knew of him as the Delhi Metro man, did not know he was also responsible for Konkan Railway and Pamban bridge), and what he has done is phenomenal and deserves every kudos from the public, it does not even come close to what someone like Norman Borlaug accomplished. Perhaps we can give credit to a person like Verghese Kurien of the Milk revolution for something similar to Borlaug. More aptly someone like C Subramaniam, the food minister in Indira Gandhi’s cabinet, or MS Swaminathan, the agricultural scientist, who enabled Borlaug’s green revolution to work is a good comparison. Whatever Sreedharan did was special mainly because it was done in India with its innumerable blocks and that too by a bureaucrat. His contribution is organizational and logistical, but he had both Indian and foreign role models to follow, most of the engineering skill was already invented (I am sure there were several Indian ingenious innovations, by engineers, bankers, bureaucrats, politicians and low-level workers who never get credit). Much messier projects of far greater engineering complexity were accomplished in other countries – the Trans Siberian railway, the recent Beijing Tibet line, the introduction of container shipping, the Manhattan project, Sputnik & Apollo, Germany and Japan’s post war miracles, the English chunnel, America’s skyscrapers, the Panama and Suez canals, Isadore Kingdom Brunel’s bridge and ship building – you get the gist. Indian history from 1857 to 1947 is merely presented as the Sepoy Mutiny and the Congress party’s struggle against the British, with 2 world wars thrown in. This is entirely a political and legislative / administrative approach to history and is horseshit, as any engineer should know, if not the general public. The British built the Indian Railways entirely with imported material (rails, engines, carriages, wheels) brought by ships and docked into the modern harbours which also the British built. Only the labour was mostly Indian, and the wooden sleepers. They had to clear forests, quell tribes, and the logistical nightmare of several kingdoms and local idiots must have been severe. And they built the macadamized road network that India has, they built the electrical and metal working infrastructure, they built colleges so that they could teach, in a foreign language, ideas, technologies and techniques that were only invented one or two decades earlier!! They electrified India, built its postal system, and last but not least built an enormous number of bridges across rivers and ravines and tunneled through hills and rocks. A large number of educated Indians (including most of our ancestors) were in deep awe and unabashed gratitude to the wonders that the Englishmen brought practically every day (these were the marvels of engineering and science, no doubt, several of whose progenitors were German, French, Swedish, American, Russian etc. – but they could have kept us as backward as the Naxalite districts of today, please remember). It took brainwashing on a Gandhian scale and the destructive power of Hitler & the specter of Stalinist Communism to make the British leave Not Gandhi’s hunger strikes and khadi, as most of us would like to believe. Was Sreedharan’s work in any way comparable to what the British did before him (in India alone) or the other feats I have mentioned? Gopu ———————————– http://globalconnections.hsbc.com/united-kingdom/en/news-insight/the-world-suddenly-got-better-but-no-one-noticed • vijaiyaprathap The britishers did the things you mentioned not for indians but for themselves.. they built bridges, tunnels and roads so that they could travel to each part of india easily. so that cotton from interior places can reach the ports like mumbai, kolkotta, and chennai from there could go to england.. even with modern technology, britishers or anyone cannot build a structure like tanjore temple.. we have countries like japan, singapore, china etc.. which were not ruled by britishers were now super powers.. the only advantage we got because of britishers is that now we are fluent in english which is an international language.. • I am probably very late to this – but isn’t the fact that you are comparing one man’s work to the work of the ‘British’ already a mark of how phenomenal his contribution has been? • ramanath very true & wise comment…i watch Extreme Engineering shows, I understand the gist & also understand about the English contribution.. bt this Civil servant here is a true & passionate engineer, who loved his job obviously & had a great desire to accomplish things & went about and did it. This is undeniably a guy for young Indian engineers to look up to, be aware of..& a bit of spot light on his feats will do only good..Some of the better ones might take up Public work, May be its d general public who r nt aware & every Mech/Civil engineers are aware of the guy (or may be they only know of his US/European counterparts.) 185. Anonymous Amazing Article on Dr. Sreedharan!! A big salute to him for the wonderful achievements and the infrastructure he has given to India. 186. vijaiya prathap i salute Dr.Sreedharan. fantastically written by you kaipulla.. 🙂 187. Ramesh Raman Outstanding. There was hardly anything known about Sri Sreedharan. Your blog has been an eye opener. Thank you. – Ramesh 188. Amit M Hear! Hear! 189. Rahul yesudas deserve Bharataratna. 190. Manu why latamangeskar got Bharata ratna?, she don’t deserve that. 191. if there is a best blogger award, it shouldnt be given to you and you know the reason!! hats offf!! 192. Thank you for this write up, it’s good to recall the accomplishment of Dr S. for he is an example for all of us. One thing I would say, is that he has been blessed with an even greater award that the BS… I mean the BR award, that of serving Mother India in the True sense of the word. I personally would feel blessed to be able to contribute so much to this great nation. 193. aurovrata Thank you for this write up, it’s good to recall the accomplishment of Dr S. for he is an example for all of us. One thing I would say, is that he has been blessed with an even greater award that the BS… I mean the BR award, that of serving Mother India in the True sense of the word. I personally would feel blessed to be able to contribute so much to this great nation. 194. ShankarNarayanan The best part of the blog was ” If Delhi Metro had benchmarked Calcutta Metro which finished work at the rate of 0.76 km per year, Delhi Metro would have been completed in the year 2083.””” Amazing stuff…Am not able to control my laughter…Do you know that in WEST BENGAL there was a clerk in PF Office who settled 7 YES …JUST SEVEN CLAIMS OF RETIRED GOVT SERVANTS IN HIS ENTIRE EFFIN CAREER….Now put that in context against what Dr.Sreedharan has achieved, one will understand and appreciate. Hats of Dr Sreedharan and TO YOU for bringing some more facts apart from the Konkan Railway and the DMRC to me. I was not aware of the Dhanushkodi and Rameswaram bridge incident. I am sure even the f….in govt would have forgotten those achievements. Only thing I did not like in this article was displeasure over awarding Bharat Ratna to some musicians (who are incidently from North India). The comment “who had negligible impact on the population of India.” is hilarious and I doubt if gyri and sulci are prsent on this authors brain. He avoids mentioning the name of M S, who probably (according to him) is ONLY musician deserving Bharata Ratna. 196. Chaitanya Hope u had written the article without defiling other legends and there by belittling urself n ur write up … 197. #RESPECT! Together with an incredible insight into things, you also possess a great sense of-for lack of a better phrase-‘music of language’. Elegant and top notch! 198. Anonymous I don’t understand why this much noise about this man there many more civil servants in this country who are quietly doing their job even more efficiently. So you mean to say we are in a country where one should be awarded Bharat Ratna for one simple reason that He got paid and did his job I pitiable writing and I don’t understand your vested interest. Krv • Ashok Kohli Dear Sir, No I don’t mean to say one should be awarded Bharat Ratna for one simple reason that He got paid and did his job but for the selfless commitment to the country and differentiating them from the ordinary citizens. I think we may also institute some infamous awards for people who do not do their job, stop others from doing their job, bribe or force to bribe and the kind. People of infamous category should be debarred from contesting elections or holding any public post. And those already holding such posts, public must have to call them back or negative voting. Regards Ashok Kohli ________________________________ 199. Anonymous Respect 200. suverthee superb…this man deserves all the kudos 201. Abhinav Excellent work… Keep up the good work…. 202. I would like to term him as a real idol for peoples in the field of project development. 203. We are a bunch of volunteers and starting a new scheme in our community. Your site provided us with valuable information to work on. You’ve done a formidable job and our whole community can be thankful to you. 204. Mahesh I respect this person Dr.Shreedhran. Even though every fact was known to me still it was awesome to read it. Liked line planning commission only plans 😛 205. May I simply just say what a relief to find someone that really knows what they’re discussing on the internet. You actually know how to bring an issue to light and make it important. More and more people really need to look at this and understand this side of your story. It’s surprising you are not more popular because 206. titus Sir, what a brilliant article….Dr. Sreedharan really deserves to be honored with Bharath Ratna…Now he is in charge of the ambitious Kochi metro initiative by Kerala government and he ias asked to complete the same with in 1000 days….. 207. I knew about his Konkan rail and Deelhi Metro and later ones and that He was INVITED to Pakistan as a consultant and HE REFUSED to go……NEVER knew that it all started from DHAUSHKODI…though in my childhood have traveeled and marvelled that BRIDGE !!!!…and I AM A NORTH INDIAN too !!!!! from Delhi…Thanks SRIDHARAN for being what u ARE !!!!! U deserve a Bharat Ratna and a NOBLE if There is ONE…U RULE in Hearts of People…EVERYONE recognises the SUPERHUMAN efforts put in…..People like us do have antics for Madrasis BUT we RECOGNISE the TALENT too !!!!! 3 CHEERS !!!!! 208. Murali Dear KS, Excellent article. So, googled Bharat Ratna this is what I got , and it goes on like ” Bharat Ratna is the Republic of India’s highest civilian award, for performance of highest order in any field of human endeavour “. Now can somebody tell me what human endeavour has Sachin done ? So before some body comes and stab me, please also google human endeavour. PS No disrespect to Sachin and I am a big.fan ,but the in-appropriation of giving the awards makes go bonkers and I wonder why are Indian people so emotional and hence stereo-typically justified. Peace . 209. Dear KS, Excellent article. So, googled Bharat Ratna this is what I got , and it goes on like ” Bharat Ratna is the Republic of India’s highest civilian award, for performance of highest order in any field of human endeavour “. Now can somebody tell me what human endeavour has Sachin done ? So before some body comes and stab me, please also google human endeavour. PS No disrespect to Sachin and I am a big.fan, but the in-appropriation of giving the awards makes go bonkers and I wonder why are Indian people so emotional and hence stereo-typically justified. Peace 210. malik Really nice article. I liked the information as well as style of writing with all the dry humour. Keep up the good work! 211. Bharath Amazing article.Thanks for making world aware of this great man.! • Bharath *the world 🙂 212. A great write-up. You have succeeded in covering all subjects concerning Mr. Sreedharan. FYI. He is presently at the helm of affairs of Kochi Metro Rail project. I wish him all the very best in his new mission. 213. Anonymous Amazing article.Need mentors like Sreedharan for young India. In a country like India where there are 1 billion people…. one or two bharath ratna’s is not important nor arguing who should get it matters…..as long as true stories of immaculate and passionate people keeps inspiring the youth to go and achieve the impossible. 214. Swetachalapathi Sonti One can appreciate the tremendous work of Dr. Sreedharan but why try to denigrate others? Just because you do not like the musicians? How can you say that these people have not touched the lives of the millions of Indians and gave them great pleasure? Coming to your anti -virus Sales man, yes, I agree that the award to him is too premature. But did he ask for it? It is those millions of fans who were clamouring for it and most of them are Indian citizens. So, how giving a Bharat ratna to a person so loved by millions of Indians needs condemnation? 215. Raghu Vamshee M Excellent. Just excellent. And even more Excellent. And just mentioning- The day this great legend died, the next day this was not the news on the opening page of TOI. It was a silly Sonia Gandhi’s and that Priyanka’s funny picture..(where one is playing with the other’s cheek) Pathetic. I am happy atleast there are people like him. I salute. 216. saisurya Simply superb…I consider you my greatest blogger. This should be ready by every kid, to understand why India is ‘still’ developing. 217. Rajendra Sharma There are so many ratnas in bharat who were ignored and completely forgotten simply for political gains:(:( 218. Incredible write up this. The name is vaguely familiar to me in connection with the Konkan railway project. I knew that Konkan railway was a phenomenal masterpiece but did not realize one man played such a big part in it. His other achievements are superhuman as well, but I haven’t heard about them. Unfortunately, Bharat Ratna award is usually based on hype, politics, mass appeal and popularity and not based on merit. There is a nobel laureate ( the only fourth Indian ever to be awarded Nobel for Science) Dr.Venkatraman Ramakrishnan, who isn’t being considered for Bharat Ratna. Our country does not value real merit. 219. Some facts: Mr. Nerhu & Mrs. G self-anointed themselves with that ratna stuff in 1971 Mr. beta G got that at the same time of Sardar Patel Be rest assured Soina G and pappu G will also receive it sooner than later That ratna stuff or whateva is being discussed is not even worth a discussion man! what a waste of time…yikes! E Sreedharan has done something really great, hope he his not demeaned by this award, seeing Sachin get it was painful enough! p.s. that ratna was withdrawn for Subhash Bose (a PIL saved his name from being tarnished), and yeah please bomb all the IITs and IIMs, replace those with Amity & Lovely universities! 220. K.M.Narendran E.Sreedharan, DR.M.Balamuralikrishna, Viswanathan Anand: all of them deserve Bharath Rathna. 221. Anonymous dear sir An excellent article. Kudos to that Great Man who has done much to our country.Living example for simplicity makes history by getting the Highest Award in India 222. Manasmitha Nyamagouda Dr Sreedharan should not be given the Bharat Ratna. Because the Bharat Ratna does not deserve a man like Dr Sreedharan. you summed it up. Next time, before I use the word Mallu, for a malayalee, I’ll remember E Sreedharan. 223. Pingback: Quora 224. srmenon The most renowned Technoscientist the country has ever produced is the Hon’ble E Sreedharan. Though this miracle worker deserved the Bharat Ratna long back even for completing the Pamban Bridge before time. Now I think he is beyond any award. But still I will pray for him to be awarded with Bharat Ratna for his unique contribution to our Country. He is even ready to complete the High Speed Rail project in India if it entrusted to him. And I think the present Government which is pro-development must consider him for the top post for this long pending project which youth of this country also wants to have high speed rail network throughout India before it is too late. Even I had earlier also commented for this renowned technoscientist for the prestigious Bharat Ratna award when the French Government can give the prestigious award of their country to this Metro Man. 225. This paragraph will assist the internet users for creating new web site or even a blog from start to end. 226. Outstanding quest there. What occurred after? Good luck! 227. They feature with the amount of features in which you will get much options for achieving much more quantity of IMVU credits for not shelling out any cash into it. It is much convenient to use the IMVU credits that you simply make in PrizeRebel. You should have a merchant account in this particular internet site and you can gain IMVU credits for any things to do one does and whenever you attain the level of points you could redeem, you are able to redeem it through winning prize segment. You should then log into IMVU and access the page for prepaid greeting cards just where you have to enter in the credit card code to be able to work with the credits you gained in IMVU. 228. What a bloody brilliant write-up on a superhero. Awesome stuff. Wow! And of course, the Metro Man is a living legend. 229. Rajan There is a small plaque in “Chawry Bazar” station on the Delhi Metro. I suggest every body reads it. Yes!!! Bharat Ratna does not deserve E.Sreedharan. How well said. 230. Dr Sridharan is a civil engineer by training and a brilliant manager. I am not sure that he is a ship designer by training. So the statement that he “designed, built and commissioned India’s first indigenously built Merchant vessel, the Rani Padmini, in 1981” needs to be verified. Also, you should avoid the temptation to pull down other people just to make the person of your article look good. It detracts from an otherwise wonderful writeup and leaves a bad taste in the mouth. 231. Anonymous I had known of Mr. Sreedharan only as a Metro man. This article has been an eye opener. It is amazing to to read about a man who has done so much. It is a shameful that we don’t recognize or highlight his acts. `Thank you for writing about him. 232. Ruhi Dude seriously, where do you research your stuff from..You are awesome..I just saw your blog by chance and now am hooked…but where are the most recent posts? I’ve been reading posts from 2011-2012. N hats off..carry on the good work! 233. Rakesh Vaisoha I’d like Dr. E. Sreedharan to be the next President of India. 234. Tanay Rajpurohit Brilliant write-up. Dr. E Sreedharan is one of the most remarkable officers India has ever had. He puts the best corporates and their efficiencies to shame. A devout karma-yogi, Dr. E is a huge inspiration. I wish more people knew about the achievements and contributions (to India) of this great man. 235. Really a ROLE MODEL ENGINEER for the upcoming new Engineering Graduates 236. Hi blogger, i’ve been reading your posts for some time and I really like coming back here. I can see that you probably don’t make money on your website. I know one cool method of earning money, I think you will like it. Search google for: dracko’s tricks	{}
# Realism of a multi-arrow bow You may have heard of this common fantasy trope, an archer firing multiple (usually 3) arrows in a single pull of the bow. Now, I am not an expert at bows or tactics, but it seems to me that the loss of accuracy is not worth even attempting this and that's assuming it could work! What problems appear with this 'Multi-arrow' bow and how can I overcome these? What are the actual realistic advantages and disadvantages of a multi-arrow bow if it could work? • Having tried this before...yeah your assumption is correct. I am no master yeoman but its a dumb idea. – James Jan 14 '16 at 6:20 • Not an answer: I am an archer, and have fired two arrows, both nocked next to each other, and they landed next to each other, if this helps at all. It required a lot of patience and bad posture (leveling out the bow a bit to the side so the arrows could rest on it, etc). It was on a modern recurve bow (similar-ish to what you see in movies). – Mikey Jan 23 '16 at 3:10 • @Mikey That must have been cool, I wish I could have seen it – TrEs-2b Jan 23 '16 at 21:58 • @Mikey, I'm an archer too and can confirm what you say: there is no accuracy or aiming problem (with two arrows, I've never tried more). The arrows fly a near identical path, right next to each other. – Jacco Dec 9 '16 at 20:46 • As someone with an invested interest in Ancient warfare, this topic caught my eye so I couldn't resist sharing that there is a very intriguing crosspost from history.stackexchange titled "Are there any historical sources that support the claim that ancient high-speed archers held multiple arrows in their hand?" that might provide further insight into what many perceive is an unrealistic topic, but there is definitely historical basis to this question! The post has – SanDiegoBookReview Jan 6 '17 at 2:12 Your main problem is that you're trying to fling multiple projectiles with a single bowstring. I'm no physics guru, but it seems that you're going to be dividing the pounds of pull across the total number of projectiles -- so you'll get correspondingly less range & target penetration than if you stuck with a single arrow (ignoring more obvious issues like aiming). Someone actually ran the numbers on how much kinetic energy from the bowstring is transferred to the arrow, if you're interested. A crossbow is a much more workable solution, as the strings can be drawn in advance, and held in place. That leads to the possibility of firing more than one projectile, in quick succession. Such weapons are indeed a trope, usually because the increased firepower is desirable from an action standpoint, but the limitations of the time period will not allow firearms. One solution to firing multiple arrows is a special crossbow, featured in a battle scene in the 2000 film The Gladiator. On first glance, it looks rather ridiculous, but I actually feel it is a workable design (albeit perhaps not terribly practical for anything other than a showy arena skirmish) for the simple reason that each crossbow bolt has its own string, rather than some sort of imaginary bolt "magazine" that ignores the necessity of re-cocking the string after each firing. Additionally, it dispenses with any sort of fanciful common trigger system, and appears to have the wielder just rotate the entire weapon on its axis. A simpler version would be the double-sided crossbow, from the 2000 film Dracula: • actualy rapid fire crossbows where a thing in real history: en.wikipedia.org/wiki/Repeating_crossbow Additionaly: while it is true that each arrow would have less force if you fire multiple from one string this might not be a problem depending on what you shoot at...penetrating a medival plate armor? bad. shooting at unarmored peasants? still more than enough force. – m.fuss Jan 6 '17 at 15:12 • Even then, I suspect that you wouldn't be dividing by 2; I suspect that only a fraction of the stored energy in the bow goes into propelling the arrow: specifically, the amount of energy that changes how fast it straightens up, between dry-firing a bow and firing it with an arrow in... which is not a lot. Seems to me, that fraction should rise as you add more mass to push out the way to straighten the bow; the bow might even straighten noticeably slower if firing a hundred arrows. – Dewi Morgan Oct 17 '17 at 14:12 ## Accuracy Let me expand on "reduced accuracy". I base this on my own experience shooting target-weight bows and observing others. Accuracy at any decent range depends on some fairly fine-grained factors. Aim a fraction of an inch high or low, or draw the string back not quite as far as usual (or just a bit farther), and you're probably going to miss your mark. You might still hit a large target somewhere, but it won't be what you were aiming for. It is not uncommon to mark the center-point of the string for this reason, so that you're nocking the arrow at the same position (and thus drawing at the same angle with respect to the bow) each time. The key to good aim is consistency. If you add arrows to the string in a single draw, then by definition those arrows are not positioned to hit your target. Movies usually depict multiple arrows being spaced an inch or so apart on the string, maybe more. And if those arrows have fletching (feathers), you probably can't get them much closer than half a inch on the string. (For any decently-heavy bow, the arrow itself is about a quarter-inch in diameter.) That's a huge difference when shooting at even 30 yards, to say nothing of 50 or 100 yards. Those extra arrows may look pretty, but they aren't hitting your target. I suppose if you're shooting at a large charging army they might hit somebody else, if you're very lucky, but probably not (see "power"). ## Power Other answers have already addressed the reduced power so I won't repeat them. Your arrows aren't going to pack the punch you need to do damage. They're also not all going to get the same amount of power; the one in the middle will have more force than the others, which will likely fall short. (Thanks to XandarTheZenon for pointing this out in a comment.) ## Practicality A competent archer can fire an aimed shot about once every five seconds; good ones are even faster. A lot of this speed comes from the ease of the load-draw-release cycle. It's a very smooth motion; you draw an arrow from the quiver by its nock and place it on the string. You don't even need to be looking at it; this is done by feel. (Instead you're looking at your target.) Now, how long is it going to take you to load three arrows onto the bowstring? I don't think you can do them all at once, and if you do them one at a time then the ones already on the string are going to slow you down. I haven't tried the experiment, but I'm going to be bold and say that it will take you longer to load and fire your trio of arrows than it would take to fire them individually. So you can shoot less-accurate under-powered arrows more slowly, or you can shoot them one at a time instead. I know which I'd choose. • I think that this sums up what would happen much better than what I said. – Xandar The Zenon Jan 14 '16 at 3:12 • @XandarTheZenon your point about the angles is interesting; I'd been assuming three "shelves" on the bow so the arrows would be parallel (though two are not at your draw length), but if you're trying to fire them all off the same shelf, no that's definitely not going to work. Also, sorry, didn't mean to ninja you there -- I hadn't seen your answer before writing mine. – Monica Cellio Jan 14 '16 at 3:18 • Don't worry about it, I admire those who can write a one page essay on a random topic and make it neat and sensible. Something I don't think has been addressed here is that the amount of force exerted on an arrow would be less for the bottom and top and more for the middle. The point with the most power would be where you pull back the bow, and then I'm pretty sure the arrows' power would exponentially decrease as you move away from it. After all, there's a reason bows were designed the way they were. For a single arrow in the center. – Xandar The Zenon Jan 14 '16 at 3:25 • I proffer my ideas that other my profit from them. I actually enjoy seeing people use my ideas, or people with similar ones. – Xandar The Zenon Jan 14 '16 at 3:51 • From my experience as a recurve and longbow archer I can wholly agree with this answer. – fgysin reinstate Monica Jan 14 '16 at 14:21 I use a recurve and longbow. Getting good at this took practice, like a lot of practice. First. There are some very expert opinions out there on why this is fake and or a bad idea but there are some problems with that. Most people aren't thinking of the bow as a weapon of war. They are thinking of it as a hunting/sport weapon. They are thinking of compound and recurve bows where it is pretty much a given that you are going to be using bow hand draw and not arrow hand draw. Second, the physics around arrow flight are really REALLY complicated. So complicated that for a long time people thought bows were an impossible mechanism leading to the coining of the "archers paradox" now I'm not going to get into the specifics but suffice it to say that the loss of acceleration is correct but not as dramatic nor as important as you'd think in the context of a battle. third thing is, Arrow hand draw is important which means owning a bow that allows you to do this is important, you can get a longbow, and if you are interested in them I highly reccommend it, or you can just get a recurve for the opposite of your dominant hand and sand the grip to shape it a bit. Professional olympic archers can fire an arrow maybe every once every 3-5 seconds an ancient archer could fire an arrow about once every second. The reason is arrow hand draw lets you rest, nock, and draw the arrow as a single motion rather than maneuvering the clumsy bow hand draw. This is going to help you keep the top arrow from going off course or worse, interfering with the flight of your other arrow. fourth thing, in sports and hunting accuracy is important, in battle they are important but the MOST important thing is how many arrows you can put down range within a specific amount of time. It may not surprise you to learn that people really don't like getting shot with an arrow, so the less of a break in the screaming bodkin-pointed death missiles there was, the less of an opening you presented. Also the average range for an archer to be shooting at individual targets in battle is 170-200 feet at most otherwise they would just be firing into the mass of people hoping to hit. So you don't actually have to get it that far. So to pull off the famed "multishot" you need to knock two arrows, I know theres a lot of ways to do that and some people use a mechanism to aid them in having a clean release. I've personally found this is the worst way to do this trick so I would avoid it. The method I've always used is one finger above the nock and two below but for this you are going to make an "arrow sandwich" with one below the first arrow one in the middle and one on top of the second. and you need to grip the bow so that your index (pointer) finger can extend and the tip can rest between the two. If you did it right both the arrows should be perfectly straight. Now here's the part that's going to take practice. There are two things i've found that generally make this trick work reliably. First, and most important, once you release the string your index finger needs to slide out from between the arrows, otherwise the hens (the two matching coloured flights) are going to sting it on the way past, your instinct is going to be to curl it back in but what usually happpens when you do that is it deflects the bottom arrow down. The best thing to do is to flick it straight out so it's out of the way, this may deflect the top arrow up a bit but over a longer distance it will drop back down and you should still get a pretty accurate shot. Second, draw the bow with only your middle finger, keep rour ring and index finger extended to stabilize the arrows but you should be pulling from the middle, the reason for this is, while the bow will deliver all of its force across the entire string it delivers the greatest acceleration from the point you drew it back, for maximum accuracy this should be directly between the two arrows. ## Reduced Accuracy As you mentioned, the accuracy of all of the arrows would be significantly degraded. ## Reduced Power A bow transfers energy from the bent wood/materials of the bow into an arrow through its bowstring. The amount of momentum imparted depends upon a number of factors but by adding multiple arrows to the bow string, you are at the least dividing the momentum by the number of arrows added. Decreased momentum means: 1. Lower range 2. Less penetrating power ## Historically Speaking I suppose the best argument about whether you gain or lose more by adding arrows to the bow string is by looking through history. The trick is an obvious one so if the people who did it greatly benefited from it, you'd expect everyone to be doing it. ## Limiting Factors The limiting factors of arrow acceleration are: • The mass to be propelled, for example the arrow(s) - $m_{arrows}$ • The mass of the limbs of the bow - $m_{limbs}$ Also there's a scaling factor, in that the firing of the bow causes the bowstring (and arrows) to pass through the entire draw length (perhaps 24-30 inches). Meanwhile the limbs of the bow (which are forcing the arrow through that distance) may only pass through a few inches. This distance might be 8-10 inches or so (see image below). Bow drawn and at rest: Because the ratio of motion between the arrow and bow limbs are not constant you actually get a relatively complex interaction between the two. Initially the force propelling arrow and limbs are high. As the string returns to rest, the amount for force decreases while the amount of motion increases. Even if you set the arrow mass at zero (no arrow notched), the bow will take time to return to rest state (FYI, never draw and release a bow with no arrow notched, this can break your bow) because the bow limbs have finite mass. • I'll have to insert a cautionary note here concerning power. I suspect that stepping up to 3 arrows will not cut arrow speed by a factor of 3, although it will have some effect. Clearly arrow resistance is not the only factor limiting bowstring velocity. If this were true, then firing an arrow with extremely low mass would provide extremely high velocity, and this is not true. – WhatRoughBeast Jan 14 '16 at 3:36 • @WhatRoughBeast, the momentum of the bow's arms plays a roll too. Because the bow arms apply the force, the total acceleration/max velocity that can be imparted is limited to speed at which the bow's arms can return to their normal position. Shoot too much mass are you're limited to the acceleration you can apply to the arrow. Shoot too little mass and you're limited to the acceleration of the bow's arms. You get the lesser of the two numbers. – Jim2B Jan 14 '16 at 4:55 Actually, it is quite possible to fire THREE arrows with a single draw. I first saw this technique in the Hindi movie, Bahubali (2), The Conclusion. Google for images and relevant yt clips. Now, while I'm sure there was liberal application of Movie Magic, especially when the hero shot past the gal's head—two past her ears, one above her crown…I've not been able to figure out how to bump one of the arrows into a higher/wider flight—I've nonetheless been able to duplicate the draw if not the loading speed and fluency. Note the draw hand's inverted grip with thumb-down, palm-away. Inverting is not necessary, but is definitely a conversation starter on ranges. When the arrows are parallel, their grouping is quite tight. I've yet to get a spread triple hit by putting angles between the shafts. But instead of me telling y'all about it, I recently found this: https://youtu.be/l6HdEqOpgzE • Please don't answer with only a link. Answers should be written in a way that will make them usable even if all links expire. Without seeing a movie or youtube clip, your answer makes no sense now. – Mołot Oct 16 '17 at 9:00 • Welcome to Worldbuilding! As the above two comments rightly say, you've answered the question in a youtube video. What the above comments fail to say is that this is actually great as it's a direct demonstration of what the question is looking for as opposed to some explanation with no back-up. However, it would be very useful if you could edit your answer to include the relevant points of the video as well in case the link stops working. Thanks! – Mithrandir24601 Oct 16 '17 at 9:56 • I think this looks okay already, but it would definitely be better if you could edit your answer to tell people what can be seen in the video. Not everyone can watch videos every time and we try to give readers all the information necessary directly. Links of all sorts can get outdated, which would leave this answer somewhat lacking. If you have a moment please take the tour and visit the help center to learn more about the site. Have fun Love Robin Miller! – Sec SE - clear Monica's name Oct 16 '17 at 10:28 This is a very interesting idea and one that I would like to look into. One factor that I would like to point out, is that all of the bows made today are for firing one arrow. so trying to fire multiple arrows will not work. If someone were to make a bow that was specifically designed for multiple arrows, the Idea might actually work. You could also work out a way for the arrows to shoot in such a way that they would fire at a wider angle and thus prove an actually useful point. New contributor Another curious archer is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.	{}
Spring constant 1. Sep 20, 2009 stpmmaths 1. The problem statement, all variables and given/known data A spring with spring constant 8 N m-1 is cut into 2 equal parts. Find the spring constant of each part? 2. Relevant equations 3. The attempt at a solution Spring constant of each part = 16 N m-1 1. Why should we times 2? 2. If the spring is cut into a ratio of 1:2:3, then what should be the spring constant of each parts? 2. Sep 20, 2009 rl.bhat Force constant is force required to produce unit extension. In the spring for a given load, extension is proportional to the length of the spring. So when the length reduces, extension reduces and k increases. 3. Sep 20, 2009 inhere Hello. Consider the spring to be like two capacitors. 1/keq = 1/k + 1/k'. So in this case we consider the entire spring with keq to be the sum of two individual spring halves that have some other spring constant k where k = k' for this problem. Doing a bit of algebra shows that k = 2*keq.[ 4. Sep 20, 2009 stpmmaths Thanks a lot.	{}
# Whay is the form of eventual total energy The total energy of the ball is the sum of kinetic energy and potential energy, kinetic energy depends upon the velocity of the body, potential energy depends upon the position of the ball (height from the surface of the earth in gravitational potential energy), they keep varying but their sum i.e. total energy always remains constant. Now eventually initially ball was at rest at a height h above the surface of the earth so form of energy was potential energy now as we know the Relation for Potential energy of an object of mass m due to earth’s gravity (g is acceleration due to gravity) held at a height h above ground is • 0 What are you looking for?	{}
In an urban area, a median is provided to separate the opposing stream of traffic. As per $IRC:86-1983$, the desirable minimum width (in $m$, expressed as integer) of the median, is _______.	{}
GR 8677927796770177 \| # Login \| Register GR9677 #92 Problem GREPhysics.NET Official Solution Alternate Solutions This problem is still being typed. Mechanics$\Rightarrow$}Potential Given $V(x)=-ax^2+bx^4$, one can find the minimum by taking the first derivative (second derivative test indicates concave up), $V'(x) =\left( -2ax+4bx^3\right)_{x_0}=0 \Rightarrow x_0=\sqrt{\frac{a}{2b}}$. The force is given by $F=-dV/dx=2ax-4bx^3$. The angular frequency of small oscilations about the minimum can be found from, $\begin{eqnarray} F(x-x_0)&=&2a(x-x_0)-4b(x-x_0)^3\\ &\approx& 2ax-4b(3x_0^2x) + O(2)\\ &=&(2a-12bx_0^2)x\\ m\ddot{x}&=&-4ax\\ \ddot{x}&=&-\omega^2 x \Rightarrow \omega^2 = \frac{4a}{m} \end{eqnarray}$ where one might recall the binomial theorem or pascal's triangle to quickly figure out the trinomial coefficients. One finds that $\omega =2\sqrt{\frac{a}{m}}$, as in choice (D). Alternate Solutions walczyk2011-04-03 02:05:24 Hi all, So I didn't want to do it using taylor expansion so I instead used a change in variables after finding the minimum. $u=x-\sqrt{\frac{a}{2b}}$ After plugging in and rearranging all the terms you get a pretty attractive potential function, obviously the new minimum is at $u=0$. $V(u)=bu^4+2\sqrt{ab}u^3+2au^2-\frac{a^2}{4b}$ Recall $F=-\frac{dV}{du}$. So $F=-4bu^3-6\sqrt{ab}u^2-4au$. Remember for very small oscillations $u \ll 1$, so higher order terms can be ignored for the ease of this problem. $F=m\ddot{x}$ and you undoubtedly know hooke's law, so $m\ddot{x}=-kx$ and $\omega^2=\frac{k}{m}$. Unfortunately you cannot drop the quartic term after finding the minimum. If you work through the change of variables you find that the quartic term expands to provide a term $3au^2$ and the quadratic term contributes a $-au^2$, which is necessary for the right angular frequency. There may be a better answer though, if there is reasoning that there must be a $\sqrt{\frac{a}{m}}$ term. That's all folks! Reply to this comment munster2009-10-06 19:13:47 First of all, you can use dimensional analysis to rule out options (a) and (c). Second of all, we know, without question, that the potential wells caused by the $bx^4$ perturbation will be sharper (or deeper, however you like to look at it) than a well described purely by $ax^2$ (keep in mind here that the $x^2$ term is NEGATIVE, which means it by itself would not be a potential well, which is why the statement above works). We can see that if the potential were $V(x)=ax^2$ then $\omega=\sqrt{\frac{2a}{m}}$, which rules out (e) because it is too small. (b) doesn't make sense because having a $\pi$ in the answer would never come up in the proper method (Taylor expansion, etc). This leaves us with (d). Reply to this comment jmason862009-09-28 18:28:03 I did this problem using test taking strategies, although I'm not sure my thoughts were sound: Since ETS says "for small oscillations" then the $x^2$ term will dominate. This means that the constant $b$ should not be in your answer. Eliminate (A) and (C). $\pi$ should only become involved if we were asked for period or normal frequency. Eliminate (B). Getting the frequency is probably going to involve taking a derivative at some point which will bring that 2 down from the $ax^2$. You'll end up with a 2 in your numerator. Eliminate (E). (D) is left. Reply to this comment ETScustomer 2017-10-05 01:52:54 It\\\\\\\'s such a twist that there\\\\\\\'s no $b$ in the answer! Yet, the $b$ matters ($b=0$ would give a square root of two rather than a square root of four). Voltsmann 2011-10-12 21:04:53 I think a lot of people have given basically this solution. But I had some trouble understanding it, so maybe my take on it will help someone. We have the functional form of the potential, V(x). We can find the minima via the usual method: taking the derivative and setting it to zero to find x0 s.t. (dV/dx)(x0)=0 (and checking that the second derivative is positive at x0). Then, since we're interested in small oscillations around xo, we can Taylor expand! We have V(dx)=V(x0)+V'(x0)dx+V''(x0)dx^2/2+O(dx^3), where dx = x-x0. Now we can throw out the first term, because it's a constant, so has no effect on the motion. The second term is zero, because x0 was determined by setting V'(x0) = 0. Then we're left with V(dx) = (1/2)V''(x0)dx^2 = 2adx^2. Now the force is F = -dV/dx = -4adx. Now we have a Hooke's Law force with k = 4a. The frequency of oscillation is then $\sqrt{k/m}$ = 2$\sqrt{a/m}$, which is (D). walczyk 2011-04-03 02:05:24 Hi all, So I didn't want to do it using taylor expansion so I instead used a change in variables after finding the minimum. $u=x-\sqrt{\frac{a}{2b}}$ After plugging in and rearranging all the terms you get a pretty attractive potential function, obviously the new minimum is at $u=0$. $V(u)=bu^4+2\sqrt{ab}u^3+2au^2-\frac{a^2}{4b}$ Recall $F=-\frac{dV}{du}$. So $F=-4bu^3-6\sqrt{ab}u^2-4au$. Remember for very small oscillations $u \ll 1$, so higher order terms can be ignored for the ease of this problem. $F=m\ddot{x}$ and you undoubtedly know hooke's law, so $m\ddot{x}=-kx$ and $\omega^2=\frac{k}{m}$. Unfortunately you cannot drop the quartic term after finding the minimum. If you work through the change of variables you find that the quartic term expands to provide a term $3au^2$ and the quadratic term contributes a $-au^2$, which is necessary for the right angular frequency. There may be a better answer though, if there is reasoning that there must be a $\sqrt{\frac{a}{m}}$ term. That's all folks! munster 2009-10-06 19:13:47 First of all, you can use dimensional analysis to rule out options (a) and (c). Second of all, we know, without question, that the potential wells caused by the $bx^4$ perturbation will be sharper (or deeper, however you like to look at it) than a well described purely by $ax^2$ (keep in mind here that the $x^2$ term is NEGATIVE, which means it by itself would not be a potential well, which is why the statement above works). We can see that if the potential were $V(x)=ax^2$ then $\omega=\sqrt{\frac{2a}{m}}$, which rules out (e) because it is too small. (b) doesn't make sense because having a $\pi$ in the answer would never come up in the proper method (Taylor expansion, etc). This leaves us with (d). flyboy6212010-11-05 23:21:29 Well done! jmason86 2009-09-28 18:28:03 I did this problem using test taking strategies, although I'm not sure my thoughts were sound: Since ETS says "for small oscillations" then the $x^2$ term will dominate. This means that the constant $b$ should not be in your answer. Eliminate (A) and (C). $\pi$ should only become involved if we were asked for period or normal frequency. Eliminate (B). Getting the frequency is probably going to involve taking a derivative at some point which will bring that 2 down from the $ax^2$. You'll end up with a 2 in your numerator. Eliminate (E). (D) is left. anmuhich 2009-04-02 11:58:57 I used a really simple way of thinking about this to get the answer in about 20 seconds. Just remember that the potential for a simple harmonic oscillator is 1/2kx^2. Since the question says small oscillations around the minima, you can just take the central minimum, which for small oscillations looks a lot like a potential for the SHO, and around which the x^4 term will not contribute hardly. Angular frequency for this SHO is (k/m)^1/2 . But since our original potential is twice that of the potential for the above SHO you know the angular frequency will be greater than (k/m)^1/2 . The extra pi factor for B doesn't really make any sense so the only answer that fits is D. furlong2009-08-06 20:39:13 how do you know a is greater than 1/2? If a is less than 1/2, then the original potential will be less than 1/2kx^2. in other words, did you just assume that a was 1 or am I confused? plapas 2009-04-01 13:31:11 In this type of exercise, the safest way to solve is the following: (1) Set dV/dx = 0 and find the roots. (2) The frequency of small oscillations can be found by the second-derivative term of the taylor expansion, i.e. (1/2) d/dx(dV/dx) (at min) = (1/2)m ω^2 and through this relation one can find ω. The method proposed in the solution is good enough, but one can make a lot of careless errors since the time available for the question is restricted. mudder2009-09-28 01:43:22 This is good. Solving for ω we see that in general, Set V'(x_o) = 0 then ω = (V''(x_0)/m)^(1/2) Bam! There you go. Simple way of solving SHO problems theodiggers 2007-10-19 19:58:27 You can also find the minimum, shift the origin of the potential by the minimum so its now centered at zero, then just generate the x^2 term in a Taylor expansion of your shifted potential and get the force from there. theodiggers2007-10-19 20:10:42 But that triangle expansion method is some sick shit sawtooth2007-10-30 07:47:35 I agree, I prefer finding the minimum $x_0$, Taylor expanding around $x_0$: $V(x-x_0)=V(x_0)-V'(x_0)(x-x_0)+V''(x_0)\frac{(x-x_0)^2}{2}+...$ dropping everything except the coefficient of $x^2$ creating something of the form: $V(\epsilon)=V(x-x_0)=...+\frac{1}{2}(4\alpha)x^2+... = \frac{1}{2}k x^2$ and we remember that usually $\omega=\sqrt{\frac{k}{m}}$. sawtooth2007-10-30 07:52:22 Ah, and ofcourse, since $x_0$ is a minimum $V'(x_0)=0$ as we have already used, so no trouble in the expansion, and simirarly, we already have the second derivative (we made sure that it was >0 so its a minimum). To be more precise, btw, I should have had $=\frac{1}{2} 4\alpha \epsilon^2 = \frac{1}{2}k \epsilon^2$ since this is the oscillating quantity... I think! carlosoctavius2008-08-28 07:48:24 How are you getting $\(1/2)4a(x-x0)^2$? From the Taylor expansion of V(x) we should have $\(1/2)2a(x-x0)^2$ Which leads to the wrong answer :( TheXDestroyer 2007-09-29 22:42:05 Can anyone explain how did we move from F(x-x0) to mx''=-4ax Thanks antithesis2007-10-05 07:19:50 Plug in the value you found for $x_0$ into the line with $2a-12bx^2$, you get $-4a$ cyberdeathreaper 2007-02-18 19:50:59 I agree - I don't see how this approximation is done. Can someone expand on the steps? alpha2007-03-30 22:29:13 As mentioned in solution, the approximation is done by using the binomial theorem or Pascal's Triangle, then dropping higher order terms. (very quick; for 3rd power, it is the 3rd row of Pascal's Triangle with coefficients 1 3 3 1, or $(a+b)^3 = a^3 + 3a^2b + 3 ab^2 + b^3$ ... in the approximation, higher powers of x are dropped out, so we end up with $\approx 2ax-12bxx_0^2$) naama99 2006-11-15 16:10:27 Can someone please elaborate a little on how to do the approcimation? Thanks. alpha 2005-11-07 14:23:08 kolndom, isn't that what's already there? kolndom 2005-11-07 07:28:01 Hi, how can a force F=2adx leads to small oscillation? The actual forced experienced by the particle can be given as following: F=2ax-4bx^3 =2a(L+dx)-4b(L+dx)^3 =2aL(1+dx/L)-4bL^3(1+dx/L)^3 =2aL-4bL^3+2adx-4bL^33dx/L =2adx-12bL^2dx =2adx-12b(a/2b)dx =2adx-6adx =-4adx :) Anastomosis2008-04-11 13:12:36 Mmmm LaTeX: $F=2ax-4bx^3$ $=2a(L+dx)-4b(L+dx)^3$ $=2aL(1+dx/L)-4bL^3(1+dx/L)^3$ $=2aL-4bL^3+2adx-4bL^33dx/L$ $=2adx-12bL^2dx$ $=2adx-12b(a/2b)dx$ $=2adx-6adx$ $=-4a*dx$ LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$	{}
# Drawing a 1 pixel thick line in Unity I'm trying to get a cross at the center of the screen (Two lines will do as well). I have a 2048x2048 Texture which I have thrown into a canvas. Filtering is set to point. The image is set to be in the middle of the screen no matter the resolution. The image is set to 2048x2048 no matter the screen resolution which works as well. However when I change the Screen Resolution I get results like the following, no matter what I do, the line refuses to stay at the same thickness. When I enable Bilinear filtering the lines constantly fade and thicken. This is what I get when I use a 16x16 image: EDIT: i tried using a black square and setting its width to 1 and height to 2048, and another one with width to 2048 and height to 1. The result i got is not what i expected: Can someone help me to fix this please? • I think your main issue is that your crosshair is not pixel-aligned, leading to some or all of those strange artifacts. But your overall approach is "wrong" too, so I'll post an answer with better ones. – XenoRo Dec 20 '17 at 21:09 • @Gabriele Vierti Appreciate the effort but your edit took all life out of my question. XenoRo Anticipating it with great enthusiasm. Nows Bedtime though. – AzulShiva Dec 20 '17 at 21:26 Textures mapped onto the surfaces of 3D geometry were developed to represent, well, surface texture. This type of detail usually doesn't align precisely with the grid of screen pixels, due to varying object sizes, distances, orientations, and camera perspectives, so it doesn't include any built-in features to snap the sampled texels to screen pixels. In fact, we more frequently want the opposite, smooth blending that hides the effect of the screen grid (which can show up as aliasing). But Unity gives us the tools to be pixel-perfect when we want to be. Here's two ways we can approach it: 1. Create a new UI Canvas. Leave it in the default Screen Space - Overlay mode 2. Check the Pixel Perfect checkbox and set the Canvas Scaler mode to "Constant Pixel Size" 3. Create two Panels as children of the canvas. Remove their sprite so they're just a flat colour of your choosing. 4. Set one to centered horizontally / stretched vertically with width 1, and the other to the opposite with height 1. Or, if you prefer to render this on other geometry, you can use a shader that clips out everything but the center cross: CGPROGRAM #pragma vertex vert #pragma fragment frag #pragma target 3.0 #include "UnityCG.cginc" void vert ( float4 vertex : POSITION, // vertex position input out float4 outpos : SV_POSITION // clip space position output ) { outpos = UnityObjectToClipPos(vertex); } fixed4 frag (UNITY_VPOS_TYPE screenPos : VPOS) : SV_Target { // Compute the pixel center of the screen. float2 center = _ScreenParams.xy/2.0f; // Keep only the first pixel on/beside the center in the positive direction. float2 onLine = (screenPos >= center) * (screenPos < (center + 1.0f)); clip(max(onLine.x, onLine.y) - 0.5f); // Draw black. (Or the colour of your choice) return 0; } ENDCG With both of these solutions, though I could verify that Unity was writing a crisp single-pixel-wide line into the buffer, I ended up seeing a fuzzy line in screenshots. It turned out to be Windows applying its scaling logic for high-resolution displays - it was forcing Unity to render at a lower-than-native scale and then upscale. Disabling that Windows feature let the lines come out crisp, as expected. There are many ways to create cross-hairs and cross-hair-like effects in Unity. Here are just a few of them: # 9-sliced image: You don't need a 2kx2k pixel texture just for a cross-hair. In fact, that would be (very) bad practice. Instead, you can have Unity scale and tile a simple, very small texture, based on it's parts. In the case of a single-pixel cross-hair, a 3x3 image is enough, through the process of 9-slicing; there the image will be sliced in up to 9 different parts, and where each edge slice will be tiled or stretched to fit/fill a certain area. # Normal image(s): You don't need a 2kx2k texture. A 1-pixel image stretched across the screen will look like a 1-pixel line. Make two of them cross at the middle, and you have a 1-pixel cross-hair. When dealing with UI, transformations, variations and combinations of small images are pretty much always better than monolithic textures. # Vector-Graphics: In engines that support this, or with the help of plugins, vector-graphics is often the best choice for screen-space UI, as it provides clean, crisp, graphics, which are easy to transform, maintain, extend, and can greatly simplify the process of making dynamic effects. Unfortunately, very few game-engines support this out-of-the-box, and Unity is not one of them. Further unfortunately still, almost all plugins that provide this functionality do so through work-arounds, like transforming the vector-graphics into normal images, which, despite allowing the engine to "support vector-graphics", doesn't really maintain most of the advantages of vector-graphics. # Scripted Graphics-Library rendering: Most game engines allow (direct or indirect) access to their underlying low-level Graphics-Library (IE OpenGL, DirectX, Vulkan, etc), and it is trivial to draw simple cross-hairs through these libraries. In Unity, this access is provided through the GL class. Note: This is usually not recommended, because it's overkill and much harder to change/maintain than an image-based approach, specially since there is no difference if you do the image-based approach correctly. # And many more... These are just the main/most common few of many, many other methods... You're strongly advised to do some research on this, as many of the methods for creating cross-hairs also provide interesting starting-points and/or clues for more advanced graphics and effects. # NOTE: To get a pixel-perfect effect, you need a pixel-aligned approach, obviously. In Unity, for screen-space UI, that's as easy as marking the "pixel-perfect" checkbox in the canvas' inspector.	{}
1. ## table, font size hello, i have got two questions about LaTeX. First i would like to change font size only one word in my text. How can I do that? For example, I have a sentence: "I love Math " word "I" and "Math" are normal size, but love is bigger than I and Math. Second question is about eumerate tables in Latex. In standard when i put table in text i have Table 1. i would like to change to Table B.1. so I would like to have the letter.dot.number.dot (not number.dot as standard ) thanks for help 2. Hi, Originally Posted by spider865 First i would like to change font size only one word in my text. How can I do that? For example, I have a sentence: "I love Math " word "I" and "Math" are normal size, but love is bigger than I and Math. This code Code: You {\Huge love} Math. gives $\text{You {\Huge love} Math.}$ (\Huge can be replaced by \huge, \LARGE, \Large or \large...) Second question is about eumerate tables in Latex. In standard when i put table in text i have Table 1. i would like to change to Table B.1. so I would like to have the letter.dot.number.dot (not number.dot as standard ) If you want "the letter" to always be a B you can use this : Code: \renewcommand{\thetable}{B.\arabic{table}} If the letter B refers to, say, the current section then Code: \renewcommand{\thetable}{\Alph{section}.\arabic{table}} is better. (Help On LaTeX Counters) 3. thanks	{}
An R package and jamovi module for equivalence testing Background Scientists should be able to provide support for the absence of a meaningful effect. Currently, researchers often incorrectly conclude an effect is absent based a non-significant result. A widely recommended approach within a frequentist framework is to test for equivalence. In equivalence tests, such as the two one-sided tests (TOST) procedure implemented in this package, an upper and lower equivalence bound is specified based on the smallest effect size of interest. The TOST procedure can be used to statistically reject the presence of effects large enough to be considered worthwhile. Extending your statistical tool kit with equivalence tests is an easy way to improve your statistical and theoretical inferences. # Installation The developmental version, and most up-to-date, can be installed from GitHub: devtools::install_github("Lakens/TOSTER") install.packages("TOSTER") ## Educational Material For educational material on setting the smallest effect size of interest and equivalence tests, see week 2 of my MOOC “Improving Your Statistical Questions”. https://www.coursera.org/teach/improving-statistical-questions. For an introduction to equivalence testing and the TOSTER package (this is recommended reading material to understand the basics of equivalence tests) see: Lakens, D. (2017). Equivalence Tests: A Practical Primer for t Tests, Correlations, and Meta-Analyses. Social Psychological and Personality Science, 8(4), 355–362. https://doi.org/10.1177/1948550617697177 For a tutorial paper (this is the recommended reading material if you want to start using equivalence testing) see: Lakens, D., Scheel, A. M., & Isager, P. M. (2018). Equivalence Testing for Psychological Research: A Tutorial. Advances in Methods and Practices in Psychological Science, 1(2), 259–269. https://doi.org/10.1177/2515245918770963 For a comparison of Bayes factors and equivalence test (they turn out to lead to very similar inferences when used well) see: Lakens, D., McLatchie, N., Isager, P. M., Scheel, A. M., & Dienes, Z. (2018). Improving Inferences about Null Effects with Bayes Factors and Equivalence Tests. The Journals of Gerontology: Series B. https://doi.org/10.1093/geronb/gby065 For a comparison of equivalence tests and second generation p-values (equivalence tests are probably a better tool) see: Lakens, D., & Delacre, M. (2019). Equivalence Testing and the Second Generation P-Value. Meta-Psychology. https://doi.org/10.31234/osf.io/7k6ay For a general introduction to the importance of being able to support ‘null’ effects, and ways to do this, including equivalence tests, bayesian estimation, and bayes factors, see: Harms, C., & Lakens, D. (2018). Making “null effects” informative: Statistical techniques and inferential frameworks. Journal of Clinical and Translational Research, (3), 382–393. https://doi.org/10.18053/jctres.03.2017S2.007	{}
LEARN ENOUGH Video Screencasts for Learn Enough Text Editor to Be Dangerous Mar 15, 2016 • posted by Michael Hartl The humble yet essential text editor is perhaps the most important item in every computer magician’s bag of tricks. Although this subject is well-served by a written tutorial, there’s nothing like video screencasts to really bring text editors to life. Today I’m pleased to announce the availability of the Learn Enough Text Editor to Be Dangerous video screencasts. The full Learn Enough Text Editor to Be Dangerous video series includes almost an hour and a half of video, covering not only all the subjects in the written tutorial but also the kind of tips & tricks that are hard to capture in print form. To get an idea of what the screencasts include, I’ve uploaded a short preview to YouTube here: Topics include: • Minimum Viable Vim • The Most Important Vim Command • Basics of the Atom, Sublime Text, and Cloud9 editors • Worked examples of moving around, selecting text, and cut/copy/paste • Saving your bacon with Undo • Finding and replacing, both in single files and project-wide • Autocomplete and tab triggers • Writing source code • How to extend your shell with a real executable Bash script The full price for the Learn Enough Text Editor to Be Dangerous screencasts is $12, which you can also buy bundled with the ebooks for$19: Although the Learn Enough Text Editor to Be Dangerous screencasts are designed for complete beginners, in the nearly 90 minutes of content there’s also lots of material that even more experienced users will find valuable, including coverage of Minimum Viable Vim and a thorough introduction to “modern” text editors like Atom and Sublime. Enjoy! Discuss on Hacker News Michael Hartl I’m Michael Hartl—author, educator, and entrepreneur. I’m probably best known as the creator of the Ruby on Rails Tutorial, a book and screencast series that together constitute one of the leading introductions to web development. Once called his “favorite book” by Wikipedia founder Jimmy Wales, the Ruby on Rails Tutorial currently has over 150 5-star reviews at Amazon. I’m also (in)famous for creating Tau Day and The Tau Manifesto, which have inspired an international movement dedicated to the proposition that “pi is wrong.” (For example, as a result of The Tau Manifesto, MIT releases their admissions decisions each year at “Tau Time” (6:28 p.m.), and typing tau/2 at Google yields 3.14159…) Finally, I’m a founder of Softcover, a publishing system and sales platform for technical authors, which among other things powers both The Tau Manifesto and the Ruby on Rails Tutorial. I’m a graduate of Harvard College and have a Ph.D. in Physics from Caltech, where I studied black hole dynamics and was an award-winning instructor in theoretical and computational physics. I’m also an alumnus of Y Combinator, the entrepreneur program that has produced companies such as Dropbox and Airbnb. (Alas, my own Y Combinator startup was neither Dropbox nor Airbnb.)	{}
All-sky search for continuous gravitational waves from isolated neutron stars using Advanced LIGO and Advanced Virgo O3 data Document #: LIGO-P2100367-v13 Document type: P - Publications Other Versions: LIGO-P2100367-v8 03 Jan 2022, 02:43 Abstract: We present results of an all-sky search for continuous gravitational waves which can be produced by spinning neutron stars with an asymmetry around their rotation axis, using data from the third observing run of the Advanced LIGO and Advanced Virgo detectors. Four different analysis methods are used to search in a gravitational-wave frequency band from 10 to 2048 Hz and a first frequency derivative from $$-10^{-8}$$ to $$10^{-9}$$ Hz/s. No statistically-significant periodic gravitational-wave signal is observed by any of the four searches. As a result, upper limits on the gravitational-wave strain amplitude $$h_0$$ are calculated. The best upper limits are obtained in the frequency range of 100 to 200 Hz and they are $${\sim}1.1\times10^{-25}$$ at 95\% confidence-level. The minimum upper limit of $$1.10\times10^{-25}$$ is achieved at a frequency 111.5 Hz. We also place constraints on the rates and abundances of nearby planetary- and asteroid-mass primordial black holes that could give rise to continuous gravitational-wave signals. Files in Document: Other Files: Topics: Author Groups: DCC Version 3.4.1, contact DCC Help	{}
Conditional in HLSL pass This topic is 2976 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. Recommended Posts Is it possible to make a conditional check in a pass? I am trying to change CullMode based on a global set from the client side.If it is not possible should I just set it from the client with SetRenderState or should I make two different techniques like ShadowOneSided and ShadowTwoSided. The following code gives the error None identifier not defined. Any suggestions? technique Shadow { pass P0 { ColorWriteEnable = RED; AlphaBlendEnable = true; SrcBlend = One; DestBlend = Zero; CullMode = (g_TwoSided)?(None):(CCW); } } Share on other sites This isn't something I've tried before so I don't know if it works, but it seems to at least compile: bool g_TwoSided;int CullType(){ return (g_TwoSided)?(0):(2); // Can't use the "None" and "CCW" constants here}technique Shadow{ pass P0 { ... CullMode = CullType(); ... }} Share on other sites MJP's solution looks great, but I don't know if it works. My suggestion is if the changes are small, go for a setrenderstate. When the changes are significant go for #ifdefs. In my deferred renderer I have a single .fx file containing the light shader, which generates 6 shaders via #ifdefs, two for each light type. What's good about #ifdefs is you can reuse the same system to decide which features must be enabled at runtime (and rebuild the shaders on the fly). The total amount of permutations is very high as #defines control features like linear color space, shadow mapping and lighting models supported. Of course the trick is to generate a few shaders, but in case a project didn't need all lighting models, I would just tell the fx compiler not to generate that code. The cost of switching shaders is high in D3D9, but in D3D10 if the vertex layout is the same (and in theory it's the same) the operation looks a lot cheaper, so #ifdefs could be a good solution... Share on other sites If you decide to use two techniques, one thing I discovered lately is that doing 'compile' within a technique always creates a shader, so if you compile the same shader in several techniques, you get duplicates. Better to compile before and assign to a variable. As for the question itself, I'd say do whatever fits your code structure the most. If your code is built around techniques, then introducing another technique might be easier than introducing SetRenderTarget into the code. If, as undead's solution suggests, this isn't dynamic, but an option that's set once, and you can compile the shader with it, what he suggests sounds like a fine solution.	{}
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. As my student Kévin Guimard had not mailed me his own Poisson slice sampler of a Poisson distribution, I could not tell why the code was not working! My earlier post prompted him to do so and a somewhat optimised version is given below: nsim = 10^4 lambda = 6 max.factorial = function(x,u){ k = x parf=1 while (parfu<1){ k = k + 1 parf = parf k } k = k - (parf*u>1) return (k) } x = rep(floor(lambda), nsim) for (t in 2:nsim){ v1 = ceiling((log(runif(1))/log(lambda))+x[t-1]) ranj=max(0,v1):max.factorial(x[t-1],runif(1)) x[t]=sample(ranj,size=1) } barplot(as.vector(rbind( table(x)/length(x),dpois(min(x):max(x), lambda))),col=c("sienna","gold")) As you can easily check by running the code, it does not work. My student actually majored my MCMC class and he spent quite a while pondering why the code was not working. I did ponder as well for a part of a morning in Warwick, removing causes for exponential or factorial overflows (hence the shape of the code), but not eliciting the issue… (This now sounds like lethal fugu sashimi! ) Before reading any further, can you spot the problem?! The corrected R code is as follows: x = rep(lambda, nsim) for (t in 2:nsim){ v1=ceiling((log(runif(1))/log(lambda))+x[t-1]) ranj=max(0,v1):max.factorial(x[t-1],runif(1)) if (length(ranj)>1){ x[t] = sample(ranj, size = 1) }else{ x[t]=ranj} } The culprit is thus the R function sample which simply does not understand Dirac masses and the basics of probability! When running > sample(150:150,1) [1] 23 you can clearly see where the problem stands…! Well-documented issue with sample that already caused me woes… Another interesting thing about this slice sampler is that it is awfully slow in exploring the tails. And to converge to the centre from the tails. This is not very pronounced in the above graph with a mean of 6. Moving to 50 makes it more apparent: This is due to the poor mixing of the chain, as shown by the raw sequence below, which strives to achieve a single cycle out of 10⁵ iterations! In any case, thanks to Kévin for an interesting morning! Filed under: Books, Kids, pictures, R, Running, Statistics, University life Tagged: convergence assessment, ENSAE, Gibbs sampling, Monte Carlo Statistical Methods, Poisson distribution, R, sample, sampling from an atomic population, slice sampling, slow convergence	{}
Is the following function uniformly continuous? • March 17th 2010, 06:38 PM Pinkk Is the following function uniformly continuous? Let $f(x)=x^{2}sin(\frac{1}{x}), x\ne 0; f(0)=0$. Is $f$ uniformly continuous on $\mathbb{R}$? Now this one I'm completely stumped on and I don't even know where to begin. • March 17th 2010, 06:46 PM Drexel28 Quote: Originally Posted by Pinkk Let $f(x)=x^{2}sin(\frac{1}{x}), x\ne 0; f(0)=0$. Is $f$ uniformly continuous on $\mathbb{R}$? Now this one I'm completely stumped on and I don't even know where to begin. Note that the function is differentiable and that $f'(x)=\begin{cases}0 & \mbox{if} \quad x=0 \\ x\sin\left(\tfrac{1}{x}\right) -\cos\left(\tfrac{1}{x}\right) & \mbox{if} \quad x\ne 0\end{cases}$ and so $\|f'(x)\|\leqslant \|x\sin\left(\tfrac{1}{x}\right)-\cos\left(\tfrac{1}{x}\right)\|\leqslant \|x\sin\left(\tfrac{1}{x}\right)\|+\|\cos\left(\tfrac {1}{x}\right)\|\leqslant 1+1=2$. Thus, your function has a bounded derivative and is thus Lipschitz and so trivially uniformly continuous. • March 17th 2010, 07:07 PM Pinkk Ah okay, makes sense because if $f'\le M, M>0$ then if $\|b-a\|<\frac{\epsilon}{M}$, then $\|f(b)-f(a)\| = \|f'(x)\|\|b-a\|\le M\|b-a\| < M\frac{\epsilon}{M}=\epsilon$, if I'm not mistaken. Thanks again, man. • March 17th 2010, 07:08 PM Drexel28 Quote: Originally Posted by Pinkk Ah okay, makes sense because if $f'\le M, M>0$ then if $\|b-a\|<\frac{\epsilon}{M}$, then $\|f(b)-f(a)\| = \|f'(x)\|\|b-a\|\le M\|b-a\| < M\frac{\epsilon}{M}=\epsilon$, if I'm not mistaken. Thanks again, man. Good call.	{}
# Testing faster: How to avoid making compile times worse by adding tests ## Background We'd like to keep cargo test, cargo build, cargo check, ... reasonably fast, and we'd really like to keep them fast if you pass -p for a specific project. Unfortunately, there are a few ways this can become unexpectedly slow. The easiest of these problems for us to combat at the moment is the unfortunate placement of dev-dependencies in our build graph. If you perform a cargo test -p foo, all dev-dependencies of foo must be compiled before foo's tests can start. This includes dependencies only used non-test targets, such as examples or benchmarks. In an ideal world, cargo could run your tests as soon as it finished with the dependencies it needs for those tests, instead of waiting for your benchmark suite, or the arg-parser your examples use, or etc. Unfortunately, all cargo knows is that these are dev-dependencies, and not which targets actually use them. Additionally, unqualified invocations of cargo (that is, without -p) might have an even worse time if we aren't careful. If I run, cargo test, cargo knows every crate in the workspace needs to be built with all dev dependencies, if places depends on fxa-client, all of fxa-clients dev-dependencies must be compiled, ready, and linked in at least to the lib target before we can even think about starting on places. We have not been careful about what shape the dependency graph ends up as when example code is taken into consideration (as it is by cargo during certain builds), and as a result, we have this problem. Which isn't really a problem we want to fix: Example code can and should depend on several different components, and use them together in interesting ways. So, because we don't want to change what our examples do, or make major architectural changes of the non-test code for something like this, we need to do something else. ## The Solution To fix this, we manually insert "cuts" into the dependency graph to help cargo out. That is, we pull some of these build targets (e.g. examples, benchmarks, tests if they cause a substantial compile overhead) into their own dedicated crates so that: 1. They can be built in parallel with each other. 2. Crates depending on the component itself are not waiting on the test/bench/example build in order for their test build to begin. 3. A potentially smaller set of our crates need to be rebuilt -- and a smaller set of possible configurations exist meaning fewer items to add pressure to caches. 4. ... Some rules of thumb for when / when not to do this: • All rust examples should be put in examples/. • All rust benchmarks should be put in testing/separated/. See the section below on how to set your benchmark up to avoid redundant compiles. • Rust tests which brings in heavyweight dependencies should be evaluated on an ad-hoc basis. If you're concerned, measure how long compilation takes with/without, and consider how many crates depend on the crate where the test lives (e.g. a slow test in support/foo might be far worse than one in a leaf crate), etc... ### Appendix: How to avoid redundant compiles for benchmarks and integration tests To be clear, this is way more important for benchmarks (which always compile as release and have a costly link phase). Say you have a directory structure like the following: mycrate ├── src │ └── lib.rs \| ... ├── benches │ ├── bench0.rs \| ├── bench1.rs │ └── bench2.rs ├── tests │ ├── test0.rs \| ├── test1.rs │ └── test2.rs └── ... When you run your integration tests or benchmarks, each of test0, test1, test2 or bench0, bench1, bench2 is compiled as it's own crate that runs the tests in question and exits. That means 3 benchmark executables are built on release settings, and 3 integration test executables. If you've ever tried to add a piece of shared utility code into your integration tests, only to have cargo (falsely) complain that it is dead code: this is why. Even if test0.rs and test2.rs both use the utility function, unless every test crate uses every shared utility, the crate that doesn't will complain. (Aside: This turns out to be an unintentional secondary benefit of this approach -- easier shared code among tests, without having to put a #![allow(dead_code)] in your utils.rs. We haven't hit that very much here, since we tend to stick to unit tests, but it came up in mentat several times, and is a frequent complaint people have) Anyway, the solution here is simple: Create a new crate. If you were working in components/mycrate and you want to add some integration tests or benchmarks, you should do cargo new --lib testing/separated/mycrate-test (or .../mycrate-bench). Delete .../mycrate-test/src/lib.rs. Yep, really, we're making a crate that only has integration tests/benchmarks (See the "FAQ0" section at the bottom of the file if you're getting incredulous). Now, add a src/tests.rs or a src/benches.rs. This file should contain mod foo; declarations for each submodule containing tests/benchmarks, if any. For benches, this is also where you set up the benchmark harness (refer to benchmark library docs for how). [[test]] name = "mycrate-test" path = "src/tests.rs" [[test]] name = "mycrate-benches" path = "src/benches.rs" harness = false Because we aren't using src/lib.rs, this is what declares which file is the root of the test/benchmark crate. Because there's only one target (unlike with tests/* / benches/* under default settings), this will compile more quickly. Additionally, src/tests.rs and src/benches.rs will behave like a normal crate, the only difference being that they don't produce a lib, and that they're triggered by cargo test/cargo run respectively. ### FAQ0: Why put tests/benches in src/* instead of disabling autotests/autobenches Instead of putting tests/benchmarks inside src, we could just delete the src dir outright, and place everything in tests/benches. Then, to get the same one-rebuild-per-file behavior that we'll get in src, we need to add autotests = false or autobenches = false to our Cargo.toml, adding a root tests/tests.rs (or benches/benches.rs) containing mod decls for all submodules, and finally by referencing that "root" in the Cargo.toml [[tests]] / [[benches]] list, exactly the same way we did for using src/. This would work, and on the surface, using tests/.rs and benches/.rs seems more consistent, so it seems weird to use src/.rs for these files. My reasoning is as follows: Almost universally, tests/.rs, examples/.rs, benches/.rs, etc. are automatic. If you add a test into the tests folder, it will run without anything else. If we're going to set up one-build-per-{test,bench}suite as I described, this fundamentally cannot be true. In this paradigm, if you add a test file named blah.rs, you must add a mod blah it to the parent module. It seems both confusing and error-prone to use tests/, but have it behave that way, however this is absolutely the normal behavior for files in src/*.rs -- When you add a file, you then need to add it to it's parent module, and this is something Rust programmers are pretty used to. (In fact, we even replicated this behavior (for no reason) in the places integration tests, and added the mod declarations to a "controlling" parent module -- It seems weird to be in an environment where this isn't required) So, that's why. This way, we make it way less likely that you add a test file to some directory, and have it get ignored because you didn't realize that in this one folder, you need to add a mod mytest` into a neighboring tests.rs.	{}
# Jul 30, 2008 correlated electron systems. In this contribution we introduce this model along with numerically exact method of diagonalization of the model. We use an exact diagonalization method to study the ionic Hubbard model in . A ground state is found by the conventional Lanczos method , and then used as an initial state for the time-dependent Schrödinger equation. Time evolution is implemented by the Krylov subspace method based on the Lanczos method [36–38]. 4. Exact diagonalization 4.1 Hamiltonian operators for strongly correlated electron systems 4.1.1The Hubbard model The Hubbard model represents interacting electrons in narrow bands. It was originally proposed to study metal-insulator transitions and ferro-magnetism of itinerant electrons in narrow bands but it has also acquired An exact-diagonalization technique on small clusters is used to study the ground states and single-particle excitations of the Hubbard model with on-site (U) and nearest-neighbor (V) Coulombic repulsive interactions. It is shown that the long-range charge-density-wave state realized in a half-filled two-dimensional square lattice for 4V\\ensuremath{\\gtrsim}U persists up to quarter filling Hubbard model is an important model in the theory of strongly correlated electron systems. Introduction to Hubbard model and exact diagonalization Exact diagonalization: The Bose–Hubbard model as an example 593 term is the interaction part (Hˆ int) and is due to the particle–particle interaction, the strength of which is characterized by the parameter U. The Bose–Hubbard model has been realized with ultracold boson atoms in an optical lattice [4]. Moreover, in this system, the parameters PHYSICAL REVIEW A 85, 065601 (2012) Exact diagonalization of the one-dimensional Bose-Hubbard model with local three-body interactions Tomasz Sowinski´ Institute of Physics of the Polish Academy of Sciences, Aleja Lotnikow 32/46, 02-668 Warsaw, Poland´ We study Mott transition in the two-dimensional Hubbard model on an anisotropic triangular lattice. We use the Lanczos exact diagonalization of finite-size clusters up to eighteen sites, and calculate Drude weight, charge gap, double occupancy and spin structure factor. , M. Jaščur b. , A. Bobák b. Feb 27, 2019 Exact diagonalization study of the Hubbard-parametrized four-spin ring exchange model on a square lattice. C. B. Larsen, A. T. Rømer, By exactly diagonalizing the Hubbard model for ten electrons on ten sites in a one-Dimensional (1D) ring, we extend the study of Jafari (2008) to more than two Exact diagonalization has played a very important role in understanding the ground state properties of quantum spin systems. The idea is to set up the Hamiltonian Jul 30, 2008 correlated electron systems. In this contribution we introduce this model along with numerically exact method of diagonalization of the model. ## DWA-01 Simulating the Bose Hubbard model using dwa QMC code (revisiting tutorial MC-05) DWA-02 Density profile of a 3D optical lattice in a harmonic trap ; DWA-03 Time-of-flight images of a 3D optical lattice in a harmonic trap ; Exact diagonalization. ED-01 Sparse Diagonalization (Lanczos) ED-02 Spin gaps of 1D quantum systems HΦ also supports the massively parallel computations. The Lanczos algorithm for obtaining the ground state and thermal pure quantum state method for finite-temperature calculations are implemented. ### Hubbard model is an important model in the theory of strongly correlated electron systems. In this contribution we introduce this model and the concepts of Introduction to Hubbard model and exact diagonalization S. Akbar Jafari; Affiliations An exact solution for the Hubbard model in the case of a one-dimensional chain exists, called the Bethe Aug 9, 2019 Wigner transformation, (ii) free particle systems: the Su-Schrieffer-Heeger (SSH) model,. (iii) the many-body localized 1D Fermi-Hubbard model Distributional exact diagonalization formalism for quantum impurity models for the single-impurity Anderson model and the two-orbital Hubbard model within Avhandlingar om EXACT DIAGONALIZATION. Sökning: "Exact Diagonalization" a Z2 symmetric model, with long range couplings and a phase gradient. Detecting quantum critical points in the t-t ' Fermi-Hubbard model via complex network within a four-by-four plaquette that is solved by exact diagonalization. Green's function, DFT, TDDFT, Hubbard model, Many-Body Perturbation in a test system against exact diagonalization, and was found to perform well in most. We use a generalized Hubbard model that takes into account all the interactions on a lattice site, and solve the many-particle problem by exact diagonalization. Distributional exact diagonalization formalism for quantum impurity models Enhanced pairing susceptibility in a photodoped two-orbital Hubbard model. , M. Jaščur b. , A. Bobák b. Feb 27, 2019 Exact diagonalization study of the Hubbard-parametrized four-spin ring exchange model on a square lattice. C. B. Larsen, A. T. Rømer, By exactly diagonalizing the Hubbard model for ten electrons on ten sites in a one-Dimensional (1D) ring, we extend the study of Jafari (2008) to more than two Exact diagonalization has played a very important role in understanding the ground state properties of quantum spin systems. Industrial design drawings It was originally proposed to study metal-insulator transitions and ferro-magnetism of itinerant electrons in narrow bands but it has also acquired An exact-diagonalization technique on small clusters is used to study the ground states and single-particle excitations of the Hubbard model with on-site (U) and nearest-neighbor (V) Coulombic repulsive interactions. It is shown that the long-range charge-density-wave state realized in a half-filled two-dimensional square lattice for 4V\\ensuremath{\\gtrsim}U persists up to quarter filling Hubbard model is an important model in the theory of strongly correlated electron systems. We use the Lanczos exact diagonalization of finite-size clusters up to eighteen sites, and calculate Drude weight, charge gap, double occupancy and spin structure factor. 2008-07-30 · Hubbard model is an important model in theory of strongly correlated electron systems. Vem betalar omvänd moms ### We study Mott transition in the two-dimensional Hubbard model on an anisotropic triangular lattice. We use the Lanczos exact diagonalization of finite-size clusters up to eighteen sites, and calculate Drude weight, charge gap, double occupancy and spin structure factor. By exactly diagonalizing the Hubbard model for ten electrons on ten sites in a one-Dimensional (1D) ring, we extend the study of Jafari (2008) to more than two electrons on two sites. We equally show the sparsity patterns of the Hamiltonian matrices for four- and eight-site problems and obtain the ground state energy eigenvalues for ten electrons on ten-sites. Exact diagonalization: the Bose-Hubbard model as an example. Lon efter skatt ekonomifakta ### Finite temperature electronic and magnetic properties of small clusters are investigated in the framework of the Hubbard model by using exact diagonalization methods and by sampling the different cluster topologies exhaustively. Results are discussed for the specific heat C(T), magnetic susceptibility χ(T), local magnetic moments μi(T), average magnetic moments $\\overline\\mu_N(T)$ and spin By exactly diagonalizing the Hubbard model for ten electrons on ten sites in a one-Dimensional (1D) ring, we extend the study of Jafari (2008) to more than two electrons on two sites. We equally show the sparsity patterns of the Hamiltonian matrices for four- and eight-site problems and obtain the ground state energy eigenvalues for ten electrons on ten-sites. A considerable amount of work has been based on the study of the single band 2D Hubbard model us- ing numerical Monte Carlo techniques or exact dia- gonalization [3]. The advantages are that taking trace of only Cu atoms, the single band Hubbard model is simpler, contains less than half the atoms of a real Cu02 cluster, and it allows larger sizes to be treatable.	{}
# All Questions 19 views ### AES decryption vs encryption speed Let's consider the CTR mode. For a faster encryption/decryption, is it preferable to use the decryption operation of AES, or its encryption ? 12 views ### Confusion regarding computing Multiplicative Inverse Modulo P? May be a silly doubt, please rectify my confusion regarding below problem: For concreteness assume $g=2, p=11, a=6$ and $x=9$ $$A = g^a \bmod p = 2^6 \bmod 11 = 9$$ X = g^x \bmod p = 2^9 \bmod 11 ... 18 views ### Deriving message from signature (RSA) Usually signatures are used for verification, and this requires that you pass both the message and the signature. I was wondering; if the message holds to a certain format (for example, valid json ... 5 views ### OpenSSL ECDH key exchang mechanism I am using FIPS based OpenSSL module for encryption of sensitive data for my desktop socket server and client applications. 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How I do this currently: ... 12 views ### Assuring Client is Authorized to use JSON Web Token JSON Web Tokens ( JWT ) are server-signed objects that the issuing server uses to identify a user, track session data and authorize requests. The fact that JWT are server-signed gives assurance that ... 214 views ### What is stopping someone from saving encrypted info, and decoding it later? When I send encrypted data to a website, using public key encryption, what is stopping someone who is listening from storing my encrypted data, and decryption it in a few years after they have ... 131 views +100 ### Key Size for Symmetric Homomorphic Encryption Over the Integers In the paper Fully Homomorphic Encryption over the Integers, it mentions a symmetric key scheme on page 1 and 2. Key Generation: Pick a random odd number $p \epsilon [2^{N-1},2^N)$ Encrypt A Bit m: ... 29 views ### Why Curve25519 for encryption but Ed25519 for signatures? 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I know that KP-ABE depends on the ... 49 views ### POODLE attack on TLS 1.2 The POODLE attack uses the way block ciphers in CBC mode are decrypted in combination with the packet's padding to determine some byte's value. From what I understand, even TLS 1.1 can be vulnerable ... 43 views ### What is the impact of a server not fully verifying the data in the Client Finished packet of a TLS connection? F5 BIG-IP devices suffer from a vulnerability in which not all bytes of the Finished packet in a TLS connection are validated. The sort-of technical description can be found here. F5 don't consider it ... 138 views ### Using a hash (like SHA-256) vs AES as the source for pseudo-random values in Feistel network? This question is in relation to Wikipedia article on Format Preserving Encryption It says the following It is also possible to make a FPE algorithm using a Feistel network. 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Is there any way to make an attacker who is successfully performing an MITM attack to actually not be able to read or change the messages? Assuming there is no ... 27 views ### In HMAC-MD5 Implementation, why “i_key_pad” and “o_key_pad” if They Are Public and Constant? [duplicate] Please Consider: English is My Second Language This Wikipedia article on HMACs shows a part of a HMAC-MD5 implementation making use of i_key_pad and ...	{}
# 1.4 Dimensional Analysis - University Physics Volume 1 _ OpenStax.pdf • 5 This preview shows page 1 - 3 out of 5 pages. The preview shows page 1 - 3 out of 5 pages. 5/21/20201.4 Dimensional Analysis - University Physics Volume 1 \| OpenStax1/5Learning ObjectivesBy the end of this section, you will be able to:Find the dimensions of a mathematical expression involving physical quantities.Determine whether an equation involving physical quantities is dimensionally consistent.The dimensionof any physical quantity expresses its dependence on the base quantities as a product of symbols (or powers of symbols)representing the base quantities. Table 1.3 lists the base quantities and the symbols used for their dimension. For example, a measurement oflength is said to have dimension L or L, a measurement of mass has dimension M or M, and a measurement of time has dimension T or T. Likeunits, dimensions obey the rules of algebra. Thus, area is the product of two lengths and so has dimension L, or length squared. Similarly, volumeis the product of three lengths and has dimension L, or length cubed. Speed has dimension length over time, L/T or LT. Volumetric mass densityhas dimension M/Lor ML, or mass over length cubed. In general, the dimension of any physical quantity can be written as for some powers and g. We can write the dimensions of a length in this form with and the remaining six powers all set equalto zero: Any quantity with a dimension that can be written so that all seven powers are zero (that is, its dimension is ) is called dimensionless(or sometimes “of dimension 1,” because anything raised to the zero power is one). Physicists oftencall dimensionless quantities pure numbers.Base QuantitySymbol for DimensionLengthLMassMTimeTCurrentIThermodynamic temperatureΘAmount of substanceNLuminous intensityJTable 1.3Base Quantities and Their DimensionsPhysicists often use square brackets around the symbol for a physical quantity to represent the dimensions of that quantity. For example, if isthe radius of a cylinder and is its height, then we write and to indicate the dimensions of the radius and height are both those11123–13–3 5/21/20201.4 Dimensional Analysis - University Physics Volume 1 \| OpenStax2/5of length, or L. Similarly, if we use the symbol for the surface area of a cylinder and for its volume, then [A] = Land [V] = L. If we use thesymbol for the mass of the cylinder and for the density of the material from which the cylinder is made, then and Course Hero member to access this document Course Hero member to access this document End of preview. Want to read all 5 pages?	{}
# Duration: Parallel shift in yield curve assumption ## General Intro I'm trying to really understand the assumptions of dollar duration for a portfolio of bonds. In particular I don't fully understand that the assumption that there are parallel shifts in the yield curve when computing the duration of a portfolio of bonds. ## Background Assume continuous compounding so that the duration of a bond $B$ is $-\frac{1}{B}\frac{\partial B}{\partial y}$ where $y$ is the yield to maturity of the bond. The duration of a portfolio of bonds $\{B_1, \ldots, B_p\}$ is defined as $$\frac{-\sum_{i=1}^p \frac{\partial B_i}{\partial y_i}}{\sum_{i=1}^t B_i}$$ which is the weighted average duration (weighted by value, assuming for simplicity all bonds have par value of 1). The above definition of weighted average duration always comes with the caveat that the definition depends on the assumption that there are parallel shifts in the yield curve. ## Question 1. What curve are the shifts talking about? Parallel shifts in the yield curve means (to me) parallel shifts in the zero-rate yield curve, aka the yield to maturity against maturity for zero-coupon bonds. If the assumption is talking about parallel shifts in this zero-rate curve then I don't understand the assumption at all since the yield (to maturity) of the coupon bond is a non-linear function of zero-rates. I don't see how parallel shifts in the zero-rate curve translate into a easy statement about moves in yield (to maturity) of bonds with different maturities. If the shifts are referring to parallel shifts in the yield to maturity curve of the bonds themselves, then this also does not make much sense to me. In this case the bonds with the same maturity may have different yields depending on their coupon rates and price so the 'yield curve' in this sense is not well defined. 2. In "The Handbook of Fixed Income Securities" by Frank Fabozzi (2005) on page 208 he claims the the assumption that each yield has moved the same amount is equivalent to the correlation between the change in the yields is 1. Is this correct? Such an assumption would say $\Delta y_i = a\Delta y_j + b$ but it seems we need the much stronger assumption that $\Delta y_i = \Delta y_j.$ • So I take it the assumption really is that there are equal changes in yield to maturity: $\Delta y_i = \Delta y_j?.$ Or more precisely $\frac{\partial y_i}{\partial y_j} = 1$ so that way the above definition of modified duration makes sense since by the chain rule $\frac{\partial B_i}{\partial y_j} = \frac{\partial B_i}{\partial y_i}\frac{\partial y_i}{\partial y_j} = \frac{\partial B_i}{\partial y_i}$ so that the derivatives can be thought of as being taken with respect to a single quantity? – moquant Aug 9 '18 at 13:16	{}
dc.creator Albert, S. en_US dc.creator Winnewisser, B. P. en_US dc.creator Winnewisser, M. en_US dc.date.accessioned 2007-11-20T17:15:33Z dc.date.available 2007-11-20T17:15:33Z dc.date.issued 1995 en_US dc.identifier 1995-FA'-06 en_US dc.identifier.uri http://hdl.handle.net/1811/29896 dc.description Author Institution: Justus-Liebig-Universität Giessen, 35392 Giessen, Germany. en_US dc.description.abstract Fulminic acid belongs to the class of quasilinear molecules. Precise ab initio methods are not available to reproduce the quasilinear behavior of the molecule. In particular, the calculations cannot predict the multitude of accidental resonance interactions. We have now analyzed the infrared spectra of HCNO and its isotopomers $H^{13}CNO$ and $H^{13}C^{15}NO$ up to $7000 cm^{-1}$. More then 30000 lines could be assigned. The low symmetry of the molecule and the anharmonic quasilinear bending mode lead to an extraordinarily rich array of interactions. Most of the interactions could be classified into a network of resonance systems. As a result of such classifications, it has been possible to identify dark states. An overview and comparison of the different network in HCNO, $H^{13}CNO$ and $H^{13}C^{15}NO$ will be discussed. The term value schemes built up with the help of the Ritz combination principle will be presented. en_US dc.format.extent 51311 bytes dc.format.mimetype image/jpeg dc.language.iso English en_US dc.publisher Ohio State University en_US dc.title OVERVIEW OF THE NETWORKS OF RESONANCE SYSTEMS IN THE SPECTRA OF FULMINIC ACID AND ITS ISOTOPOMERS en_US dc.type article en_US	{}
Click HERE to return to the list of problems. The Inverse Function Rule • Examples If x = f(y) then dy dx dx dy 1 = i) … 57 0 obj <> endobj 85 0 obj <>/Filter/FlateDecode/ID[<01EE306CED8D4CF6AAF868D0BD1190D2>]/Index[57 95]/Info 56 0 R/Length 124/Prev 95892/Root 58 0 R/Size 152/Type/XRef/W[1 2 1]>>stream Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. This rule is obtained from the chain rule by choosing u … dv dy dx dy = 18 8. Substitute into the original problem, replacing all forms of , getting . In Leibniz notation, if y = f (u) and u = g (x) are both differentiable functions, then Note: In the Chain Rule, we work from the outside to the inside. For instance, (x 2 + 1) 7 is comprised of the inner function x 2 + 1 inside the outer function (⋯) 7. Differentiation Using the Chain Rule. Revision of the chain rule We revise the chain rule by means of an example. If f(x) = g(h(x)) then f0(x) = g0(h(x))h0(x). H��TMo�0��W�h'��r�zȒl�8X��+NҸk�!"�O�\|��Q��&�ʨ��C�%�BR��Q��z;�^_ڨ! Learn its definition, formulas, product rule, chain rule and examples at BYJU'S. From there, it is just about going along with the formula. h�bf��A��b�,;>��1Y��Z�b��k��V��Y��4bk�t�n W�h��}b�D��I5��mM꺫�g-��w�Z�l�5��G�t� ��t�c�:��bY��0�10H+$8�e��˦0]��#��%llRG�.�,��1��/]�K�ŝ�X7@�&��X�� %�bl endstream endobj 58 0 obj <> endobj 59 0 obj <> endobj 60 0 obj <>stream Now we’re almost there: since u = 1−x2, x2 = 1− u and the integral is Z − 1 2 (1−u) √ udu. Chain Rule Examples (both methods) doc, 170 KB. We ﬁrst explain what is meant by this term and then learn about the Chain Rule which is the technique used to perform the diﬀerentiation. You must use the Chain rule to find the derivative of any function that is comprised of one function inside of another function. Now apply the product rule twice. Step 1. •Prove the chain rule •Learn how to use it •Do example problems . doc, 90 KB. For example, all have just x as the argument. SOLUTION 6 : Differentiate . Class 1 - 3; Class 4 - 5; Class 6 - 10; Class 11 - 12; CBSE. Scroll down the page for more examples and solutions. Definition •In calculus, the chain rule is a formula for computing the derivative of the composition of two or more functions. Example 1 Find the rate of change of the area of a circle per second with respect to its … �x$�V �L�@na%�'�3� 0 �0S endstream endobj startxref 0 %%EOF 151 0 obj <>stream The Chain Rule for Powers 4. 5 0 obj Title: Calculus: Differentiation using the chain rule. Written this way we could then say that f is diﬀerentiable at a if there is a number λ ∈ R such that lim h→0 f(a+h)− f(a)−λh h = 0. If our function f(x) = (g◦h)(x), where g and h are simpler functions, then the Chain Rule may be stated as f′(x) = (g◦h) (x) = (g′◦h)(x)h′(x). Work through some of the examples in your textbook, and compare your solution to the detailed solution o ered by the textbook. In fact we have already found the derivative of g(x) = sin(x2) in Example 1, so we can reuse that result here. Example Suppose we wish to diﬀerentiate y = (5+2x)10 in order to calculate dy dx. Find the derivative of $$f(x) = (3x + 1)^5$$. 2 1 0 1 2 y 2 10 1 2 x Figure 21: The hyperbola y − x2 = 1. (b) We need to use the product rule and the Chain Rule: d dx (x 3 y 2) = x 3. d dx (y 2) + y 2. d dx (x 3) = x 3 2y. Even though we had to evaluate f′ at g(x)=−2x+5, that didn't make a difference since f′=6 not matter what its input is. Solution Again, we use our knowledge of the derivative of ex together with the chain rule. x + dx dy dx dv. Implicit Diﬀerentiation and the Chain Rule The chain rule tells us that: d df dg (f g) = . We always appreciate your feedback. Show all files. Final Quiz Solutions to Exercises Solutions to Quizzes. For example: 1 y = x2 2 y =3 √ x =3x1/2 3 y = ax+bx2 +c (2) Each equation is illustrated in Figure 1. y y y x x Y = x2 Y = x1/2 Y = ax2 + bx Figure 1: 1.2 The Derivative Given the general function y = f(x) the derivative of y is denoted as dy dx = f0(x)(=y0) 1. Tree diagrams are useful for deriving formulas for the chain rule for functions of more than one variable, where each independent variable also depends on other variables. The rules of di erentiation are straightforward, but knowing when to use them and in what order takes practice. �ʆ�f��7w��ٴ"L��,��Jڜ �X��0�mm�%�h�tc� m�p}��J�b�f�4Q��XXЛ�p0��迒1�A�� eܟN�{P��1��\XL�O5M�ܑw��q��)D0��a�\�R(y�2s�B� ��\|0�e��'��V�?��꟒��d� a躆�i�2�6�J�=��2�iW;�Mf��B=�}T�G�Y�M�. Click HERE to return to the list of problems. /� �؈L@'ͱ݌�z��X�0�d\�R��9��y~c The Chain Rule (Implicit Function Rule) • If y is a function of v, and v is a function of x, then y is a function of x and dx dv. The problem that many students have trouble with is trying to figure out which parts of the function are within other functions (i.e., in the above example, which part if g(x) and which part is h(x). View Notes - Introduction to Chain Rule Solutions.pdf from MAT 122 at Phoenix College. For functions f and g d dx [f(g(x))] = f0(g(x)) g0(x): In the composition f(g(x)), we call f the outside function and g the inside function. As we apply the chain rule, we will always focus on figuring out what the “outside” and “inside” functions are first. D(y ) = 3 y 2. y '. Find it using the chain rule. This is called a tree diagram for the chain rule for functions of one variable and it provides a way to remember the formula (Figure $$\PageIndex{1}$$). After having gone through the stuff given above, we hope that the students would have understood, "Chain Rule Examples With Solutions"Apart from the stuff given in "Chain Rule Examples With Solutions", if you need any other stuff in math, please use our google custom search here. In examples such as the above one, with practise it should be possible for you to be able to simply write down the answer without having to let t = 1 + x² etc. Created: Dec 4, 2011. Both use the rules for derivatives by applying them in slightly different ways to differentiate the complex equations without much hassle. dy dx + y 2. %PDF-1.4 %�� Target: On completion of this worksheet you should be able to use the chain rule to differentiate functions of a function. rule d y d x = d y d u d u d x ecomes Rule) d d x f ( g ( x = f 0 ( g ( x )) g 0 ( x ) \outer" function times of function. Write the solutions by plugging the roots in the solution form. For this problem the outside function is (hopefully) clearly the exponent of 4 on the parenthesis while the inside function is the polynomial that is being raised to the power. That is, if f is a function and g is a function, then the chain rule expresses the derivative of the composite function f ∘ g in terms of the derivatives of f and g. Definition •If g is differentiable at x and f is differentiable at g(x), … Differentiating using the chain rule usually involves a little intuition. Since the functions were linear, this example was trivial. Hyperbolic Functions - The Basics. Differentiate algebraic and trigonometric equations, rate of change, stationary points, nature, curve sketching, and equation of tangent in Higher Maths. Although the chain rule is no more com-plicated than the rest, it’s easier to misunderstand it, and it takes care to determine whether the chain rule or the product rule is needed. 6 f ' x = 56 2x - 7 1. f x 4 2x 7 2. f x 39x 4 3. f x 3x 2 7 x 2 4. f x 4 x 5x 7 f ' x = 4 5 5 -108 9x - For the matrices that are stochastic matrices, draw the associated Markov Chain and obtain the steady state probabilities (if they exist, if In this presentation, both the chain rule and implicit differentiation will 'ɗ�m��sq'�"d��a�n��R�� S>��X��j��e�\�i'�9��hl�֊�˟o��[1dv�{� g�?�h��#H��G��~�1�yӅOXx�. If and , determine an equation of the line tangent to the graph of h at x=0 . We are nding the derivative of the logarithm of 1 x2; the of almost always means a chain rule. To avoid using the chain rule, recall the trigonometry identity , and first rewrite the problem as . {�F?р��F��㸝.�7�FS��V��zΑm��%a=;^�K��v_6��ft�8CR�,�vy>5d륜��UG�/��A�FR0ם�'�,u#K �B7~��1&��\|��J�ꉤZ��GV�q��T��{��70"cU��1j�V_U('u�k��ZT. There is a separate unit which covers this particular rule thoroughly, although we will revise it brieﬂy here. The Rules of Partial Diﬀerentiation Since partial diﬀerentiation is essentially the same as ordinary diﬀer-entiation, the product, quotient and chain rules may be applied. Study the examples in your lecture notes in detail. The Chain Rule for Powers The chain rule for powers tells us how to diﬀerentiate a function raised to a power. Solution. Ok, so what’s the chain rule? If and , determine an equation of the line tangent to the graph of h at x=0 . Moveover, in this case, if we calculate h(x),h(x)=f(g(x))=f(−2x+5)=6(−2x+5)+3=−12x+30+3=−12… A good way to detect the chain rule is to read the problem aloud. Section 1: Basic Results 3 1. 13) Give a function that requires three applications of the chain rule to differentiate. d dx (ex3+2x)= deu dx (where u = x3 +2x) = eu × du dx (by the chain rule) = ex3+2x × d dx (x3 +2x) =(3x2 +2)×ex3+2x. Chain Rule Examples (both methods) doc, 170 KB. General Procedure 1. To avoid using the chain rule, first rewrite the problem as . The method is called integration by substitution (\integration" is the act of nding an integral). The outer layer of this function is the third power'' and the inner layer is f(x) . It’s also one of the most used. If , where u is a differentiable function of x and n is a rational number, then Examples: Find the derivative of each function given below. It is often useful to create a visual representation of Equation for the chain rule. We must identify the functions g and h which we compose to get log(1 x2). The following chain rule examples show you how to differentiate (find the derivative of) many functions that have an “ inner function ” and an “ outer function.” For an example, take the function y = √ (x 2 – 3). Solution: This problem requires the chain rule. dx dy dx Why can we treat y as a function of x in this way? The Chain Rule is a formula for computing the derivative of the composition of two or more functions. NCERT Books. Example: a) Find dy dx by implicit di erentiation given that x2 + y2 = 25. A function of a … Chain rule. There is also another notation which can be easier to work with when using the Chain Rule. This 105. is captured by the third of the four branch diagrams on the previous page. Section 3-9 : Chain Rule. If , where u is a differentiable function of x and n is a rational number, then Examples: Find the derivative of … The following examples demonstrate how to solve these equations with TI-Nspire CAS when x >0. Show Solution For this problem the outside function is (hopefully) clearly the exponent of 4 on the parenthesis while the inside function is the polynomial that is being raised to the power. dv dy dx dy = 17 Examples i) y = (ax2 + bx)½ let v = (ax2 + bx) , so y = v½ ()ax bx . Solution. Click HERE to return to the list of problems. Scroll down the page for more examples and solutions. ()ax b dx dy = + + − 2 2 1 2 1 2 ii) y = (4x3 + 3x – 7 )4 let v = (4x3 + 3x – 7 ), so y = v4 4()(4 3 7 12 2 3) = x3 + x − 3 . Solution: The derivatives of f and g aref′(x)=6g′(x)=−2.According to the chain rule, h′(x)=f′(g(x))g′(x)=f′(−2x+5)(−2)=6(−2)=−12. Some examples involving trigonometric functions 4 5. by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²)². Make use of it. 2. by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²)². The following figure gives the Chain Rule that is used to find the derivative of composite functions. u and the chain rule gives df dx = df du du dv dv dx = cosv 3u2=3 1 3x2=3 = cos 3 p x 9(xsin 3 p x)2=3: 11. 2 1 0 1 2 y 2 10 1 2 x Figure 21: The hyperbola y − x2 = 1. Now apply the product rule. Solution Again, we use our knowledge of the derivative of ex together with the chain rule. Let f(x)=6x+3 and g(x)=−2x+5. The chain rule is probably the trickiest among the advanced derivative rules, but it’s really not that bad if you focus clearly on what’s going on. The inner function is the one inside the parentheses: x 2 -3. Section 3: The Chain Rule for Powers 8 3. The symbol dy dx is an abbreviation for ”the change in y (dy) FROM a change in x (dx)”; or the ”rise over the run”. For instance, if f and g are functions, then the chain rule expresses the derivative of their composition. d dx (e3x2)= deu dx where u =3x2 = deu du × du dx by the chain rule = eu × du dx = e3x2 × d dx (3x2) =6xe3x2. Example. Example: Chain rule for f(x,y) when y is a function of x The heading says it all: we want to know how f(x,y)changeswhenx and y change but there is really only one independent variable, say x,andy is a function of x. dx dg dx While implicitly diﬀerentiating an expression like x + y2 we use the chain rule as follows: d (y 2 ) = d(y2) dy = 2yy . %�쏢 Implicit Diﬀerentiation and the Chain Rule The chain rule tells us that: d df dg (f g) = . We are nding the derivative of the logarithm of 1 x2; the of almost always means a chain rule. has solution: 8 >> >< >> >: ˇ R = 53 1241 ˇ A = 326 1241 ˇ P = 367 1241 ˇ D = 495 1241 2.Consider the following matrices. Example: Find the derivative of . 2. To write the indicial equation, use the TI-Nspire CAS constraint operator to substitute the values of the constants in the symbolic … Solution 4: Here we have a composition of three functions and while there is a version of the Chain Rule that will deal with this situation, it can be easier to just use the ordinary Chain Rule twice, and that is what we will do here. 4 Examples 4.1 Example 1 Solve the differential equation 3x2y00+xy0 8y=0. If f(x) = g(h(x)) then f0(x) = g0(h(x))h0(x). functionofafunction. Find the derivative of y = 6e7x+22 Answer: y0= 42e7x+22 3x 2 = 2x 3 y. dy … (b) For this part, T is treated as a constant. y = x3 + 2 is a function of x y = (x3 + 2)2 is a function (the square) of the function (x3 + 2) of x. The chain rule gives us that the derivative of h is . Example: Find d d x sin( x 2). It states: if y = (f(x))n, then dy dx = nf0(x)(f(x))n−1 where f0(x) is the derivative of f(x) with respect to x. For this equation, a = 3;b = 1, and c = 8. The derivative is then, \[f'\left( x \right) = 4{\left( {6{x^2} + 7x} \right)^3}\left( … dx dg dx While implicitly diﬀerentiating an expression like x + y2 we use the chain rule as follows: d (y 2 ) = d(y2) dy = 2yy . Basic Results Diﬀerentiation is a very powerful mathematical tool. Examples using the chain rule. The rule is given without any proof. !w�@��Bab��JIDу>�Y"�Ĉ2FP;=�E��Y��Ɯ��M��3f�+o��DLp�[BEGG��#�=a��G�f�l��ODK��oDǟp&�)�8>ø�%�:�>��e �i":��&�_�J�\�\|�h��xH�=��e"}�,*��N8l��y��:ZY�S�b{�齖\|�3[�Zk��U�6H��h��%�T68N��o�� This section shows how to differentiate the function y = 3x + 1 2 using the chain rule. Let Then 2. Most of the basic derivative rules have a plain old x as the argument (or input variable) of the function. 1. Updated: Mar 23, 2017. doc, 23 KB. differentiate and to use the Chain Rule or the Power Rule for Functions. Use u-substitution. About this resource. dx dy dx Why can we treat y as a function of x in this way? Then . In this unit we will refer to it as the chain rule. (medium) Suppose the derivative of lnx exists. A good way to detect the chain rule is to read the problem aloud. Take d dx of both sides of the equation. Info. … <> Solution: Step 1 d dx x2 + y2 d dx 25 d dx x2 + d dx y2 = 0 Use: d dx y2 = d dx f(x) 2 = 2f(x) f0(x) = 2y y0 2x + 2y y0= 0 Step 2 Then (This is an acceptable answer. SOLUTION 6 : Differentiate . Chain Rule Worksheets with Answers admin October 1, 2019 Some of the Worksheets below are Chain Rule Worksheets with Answers, usage of the chain rule to obtain the derivatives of functions, several interesting chain rule exercises with step by step solutions and quizzes with answers, … To differentiate this we write u = (x3 + 2), so that y = u2 Solution: Using the above table and the Chain Rule. BOOK FREE CLASS; COMPETITIVE EXAMS. Usually what follows BNAT; Classes. du dx Chain-Log Rule Ex3a. Example 3 Find ∂z ∂x for each of the following functions. A simple technique for diﬀerentiating directly 5 www.mathcentre.ac.uk 1 c mathcentre 2009. Example 2. The outer function is √ (x). The Chain Rule 4 3. Now apply the product rule twice. Multi-variable Taylor Expansions 7 1. Then (This is an acceptable answer. For example, they can help you get started on an exercise, or they can allow you to check whether your intermediate results are correct Try to make less use of the full solutions as you work your way ... using the chain rule, and dv dx = (x+1) This might … Example: Differentiate . The chain rule for functions of more than one variable involves the partial derivatives with respect to all the independent variables. %PDF-1.4 SOLUTION 8 : Integrate . Solution This is an application of the chain rule together with our knowledge of the derivative of ex. Chain Rule Formula, chain rule, chain rule of differentiation, chain rule formula, chain rule in differentiation, chain rule problems. The difficulty in using the chain rule: Implementing the chain rule is usually not difficult. The more times you apply the chain rule to different problems, the easier it becomes to recognize how to apply the rule. Now apply the product rule. SOLUTION 9 : Integrate . Example Diﬀerentiate ln(2x3 +5x2 −3). In examples such as the above one, with practise it should be possible for you to be able to simply write down the answer without having to let t = 1 + x² etc. Does your textbook come with a review section for each chapter or grouping of chapters? Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. Chain rule examples: Exponential Functions. 1.3 The Five Rules 1.3.1 The … if x f t= ( ) and y g t= ( ), then by Chain Rule dy dx = dy dx dt dt, if 0 dx dt ≠ Chapter 6 APPLICATION OF DERIVATIVES APPLICATION OF DERIVATIVES 195 Thus, the rate of change of y with respect to x can be calculated using the rate of change of y and that of x both with respect to t. Let us consider some examples. Solution: This problem requires the chain rule. Then if such a number λ exists we deﬁne f′(a) = λ. Example 1: Assume that y is a function of x . The chain rule provides a method for replacing a complicated integral by a simpler integral. 2.Write y0= dy dx and solve for y 0. x��\Y��uN^��y�L�۪}1�-A�Al_�v S�D�u). ߼8\|~�! � ��5��n�J_�� .��w/n�x��t��c��VΫ�/Nb��h��AZ��o�ga��O�Vy\|K_J��LOO�\hỿ��: bچ1��ӭ�� [ =X�\|��5R��4nܶ3��4��t+u�� just about going along with the rule... Substitution ( \integration '' is the act of nding an integral ) detect chain. 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Regularity of the solutions of second order evolution equations and their attractors Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, Serie 4, Volume 14 (1987) no. 3, pp. 485-511. @article{ASNSP_1987_4_14_3_485_0, author = {Ghidaglia, J. M. and Temam, R.}, title = {Regularity of the solutions of second order evolution equations and their attractors}, journal = {Annali della Scuola Normale Superiore di Pisa - Classe di Scienze}, pages = {485--511}, publisher = {Scuola normale superiore}, volume = {Ser. 4, 14}, number = {3}, year = {1987}, zbl = {0666.35062}, mrnumber = {951230}, language = {en}, url = {http://archive.numdam.org/item/ASNSP_1987_4_14_3_485_0/} } TY - JOUR AU - Ghidaglia, J. M. AU - Temam, R. TI - Regularity of the solutions of second order evolution equations and their attractors JO - Annali della Scuola Normale Superiore di Pisa - Classe di Scienze PY - 1987 DA - 1987/// SP - 485 EP - 511 VL - Ser. 4, 14 IS - 3 PB - Scuola normale superiore UR - http://archive.numdam.org/item/ASNSP_1987_4_14_3_485_0/ UR - https://zbmath.org/?q=an%3A0666.35062 UR - https://www.ams.org/mathscinet-getitem?mr=951230 LA - en ID - ASNSP_1987_4_14_3_485_0 ER - %0 Journal Article %A Ghidaglia, J. M. %A Temam, R. %T Regularity of the solutions of second order evolution equations and their attractors %J Annali della Scuola Normale Superiore di Pisa - Classe di Scienze %D 1987 %P 485-511 %V Ser. 4, 14 %N 3 %I Scuola normale superiore %G en %F ASNSP_1987_4_14_3_485_0 Ghidaglia, J. M.; Temam, R. Regularity of the solutions of second order evolution equations and their attractors. Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, Serie 4, Volume 14 (1987) no. 3, pp. 485-511. http://archive.numdam.org/item/ASNSP_1987_4_14_3_485_0/ [1] S. Agmon, A. Douglis and L. Nirenberg, Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions I, Comm. Pure Appl. Math., 12 (1959), 623-727. \| MR \| Zbl [2] A.V. Babin and M.I. Vishik, Regular attractors of semigroups and evolution equations, J. Math Pures Appl., 62 (1983), 441-491. \| MR \| Zbl [3] A.V. Babin et M.I. Vishik, Attracteurs maximaux dans les équations aux dérivées partielles, Séminaire du Collège de France, J.L. Lions Ed., Pitman, Boston, 1985. \| MR \| Zbl [4] C. Bardos, A regularity theorem for parabolic equations, J. Funct. Anal., 7 (1971), 311-322. \| MR \| Zbl [5] L. Comtet, Advanced Combinatorics, D. Reidel, Dordrecht, 1978. \| Zbl [6] N. Dunford and J. Schwartz, Linear Operators Part II, Self adjoint Operators in Hilbert Spaces, Interscience Publishers, 1963. \| Zbl [7] H. Federer, Geometric Measure Theory, Springer-Verlag, New York, 1969. \| MR \| Zbl [8] J.M. Ghidagua and R. Temam, Attractors for damped nonlinear hyperbolic equations, J. Math. Pures et Appl., 66(3) 1987, pp. 273-319, and C.R. Acad. Sci. Paris. t. 300, série I, n° 7, 1985, pp. 185-188. \| MR \| Zbl [9] J.M. Ghidaglia and R. Temam, Dimension of the universal attractor describing time periodically driven Sine-Gordon equation, Transport Theory and Statistical Physics, 16, 1987, pp. 253-265. \| MR \| Zbl [10] J. Hale, Asymptotic behavior and dynamics in infinite dimensions, Lectures in Granada, Pitman, Boston, 1985. \| MR \| Zbl [11] J.K. Hale and J. Scheurle, Smoothness of bounded solutions of nonlinear evolution equations, J. Differential Equations, 56 (1985), 142-163. \| MR \| Zbl [12] A. Haraux, Two remarks on dissipative hyperbolic problems, Séminaire du Collège de France, J.L. Lions Ed., Pitman, Boston, 1985. \| MR [13] O.A. Ladyzhenksaya, V.A. Solonikov and N.N. Ural, Linear and quasilinear equations of parabolic type, Translation of Math. Monographs, n° 23, Amer. Math. Soc., Providence R.I. [14] N. Levinson, Transformation theory of nonlinear differential equations of the second order, Annals of Math., 45, 4, (1944), 723-737. \| MR \| Zbl [15] J.L. Lions, Quelques méthodes de résolution des problèmes aux limites non linéaires, Dunod, Paris, 1969. \| MR \| Zbl [16] J.L. Lions and E. Magenes, Nonhomogeneous boundary value problems and applications, Springer, Berlin, 1972 (translated from Dunod, Paris, 1968). \| Zbl [17] J.L. Lions and W.A. Strauss, Some nonlinear evolution equations, Bull. Soc. Math. France, 93 (1965), 43-96. \| Numdam \| MR \| Zbl [18] B. Mandelbrot, Fractals: Form, Chance and Dimension, Freeman, San Francisco, 1977. \| MR \| Zbl [19] J.B. Rauch and F. Masseyiii, Differentiability of solutions to hyperbolic initial-boundary value problem, Trans. Amer. Math. Soc., 189 (1974), 303-318. \| MR \| Zbl [20] J. Sather, The existence of global classical solution of the initial-boundary value problem for □ u + u3 = f. Arch. Rational Mech. Anal., 22 (1966), 292-307. \| Zbl [21] I. Segal, Nonlinear semi-groups, Ann. of Math., 78, n° 2. 1963. \| MR \| Zbl [22] J. Smale, Smooth solutions of the heat and wave equations, Comment. Math. Helvetici, 55 (1980), 1-12. \| MR \| Zbl [23] R. Temam, Behaviour at time t = 0 of the solutions of semi-linear evolution equations, J. Differential Equations, 43 (1982), 73-92. \| MR \| Zbl [24] J.M. Bony, Calcul symbolique et propagation des singularités pour les équations aux dérivées partielles non linéaires, Ann. Scient. Ecole Norm. Sup., 4e Série, 14 (1981), 209-246. \| Numdam \| MR \| Zbl	{}
# If $\text{ }{{\text{K}}_{\text{sp}}}$ of $\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$is $\text{ 8 }\times \text{1}{{\text{0}}^{-12}}\,\text{ }$, the molar solubility of $\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$in $\text{ 0}\text{.1 M AgN}{{\text{O}}_{\text{3}}}$ is :A) $\text{ 8}\times \text{1}{{\text{0}}^{-12}}\text{M}$B) $\text{ 8}\times \text{1}{{\text{0}}^{-10}}\text{M}$C) $\text{ 8}\times \text{1}{{\text{0}}^{-11}}\text{M}$D) $\text{ 8}\times \text{1}{{\text{0}}^{-13}}\text{M}$ Verified 119.7k+ views Hint: The molar solubility is the number of moles of a substance that can be dissolved per litre of the solution after achieving saturation. The molar solubility for salt $\text{ AxBy }$ is written as, $\text{ AxBy }\rightleftharpoons \text{ x}{{\text{A}}^{\text{+}}}\text{ (aq) + y}{{\text{B}}^{-}}\text{(aq) }$ Solubility product is, $\text{ }{{\text{K}}_{\text{sp}}}\text{ = }{{\left[ {{\text{A}}^{\text{+}}} \right]}^{\text{x}}}{{\left[ {{\text{B}}^{-}} \right]}^{\text{y}}}\text{ }$ The presence of the same ion in the solution reduces the solubility of the salt. The increased concentration of ions is taken into consideration while solving for the molar solubility. We have given the following data on the problem. Solubility product of silver carbonate$\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ is $\text{ }{{\text{K}}_{\text{sp}}}\text{ = 8 }\times \text{1}{{\text{0}}^{-12}}\,\text{ }$ The $\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$is in the $\text{ 0}\text{.1 M AgN}{{\text{O}}_{\text{3}}}$ solution. We are interested to find out the molar solubility of $\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$. The solute like silver carbonate dissolves in the solution and forms corresponding silver and carbonate ions. The equilibrium reaction between the ions and solute is as shown below, $\text{ }\begin{matrix} {} & \text{A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{(s)} & \rightleftharpoons & \text{2A}{{\text{g}}^{\text{+}}}\text{(aq)} & \text{+} & \text{CO}_{\text{3}}^{\text{2-}} \\ \text{Before} & \text{S} & {} & \text{0} & {} & \text{0} \\ \text{After} & \text{-} & {} & \text{2S+0}\text{.1} & {} & \text{S} \\ \end{matrix}$ This $\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$is dissolved in the $\text{ 0}\text{.1 M AgN}{{\text{O}}_{\text{3}}}$ solution. Both salts have the common$\text{ A}{{\text{g}}^{+}}\text{ }$. Thus, the silver ion from both the salt contributes towards the molar solubility.The solubility product for the $\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$is written as follows, $\text{ }{{\text{K}}_{\text{sp}}}\text{ = }{{\left[ \text{A}{{\text{g}}^{\text{+}}} \right]}^{2+}}\left[ \text{CO}_{3}^{2-} \right]\text{ }$ Let’s substitute the values in the above equation we have, $\text{ 8}\times \text{1}{{\text{0}}^{-12}}\text{ = }{{\left( 0.1+2S \right)}^{2+}}\left( S \right)$ The value of solubility product is very small, thus we neglect it $\text{ 0}\text{.1 + 2S = 0}\text{.1 }$.The equation now becomes, \begin{align} & \text{ 8}\times \text{1}{{\text{0}}^{-12}}\text{ = }{{\left( 0.1 \right)}^{2}}\left( S \right) \\ & S=\dfrac{\text{8}\times \text{1}{{\text{0}}^{-12}}}{{{\left( 0.1 \right)}^{2}}}=\text{8}\times \text{1}{{\text{0}}^{-10}}\text{M} \\ \end{align} Therefore, molar solubility of a silver carbonate in presence of silver nitrate is equal to $\text{ 8}\times \text{1}{{\text{0}}^{-10}}\text{M}$ . Hence, (B) is the correct option. Note: The common ion effect is the addition of the salt to a solution such that the two slats have the common ion.for example, silver carbonate and silver nitrate. Both salts have the common ion as silver. The solubility of the salt is measuredly affected by the common ion. The silver carbonate is less soluble as compared to silver nitrate, but the presence of silver nitrate increases the molar solubility of less soluble salt.	{}
## Gradient methods for minimizing composite objective function In this paper we analyze several new methods for solving optimization problems with the objective function formed as a sum of two convex terms: one is smooth and given by a black-box oracle, and another is general but simple and its structure is known. Despite to the bad properties of the sum, such problems, both … Read more ## Gradient methods for minimizing composite objective function In this paper we analyze several new methods for solving optimization problems with the objective function formed as a sum of two convex terms: one is smooth and given by a black-box oracle, and another is general but simple and its structure is known. Despite to the bad properties of the sum, such problems, both … Read more ## A secant method for nonsmooth optimization The notion of a secant for locally Lipschitz continuous functions is introduced and a new algorithm to locally minimize nonsmooth, nonconvex functions based on secants is developed. We demonestrate that the secants can be used to design an algorithm to find descent directions of locally Lipschitz continuous functions. This algorithm is applied to design a … Read more ## Max-min separability: incremental approach and application to supervised data classification A new algorithm for the computation of a piecewise linear function separating two finite point sets in $n$-dimensional space is developed and the algorithm is applied to solve supervised data classification problems. The algorithm computes hyperplanes incrementally and it finds as many hyperplanes as necessary to separate two sets with respect to some tolerance. An … Read more ## The Speed of Shor’s R-Algorithm Shor’s r-algorithm is an iterative method for unconstrained optimization, designed for minimizing nonsmooth functions, for which its reported success has been considerable. Although some limited convergence results are known, nothing seems to be known about the algorithm’s rate of convergence, even in the smooth case. We study how the method behaves on convex quadratics, proving … Read more ## A Proximal Cutting Plane Method Using Chebychev Center for Nonsmooth Convex Optimization An algorithm is developed for minimizing nonsmooth convex functions. This algorithm extends Elzinga-Moore cutting plane algorithm by enforcing the search of the next test point not too far from the previous ones, thus removing compactness assumption. Our method is to Elzinga-Moore’s algorithm what a proximal bundle method is to Kelley’s algorithm. Instead of lower approximations … Read more ## Discrete gradient method: a derivative free method for nonsmooth optimization In this paper a new derivative-free method is developed for solving unconstrained nonsmooth optimization problems. This method is based on the notion of a discrete gradient. It is demonstrated that the discrete gradients can be used to approximate subgradients of a broad class of nonsmooth functions. It is also shown that the discrete gradients can … Read more ## A conic interior point decomposition approach for large scale semidefinite programming We describe a conic interior point decomposition approach for solving a large scale semidefinite programs (SDP) whose primal feasible set is bounded. The idea is to solve such an SDP using existing primal-dual interior point methods, in an iterative fashion between a {\em master problem} and a {\em subproblem}. In our case, the master problem … Read more ## Computing Proximal Points on Nonconvex Functions The proximal point mapping is the basis of many optimization techniques for convex functions. By means of variational analysis, the concept of proximal mapping was recently extended to nonconvex functions that are prox-regular and prox-bounded. In such a setting, the proximal point mapping is locally Lipschitz continuous and its set of fixed points coincide with … Read more ## An incremental method for solving convex finite minmax problems We introduce a new approach to minimizing a function defined as the pointwise maximum over finitely many convex real functions (next referred to as the “component functions”), with the aim of working on the basis of “incomplete knowledge” of the objective function. In fact, a descent algorithm is proposed which does not necessarily require at … Read more	{}
One of the most fundamental inequalities in math is the Trivial Inequality, which says: If x is a real number, then $x^2 \geq 0$. $x^2 = 0$ if and only if $x = 0$. This seemingly innocuous statement, which most of us may already know as ‘all squares are nonnegative’, can actually be used to prove much more complex inequalities. Let’s start with some examples. Also please note the equality case – that in order for a squared expression to be equal to 0, the expression itself must equal 0. This is a hint that it will be necessary in some of the exercises at the end. Example 1: Prove that $a^2 + b^2 \geq 2ab$ for real numbers $a, b$ We see that there are some squares, which is what we want, however that’s not enough. All we can say about those is that $a^2 \geq 0$, and the same is true of $b^2$. The issue is that $2ab$ could be positive and bigger than the two squares, so we try again. Moving the $2ab$ to the left side, we get a familiar looking expression, $a^2 - 2ab + b^2$, which looks like it can be factorized. Sure enough, we get $(a-b)^2 \geq 0$. Now, we have a square! And from the Trivial Inequality, it is true, as all squares are nonnegative. In an actual proof, we would have to start with the factorized expression and work our way back, as this makes more logical sense, but if you’re not bothered about, then don’t worry. Example 2: Prove that $b^2 / a^2 \geq 4b/a - 4$ for real numbers $a, b$ This looks slightly trickier, because the algebra doesn’t look as nice. So let’s clean it up! Because our inequality has a square one side and 0 on the other, we try and replicate this by moving everything to the same side. Moving all of the terms to the left side, we get $b^2 / a^2 - 4b/a + 4 \geq 0$. Denominators are hard to work with, so we multiply by $a^2$ on both sides, to get $b^2 -4ab + 4a^2 \geq 0$. This is much more manageable and familiar too, as it looks like we can factorize it. Sure enough, we can factorize to get $(b-2a)^2 \geq 0$, which is again true by the Trivial Inequality. Example 3 (a slightly trickier one): Prove that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for real numbers $a, b, c$ We have squares and terms on both sides, so we move everything to one side. We get $a^2 + b^2 + c^2 - ab - bc - ca \geq 0$. We’re kind of at a loss now. We have squares, but we also have the subtracted terms. If only we had $-2ab$, so we could factorize… We try this approach, subtracting $-ab$ from both sides and factorizing to get $(a-b)^2 + c^2 - bc - ca \geq -ab$. This seems promising, as we have a nonnegative quantity now, but now we’re still stuck with the $bc$ and $ca$. Also, as $a, b, c$ are not necessarily always positive, the right side could be positive as well, making things more tricky, so we try again. Because we want $-2ab, -2bc$ and $-2ac$, we multiply both sides by two. $2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac \geq 0$. We try to factorize now, taking $a^2 - 2ab + b^2$, to get $(a-b)^2 + a^2 + b^2 + 2c^2 - 2ac - 2bc \geq 0$. We don’t really see anything which indicates this won’t work, or will get messy and so we keep going, to get $(a-b)^2 + (a-c)^2 + (b-c)^2 \geq 0$! And now we are done, as each of the squared expressions is nonnegative, so their sum is also nonnegative! Note: for ‘is something true’ questions, either show it is not by providing a counterexample (a value which makes the inequality false) or prove that it is indeed true. Exercises (all variables represent real numbers unless otherwise specified) 1. Is $a^2 + b^2 + ab \geq 0$ true for real numbers $a, b$ where $a, b$ are of the same sign (if one is positive, the other is negative and vice versa)? 2. Is the problem above true if $a, b$ are of opposite sign? 3. Prove $a/b + b/a \geq 2$ 4. Prove $a/2 + b/2 \geq \sqrt{ab}$ (AM-GM inequality in two variables) 5. Find all solutions to the equation $34y + z = 2\sqrt{34yz}$ 6. Find all integer solutions $x, y, z$ to the equation $x^2 + 5y^2 + 10z^2 = 4xy + 6yz + 2z - 1$ (AoPS) When you’re done, email your answers to Zach Marinov, Paris Suksmith, or Aarit Bhattacharya for feedback, and don’t forget to post any questions in the comments section below!	{}
17th International Conference on Ion Sources 15-20 October 2017 CERN Europe/Zurich timezone Power Transfer Efficiency in Inductively Coupled Radio-Frequency Ion Source: Case Study for the NIO1 17 Oct 2017, 16:30 2h 30m CERN CERN Centre international de Conférence Genève (CICG). http://www.cicg.ch/ Poster presentation Ion sources for fusion Speaker Ms Palak Jain (Consorzio RFX & Università degli studi di Padova) Description An ion source called NIO1 (Negative Ion Optimization, phase 1) has been developed by Consorzio RFX and INFN-LNL and is currently in operation in the Consorzio RFX premises in Padova. NIO1 has a radio frequency (RF) inductively coupled (IC) ion source designed to produce a total of 130 mA $\text{H}^-$ current and to accelerate the ions up to energy of 60 keV; it operates at a frequency of 2 ± 0.2 MHz with maximum RF power of 2.5 kW. In the IC RF ion source, plasma is generated and heated in a component called driver by the RF field induced by a coil. In a preliminary work, a methodology was developed to evaluate the equivalent electrical parameters for IC RF ion sources and power transfer efficiency to the plasma. This methodology takes into account various relevant mechanisms of plasma particle dynamics like ohmic heating, stochastic heating and classical skin depth, which are modelled based on the available literature. These formulations are then integrated with a simplified scheme describing the magnetic coupling between plasma and the RF coil (2-windings transformer model). In order to develop the methodology further, this work provides an improved equivalent electrical model of the driver based on a multiple winding transformer scheme accounting for the mutual coupling between the RF coil, plasma and the surrounding metallic structures. The developed methodology is then applied to NIO1 ion source. The results in terms of equivalent electrical parameters and power transfer efficiency will be presented based on both the simplified and the improved approach. Acknowledgements: This work has been carried out within the framework of the EUROfusion Consortium and has received funding from the Euratom research and training programme 2014-2018 under grant agreement No 633053. The views and opinions expressed herein do not necessarily reflect those of the European Commission. Primary author Ms Palak Jain (Consorzio RFX & Università degli studi di Padova) Co-authors Dr Mauro Recchia (Consorzio RFX) Mr Alberto Maistrello (Consorzio RFX & Università degli studi di Padova) Dr Elena Gaio (Consorzio RFX) Dr Pierluigi Veltri (Consorzio RFX & INFN-LNL)	{}
International Tables for Crystallography Volume D Physical properties of crystals Edited by A. Authier International Tables for Crystallography (2013). Vol. D, ch. 2.3, pp. 336-337 ## Section 2.3.3.2. Symmetry properties of the scattering cross section I. Gregoraa* aInstitute of Physics, Academy of Sciences of the Czech Republic, Na Slovance 2, CZ-18221 Prague 8, Czech Republic Correspondence e-mail: gregora@fzu.cz #### 2.3.3.2. Symmetry properties of the scattering cross section \| top \| pdf \| The quantity that controls the symmetry properties of the scattering cross section due to excitation is the squared modulus of the corresponding second-order susceptibility (second-rank tensor), contracted with the two polarization vectors of the incident and scattered light: The nonlinear susceptibility tensor is usually referred to as the first-order Raman tensor (defined in the literature to within a factor). Before discussing the consequences of the crystal symmetry on the form of the Raman tensor, let us mention two important approximations on which conventional analysis of its symmetry properties is based. In a general case, the second-order susceptibilities are not necessarily symmetric. However, they fulfil a general symmetry property which follows from the symmetry of the scattering with respect to time inversion. Since the anti-Stokes process can be regarded as a time-inverted Stokes process (exchanging the role of the incident and scattered photons), it can be shown that in non-magnetic materials the susceptibilities obey the relation In the quasi-static limit, i.e. if the scattering frequency is negligibly small compared with the incident photon frequency (), it follows that the susceptibilities of non-magnetic materials become symmetric in the Cartesian indices α, β. This symmetry is very well fulfilled in a great majority of cases. Appreciable antisymmetric contributions are known to occur, e.g. under resonant conditions, where the quasi-static approximation breaks down as the energy of the incident (or scattered) photon approaches those of electronic transitions. Thus, in the first approximation, we set equal to zero and remove the time dependence in the phonon amplitudes, treating the normal coordinates as static. Then the nonlinear susceptibilities correspond to susceptibility derivatives, where we suppressed the explicit dependence on and introduced a simplified notation for the Raman tensor , still keeping the dependence on the scattering wavevector. In deriving the symmetry properties of the Raman tensor that follow from the crystal lattice symmetry, the main point is thus to determine its transformation properties under the symmetry operation of the crystal space group. Since the magnitude of the scattering vector is very small compared with the Brillouin-zone dimensions, another conventional approximation is to neglect the q dependence of the susceptibilities. Setting enables us to analyse the symmetry of the Raman tensor in terms of the factor group , which is isomorphous to the point group of the crystal lattice. This approach is, again, appropriate for the vast majority of cases. An important exception is, for instance, the scattering by acoustic modes (Brillouin scattering) or scattering by longitudinal plasma waves in semiconductors (plasmons): in these cases the Raman tensor vanishes for , since this limit corresponds to a homogeneous displacement of the system. Possible q-dependent effects can be treated by expanding the Raman tensor in powers of q and using compatibility relations between the symmetries at and at the full symmetries applicable in the case. Let us mention that another notation is sometimes used in the literature for the Raman tensor. Since the square modulus of a second-rank tensor contracted with two vectors can be written as a fourth-rank tensor contracted with four vectors, one can introduce a fourth-rank tensor I(j), so that the scattering cross section of the jth mode is If there are no antisymmetric components in the susceptibility derivatives, it can be shown that the fourth-rank tensor has at most 21 independent components, as for the elastic constants tensor.	{}
# Electrodynamics and the Lagrangian density [duplicate] Could anyone tell me what equations can I obtain from the Lagrangian density $${\cal L}(\phi,\,\,\phi_{,i},\,\,A_i, \dot A_i,\,\,A_{i,j})~=~\frac{1}{2}\|\dot A+\nabla\phi\|^2-\frac{1}{2}\|\nabla \times A\|^2-\rho\phi+J\cdot A$$ by the Euler-Lagrange equations? ## marked as duplicate by Qmechanic♦ lagrangian-formalism StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 12 at 19:02 If your $A$ is a 3D vector potential $\mathbf{A}$, then your Lagrangian is a Lagrangian for electromagnetic field potentials $(\phi,\mathbf{A})$ created with a charge $\propto c$ and the 3D charge velocity $\propto d$. The exact proportionality coefficients depend on units.	{}
CPL - Chalmers Publication Library # Stochastic domination for the Ising and fuzzy Potts models Marcus Warfheimer (Institutionen för matematiska vetenskaper, matematik) (2010) [Preprint] We discuss various aspects concerning stochastic domination for the Ising model and the fuzzy Potts model. We begin by considering the Ising model on the homogeneous tree of degree $d$, $\Td$. For given interaction parameters $J_1$, $J_2>0$ and external field $h_1\in\RR$, we compute the smallest external field $\tilde{h}$ such that the plus measure with parameters $J_2$ and $h$ dominates the plus measure with parameters $J_1$ and $h_1$ for all $h\geq\tilde{h}$. Moreover, we discuss continuity of $\tilde{h}$ with respect to the three parameters $J_1$, $J_2$, $h$ and also how the plus measures are stochastically ordered in the interaction parameter for a fixed external field. Next, we consider the fuzzy Potts model and prove that on $\Zd$ the fuzzy Potts measures dominate the same set of product measures while on $\Td$, for certain parameter values, the free and minus fuzzy Potts measures dominate different product measures. For the Ising model, Liggett and Steif proved that on $\Zd$ the plus measures dominate the same set of product measures while on $\T^2$ that statement fails completely except when there is a unique phase. CPL Pubid: 118145 # Läs direkt! Länk till annan sajt (kan kräva inloggning) # Institutioner (Chalmers) Institutionen för matematiska vetenskaper, matematik (2005-2016)	{}
# Insights Why Is Quantum Mechanics So Difficult? - Comments 1. Aug 12, 2014 ### Greg Bernhardt Last edited: Aug 24, 2016 2. Aug 12, 2014 ### Staff: Mentor Agree entirely. The mathematical formalism is required to understand the concepts. That is exactly the process taken in my favourite QM book, Ballentine, and is much more rational than the semi historical approach usually taken. The only problem with Ballentine is it is at graduate level. I have always thought a book like Ballentine, but accessible to undergraduate students, would be the ideal introduction. In particular it would have a 'watered' down version of the very important chapter 3 that explains the dynamics of QM from symmetry. Its a long hard slog even for math graduates like me - definitely not for undergraduates. But the key results and theorems can be stated, and their importance explained, without the proofs. I think its very important for beginning students to understand the correct foundation of Schroedinger's equation etc from the start - if the not the mathematical detail. Thanks Bill Last edited: Aug 12, 2014 3. Aug 12, 2014 ### bohm2 I don't think there's one best way. Some learn better using the approach advocated here. Others learn better the other way around. 4. Aug 12, 2014 ### WannabeNewton Honestly I think at the undergraduate level QM is the easiest physics class one has to take. It is just a cookbook on calculations. Every book is uninspired and my class was certainly uninspired. It is an incredibly boring subject at this level. So I don't think difficulty is the issue. It is simply the lack of physical concepts and a healthy dose of philosophy that is avoided when teaching QM at the undergraduate level. Indeed one of the professors I know basically called Griffiths' book a cookbook in differential equations. A good book can go a long way. For me the saving grace was Landau and Lifshitz. It is the sole reason I started liking QM. Seriously the way undergrad QM is taught really isn't fun for the students. Boredom from a lack of intellectusl stimulation really isn't how a physics class should be. 5. Aug 12, 2014 ### atyy Hmmm, I still can't derive the Stefan-Boltzmann whatever - chills down my spine. How is that easy? 6. Aug 12, 2014 ### WannabeNewton What 7. Aug 12, 2014 ### atyy Is it easy? 8. Aug 12, 2014 ### WannabeNewton Im actually not sure what youre referring to. Are you talking about the Stefan Boltzmann law of radiation? Im not sure what that has to do with undergrad QM apart from historical impetus but there is a particularly lucid derivation in section 9.13 of Reif if youre interested. It's more of a statistical mechanics derivation. Which is good because statistical mechanics, both classical and quantum, is actually extremely interesting at the undergrad level. 9. Aug 12, 2014 ### Staff: Mentor That I agree with. I gave it away for health reasons no need to go into here. But I did enrol in a Masters in Applied Math at my old alma mater that included a good dose of QM. When mapping out the course structure with my adviser he said forget the intro QM course - since you have taken courses on advanced linear algebra, Hilbert spaces, partial differential equations etc it's completely redundant. Other students with a similar background to mine were totally bored. He suggested I start on the advanced course right away. Really I think it points to doing a math of QM course before the actual QM course where you study the Dirac notation etc - basically the first and a bit of the second chapter of Ballentine. You can then get stuck into the actual QM. And yes - I like Landau and Lifshitz too. Their Mechanics book was a revelation; QM, while good and better than most, wasn't quite as impressive to me as Ballintine. But like all books in that series it's, how to put it, terse, and the problems are, again how to put it, challenging, but to compensate actually relevant. Thanks Bill Last edited: Aug 13, 2014 10. Aug 12, 2014 ### atyy Actually, I only dimly remember what it is, although it was very exciting. It sounds right that it should be in a stat mech book, because the whole point IIRC was that classical thermodynamics was able to derive all sorts of completely correct things about blackbody radiation, yet classical stat mech could not. Then miraculously when one switched to quantum stat mech everything fell in place with classical thermo. I remember the narrative, but none of the calculations except Planck's. The text we used was Gasiorowicz, and I think his chapter 1 is all about this. Apart from the Stefan-Boltzmann law, the other amazing derivation was Wien's displacement law. IIRC, these were all from classical thermodynamics, with no quantum mechanics, yet they are correct! Last edited: Aug 12, 2014 11. Aug 13, 2014 ### vanhees71 Well, the historic approach is bad. You are taught "old quantum mechanics" a la Einstein and Bohr only to be adviced to forget all this right away when doing "new quantum mechanics". I've never heard that it is a good didactical approach to teach something you want the students to forget. They always forget inevitably most important things you try to teach them anyway, but in a kind of Murphy's Law they remember all the wrong things being taught in the introductory QM lecture. You see it in this forum: Most people remember the utmost wrong picture about photons, and it is very difficult to make them forget these ideas, because they are apparently simple. The only trouble is they are also very wrong. As Einstein said, you should explain things as simple as possible but not simpler. Concerning philosophy, I think the healthy dose is 0! Nobody tends to introduce some philosophy in the introductory mechanics or electrodynamics lecture. Why should one need to do so in introdutory QM? If you want to rise interpretational problems at all, you shouldn't do this in QM 1 or at least not too early. First you should understand the pure physics, and that's done with the minimal statistical interpretation. If you like Landau/Lifshits (all volumes are among the most excellent textbooks ever written, but they are for sure not for undergrads; this holds also true for the also very excellent Feynman lectures which are clearly not a freshmen course but benefit advanced students a lot), I don't understand why you like to introduce philosophy into a QM lecture. This book is totally void of it, and that's partially what it makes so good ;-)). 12. Aug 13, 2014 ### Staff: Mentor Abso-friggen-lutely. And to make matters worse they do not go back and show exactly how the correct theory accounts for the historical stuff and students are left with a sort of hodge podge, not knowing whats been replaced and what changed or the why of things like the double slit experiment. Thanks Bill 13. Aug 13, 2014 ### microsansfil May be QM is primarily predictive. Quantum mechanics construed as a predictive structure. After we try to interpret it with épistemic or ontological human sense. For example "The debate on the interpretation of quantum mechanics has been dominated by a lasting controversy between realists and empiricists" : http://michel.bitbol.pagesperso-orange.fr/transcendental.html Patrick 14. Aug 13, 2014 ### Staff: Mentor I think philosophers worry more about that sort of thing more than physicists or mathematicians. An axiomatic development similar to what Ballentine does is all that's really required, with perhaps a bit of interpretational stuff thrown in just to keep the key idea behind the principles clear. And I really do mean IDEA - not ideas - see post 137: It always amazes me exactly the minimal assumptions that goes into QM and what needs 'interpreting'. Thanks Bill 15. Aug 13, 2014 ### stevendaryl Staff Emeritus When discussing the best approach to teaching something like quantum mechanics, I think you really have to consider the purpose in teaching it. Some of the people studying quantum mechanics are going to go on to become physics researchers, but my guess is that that is a tiny, tiny fraction. A small fraction of those who learn QM go on to get undergrad physics degrees, and a small fraction of them go on to get postgraduate physics degrees, and a small fraction of them go on to get jobs as physics researchers. So for the majority (I'm pretty sure it's a majority) who are not going to become physics researchers, what do we want them to know about quantum mechanics? I'm not asking these as rhetorical questions, I really don't know. But I think that if we want people to be able to solve problems in QM, there might be a best way to teach it to get them up to speed in solving problems. If we want them to understand the mathematical foundations, there might be a different way to teach it. If we want them to be able to apply QM to problems arising in other fields--say chemistry or biology or electronics--there might be another best way to teach it. So when people say things like "You shouldn't bring up X, because that will just confuse the student" or "The historical approach, with all of its false starts and blunders, is just not relevant to today's students", they need to get clear what, exactly, they want the student to get out of their course in QM. And I think that the answer to that question isn't always the same for all students. 16. Aug 13, 2014 ### stevendaryl Staff Emeritus I assume you mean the idea expressed by the sentence: I would say that that's a single sentence, but I'm not sure I would call it a single idea. There are many other ideas involved in understanding why we would want basis-independence, why we are looking for probabilities in the first place, why we want the outcome probabilities to be determined by $E_i$ (as opposed to depending on both the system being measured and the device doing the measurement), what is an "observation" or "measurement", why should it have a discrete set of possible results, etc. 17. Aug 13, 2014 ### microsansfil probably not theory, but the people : Erwin Schrodinger : Mind and matter - What Is Life? - My View of the World - ... Werner Heisenberg : Physics and Philosophy: The Revolution in Modern Science - Mind and Matter - The physicist's conception of nature - ... ... Patrick 18. Aug 13, 2014 ### Staff: Mentor Know both those books - but they are old mate. These days the following is much better at that sort of level: https://www.amazon.com/Understanding-Quantum-Mechanics-Roland-Omnès/dp/0691004358 But of relevance to this thread you will get a lot more out of that book if you know some of the real deal detail. Thanks Bill Last edited by a moderator: May 6, 2017 19. Aug 13, 2014 ### microsansfil To understand the quantum theory in terms of mathematical language, we have in "France" some good free lecture like this one from "Ecole polytechnique" : http://www.phys.ens.fr/~dalibard/Notes_de_cours/X_MQ_2003.pdf on the other side there is not a unique look on its interpretation. Patrick 20. Aug 13, 2014 ### atyy In fact Landau and Lifshitz introduce philosophy early and correctly in their QM book, which is what makes it so wonderful. 21. Aug 13, 2014 ### stevendaryl Staff Emeritus I just wanted to add that, whether or not the student is going to go on to become a physicist, there are certain ways to teach quantum mechanics that I think are just bad. There might be ways to teach a little bit of the feel of what quantum mechanics is about without getting into the mathematics that would be necessary to solve actual problems. But what is worse than useless is to skip the actual facts about quantum mechanics and instead teach people sound bites about how "Quantum mechanics teaches us that the mind creates its own reality" or whatever Deepak Chopra might say about it. However, the goal of giving the layman a flavor of quantum mechanics without being misleading is very difficult to pull off. 22. Aug 13, 2014 ### RUTA I start my undergrad QM course with Quantum Mechanics and Experience, David Z. Albert, Harvard Univ Press, 1992, ISBN 0-674-74113-7. It's not the dry math start that you find in, say, Principles of Quantum Mechanics, 2nd Ed., R. Shankar, Plenum Press, 1994, ISBN 0-306-44790-8. Don't get me wrong, I like Shankar and use it after the students do the calculations in Albert and some AJP papers cited below. I choose this intro because it involves some interesting phenomena that we can easily model mathematically. The phenomena is electron spin to include entanglement, so its "weirdness" tends to motivate the students to work on the matrix algebra needed to model it. And, the parameters in the matrix algebra correspond directly to Stern-Gerlach orientations and spatial locations of detector outcomes which are easy to visualize. Thus, while the outcomes are "mysterious," the modeling of the experiment is intuitive. I then have them reproduce the quantum calculations for each of Mermin's AJP papers on "no instruction sets": "Bringing home the quantum world: Quantum mysteries for anybody," N.D. Mermin, Am. J. Phys. 49, Oct 1981, 940-943. “Quantum mysteries revisited,” N.D. Mermin, Am. J. Phys. 58, Aug 1990, 731-734. “Quantum mysteries refined,” N.D. Mermin, Am. J. Phys. 62, Oct 1994, 880-887. Again, in each case, there is an easy-to-understand counterintuitive outcome that motivates the students to work with the simple, intuitive matrix modeling. We finish this intro by reproducing all the calculations in: “Entangled photons, nonlocality, and Bell inequalities in the undergraduate laboratory,” D. Dehlinger and M.W. Mitchell, Am. J. Phys. 70, Sep 2002, 903-910 to include the error analysis. That gives them a grounding in an actual experiment. Only after all that do we proceed to Shankar. 23. Aug 14, 2014 ### Barry911 However; There seems to be a Platonic trend among the "speculative" types including all the string theorists, i.e.-no observables, no predictions....sounds like an elegant theory of pure mathematics. Multiverses, "anthropic principle, demanding multiverses, Maldecena's conjecture ADS/cft also elegant but lacking physical relevance. His holographic universe came about because he felt that information is conserved in two dimensions inside black holes! Have these people no humility? Q.M. requires more than analysis unless your limited to applied physics and just don't care. The power of QM of course lies in its mathematical formalism but it is a physical theory and requires interpretation. At this time, however (I'll say it again) all interpretation is premature. but even the extraordinarily inelegant interpretations are better than a strictly analytical i.e.-Platonic approach. Respectfully, Barry911 24. Aug 14, 2014 ### Staff: Mentor M Pure math is the last thing QM is. At the axiomatic level the primitive of the theory is an observation eg see post 137: The fundamental axiom is: An observation/measurement with possible outcomes i = 1, 2, 3 ..... is described by a POVM Ei such that the probability of outcome i is determined by Ei, and only by Ei, in particular it does not depend on what POVM it is part of. When you get right down to it much of the difficulty of QM boils down to exactly what is an observation? Its generally taken to be something that occurs here in an assumed classical common-sense world. But QM is supposed to be the theory that explains that world - yet assumes its existence from the get-go. Much of the modern research into the foundations of QM has been how to resolve that tricky issue - with decoherence playing a prominent role. A lot of progress has been made - but issues still remain - although opinions vary as to how serious they are. Thanks Bill 25. Aug 15, 2014 ### Barry911 Problem: The outcome of single quantun outcomes does not yield a meaningful outcome. Iteration of "identical experiments" yields probability densities. P-densities do not predict where a quantum event will occur only statistical weightings	{}
# Understanding the pdf of a truncated normal distribution Suppose $$\boldsymbol{x} = (x_1, \ldots, x_m)^T$$ follows a multivariate normal distribution with 2-sided truncation $$a_i \leq x_i \leq b_i$$. This is a truncated multivariate normal defined by $$TN(\mu, \Sigma, a, b)$$ where $$a = (a_1, \ldots, a_m)^T$$ and $$b = (b_1, \ldots, b_m)^T$$. The probability density function for $$TN(\mu, \Sigma, a, b)$$ can be expressed as $$f(x, \mu, \Sigma, a, b) = \frac{\exp\left\{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)\right\}}{\int_{a_1}^{b_1} \int_{a_2}^{b_2}\cdots \int_{a_m}^{b_m} \exp\left\{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)\right\}dx_m \cdots dx_1}$$ for $$a \leq x \leq b$$ and 0 otherwise. My question is: suppose $$x = (x_1, x_2)$$, is it possible to have the following pdf for $$TN\left(\mu, \Sigma, a = (a_1, a_2), b = (b_1, 2x_1)\right)$$ $$f(x, \mu, \Sigma, a, b) = \frac{\exp\left\{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)\right\}}{\int_{a_1}^{b_1} \int_{a_2}^{2x_1} \exp\left\{-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu)\right\}dx_2dx_1}$$ where the upper truncation for $$x_2$$ depends on $$x_1$$? Is this valid? • Don't let the notation confuse you: you can truncate any distribution to any measurable set of positive measure. The procedure is the same: all values outside the set get zero probability and the chances of all values inside the set are inflated to make the total truncated probability equal to $1.$ Here's an example of truncation to the complement of a disk: stats.stackexchange.com/questions/157963 and here is an example of arbitrarily complex truncation in one dimension: stats.stackexchange.com/a/491045/919. – whuber May 25 at 18:56 • @whuber, I see. So to double-check, it is valid to have something like $TN(\mu, \Sigma, a = (a_1, a_2), b = (b_1, 2x_1))$ where the truncation for $x_2$ depends on $x_1$? – Adrian May 25 at 18:58 • I don't understand what that notation means. But unless you are truncating to a vertical rectangle in $(x_1,x_2)$ coordinates, along the boundary of the truncation region $x_2$ must vary with $x_1.$ – whuber May 25 at 19:00 • I'm writing $TN(\mu, \Sigma, a = (a_1, a_2), b = (b_1, 2x_1))$ to denote a truncated bivariate normal distribution where $x_1$ lies between $a_1$ and $b_1$, and $x_2$ lies between $a_2$ and $2x_1$. That is, the truncation region of $x_2$ varies with $x_1$. – Adrian May 25 at 19:02 • Just to understand your comment better @whuber, if I truncated to $a_1 \leq x_1 \leq b_1$, $a_2 \leq x_2 \leq b_2$, this is equivalent to truncating to a rectangle in $(x_1, x_2)$ coordinates. In my case, having $a_1 \leq x_1 \leq b_1, a_2 \leq x_2 \leq 2x_1$ is equivalent to truncating to something other than a rectangle, correct? My question is: does the multivariate truncated normal distribution ALWAYS truncate to a rectangle, i.e., do lower limits $a$ and upper limits $b$ always have to be constants? Thank you. – Adrian May 25 at 19:11 Perhaps a more general notation will uncover the basic concepts and help you answer your question. There's little more to the following analysis than using mathematical notation carefully. Because that makes it abstract, I rephrase the key results in English: look for the quoted passages. When there are $$m$$ random variables $$X=(X_1, \ldots, X_m)$$ they have a distribution. This distribution gives the chance that $$X$$ lies in any Borel measurable set $$\mathcal{R},$$ written $$F_X(\mathcal{R}) = \Pr(X\in\mathcal{R}).$$ The distribution is (absolutely) continuous when there is a density function $$f_X$$ defined on all of $$\mathbb{R}^m$$ whose integral gives the probability. That is, for all $$\mathcal R,$$ $$\Pr(X\in\mathcal{R}) = F_X(\mathcal{R})= \iint_\mathcal{R} f_X(x)\mathrm{d}x = \iint_{\mathbb{R}^m} \mathcal{I}_{\mathcal{R}}(x)f_X(x)\,\mathrm{d}x.\tag{*}$$ The latter expression involves the indicator function $$\mathcal{I}_{\mathcal{R}}$$ (which by definition takes the value $$1$$ at all points in $$\mathbb{R}$$ and otherwise is zero). It enables us to express the integral over any region in terms of an integral over the entire space $$\mathbb{R}^m.$$ Suppose $$\mathcal{E} \subset \mathbb{R}^m$$ is a measurable set. Then the truncation of $$F_X$$ to $$\mathcal{E}$$ is a distribution function that arises when we "throw out all outcomes where $$X$$ is not in $$\mathcal{E}.$$" It is obtained in the simplest possible manner: just "limit the probability to the part of $$\mathcal{R}$$ lying in $$\mathcal{E}:$$" $$F_X^{\mathcal{E}}(\mathcal{R})\, \propto\, F_X(\mathcal{E}\cap\mathcal{R}).$$ The implicit multiple $$\lambda$$ in this proportion has to be such to make the total probability $$1,$$ leading to the equation $$1 =F^{\mathcal{E}}_X(\mathbb{R}^m) = \lambda F_X(\mathcal{E}\cap\mathbb{R}^m)= \lambda F_X(\mathcal{E})$$ yielding a unique (and obvious) value for $$\lambda$$ which we may plug into the foregoing to yield $$F^{\mathcal{E}}_X(\mathcal{R}) = \frac{F_X(\mathcal{E}\cap\mathcal{R})}{F_X(\mathcal{E})}.$$ This reads as "the chance $$X$$ is in the part of $$\mathcal{R}$$ lying in $$\mathcal{E}$$ relative to the chance $$X$$ is in $$\mathcal{E}.$$" When $$F_X$$ is absolutely continuous the identity $$\mathcal{I}_{\mathcal{E}\cap\mathcal{R}} = \mathcal{I}_{\mathcal{E}}\,\mathcal{I}_{\mathcal{R}}$$ (a consequence of the arithmetic facts $$1\times1=1$$ and $$1\times 0 = 0\times 0 = 0$$) produces $$F^{\mathcal{E}}_X(\mathcal{R})\, \propto\, \iint_{\mathcal{E} \cap \mathcal{R}} f_X(x)\,\mathrm{d}x = \iint_{\mathbb{R}^m} \mathcal{I}_{\mathcal{E}}(x)\mathcal{I}_{\mathcal{R}}(x)f_X(x)\,\mathrm{d}x = \iint_{\mathcal{R}} \mathcal{I}_{\mathcal{E}}(x)f_X(x)\,\mathrm{d}x$$ and therefore $$F^{\mathcal{E}}_X(\mathcal{R}) = \frac{\iint_{\mathcal{R}} \mathcal{I}_{\mathcal{E}}(x)f_X(x)\,\mathrm{d}x}{\iint_{\mathcal{E}} f_X(x)\,\mathrm{d}x} = \iint_{\mathcal{R}}\frac{ \mathcal{I}_{\mathcal{E}}(x)f_X(x)}{\iint_{\mathcal{E}} f_X(x)\,\mathrm{d}x}\,\mathrm{d}x = \iint f_X^{\mathcal{E}}(x)\,\mathrm{d}x.$$ This exhibits the density of the truncated variable as $$f_X^{\mathcal{E}}(x) = \frac{ \mathcal{I}_{\mathcal{E}}(x)f_X(x)}{\iint_{\mathcal{E}} f_X(x)\,\mathrm{d}x}.$$ It is "the density $$f_X,$$ zeroed beyond $$\mathcal{E},$$ as renormalized to integrate to unity." Here is an application. Let $$m=2$$ and suppose $$\mathcal{E}$$ is the region defined by $$\mathcal{E} = \{(x_1,x_2)\mid a_1\le x_1\le b_1,\, a_2 \le x_2 \le 2x_1\}.$$ It is either empty, a point, a triangle, or (generically) a trapezoid. Applying the preceding analysis shows that for any density $$f_{X_1,X_2},$$ the truncated density is given by Fubini's Theorem as \begin{aligned} f_{X_1,X_2}^\mathcal{E}(x_1,x_2) &= \frac{ \mathcal{I}_{\mathcal{E}}(x)f_X(x)}{\iint_{\mathcal{E}} f_X(x)\,\mathrm{d}x} \\ &= \frac{ \mathcal{I}_{\mathcal{E}}(x_1,x_2) f_{X_1,X_2}(x_1,x_2)}{\int_{a_1}^{b_1}\int_{a_2}^{2x_1} f_{X_1,X_2}(x_1,x_2)\,\mathrm{d}x_2\mathrm{d}x_1} \\ &= \frac{ f_{X_1,X_2}(x_1,x_2)}{\int_{a_1}^{b_1}\int_{a_2}^{2x_1} f_{X_1,X_2}(x_1,x_2)\,\mathrm{d}x_2\mathrm{d}x_1} \end{aligned} for $$(x_1,x_2)\in\mathcal{E}$$ (and zero otherwise). When you use the binormal density for $$f_X$$ you have the answer to the question.	{}
# Site condition and design Authors: Alex Chandel, Eric Jiang, Minwook Kim, Todor Kukushliev, William Lassman (ChE 352 in Winter 2014) Sheridan Lichtor (ChBE 352 in Winter 2016) Steward: Daniel Garcia, David Chen, Fengqi You Date Presented: 2/21/2016 & 1/17/2014 ## Introduction One of the first and most important decisions when designing a chemical plant is plant location. If a new plant is being built, a suitable site must be found and a plant layout considered. However, if the chemical plant is being built on the site of an old plant (possible upgrade or expansion) the existing site’s infrastructure must be considered. Of incredible importance are local laws and ordinances concerning chemical disposal, safety of the local population, and considerations for the employed operators. Suitable locations for chemical plants often have several plants in close proximity. The existence of these locations is often beneficial as there are often living infrastructure nearby to support the labor. Figure 1 shows the distribution of labor across the US and implicitly the common locations of many chemical plants. Figure 1: Employment of Chemical Engineers by Area as of May 2012 Above is shown the occupational employment density of chemical engineers separated county. It is noticeable that the coastal areas of the United States are most attractive for chemical process industries due, no doubt, to the easy access to water transportation routes, which are cheaper and faster than land transportation. Building a process plant in any of the “240-3,740” density shaded regions would capture the additional benefit of having the process plant built in an area where supporting industries already thrive, therefore making repairs and operational costs as a whole as low as possible as determined by location. The states of greatest chemical process plant density are California, Arizona, Texas, Louisiana, Mississippi, Illinois, Michigan, Indiana, Ohio, and the majority of the east coast states. The local corporate tax rates of these locations are highly dependent on income bracket as per the information found at http://www.taxadmin.org/fta/rate/corp_inc.pdf ## Geographical selection The location surrounding a chemical plant can substantially influence its construction costs and operating costs, and may affect long-term profitability. Thus it is important to choose an appropriate location for every facility. ### Factors considered #### Natural resources Transporting materials to and from the plant is a huge operational cost that is heavily factored into when selecting a plant location. Therefore, choosing to build a plant near natural resources reduce the operational cost of the plant tremendously. Natural resources such as river, lake, sea, and oil well near operating plants can be a huge bonanza for them. Major chemical plants processes need cooling system, which require immense amount of water. If river, lake, or sea is in close proximity, plants can utilize the water readily and relatively cheap. Plant needing of a great energy may build a dam on a river to resolve power issue. Upstream oil sectors look for oil wells to operate and drill out the oil and gas. Companies find themselves saving or making money when they build a plant near natural resources which they can take advantage. #### Weather Local climate conditions should be a strong consideration in the selection of a site for (chemical) processing or production facilities. As previously discussed, a variety of considerations are involved in site selection, including: the availability of raw materials, transportation capability, availability of labor, waste capacity and regulations, and local community and environmental considerations. In some instances, the aforementioned considerations will be prioritized over considerations regarding the local climate of the site; in other instances, a company is limited to the geographic locations in which they already own land or are involved in manufacturing and production. In the case where a company cannot select a site whose climate is optimized to meet production needs, there are several design considerations that need to be taken into account to accommodate the local climate conditions when setting up a facility. A reality of large chemical processing and production facilities is that it is oftentimes difficult to control the ambient environmental conditions in which manufacturing occurs. In industry, it is common to use open, structural steelwork buildings to house processing equipment (Towler 511). Oftentimes, this type of setup provides little protection from the weather and local climate. In some cases closed buildings house processing equipment in operations that can be particularly sensitive to disturbances (such as the disturbances that adverse weather conditions might present), in small plants, or in processes that have ventilation components for which the vent gas scrubbing is necessary (Towler 511). It is generally cheaper, however, to use open setups for production given their lower capital costs of construction. The usage of open setups also has downsides, particularly with respect to local climate. A site in which adverse climate conditions are commonplace will logically result in higher costs for the setup and upkeep of the facility. Examples of climate conditions that should be factored into the design process for plant location and site selection include: temperature, humidity, wind, precipitation, and natural disaster frequency (hurricanes, tornadoes, earthquakes, tsunamis). Strong, reinforced structures are required in locations that are subjected to high winds and in climates that receive hurricanes, tornadoes, earthquakes, and tsunamis (Towler 507). This section will specifically focus in depth on the implications of two critical climate conditions, temperature and humidity, as they relate to chemical processing and site selection. ##### Temperature In many geographic locations, temperature can fluctuate significantly depending on the time of year. In these cases, processing equipment should be able to withstand the stresses of gradual annual shifts in temperature, as well as faster day-to-day changes. In areas where the climate crosses 0 ºC, cycles of freezing and thawing may weaken the structural integrity of pipes and other processing equipments. Abnormally low temperatures may necessitate the addition of heating and added insulation, whereas abnormally high temperatures may require the provision of additional cooling systems to control the process temperature (Booth 154). Furthermore, the potential for a catastrophic burst or leakage is possible in cases where freezing water has the possibility of touching or interacting with pipelines or processing equipment. Specifically, in some circumstances a valve or joint might have a defect or crack that could propagate and cause a catastrophic failure from the constant freezing and thawing cycles on the equipment (Booth 154). Extreme temperatures are known to lower productivity of laborers and machinery. Heat, for example, can impact machinery that uses belts; warm temperatures loosen belts and can lower the product output due to processing irregularities stemming from belt slippage (Booth 157). Another general concern with temperature is that worker labor and productivity is adversely affected by extreme cold and hot; this may occur either in instances where production is not shielded from extreme outside climates or when production itself necessitates extreme temperature climates. Extreme heat, in particular, can hinder the mental and physical capability of workers; as a result, many companies give workers enforced vacation and additional mandatory break times. While this is good for the health and safety of the workers, it is also at the company’s expense. Local climate temperature should not be overlooked in the site selection process for a chemical plant. Figure 2 demonstrates the drastic differences that can exist in minimum and maximum temperature for one location (Chicago, IL) based on seasonality. According the graph, the minimum and maximum temperature over the year 2015 in Chicago fluctuates on average 11 ºC within each month ("World Weather"). Furthermore, the average temperature in Chicago during July is 23 ºC and the average temperature in January is -6 ºC; this is a 29 ºC range in the average temeperature throughout the year; the equipment used in a chemical plant, and also the materials being processed, need to withstand this large annual fluctuatiaon in temperature. Figure 2: Minimum and Maximum Temperature Throughout the Year in Chicago ##### Humidity Like temperature, humidity can fluctuate significantly depending on the season and even time of day. Unlike temperature, however, humidity is less so a problem for processing equipment as it is for the chemicals and substances being processed. Namely, hygroscopic effects become significant factors associated with high humidity processing environments (Booth 156). Hygroscopy concerns itself with a material’s affinity to pull in and store moisture from the environment, either via absorption or adsorption. Moisture uptake and hygroscopic effects are a major problem in cases where knowing the weight fractions of different materials is critical. For example, reactions usually call for specific amounts and weight fractions of reactants in order to get the desired product and meet detailed specifications. If one is not aware of the water fraction of the materials going into the reaction, then there may be unforeseen (and potentially very dangerous) consequences associated with either having an incorrect weight fraction reactant entering the reactor or having water involved in the reaction. Powders are also very susceptible to hygroscopic effects. Many food products, such as baked goods, use powder ingredients that are sensitive to moisture effects; moisture content of packaged foods is critical to shelf life and preventing the growth of bacteria. Outside of food applications, powders are also used in making glass, composites, ceramics, and pharmacological drugs. In their processing, it is critical to prevent caking by limiting the moisture uptake. This can impact the rheology (flow behavior) of the materials, which ultimately will have implications on the quality and purity of the final product. By contrast, some materials, such as cotton, linen, jute, and hemp, need to spun and woven in humid environments. Examples of hygroscopic materials include: glucose, flour, starch, glycerin, calcium chloride, and sodium chloride (Booth 157). A psychrometric diagram, such as the one in Figure 3, can explain the correlation between dry-bulb temperature, wet-bulb temperature, and relative humidity (NSF). Dry-bulb temperature is simply the ambient air temperature and is unaffected by the moisture content of the air; dry-bulb temperature is measured by a standard thermometer (NSF). Wet-bulb temperature is the saturation temperature of air, and it is the lowest temperature at which water can evaporate into the air; this is measured using a thermometer with a moist cloth wrapped around the bulb (Padfield). Relative humidity is the amount of water vapor present in air, and is calculated as the ratio of partial pressure of water vapor to the equilibrium vapor pressure of water (Padfield). Figure 3 suggests strong correlations between air temperature moisture content in the air, measured as relative humidity. For example, at constant dry-bulb temperatures, the relative humidity increases as the wet-bulb temperature increases. Likewise, at constant wet-bulb temperatures, the relative humidity decreases as the dry-bulb temperature increases. In selecting the location of a chemical processing facility, especially for facilities where there is little control over the ambient temperature, it is important to understand that small changes in temperature can have exponential consequences on the moisture content/ relative humidity of the air; these temperature and humidity changes can affect both processing equipment and the materials being processed. Figure 3: Relative Humidity as a Function of Wet-Bulb and Dry-Bulb Temperatures ##### Weather Example A variety of case studies have looked at weather effects on chemical processing. One such case explored the effects of temperature and humidity on phenol-formaldehyde resin bonding (Wang 253). Phenol-formaldehyde resin is a thermosetting adhesive that polymerizes and reacts with wood as part of the curing process in wood composite manufacturing. The strength of the resin bond is thought to be influenced by a variety of factors related to processing environment, including temperature and humidity. Figure 4 depicts the results from a study that compared the bond strength as a function of temperature, relative humidity, and bonding time (Wang 258-259). Figure 4: Effects of Humidity, Temperature, and Bonding Time on Bond Strength As the results suggest, drastically different resin strength profiles are expected depending on relative humidity. Considering just the samples that were bonded at 110 ºC, the resins that were cured at 41% relative humidity overall cured stronger than their counterparts that were cured at the same time but at higher relative humidities. An interesting feature that is prevalent in the 110 ºC bonding samples is that processing conditions at higher relative humidities is not always indicative of a depreciated bond strength. As the graph suggests between the 75% and 90% relative humidity samples, when bonded for less than 10 minutes, the 90% relative humidity samples actually are stronger than the 75% samples. However, after about 10 minutes the trend exists such that samples cured at higher relative humidities overall bond more weakly. There also appear to be striking differences between the 110 ºC and 120 ºC samples. In fact, it appears that less bonding time is required for the 120 ºC as is the time required to get comparable strengths for the 110 ºC samples. Also, at higher processing temperature of 120 ºC, it is evident that the samples at 75% relative humidity now can have comparable bond strengths as the samples at 41% relative humidity. Additionally, the samples at the 90% relative humidity also have somewhat higher binding strengths for the 120 ºC bonding temperatures compared to the 110 ºC bonding temperatures. Thus, this study indicates the appreciable differences that can exist in the product quality based on humidity and temperature effects. Thus, depending on the desired product qualities (bond strength in this resin study), humidity and temperature are critical metrics in defining the process environment. This phenol-formaldehyde resin study is particularly useful in demonstrating the effects of ambient relative humidity on the mechanical strength of the product, and relative humidity is definitely a parameter that could fluctuate depending on the weather patterns of the processing environment. Additionally, 10 ºC (the difference between bonding at 110 ºC and 120 ºC) is well within the monthly and seasonal temperature fluctuations of different locations; whether or not the weather could be attributed to such processing differences at these high temperatures is a possibility. #### Proximity to related chemical operations The availability and price of raw materials for feed streams often play a large part in determining the plant location. For example, many ethylene plants are built in the Middle East near supplies of natural gas. If building a plant near raw materials is not possible, often the next determining factor is ease of transportation. For most chemicals, proximity to major road, rail, waterway, or ports are desired. For high-value products such as pharmaceuticals, proximity to air ports can be used to prevent degradation of product during transport. Ease of transportation results in cheaper logistics cost for transport between both suppliers and buyers. Proximity to utilities are important in chemical process. Water is ubiquitous in chemical plants and are often require in substantial amounts. Construction of plants near rivers and lakes are often desired to reduce the cost of process water. If drawing from local water is not possible, cooling towers will need to be used. Electrical power is required in all plants, often requiring plants to be built on available power grids. #### Laws and regulations Federal laws will be listed as it serves as a baseline for the entire country. State and local laws sometimes are stricter than the established federal laws resulting. Property costs, property taxes, corporate income taxes, and fines also vary between states. Therefore, further consultation of the state and local laws must also be done beyond the laws listed in this text to ensure adherence to all laws required for the location of the plant. Below are several hallmark federal laws which proper treatment and disposal of waste in the air, ground, and water (Towler and Sinnott, 2013). ##### The Clean Air Act The Clean Air Act (CAA) was first passed in 1970 and was amended in 1990. The CAA empowers the EPA to set National Ambient Air Quality Standards (NAAQS)for the following seven contaminants: • Ozone • Carbon Monozide • Nitrogen Dioxide • Sulfur Dioxide • PM10: Partulate matter with mean diameter less than 10 μm. • PM2.5: Partulate matter with mean diameter less than 2.5 μm. In addition to these seven contaminatns, the CAA also empowered EPA to regulate an additional 189 hazardous air pollutants listed in the National Emission Standards for Hazardous Air Pollutants. Failure to meet NAAQS levels will result in the requirement of remediation steps to be taken to lower emissions before the plant is allowed to be operational. ##### The Clean Water Act The Clean Water Act was first passed in 1972 and was amended in 1977 and 1987. It seeks to achieve clean water for swimming, boating, and protecting fish and water life. This act mandates that a permit from the EPA must be issued for discharge of pollutants into navigable waters. ##### The Safe Drinking Water Act The Safe Drinking Water Act was passed in 1974. It empowers the EPA to set standards on the required purity of any water which could be used for drinking. ##### The Resource Conservation and Recovery Act The Resource Conservation and Recovery Act was passed in 1976 to protect groundwater from contamination. This Act states that all waste producers are legally liable at any time from waste production to final disposal. Hazardous waste must be clearly labeled and tracked in transport and treated to levels specified by the EPA. #### Waste Minimization and Management ##### Waste Minimization Production of waste is arises naturally in any plant and require a noticeable amount of resources to take care of. Before even considering methods of managing ways, cost can significantly be reduced by efficient management by source reduction. Below is a five-step review often conducted to minimize waste production (Towler and Sinnott, 2013): 1. Identify waste: Identify what waste products are produced. 2. Economical Impact: Determine the size of the waste stream and the cost of treatment 3. Root causes: Determine the root cause of the waste streams. 4. Modifications: Analyze the effectiveness and cost of potential solutions. 5. Implement: Implement and optimize solutions of waste management. Below are some source reduction strategies which can be employed (Towler and Sinnott, 2013): • Purification of feeds: Impurities in feed streams can lead to side reactions and formation of waste. Either purchase of purer feeds or employment of purification techniques which do not generate more waste can be used. Purification of feeds will also lead to the reduction of purge and vent streams. • Protect catalysis and adsorbents: Catalysts are sensitive to containment in the feed and be deactivated. One method of protecting the catalyst or adsorbent is by employing a guard bed of material to absorb or filter out contaminants. • Eliminate use of extraneous materials: Limiting the diversity of solvents is beneficial. The mixing of different solvents can result in waste formation when solvents are degraded. • Increase recovery from separations: Higher product recovery results in lower concentrations of products in the the waste streams and less waste formation. • Improve fuel quality: Cleaner-burning fuel can have less harmful emissions. • Recycle or sell side-products: Rather than processing side-products, attempt to identify companies which might use the waste as raw material for another process. Keep in mind for all the strategies which can be employed to minimize waste production and therefore waste treatment, the overall cost must be considered. The savings from minimizing waste must be more than the additional cost implementing minimization. ##### Waste Management There are many methods of waste treatment and safe disposal. The availability and efficiency of these methods depend heavily on location. Adherence to federal, state, and local laws may further restrict the availability, of some of these techniques. Common techniques include: • Dilution and dispersion. • Discharge into municipal sewer. • Chemical treatment: precipitation, neutralization. • Biological treatment: composting, anaerobic digestion. • Incineration. • Landfill at controlled sites. • Sea dumping. ## Economical definition of cost ### Cost All of the above criteria ultimately influence the capital and operating costs of a plant, and its expected lifespan. $TC = r K + w L$ Local wages, prices of chemical feedstock, shipping costs, and utilities all contribute to total operating costs. Property prices, rental fees, taxes, and existing company property in the area contribute to recurring investment costs. ## Site Layout: Design and Construction ### Site Layout The needs of a site for a chemical process vary considerably from process to process. In general, however, all chemical plants require the following (Towler and Sinnott, 2013): • Shipping and receiving of products and raw materials • Storage • Utilities • Offices and laboratories for management and quality control personnel • Medical and fire services for emergency management • Cafeterias, parking lots, and other amenities for employees. These auxiliary buildings are often referred to as ancillary structures and they are placed within a chemical process to minimize transportation of goods and personnel, and to maximize safety. The following procedure is followed when determining the site layout of a chemical process (Mecklenburgh, 1985): 1. Major process equipment is placed in a logical order to minimize transportation of process streams. Extra emphasis is placed on the separation and treatment of hazardous materials as quickly as possible. 2. Utilities such as boilers and power plants are placed to minimize transportation of utility to its use within the process. Utilities are usually consolidated into one section of the chemical plant because they are usually generated together. For example, a boiler produces high pressure steam; half the steam is sent through a turbine to generate electricity and to expand the steam into low pressure steam. 3. Shipping and receiving are placed wherever there is a need to conform to preexisting infrastructure. For example, if the plant is located on a harbor, shipping and receiving for all barge shipments are located by the water. If the plant is built next to a railway, shipping by rail is located next to the tracks. 4. Storage tanks and warehouses are consolidated as much as possible. Storage of raw materials and products are stored between where they enter or exit the process and where they are shipped or received. 5. Ancillary buildings are placed to minimize time personnel spend traveling around the site. 6. Offices and labs are located as far away from hazardous processes as possible. 7. Walkways and roadways are added as needed to assist with construction and transportation during plant operation. ### Site Construction Site construction, along with process design, is an iterative process that follows a multi-step procedure (Mecklenburgh, 1985). #### Stage One Layout The "Proposal" or Stage One layout is the first step towards designing a site layout. The purpose of the Stage One layout is to assess the feasibility of the process according to the cost, hazard, risk, and environmental standards set by the interested parties. The information included in a Stage One layout is the relative position of buildings and process equipment, and any other data that may come from a preliminary case study of a particular process. Additionally, preliminary estimates by manufacturers and contractors for process equipment and ancillary structures, as well as local building codes and regulations are used in generating the Stage One Layout. Usually, different layouts for the same process may produce different costs. At this stage in development, many different layouts should be generated and the different layouts should be compared in a systematic way. It is usually very difficult to tell which layout is superior based purely on inspection. Once a Stage One design is finalized, the layout can move on to the next stage. #### Stage Two Layout Stage Two Layouts are generated based on the finalized Stage One design. Changes to the Stage One design are minimized. The purpose of the Stage Two Layout is to determine an accurate detailed cost of the entire process. Additionally, detailed hazard and environmental information is determined and submitted to all involved regulatory parties at this stage. #### Final Stage Layout After the detailed cost information, hazard, and environmental information are approved, the Final Design layout commences. At this stage, detailed drawings of all equipment, piping, and layouts are finalized. At the conclusion of the Final Stage layout, orders with contractors are placed and fabrication of process equipment begins, and the site land is purchased. Essentially, this is the "point of no return." #### Construction The first step in constructing the plant is remediation and preparation of the land for construction of a chemical plant. This can include clearing the land of trees and vegetation, removing other natural obstacles such as boulders and ditches, implementing a drainage system, landscaping, grading to remove difficult topography, and anything else that is necessary. Site selection attempts to minimize costs associated with this step, but there is invariably some form of preparation required for every site. The second step is to construct all roadways, sidewalks, and fences required for both plant operation and plant construction. Costs associated with this step can range from 2 to 10 percent of the total capital investment for a chemical plant (Peters et al., 2002). Process equipment and buildings are then constructed as soon as they are available. While construction schedules vary considerably from process to process, in some cases it is possible to perform the final construction steps once the process has already begun to operate, and the construction schedule is designed with this in mind (Mecklenburgh, 1985). A new aspect of construction of process equipment is a modular approach, where process equipment is assembled as completely as possible by the manufacturer and shipped while assembled. The advantage to this approach is a more comprehensive testing of the equipment by the manufacturer and less installation time once the equipment has arrived on site (Towler and Sinnott, 2013). ## References Booth WM. Transactions of the American Institute of Chemical Engineers. Volume VI. New York: D Van Nostrond Co; 1913. Mecklenburgh JC. Process plant layout. New York: Halsted Press; 1985. National Science Foundation. "Hands-on Activity: Swamp Cooler." Teach Engineering. College of Engineering at University of Colorado Boulder and Integrated Teaching and Learning, 19 Feb. 2016. Web. 21 Feb. 2016. <https://www.teachengineering.org/view_activity.php?url=collection/cub_/activities/cub_housing/cub_housing_lesson01_activity2.xml>. Padfield, Tim. "Glossary of the Microclimate Variables and Units Used in Conservation Physics." Conservation Physics. National Museum of Denmark, 14 June 2012. Web. 21 Feb. 2016. <http://www.conservationphysics.org/cpw/Storage/Fundamentals>. Peters MS, Timmerhaus KD, West RE. Plant Design and Economics for Chemical Engineers. 5th ed. New York: McGraw-Hill; 2002. Towler G, Sinnott R. General Site Considerations. In: Chemical Engineering Design: Principles, Practice and Economics of Plant and Process Design. 2nd ed. Boston: Elsevier; 2013. Wang, Xiang-Ming, Bernard Riedl, Alfred Christiansen, and W. Geimer. "The Effects of Temperature and Humidity on Phenol-formaldehyde Resin Bonding." Wood Science and Technology 29.4 (1995): 253-66. Web. "World Weather & Climate Information." Weather and Climate: Chicago, United States of America. World Weather and Climate Information, 2015. Web. 20 Feb. 2016. <https://weather-and-climate.com/average-monthly-Rainfall-Temperature-Sunshine,Chicago,United-States-of-America>.	{}
# Congruence lattices of finite diagram monoids Published in Advances in Mathematics, 2018 Recommended citation: J. East, J. D. Mitchell, N. Ruškuc and M. Torpey. Congruence lattices of finite diagram monoids, Advances in Mathematics 333 (Jul 2018) 931–1003. https://doi.org/10.1016/j.aim.2018.05.016 We give a complete description of the congruence lattices of the following finite diagram monoids: the partition monoid, the planar partition monoid, the Brauer monoid, the Jones monoid (also known as the Temperley–Lieb monoid), the Motzkin monoid, and the partial Brauer monoid. All the congruences under discussion arise as special instances of a new construction, involving an ideal $I$, a retraction $I \to M$ onto the minimal ideal, a congruence on $M$, and a normal subgroup of a maximal subgroup outside $I$.	{}
### Iron Ore Smelting Process - Brighthub Engineering The ore is loaded into a blast furnace along with measured quantities of coke and limestone Hot combustion air is supplied to the furnace and some form of fuel used to raise the temperature The iron is reduced from the ore by carbon in the coke, the limestone aiding slag separation from the molten iron The slag and molten iron are tapped off from the bottom of the furnace, the slag being .... Know More ### Blast furnace - Simple English Wikipedia, the free , Ore, limestone and carbon in the form of coke are put into the top of the blast furnace in layersAt the same time, hot air called "wind" is blown inside the furnace Special nozzles called "tuyeres" are used to put the air in the furnace The nozzles are at the bottom of the furnace This process is called "blasting" It is why it is called a "blast furnace"... Know More ### Why limestone is used in the blast furnace - Answers Limestone is used mainly in the iron making part of steel production where it is added to the iron ore before entering the blast furnace The limestone reacts with acidic impurities called slag .... Know More ### What is the purpose of limestone in a blast furnace? - Quora Nov 13, 2017· limestone is used to remove impurities in the furnace and the main impurity of the blast furnace is silica (sand and rock) which is silicon dioxide silicon dioxide is a solid at furnace temperatures so it reacts with calcium oxide from the decomp.... Know More ### Creating Iron \| HowStuffWorks The more advanced way to smelt iron is in a blast furnace A blast furnace is charged with iron ore, charcoal or coke (coke is charcoal made from coal) and limestone (CaCO 3 ) Huge quantities of air blast in at the bottom of the furnace, and the calcium in the limestone combines with the silicates to ,... Know More ### What is the use of limestone in a blast furnace? \| Yahoo , Sep 16, 2008· The formula is CaO + SiO(2) = CaSiO(3) This is molten at blast furnace temperatures and floats on top of the molten iron It is then removed and cooled into hardened slag Thus, the lime from limestone combines with the silica and allows the separation of the iron from the silica impuriti... Know More ### Why is calcium carbonate added to the blast furnace , The purpose of a blast furnace is to chemically reduce and physically convert iron oxides into liquid iron called "hot metal" The blast furnace is a huge, steel stack lined with refractory brick, where iron ore, coke and limestone are dumped into the top, and preheated air is blown into the bottom... Know More ### Blast Furnace Slag - Material Description - User , Granulated blast furnace slag and vitrified pelletized blast furnace slag are also used in the manufacture of blended hydraulic cements (AASHTO M240) (3) When used in blended cements, granulated blast furnace slag or vitrified pelletized slag are milled to a fine particle size in accordance with AASHTO M302 requirements... Know More ### The Extraction of Iron - Chemistry LibreTexts The heat of the furnace decomposes the limestone to give calcium oxide $CaCO_3 \rightarrow CaO + CO_2 \tag{5}$ This is an endothermic reaction, absorbing heat from the furnace It is therefore important not to add too much limestone because it would otherwise cool the furnace , Most of the molten iron from a Blast Furnace is used to make .... Know More ### -Limestone Consumption In Blast Furnace- limestone used in blast furnace - vhf-venwnl limestone used in blast furnace - chicshocoza limestone used in blast furnace Limestone fillers cement based composites Effects of Limestone filler is a raw material that is already used in several applications like paints bricks and bituminous mixtures Moreover and particularly in Belgium .... Know More ### The Making of Iron & Steel - SEAISI In modern practice, the combined iron plus coke burden is self fluxing and virtually no limestone is required at the blast furnace When the flux is added at the sinter plant, the limestone is precalcined, the sinter strength and chemical reactivity are improved and ,... Know More ### GCSE CHEMISTRY: The Blast Furnace Flashcards \| Quizlet What are the raw materials used in the Blast furnace? Haematite (iron ore), Coke (impure carbon), limestone (calcium carbonate) Write the equation of the first reaction... Know More ### Blast furnace - Wikipedia A blast furnace is a type of metallurgical furnace used for smelting to produce industrial metals, generally pig iron, but also others such as lead or copper Blast refers to the combustion air being "forced" or supplied above atmospheric pressure In a blast furnace, fuel (), ores, and flux are continuously supplied through the top of the furnace, while a hot blast of air (sometimes with .... Know More ### Blast furnace - Simple English Wikipedia, the free , Process Ore, limestone and carbon in the form of coke are put into the top of the blast furnace in layersAt the same time, hot air called "wind" is blown inside the furnace Special nozzles called "tuyeres" are used to put the air in the furnace The nozzles are at the bottom of the furnace This process is called "blasting" It is why it is called a "blast furnace"... Know More ### Limestone - Wikipedia Used in blast furnaces, limestone binds with silica and other impurities to remove them from the iron It is used in sculptures because of its suitability for carving Occupational safety and health People can be exposed to limestone in the workplace by inhalation of and eye contact with the dust... Know More ### How to Make Cement From Blast-Furnace Slag Probably the first use made of blast-furnace slag in the cement industry was not of an honourable nature, to use an almost worthless and cumbersome by-product as a not readily detectable adulterant in Portland cement, which was at that time very costly to manufacture... Know More ### Iron Ore Processing For The Blast Furnace - t-gende Blast furnace gas is produced during the iron oxide reduction in blast furnace iron making in which iron ore, coke and limestone are heated and melted in a blast furnace and is an indigenous process gas of the steelworks industry pugh et al2013 Blast Furnace And Stoves Eurotherm By Schneider... Know More ### Blast furnace \| metallurgy \| Britannica Basically, the blast furnace is a countercurrent heat and oxygen exchanger in which rising combustion gas loses most of its heat on the, The stack is kept full with alternating layers of coke, ore, and limestone admitted at the top during continuous operation Coke is ignited at the bottom and .... Know More ### EXTRACTION OF IRON IN A BLAST FURNACE reduction , EXTRACTION OF IRON IN A BLAST FURNACE Iron is also below carbon in the reactivity series, so therefore carbon can be used to reduce it (note: reduction can also mean to remove the oxygen from a compound) This is done in a blast furnace... Know More ### How it works: The Blast Furnace - Stoke-on-Trent The purpose of a blast furnace is to chemically reduce and physically convert iron oxides into liquid iron called "hot metal" The blast furnace is a huge, steel stack lined with refractory brick, where iron ore, coke and limestone are dumped into the top, and preheated air is blown into the bottom... Know More ### What is a Blast Furnace? (with picture) - wisegeek Oct 17, 2019· Furnaces of this type are often used in the steelmaking process A blast furnace is ideal for combining charcoal and iron ore together The extreme heat in this type of furnace makes it possible to melt both substances into an integrated liquid metal that ultimately forms what is known as pig iron The pig iron is removed from the base of the .... Know More ### Metal extraction - MYP Chem Limestone; Air; Each of the raw materials is necessary for the production - click to see its purpose The blast furnace The iron ore, coke and limestone ("charge") is fed into the blast furnace from the top Once inside the hot furnace the coke reacts with the blast of ,... Know More ### blast furnace Flashcards and Study Sets \| Quizlet Learn blast furnace with free interactive flashcards Choose from 252 different sets of blast furnace flashcards on Quizlet... Know More ### furnace for limestone - YouTube Aug 09, 2019· This video is unavailable Watch Queue Queue Watch Queue Queue... Know More ### What is the purpose of coke in a blast furnace? \| AnswersDrive The purpose of a blast furnace is to chemically reduce and physically convert iron oxides into liquid iron called "hot metal" The blast furnace is a huge, steel stack lined with refractory brick, where iron ore, coke and limestone are dumped into the top, and preheated air is blown into the bottom... Know More ### Blast Furnace - an overview \| ScienceDirect Topics Blast furnace gas is produced during the iron oxide reduction in blast furnace iron making in which iron ore, coke and limestone are heated and melted in a blast furnace and is an indigenous process gas of the steelworks industry (Pugh et al, 2013)... Know More ### why limestone and dolomite using during iron ore pellets , Jun 30, 2013· Iron making furnaces like the blast furnace or shaft furnace need burden that form a , the mechanized mining and close sizing of the iron ore, the quantity of fines generated , has been given in recent years to the use of fluxed pellets in the blast furnace due to their , like limestone, dolomite, pyroxenite and magnesite III ... Know More ### Blast Furnace operation: HOW A BLAST FURNACE WORKS Aug 04, 2011· The purpose of a blast furnace is to chemically reduce and physically convert iron oxides into liquid iron called "hot metal" The blast furnace is a huge, steel stack lined with refractory brick, where iron ore, coke and limestone are dumped into the top, and preheated air is blown into the bottom... Know More ### Iron Ore Smelting Process - Brighthub Engineering The ore is loaded into a blast furnace along with measured quantities of coke and limestone Hot combustion air is supplied to the furnace and some form of fuel used to raise the temperature The iron is reduced from the ore by carbon in the coke, the limestone aiding slag separation from the molten iron The slag and molten iron are tapped off from the bottom of the furnace, the slag being .... Know More ### Blast Furnace operation: HOW A BLAST FURNACE WORKS Aug 04, 2011· The purpose of a blast furnace is to chemically reduce and physically convert iron oxides into liquid iron called "hot metal" The blast furnace is a huge, steel stack lined with refractory brick, where iron ore, coke and limestone are dumped into the top, and preheated air is blown into the bottom... Know More	{}
# 2.6: Unconstrained Optimization- Numerical Methods The types of problems that we solved in the previous section were examples of unconstrained optimization problems. That is, we tried to find local (and perhaps even global) maximum and minimum points of real-valued functions $$f (x, y)$$, where the points $$(x, y)$$ could be any points in the domain of $$f$$. The method we used required us to find the critical points of $$f$$, which meant having to solve the equation $$\nabla f = \textbf{0}$$, which in general is a system of two equations in two unknowns ($$x \text{ and }y$$). While this was relatively simple for the examples we did, in general this will not be the case. If the equations involve polynomials in $$x \text{ and }y$$ of degree three or higher, or complicated expressions involving trigonometric, exponential, or logarithmic functions, then solving even one such equation, let alone two, could be impossible by elementary means. For example, if one of the equations that had to be solved was $\nonumber x^3 +9x−2 = 0 ,$ you may have a hard time getting the exact solutions. Trial and error would not help much, especially since the only real solution turns out to be $\nonumber \sqrt[3]{\sqrt{28}+1-\sqrt[3]{\sqrt{28}-1}}.$ In a situation such as this, the only choice may be to find a solution using some numerical method which gives a sequence of numbers which converge to the actual solution. For example, Newton’s method for solving equations $$f (x) = 0$$, which you probably learned in single-variable calculus. In this section we will describe another method of Newton for finding critical points of real-valued functions of two variables. Let $$f (x, y)$$ be a smooth real-valued function, and define $\nonumber D(x, y) = \dfrac{∂^2 f}{ ∂x^2} (x, y) \dfrac{∂^2 f}{ ∂y^2} (x, y)− \left (\dfrac{∂^2 f}{∂y∂x} (x, y)\right )^2$ Newton’s algorithm: Pick an initial point $$(x_0 , y_0)$$. For $$n$$ = 0,1,2,3,..., define: $x_{n+1}=x_n - \dfrac{\begin{vmatrix} \dfrac{∂^2 f}{ ∂y^2} (x_n, y_n) & \dfrac{∂^2 f}{∂x∂y} (x_n, y_n) \\ \dfrac{∂f}{∂y} (x_n, y_n) & \dfrac{∂f}{∂x} (x_n, y_n)\\ \end{vmatrix}}{D(x_n, y_n)},\quad y_{n+1}=y_n - \dfrac{\begin{vmatrix} \dfrac{∂^2 f}{ ∂x^2} (x_n, y_n) & \dfrac{∂^2 f}{∂x∂y} (x_n, y_n) \\ \dfrac{∂f}{∂x} (x_n, y_n) & \dfrac{∂f}{∂y} (x_n, y_n)\\ \end{vmatrix}}{D(x_n, y_n)}\label{Eq2.14}$ Then the sequence of points $$(x_n, y_n)_{n=1}^{\infty}$$ converges to a critical point. If there are several critical points, then you will have to try different initial points to find them. $$f (x, y) = x^ 3 − x y− x+ x y^3 − y^ 4\text{ for }−1 ≤ x ≤ 0 \text{ and }0 ≤ y ≤ 1$$ The derivation of Newton’s algorithm, and the proof that it converges (given a “reasonable” choice for the initial point) requires techniques beyond the scope of this text. See RALSTON and RABINOWITZ for more detail and for discussion of other numerical methods. Our description of Newton’s algorithm is the special two-variable case of a more general algorithm that can be applied to functions of $$n \ge 2$$ variables. In the case of functions which have a global maximum or minimum, Newton’s algorithm can be used to find those points. In general, global maxima and minima tend to be more interesting than local versions, at least in practical applications. A maximization problem can always be turned into a minimization problem (why?), so a large number of methods have been developed to find the global minimum of functions of any number of variables. This field of study is called nonlinear programming. Many of these methods are based on the steepest descent technique, which is based on an idea that we discussed in Section 2.4. Recall that the negative gradient $$- \nabla f$$ gives the direction of the fastest rate of decrease of a function $$f$$. The crux of the steepest descent idea, then, is that starting from some initial point, you move a certain amount in the direction of $$-\nabla f$$ at that point. Wherever that takes you becomes your new point, and you then just keep repeating that procedure until eventually (hopefully) you reach the point where f has its smallest value. There is a “pure” steepest descent method, and a multitude of variations on it that improve the rate of convergence, ease of calculation, etc. In fact, Newton’s algorithm can be interpreted as a modified steepest descent method. For more discussion of this, and of nonlinear programming in general, see BAZARAA, SHERALI and SHETTY.	{}
# Sari has 3 times as many pencil eraser as sam they have 28 eraser how many eraser does sari have? Sari has 3 times as many pencil eraser as sam they have 28 eraser how many eraser does sari have? ## This Post Has 3 Comments 1. Expert says: he kicked it up? is this the answer ? step-by-step explanation: 2. Expert says: what are you doing do you need examples of variable questions? step-by-step explanation: 5a-1=9 bring the one over the = a=2 brainliest? ? here try this one 3a-24=0 3. coreyscott1281 says: The equation is 3x + x = 28 4x = 28 Divide both sides by 4 and you get x = 7 ( Sam's amount) Because Sari has 3 times as much multiply 7 by 3 Sari = 21	{}
## College Algebra (6th Edition) fill the blanks with $b^{x}$ ... $(-\infty, \infty)$ ... $(0, \infty)$ Book references: Definition of the Exponential Function (just below Figure 4.1) and Characteristics of Exponential Functions of the Form $f(x)=b^{x}$ (below Figure 4.4) The exponential function with base $b$ is defined by $f(x)=b^{x},\ b>0$ and $b\neq 1$. The domain is: all real numbers x. An image with examples for both cases , $b>1$ and \$0	{}
# when will page numbers be with section styles for 7 years now i've been trying assorted hacks to get proper page numbering in writer with multisection docs.And seen hundreds of very frustrated people on the web who also can't hack out how to do page numbering. And nobody has any solution. of course your non-intuitive approach to page numbers (attached to paragraph styles) works well for simple things like letters, essays, but NOT for multi-section books with different sections like frontmatter and main bookmatter and backmatter You don't understand???? Please listen. each section front, main, back RESTARTs renumbering the pages at page 1 Did you read that??? Three sections each starting at page 1.In the same document. NOT unusual. Common. Common. Common. The usual thing in books. And libreoffice just cannot handle this. if in the main book section you have 10 chapters (all starting with same paragraph style) you don't want them all starting at page 1, just the chapter 1. So useless to tell a paragraph style to restart numbering pages. Don't get it???? Reread what i just wrote. Reread it again. and again and again. until you get it. Don't bother to answer me if you haven't read the problem at least 10 times. In the past i've only gotten replies from people who never read the question. I'm so frustrated, angry and irritated by people who never read the problem and give CLICHE answers that don't apply to the problem. So please don't give pat but WRONG answers about page numbers in paragraph styles. Just say simply, yes or no. Just say yes you will or no you never will ever offer proper page numbering in sections. so we can all know finally to just give up on libreoffice, go elsewhere (and warn our friends to do the same). edit retag close merge delete 2 I feel frustrated by people not reading the manual. Just one example Sections are not Chapters. I have many documents where the numbering restarts as you describe, some have roman numerals, some have arabic numerals, some have letters; all work perfectly, thank you. ( 2021-05-03 08:14:38 +0200 )edit easy to say things like i've been to the moon, but no way to do it libreoffice can do anything..... if you do it by hand.... so what ( 2021-05-03 19:26:36 +0200 )edit Sort by » oldest newest most voted First of all, we aren't developers but users helping other users with the tricks we've discovered on how to make things work the way we want. So we "can't offer proper X, Y or Z". We only configure LO so that it achieves the needed formatting. Second, have you read the Writer Guide? Relying on one's own intuition (a masquerading word for "routine") to use any software application is not the best way learn it and use efficiently. There a vocabulary ambiguity in your question. In Writer, a section is a part of a page with a different layout. Consequently, a section can't have a page number of its own. You are likely talking about "parts" of a book with a distinctive formatting and layout (header, footer, margins, background, columns, page number style, ...). All these attributes are recorded in a page style. Every "part" has a dedicated page style (you can build "dynamic" page styles fitting several parts but this requires advance knowledge; start with simple "static" ones). You switch from one page style to another one at the boundary between "parts" with a special page break (not the usual Ctrl+Enter which only adds a page break without changing the page style). This page break can be created • either with Insert>Manual Break>Other Break (wording not guaranteed because I have not presently the latest version on this computer) • or with a dedicated paragraph style configuration in Text Flow tab In both cases, you choose the page style to switch to. Optionally, you can reset the page number. If nothing is entered in the page number field, page number increments in sequence. It is reset only if an explicit non-zero value is given. Just leave the field blank to keep the usual page number sequence. If this doesn't solve your problem, make a skeleton sample file (5 pages max) and edit your question to attach it. A file is attached with the paperclip tool which is provided in a question, not in a comment. Don't use "Add answer" because this site is not a forum and answers are reserved for solutions. To show the community your question has been answered, click the ✓ next to the correct answer, and "upvote" by clicking on the ^ arrow of any helpful answers. These are the mechanisms for communicating the quality of the Q&A on this site. Thanks! more 1 Optionally, you can reset the page number. If nothing is entered in the page number field, page number increments in sequence. It is reset only if an explicit non-zero value is given. @bookwriterman: don't get it???? Reread what @ajlittoz just wrote. Reread it again. and again and again. until you get it. ( 2021-05-03 08:48:13 +0200 )edit same old same old ... blame the questioner your ideas are old... 7 years old.... still don't work try it before you speak manual page break drops eventually via page inserted to first paragraph style of inserted page... it all goes back to paragraph style read the manual please if you don't understand ( 2021-05-03 19:33:38 +0200 )edit lol :-) it's easy to tell "nothing works" and then reply to anything "been there". Just describe what you try to do, then see how easy people implement that, then learn that you simply are incapable to learn. And I bet that you replied without having read the answer at least 10 times ( 2021-05-03 19:39:36 +0200 )edit same old same old.... insult the user easy to do after 7 years i probably know twice what you do about libreoffice and have probably read the manual much more than you have of course things can be done.... if you do it by hand or manually.... or semi-manually but you don't use libreoffice so you can get around its idiosyncracies by doing stuff by hand then you might as well just use a text editor ( 2021-05-03 19:46:13 +0200 )edit 1 Aha, so a troll unable to describe any task that they complain about, only mumble. Miserable. Present a task, or get lost. And stop talking about something you don't know. Even if you saw the manual (you can't even describe something from it), you didn't understand nothing. "i probably know twice what you do about libreoffice" - lol, after you develop for LibreOffice at least half I did, I would start thinking you learned something. ( 2021-05-03 19:50:42 +0200 )edit Then for 7 years you missed the point since the page numbering is done by the Page style. See for example that FAQ about page numbering, there are other sections in this FAQ that should answer your question. Basically, you need to insert Page breaks, either manually or with a heading style for example, see this page, it applies to restarting the page numbering at 1, too. more go look at page styles.... nowhere is there in it is there a way to set starting page numbers page numbers in libreoffice are set by paragraph style.... please read the manual ( 2021-05-03 19:29:49 +0200 )edit @bookwriterman: first thing first, could you use some form of syntactically correct simplified English so that we can understand your plights? You have duplicated words, nearly no punctuation, many grammar mistakes … It is not a fault not being fluent in English, many questioners are in the same case, but they try to circumvent their (un)fluency by explaining with simple terms. Help us to help you. This means, explain what you try to achieve (quite clear: get continuous page numbering across page styles), how you do it (this is not explained), eventually attach a small sample file to your question, not to a comment. ( 2021-05-03 20:13:27 +0200 )edit Well, last post in this discussion. If there is one thing to understand, it's this screenshot: Be it in a paragraph style or a direct formatting of a paragraph, the trick is to activate the page break and defining a page style (that can be the same as the previous page). Then, you have access to the numbering to be applied for the new page. ( 2021-05-03 21:18:44 +0200 )edit	{}
# zbMATH — the first resource for mathematics Root-N-consistent semiparametric regression. (English) Zbl 0647.62100 Summary: One type of semiparametric regression on an $${\mathcal R}$$ $$p\times {\mathcal R}$$ q-valued random variable (X,Z) is $$\beta 'X+\theta (Z)$$, where $$\beta$$ and $$\theta$$ (Z) are an unknown slope coefficient vector and function, and X is neither wholly dependent on Z nor necessarily independent of it. Estimators of $$\beta$$ based on incorrect parameterization of $$\theta$$ are generally inconsistent, whereas consistent nonparametric estimators deviate from $$\beta$$ by a larger probability order than $$N^{-}$$, where N is sample size. An estimator generalizing the ordinary least squares estimator of $$\beta$$ is constructed by inserting nonparametric regression estimators in the nonlinear orthogonal projection on Z. Under regularity conditions $${\hat \beta}$$ is shown to be $$N^{1/2}$$-consistent for $$\beta$$ and asymptotically normal, and a consistent estimator of its limiting covariance matrix is given, affording statistical inference that is not only asymptotically valid but has nonzero asymptotic first-order efficiency relative to estimators based on a correctly parameterized $$\theta$$. We discuss the identification problem and $${\hat \beta}$$’s efficiency, and report results of a Monte Carlo study of finite-sample performance. While the paper focuses on the simplest interesting setting of multiple regression with independent observations, extensions to other econometric models are described, in particular seemingly unrelated and nonlinear regressions, simultaneous equations, distributed lags, and sample selectivity models. ##### MSC: 62P20 Applications of statistics to economics 62G05 Nonparametric estimation 62J02 General nonlinear regression 62J05 Linear regression; mixed models Full Text:	{}
# A light string of mass m and length L has its ends tied to two walls that are separated by the distance D . Two objects each of mass M are suspended from the string as shown in figure given . if a wave pulse is sent from point A find speed is AB string $(a)\;\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}\qquad(b)\;\sqrt{\large\frac{(2D-L)}{4D^2-4DL}}\qquad(c)\;\sqrt{2l(2D-L)}\qquad(d)\;None$ Answer : (a) $\;\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}$ Explanation : $2 \times \large\frac{L}{4} cos \theta + \large\frac{L}{2}=D$ $1+cos \theta =\large\frac{2D}{L}$ $cos \theta=\large\frac{2D-L}{L}$ $Tsin \theta =mg$ $T=\large\frac{mg}{sin \theta}$ $T cos \theta =T_{1} \qquad \;$ AB string mass = $\large\frac{m}{2}$ $v=\sqrt{\large\frac{T_{1}}{\mu}}= \qquad \; \mu=\large\frac{mg}{2l}$ $v=\sqrt{\large\frac{T cos \theta 2l}{mg}}$ $v=\sqrt{\large\frac{mg}{sin \theta} \times \large\frac{cos 2 \theta 2l}{mg}}$ $v=\sqrt{cos \theta 2l}$ $=\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}\;.$ Answer : (a) $\;\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}$ Explanation : $2 \times \large\frac{L}{4} cos \theta + \large\frac{L}{2}=D$ $1+cos \theta =\large\frac{2D}{L}$ $cos \theta=\large\frac{2D-L}{L}$ $Tsin \theta =mg$ $T=\large\frac{mg}{sin \theta}$ $T cos \theta =T_{1} \qquad \;$ AB string mass = $\large\frac{m}{2}$ $v=\sqrt{\large\frac{T_{1}}{\mu}}= \qquad \; \mu=\large\frac{mg}{2l}$ $v=\sqrt{\large\frac{T cos \theta 2l}{mg}}$ $v=\sqrt{\large\frac{mg}{sin \theta} \times \large\frac{cos 2 \theta 2l}{mg}}$ $v=\sqrt{cos \theta 2l}$ $=\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}\;.$ Answer : (a) $\;\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}$ Explanation : $2 \times \large\frac{L}{4} cos \theta + \large\frac{L}{2}=D$ $1+cos \theta =\large\frac{2D}{L}$ $cos \theta=\large\frac{2D-L}{L}$ $Tsin \theta =mg$ $T=\large\frac{mg}{sin \theta}$ $T cos \theta =T_{1} \qquad \;$ AB string mass = $\large\frac{m}{2}$ $v=\sqrt{\large\frac{T_{1}}{\mu}}= \qquad \; \mu=\large\frac{mg}{2l}$ $v=\sqrt{\large\frac{T cos \theta 2l}{mg}}$ $v=\sqrt{\large\frac{mg}{sin \theta} \times \large\frac{cos 2 \theta 2l}{mg}}$ $v=\sqrt{cos \theta 2l}$ $=\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}\;.$ Answer : (a) $\;\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}$ Explanation : $2 \times \large\frac{L}{4} cos \theta + \large\frac{L}{2}=D$ $1+cos \theta =\large\frac{2D}{L}$ $cos \theta=\large\frac{2D-L}{L}$ $Tsin \theta =mg$ $T=\large\frac{mg}{sin \theta}$ $T cos \theta =T_{1} \qquad \;$ AB string mass = $\large\frac{m}{2}$ $v=\sqrt{\large\frac{T_{1}}{\mu}}= \qquad \; \mu=\large\frac{mg}{2l}$ $v=\sqrt{\large\frac{T cos \theta 2l}{mg}}$ $v=\sqrt{\large\frac{mg}{sin \theta} \times \large\frac{cos 2 \theta 2l}{mg}}$ $v=\sqrt{cos \theta 2l}$ $=\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}\;.$	{}
Question #64d80 Dec 16, 2017 There are two types of wave, but all transfer energy (or information) from source to detector. Explanation: You could also say they are all repeated periodic oscillations in an elastic medium (so the particles in the medium do not translate, only oscillate), they all obey the wave equation ($v = f \times \lambda$) and all (I think all) decrease in intensity with distance squared.	{}
Suggested languages for you: Americas Europe Q7 CQ Expert-verified Found in: Page 186 ### College Physics (Urone) Book edition 1st Edition Author(s) Paul Peter Urone Pages 1272 pages ISBN 9781938168000 # As cars travel, oil and gasoline leaks onto the road surface. If a light rain falls, what does this do to the control of the car? Does a heavy rain make any difference? A vehicle can lose control. Heavy rain does not affect the movement of cars as much. See the step by step solution ## Definition of Gasoline Gasoline or petrol is a clear, flammable liquid derived from petroleum and is used primarily as a fuel in most spark-ignition internal combustion engines (also known as gasoline engines). ## Determining status of car and effect of heavy rain Car motion is caused by the friction force. Static friction works when there is no slipping between the point of contact and road. When the breaks are applied the nature of friction force converts from static friction to kinetic friction. In the absence of friction force motion of car will not possible because the car moves due to the force which road applies on the car which is a result of friction force acting between the surfaces. Because oil is less dense than water, it rises to the top and settles on the road when a light rain falls. This produces a potentially dangerous condition because the friction between the surfaces is considerably reduced. As a result, a vehicle can lose control. The oil is distributed in a strong rain, which keeps the friction force low. Therefore, they do not affect the movement of cars as much.	{}
# Outputting a fraction on one line with an oblique division sign How can I make Mathematica output a fraction as x/2 as opposed to the x on top with a horizontal bar and 2 on the bottom? Perhaps I need to work more with Riffle but so far I cannot get it. I have tried concatenating strings, but Mathematica reverses the order of the two elements. Specifically, I am trying to put the symbol theta with a subscript i and then a division by 2. It is going onto a figure, and is very difficult to read with the vertical division. My code is: Do[ alabel[i] = Text[StyleForm[Subscript["\[Theta]", i]/2, FontSize -> 10, FontColor -> Black, FontFamily -> "Times"], {Re[ww[tt[[3]]]], Im[ww[tt[[3]]]]} + .2 {Cos[aloc3[[i]]], Sin[aloc3[[i]]]}, ioffset[i]] , {i, 1, 6}]; th3 = Graphics[Table[alabel[j], {j, 1, 6}]]; (The coordinate ww[[tt[[33]]]] is just some point in the plane). - I can't understand what you are trying to accomplish by inserting text with 3D coordinates into a 2D graphics object. Could you clarify your Question regarding that. – m_goldberg Oct 7 '13 at 21:38 Assuming you want your labels inserted into a 2D Graphic, it can be done like this: With[{n = 3}, Module[{pts, lbls}, SeedRandom@1; pts = RandomReal[1., {n, 2}]; lbls = Table[ Text[Row[{Subscript[θ, i], "/\[ThinSpace]2"}, "", BaseStyle -> {"TR", 10}], pts[[i]]], {i, Range @ n}]; Framed @ Graphics[lbls]]] The salient points in the above example are in the use of Row with the option BaseStyle and in adding a \[ThinSpace] to adjust the spacing between the slash and the denominator 2. Without the \[ThinSpace] the text looks too tight. - \$PrePrint = RawBoxes[ToBoxes[#] //. FractionBox[x_, y_] :> RowBox[{If[Head[x] === RowBox, RowBox[{"(", x, ")"}], x], "/", If[Head[y] === RowBox, RowBox[{"(", y, ")"}], y]}]] &; Then every fraction in your output will be of the form x/y. For example: - Here's a way to get the slash to not be written in the standard form: Plot[Sin[2 Pi t], {t, 0, 2 Pi}, PlotLabel -> Row[{Subscript[θ, i], "/", 2}]] -	{}
Measurements of the second Fourier harmonic coefficient ($v_2$) of the azimuthal distributions of prompt and nonprompt D$^0$ mesons produced in pp and pPb collisions are presented. Nonprompt D$^0$ mesons come from beauty hadron decays. The data samples are collected by the CMS experiment at nucleon-nucleon center-of-mass energies of 13 and 8.16 TeV, respectively. In high multiplicity pp collisions, $v_2$ signals for prompt charm hadrons are reported for the first time, and are found to be comparable to those for light-flavor hadron species over a transverse momentum ($p_\mathrm{T}$) range of 2-6 GeV. Compared at similar event multiplicities, the prompt D$^0$ meson $v_2$ values in pp and pPb collisions are similar in magnitude. The $v_2$ values for open beauty hadrons are extracted for the first time via nonprompt D$^0$ mesons in pPb collisions. For $p_\mathrm{T}$ in the range of 2-5 GeV, the results suggest that $v_2$ for nonprompt D$^0$ mesons are smaller than those for prompt D$^0$ mesons. These new measurements indicate a positive charm hadron $v_2$ in pp collisions and suggest a mass dependence in $v_2$ between charm and beauty hadrons in the pPb system. These results provide insights into the origin of heavy-flavor quark collectivity in small systems. ### Studies of charm and beauty hadron long-range correlations in pp and pPb collisions at LHC energies #### Abstract Measurements of the second Fourier harmonic coefficient ($v_2$) of the azimuthal distributions of prompt and nonprompt D$^0$ mesons produced in pp and pPb collisions are presented. Nonprompt D$^0$ mesons come from beauty hadron decays. The data samples are collected by the CMS experiment at nucleon-nucleon center-of-mass energies of 13 and 8.16 TeV, respectively. In high multiplicity pp collisions, $v_2$ signals for prompt charm hadrons are reported for the first time, and are found to be comparable to those for light-flavor hadron species over a transverse momentum ($p_\mathrm{T}$) range of 2-6 GeV. Compared at similar event multiplicities, the prompt D$^0$ meson $v_2$ values in pp and pPb collisions are similar in magnitude. The $v_2$ values for open beauty hadrons are extracted for the first time via nonprompt D$^0$ mesons in pPb collisions. For $p_\mathrm{T}$ in the range of 2-5 GeV, the results suggest that $v_2$ for nonprompt D$^0$ mesons are smaller than those for prompt D$^0$ mesons. These new measurements indicate a positive charm hadron $v_2$ in pp collisions and suggest a mass dependence in $v_2$ between charm and beauty hadrons in the pPb system. These results provide insights into the origin of heavy-flavor quark collectivity in small systems. ##### Scheda breve Scheda completa Scheda completa (DC) 2021 File in questo prodotto: Non ci sono file associati a questo prodotto. I documenti in IRIS sono protetti da copyright e tutti i diritti sono riservati, salvo diversa indicazione. Utilizza questo identificativo per citare o creare un link a questo documento: https://hdl.handle.net/11563/159089 ##### Attenzione Attenzione! I dati visualizzati non sono stati sottoposti a validazione da parte dell'ateneo • ND • 8 • ND	{}
# Increase spacing between equations \documentclass[11pt]{article} \usepackage{siunitx} \usepackage{amsmath} \begin{document} \begin{align} \text{Concentration of unknown sample solution} &= \frac{303 - 0.00227}{0.215} \\ &= 1.400 \si{\mg \per \L} \end{align} \end{document} Is it possible to have a line break like this? concentration of unknown = sample solution And increase the spacing between the equations? • It should be \SI{1.400}{\mg\per\L} Nov 10 '14 at 23:29 • @egreg So that's why my units were touching the numbers. I thought it was meant to be like that. – Jack Nov 10 '14 at 23:39 • A thin space should separate the number from the unit, which is what \SI does. Nov 10 '14 at 23:40 \documentclass[11pt]{article} \usepackage{siunitx} \usepackage{amsmath} \begin{document} \begin{align} \parbox{5cm}{\centering Concentration of unknown sample solution} &= \frac{303 - 0.00227}{0.215} \\[\jot] &= \SI{1.400}{\mg \per \L} \end{align} \end{document} • varwidth or tabular instead of \parbox? Nov 10 '14 at 23:33	{}
## Src/config plan for installed software This is a preliminary document describing how software configuration and source management should be handled for software that is installed in /s. Note: This is a very preliminary document; suggestions are welcome and appreciated on all points discussed. ## Goals The goals of this Src/configuration system are to: • allow us to determine precisely what source pool was used to build any of the installed software. This includes information about both: • where the sources originally came from, and • what local modifications were made to the sources. • allow us to determine precisely what binaries are installed. This includes information such as: • the set of configuration options used in compiling the software • where the binaries came from, if not built locally. • any changes made to binaries as installed (such as binary edits) ## Assumptions This document assumes that software is already being installed according to the recommendations made in the /s document. When reference is made to "the package's directory", we're referring to the directory /s/package-name. ## Plan of Action The set of goals above will be discussed one-by-one, with a suggested plan of action for each. ### Determining where sources originally came from • A URL pointing to the original sources for the package should be placed in a well-known canonical place. I recommend that a file named ChangeLog be created in each package directory for this purpose, and contain such a pointer. (The ChangeLog file will also contain other information, which is detailed below.) Some other name for the URL-containing file is fine; I first considered a file (called URL or ANCESTRY) containing only this pointer, but saw no advantage over combining it with other information in the ChangeLog. The primary distribution site is the preferred one to use as the URL for the package; by "primary", I mean the site at which a new version of the software package will normally be released. This will simplify in the future the task of checking whether a new version of a package is available. ### Determining what local modifications have been made to sources It is very useful for local modification information be available on two different levels of granularity. For example, sometimes a simple "executive summary" type of information is desirable: a couple of sentences describing what's changed. It's also important that a set of local patches be simple to obtain. These two different needs are handled by two different solutions: • Executive summary: Each package should have a ChangeLog type of file; this should detail any changes that are made to a package at a relatively high level of abstraction. The sorts of messages that are appropriate are along the same lines as what would be reported in a weekly staff meeting. Here's a sample of what a ChangeLog file might contain: Wed May 24 10:00:09 1995 Jon Cargille * Applied patches found on site <URL> rebuilt for all architectures Mon May 22 11:34:55 1995 Jon Cargille * Fetched version XX.YY.ZZ from <URL> built for all supported architectures • Use CVS to track details of local modifications: There are several reasons to choose CVS over RCS; the most obvious such is that CVS is much better at handling software trees, as opposed to single directories of sources. This translates into a higher probability that people will correctly save away an entire unmodified copy of a package. CVS should be used in the following way: • An unmodified copy of the entire package's distribution tree should be stored before any modifications are made. (This corresponds to an import command in CVS-speak.) This guarantees that we can easily see how the package was originally distributed, without having to grab a fresh distribution tar-ball and compare. • Each significant change to the package should be saved in the CVS repository (called a commit in CVS lingo). This includes changes to the configuration of the packages, as well as source changes. This guarantees that we can easily generate a set of all the local changes that have been made. • Each time a "commit" is done, an entry should also be added to the package's ChangeLog. The same entry can generally be used both for CVS's logs, and the ChangeLog. Note that various versions of the same software package will be imported into the CVS tree as different packages. This will cause some extra space usage, but will pay off significantly in the long run. The advantages are: • importing a newer packages is much simpler (one command), as opposed to a potential rats-nest of dealing with filenames that have changed, new files, files that have been removed, etc. This is one of the area that CVS is weak in, and by treating each new version as a separate package in the CVS tree, we avoid its area of weakness. • removing old versions of sources from the CVS repository becomes possible and easy (one command); if the sources are imported on top of one another (i.e. over-layed), removing old versions could be a nightmare. So in spite of the higher disk-space utilization in the short term, the long-term disk-space utilization should be lower. ### Determining what configuration was used in compiling a package There are a number of common but different configuration methods; each of these will have to be dealt with somewhat differently. • Packages configured with "configure" (generated by GNU autoconf) GNU configure are probably the easiest to deal with; if a package configured automatically once, it is likely to do so again. If a non-standard configure command is required (i.e. any command-line options to configure), they should be detailed in the ChangeLog file. If any of the configure-generated scripts are hand-edited, they should be committed to the CVS repository (both unmodified and modified versions) as well as described in the ChangeLog. • Packages with human-generated configure scripts Any package that comes with its own custom configure scripts should be handled in the same way as packages using GNU configure. Any configuration file that is hand-edited should be committed to the repository (both modified and unmodified versions), and described in the ChangeLog. Any command-line options that are passed to the configuration scripts should be described in the ChangeLog. If a package requires interactive input during the configuration process (such as prompting for the desired locations of various files, as news software does), a typescript file should be generated of the configuration process using the script command. This should be committed to the CVS repository as well. • Packages with no auto-configuration Packages that must be configured by hand-editing the source files (i.e. "config.h") and Makefiles should just have their modified Makefiles and source files committed to the CVS repository. Can anyone point out common configuration methods that I've missed? Are there packages for which the above would not capture enough information to repeat the build? You guys have more experience installing packages than I, so I'd be happy to have some feedback on this... ### Determining where binaries came from, if not built locally • A URL for the original binaries on the distribution site should be saved, in the same way a URL for the original sources would be handled. ### Determining what changes have been made to binaries as installed • Changes to a binary should be detailed in the ChangeLog file. ## Using CVS for source control ### Creating the CVSROOT Each package will have its own CVSROOT area, located in /s/pkgname/src/CVSROOT. The CVSROOT can be created and initialized with the following command: > setenv CVSROOT /s/pkgname/src/CVSROOT > cd /s/cvs/src/cvs-1.3 > ./cvsinit Note: The requirement of having your current directory be in the cvs src tree will go away when cvsinit gets fixed. I intend to do this, though it probably won't be right away. ### Importing the distribution It is important that each package's original distribution be imported into CVS, so that we can track all modifications that are made locally to each package. The initial distribution should be imported immediately after the distribution is extracted, as demonstrated by the following example: > echo \$CVSROOT /s/pkgname-x.y.z/src/CVSROOT > cd /s/pkgname-x.y.z/src > tar xzvf pkgname-x.y.z.tgz > cd pkgname-x.y.z > cvs import -I ! -m "pkgname x.y.z distribution" pkgname-x.y.z VENDOR PKGNAME_X_Y_Z Thus, the import line I used to import GNU-make was: > cvs import -I ! -m "GNU-make 3.74 distribution" make-3.74 GNU GNUMAKE_3_74 where: make-3.74 was the directory that I wanted to import GNU was the vendor GNUMAKE_3_74 was the version tag Note that the "import" command does update the imported tree at all. In other words, the equivalent of the "checkout" is not performed. So the first thing you want to do after the import is blow away the tree, and extract the version you just imported, so that additions and changes can be tracked. So, to continue the above example: > cd .. > \rm -rf pkgname-x.y.z > cvs co pkgname-x.y.z	{}
# Contest: fastest way to sort a big array of Gaussian-distributed data Following the interest in this question, I thought it would be interesting to make answers a bit more objective and quantitative by proposing a contest. The idea is simple: I have generated a binary file containing 50 million gaussian-distributed doubles (average: 0, stdev 1). The goal is to make a program that will sort these in memory as fast as possible. A very simple reference implementation in python takes 1m4s to complete. How low can we go? The rules are as follows: answer with a program that opens the file "gaussian.dat" and sorts the numbers in memory (no need to output them), and instructions for building and running the program. The program must be able to work on my Arch Linux machine (meaning you can use any programming language or library that is easily installable on this system). The program must be reasonably readable, so that I can make sure it is safe to launch (no assembler-only solution, please!). I will run the answers on my machine (quad core, 4 Gigabytes of RAM). The fastest solution will get the accepted answer and a 100 points bounty :) The program used to generate the numbers: #!/usr/bin/env python import random from array import array from sys import argv count=int(argv[1]) a=array('d',(random.gauss(0,1) for x in xrange(count))) f=open("gaussian.dat","wb") a.tofile(f) The simple reference implementation: #!/usr/bin/env python from array import array from sys import argv count=int(argv[1]) a=array('d') a.fromfile(open("gaussian.dat"),count) print "sorting..." b=sorted(a) EDIT: only 4 GB of RAM, sorry EDIT#2: Note that the point of the contest is to see if we can use prior information about the data. it is not supposed to be a pissing match between different programming language implementations! • Take each value and move it directly to its "expected" position, repeat for the displaced value. Not sure how to resolve a couple issue with that. When done, bubble sort until complete (a couple passes should do). – phkahler Jun 7 '11 at 20:29 • I will post a bucket sort solution tomorrow evening if this hasn't been closed by then :) – jonderry Jun 7 '11 at 21:22 • @static_rtti - as a heavy CG user, this is exactly the sort of thing "we" like to hack on at CG.SE. To any reading mods, move this to CG, don't close it. Jun 8 '11 at 0:44 • Welcome to CodeGolf.SE! I've cleared a lot of the commentary from the SO original concerning where this does or does not belong, and re-tagged to be closer to the CodeGolf.SE mainstream. Jun 8 '11 at 13:26 • The one tricky issue here is that we look for objective winning criteria, and "fastest" introduces platform dependencies...does a O(n^{1.2}) algorithm implemented on the cpython virtual machine beat a O(n^{1.3}) algorithm with a similar constant implemented in c? I generally suggest some discussion on the performance characteristics of each solution, as this may help people to judge what is going on. Jun 8 '11 at 13:26 Here is a solution in C++ which first partitions the numbers into buckets with same expected number of elements and then sorts each bucket separately. It precomputes a table of the cumulative distribution function based on some formulas from Wikipedia and then interpolates values from this table to get a fast approximation. Several steps run in multiple threads to make use of the four cores. #include <cstdlib> #include <math.h> #include <stdio.h> #include <algorithm> #include <tbb/parallel_for.h> using namespace std; typedef unsigned long long ull; double signum(double x) { return (x<0) ? -1 : (x>0) ? 1 : 0; } const double fourOverPI = 4 / M_PI; double erf(double x) { double a = 0.147; double x2 = xx; double ax2 = ax2; double f1 = -x2 * (fourOverPI + ax2) / (1 + ax2); double s1 = sqrt(1 - exp(f1)); return signum(x) * s1; } const double sqrt2 = sqrt(2); double cdf(double x) { return 0.5 + erf(x / sqrt2) / 2; } const int cdfTableSize = 200; const double cdfTableLimit = 5; double* computeCdfTable(int size) { double* res = new double[size]; for (int i = 0; i < size; ++i) { res[i] = cdf(cdfTableLimit * i / (size - 1)); } return res; } const double* const cdfTable = computeCdfTable(cdfTableSize); double cdfApprox(double x) { bool negative = (x < 0); if (negative) x = -x; if (x > cdfTableLimit) return negative ? cdf(-x) : cdf(x); double p = (cdfTableSize - 1) * x / cdfTableLimit; int below = (int) p; if (p == below) return negative ? -cdfTable[below] : cdfTable[below]; int above = below + 1; double ret = cdfTable[below] + (cdfTable[above] - cdfTable[below])(p - below); return negative ? 1 - ret : ret; } void print(const double arr, int len) { for (int i = 0; i < len; ++i) { printf("%e; ", arr[i]); } puts(""); } void print(const int* arr, int len) { for (int i = 0; i < len; ++i) { printf("%d; ", arr[i]); } puts(""); } void fillBuckets(int N, int bucketCount, double* data, int* partitions, double* buckets, int* offsets) { for (int i = 0; i < N; ++i) { ++offsets[partitions[i]]; } int offset = 0; for (int i = 0; i < bucketCount; ++i) { int t = offsets[i]; offsets[i] = offset; offset += t; } offsets[bucketCount] = N; int next[bucketCount]; memset(next, 0, sizeof(next)); for (int i = 0; i < N; ++i) { int p = partitions[i]; int j = offsets[p] + next[p]; ++next[p]; buckets[j] = data[i]; } } class Sorter { public: Sorter(double* data, int* offsets) { this->data = data; this->offsets = offsets; } static void radixSort(double* arr, int len) { ull* encoded = (ull)arr; for (int i = 0; i < len; ++i) { ull n = encoded[i]; if (n & signBit) { n ^= allBits; } else { n ^= signBit; } encoded[i] = n; } const int step = 11; const ull mask = (1ull << step) - 1; int offsets[8][1ull << step]; memset(offsets, 0, sizeof(offsets)); for (int i = 0; i < len; ++i) { for (int b = 0, j = 0; b < 64; b += step, ++j) { int p = (encoded[i] >> b) & mask; ++offsets[j][p]; } } int sum[8] = {0}; for (int i = 0; i <= mask; i++) { for (int b = 0, j = 0; b < 64; b += step, ++j) { int t = sum[j] + offsets[j][i]; offsets[j][i] = sum[j]; sum[j] = t; } } ull copy = new ull[len]; ull* current = encoded; for (int b = 0, j = 0; b < 64; b += step, ++j) { for (int i = 0; i < len; ++i) { int p = (current[i] >> b) & mask; copy[offsets[j][p]] = current[i]; ++offsets[j][p]; } ull* t = copy; copy = current; current = t; } if (current != encoded) { for (int i = 0; i < len; ++i) { encoded[i] = current[i]; } } for (int i = 0; i < len; ++i) { ull n = encoded[i]; if (n & signBit) { n ^= signBit; } else { n ^= allBits; } encoded[i] = n; } } void operator() (tbb::blocked_range<int>& range) const { for (int i = range.begin(); i < range.end(); ++i) { double* begin = &data[offsets[i]]; double* end = &data[offsets[i+1]]; //std::sort(begin, end); } } private: double* data; int* offsets; static const ull signBit = 1ull << 63; static const ull allBits = ~0ull; }; void sortBuckets(int bucketCount, double* data, int* offsets) { Sorter sorter(data, offsets); tbb::blocked_range<int> range(0, bucketCount); tbb::parallel_for(range, sorter); //sorter(range); } class Partitioner { public: Partitioner(int bucketCount, double* data, int* partitions) { this->data = data; this->partitions = partitions; this->bucketCount = bucketCount; } void operator() (tbb::blocked_range<int>& range) const { for (int i = range.begin(); i < range.end(); ++i) { double d = data[i]; int p = (int) (cdfApprox(d) * bucketCount); partitions[i] = p; } } private: double* data; int* partitions; int bucketCount; }; const int bucketCount = 512; int offsets[bucketCount + 1]; int main(int argc, char** argv) { if (argc != 2) { printf("Usage: %s N\n N = the size of the input\n", argv[0]); return 1; } puts("initializing..."); int N = atoi(argv[1]); double* data = new double[N]; double* buckets = new double[N]; memset(offsets, 0, sizeof(offsets)); int* partitions = new int[N]; FILE* fp = fopen("gaussian.dat", "rb"); if (fp == 0 \|\| fread(data, sizeof(data), N, fp) != N) { return 1; } //print(data, N); puts("assigning partitions..."); tbb::parallel_for(tbb::blocked_range<int>(0, N), Partitioner(bucketCount, data, partitions)); puts("filling buckets..."); fillBuckets(N, bucketCount, data, partitions, buckets, offsets); data = buckets; puts("sorting buckets..."); sortBuckets(bucketCount, data, offsets); puts("done."); / for (int i = 0; i < N-1; ++i) { if (data[i] > data[i+1]) { printf("error at %d: %e > %e\n", i, data[i], data[i+1]); } } / //print(data, N); return 0; } To compile and run it, use this command: g++ -O3 -ltbb -o gsort gsort.cpp && time ./gsort 50000000 EDIT: All buckets are now placed into the same array to remove the need to copy the buckets back into the array. Also the size of the table with precomputed values was reduced, because the values are accurate enough. Still, if I change the number of buckets above 256, the program takes longer to run than with that number of buckets. EDIT: The same algorithm, different programming language. I used C++ instead of Java and the running time reduced from ~3.2s to ~2.35s on my machine. The optimal number of buckets is still around 256 (again, on my computer). By the way, tbb really is awesome. EDIT: I was inspired by Alexandru's great solution and replaced the std::sort in the last phase by a modified version of his radix sort. I did use a different method to deal with the positive/negative numbers, even though it needs more passes through the array. I also decided to sort the array exactly and remove the insertion sort. I will later spend some time testing how do these changes influence performance and possibly revert them. However, by using radix sort, the time decreased from ~2.35s to ~1.63s. • Nice. I got 3.055 on mine. Lowest I was able to get mine to was 6.3. I'm picking through yours to get the stats better. Why did you choose 256 as the number of buckets? I tried 128 and 512, yet 256 worked the best. Jun 9 '11 at 14:07 • Why did I choose 256 as the number of buckets? I tried 128 and 512, yet 256 worked the best. :) I found it empirically and I am not sure why increasing the number of buckets slows down the algorithm - memory allocation should not take that long. Maybe something related to cache size? – k21 Jun 9 '11 at 17:08 • 2.725s on my machine. Pretty nice for a java solution, taking into account the loading time of the JVM. Jun 9 '11 at 19:17 • I switched your code over to use the nio packages, per my and Arjan's solution(used his syntax, since it was cleaner than mine) and was able to get it .3 seconds faster. I've got an ssd, I wonder what the implications might be if not. It also gets rid of some of your bit twiddling. Modded sections are here. Jun 9 '11 at 20:06 • This is the fastest parallel solution on my tests (16core cpu). 1.22s far from 1.94s second place. Jun 10 '11 at 11:46 Without getting smart, just to provide a much faster naïve sorter, here's one in C which should be pretty much equivalent to your Python one: #include <stdio.h> #include <stdlib.h> #include <string.h> int cmp(const void av, const void* bv) { double a = (const double)av; double b = (const double)bv; return a < b ? -1 : a > b ? 1 : 0; } int main(int argc, char** argv) { if (argc <= 1) return puts("No argument!"); unsigned count = atoi(argv[1]); double a = malloc(count sizeof a); FILE f = fopen("gaussian.dat", "rb"); if (fread(a, sizeof a, count, f) != count) fclose(f); puts("sorting..."); double b = malloc(count * sizeof b); memcpy(b, a, count sizeof b); qsort(b, count, sizeof b, cmp); return 0; } Compiled with gcc -O3, on my machine this takes more than a minute less than the Python: about 11 s compared to 87 s. • Took 10.086s on my machine, which makes you the current leader! But I'm pretty sure we can do better :) – static_rtti Jun 7 '11 at 17:17 • Could you try to remove the second ternary operator and simply return 1 for that case because random doubles are like not equal to each other in these amount of data. Jun 7 '11 at 17:39 • @Codism: I would add that we don't care about swapping locations of equivalent data so therefore even if we could get equivalent values it would be an appropriate simplification. – Guvante Jun 7 '11 at 18:43 I partitioned into segments based on the standard deviation that should best break it down into 4ths. Edit: Rewritten to partition based on x value in http://en.wikipedia.org/wiki/Error_function#Table_of_values http://www.wolframalpha.com/input/?i=percentages+by++normal+distribution I tried using smaller buckets, but it seemed to have little effect once 2 * beyond the number of cores available. Without any parallel collections, it would take 37 seconds on my box and 24 with the parallel collections. If partitioning via distribution, you can't just use an array, so there is some more overhead. I'm not clear on when a value would be boxed/unboxed in scala. I'm using scala 2.9, for the parallel collection. You can just download the tar.gz distribution of it. To compile: scalac SortFile.scala (I just copied it directly in the scala/bin folder. To run: JAVA_OPTS="-Xmx4096M" ./scala SortFile (I ran it with 2 gigs of ram and got about the same time) Edit: Removed allocateDirect, slower than just allocate. Removed priming of initial size for array buffers. Actually made it read the whole 50000000 values. Rewrote to hopefully avoid autoboxing issues (still slower than naive c) import java.io.FileInputStream; import java.nio.ByteBuffer import java.nio.ByteOrder import scala.collection.mutable.ArrayBuilder object SortFile { //used partition numbers from Damascus' solution val partList = List(0, 0.15731, 0.31864, 0.48878, 0.67449, 0.88715, 1.1503, 1.5341) val listSize = partList.size * 2; val posZero = partList.size; val neg = partList.map( _ * -1).reverse.zipWithIndex val pos = partList.map( _ * 1).zipWithIndex.reverse def partition(dbl:Double): Int = { //for each partition, i am running through the vals in order //could make this a binary search to be more performant... but our list size is 4 (per side) if(dbl < 0) { return neg.find( dbl < _._1).get._2 } if(dbl > 0) { return posZero + pos.find( dbl > _._1).get._2 } return posZero; } def main(args: Array[String]) { var l = 0 val dbls = new Array[Double](50000000) val partList = new Array[Int](50000000) val pa = Array.fill(listSize){Array.newBuilder[Double]} val channel = new FileInputStream("../../gaussian.dat").getChannel() val bb = ByteBuffer.allocate(50000000 * 8) bb.order(ByteOrder.LITTLE_ENDIAN) bb.rewind var dbl = 0.0 while(bb.hasRemaining) { dbl = bb.getDouble dbls.update(l,dbl) l+=1 } for( i <- (0 to 49999999).par) { partList.update(i, partition(dbls(i)))} println("Partition computed" + System.currentTimeMillis() ) for(i <- (0 to 49999999)) { pa(partList(i)) += dbls(i) } println("Partition completed " + System.currentTimeMillis()) val toSort = for( i <- pa) yield i.result() println("Arrays Built" + System.currentTimeMillis()); toSort.par.foreach{i:Array[Double] =>scala.util.Sorting.quickSort(i)}; } } • 8.185s! Nice for a scala solution, I guess... Also, bravo for providing the first solution that actually uses the Gaussian distribution in some way! – static_rtti Jun 8 '11 at 6:12 • I was only aiming to compete with the c# solution. Didn't figure I'd beat c/c++. Also.. it's behaving much differently for you than for me. I'm using openJDK on my end and it's a lot slower. I wonder if adding more partitions would help in your env. Jun 8 '11 at 13:00 Just put this in a cs file and compile it with csc in theory: (Requires mono) using System; using System.IO; namespace Sort { class Program { const int count = 50000000; static double[][] doubles; static WaitHandle[] waiting = new WaitHandle[4]; static AutoResetEvent[] events = new AutoResetEvent[4]; static double[] Merge(double[] left, double[] right) { double[] result = new double[left.Length + right.Length]; int l = 0, r = 0, spot = 0; while (l < left.Length && r < right.Length) { if (right[r] < left[l]) result[spot++] = right[r++]; else result[spot++] = left[l++]; } while (l < left.Length) result[spot++] = left[l++]; while (r < right.Length) result[spot++] = right[r++]; return result; } { int index = (int)data; Array.Sort(doubles[index]); events[index].Set(); } static void Main(string[] args) { System.Diagnostics.Stopwatch watch = new System.Diagnostics.Stopwatch(); watch.Start(); doubles = new double[][] { new double[count / 4], new double[count / 4], new double[count / 4], new double[count / 4] }; for (int i = 0; i < 4; i++) { for (int j = 0; j < count / 4; j++) { doubles[i][j] = BitConverter.ToDouble(bytes, i * count/4 + j * 8); } } for (int i = 0; i < 4; i++) { waiting[i] = events[i] = new AutoResetEvent(false); } WaitHandle.WaitAll(waiting); double[] left = Merge(doubles[0], doubles[1]); double[] right = Merge(doubles[2], doubles[3]); double[] result = Merge(left, right); watch.Stop(); Console.WriteLine(watch.Elapsed.ToString()); } } } • Can I run your solutions with Mono? How should I do it? – static_rtti Jun 7 '11 at 17:21 • Haven't used Mono, didn't think of that, you should be able to compile the F# and then run it. – Guvante Jun 7 '11 at 17:38 • Updated to use four threads to improve performance. Now gives me 6 sec. Note that this could be significantly improved (5 sec likely) if you use only one spare array and avoid initializing a ton of memory to zero, which is done by the CLR, since everything is being written to at least once. – Guvante Jun 7 '11 at 18:41 • 9.598s on my machine! You're the current leader :) – static_rtti Jun 7 '11 at 18:55 • My mother told me to stay away from guys with Mono! – Mark Canlas Jun 8 '11 at 1:54 Since you know what the distribution is, you can use a direct indexing O(N) sort. (If you're wondering what that is, suppose you have a deck of 52 cards and you want to sort it. Just have 52 bins and toss each card into it's own bin.) You have 5e7 doubles. Allocate a result array R of 5e7 doubles. Take each number x and get i = phi(x) * 5e7. Basically do R[i] = x. Have a way to handle collisions, such as moving the number it may be colliding with (as in simple hash coding). Alternatively, you could make R a few times larger, filled with a unique empty value. At the end, you just sweep up the elements of R. phi is just the gaussian cumulative distribution function. It converts a gaussian distributed number between +/- infinity into a uniform distributed number between 0 and 1. A simple way to calculate it is with table lookup and interpolation. • Be careful: you know the approximate distribution, not the exact distribution. You know the data was generated using a Gaussian law, but since it is finite, it doesn't exactly follow a Gaussian. – static_rtti Jun 7 '11 at 18:48 • @static_rtti: In this case the necessary approximation of phi would create a bigger hassle than any irregularities in the data set IMO. – Guvante Jun 7 '11 at 18:54 • @static_rtti: it doesn't have to be exact. It only has to spread out the data so it is approximately uniform, so it doesn't bunch up too much in some places. – Mike Dunlavey Jun 8 '11 at 4:46 • Suppose you have 5e7 doubles. Why not just make each entry in R a vector of, say, 5e6 vectors of double. Then, push_back each double in its appropriate vector. Sort the vectors and you're done. This should take linear time in the size of the input. Jun 8 '11 at 7:32 • Actually, I see that mdkess already came up with that solution. Jun 8 '11 at 7:33 Here is another sequential solution: #include <stdio.h> #include <stdlib.h> #include <algorithm> #include <ctime> typedef unsigned long long ull; int size; double dbuf, copy; int cnt[8][1 << 16]; void sort() { const int step = 10; const int start = 24; ull mask = (1ULL << step) - 1; ull ibuf = (ull ) dbuf; for (int i = 0; i < size; i++) { for (int w = start, v = 0; w < 64; w += step, v++) { int p = (~ibuf[i] >> w) & mask; cnt[v][p]++; } } int sum[8] = { 0 }; for (int i = 0; i <= mask; i++) { for (int w = start, v = 0; w < 64; w += step, v++) { int tmp = sum[v] + cnt[v][i]; cnt[v][i] = sum[v]; sum[v] = tmp; } } for (int w = start, v = 0; w < 64; w += step, v++) { ull ibuf = (ull ) dbuf; for (int i = 0; i < size; i++) { int p = (~ibuf[i] >> w) & mask; copy[cnt[v][p]++] = dbuf[i]; } double tmp = copy; copy = dbuf; dbuf = tmp; } for (int p = 0; p < size; p++) if (dbuf[p] >= 0.) { std::reverse(dbuf + p, dbuf + size); break; } // Insertion sort for (int i = 1; i < size; i++) { double value = dbuf[i]; if (value < dbuf[i - 1]) { dbuf[i] = dbuf[i - 1]; int p = i - 1; for (; p > 0 && value < dbuf[p - 1]; p--) dbuf[p] = dbuf[p - 1]; dbuf[p] = value; } } } int main(int argc, char argv) { size = atoi(argv[1]); dbuf = new double[size]; copy = new double[size]; FILE f = fopen("gaussian.dat", "r"); fclose(f); clock_t c0 = clock(); sort(); printf("Finished after %.3f\n", (double) ((clock() - c0)) / CLOCKS_PER_SEC); return 0; } I doubt it beats the multi-threaded solution, but the timings on my i7 laptop are (stdsort is the C++ solution provided in another answer): $g++ -O3 mysort.cpp -o mysort && ./mysort 50000000 Finished after 2.10$ g++ -O3 stdsort.cpp -o stdsort && ./stdsort Finished after 7.12 Note that this solution has linear time complexity (because it uses the special representation of doubles). EDIT: Fixed the order of elements to be increasing. EDIT: Improved speed by almost half a second. EDIT: Improved speed by another 0.7 seconds. Made the algorithm more cache friendly. EDIT: Improved speed by another 1 second. Since there only 50.000.000 elements I can partially sort the mantissa and use insert sort (which is cache friendly) to fix out-of-place elements. This idea removes around two iterations from the last radix sorting loop. EDIT: 0.16 less seconds. First std::reverse can be eliminated if the sorting order is reversed. • Now that is getting interesting! What kind of sort algorithm is it? Jun 8 '11 at 11:29 • Least significant digit radix sort. You can sort the mantissa, then the exponent, then the sign. The algorithm presented here takes this idea one step further. It can be parallelized using a partitioning idea provided in a different answer. Jun 8 '11 at 11:37 • Pretty fast for a single threaded solution: 2.552s! Do you think you could change your solution to make use of the fact that the data is normally distributed? You could probably do better than the current best multi-threaded solutions. Jun 8 '11 at 18:13 • @static_rtti: I see that Damascus Steel already posted a multithreaded version of this implementation. I improved caching behavior of this algorithm, so you should get better timings now. Please test this new version. Jun 9 '11 at 11:29 • 1.459s in my latests tests. While this solution is not the winner per my rules, it really deserves big kudos. Congratulations! Jun 15 '11 at 22:02 Taking Christian Ammer's solution and parallelizing it with Intel's Threaded Building Blocks #include <iostream> #include <fstream> #include <algorithm> #include <vector> #include <ctime> #include <tbb/parallel_sort.h> int main(void) { std::ifstream ifs("gaussian.dat", std::ios::binary \| std::ios::in); std::vector<double> values; values.reserve(50000000); double d; values.push_back(d); clock_t c0 = clock(); tbb::parallel_sort(values.begin(), values.end()); std::cout << "Finished after " << static_cast<double>((clock() - c0)) / CLOCKS_PER_SEC << std::endl; } If you have access to Intel's Performance Primitives (IPP) library, you can use its radix sort. Just replace #include <tbb/parallel_sort.h> with #include "ipps.h" and tbb::parallel_sort(values.begin(), values.end()); with std::vector<double> copy(values.size()); On my dual core laptop, the timings are C 16.4 s C# 20 s C++ std::sort 7.2 s C++ tbb 5 s C++ ipp 4.5 s python too long • 2.958s! TBB seems pretty cool and easy to use! – static_rtti Jun 8 '11 at 6:06 • TBB is absurdly awesome. It's exactly the right level of abstraction for algorithmic work. Jun 8 '11 at 8:16 How about an implementation of parallel quicksort that chooses its pivot values based on the statistics of the distribution, thereby ensuring equal sized partitions? The first pivot would be at the mean (zero in this case), the next pair would be at the 25th and 75th percentiles (+/- -0.67449 standard deviations), and so on, with each partition halving the remaining data set more or less perfectly. • That's effectively what I did on mine.. of course you got this post up before I could finish my writeup. – Scott Jun 7 '11 at 21:12 Very ugly (why use arrays when I can use variables ending with numbers), but fast code (my first try to std::threads), whole time (time real) on my system 1,8 s (comparing to std::sort() 4,8 s), compile with g++ -std=c++0x -O3 -march=native -pthread Just pass data through stdin (works only for 50M). #include <cstdio> #include <iostream> #include <algorithm> using namespace std; const size_t size=50000000; void pivot(double* start,double * end, double middle,size_t& koniec){ double * beg=start; end--; while (start!=end){ if (start>middle) swap (start,end--); else start++; } if (end<middle) start+=1; koniec= start-beg; } void s(double * a, double* b){ sort(a,b); } int main(){ double data=new double[size]; FILE f = fopen("gaussian.dat", "rb"); size_t end1,end2,end3,temp; pivot(data, data+size,0,end2); pivot(data, data+end2,-0.6745,end1); pivot(data+end2,data+size,0.6745,end3); end3+=end2; ts1.join(),ts2.join(),ts3.join(),ts4.join(); //for (int i=0; i<size-1; i++){ //} fclose(f); //fwrite(data,8,size,stdout); } //Edit changed to read gaussian.dat file. • Could you change it to read gaussian.dat, as the above C++ solutions do? – static_rtti Jun 8 '11 at 6:07 • I'll try later when I come home. Jun 8 '11 at 8:12 • Very nice solution, you're the current leader (1.949s)! And nice use of the gaussian distribution :) Jun 8 '11 at 18:11 A C++ solution using std::sort (eventually faster than qsort, regarding Performance of qsort vs std::sort) #include <iostream> #include <fstream> #include <algorithm> #include <vector> #include <ctime> int main(void) { std::ifstream ifs("C:\\Temp\\gaussian.dat", std::ios::binary \| std::ios::in); std::vector<double> values; values.reserve(50000000); double d; values.push_back(d); clock_t c0 = clock(); std::sort(values.begin(), values.end()); std::cout << "Finished after " << static_cast<double>((clock() - c0)) / CLOCKS_PER_SEC << std::endl; } I can't reliable say how long it takes because I only have 1GB on my machine and with the given Python code I could only make a gaussian.dat file with only 25mio doubles (without getting a Memory error). But I'm very interested how long the std::sort algorithm runs. • 6.425s! As expected, C++ shines :) – static_rtti Jun 8 '11 at 6:04 • @static_rtti: I've tried swensons Timsort algorithm (as suggested from Matthieu M. in your first question). I had to make some changes to the sort.h file to compile it with C++. It was about twice as slow than std::sort. Don't know why, maybe because of compiler optimizations? Jun 8 '11 at 19:39 Here is a mix of Alexandru's radix sort with Zjarek's threaded smart pivoting. Compile it with g++ -std=c++0x -pthread -O3 -march=native sorter_gaussian_radix.cxx -o sorter_gaussian_radix You can change the radix size by defining STEP (e.g. add -DSTEP=11). I found the best for my laptop is 8 (the default). By default, it splits the problem into 4 pieces and runs that on multiple threads. You can change that by passing a depth parameter to the command line. So if you have two cores, run it as sorter_gaussian_radix 50000000 1 and if you have 16 cores sorter_gaussian_radix 50000000 4 The max depth right now is 6 (64 threads). If you put too many levels, you will just slow the code down. One thing I also tried was the radix sort from the Intel Performance Primitives (IPP) library. Alexandru's implementation soundly trounces IPP, with IPP being about 30% slower. That variation is also included here (commented out). #include <stdlib.h> #include <string.h> #include <algorithm> #include <ctime> #include <iostream> #include <vector> #include <boost/cstdint.hpp> // #include "ipps.h" #ifndef STEP #define STEP 8 #endif const int step = STEP; const int start_step=24; const int num_steps=(64-start_step+step-1)/step; int size; double dbuf, copy; clock_t c1, c2, c3, c4, c5; const double distrib[]={-2.15387, -1.86273, -1.67594, -1.53412, -1.4178, -1.31801, -1.22986, -1.15035, -1.07752, -1.00999, -0.946782, -0.887147, -0.830511, -0.776422, -0.724514, -0.67449, -0.626099, -0.579132, -0.53341, -0.488776, -0.445096, -0.40225, -0.36013, -0.318639, -0.27769, -0.237202, -0.197099, -0.157311, -0.11777, -0.0784124, -0.0391761, 0, 0.0391761, 0.0784124, 0.11777, 0.157311, 0.197099, 0.237202, 0.27769, 0.318639, 0.36013, 0.40225, 0.445097, 0.488776, 0.53341, 0.579132, 0.626099, 0.67449, 0.724514, 0.776422, 0.830511, 0.887147, 0.946782, 1.00999, 1.07752, 1.15035, 1.22986, 1.31801, 1.4178, 1.53412, 1.67594, 1.86273, 2.15387}; class Distrib { const int value; public: Distrib(const double &v): value(v) {} bool operator()(double a) { return a<value; } }; void recursive_sort(const int start, const int end, const int index, const int offset, const int depth, const int max_depth) { if(depth<max_depth) { Distrib dist(distrib[index]); const int middle=std::partition(dbuf+start,dbuf+end,dist) - dbuf; // const int middle= // std::partition(dbuf+start,dbuf+end,[&](double a) // {return a<distrib[index];}) // - dbuf; depth+1,max_depth); depth+1,max_depth); lower.join(), upper.join(); } else { c1=clock(); double dbuf_local(dbuf), copy_local(copy); boost::uint64_t mask = (1 << step) - 1; boost::uint64_t ibuf = reinterpret_cast<boost::uint64_t > (dbuf_local); for(int i=0;i<num_steps;++i) cnt[i][j]=0; for (int i = start; i < end; i++) { for (int w = start_step, v = 0; w < 64; w += step, v++) { int p = (~ibuf[i] >> w) & mask; (cnt[v][p])++; } } c2=clock(); std::vector<int> sum(num_steps,0); for (uint i = 0; i <= mask; i++) { for (int w = start_step, v = 0; w < 64; w += step, v++) { int tmp = sum[v] + cnt[v][i]; cnt[v][i] = sum[v]; sum[v] = tmp; } } c3=clock(); for (int w = start_step, v = 0; w < 64; w += step, v++) { ibuf = reinterpret_cast<boost::uint64_t >(dbuf_local); for (int i = start; i < end; i++) { int p = (~ibuf[i] >> w) & mask; copy_local[start+((cnt[v][p])++)] = dbuf_local[i]; } std::swap(copy_local,dbuf_local); } // Do the last set of reversals for (int p = start; p < end; p++) if (dbuf_local[p] >= 0.) { std::reverse(dbuf_local+p, dbuf_local + end); break; } c4=clock(); // Insertion sort for (int i = start+1; i < end; i++) { double value = dbuf_local[i]; if (value < dbuf_local[i - 1]) { dbuf_local[i] = dbuf_local[i - 1]; int p = i - 1; for (; p > 0 && value < dbuf_local[p - 1]; p--) dbuf_local[p] = dbuf_local[p - 1]; dbuf_local[p] = value; } } c5=clock(); } } int main(int argc, char argv) { size = atoi(argv[1]); copy = new double[size]; dbuf = new double[size]; FILE f = fopen("gaussian.dat", "r"); fclose(f); clock_t c0 = clock(); const int max_depth= (argc > 2) ? atoi(argv[2]) : 2; recursive_sort(0,size,31,16,0,max_depth); if(num_steps%2==1) std::swap(dbuf,copy); // for (int i=0; i<size-1; i++){ // if (dbuf[i]>dbuf[i+1]) // << i << " " // << dbuf[i] << " " // << dbuf[i+1] << " " // << "\n"; // } std::cout << "Finished after " << (double) (c1 - c0) / CLOCKS_PER_SEC << " " << (double) (c2 - c1) / CLOCKS_PER_SEC << " " << (double) (c3 - c2) / CLOCKS_PER_SEC << " " << (double) (c4 - c3) / CLOCKS_PER_SEC << " " << (double) (c5 - c4) / CLOCKS_PER_SEC << " " << "\n"; // delete [] dbuf; // delete [] copy; return 0; } EDIT: I implemented Alexandru's cache improvements, and that shaved off about 30% of the time on my machine. EDIT: This implements a recursive sort, so it should work well on Alexandru's 16 core machine. It also use's Alexandru's last improvement and removes one of the reverse's. For me, this gave a 20% improvement. EDIT: Fixed a sign bug that caused inefficiency when there are more than 2 cores. EDIT: Removed the lambda, so it will compile with older versions of gcc. It includes the IPP code variation commented out. I also fixed the documentation for running on 16 cores. As far as I can tell, this is the fastest implementation. EDIT: Fixed a bug when STEP is not 8. Increased the maximum number of threads to 64. Added some timing info. • Nice. Radix sort is very cache unfriendly. See if you can get better results by changing step (11 was optimal on my laptop). Jun 9 '11 at 9:26 • You have a bug: int cnt[mask] should be int cnt[mask + 1]. For better results use a fixed value int cnt[1 << 16]. Jun 9 '11 at 9:38 • I'll try all these solutions later today when I get home. Jun 9 '11 at 9:40 • 1.534s!!! I think we have a leader :-D Jun 9 '11 at 19:27 • @static_rtti: Could you try this again? It has gotten significantly faster than the last time you tried it. On my machine, it is substantially faster than any other solution. Jun 16 '11 at 2:08 I guess this really depends on what you want to do. If you want to sort a bunch of Gaussians, then this won't help you. But if you want a bunch of sorted Gaussians, this will. Even if this misses the problem a bit, I think it will be interesting to compare vs actual sorting routines. If you want to something to be fast, do less. Instead of generating a bunch of random samples from the normal distribution and then sorting, you can generate a bunch of samples from the normal distribution in sorted order. You can use the solution here to generate n uniform random numbers in sorted order. Then you can use the inverse cdf (scipy.stats.norm.ppf) of the normal distribution to turn the uniform random numbers into numbers from the normal distribution via inverse transform sampling. import scipy.stats import random # slightly modified from linked stackoverflow post def n_random_numbers_increasing(n): """Like sorted(random() for i in range(n))), but faster because we avoid sorting.""" v = 1.0 while n: v = random.random() * (1.0 / n) yield 1 - v n -= 1 def n_normal_samples_increasing(n): return map(scipy.stats.norm.ppf, n_random_numbers_increasing(n)) If you want to get your hands dirtier, I'd guess that you might be able to speed up the many inverse cdf calculations by using some kind of iterative method, and using the previous result as your initial guess. Since the guesses are going to be very close, probably a single iteration will give you great accuracy. • Nice answer, but that would be cheating :) The idea of my question is that while sorting algorithms have been given enormous attention, there is almost no literature on the use of prior knowledge about the data for sorting, even though the few papers that have addressed the issue have reported nice gains. So let's see what's possible! – static_rtti Jun 7 '11 at 16:50 Try this changing Guvante's solution with this Main(), it starts sorting as soon as 1/4 IO reading is done, it's faster in my test: static void Main(string[] args) { FileStream filestream = new FileStream(@"..\..\..\gaussian.dat", FileMode.Open, FileAccess.Read); doubles = new double[][] { new double[count / 4], new double[count / 4], new double[count / 4], new double[count / 4] }; for (int i = 0; i < 4; i++) { byte[] bytes = new byte[count * 4]; for (int j = 0; j < count / 4; j++) { doubles[i][j] = BitConverter.ToDouble(bytes, i * count/4 + j * 8); } waiting[i] = events[i] = new AutoResetEvent(false); } WaitHandle.WaitAll(waiting); double[] left = Merge(doubles[0], doubles[1]); double[] right = Merge(doubles[2], doubles[3]); double[] result = Merge(left, right); } } • 8.933s. Slightly faster :) – static_rtti Jun 8 '11 at 6:02 Since you know the distribution, my idea would be to make k buckets, each with the same expected number of elements (since you know the distribution, you can compute this). Then in O(n) time, sweep the array and put elements into their buckets. Then concurrently sort the buckets. Suppose you have k buckets, and n elements. A bucket will take (n/k) lg (n/k) time to sort. Now suppose that you have p processors that you can use. Since buckets can be sorted independently, you have a multiplier of ceil(k/p) to deal with. This gives a final runtime of n + ceil(k/p)(n/k) lg (n/k), which should be a good deal faster than n lg n if you choose k well. • I think this is the best solution. Jun 8 '11 at 7:33 • You don't exactly know the number of elements who will end up in a bucket, so the math is actually wrong. That being said, this is a good answer I think. Jun 8 '11 at 19:43 • @pouejapon: You're right. Jun 9 '11 at 6:18 • This answer sounds really nice. The problem is -- it isn't really fast. I implemented this in C99 (see my answer), and it certainly easily beats std::sort(), but it is way slower than Alexandru's radixsort solution. Jun 9 '11 at 17:54 One low-level optimization idea is to fit two doubles in a SSE register, so each thread would work with two items at a time. This might be complicated to do for some algorithms. Another thing to do is sorting the array in cache-friendly chunks, then merging the results. Two levels should be used: for example first 4 KB for L1 then 64 KB for L2. This should be very cache-friendly, since the bucket sort will not go outside the cache, and the final merge will walk memory sequentially. These days computation is much cheaper than memory accesses. However we have a large number of items, so it's hard to tell which is the array size when dumb cache-aware sort is slower than a low-complexity non-cache-aware version. But I won't provide an implementation of the above since I would do it in Windows (VC++). Here's a linear scan bucket sort implementation. I think it's faster than all current single-threaded implementations except for radix sort. It should have linear expected running time if I'm estimating the cdf accurately enough (I'm using linear interpolation of values I found on the web) and haven't made any mistakes that would cause excessive scanning: #include <stdio.h> #include <stdlib.h> #include <string.h> #include <algorithm> #include <ctime> using std::fill; const double q[] = { 0.0, 9.865E-10, 2.8665150000000003E-7, 3.167E-5, 0.001349898, 0.022750132, 0.158655254, 0.5, 0.8413447460000001, 0.9772498679999999, 0.998650102, 0.99996833, 0.9999997133485, 0.9999999990134999, 1.0, }; int main(int argc, char* argv) { if (argc <= 1) return puts("No argument!"); unsigned count = atoi(argv[1]); unsigned count2 = 3 * count; bool ba = new bool[count2 + 1000]; fill(ba, ba + count2 + 1000, false); double a = new double[count]; double c = new double[count2 + 1000]; FILE f = fopen("gaussian.dat", "rb"); if (fread(a, 8, count, f) != count) fclose(f); int i; int j; bool s; int t; double z; double p; double d1; double d2; for (i = 0; i < count; i++) { s = a[i] < 0; t = a[i]; if (s) t--; z = a[i] - t; t += 7; if (t < 0) { t = 0; z = 0; } else if (t >= 14) { t = 13; z = 1; } p = q[t] * (1 - z) + q[t + 1] * z; j = count2 * p; while (ba[j] && c[j] < a[i]) { j++; } if (!ba[j]) { ba[j] = true; c[j] = a[i]; } else { d1 = c[j]; c[j] = a[i]; j++; while (ba[j]) { d2 = c[j]; c[j] = d1; d1 = d2; j++; } c[j] = d1; ba[j] = true; } } i = 0; int max = count2 + 1000; for (j = 0; j < max; j++) { if (ba[j]) { a[i++] = c[j]; } } // for (i = 0; i < count; i += 1) { // printf("here %f\n", a[i]); // } return 0; } • I'll try this later today when I get home. In the meantime, can I say your code is very ugly? :-D Jun 9 '11 at 9:41 • 3.071s! Not bad for a single-threaded solution! Jun 9 '11 at 19:02 I don't know, why I can't edit my previous post, so here's new version, 0,2 seconds faster (but about 1,5 s faster in CPU time (user)). This solution have 2 programs, first precalculates quantiles for normal distribution for bucket sort, and stores it in table, t[double * scale] = bucket index, where scale is some arbitrary number which makes casting to double possible. Then main program can use this data to put doubles in correct bucket. It has one drawback, if data is not gaussian it will not work correctly (and also there is almost zero chance to work incorrectly for normal distribution), but modification for special case is easy and fast (only number of buckets checks and falling to std::sort()). Compiling: g++ => http://pastebin.com/WG7pZEzH helper program g++ -std=c++0x -O3 -march=native -pthread => http://pastebin.com/T3yzViZP main sorting program • 1.621s! I think you're the leader, but I'm rapidly losing track with all these answers :) Jun 9 '11 at 19:08 Here is another sequential solution. This one uses the fact that the elements are normal distributed, and the I think the idea is generally applicable to get sorting close to linear time. The algorithm is like this: • Approximate CDF (see phi() function in the implementation) • For all elements compute the approximate position in the sorted array: size * phi(x) • Put elements in a new array close to their final position • In my implementation destination array has some gaps in it so I don't have to shift too many elements when inserting. • Use insertsort to sort the final elements (insertsort is linear if distance to final position is smaller than a constant). Unfortunately, the hidden constant is quite large and this solution is twice as slow as the radix sort algorithm. • 2.470s! Very nice ideas. It doesn't matter that the solution isn't the fastest if the ideas are interesting :) Jun 9 '11 at 19:35 • This is the same as mine, but grouping the phi computations together and the shifts together for better cache performance, right? Jun 10 '11 at 1:13 • @jonderry: I upvoted your solution, now that I understand what it does. Didn't mean to steal your idea. I included your implementation in my (unofficial) set of tests Jun 10 '11 at 10:58 My personal favorite using Intel's Threaded Building Blocks has already been posted, but here's a crude parallel solution using JDK 7 and its new fork/join API: import java.io.FileInputStream; import java.nio.channels.FileChannel; import java.util.Arrays; import java.util.concurrent.; import static java.nio.ByteOrder.LITTLE_ENDIAN; /* * * Original Quicksort: https://github.com/pmbauer/parallel/tree/master/src/main/java/pmbauer/parallel * / public class ForkJoinQuicksortTask extends RecursiveAction { public static void main(String[] args) throws Exception { double[] array = new double[Integer.valueOf(args[0])]; FileChannel fileChannel = new FileInputStream("gaussian.dat").getChannel(); ForkJoinPool mainPool = new ForkJoinPool(); System.out.println("Starting parallel computation"); } private static final long serialVersionUID = -642903763239072866L; private static final int SERIAL_THRESHOLD = 0x1000; private final double a[]; private final int left, right; public ForkJoinQuicksortTask(double[] a) {this(a, 0, a.length - 1);} private ForkJoinQuicksortTask(double[] a, int left, int right) { this.a = a; this.left = left; this.right = right; } @Override protected void compute() { if (right - left < SERIAL_THRESHOLD) { Arrays.sort(a, left, right + 1); } else { int pivotIndex = partition(a, left, right); if (left < pivotIndex) t1 = new ForkJoinQuicksortTask(a, left, pivotIndex).fork(); if (pivotIndex + 1 < right) new ForkJoinQuicksortTask(a, pivotIndex + 1, right).invoke(); if (t1 != null) t1.join(); } } public static int partition(double[] a, int left, int right) { // chose middle value of range for our pivot double pivotValue = a[left + (right - left) / 2]; --left; ++right; while (true) { do ++left; while (a[left] < pivotValue); do --right; while (a[right] > pivotValue); if (left < right) { double tmp = a[left]; a[left] = a[right]; a[right] = tmp; } else { return right; } } } } Important disclaimer: I took the quick-sort adaption for fork/join from: https://github.com/pmbauer/parallel/tree/master/src/main/java/pmbauer/parallel To run this you need a beta build of JDK 7 (http://jdk7.java.net/download.html). On my 2.93Ghz Quad core i7 (OS X): Python reference time python sort.py 50000000 sorting... real 1m13.885s user 1m11.942s sys 0m1.935s Java JDK 7 fork/join time java ForkJoinQuicksortTask 50000000 Starting parallel computation real 0m2.404s user 0m10.195s sys 0m0.347s I also tried to do some experimenting with parallel reading and converting the bytes to doubles, but I did not see any difference there. Update: If anyone wants to experiment with parallel loading of the data, the parallel loading version is below. In theory this could make it go a little faster still, if your IO device has enough parallel capacity (SSDs usually do). There's also some overhead in creating Doubles from bytes, so that could potentially go faster in parallel as well. On my systems (Ubuntu 10.10/Nehalem Quad/Intel X25M SSD, and OS X 10.6/i7 Quad/Samsung SSD) I did not see any real difference. import static java.nio.ByteOrder.LITTLE_ENDIAN; import java.io.FileInputStream; import java.nio.DoubleBuffer; import java.nio.channels.FileChannel; import java.util.Arrays; import java.util.concurrent.ForkJoinPool; import java.util.concurrent.RecursiveAction; /* * * Original Quicksort: https://github.com/pmbauer/parallel/tree/master/src/main/java/pmbauer/parallel * / public class ForkJoinQuicksortTask extends RecursiveAction { public static void main(String[] args) throws Exception { ForkJoinPool mainPool = new ForkJoinPool(); double[] array = new double[Integer.valueOf(args[0])]; FileChannel fileChannel = new FileInputStream("gaussian.dat").getChannel(); DoubleBuffer buffer = fileChannel.map(READ_ONLY, 0, fileChannel.size()).order(LITTLE_ENDIAN).asDoubleBuffer(); } private static final long serialVersionUID = -642903763239072866L; private static final int SERIAL_THRESHOLD = 0x1000; private final double a[]; private final int left, right; public ForkJoinQuicksortTask(double[] a) {this(a, 0, a.length - 1);} private ForkJoinQuicksortTask(double[] a, int left, int right) { this.a = a; this.left = left; this.right = right; } @Override protected void compute() { if (right - left < SERIAL_THRESHOLD) { Arrays.sort(a, left, right + 1); } else { int pivotIndex = partition(a, left, right); if (left < pivotIndex) t1 = new ForkJoinQuicksortTask(a, left, pivotIndex).fork(); if (pivotIndex + 1 < right) new ForkJoinQuicksortTask(a, pivotIndex + 1, right).invoke(); if (t1 != null) t1.join(); } } public static int partition(double[] a, int left, int right) { // chose middle value of range for our pivot double pivotValue = a[left + (right - left) / 2]; --left; ++right; while (true) { do ++left; while (a[left] < pivotValue); do --right; while (a[right] > pivotValue); if (left < right) { double tmp = a[left]; a[left] = a[right]; a[right] = tmp; } else { return right; } } } } private static final long serialVersionUID = -3498527500076085483L; private final DoubleBuffer buffer; private final double[] array; private final int low, high; public ReadAction(DoubleBuffer buffer, double[] array, int low, int high) { this.buffer = buffer; this.array = array; this.low = low; this.high = high; } @Override protected void compute() { if (high - low < 100000) { buffer.position(low); buffer.get(array, low, high-low); } else { int middle = (low + high) >>> 1; } } } Update2: I executed the code on one of our 12 core dev machines with a slight modification to set a fixed amount of cores. This gave the following results: Cores Time 1 7.568s 2 3.903s 3 3.325s 4 2.388s 5 2.227s 6 1.956s 7 1.856s 8 1.827s 9 1.682s 10 1.698s 11 1.620s 12 1.503s On this system I also tried the Python version which took 1m2.994s and Zjarek's C++ version which took 1.925s (for some reason Zjarek's C++ version seems to run relatively faster on static_rtti's computer). I also tried what happened if I doubled the file size to 100,000,000 doubles: Cores Time 1 15.056s 2 8.116s 3 5.925s 4 4.802s 5 4.430s 6 3.733s 7 3.540s 8 3.228s 9 3.103s 10 2.827s 11 2.784s 12 2.689s In this case, Zjarek's C++ version took 3.968s. Python just took too long here. 150,000,000 doubles: Cores Time 1 23.295s 2 12.391s 3 8.944s 4 6.990s 5 6.216s 6 6.211s 7 5.446s 8 5.155s 9 4.840s 10 4.435s 11 4.248s 12 4.174s In this case, Zjarek's C++ version was 6.044s. I didn't even attempt Python. The C++ version is very consistent with its results, where Java swings a little. First it gets a little more efficient when the problem gets larger, but then less efficient again. • This code does not parse the double values correctly for me. Is Java 7 required to correctly parse the values from the file? Jun 9 '11 at 2:55 • Ah, silly me. I forgot to set the endianness again after I locally refactored the IO code from multiple lines to one. Java 7 would normally be needed, unless you added fork/join separately to Java 6 of course. Jun 9 '11 at 17:57 • 3.411s on my machine. Not bad, but slower than koumes21's java solution :) Jun 9 '11 at 19:21 • I'll try koumes21's solution here too locally to see what the relative differences are on my system. Anyway, no shame in 'loosing' from koumes21 since it's a much more clever solution. This is just an almost standard quick-sort thrown into a fork/join pool ;) Jun 9 '11 at 21:52 A version using traditional pthreads. Code for merging copied from Guvante's answer. Compile with g++ -O3 -pthread. #include <stdio.h> #include <stdlib.h> #include <algorithm> static unsigned int nthreads = 4; static unsigned int size = 50000000; typedef struct { double array; int size; } array_t; void merge(double left, int leftsize, double right, int rightsize, double result) { int l = 0, r = 0, insertat = 0; while (l < leftsize && r < rightsize) { if (left[l] < right[r]) result[insertat++] = left[l++]; else result[insertat++] = right[r++]; } while (l < leftsize) result[insertat++] = left[l++]; while (r < rightsize) result[insertat++] = right[r++]; } void { array_t numbers = (array_t )input; std::sort(numbers.array, numbers.array+numbers.size); } int main(int argc, char *argv) { double numbers = (double ) malloc(size sizeof(double)); FILE f = fopen("gaussian.dat", "rb"); if (fread(numbers, sizeof(double), size, f) != size) fclose(f); int worksetsize = size / nthreads; for (int i = 0; i < nthreads; i++) { worksets[i].array=numbers+(iworksetsize); worksets[i].size=worksetsize; } for (int i = 0; i < nthreads; i++) { } for (int i = 0; i < nthreads; i++) { } double tmp = (double ) malloc(size * sizeof(double)); merge(numbers, worksetsize, numbers+worksetsize, worksetsize, tmp); merge(numbers+(worksetsize2), worksetsize, numbers+(worksetsize3), worksetsize, tmp+(size/2)); merge(tmp, worksetsize2, tmp+(size/2), worksetsize2, numbers); /* printf("Verifying result..\n"); for (int i = 0; i < size - 1; i++) { if (numbers[i] > numbers[i+1]) printf("Result is not correct\n"); } / return 0; } On my laptop I get the following results: real 0m6.660s user 0m9.449s sys 0m1.160s Here is a sequential C99 implementation that tries to really make use of the known distribution. It basically does a single round of bucket sort using the distribution information, then a few rounds of quicksort on each bucket assuming a uniform distribution within the limits of the bucket and finally a modified selection sort to copy the data back to the original buffer. The quicksort memorizes the split points, so selection sort only needs to operate on small cunks. And in spite (because?) of all that complexity, it isn't even really fast. To make evaluating Φ fast, the values are sampled in a few points and later on only linear interpolation is used. It actually does not matter if Φ is evaluated exactly, as long as the approximation is strictly monotonic. The bin sizes are chosen such that the chance of an bin overflow is negligible. More precisely, with the current parameters, the chance that a dataset of 50000000 elements will cause a bin overflow is 3.65e-09. (This can be computed using the survival function of the Poisson distribution.) gcc -std=c99 -msse3 -O3 -ffinite-math-only Since there is considerably more computation than in the other solutions, these compiler flags are needed to make it at least reasonably fast. Without -msse3 the conversions from double to int become really slow. If your architecture does not support SSE3, these conversions can also be done using the lrint() function. The code is rather ugly -- not sure if this meets the requirement of being "reasonably readable"... #include <stdio.h> #include <stdlib.h> #include <assert.h> #include <math.h> #define N 50000000 #define BINSIZE 720 #define MAXBINSIZE 880 #define BINCOUNT (N / BINSIZE) #define SPLITS 64 #define PHI_VALS 513 double phi_vals[PHI_VALS]; int bin_index(double x) { double y = (x + 8.0) ((PHI_VALS - 1) / 16.0); int interval = y; y -= interval; return (1.0 - y) * phi_vals[interval] + y * phi_vals[interval + 1]; } double bin_value(int bin) { int left = 0; int right = PHI_VALS - 1; do { int centre = (left + right) / 2; if (bin < phi_vals[centre]) right = centre; else left = centre; } while (right - left > 1); double frac = (bin - phi_vals[left]) / (phi_vals[right] - phi_vals[left]); return (left + frac) * (16.0 / (PHI_VALS - 1)) - 8.0; } void gaussian_sort(double restrict a) { double b = malloc(BINCOUNT * MAXBINSIZE * sizeof(double)); double *pos = malloc(BINCOUNT sizeof(double)); for (size_t i = 0; i < BINCOUNT; ++i) pos[i] = b + MAXBINSIZE i; for (size_t i = 0; i < N; ++i) pos[bin_index(a[i])]++ = a[i]; double left_val, right_val = bin_value(0); for (size_t bin = 0, i = 0; bin < BINCOUNT; ++bin) { left_val = right_val; right_val = bin_value(bin + 1); double splits[SPLITS + 1]; splits[0] = b + bin * MAXBINSIZE; splits[SPLITS] = pos[bin]; for (int step = SPLITS; step > 1; step >>= 1) for (int left_split = 0; left_split < SPLITS; left_split += step) { double left = splits[left_split]; double right = splits[left_split + step] - 1; double frac = (double)(left_split + (step >> 1)) / SPLITS; double pivot = (1.0 - frac) * left_val + frac * right_val; while (1) { while (left < pivot && left <= right) ++left; while (right >= pivot && left < right) --right; if (left >= right) break; double tmp = left; left = right; right = tmp; ++left; --right; } splits[left_split + (step >> 1)] = left; } for (int left_split = 0; left_split < SPLITS; ++left_split) { double left = splits[left_split]; double right = splits[left_split + 1] - 1; while (left <= right) { double min = left; for (double tmp = left + 1; tmp <= right; ++tmp) if (tmp < min) min = tmp; a[i++] = min; min = right--; } } } free(b); free(pos); } int main() { double a = malloc(N * sizeof(double)); FILE f = fopen("gaussian.dat", "rb"); assert(fread(a, sizeof(double), N, f) == N); fclose(f); for (int i = 0; i < PHI_VALS; ++i) { double x = (i (16.0 / PHI_VALS) - 8.0) / sqrt(2.0); phi_vals[i] = (erf(x) + 1.0) * 0.5 * BINCOUNT; } gaussian_sort(a); free(a); } • 4.098s! I had to add -lm to compile it (for erf). Jun 9 '11 at 19:25 #include <stdio.h> #include <math.h> #include <stdlib.h> #include <memory.h> #include <algorithm> // maps [-inf,+inf] to (0,1) double normcdf(double x) { return 0.5 * (1 + erf(x * M_SQRT1_2)); } int calcbin(double x, int bins) { return (int)floor(normcdf(x) * bins); } int docensus(int bins, int n, double arr) { int hist = calloc(bins, sizeof(int)); int i; for(i = 0; i < n; i++) { hist[calcbin(arr[i], bins)]++; } return hist; } void partition(int bins, int orig_counts, double arr) { int counts = malloc(bins * sizeof(int)); memcpy(counts, orig_counts, binssizeof(int)); int starts = malloc(bins * sizeof(int)); int b, i; starts[0] = 0; for(i = 1; i < bins; i++) { starts[i] = starts[i-1] + counts[i-1]; } for(b = 0; b < bins; b++) { while (counts[b] > 0) { double v = arr[starts[b]]; int correctbin; do { correctbin = calcbin(v, bins); int swappos = starts[correctbin]; double tmp = arr[swappos]; arr[swappos] = v; v = tmp; starts[correctbin]++; counts[correctbin]--; } while (correctbin != b); } } free(counts); free(starts); } void sortbins(int bins, int counts, double arr) { int start = 0; int b; for(b = 0; b < bins; b++) { std::sort(arr + start, arr + start + counts[b]); start += counts[b]; } } void checksorted(double arr, int n) { int i; for(i = 1; i < n; i++) { if (arr[i-1] > arr[i]) { printf("out of order at %d: %lf %lf\n", i, arr[i-1], arr[i]); exit(1); } } } int main(int argc, char argv[]) { if (argc == 1 \|\| argv[1] == NULL) { printf("Expected data size as argument\n"); exit(1); } int n = atoi(argv[1]); const int cachesize = 128 * 1024; // a guess int bins = (int) (1.1 * n * sizeof(double) / cachesize); if (argc > 2) { bins = atoi(argv[2]); } printf("Using %d bins\n", bins); FILE f = fopen("gaussian.dat", "rb"); if (f == NULL) { printf("Couldn't open gaussian.dat\n"); exit(1); } double arr = malloc(n * sizeof(double)); fclose(f); int *counts = docensus(bins, n, arr); partition(bins, counts, arr); sortbins(bins, counts, arr); checksorted(arr, n); return 0; } This uses erf() to place each element appropriately into a bin, then sorts each bin. It keeps the array entirely in-place. First pass: docensus() counts the number of elements in each bin. Second pass: partition() permutes the array, placing each element into its proper bin Third pass: sortbins() performs a qsort on each bin. It's kind of naive, and calls the expensive erf() function twice for every value. The first and third passes are potentially parallelizable. The second is highly serial and is probably slowed down by its highly random memory access patterns. It also might be worthwhile to cache each double's bin number, depending on the CPU power to memory speed ratio. This program lets you choose the number of bins to use. Just add a second number to the command line. I compiled it with gcc -O3, but my machine's so feeble I can't tell you any good performance numbers. Edit: Poof! My C program has magically transformed into a C++ program using std::sort! • You can use phi for a faster stdnormal_cdf. Jun 9 '11 at 18:11 • How many bins should I put, approximately? Jun 9 '11 at 19:09 • @Alexandru: I added a piecewise linear approximation to normcdf and only gained about 5% speed. – frud Jun 9 '11 at 19:14 • @static_rtti: You don't have to put any. By default, the code chooses the bins count so the average bin size is 10/11 of 128kb. Too few bins and you don't get the benefit of partitioning. Too many and the partition phase bogs down due to overflowing the cache. – frud Jun 9 '11 at 19:19 • 10.6s! I tried playing a bit with the number of bins, and I got the best results with 5000 (slightly over the default value of 3356). I must say I was expected to see a much better performance for your solution... Maybe it's the fact that you're using qsort instead of the potentially faster std::sort of the C++ solutions? Jun 9 '11 at 19:43 Have a look at the radix sort implementation by Michael Herf (Radix Tricks). On my machine sorting was 5 times faster compared to the std::sort algorithm in my first answer. The name of the sorting function is RadixSort11. int main(void) { std::ifstream ifs("C:\\Temp\\gaussian.dat", std::ios::binary \| std::ios::in); std::vector<float> v; v.reserve(50000000); double d;	{}
(Non-)formality of the extended Swiss Cheese operads Friday, 9 June, 2017 Published in: arXiv:1706.02945 We study two colored operads of configurations of little n-disks in a unit n-disk, with the centers of the small disks of one color restricted to an m-plane, m<n. We compute the rational homotopy type of these \emph{extended Swiss Cheese operads} and show how they are connected to the rational homotopy types of the inclusion maps from the little m-disks to the little n-disks operad. Author(s): Thomas Willwacher	{}
## Archive for January, 2009 ### Electrifying Gowdy January 27, 2009 In a previous post I wrote about the polarized Gowdy equations. If the condition of polarization is dropped the full Gowdy equations are obtained. They form a system of two semilinear wave equations: $\displaystyle P_{tt}+t^{-1}P_t-P_{\theta\theta}=e^{2P}(Q_t^2-Q_\theta^2)$, $\displaystyle Q_{tt}+t^{-1}Q_t-Q_{\theta\theta}=-2(P_tQ_t-P_\theta Q_\theta)$. These equations are very similar to those for a wave map from two-dimensional Minkowski space to the hyperbolic plane with metric $dP^2+e^{2P}dQ^2$ with the difference of the singular term involving $t^{-1}$. (It fact it is known that these equations can be given a formulation in terms of a wave map. The relevant wave map is on the domain with metric $-dt^2+d\theta^2+t^2 dx^2$ and is asssumed to be independent of the additional coordinate $x$.) The boundary conditions are as in the polarized case. Given initial data for $t=t_0$ there exists a unique solution on the time interval $(0,\infty)$, as was proved by Vincent Moncrief in 1981. The asymptotics for $t\to 0$ is significantly more complicated than in the polarized case. There is a large class of solutions with rather simple asymptotics. Their asymptotic form in the limit $t\to 0$ is: $P(t,\theta)=-k(\theta)\log t+\phi(\theta)+\ldots$,$Q(t,\theta)=q(\theta)+t^{2k(\theta)}\psi(\theta)+\ldots$. In 1998 Satyanad Kichenassamy and I proved the existence of solutions of this type for prescribed functions $k$, $\phi$, $q$ and $\psi$, subject to the condition $0. I will say no more about he positivity condition on $k$ here but instead concentrate on the inequality $k<1$. The quantity $k$ is sometimes called the asymptotic velocity and the condition $k<1$ the low velocity condition. It turns out that this is not just a technical restriction. It has been shown by Hans Ringström that the quantity $-tP_t(t,\theta)$ always has a limit as $t\to 0$. Call its modulus the asymptotic velocity $v_\infty (\theta)$. In the low velocity case described above it coincides with $k(\theta)$. More generally, when it is allowed to exceed one other phenomena occur. They are associated to the phenomenon of spikes. At a low velocity point $P_\theta/P$ converges to a smooth limit as $t\to 0$. A spike is a value of $\theta$ where $P_\theta/P$ becomes unbounded. In general $v_\infty$ has a discontinuity at a point of this type. These phenomena have been analysed in detail under a genericity assumption by Ringström. He used these results in his proof of strong cosmic censorship for Gowdy spacetimes. There are also subtle phenomena which occur in the limit $t\to\infty$. (For more information about many of the results mentioned in this post see the papers on Ringström’s web page.) The Gowdy spacetimes are solutions of the Einstein vacuum equations. They can be generalized by adding an electromagnetic field. Recently Ernesto Nungesser completed his diploma thesis under my supervision on the subject of these electromagnetic generalizations of Gowdy solutions. He and I just produced a paper where the results of his thesis are described and extended somewhat. One key element of this work is that there is a class of polarized ‘electro-Gowdy’ solutions which can be defined by symmetry conditions. Another is that there is a change of variables which transforms the polarized electro-Gowdy equations to the full (i.e. non-polarized) Gowdy equations. This allows the extensive results available on Gowdy solutions to be applied to the electromagnetic case. On the analytical level nothing happens but the geometrical interpretation of the variables is different in the two cases. It turns out that in this way strong cosmic censorship can be proved for polarized electro-Gowdy spacetimes. As discussed in the paper, much less is known in the non-polarized electro-Gowdy case . The principle of competitive exclusion arose in ecology. Roughly speaking it says that in an ecosystem with only $k$ different niches (or only $k$ different resources) no more than $k$ different species can coexist on a long-term basis. If there are originally $n$ different species with $n>k$ then at least $n-k$ of them must die out. This principle was introduced by Georgii Gause in his 1934 book ‘The struggle for existence’. On a mathematical level it leads to questions about the behaviour of the solutions of certain systems of ODE. In fact many related questions may be formulated. The ODE systems of relevance here are generalizations of the classical chemostat. This models a vessel containing fluid in which one species of bacteria lives. A nutrient is introduced continuously from a reservoir with fixed concentration in such a way that the rate of inflow is constant. Fluid is removed at the same rate. It is assumed that everything else the bacteria need to live and proliferate is present in sufficient quantities so as to present no limit to the population growth. The unknowns in the system are the concentrations of bacteria and nutrient. Call them $y$ and $C$ respectively. For an introduction to the subject see for instance the book “Mathematical models in population biology and epidemiology” by Brauer and Castillo-Chavez. This kind of system was considered by Novick and Szilard and by Monod in 1950. In the simplest case, which may be called the ‘classical’ chemostat, it is assumed that the proliferation rate of the bacteria is equal to the population density times an uptake function $r(C)=\frac {aC}{C+A}$ where $a$ and $A$ are constants. It can be proved that in this case there is a threshold value of the flow rate such that above the threshold the population dies out while below the threshold it tends to a constant value. It is very helpful that the system is two-dimensional and admits a Dulac function given by $C\log y$. This rules out periodic solutions. The conclusion also holds for rather general uptake functions $r(C)$.	{}
Posted by: atri \| September 21, 2010 ## Clarification on Q 3 on HW 2 One of you asked about what does it mean for an algorithm to have a running time of $T(n)$? I’ll talk about this in the lecture tomorrow but I thought it would be good to give some clarification now rather than later. We say an algorithm has running time of $T(n)$ if for every $n\ge 1$ and every input of size $n$, the algorithm runs for at most $T(n)$ steps. In other words, $T(n)$ is the maximum number of steps an algorithm can spend on any input of size $n$. In the above, $T(n)$ is an exact count. Let us see what this means when we talk about $T(n)$ in the asymptotic sense: • If we say $T(n)$ is $O(f(n))$, then (by the definition of Big-Oh and $T(n)$ above) this means that for some absolute constant $c>0$ and large enough $n$, for every input of size $n$, the algorithm uses up at most $c\cdot f(n)$ steps. • If we say $T(n)$ is $\Omega(g(n))$, then this means that for some absolute constant $\epsilon>0$ and every large enough $n$, for some input of size $n$, the algorithm makes at least $\epsilon\cdot g(n)$ steps.	{}
## Simply connected projective manifolds in characteristic $$p>0$$ have no nontrivial stratified bundles.(English)Zbl 1203.14029 Let $$X$$ be a smooth projective variety defined over an algebraically closed field of characteristic $$p>0$$. Let $$F$$ denote the absolute Frobenius morphism on $$X$$. A stratified bundle on $$X$$ is a sequence of locally free sheaves $$\{E_n\} _{n\in {\mathbb N}}$$ together with isomorphisms $$\sigma _n: F^*E_{n+1}\to E_n$$. D. Gieseker [Ann. Sc. Norm. Super. Pisa 2, 1–31 (1975; Zbl 0322.14009)] conjectured that if the étale fundamental group of $$X$$ vanishes then $$X$$ has only trivial stratified bundles. The main aim of the paper under review is proof of this conjecture. In characteristic zero the above conjecture is analogous to the fact that if étale fundamental group of $$X$$ vanishes then a pro-algebraic completion of $$X$$ also vanishes (this was known due to Grothendieck and, independently, Malcev). Roughly speaking, the proof follows from application of E. Hrushovski’s theorem on periodic points [arXiv:math/0406514]) to a Verschiebung morphism on the moduli space of semistable vector bundles on $$X$$ (see the reviewer’s paper in [Ann. Math. 159, 251–276 (2004; Zbl 1080.14014)]). ### MSC: 14G17 Positive characteristic ground fields in algebraic geometry 14J60 Vector bundles on surfaces and higher-dimensional varieties, and their moduli ### Keywords: fundamental group; stratified bundle; Verschiebung ### Citations: Zbl 0322.14009; Zbl 1080.14014 Full Text: ### References: [1] Brenner, H., Kaid, A.: On deep Frobenius descent and flat bundles. Math. Res. Lett. 15, 10001–10015 (2008) · Zbl 1200.14061 [2] Deligne, P.: Équations Différentielles à Points Singuliers Réguliers. Lecture Notes in Mathematics, vol. 163. Springer, Berlin-New York (1970) · Zbl 0244.14004 [3] dos Santos, J.-P.-S.: Fundamental group schemes for stratified sheaves. J. Algebra 317(2), 691–713 (2007) · Zbl 1130.14032 [4] Gieseker, D.: Stable vector bundles and the Frobenius morphism. Ann. Sc. Norm. Super. 4-ième Sér. 6(1), 95–101 (1973) · Zbl 0281.14013 [5] Gieseker, D.: Flat vector bundles and the fundamental group in non-zero characteristics. Ann. Sc. Norm. Super. Pisa, 4 Sér. 2(1), 1–31 (1975) · Zbl 0322.14009 [6] Gieseker, D.: On the moduli of vector bundles on an algebraic surface. Ann. Math. 106(1), 45–60 (1977) · Zbl 0381.14003 [7] Grothendieck, A.: Représentations linéaires et compactifications profinies des groupes discrets. Manuscr. Math. 2, 375–396 (1970) · Zbl 0239.20065 [8] Hrushovski, E.: The elementary theory of Frobenius automorphisms. http://de.arxiv.org/pdf/math/0406514v1 [9] Huybrechts, D., Lehn, M.: The Geometry of Moduli Spaces of Sheaves. Aspects of Mathematics, vol. E31. Vieweg, Braunschweig (1997) · Zbl 0872.14002 [10] Huppert, B.: Endliche Gruppen. Die Grundlehren der mathematischen Wissenschaft, vol. 134. Springer, Berlin (1967) · Zbl 0217.07201 [11] Lange, H., Stuhler, U.: Vektorbündel auf Kurven und Darstellungen der algebraischen Fundamentalgruppe. Math. Z. 156, 73–83 (1977) · Zbl 0349.14018 [12] Langer, A.: Semistable sheaves in positive characteristic. Ann. Math. 159, 251–276 (2004) · Zbl 1080.14014 [13] Langer, A.: Moduli spaces of sheaves in mixed characteristics. Duke Math. J. 124(3), 571–586 (2004) · Zbl 1086.14036 [14] Malcev, A.: On isomorphic matrix representations of infinite groups. Mat. Sb. N.S. 8(50), 405–422 (1940) · JFM 66.0088.03 [15] Maurischat, A.: Galois theory for iterative connections and nonreduced Galois groups. http://de.arxiv.org/pdf/math/0712.3748v3 · Zbl 1250.13009 [16] Simpson, C.: Moduli of representations of the fundamental group of a smooth projective variety. Publ. Math. I.H.E.S. 79, 47–129 (1994) · Zbl 0891.14005 [17] SGA1: Revêtements étales et groupe fondamental. SGA 1 [18] SGA2: Cohomologie locale des faisceaux cohérents et théorèmes de Lefschetz locaux et globaux. SGA2 This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	{}
## Problem A. Advanced drag racing ≡ • problems Author: I. Ludov Time limit: 1 sec Input file: input.txt Memory limit: 256 Mb Output file: output.txt ### Statement There is a long discussion about drag racing as a motorsport. Some critics argue that it is too technically conditioned and hence everything there depends on a car preparation and tuning, not on driver skills. Enthusiasts may note that these things are said in such a way, as if they were something bad. Organizers of one drag racing contest try to make their event interesting to both sides by changing some rules. The main principle is essentially the same: two participants start at the same time along parallel lines and try to pass a given range as fast as their cars allow. But still there is one very notable change. Whereas the standard drag racing track is a 400 meters straight line, this contest is conducted on a track with a single turn, connecting two straight segments of lengths L0 and L1 meters with an angle A radians between them. Since the driver must turn, and he cannot do this instantly, he must leave the first straight segment before it ends, turn and arrive into some point of second segment with velocity parallel to this segment. During this process he should not deviate for more than D meters from segments, or else he risks to hit the border of the track. Also, rules of the contest state that there should be only one such turn. Your car is equipped with an engine so powerful, that available acceleration is limited only by driving wheels traction. This maximum acceleration is G meters / second2 and it can push car along a straight line, decelerate car in straight line and turn car's velocity direction, leaving its amplitude unchanged, acting as a centripetal acceleration. To win the contest you must determine an optimal strategy of passing a track. It is rather obvious, that after the turn you should accelerate as hard as possible. But the question is how to approach turn and when to start turning. Such approach will consist of maximum acceleration which may or may not be followed by maximum braking. So your task is to write a program that will find values of time durations of accelerating and braking which make your car do its best on a given track, i.e. finishes in time that is at most 0.01 seconds more than minimum possible. ### Input file format Input file contains real numbers L0 L1 A D G, length of a straight segment before turn point, length of a straight segment after turn point, turn angle in radians, maximum allowed side deviation and maximum car acceleration respectively. All lengths are given in meters. ### Output file format Output file should contain two real numbers Ta Tb, duration of acceleration and braking respectively. If braking is not needed, Tb should be equal to 0. All durations must be specified in seconds. ### Constraints 0.02 ≤ A ≤ π  / 2; 1 ≤ L0, L1 ≤ 1000; 1 ≤ D ≤ 3; 1 ≤ G ≤ 20 ### Sample tests No. Input file (input.txt) Output file (output.txt) 1 50 100 1 1 10 2.22863 1.38195 2 50 100 0.1 1 10 3.015 0 0.125s 0.015s 15	{}
# Is there a closed form for the sum of the cubes of the binomial coefficients? We know that $$\sum_{k=0}^n \binom{n}{k} = 2^n\;\; \text{ and }\;\; \sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}$$ hold for all $$n\in \mathbb{N}_0$$. Now I tried to find a similar expression for $$\sum_{k=0}^n \binom{n}{k}^3$$ but didn't get anywhere at all. What I found were only asymptotic estimates (see Sum of cubes of binomial coefficients or Asymptotics of $$\sum_{k=0}^{n} {\binom n k}^a$$). Now is there a closed form for this sum or, what would be even better, for $$\sum_{k=0}^n \binom{n}{k}^\alpha$$ with any $$\alpha \in \mathbb{N}_0$$? These numbers are called the Franel Numbers. It's proven in (Petkovšek, M., Wilf, H. and Zeilberger, D. (1996). A=B. Wellesley, MA: A K Peters. p. 160) that there is no closed form for these numbers, in terms of the sum of a fixed number of hypergeometric terms. However, as @Robert_Israel points out, the expression could possibly be represented by different types of closed form. • ... if "closed form" is defined as "the sum of a fixed number of hypergeometric terms". There could be other types of "closed form". – Robert Israel Nov 28 '18 at 18:24 • @Robert Was just thinking that. Thanks for the suggestion. – Jam Nov 28 '18 at 18:26 The binomial coefficient for a given pair of $$n \geq k \geq 0$$ integers can be expressed in terms of a Pochhammer symbol as the following. $$\binom n k = \frac{(-1)^k(-n)_k} {k!}.$$ The expression is valid even if $$n$$ is an arbitrary real number. Here we note two things. 1. The Pochhammer symbol $$(-n)_k$$ is zero, if $$n \geq 0$$ and $$k > -n$$. 2. The factorial $$k!$$ can be written as $$(1)_k$$. Using these observations, we can express your sums in terms of a generalized hypergeometric function $$_pF_q$$ as the following. For the sum of the binomial coefficients, we have $$\sum_{k=0}^n \binom n k = \sum_{k=0}^n \frac{(-1)^k(-n)_k}{k!} = \sum_{k=0}^\infty (-n)_k{\frac{(-1)^k}{k!}} = {_1F_0}\left({{-n}\atop{-}}\middle\|\,-1\right).$$ For the sum the square of the binomial coefficients, we have $$\sum_{k=0}^n {\binom n k}^2 = \sum_{k=0}^n \left(\frac{(-1)^k(-n)_k}{k!}\right)^2 = \sum_{k=0}^\infty \frac{\left((-n)_k\right)^2}{k!} \cdot \frac{1}{k!} = {_2F_1}\left({{-n, -n}\atop{1}}\middle\|\,1\right).$$ And for the sum of the cube of the binomial coefficients $$-$$ also known as Franel numbers $$-$$, we have $$\sum_{k=0}^n {\binom n k}^3 = \sum_{k=0}^n \left(\frac{(-1)^k(-n)_k}{k!}\right)^3 = \sum_{k=0}^\infty \frac{\left((-n)_k\right)^3}{(k!)^2} \cdot \frac{(-1)^k}{k!} = {_3F_2}\left({{-n, -n, -n}\atop{1, 1}}\middle\|\,-1\right).$$ In general, for a positive integer $$r$$, we have the binomial sum \begin{align} \sum_{k=0}^n {\binom n k}^r &= \sum_{k=0}^n \left(\frac{(-1)^k(-n)_k}{k!}\right)^r = \sum_{k=0}^\infty \frac{\left((-n)_k\right)^r}{(k!)^{r-1}} \cdot \frac{(-1)^{rk}}{k!} \\ &= {_rF_{r-1}}\left({{-n, -n, \dots, -n}\atop{1, \dots, 1}}\middle\|\,(-1)^r\right). \end{align}	{}
💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! Oh no! Our educators are currently working hard solving this question. In the meantime, our AI Tutor recommends this similar expert step-by-step video covering the same topics. # (a) $$2 \pi\left[b^{2}+\frac{a^{2} b \sin ^{-1} \frac{\sqrt{a^{2}-b^{2}}}{a}}{\sqrt{a^{2}-b^{2}}}\right]$$(b) $$2 \pi a^{2}+\frac{2 \pi a b^{2}}{\sqrt{a^{2}-b^{2}}} \ln \frac{\sqrt{a^{2}-b^{2}}+a}{b}$$ ## a) $$2 \pi\left[\frac{a^{2} b}{\sqrt{a^{2}-b^{2}}} \arcsin \left(\frac{\sqrt{a^{2}-b^{2}}}{a}\right)+b^{2}\right]$$b) $$2 \pi a^{2}+\frac{2 \pi a b^{2}}{\sqrt{a^{2}-b^{2}}} \ln \left(\frac{a+\sqrt{a^{2}-b^{2}}}{b}\right)$$ #### Topics Applications of Integration ### Discussion You must be signed in to discuss. Lectures Join Bootcamp ### Video Transcript pants, Clara. So when he married here. So we're going to use the parametric equations, which is X is equal to a co sign of data. 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Be divided by becomes equal to two pine terms a square past. Be square over E Oh, um of one plus over the square root of farm miners E square. #### Topics Applications of Integration Lectures Join Bootcamp	{}
# The Lyapunov spectrum of the Lorenz system¶ Here, we present a calculation of the Lyapunov spectrum of the Lorenz system, using TaylorIntegration.jl. Our calculation involves evaluating the 1st order variational equations $\dot \xi = J \cdot \xi$ for this system, where $J = \operatorname{D}f$ is the Jacobian. By default, the numerical value of the Jacobian is computed using automatic differentiation techniques implemented in TaylorSeries.jl. Automatic differentation helps us to avoid writing down manually the Jacobian; instead, TaylorIntegration.jl does that for us! For complex problems, or when the number of dependent variables is large, this may help save typos, which sometimes are hard to trace. Otherwise, if performance is critical, then the user may provide the Jacobian function, as will be shown below. The Lorenz system is the ODE defined as: \begin{align} \begin{split} \dot x_1 &= \sigma(x_2-x_1) \\ \dot x_2 &= x_1(\rho-x_3)-x_2 \\ \dot x_3 &= x_1x_2-\beta x_3 \end{split} \end{align} where $\sigma$, $\rho$ and $\beta$ are constant parameters. ## The setup¶ First, we write a Julia function which evaluates (in-place) the Lorenz system equations: In [1]: #Lorenz system ODE: function lorenz!(t, x, dx) dx[1] = σ(x[2]-x[1]) dx[2] = x[1](ρ-x[3])-x[2] dx[3] = x[1]x[2]-βx[3] nothing end Out[1]: lorenz! (generic function with 1 method) Next, we define the parameters: In [2]: #Lorenz system parameters #we use the const prefix in order to help the compiler speed things up const σ = 16.0 const β = 4.0 const ρ = 45.92 Out[2]: 45.92 The initial conditions, the initial and final time are: In [3]: const x0 = [19.0, 20.0, 50.0] #the initial condition const t0 = 0.0 #the initial time const tmax = 100.0 #final time of integration Out[3]: 100.0 We know that the sum of the Lyapunov spectrum has to be equal to the trace of the Jacobian of the equations of motion. We will calculate this trace using TaylorSeries.jl, and after the numerical integration, we will come back to this value to check that this is indeed the case: In [4]: # Note that TaylorSeries.jl exported variables and methods are @reexport-ed by TaylorIntegration.jl #Calculate trace of Lorenz system Jacobian via TaylorSeries.jacobian: import LinearAlgebra: tr using TaylorIntegration xi = set_variables("δ", order=1, numvars=length(x0)) x0TN = [ x0[1]+xi[1], x0[2]+xi[2], x0[3]+xi[3] ] dx0TN = similar(x0TN) lorenz!(t0, x0TN, dx0TN) lorenztr = tr(TaylorSeries.jacobian(dx0TN)) #trace of Lorenz system Jacobian matrix Out[4]: -21.0 # The integration¶ Now, we are ready to perform the integration, using TaylorIntegration.jl. Usually, we would use the taylorinteg method in order to integrate the equations of motion; but, since we are interested in evaluating the Lyapunov spectrum, we will use the lyap_taylorinteg method, which calculates the Lyapunov spectrum via the variational equations and Oseledets' theorem. The expansion order will be $28$; the local absolute tolerance will be $10^{-20}$. lyap_taylorinteg will return three arrays: one with the evaluation times, one with the value of the dependent variables at the time of evaluation, and another one with the vaues of the Lyapunov spectrum at each time of the evaluation. In [5]: @time tv, xv, λv = lyap_taylorinteg(lorenz!, x0, t0, tmax, 28, 1e-20; maxsteps=2000000); 3.038651 seconds (15.28 M allocations: 1.371 GiB, 13.00% gc time) As explained above, instead of using automatic differentiation to compute the numerical value of the Jacobian, the user may provide a function which computes the Jacobian in-place, as follows: In [6]: #Lorenz system Jacobian (in-place): function lorenz_jac!(jac, t, x) jac[1,1] = -σ+zero(x[1]); jac[1,2] = σ+zero(x[1]); jac[1,3] = zero(x[1]) jac[2,1] = ρ-x[3]; jac[2,2] = -1.0+zero(x[1]); jac[2,3] = -x[1] jac[3,1] = x[2]; jac[3,2] = x[1]; jac[3,3] = -β+zero(x[1]) nothing end Out[6]: lorenz_jac! (generic function with 1 method) Then, we may perform the same integration as before using lorenz_jac!: In [7]: @time tv_, xv_, λv_ = lyap_taylorinteg(lorenz!, x0, t0, tmax, 28, 1e-20, lorenz_jac!; maxsteps=2000000); 0.856854 seconds (4.35 M allocations: 575.767 MiB, 20.75% gc time) In general, we may check the consistency of our integration checking that both solutions are either equal, or at least approximately equal with differences comparable to roundoff errors. In the particular case of the integrations performed above, we obtain the same solution, exactly: In [8]: tv == tv_, xv == xv_, λv == λv_ Out[8]: (true, true, true) The number of steps taken is: In [9]: length(tv) Out[9]: 4093 Is the final time actually the requested value? In [10]: tv[end] == tmax Out[10]: true What is the minimum, maximum and mean time-step? In [11]: import Statistics: mean minimum(diff(tv)), maximum(diff(tv)), mean(diff(tv)) Out[11]: (0.009471209659452029, 0.0608034582365633, 0.024437927663734114) What is the standard deviation of the time-step size distribution? In [12]: import Statistics: std std(diff(tv)) Out[12]: 0.007428238447533755 What is the final value for each of the exponents? In [13]: λv[end,:] Out[13]: 3-element Array{Float64,1}: 1.4856676067102397 -0.0012149610701265275 -22.484452645640157 What is the value of the sum of the value we obtained for the spectrum? In [14]: sum(λv[end,:]) Out[14]: -21.000000000000043 Now, is the sum of the exponents exactly equal to the trace of the Jacobian? In [15]: sum(λv[end,:]) == lorenztr Out[15]: false So the sum is not exactly equal, but is it approximately equal? What is the relative error in the computed vs the expected value? In [16]: rel_error = (sum(λv[end,:])-lorenztr)/lorenztr Out[16]: 2.0301221021717147e-15 How does this error compare to the machine epsilon? In [17]: rel_error/eps() Out[17]: 9.142857142857142 So the relative error in our computation for the sum of the Lyapunov spectrum is comparable to the machine epsilon. # Visualization¶ Finally, we plot our results: In [18]: using Plots #gr() #using LaTeXStrings #plotlyjs() ### Lyapunov exponents vs time plot¶ In [19]: plot(tv, λv[:,1], label="L_1") plot!(tv, λv[:,2], label="L_2") plot!(tv, λv[:,3], label="L_3") xlabel!("time") ylabel!("L_i, i=1,2,3") title!("Lyapunov exponents vs time") Out[19]: ### Lyapunov exponents vs time plot (semi-log)¶ In [20]: plot(tv[2:end], λv[:,1][2:end], xscale=:log10, label="L_1") plot!(tv[2:end], λv[:,2][2:end], label="L_2") plot!(tv[2:end], λv[:,3][2:end], label="L_3") xlabel!("time") ylabel!("L_i, i=1,2,3") title!("Lyapunov exponents vs time (semi-log)") Out[20]: ### Convergence of Lyapunov exponents vs time plot (log-log)¶ In [21]: plot(tv[2:end], abs.(diff(λv[:,1])), yscale=:log10, xscale=:log10, label="dL_1") plot!(tv[2:end], abs.(diff(λv[:,2])), label="dL_2") plot!(tv[2:end], abs.(diff(λv[:,3])), label="dL_3") xlabel!("time") ylabel!("dL_i, i=1,2,3") title!("Convergence of Lyapunov exponents vs time (log-log)") Out[21]: ### Absolute difference wrt trace of Jacobian vs time plot (log-log)¶ In [22]: plot(tv[2:end], log10.(abs.(λv[2:end,1]+λv[2:end,2]+λv[2:end,3].-lorenztr)), xscale=:log10, leg=false) #ylims!(-15,-12) xlabel!("time (log10)") ylabel!("Absolute difference wrt Tr(J) (log10)") Out[22]: ## The Lorenz attractor¶ In [23]: plot(xv[:,1], xv[:,2], xv[:,3], leg=false) Out[23]: In [ ]:	{}
# Choose the correct alternative. The order and degree of (dydx)3-d3ydx3+yex=0 are respectively. - Mathematics and Statistics MCQ Sum Choose the correct alternative. The order and degree of (dy/dx)^3 - (d^3y)/dx^3 + ye^x = 0 are respectively. • 3, 1 • 1, 3 • 3, 3 • 1, 1 #### Solution The order and degree of (dy/dx)^3 - (d^3y)/dx^3 + ye^x = 0are respectively - 3, 1 Is there an error in this question or solution? #### APPEARS IN Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 8 Differential Equation and Applications Miscellaneous Exercise 8 \| Q 1.01 \| Page 171	{}
# Voice Conversion using GANs: An Extensive Review Published: For my project on code-mixed speech recognition with Prof. Preethi Jyothi, I did a literature review of voice conversion and found a lot of recent papers that used GANs for the problem. So I decided to write this post to summarize my review. I’ve also attached the slides and references I used in my group meeting at the end of this post. Back in 2015, before GANs became popular for conversion problems, voice conversion models were trained for a particular target speaker and models were trained simply on regression loss. One good example is Lifa et al, 2015 where a deep BiLSTM network is used. They use STRAIGHT vocoder system to convert an utterance into 3 types of features viz. Mel-cepstral coefficients, aperiodicity measures, and $\log F_0$. Only the Mel-cepstral coefficients are transformed by BiLSTM for voice conversion. $\log F_0$ is linearly transformed using input speaker information and aperiodicity measures were simply forwarded to the synthesis system. The synthesis system took the transformed Mel-cepstral coefficients, aperiodicity measures and transformed $log F_0$ to generate output utterance. This system had several limitations including a single target speaker and need for parallel utterances which also had to be aligned using dynamic-time warping. Before GANs were used for voice conversion, they were explored in the image domain, very notably by Zhu et al, 2017 and Kim et al, 2017. In Zhu et al, 2017, the idea of cycle-consistency loss is introduced which makes the whole thing possible. It is best explained from the figure shown below from the same paper. The idea is that there are two transformation networks $G$ and $F$, which transform from domain $X$ to $Y$ and $Y$ to $X$ respectively. In addition to the discriminative loss for real-fake classification, a consistency loss is used to ensure that $\hat{x} = F(G(x))$ is close to $x$ itself and $\hat{y} = G(F(y)$ is close to $y$ itself. The idea is totally life-changing! In case one is not impressed enough by the elegance of this idea, their mind will surely be blown away by the amazing results they’ve managed to achieve. An example from the paper is shown below where paintings from different painters were treated as different domains. Even if these results don’t satisfy you (who are you?), think about the big advantage that this model has: there is no need for parallel data. Just get a set of images from different domains, and transform them to impress your friends. The idea proposed in Kim et al, 2017 is very similar except that they use it to find relationships among datapoints different domains. An example is shown below where the model learns to identify the orientation of the input image of a chair and give out an image of a car with the same orientation. Another example shown below is for the transformation between a male face and a female face. Extending these ideas to voice conversion is pretty straightforward. Kaneko et al, 2017 used the same idea of cycle-consistency loss for training a voice conversion system where they treated two different speakers as two domains. For details on their gated-CNN based model and how they use features from WORLD vocoder, one should refer to the paper. They have also provided samples here. Another very similar work is by Fang et al, 2018 where they’ve only transformed the lower order Mel-cepstrum using the GAN and copied the higher-order information for synthesis owing to the argument that speaker-specific information is only present in lower-order cepstrum. Both of the above systems still have the disadvantage that only a pair of speakers can be used for training and inference in voice conversion. This was resolved by Kameoka et al, 2018 in their model called the StarGAN. This model has 3 parts: a generator (or more aptly called a transformer), a discriminator and a domain classifier. Here the generator and discriminator are conditioned on the domain by an additional input specifying the speaker. The simplest one to train is the classifier which is trained on real data to predict the domain of the input, or simply the speaker in case of voice conversion. The discriminator is, as usual, trained to identify real and fake data points. The loss for training generator has four terms with scaling parameters: an adversarial loss to fool discriminator, a domain classifier loss to ensure that output belongs to the domain the generator is conditioned on, a cyclic loss to ensure reliable conversion and an identity loss for identity conversion as a regularization. The demo can be seen here. They’ve also proposed other systems which can all be seen in the list given here. Very similar work is by Gao et al. where they’ve treated genders as the two domains. The applications of the idea of cycle-consistency have been pretty impressive. Another important papers in this line is the work by Meng et al, 2018 on speech enhancement by learning noise using cycle-consistency GANs where the two domains are that of clean speech and noisy speech which is also an extension to their previous work in Meng et al, 2018 that didn’t use cycle-consistency. Another very interesting application is presented by Tanaka et al, 2018 where they train a cycle-consistent GAN to make synthetic speech sound more natural. As one can guess, the two domains are that of synthetic speech and natural speech. They propose that this transformation can be used as an additional processing step in the vocoder pipelines after synthesis. The rest of the post is about other related stuff that I found interesting. Since I’m talking about a generative model, it is important to mention that variational auto-encoders (VAEs) have also been used for voice conversion. An example is the work by Kameoka et al, 2018 where they maximize the mutual information between the output and domain in addition to the usual VAE loss. This maximization ensures that the encoder and decoder actually account for the conditioning on the domain and maximizes the usage of the bottleneck to preserve content. They show, using variational information maximization, that maximizing this mutual information reduces down to adding an auxiliary classifier and using to train VAE for minimizing classification loss. The demo can be seen here. One interested in their fully convolutional VAE for voice conversion should check out the paper in more detail. Just to end this post on a positive note about GANs, I wish to mention two very popular papers that have explored the stability of GANs which have been applied in the image domain. I think that it would be very interesting to see this work in improving systems for voice conversion. The first one I wish to mention is the work by Karras et al, 2018 where they progressively increase layers in the network over time while ensuring a smooth change in dependence on new parameters, and fading in the new layers. The resolution of the image improves over each additional layer, finally leading to amazingly realistic fake images as the ones shown below. Second is the paper by Mao et al, 2017 where they show that using least square error on discriminator output is more stable than using cross-entropy which is based on minimizing Jenson-Shannon divergence between the generator output distribution $P_G$ and the real data $P_D$. They show that minimizing the square error is instead equivalent to minimizing the Pearson $\chi^2$ divergence between $P_D + P_G$ and $2P_G$. All the references and some further reading (which I’ll update over time) are given below. The slides from my presentation can be found here. ## References Voice conversion using deep Bidirectional Long Short-Term Memory based Recurrent Neural Networks L. Sun et al., ICASSP 2015 Unpaired Image-to-Image Translation using Cycle-Consistent Adversarial Networks J. Zhu et al., ICCV 2017 Learning to Discover Cross-Domain Relations with Generative Adversarial Networks T. Kim et al., ICML 2017 Parallel-Data-Free Voice Conversion Using Cycle-Consistent Adversarial Networks T. Kaneko et al., arXiv preprint arXiv:1711.11293, 2017 High-Quality Nonparallel Voice Conversion Based on Cycle-Consistent Adversarial Network F. Fang et al., ICASSP 2018 StarGAN-VC: non-parallel many-to-many Voice Conversion Using Star Generative Adversarial Network H. Kameoka et al., IEEE SLT 2018 Voice Impersonation using Generative Adversarial Networks Y. Gao et al., arXiv preprint arXiv:1802.06840, 2018 Cycle-Consistent Speech Enhancement Z. Meng et al., Interspeech 2018 Z. Meng et al., Interspeech 2018 WaveCycleGAN: Synthetic-to-natural speech waveform conversion using cycle-consistent adversarial networks K. Tanaka et al., arXiv preprint, 2018 ACVAE-VC: Non-parallel Many-to-many VC with Auxiliary Classifier VAE H. Kameoka et al., arXiv preprint, 2018 Progressive Growing of GANs for Improved Quality, Stability and Variation T. Karras et al., ICLR 2018 X. Mao et al., ICCV 2017	{}
# LQG makes predictions? 1. Feb 2, 2006 ### ZapperZ Staff Emeritus At least LQG tries to make measurable predictions! When is String theory going to even try? http://physicsweb.org/articles/news/10/2/2/1 Zz. 2. Feb 2, 2006 Staff Emeritus It's good to see the beginning of some empirical tests in quantum gravity, but I have to say that string theorists have been trying to come up with checkable predictions - trying very hard, too - for nigh on to thutty years now. It's just that they have always failed. 3. Feb 2, 2006 ### marcus I remember that Parampreet Singh gave a seminar talk about this very thing last Fall, at Penn State. the slides and/or the audio recording were online. I will see if I can find some links. I dont know if this can be put to observational use NOW. But it is at least a serious try at getting an observable signature of some QG. Here is a free version of their article: http://arxiv.org/abs/gr-qc/0506129 Quantum evaporation of a naked singularity Authors: Rituparno Goswami, Pankaj S. Joshi, Parampreet Singh Comments: 4 pages, 2 figures. Minor changes to match published version in Physical Review Letters Report-no: IGPG-05/6-8 We investigate here quantum effects in gravitational collapse of a scalar field model which classically leads to a naked singularity. We show that non-perturbative semi-classical modifications near the singularity, based on loop quantum gravity, give rise to a strong outward flux of energy. This leads to the dissolution of the collapsing cloud before the singularity can form. Quantum gravitational effects thus censor naked singularities by avoiding their formation. Further, quantum gravity induced mass flux has a distinct feature which may lead to a novel observable signature in astrophysical bursts. Last edited: Feb 3, 2006 4. Feb 2, 2006 ### marcus Here is Parampreet Singh's October seminar talk http://www.phys.psu.edu/events/index.html?event_id=1302;event_type_ids=0;span=2005-08-20.2005-12-25 Both the audio and the lecture-note/slides are available online at that link. the talk was given 21 October several months after the article was submitted to Phys. Rev. Letters. so it must be reasonably up-to-date. ============== I think this signature "flicker" in the Gammaray burst of a certain type of stellar collapse is only ONE OF SEVERAL ways that Loop researchers are getting into position to test. It may not even be the best or farthest along. I don't mean to criticize their efforts---I think it's great. but these things all seem tentative to me and nothing has reached the stage of a cut-and-dried ironclad test that you can actually do next year. it's "getting there" but not there yet. my personal opinion. Parampreet Singh may say different. 5. Feb 3, 2006 ### Nucleonics Zapper, I was under the impression that some models of string theory HAD made measurable predictions -- namely the deviation from inverse-square-law gravity at the sub-millimeter scale. Is this not generally accepted as a measurable prediction of ST? physicsweb.org/articles/news/7/2/14/1 Also, "split symmetry" models of ST predict gauginos and higgsinos at the TeV scale. There are also other models of ST at the TeV scale. They make predictions that would be testable at the LHC. So, am I missing something when I think that ST does actually make testable predictions? Best wishes, Nucleonics 6. Feb 3, 2006 ### josh1 Hi guys, My understanding of LQG is that they dont know whether it can describe the universe at the low energies we see it today. So Im wondering what you guys mean when you say that LQG makes testable predictions. For example, even if somebody did find evidence of area quantization or some effect depending on it, they cant claim that theyve verified LQG since they cant say that LQG describes the universe in which they did their experiment. In fact, from what I understand, detecting this kind of discreteness in the spectrum of whatever you are measuring doesnt necessarily imply quantization. It could simply be revealing a fundamental length in the theory rather than a genuine quantization of the observables spectrum. So can someone give me an example of an experiment that can be performed that could verify LQG? Maybe someone knows some way that LQG is falsifiable. What you guys are claiming just doesnt make sense to me since the only universe experimentally accessible to us is one which LQG hasnt been able to describe. By the way, please dont post some quotes from or links to a paper. Just explain it to me yourself. If I wanted to go read papers, I wouldn`t have posted this. 7. Feb 4, 2006 ### marcus This is a good point. At least it tries, and I think it's fair to say that SOME VARIANTS of LQG do make testable predictions or are coming close to that goal. One thing to mention is that there is no one unique LQG. People use LQG as a catch-all term for a number of leading QG alternatives to string. These include canonical LQG, spinfoam, LQC (loop quantum cosmology, a symmetry-reduced version of canonical LQG), Thiemann's masterconstraint, Gambini's discretized LQG. One can't say which of these are equivalent to which others. In some the dynamics have been worked out and the semiclassical or largescale limit has been checked. In others not. LQC is an example of where the dynamics have been worked out and low energy limit checked. It gives the same answers as ordinary Gen Rel except around the big bang ('big bounce'). LQC can make predictions IN ITS OWN RIGHT and needs to be tested in its own right. Since the logical relation with canonical LQG is not clear, you have people working on LQC phenomenology. Parampreet Singh has given some seminar talks about this at Penn State---audio and visuals are online. There is also an attempt to make GENERIC predictions that would be valid consequences of several different LQG models regardless of whether the details of dynamics and classical limit have been worked out in each separate case. An example of this is DSR. Some people argue that a variety of LQG models will not work without some modification of SR symmetry. Unfortunately there is disagreement about this. There is time pressure to get this issue resolved by 2007, or whenever the GLAST satellite is launched. It is argued that a number of LQG models require a slight energy-dependence of the speed of light. This gives reason to hope that one might RULE OUT SUCH MODELS regardless of whether details have been established and checked in each case---by finding gammaray dispersion in the opposite sense to that predicted. But this issue is still unresolved. Zapper's main point is a sound one. Researchers pursuing various LQG approaches are indeed making an effort to arrive at tests which could falsify some or all of the different LQG models. Last edited: Feb 4, 2006	{}
# quick lathe question ### Help Support The Rocketry Forum: #### rabidsheeep ##### Well-Known Member i remember the topic of a cheap lathe coming up for nosecones and other things awhile ago... im just wondering where that lathe was or which i should get? basically what do i need to get started? a small one would be best... one that isn't anything stationary... thanks #### rocket trike ##### Well-Known Member A couple weeks ago I pick up one from Menards. It is a 12" with a 8" swing.I got it for $79.00 they ussally sell for$99.00. If you got a menards by you go there and check it out to see if they got one. Here is a picture of the lathe and things that come with them. #### rabidsheeep ##### Well-Known Member umm anyone know anywhere in new york or one i could order offline? #### rabidsheeep ##### Well-Known Member thanks and umm... im still lookin for that other post you put up those (this is probably the wrong word for it) chisel things... #### sandman ##### Well-Known Member Lathe turning tools! here! https://www.harborfreight.com/cpi/ctaf/Displayitem.taf?itemnumber=3793 Actually they are kinda useless on balsa wood...they tear it up! They are designed for hardwoods. I make "emery boards" out of 40 grit, 60 grit all the way up to 220 grit. I glue it to blocks of wood with contact cement. It last an unbelievably long time too. The 40 grit is ....well it's sorta like driveway gravel glued onto paper...really hogs the wood of fast! So be careful with it! Where a respirator!!! They don't call me "sandman" for nothing! #### rabidsheeep ##### Well-Known Member lol ya know i figured you be the one to respond to this post first... like you have some special "nosecone topic sensing abilites" anyway since your a genius what kinda would should i use on these things? try it with balsa and see if it blows up first? or like chop down the oak tree in the backyard and have enough nosecones for a lifetime #### sandman ##### Well-Known Member What kind of wood??? Balsa is yer best bet, easy to shape and readily available. Lots of different weights and densitys around from a couple of pounds per cubic foot to almost 15 pounds per cubic foot Basswood is a second choice, very tight grained and "fairly light". Available in up to 4"x4" but expensive! Poplar is pretty touch, medium heavy and a real pretty grain. Hard to find now in anything thicker than 3". Oak and maple...well only for little stuff otherwise they get really heavy! Gonna do some black walnut nose cones soon (as soon as it warms up a bit more! Too darn cold out!) tight grained and pretty light. Heavier than basswood but lighter than poplar. And very very pretty! #### astronboy ##### Well-Known Member Sandman and Rabidsheep: Thanks for this timely thread!! My B-day is in one month, and I think I will ask my wife to get me this lathe. I have been turning comes using my drill press. It is not bad, but a bit awkward. For 'tools' I use the various emery boards that are available for fingernails. They are pretty cheap at the local beauty supply shop. I have been tossing the idea around of turning a cone out of a nice chunk of ebony... I know that this is a REALLY hard wood, and I would need to use real woodturning tools and a lathe. But I think it would work for a certain design I am pondering that would need a weighted nosecone. #### sandman ##### Well-Known Member I have been tossing the idea around of turning a cone out of a nice chunk of ebony astronboy, DON'T DO IT! Be warned! 1. Ebony is extreamly hard almost to the hardness of some metals! Keep your lathe tools very very sharp! Sharpen constantly! You may even break an edge off of a tool...be careful! Yes, it is THAT hard! 2. Ebony is absolutely BLACK almost no grain will show (look at a clarinet they are made of ebony!) 3. When you finish you will have a very hard very heavy nose cone that will look like a black plastic nose cone! 4. Although they are very pretty be carful around some exotic woods, they can be toxic. Some are even protected. You cannot get any more Cocoabolo (a beautiful brownish red rosewood). Start out slowly. If you want exotic try black walnut for now. sandman #### teflonrocketry1 ##### Well-Known Member Sandman, Did you ever try using drywall sanding sceens to shape balsa? I use a "fine" grit drywall screen to sand balsa fin sets clamped together to same shape. The screens work fast and don't clog with balsa dust. The sceens don't grab the wood either, especially against the grain. Did you ever work with Paulownia wood? It supposedly has almost twice the strength to weight ratio as balsa. Visit: https://www.paulowniawood.com/ Bruce S. Levison, NAR #69055 #### astronboy ##### Well-Known Member OK Sandman, I will take it slowly. Although I do not have experience in turning woods, I was an apprentice to a guitar builder in the mid 1980's. (That made my Dad REAL happy ) Anyway, we built a custom order Electric Bass for a client out of Purple Heart, and it darned near ruined every single tool we used on it! This stuff was like steel. Based on that experience alone, I will take your warning, and proceed slowly up the hardwood charts in my turning. I have used ebony on some ship models (1/8"x3" stock), so I have some experience with it. I would actuallly say that sanding it was harder than sanding Iron!! (I have sanded Iron doing some gunsmith repairs on my Brown Bess Musket... but that is another hobby, and another story altogether.....) Cocobolo is beautiful!! I have a Fife that is made of Coco, and it is stunning!! #### powderburner ##### Well-Known Member Originally posted by astronboy For 'tools' I use the various emery boards that are available for fingernails. They are pretty cheap at the local beauty supply shop. The cheapo emery boards are EXTREMELY cheap at the dollar store near me. I think I get about three dzn of the things (in various lengths) in one pkg for a buck. Now if I could just figure out how to get all that balsa dust out of my Dremel. #### Micromeister ##### Micro Craftman/ClusterNut TRF Supporter A vacuum works wonders on removing dust You might also want to check out Grizzly.com for a 40.00 alum bed handdrill lathe setup I'm told is very nice. Or you can make your own lathe from scrap wood and your 3/8" or 1/2" hand drill very cheaply. heres a pic of a homemade lathe the cost 1.86 to make excluding the drill and the 6" c clamps. 4.5" swing over bed 22" between centers, OA 32" long. Stevem also made a homemade handdrill lathe with a harddrive motor bearing as the tailstock live center some pic in the scratch built section I believe. #### powderburner ##### Well-Known Member Micro, I appreciate the suggestions but am currently dabbling with trying to adapt an orphaned kitchen blender base to be an upside-down 'drill press' machine for turning balsa cones. I even have an extra chuck (from an old dead hand drill) that I could use to mount on top of the blender to hold my stock. If I can make it work, should be a powerful, fast way to spin balsa. Oh yeah, so far it's made from junk (free junk). We'll see.... #### Micromeister ##### Micro Craftman/ClusterNut TRF Supporter Good grief guys: Go to Sears. your can pick up a skewed chisel and parting tool, either individually or as a set for around \$25.00 or so buck. Harbors turning tool set is fine if you can wait for it Astroboy #### astronboy ##### Well-Known Member Well, I do not want to wait that long, but my Birthday is a month away..... #### rabidsheeep ##### Well-Known Member lathe purchase was put on hold cause of this: new computer well it cant help me out with nosecones but atleast i can play battlefield #### astronboy ##### Well-Known Member Well, I did receive the lathe stand, and the chisels, but the LATHE is out of stock for a few more weeks!!. BTW that set of Chisels looks VERY nice. #### sandman ##### Well-Known Member What is the brand name on the chisel set? It should be stanped on the chisels. It isn't Sorby is it? sandman #### rkt2k1 ##### Well-Known Member Sandman, I finally received my first set of wood turning tools from Harbor Freight. They were on back ordered since Xmas. (Must be all the new TRF lathe owners! ) The brand stamped on the chisels is Pittsburgh.	{}
# How do you integrate int 1/(x^4 - 16) using partial fractions? Feb 1, 2017 $\int \frac{1}{{x}^{4} - 16} \mathrm{dx} = \frac{1}{32} \ln \| x - 2 \| - \frac{1}{32} \ln \| x + 2 \| - \frac{1}{16} \arctan \left(\frac{x}{2}\right) + C$ #### Explanation: ${x}^{4} - 16 = \left({x}^{2} - 4\right) \left({x}^{2} + 4\right) = \left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 4\right)$ The partial fraction decomposition will therefore be of the form $\frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C x + D}{{x}^{2} + 4}$. $\frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C x + D}{{x}^{2} + 4} = \frac{1}{\left(x + 2\right) \left(x - 2\right) \left({x}^{2} + 4\right)}$ $A \left(x - 2\right) \left({x}^{2} + 4\right) + B \left(x + 2\right) \left({x}^{2} + 4\right) + \left(C x + D\right) \left(x + 2\right) \left(x - 2\right) = 1$ $A \left({x}^{3} - 2 {x}^{2} + 4 x - 8\right) + B \left({x}^{3} + 2 {x}^{2} + 4 x + 8\right) + \left(C x + D\right) \left({x}^{2} - 4\right) = 1$ $A {x}^{3} - 2 A {x}^{2} + 4 A x - 8 A + B {x}^{3} + 2 B {x}^{2} + 4 B x + 8 B + C {x}^{3} + D {x}^{2} - 4 C x - 4 D = 1$ $\left(A + B + C\right) {x}^{3} + \left(2 B - 2 A + D\right) {x}^{2} + \left(4 A + 4 B - 4 C\right) x + \left(8 B - 8 A - 4 D\right) = 1$ Write a system of equations now: {(A + B + C = 0),(2B - 2A + D = 0), (4A + 4B - 4C = 0), (8B - 8A - 4D = 1 ):} First of all, the third equation can be simplified to $A + B - C = 0$. We can get rid of the $C$ variable by adding the first and the third equations, to get $2 A + 2 B = 0 \to A + B = 0$. This means that $B = - A$. Also, from the second equation, we deduce that $D = 2 A - 2 B$. Substitute into the fourth equation: $8 \left(- A\right) - 8 A - 4 \left(2 A - 2 \left(- A\right)\right) = 1$ $- 8 A - 8 A - 8 A - 8 A = 1$ $- 32 A = 1$ $A = - \frac{1}{32}$ Solving for the other variables is now relatively simple. You should get $A = - \frac{1}{32}$, $B = \frac{1}{32}$,$C = 0$ and $D = - \frac{1}{8}$ as final answers. $\therefore$The partial fraction decomposition is $- \frac{1}{32 \left(x + 2\right)} + \frac{1}{32 \left(x - 2\right)} - \frac{1}{8 \left({x}^{2} + 4\right)}$ Let's look closely at the integration of $\int - \frac{1}{8 \left({x}^{2} + 4\right)} \mathrm{dx}$. This will be of the arctangent form, but first we need to perform a u-substitution. Let $u = \frac{x}{2}$. Then $\mathrm{du} = \frac{1}{2} \mathrm{dx}$. $- \frac{1}{8} \int \frac{1}{{u}^{2} + 1} \cdot \frac{1}{2} \mathrm{du}$ $- \frac{1}{16} \int \frac{1}{{u}^{2} + 1} \mathrm{du}$ This is a standard integral. $- \frac{1}{16} \arctan u$ $- \frac{1}{16} \arctan \left(\frac{x}{2}\right)$ Now deal with the other parts of the problem. The other two terms can be integrated using $\int \frac{1}{x} \mathrm{dx} = \ln \| x \| + C$. $\int \frac{1}{32 \left(x - 2\right)} \mathrm{dx} = \frac{1}{32} \ln \| x - 2 \|$ $\int - \frac{1}{32 \left(x + 2\right)} \mathrm{dx} = - \frac{1}{32} \ln \| x + 2 \|$ Therefore, $\int \frac{1}{{x}^{4} - 16} \mathrm{dx} = \frac{1}{32} \ln \| x - 2 \| - \frac{1}{32} \ln \| x + 2 \| - \frac{1}{16} \arctan \left(\frac{x}{2}\right) + C$ Hopefully this helps!	{}
# Browsing School of Physics (Scholarly Articles) by Title Sort by: Order: Results: • (2007) The stability of the optical pulse of the Crab pulsar is analyzed based on the 1 $\mu$s resolution observations with the Russian 6-meter and William Hershel telescopes equipped with different photon-counting detectors. The ... • (2009) The accretion of matter onto a massive black hole is believed to feed the relativistic plasma jets found in many active galactic nuclei (AGN). Although some AGN accelerate particles to energies exceeding 10^12 electron ... • (2002) A decade after the discovery of TeV gamma-rays from the blazar Mrk 421 (Punch et al. 1992), the list of TeV blazars has increased to five BL Lac objects: Mrk 421 (Punch et al. 1992; Petry et al. 1996; Piron et al. 2001), ... • (2007) We find periodic I-band variability in two ultracool dwarfs, TVLM 513-46546 and 2MASS J00361617+1821104, on either side of the M/L dwarf boundary. Both of these targets are short-period radio transients, with the detected ... • (2007) We have carried out a systematic search for the molecular ion CO+ in a sample of 8 protoplanetary and planetary nebulae in order to determine the origin of the unexpectedly strong HCO+ emission previously detected in these ... • (2011) The Rotating RAdio Transient (RRAT) J1819-1458 exhibits ~3 ms bursts in the radio every ~3 min, implying that it is visible for only ~1 per day. Assuming that the optical light behaves in a similar manner, long exposures ... • (1999) We present the results of a search for pulsed TeV emission from the Crab pulsar using the Whipple Observatory's 10m gamma-ray telescope. The direction of the Crab pulsar was observed for a total of 73.4 hours between 1994 ... • (1999) We present the results of a search for pulsed TeV emission from the Crab pulsar using the Whipple Observatory's 10m gamma-ray telescope. The direction of the Crab pulsar was observed for a total of 73.4 hours between 1994 ... • (1999) We present the results of a search for pulsed TeV emission from the Crab pulsar using the Whipple Observatory's 10 m gamma-ray telescope. The direction of the Crab pulsar was observed for a total of 73.4 hours between 1994 ... • (1999) We present the results of a search for pulsed TeV emission from the Crab pulsar using the Whipple Observatory's 10 m gamma-ray telescope. The direction of the Crab pulsar was observed for a total of 73.4 hours between 1994 ... • (2009) We report on the results of two coordinated multiwavelength campaigns that focused on the blazar Markarian 421 during its 2006 and 2008 outbursts. These campaigns obtained UV and X-ray data using the XMM-Newton satellite, ... • (2008) VERITAS, an Imaging Atmospheric Cherenkov Telescope (IACT) system for gammma-ray astronomy in the GeV-TeV range, has recently completed its first season of observations with a full array of four telescopes. A number of ... • (2009) An example of a cold massive core, JCMT 18354-0649S, a possible high mass analogue to a low mass star forming core is studied. Line and continuum observations from JCMT, Mopra Telescope and Spitzer are presented and modelled ... • (2011) We report on TeV gamma-ray observations of the blazar Mrk 421 (redshift of 0.031) with the VERITAS observatory and the Whipple 10m Cherenkov telescope. The excellent sensitivity of VERITAS allowed us to sample the TeV ... • (2006) Galaxy clusters might be sources of TeV gamma rays emitted by high-energy protons and electrons accelerated by large scale structure formation shocks, galactic winds, or active galactic nuclei. Furthermore, gamma rays may ... • (2010) • (2010) M87 is a nearby radio galaxy that is detected at energies ranging from radio to VHE gamma-rays. Its proximity and its jet, misaligned from our line-of-sight, enable detailed morphological studies and extensive modeling at ... • (2008) We report the detection of very high-energy gamma-ray emission from the intermediate-frequency-peaked BL Lacertae object W Comae (z=0.102) by VERITAS. The source was observed between January and April 2008. A strong outburst ... • (2009) The intermediate-frequency peaked BL Lacertae (IBL) object 3C 66A is detected during 2007 - 2008 in VHE (very high energy: E > 100 GeV) gamma-rays with the VERITAS stereoscopic array of imaging atmospheric Cherenkov ... • (2009) The VERITAS collaboration reports the detection of very-high-energy (VHE) gamma-ray emission from the high-frequency-peaked BL Lac object 1ES 1218+304 located at a redshift of z=0.182. A gamma-ray signal was detected with ...	{}
## anonymous 4 years ago 5√3+√6/√6+√3? would the numerators cancel out leaving only the five? or is there more to this? 1. saifoo.khan $\Large \frac{5 \sqrt 3 + \sqrt6}{\sqrt6+\sqrt3} \to -3 +4\sqrt2$ 2. anonymous hmm... could you explain in in further detail? 3. saifoo.khan Can u solve now? 4. anonymous ill try 5. saifoo.khan wait i did a mistake, 6. anonymous D: 7. saifoo.khan $\Large \frac{5 \sqrt 3 + \sqrt6}{\sqrt6+\sqrt3} \times \frac{\sqrt6-\sqrt3}{\sqrt6-\sqrt3}$Now it's Accurate. 8. anonymous i dont know where you got the second half of the problem... its only the first set of numbers 9. saifoo.khan It's a method to solve these type of questions. 10. anonymous oh neat 11. saifoo.khan You have multiply and divide with the denominator but with different sign. 12. anonymous im trying to get it worked out 13. anonymous how do you multiply a √3 X -√3? thats what i dont understand 14. saifoo.khan $\sqrt 3 \times - \sqrt3 = -3$ 15. saifoo.khan The roots cancels out. 16. anonymous ahhh. brilliant! 17. saifoo.khan Thank you. heading to bed. C ya! :) 18. anonymous thanks 19. saifoo.khan	{}
# Antiderivative Practice Khan org are unblocked. Integration : Edexcel Core Maths C4 June 2012 Q7(b) : ExamSolutions Maths Revision - youtube Video Part (c): Edexcel Core Maths C4 June 2012 Q7(c) : ExamSolutions Maths Revision - youtube Video. Quotient Identities. See the complete profile on LinkedIn and discover Umair’s connections and jobs at similar companies. I don't know if there's enough space, but they are working hard to make it work with maybe hybrid or staggered schedules. Furthermore, the product of antiderivatives, $$x^2e^x/2$$ is not an antiderivative of $$xe^x$$ since $$\dfrac{d}{dx}(\dfrac{x^2e^x}{2})=xe^x+\dfrac{x^2e^x}{2}≠xe^x$$. Integration by Trigonometric Substitution. Line Integral Example 1 lesson plan template and teaching resources. Vajih has 10 jobs listed on their profile. Geothermal Energy. لدى Zeeshan2 وظيفة مدرجة على الملف الشخصي عرض الملف الشخصي الكامل على LinkedIn وتعرف على زملاء Zeeshan والوظائف في الشركات المماثلة. If we divide everything on the numerator and everything on the denominator by x 2, we get: = ∫ (3x – 4x-1 – 5x-2) dx. Inside the double integral, we still need to include a single integral in the third "vertical" variable, where this variable ranges from the bottom of $\dlv$ to its top. An introduction to antiderivatives, or indefinite integrals, the process that 'undoes' the derivative. عرض ملف Imtiyaz Khan الشخصي على LinkedIn، أكبر شبكة للمحترفين في العالم. The paper studied the integration of sustainability through the use of computers and technology in the examination of the universities viz. Available for Pre-Algebra, Algebra 1, Geometry, Algebra 2, Precalculus, and Calculus. Practice questions. If you’d like a pdf document containing the solutions the download tab above contains links to pdf’s containing the solutions for the full book, chapter and section. We will assume knowledge of the following well-known, basic indefinite integral formulas : , where is a constant , where is a constant Most of the following problems are average. 3, and remains large if inflows of foreign capital are limited to a fraction of the. We answer your questions on math, physics & computer science. Prepare for the SAT, GMAT, LSAT, MCAT, NCLEX-RN, and AP exams. The following practice questions ask you to find values that satisfy the Mean Value Theorem in a given interval. COM TOMORROW September 11,2001. Vertical integration is a strategy where a company expands its business operations into different steps on the same production path, such as when a manufacturer owns its supplier and/or. In general, the product of antiderivatives is not an antiderivative of a product. Commenting on U. Integration is the inverse operation of differentiation. The common-sense way of thinking about continuity is that a curve is continuous wherever you can draw the curve without taking. Four full-length practice tests Sample test questions. Described are the origin and nature of geothermal energy. Our mission is to end the stress and burnout epidemic by offering sustainable, science-based solutions to enhance well-being, performance, and purpose, and create a healthier relationship with technology. It is assumed that you are familiar with the following rules of differentiation. Account for the magnitude and speed of the Mongol conquests. Join today to get access to thousands of courses. I have read different topics of integration and differentiation, i mean different ways to integrate or differentiate. D (LO) , FUN‑6. All India Bank of Baroda Officers' Association. For example, if you tried to evaluate \begin{align} \int_0^1 \int_x^1 e^{y^2}dy\,dx \end{align} directly, you would run into trouble. Find displacement by find area under a derivative graph; integration notation and properties (video to the right) Notes and practice: notes_displacement_def_integral. We will assume knowledge of the following well-known differentiation formulas : , where , and , where a is any positive constant not equal to 1 and is the natural (base e) logarithm of a. The Add 1 or 10 exercise appears under the Early math Math Mission. CORE Group GHPC15 October 8, 2015 Concurrent Session: New Trends for HIV Clinical Platforms: Prevention, Care, and Treatment of NCDs. Over the course of the 20th and 21st centuries, across the Muslim world (historical and contemporary), the roles and contexts of. Test your skills at finding antiderivatives. The first type of integral equations which involve constants as both the limits — are called Fredholm Type Integral equations. Canvas supports an LTI integration with Microsoft Office 365 in Assignments, Collaborations, Modules, Course Navigation, and the Rich Content Editor. This page can show you how to do some very basic integrals. Integration by doing the chain rule in reverse. Examples of taking the indefinite integral (or anti-derivative) of polynomials. Integration is one of the two main operations of calculus; its inverse operation, differentiation, is the other. If the integrand (the expression after the integral sign) is in the form of an algebraic fraction and the integral cannot be evaluated by simple methods, the fraction needs to be expressed in partial fractions before integration takes place. By using this website, you agree to our Cookie Policy. dx x xx 1 5. In other words, the derivative of is. 2 Participants were encouraged to provide relevant information to include in the database. His effective articulation and clarity of thought gives him the rare ability to break down complex technologies into simple comprehensible frameworks. So if you wish to fine tune your Technique or desire to upgrade your skills, enroll yourself at 9 Aces! Driving Range:. Navy are to declare the ability to operate and deploy the F-35 in 2016 and 2018 respectively, and full-rate production of the aircraft is to begincapability”) in 2016 and 2018 respectively, and full-rate production decision of the program is planned for 2019. Integration by parts is a method of breaking down equations to solve them more easily. I usually do about 1 item a day, such as 1 video, 1 practice, 1 article, etc. Khan Academy is a 501(c)(3) nonprofit organization. These practice questions will give you a better idea of what to study on your TEAS exam. Detailed step by step solutions to your Integration by trigonometric substitution problems online with our math solver and calculator. For g(x) = x 3 + x 2 – x, find all the values c in the interval (–2, 1) that satisfy the Mean Value Theorem. Issuu is a digital publishing platform that makes it simple to publish magazines, catalogs, newspapers, books, and more online. Sample Quizzes with Answers Search by content rather than week number. The app includes the same information and practice questions found in the CLEP Official Study Guide and subject-specific Examination Guide but offers the convenience of answering sample questions on your mobile device. Continuous integration (CI) and continuous delivery (CD) embody a culture, set of operating principles, and collection of practices that enable application development teams to deliver code. Take the integral of y = -1. Integration by substitution (or the reverse-chain-rule). The Integral Calculator solves an indefinite integral of a function. Oracle Data Integration Cloud – The Future of Data Integration Now! Room 3 LIMITED Chaitanya Pydimukkala. You can also get a better visual and understanding of the function and area under the curve using our graphing tool. 01) FILLING Basheer Khan. Math 104: Calculus has been evaluated and recommended for up to 6 semester hours and may be transferred to over 2,000 colleges and universities. Example problem 1: Find the area between the curves y = x and y = x 2 between x = 0 and x = 1. Fubini's theorem 1 Fubini's theorem In mathematical analysis Fubini's theorem, named after Guido Fubini, is a result which gives conditions under which it is possible to compute a double integral using iterated integrals. In this case, it is called an indefinite integral and is written:. The integration-by-parts formula tells you to do the top part of the 7. 55 m from the natural length, so the lower limit of the integral is 0. You can also get a better visual and understanding of the function and area under the curve using our graphing tool. To denote the heights of the rectangles we let y i = f(x. Wiesser works with a team of over 300 planners and …. When you learn about the fundamental theorem of calculus, you will learn that the antiderivative has a very, very important property. Khan Academy is a 501(c)(3) nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. The reason is because integration is simply a harder task to do - while a derivative is only concerned with the behavior of a function at a point, an integral, being a glorified sum, integration requires global knowledge of the function. Collaborative Team Teaching Through Arts Integration Zuni NMTEACH Domain 2:. An indefinite integral is a function that takes the antiderivative of another function. The ones marked * may be different from the article in the profile. We can often find that area just by sketching and using basic calculations, but other times we may need to use the Integration Rules. Teledermatology, originating in 1995, has been one of the first telemedicine services to see the light of day. Z xsin 1 xdx 3. This exercise shows how basic antiderivatives are graphed in respect to their constants. Khan Academy being integrated in the same way for exercises and practice (not just videos as it is currently) would be perfect! I understand the argument that there isn't much of an incentive to do this for a product that is available for free, but with so many instructors utilizing Khan Academy, it seems that it makes sense for Canvas to be. Rosetta Stone Classroom. For example, look at the sum. 2 Worksheet by Kuta Software LLC. About the journal. Inside the double integral, we still need to include a single integral in the third "vertical" variable, where this variable ranges from the bottom of $\dlv$ to its top. Here is a set of practice problems to accompany the Computing Indefinite Integrals section of the Integrals chapter of the notes for Paul Dawkins Calculus I course at Lamar University. See the complete profile on LinkedIn and discover Usman's. In fact, as you learn more advanced techniques, you will still probably use this one also, in addition to the more advanced techniques, even on the same problem. Renaissance Learning. I have just written an answer about this anti-derivative, which can be read here. ESPDIC (Esperanto – English Dictionary) – 2 August 2015 - Paul Denisowski (www. Review the material from this module by completing the practice test below:. So you would probably have to try something else. For this reason, when we integrate, we have to add a constant. View Shabu Khan’s profile on LinkedIn, the world's largest professional community. The Aga Khan Award for Architecture (AKAA) is an architectural prize established by Aga Khan IV in 1977. Practice unlimited questions for JEE Main and other Engineering Exams, JEE Main Practice Tests, Solved Questions for JEE Main, JEE Main Previous Year papers. Explanations are given when you click on the correct answer. Our Windows 8 app is the best way to view Khan Academy's complete library of over 3,400 videos. Khan Academy allows you to learn almost anything for free. Types of Problems There are two types of problems in this exercise: Find the definite integral: This problem has an integral that could use a trigonometric identity to be made easier. Use substitution of variables to calculate antiderivatives (including changing limits for definite integrals). Here are a set of practice problems for the Derivatives chapter of the Calculus I notes. Curriculum Pathways provides interactive, standards-based resources in English language arts, math, science, social studies, and Spanish (grades K-12). More on why the anti derivative and the area under a curve are essentially the same thing. Co-Function Identities. Commenting on U. Another example of a tutorial is Come Live with Me, a guided tutorial (by Herman the Worm) on how to create a compost box. Students can practice maths skills there and are given assistance when they have difficulty. The phrase heading toward is emphasized here because what happens precisely at the given x value isn't relevant to this limit inquiry. The definite integral of on the interval can now be alternatively defined by. Take the integral of y = -1. See the complete profile on LinkedIn and discover Mariam’s connections and jobs at similar companies. Official website of Mayor of London, Sadiq Khan, and the 25 London Assembly Members. ©T M2G0j1f3 F XKTuvt3a n iS po Qf2t9wOaRrte m HLNL4CF. If you are asked to integrate a fraction, try multiplying or dividing the top and bottom of the fraction by a number. EXPONENT RULES & PRACTICE 1. What is Integration?. This activity contains 25 questions. 1992-01-01. If it is convergent, nd which value it converges to. The function you input will be shown in blue underneath as ; The Density slider controls the number of vector lines. Furthermore, a substitution which at ﬁrst sight might seem sensible, can lead nowhere. لدى Imtiyaz4 وظيفة مدرجة على الملف الشخصي عرض الملف الشخصي الكامل على LinkedIn وتعرف على زملاء Imtiyaz والوظائف في الشركات المماثلة. The idea it is based on is very simple: applying the product rule to solve integrals. org, join us on Facebook or. We'll review your answers and create a Test Prep Plan for you. Integration by parts formula: ? u d v = u v-? v d u. Listed are some common derivatives and antiderivatives. Rehan has 7 jobs listed on their profile. org are unblocked. Your practice problems for this module will be on Khan Academy. The app includes the same information and practice questions found in the CLEP Official Study Guide and subject-specific Examination Guide but offers the convenience of answering sample questions on your mobile device. Integration: The Basic Logarithmic Form. It is visually represented as an integral symbol, a function, and then a dx at the end. Precalculus \| Khan Academy by Khan Academy. Evaluate an Integral Step 1: Enter an expression below to find the indefinite integral, or add bounds to solve for the definite integral. View Rehan Khan🚗🌍's profile on LinkedIn, the world's largest professional community. To find antiderivatives of basic functions, the following rules can be used:. MuleSoft’s Anypoint Platform™ is the world’s leading integration platform for SOA, SaaS, and APIs. Khan (Editor). Click here Anna University Syllabus. Parveen has 4 jobs listed on their profile. •So by substitution, the limits of integration also change, giving us new Integral in new Variable as well as new limits in the same variable. Explore math with Desmos. View Sameer Khan's profile on LinkedIn, the world's largest professional community. Curriculum Pathways provides interactive, standards-based resources in English language arts, math, science, social studies, and Spanish (grades K-12). Step 2: Click the blue arrow to compute the integral. Khan Abdul Gaffar Khan (popularly known as "Frontier Gandhi") was a noted nationalist who spent 45 of his 95 years of life in jail; Barakatullah of Bhopal was one of the founders of the Ghadar Party, which created a network of anti-British organisations; Syed Rahmat Shah of the Ghadar Party worked as an underground revolutionary in France and. MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum. Khan Academy is a 501(c)(3) nonprofit organization. Free definite integral calculator - solve definite integrals with all the steps. We're building a library of world-class instructional and practice resources that empowers learners. Learning Math has never been such fun!. THE INTEGRATION OF EXPONENTIAL FUNCTIONS The following problems involve the integration of exponential functions. Each Practice Test consists of 10 to 12 AP Calculus (AB) problems; you can think of Practice Tests as being like little quizzes which you can use to hone your skills. It was a pleasure of working with Morshed Heyder Khan for the past 3 years at Square Toiletries Limited. Khan Academy is a 501(c)(3) nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. Answer: The given limits are (inner) y from x to π/2; (outer) x from 0 to π/2. Get a head start on next semester's geometry fundamentals. This method is based on the simple concept of adding fractions by getting a common denominator. Definite Integrals and Indefinite Integrals. Integration by Parts: Knowing which function to call u and which to call dv takes some practice. u is the function u(x) v is the function v(x). The first and most vital step is to be able to write our integral in this form:. For more information, visit www. 2014, Khulood Salem Albeladi, Usman A. Imran has 6 jobs listed on their profile. Sometimes this is a simple problem, since it will be apparent that the function you wish to integrate is a derivative in some straightforward way. Integration can be used to find areas, volumes, central points and many useful things. Practice - MAP Item Sampler. The familiar trigonometric identities may be used to eliminate radicals from integrals. REMINDER! WE ARE CLOSING JSPEK. Find more online open access dissertations / theses in this subject. Then du= cosxdxand v= ex. Navy are to declare the ability to operate and deploy the F-35 in 2016 and 2018 respectively, and full-rate production of the aircraft is to begincapability”) in 2016 and 2018 respectively, and full-rate production decision of the program is planned for 2019. For example, ln(x)e x. Over 100,000 Manuscripts, Books Burnt By ISIL across Iraq's Anbar Photo: Over 100,000 Manuscripts, Books Burnt By ISIL across Ira. Take this Practice Quiz as though it were a real test. Step 2: Click the blue arrow to compute the integral. Changing the order of integration 1. Differentiating using multiple rules: strategy. View Omar Khan’s profile on LinkedIn, the world's largest professional community. Types of Problems There are two types of problems in this exercise: Find the definite integral: This problem has an integral that could use a trigonometric identity to be made easier. View Sajid Ahmed Khan’s profile on LinkedIn, the world's largest professional community. Collaborative Team Teaching Through Arts Integration Zuni NMTEACH Domain 2:. View Saima Zeb-Khan's profile on LinkedIn, the world's largest professional community. Having been involved with Specsavers for most of the 90’s, our director Soharab Khan decided he would like to open an independent practice. 2nd Semester Review Packet. Integration and Integration Techniques Chapter Exam Take this practice test to check your existing knowledge of the course material. Notice that not only is x 5 an antiderivative of f, but so are. An arbitrary domain value, x i, is chosen in each subinterval, and its subsequent function. MuleSoft’s Anypoint Platform™ is the world’s leading integration platform for SOA, SaaS, and APIs. It is not very "smart" though, so do not be surprised if it cannot do your integral. After you have chosen the answer, click on the button Check Answers. Prepare your students for success with meticulously researched ELA, math, and science practice for grades 5-8. Being able to do an integral is a key skill for any Calculus student. 9 years ago \| 277 views. The number the student is adding to the two. Parveen has 4 jobs listed on their profile. This integration allows students and instructors to use, create, share, and collaborate on Office 365 files within Canvas. I go through 30 AP Calculus AB Practice Exam Problems and Solutions (Section 1, AP Calculus AB & AP Calculus BC 2018 Exam FRQ #3 AP Calculus AB & AP Calculus BC 2018 Exam Free. View Usman Khan's profile on LinkedIn, the world's largest professional community. The problem is i don't understand that, is there any specific way to solve a particular problem. Integration by Trigonometric Substitution. Described are the origin and nature of geothermal energy. 2016-04-01. In what follows, C is the constant of integration. Furthermore, the product of antiderivatives, $$x^2e^x/2$$ is not an antiderivative of $$xe^x$$ since $$\dfrac{d}{dx}(\dfrac{x^2e^x}{2})=xe^x+\dfrac{x^2e^x}{2}≠xe^x$$. (see: calculating definite integrals). Khan Academy - Connecting Free Practice to the Classroom Zuni. Practice for this exam with the Official CLEP Study Guide App from examIam. ) But after so many discussions, nothing came out to solve the irritating situation in the. The enacting of…. Another example of a tutorial is Come Live with Me, a guided tutorial (by Herman the Worm) on how to create a compost box. Rubayat Khan is an extraordinary technology advisor and evangelist. Practice - MAP Item Sampler. U-Substitution Integration Problems. So, we are going to begin by recalling the product rule. We will assume knowledge of the following well-known differentiation formulas : , where , and , where a is any positive constant not equal to 1 and is the natural (base e) logarithm of a. Listed are some common derivatives and antiderivatives. ERIC is an online library of education research and information, sponsored by the Institute of Education Sciences (IES) of the U. When you learn about the fundamental theorem of calculus, you will learn that the antiderivative has a very, very important property. Make this ad disappear by upgrading to Symbaloo PRO. (b) Decide if the integral is convergent or divergent. Developed an appreciation for the practice of law in addition to skills integral to that practice such as working independently, time management skills and interdisciplinary analysis. LinkedIn is the world's largest business network, helping professionals like Sobia Khan discover inside connections to recommended job candidates, industry experts, and business partners. Explanations are given when you click on the correct answer. Quotient Identities. Need Math Homework Help? Read free Math courses, problems explained simply and in few words. Math video on how to determine the position of an object by solving a differential equation that describes it acceleration. But it is often used to find the area underneath the graph of a function like this: The integral of many functions are well known, and there are useful rules to work out the integral of more complicated functions, many of which. I say that because this free prep website only offers a program for the SAT, so you won't find any official ACT practice questions here. Important Formulae. The reason is because integration is simply a harder task to do - while a derivative is only concerned with the behavior of a function at a point, an integral, being a glorified sum, integration requires global knowledge of the function. The Khan Academy has a mission to provide free education of world-class quality. Using upper sums to evaluate a definite integral. Khan Academy has been translated into dozens of languages, and 100 million people use our platform worldwide every year. khanacademy. Examples of taking the indefinite integral (or anti-derivative) of polynomials. We can often find that area just by sketching and using basic calculations, but other times we may need to use the Integration Rules. Join today to get access to thousands of courses. These results have been interpreted as evidence that motor learning is specific to the sources of afferent information. For example, faced with Z x10 dx. Indefinite Integration (part V) lesson plan template and teaching resources. The derivative is the function slope or slope of the tangent line at point x. The basic idea of Integral calculus is finding the area under a curve. Choose a test from the listing below to start your AP Calc review. Check out all of our online calculators here!. Khan Academy, found at www. It is assumed that you are familiar with the following rules of differentiation. You just clipped your first slide! Clipping is a handy way to collect important slides you want to go back to later. A few are somewhat challenging. For g(x) = x 3 + x 2 - x, find all the values c in the interval (-2, 1) that satisfy the Mean Value Theorem. It is not very "smart" though, so do not be surprised if it cannot do your integral. Khan Academy Practice. More on why the anti derivative and the area under a curve are essentially the same thing. (this number is conflicting, forgive any errors. Flawed advice and the management trap: How managers can know they’re getting good advice, and when they’re not. An arbitrary domain value, x i, is chosen in each subinterval, and its subsequent function. Solve Math problems online. Problem 5 (10 points) Find. For sqrt(a^2-x^2), use x =a sin theta. The development of the definition of the definite integral begins with a function f( x), which is continuous on a closed interval [ a, b]. 0m to its left. Leadership & Organization Development Journal, 29(3), 212-234. Software for math teachers that creates exactly the worksheets you need in a matter of minutes. Cantrell, Mary Lynn. CIVIL ENGIN. The importance of integrated approaches and awareness of linkages between different intervention levels were pointed out. A Leibniz integral rule for a two dimensional surface moving in three dimensional space is ∬ (,) ⋅ = ∬ ((,) + [∇ ⋅ (,)]) ⋅ − ∮ ∂ ⁡ [× (,)] ⋅, where: F(r, t) is a vector field at the spatial position r at time t, Σ is a surface bounded by the closed curve ∂Σ, dA is a vector element of the surface Σ, ds is a vector element of the curve ∂Σ,. We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. Typical (straight sided) Problem. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. It has been a pleasure to engage with him to seek his advice on disruptive technologies and project execution. ) But after so many discussions, nothing came out to solve the irritating situation in the. Renaissance Learning. Several Examples with detailed solutions are presented. See the complete profile on LinkedIn and discover Sameer's connections and jobs at similar companies. Choose "Evaluate the Integral" from the topic selector and click to. 1 AFFILIATED INSTITUTIONS ANNA UNIVERSITY CHENNAI : : CHENNAI 600 025 REGULATIONS - 2008 VI TO VIII SEMESTERS AND ELECTIVES B. For more information please also view the AAEA press announcement here, or visit AEPP's new website. This exercise has the Khan Academy user practice adding one or ten to a two-digit number. Khan Study Group (KSG India) Khan Study Group is one of the leading premier institute that provides the IAS Examination study. 2nd Semester Review Packet. Integral definition is - essential to completeness : constituent. Integration by parts formula: ? u d v = u v-? v d u. Compare and contrast the effects of Mongol rule on Russia and the lands of Islam with the effects on East Asia. I say that because this free prep website only offers a program for the SAT, so you won't find any official ACT practice questions here. Bristlecone is a supply chain consulting and system Integration multinational company, founded in 1998, with its headquarters situated at San Jose, California. f (x)dx means the antiderivative of f with respect to x. The first and most vital step is to be able to write our integral in this form:. The company is a part of $20. See the complete profile on LinkedIn and discover Vajih's. Muhammad Bilal has 7 jobs listed on their profile. Important Formulae. Z 1 0 1 4 p 1 + x dx Solution: (a) Improper because it is an in nite integral (called a Type I. Integral definition is - essential to completeness : constituent. Khan Academy. Shabu has 7 jobs listed on their profile. Leaders as facilitators of individual and organizational learning. So, we are going to begin by recalling the product rule. Q&A for Work. View Vajih Khan's profile on LinkedIn, the world's largest professional community. Elementary Trigonometric Functions. Definite Integral. Described are the origin and nature of geothermal energy. Integration and Integration Techniques Chapter Exam Take this practice test to check your existing knowledge of the course material. y c CA9l5l W ur Yimgh1tTs y mr6e Os5eVr3vkejdW. Barring some drastic downturn, they will be there in some form, they know that school can't really work for these grades unless they get the kids in class with the teacher. This conference explores three major tracks: information reuse, information integration, and reusable systems. You click on the circle next to the answer which you believe that is correct. Unit 6: Integration and Accumulation of Change You’ll learn to apply limits to define definite integrals and how the Fundamental Theorem connects integration and differentiation. The Fundamental Theorem of Calculus states the relation between differentiation and integration. Step 1: Find the definite integral for each equation over the range x = 0 and x = 1, using the usual integration rules to integrate each term. Here are a set of practice problems for the Integration Techniques chapter of the Calculus II notes. In other words, the derivative of is. Here is a set of practice problems to accompany the Newton's Method section of the Applications of Derivatives chapter of the notes for Paul Dawkins Calculus I course at Lamar University. A differential equation is an equation that relates an unknown function and one or more of its derivatives. If you are asked to integrate a fraction, try multiplying or dividing the top and bottom of the fraction by a number. Pythagorean Identities. khanacademy. 1 online graduate program in Texas. Good luck with your TEAS test studying. It is an integral, but in practice it just means to find the net area of. About the journal. Problems for Fun and Practice 1. Integral definition is - essential to completeness : constituent. It justifies our procedure of evaluating an antiderivative at the. Students, teachers, parents, and everyone can find solutions to their math problems instantly. لدى Imtiyaz4 وظيفة مدرجة على الملف الشخصي عرض الملف الشخصي الكامل على LinkedIn وتعرف على زملاء Imtiyaz والوظائف في الشركات المماثلة. Setting: A general dental practice and a general medical practice. Available for Pre-Algebra, Algebra 1, Geometry, Algebra 2, Precalculus, and Calculus. The second derivative is given by: Or simply derive the first derivative: Nth derivative. The first type of integral equations which involve constants as both the limits — are called Fredholm Type Integral equations. Techniques of Integration Over the next few sections we examine some techniques that are frequently successful when seeking antiderivatives of functions. Practice finding indefinite integrals using the method of integration by parts. It has almost 2,500 students and 14,000 staff. Definite Integral Calculus Examples, Integration - Basic Introduction, Practice Problems by The Organic Chemistry Tutor. This is a full-length test with a complete set of multiple choice and free response questions. Easily share your publications and get them in front of Issuu’s. Antiderivative of xcosx using integration by parts. Reciprocal identities. View Umair Khan’s profile on LinkedIn, the world's largest professional community. Math and Probability for Life Sciences. Get involved and shape London’s future. I usually do about 1 item a day, such as 1 video, 1 practice, 1 article, etc. y The given limits have inner variable y. Step 1: Find the definite integral for each equation over the range x = 0 and x = 1, using the usual integration rules to integrate each term. View Dr Mohammad Zahidul Islam Khan’s profile on LinkedIn, the world's largest professional community. Masood has 7 jobs listed on their profile. Being able to do an integral is a key skill for any Calculus student. See the complete profile on LinkedIn and discover Khalid’s connections and jobs at similar companies. Strategy: Use Integration by Parts. Our mission is to provide a free, world-class education to anyone, anywhere. Explore degrees available through the No. During this time, we worked together on multiple projects related to Compliance, Project Management, Analytics etc. Divergence theorem examples. It is an integral, but in practice it just means to find the net area of. Integration by parts is a "fancy" technique for solving integrals. 55 16x\ dx`. Integration is the inverse operation of differentiation. Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step. The integration-by-parts formula tells you to do the top part of the 7, namely. Partial Fractions Calculator - find the partial fractions of a fractions step-by-step This website uses cookies to ensure you get the best experience. As part of the knowledge sharing initiative, a new informal economy resource database was presented. Take the integral of y = -1. 01 Single Variable Calculus, Fall 2005 Prof. Table of Basic Integrals1 (1) Z xn dx = 1 n+1 xn+1; n 6= 1 (2) Z 1 x dx = lnjxj (3) Z u dv = uv Z vdu (4) Z e xdx = e (5) Z ax dx = 1 lna ax (6) Z lnxdx = xlnx x (7) Z sinxdx = cosx (8) Z cosxdx = sinx (9) Z. Definite integration finds the accumulation of quantities, which has become a basic tool in calculus and has numerous applications in science and engineering. Integration is often introduced as the reverse process to differentiation, and has wide applications, for example in finding areas under curves and volumes of solids. Antiderivatives. His effective articulation and clarity of thought gives him the rare ability to break down complex technologies into simple comprehensible frameworks. View Umair Khan’s profile on LinkedIn, the world's largest professional community. List of games that practice all of your exponent rules. If you’re looking to get into nursing school, there’s a very good chance you will have to take the Test of Essential Academic Skills (TEAS), which tests your reading, math, science, and English comprehension. Find materials for this course in the pages linked along the left. 0s with a constant acceleration of 20. Integration by substitution (or the reverse-chain-rule). Examples: A. Examples of taking the indefinite integral (or anti-derivative) of polynomials. When you learn about the fundamental theorem of calculus, you will learn that the antiderivative has a very, very important property. Considers cults as gangs, but also distinguishes cults from gangs by the cult's reference to and insistence on allegiance to single higher authority, usually spirit figure or spiritual leader. Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways. Problems for Fun and Practice 1. See the complete profile on LinkedIn and discover Shabu's. Definite Integral Using U-Substitution •When evaluating a definite integral using u-substitution, one has to deal with the limits of integration. Saad has 6 jobs listed on their profile. Sameer has 11 jobs listed on their profile. There are several videos and some other helpul tools on the site if you have questions or get stuck. BASIC ANTIDERIVATIVE FORMULAS YOU REALLY NEED TO KNOW !! ex dx = ex +C ax dx = ax lna +C 1 x dx =ln\|x\| +C cosxdx=sinx+C sec2 xdx=tanx+C sinxdx= −cosx+ C csc2 xdx= −cotx +C secxtanxdx=secx+ C 1 1+x2 dx =arctanx+C 1 √ 1− x2 dx =arcsinx+C cscxcotxdx= −cscx+ C secxdx=ln\|secx+tanx\|+ C cscxdx= −ln\|cscx+cotx\|+ C xn dx = xn+1 n+1 +C, when n. Antiderivative Rules. About Khan Academy Khan Academy is a 501(c)(3) nonprofit organization with the mission of providing a free, world-class education for anyone, anywhere. 5 triple integrals. Oracle Data Integration Cloud – The Future of Data Integration Now! Room 3 LIMITED Chaitanya Pydimukkala. Chapter 1 : Integration Techniques. Centroid of an Area by Integration. For example, ln(x)e x. we showed that an antiderivative of the sum x + e x x + e x is given by the sum (x 2 2) + e x (x 2 2) + e x —that is, an antiderivative of a sum is given by a sum of antiderivatives. Integrals involving u-substitution and strange-looking bounds. Integration by substitution (or the reverse-chain-rule). Integration is an important tool in calculus that can give an antiderivative or represent area under a curve. Then where C is a constant of integration. (b) Decide if the integral is convergent or divergent. Following the table of contents in Applied Calculus 7e by Stefan Waner and Steven R. As a consequence it allows the order of integration to be changed in iterated integrals. Integral; Description Draw a graph of any function and see graphs of its derivative and integral. Indefinite integration is implemented in the Wolfram Language as Integrate[f, z]. ERIC is an online library of education research and information, sponsored by the Institute of Education Sciences (IES) of the U. Choose from 279 different sets of introduction ap ohs chapter 10 cultural landscape flashcards on Quizlet. A free online AP Calc practice test from Barron's. Bristlecone is a supply chain consulting and system Integration multinational company, founded in 1998, with its headquarters situated at San Jose, California. This "Cited by" count includes citations to the following articles in Scholar. (8 SEMESTER) ELECTRONICS AND COMMUNICATION ENGINEERING CURRICU. Department of Education. Basic Functions. Please sign in to leave a comment. It is visually represented as an integral symbol, a function, and then a dx at the end. When you learn about the fundamental theorem of calculus, you will learn that the antiderivative has a very, very important property. Techniques of Integration Over the next few sections we examine some techniques that are frequently successful when seeking antiderivatives of functions. Use substitution of variables to calculate antiderivatives (including changing limits for definite integrals). Learning Math has never been such fun!. Integration by trigonometric substitution Calculator online with solution and steps. Developer API & Integration; Customers. Watch fullscreen. COM TOMORROW September 11,2001. Numerically evaluate triple integral matlab integral3. Learn introduction ap ohs chapter 10 cultural landscape with free interactive flashcards. Changing the order of integration 1. After you have chosen the answer, click on the button Check Answers. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. •So by substitution, the limits of integration also change, giving us new Integral in new Variable as well as new limits in the same variable. Let u(x) and v(x) be two differentiable functions. Note: use your eyes and common sense when using this! Some curves don't work well, for example tan(x), 1/x near 0, and functions with sharp changes give bad results. The basic idea of integration by parts is to transform an integral you can't do into a simple product minus an integral you can do. A partner of the College Board (the creators of the SAT—not ACT), Khan Academy is a great website to use for ACT Math practice and review, as long as you know how to use it effectively. These formulas lead immediately to the. لدى Imtiyaz4 وظيفة مدرجة على الملف الشخصي عرض الملف الشخصي الكامل على LinkedIn وتعرف على زملاء Imtiyaz والوظائف في الشركات المماثلة. For example, so that we can now say that a partial fractions decomposition for is. View Mehdi Hasan Khan’s profile on LinkedIn, the world's largest professional community. Magnolia - MS Alliance. Easily share your publications and get them in front of Issuu’s. Created by Sal Khan. Divergence theorem examples. Urooj Reza Khan for a period of one year from September 2005 to September 2006. There are several videos and some other helpul tools on the site if you have questions or get stuck. Don't show me this again. Explore math with Desmos. Solve an Indefinite Integral - powered by WebMath. Spend an afternoon brushing up on statistics. See the complete profile on LinkedIn and discover Umair’s connections and jobs at similar companies. In 2 experiments, the authors investigated a potential interaction involving the processing of concurrent feedback using design features from the specificity of practice literature and the processing of terminal feedback using a manipulation from the guidance hypothesis literature. Our mission is to provide a free, world-class education to anyone, anywhere. Choose "Evaluate the Integral" from the topic selector and click to. Setting: A general dental practice and a general medical practice. Our practice name is slightly quirky, but this was done so that we stood out from any other optician on the high street. Bristlecone is a supply chain consulting and system Integration multinational company, founded in 1998, with its headquarters situated at San Jose, California. INTEGRAL CALCULUS - EXERCISES 43 Homework In problems 1 through 13, ﬁnd the indicated integral. 20 free AP calculus bc practice tests. Explore math with Desmos. They do not cover everything so a careful review of the Chapter and your class notes is also in order. com, a math practice program for schools and individual families. Practice Problems (solutions follow) For each of the following, set up the triple integral: ZZZ E f(x;y;z) dV. Controversial London mayor Sadiq Khan recently toured the United States in support of presidential candidate Hillary Clinton. ( ) 3 x dx. (see: calculating definite integrals). AP Calculus AB Exam Review: Practice Exam Problems & Solutions (Multiple Choice, No Calculator) AP Calculus AB Exam Review (Exam Prep). Integration is one of the two main operations of calculus; its inverse operation, differentiation, is the other. Four full-length practice tests Sample test questions. Imran has 6 jobs listed on their profile. You can represent the entire family of antiderivatives of a function by adding a constant to a known antiderivative. Jason Starr. Try for free. This section explains what is meant by integration and provides many standard integration techniques. For more information, visit www. The importance of integrated approaches and awareness of linkages between different intervention levels were pointed out. We redesigned Thinking Blocks and packed it full of new features! read aloud word problems - visual prompts - better models - engaging themes - mobile friendly Thinking Blocks works well on all devices. On loading of the question, it gives me a hint. The antiderivative xex−ex is not a product of the antiderivatives. If you are asked to integrate a fraction, try multiplying or dividing the top and bottom of the fraction by a number. Given: v i = 10 m/s, a = 2 m/s 2, Δt = 4 s Unknown: Δx = ? The value that is not given or asked for is v, so we must use the equation that does not contain v. Aprenda cálculo integral—integrais indefinidas, somas de Riemann, integrais definidas, problemas de aplicação e muito mais. View Zahir Khan AFRIDI’S profile on LinkedIn, the world's largest professional community. Please use your browsers back button or navigate to the home pagehome page. During this time, we worked together on multiple projects related to Compliance, Project Management, Analytics etc. Bristlecone is a supply chain consulting and system Integration multinational company, founded in 1998, with its headquarters situated at San Jose, California. Study online to earn the same quality degree as on campus. See the complete profile on LinkedIn and discover Saima’s connections and jobs at similar companies. We have seen from finding the area that the definite integral of a function can be interpreted as the area under the graph of a function. IRI plays a pivotal role in the capture, representation, maintenance, integration, validation, and extrapolation of information; and applies both information and knowledge for enhancing decision-making in various application domains. This integration allows students and instructors to use, create, share, and collaborate on Office 365 files within Canvas. General Editors: David Bourget (Western Ontario) David Chalmers (ANU, NYU) Area Editors: David Bourget Gwen Bradford. For example, find the indefinite integral of 14t². Integration by Parts is yet another integration trick that can be used when you have an integral that happens to be a product of algebraic, exponential, logarithm, or trigonometric functions. I was actually looking for a khan academy video specifically on root test and couldn't find it. How do we find the center of mass for such an uneven shape?. The University is both a model. If x is restricted to lie on the real line, the definite integral is known as a Riemann integral (which is the usual definition encountered in elementary textbooks). So if F(x) is the antiderivative of f(x), then the family of the antiderivatives would be F(x) + C. So the integral of 2 can be 2x + 3, 2x + 5, 2x, etc. The Fundamental Theorem of Calculus states the relation between differentiation and integration. Free Integration Online Practice Tests 71 Tests found for Integration : XII_Maths_Integrations_1 10 Questions \| 1939 Attempts AIEEE Mathematics: Integral Calculus INTEGRATION 1 [ FOR CBSE PLUS 2 , GCE A - LEVEL , BI , AIEEE , JCE ]. It is subject to change at any time. E lies under the plane z = 1+x+y and above the region in the xy-plane bounded by the curves y = p x, y = 0 and x = 1. Math and Probability for Life Sciences. See the complete profile on LinkedIn and discover Umair’s connections and jobs at similar companies. khanacademy. Fubini's theorem 1 Fubini's theorem In mathematical analysis Fubini's theorem, named after Guido Fubini, is a result which gives conditions under which it is possible to compute a double integral using iterated integrals. Graph functions, plot data, evaluate equations, explore transformations, and much more – for free! Products Classroom Activities. Using upper sums to evaluate a definite integral. View Saeem Khan's profile on LinkedIn, the world's largest professional community. 0s with a constant acceleration of 20. View Parveen Khan’s profile on LinkedIn, the world's largest professional community. View Rehan Khan’s profile on LinkedIn, the world's largest professional community. U-Substitution Integration, Indefinite & Definite Integral - Fractions & Trig Functions Calculus - Duration: 42:17. On this page, several problems related to kinematics are given. Take the integral of y = -1. 01 Single Variable Calculus, Fall 2005 Prof. See the complete profile on LinkedIn and discover Shabu's. View Raheel Khan's profile on LinkedIn, the world's largest professional community. Software for math teachers that creates exactly the worksheets you need in a matter of minutes. Available for Pre-Algebra, Algebra 1, Geometry, Algebra 2, Precalculus, and Calculus. We have seen from finding the area that the definite integral of a function can be interpreted as the area under the graph of a function. 1995-12-01. The students used the website’s question bank to practice and answer exercises on certain lessons, mainly on techniques of integration. عرض ملف Imtiyaz Khan الشخصي على LinkedIn، أكبر شبكة للمحترفين في العالم. Recall that the definition of an integral requires the function f(x) to be bounded on the bounded interval [a,b] (where a and b are two real numbers). Commenting on U. A compilation of selected projects from my Masters course in the Building Technology track at TU Delft. Practice - SATP2 & MST2. Integration is the reverse of differentiation. We’ll discuss integration of mobility, energy, water and waste disposal into planning practice, the role of educational institutions in the life of a city and the effects of art, culture and sport on planning strategies. Renaissance Learning. Try for free. If you're seeing this message, it means we're having trouble loading external resources on our website. Usman has 10 jobs listed on their profile. An Introduction to Technology Integration Integrating technology with classroom practice can be a great way to strengthen engagement by linking students to a global audience, turning them into creators of digital media, and helping them practice collaboration skills that will prepare them for the future. • Created, developed, and implemented an innovative HR Time Shop. To evaluate the integral we may consider choosing u = 3 sin 2 (x) + e 8 du = 6 sin(x) cos(x) dx. f(x) sin(nx π /L) between −L and L. com, a math practice program for schools and individual families. As a consequence it allows the order of. These are automatic, one-step antiderivatives with the exception of the reverse power rule, which is only slightly harder. Khan Academy problems work on a "mastery" system, which means you keep doing the problem until you get it right. F-35 Sustainment: DOD Needs a Plan to Address Risks Related to Its Central Logistics System. Example problem 1: Find the area between the curves y = x and y = x 2 between x = 0 and x = 1. Navy are to declare the ability to operate and deploy the F-35 in 2016 and 2018 respectively, and full-rate production of the aircraft is to begincapability”) in 2016 and 2018 respectively, and full-rate production decision of the program is planned for 2019. Khan Academy being integrated in the same way for exercises and practice (not just videos as it is currently) would be perfect! I understand the argument that there isn't much of an incentive to do this for a product that is available for free, but with so many instructors utilizing Khan Academy, it seems that it makes sense for Canvas to be. We will assume knowledge of the following well-known differentiation formulas : , where , and , where a is any positive constant not equal to 1 and is the natural (base e) logarithm of a. Let's do some problems and set up the $$u$$-sub. Khan Abdul Gaffar Khan (popularly known as "Frontier Gandhi") was a noted nationalist who spent 45 of his 95 years of life in jail; Barakatullah of Bhopal was one of the founders of the Ghadar Party, which created a network of anti-British organisations; Syed Rahmat Shah of the Ghadar Party worked as an underground revolutionary in France and. Vertical integration is a strategy where a company expands its business operations into different steps on the same production path, such as when a manufacturer owns its supplier and/or. This activity contains 25 questions. Integration is then carried out with respect to u, before reverting to the original variable x. Review Albert's AP® Calculus math concepts, from limits to infinity, with exam prep practice questions on the applications of rates of change and the accumulation of small quantities. Get Started. In Example 4. Magnolia - MS Alliance. We can often find that area just by sketching and using basic calculations, but other times we may need to use the Integration Rules. khanacademy. Precalculus \| Khan Academy by Khan Academy. Definition by ISTQB integration testing: Testing performed to expose… Read More »Integration Testing. THE INTEGRATION OF EXPONENTIAL FUNCTIONS The following problems involve the integration of exponential functions. Step 2: Click the blue arrow to compute the integral. • Created, developed, and implemented an innovative HR Time Shop. 5 Centroids by Integration Centroidal Axis. E lies under the plane z = 1+x+y and above the region in the xy-plane bounded by the curves y = p x, y = 0 and x = 1. Definite Integral. His role encompasses utilising Fusion tools and processes for development and activation. If you're seeing this message, it means we're having trouble loading external resources on our website. In this video, Mr. عرض ملف Qasim Khan الشخصي على LinkedIn، أكبر شبكة للمحترفين في العالم. "He also stated that he believes in "integration" rather than "assimilation. Cantrell, Mary Lynn. These formulas lead immediately to the. Find displacement by find area under a derivative graph; integration notation and properties (video to the right) Notes and practice: notes_displacement_def_integral. One function has many antiderivatives, but they all take the form of a function plus an arbitrary constant. Curriculum Pathways provides interactive, standards-based resources in English language arts, math, science, social studies, and Spanish (grades K-12). The Graphs of antiderivatives exercise appears under the Integral calculus Math Mission. org are unblocked. Integration and Integration Techniques Chapter Exam Take this practice test to check your existing knowledge of the course material. The force required to lift the water is the water's weight. We redesigned Thinking Blocks and packed it full of new features! read aloud word problems - visual prompts - better models - engaging themes - mobile friendly Thinking Blocks works well on all devices. A student with a computer and a broadband connection at home can watch a lesson from one of America’s most experienced teachers. Join over 8 million developers in solving code challenges on HackerRank, one of the best ways to prepare for programming interviews. On 19-21 July Thomas is invited to the Aga Khan Award for Architecture Seminar on «Emerging Models of Planning Practices» in Singapore. Find displacement by find area under a derivative graph; integration notation and properties (video to the right) Notes and practice: notes_displacement_def_integral. Divergent improper integral. You’ll apply properties of integrals and practice useful integration techniques. NASA Astrophysics Data System (ADS) Maharana, Pyarimohan; Abdel-Lathif, Ahmat Younous; Pattnayak, Kanhu Charan. Invaluable in and out of the classroom Designed to develop deep mathematical understanding and all the skills students need for their AS/A level studies and beyond. We referred the GOI, DFS, order dated 08-09-2011(GOI no. "Integrating Learning. The symbol dx represents an infinitesimal. The development of the definition of the definite integral begins with a function f( x), which is continuous on a closed interval [ a, b]. The equation. Benoit Wiesser leads strategy integration across all disciplines and Asia-Pacific markets. Aprenda Matemática, Artes, Programação de Computadores, Economia, Física, Química, Biologia, Medicina, Finanças, História e muito mais, gratuitamente. 5 * 2 * 4 2 + 10 * 4 = 16 + 40 = 56 m Note that since we are not given an initial position and are asked for the displacement, not the final position, we simply. Solution: The region can be described simply if we change to coordinates$\cvarfv$and$\cvarsvwhere \begin{align} \cvarfv &= y-x\notag\\ \cvarsv&=xy \label{thechangevar} \end{align} With this change of variables, our new region of integration\dlr^*$is$0 \le \cvarfv. Basic Functions. com, a math practice program for schools and individual families. Kinematics Practice Problems. Here is a set of practice problems to accompany the Substitution Rule for Indefinite Integrals section of the Integrals chapter of the notes for Paul Dawkins Calculus I course at Lamar University. 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# 11.3 Bohr’s theory of the hydrogen atom (Page 2/7) Page 2 / 7 ## Bohr’s solution for hydrogen Bohr was able to derive the formula for the hydrogen spectrum using basic physics, the planetary model of the atom, and some very important new proposals. His first proposal is that only certain orbits are allowed: we say that the orbits of electrons in atoms are quantized . Each orbit has a different energy, and electrons can move to a higher orbit by absorbing energy and drop to a lower orbit by emitting energy. If the orbits are quantized, the amount of energy absorbed or emitted is also quantized, producing discrete spectra. Photon absorption and emission are among the primary methods of transferring energy into and out of atoms. The energies of the photons are quantized, and their energy is explained as being equal to the change in energy of the electron when it moves from one orbit to another. In equation form, this is $\Delta E=\text{hf}={E}_{i}-{E}_{f}.$ Here, $\Delta E$ is the change in energy between the initial and final orbits, and $\text{hf}$ is the energy of the absorbed or emitted photon. It is quite logical (that is, expected from our everyday experience) that energy is involved in changing orbits. A blast of energy is required for the space shuttle, for example, to climb to a higher orbit. What is not expected is that atomic orbits should be quantized. This is not observed for satellites or planets, which can have any orbit given the proper energy. (See [link] .) [link] shows an energy-level diagram , a convenient way to display energy states. In the present discussion, we take these to be the allowed energy levels of the electron. Energy is plotted vertically with the lowest or ground state at the bottom and with excited states above. Given the energies of the lines in an atomic spectrum, it is possible (although sometimes very difficult) to determine the energy levels of an atom. Energy-level diagrams are used for many systems, including molecules and nuclei. A theory of the atom or any other system must predict its energies based on the physics of the system. Bohr was clever enough to find a way to calculate the electron orbital energies in hydrogen. This was an important first step that has been improved upon, but it does correctly describe many characteristics of hydrogen. Bohr proposed that only very specific circular orbits were allowed. The radius of these orbits is calculated to be ${r}_{\mathrm{n}}=\frac{{n}^{2}}{Z}{a}_{\mathrm{B}}$ where $n=1,2,3,\dots ,$ $Z$ is the atomic number of the element, and ${a}_{\mathrm{B}}=0.529×{10}^{-10}\phantom{\rule{.1667em}{0ex}}\mathrm{m},$ is known as the Bohr radius. The value of $n$ largely determines the energy level of the atom and is called the principal quantum number . This equation can be used to calculate the radii of the allowed (quantized) electron orbits in any hydrogen-like atom . It is impressive that the formula gives the correct size of hydrogen, which is measured experimentally to be very close to the Bohr radius. This equation also tells us that the orbital radius is proportional to ${n}^{2}$ , as illustrated in [link] . How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe anyone know any internet site where one can find nanotechnology papers? research.net kanaga sciencedirect big data base Ernesto Introduction about quantum dots in nanotechnology what does nano mean? nano basically means 10^(-9). nanometer is a unit to measure length. Bharti do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment? absolutely yes Daniel how to know photocatalytic properties of tio2 nanoparticles...what to do now it is a goid question and i want to know the answer as well Maciej Abigail for teaching engĺish at school how nano technology help us Anassong How can I make nanorobot? Lily Do somebody tell me a best nano engineering book for beginners? there is no specific books for beginners but there is book called principle of nanotechnology NANO how can I make nanorobot? Lily what is fullerene does it is used to make bukky balls are you nano engineer ? s. fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball. Tarell what is the actual application of fullerenes nowadays? Damian That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes. Tarell what is the Synthesis, properties,and applications of carbon nano chemistry Mostly, they use nano carbon for electronics and for materials to be strengthened. Virgil is Bucky paper clear? CYNTHIA carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc NANO Got questions? Join the online conversation and get instant answers!	{}
My Math Forum logarithm help algebra 2 Algebra Pre-Algebra and Basic Algebra Math Forum April 30th, 2014, 06:43 PM #1 Newbie Joined: Apr 2014 From: Boulder, Colorado Posts: 3 Thanks: 0 logarithm help algebra 2 IMG_2862.jpgI am doing an online Algebra II course and am having a hard time understanding an example given, which I have attached a picture of to this post (at least I think I'm pretty sure I was able to). The part that is confusing me is step three. I understand that ten is the common logarithm but I am failing to see how they got from log ((n+9)/4n) = 0 to 10^0 = (n+9/4n. If someone could explain this to me that would be great. Thanks. By the way I realize this is my second post about this but I forgot to attach the link to the first one and have no idea how to delete it. Last edited by FinlayB; April 30th, 2014 at 06:47 PM. April 30th, 2014, 07:01 PM #2 Math Team Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra This is one definition of the logarithm. $$\log_a{b} = c \iff a^c = b$$ April 30th, 2014, 07:05 PM #3 Newbie Joined: Apr 2014 From: Boulder, Colorado Posts: 3 Thanks: 0 thank you so much! Just as a side note would I use those same steps in solving the equation log(5x) = log(2x+9)? April 30th, 2014, 07:27 PM #4 Math Team Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra You could do, yes. Tags algebra, logarithm, logarithms, logs Thread Tools Display Modes Linear Mode Similar Threads Thread Thread Starter Forum Replies Last Post anco995 Algebra 5 December 16th, 2012 03:36 AM Algebra 4 February 16th, 2012 06:40 AM sea_wave Applied Math 2 October 24th, 2009 03:45 AM HHH Algebra 5 September 27th, 2009 05:05 PM anco995 Calculus 0 December 31st, 1969 04:00 PM Contact - Home - Forums - Cryptocurrency Forum - Top	{}
# WorkLock¶ Important Participation in WorkLock is now closed, but since NU is an ERC-20 token, it can be purchased through exchanges. ## Overview¶ WorkLock is a novel, permissionless decentralized network node setup mechanism, developed at NuCypher, which requires participants to temporarily escrow ETH and operate NuCypher nodes in order to be allocated NU, the native cryptocurrency of the NuCypher network that enables node operation. WorkLock is designed to onboard threshold cryptography nodes to the live NuCypher network using a process that selects participants who are most likely to strengthen and sustain the network by committing to staking and running active nodes. The WorkLock begins with an escrow period, during which anyone seeking to operate a NuCypher node can send ETH to the WorkLock contract to be temporarily escrowed on-chain. At any time during the escrow period, WorkLock participants can cancel their participation to forgo NU and recoup their escrowed ETH immediately. Once the escrow period closes, the WorkLock contract does not accept more ETH, but will still accept cancellations during an additional time window (the “cancellation window”). At the end of this cancellation period, the allocation period begins and stake-locked NU are allocated to participants in the form of non-transferable stakes. These stakes are designed for the limited and explicit purpose of running threshold cryptography nodes on the live NuCypher network. Stake-locked NU will be allocated at network launch according to the following principles: • All of the NU held by WorkLock will be allocated to network nodes in non-transferable stakes designed to enable node operation. • All ETH escrows must be greater than or equal to the minimum allowed escrow. • For each escrow, the surplus above the minimum allowed escrow is called the bonus; all escrows are composed of a base escrow (fixed minimum) and a bonus escrow (variable amount). • Each participant will receive at least the minimum amount of staked NU needed to operate a network node. • Once all participants have been allocated the minimum amount of NU, each participant with a bonus will be allocated a portion of the remaining NU, allocated pro rata across all participants, taking into consideration only their bonus ETH amounts. • If the resulting NU allocated to a participant is above the maximum allowed NU needed to operate a network node, then such a participant has their escrow partially refunded until the corresponding amount of NU is within the allowed limits for node operation. Finally, if WorkLock participants use that stake-locked NU to run a node and provide threshold cryptography services on the live network for a minimum of six months, the NU will subsequently unlock and their temporarily escrowed ETH will be returned in full. Participants that fail to successfully run a NuCypher network node for six months of the live network will not receive stake-rewarded NU and their ETH will remain escrowed in the WorkLock contract. ## Hypothetical WorkLock Scenarios¶ Important These examples use hypothetical values that allow for easier illustration and calculations. The specified WorkLock properties in the examples are not intended to be reflective of what actual values will be. For each scenario, assume the following hypothetical WorkLock properties: 1. WorkLock holds 225,000,000 NU and the minimum escrow is 5 ETH. 2. The minimum amount of NU required to stake is 15,000 NU and the maximum stake size is 30,000,000 NU. 3. The total number of participants is 1000 (including you) with a total of 50,000 ETH escrowed (including your escrow). 4. For our purposes, a whale escrow is an escrow that causes the calculated stake size to be larger than the maximum stake size (30,000,000 NU). Note To reduce complexity, calculations are performed in a step-wise manner which may lead to minor rounding differences in the determined values. ### Scenario 1: Resulting stake size does not exceed maximum stake size (no whale escrows)¶ You submit an escrow of 32 ETH i.e. 5 ETH minimum + 27 bonus ETH. How many NU would be allocated to your network node? • Each of the 1000 participants (including you) would be allocated at least the minimum NU to stake = 15,000 NU • Remaining NU in WorkLock after minimum allocation is $225,000,000 NU - (15,000 NU \times 1000 \,participants) = 210,000,000 NU$ • Bonus ETH supply (i.e. total ETH not including minimum escrows) is $50,000 ETH - (5 ETH \times 1000 \,participants) = 45,000 ETH$ • Your bonus portion of the bonus ETH supply is $\frac{27 ETH}{45,000 ETH} = 0.06\%$ • Your allocation of the remaining NU is $0.06\% \times 210,000,000 NU = 126,000 NU$ Total NU received = 15,000 NU + 126,000 NU = 141,000 NU ### Scenario 2: Resulting stake size exceeds maximum stake size (1 whale escrow)¶ You submit an escrow of 6,755 ETH i.e. 5 ETH minimum + 6,750 bonus ETH. How many NU would be allocated to your network node? • Each of the 1000 participants (including you) would be allocated at least the minimum NU to stake = 15,000 NU • Remaining NU in WorkLock after minimum allocation is $225,000,000 NU - (15,000 NU \times 1000 \,participants) = 210,000,000 NU$ • Bonus ETH supply (i.e. total ETH not including minimum escrows) is $50,000 ETH - (5 ETH \times 1000 \,participants) = 45,000 ETH$ • Your bonus allocation of the bonus ETH supply is $\frac{6,750 ETH}{45,000 ETH} = 15\%$ • Your allocation of the remaining NU is $15\% \times 210,000,000 NU = 31,500,000 NU$ However, the total amount of NU to be allocated is 15,000 NU + 31,500,000 NU = 31,515,000 NU which is greater than the maximum stake amount (30,000,000 NU). Therefore, the amount of NU allocated to you needs to be reduced, and some of your bonus ETH refunded. • Typically the calculation for the NU allocated from the bonus portion is $\frac{\text{your bonus ETH}}{\text{bonus ETH supply}} \times \text{remaining NU bonus supply}$ • The additional complication here is that refunding bonus ETH reduces your bonus ETH AND the bonus ETH supply since the bonus ETH supply includes the bonus ETH portion of your escrow. • A more complicated equation arises for the bonus part of the calculation, where x is the refunded ETH: $\text{stake size} = \frac{\text{(your bonus ETH - x)}}{\text{(bonus ETH supply - x)}} \times \text{remaining NU}$ • Since you will be allocated a 15,000 NU minimum, and the maximum stake size is 30,000,000 NU, the most you can be allocated from the remaining NU is $30,000,000 NU - 15,000 NU = 29,985,000 NU$ • Therefore using values in the equation above yields $29,985,000 NU = \frac{6,750 ETH - x ETH}{45,000 ETH - x ETH} \times 210,000,000 NU$ • Reorganizing the equation $\begin{split}x &= \frac{6,750 ETH \times 210,000,000 NU - 45,000 ETH \times 29,985,000 NU}{210,000,000 NU - 29,985,000 NU} \\ &\approx 378.72 ETH\end{split}$ • Therefore, your final bonus escrow is $6,750 ETH - 378.72 ETH \approx 6,371.28 ETH$ • Your portion of the bonus ETH supply is $\frac{6,371.28}{(45,000 ETH - 378.72 ETH)} \approx 14.279\%$ • Your allocation of the remaining NU is $14.279\% \times 210,000,000 NU \approx 29,985,000 NU$ Total NU allocated ~ 15,000 NU + 29,985,000 NU (rounding) ~ 30,000,000 NU, and refunded ETH ~ 378.72 ETH ### Scenario 3: Resulting stake size exceeds maximum stake size (2 whale escrows)¶ Someone else submitted an escrow of 6,108 ETH i.e. 5 ETH + 6,100 bonus ETH; we’ll call them “whale_1”. You submit an escrow of 6,755 ETH i.e. 5 ETH minimum + 6,750 bonus ETH; you are “whale_2”. How many NU would be allocated to your network node? • Each of the 1000 participants (including you) would receive at least the minimum NU to stake = 15,000 NU • Remaining NU in WorkLock after minimum allocation is $225,000,000 NU - (15,000 NU \times 1000 \,participants) = 210,000,000 NU$ • Bonus ETH supply (i.e. total ETH not including minimum escrows) is $50,000 ETH - (5 ETH \times 1000 \,participants) = 45,000 ETH$ • Your portion of the bonus ETH supply is $\frac{6,750 ETH}{45,000 ETH} = 15\%$ • Your allocation of the remaining NU is $15\% \times 210,000,000 NU= 31,500,000 NU$ However, the total amount of NU to be allocated to receive is 15,000 NU + 31,500,000 NU = 31,515,000 NU which is greater than the maximum stake amount (30,000,000 NU). • From the previous scenario, the equation for the bonus part of the calculation is as follows, where x is the refunded ETH $\text{stake size} = \frac{\text{(your bonus ETH - x)}}{\text{(bonus ETH supply - x)}} \times \text{remaining NU}$ • Additionally, there is more than one whale escrow, which would also cause the bonus ETH supply to reduce as well. • Instead the following whale resolution algorithm is employed: 1. Select the smallest whale bonus ETH escrow; in this case 6,390 ETH from whale_1 < 6,750 ETH from whale_2 2. Equalize the bonus ETH whale escrows for all other whales (in this case, just whale_2 i.e. just you) to be the smallest whale bonus escrow i.e. 6,390 ETH in this case 3. Since your bonus ETH escrow is > 6,390 ETH, you will be refunded $6,750 ETH - 6,390 ETH = 360 ETH$ 4. This reduces the resulting bonus ETH supply which will now be $45,000 ETH - 360 ETH = 44,640 ETH$ 5. We now need to calculate the bonus ETH refunds based on the updated bonus ETH supply, and the maximum stake size. 6. Remember that everyone is allocated a 15,000 NU minimum, and the maximum stake size is 30,000,000 NU, so the most that can be allocated to you from the remaining NU is $30,000,000 NU - 15,000 NU = 29,985,000 NU$ 7. Since we have multiple whales, our equation is the following , where n is the number of whale escrows $x = \frac{\text{(min whale escrow} \times \text{NU supply - eth_supply} \times \text{max stake)}}{\text{(NU supply - n} \times \text{max stake)}}$ 8. Plugging in values $\begin{split}x &= \frac{(6,390 ETH \times 210,000,000 NU - 44,640 ETH \times 29,985,000 NU)}{(210,000,000 NU - 2 \times 29,985,000 NU)} \\ &\approx 22.46 ETH\end{split}$ • hence each whale gets additionally refunded ~ 22.46 ETH 9. Therefore, • whale_1 is refunded ~ 22.46 ETH • whale_2 (i.e. you) is refunded ~ 22.46 ETH + 360 ETH (from Step 3) ~ 382.46 ETH 10. Based on the refunds • The bonus escrows for the whales will now be equalized: • whale_1 bonus ~ 6,390 ETH - 22.46 ETH ~ 6,367.54 ETH • whale_2 bonus ~ 6,750 ETH - 382.46 ETH ~ 6,367.54 ETH • The updated bonus ETH supply will be $45,000 ETH - (22.46 ETH + 382.46 ETH) \approx 44,595.08 ETH$ 11. Each whale’s portion of the bonus ETH supply is therefore $\frac{6,367.54 ETH}{44,595.08 ETH} \approx 14.279\%$ 12. And each whale’s allocation of the remaining NU is $14.279\% \times 210,000,000 NU \approx 29,985,900 NU$ Total NU allocated ~ 15,000 NU + 29,985,900 NU (rounding) ~ 30,000,000 NU, and refunded ETH ~ 382.46 ETH Note In Scenarios 2 and 3, you will notice that the bonus ETH supply was reduced. This produces a very subtle situation - for previous non-whale participants (escrows in the original bonus ETH supply that did not produce a stake larger than the maximum stake) their escrows remained unchanged, but the bonus ETH supply was reduced. This means that some participants that were not originally whales, may become whales once the bonus ETH supply is reduced since their proportion of the bonus pool increased. Therefore, the whale resolution algorithm described in Scenario 3 may be repeated for multiple rounds until there are no longer any whales. To keep the explanation simple, both Scenarios 2 and 3 ignore such a situation since the calculations become even more complex.	{}
Article \| Open \| Published: # Exposure to genetically engineered olive fly (Bactrocera oleae) has no negative impact on three non-target organisms ## Abstract Bactrocera oleae (Diptera: Tephritidae) remains a major pest of olive fruit production worldwide. Current pest management programs largely depend on chemical insecticides, resulting in high economic and environmental costs. Alternative pest control approaches are therefore highly desirable. We have created a conditional female-specific self-limiting strain of B. oleae (OX3097D-Bol) that could be applied for sustainable pest control. OX3097D-Bol olive fly carries a fluorescent marker (DsRed2) for identification and a self-limiting genetic trait that is repressed by tetracycline. In the absence of tetracycline, the tetracycline transactivator (tTAV) accumulates, resulting in female death at larvae and early pupal stages. The aim of this study was to evaluate the impact of genetically engineered OX3097D-Bol olive fly on three non-target organisms that either predate or parasitize olive flies, one from the guild of parasitoids (Psyttalia concolor) and two from the guild of predators (Pardosa spider species and the rove beetle Aleochara bilineata). No significant negative effect was observed on life history parameters, mortality and reproductive capacity of the non-target organisms studied. These results suggest that potential exposure to DsRed2 and tTAV gene products (e.g. mRNA and encoded proteins) would have a negligible impact on on-target organisms in the guilds or predators and parasitoids. ## Introduction The olive fruit fly, Bactrocera oleae (Rossi) (Diptera: Tephritidae), is a major agricultural pest of olive fruit production across the world. The female olive fly lays eggs inside the olive fruit and larvae feed on fruit flesh causing extensive crop damage1. Current pest control programs rely heavily on chemical insecticides, which not only have high economic and environmental costs, but have resulted in the appearance of insecticide-resistant pest populations2. In addition, these chemistries not only act on pests but also affect beneficial insects. Thus, integrated pest management strategies that allow the use of control measures and have no adverse effect on beneficial insects offer further advantages to growers. Alternative pest control methods, such as the sterile insect technique (SIT), have been previously described for olive fly3. However, these have consistently yielded poor results, most likely due to the fitness penalties acquired by the radiation-sterilized flies3,4,5,6. SIT male-only releases are preferable and have been proposed as a new approach to improve pest control methods4, 5. Releases of males are cost-effective and ensure the reduction of the next generation’s population since they mate with wild females. To address this, we have generated a conditional female-specific self-limiting strain (OX3097D-Bol) of the olive fly B. oleae that could be used in pest control programs. The genetically engineered OX3097D-Bol strain carries a fluorescent protein marker gene (DsRed2) that facilitates quick identification of all genetically engineered individuals at all life-stages. OX3097D-Bol also carries a conditional female-specific self-limiting gene that is dominantly inherited7. The conditional female-specific self-limiting gene is repressed by tetracycline, present in the larval dietary medium8. In the presence of tetracycline, the system is inactivated and both females and males can be reared in the laboratory. When tetracycline is absent from the rearing process, the tetracycline transactivator (tTAV) is accumulated in females leading to male-only survival7. Data from laboratory and greenhouse studies demonstrated that the release of OX3097D-Bol males could provide sustainable control of pest olive fly as they will mate with wild females and produce progeny that will die at larval and pupal stages7. The assessment of the environmental impact of genetically engineered insects is an important safety aspect and a prerequisite for their commercialization as vector/pest control agents9. Risk of exposure to non-target organisms could occur through parasitizing the B. oleae larvae or direct feeding on B. oleae larvae or eggs10. Our study has tested the potential impact of genetically engineered olive fly B. oleae on three relevant non-target organisms, one from the guild of parasitoids (Psyttalia concolor (Szépligeti), Hymenoptera: Braconidae) and two from the guild of predators (Pardosa spiders (Koch) (Araneae, Lycosidae) and the rove beetle Aleochara bilineata (Gyllenhal) (Choleoptera: Staphylinidae)). P. concolor is a larval-pupal endoparasitoid that develops within a range of Tephritid larvae including medfly (Ceritatis capitate (Wiedemann) (Diptera: Tephritidae)) and olive fly (B. oleae)11. P. concolor has a global distribution spanning Africa, Europe, America, Central America and Pakistan as a result of accidental or intentional introduction for use as a biological control agent12, 13. Female P. concolor detect olive fly larvae by the vibrations made when burrowing through the fruit. The female parasitoid wasp lays an egg inside the larvae, through the fruit, by using her ovipositor12. Once the parasitoid egg is within its host, the host larvae continues to feed on the fruit until it is ready to pupate. Whilst the host is pupating, the parasitoid egg hatches and the parasitoid larvae feeds on the host, causing its death. Pardosa species (wolf spiders) are generalist predators, attacking a range of live prey species, including Diptera. They can be found in large numbers in agricultural fields in the spring and summer months. Spiders from the Pardosa genus have been identified in citrus orchard in Spain14, 15 and in olive groves in the Mediterranean16. Pardosa species are often the organism of choice for testing the effect of plant protection products on non-target organisms17. In such experiments, Pardosa are caught from the wild as there is no reliable laboratory breeding protocol17. Mature females and males mate in the early spring and females can be seen carrying an egg sac from May onwards. Species of the genus Aleochara (Coleoptera: Staphylinidae) are predators of olive fly. A. bilineata is normally reared of the pupae of cyclorrhaphous Diptera. First instar larvae enter the host puparium, and after having consumed the host and completed the three larval stages, they pupate inside the host puparium. Beetles then emerge from the host pupae. In this study, pupae of the onion fly Delia antiqua (Meigen) (Diptera: Anthomyiidae) were used as host organism for the larvae of A. bilineata. Reproductive success of beetles is usually measured by counting the number of second generation beetles (offspring) emerging from the onion fly pupae. Here we assessed the effect of the exposure to genetically engineered OX3097D-Bol olive fly, compared to wild-type, on: (i) pupal emergence rate of the olive fly parasitized by P. concolor and life history parameters of the parasitoid, (ii) mortality and behavior of Pardosa spiders when fed olive fly larvae and (iii) mortality and reproductive success of A. bilineata fed olive fly larvae. ## Results The development of the genetically engineered OX3097D-Bol olive fly strain tested in this study has been described elsewhere7. OX3097D-Bol contains a fluorescent protein marker gene (DsRed2) and a conditional female-specific self-limiting genetic trait which in the absence of tetracycline allows the accumulation of the tetracycline transactivator (tTAV) gene products resulting in female death at larval/pupal stages7. There are currently no known methods for accurately sexing olive fly larvae whilst alive. Therefore, in this study we used mixed sex OX3097D-Bol and wild-type larvae reared on tetracycline. To expose the non-target organisms to both DsRed2 and tTAV gene products (e.g. mRNA and encoded proteins) olive fly larvae or adults were then fed fresh diet without tetracycline. ### Wild-type and OX3097D-Bol olive fly are equally adequate hosts of the parasitoid P. concolor To determine whether exposure to OX3097D-Bol olive fly has any effect on the parasitoid P. concolor, female wasps were presented with third instar olive fly for oviposition. Predictably, olive fly larvae exposed to P. concolor did not survived parasitism. The percentage of parasitized olive fly adults that emerged was approximately 0.35% and 0.17% for wild-type and OX3097D-Bol larvae, respectively, in contrast to 73.33% and 62.50% emergence of non-parasitized wild-type and OX3097D-Bol larvae (Fig. 1A). The difference in emergence in non-parasitized and parasitized wild-type versus OX3097D-Bol olive fly larvae was not statistically significant (P = 0.274). There was no significant difference between the mean number of P. concolor eggs laid into OX3097D-Bol (n = 144) or into wild-type (n = 144) olive fly larvae (P = 0.931, Fig. 1B). There was also no significant difference between the percentage of P. concolor adults emerging from wild-type larvae (33.33%; n = 192 out of 576) and that emerging from OX3097D-Bol larvae (32.99%; n = 190 out of 576) (P = 0.95, Fig. 1C). Furthermore, no significant difference in the length of P. concolor developmental time between the treatments was observed when males (wild-type, n = 80 and OX3097D-Bol, n = 55; P = 0.565) and females (wild-type, n = 112 and OX3097D-Bol, n = 135; P = 0.503) were directly compared to each other (Fig. 1D and E). The reproductive potential of newly eclosed P. concolor females, as determined by the number of fully developed eggs in both ovaries, did not vary significantly (P = 0.709) between females that parasitized wild-type larvae (n = 112) and those that parasitized OX3097D-Bol larvae (n = 135) (Fig. 1F). Taken together, these data suggest that both olive fly strains, wild-type and OX3097D-Bol, were perceived as equally adequate hosts by P. concolor females. ### Feeding Pardosa spiders with OX3097D-Bol has no detrimental effects To assess the effect of a diet consisting of genetically engineered compared to wild-type olive fly on Pardosa species food consumption, mortality and reproductive potential, spiders were fed exclusively on olive OX3097D-Bol or wild-type flies. Most spiders captured were females, each carrying an egg sac. Only 4/70 spiders captured were males, most likely because male spiders would have died, once mating has occurred early in the season17,18,19. The four male spiders collected were randomly allocated to the three study groups; 1 male was allocated to the OX3097D-Bol group (Treatment group), 1 male to the wild-type group (Control 1 group) and 2 males to the starved group (Control 2 group). The female spiders had their egg sac removed seven days before the start of the study (as experimental design in Supplementary Figure S1). All spiders were fed adult OX3097D-Bol or wild-type olive flies for 14 days. Food consumption was consistently high over the duration of the study (Table 1), and there was no sign of avoidance or preference behavior towards either the wild-type (n = 30) or OX3097D-Bol (n = 30) olive flies. There were no significant differences in the mean number of flies consumed per spider per feeding session between the Treatment and Control 1 groups (see specific P-values in Table 1). Similarly, there was no significant difference (P = 0.841) in the total number of flies consumed by spiders fed with wild-type (total = 10.92 ± 1.83) or OX3097D-Bol (total = 10.97 ± 1.98) olive flies. There were five dead spiders in total by the end of the study, three of which were males (one from the Treatment group and two from the Control 2 group). There was no significant difference between mortality rates of spiders fed a diet consisting of OX3097D-Bol olive fly (n = 30) compared to those fed wild type olive fly (n = 30; P > 0.999) (Table 1). As expected, there was a significant difference between mortality rates of Pardosa spiders fed with OX3097D-Bol olive fly (n = 30) compared to starved spiders (n = 10; P = 0.019). Although female spiders had their egg sac removed at the beginning of the study, a number of females (seven spiders from the Control 1 group and nine spiders from the Treatment group) had produced a new egg sac during the course of the study. None of the spiders in the starved group (Control 2 group) produced a new egg sac. This implies that both wild-type and OX3097D-Bol olive fly diets are equally adequate for female spiders to produce potential off-spring. These data suggest that there is no significant difference in survival or behavior of Pardosa spiders when they are fed exclusively OX3097D-Bol flies or wild-type flies. ### Feeding rove beetles with OX3097D-Bol has not effect on their reproductive capacity To assess the effect of a diet of genetically engineered olive fly on the full reproductive cycle of the predator A. bilineata, rove beetles were fed OX3097D-Bol or wild-type olive flies. Rove beetles were exposed to either genetically engineered OX3097D-Bol (Treatment group; n = 20) or wild-type (Control 1 group; n = 20) olive flies. As additional controls, beetles were also exposed to a diet of Chironomus species in the absence (Control 2 group; n = 20) and presence (Control 3 group; n = 20) of a toxic reference substance (ROGOR PLUS) known to reduce their reproductive capacity drastically at the dose used (440 g active substance/ha). An average of 450.3 beetles out of 6000 (approximately 26.7%) emerged from onion fly pupae when fed wild-type olive fly larvae (Control 1 group). In the toxic reference substance group (Control 3 group), the reproductive capacity of A. bilineata was reduced by >50%. Therefore, and according to the criteria selected a priori 19, the study was considered valid. In this study, 28-day mortality data and cumulative fecundity data of adult rove beetles fed with a diet of OX3097D-Bol (Treatment group) or wild-type (Control 1group) olive fly larvae were compared. Mortality after 28 days was 6.25% in the Treatment group, 12.5% in the Control 1 group and 10.0% in the Control 2 group whilst it was 83.8% in the Control 3 group (Table 2). There was no significant difference between mortality in the Control 1 group (beetles fed with 0.10 g wild-type olive fly larvae) and the Treatment group (beetles fed with 0.10 g of OX3097D-Bol olive fly larvae) after 28 days (P = 0.222). By contrast, there was a significant difference between mortality in the Control 1 group and Control 3 group (P = 0.004). Fecundity effects were estimated by comparing the mean number of offspring produced per beetle in the Treatment group with controls (Table 2). Fecundity in the Control 1 group was 40.5 offspring per beetle, whilst in the Control 2 group fecundity was 25.3 offspring per beetle. The reduction of the reproductive capacity in the Control 3 group (toxic reference substance) was −95.6%. There was no significant difference between the cumulative fecundity in the Treatment group and Control 1 group (P = 0.079). As expected, there was a significant difference in fecundity between the Control 1 group and the Control 3 group (P = 0.002). No other behavioural effects were observed with the test item treatment. These results suggest that there is no significant difference in rove beetles’ life cycle when they are fed exclusively OX3097D-Bol flies or wild-type flies. ## Discussion B. oleae is a major pest of olive fruit that is currently controlled through the application of chemical insecticides. However, it is acknowledged that the use of these insecticides poses several risks to pest control management. On one hand, these pesticides are suffering from loss of registration (e.g. the neonicotinoids in the European Union)20 and the development of resistance in the target pest insects (including the synthetic pyrethroids, carbamates)21,22,23,24. On the other hand, these chemistries have a broad spectrum and not only act on pest insects but also affect beneficial insects such as pollinators, predators and parasitoids. Therefore, and to offer further advantages to growers, it is crucial to develop integrated pest management strategies that are not harmful to beneficial insects. Sterile insect Technique (SIT) is an alternative pest control method unsuccessfully trialed for olive fly3 and thus, male-only release has been proposed as a new approach to improve pest control programs5. We have generated a conditional female-specific self-limiting strain (OX3097D-Bol) of the olive fly B. oleae that could be used in pest control programs, as the female progeny do not survive to adulthood, in the absence of tetracycline. Release of self-limiting olive flies would be an effective and direct, species-specific pest management approach by preventing survival of female progeny. OX3097D-Bol males released into olive groves could subsequently mate with wild females. Those wild female flies that had mated with OX3097D-Bol males would produce progeny that would die at the larval/pupal stages. Exposure of parasitoids and predators to the transgenes and/or their products (e.g. mRNA and encoded proteins), could potentially have adverse effects on such non-target organisms. In this study, we have demonstrated for the first time that exposure (via diet or predation routes) to genetically engineered olive fly B. oleae has no detrimental impact on three non-target organisms, one from the guild of parasitoids (P. concolor) and two from the guild of predators (Pardosa spiders and rove beetle A. bilineata). The Braconid wasp P. concolor was selected for this study as it can develop within a range of Tephritid larvae11 and because of its global distribution throughout four different continents12, 13. Mass rearing of P. concolor in laboratory conditions relies on a delicate balance of three factors: a) the parasitoid/host ratio, b) host exposure time and c) female status (naïve or experienced)25. Fruit fly larvae parasitized by P. concolor do not generally survive. It is possible for the host to encapsulate the parasitoid egg and survive to adulthood, however, the frequency of this occurrence for P. concolor parasitoids and their hosts is unknown26. In our study, most olive fly larvae (>99.5%), whether genetically engineered OX3097D-Bol or wild-type larvae, did not survive following exposure to P. concolor. In agreement with previous studies, there was also very low level of hyperparasitism or superparasitism11. Levels of P. concolor emergence >35% are rarely reported13, 27. A study using wild olive fly larvae, collected from the field and parasitized in the laboratory, reported 6.2% P. concolor emergence28, whilst 28.2% emergence has been observed from lab reared medfly28. In the present study, emergence of P. concolor from laboratory reared OX3097D-Bol and wild-type olive fly is close to the emergence level detected for laboratory reared medfly. No changes in life history parameters of P. concolor where observed when exposed to genetically engineered OX3097D-Bol olive fly compared to wild-type olive fly. In addition, exposure to genetically engineered OX3097D-Bol olive fly compared to wild-type does not affect the reproductive potential of P. concolor females. Pardosa spiders captured in May were overwhelmingly females, carrying an egg sac. In the wild, the first spiders to die tend to be male, as their natural mortality radically increases after reproduction19. Given that most Pardosa females in our study were carrying an egg sac on capture, we can safely assume males have reached their reproductive phase. Removal of the first egg sac from the females appears to have advanced their development, as even though Pardosa are known to produce a second egg sac, this normally occurs from mid-July to September17. These results suggest that Pardosa spiders fed OX3097D-Bol olive flies could potentially produce as much offspring as those fed with wild-type olive flies. Wild caught Pardosa fed 100% OX3097D-Bol or wild-type flies did not show any avoidance behavior towards the genetically engineered olive fly compared to wild-type. This study falls within the limits of validity19, therefore, we accept the null hypothesis that there is no significant difference between the levels of mortality or behavior of spiders fed a diet consisting of genetically engineered OX3097D-Bol olive fly compared to those fed a diet consisting of wild-type olive fly. Consequently, we can conclude that DsRed2 and tTAV gene products are unlikely to be harmful to Pardosa spiders at 100% of diet. As Pardosa spiders are generalist predators and consume a wide variety of prey29, our results are likely to represent a significant safety margin. Here, we have also assessed the effect of ingestion of B. oleae OX3097D-Bol on the mortality and reproductive capacity of A. bilineata. In the Pardosa studies, we concluded that female spiders fed genetically engineered OX3097D-Bol olive flies had similar reproductive potential as those fed wild-type olive flies. The study with A. bilineata is important as the test covered the entire life cycle of this predator and its overall reproductive capacity was used as endpoint. A. bilineata is a parasitoid of the pupae of cyclorraphous Diptera. Rove beetles fed with genetically engineered OX3097D-Bol olive fly did not show increased mortality or reduced reproductive capacity compared to those fed with wild-type olive fly or Chironomus species. No other effects were observed. Having assessed changes in the whole life cycle of this predator, we can conclude that any residues of DsRed2 and tTAV gene products from the genetically engineered OX3097D-Bol olive fly are unlikely to affect the reproductive capacity of A. bilineata. Our results are in agreement with previously published data on safety of genetically engineered mosquitoes carrying similar transgenes and their exposure to non-target organisms9. Importantly, the DsRed2 marker gene and its protein have been widely used in a variety of species, including insects, without any adverse effect30. In mosquitoes, the tTAV expression system is repressed by tetracycline, but the corresponding mRNA has been shown to be expressed in the absence of the antibiotic31. In that previous study, no adverse effects were detected in Toxorhynchites’ development fed with mosquito larvae reared either in the presence or absence of tetracycline9. Non-target organisms such as parasitoids and predators may be exposed to the transgenes and their products directly from consuming insects that express and/or accumulate those products. The results presented here suggest that potential exposure to OX3097D-Bol olive flies carrying DsRed2 and tTAV genes, have no detectable impact on the lethal and sub-lethal life history traits of non-target organisms included in this study and by extrapolation, on other members of ecological guilds of predators and parasitoids as relevant sensitive indicator species were chosen for this study. ## Materials and Methods ### Olive fly strains The B. oleae Argov strain was used as wild-type (kindly provided by the Israeli Agricultural Research Organisation (ARO) and Bio-Fly Ltd). B. oleae Dacus (Democritus laboratory, Greece) was used as a background strain to create the OX3097D-Bol strain, as previously described7. Briefly, OX3097D-Bol contains a fluorescent protein genetic marker (DsRed2), regulated by the immediate early gene 1 (ie1) promoter and homologous region 5 (Hr5) enhancer (Hr5/ie1) from the Autographa californica multicapsid nucleopolyhedrovirus (AcMNPV)32,33,34. OX3097D-Bol also contains a conditional female-specific tetracycline transactivator (tTAV) expression system linked to the fluorescent marker8. The sex-specific splicing module of the Mediterranean fruit fly C. capitata transformer (Cctra) gene was linked to the tTAV gene coding region (Ant et al., 2012). The tTAV gene is under the control of a minimal promoter linked to the tetracycline responsive operator (TetO). In the absence of tetracycline, intracellular levels of tTAV gene products (e.g. mRNA and encoded proteins) accumulate in sufficient quantities that OX3097D-Bol females die at larval or early pupal stages. In the presence of tetracycline (100 μg/ml), tTAV expression is repressed allowing for normal female development8, 31. ### Olive fly rearing Olive fly larvae (OX3097D-Bol and wild-type) were reared in 30mm Petri dishes on an artificial diet as previously described7. For the Pardosa and P. concolor experiments, olive fly larvae were reared in the presence of tetracycline (final concentration 100 μg/ml) added during the rearing process. For the Pardosa study, adults eclosed from larvae reared on tetracycline were then reared without tetracycline and therefore, spiders fed on OX3097D-Bol adult female olive flies should be exposed to tTAV gene products. Adult olive fly lines (OX3097D-Bol and wild-type) were housed in BugDorm cages (Model DP1000B) and reared as previously described7. OX3097D-Bol and wild-type olive flies were reared at 23 °C (±2 °C), 50% (±5%) relative humidity and 12 h:12 h Light/Dark cycle. Olive fly larvae for the A. bilineata study were reared without tetracycline. ### P. concolor rearing P. concolor infested Mediterranean fruit fly (medfly) pupae were supplied by the Department of Agriculture, Food and Environment, University of Pisa, Italy. P. concolor adults emerged into a cage (length 50 cm, width 20 cm, height 25 cm) which housed around 300–400 individuals. Adult P. concolor were fed a honey (Rowse): pollen (100% pure Bee Pollen) mixture (1:1 ratio) and fresh water to drink, both ad libitum, and maintained in a room at 25 °C (±2 °C), uncontrolled humidity 35–50%, with a 16:8 light:dark cycle. Males P. concolor, emerged on average 2 days before females27. To ensure all females had equal opportunity to mate, female P. concolor were selected for experimental cages between 5–12 days following the first male emergence. Prior to the start of the experiment, all females were naïve (no previous experience of egg laying). ### P. concolor study Each treatment cage (3.7 L, see Supplementary Figure S2) contained 30 P. concolor females and 10 P. concolor males. For each of the eight replicates the following cages were set up: 1. 1) Treatment 1: P. concolor presented with 90 OX3097D-Bol third instar olive fly larvae 2. 2) Treatment 2: P. concolor presented with 90 wild type third instar olive fly larvae 3. 3) Control 1: 50 OX3097D-Bol third instar olive fly larvae 4. 4) Control 2: 50 wild type third instar olive fly larvae The control cages contained no parasitoid wasps. OX3097D-Bol and wild-type larvae were handled in the same manner and placed in a nylon exposure net bag which was wrapped around pipe insulation (as seen in Supplementary Figure S2) and hung into the cage without any parasitoids present. This provided a baseline for the survival of olive fly larvae after being handled via this method. Third instar olive fly larvae (9–10 days old) were collected from artificial diet or sand and placed in a single layer inside a nylon fine mesh bag which was then wrapped around pipe insulation and secured in place by elastic bands (Supplementary Figure S2). Parasitoids were exposed to olive fly larvae for oviposition at a 1:3 ratio (parasitoid females: olive fly larvae) for 30 min, after which P. concolor females were encouraged to fly away from the nylon bag by lifting the tubing up a short distance and blowing them off the nylon mesh bag into the cage. The nylon mesh bag, with larvae inside, was then removed from the test cages and the hole was covered with a Petri dish lid. Larvae were then tipped into a Petri dish where a representative subsample (approximately 20%) of the larvae from the test cages were collected. To determine the level of parasitism and check that the exposure period and larvae to parasitoid ratio, was appropriate, these larvae were dissected to count the number of parasitoid eggs within their bodies. The remaining larvae (approximately 80%) were placed onto a Petri dish of fresh diet in a plastic box with sand lining the bottom and allowed to feed or pupate. Pupae were collected 2 days later from the boxes and placed in Petri dishes until emergence. At 25 °C P. concolor were expected to emerge in 17–18 days27. Any pupae not emerged on day 27 post-larval exposure to parasitoids were considered dead. Olive fly were allowed to emerge and die in the Petri dish and counted/sexed once all had eclosed. To record the developmental time of each eclosed parasitoid wasp, P. concolor were collected on the same day they eclosed (from day 17–27) post-parasitoid exposure and stored in 70% ethanol in a 1.5 ml microtube (Eppendorf) at room temperature. All newly eclosed females from each cage were dissected to identify the number of mature eggs in both ovaries. ### Pardosa spider collection and rearing Pardosa spiders were collected in May 2014 from an allotment (compost heap, greenhouse, vegetable patch) in Faringdon (51o 39′ 28″ N, 1o 34′ 57″ W), Oxfordshire, UK. This study did not interfere with any endangered or protected species. Spiders were collected within a 10 m2 area on the same collection day and placed individually into large Drosophila vials (diameter x height = 2 × 6 cm) sealed with a ball of cotton wool. Attention was taken to avoid spiders losing limbs during the process. All captured Pardosa spiders had a mottled brown and black appearance with a beige stripe down their cephalothorax (top body segment) through the black coloration and the characteristic eye pattern of Lycosidae (wolf spiders). Spiders were caught 10 days prior to the start of the experiment as previously described19, acclimated to the laboratory conditions and deprived of food. Spiders were reared in a temperature controlled room at 23 °C (±1 °C), 50% humidity and 12 h:12 h light/dark cycle. Humidity within the vials was recorded to be around 70% using a digital recorder. Spiders were transferred to fresh vials with fresh water and sand weekly and they were fed once every two days on living OX3790D-Bol or wild-type adult olive flies. ### Pardosa study The study was designed to include three groups: 1. 1) Treatment: 30 Pardosa spiders fed OX3097D-Bol olive flies 2. 2) Control 1: 30 Pardosa spiders fed wild-type olive flies 3. 3) Control 2: 10 Starved Pardosa spiders The weight of the spiders was measured prior to the start of the study and spiders were allocated a number so that they could be traced for the duration of the study. Spiders were ranked by their weight and randomly assigned to Treatment, Control 1 and Control 2 groups, as spiders were also allocated to the Control 2 (starved) group, a spider for the starved group was selected every third iteration. Spiders were kept individually throughout the course of the study. Only female adult olive flies were fed to the spiders. Females were individually collected in 1.5 ml microtubes (Eppendorf) and stored on ice to slow down their metabolism. Each spider in the wild-type control and OX3097D-Bol treatment groups received 6 flies per week (2 flies/feeding) following the schedule in Supplementary Figure S2. Briefly, on day −7 spiders carrying an egg sac were anesthetized with 0.2 bar C02, applied for less than 10 sec, and the egg sac carefully removed using tweezers. This method caused no spider mortality. Food uptake was recorded 24 hours post-feeding and uptake was measured as either; 0, 0.5, 1, 1.5 or 2 flies. Any remaining flies were removed at this point and debris was collected to prevent mold growth and further consumption. Notable behavioral changes such as withdrawal of legs under body, uncoordinated movement or curling of the legs were recorded as described elsewhere19. Death was recorded when the spider showed no response to prodding. At the end of the study, all spiders were killed by freezing at −20 °C for 24 h and then stored in 70% ethanol at room temperature. The study was subjected to the validity criteria described previously19 and was deemed acceptable if a mortality rate of ≤6.7% in the wild-type fed control group over two weeks feeding experiment was not exceeded. Mortality rates over 6.7% in the wild-type control group would render the experiment invalid. ### Rove beetle A. bilineata rearing The study was conducted at the SynTech Research laboratories (SynTech Research France SAS, 613 route du Bois de Loyse, 71570 La Chapelle de Guinchay, France). The non-target arthropods used were adults of A. bilineata. Beetles were reared at 20 °C ± 2 °C, relative humidity 60–90% and 16 h:8 h light:dark cycle using onion fly pupae (D. antiqua; Diptera: Anthomyiidae) as hosts for the larvae. OX3097D-Bol or wild-type olive fly larvae or Chironomus species larvae (control) were used as food for the adult rove beetles. Onion fly pupae and the emerging rove beetles were reared in moist standard sandy soil (LUFA 2.1, German BBA-standard test soil; LUFA, D-67346 Speyer) at 35% (±5%) maximum water-holding capacity and remoistening once a week with distilled water. Weight of the sandy soil will be used as an indirect measure of water loss. Onion fly pupae and rove beetles were reared in plastic containers (surface 150 cm2, volume 750 cm3) containing a layer of moist standard sandy soil (minimum volume 600 cm3/test container). Containers were sealed with a lid with an opening covered with fine mesh nylon netting (test units). Onion fly pupae were removed from the moist soil and placed in hatching test units consisting of vessels with a sieve bottom (mesh 2 mm) placed above a second vessel. The pupae were placed onto the sieve and emerging rove beetles fell through the holes and were collected in the vessel below. ### Rove beetle study The study included four groups: 1. 1. Treatment 1: 20 rove beetles fed OX3097D-Bol olive fly larvae at 0.10 g larvae/test unit 2. 2. Control 1: 20 rove beetles fed wild-type olive fly larvae at 0.10 g larvae/test unit 3. 3. Control 2: 20 rove beetles fed Chironomus species larvae at 0.10 g larvae/test unit 4. 4. Control 3: 20 rove beetles fed Chironomus species larvae at 0.10 g larvae/test unit in the presence of toxic reference substance ROGOR PLUS at 440 g active substance/ha. Four replicates were set-up for each treatment group. The study was design to include 28 days exposure phase to assess mortality and 49 days hatching phase to assess fecundity. Beetles were fed 3 times per week with the genetically engineered olive fly or the toxic substance during the exposure phase. The toxic reference substance (ROGOR PLUS, Plant Remedies PVT. Ltd., Patna, India) was included to demonstrate both the susceptibility of the test organism to the substance and the sensitivity of the test system. ROGOR PLUS inhibits acetylcholinesterase, an enzyme essential for the nervous system function. The toxic substance was applied once to the moist standard sandy soil using a laboratory spraying apparatus (DeVries Spray Booth Generation III, propeller application, nozzle Teeject 8002E), calibrated to deliver the exact dose. The dose used (440 g active substance/ha) should result in a minimum reduction of 50% of the reproductive capacity of Chironomus species compared to the untreated control (Control 2). The test units were maintained in control conditions of 20 °C ± 2 °C, relative humidity 60–90% and 16 h:8 h light:dark cycle. Rove beetle mortality was assessed over 28 days (exposure phase). Fecundity was assessed in the next generation (following exposure and during the hatching phase) by measuring the mean number of offspring produced per beetle and calculated for each group. Fecundity effect was expressed as percent reduction of the reproductive performance (mean numbers of offspring per beetle) in the treatment group in comparison to the control group and using the following expression: $$( \% )=100\times -[1-(\text{Rt}/\text{Rc})]$$ where Rt = reproduction (number of offspring/beetle) in the treated group, Rc = reproduction in the control group. The test was subjected to the following validity criteria: the average number of emerging rove beetles from the onion fly pupae in the Control 2 group should not be lower than 400 out of the 6000 (26.7%) flies introduced in the test units or the toxic reference substance in the Control 3 group should not result in a reduction of reproductive capacity ≤50%. Deviation from these criteria would render the study invalid as previously described19. The study was conducted in accordance with the recommendations of the ESCORT workshops35. The method was based on the guidelines proposed by Grimm et al.36, Environmental conditions were monitored throughout the study at regular intervals. The substrate was maintained at its original water content (not more than 50% of the water content evaporated). ### Statistical tests A Fisher’s exact test (two-tailed) was performed on the emergence of parasitized and non-parasitized olive fly larvae. Generalized linear models (Glm’s) with Poisson errors or quasi Poisson errors (where over dispersion occurs) were used to determine whether there was a significant difference between life history parameters of parasitoids reared on OX3097D-Bol olive fly compared to wild-type olive fly. A Glm with Poisson/quasi Poisson errors was required to take into account that the data is count data and therefore have errors that are not normally distributed and variance that is not constant. A Student’s t-test (two-tailed) was carried out on the subsample of larvae that were dissected to identify the number of parasitoid eggs per olive fly larvae. Wilcox rank sum test was used to compare Pardosa’s food consumption. Fisher’s exact test (two-tailed) was performed on Pardosa mortality rates. In the A. bilineata study, a two-tailed Student’s t-test was used for comparison of mortality and fecundity between beetles fed OX3097D-Bol olive fly and wild-type olive fly and between beetles fed Chironomus species in the presence of ROGOR PLUS and a diet of wild-type olive flies. P-values ≤ 0.05 were considered statistically significant. ### Data availability Data and associated protocols described in this study are available to readers upon request from the authors (enca.martin-rendon@oxitec.com). As the study was conducted by Oxitec Ltd. in 2014, biological material is no longer available. Publisher's note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. ## References 1. 1. Daane, K. M. & Johnson, M. W. Olive Fruit Fly: Managing an Ancient Pest in Modern Times. Annu Rev Entomol 55, 151–169, https://doi.org/10.1146/annurev.ento.54.110807.090553 (2010). 2. 2. Vontas, J. et al. Insecticide resistance in Tephritid flies. Pesticide Biochemistry and Physiology 100, 199–205, https://doi.org/10.1016/j.pestbp.2011.04.004 (2011). 3. 3. Economopoulos, A. P. et al. Experiments on Control of Olive Fly, Dacus Oleae (Gmel), by Combined Effect of Insecticides and Releases of Gamma-Ray Sterilized Insects. 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Aspects of the ecology and energetics of the egg sac parasites of the wolf spider Pardosa lugubris (Walckenaer). Oecologia 7, 155–163, https://doi.org/10.1007/BF00346357 (1971). 19. 19. Heimbach, U., Kral, G. & Niemann, P. Implementation of resistance risk analysis of plant protection products in the German authorisation procedure. Bcpc Conference: Pests & Diseases 2000, Vols 1–3, Proceedings 1-3, 771–776 (2000). 20. 20. Walters, K. F. A. N. bees and opportunity costs for conservation. Insect Conserv Diver 9, 375–383, https://doi.org/10.1111/icad.12177 (2016). 21. 21. Bass, C. et al. The evolution of insecticide resistance in the peach potato aphid, Myzus persicae. Insect Biochem Mol Biol 51, 41–51, https://doi.org/10.1016/j.ibmb.2014.05.003 (2014). 22. 22. Chen, Y. et al. Evidence of superclones in Australian cotton aphid Aphis gossypii Glover (Aphididae: Hemiptera). Pest Manag Sci 69, 938–948, https://doi.org/10.1002/ps.3455 (2013). 23. 23. Charaabi, K., Boukhris-Bouhachem, S., Makni, M., Fenton, B. & Denholm, I. Genetic variation in target-site resistance to pyrethroids and pirimicarb in Tunisian populations of the peach potato aphid, Myzus persicae (Sulzer) (Hemiptera: Aphididae). Pest Manag Sci 72, 2313–2320, https://doi.org/10.1002/ps.4276 (2016). 24. 24. Corbel, V., Raymond, M., Chandre, F., Darriet, F. & Hougard, J. M. Efficacy of insecticide mixtures against larvae of Culex quinquefasciatus (Say) (Diptera: Culicidae) resistant to pyrethroids and carbamates. Pest Manag Sci 60, 375–380, https://doi.org/10.1002/ps.809 (2004). 25. 25. Loni, A. Impact of host exposure time on mass-rearing of Psyttalia concolor (Szepligeti) (Hymenoptera: Braconidae) on Ceratitis capitata (Diptera: Tephritidae). Bulletin of Insectology 56, 277–282 (2003). 26. 26. Hegazi, E. & Khafagi, W. The effects of host age and superparasitism by the parasitoid, Microplitis rufiventris on the cellular and humoral immune response of Spodoptera littoralis larvae. J Invertebr Pathol 98, 79–84, https://doi.org/10.1016/j.jip.2008.02.009 (2008). 27. 27. Loni, A. Developmental rate of Opius concolor (Hym: Braconidae) at vaious constant temperatures. Entomophaga 42, 359–366 (1997). 28. 28. Loni, A. & Canale, A. Reproductive success of Psyttalia concolor (Szepligeti) (Hymenoptera: Braconidae) on different hosts. Frustula Entomologica 28-29, 166–171 (2006). 29. 29. Riechert, S. E. & Lawrence, K. Test for predation effects of single versus multiple species of generalist predators: spiders and their insect prey. Entomologia Experimentalis at Applicata 84, 147–155, https://doi.org/10.1046/j.1570-7458.1997.00209.x (1997). 30. 30. Pavely C & M, F. Early Food Safety Evaluation for a Red Fluorescent Protein:DsRed2. Pioneer Hi-Bred International. (2006). 31. 31. Gong, P. et al. A dominant lethal genetic system for autocidal control of the Mediterranean fruitfly. Nat Biotechnol 23, 453–456, https://doi.org/10.1038/nbt1071 (2005). 32. 32. Rodems, S. M. & Friesen, P. D. The hr5 transcriptional enhancer stimulates early expression from the Autographa californica nuclear polyhedrosis virus genome but is not required for virus replication. J Virol 67, 5776–5785 (1993). 33. 33. Guarino, L. A. & Dong, W. Expression of an enhancer-binding protein in insect cells transfected with the Autographa californica nuclear polyhedrosis virus IE1 gene. J Virol 65, 3676–3680 (1991). 34. 34. Guarino, L. A. & Summers, M. D. Interspersed Homologous DNA of Autographa californica Nuclear Polyhedrosis Virus Enhances Delayed-Early Gene Expression. J Virol 60, 215–223 (1986). 35. 35. Candolfi, M. P. & Blümel, S. Guidelines to Evaluate Side-effects of Plant Protection Products to Non-target Arthropods: IOBC, BART and EPPO Joint Initiative. 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(ed Candolfi) 1–12 (IOBC/WPRS, 2000). ## Author information ### Author notes • Clare Cassidy Present address: Unilever, 3 St. James Rd, Kingston upon Thames, KT1 2BA, UK • Camilla Beech Present address: Cambea Consulting Ltd., 10 Beech Court, Wokingham Road, Hurst, Berkshire, RG10 0RQ, UK ### Affiliations 1. #### Oxitec Ltd., 71 Innovation Drive, Abingdon, Oxfordshire, OX14 4RX, United Kingdom • Thea Marubbi • , Clare Cassidy • , Esther Miller • , Martha Koukidou • , Enca Martin-Rendon • , Simon Warner • & Camilla Beech 2. #### Department of Agriculture, Food and Environment, University of Pisa, Via del Borghetto 80, 56124, Pisa, Italy • Augusto Loni ### Contributions Thea Marubbi: Conducted experiments, analyzed data, reviewed final draft of the manuscript. Clare Cassidy: Conducted experiments, analyzed data, reviewed final draft of the manuscript. Esther Miller: Designed and conducted experiments, reviewed the final draft of the manuscript. Martha Koukidou: Designed and supervised experiments, analyzed data, contributed to the writing of the manuscript and reviewed the final draft of the manuscript. Enca Martin-Rendon: Analyzed data and wrote the manuscript. Simon Warner: Analyzed data and wrote the manuscript. Augusto Loni: Provided essential reagents for the project, designed experiments, analyzed data, contributed to the writing of the manuscript and reviewed final draft of the manuscript. Camilla Beech: Designed and supervised experiments, analyzed data, contributed to the writing of the manuscript and reviewed final draft of the manuscript. ### Competing Interests This study was funded by Oxitec. Ltd. T.M., M.K., E.M.R. and S.W. are employed by Oxitec Ltd. C.C., E.M. and C.B. areformer Oxitec employees. Oxitec is a wholly owned subsidiary of Intrexon. E.M. and C.B. hold shares in Intrexon.Syntech Research (France) was contracted by Oxitec Ltd to conduct the A. bilineata study. A.L. have no competing financial interest to declare. ### Corresponding author Correspondence to Enca Martin-Rendon.	{}
## Soft and hard classification by reproducing kernel Hilbert space methods.(English)Zbl 1106.62338 Summary: Reproducing kernel Hilbert space (RKHS) methods provide a unified context for solving a wide variety of statistical modelling and function estimation problems. We consider two such problems: We are given a training set $$\{y_i,t_i$$, $$i=1,\dots,n\}$$, where $$y_i$$ is the response for the $$i$$th subject, and $$t_i$$ is a vector of attributes for this subject. The value of $$y_i$$ is a label that indicates which category it came from. For the first problem, we wish to build a model from the training set that assigns to each $$t$$ in an attribute domain of interest an estimate of the probability $$p_j(t)$$ that a (future) subject with attribute vector $$t$$ is in category $$j$$. The second problem is in some sense less ambitious; it is to build a model that assigns to each $$t$$ a label, which classifies a future subject with that $$t$$ into one of the categories or possibly “none of the above”. The approach to the first of these two problems discussed here is a special case of what is known as penalized likelihood estimation. The approach to the second problem is known as the support vector machine. We also note some alternate but closely related approaches to the second problem. These approaches are all obtained as solutions to optimization problems in RKHS. Many other problems, in particular the solution of ill-posed inverse problems, can be obtained as solutions to optimization problems in RKHS and are mentioned in passing. We caution the reader that although a large literature exists in all of these topics, in this inaugural article we are selectively highlighting work of the author, former students, and other collaborators. ### MSC: 62H30 Classification and discrimination; cluster analysis (statistical aspects) 46N30 Applications of functional analysis in probability theory and statistics gss Full Text: ### References: [1] ADV COMPUT MATH 13 pp 1– (2000) · Zbl 0939.68098 [2] 33 pp 82– (1971) · Zbl 0201.39702 [3] TRANS AM MATH SOC 68 pp 337– (1950) [4] 92 pp 107– (1997) [5] 81 pp 96– (1986) [6] ANN STATIST 23 pp 1865– (1995) · Zbl 0854.62042 [7] Klein, Archives of Ophthalmology 102 (4) pp 520– (1984) [8] COMMUN STAT SIMUL COMPUT 26 pp 765– (1997) [9] 1 pp 169– (1992) [10] STAT SIN 6 pp 675– (1996) [11] ANN STATIST 18 pp 1676– (1990) · Zbl 0719.62051 [12] ANN STATIST 28 pp 734– (2000) · Zbl 1105.62329 [13] DATA MINING KNOWL DISCOV 6 pp 259– (2002) · Zbl 05660804 [14] PNAS 98 (26) pp 15149– (2001) [15] MACH LEARN 46 pp 191– (2002) · Zbl 0998.68103 [16] NUMER MATH 31 pp 377– (1979) [17] Technometrics 21 pp 215– (1979) [18] ANN STATIST 14 pp 1101– (1986) · Zbl 0629.62043 [19] ANN STATIST 13 pp 970– (1985) · Zbl 0585.62074 [20] COMMUN STAT SIMUL 18 pp 1059– (1989) · Zbl 0695.62113 [21] NUMER MATH 56 pp 1– (1989) · Zbl 0665.65010 [22] MACH LEARN 46 pp 131– (2002) · Zbl 0998.68101 [23] CONSTR APPROXIMATION 2 pp 11– (1986) · Zbl 0625.41005 [24] MON WEATHER REV 123 pp 3358– (1995) [25] 79 pp 832– (1984) [26] J COMPUT PHYS 59 pp 441– (1985) · Zbl 0626.65053 [27] SIAM J NUMER ANAL 14 pp 651– (1977) · Zbl 0402.65032 This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	{}

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