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# Descriptive Set Theory by Y. N. Moschovakis

By Y. N. Moschovakis

Read Online or Download Descriptive Set Theory PDF

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Prove that if f : R 'X is continuous and F is a closed set of reals, then f [ F ] is 2 : . Hint. R is a countable union of compact sets. -I Practically every specific pointset which comes up in the usual constructions of analysis and topology is easily shown to be projective-in fact, almost always, it is 2; or n:. We only mention a couple of simple examples here, since we will meet several interesting projective pointsets later on. 8. On the space C[O, 11 of continuous real functions on the unit interval, put Q ( f )w f is differentiable on [0, 11, Rcf)H f is continuously differentiable on [0, 11, where a t the endpoints we naturally take the one-sided derivatives.

We refer to any such g as a Borel inverse of f. It will turn out that every Borel injection is a good Borel injection. This is a special case of a fairly difficult theorem which we will prove in 2E and again in Chapter 4. Here we only need show that enough good Borel injections exist. Notice that if f : X y is a good Borel injection, then - Y f m1- f(dY)) = Y with g any Borel inverse of f, so f[X] is a Borel set.

4. 3 Clearly T and T are Zz by the closure properties of this pointclass. T” is X:+l. 1and since i Q is also is Z;,, by two applications of closure of this pointclass under 3”. This kind of computation is so simple that we will not usually bother to put it down. 4 above which shows step-by-step the properties of the relevant pointclasses that we use. 3. Let f :63 4(R be a continuous function on the line. Prove that the relations Y) f’(x> = Y, m, @ Q(x) e f ’ ( x )exists are both @. Hint. Let r,, r I ,...

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# Linear function maybe?

Algebra Level 2

Given the $$f(x)$$ is a linear function such that this function is satisfy for all real $$x$$.

If $$f(2016)-f(2004)=5$$, find the value of $$f(2012)-f(1184)$$.

×

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# Kerbal Space Program Wiki:Migration problems

After the recent update of the MediaWiki software to 1.24.2 some problems have been introduced and not addressed yet but please don't annoy Squad employees.

## Current status

Apparently these issues are due to the fact that the wiki currently resides on a temporary server. Unfortunately the person employed by Squad (aLeXmOrA) left over a year ago and thus it's unknown when these issues will be fixed. It was originally planned to move the wiki back onto a permanent server and with that move these issues here should be resolved. As soon as more information comes available this section will be updated.

## Issues

### File namespace

There were previously several problems with the File namespace: uploading and moving files was impossible, and images would fail to display except in resolutions that had been used previously. These problems have been mostly fixed, however, the Mediawiki software still fails to convert .svg images into .png thumbnails, displaying the following error:

Original source: `[[File:Image_needed.svg|91px]]`

The ability to upload files is working for new files just fine, but if you try to upload a new version of a file, it gives the following error:

• Could not read or write file "mwstore://local-backend/local-public/5/5d/TinyKerbin.png" due to insufficient permissions or missing directories/containers.
• Could not store file "/tmp/phpg9bLeI" at "mwstore://local-backend/local-public/5/5d/TinyKerbin.png".

I tested around with this and noticed that if you are the uploader of the first image it works just fine. It only gives the above error when youre not the original uploader.

### Math tag

Previously a [itex]-tag could be used to nicely print formulas using TeX but since the update an error appears: ${\displaystyle y=x^{3}}$ Original source: `[itex]y = x^3[/itex]`

Update 04-17-2018: Math tags are working again.

### Job queue

There seems to be an issue with the job queue not running.

The number of jobs doesn't seem to be counting down as it should be. Subtle things seem to not catch up properly, such as category membership and "what links here". It's not something that will be immediately noticed in terms of viewing the site, as it doesn't block every category update, or every update of "what links here", just things which only ever occur as a background task and will never get triggered via other mechanisms. It is stuck at `"jobs": 5662` at the time of posting this issue, and has not changed for a good 5–10 minutes, where the expected behaviour is that it should be steadily counting downwards (except when an edit causes new tasks to be added to the queue). It's not an issue of the API page getting cached, as "views" is correctly updating with every reload of that URL.

### Vector extension

According to Special:Version, the discontinued Vector extension is still installed. There are not any obvious specific issues caused by this, but it is an invalid configuration for MediaWiki 1.24.2, and could easily be responsible for odd behaviour somewhere. A fresh install of 1.24.2 does not include it, and the official docs say that it should have been uninstalled. mw:MediaWiki 1.22#Upgrades to Vector and other skins:

If you were previously using the Vector extension, you must uninstall it (the extension, not the skin) before upgrading to 1.22.

Prior to uninstalling the discontinued extension, it would be prudent to do a quick check to confirm that the installed Vector skin is actually fully up to date with the version supplied as part of 1.24.2 (which includes all of the functionality previously supplied by the extension, either in the MW core or in the skin).

The ExpandTemplates extension should also be uninstalled, as it was merged into the MediaWiki core in 1.23. This is less likely to be responsible for visible problems, as it is just a diagnostic/development tool; but bad practice all the same to keep the outdated extension around after it has been merged into the core, and may break something.


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This topic is 2570 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

My game has classes called chunks.
Each chunk has 2 16x16x16 unsigned char arrays called position and data.

I have two functions below a read and write.

Somehow between them the data becomes corrupt.
Is "(char)Chunk->position[j][k];" truncating my data?
The data array contains values from 0 to 255.

When it converts them from unsigned chars to chars is it losing the data? (even though thier converted back " newChunk->data[j][k] = (unsigned char)*pLoc;" )

After the readChunk routine runs I can look at the data[] and see that all of the values are 205.

Someone please point out my mistake.

Thanks,
Walker

 bool file::writeChunk(chunk* Chunk) { if (Chunk->requiresWriting == false) { return false; } std::stringstream name; std::string temp; signed int x = Chunk->absolutePos.x; signed int y = Chunk->absolutePos.y; signed int z = Chunk->absolutePos.z; ofstream fileStream; ////Open File to save in name.str(std::string()); name << "save/" << world0->name << "x" << x << "x" << y << "x" << z << ".txt"; temp = name.str(); fileStream.open(temp.c_str(), ios::out|ios::binary); if (fileStream.is_open()) { ////Make the 2 pointers char *pStoreMemory; char *pLoc; ////Allocate memory pStoreMemory = new char [16*16*16*2]; pLoc = pStoreMemory; ////Write needed data to memory for (int i = 0; i < 16; i++) { for (int j = 0; j < 16; j++) { for (int k = 0; k < 16; k++) { *pLoc = (char)Chunk->position[j][k]; pLoc++; } } } for (int i = 0; i < 16; i++) { for (int j = 0; j < 16; j++) { for (int k = 0; k < 16; k++) { *pLoc = (char)Chunk->data[j][k]; pLoc++; } } } ////Copy memory to file fileStream.write(pStoreMemory,16*16*16*2); fileStream.close(); delete[] pStoreMemory; //cout << "Done!\n"; } else { //cout << "/!\\ Cannot open " << temp.c_str() << "\n"; } delete Chunk; return true; } chunk* file::readChunk(signed int x,signed int y,signed int z) { std::stringstream name; std::string temp; chunk* newChunk; ifstream fileStreamRead; ////Open File to load from name.str(std::string()); name << "save/" << world0->name << "x" << x << "x" << y << "x" << z << ".txt"; temp = name.str(); fileStreamRead.open(temp.c_str(), ios::in|ios::binary); if (fileStreamRead.is_open()) { iVertex3d position; position.x = x; position.y = y; position.z = z; newChunk = new chunk(position); ////Make the 2 pointers char *pStoreMemory; char *pLoc; ////Allocate memory pStoreMemory = new char [16*16*16*2]; pLoc = pStoreMemory; ////Copy memory to file fileStreamRead.read(pStoreMemory,16*16*16*2); ////read needed data from memory //if ((unsigned char)*pLoc != 48) { //newChunk->empty = false; for (int i = 0; i < 16; i++) { for (int j = 0; j < 16; j++) { for (int k = 0; k < 16; k++) { newChunk->position[j][k] = (unsigned char)*pLoc; // /!\ Add to watch Queue if block has potential to move switch (*pLoc) { case steam: case smoke: case water1: case water2: case water3: case water4: case water5: case lava1: case lava2: case lava3: case lava4: case lava5: case oil1: case oil2: case oil3: case oil4: case oil5: case fire: newChunk->liquidBufferAdd(i,j,k); break; } pLoc++; } } } } for (int i = 0; i < 16; i++) { for (int j = 0; j < 16; j++) { for (int k = 0; k < 16; k++) { newChunk->data[j][k] = (unsigned char)*pLoc; } } } fileStreamRead.close(); delete[] pStoreMemory; return newChunk; } else { //cout << "/!\\ Cannot open " << temp.c_str() << "\n"; iVertex3d position; position.x = x; position.y = y; position.z = z; newChunk = new chunk(position); generateChunk(newChunk); return newChunk; } }

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You are forgetting to increase pLoc in the second 3d loop in readChunk().

Of course, if position and data are 3d (unsigned) char arrays, why not just write (char *)position and (char *)data rather than copying them to another array?

##### Share on other sites
Thanks a ton!!!

(I just knew it would be the unsigned char to char problem)

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Article

# CLASH: New Multiple-Images Constraining the Inner Mass Profile of MACS J1206.2-0847

The Astrophysical Journal (Impact Factor: 6.73). 07/2011; DOI:10.1088/0004-637X/749/2/97
Source: arXiv

ABSTRACT We present a strong-lensing analysis of the galaxy cluster MACS J1206.2-0847
($z$=0.44) using UV, Optical, and IR, HST/ACS/WFC3 data taken as part of the
CLASH multi-cycle treasury program, with VLT/VIMOS spectroscopy for some of the
multiply-lensed arcs. The CLASH observations, combined with our mass-model,
allow us to identify 47 new multiply-lensed images of 12 distant sources. These
images, along with the previously known arc, span the redshift range $1\la z\la5.5$, and thus enable us to derive a detailed mass distribution and to
accurately constrain, for the first time, the inner mass-profile of this
cluster. We find an inner profile slope of $d\log \Sigma/d\log \theta\simeq -0.55\pm 0.1$ (in the range [1\arcsec, 53\arcsec], or $5\la r \la300$ kpc), as
commonly found for relaxed and well-concentrated clusters. Using the many
systems uncovered here we derive credible critical curves and Einstein radii
for different source redshifts. For a source at $z_{s}\simeq2.5$, the critical
curve encloses a large area with an effective Einstein radius of
$\theta_{E}=28\pm3\arcsec$, and a projected mass of $1.34\pm0.15\times10^{14} M_{\odot}$. From the current understanding of structure formation in
concordance cosmology, these values are relatively high for clusters at
$z\sim0.5$, so that detailed studies of the inner mass distribution of clusters
such as MACS J1206.2-0847 can provide stringent tests of the $\Lambda$CDM

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text: A complete group is a centerless group that has only inner automorphisms ($\textrm{Aut}(G) = \textrm{Inn}(G)$). Equivalently a centerless group is complete iff $\textrm{Aut}(G)$ is isomorphic to $G$. To be used with the tag [group-theory].


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# Volcano Watch — Unusual events of March-April 1868

Release Date:

On Friday, March 27, 1868, at 5:30 a.m., several whaling ships anchored in Kawaihae Harbor noticed a dense column of fume reflected by a bright light southwest of the summit of Mauna Loa.

On Friday, March 27, 1868, at 5:30 a.m., several whaling ships anchored in Kawaihae Harbor noticed a dense column of fume reflected by a bright light southwest of the summit of Mauna Loa. An eruption near Mokuaweoweo had taken place, lasting several hours before subsiding. Pele's hair had drifted down upon the residents of Kau and South Kona, indicating the presence of lava fountains above.

In near synchrony with its larger neighbor, Kīlauea Volcano began to shake at 10:00 a.m. the next day with a series of earthquakes that increased in intensity for several days. Cracks appeared around the summit of Kīlauea; the level of the lava lake in Halemaumau fluctuated rapidly. Stone walls collapsed, houses shook, and trees vibrated.

The largest historic earthquake in the Hawaiian Islands happened at 3:40 p.m. on April 2. The earthquake had an estimated magnitude of 7.9. Clocks stopped on Oahu. At Kohala, the shock was so strong that it stopped all of the engines at the Kohala Mill, including the 75-horsepower main engine. All over the Island of Hawaii, the effects were felt as a large portion of the coastline from Honuapo to Kapoho subsided 1.2 to 2.1 meters (4 to 7 feet).

Coastal villages were inundated by a huge tsunami, at places over 15 meters (50 feet) high. The coastal villages of Apua, Keauhou, Punaluu, Ninole, Kawaa, Honuapo and Kaalualu were destroyed. There were five large waves in succession. The largest came first. A total of 75 people and numerous animals were swept out to sea and drowned.

Along with the coastal subsidence and tsunami, the earth at Kiolakaa, near Wood Valley, broke loose and slid 300 meters (1,000 feet) down from the summit and southeast side of the hill onto the valley below, covering houses and trees. This event, known as the mud flow, buried people, horses, cattle, goats and sheep under a thick layer of mud. The mud flow was 5 km (3 miles) long, as wide as 1.5 km (1 mile), and varied in thickness from 1 meter (3 feet) to 27 meters (90 feet).

All stone walls and dwellings in Kau were flattened by the earthquake; people and animals were thrown down, and ground cracks appeared throughout Kīlauea and in the district of Kau. Landslides occurred island-wide, and plantation chimneys in the Hilo area fell down. Aftershocks almost as violent as the April 2 event hit at 12:20 a.m. and again at 12.45 a.m. on April 4.

At about 5:00 p.m. on April 7, a great crack opened along Mauna Loa's southwest rift zone above the Captain Robert Brown ranch in Kahuku. Lava gushed from the earth and flowed directly toward the ranch house. Captain Brown, his wife, and nine children ran for their lives as the molten flood engulfed their home. Within three hours, the flow reached the sea, a distance of 17 kilometers (10 miles) from the vent.

### Volcano Activity Update

Eruptive activity of Kīlauea Volcano continued unabated during the past week. Lava is erupting from Puu Oo and flowing through a network of tubes from the vent to the sea. Lava is entering the ocean near Kamokuna and enlarging the bench. Occasional phreatic activity was observed at the entry. The public is reminded that the ocean-entry area is extremely hazardous, with explosions accompanying unpredictable collapses of the new land. The steam clouds are highly acidic and laced with glass particles.

There were no earthquakes reported felt during the week ending on July 22, 1999.


text:
# associator

## Primary tabs

\documentclass{article}
% this is the default PlanetMath preamble.  as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.

% almost certainly you want these
\usepackage{amssymb,amscd}
\usepackage{amsmath}
\usepackage{amsfonts}

% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
%\usepackage{amsthm}
% making logically defined graphics
%%%\usepackage{xypic}

% there are many more packages, add them here as you need them

% define commands here
\begin{document}
Let $A$ be a non-associative algebra over a field.  The \emph{associator} of $A$, denoted by $[\ , , ]$, is a \PMlinkname{trilinear}{multilinear} map from $A\times A\times A$ to $A$ given by:
$$[\ a,b,c\ ]=(ab)c-a(bc).$$
Just as the commutator measures how close an algebra is to being commutative, the associator measures how close it is to being associative.  $[\ , , ]=0$ identically iff $A$ is associative.


text:
# Complex conjugates of functions

brmath
I am looking at a complex function f(z) and want to know something about f(z)f*(1/z*). We could assume for now that f(z) is analytic or at least meromorphic. Are there any identities involving this product? Is there any way to decompose the f*(1/z*) into a function of f?

Mandelbroth
I am looking at a complex function f(z) and want to know something about f(z)f*(1/z*). We could assume for now that f(z) is analytic or at least meromorphic. Are there any identities involving this product? Is there any way to decompose the f*(1/z*) into a function of f?
Caveat: You mean "a complex function ##f##." ##f(z)## is the image (or "output") of ##z## under the function ##f##. This is a common abuse of notation.

Suppose ##f(x+iy)= u(x,y)+iv(x,y)##, where ##u## and ##v## are real-valued functions. Then, we have that ##\bar{f}(\frac{1}{x-iy})=\bar{f}(\frac{x}{x^2+y^2}+i\frac{y}{x^2+y^2})=u(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})-iv(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})##.

Thus, we have ##f(z)\bar{f}(\bar{z}^{-1})=u(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})u(x,y)+v(\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2})v(x,y)+i (u(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}) v(x,y)-v(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}) u(x,y))##.

brmath
Hi, I'll try not to write f(z) when I mean f. I already know that 1/z* = z/|z|. I had hoped there are some standard indentities or inequalities involving f(z)f*(1/z*).


text:
THE CLASS OF MODULES WITH PROJECTIVE COVER

Title & Authors
THE CLASS OF MODULES WITH PROJECTIVE COVER
Guo, Yong-Hua;

Abstract
Let R be a ring. A right R-module M is called perfect if M possesses a projective cover. In this paper, we consider the relationship between the class of perfect modules and other classes of modules. Some known rings are characterized by these relationships.
Keywords
projective cover;$\small{C_p}$-hereditary ring;perfect ring;small submodule;
Language
English
Cited by
1.
Rings Whose Nonsingular Modules Have Projective Covers, Ukrainian Mathematical Journal, 2016, 68, 1, 1
2.
ON A SPECIAL CLASS OF EXACT SEQUENCES, Journal of Algebra and Its Applications, 2011, 10, 05, 915
3.
EXISTENCE OF ALMOST SPLIT SEQUENCES VIA REGULAR SEQUENCES, Bulletin of the Australian Mathematical Society, 2013, 88, 02, 218
References
1.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, Second edition. Graduate Texts in Mathematics, 13. Springer-Verlag, New York, 1992.

2.
G. Azumaya and A. Facchini, Rings of pure global dimension zero and Mittag-Leffler modules, J. Pure Appl. Algebra 62 (1989), no. 2, 109-122.

3.
H. Bass, Finitistic dimension and a homological generalization of semi-primary rings, Trans. Amer. Math. Soc. 95 (1960), 466-488.

4.
H. Cartan and S. Eilenberg, Homological Algebra, Princeton Univ. Press, Princeton, 1956.

5.
J. Clark, C. Lomp, N. Vanaja, and R. Wisbauer, Lifting Modules, Supplements and projectivity in module theory. Frontiers in Mathematics. Birkhauser Verlag, Basel, 2006.

6.
D. J. Fieldhouse, Characterizations of modules, Canad. J. Math. 23 (1971), 608-610.

7.
T. H. Fay and S. V. Joubert, Relative injectivity, Chinese J. Math. 22 (1994), no. 1, 65-94.

8.
K. R. Goodeal, Ring Theory: Nonsingular Rings and Modules, Pure and Applied Math-ematics, No. 33. Marcel Dekker, Inc., New York-Basel, 1976.

9.
P. A. Guil Asensio and I. Herzog, Sigma-cotorsion rings, Adv. Math. 191 (2005), no. 1, 11-28.

10.
S.-T. Hu, Introduction to Homological Algebra, Holden-Day, Inc., San Francisco, Calif.- London-Amsterdam 1968.

11.
T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics, 131. Springer-Verlag, New York, 1991.

12.
T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics, 189. Springer- Verlag, New York, 1999.

13.
L. X. Mao and N. Q. Ding, Notes on cotorsion modules, Comm. Algebra 33 (2005), no.1, 349-360.

14.
J. J. Rotman, An Introduction to Homological Algebra, Pure and Applied Mathematics, 85. Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], New York-London, 1979.

15.
F. L. Sandomierski, Homological dimensions under change of rings, Math. Z. 130(1973), 55-65.

16.
F. L. Sandomierski, On semiperfect and perfect ring, Proc. Amer. Math. Soc. 21 (1969), no. 1, 205-207.

17.
L. Shen and J. L. Chen, New characterizations of quasi-Frobenius rings, Comm. Algebra 34 (2006), no. 6, 2157-2165.

18.
U. Shukla, On the projective cover of a module and related results, Pacific J. Math. 12 (1962), 709-717.

19.
J. Xu, Flat Covers of Modules, Lecture Notes in Mathematics, 1634. Springer-Verlag, Berlin, 1996.

20.
D. Zhou, Rings characterized by a class of modules, Comm. Algebra 33 (2005), no. 9, 2941-2955.


text:
Optimize this computation of Liapunov exponent

Introduction

In dynamical systems there is an important exponent called Liapunov exponent. I'm trying to create a visual representation of Liapunov exponent for one dimension finite-difference equations. For example, the most common equation is the logistic map: $$(1)\quad x_{n + 1} = rx_{n}(1 - x_{n})$$, where $$x_{n}$$ is the $$n$$-th iteration of (1), we define $$x_0 = x$$, $$n$$ goes from 0 to $$\infty$$, and $$r$$ goes from 0 to 4. Equation (1) is discrete, but we can consider the continuous function $$(2)\quad f(x) = rx(1 - x)$$ in order to do all the computations (I will not prove nothing here), then $$(3)\quad f'(x) = r(1 - 2x)$$.

Finally, Liapunov exponent is

$$$$(4) \hspace{1em} \lambda(r) = \lim_{n \to \infty} \left\{\frac{1}{n} \sum_{i = 0}^{n - 1} \ln \left|f'(x_i)\right|\right\}$$$$

and we can define any other $$f(x)$$ depending the finite-difference equations. For (1) the graph is:

The code

I want a good approximations of this exponent and the following code plots is that way, but I want a smooth graph and it takes a lot of time.

ClearAll["Global*"]

ITERATIONS = 100; (* Fidelity *)
F[x_, r_] := r*x*(1 - x); (* Logistic map *)
f[x_, r_] := r*(1 - 2*x); (* Derivate *)
λ[r_, u_] := (1/ITERATIONS)*
Sum[Log[Abs[f[Nest[F[#, r] &, u, i], r]]], {i, 0,
ITERATIONS - 1}]; (* Liapunov exponent approximation *)

Plot[
Legended[
Style[λ[r, RandomReal[]], Black, Thickness[0.001]],
Style["Fidelity = 100", FontSize -> 20]
],
{r, 3 , 4},
PlotTheme -> "Scientific",
FrameStyle -> Directive[Thickness[0.003], FontSize -> 20, Black],
ImageSize -> Large,
AspectRatio -> 1/2,
PlotRange -> {{3, 4}, {-1, 1}},
Axes -> True,
AxesStyle -> Directive[Thickness[0.002], Blue]
]

I know this problem can be solved with a "manual Nest" because Nest itself takes a lot of time with large number of iterations. How can I improve this code? How can I use a function like a counter, but, a Nest counter?

You are using symbolic Nest which is extremely slow. Anyway, it is more efficient to use RecurrenceTable. The plot below generates in less than 2 seconds with 20000 iterations.

ClearAll["Global*"]

ITERATIONS = 20000; (*Fidelity*)
F[x_, r_] := r*x*(1. - x); (*Logistic map*)
f[x_, r_] := r*(1 - 2*x);
xlist[x0_?NumericQ, r_] := RecurrenceTable[{x[n + 1] == r*x[n]*(1 - x[n]), x[0] == x0},
x, {n, 0, ITERATIONS - 1}]
lambda[r_, xx_?NumericQ] := Mean[xlist[xx, r] // f[#, r] & // Abs // Log]

Plot[lambda[r, RandomReal[]], {r, 3, 4}]

You can superimpose with the original plot (see 18767):

img = ImageCrop@Import["https://i.stack.imgur.com/dBrkG.png"]
img = ImageTrim[img, {{60, 65}, {475, 420}}]
Plot[lambda[r, RandomReal[]], {r, 3, 4}, AspectRatio -> 1/2,
ImageSize -> 1000, PlotRange -> {{3, 4}, {-1, 1}},
PlotStyle -> Thickness[0.005],
Prolog -> {Texture[img],
Polygon[{Scaled[{0, 0}], Scaled[{1, 0}], Scaled[{1, 1}],
Scaled[{0, 1}]},
VertexTextureCoordinates -> {{0, 0}, {1, 0}, {1, 1}, {0, 1}}]}]

• I had never seen before the notation ?NumericQ, can you explain it to me? I found this Reference – tajiri_numero_1 Oct 30 '20 at 17:57
• It's a Pattern, it means that the function is only defined if NumericQ returns True for the argument (= the argument is numerical value). Try for instance f[x_NumericQ] := x^2 and f[1] then f[abc]. The latter returns unevaluated. – anderstood Oct 30 '20 at 18:11
• @tajiri_numero_1 Is it correct that the system is stable iff $\lambda <0$? $r$ would be a parameter of a dynamical system (such as mass or length)? – anderstood Oct 30 '20 at 18:15
• Yes, if $\lambda < 0$ represents a stable behavior, $\lambda > 0$ represents a chaotic behavior and $\lambda = 0$ is a marginal value. $r$ in Logistic map could represent a lot of different things and it is given by the problem. In biology, for example, $r$ could represent a parameter of birth and dead of a population. – tajiri_numero_1 Oct 30 '20 at 18:28


text:
# PyLops¶

This Python library is inspired by the MATLAB Spot – A Linear-Operator Toolbox project.

Linear operators and inverse problems are at the core of many of the most used algorithms in signal processing, image processing, and remote sensing. When dealing with small-scale problems, the Python numerical scientific libraries numpy and scipy allow to perform most of the underlying matrix operations (e.g., computation of matrix-vector products and manipulation of matrices) in a simple and expressive way.

Many useful operators, however, do not lend themselves to an explicit matrix representation when used to solve large-scale problems. PyLops operators, on the other hand, still represent a matrix and can be treated in a similar way, but do not rely on the explicit creation of a dense (or sparse) matrix itself. Conversely, the forward and adjoint operators are represented by small pieces of codes that mimic the effect of the matrix on a vector or another matrix.

Luckily, many iterative methods (e.g. cg, lsqr) do not need to know the individual entries of a matrix to solve a linear system. Such solvers only require the computation of forward and adjoint matrix-vector products as done for any of the PyLops operators.

Here is a simple example showing how a dense first-order first derivative operator can be created, applied and inverted using numpy/scipy commands:

import numpy as np
from scipy.linalg import lstsq

nx = 7
x = np.arange(nx) - (nx-1)/2

D = np.diag(0.5*np.ones(nx-1), k=1) - \
np.diag(0.5*np.ones(nx-1), k=-1)
D[0] = D[-1] = 0 # take away edge effects

# y = Dx
y = np.dot(D, x)
# x = D'y
# xinv = D^-1 y
xinv = lstsq(D, y)[0]

and similarly using PyLops commands:

from pylops import FirstDerivative

nx = 7
x = range(-(nx // 2), nx // 2 + (1 if nx % 2 else 0))

Dlop = FirstDerivative(nx, dtype='float64')

# y = Dx
y = Dlop*x
# x = D'y
# xinv = D^-1 y
xinv = Dlop / y

Note how this second approach does not require creating a dense matrix, reducing both the memory load and the computational cost of applying a derivative to an input vector $$\mathbf{x}$$. Moreover, the code becomes even more compact and espressive than in the previous case letting the user focus on the formulation of equations of the forward problem to be solved by inversion.

## Terminology¶

A common terminology is used within the entire documentation of PyLops. Every linear operator and its application to a model will be referred to as forward model (or operation)

$\mathbf{y} = \mathbf{A} \mathbf{x}$

while its application to a data is referred to as adjoint modelling (or operation)

$\mathbf{x} = \mathbf{A}^H \mathbf{y}$

where $$\mathbf{x}$$ is called model and $$\mathbf{y}$$ is called data. The operator $$\mathbf{A}:\mathbb{F}^m \to \mathbb{F}^n$$ effectively maps a vector of size $$m$$ in the model space to a vector of size $$n$$ in the data space, conversely the adjoint operator $$\mathbf{A}^H:\mathbb{F}^n \to \mathbb{F}^m$$ maps a vector of size $$n$$ in the data space to a vector of size $$m$$ in the model space. As linear operators mimics the effect a matrix on a vector we can also loosely refer to $$m$$ as the number of columns and $$n$$ as the number of rows of the operator.

Ultimately, solving an inverse problems accounts to removing the effect of $$\mathbf{A}$$ from the data $$\mathbf{y}$$ to retrieve the model $$\mathbf{x}$$.

For a more detailed description of the concepts of linear operators, adjoints and inverse problems in general, you can head over to one of Jon Claerbout’s books such as Basic Earth Imaging.

## Implementation¶

PyLops is build on top of the scipy class scipy.sparse.linalg.LinearOperator.

This class allows in fact for the creation of objects (or interfaces) for matrix-vector and matrix-matrix products that can ultimately be used to solve any inverse problem of the form $$\mathbf{y}=\mathbf{A}\mathbf{x}$$.

As explained in the scipy LinearOperator official documentation, to construct a scipy.sparse.linalg.LinearOperator, a user is required to pass appropriate callables to the constructor of this class, or subclass it. More specifically one of the methods _matvec and _matmat must be implemented for the forward operator and one of the methods _rmatvec or _adjoint may be implemented to apply the Hermitian adjoint. The attributes/properties shape (pair of integers) and dtype (may be None) must also be provided during __init__ of this class.

Any linear operator developed within the PyLops library follows this philosophy. As explained more in details in Implementing new operators section, a linear operator is created by subclassing the scipy.sparse.linalg.LinearOperator class and _matvec and _rmatvec are implemented.

## History¶

PyLops was initially written by Equinor It is a flexible and scalable python library for large-scale optimization with linear operators that can be tailored to our needs, and as contribution to the free software community. Since June 2021, PyLops is a NUMFOCUS Affiliated Project.


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# In a group of $950$ persons, $750$ can speak Hindi and $460$ can speak English. Find how many can speak both Hindi and English.

Last updated date: 20th Mar 2023
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Hint: This question is based on the concept of set theory. Try to recall all the formula of set theory which relate the union and intersection of the two sets. In this question, one set is of those persons who can speak Hindi and the other set is of those persons who can speak English.

Before proceeding with this question, we must know the formula that is required to solve this question. Let us consider two sets, set $A$ and set $B$. In set theory, we have a formula that relates the number of terms in the intersection and the union of the two sets $A$ and $B$. The formula is,
$n\left( A\cup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\cap B \right)...............\left( 1 \right)$
For this question, let us denote set $A$ for those persons who can speak Hindi and let us denote set $B$ for those persons who can speak English.
It is given that there are a total $950$ persons in the group. Hence, we can say,
$n\left( A\cup B \right)=950..........\left( 2 \right)$
It is given that $750$ persons can speak Hindi. Hence, we can say,
$n\left( A \right)=750.............\left( 3 \right)$
Also, it is given that $460$ persons can speak English. Hence, we can say,
$n\left( B \right)=460.............\left( 4 \right)$
In the question, we have to find the number of persons that can speak both Hindi and English. This means, we have to calculate the number of elements in the intersection of the sets $A$ and $B$. Substituting $n\left( A\cup B \right)$ from equation $\left( 2 \right)$, $n\left( A \right)$ from equation $\left( 3 \right)$ and $n\left( B \right)$ from equation $\left( 4 \right)$ in equation $\left( 1 \right)$, we get,
\begin{align} & 950=750+460-n\left( A\cap B \right) \\ & \Rightarrow n\left( A\cap B \right)=750+460-950 \\ & \Rightarrow n\left( A\cap B \right)=260 \\ \end{align}
Hence, the number of persons that can speak both Hindi and English is $260$.

Note: In this question, it may become easier for a student to analyze the data given in the question if he/she analyzes this question with the help of a venn diagram. So, one can also do this question with the help of a venn diagram to avoid any small mistakes.


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# [SOLVED] Help! - Values

• Feb 7th 2008, 08:21 AM
weasley74
[SOLVED] Help! - Values
I need help finding out the biggest and smallest values of;

f(x)=2(cosx)^2 - 2sinx + 1

without using the derivative.

Please, need help as soon as possible!
• Feb 7th 2008, 08:25 AM
colby2152
Quote:

Originally Posted by weasley74
I need help finding out the biggest and smallest values of;

f(x)=2(cosx)^2 - 2sinx + 1

without using the derivative.

Please, need help as soon as possible!

You want values where sine is small and cosine is large for your maximum of the function. Sine is small (zero) and cosine is large (one) when x = 0, pi, 2pi, ....
• Feb 7th 2008, 08:37 AM
weasley74
what about the smallest value..? is that just the other way around?
• Feb 7th 2008, 09:03 AM
colby2152
Quote:

Originally Posted by weasley74
what about the smallest value..? is that just the other way around?

Correct!(Yes)
• Feb 7th 2008, 11:40 AM
weasley74
so for minimum x=pi/2 +2npi

and for maximum x=0, x=pi+2npi ?
• Feb 7th 2008, 04:17 PM
Soroban
Hello, weasley74!

Quote:

Find the biggest and smallest values of: . $f(x)\:=\:2\cos^2\!x - 2\sin x + 1$
without using the derivative.

Replace $\cos^2\!x$ with $1 - \sin^2\!x$

We have: . $f(x) \:=\:2(1-\sin^2\!x) - 2\sin x + 1 \quad\Rightarrow\quad f(x) \:=\:-2\sin^2\!x - 2\sin x + 3$

Let $u \,=\,\sin x\!:\;\;f(u) \:=\:-2u^2 - 2u + 3$

This is a down-opening parabola; its maximum is at its vertex.
. . The vertex is at: . $u\:=\:\frac{-b}{2a} \:=\:\frac{-(-2)}{2(-2)} \:=\:-\frac{1}{2}$

Hence, the maximum is: . $f\left(-\frac{1}{2}\right) \:=\:-2\left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) + 3 \;=\;\boxed{\frac{7}{2}}$

Normally, there is no minimum for this parabola.
. . However, the values of $u$ are limited: . $-1 \leq u \leq 1$

The minimum occurs at an endpoint of the interval.

. . And we find that the minimum is: . $f(1) \:=\:\boxed{-1}$

• Feb 7th 2008, 11:24 PM
weasley74
Quote:

Originally Posted by Soroban

This is a down-opening parabola; its maximum is at its vertex.
. . The vertex is at: . $u\:=\:\frac{-b}{2a} \:=\:\frac{-(-2)}{2(-2)} \:=\:-\frac{1}{2}$

how do you get to this point? what is a and b?
• Feb 8th 2008, 02:31 AM
mr fantastic
Quote:

Originally Posted by weasley74
how do you get to this point? what is a and b?

Are you familiar with parabolas? Do you know how to find the turning point of a parabola?

There are many ways to get the turning point - Soroban gave one which clearly you're not familiar with. What about completing the square to get turning point form? You've gotta have met that one ....?

$f(u) = -2u^2 - 2u + 3 = -2 \left( u^2 + u - \frac{3}{2} \right) = -2 \left( \left[ u + \frac{1}{2} \right]^2 - \frac{7}{4} \right) = -2\left(u + \frac{1}{2} \right)^2 + \frac{7}{2}$.

Maximum turning point at $\left(-\frac{1}{2}, \, \frac{7}{2}\right)$, that is, f(u) has a maximum value of $\, \frac{7}{2} \,$ (corresponding to $u = -\frac{1}{2} \Rightarrow \sin x = -\frac{1}{2}$ .....)
• Feb 8th 2008, 02:39 AM
mr fantastic
Quote:

Originally Posted by colby2152
You want values where sine is small and cosine is large for your maximum of the function. Sine is small (zero) and cosine is large (one) when x = 0, pi, 2pi, ....

Not wanting to rub it in ....... The error in this argument is that, unfortunately, it forgets that $\sin x$ can be negative (and hence $- 2 \sin x$ can also make a positive contribution).
• Feb 8th 2008, 04:36 AM
colby2152
Quote:

Originally Posted by mr fantastic
Not wanting to rub it in ....... The error in this argument is that, unfortunately, it forgets that $\sin x$ can be negative (and hence $- 2 \sin x$ can also make a positive contribution).

Yes, that thinking only works best in the 1st quadrant.
• Feb 8th 2008, 05:48 AM
Krizalid
\begin{aligned}
f(u) &= - 2u^2 - 2u + 3\\
&= - \frac{{\left( {4u^2 + 4u - 6} \right)}}
{2}\\
&= - \frac{{(2u + 1)^2 - 7}}
{2}.
\end{aligned}

May be this could be another way to get minimum point.


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# MATLAB: Hello everyone, I am stuck in a question. Can anyone there help me out? The question is as follows-

row selection from a matrix

I have a 3D matrix in which each 2D submatrix is of size 7×4.I have to randomly select any 4 out of these 7 rows(for various 3D indexes) except 1 row which will be decided by user input.That means, I can't select that particular row while selecting 4 rows. Thanks in advance.

• A = rand(7,4) ;NS = 2 ;    %Don't select his row idx = randperm(7,5) ;  % random indices of row idx = setdiff(idx,NS) ; % remove NS if presentS = idx(1:4) ;   % select these rowsiwant = A(S,:)


text:
# Boundary Conditions for an inviscid fluid at a fixed boundary

1. Mar 24, 2014

### Type1civ

This is my first post so I hope this in the right place. I am fairly sure this is quite a straight forward question but I having trouble working out the details of it.

"State the boundary conditions for an inviscid fluid at an impermeable fixed boundary

$$x_3-h(x_1,x_3)=0$$

where we do not permit normal velocity but permit non-zero tangential velocity"

So my attempt is quite limited as I feel if I could get started I'd be able to make some progress. But I believe that since the wall is stationary then the velocity of the fluid at the wall must also be stationary. And since it says that normal velocity is zero i guess this means that in the $x_3$ direction the velocity is always zero.
so:
$$\mathbf{u}\cdot \mathbf{\hat x}_3 =0$$
Then the velocity of the fluid only has components in the other two directions...
I was then also thinking that the $\hat x_1$ component of velocity wouldn't change with $x_1$ but I am not really sure why I think that. I am just getting a bit confused as you might be able to tell, any help would be really appreciated!

Edit: I am also not sure why my tex commands aren't working so if somebody could point out what I am doing wrong that would also be great...

Last edited: Mar 24, 2014
2. Mar 24, 2014

### pasmith

That boundary seems an unnecessary complicated way to write $g(x_1,x_3) = 0$. Does the question actually say
$$x_3 - h(x_1, x_2) = 0$$
which would make sense as the upper boundary of a fluid layer of variable depth?

For an inviscid fluid only the normal component must be stationary relative to the wall.

If the normal to the boundary were in the $x_3$ direction this would be correct, but that's not necessarily the case.

Following the question as written, the first thing is to define
$$g(x_1,x_3) = x_3 - h(x_1,x_3)$$
so that the boundary is at $g(x_1,x_3) = 0$. The boundary is then a contour of $g$, so the direction normal to the boundary is given by
$$\nabla g = \frac{\partial g}{\partial x_1} \hat{\mathbf{x}}_1 + \frac{\partial g}{\partial x_3} \hat{\mathbf{x}}_3.$$

The closing tag is [/tex] not [\tex].

3. Mar 24, 2014

### Type1civ

Yeah, sorry! It is:
$$x_3-h(x_1,x_2)=0$$

So then in this case my direction normal would be:
$$\nabla_h=\hat x_3 -\frac{\partial h}{\partial x_1} \hat x_1 -\frac{\partial h}{\partial x_2} \hat x_2$$
and I can dot this with the velocity of the fluid to get zero. So that:
$$\mathbf{u}\cdot\nabla_h= \mathbf{u}\cdot\hat x_3 -\mathbf{u}\cdot \frac{\partial h}{\partial x_1} \hat x_1 - \mathbf{u}\cdot\frac{\partial h}{\partial x_2} \hat x_2=0$$
This surely isn't the whole whole answer though is it?


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# AGHmatrix Tutorial

## Contact

Horticultural Sciences Department
University of Florida

## Overview

AGHmatrix software is an R-package to build relationship matrices using pedigree (A matrix) and/or molecular markers (G matrix) with the possibility to build a combined matrix of Pedigree corrected by Molecular (H matrix). The package works with diploid and autopolyploid data.

## Matrices computation implemented in the AGHmatrix

Currently the package computes the following 17 different relationship matrices:

### Pedigree-based relationship matrix (A matrix)

Diploid Henderson (1976) Cockerham (1954)
Autopolyploid Kerr (2012), Slater (2013)

### Molecular-based relationship matrix (G matrix)

Diploid Yang (2010), VanRaden (2012), Liu (2020) Su (2012), Vitezica (2013)
Polyploid Slater (2016), VanRaden (2021) Slater (2016), Endelman (2018)

### Combined pedigree and molecular-based relationship matrix (H matrix)

Any ploidy/effect
Legarra (2009), Munoz (2014), Martini (2018)

Additionally there is a beta implementation to compute A matrix when parentage is not deterministic as in a polycross design. See ?AmatrixPolycross.

## Citation

To cite this R package:

Amadeu, RR, C Cellon, JW Olmstead, AAF Garcia, MFR Resende, and PR Munoz. 2016. AGHmatrix: R package to construct relationship matrices for autotetraploid and diploid species: a blueberry example. The Plant Genome 9(3).

Within R:

## Install stable version
install.packages("AGHmatrix")

## Install development version
install.packages("devtools")

library(AGHmatrix)

## Relationship matrices using pedigree data - A matrix

Amatrix process the pedigree and build the A-matrix related to that given pedigree. The matrix is built based in the recursive method presented in Mrode (2014) and described by Henderson (1976). This method is expanded for higher ploidies (n-ploidy) as detailed in Kerr et al. (2012). After loading the package you have to load your data file into the software. To do this, you can use the function read.data() or read.csv() for example. Your data should be available in R as a data.frame structure in the following order: column 1 must be the individual/genotype names (id), columns 2 and 3 must be the parent names. For the algorithm, it does not matter who is the mother and who is the father (so, no sex column). There is a pedigree data example (ped.mrode) that can be used to look at the structure and order the data.

To load ped.mrode:

data(ped.mrode)
ped.mrode
#>    Ind Par1 Par2
#> 1 Anc1    0    0
#> 2 Anc2    0    0
#> 3 Var1 Anc1 Anc2
#> 4 Var2 Anc1    0
#> 5 Var3 Var2 Var1
#> 6 Var4 Var3 Anc2
str(ped.mrode) #check the structure
#> 'data.frame':    6 obs. of  3 variables:
#>  $Ind : Factor w/ 6 levels "Anc1","Anc2",..: 1 2 3 4 5 6 #>$ Par1: Factor w/ 4 levels "0","Anc1","Var2",..: 1 1 2 2 3 4
#>  $Par2: Factor w/ 3 levels "0","Anc2","Var1": 1 1 2 1 3 2 The example ped.mrode has 3 columns, column 1 contains the names of the individual/genotypes, column 2 contains the names of the first parent, column 3 contains the names of the second parental (example from Table 2.1 of Mrode 2014). There is no header template, and the unknown value must be equal 0. Your data has to be in the same format of ped.mrode. Internally the algorithm first pre-process the pedigree: the individuals are numerated $$1$$ to $$n$$. Then, it is verified whether the genotypes in the pedigree are in chronological order (i.e. if the parents of a given individual are located before to this individual in the pedigree data set). If this order is not followed, the algorithm performs the necessary changes to correct them in a iterative way. After this pre-processing, the matrix computation proceeds as in Henderson (1976) for diploid - for additive or dominance relationship - and as in Kerr et al. (2012) for autotetraploids - for additive relationship. For autotetraploids, there is the option to include double-reduction fraction. For diploids there is the option to compute the dominant relationship matrix (Cockerham, 1954). It follows some usage examples with the ped.mrode. #Computing additive relationship matrix for diploids (Henderson 1976): Amatrix(ped.mrode, ploidy=2) #Computing dominant relationship matrix for diploids (Cockerham 1954): Amatrix(ped.mrode, ploidy=2, dominance=TRUE) #Computing additive relationship matrix for autotetraploids (Kerr 2012): Amatrix(ped.mrode, ploidy=4) #Computing additive relationship matrix for autooctaploids (Kerr 2012): Amatrix(ped.mrode, ploidy=8) #Computing additive relationship matrix for autotetraploids # and double-reduction of 0.1 (Kerr 2012): Amatrix(ped.mrode, ploidy=4, w=0.1) #Computing additive relationship matrix for autotetraploids # and double-reduction of 0.1 as in Slater et al. (2014): Amatrix(ped.mrode, ploidy=4, w=0.1, slater = TRUE) #not recommended, but kept in the package to reproduce some former analysis #Computing additive relationship matrix for autohexaploids # and double-reduction of 0.1 (Kerr 2012): Amatrix(ped.mrode, ploidy=6, w=0.1) More information about Amatrix can be found with: ?Amatrix ## Diploid G matrix: relationship matrices using the molecular data Gmatrix handles the molecular-marker matrix and builds the relationship matrix. Molecular markers data should be organized in a matrix format (individuals in rows and markers in columns) coded as 0, 1, 2 and missing data value (numeric or NA). Import your molecular marker data into R with the function read.table() or read.csv() and convert to a matrix format with the function as.matrix(). The function Gmatrix can be used to construct the additive relationship either as proposed by Yang et al. (2010) or the proposed by VanRaden (2008). The function can also construct the dominance relationship matrix either as proposed by Su et al. (2012) or as proposed by Vitezica et al. (2013). As an example, here we build the four matrices using real data from Resende et al. (2012). To load snp.pine and to check its structure: data(snp.pine) snp.pine[1:5,1:5] #> 0-10024-01-114 0-10037-01-257 0-10040-02-394 0-10040-02-41 #> 1087120 -9 1 -9 -9 #> 1085618 2 2 2 2 #> 1091040 2 1 2 2 #> 1091686 2 1 2 2 #> 1082624 2 2 0 1 #> 0-10044-01-392 #> 1087120 -9 #> 1085618 2 #> 1091040 2 #> 1091686 2 #> 1082624 1 str(snp.pine) #> int [1:926, 1:4853] -9 2 2 2 2 2 2 2 2 2 ... #> - attr(*, "dimnames")=List of 2 #> ..$ : chr [1:926] "1087120" "1085618" "1091040" "1091686" ...
#>   ..\$ : chr [1:4853] "0-10024-01-114" "0-10037-01-257" "0-10040-02-394" "0-10040-02-41" ...

In this dataset, we have 926 individuals with 4853 markers and the missing data value is -9.

It follows some examples with the snp.pine data where the unknown value (missingValue) is -9. Here we set minimum allele frequency to 0.05, so markers with minor allele frequency lower than 0.05 are removed from the dataset prior to the G matrix construction.

#Computing the additive relationship matrix based on VanRaden 2008

#Computing the additive relationship matrix based on Yang 2010
G_YangPine <- Gmatrix(SNPmatrix=snp.pine, missingValue=-9,
maf=0.05, method="Yang")

#Computing the dominance relationship matrix based on Su 2012
G_SuPine <- Gmatrix(SNPmatrix=snp.pine, missingValue=-9,
maf=0.05, method="Su")

#Computing the dominance relationship matrix based on Vitezica 2013
G_VitezicaPine <- Gmatrix(SNPmatrix=snp.pine, missingValue=-9,
maf=0.05, method="Vitezica")

More information about Gmatrix can be found with:

?Gmatrix

## Autopolyploid G matrix: relationship matrices using the molecular data

Molecular markers data should be organized in a matrix format (individual in rows and markers in columns) coded according to the dosage level: 0, 1, 2, …, ploidy level, and missing data value (numeric user-defined or NA). As an example, an autotetraploid should be coded as 0, 1, 2, 3, 4, and a missing data value. In autopolyploids, the function Gmatrix can be used to construct: i) the additive relationship based on VanRaden (2008) and extended by Ashraf (2016); ii) the full-autopolyploid including additive and non-additive model as equations 8 and 9 in Slater et al. (2016); iii) the pseudo-diploid model as equations 5, 6, and 7 Slater et al. (2016). iv) the digenic-dominant model based on Endelman et al. (2018). As an example, here we build the matrices using data from Endelman et al. (2018) (snp.sol). There is also an option to build weighted relationship matrices as in Liu et al. (2020).

The argument ploidy.correction defines the denominator of the formula for the VanRaden method. If ploidy.correction=TRUE, it uses the parametric correction as $$\sum_i p f_i(1-f_i)$$, where $$p$$ is the ploidy level and $$f_i$$ is the minor allele frequency of the $$i_th$$ marker. If ploidy.correction=FALSE, it uses the sampling variance correction as $$\sum_i \frac{1}{p} s^2(m_i)$$, where $$s^2(m_i)$$ is the sampling variance of the $$i_th$$ marker. Both corrections are equivalent when sampling size goes to the infinity. The default is to use the sampling variance as the correction (i.e., ploidy.correction=FALSE).

#Loading the data
data(snp.sol)
str(snp.sol)

#Computing the dominance (digenic) matrix based on Endelman 2018 (Eq. 19)
G_Dominance <- Gmatrix(snp.sol, method="Endelman", ploidy=4)

#Computing the full-autopolyploid matrix based on Slater 2016 (Eq. 8
#and 9)
G_FullAutopolyploid <- Gmatrix(snp.sol, method="Slater", ploidy=4)

#Computing the pseudodiploid matrix based on Slater 2016 (Eq. 5, 6,
#and 7)
G_Pseudodiploid <- Gmatrix(snp.sol, method="VanRaden", ploidy=4, pseudo.diploid=TRUE)

#Computing G matrix with specific weight for each marker as
# in Liu et al. (2020).
Gmatrix_weighted <- Gmatrix(snp.sol, method="VanRaden", weights = runif(3895,0.001,0.1), ploidy=4)

More information about Gmatrix can be found with:

?Gmatrix

## Ratio (non-dosage) G matrix: relationship matrices using the molecular data without dosage calling

Molecular markers data should be organized in a matrix format (individual in rows and markers in columns) coded according to a fraction that represents its molecular information, this can be any number between 0 to 1. Such ratio can represent the count of alternative alleles over the read depth for each individual-marker combination (GBS-like technique). It can be the signal of the alternative allele over the sum of the signals of the alternative and reference alleles (GCMS-like technique). It can also be used for family-pool genotypes.

#Loading the data
library(AGHmatrix)
data(snp.sol)
snp.sol.ratio = snp.sol/4 #transforming it in a ratio of the minor allele frequency
Gmatrix <- Gmatrix(snp.sol, method="VanRaden", ploidy=4, ratio=FALSE)
Gmatrix.ratio <- Gmatrix(snp.sol.ratio, method="VanRaden", ploidy=4, ratio=TRUE)
Gmatrix[1:5,1:5]==Gmatrix.ratio[1:5,1:5]

## it also has the ploidy.correction option
Gmatrix.alternative <- Gmatrix(snp.sol,
ploidy=4,
ratio=FALSE,
ploidy.correction=TRUE)

Gmatrix.ratio.alternative <- Gmatrix(snp.sol.ratio,
ploidy=4,
ratio=TRUE,
ploidy.correction=TRUE)
Gmatrix[1:5,1:5]==Gmatrix.alternative[1:5,1:5]
Gmatrix.alternative[1:5,1:5]==Gmatrix.ratio.alternative[1:5,1:5]

## Combined relationship matrix - H matrix

H matrix is the relationship matrix using combined information from the pedigree and genomic relationship matrices. First, you need to compute the matrices separated and then use them as input to build the combined H matrix. Two methods are implemented: Munoz shrinks the G matrix towards the A matrix scaling the molecular relatadness by each relationship classes; Martini is a modified version from Legarra et al. 2009 where combines A and G matrix using scaling factors. As an example, here we build the matrices using data from Endelman et al. (2018) (ped.sol and snp.sol).

data(ped.sol)
data(snp.sol)

#Computing the numerator relationship matrix 10% of double-reduction
Amat <- Amatrix(ped.sol, ploidy=4, w = 0.1)
Gmat <- Gmatrix(snp.sol, ploidy=4,
Gmat <- round(Gmat,3) #see appendix

#Computing H matrix (Martini)
Hmat_Martini <- Hmatrix(A=Amat, G=Gmat, method="Martini",
ploidy=4, missingValue=-9, maf=0.05)

#Computing H matrix (Munoz)
Hmat_Munoz <- Hmatrix(A=Amat, G=Gmat, markers = snp.sol,
ploidy=4, method="Munoz",
missingValue=-9, maf=0.05)

## Covariance matrices due to epistatic terms

Here we present how to compute the epistasis relationship matrices using Hadamard products (i.e. cell-by-cell product), denoted by *. For more information please see Munoz et al. (2014). In this example we are using the molecular-based relationship matrices. First, build the additive and dominance matrices:

data(snp.pine)
A <- Gmatrix(SNPmatrix=snp.pine, method="VanRaden", missingValue=-9, maf=0.05)
D <- Gmatrix(SNPmatrix=snp.pine, method="Vitezica", missingValue=-9,maf=0.05)

For the first degree epistatic terms:

#Additive-by-Additive Interactions
A_A <- A*A
D_A <- D*A
#Dominance-by-Dominance Interactions
D_D <- D*D

For the second degree epistatic terms:

#Additive-by-Additive-by-Additive Interactions
A_A_A <- A*A*A
A_A_D <- A*A*D
A_D_D <- A*D*D
#Dominance-by-Dominance-by-Dominance Interactions
D_D_D <- D*D*D

And so on…

## Exporting your matrix as three columns and sparse format (ASReml - csv format)

That is the lower diagonal matrix formatted in three columns in .csv format (other ASCII extension could be used as well). In order to do this, we need to build a matrix, its inverse, and export it using formatmatrix function. ASReml can invert the relationship matrix as well, probably more efficiently than R for large matrices (i.e. solve() function), so no need to invert the matrix in R if matrix is large. This function has as options: round.by, which let you decide the number of decimals you want; exclude.0, if TRUE, remove all the zeros from your data (i.e., transforms into sparse); and, name that defines the name to be used in the exported file. Use the default if not sure what parameter use in these function. Here an example using ped.mrode data:

#Loading the data example
data(ped.mrode)

#Computing the matrix
A <- Amatrix(data=ped.mrode, ploidy=4, w=0.1)

#Building its inverse
Ainv <- solve(A)

#Exporting it. The function "formatmatrix"
# will convert it and save in your working directory
formatmatrix(Ainv, round.by=12, exclude.0=TRUE, name="Ainv")

## Relationship matrices using pedigree data for polycrosses - A matrix (beta)

Creates an additive relationship matrix A based on a non-deterministic pedigree with 4+ columns where each column represents a possible parent. This function was built with the following designs in mind. 1) A mating design where you have equally possible parents. For example, a generation of insects derived from the mating of three insects in a cage. All the insects in this generation will have the same expected relatedness with all the possible parents (1/3). If there are only two parents in the cage, the function assumes no-inbreeding and the pedigree is deterministic (the individual is offspring of the cross between the two parents). Another example, a population of 10 open-pollinated plants where you harvest the seeds without tracking the mother. 2) When fixedParent is TRUE: a mating design where you know one parent and might know the other possible parents. For example, a polycross design where you have seeds harvested from a mother plant and possible polén donors.

The following pedigree has the id of the individual followed by possible parents. The possible parents are filled from left to right, in the pedigree data frame: id 1,2,3,4 have unknown parents and are assumed unrelated; id 5 has three possible parents (1,2,3); id 6 has three possible parents (2,3,4); id 7 has two parents (deterministic case here, the parents are 3 and 4); id 8 has four possible parents (5,6,7,1).

pedigree = data.frame(id=1:8,
parent1 = c(0,0,0,0,1,2,3,5),
parent2 = c(0,0,0,0,2,3,4,6),
parent3 = c(0,0,0,0,3,4,0,7),
parent4 = c(0,0,0,0,0,0,0,1),
parent5 = 0)

print(pedigree)
#>   id parent1 parent2 parent3 parent4 parent5
#> 1  1       0       0       0       0       0
#> 2  2       0       0       0       0       0
#> 3  3       0       0       0       0       0
#> 4  4       0       0       0       0       0
#> 5  5       1       2       3       0       0
#> 6  6       2       3       4       0       0
#> 7  7       3       4       0       0       0
#> 8  8       5       6       7       1       0
AmatrixPolyCross(pedigree)
#> Your data was chronologically organized with success.
#> Your data was chronologically organized with success.
#> Your data was chronologically organized with success.
#> Your data was chronologically organized with success.
#> Constructing matrix A using ploidy = 2
#> Completed! Time = 3.333333e-05  minutes
#>           1         2         3         4         5         6         7
#> 1 1.0000000 0.0000000 0.0000000 0.0000000 0.3333333 0.0000000 0.0000000
#> 2 0.0000000 1.0000000 0.0000000 0.0000000 0.3333333 0.3333333 0.0000000
#> 3 0.0000000 0.0000000 1.0000000 0.0000000 0.3333333 0.3333333 0.5000000
#> 4 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 0.3333333 0.5000000
#> 5 0.3333333 0.3333333 0.3333333 0.0000000 1.0000000 0.2222222 0.1666667
#> 6 0.0000000 0.3333333 0.3333333 0.3333333 0.2222222 1.0000000 0.3333333
#> 7 0.0000000 0.0000000 0.5000000 0.5000000 0.1666667 0.3333333 1.0000000
#> 8 0.3333333 0.1666667 0.2916667 0.2083333 0.4305556 0.3888889 0.3750000
#>           8
#> 1 0.3333333
#> 2 0.1666667
#> 3 0.2916667
#> 4 0.2083333
#> 5 0.4305556
#> 6 0.3888889
#> 7 0.3750000
#> 8 1.0879630

If fixedParent=TRUE, the above pedigree will be interpreted with the possible parents are filled from left to right after the known parent, in the pedigree data frame: id 1,2,3,4 have unknown parents and are assumed unrelated; id 5 is offspring of parent 1 in a deterministic way and two other possible parents (2,3); id 6 is offspring of parent 2 in a deterministic way and two other possible parents (3,4); id 7 has two parents (deterministic case here, the parents are 3 and 4); as before; id 8 is offspring of parent 5 in a deterministic way and has three other possible parents (6,7,1).

AmatrixPolyCross(pedigree,fixedParent=TRUE)
#> Your data was chronologically organized with success.
#> Your data was chronologically organized with success.
#> Your data was chronologically organized with success.
#> Your data was chronologically organized with success.
#> Completed! Time = 1.666667e-05  minutes
#>           1         2    3     4         5         6         7         8
#> 1 1.0000000 0.0000000 0.00 0.000 0.5000000 0.0000000 0.0000000 0.4166667
#> 2 0.0000000 1.0000000 0.00 0.000 0.2500000 0.5000000 0.0000000 0.2083333
#> 3 0.0000000 0.0000000 1.00 0.000 0.2500000 0.2500000 0.5000000 0.2500000
#> 4 0.0000000 0.0000000 0.00 1.000 0.0000000 0.2500000 0.5000000 0.1250000
#> 5 0.5000000 0.2500000 0.25 0.000 1.0000000 0.1875000 0.1250000 0.6354167
#> 6 0.0000000 0.5000000 0.25 0.250 0.1875000 1.0000000 0.2500000 0.3020833
#> 7 0.0000000 0.0000000 0.50 0.500 0.1250000 0.2500000 1.0000000 0.2708333
#> 8 0.4166667 0.2083333 0.25 0.125 0.6354167 0.3020833 0.2708333 1.1354167

## Amatrix() benchmark

It follows a small memory and computational time profiling for the Amatrix() function. The required RAM was computed based on the peak of the process for different pedigree sizes (based on /usr/bin/time -v output). The time profiling was done using AMD Milan 2.95GHz, so it might be an underestimated value when compared with lower speed processors. Numerator relationship matrices for pedigrees with less than 20,000 rows can built with low-specs user-end machines (<8GB RAM) using our package.

## Bibliography

Amadeu, RR, et al. 2016 AGHmatrix: R package to construct relationship matrices for autotetraploid and diploid species: a blueberry example. The Plant Genome 9, 4.

Ashraf, BH, et al.. 2016 Estimating genomic heritabilities at the level of family-pool samples of perennial ryegrass using genotyping-by-sequencing. Theoretical and Applied Genetics 129, 45-52.

Cockerham, CC. 1954. An extension of the concept of partitioning hereditary variance for analysis of covariances among relatives when epistasis is present. Genetics 39, 859–882

Endelman, JB, et al. 2018. Genetic variance partitioning and genome-wide prediction with allele dosage information in autotetraploid potato. Genetics 209, 77-87.

Hamilton, MG, et al. 2017 Computation of the inverse additive relationship matrix for autopolyploid and multiple-ploidy populations. Theoretical and Applied Genetics 131, 851-890.

Henderson, C. 1976 A simple method for computing the inverse of a numerator relationship matrix used in prediction of breeding values. Biometrics 32, 69–83.

Kerr, RJ, et al., 2012 Use of the numerator relation ship matrix in genetic analysis of autopolyploid species. Theoretical and Applied Genetics 124, 1271–1282.

Legarra, A, et al. 2009 A relationship matrix including full pedigree and genomic information. Journal of Dairy Science 92, 4656–4663.

Liu, A, et al. 2020. Weighted single-step genomic best linear unbiased prediction integrating variants selected from sequencing data by association and bioinformatics analyses. Genet Sel Evol 52, 48.

Martini, JW, et al., 2018, The effect of the H$$^{1}$$ scaling factors $$\tau$$ and $$\omega$$ on the structure of H in the single-step procedure. Genetics Selection Evolution 50(1), 16.

Mrode, RA. 2014 Linear models for the prediction of animal breeding values. Cabi. 3rd ed.

Munoz, PR, et al., 2014 Unraveling additive from nonadditive effects using genomic relationship matrices. Genetics 198, 1759–1768.

R Core Team, 2016 R: A Language and Environment for Statistical Computing. R Foundation for Statistical Computing, Vienna, Austria.

Resende, MF, et al., 2012 Accuracy of genomic selection methods in a standard data set of loblolly pine (Pinus taeda l.). Genetics 190, 1503–1510.

Slater, AT, et al. 2014 Improving the analysis of low heritability complex traits for enhanced genetic gain in potato. Theoretical and applied genetics 127, 809–820.

Slater, AT, et al. 2016 Improving genetic gain with genomic selection in autotetraploid potato. The Plant Genome 9.

Su, G, et al. 2012 Estimating additive and non-additive genetic variances and predicting genetic merits using genome-wide dense single nucleotide polymorphism markers. PloS one 7, e45293.

VanRaden, P. 2008 Efficient methods to compute genomic predictions. Journal of dairy science 91, 4414–4423.

Vitezica, ZG, et al. 2013 On the additive and dominant variance and covariance of individuals within the genomic selection scope. Genetics 195, 1223–1230.

Yang, J, et al. 2010 Common snps explain a large proportion of the heritability for human height. Nature genetics 42, 565–569.

sessionInfo()
#> R version 4.1.0 (2021-05-18)
#> Platform: x86_64-pc-linux-gnu (64-bit)
#> Running under: Ubuntu 20.04.2 LTS
#>
#> Matrix products: default
#>
#> locale:
#>  [1] LC_CTYPE=en_US.UTF-8       LC_NUMERIC=C
#>  [3] LC_TIME=en_US.UTF-8        LC_COLLATE=C
#>  [5] LC_MONETARY=en_US.UTF-8    LC_MESSAGES=en_US.UTF-8
#>  [7] LC_PAPER=en_US.UTF-8       LC_NAME=C
#> [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
#>
#> attached base packages:
#> [1] stats     graphics  grDevices utils     datasets  methods   base
#>
#> other attached packages:
#> [1] AGHmatrix_2.0.4 rmarkdown_2.8   knitr_1.33
#>
#> loaded via a namespace (and not attached):
#>  [1] codetools_0.2-18  lattice_0.20-44   zoo_1.8-9         digest_0.6.27
#>  [5] grid_4.1.0        R6_2.5.0          jsonlite_1.7.2    magrittr_2.0.1
#>  [9] evaluate_0.14     highr_0.9         rlang_0.4.11      stringi_1.6.2
#> [13] jquerylib_0.1.4   Matrix_1.3-4      bslib_0.2.5.1     tools_4.1.0
#> [17] stringr_1.4.0     xfun_0.23         yaml_2.2.1        compiler_4.1.0
#> [21] htmltools_0.5.1.1 sass_0.4.0


text:
## Need help on my understanding of gravitational waves

Hi, Guys. So i am doing a research paper on gravitational waves. As I read through review articles on gravitational waves and LIGO, I don't quite get a few points. The materials are really dense and I don't have the math and physics level to back it up. But I am really interested in the topic.

1) So first of all, I think one article is saying that gravitational waves have this quadrupole moment which allows it to occur. But how does quadrupole moment plays a part in binary star systems such as the neutron stars spin down. What is the connection? I can see energy loses in the system is being emitted as gravitational waves.

2) "LIGO should be able to detect gravitational waves as small as h \approx 5\times 10^{-22}"(WIKI) Using the h strength of gws equation, we approximate the magnitude of the gws to be detected in galaxy. But where in the galaxy? What is the hubble distance? I know this h equation approximates the order of accuracy in LIGO. But how did it come about?

3) Why GWs are detected in amplitude instead of frequency? Aren't we detecting gws base on the strength(as in magnitude of energy) of it.

4)How do we know if we pick up the GWs signal? This is very ambiguous. One article i read said they derived equation to come up with the GWs signal in form of radio-wave. I know they detect the waves in the photodiode where the two beams in LIGO are measured to see if they are out of phrase.

Thank you guys so much.
 PhysOrg.com science news on PhysOrg.com >> Leading 3-D printer firms to merge in $403M deal (Update)>> LA to give every student an iPad;$30M order>> CIA faulted for choosing Amazon over IBM on cloud contract

Recognitions:
 Quote by peacewar1 1) So first of all, I think one article is saying that gravitational waves have this quadrupole moment which allows it to occur. But how does quadrupole moment plays a part in binary star systems such as the neutron stars spin down. What is the connection? I can see energy loses in the system is being emitted as gravitational waves.
What is meant is that gravitational radiation is emitted by changes in the quadrupole moment of a mass distribution, NOT the dipole moment as is the case in the electromagnetic case. This means that if you have, for example, a spherical star pulsating radially that it will not radiate gravitational waves! On the other hand, two orbiting bodies produce a changing quadrupole moment.

 2) "LIGO should be able to detect gravitational waves as small as h \approx 5\times 10^{-22}"(WIKI) Using the h strength of gws equation, we approximate the magnitude of the gws to be detected in galaxy. But where in the galaxy? What is the hubble distance? I know this h equation approximates the order of accuracy in LIGO. But how did it come about?
It comes about through complicated sensitivity curves that look like this: http://www.ligo-wa.caltech.edu/ligo_...G060293-00.jpg These are generated by the complicated noise models of the actual detectors.

Hubble distance? That doesn't have anything to do with this.

 3) Why GWs are detected in amplitude instead of frequency? Aren't we detecting gws base on the strength(as in magnitude of energy) of it.
Because the GW changes the length of the arms of the interferometer, which is an effect proportional to the amplitude of the wave. The frequency just tells you how often it's doing this. The energy in the wave is actually irrelevant for LIGO's purposes (it's not like in EM where photons have energy hbar*omega, and that's the relevant quantity. Rather, if you think of looking at a test charge as an electromagnetic wave passes by, the E-field will cause the charge to oscillate with a magnitude equal to the amplitude of the EM wave, and the same frequency as well.)

 4)How do we know if we pick up the GWs signal? This is very ambiguous. One article i read said they derived equation to come up with the GWs signal in form of radio-wave. I know they detect the waves in the photodiode where the two beams in LIGO are measured to see if they are out of phrase.
What do you mean? You measure the distance in the two arms of the interferometer. If you see this distance changing by larger amounts than your noise threshold, that's a signal. If it has the right frequency characteristics, then it's possibly a GW of astrophysical origin (other things are of course possible, like falling trees nearby or construction equipment). The characteristic 'chirp' of a binary inspiral is actually an unmistakable signal, you can hear it here: http://www.einstein-online.info/spot..._neutron_stars ).
 Thank you so much for your time. You are a great help!

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# Which is greater 105^19 or 100^20?

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Kudos [?]: 21 [1] , given: 115

Which is greater 105^19 or 100^20? [#permalink]

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Dear fellowmembers,

I have come across a question that I am not able to either solve it or find a suitable explanation; the question is: -

Q). Which is greater 105^19 or 100^20?

Please help!
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Re: Question [#permalink]

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11 Jan 2013, 04:01
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$$100^{20}$$ = $$100^{19} * 100$$ i.e 100 times $$100^{19}$$

= $$100^{19} + 100^{19} + 100^{19} +$$ ...... and so on till hundred times --> (1)

now,

$$105^{19}$$ = $$(100 + 5) ^{19}$$

Now, as we know $$(a + b)^n = a^{19} + ^nC1*a^{(n-1)}*b^1 + ^nC2*a^{(n-2)}*b^2 +$$ ........ $$+ ^nCn-1*a^1*b^{(n-1)} + b^n$$

$$(100 + 5) ^{19} = 100^{19} + 19*100^{18}*5^1 + 171*100^{17}*5^2 + .... + 5^{19}$$

$$= 100^{19} + 85*100^{18} +$$ ... preceding all the terms will be $$< 100^{19}$$ --> (2)

so compare (1) and (2) now, we get

$$100^{20} > 105^{19}$$
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Last edited by goutamread on 12 Jan 2013, 01:04, edited 3 times in total.
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Which is greater 105^19 or 100^20? [#permalink]

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11 Jan 2013, 04:56
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Let me try :

1st expression : $$105^{19}$$ $$= (100 *1.05)^{19}$$ $$= 100^{19}*1.05^{19}$$ --- [ as $$(a*b)^n = (a^n)*(b^n)$$ ]

2nd expression : $$100^{20}$$ $$= 100^{19} * 100^1$$ $$= 100^{19}*100$$ --- [ as $$a^n = (a^1)*(a^{n-1})$$ ]

now as $$100^{19}$$ is common,

Compare $$1.05^{19}$$ with $$100$$.

as 1.05 is slightly greater than 1

surely, $$1.05^{19} < 100$$

thus

$$105^{19} < 100^{20}$$

Let me know if this worked for you.
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Last edited by goutamread on 12 Jan 2013, 01:20, edited 2 times in total.
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Re: Question [#permalink]

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11 Jan 2013, 22:18
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Expert's post
ObsessedWithGMAT wrote:
Dear fellowmembers,

I have come across a question that I am not able to either solve it or find a suitable explanation; the question is: -

Q). Which is greater 105^19 or 100^20?

Please help!
Thanks in advance.

goutamread has given you a great approach. I would use a similar one though I just try to simplify first. It helps you think faster in some cases.

$$105^{19}$$ and $$100^{20}$$ have $$5^{19}$$ common. Cancel out the common $$5^{19}$$ and you are left with

$$105^{19} = 5^{19}*21^{19}$$
and $$100^{20} = 5^{19}*20^{19}*100$$

Now compare $$21^{19}$$ and $$20^{19}*100$$

It is easy to make some sense out of $$20^{19}*100$$. It is 100 times $$20^{19}$$ i.e. you can get this expression if you add 100 terms of $$20^{19}$$ together

$$20^{19}*100 = 20^{19} + 20^{19} + 20^{19}$$ ... (100 terms) ..............(I)

To compare, we can use binomial for $$21^{19}$$. If you do not know how to use binomial, check out this link:
http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

$$21^{19} = (20 + 1)^{19}$$

Let's try to expand $$(20 + 1)^{19}$$ using binomial. We will get 19+1 = 20 terms when we expand.

$$(20 + 1)^{19} = 20^{19} + 19*20^{18} + 19* 18/2 * 20^{17} + ... + 1^{19}$$ ............... (II)

Notice that this expression has 20 terms and none of the terms will be greater than $$20^{19}$$.

Compare (I) with (II).
(I) has 100 terms, all of them $$20^{19}$$
(II) has 20 terms, all of them equal to or less than $$20^{19}$$.

Hence, (I) will be greater than (II). Or we can say that $$100^{20}$$ will be greater than $$105^{19}$$.
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Re: Which is greater 105^19 or 100^20? [#permalink]

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18 Jan 2013, 08:50
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ObsessedWithGMAT wrote:
Thanks a lot goutamread! This one is really better. But, I want to know that what made you to split 105 into 100*1.05. This is some thing that is really very important to know because I believe that these are the tactics that one has to think of while attacking a question.

I'm not an expert, rather I'm just another aspirant on the forum. I could try to explain:

See, whenever we deal with a big number/expression, we try to simplify the expression to make it realistic

There could be numerous ways to solve a question, what matters is which one you feel at your ease... now as you mentioned inequality,

assume $$105^{19} > 100^{20}$$ lets try to prove whther our assumption is correct or not. If yes, then $$105^{19} > 100^{20}$$ and if not then, $$105^{19} < 100^{20}$$

$$105^{19} > 100^{20}$$
Now, don't these expressions scare us. yes they are scary.. so lets try and simplify.
lets divide both side by $$100^{20}$$, (as $$100^{20}$$ is +ve --> inequality sign wont change)
$$\frac{105^{19}}{100^{20}} > 1$$
$$\frac{105^{19}}{{100^{19}*100}} > 1$$
$$(\frac{105}{100})^{19}*\frac{1}{100} > 1$$
$$\frac{(1.05) ^{19}}{100} > 1$$ OR $$(\frac{26}{25}) ^{19}* \frac{1}{100} > 1$$ ==> $$26 ^{19} > 100 * 25^{19}$$

From here its already discussed earlier how to solve the problem. Take away could be : There are various ways to solve a problem, what matters is what suits you, but you should be aware of more than one trick.
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Last edited by goutamread on 20 Jan 2013, 03:23, edited 1 time in total.
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Re: Question [#permalink]

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11 Jan 2013, 22:05
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ObsessedWithGMAT wrote:
Dear fellowmembers,

I have come across a question that I am not able to either solve it or find a suitable explanation; the question is: -

Q). Which is greater 105^19 or 100^20?

Please help!
Thanks in advance.

Could you change the main subject as 'Which is greater 105^19 or 100^20?' So, It may be helpful for others too, in case they need to dive into the problem.
_________________

Thanks and Regards!

P.S. +Kudos Please! in case you like my post.

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Re: Which is greater 105^19 or 100^20? [#permalink]

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17 Jan 2013, 11:52
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Here is the smart solution. Lets assume both are equal. If after some manipulation we can show LHS > RHS or LHS < RHS we would have got the answer.

105^19 = 100^20

Take 100^19 to the LHS (Divide both sides by 100^19)

We have 1.05^19 = 100

From Binomial theorem we know than (1+x)^n = 1 + nx when x<<1

Hence 1.05^19 = (1 + 0.05*19) = 1.95 < 100

Hence 105^19 < 100^20
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Re: Which is greater 105^19 or 100^20? [#permalink]

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18 Jan 2013, 09:39
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For more practice : which-is-greater-54-200-or-5145.html
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Re: Question [#permalink]

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11 Jan 2013, 04:38
goutamread wrote:
100^20 = 100^19 * 100 i.e 100 times 100^19
= 100^19 + 100^19 + 100^19 + ...... and so on till hundred times --> (1)

now,

105^19 = (100 + 5) ^19
Now, as we know (a + b)^n = a^19 + (nC1)(a^(n-1))(b^1) + (nC2)(a^(n-2))(b^2) + .... + (nCn-1)(a^1)(b^(n-1)) + b^n
(100 + 5) ^19 = 100^19 + (19)(100^18)(5^1) + (171)(100^17)(5^2) + .... + 5^19
= 100^19 + (85)(100^18) + ... preceding all the terms will < 100^19 --> (2)

so compare (1) and (2) now, we get

100^20 > 105^19

Hi Goutamread,

Thanks for the response and effort! Although it seems that you have tried to explain in the best possible way that you are aware of, the explanation has just gone over my head
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Re: Which is greater 105^19 or 100^20? [#permalink]

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16 Jan 2013, 07:46
goutamread wrote:
Let me try :

1st expression : $$105^{19}$$ $$= (100 *1.05)^{19}$$ $$= 100^{19}*1.05^{19}$$ --- [ as $$(a*b)^n = (a^n)*(b^n)$$ ]

2nd expression : $$100^{20}$$ $$= 100^{19} * 100^1$$ $$= 100^{19}*100$$ --- [ as $$a^n = (a^1)*(a^{n-1})$$ ]

now as $$100^{19}$$ is common,

Compare $$1.05^{19}$$ with $$100$$.

as 1.05 is slightly greater than 1

surely, $$1.05^{19} < 100$$

thus

$$105^{19} < 100^{20}$$

Let me know if this worked for you.

Thanks a lot goutamread! This one is really better. But, I want to know that what made you to split 105 into 100*1.05. This is some thing that is really very important to know because I believe that these are the tactics that one has to think of while attacking a question.
_________________

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Re: Question [#permalink]

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16 Jan 2013, 07:47
goutamread wrote:
ObsessedWithGMAT wrote:
Dear fellowmembers,

I have come across a question that I am not able to either solve it or find a suitable explanation; the question is: -

Q). Which is greater 105^19 or 100^20?

Please help!
Thanks in advance.

Could you change the main subject as 'Which is greater 105^19 or 100^20?' So, It may be helpful for others too, in case they need to dive into the problem.

Done!
_________________

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Re: Question [#permalink]

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16 Jan 2013, 07:50
VeritasPrepKarishma wrote:
ObsessedWithGMAT wrote:
Dear fellowmembers,

I have come across a question that I am not able to either solve it or find a suitable explanation; the question is: -

Q). Which is greater 105^19 or 100^20?

Please help!
Thanks in advance.

goutamread has given you a great approach. I would use a similar one though I just try to simplify first. It helps you think faster in some cases.

$$105^{19}$$ and $$100^{20}$$ have $$5^{19}$$ common. Cancel out the common $$5^{19}$$ and you are left with

$$105^{19} = 5^{19}*21^{19}$$
and $$100^{20} = 5^{19}*20^{19}*100$$

Now compare $$21^{19}$$ and $$20^{19}*100$$

It is easy to make some sense out of $$20^{19}*100$$. It is 100 times $$20^{19}$$ i.e. you can get this expression if you add 100 terms of $$20^{19}$$ together

$$20^{19}*100 = 20^{19} + 20^{19} + 20^{19}$$ ... (100 terms) ..............(I)

To compare, we can use binomial for $$21^{19}$$. If you do not know how to use binomial, check out this link:
http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

$$21^{19} = (20 + 1)^{19}$$

Let's try to expand $$(20 + 1)^{19}$$ using binomial. We will get 19+1 = 20 terms when we expand.

$$(20 + 1)^{19} = 20^{19} + 19*20^{18} + 19* 18/2 * 20^{17} + ... + 1^{19}$$ ............... (II)

Notice that this expression has 20 terms and none of the terms will be greater than $$20^{19}$$.

Compare (I) with (II).
(I) has 100 terms, all of them $$20^{19}$$
(II) has 20 terms, all of them equal to or less than $$20^{19}$$.

Hence, (I) will be greater than (II). Or we can say that $$100^{20}$$ will be greater than $$105^{19}$$.

Dear Karishma,

Thanks a lot for taking out time to reply to my question! I am still struggling with the fact that this question is from Inequalities and neither you nor goutamread applied any concept of Inequalities or is it that I am not able to identify?
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Re: Question   [#permalink] 16 Jan 2013, 07:50
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