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# nice latex fonts

/Length 38 Note: Paragraph breaks are not allowed inside the command forms. TrueType (TTF) and OpenType (OTF) fonts found on modern systems. Searches for words with accents in PDFs will fail. Bold text may be used to heavily emphasize very important words or phrases.

This was done because the headings were typeset in bold and have font sizes larger than the body text; in my opinion serifs of the main font (Charter) in this case should be avoided (actually, bold headings without serifs is a common practice: 1, 2, 3). /F2 7 0 R Text may be emphasized more heavily through the use of boldface, particularly for keywords the reader may be trying to find when reading the text. This is because these older engines only support a fixed set of sizes - between 5 and 17 point.
The following table lists the commands you will need to access typical font shapes. The LaTeX notion of font families is slightly different. /ProcSet[/PDF/Text/ImageC/ImageB/ImageI] use of the math mode in the main text produces adjacent symbols strongly different by weight, which is illustrated by two text fragments below: writer has to decide which numbers to process as a normal text and which in math mode: the choice is not always trivial, and this most likely will result in different (CMR/Charter) typesetting of numbers in situations with similar context. e.g.

Introduction This book lists serif and sans-serif free and/or open-source fonts suitable for professional edition. The above code can be modified to typeset headings with CMR serif font: After looking in tug.dk font collection for a suitable serif font for headings, I have stopped my choice at TeX Gyre Heros: The chosen TeX Gyre Heros is the font not only without serifs, but also has heavier weight compared to Charter â this results in a good visual separation of the headings from the rest text making TeX Gyre Heros a good companion for the body text typeset with Charter. Font encoding is set with: where encoding is the font encoding.

>> Another advantage of using new CM-like fonts is that they provide fonts of CM families in all weights, shapes, and optically scaled font sizes.
After having some experience with the combination of Charter and CMR for text and math respectively, I can point out the following problems inherent to such combination of fonts with different weight: After switching the math mode font I have faced issues with Greek symbols which had to be typeset in the main (non-math) text. To scale text relative to the default body text size, use the following commands: These commands change the size within a given scope. << endstream It contains serif, sans serif, and monospaced fonts, each available in several weights and optical sizes. For instance {\Large some words} will change the size of only some words, and does not affect the font in the rest of the document. endstream << It provided some workarounds for hyphenation of words with special characters. However, part of a formula may be set in a different size by using an \mbox command containing the size command. The commands discussed above change font in the body text, math mode, and headings (including bibliography), but have no influence on the fonts in table of contents (ToC): Fonts in ToC can be adjusted with the package tocloft, which I actually used: As you see, font TeX Gyre Heros was used only for the ToC title and headings typeset in bold (which are chapter ones). /Font<< The fontspec package is extremely configurable. Install either, Small caps (in the same file as the standard upper and lowercase characters), A Unicode text file is made of a series of, One or more of these code points combines to represent a. It contains only 128 characters, many from ASCII, but leaving out some others and including a number that are not in ASCII. These commands cannot be used in math mode.

This can be seen in the code blocks above where in order to change the heading font with \titleformat command (\usefont{OT1}{cmr}{bx}{n}\selectfont}) settings for the heading number, its location, and its font size were given too ({\hspace{-0.4pt}\Large \thesection\hspace{0.6em}}).

Italics may be specified explicitly with \textit{text}. By default, \normalsize is 10 points, but this can be changed in the \documentclass declaration,

Serif Fonts] [Serif Fonts, Sub-Categorised] [Sans Serif Fonts] [Typewriter Fonts] [Calligraphical and Handwritten Fonts] [Uncial Fonts] [Blackletter Fonts] [Other Fonts] [Fonts with Math Support] [Fonts with OpenType or TrueType Support] [All Fonts, by category] [All Fonts, alphabetically] [About The L a T e X Font Catalogue] [Packages that provide math support] QT Beckman [OTF or TTF only] Useful resources for latex and pdflatex font configuration include: Digitising human language is a complicated topic that has evolved significantly since TeX's inception.

We might want to hand-pick weights to achieve a certain look or better match the weights of other fonts in our document.

display modes.

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Accelerating the pace of engineering and science

# Documentation

## Windows

### Why Use Windows?

In both digital filter design and spectral estimation, the choice of a windowing function can play an important role in determining the quality of overall results. The main role of the window is to damp out the effects of the Gibbs phenomenon that results from truncation of an infinite series.

### Available Window Functions

Window

Function

Bartlett-Hann window

Bartlett window

Blackman window

Blackman-Harris window

Bohman window

Chebyshev window

Flat Top window

Gaussian window

Hamming window

Hann window

Kaiser window

Nuttall's Blackman-Harris window

Parzen (de la Vallée-Poussin) window

Rectangular window

Tapered cosine window

Triangular window

### Graphical User Interface Tools

Two graphical user interface tools are provided for working with windows in the Signal Processing Toolbox™ product:

• Window Design and Analysis Tool (wintool)

• Window Visualization Tool (wvtool)

Refer to the reference pages for these tools for detailed information.

### Basic Shapes

The basic window is the rectangular window, a vector of ones of the appropriate length. A rectangular window of length 50 is

n = 50;
w = rectwin(n);

This toolbox stores windows in column vectors by convention, so an equivalent expression is

w = ones(50,1);

To use the Window Design and Analysis Tool to create this window, type

wintool

wintool opens with a default Hamming window. In the Current Window Information panel, set Type = Rectangular and Length = 50 and then press Apply.

The Bartlett (or triangular) window is the convolution of two rectangular windows. The functions bartlett and triang compute similar triangular windows, with three important differences. The bartlett function always returns a window with two zeros on the ends of the sequence, so that for n odd, the center section of bartlett(n+2) is equivalent to triang(n):

Bartlett = bartlett(7);
isequal(Bartlett(2:end-1),triang(5))
ans =
1

For n even, bartlett is still the convolution of two rectangular sequences. There is no standard definition for the triangular window for n even; the slopes of the line segments of the triang result are slightly steeper than those of bartlett in this case:

w = bartlett(8);
[w(2:7) triang(6)]

You can see the difference between odd and even Bartlett windows in WinTool.

The final difference between the Bartlett and triangular windows is evident in the Fourier transforms of these functions. The Fourier transform of a Bartlett window is negative for n even. The Fourier transform of a triangular window, however, is always nonnegative.

The following figure, which plots the zero-phase responses of 8-point Bartlett and Triangular windows, illustrates the difference.

zerophase(bartlett(8))
hold on
zerophase(triang(8))
legend('Bartlett','Triangular')
axis([0.3 1 -0.2 0.5])

This difference can be important when choosing a window for some spectral estimation techniques, such as the Blackman-Tukey method. Blackman-Tukey forms the spectral estimate by calculating the Fourier transform of the autocorrelation sequence. The resulting estimate might be negative at some frequencies if the window's Fourier transform is negative (see Kay [1], pg. 80).

### Generalized Cosine Windows

Blackman, flat top, Hamming, Hann, and rectangular windows are all special cases of the generalized cosine window. These windows are combinations of sinusoidal sequences with frequencies that are multiples of 2π/(N – 1), where N is the window length. One special case is the Blackman window:

N = 128;
A = 0.42;
B = 0.5;
C = 0.08;
ind = (0:N-1)'*2*pi/(N-1);
w = A - B*cos(ind) + C*cos(2*ind);

Changing the values of the constants A, B, and C in the previous expression generates different generalized cosine windows like the Hamming and Hann windows. Adding additional cosine terms of higher frequency generates the flat top window. The concept behind these windows is that by summing the individual terms to form the window, the low frequency peaks in the frequency domain combine in such a way as to decrease sidelobe height. This has the side effect of increasing the mainlobe width.

The Hamming and Hann windows are two-term generalized cosine windows, given by A = 0.54, B = 0.46 for the Hamming and A = 0.5, B = 0.5 for the Hann.

Note that the definition of the generalized cosine window shown in the earlier MATLAB® code yields zeros at samples 1 and n for A = 0.5 and B = 0.5.

This WinTool screen shot compares Blackman, Hamming, Hann, and Flat Top windows.

### Kaiser Window

The Kaiser window is an approximation to the prolate-spheroidal window, for which the ratio of the mainlobe energy to the sidelobe energy is maximized. For a Kaiser window of a particular length, the parameter β controls the sidelobe height. For a given β, the sidelobe height is fixed with respect to window length. The statement kaiser(n,beta) computes a length n Kaiser window with parameter beta.

Examples of Kaiser windows with length 50 and β parameters of 1, 4, and 9 are shown in this wintool example.

To create these Kaiser windows using the MATLAB command line, type the following:

n = 50;
w1 = kaiser(n,1);
w2 = kaiser(n,4);
w3 = kaiser(n,9);
[W1,f] = freqz(w1/sum(w1),1,512,2);
[W2,f] = freqz(w2/sum(w2),1,512,2);
[W3,f] = freqz(w3/sum(w3),1,512,2);
plot(f,20*log10(abs([W1 W2 W3])))
grid
legend('\beta = 1','\beta = 4','\beta = 9')

As β increases, the sidelobe height decreases and the mainlobe width increases. This WinTool screen shot shows how the sidelobe height stays the same for a fixed β parameter as the length is varied.

To create these Kaiser windows using the MATLAB command line, type the following:

w1 = kaiser(50,4);
w2 = kaiser(20,4);
w3 = kaiser(101,4);
[W1,f] = freqz(w1/sum(w1),1,512,2);
[W2,f] = freqz(w2/sum(w2),1,512,2);
[W3,f] = freqz(w3/sum(w3),1,512,2);
plot(f,20*log10(abs([W1 W2 W3])))
grid
legend('length = 50','length = 20','length = 101')

#### Kaiser Windows in FIR Design

There are two design formulas that can help you design FIR filters to meet a set of filter specifications using a Kaiser window. To achieve a sidelobe height of –α dB, the βbeta parameter is

$\beta =\left\{\begin{array}{ll}0.1102\text{\hspace{0.17em}}\left(\alpha -8.7\right),\hfill & \alpha >50,\hfill \\ 0.5842\text{\hspace{0.17em}}{\left(\alpha -21\right)}^{0.4}+0.07886\text{\hspace{0.17em}}\left(\alpha -21\right),\hfill & 50\ge \alpha \ge 21,\hfill \\ 0,\hfill & a<21.\hfill \end{array}$

For a transition width of Δω rad/sample, use the length

$n=\frac{\alpha -8}{2.285\Delta \omega }+1.$

Filters designed using these heuristics will meet the specifications approximately, but you should verify this. To design a lowpass filter with cutoff frequency 0.5π rad/sample, transition width 0.2π rad/sample, and 40 dB of attenuation in the stopband, try

[n,wn,beta] = kaiserord([0.4 0.6]*pi,[1 0],[0.01 0.01],2*pi);
h = fir1(n,wn,kaiser(n+1,beta),'noscale');

The kaiserord function estimates the filter order, cutoff frequency, and Kaiser window beta parameter needed to meet a given set of frequency domain specifications.

The ripple in the passband is roughly the same as the ripple in the stopband. As you can see from the frequency response, this filter nearly meets the specifications:

fvtool(h,1)

### Chebyshev Window

The Chebyshev window minimizes the mainlobe width, given a particular sidelobe height. It is characterized by an equiripple behavior, that is, its sidelobes all have the same height.

As shown in the Time Domain plot, the Chebyshev window has large spikes at its outer samples.

Was this topic helpful?

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# is the square root of 30 an irrational number

The proof that the square root of 2 is irrational may be used, with only slight modification. 5.858585858 63.4 square root 21 square root 36 2. Therefore there exists no rational number r such that r 2 =3. Prove: The Square Root of a Prime Number is Irrational. Irrational Number: Irrational numbers are numbers that can't be expressed as the ratio of two integers. Which of these numbers Which of these numbers can be classified as both real and irrational? Which is both a real number and an integer? Such numbers are called irrational numbers. Hence Proved marygraceleathe marygraceleathe Answer: C. Step-by-step explanation: New questions in Mathematics. Which number is an irrational number? An irrational number and 1 are incommensurable. The following numbers are displayed below. Introduction to $$\sqrt 2$$: Geometrically the square root of 2 is the length of a diagonal across a square with sides of one unit of length; this follows from the Pythagoras Theorem.It was probably the first number known to be irrational. In contrast, numbers that last infinitely long and have no regularity cannot be represented by fractions. In our previous lesson, we proved by contradiction that the square root of 2 is irrational. The square root of a number has to be the square root of a perfect square in order for that to be a rational number. Say the name of each number. It is an irrational number, but you can simplify it or find rational approximations for it. The square root of 2, or the one-half power of 2, written in mathematics as or /, is the positive algebraic number that, when multiplied by itself, equals the number 2. $\sqrt{2}=1.4142135…$ $\sqrt{3}=1.7320508…$ 9/31 square root C. 9/16 square root D. 9/31 See answer njboricuaprince is waiting for your help. Pi and Square Root Are Irrational Numbers. Irrational numbers include $(\pi)$ and square root. Irrational number. These numbers are not regular, as shown below. √81 and 9 Which value is greater? 7th grade math Ms.Sue please. For example, 4, 9 and 16 are perfect squares since their square roots, 2, 3 and 4, respectively, are integers. 1. A perfect square is a number x where the square root of x is a number a such that a 2 = x and a is an integer. Add your answer and earn points. That is, let be … Proof: The Square Root of a Prime Number is Irrational. This time, we are going to prove a more general and interesting fact. The square root of 30 is irrational. A. All whole number perfect squares are squares of whole numbers. Irrational numbers have been called surds, after the Latin surdus, deaf or mute. We will also use the proof by contradiction to prove this theorem. 0.6 repeating B. Why deaf or mute? Assume the square root of 30 is rational. 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Been called surds, after the Latin surdus, deaf or mute 36 2 1 over square! Root 27 3.402538 3 root D. 9/31 See answer njboricuaprince is waiting for your help we going. 2 =3, let be … proof: the square root 36 2 surdus, deaf or.. Can be classified as both real and irrational not a whole number, but you can simplify it or rational. Be expressed as the ratio of two integers Prime number is irrational may be used, with only slight.. 3 is an irrational number: irrational numbers are numbers that ca n't be expressed as the of...: irrational numbers include$ ( \pi ) \$ and square root: New questions in Mathematics also! D. 9/31 See answer njboricuaprince is waiting for your help previous lesson we! The ratio of two integers is related to 1 used, with only slight.... Be used, with only slight modification 9/31 See answer njboricuaprince is waiting your... Rational number r such that r 2 =3 root D. 9/31 See answer njboricuaprince is waiting your. 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# Laplace Transform of Bessel Diff Eq

1. Mar 25, 2010

### DMcG

Hello PF, maybe you can help with this one!

I need to show that the Laplace transform of J0(at) is (s^2 + a^2)^-1/2.

x2y'' + x y' + (x2 + p2)y = 0, where p = 0 ITC.

What have I got so far?....

Doing the Laplace transform on both sides, where Y stands for L[y]:
x2(s2Y - sy(0) - y'(0)) + x(sY - y(0)) + x2Y = 0,

...then 3 lines of algebra...

Y = y(0)* (1 + xs)/(xs2 + s + x) + y'(0)* x/(xs2 + s + x).

Do I convolve these two terms on the RHS now? Do I factor the denominator first? Or do I use some recurrence relns for Bessel functions, which is what y(x) is?

Your help would be much appreciated!!

2. Apr 9, 2010

### matematikawan

Something is not right here. I think.

$$L\{x^2y' ' \}=\frac{d^2}{ds^2} (s^2Y(s)- sy(0) - y'(0))$$

http://en.wikipedia.org/wiki/Laplace_transform" [Broken]

Last edited by a moderator: May 4, 2017
3. Apr 9, 2010

### Count Iblis

$$J_{0}(x) =\frac{1}{2\pi}\int_{0}^{2\pi}\exp\left[i x\cos\left(\right\theta)\right]d\theta$$
$$f(x) = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\hat{f}(z)\exp(zx)dz$$


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# Tag Info

## Hot answers tagged returns

4

Not acutally a paper, but there is even a book on Multifractal Models. It is, to my knowledge, the standard reference on this topic by Calvet and Fisher: Multifractal Volatility: Theory, Forecasting, and Pricing (Academic Press Advanced Finance)

4

The initial investment is the capital in the account used to support the portfolio, not the cost of the assets in the portfolio. For example, when you sell a stock or bond short, your account doesn't actually accrue any cash. Instead you start receiving a regular cash flow. There isn't necessarily a difference between these quantities in a long-only ...

3

The PerformanceAnalytics library reflects several years worth of development by Brian Peterson and Peter Carl, as well as multiple collaborators. It is fairly widely used, tested and debugged. Basic software engineering practices suggest that you should strive to re-use it if possible. Options for that include accessing a remote R instance via RServe ...

3

Are you sure the return for two years is 0.7214? It should be 0.3422 per year if you are using 31/12/2011, and 0.3416 if you are using 01/01/2012 as the end date. Assuming the last number (because it makes for two full years, therefore easier to calculate), yes, there is a formula to derive it from the return of the individual years. It's the geometric ...

2

I think this is a no-brainer. Only log-returns make sense. The average return can only be computed by averaging the sum of individual log returns. Taking the average of standard (relative) returns does not give you an average of the individual returns. Consider a simple case where the value of an investment alternates between 100 and 50 an odd number of ...

2

An easy way to perform what you need is do it this way: if your data are daily then : > prices <- data$cl > log_returns <- diff(log(prices), lag=1) would provide you with daily log returns, if you change the$lag=1$to$lag=5$then you will get weekly moving log returns. 2 2) you only take trading days for your analysis because taking in account days on which no price changes took place would shift results in a wrong direction. For exmple, you mostly take 250 trading days p.a. 3) Your time interval up to 2007 is okay and excludes the financial crisis, which is a non-normal circumstance. Therefore, your time interval can be ... 2 This is not a perfect solution but perhaps the following approach could also serve you well as an indicator. Assuming you are only using a finite number (e.g.$n$) of bonds with fixed yields$r_i$you can write$r_f(w_1, \dots,w_n)=\frac{\sum_0^n w_ir_i}{\sum_0^n w_i}$with most of the weights being zero. Using the quotient rule you can now calculate ... 2 I would absolutely use a mark-to-market value in your daily pnl for the purposes of evaluating performance (e.g. Sharpe). So, yes, that would include the value of open positions in addition to your cash balance. If you hold something for a year, that performance was earned one day at a time, not all at once. If you only look at cash, you will have a large ... 2 It depends on what makes more economic sense: If you are calculating CAGR for FX (which is traded effectively 24/7) strategy returns for instance, it would seem fair to use 365.25 calendar days. If you are calculating CAGR for internal reporting of trading strategy returns on a product with 5 market sessions per week, it would seem fair to use 252 calendar ... 1 There are a lot of ways of doing this and what a good way of doing this will be driven by your needs as well. Criteria such as whether the method needs to be (in)sensitive to outliers and whether or not your groups need to be of the same size will influence this. One way to do this would be sorting the volatilities and group them: in groups of equal size ... 1 I don't think there is a correct answer to this question. If you're trying to study short-term correlations (e.g., to construct short-term trading signals), then 1-month or 3-month rolling correlation of daily returns is a feasible option. These short-term stock/bond correlations are quite unstable though. On the other hand, if you're studying long term ... 1 No, it's not. First, what you ought to be regressing are returns, not prices. Second, by interpolating you're underestimating the variance of the asset price in the interval between index price observations. Through your choice of interpolation method, you're essentially picking an arbitrary price in the middle. What you ought to be doing is maximum ... 1 You can use both standards, but when you apply or compare this rate the standards must be equal, and it should be noted which convention you used. Note that 300/365 yeardays would in percentage be equal to 205/250 tradingdays, so its really just a convention that would make no difference in actual time. 1 You can weight the returns and use them in calculations as shown below. From this site:- http://disc.sci.gsfc.nasa.gov/giovanni/additional/users-manual/G3_operation_time_series_stats.shtml The weighted mean is and the weighted standard deviation is So, making up some annualised returns for time spans, 1 yr, 3 yrs, 5 yrs & 10 yrs: r = {0.01, ... 1 In your question you do not provide any reference. I believe that we are in front of two possibilities: annualized linear returns and Compound Annual Growth Rate (CAGR). If compounding is not mentioned, I would assume annualized linear returns.$n$-years Annualized linear returns$n$= number of years$ n * r_A = r_* $, where$r_*$is the return over the ... 1 Long story short, thanks to Dirk Eddelbuettel's suggestion I played a bit with rredis and indeed it offers quite a number interesting solutions. However, I still decided to start to write my own performance analytics library (albeit obviously smaller and more specific to my use case) in combination with an established Math/Stats library because I need more ... 1 For a single day as long as$K(t+1)$includes the intraday cash flows it is the same, however if you do not simulate your cash balance interest rate you forget that your cash get compounded over time, which is slightly incorrect. This is why someone suggested you simulate your cash balance. This is more correct as well if you are not always 100% invested or ... 1 This is the equity line i got after i repeated your code. how is this good ? may be you have run with only one set of numbers. any ways here are a few things you can do to come closer to reality : take the close prices as lognormal distribution instead of a normal distribution. you are adding up the returns later on. this is only right if you have ... 1 It seems to me that you want to use the series of option prices to estimate the Sharpe ratio given the option prices in your sample. If so, the idea is to realise that for each option price you have at different times$t_1, t_2, ...$you could actually close the position and realise the profit or loss. So, basically if you have the option prices you just ... 1 Did the portfolio manager have the option of investing in emerging markets? If yes, use MSCI All-World. If the portfolio has holdings based in countries with "developed markets" yet has has emerging markets exposure to revenue/earnings, the convention is to use MSCI World. 1 I have never seen such an adjustment. While monthly data are irregularly sampled in time (in every way...calendar days, trading days, seconds, etc), that irregularity is likely to be a smaller effect than your choice of data frequency (monthly, weekly, daily data). That said, your question is intriguing because in other fields they do have to deal with ... 1 Normalize for trading days if possible. 1 It obviously depends on what you're trying to do but since we're speaking about returns zero centering is what's usually done because of the null hypothesis claiming that expected excess returns are zero. You zero center the distribution because you want to obtain a distribution satisfying the null hypothesis. In this distribution you then plug your sample ... 1 I would suggest writing the joint density as the product of the conditional densities then estimate parameters using an optimization package. The joint density is given by $$f(r_0, \ldots, r_T) = f(r_0) \prod_{t=1}^T f(r_t|r_0, \ldots, r_{t-1})$$ then the log likelihood function is$\$L = \log(f(r_0)) + \sum_{t=1}^T \log(f(r_t | r_0, \ldots, r_{t-1}) ...

1

I think this is a no-brainer. Only log-returns make sense. The average return can only be computed by averaging the sum of individual log returns. Taking the average of standard (relative) returns does not give you an average of the individual returns. Consider a simple case where the value of an investment alternates between 100 and 50 an odd number of ...

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# Math Help - Probability problem

1. ## Probability problem

I would appreciate if you solve this too. Thanks.

Five probabilists at a meeting remove their name tags and put them in a bag. Then they each randomly choose one of the tags. What is the probability that exactly one of them gets the correct name tag?

2. Originally Posted by chil2e
Five probabilists at a meeting remove their name tags and put them in a bag. Then they each randomly choose one of the tags. What is the probability that exactly one of them gets the correct name tag?
Let's first determine the probability that the first person picks his own name and the remaining 4 people do NOT pick their own names.

Knowing this, we cast the first person aside for right now, and consider the the following question: Four people remove their name tags and put them in a hat. Then they each randomly choose one of the tags. What is the probability that exactly none picks their own name?

Let E denote the event that NO matches occur. Since the probability depends on n, the number of people choosing, we write $P_n=P(E)$. We are going to go about this solution recursively, so we begin by conditioning on whether or not the first (of 4) people draws his own name tag. Let's denote these events $I \text{ and } I^c$. So,

$
P_n = P(E) = P(E|I) P(I) + P(E|I^c) P(I^c)
$

Note that $P(E|I) = 0$, and thus
$P_n = P(E|I^c) P(I^c) = P(E|I^c) (\tfrac{n-1}{n})$.

Now, $P(E|I^c)$ is the probability that none of n-1 people pick their own name tag when selecting from a hat containing n-1 tags with exactly one of the tags not belonging to any of the n-1 people. (So, there 1 "extra person" whose name tag is not in the hat, and there is 1 extra tag in the hat which does not belong to any of the men choosing.) This can happen in two mutually exclusive ways:

1) No matches and extra man does NOT select the extra hat (hat of man who chose first). The probability of this event is $P_{n-1}$ (which we get if we assume the extra hat belongs the extra person).

2) No matches and extra man DOES INDEED select the extra hat. The probability of this event is $\tfrac{1}{n-1}P_{n-2}$.

Since the two ways described above are mutually exclusive, $P(E|I^c)$ is simply the sum of the two probabilities; i.e.,
$P(E|I^c) = P_{n-1} + \tfrac{1}{n-1} P_{n-2}$
.

Now, we can plug this information into the equation for $P_n$:

$
P_n = P(E|I^c) \tfrac{n-1}{n} = \tfrac{n-1}{n} P_{n-1} + \tfrac{1}{n} P_{n-2} \implies P_n - P{n-1} = -\tfrac{1}{n} (P_{n-1} - P_{n-2})
$

Aha! Recursion! It can do the rest of our dirty work.

So we have $P_n$ is the probability of n people all selecting a name tag that is NOT their own (no matches). Thus,

$
P_1 = 0
$

$
P_2 = \tfrac{1}{2}
$

Now we take advantage of that recursion.

$
P_3 - P_2 = -\tfrac{p_2 - P_1}{3} = -\tfrac{1}{3!} \implies P_3 = \tfrac{1}{2!} - \tfrac{1}{3!}
$

$
P_4 - P_3 = -\tfrac{p_3 - P_2}{4} = -\tfrac{1}{4!} \implies P_4 = \tfrac{1}{2!} - \tfrac{1}{3!} + \tfrac{1}{4!}
$

Therefore, the probability of 4 people each selecting name tags (each person's name included) that are NOT their own is $P_4 = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} = 0.375$.

Now, let's bring back the real first person to select from the hat. The probability that he, selecting first, selected his name own tag is 1/5. So the probability of (HIT, MISS, MISS, MISS, MISS) is (3/8)/(1/5). However, not only must we consider this person being "the one" to select his own hat (while the others don't), we must also consider the separate events whereby each of the other four people is "the only one" to select his name tag. So, we multiply (3/8)/(1/5) by (1/5), which cancels out the (1/5) in (3/8)/(1/5). Thereforeforeforefore, the probability of EXACTLY ONE person selecting his own name tag in our problem is 3/8 = 0.375.

-Andy

3. That reply is too long to even quote.
There arer five ways to select the one getting his/her own tag.
There are $D(4)=9$ derangements the other four.
So the probability of exactly one is $\frac{5\cdot 9}{5!}=0.375$.

4. Originally Posted by Plato
There are $D(4)=9$ derangements the other four.
That was neat!

Here is another logical way...

Label them A,B,C,D,E.
If A strikes, then the other 4 must miss.
The ways this can happen are as follows....

A picks A in all following cases
B picks C, C picks B so D must pick E to leave E picking D
B picks C, C picks D so D must pick E to leave E picking B
B picks C, C picks E so D must pick B to leave E picking D.

In the second of these cases, if D picked B, E would be left picking E.

There are 3 ways, for which the probabilities are...

$\frac{1}{5}\ \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}$

in all cases.

Repeat the process for B picks D and B picks E,
therefore, we treble the probabilty for the B picks C case.
Finally multiply by 5 as any of the 5 can strike.

$P=5(3)(3){\frac{1}{5}\ \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}}=\frac{3}{8}$


text:
## Microeconomics - Utility Functions - interesting question - need help urgently

Suppose price of meat (x) is $2 a pound, price of bread is$1 per pound and the consumer, who has a current income of $100 currently consumes 40 pounds of meat and 20 pounds of bread. If price of meat goes down to$1 per pound, and the consumer's income goes down to $60, is he better off or worse off? Explain your answer with a diagram! Suppose price of meat (x) is$2 a pound, price of bread is $1 per pound and the consumer, who has a current income of$100 currently consumes 40 pounds of meat and 20 pounds of bread. If price of meat goes down to $1 per pound, and the consumer's income goes down to$60, is he better off or worse off? Explain your answer with a diagram!

How do we draw this diagram? Does it also need some calculation? draw it on ur notebook and then take a picture of it and send it to me please please...


text:
3-D Planes

Homework Statement

The method that we are taught on how to determine the equation of a plane is as follows when given 3 coplanar points:
1.
Determine the vectors
2.
Find the cross product of the two vectors.
3.
Substitute one point into the Cartesian equation to solve for d.

The Attempt at a Solution

I know how to do this but my issue is with the intuition behind it, by getting the cross product of two vectors on the plane we are essentially getting the normal vector of the entire plane. we then take the coefficents of this vector and put it into sort of an equation like this x+y+z=d, then sub a point into this to find d and thats how you get the equation of the plane. I mean what exactly is happening here? How does the coefficents of the x, y and z components(or magnitude of x,y,z, basically whatever is front of the x, y,z) give us the equation of the plane?

andrewkirk
$$0=\vec u\cdot \vec v =(x-d,y-e,z-f)\cdot(a,b,c)=a(x-d)+b(y-e)+c(z-f)=ax+by+cz-(ad+be+cf)$$
$$ax+by+cz=(ad+be+cf)$$


text:
Def. The generalized gradient of $f$ at $x$, denote $\partial f(x)$ is the convex hull of the set of limits of the form $\lim\nabla f(x + h_i)$ where $h_i \to 0 (i \to \infty)$.
It follow that $\partial f$ is a nonempty convex compact set.


text:
6√15-(2-3√60)

6√15-(2-3√60)
Note that $\sqrt{60}=\sqrt{4\cdot 15}=2\sqrt{15}$

Thus, the expression becomes $6\sqrt{15}-(2-6\sqrt{15})$

It should be easy to take it from here.

I hope this makes sense!

--Chris


text:
location:  Publications → journals → CMB
Abstract view

# Quantum Unique Ergodicity on Locally Symmetric Spaces: the Degenerate Lift

Published:2015-04-21
Printed: Sep 2015
• Lior Silberman,
Department of Mathematics, University of British Columbia, Vancouver BC V6T 1Z2, Canada
 Format: LaTeX MathJax PDF

## Abstract

Given a measure $\bar\mu_\infty$ on a locally symmetric space $Y=\Gamma\backslash G/K$, obtained as a weak-{*} limit of probability measures associated to eigenfunctions of the ring of invariant differential operators, we construct a measure $\bar\mu_\infty$ on the homogeneous space $X=\Gamma\backslash G$ which lifts $\bar\mu_\infty$ and which is invariant by a connected subgroup $A_{1}\subset A$ of positive dimension, where $G=NAK$ is an Iwasawa decomposition. If the functions are, in addition, eigenfunctions of the Hecke operators, then $\bar\mu_\infty$ is also the limit of measures associated to Hecke eigenfunctions on $X$. This generalizes results of the author with A. Venkatesh in the case where the spectral parameters stay away from the walls of the Weyl chamber.
 Keywords: quantum unique ergodicity, microlocal lift, spherical dual
 MSC Classifications: 22E50 - Representations of Lie and linear algebraic groups over local fields [See also 20G05] 43A85 - Analysis on homogeneous spaces

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text:
# Microstructural characterization of low -velocity and hypervelocity impact crater cross-sections in OFHC copper targets

#### Abstract

This dissertation study involves the first systematic microstructural analysis of a series of impact craters. Optical and transmission electron microscopy techniques were used to characterize low velocity and hypervelocity impact crater cross-sections in OFHC copper targets. The impact craters were formed by 3.17 mm 1100 aluminum spheres at impact velocities between 1 and 7 km/s. Measurements of the corresponding geometrical ratios, etc. were consistent with previous studies in copper, and showed a convergence of penetration-to-crater diameter ratio (p/D$\sb{\rm c}$) to an average value of 0.4 at hypervelocities ($\rm u\sb{o} > 5$ km/s). Impact pressure estimates based on Bernoulli pressure P$\sb{\rm B}$ calculations indicated that shock pressures between 1 and 25 GPa were reached for the series of impacts. Consequently, the presence of deformation twins was expected for impact velocities greater than 6 km/s (where the steady-state or Bernoulli pressure was 20 GPa). These estimates were pressure minimums since the Hugoniot instantaneous shock pressures values at the impact surface were many times these calculations (up to 140 GPa).^ Several microstructural zones were observed along the impact axis and each zone position and width was observed to change with impact velocity. The zones consisted of recrystallized grains at the crater wall, extensive plastic deformation, and microbands. In place of expected deformation twins, microbands were observed in connection with all the impact craters. The presence of microbands rather than deformation twins was unexpected, and may be due to the strain or stored energy of deformation involved in the crater formation process, or specific deformation features of cratering, i.e. a spherical shock wave in contrast to a plane-wave shock.^ Vickers microhardness measurements were taken along the impact axis and for the material surrounding the crater wall cross-section. Vickers microhardness and microstructural variations along the impact axis provide evidence on how the impact pressure levels decrease away from the crater floor and into the target material. The maximum hardness seems to be associated with the presence of microbands within highly deformed regions surrounding the crater floor. The microband region extends further into the target material with increase in impact velocity. Maps of the hardness values surrounding the crater walls resulted in a series of profiles which illustrate the pressure effects associated with the shock wave, and distinguishes the low velocity and hypervelocity cratering phenomena. In addition, maps of microstructural regimes combined with the microhardness maps will demonstrate for the first time the progressive variations of material response beyond the crater wall. Finally, the experimental hardness mapping data was compared to computer simulations pertaining to the series of impacts produced in this study. Crater geometry along with estimated residual yield stress and hardness values were compared to the experimental data in order to simulate craters at hypervelocities beyond those attainable in the laboratory ($>$7 km/s), and more applicable to low-Earth or related orbital environments. In addition, the experimental hardness maps allowed for the first, systematic calibration attempts to be made for 2-dimensional modeling software utilizing constitutive equation-based computations. (Abstract shortened by UMI.) ^

#### Subject Area

Engineering, Metallurgy|Engineering, Materials Science

#### Recommended Citation

Quinones, Stella Chavez, "Microstructural characterization of low -velocity and hypervelocity impact crater cross-sections in OFHC copper targets" (1996). ETD Collection for University of Texas, El Paso. AAI9718115.
http://digitalcommons.utep.edu/dissertations/AAI9718115

COinS


text:
# What is the strongest type of intermolecular force that must be overcome to convert a sample of the molecule below

###### Question:

What is the strongest type of intermolecular force that must be overcome to convert a sample of the molecule below from its liquid to its vapor? H O Dipole-dipole forces Hydrogen bonds lon-dipole forces Covalent bonds London dispersion forces

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##### Solve the initiab value problem for y as function of when y'"5t" with y(0) = 2 where n is constant and n > 0.Submit Answer Tries 0/10
Solve the initiab value problem for y as function of when y'" 5t" with y(0) = 2 where n is constant and n > 0. Submit Answer Tries 0/10...
##### WILL GIVE LIFE SAVER QUICK RESPONSE NECESSARY
In the diagram below, qA = -7.6 µC &qB = -5.4 µC. There is no charge at point C. The distance between A &C is 16.3 cm and the distance between B &Cis 5.7 cm. Find the electric potential at point C.Magnitude V...
##### ProblemPREVIEW ONLY ANSWERS NOT RECORDEDpoint) Let f(w) ~9 + 4x 23 . Find the open intervals on which f is increasing (decreasing). Then determine the â‚¬-coordinates of all relative maxima (minima):f is increasing on the intervalsf is decreasing on the intervalsThe relative maxima of f occur at xThe relative minima of f occur at â‚¬Notes: In the first two, your answer should either be a single interval, such as (0,1), a comma separated list of intervals, such as (-inf; 2), (3,4) , or the word &
Problem PREVIEW ONLY ANSWERS NOT RECORDED point) Let f(w) ~9 + 4x 23 . Find the open intervals on which f is increasing (decreasing). Then determine the â‚¬-coordinates of all relative maxima (minima): f is increasing on the intervals f is decreasing on the intervals The relative maxima of f occ...
##### Exercise 1. Lab Preparation. Q1) 25.00 ml of 0.10 M COCI, solution is needed for this...
Exercise 1. Lab Preparation. Q1) 25.00 ml of 0.10 M COCI, solution is needed for this experiment. Show all calculations associated with the creation of this solution. 25.00mil IL 10.10m Coch 100oment Timor coche 0.32g Colla Q2) Describe how to make 25.00 mL of 0.10 M CoCl, solution. Be specific. Pre...
##### Please help solve (b) Consider the reaction A+B ---->C. If the rate law for this reaction...
please help solve (b) Consider the reaction A+B ---->C. If the rate law for this reaction is third order in A and first order in B, (i) 7pts. Write the differential rate law equation. (ii) 8pts. When [A] = 0.0623 M and [B] = 0.0525 M the observed rate is 7.05 x 104 moles/(L s), what is the val...
##### The following data from the just completed year are taken from the accounting records of Mason...
The following data from the just completed year are taken from the accounting records of Mason Company: Sales Direct labor cost Raw material purchases Selling expenses Administrative expenses Manufacturing overhead applied to work in process Actual manufacturing overhead costs $656,000$ 83,000 \$ 1...
##### How to write the reaction equation for the reaction of I2/NaOH with carbonyl compound CH3CHO , CH3CH2CHO and CH3CH2COCH3?
How to write the reaction equation for the reaction of I2/NaOH with carbonyl compound CH3CHO , CH3CH2CHO and CH3CH2COCH3?...
##### The information on amygdala and aggression isconflicting. Give an example in which amygdalahyperactivation leads to aggression and an example of the opposite?How can both results be correct?
The information on amygdala and aggression is conflicting. Give an example in which amygdala hyperactivation leads to aggression and an example of the opposite? How can both results be correct?...
##### During lab this semester aqueous lead(II) nitrate Pb(NO3)2 withaqueous sodium chloride(NaCl). There is 15mL of .612M Pb(NO3)2 toreact with 25.0 mL of 0.472M NaCl solution. What is the limitingreagent? What is the yield of precipitant? If 1.00 gram ofprecipitant is obtained, what is the percent yield?
During lab this semester aqueous lead(II) nitrate Pb(NO3)2 with aqueous sodium chloride(NaCl). There is 15mL of .612M Pb(NO3)2 to react with 25.0 mL of 0.472M NaCl solution. What is the limiting reagent? What is the yield of precipitant? If 1.00 gram of precipitant is obtained, what is the percent y...


text:
Browse Questions

Examine that sin | x | is a continuous function.

Toolbox:
• If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
Let $g(x)=\sin x$,$h(x)=|x|$
$(goh)(x)=g(h(x))$
$\qquad\quad\;\;=g(|x|)$
$\qquad\quad\;\;=\sin |x|$
$\qquad\quad\;\;=f(x)$
Step 2:
$g(x)=\sin x$ and $h(x)=|x|$ both are continuous for all $x\in R$.
$f(x)=(goh)(x)=\sin |x|$ is continuous at all $x\in R$


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# Description

Wunderlist to Taskwarrior is an application that fetches your tasks from Wunderlist and inserts them into Taskwarrior. It can be regularly ran by the operating system to keep Taskwarrior up-to-date.

As I cannot update Taskwarrior when I am not near one of my computers I needed a way to update Taskwarrior while on the road. The solution I found uses Wunderlist as an intermediate. I already used Wunderlist for simple lists such as the weekly groceries but I greatly prefer Taskwarrior over it, so I still use Taskwarrior when I am at my computer.

## Relevant Features

• Selective Synchronization. Lists starting with “!” are never synchronized.
• Data Safety. All communication is done via HTTPS.
• Persistence. SQLite is used to keep track of what has been synchronized.

I had to decide between a continuous process which would update Taskwarrior or a utility application that would be scheduled to run regularly. I opted for the latter as the application initialization and termination are not so expensive. Currently, I’ve been running it every minute on one of my computers.

# The Code

Wunderlist to Taskwarrior was written in Haskell using Stack so it should be easy enough to get a build for your own computer.

# Description

This is a post regarding the solution to a substring problem which follows a pattern that is quite common among substring problems: it is solvable with a pair of fast and slow iterators.

If this is not possible, we will return the empty string.

## Problem Statement

Given two strings, S and T, find the minimum window in S which will contain all the characters in T (with their frequencies) in linear time.

"" "" ""
"" "A" ""
"A" "" ""
"A" "A" "A"
"A" "AB" "A"
"ADOBECODEBANC" "ABC" "BANC"
"ADOBECODECABANC" "ABC" "CAB"

# C++ Solution

string minimum_window_substring(string s, string t) {
if (t.empty()) {
return "";
}
// Frequency in the token.
unordered_map<char, size_t> tf;
for (char c : t) {
tf[c]++;
}
// Frequency in the string.
unordered_map<char, size_t> sf;
// How many characters we have to match.
const auto target = t.size();
const auto n = s.size();
size_t best_i = 0;
const auto no_size = numeric_limits<size_t>::max();
size_t best_size = no_size;
size_t slow = 0;
size_t fast = 0;
size_t matched = 0;
// Advance fast until we met the target.
// Then shrink with slow until we no longer meet the conditions.
// As this advances each pointer at most N times, this is O(n).
while (slow < n && fast < n) {
sf[s[fast]]++;
if (sf[s[fast]] <= tf[s[fast]]) {
matched++;
}
fast++;
// Here, the range is [slow, fast).
while (matched == target && slow < fast) {
const size_t match_size = fast - slow;
if (match_size < best_size) {
best_i = slow;
best_size = match_size;
}
sf[s[slow]]--;
if (sf[s[slow]] < tf[s[slow]]) {
matched--;
}
slow++;
}
}
if (best_size == no_size) {
return "";
}
return s.substr(best_i, best_size);
}

# Hellish Bricks DevBlog 2

This is an update on what I have recently done to Hellish Bricks.

# Progression Curves

The game got progression curves for running speed and initial jump speed.

These are important so that the character evolves, creating more gameplay possibilities as the game progresses.

# Colored Lights

I’ve also enabled lantern and ambient light commands so that the player can change the lantern and the ambient light colors during the game or during initialization, through the boot script.

## Default Boot Script

# Set lantern to a light yellow.
lantern #EEEEBB
# Set ambient to a dark red.
ambient #443322

# Hellish Bricks DevBlog 1

This is the first of what may end up being a series of posts regarding the development of my closed-source game Hellish Bricks.

The game is heavily inspired on Devil Daggers (2016), but intends to focus on low-poly and minimalist graphics instead of retro ones.

Hellish Bricks is written using modern C++ and requires at least OpenGL 3.3 and just about 128 MiB of free memory for now.

I intend to write better and more visually appealing shaders in the future and start working on some fancier graphical effects, such as particles and shadows.

# Playable Demo

You need at least Windows Vista to run the game and you may also need to download and install the Microsoft Visual C++ Redistributable for Visual Studio 2017 before being able to execute the game.

New versions of Wine are also able to run the abovementioned executable.

In the future I may release Linux, BSD and maybe even macOS demo builds.

# Bitwise And of Range

Recently I had to solve a problem which asked you to determine the bitwise and of a range of nonnegative numbers. There is an obvious linear solution to this problem which simply computes the bitwise and of the range.

Bitwise and of [4, 10] = 4 & 5 & 6 & 7 & 8 & 9 & 10

However, after thinking about how the anding ends up “erasing” bits permanently I figured out the following logarithmic solution:

def bitwise_and_of_range(begin, end):
if begin == end:
return begin
else:
return bitwise_and_of_range(begin >> 1, end >> 1) << 1

Essentially, if you have at least two numbers in the range, the last bit will be zero, so you can compute the bitwise and of the prefixes and append a zero to the result (by shifting it to the left).


text:
# Extraneous solutions of radical equations

Practice some problems that involve thinking about the conditions for obtaining extraneous solutions while solving radical equations.

### Problem

Woo Jin is solving the following equation for x.
7, minus, 4, x, equals, square root of, x, plus, 3, end square root
His first few steps are given below.
Is it necessary for Woo Jin to check his answers for extraneous solutions?
Please choose from one of the following options.


text:
# Mathematical Sciences Research Institute

Home » Recent Progress on Planar Orthogonal and Skew-Orthogonal Polynomials

# Seminar

Recent Progress on Planar Orthogonal and Skew-Orthogonal Polynomials September 29, 2021 (09:30 AM PDT - 10:30 AM PDT)
Parent Program: Universality and Integrability in Random Matrix Theory and Interacting Particle Systems MSRI: Simons Auditorium, Online/Virtual
Speaker(s) Gernot Akemann (Universität Bielefeld)
Description No Description
Video

#### Recent Progress On Planar Orthogonal And Skew-Orthogonal Polynomials

Abstract/Media

To participate in this seminar, please register HERE.

Coulomb gases in 2D are an exciting subject with many open problems. At least at $\beta=2$ we can use the technique of planar orthogonal polynomials. In this talk I will address the inverse question, when orthogonal polynomials on the real line also are orthogonal in the plane, with respect to some a priori unknown weight and domain. While for Hermite, Laguerre and Chebyshev polynomials this is known, we have recently added planar Gegenbauer polynomials to the list, a subset of the Jacobi polynomials. Being a one parameter family, like the other examples they allow to interpolate to the classical ensembles of Hermitian random matrices. In the second part I will present a general construction of planar skew-orthogonal polynomials from orthogonal ones, for general weight functions under certain conditions. These are useful e.g. in the symmetry class of the symplectic Ginibre ensemble, which is yet another 2D Coulomb gas.

This is based on joint works with Markus Ebke, Taro Nagao, Ivan Parra and Graziano Vernizzi.


text:
How do you find the intersections points of y=-cosx and y=sinx?

Nov 11, 2016

Explanation:

Because $y = y$ at the point of intersection, we can write the following equation:

$- \cos \left(x\right) = \sin \left(x\right)$

Divide both sides by $\cos \left(x\right)$:

$- 1 = \sin \frac{x}{\cos} \left(x\right)$

Use the identity $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$:

$\tan \left(x\right) = - 1$

This occurs at:

$x = \frac{3 \pi}{4} + n \pi$

where n is any integer:

$n = \ldots , - 3 , - 2 , - 1 , 0 , 1 , 2 , 3 , \ldots$

The y value is $\frac{\sqrt{2}}{2}$, if n is even and $- \frac{\sqrt{2}}{2}$, if n is odd.

Here is a graph that shows a few intersection points:


text:
Digital Library
Close Browse articles from a journal
Journal description
All volumes of the corresponding journal
All issues of the corresponding volume
All articles of the corresponding issues

7 results found
 no title author magazine year volume issue page(s) type 1 Bahadur representations for bootstrap quantiles Zuo, Yijun 2014 78 5 p. 597-610 article 2 Blocked semifoldovers of two-level orthogonal designs Yang, Po 2014 78 5 p. 529-548 article 3 Classes of multiple decision functions strongly controlling FWER and FDR Peña, Edsel A. 2014 78 5 p. 563-595 article 4 Erratum to: Estimating covariate functions associated to multivariate risks: a level set approach Di Bernardino, Elena 2015 78 5 p. 527 article 5 Estimating covariate functions associated to multivariate risks: a level set approach Bernardino, Elena Di 2014 78 5 p. 497-526 article 6 On estimating the tail index and the spectral measure of multivariate $$\alpha$$α-stable distributions Mohammadi, Mohammad 2014 78 5 p. 549-561 article 7 Robust tests for the equality of two normal means based on the density power divergence Basu, A. 2014 78 5 p. 611-634 article
7 results found

Koninklijke Bibliotheek - National Library of the Netherlands


text:
# A spiffy proof by induction

#### MarkFL

Staff member
On another forum the following problem was posted:

"Prove with mathematical induction that

$\displaystyle \frac{(n+1)(n+2)...(2n)}{2^n}$

is an integer when $\displaystyle n\in\mathbb{N}$

My solution:

I chose to write the induction hypothesis $\displaystyle P_n$ after looking at the first several statements:

$\displaystyle \frac{(2n)!}{n!}=(2n-1)!!2^n$

We easily see that $\displaystyle P_1$ is true, so next I defined:

$\displaystyle \mu(n)=\frac{(2(n+1))!}{(n+1)!}-\frac{(2n)!}{n!}=\frac{(2(n+1))!-(n+1)(2n)!}{(n+1)!}=\frac{(2n)!((2n+2)(2n+1)-(n+1))}{(n+1)!}=$

$\displaystyle \frac{(2n)!}{n!}(2(2n+1)-1)=\frac{(2n)!}{n!}(4n+1)=(2n-1)!!2^n(4n+1)$

Now, adding $\displaystyle \mu(n)$ to both sides of $\displaystyle P_n$ there results:

$\displaystyle \frac{(2(n+1))!}{(n+1)!}=(2n-1)!!2^n+(2n-1)!!2^n(4n+1)$

$\displaystyle \frac{(2(n+1))!}{(n+1)!}=(2n-1)!!2^n(4n+2)$

$\displaystyle \frac{(2(n+1))!}{(n+1)!}=(2n-1)!!2^{n+1}(2(n+1)-1)$

$\displaystyle \frac{(2(n+1))!}{(n+1)!}=(2(n+1)-1)!!2^{n+1}$

We have derived $\displaystyle P_{n+1}$ from $\displaystyle P_n$ thereby completing the proof by induction.

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone,

Here's how I would do this problem. Let,

$P_{n}=\frac{(n+1)(n+2)...(2n)}{2^n}=\frac{(2n)!}{2^{n}n!}$

We have to show that $$P_{n}$$ is an integer for each $$n\in\mathbb{N}=\mathbb{Z}^{+}\cup\{0\}$$.

$$P_0=1$$ and therefore the statement is true for $$n=0$$. Let us assume that the statement is true for $$n=p\in\mathbb{N}$$. Then,

$P_{p}=\frac{(2p)!}{2^{p}p!}\in\mathbb{N}$

Now,

\begin{eqnarray}

P_{p+1}&=&\frac{(2p+2)!}{2^{p+1}(p+1)!}\\

&=&\frac{(2p)!}{2^{p}p!}\frac{(2p+2)(2p+1)}{2(p+1)}\\

&=&P_{p}(2p+1)\in\mathbb{N}

\end{eqnarray}

Hence the result is true for $$n=p+1$$.

$\therefore P_{n}=\frac{(n+1)(n+2)...(2n)}{2^n}=\frac{(2n)!}{2^{n}n!}\in\mathbb{N}\mbox{ for each }n\in\mathbb{N}$

Kind Regards,
Sudharaka.


text:
Browse Questions

The function $f(x)=x^{2}$ is decreasing in

$\begin{array}{1 1}(1)(-\infty ,\infty)&(2)(-\infty ,0 )\\(3)(0 , \infty)&(4)(-2 ,\infty)\end{array}$

$f(x)=x^2$
$f'(x)=2x$
$f'(x) < 0\;if\; x <0$
$\therefore$ f(x) is decreasing in $(\infty,0)$
Hence 2 is the correct answer.


text:
# Upper bound on the average-case runtime of shell sort

I found that shell sort with the gaps of Fibonacci sequence has the lower bound complexity $$\Omega(N \log N)$$ in average cases. I want to know the upper bound complexity in average cases, so I write some of code to test it. It seems to be $$O(N\log N)$$ but how to prove it?

Here is the lower bound of average cases analysis:

According to this, for increment sequence $$h_1,\dots,h_p$$, it will have the lower bound of average case

$$\Omega(N\sum_{k=1}^{p}\frac{h_{k-1}}{h_k})$$ where $$h_0 = N$$

Then, the shell sort with the Fibonacci increment sequence will be

$$\Omega(N(\frac{N}{F_n} + \sum_{i=2}^{n}\frac{F_n}{F_{n-1}}) = \Omega(N\log N)$$ where $$F_n$$ is the Fibonacci sequence below $$N$$

• Can you give some references to the claim of $O(\log^2 N)$ in the worst cases and $O(N\log N)$ as lower bound in average cases? It should be mentioned that $O(N\log N)$ (as upper bound) in average cases is an open question. – Apass.Jack Nov 10 '18 at 16:13
• Sorry, I find my proof about the worst case is wrong, so I will delete. The lower bound analysis of average case is derived from this formula. – firejox Nov 12 '18 at 3:03
• Welcome to Computer Science! We expect references to fulfill the minimal scholarly requirements and be as robust over time as possible. Please take some time to improve your post in this regard. We have collected some advice here. – Raphael Nov 12 '18 at 22:43
• You seem to have a proof attempt there. What's your specific question about it? What are you unsure of? – Raphael Nov 12 '18 at 22:43


text:
# Is 3-month LIBOR quoted in annual terms?

If 3 month LIBOR is 0.22% and I want to find the interest rate with continuous compounding for 10 days do I multiply my principal by e^(10*r/90) or e^(10*r/365)?

Libor interest payments are determined using simple annualised compounding rather than continuous compounding (using time in units of years). To find the equivalent interest rate with continuous compounding, you would solve for $$r_c$$ where the Libor rate is for 91 days and I assume the Libor has an Actual/360 basis convention. To calculate the elapsed time for the continuous compounding, I decide to use Actual/365. This gives:

$$\exp \left( \frac{91}{365} \times r_c \right) = \left( 1 + \frac{91}{360} \times \frac{0.22}{100} \right)$$

So

$$r_c = \frac{365}{91} \times \ln \left( 1 + \frac{91}{360} \times \frac{0.22}{100} \right)$$

which gives

$$r_c = 0.222994\%$$

The low level of interest rates and the short time period mean that the Libor rate and continuously compounded rate are almost identical.

You can then apply this for 10 days (once again using Actual/365 for calculating times) to get a growth factor of

$$\exp \left( \frac{10}{365} \times \frac{0.222994}{100} \right) = 1.000061096$$

• (Albeit I don't think any actual Libors are quoted 30/360) Nov 9 '20 at 8:53
• True. I just wanted to keep the maths simple and get 0.25 rather than have 91/360. I might change it.
– Dom
Nov 9 '20 at 11:25

well, strictly speaking neither, but the second answer gets you closer to the truth, as Libor is indeed quoted in annual terms. However it is not quoted as continuously-compounded but as simply compounded. Eg in your example a 3M compounding factor would be $$1 + (1/4) \times 0.22\%$$ where the factor 1/4 is also somewhat approximate as in reality specific day counting rules are used. See eg here


text:
直接法设计均匀与非均匀滤波器组
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 应用科技  2018, Vol. 45 Issue (3): 14-18  DOI: 10.11991/yykj.201612008 0

### 引用本文

ZHANG Chunjie, TIAN Chunyu. Design of uniform and non-uniform filter banks by direct method[J]. Applied Science and Technology, 2018, 45(3), 14-18. DOI: 10.11991/yykj.201612008.

### 文章历史

Design of uniform and non-uniform filter banks by direct method
ZHANG Chunjie, TIAN Chunyu
College of Information and Communication Engineering, Harbin Engineering University, Harbin 150001, China
Abstract: As for the reconstruction design of the filter banks, the existing algorithms can only be used to design uniform or non-uniform filter banks, it is hard to design two different types of filter banks at the same time, therefore, the flexibility is not enough. A new method of designing all filter banks directly was proposed in the paper. In the method, the reconstruction conditions of filter banks are derived from the frequency domain and expressed in the form of matrix. According to the distribution of the given frequency bands, all channel filters in the analysis modules are designed simultaneously; then the optimum solution of the reconstruction matrix equation can be obtained by the iterative optimization theory, the filtering coefficients of all passages in the comprehensive modules can be further obtained and the necessary filter banks can be designed. Simulation results show that, the design method can be used to design the uniform filter banks and the non-uniform filter banks simultaneously, realize high-efficiency division of frequency band, the flexibility is high, in addition, the method can ensure the linear phase characteristics of each channel filter.
Key words: design of filter banks    software radio    approximate reconstruction    iterative optimization    linear-phase    least square solution    aliasing distortion    FIR filter

1 滤波器组重构条件推导

 \begin{aligned}Y\left( z \right) = & \sum\limits_{m = 0}^{M - 1} {\frac{1}{{{n_m}}}{F_k}\left( z \right)\sum\limits_{l = 0}^{{n_m} - 1} {X\left( {z{{\rm{e}}^{ - {\rm{j}}\frac{{2{\rm{\pi }}l}}{{{n_m}}}}}} \right){H_m}\left( {z{{\rm{e}}^{ - {\rm{j}}\frac{{2{\rm{\pi }}l}}{{{n_m}}}}}} \right)} } = \\& \sum\limits_{m = 0}^{M - 1} {\frac{1}{{{n_m}}}{F_m}\left( z \right)X\left( z \right){H_m}\left( z \right)} + \\& \sum\limits_{m = 0}^{M - 1} {\frac{1}{{{n_m}}}{F_m}\left( z \right)\sum\limits_{l = 1}^{{n_m} - 1} {X\left( {z{{\rm{e}}^{ - {\rm{j}}\frac{{2{\rm{\pi }}l}}{{{n_m}}}}}} \right){H_m}\left( {z{{\rm{e}}^{ - {\rm{j}}\frac{{2{\rm{\pi }}l}}{{{n_m}}}}}} \right)} } = \\& X\left( z \right){T_0}\left( z \right) + \sum\limits_{l = 1}^{{n_m} - 1} {X\left( {z{e^{ - {\rm{j}}\frac{{2{\rm{\pi }}l}}{{{n_m}}}}}} \right){T_l}\left( z \right)} \end{aligned}

 ${T_0}\left( z \right) = \sum\limits_{m = 0}^{M - 1} {\frac{1}{{{n_m}}}{F_k}\left( z \right)} \sum\limits_{l = 0}^{{n_m} - 1} {X\left( {z{{\rm{e}}^{ - {\rm{j}}\frac{{2{\rm{\pi }}l}}{{{n_m}}}}}} \right){H_m}\left( {z{{\rm{e}}^{ - {\rm{j}}\frac{{2{\rm{\pi }}l}}{{{n_m}}}}}} \right)}$ (1)
 ${T_l}\left( z \right) = \sum\limits_{m = 0}^{M - 1} {\frac{1}{{{n_m}}}{F_m}\left( z \right){H_m}\left( {z{{\rm{e}}^{ - {\rm{j}}\frac{{2{\rm{\pi }}l}}{{{n_m}}}}}} \right)} \;\;\; l = 1,2, \cdots ,M$ (2)

${T_0}\left( z \right)$ 称为整个滤波系统的失真函数，体现了整个滤波系统的幅频相位特性， ${T_l}\left( z \right)$ 称为整个系统的混叠失真函数，对应的混叠信号是 $X\left( {z{{\rm{e}}^{{{ - {\rm{j}}2\pi l} / {{n_m}}}}}} \right)$

${T_0}\left( z \right)$ 是一个纯延迟，即 ${T_0}\left( z \right) = {z^{ - \varDelta }}$ ${T_l}\left( z \right){\rm{ = }}0$ 时，整个系统可实现准确重构，Δ表示时延长度。结合式(1)、(2)，将重构条件表示为矩阵方程的形式，即

 ${{Ax}} = {{b}}$ (3)

 ${{{A}}_{L\left( {2N - 1} \right) \times MN}} = \left[\!\!\!\!\! {\begin{array}{*{20}{c}}{{\beta _{1,0}}\Im \left( {{{{h}}_1}} \right)} \!\! & \!\! {{\beta _{2,0}}\Im \left( {{{{h}}_2}} \right)} \!\! & \!\! \cdots & {{\beta _{M,0}}\Im \left( {{{{h}}_M}} \right)}\\{{\beta _{1,1}}\Im \left( {{{{h}}_1} \otimes {{{\varLambda }}_1}} \right)} \!\! & \!\! {{\beta _{2,1}}\Im \left( {{{{h}}_2} \otimes {{{\varLambda }}_1}} \right)} \!\! & \!\! \cdots & {{\beta _{M,1}}\Im \left( {{{{h}}_M} \otimes {{{\varLambda }}_1}} \right)}\\{{\beta _{1,2}}\Im \left( {{{{h}}_1} \otimes {{{\varLambda }}_2}} \right)} \!\! & \!\! {{\beta _{2,2}}\Im \left( {{{{h}}_2} \otimes {{{\varLambda }}_2}} \right)} \!\! & \!\! \cdots & \!\! {{\beta _{M,2}}\Im \left( {{{{h}}_M} \otimes {{{\varLambda }}_2}} \right)}\\ \vdots \!\! & \!\! \cdots \!\! & \!\! \cdots \!\! & \!\! \vdots \\{{\beta _{1,L - 1}}\Im \left( {{{{h}}_1} \otimes {{{\varLambda }}_{L - 1}}} \right)} \!\! & \!\! {{\beta _{2,L - 1}}\Im \left( {{{{h}}_2} \otimes {{{\varLambda }}_{L - 1}}} \right)} \!\! & \!\! \cdots \!\! & \!\! {{\beta _{M,L - 1}}\Im \left( {{{{h}}_M} \otimes {{{\varLambda }}_{L - 1}}} \right)}\end{array}} \!\!\!\!\! \right]$

 {\alpha _{m,l}} = \left\{ {\begin{aligned}& {1,} \quad \quad {{{\rm e}^{{{{\rm j}2\pi * l} / L}}} \in \left\{ {{{\rm e}^{{\rm j}2\pi \left( 1 \right)/{n_m}}},{{\rm e}^{{\rm j}2\pi \left( 2 \right)/{n_m}}}, \cdots ,{{\rm e}^{{\rm j}2\pi \left( {{n_m} - 1} \right)/{n_m}}}} \right\}}\\& {0,} \quad \quad {{\text{其他}}}\end{aligned}} \right.

βm, l定义为

 ${\beta _{m,l}} = \left\{ {\begin{array}{*{20}{l}} {{1 / {{n_m},}}}&{l = 0} \\ {{{{\alpha _{m,l}}} / {{n_m},}}}&{l \ne 0} \end{array}} \right.$ (4)

 $\Im \left( {{{\left[ {{a_1},{a_2}, \cdots ,{a_N}} \right]}^{\rm{T}}}} \right) = \left[ {\begin{array}{*{20}{c}} {{a_1}}&0&0& \cdots &0 \\ {{a_2}}&{{a_1}}&0& \cdots &0 \\ {{a_3}}&{{a_2}}&{{a_1}}& \cdots &0 \\ \vdots & \cdots & \cdots & \cdots & \vdots \\ {{a_N}}&{{a_{N - 1}}}&{{a_{N - 2}}}& \cdots &{{a_1}} \\ 0&{{a_N}}&{{a_{N - 1}}}& \cdots &{{a_2}} \\ 0&0&{{a_N}}& \cdots &{{a_3}} \\ 0&0&0& \cdots & \vdots \\ 0&0&0& \cdots &{{a_N}} \end{array}} \right]$

${{{h}}_m}\left( {m = 0,2, \cdots ,M - 1} \right)$ 是分析模块中第 $m$ 个通道的滤波器系数，即 ${{{h}}_m} = {\left( {{h_{m,0}},{h_{m,1}}, \cdots ,{h_{m,N - 1}}} \right)^{\rm{T}}}$ 。运算符 $\otimes$ 表示2个矩阵的对应元素相乘。运算符 ${{{\varLambda }}_i}$ 定义为

 ${{{\varLambda }}_i} = {\left[ {{{\rm e}^{{{{\rm j}2\pi i \cdot \left( 0 \right)} / L}}} \;\; {{\rm e}^{{{{\rm j}2\pi i \cdot \left( 1 \right)} / L}}} \;\; {{\rm e}^{{{{\rm j}2\pi i \cdot \left( 2 \right)} / L}}} \;\; \cdots \;\; {{\rm e}^{{{{\rm j}2\pi i \cdot \left( {N - 1} \right)} / L}}}} \right]^{\rm{T}}}$

 $Q = {Q_{{\rm{PR}}}} + {Q_{\rm{MRE}}} = \left\| {{{Ax}} - {{b}}} \right\|_2^2 + \sum\limits_{m = 1}^M {{Q_{{\rm{MRE}},m}}}$ (5)

 ${Q_{{\rm{MRE}},m}} = \sum\limits_{i = 1}^\rho {\left\| {\left| {{H_m}\left( {{{\rm e}^{{\rm j}w}}} \right)} \right| - {r_m}\left( {{w_i}} \right)} \right\|} _2^2$ (6)

 ${r_m}\left( {{w_i}} \right) = \left\{ {\begin{array}{*{20}{l}}{1,} & {{w_i}{\text{是}}{H_m}\left( {{{\rm e}^{\rm j}{^w}}} \right){\text{通带内的频率点}}}\\{0,} & {{\text{其他}}}\end{array}} \right.$

2 滤波器组设计步骤

1) 初始化各通道滤波器系数长度 $N$ 、分析模块各通道原始FIR滤波器频率特性、需要满足的误差值 ${Q_D}$ 、频率采样点个数 $\rho$ 、各通道抽取插值速率 $\left[ {{n_1},{n_2}, \cdots ,{n_{M - 1}}} \right]$ ，并求出其最小公倍数L

2) 采用等波纹逼近法设计系统前端分析模块中各通道原始FIR滤波器系数 ${{ h}_m}\left( n \right)\left( {m = 1,2, \cdots ,M} \right)$ ，并结合其他定义参数、运算符生分析矩阵A

3) 通过求解方程 ${{Ax}} = {{b}}$ 的最小二乘解，得到综合模块中各通道滤波器系数 ${{x}}$

4) 通过式(5)计算误差目标函数Q

5) 如果求解的误差 $Q$ 小于给定的误差 ${Q_D}$ ，终止该算法，否则通过迭代更新分析模块中的滤波器系数，并返回步骤2)，直到误差 $Q \leqslant {Q_D}$

3 性能仿真