text:
## Elementary Algebra

Published by Cengage Learning

# Chapter 4 - Proportions, Percents, and Solving Inequalities - 4.2 - Percents and Problem Solving - Problem Set 4.2: 63

#### Answer

The interest charged will be 510 dollars.

#### Work Step by Step

To find the interest charged, we use the interest formula. The interest formula is: I = P $\times$ r $\times$ t P = 8000 r = 8.5% = 0.085 t = 9 months = 9 $\times$ $\frac{1}{12}$ = 0.75 (Because 1 month = $\frac{1}{12}$ year.) Substitute these values in the formula to obtain: I = 8000 $\times$ 0.085 $\times$ 0.75 I = 510 The interest charged will be 510 dollars.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.

format: plain


text:
## Isometries on the $\ell^1$-sum of $C_0(Î©,E)$ type spaces

J. Math. Sci. Univ. Tokyo
Vol. 2 (1995), No. 1, Page 131--154.

Wang, Risheng ; Orihara, Akio
Isometries on the $\ell^1$-sum of $C_0(Î©,E)$ type spaces
[Full Article (PDF)] [MathSciNet Review (HTML)] [MathSciNet Review (PDF)]

Abstract:
In this paper, we study the representations of isometries on the unit sphere of the $\ell\sp 1$-sum of $C\sb 0 (Î©,E)$ type spaces and give an affirmative answer to the corresponding Tingley's problem. Especially, we can get the results of [1].

Keywords: Isometry; $\ell\sp p$-sum; Tingley's problem; Mazur-Ulam's theorem

Mathematics Subject Classification (1991): Primary 46A22; Secondary 46B20, 46E15, 46E40
Mathematical Reviews Number: MR1348025


text:
# Can someone explain this summation definition to me?

1. Nov 18, 2014

A=limn→∞Rn=limn→∞[f(x1)Δx+f(x2)Δx+...+f(xn)Δx]

Consider the function f(x)=4√x, 1≤x≤16. Using the above definition, determine which of the following expressions represents the area under the graph of f as a limit.

I knew the correct answer was $$\sum \frac{15}{n} (4√x+\frac{15i}{n})$$

I figured out most of this, but the only thing I don't get is how you figure out is basically since you start the sum from i=1 to n, that you have to shift the sum up. If i=0, then I think there would be no "1+" term, right?
But let's say it was.... i =5 to n. I have no idea why, or where you would put the added terms in order for the sum to work. I thought that it would be outside of the function 4√x but inside of the summnation, because, well, you're just adding values, right?
I don't see or understand intuitively why 1+ would go inside of 4√x.

Similarly, I got this right but didn't understand the idea.
A=limn→∞Rn=limn→∞[f(x1)Δx+f(x2)Δx+...+f(xn)Δx]

Consider the function $$f(x) = \frac{ln(x)}{x}$$,3≤x≤10. Using the above definition, determine which of the following expressions represents the area under the graph of f as a limit.

Of course, the answer was Δx and $$\frac{ln 3+\frac{7i}{n}}{3+\frac{7i}{n}}$$, but just like above, wasn't sure why the $$\frac{ln3}{3}$$ went there. I thought if anything, it should be an added term, not mixed up with the main fraction.

Last edited by a moderator: Nov 19, 2014
2. Nov 19, 2014

### lurflurf

The area is

$$\lim_{n\rightarrow\infty}\sum_{i=1}^n \dfrac{b-a}{n}\mathrm{f}\left(a+i\dfrac{b-a}{n}\right)$$

The function values are determined by dividing the interval into equal parts. a+i(b-a)/n are the points that that divide the interval evenly.

3. Nov 20, 2014

Thanks, I understand that, but just about the notation- Do you mean to say f(x)? And the X being all the stuff in parentheses?

But sorry, I have another confusion- if a is on the inside of those parentheses, isn't A the starting value? Then why, for example is the answer for that lnx/x one not the A value from, 3≤x≤10, which would make it 3, right? Instead of ln3?

4. Nov 20, 2014

### Staff: Mentor

I'm pretty sure he didn't mean f(x) or he would have written that. To understand what that notation means, try it out with a simple function like f(x) = x2 and an interval [0, 2].

See what you get with, say, n = 4. The summation will have 4 terms.
a is the left endpoint of the interval.
This doesn't make much sense, so I don't know what you're asking.

5. Nov 24, 2014

### vela

Staff Emeritus
It would have been clearer with parentheses. You should have written
$$\frac{\ln \left(3 + \frac{7i}n\right)}{3 + \frac{7i}n}$$ which I hope you recognize as $f\left(3 + \frac{7i}n\right)$ when $f(x) = \frac{\ln x}{x}$.


text:
# The product/Service project: secunity drones for which monitors who in and out. farms goes this. o...

###### Question:

The product/Service project: secunity drones for which monitors who in and out. farms goes this. o what are the key steps to develop prodhet/senice @ what key relationship needs to be established ? & what are the funding requirements? what are the they challenges ahead?

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text:
# Minimum Spanning Tree Clustering¶

This is a short notebook describing the Minimum Spanning Tree (MST) Clustering algorithm implemented at http://github.com/jakevdp/mst_clustering/. The API of the clustering estimator is compatible with scikit-learn.

Note that MST clustering is effectively identical to single-linkage Agglomerative Clustering, but uses a top-down approach rather than a bottom-up approach. Similar methods are implemented in scikit-learn's Agglomerative Clustering estimator.

We'll start with some initial imports and definitions; you can ignore this for now:

In [1]:
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns; sns.set()

# matplotlib 1.4 + numpy 1.10 produces warnings; we'll filter these
import warnings; warnings.filterwarnings('ignore', message='elementwise')

def plot_mst(model, cmap='rainbow'):
"""Utility code to visualize a minimum spanning tree"""
X = model.X_fit_
fig, ax = plt.subplots(1, 2, figsize=(16, 6), sharex=True, sharey=True)
for axi, full_graph, colors in zip(ax, [True, False], ['lightblue', model.labels_]):
segments = model.get_graph_segments(full_graph=full_graph)
axi.plot(segments[0], segments[1], '-k', zorder=1, lw=1)
axi.scatter(X[:, 0], X[:, 1], c=colors, cmap=cmap, zorder=2)
axi.axis('tight')

ax[0].set_title('Full Minimum Spanning Tree', size=16)
ax[1].set_title('Trimmed Minimum Spanning Tree', size=16);

## Example and Explanation of the Algorithm¶

Minimum Spanning Tree clustering is a very intuitive algorithm. Consider the following data:

In [2]:
from sklearn.datasets import make_blobs
X, y = make_blobs(200, centers=4, random_state=42)
plt.scatter(X[:, 0], X[:, 1], c='lightblue');

By eye we can see that it contains four well-separated clusters, and a clustering algorithm should be able to find these. The MSTClustering estimator can accomplish this quite well:

In [3]:
from mst_clustering import MSTClustering
model = MSTClustering(cutoff_scale=2, approximate=False)
labels = model.fit_predict(X)
plt.scatter(X[:, 0], X[:, 1], c=labels, cmap='rainbow');

Notice that the algorithm finds five clusters: the one outlier near the yellow cluster is put in its own group.

This clustering result is computed by first constructing a graph from the input data, with nodes given by points in the dataset and edges given by distances between pairs of points. Then we find a Minimum Spanning Tree (MST) over this graph: an MST is a subgraph that connects all nodes such that the sum of the graph edges is minimized.

We can see a representation of the minimum spanning tree over this data in the left panel of the following plot:

In [4]:
plot_mst(model)

The right panel shows the clusters derived from this minimum spanning tree, by removing all graph edges larger than the specified cutoff_scale parameter (here, cutoff_scale = 2). Points which remain connected after the truncation are given the same label.

## Strengths of MST clustering¶

MST clustering has several advantages over other clustering algorithms:

1. MST clustering is capable of detecting clusters with very irregular boundaries. This compares favorably to K Means, Meanshift, GMM, etc.
2. The number of clusters is not decided a priori as in many other algorithms. This compares favorably to K Means, some Spectral methods, etc.
3. A suitably tuned algorithm can detect clusters within background noise. This compares favorably to most other methods.
4. Perhaps most importantly, the tuning parameters of the clustering have very intuitive meanings, and thus MST clustering can be a useful method for scientific data in which the user has a good prior knowledge regarding the meaningful length scale and/or cluster size.

Many of these advantages are shared with, e.g. DBSCAN which uses a similar but conceptually distinct model.

## Clusters In Background Noise¶

Let's take a look at the results of MST clustering in the presense of background noise. Consider this data:

In [5]:
rng = np.random.RandomState(int(100 * y[-1]))
noise = -14 + 28 * rng.rand(200, 2)

X_noisy = np.vstack([X, noise])
y_noisy = np.concatenate([y, np.full(200, -1, dtype=int)])

plt.scatter(X_noisy[:, 0], X_noisy[:, 1], c='lightblue', cmap='spectral_r')
plt.xlim(-15, 15)
plt.ylim(-15, 15);

We can see that there are four distinct overdensities in this data. Let's take a look at what the minimum spanning tree does in this case:

In [6]:
noisy_model = MSTClustering(cutoff_scale=1)
noisy_model.fit(X_noisy)
plot_mst(noisy_model)

Cutting the long edges does separate the four large clusters, but it also leads to a large number of isolated clusters with just a handful of points. We can easily filter these out by specifying a minimum cluster size; all points which don't belong to a large enough cluster are grouped together with the label -1:

In [7]:
noisy_model = MSTClustering(cutoff_scale=1, min_cluster_size=10)
noisy_model.fit(X_noisy)
plot_mst(noisy_model, cmap='spectral_r')

Thus, by appropriately tuning the cutoff scale and the minimum number of points per cluster, you can recover even clusters within a noisy background.

## Future Work¶

There are methods in the literature that discuss how to tune these parameters automatically; I hope to implement some of them in this package in the future.


text:
# Revision history [back]

### How to efficiently access vector of Mat ?

I read this doc here http://docs.opencv.org/2.4/doc/tutorials/core/mat_the_basic_image_container/mat_the_basic_image_container.html that teaches me how I can efficiently scan an image.

Mat& ScanImageAndReduceC(Mat& I, const uchar* const table)
{
// accept only char type matrices
CV_Assert(I.depth() == CV_8U);

int channels = I.channels();

int nRows = I.rows;
int nCols = I.cols * channels;

if (I.isContinuous())
{
nCols *= nRows;
nRows = 1;
}

int i,j;
uchar* p;
for( i = 0; i < nRows; ++i)
{
p = I.ptr<uchar>(i);
for ( j = 0; j < nCols; ++j)
{
p[j] = table[p[j]];
}
}
return I;
}

What if as input I have an additional vector of Mat and I am trying to copy the value of an element from Mat angle to a certain Mat angle_ii depending on the value is has ? In the code below, am I forced to create multiple angle_ii_row so that each variable points to each Mat of angle_ii ?

void construct_ii( cv::Mat& angle, std::vector<cv::Mat>& angle_ii)
{
int rows = angle.rows;
int cols = angle.cols;

if (angle.isContinuous())
{
cols *= rows;
rows = 1;
}

for (int i = 0; i < angle.rows; i++)
{
float* angle_row = angle.ptr<float>(i);
float** angle_ii_row = angle_ii.ptr<float>(i); // wrong

for (int j = 0; j < angle.rows; j++)
{
float angle_temp = angle_row[j];
int which = floor(angle_temp/20);

switch (which)
{
case 0 :  angle_ii[0].at<float>(row,col) = angle_temp; // also wrong
}

}
}

}

### How to efficiently access vector of Mat ?

I read this doc here http://docs.opencv.org/2.4/doc/tutorials/core/mat_the_basic_image_container/mat_the_basic_image_container.html that teaches me how I can efficiently scan an image.

Mat& ScanImageAndReduceC(Mat& I, const uchar* const table)
{
// accept only char type matrices
CV_Assert(I.depth() == CV_8U);

int channels = I.channels();

int nRows = I.rows;
int nCols = I.cols * channels;

if (I.isContinuous())
{
nCols *= nRows;
nRows = 1;
}

int i,j;
uchar* p;
for( i = 0; i < nRows; ++i)
{
p = I.ptr<uchar>(i);
for ( j = 0; j < nCols; ++j)
{
p[j] = table[p[j]];
}
}
return I;
}

What if as input I have an additional vector of Mat and I am trying to copy the value of an element from Mat angle to a certain Mat angle_ii depending on the value is has ? In the code below, am I forced to create multiple angle_ii_row so that each variable points to each Mat of angle_ii ?

void construct_ii( cv::Mat& angle, std::vector<cv::Mat>& angle_ii)
{
int rows = angle.rows;
int cols = angle.cols;

if (angle.isContinuous())
{
cols *= rows;
rows = 1;
}

for (int i = 0; i < angle.rows; i++)
{
float* angle_row = angle.ptr<float>(i);
float** angle_ii_row = angle_ii.ptr<float>(i); // wrong

for (int j = 0; j < angle.rows; j++)
{
float angle_temp = angle_row[j];
int which = floor(angle_temp/20);

switch (which)
{
case 0 :  angle_ii[0].at<float>(row,col) = angle_temp; // also wrong
wrong. and not efficient
}

}
}

}


text:
Announcing the TensorFlow Dev Summit 2020

# TensorFlow code style guide

## Python style

Follow the PEP 8 Python style guide, except TensorFlow uses 2 spaces instead of 4. Please conform to the Google Python Style Guide, and use pylint to check your Python changes.

### pylint

To install pylint and retrieve TensorFlow's custom style definition:

$pip install pylint$ wget -O /tmp/pylintrc https://raw.githubusercontent.com/tensorflow/tensorflow/master/tensorflow/tools/ci_build/pylintrc

To check a file with pylint:

$pylint --rcfile=/tmp/pylintrc myfile.py ### Supported Python versions TensorFlow supports Python 2.7 and Python >= 3.4. See the installation guide for details. See the TensorFlow continuous build status for official and community supported builds. #### Legacy Python compatibility TensorFlow will support Legacy Python (Python 2.7) until January 1, 2020. Until that time, all code will need to be compatible with the Python versions listed above. These lines should be present in every Python file: from __future__ import absolute_import from __future__ import division from __future__ import print_function Use six to write compatible code (for example, six.moves.range). ## C++ coding style Changes to TensorFlow C++ code should conform to the Google C++ Style Guide. Use clang-tidy to check your C/C++ changes. To install clang-tidy on Ubuntu 16+, do:$ apt-get install -y clang-tidy

You can check a C/C++ file by using the following:

$clang-format <my_cc_file> --style=google > /tmp/my_cc_file.cc$ diff <my_cc_file> /tmp/my_cc_file.cc

## TensorFlow conventions and special uses

### Tensors

• Operations that deal with batches may assume that the first dimension of a Tensor is the batch dimension.
• In most models, the last dimension is the number of channels.
• Dimensions excluding the first and last usually make up the space dimensions: sequence-length, or image-size.
• When possible, use a Tensor's overloaded operators rather than TensorFlow functions. For example, we prefer **, +, /, *, -, and/or over tf.pow, tf.add, tf.divide, tf.multiply, tf.subtract, and tf.logical_* — unless a specific name for the operation is desired.

### Python operations

A Python operation is a function that, given input tensors and parameters, creates a part of the graph and returns output tensors.

• The first argument should be tensors, followed by basic Python parameters. The last argument is name with a default value of None.
• Tensor arguments should be either a single tensor or an iterable of tensors. That is, a "Tensor or list of Tensors" is too broad. See assert_proper_iterable.
• Operations that take tensors as arguments should call convert_to_tensor to convert non-tensor inputs into tensors if they are using C++ operations. Note that the arguments are still described as a Tensor object of a specific dtype in the documentation.
• Each Python operation should have a name_scope. As seen below, pass the name of the op as a string.
• Operations should contain an extensive Python comment with Args and Returns declarations that explain both the type and meaning of each value. Possible shapes, dtypes, or ranks should be specified in the description. See documentation details.
• For increased usability, include an example of usage with inputs / outputs of the op in Example section.
• Avoid making explicit use of tf.Tensor.eval or tf.Session.run. For example, to write logic that depends on the Tensor value, use the TensorFlow control flow. Alternatively, restrict the operation to only run when eager execution is enabled (tf.executing_eagerly()).

Example:

def my_op(tensor_in, other_tensor_in, my_param, other_param=0.5,
output_collections=(), name=None):
"""My operation that adds two tensors with given coefficients.

Args:
tensor_in: Tensor, input tensor.
other_tensor_in: Tensor, same shape as tensor_in, other input tensor.
my_param: float, coefficient for tensor_in.
other_param: float, coefficient for other_tensor_in.
output_collections: tuple of strings, name of the collection to
collect result of this op.
name: string, name of the operation.

Returns:
Tensor of same shape as tensor_in, sum of input values with coefficients.

Example:
>>> my_op([1., 2.], [3., 4.], my_param=0.5, other_param=0.6,
[2.3, 3.4]
"""
with tf.name_scope(name or "my_op"):
tensor_in = tf.convert_to_tensor(tensor_in)
other_tensor_in = tf.convert_to_tensor(other_tensor_in)
result = my_param * tensor_in + other_param * other_tensor_in
return result

Usage:

output = my_op(t1, t2, my_param=0.5, other_param=0.6,


text:
## SectionAlgebra of Polynomials

In the previous section, we defined functions and practiced using the compact function notation. In this section, we will explore a particularly useful kind of function.

###### In this section, you will...
• define polynomials and explore their characteristics

• evaluate polynomials

• perform the four basic operations on polynomials

A polynomial is an expression made up of a single term or sum of terms with only one variable in which each exponent is a whole number. For example,

\begin{equation*} 2x^3, \hspace{.5cm} 8x^4-x^2+9, \hspace{.5cm} x^9, \hspace{.5cm} x+4 \end{equation*}

are all polynomials, while

\begin{equation*} x^{-3}, \hspace{.5cm} 4x^{0.5}+2, \hspace{.5cm} \dfrac{1}{x^4} \end{equation*}

are not polynomials. Polynomials define functions of the form

\begin{gather*} f(x)=a_n x^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0. \end{gather*}

Here $a_n$ represents any real number and $n$ represents any whole number. The degree of a polynomial with one variable is the largest exponent of all the terms. Typically, we arrange terms of polynomials in descending order based on their degree and classify them as follows:

\begin{align*} \begin{aligned} f(x)\amp =2 \ \alert{\text{ Constant function (degree }0\text{)}}\\ g(x)\amp = 3x+2 \ \alert{\text{ Linear function (degree }1\text{)}}\\ h(x)\amp = 4x^2+3x+2 \ \alert{\text{ Quadratic function (degree }2\text{)}}\\ p(x)\amp = 5x^3+4x^2+3x+2 \ \alert{\text{ Cubic function (degree }3\text{)}} \end{aligned} \end{align*}

In this course, we call any polynomial with degree higher than 3 an $n$th-degree polynomial. For example, if the degree is 4, we call it a fourth-degree polynomial; if the degree is 5, we call it a fifth-degree polynomial, and so on.

When a polynomial is arranged in descending order based on their degree, we call the first term of the sum the leading term, and the coefficient part of this term is called the leading coefficient.

###### Example130

Given $f(x)=3x^2+5x+17\text{,}$ $g(x)=15x^{24}+36x^{22}+17x^4+31\text{,}$ $h(x)=11x^2+17x^3+121x\text{,}$ $j(x)=5x^3+2x+4x^3+3x+5\text{,}$ and $k(x)=25\text{,}$ state the degree, leading term, and leading coefficient of each polynomial.

Solution

Looking at $f(x)\text{,}$ we see that the highest power of $x$ is $2\text{,}$ so the degree is $2\text{.}$ The term with the highest degree is $3x^2\text{,}$ meaning this is our leading term, and the coefficient of this term is $3\text{,}$ which means this is our leading coefficient.

Looking at $g(x)\text{,}$ we see that the highest power of $x$ is $24\text{,}$ so the degree is $24\text{.}$ The term with the highest degree is $15x^{24}\text{,}$ meaning this is our leading term, and the coefficient of this term is $15\text{,}$ which means this is our leading coefficient.

Looking at $h(x)\text{,}$ we see that the polynomial is not arranged in descending order of degree, so we will rearrange first.

\begin{equation*} h(x)=17x^3+11x^2+121x \end{equation*}

Now we see that the highest power of $x$ is $3\text{,}$ so the degree is $3\text{.}$ The term with the highest degree is $17x^{3}\text{,}$ meaning this is our leading term, and the coefficient of this term is $17\text{,}$ which means this is our leading coefficient.

Looking at $j(x)\text{,}$ we see that the polynomial is not arranged in descending order of degree, so we will rearrange first.

\begin{equation*} j(x)=5x^3+4x^3+2x+3x+5=9x^3+5x+5 \end{equation*}

Now we see that the highest power of $x$ is $3\text{,}$ so the degree is $3\text{.}$ The term with the highest degree is $9x^{3}\text{,}$ meaning this is our leading term, and the coefficient of this term is $9\text{,}$ which means this is our leading coefficient.

Looking at $k(x)\text{,}$ we have just a constant. This is the same as $k(x)=25\cdot 1=25x^0\text{,}$ so we see that the highest power of $x$ is $0\text{,}$ so the degree is $0\text{.}$ The term with the highest degree is $25x^0=25\text{,}$ meaning this is our leading term, and the coefficient of this term is $25\text{,}$ which means this is our leading coefficient.

###### Example131

Given $f(x)=x^2-8x+17\text{,}$ find $f(2)$ and $f(4)\text{.}$

Solution

Replace each instance of $x$ with the value given inside the parentheses.

We can write $f(2)=5$ and $f(4)=1\text{.}$ Remember that $f(x)=y$ and so we can interpret these results on the graph as follows:

We find that $f(2)=5$ and $f(4)=1\text{.}$

One useful application of polynomials appears in physics. The height of an object launched upward, ignoring the effects of air resistance, can be modeled with the following quadratic function:

\begin{gather*} h(t)=-\frac{1}{2}gt^2+v_0t+s_0 \end{gather*}

With this formula, the height $h(t)$ can be calculated at any given time $t$ after the object is launched. The letter $g$ represents acceleration due to gravity on the surface of the Earth, which is 32 feet per second squared (or, using metric units, $g = 9.8$ meters per second squared). The parameter $v_0$ pronounced v-naught, or sometimes v-zero, represents the initial velocity of the object. The parameter $s_0$ represents the initial height from which the object was launched.

###### Example132

An object is launched from the ground at a speed of $64$ feet per second. Write a function that models the height of the object and use it to calculate the object's height at $1$ second and at $3.5$ seconds.

Solution

We know that the acceleration due to gravity is $g=32$ feet per second squared and we are given the initial velocity $v_0=64$ feet per second. Since the object is launched from the ground, the initial height is $s_0=0$ feet. Create the mathematical model by substituting these coefficients into the following formula:

Use this model to calculate the height of the object at $1$ second and $3.5$ seconds.

Our function is $h(t)=-16t^2+64t\text{.}$ This tells us that at $1$ second the object is at a height of $48$ feet, and at $3.5$ seconds it is at a height of $28$ feet.

###### Example133

An object is dropped from a height of $6$ meters. Write a function that models the height of the object and use it to calculate the object's height $1$ second after it is dropped.

Solution

Our function is $h(t)=-4.9t^2+6\text{.}$ Since

\begin{equation*} \begin{aligned} h(1) \amp= -4.9(1)^2+6 =1.1, \end{aligned} \end{equation*}

at $1$ second the object is at a height of $1.1$ meters.

On occasion a special notation is used to indicate addition and subtraction of functions. This mentioned here in the book so you have some exposure to the notation. The notation is as follows:

\begin{gather*} \text{Addition of functions: }(f+g)(x)=f(x)+g(x)\\ \text{Subtraction of functions: }(f-g)(x)=f(x)-g(x). \end{gather*}

When using function notation, be careful to group the entire function and add or subtract accordingly.

###### Example134

Given $f(x)=x^3-5x-7$ and $g(x)=3x^2+7x-2\text{,}$ find $f(x)+g(x)$ and $f(x)-g(x)\text{.}$

Solution

The notation $f(x)+g(x)$ indicates that we should add the given expressions.

\begin{equation*} \begin{aligned} f(x)+g(x)\amp = (x^3-5x-7)+(3x^2+7x-2)\\ \amp = x^3-5x-7+3x^2+7x-2\\ \amp = x^3+3x^2+2x-9 \end{aligned} \end{equation*}

The notation $f(x)-g(x)$ indicates that we should subtract the given expressions. When subtracting, the parentheses become very important. We can eliminate them after applying the distributive property.

\begin{equation*} \begin{aligned} f(x)-g(x) \amp = (x^3-5x-7)-(3x^2+7x-2)\\ \amp = x^3-5x-7-3x^2-7x+2\\ \amp = x^3-3x^2-12x-5 \end{aligned} \end{equation*}

Our solutions are $f(x)+g(x)=x^3+3x^2+2x-9$ and $f(x)-g(x)= x^3-3x^2-12x-5\text{.}$

We may be asked to evaluate the sum or difference of two functions. We have the option to first find the sum or difference in general and then use the resulting function to evaluate for the given variable, or evaluate each first and then find the sum or difference.

###### Example135

Evaluate $f(3)-g(3)$ given $f(x)=5x^2-x+4$ and $g(x)=x^2+2x-3\text{.}$

Solution

First, find $(f-g)(x)\text{.}$

\begin{equation*} \begin{aligned} f(x)-g(x)\amp = (5x^2-x+4)-(x^2+2x-3)\\ \amp = 5x^2-x+4-x^2-2x+3\\ \amp = 4x^2-3x+7 \end{aligned} \end{equation*}

Now we substitute $3$ in for the variable $x\text{.}$

Our solution is $f(3)-g(3)=34\text{.}$

Alternatively, we could have found $f(3)$ and $g(3)$ and then subtracted the results to obtain our answer.

Note: If multiple values are to be evaluated, it is best to find the sum or difference in general first and then use it to evaluate.

###### Example136

Given $f(x)=x^3+x-8$ and $g(x)=2x^2-x+9\text{,}$ evaluate $f(2)+g(2)$ and $f(-2)+g(-2)\text{.}$

Solution

We will compute $f(x)+g(x)$ on its own first.

\begin{equation*} \begin{aligned} f(x)+g(x) \amp= (x^3+x-8)+(2x^2-x+9)=x^3+2x^2+1 \end{aligned} \end{equation*}

Using this, we see that

and

### SubsectionMultiplying and Dividing Polynomials

The notation that is sometimes used to indicate multiplication and division of functions follows:

\begin{gather*} \text{Multiplication of functions: }(f\cdot g)(x)=f(x)\cdot g(x)\\ \text{Division of functions: }\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}, \text{ where }g(x)\neq 0. \end{gather*}

In this book we will simply consider multiplication notated by $f(x)\cdot g(x)$ and division notated by $\frac{f(x)}{g(x)}, \text{ where }g(x)\neq 0.$

###### Example137

Given $f(x)=15x^4-9x^3+6x^2$ and $g(x)=3x^2\text{,}$ find $f(x)\cdot g(x)$ and $\frac{f(x)}{g(x)}\text{.}$

Solution

Apply the distributive property and simplify.

For this quotient, assume $x\neq 0\text{.}$

\begin{equation*} \begin{aligned} \frac{f(x)}{g(x)}\amp = \frac{15x^4-9x^3+6x^2}{3x^2}\\ \amp = \frac{15x^4}{3x^2}-\frac{9x^3}{3x^2}+\frac{6x^2}{3x^2}\\ \amp = 5x^2-3x+2 \end{aligned} \end{equation*}

Our solutions are $f(x)\cdot g(x)=45x^6-27x^5+18x^4$ and $\frac{f(x)}{g(x)}= 5x^2-3x+2\text{.}$

###### Example138

Given $f(x)=6x-5$ and $g(x)=3x^2-2x-1\text{,}$ find $f(0)\cdot g(0)$ and $f(-1)\cdot g(-1)\text{.}$

Solution

To find $f(x)\cdot g(x)\text{,}$ we need to apply the distributive property multiple times. First, we distribute $6x-5$ to each of the terms in $g(x)$ and then distribute the terms to each of the $6x-5\text{.}$

\begin{equation*} f(x)\cdot g(x)\amp = (6x-5)(3x^2-2x-1) \\ \amp= (6x-5)(3x^2)-(6x-5)(2x)-(6x-5)(1) \\ \amp= (6x)(3x^2)-(5)(3x^2)-(6x)(2x)+(5)(2x)-(6x)(1)+(5)(1) \\ \amp= 18x^3-15x^2-12x^2+10x-6x+5 \\ \amp= 18x^3-27x^3+4x+5 \end{aligned} \end{equation*}

Now we substitute $0$ and $-1$ in for the variable $x\text{.}$

\begin{equation*} \begin{aligned} f(0)\cdot g(0)\amp = 18(0)^3-27(0)^2+4(0)+5\\ \amp =5 \end{aligned} \end{equation*}
\begin{equation*} \begin{aligned} f(-1)\cdot g(-1)\amp = 18(-1)^3-27(-1)^2+4(-1)+5\\ \amp = -18-27-4+5\\ \amp=-44 \end{aligned} \end{equation*}

Our solutions are $f(0)\cdot g(0)=5$ and $f(-1)\cdot g(-1)=-44\text{.}$

###### Example139

Given $f(x)=x^3+x-8$ and $g(x)=2x^2-x+9\text{,}$ find $f(x)\cdot g(x)\text{.}$

Solution
\begin{equation*} \begin{aligned} f(x)\cdot g(x) \amp= (x^3+x-8)(2x^2-x+9) \\ \amp= (x^3+x-8)(2x^2)-(x^3+x-8)(x)+(x^3+x-8)(9) \\ \amp= (x^3)(2x^2)+(x)(2x^2)-(8)(2x^2)-(x^3)(x)-(x)(x)+(8)(x)+(x^3)(9)+(x)(9)-(8)(9) \\ \amp= 2x^5+2x^3-16x^2-x^4-x^2+8x+9x^3+9x-72 \\ \amp= 2x^5-x^4+11x^3-17x^2+17x-72 \end{aligned} \end{equation*}


text:
# Thread: why is this equal to zero?

1. ## why is this equal to zero?

sorry it's not equal to zero, I was looking at the wrong answer:

I tried using the f'(t)h(f(t) - g'(t)h(g(t)) method but I got

$2sinx cosx(csc^2x+sec^x)$

2. ## Re: why is this equal to zero?

You can solve it like this:

\begin{align*} \frac{d}{dy}\int_{\cos^{2}(y)}^{\sin^{2}(y)}\frac{ 1}{t} dt =& \frac{d}{dy}\left( \ln(t) \Big{|}^{\sin^{2}(y)}_{\cos^{2}(y)}\right) \\ =& \frac{d}{dy}[\ln{(\sin^{2}(y))} - \ln{(\cos^{2}(y))}] \\ =& \frac{2 \sin(y) \cos(y)}{\sin^{2}(y)} + \frac{2 \sin(y) \cos(y)}{\cos^{2}(y)} \\ =& \frac{2\sin(y)\cos^{3}(y) + 2\sin^{3}(y)\cos(y)}{\sin^{2}(y)\cos^{2}(y)} \\ =& \frac{2\sin(y)\cos(y)(\sin^{2}(y) + \cos^{2}(y))}{\sin^{2}(y)\cos^{2}(y)} \\ =& \frac{2\sin(y)\cos(y)}{\sin^{2}(y)\cos^{2}(y)} \\ =& 2\csc(y)\sec(y) \end{align*}


text:
Private GMAT Tutoring in NYC & Online

# GMAT Official Guide 2016 Breakdown!

The GMAT Official Guide for 2016 (or OG 2016 as we'll be calling it throughout this review) is the book to beat for official GMAT practice questions. It represents the newest set of available official GMAT practice questions and seems to align a bit more with what I have seen on recent Official GMAT CATs. No surprise: the guide’s explanations still need improvement as does the online question bank. Read on for an ultra detailed review of this new Official Guide for 2016 along with a complete breakdown of the new questions. Here is the corrected answer key for the kindle version of OG 2016.

## an Official Context

The GMAT Official Guide 2016 comes nipping at the heels of the much derided 2015 edition. With the latter you paid for a snazzy new cover and a ghetto question bank filled with the exact same questions as the GMAT Official Guide 13th Edition. The GMAT Official Guide 2015 guide seemed like a thinly veiled cash grab on the part of GMAC and was a downgrade from the Official Guide 13th editions considering that the 13th could be purchased as a Kindle/Ebook. Needless to say I was skeptical of OG 2016 but being the curious GMAT tutor that I am I bit the bullet and trucked down to Barnes and Nobles in Tribeca (the only store that I could find that had the GMAT Official Guide 2016 in stock) to see what all of the fuss was about. I feared that the questions would be re-releases from some older, outdated sources but to my pleasant surprise, I can confirm that the questions  are, for the most part, shiny and new.

"The GMAT Official Guide 2015 seemed like a thinly veiled cash grab on the part of GMAC."

## GMAT Official Guide 2016 Quant

The OG 2016 problem solving seems to align more with the quant that I’ve seen on recent GMATs that I’ve taken. My feeling is that there is a bit more combinatorics on the real test and in line with that we’ve got a few extra combinatorics questions here in the new guide including a tricky code question (PS 117) with a bunch of conditions. Overlapping sets also seem to have to received a boost. A couple of oddities:

1. Over 20% of the new questions are Geometry questions. The re-shuffle isn’t proportionate with the amount of Geometry in the problem solving. Also - there doesn’t seem to be anything special about these geometry questions or anything particularly difficult about them. I’m not quite sure why there was this big infusion.
1. The new questions stop at #153. So 154-230 are all "old" questions. The questions are in order of difficulty so we can safely assume that we weren’t treated to any new super tough questions. Having gone through all of the new questions I can confirm this. Does this make the book less useful for top scorers? A little. Still the medium level questions have been bulked up and these questions are probably more important for most people.

"Medium level problem solving questions have been bulked up in this new edition of the GMAT Official Guide"

After seeing the lopsidedness in the problem solving I was expecting some weirdness in the Data Sufficiency but the OG 2016 additions seem evenly distributed in terms of content and level of difficulty. Here is a chart from the graphics maestros at Bob and Uncle Design showing most of the major changes in the GMAT Official guide 2016:

### PROBLEM SOLVING 57 New Questions - 24.7%:1, 2, 3, 11, 13, 14, 15, 17, 18, 21, 25, 27, 30, 31, 33, 38, 39, 43, 44, 45, 55, 56, 57, 59, ,60, 61, 63, 66, 67, 68, 69, 70, 73, 76, 82, 83, 85, 86, 87, 88, 89, 91, 97, 99, 107, 108, 111, 114, 117, 122, 125, 134, 136, 137, 139, 142, 153

Percents  1 (Percents), 11 (Percents), 55 (Percent/Word Problem),  30 (Percents/Arithmetic/Word Problem), 99 (Word Problem/Percent)

Work/Rate  59 (Work/Rate), 76 (Work/Rate),

Ratio/Proportions 134 (Proportions/Word Problem), 114 (Mixture)

Algebra/Arithmetic 2 (Arithmetic/Fractions), 3 (Algebraic Translation), 14 (Arithmetic), 21(Algebra/Arithmetic), 25 (Algebra/Multiple Equations), 61 (Inequalities), 66 (System of Equations)

Exponents 56 (Exponents/Word Problem), 57 (Exponents/Roots/Number Properties), 83 (Exponents), 85 (Exponents/Roots/Arithmetic), 67 (Arithmetic/Exponents), 68 (System of Equations/Exponents),

Number Properties  38 (Number Properties/Digits), 107 (Number Properties), 108 (Number Properties), 17 (Number Properties/Absolute value/Exponents), 43 (Number Properties), 87 (Consecutive Integers)

Overlapping Sets 69 (Overlapping Sets/Percents/Word Problem), 89 (overlapping sets), 91 (overlapping sets)

Combinatorics 117 (Combinatorics), 136 (Probability)

Statistics  137 (Average), 15 (Average), 88 (Average), 111 (Average)

Functions/Sequences/Formula 39 (sequence),63 (Functions/Exponents), 82 (Formula/Arithmetic), 97 (formula), 27 (Data Interpretation)

Geometry 13 (Geometry/Quadrilaterals/Word Problem), 18 (Geometry/Triangles), 31 (Geometry/Mixed Shapes), 33 (Geometry/Triangles), 44 (geometry/coordinate), 45 (Geometry/Quadrilaterals), 60 (Geometry/Coordinate), 70 (Geometry/Mixed Shapes), 73 (Geometry/Quadrilaterals), 86 (Geometry/3D Shapes), 122 (Geometry/Mixed Shapes), 125 (Geometry/Quadrilateral/Word Problem), 139 (Geometry/Quadrilateral/ratio), 142, 153 (Geometry/Coordinate),

### GMAT Official guide DATA SUFFICIENCY - 44 new questions / 25.2%

7, 10, 12, 14, 15, 16, 18, 19, 20, 21, 23, 26, 27, 29, 34, 37, 38, 39, 40, 41, 46, 48, 49, 51, 54, 55, 60, 64, 67, 73, 77, 80, 83, 84, 86, 92, 94, 115, 121, 134, 135, 136, 137, 150

Combinatorics 7 (probability), 73 (Geometry/Combinatorics)

Exponents 20 (Exponents), 37 (Square Root/Match the Expression/Special Quadratic), 39 (special quadratics), 41 (Exponents/Inequalities)

Geometry 136 (Coordinate), 137 (Triangles), 26 (Mixed Shapes/Quadrilateral), 27 (Coordinate), 49 (Quadrilaterals), 51 (Quadrilaterals), 115 (Mixed Shapes), 86 (3d Shapes), 77 (Geometry/Number Line)

Word/Rate 134 (Threshold/Rate/Word Problem), 84 (Average Rate)

Counting Equations 10 (Translation), 21 (Interest/Word Problem), 23 (Word Problem), 64 (Word Problem), 67 (Word Problem)

Number Properties 16 (# properties/Inequalities), 94 (Number Properties/Digits), 54 (Exponents/# Properties), 60 (# Properties),

Overlapping Sets 150 (Overlapping Sets), 29 (Overlapping Sets), 19 (Overlapping Sets)

Statistics  135 (Statistics), 14 (Statistics), 15 (Statistics), 92 (Average)

Threshold 48 (Threshold/Work), 46 (Threshold/Word Problem), 34 (Threshold), 80 (Threshold)

Translation 55 (Ratio/Word Problems), 83 (Threshold), 18 (Algebra),

Formula/Function/Sequence  12 (Sequence), 38 (Formula),  121(Function/Number Properties/Exponents)

## GMAT Official Guide Verbal

The Official Guide 2016 adds more difficulty to the reading comprehension. Unfortunately the quality does not come with quantity. I feel a little burned with only six new reading comprehension passages. Is it sounding entitled wanting an extra couple of tough ones? Maybe it is. But I did pay the full 45 cover price at Barnes and Noble. Hint: the Amazon price for this book is way cheaper (I bought my second copy from them). Back on topic: happy to see more tough passages. Happy to see more long passages. Still the Official Guide doesn't have enough really tough reading comprehension so if you really want to knock the verbal out of the park then I would recommend supplementing with LSAT work. Here is an article on how use LSAT for GMAT verbal practice. The OG 2016 is awash with new critical reasoning and I am happy to report that this CR aligns more with what I’ve seen on recent tests. Not only do the questions seem a bit trickier but some the question prompts hide the real question. For instance, you’ll be asked to provide a secondary explanation for a set of facts rather than weaken the argument. Those two actions are really the same but most people feel much more comfortable with the simple, straightforward language of “weaken”, “strengthen”. This is a positive move and should help prepare you for the real GMAT. One oddity here in the critical reasoning: about 75% of the updates are in the assumption based family. That’s a major shift. The official guide for GMAT review 2016 has 29 new assumption based questions. Wow. On the flipside, the content family has seen only six new additions. That seems strange considering that the content questions account for about 40% of the section. Why this lumpiness? The sentence correction update reminds me of the Data Sufficiency update. I’m happy about it but there’s nothing terribly interesting to report. If I find anything as I dig deeper into this new OG I will report back. "GMAT Official Guide 2016 is awash with new critical reasoning." ### Official Guide for GMAT Review 2016 Reading Comprehension - 6 new passages, 31 new Questions /22.3% Page 370 (Short)/Questions 8-10 Page 386 (Long)/Questions 42-47 Page 404 (Long)/Questions 96-101 Page 408 (Long)/Questions 107-112 Page 416 (Long)/Questions 130-136 Page 418 (Short)/Questions 137-139 ### Official Guide for GMAT Review 2016 CRITICAL REASONING - 34 new questions / 28.2% 1, 2, 3, 4, 8, 21, 30, 37, 39, 41, 43, 48, 49, 53, 55, 57, 58, 60, 63, 65, 67, 70, 76, 77, 79, 81, 84, 94, 97, 100, 106, 109, 121, 124 Strengthen/Weaken: 2, 3, 4, 21, 30, 37, 39 (oddball), 48, 55, 57, 60, 63, 65, 67 (oddball), 76, 77, 79, 94, 97, 106, 121, 124 Assumption: 8, 37, 41, 49, 70, 84, 109 | Evaluate: 100 | Inference: 1, 53 | Paradox: 58 | Boldface: 81 | Fill in the Blank: 43 Assumption Based: 29 Strengthen/Weaken - 21 | Assumption - 7 | Evaluate - 1 Content Based: 5 Paradox - 1 | Boldface - 1 | Inference - 2 | Fill in the Blank - 1 ### GMAT Official guide 2016 SENTENCE CORRECTION - 35 new questions / 25% 2, 4, 7, 10, 13, 18, 20, 23, 27, 28, 31, 35, 37, 42, 46, 52, 63, 66, 73, 80, 86, 90, 97, 101, 111, 116, 120, 124, 126, 128, 132, 133, 136, 137, 138 Nothing terribly interesting to report on the SC. The difficulty of the questions is evenly distributed and the types of rules tested are familiar. ### The Online Question Bank powered by windows 95 It took me a second to find the GMAT official guide question bank. I even googled “Official Guide 2016 Online Question Bank” and found pretty much nothing useful. I felt the same as I had felt with other GMAC digital products: annoyed. I’ve had issues with pretty much every piece of software that I’ve bought from GMAC. Alas, flipping to the back of the OG 2016, I spied a plastic pouch with my access code along with the OG 2016 question bank website: GMAT.Wiley.com. Eureka! ### What you can do with the GMAT online question bank: 1. Set up a practice set mixing and matching order of difficulty and content. Although you have you have to pick one order of difficulty for all of the content. If you are doing problem solving hard then so must you do critical reasoning hard. This is a bug! 2. Save your sessions so that you can review at a later date. Considering that GMAC already has these questions tagged by type, it would have been helpful to have been able to set up quizzes by content type. Is that really so hard? C’mon! We paid the45 ($50+ with tax and subway ride) throw us a GMAT bone! Also - no internet means no question bank. This wouldn’t be a problem in a major city in say, South Korea but unfortunately here in the USA public wifi is spotty. That means if you are not at home and want to study you will probably have to lug around the book. Annoying not only because the GMAT Official Guide is a heavy ass book but also because you ought to be doing GMAT practice ogling a computer screen as you will on your actual GMAT (here are a few more good GMAT study habits) . Practicing on the screen can make a big difference. Oh, almost forgot the insult to injury: you only have access to the Official Guide 2016 online question bank for 1 year. ## an Official Conclusion I'm conflicted about the fact that I gave this book five stars on Amazon. My feeling is that the Official Guide for GMAT review 2016 could be a whole lot better. There's a ton of room for improvement in the explanations and in the online question bank. In addition it would be nice to just have a list of what questions are unique to each version of the Official Guide. Is that some sort of secret? And further why not make it clear why certain questions are getting removed? Right? That said, the 13th edition of the official guide was the best resource in GMAT land and now the GMAT Official Guide for 2016 is king (or queen). If you are just starting out with your GMAT preparation buy this book. If you need extra questions I’d also recommend buying this book (provided that you have exhausted Question pack 1). Happy studies! Here are some Official GMAT resources that you might also want to have: 1. Question Pack 1. This is a question bank of official GMAT Questions. This is a very close second to the Official Guide 2016 in terms of usefulness and a no brainer at$29.99
2. Exam Pack 1. Two extra Official GMAT Practice Tests. These are awesome!

These first three (OG 2016, QP1, and the Exam Pack) I would consider critical. The last three are not critical but can be very helpful.

1. The Official Guide for Quant Review. The questions here are slightly easier than the ones in the Official Guide. This is a good “skill builder” book.
2. The Official Guide for Verbal Review. Similar to the Quant review in difficulty and usefulness.
3. GMAT Focus Quizzes. Computer adaptive Quant quizzes made up of official GMAT questions. Stellar for the last week of preparation. There is some controversy with these because they are expensive and there is a bit of overlap with other resources. You can buy them here.

### One Response to GMAT Official Guide 2016 Review

1. Josh says:

Hi

Do you have a breakout of the new questions by difficulty? You have an infographic on it but it would be helpful to know which of the quant questions were Medium and Hard so I can skip the easy ones.

Thank you for putting this together!

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# HD 124953

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 Photoelectric Minima of Selected Eclipsing BinariesNot Available Rotational velocities of A-type stars in the northern hemisphere. II. Measurement of v sin iThis work is the second part of the set of measurements of v sin i forA-type stars, begun by Royer et al. (\cite{Ror_02a}). Spectra of 249 B8to F2-type stars brighter than V=7 have been collected at Observatoirede Haute-Provence (OHP). Fourier transforms of several line profiles inthe range 4200-4600 Å are used to derive v sin i from thefrequency of the first zero. Statistical analysis of the sampleindicates that measurement error mainly depends on v sin i and thisrelative error of the rotational velocity is found to be about 5% onaverage. The systematic shift with respect to standard values fromSlettebak et al. (\cite{Slk_75}), previously found in the first paper,is here confirmed. Comparisons with data from the literature agree withour findings: v sin i values from Slettebak et al. are underestimatedand the relation between both scales follows a linear law ensuremath vsin inew = 1.03 v sin iold+7.7. Finally, thesedata are combined with those from the previous paper (Royer et al.\cite{Ror_02a}), together with the catalogue of Abt & Morrell(\cite{AbtMol95}). The resulting sample includes some 2150 stars withhomogenized rotational velocities. Based on observations made atObservatoire de Haute Provence (CNRS), France. Tables \ref{results} and\ref{merging} are only available in electronic form at the CDS viaanonymous ftp to cdsarc.u-strasbg.fr (130.79.125.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/393/897 Diffraction-limited 800 nm imaging with the 2.56 m Nordic Optical TelescopeA quantitative assessment is presented of diffraction-limited stellarimages with Strehl ratios of 0.25-0.30 obtained by selection ofshort-exposure CCD images of stars brighter than +6 m at 810 nm with theNordic Optical Telescope. A revised catalogue of delta Sct starsAn extensive and up-dated list of delta Sct stars is presented here.More than 500 papers, published during the last few years, have beenrevised and 341 new variables have been added to our last list, sixyears ago. This catalogue is intended to be a comprehensive review onthe observational characteristics of all the delta Sct stars known untilnow, including stars contained in earlier catalogues together with othernew discovered variables, covering information published until January2000. In summary, 636 variables, 1149 references and 182 individualnotes are presented in this new list. Tables 1 and 2 will be accessibleonly in electronic form at the CDS via anonymous ftp tocdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/Abstract.html X-ray emission from A0-F6 spectral type starsWe use the ROSAT public data archive to study the X-ray emission of asample of supposedly single A0-F6 spectral type stars from the BrightStar Catalogue. We detected X-ray emission from 19 A- and 33 F-typestars. However, our results are not sufficient to associate withcertainty the X-ray emission to the A-type stars themselves, since theusual argument that it may originate from a binary companion cannot beexcluded. A spectral analysis was conducted for 14 sources (3 A and 11F), finding that for 12 of them a two temperature thermal plasma modelis needed to reproduce the observed spectra. The two temperatures arecentered at 0.13 and 0.54 keV, respectively. The values found for thehigher temperature are lower than that ones of X-ray selected singlelate-type stars. The X-ray luminosities are in the range LX ~10(28}-10({30)) erg s(-1) , with a distribution similar to that ofactive late-type stars. No correlation is found between LXand B-V color, V sin i and Lbol, while a positive correlationis found between the X-ray luminosity and the hardness ratio. The Angular Momentum of Main Sequence Stars and Its Relation to Stellar ActivityRotational velocities are reported for intermediate-mass main sequencestars it the field. The measurements are based on new, high S/N CCDspectra from the Coudé Feed Telescope of the Kitt Peak NationalObservatory. We analyze these rotation rates for a dependence on bothmass and age. We compare the average rotation speeds of the field starswith mean velocities for young stars in Orion, the Alpha Persei cluster,the Pleiades, and the Hyades. The average rotation speeds of stars moremassive than $\sim1.6$ \msun\experience little or no change during theevolutionary lifetimes of these stars on the zero age main sequence orwithin the main sequence band. Less massive stars in the range betwee n1.6\msun\ and 1.3\msun\ also show little decline in mean rotation ratewhile they are on the main sequence, and at most a factor of 2 decreasein velocity as they evolve off the main sequence. The {\it e}-foldingtime for the loss of angular momentum b y the latter group of stars isat least 1--2 billion years. This inferred characteristic time scale forspindown is far longer than the established rotational braking time forsolar-type stars with masses below $\sim1.3$ \msun. We conclude from acomparison of the trends in rotation with trends in chromospheric andcoronal activity that the overall decline in mean rotation speed alongthe main sequence, from $\sim2$ \msun\ down to $\sim1.3$ \msun, isimposed during the pre-main sequence phase of evolution, and that thispattern changes little thereafter while the star resides on the mainsequence. The magnetic activity implicated in the rotational spindown ofthe Sun and of similar stars during their main sequence lifetimes mus ttherefore play only a minor role in determining the rotation rates ofthe intermediate mass stars, either because a solar-like dynamo is weakor absent, or else the geometry of the magnetic field is appreciablyless effective in removing angular momentu m from these stars. (SECTION:Stars) Classification of Population II Stars in the Vilnius Photometric System. II. ResultsThe results of photometric classification of 848 true and suspectedPopulation II stars, some of which were found to belong to Population I,are presented. The stars were classified using a new calibrationdescribed in Paper I (Bartkevicius & Lazauskaite 1996). We combinethese results with our results from Paper I and discuss in greaterdetail the following groups of stars: UU Herculis-type stars and otherhigh-galactic-latitude supergiants, field red horizontal-branch stars,metal-deficient visual binaries, metal-deficient subgiants, stars fromthe Catalogue of Metal-deficient F--M Stars Classified Photometrically(MDPH; Bartkevicius 1993) and stars from one of the HIPPARCOS programs(Bartkevicius 1994a). It is confirmed that high galactic latitudesupergiants from the Bartaya (1979) catalog are giants or even dwarfs.Some stars, identified by Rose (1985) and Tautvaisiene (1996a) as fieldRHB stars, appear to be ordinary giants according to our classification.Some of the visual binaries studied can be considered as physical pairs.Quite a large fraction of stars from the MDPH catalog are found to havesolar metallicity. A number of new possible UU Herculis-type stars, RHBstars and metal-deficient subgiants are identified. The Relation between Rotational Velocities and Spectral Peculiarities among A-Type StarsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1995ApJS...99..135A&db_key=AST Vitesses radiales. Catalogue WEB: Wilson Evans Batten. Subtittle: Radial velocities: The Wilson-Evans-Batten catalogue.We give a common version of the two catalogues of Mean Radial Velocitiesby Wilson (1963) and Evans (1978) to which we have added the catalogueof spectroscopic binary systems (Batten et al. 1989). For each star,when possible, we give: 1) an acronym to enter SIMBAD (Set ofIdentifications Measurements and Bibliography for Astronomical Data) ofthe CDS (Centre de Donnees Astronomiques de Strasbourg). 2) the numberHIC of the HIPPARCOS catalogue (Turon 1992). 3) the CCDM number(Catalogue des Composantes des etoiles Doubles et Multiples) byDommanget & Nys (1994). For the cluster stars, a precise study hasbeen done, on the identificator numbers. Numerous remarks point out theproblems we have had to deal with. A catalogue of variable stars in the lower instability strip.Identifications, positions, photometry, spectra, some pulsationalfeatures, other astrophysical parameters and literature for 302pulsating variable stars in the lower instability strip, near the ZAMS,are given. About 185 stars have near homogeneous photometric informationin the Stroemgren's uvby-β photometric system. Thiscatalogue/database covers information published until November 1993. δ Scuti stars: a new revised listAn extensive and up to date list of δ Sct stars is presented. Thiscatalogue is intended to be a comprehensive review of observationalcharacteristics of all the δ Sct stars known until now, includingstars contained in earlier catalogues together with other new discoveredvariables, covering information published until November 1993. Globalinformation in the form of histograms and diagrams are also shown. Catalogue of CP stars with references to short time scale variabilityA catalogue was compiled which contains all references in the literaturesince 1962 related to variations of CP stars on time scales shorter thanthe rotation period. The role of this catalogue lies in the unbiasedlisting of all available references, and not in a critical evaluation. A seven-year northern sky survey of AP stars for rapid variabilityA high-speed photometric survey of 120 Ap stars in the northern sky, hasbeen conducted, between 1985 and 1991, in order to search for rapidvariability. Stars of spectral types, namely from B8 to F4, have beenselected for the survey. The selected pulsational variable stars occupythe hotter regions of the instability strip of the Hertzsprung-Russeldiagram. Noted is the absence of pulsations in the hotter B8-A3 Apstars; this does not, however, preclude the existence of pulsations,since HD 218495 was recently discovered to be a rapidly oscillating Ap(roAp) star. The primary result of this study is that variouscombinations of photometric indices, while pointing towards roAp starshaving the characteristic signatures of cool, SrCrEu stars, still failto isolate the roAp phenomenon from similar nonpulsating Ap stars.Color-magnitude and color-color diagrams are presented in order tocomplete this survey. The radial velocity curve and the two-dimensional spectral classification of the Delta SCT variable HD 18878The radial velocities of HD 18878 were measured from spectra of this newDelta Sct type variable. The radial velocity curve with elementsdetermined from the photoelectric observations of Frolov et al. (1990)is derived. The mean radial velocity is -20 km/s, and the amplitude ofits variation is 34 km/s. On the basis of spectral criteria and thelocation of this star on a two-color diagram, it is argued that HD 18878is an A9III star. Digging in the coronal graveyard - A ROSAT observation of the red giant ArcturusA deep exposure of the bright star Arcturus (Alpha Bootis: K1 III) withthe Roentgensatellit (Rosat) failed to detect soft X-ray emission fromthe archetype 'noncoronal' red giant. The 3-sigma upper limit in theenergy band 0.1-2.4 keV corresponds to an X-ray luminosity of less than3 x 10 to the 25th erg/s, equivalent to a coronal surface flux densityof less than 0.0001 solar. The nondetection safely eliminates coronalirradiation as a possible mechanism to produce the highly variable He I10830 feature and emphasizes the sharp decline in solarlike coronalactivity that accompanies the evolution of low-mass single stars awayfrom the main sequence. While the most conspicuous object in the Rosatfield of view was not visible in X-rays, at least one fainter star isamong the about 60 sources recorded: the Sigma Sct variable CN Boo, anA8 giant in the UMa Stream. X-ray studies of coeval star samples. III - X-ray emission in the Ursa Major streamResults are reported from a comprehensive survey of X-ray emission fromstars known or suspected to be members of the UMa cluster and/or stream.Of the 42 UMa member stars surveyed, 18 were detected as X-ray sources,and spectral analysis was performed for 10 stars with sufficient X-raycounts. Consideration is given to relations between X-ray luminosity,color, and kinematics of the sample stars, and the X-ray spectra of theUMa stars are discussed in the context of the general problem of stellarX-ray temperatures. Also confirmed is the lack of X-ray-emitting Adwarfs among UMa members; among stars of later spectra type there is arather large dispersion in X-ray luminosity. This dispersion cannotreadily be explained by contamination with field star interlopers andappears rather to be a property of the UMa X-ray luminosity distributionfunction. Empirical P-L-C relation for Delta Scuti stars - A catalogueAn extensive and up-to-date list of 192 Delta Scuti stars is presented.Empirical period - luminosity - color (P-L-C) relations are obtained forthe four lowest modes corresponding to radial pulsations. Agreement withpredicted values indicates that, in general, both Stroemgren photometriccalibration and pulsation theory work well for these stars. Third preliminary catalogue of stars observed with the photoelectric astrolabe of the Beijing Astronomical Observatory.Not Available ICCD speckle observations of binary stars. IV - Measurements during 1986-1988 from the Kitt Peak 4 M telescopeOne thousand five hundred and fifty measurements of 1006 binary starsystems observed mostly during 1986 through mid-1988 by means of speckleinterferometry with the KPNO 4-m telescope are presented. Twenty-onesystems are directly resolved for the first time, including newcomponents to the cool supergiant Alpha Her A and the Pleiades shellstar Pleione. A continuing survey of The Bright Star Catalogue yieldedeight new binaries from 293 bright stars observed. Corrections tospeckle measures from the GSU/CHARA ICCD speckle camera previouslypublished are presented and discussed. Quasi-160-minute oscillation period of Delta Scuti starsThe resonance power spectrum (or commensurability spectrum) computed for217 Delta Scuti stars indicates that the dominant (most commensurate)period for the total set of oscillation periods of these stars is 162.2+ or - 2.8 min. Within the error limits, this period coincidesapproximately with the 160-min period of global oscillations of the sun. The Sirius superclusterPhotometric data on the chemical composition of 927 A stars in the UrsaMajor stream, called the Sirius supercluster, were used to estimate theage and place of formation of the objects. The stars studied are in thesolar neighborhood and have been observed to be co-moving in a velocityellipsoid with a (U, V) velocity of 10.3 km/sec and concentrated in aspatial volume less than 10 pc across. The Stromgren and Geneva systemphotometric data show that the supercluster is homogeneous in chemicalcontent, although the value of the forbidden Fe/H ratio could not beprecisely determined. The supercluster age is projected to be from260-620 Myr, with the origin having been in the Carina spiral arm of theGalaxy. Three-dimensional calssification of F-M type halo stars in the Vilnius photometric systemNot Available Revised list of pulsating stars with ultra-short periodsA comprehensive list of 178 known Delta Scuti and RR Lyrae-stars ispresented. Using this revised list a HR diagram for these ultra shortperiod pulsating stars is plotted and the blue and red edges of theresultant instability strip are determined. Selection effects arediscussed, and the PLC relationship of Breger (1979) is tested usingdata from this list. Stars lying outside the defined instability regionare discussed. Classification of intrinsic variables. IX - The Cepheid domainThe present H-beta, RI and intermediate band observations obtained for135 high mass, long- and short-period Cepheids (LPC and SPC) throughoutthe cycle of variation are combined with observations for 100 ultrashortperiod Cepheids in order to delineate the Cepheid domain in the (beta,/C1/) and (M-bol, log Te) planes. The independently derived luminositycalibrations for bright giants and supergiants previously published arefound to closely reproduce the luminosities derived from theSandage-Tammann (1968) PL relation for most variables. The majority ofthe LPC appear to be overabundant in metals by comparison to the sun,and reddening determinations for the LPC suggest a small systematiccorrection to previous results. The distribution of pulsation constants Q of Delta Scuti starsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1981PASP...93..237A&db_key=AST 65th Name-List of Variable StarsNot Available Properties of AM stars in the Geneva photometric systemAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1980A&A....92..289H&db_key=AST Prediction of spectral classification from photometric observations - Application of the UVBY beta photometry and the MK spectra classification. II - General caseAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1980A&A....85...93M&db_key=AST Prediction of spectral classification from photometric observations-application to the UVBY beta photometry and the MK spectral classification. I - Prediction assuming a luminosity classAn algorithm based on multiple stepwise and isotonic regressions isdeveloped for the prediction of spectral classification from photometricdata. The prediction assumes a luminosity class with reference touvbybeta photometry and the MK spectral classification. The precisionattained is about 90 percent and 80 percent probability of being withinone spectral subtype respectively for luminosity groups I and V and forluminosity groups III and IV. A list of stars for which discrepanciesappear between photometry and spectral classification is given. Photometric variability on the lower part of the Cepheid instability strip. I - Evolved AM starsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1980A&A....83..114G&db_key=AST
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# The Top-to-Random Shuffle II

In the last post, I introduced the top-to-random shuffle. In particular, we considered why this sort of procedure was important as an alternative to choosing an ordering afresh, and how to we would go about measuring how close we had got to randomness.

In this post, I want to develop the second of these points. The intuition might be that we can get very close to uniform randomness if we repeat the shuffle often enough. Recall this means that even if we choose our bet in a really complicated and careful way, we still couldn’t make much profit by knowing the actual distribution of the ordering. But we might also suspect that the pack will never be exactly random, in the same way that the distribution of the proportion of heads seen on a repeatedly-flipped coin will eventually get very close to 1/2, but will not be exactly 1/2.

This intuition is extremely sensible, and in general is true. It is a nice fact, however, that it fails for the top-to-random shuffle, where we do in fact get to a uniformly random deck. Recall that we approximated how long it would take to get to a state that was roughly random by calculating the time taken for the original bottom card to rise to the top of the deck. This time was:

$n+\frac{n}{2}+\frac{n}{3}+\ldots+\frac{n}{n-1},$

where n is the total number of cards. A further shuffle is required to send the original card back into the pack somewhere. The claim is that the pack is now uniformly random. Note that if we ever actually use this, we have to be careful because although we can calculate the average time at which this event happens, the time itself is random. Rather than worry about that, let’s see why it is true.

As a motivating example, let’s assume that the pack was originally in the order 1,2,3,…,n, and consider the final relative order of the cards numbered 1 and 2. There is some small probability that on the first go (and then again on the second go possibly also) that the card number 1 stays put. Let’s ignore that possibility and progress to the first time that the 1 has been moved into the interior of the pack, so the 2 is now on top. When we do this, we are choosing a number between 2 and n uniformly at random, and moving the card numbered 1 to this position. Let’s call this number X.

Now, when we try to shuffle the 2, we are choosing a number between 1 and n uniformly at random, and moving the card numbered 2 to this position. Let’s call this number Y. Under exactly what circumstances does 2 end up above 1? The clearest example is if Y=X. Then card 2 has been moved to the position previously occupied by card 1. So card 1 moves up a position (since the card 2 is no longer on the top). The final configuration therefore includes card 1 directly above card 2. So we can say

$\mathbb{P}(\text{2 above 1})=\mathbb{P}(Y

where the fact that the inequality in the second probability is strict is important. To calculate this second probability, we want to exploit the symmetry of the situation. The only problems are that the case X=Y is not symmetric as then 1 ends up above 2 as described above, and also that X cannot be 1. So we account for these separately. Note

$\mathbb{P}(Y=1)=\frac{1}{n},\quad \mathbb{P}(Y=X)=\frac{1}{n}.$

The second result holds overall since it holds whenever we condition on a particular value of X. These events are also disjoint. Then

$\mathbb{P}(Y1, X\neq Y)\mathbb{P}(Y>1,X\neq Y)$

$= \frac{1}{n}+\frac{1}{2}(1-\mathbb{P}(Y>1,Y\neq X))$

$=\frac{1}{n}+\frac{1}{2}(1-\frac{1}{n}-\frac{1}{n}) = \frac{1}{2}.$

In summary, the cards 1 and 2 will be in uniformly random order. We might like to extend this idea, but it will get complicated when we add card 3 to the mix, as it is possible (if unlikely) that 1 and 2 will be further mixed before this. This shouldn’t affect the result much, but it will get complicated to define the notation required to carry this sort of argument all the way up to the nth card.

Even using induction is not going to make life substantially easier. Knowing that once we have inserted cards 1,2,…,k into the pack they are in uniformly random order is not enough to make inference about what happens once we put k+1 into the pack. We have to know something about the current positions of 1,2,…,k. For example, if one of these cards is definitely on the bottom of the pack, then the probability that k+1 ends up last among 1,2,…,k+1 is 1/n rather than 1/k+1 as it should be. So in fact we would have to control an annoying amount of information jointly.

In the argument we attempted above, we were looking at the first times some card k got folded back into the pack. Note that this division of time is different to the one we were using for the coupon collector approach to the mixing time in the previous post. Let’s try to use that instead here.

Now we consider the times at which a card is moved below card n. We deliberately decline to say what these cards are. But rather, we want to prove that, conditional on the cards below n being $A_k=\{a_1,\ldots,a_k\}$, the ordering of these is uniform on $S_{A_k}$, that is, every possibility is equally likely. Now this is easy to prove by induction. For, by conditioning on $A_k$ and $a_{k+1}$ being the new card to be moved below n, we are conditioning on the set of cards below n now being $A_{k+1}=A_k\cup\{a_{k+1}\}$. The position of the new card is uniformly random within this, by construction of the top-to-random shuffle, and so the new arrangement is uniformly random on the (k+1)! possibilities.

To see why we have proved the original result we wanted, note that this argument works at the time when the original bottom card is now at the top. So the remaining cards are uniformly randomly ordered. Inserting card n at random gives an arrangement that is uniformly random overall. So as we suggested before, working out how long it takes to get close to randomness in this case reduces to working out how long it is before the original bottom card hits the top and is re-inserted, as at that point, the pack genuinely is uniformly random.


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# How can we divide 2 vectors?

1. Apr 1, 2005

### TheDestroyer

How can We Devide 2 vectors? and what is the result? is it a scalar or another vector?

if we said:

$$\frac{\vec{E_1}}{\vec{E_2}}$$

What the result would be if the unit vectors or not equal or opposite? if they are then we'll get a scalar,

Can we say:

$$\frac{\vec{E_1}}{\vec{E_2}} = \vec{E_1} \cdot \frac{1}{\vec{E_2}}$$

Can this be a dot product? then we'll get a scalar also !! and if it's like this, what's the defenition of a vector inverse?

2. Apr 1, 2005

### matt grime

If you want to define division then you need to define multiplication. So how do you define the product of two (arbitrary) vectors?

3. Apr 1, 2005

### cronxeh

there are infinitely many answers, because there are that many possibile vectors, therefore vector division is not defined

4. Apr 1, 2005

### TheDestroyer

How come??? can you explain??

In electromagnetic we define the freenels coeffecients with the ratio of to electrical fields,

can you explain why there are infinite answers?

and about the definition of 2 dot product vectors simply:

a.b = ax.bx + ay.by + az.bz
or
a.b = ab cos(a,b)

5. Apr 1, 2005

6. Apr 1, 2005

### matt grime

The dot product of two vectors isn't a vector so doesn't define a division operation.

If you're going to ahve a sensible notion of division then you need a sensible notion of multiplication. 1/b is *define* to be the (unique) element (of whatever objects you're thinking about) such that b*1/b=1. This requires a *well defined* mutliplication, and an identity element. You've so far got neither of these. well defined means something like a*b=a*c implies b=c.

7. Apr 1, 2005

### TheDestroyer

I didn't like this !!! or we can say this is not logic !!! because as I said, some physical quantities are defined as the ratio of 2 vectors, that means it must have a value, even if it's not single,

I'll work on this a little, and i'll repost the result, thanks everybody for caring about answering,

TheDestroyer

8. Apr 1, 2005

### HallsofIvy

Staff Emeritus
WHAT physical quantities are defined as the ratio of two vectors? I've never seen such a thing! (The Fresnel coefficients, in particular, are NOT defined as the ratio of two vectors.)

9. Apr 1, 2005

### dextercioby

The Fresnel coefficents are defined with absolute values of the 2 electric field vectors.

I guess one could say that euclidean vectors form a vector/linear space over R,and not an algebra...

Daniel.

10. Apr 1, 2005

### Data

Well, in special cases, they do form a division algebra with an appropriate multiplication. For example, we can look at $\mathbb{R}^2$ and define the operation

$$(a, b)(c, d) = (ac-bd, ad+cb)$$

which does permit meaningful division using the definition

$$(a, b)/(c, d) = \left(\frac{ac+bd}{c^2+d^2}, \frac{cb-ad}{c^2+d^2}\right), \ \forall (c, d) \neq (0, 0)$$

Of course, these definitions just stem from the fact that $\mathbb{C}$ is a division algebra. The same thing can clearly be done for $\mathbb{R}^4$ and $\mathbb{R}^8$ using the multiplication rules for quaternions and octonions.

It doesn't work for $\mathbb{R}^3$, though.

Last edited: Apr 1, 2005
11. Apr 1, 2005

### matt grime

They do form an algebra in many non-isomorphic ways. Just almost never division algebra.

12. Apr 1, 2005

### dextercioby

Okay.However,there are more definitions of a vector...

Daniel.

13. Apr 1, 2005

### matt grime

Given any (finite dimensional) vector space I can create an algebra from it. And probably from most infinite ones too, I shouldn't wonder.

Last edited: Apr 1, 2005


text:
## Decay bounds for Bernstein functions of Hermitian matrices with applications to the fractional graph Laplacian

Marcel Schweitzer

### Abstract

For many functions of matrices $f(A)$, it is known that their entries exhibit a rapid—often exponential or even superexponential—decay away from the sparsity pattern of the matrix $A$. In this paper, we specifically focus on the class of Bernstein functions, which contains the fractional powers $A^\alpha$, $\alpha \in (0,1)$, as an important special case, and derive new decay bounds by exploiting known results for the matrix exponential in conjunction with the Lévy-Khintchine integral representation. As a particular special case, we find a result concerning the power law decay of the strength of connection in nonlocal network dynamics described by the fractional graph Laplacian, which improves upon known results from the literature by doubling the exponent in the power law.

Full Text (PDF) [985 KB], BibTeX

### Key words

matrix functions, Bernstein functions, off-diagonal decay, graph Laplacian, fractional powers, nonlocal dynamics

### AMS subject classifications

05C82, 15A16, 65F50, 65F60

### Links to the cited ETNA articles

 [7] Vol. 28 (2007-2008), pp. 16-39 Michele Benzi and Nader Razouk: Decay bounds and $O$($n$) algorithms for approximating functions of sparse matrices [22] Vol. 48 (2018), pp. 362-372 Andreas Frommer, Claudia Schimmel, and Marcel Schweitzer: Non-Toeplitz decay bounds for inverses of Hermitian positive definite tridiagonal matrices

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text:
# Permutation and Combination Quiz 1

Question 1

Time: 00:00:00
Manjistha is looking for ways to make groupings with all of the letters in the word "NATION." How many N's appear in each of these arrangements?

5!/2! , 5!

5!/2! , 5!

6!/2! , 5!

6!/2! , 5!

4!/3!, 3!

4!/3!, 3!

None of the above

None of the above

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Question 2

Time: 00:00:00
Manish wants to know how many different ways the characters in the word "PARAGLIDING" may be organized so that all of the vowels appear at the same time.

132005

132005

120452

120452

120960

120960

115639

115639

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Question 3

Time: 00:00:00
Lima and her pals were on a long drive from Bangalore to Gokarna. Five pals, only two of whom can drive, will be seated in a five-seater car with two seats in front and three seats in the back. Friends who know how to drive don't sit next to each other. The driver's seat can only be occupied by someone who knows how to drive. Determine how many different ways the five people can be seated.

25

25

21

21

36

36

42

42

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Question 4

Time: 00:00:00
Rama, a small child, is playing a Snake & Ladder match with his friends; he is on level 91 and must reach 100 to win. On 93 and 96, there is a snake. If he doesn't want to roll the dice more than three times, how many different ways can he finish the game?

14

14

16

16

13

13

10

10

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Question 5

Time: 00:00:00
For the inter-college fashion show, 8 buddies will be chosen from a group of 9 boys and 7 girls. How many different ways will the candidates with a maximum of three girls and a minimum of four boys be chosen?

5432

5432

6431

6431

7653

7653

6435

6435

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Question 6

Time: 00:00:00
Deepika must form a five-person team for an inter-school contest. The debate club has a group of seven males and eight girls, and the squad must be made up entirely of these individuals. According to the tournament's rules, each team must have at least one female member. As a result, Deepika's team must include at least one female member. In how many ways can Deepika select her team?

2500

2500

2643

2643

3125

3125

2982

2982

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Question 7

Time: 00:00:00
Manisha went grocery shopping at Spencers. Six different Lays wrappers, four different cookies boxes, and two different chocolate packages are to be positioned on a rack in such a way that the Lays wrappers, cookies boxes, and chocolate packages all stand together. What is the maximum number of such arrangements that can be made?

254604

254604

263410

263410

207360

207360

314256

314256

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Question 8

Time: 00:00:00
Reema went to see a performance with her companions. When they arrived at their location, they discovered that five individuals were to be seated in five chairs for the concert, and three of them did not want to sit next to each other. Find out how many different ways this can be accomplished.

12

12

14

14

16

16

15

15

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Question 9

Time: 00:00:00
Monica invited a few of her coworkers to supper. In her home, there is a square table with two seats on each side. There are a total of ten persons present. How many different ways can the table's chairs be filled?

12!/4!

12!/4!

3!/5!

3!/5!

10!/8!

10!/8!

None of the above

None of the above

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Question 10

Time: 00:00:00
Manoj is assigned the task of figuring out how many three-letter words can be made from the letters in the term "PRACTICES."

432

432

357

357

421

421

None of the above

None of the above

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## Campisi

The Campisi model was first proposed in a paper published by Stephen Campisi (Primer on Fixed Income Performance Attribution, Journal of Performance Measurement, June 2000). It is perhaps the simplest possible security-level model for fixed income attribution.

Breakdown of effects in the Campisi model

Notes

• Campisi's original paper describes income return as generated by coupons; specifically, as the average annual coupon divided by the average beginning market price. This is actually running yield, as it omits any pull-to-par effects.
• Treasury effect combines the returns from parallel and non-parallel curve movements into one source of return, given by . The change in treasury yield is given by the change in the treasury curve at the tenor point equal to the duration of the portfolio.
• Spread effect is the effect of changes in the average spread , given by . Treasury and spread durations are assumed to be equivalent.
• Selection, or security-specific, return is given by total return, minus the sum of the previous three effects.
• There is no residual term. Any unexplained returns is aggregated into selection effect.

The Campisi model can be applied at the portfolio, sector or security levels.

To implement the single-currency Campisi model in FIA, set the following parameters in the configuration file:

# Definition of effects
CouponDecomposition=AGGREGATED
SovereignCurveDecomposition=AGGREGATED

# Labels of effects
TimeReturnLabel= IncomeReturn
SovereignCurveReturnLabel = Treasury effect
ResidualReturnLabel = Selection effect

## Duration

The duration model provides slightly more information than the Campisi model. Treasury effects, which are returns generated by changes in the level of the sovereign curve, are decomposed into returns made by parallel movements in the curve (duration return), and those from non-parallel movements (curve return). Modified duration measures sensitivity to both types of yield movement, so this effect gives its name to the model.

The definition of how parallel movements should be calculated varies widely.

There is no formal definition of the duration model in the attribution literature, but it is one of the most widely used.

To implement the Duration model in FIA, set the following parameters in the configuration file:

# Definition of effects
CouponDecomposition = AGGREGATED
SovereignCurveDecomposition = DURATION

# Labels of effects
TimeReturnLabel = IncomeReturn
DurationCurveReturnLabel = Duration return
NonParallelCurveReturnLabel = Curve return
ResidualReturnLabel = Selection effect

## Tim Lord

The Tim Lord model is similar to the Campisi and Duration models, but includes a number of more detailed effects ('The attribution of portfolio and index returns in fixed income, Journal of Performance Measurement, Fall 1997).

Notes

• Income return is given by change in accrued interest and other income cash flows. As with the Campisi model, this is more correctly described as the return due to running yield. Unlike the Campisi model, pull-to-par and roll-down effects are measured here as calendar returns (see below).
• Duration return is defined as price return due to changes in the Treasury par curve, measured as the product of the negative modified duration of each security, times the change in yield of its duration matched Treasury (a synthetic Treasury bond with the same modified duration as the target bond).
• Shift return is generated by parallel shifts in the Treasury curve. The shift is measured at the 5-year maturity point on the curve.
• Twist return is the portion of return generated by non-parallel shifts in the Treasury curve.
• Spread return is further decomposed into sector return and issuer-specific return. Sector return is calculated by recording the change in spread for the security's defined sector, which is the average OAS change for all benchmark securities in that category.
• Issuer-specific return is the difference between spread return and sector return.
• Residual return is the price return that is unexplained by the sum of the previous effects.

The author notes that the model can generate substantial residuals if a security's pricing model is wrongly specified. However, this is easily addressed if the attribution system can use specialized pricing models, rather than a one-size-fits-all approach.

Most security-based models are variants of the Tim Lord approach.

To implement the Tim Lord model in FIA, set the following parameters in the configuration file:

# Definition of effects
CouponDecomposition = AGGREGATED
SovereignCurveDecomposition = DURATION

# Labels of effects
TimeReturnLabel = IncomeReturn
DurationCurveReturnLabel = Duration return
NonParallelCurveReturnLabel = Curve return
ResidualReturnLabel = Selection effect

In addition, assign a sector curve to each security where appropriate.

## Key rate duration

Modified duration measures the sensitivity of price to changes in overall levels of interest rates. For instance, if the yield curve moves downwards by 10 basis points, a bond with a ten year modified duration will generate a return of 100 basis points, but a bond with a two year modified duration will generate only 20 basis points.

A key rate duration measures the price sensitivity of a security to a change in its yield curve at a single maturity, rather than to movements in the curve as a whole. It is therefore suited to detailed analysis of the return of securities with cash flows spread over a range of maturities, such as amortizing bonds and MBS.

Key rate duration attribution usually follows the same pattern as Campisi and Tim Lord models. The difference is that returns from changes in the yield curve are represented as returns from different maturities along the yield curve, rather than from global movements such as shift and twist.

To implement the above KRD model in FIA, set the following parameters in the configuration file:

# Definition of effects
CouponDecomposition = AGGREGATED
SovereignCurveDecomposition = KRD

# Labels of effects
TimeReturnLabel = Income Return
ResidualReturnLabel = Selection effect

# Specific key rate durations
KRDList = 0,2,5,10,30

Top-down attribution forms a useful approach to attribution on fixed income portfolios where over- or under-weighting of sectors forms part of the investment process, in addition to taking fixed income investment decisions. Top-down attribution forms a combination of the successive portfolio methodology, of which Brinson attribution is the simplest example, and the successive spread methodology, which underlies the various models shown above. The approach is described in the GRAP paper 'Fixed income attribution: proposal of a new methodology' (Gillet and Homolle, undated).

Unlike the other types of attribution model shown here, top-down attribution requires a benchmark, just as for the Brinson models.

The first stage of a top-down attribution analysis is to define a partition on the securities in the portfolio and the benchmark, and to categorize each security into a sector. The excess asset allocation return r_a_AA generated by each sector is then given by

where and are the weights of the portfolio and benchmark sectors, and is the return of the sector in the benchmark.

The second stage is to decompose the returns of the portfolio and benchmark sectors according to a particular fixed income scheme, which can (but need not) be one of those listed above. The excess return from each source of fixed income risk is then given by

The typical user of this type of strategy has a two-fold investment strategy. Firstly, a decision is made about which sectors to overweight. For instance, if mortgage-backed debt looks attractive relative to Treasury bills, the former will be overweighted relative to the benchmark. Next, the managers for each type of asset will make a range of fixed-income specific investment decisions, and the value added by each will be calculated using the above expression.

It is of course possible to run such an analysis on a portfolio containing a large number of different asset types even when no format asset allocation decision has been made. The results are arguably of little use, since they do not reflect the underlying investment process. The reader should beware of reverse fitted analyses, in which a range of attribution analyses have been run to cherry-pick the particular strategy that generated positive returns.

To implement a top-down model in FIA that uses currency for its sectors and Campisi attribution for fixed income effects, set the following parameters in the configuration file:

# Definition of effects
CouponDecomposition = AGGREGATED
SovereignCurveDecomposition = DURATION
AssetAllocation = TRUE
AssetAllocationVariable = CURRENCY

# Labels of effects
TimeReturnLabel = IncomeReturn
DurationCurveReturnLabel = Duration return
NonParallelCurveReturnLabel = Curve return
ResidualReturnLabel = Selection effect


text:
+0

# Cosine equation?

0
134
2

How do I find the equation of a cosine function with a maximum of 3, a minimum of -3, and a period of 4?

Guest May 6, 2017
Sort:

#1
+90590
+1

How do I find the equation of a cosine function with a maximum of 3, a minimum of -3, and a period of 4?

lets see

y=3cos b(x+c)

c can be any constant becasue you have not mentioned phase shift.

where period = 2pi/b

$$\frac{2\pi}{b}=4\\ \frac{\pi}{b}=2\\ b=\frac{\pi}{2}$$

so it is

$$y=3cos[\frac{\pi }{2}(x+c)]$$

check (here is your graph )

https://www.desmos.com/calculator/nuopbcsgsn

Melody  May 6, 2017
edited by Melody  May 6, 2017
#2
+90590
+1

This is what you must remember.

$$y=a*cos[b(x+c)]+d\\~\\ \text{a is the amplitude}\\ \text{d is the vertical shift}\\ \text{c is the horizontal, or phase, shift}\\ \text{and last but not least, the period is}\frac{2\pi}{b}\\ \text{The angle is in radians}\\$$

If you remember that then these questions become almost trivia.

Sine functions work the same way.

Melody  May 6, 2017

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