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# Pysound buffer module

Martin McBride
2021-09-11

The Pysound buffer module contains some functions for working with sound buffers, and the BufferParams object.

Sound buffers are pure NumPy arrays, with a type of np.float. This means that we don't need a lot of extra functions to handle sound buffers, we just use the existing NumPy functions.

A sound buffer always requires an associated BufferParams object to specify the sample rate of the sound, but this is usually passed around separately. In most cases every component uses the same sample rate, and in many cases they all use the same length too. This means that you can get away with using just a single BufferParams object for every component.

The buffer module defines:

• The BufferParams object.
• The create_buffer function that creates a new sound buffer by either an existing sound buffer or filling a buffer with a constant value.
• The mix_buffer function that adds the content of one sound buffer into another.

## BufferParams

BufferParams is a simple object that stores the length and sample rate of a sound buffer.

BufferParams is immutable. If you want a BufferParams object with different values, you must create a new object. There are factory methods to do this, see below.

### Constructor

Creates a new BufferParams with specified values.

BufferParams(length=44100, sample_rate=44100)

Parameter Type Description
length integer The required length of the buffer, in samples.
sample_rate integer The required sample rate of the buffer.

Creates a new BufferParams object.

### get_length

Returns the length of this BufferParams object.

get_length()

Returns the length of the buffer, in samples, as an integer.

### get_sample_rate

Returns the sample rate of this BufferParams object.

get_sample_rate()

Returns the sample rate of the buffer, in samples per second, as an integer.

### t2s

Converts a time to a sample count using the buffer's sample rate.

t2s(length=44100, sample_rate=44100)

Parameter Type Description
time float The time in seconds.

For a given duration time, this function resturns the corresponding sample count. For example, if the BufferParams object had a sample rate of 5000 samples per second, and we called t2s with a time of 0.5 seconds, the result would be 2500 (there are 2500 samples in 0.5 seconds).

Note that there is potential for rounding error in this call, because time is a float.

### with_length

Creates a new BufferParams with the same sample rate but a new length.

with_length(length)

Parameter Type Description
length integer The required length of the buffer, in samples.

Returns a new BufferParams object based on an existing object.

### with_duration

Creates a new BufferParams with the same sample rate but a new length.

with_duration(duration)

Parameter Type Description
duration float The required length of the buffer, in seconds.

Returns a new BufferParams object based on an existing object. This is similar to with_length but the buffer length is expressed as a time in seconds rather than a length in samples.

For an original BufferParams object params, this is equivalent to:

params.with_length(params.t2s(duration))

As with the t2s method, this method may be subject to rounding errors.

### with_sample_rate

Creates a new BufferParams with the same length but a new sample rate.

with_sample_rate(sample_rate)

Parameter Type Description
sample_rate integer The required sample rate of the buffer, in samples per second.

Returns a new BufferParams object based on an existing object.

## create_buffer

Utility function to create a new buffer based on a value that can either be a float or an existing buffer.

create_buffer(params, value=0.0)

Parameter Type Description
params BufferParams The parameters for the new buffer.
value float or np array The data for the new buffer.

Returns a new BufferParams object. The length and sample rate of the new buffer will be set by params.

If the value is a float, the resulting buffer will be filled with the float value. The default is 0.0, creating an array of zeros,

If the value is a NumPy array, the resulting buffer will contain a copy of the supplied array.

This function assumes that the value buffer has the same sample rate as the output buffer. However, the buffers don't have to be the same length. The output array will always have the length specified in params. If value is too short, the output array will be padded with zeros. If value is too long, the data will be truncated to fit the output array.

## mix_buffer

Utility function to mix data into an existing buffer.

mix_buffer(dest, src, at)

Parameter Type Description
dest np array The destination NumPy array.
src np array The source NumPy array.
at int The position within the destination buffer where the source will be included.

This function performs simple mixing of a source buffer into a destination buffer.

For example assume the destination buffer is 1000 samples long, the source buffer is 200 sample long, and the at parameter is 100.

In that case, the source buffer will be mixed in to the destination buffer at position 100. It will affect destination samples 100 to 299.

The source sample will be added to whatever values are already in the destination samples in teh affected area.

There are a few limitations:

• It is assumed that the source and destination buffers have the same sample rate.
• The source buffer must completely fit within the destination buffer. That is to say, at cannot be less than zero, and at plus the length of the soiurce buffer cannot exceed the length of the destination buffer.


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# 2.14.1 Lifting equivalences

When working loosely, one might say that a bijection between sets $A$ and $B$ “obviously” induces an isomorphism between semigroup structures on $A$ and semigroup structures on $B$. With univalence, this is indeed obvious, because given an equivalence between types $A$ and $B$, we can automatically derive a semigroup structure on $B$ from one on $A$, and moreover show that this derivation is an equivalence of semigroup structures. The reason is that $\mathsf{SemigroupStr}$ is a family of types, and therefore has an action on paths between types given by $\mathsf{transport}$:

 $\mathsf{transport}^{\mathsf{SemigroupStr}}{(\mathsf{ua}(e))}:\mathsf{% SemigroupStr}(A)\to\mathsf{SemigroupStr}(B).$

Moreover, this map is an equivalence, because $\mathsf{transport}^{C}(\alpha)$ is always an equivalence with inverse $\mathsf{transport}^{C}{(\mathord{{\alpha}^{-1}})}$, see Lemma 2.3.9 (http://planetmath.org/23typefamiliesarefibrations#Thmprelem6),Lemma 2.1.4 (http://planetmath.org/21typesarehighergroupoids#Thmprelem3).

While the univalence axiom ensures that this map exists, we need to use facts about $\mathsf{transport}$ proven in the preceding sections to calculate what it actually does. Let $(m,a)$ be a semigroup structure on $A$, and we investigate the induced semigroup structure on $B$ given by

 $\mathsf{transport}^{\mathsf{SemigroupStr}}(\mathsf{ua}(e),(m,a)).$

First, because $\mathsf{SemigroupStr}$(X) is defined to be a $\Sigma$-type, by Theorem 2.7.4 (http://planetmath.org/27sigmatypes#Thmprethm2),

 \mathsf{transport}^{\mathsf{SemigroupStr}}(\mathsf{ua}(e),(m,a))=\begin{% aligned} \displaystyle\big{(}&\displaystyle\mathsf{transport}^{X\mapsto(X\to X% \to X)}(\mathsf{ua}(e),m),\\ &\displaystyle\mathsf{transport}^{(X,m)\mapsto\mathsf{Assoc}(X,m)}((\mathsf{% pair}^{\mathord{=}}(\mathsf{ua}(e),\mathsf{refl})),a)\big{)}\end{aligned} (2.14.2)

where $\mathsf{Assoc}(X,m)$ is the type $\mathchoice{\prod_{x,y,z:X}\,}{\mathchoice{{\textstyle\prod_{(x,y,z:X)}}}{% \prod_{(x,y,z:X)}}{\prod_{(x,y,z:X)}}{\prod_{(x,y,z:X)}}}{\mathchoice{{% \textstyle\prod_{(x,y,z:X)}}}{\prod_{(x,y,z:X)}}{\prod_{(x,y,z:X)}}{\prod_{(x,% y,z:X)}}}{\mathchoice{{\textstyle\prod_{(x,y,z:X)}}}{\prod_{(x,y,z:X)}}{\prod_% {(x,y,z:X)}}{\prod_{(x,y,z:X)}}}m(x,m(y,z))=m(m(x,y),z)$. That is, the induced semigroup structure consists of an induced multiplication operation on $B$

 $\displaystyle m^{\prime}:B\to B\to B$ $\displaystyle m^{\prime}(b_{1},b_{2}):\!\!\equiv\mathsf{transport}^{X\mapsto(X% \to X\to X)}(\mathsf{ua}(e),m)(b_{1},b_{2})$

together with an induced proof that $m^{\prime}$ is associative. By function extensionality, it suffices to investigate the behavior of $m^{\prime}$ when applied to arguments $b_{1},b_{2}:B$. By applying (LABEL:eq:transport-arrow) twice, we have that $m^{\prime}(b_{1},b_{2})$ is equal to

 $\mathsf{transport}^{X\mapsto X}\big{(}\mathsf{ua}(e),m(\mathsf{transport}^{X% \mapsto X}(\mathord{{\mathsf{ua}(e)}^{-1}},b_{1}),\mathsf{transport}^{X\mapsto X% }(\mathord{{\mathsf{ua}(e)}^{-1}},b_{2}))\big{)}.$

Then, because $\mathsf{ua}$ is quasi-inverse to $\mathsf{transport}^{X\mapsto X}$, this is equal to

 $e(m(\mathord{{e}^{-1}}(b_{1}),\mathord{{e}^{-1}}(b_{2}))).$

Thus, given two elements of $B$, the induced multiplication $m^{\prime}$ sends them to $A$ using the equivalence $e$, multiplies them in $A$, and then brings the result back to $B$ by $e$, just as one would expect.

Moreover, though we do not show the proof, one can calculate that the induced proof that $m^{\prime}$ is associative (the second component of the pair in (2.14.2)) is equal to a function sending $b_{1},b_{2},b_{3}:B$ to a path given by the following steps:

 $\displaystyle m^{\prime}(m^{\prime}(b_{1},b_{2}),b_{3})$ $\displaystyle=e(m(\mathord{{e}^{-1}}(m^{\prime}(b_{1},b_{2})),\mathord{{e}^{-1% }}(b_{3})))$ (2.14.3) $\displaystyle=e(m(\mathord{{e}^{-1}}(e(m(\mathord{{e}^{-1}}(b_{1}),\mathord{{e% }^{-1}}(b_{2})))),\mathord{{e}^{-1}}(b_{3})))$ $\displaystyle=e(m(m(\mathord{{e}^{-1}}(b_{1}),\mathord{{e}^{-1}}(b_{2})),% \mathord{{e}^{-1}}(b_{3})))$ $\displaystyle=e(m(\mathord{{e}^{-1}}(b_{1}),m(\mathord{{e}^{-1}}(b_{2}),% \mathord{{e}^{-1}}(b_{3}))))$ $\displaystyle=e(m(\mathord{{e}^{-1}}(b_{1}),\mathord{{e}^{-1}}(e(m(\mathord{{e% }^{-1}}(b_{2}),\mathord{{e}^{-1}}(b_{3}))))))$ $\displaystyle=e(m(\mathord{{e}^{-1}}(b_{1}),\mathord{{e}^{-1}}(m^{\prime}(b_{2% },b_{3}))))$ $\displaystyle=m^{\prime}(b_{1},m^{\prime}(b_{2},b_{3})).$

These steps use the proof $a$ that $m$ is associative and the inverse laws for $e$. From an algebra perspective, it may seem strange to investigate the identity of a proof that an operation is associative, but this makes sense if we think of $A$ and $B$ as general spaces, with non-trivial homotopies between paths. In http://planetmath.org/node/87576Chapter 3, we will introduce the notion of a set, which is a type with only trivial homotopies, and if we consider semigroup structures on sets, then any two such associativity proofs are automatically equal.

Title 2.14.1 Lifting equivalences
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Twenty — minutes — maybe — more.choose four wordsmore quotes

poetry: fun

EMBO Practical Course: Bioinformatics and Genome Analysis, 5–17 June 2017.

daily quotation server archives

In the late 90’s I started (a good decade for starts) a daily quotation server project at www.quoteserver.ca. The domain is now defunct—some pages are partially viewable at the Way Back Machine.

Below is the list of quotes I had collected by the end of the life of the project. Most are about love—duh—and a few are jolly jests from funny trenches. You know, that place where mustard gas makes your eyes water.

The quotes weren’t scraped from quote archives—each is meaningful and hand-picked.

the quote archive

And now for full list of 1,600 other things worth reading. Such as everything Dorothy Parker has written and ... yes, even the Pinky and Brain quotes, which are a special kind of special.

Quote collections about love, heart, desire, life, death, god, mind, science.

Feeling lucky? Read 10 random quotes. Well, will you, punk?

280
Science is a cemetery of dead ideas.
735
As soon as questions of will or decision or reason or choice of action arise, human science is at a loss.
870
The great tragedy of Science—the slaying of a
beautiful hypothesis by an ugly fact.
Biogenesis and Abiogenesis
1466
Even if there is only one possible unified theory, it is just a set of rules and equations. What is it that breathes fire into the equations and makes a universe for them to describe? The usual approach of science of constructing a mathematical model cannot answer the questions of why there should be a universe for the model to describe. Why does the universe go to all the bother of existing?
A Brief History of Time
1467
Science can purify religion from error and superstition. Religion can purify science from idolatry and false absolutes.
1561
In science one tries to tell people, in such a way as to be understood by everyone, something that no one ever knew before. But in poetry, it’s the exact opposite.
Mathematical Circles Adieu by H. Eves [quoted]
1607
Every attempt to employ mathematical methods in the study of chemical questions must be considered profoundly irrational and contrary to the spirit of chemistry... if mathematical analysis should ever hold a prominent place in chemistry — an aberration which is happily almost impossible — it would occasion a rapid and widespread degeneration of that science.
Cours de Philosophie Positive (1830)
1609
Biology is the only science in which multiplication means the same thing as division.
1610
Science has proof without any certainty. Creationists have certainty without any proof.
1625
This isn’t a rare scene in science or technology. This is the commonest scene of all. Just plain stuck. In traditional maintenance this is the worst of all moments, so bad that you have avoided even thinking about it before you come to it.
Zen and the Art of Motorcycle Maintenance
1630
Art is Science in Love.
1662
A poet is, after all, a sort of scientist, but engaged in a qualitative science in which nothing is measurable. He lives with data that cannot be numbered, and his experiments can be done only once. The information in a poem is, by definition, not reproducible. He becomes an equivalent of scientist, in the act of examining and sorting the things popping in [to his head], finding the marks of remote similarity, points of distant relationship, tiny irregularities that indicate that this one is really the same as that one over there only more important. Gauging the fit, he can meticulously place pieces of the universe together, in geometric configurations that are as beautiful and balanced as crystals.
The Medusa and the Snail: More Notes of a Biology Watcher
1665
How often people speak of art and science as though they were two entirely different things, with no interconnection. An artist is emotional, they think, and uses only his intuition; he sees all at once and has no need of reason. A scientist is cold, they think, and uses only his reason; he argues carefully step by step, and needs no imagination. That is all wrong. The true artist is quite rational as well as imaginative and knows what he is doing; if he does not, his art suffers. The true scientist is quite imaginative as well as rational, and sometimes leaps to solutions where reason can follow only slowly; if he does not, his science suffers.
The Roving Mind (Ch 25)
VIEW ALL

Tabular Data

Tue 11-04-2017
Tabulating the number of objects in categories of interest dates back to the earliest records of commerce and population censuses.

After 30 columns, this is our first one without a single figure. Sometimes a table is all you need.

In this column, we discuss nominal categorical data, in which data points are assigned to categories in which there is no implied order. We introduce one-way and two-way tables and the $\chi^2$ and Fisher's exact tests.

Altman, N. & Krzywinski, M. (2017) Points of Significance: Tabular data. Nature Methods 14:329–330.

Happy 2017 $\pi$ Day—Star Charts, Creatures Once Living and a Poem

Tue 14-03-2017

on a brim of echo,

capsized chamber
drawn into our constellation, and cooling.
—Paolo Marcazzan

Celebrate $\pi$ Day (March 14th) with star chart of the digits. The charts draw 40,000 stars generated from the first 12 million digits.

12,000,000 digits of $\pi$ interpreted as a star catalogue. (details)

The 80 constellations are extinct animals and plants. Here you'll find old friends and new stories. Read about how Desmodus is always trying to escape or how Megalodon terrorizes the poor Tecopa! Most constellations have a story.

Find friends and stories among the 80 constellations of extinct animals and plants. Oh look, a Dodo guardings his eggs! (details)

This year I collaborate with Paolo Marcazzan, a Canadian poet, who contributes a poem, Of Black Body, about space and things we might find and lose there.

Check out art from previous years: 2013 $\pi$ Day and 2014 $\pi$ Day, 2015 $\pi$ Day and and 2016 $\pi$ Day.

Data in New Dimensions: convergence of art, genomics and bioinformatics

Tue 07-03-2017

Art is science in love.
— E.F. Weisslitz

A behind-the-scenes look at the making of our stereoscopic images which were at display at the AGBT 2017 Conference in February. The art is a creative collaboration with Becton Dickinson and The Linus Group.

Its creation began with the concept of differences and my writeup of the creative and design process focuses on storytelling and how concept of differences is incorporated into the art.

Oh, and this might be a good time to pick up some red-blue 3D glasses.

A stereoscopic image and its interpretive panel of single-cell transcriptomes of blood cells: diseased versus healthy control.

Interpreting P values

Thu 02-03-2017
A P value measures a sampleâs compatibility with a hypothesis, not the truth of the hypothesis.

This month we continue our discussion about $P$ values and focus on the fact that $P$ value is a probability statement about the observed sample in the context of a hypothesis, not about the hypothesis being tested.

Nature Methods Points of Significance column: Interpreting P values. (read)

Given that we are always interested in making inferences about hypotheses, we discuss how $P$ values can be used to do this by way of the Benjamin-Berger bound, $\bar{B}$ on the Bayes factor, $B$.

Heuristics such as these are valuable in helping to interpret $P$ values, though we stress that $P$ values vary from sample to sample and hence many sources of evidence need to be examined before drawing scientific conclusions.

Altman, N. & Krzywinski, M. (2017) Points of Significance: Interpreting P values. Nature Methods 14:213–214.

Krzywinski, M. & Altman, N. (2017) Points of significance: P values and the search for significance. Nature Methods 14:3–4.

Krzywinski, M. & Altman, N. (2013) Points of significance: Significance, P values and tâtests. Nature Methods 10:1041–1042.

Snellen Charts—Typography to Really Look at

Sat 18-02-2017

Another collection of typographical posters. These ones really ask you to look.

Snellen charts designed using physical constants, Braille and elemental abundances in the universe and human body.

The charts show a variety of interesting symbols and operators found in science and math. The design is in the style of a Snellen chart and typset with the Rockwell font.


text:
MathSciNet bibliographic data MR329046 60J20 (60F20) Iosifescu, Marius On finite tail $\sigma$$\sigma$-algebras. Z. Wahrscheinlichkeitstheorie und Verw. Gebiete 24 (1972), 159–166. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


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# tf.contrib.layers.layer_norm(args, *kwargs)

### tf.contrib.layers.layer_norm(*args, **kwargs)

Adds a Layer Normalization layer from https://arxiv.org/abs/1607.06450.

"Layer Normalization"

Jimmy Lei Ba, Jamie Ryan Kiros, Geoffrey E. Hinton

Can be used as a normalizer function for conv2d and fully_connected.

#### Args:

• inputs: a tensor with 2 or more dimensions. The normalization occurs over all but the first dimension.
• center: If True, add offset of beta to normalized tensor. If False, beta is ignored.
• scale: If True, multiply by gamma. If False, gamma is not used. When the next layer is linear (also e.g. nn.relu), this can be disabled since the scaling can be done by the next layer.
• activation_fn: activation function, default set to None to skip it and maintain a linear activation.
• reuse: whether or not the layer and its variables should be reused. To be able to reuse the layer scope must be given.
• variables_collections: optional collections for the variables.
• outputs_collections: collections to add the outputs.
• trainable: If True also add variables to the graph collection GraphKeys.TRAINABLE_VARIABLES (see tf.Variable).
• scope: Optional scope for variable_scope.

#### Returns:

A Tensor representing the output of the operation.

#### Raises:

• ValueError: if rank or last dimension of inputs is undefined.


text:
# Homework Help: Satellite in an orbit around Eart

1. Dec 6, 2006

### HHveluj

1. The problem statement, all variables and given/known data
A satellite in an orbit around Earth has velocities between Vmin=V-V0 and Vmax=V+V0. Find the eccentricity of the orbit.

2. Relevant equations
rmin=l^2/(Gm1m2*mu*(1+e))
E=(Gm1m2)^2*mu*(e^2-1)/2l^2
where mu = reduced mass = m1m2/m1+m2

3. The attempt at a solution
I know all this formulas for eccentricity, but they all involve disctances/or energy, but not velocities! I don't even know how to start solving this.... Any help/hints?

Thank you very much in advance!

2. Dec 6, 2006

### nrqed

You may use conservation of angular momentum to relate the speed at aphelion to the speed at perihelion. Recall, ${\vec L} = {\vec r} \times {\vec p}$. Since at aphelion and at perihelion the motion is perpendicular to the position vector, you get that $r_{min} v_{max} = r_{max} v_{min}$. Since $r_{min} = a(1-e)$ (*if* I recall correctly) and $r_{max} = a(1+e)$ , you get a simple equation relating the max and minimum speeds. You will get a simple expression for the eccentricity in terms of V and V_0.

Hope this helps.

Patrick


text:
# 9.1 Solving trigonometric equations with identities  (Page 5/9)

 Page 5 / 9

## Simplify by rewriting and using substitution

Simplify the expression by rewriting and using identities:

${\mathrm{csc}}^{2}\theta -{\mathrm{cot}}^{2}\theta$

$1+{\mathrm{cot}}^{2}\theta ={\mathrm{csc}}^{2}\theta$

Now we can simplify by substituting $\text{\hspace{0.17em}}1+{\mathrm{cot}}^{2}\theta \text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}{\mathrm{csc}}^{2}\theta .\text{\hspace{0.17em}}$ We have

$\begin{array}{ccc}\hfill {\mathrm{csc}}^{2}\theta -{\mathrm{cot}}^{2}\theta & =& 1+{\mathrm{cot}}^{2}\theta -{\mathrm{cot}}^{2}\theta \hfill \\ & =& 1\hfill \end{array}$

Use algebraic techniques to verify the identity: $\text{\hspace{0.17em}}\frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{1+\mathrm{sin}\text{\hspace{0.17em}}\theta }=\frac{1-\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }.$

(Hint: Multiply the numerator and denominator on the left side by $\text{\hspace{0.17em}}1-\mathrm{sin}\text{\hspace{0.17em}}\theta .\right)$

$\begin{array}{ccc}\hfill \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{1+\mathrm{sin}\text{\hspace{0.17em}}\theta }\left(\frac{1-\mathrm{sin}\text{\hspace{0.17em}}\theta }{1-\mathrm{sin}\text{\hspace{0.17em}}\theta }\right)& =& \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta \left(1-\mathrm{sin}\text{\hspace{0.17em}}\theta \right)}{1-{\mathrm{sin}}^{2}\theta }\hfill \\ & =& \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta \left(1-\mathrm{sin}\text{\hspace{0.17em}}\theta \right)}{{\mathrm{cos}}^{2}\theta }\hfill \\ & =& \frac{1-\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }\hfill \end{array}$

Access these online resources for additional instruction and practice with the fundamental trigonometric identities.

## Key equations

 Pythagorean identities $\begin{array}{l}{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1\\ 1+{\mathrm{cot}}^{2}\theta ={\mathrm{csc}}^{2}\theta \\ 1+{\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta \end{array}$ Even-odd identities $\begin{array}{ccc}\mathrm{tan}\left(-\theta \right)& =& -\mathrm{tan}\text{\hspace{0.17em}}\theta \\ \mathrm{cot}\left(-\theta \right)& =& -\mathrm{cot}\text{\hspace{0.17em}}\theta \\ \mathrm{sin}\left(-\theta \right)& =& -\mathrm{sin}\text{\hspace{0.17em}}\theta \\ \mathrm{csc}\left(-\theta \right)& =& -\mathrm{csc}\text{\hspace{0.17em}}\theta \\ \mathrm{cos}\left(-\theta \right)& =& \mathrm{cos}\text{\hspace{0.17em}}\theta \\ \mathrm{sec}\left(-\theta \right)& =& \mathrm{sec}\text{\hspace{0.17em}}\theta \end{array}$ Reciprocal identities $\begin{array}{ccc}\mathrm{sin}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{csc}\text{\hspace{0.17em}}\theta }\\ \mathrm{cos}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{sec}\text{\hspace{0.17em}}\theta }\\ \mathrm{tan}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{cot}\text{\hspace{0.17em}}\theta }\\ \mathrm{csc}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{sin}\text{\hspace{0.17em}}\theta }\\ \mathrm{sec}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{cos}\text{\hspace{0.17em}}\theta }\\ \mathrm{cot}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{tan}\text{\hspace{0.17em}}\theta }\end{array}$ Quotient identities $\begin{array}{ccc}\mathrm{tan}\text{\hspace{0.17em}}\theta & =& \frac{\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }\\ \mathrm{cot}\text{\hspace{0.17em}}\theta & =& \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{\mathrm{sin}\text{\hspace{0.17em}}\theta }\end{array}$

## Key concepts

• There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem.
• Graphing both sides of an identity will verify it. See [link] .
• Simplifying one side of the equation to equal the other side is another method for verifying an identity. See [link] and [link] .
• The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation. See [link] .
• We can create an identity and then verify it. See [link] .
• Verifying an identity may involve algebra with the fundamental identities. See [link] and [link] .
• Algebraic techniques can be used to simplify trigonometric expressions. We use algebraic techniques throughout this text, as they consist of the fundamental rules of mathematics. See [link] , [link] , and [link] .

## Verbal

We know $\text{\hspace{0.17em}}g\left(x\right)=\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is an even function, and $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h\left(x\right)=\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ are odd functions. What about $\text{\hspace{0.17em}}G\left(x\right)={\mathrm{cos}}^{2}x,F\left(x\right)={\mathrm{sin}}^{2}x,$ and $\text{\hspace{0.17em}}H\left(x\right)={\mathrm{tan}}^{2}x?\text{\hspace{0.17em}}$ Are they even, odd, or neither? Why?

All three functions, $\text{\hspace{0.17em}}F,G,$ and $H,$ are even.

This is because $\text{\hspace{0.17em}}F\left(-x\right)=\mathrm{sin}\left(-x\right)\mathrm{sin}\left(-x\right)=\left(-\mathrm{sin}\text{\hspace{0.17em}}x\right)\left(-\mathrm{sin}\text{\hspace{0.17em}}x\right)={\mathrm{sin}}^{2}x=F\left(x\right),G\left(-x\right)=\mathrm{cos}\left(-x\right)\mathrm{cos}\left(-x\right)=\mathrm{cos}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x={\mathrm{cos}}^{2}x=G\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}H\left(-x\right)=\mathrm{tan}\left(-x\right)\mathrm{tan}\left(-x\right)=\left(-\mathrm{tan}\text{\hspace{0.17em}}x\right)\left(-\mathrm{tan}\text{\hspace{0.17em}}x\right)={\mathrm{tan}}^{2}x=H\left(x\right).$

Examine the graph of $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sec}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ on the interval $\text{\hspace{0.17em}}\left[-\pi ,\pi \right].\text{\hspace{0.17em}}$ How can we tell whether the function is even or odd by only observing the graph of $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sec}\text{\hspace{0.17em}}x?$

After examining the reciprocal identity for $\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}t,$ explain why the function is undefined at certain points.

When $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t=0,$ then $\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}t=\frac{1}{0},$ which is undefined.

All of the Pythagorean identities are related. Describe how to manipulate the equations to get from $\text{\hspace{0.17em}}{\mathrm{sin}}^{2}t+{\mathrm{cos}}^{2}t=1\text{\hspace{0.17em}}$ to the other forms.

## Algebraic

For the following exercises, use the fundamental identities to fully simplify the expression.

$\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x$

$\mathrm{sin}\text{\hspace{0.17em}}x$

$\mathrm{sin}\left(-x\right)\text{\hspace{0.17em}}\mathrm{cos}\left(-x\right)\text{\hspace{0.17em}}\mathrm{csc}\left(-x\right)$

$\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x+\mathrm{sec}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}{\mathrm{cos}}^{2}x$

$\mathrm{sec}\text{\hspace{0.17em}}x$

$\mathrm{csc}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cot}\left(-x\right)$

$\frac{\mathrm{cot}\text{\hspace{0.17em}}t+\mathrm{tan}\text{\hspace{0.17em}}t}{\mathrm{sec}\left(-t\right)}$

$\mathrm{csc}\text{\hspace{0.17em}}t$

$3\text{\hspace{0.17em}}{\mathrm{sin}}^{3}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}t+{\mathrm{cos}}^{2}\text{\hspace{0.17em}}t+2\text{\hspace{0.17em}}\mathrm{cos}\left(-t\right)\mathrm{cos}\text{\hspace{0.17em}}t$

$-\mathrm{tan}\left(-x\right)\mathrm{cot}\left(-x\right)$

$-1$

$\frac{-\mathrm{sin}\left(-x\right)\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x}{\mathrm{cot}\text{\hspace{0.17em}}x}$

$\frac{1+{\mathrm{tan}}^{2}\theta }{{\mathrm{csc}}^{2}\theta }+{\mathrm{sin}}^{2}\theta +\frac{1}{{\mathrm{sec}}^{2}\theta }$

${\mathrm{sec}}^{2}x$

$\left(\frac{\mathrm{tan}\text{\hspace{0.17em}}x}{{\mathrm{csc}}^{2}x}+\frac{\mathrm{tan}\text{\hspace{0.17em}}x}{{\mathrm{sec}}^{2}x}\right)\left(\frac{1+\mathrm{tan}\text{\hspace{0.17em}}x}{1+\mathrm{cot}\text{\hspace{0.17em}}x}\right)-\frac{1}{{\mathrm{cos}}^{2}x}$

$\frac{1-{\mathrm{cos}}^{2}\text{\hspace{0.17em}}x}{{\mathrm{tan}}^{2}\text{\hspace{0.17em}}x}+2\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\text{\hspace{0.17em}}x$

${\mathrm{sin}}^{2}x+1$

For the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression.

$\frac{\mathrm{tan}\text{\hspace{0.17em}}x+\mathrm{cot}\text{\hspace{0.17em}}x}{\mathrm{csc}\text{\hspace{0.17em}}x};\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x$

$\frac{\mathrm{sec}\text{\hspace{0.17em}}x+\mathrm{csc}\text{\hspace{0.17em}}x}{1+\mathrm{tan}\text{\hspace{0.17em}}x};\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

$\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}x}$

$\frac{\mathrm{cos}\text{\hspace{0.17em}}x}{1+\mathrm{sin}\text{\hspace{0.17em}}x}+\mathrm{tan}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x$

$\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x}-\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x$

$\frac{1}{\mathrm{cot}\text{\hspace{0.17em}}x}$

$\frac{1}{1-\mathrm{cos}\text{\hspace{0.17em}}x}-\frac{\mathrm{cos}\text{\hspace{0.17em}}x}{1+\mathrm{cos}\text{\hspace{0.17em}}x};\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x$

$\left(\mathrm{sec}\text{\hspace{0.17em}}x+\mathrm{csc}\text{\hspace{0.17em}}x\right)\left(\mathrm{sin}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}x\right)-2-\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x$

$\mathrm{tan}\text{\hspace{0.17em}}x$

$-4\mathrm{sec}\text{\hspace{0.17em}}x\mathrm{tan}\text{\hspace{0.17em}}x$

$\mathrm{tan}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x$

$\mathrm{sec}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x$

$±\sqrt{\frac{1}{{\mathrm{cot}}^{2}x}+1}$

$\mathrm{sec}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

$\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

$\frac{±\sqrt{1-{\mathrm{sin}}^{2}x}}{\mathrm{sin}\text{\hspace{0.17em}}x}$

$\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x$

For the following exercises, verify the identity.

$\mathrm{cos}\text{\hspace{0.17em}}x-{\mathrm{cos}}^{3}x=\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\text{\hspace{0.17em}}x$

$\begin{array}{ccc}\hfill \mathrm{cos}\text{\hspace{0.17em}}x-{\mathrm{cos}}^{3}x& =& \mathrm{cos}\text{\hspace{0.17em}}x\left(1-{\mathrm{cos}}^{2}x\right)\hfill \\ & =& \mathrm{cos}\text{\hspace{0.17em}}x{\mathrm{sin}}^{2}x\hfill \end{array}$

$\mathrm{cos}\text{\hspace{0.17em}}x\left(\mathrm{tan}\text{\hspace{0.17em}}x-\mathrm{sec}\left(-x\right)\right)=\mathrm{sin}\text{\hspace{0.17em}}x-1$

$\frac{1+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=\frac{1}{{\mathrm{cos}}^{2}x}+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=1+2\text{\hspace{0.17em}}{\mathrm{tan}}^{2}x$

$\frac{1+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=\frac{1}{{\mathrm{cos}}^{2}x}+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}={\mathrm{sec}}^{2}x+{\mathrm{tan}}^{2}x={\mathrm{tan}}^{2}x+1+{\mathrm{tan}}^{2}x=1+2{\mathrm{tan}}^{2}x$

${\left(\mathrm{sin}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}x\right)}^{2}=1+2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x$

${\mathrm{cos}}^{2}x-{\mathrm{tan}}^{2}x=2-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x$

${\mathrm{cos}}^{2}x-{\mathrm{tan}}^{2}x=1-{\mathrm{sin}}^{2}x-\left({\mathrm{sec}}^{2}x-1\right)=1-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x+1=2-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x$

## Extensions

For the following exercises, prove or disprove the identity.

$\frac{1}{1+\mathrm{cos}\text{\hspace{0.17em}}x}-\frac{1}{1-\mathrm{cos}\left(-x\right)}=-2\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x$

${\mathrm{csc}}^{2}x\left(1+{\mathrm{sin}}^{2}x\right)={\mathrm{cot}}^{2}x$

False

$\left(\frac{{\mathrm{sec}}^{2}\left(-x\right)-{\mathrm{tan}}^{2}x}{\mathrm{tan}\text{\hspace{0.17em}}x}\right)\left(\frac{2+2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x}{2+2\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x}\right)-2\text{\hspace{0.17em}}{\mathrm{sin}}^{2}x=\mathrm{cos}\text{\hspace{0.17em}}2x$

$\frac{\mathrm{tan}\text{\hspace{0.17em}}x}{\mathrm{sec}\text{\hspace{0.17em}}x}\mathrm{sin}\left(-x\right)={\mathrm{cos}}^{2}x$

False

$\frac{\mathrm{sec}\left(-x\right)}{\mathrm{tan}\text{\hspace{0.17em}}x+\mathrm{cot}\text{\hspace{0.17em}}x}=-\mathrm{sin}\left(-x\right)$

$\frac{1+\mathrm{sin}\text{\hspace{0.17em}}x}{\mathrm{cos}\text{\hspace{0.17em}}x}=\frac{\mathrm{cos}\text{\hspace{0.17em}}x}{1+\mathrm{sin}\left(-x\right)}$

Proved with negative and Pythagorean identities

For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression.

$\frac{{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta }{1-{\mathrm{tan}}^{2}\theta }={\mathrm{sin}}^{2}\theta$

$3\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta +4\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta =3+{\mathrm{cos}}^{2}\theta$

True $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta +4\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta =3\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta +3\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta +{\mathrm{cos}}^{2}\theta =3\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)+{\mathrm{cos}}^{2}\theta =3+{\mathrm{cos}}^{2}\theta$

$\frac{\mathrm{sec}\text{\hspace{0.17em}}\theta +\mathrm{tan}\text{\hspace{0.17em}}\theta }{\mathrm{cot}\text{\hspace{0.17em}}\theta +\mathrm{cos}\text{\hspace{0.17em}}\theta }={\mathrm{sec}}^{2}\theta$

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text:
tools for parallel programming

Description

 • The Threads package provides user-level commands for writing parallel code in Maple.  This allows multiple paths of execution to run concurrently. This can allow Maple to perform a greater number of computations per second when running on a multi-core machine.  See multithreaded for more information.
 • The best tool for parallel programming in Maple is the Task Programming Model.
 • Each command in the Threads package can be accessed by using either the long form or the short form of the command name in the command calling sequence.  For more information, see the UsingPackages help page.


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 J. Ocean Eng. Technol. > Volume 26(6); 2012 > Article
 J. Ocean Eng. Technol. 2012;26(6):19-26. DOI: https://doi.org/10.5574/KSOE.2012.26.6.019
 Experimental Study on Unconfined Compression Strength and Split Tensile Strength Properties in relation to Freezing Temperature and Loading Rate of Frozen Soil Seo Young-Kyo,Choi Heon-Woo Department of Ocean Engineering Korea Maritime University,Department of Ocean Engineering Korea Maritime University 동결 온도와 재하속도에 따른 동결토의 일축압축 및 쪼갬인장 강도특성 서영교,최헌우 한국해양대학교 해양공학과,한국해양대학교 해양공학과 Key Words: Frozen soil, Uniaxial compression test, Indirect tensile test, Freezing temperature, Axial strain rate, Initial tangent modulus 핵심용어: 동결토, 일축압축시험, 간접인장시험, 동결온도, 축 변형 속도, 초기탄성계수 Abstract Recently the world has been suffering from difficulties related to the demand and supply of energy due to the democratic movements sweeping across the Middle East. Consequently, many have turned their attention to never-developed extreme regions such as the polar lands or deep sea, which contain many underground resources. This research investigated the strength and initial elastic modulus values of eternally frozen ground through a uniaxial compression test and indirect tensile test using frozen artificial soil specimens. To ensure accurate test results, a sandymud mixture of standard Jumunjin sand and kaolinite (20% in weight) was used for the specimens in these laboratory tests. Specimen were prepared by varying the water content ratio (7%, 15%, and 20%). Then, the variation in the strength value, depending on the water content, was observed. This research also established three kinds of environments under freezing temperatures of $-5^{circ}C$, $-10^{circ}C$, and $-15^{circ}C$. Then, the variation in the strength value was observed, depending on the freezing environment. In addition, the tests divided the loading rate into 6 phases and observed the variation in the stress-strain ratio, depending on the loading rate. The test data showed that a lower freezing temperature resulted in a larger strength value. An increase in the ice content in the specimen with the increase in the water content ratio influenced the strength value of the specimen. A faster load rate had a greater influence on the uniaxial compression and indirect tensile strengths of a frozen specimen and produced a different strength engineering property through the initial tangential modulus of elasticity. Finally, the long-term strength under a constant water content ratio and freezing temperature was checked by producing stress-strain ratio curves depending on the loading rate.
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# Energy-momentum tensor for dust

We all know that the energy-momentum tensor for dust is just $T^{\alpha\beta}=\rho_0v^\alpha v^\beta,$ where $\rho_0$ is the mass density in the dust's rest frame and $v^α$ is the dust's four-velocity. I'm trying to derive the dust energy momentum tensor from the equation $T_{αβ}=-\frac{2}{\sqrt{-g}}\frac{δS_M}{δg_{αβ}}$ but I'm getting the wrong answer.

The action for dust is

$$S=\int -\rho_0\sqrt{-g}d^4x.$$

Thus

$$\frac{\delta S}{\delta g^{\alpha\beta}}=\frac{\delta (-\rho_0)}{\delta g^{\alpha\beta}}\sqrt{-g}-\rho_0\frac{\delta \sqrt{-g}}{\delta g^{\alpha\beta}}.$$

To evaluate $\frac{\delta (-\rho_0)}{\delta g^{\alpha\beta}}$, I define $K_\alpha=\rho_0 v_\alpha$. Then $\rho_0=\sqrt{g^{\alpha\beta}K_\alpha K_\beta}$ and thus $\delta \rho_0=\frac{1}{2\rho_0}K_\alpha K_\beta \delta g^{\alpha \beta}.$ It follows that $$T_{\alpha\beta}=\rho_0 v_\alpha v_\beta-\rho_0 g_{\alpha\beta}.$$

There's an extra term that I can't get rid of. Any idea where I went wrong?

• I do not think that your Lagrangian is OK. What are the variables describing the dust therein and what are the equation of motion of dust produced that way? – Valter Moretti Feb 20 '14 at 20:46
• Not that this answers your question, but the easy way to get $T^{\alpha\beta}$ for dust is to slip into the frame comoving with the fluid; note that $\vec{v}\to(-1,0,0,0)$, note that there is no momentum flux, pressure, or shear but only energy density $\rho_0$, and say $T^{\alpha\beta} = \rho_0 v^\alpha v^\beta$. As you have a true tensor equation, it must hold in all frames. QED. – user10851 Feb 20 '14 at 23:27

Finding an action that gives you dust when varied is actually kind of tricky. Part of the reason it is hard is that for scalar matter, the action is usually proportional to the pressure,and the pressure vanishes for dust. I'm not sure where you found your action for dust, but I think you didn't make any calculational errors, it just isn't the right action.

There are two ways I know to get dust from an action. You have to make sure to specify the fields in your theory, and then take the variations of all of them and enforce their equations of motion. The first way is to choose a nondynamical vector field $v^a$ (which means we do not enforce it's equation of motion) and a dynamical scalar field $\rho$, and write the action $$S = - \int d^4x\sqrt{-g} \frac\rho2(g_{ab}v^av^b+1)$$

Varying with respect to $\rho$ enforces that $v^a$ be unit timelike, $v^a v_a = -1$. Now when you vary with respect to the metric, the term you get from varying $\sqrt{-g}$ is proportional to $g_{ab}v^a v^b + 1$, which vanishes since $v^a$ is unit. The only remaining term comes from varying $g_{ab}$ contracting with $v^a v^b$, which gives $$\frac{\delta S}{\delta g_{ab} }= -\frac\rho2 v^a v^b$$ and hence the stress tensor is $$T^{ab} = \rho v^a v^b.$$

The second method is similar and has the advantage of not using a nondynamical field. It is described in a paper found here. This consists of introducing two scalar fields $\lambda$ and $\varphi$. The action is $$S=\int d^4x \sqrt{-g} \lambda\left(\frac12g^{ab}\partial_a\varphi \partial_b\varphi - \frac12\mu^2(\varphi)\right)$$

Varying with respect to $\lambda$ will give $$\partial_a\varphi \partial^a \varphi = \mu^2(\varphi)$$ Again, we see that when this equation is enforced it will cause the variation of $\sqrt{-g}$ in the stress tensor to drop out. Then the stress tensor is

$$T_{ab} = \lambda\partial_a\varphi \partial_b \varphi.$$

Then defining the unit $4$-velocity by $v_a = \dfrac{\partial_a\varphi}{(\partial_a\varphi \partial^a\varphi)^{1/2}}=\dfrac{\partial_a\varphi}{\mu(\varphi)}$, we see that the stress tensor take the form $$T_{ab} = \lambda \mu^2 v_a v_b$$ which is pressureless dust with energy density $\lambda\mu^2$. The final $\varphi$ equation of motion determines the evolution of $\lambda$ and $\varphi$.


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# Tag Info

17

My question is, when I run a simulation with $N$ particles and I track the Hamiltonian per particle $(H/N)$ and the magnetization per particle $\left(\sum _i s_i /N\right)$, with $K$ values going from $0.1$ to $0.7$ in increments of $0.1$, how do I spot the region of the critical coupling constant? There has to be a signature of the critical point that is ...

14

As Anyon correctly pointed out, there is no phase transition at finite temperature in 1D. In 2D there are a number of different ways to identify the phase transition (I'm assuming you're using Monte Carlo). You could directly look at the magnetization, but a more reliable signature is the magnetic susceptibility, $\chi_m (K)$, which is strongly peaked around ...

11

A locally interacting system displaying a continuous phase transition belongs to a universality class that is determined solely by the system symmetries and dimensionality. Drawing from Wikipedia's list (itself mostly based on Ódor's paper) and this answer from Physics SE, here's a partial list of universality classes and critical exponents: \begin{array}{|...

8

I think Anyon's and taciteloquence's answers are perfect. I just want to add an emphasis on the following fact that frequently leads to confusion for beginners. The formal definition of the magnetization $$m = \frac{\sum_i s_i}{N}$$ has a symmetry that $\mathrm{Prob}[m=+m_0]=\mathrm{Prob}[m=-m_0]$, since the energy of a particular ...

5

Off the top of my head, I can think of two reasons to be concerned about simulations near a phase transition, although I'm only really well-qualified to talk about one of them in detail. The one I know less about is finite size effects. There are many cases where the limited size of the system prevents it from behaving as it would at the thermodynamic limit. ...

5

Enthalpy and entropy are the temperature dependent terms in free energy from which enthalpy is a dominant term mainly at lower temperature and entropy is dominant at higher temperature (because randomness is more at higher temperature). This temperature effect can be considered by utilizing the heat capacity at constant pressure (Cp). Phonopy code is there ...

4

I felt that it's important to note that not all universality classes are determined solely by spacial dimension and symmetry, even when the interactions are local. The easiest model that exemplifies this is probably the Ashkin-Teller model in 2D. It has a continuously varying (thus infinite variety of) critical exponents (universality classes) depending on ...

4

Thermo-Calc's DGM is the normalized parallel tangent driving force per mole of components. To be able to compute the driving force of a phase, it must be set to "dormant" status so that it cannot become stable, but the driving force can still be computed. Here's an example macro of how this could be done in Thermo-Calc: @@ These commands can be ...

4

Thermodynamics There is no problem with simulating a phase transition. There is only a problem with calculating thermodynamic averages during a phase transition. Short simple answer: I don't know anyone who has ever had a use for a property that corresponds to part one phase, part another phase. There are plenty of needs for the properties of mixed phase ...

3

This seems to be a typo in the paper. As they mention in the introduction, austenite is the "hot" phase of this compound, so it should form on heating martensite. In their experiments, they heat samples with varying compositions to determine at what temperature the austensitic transformation starts/finishes. The martensite temperatures should ...

Only top voted, non community-wiki answers of a minimum length are eligible


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# rectangle: L decr 2.4cm W incr 30%, result: area 4% bigger. find original L and W

• September 7th 2012, 09:48 PM
rectangle: L decr 2.4cm W incr 30%, result: area 4% bigger. find original L and W
Hi.

So I'm looking for help with this geometry problem. It is:

the L of the rectangle is decreased by 2.4cm . The W of the rectangle is increased by 30%. As the result, the area of the rectangle grew 4%. Find the original L and W.

First i did an equation (L - 2,4cm) x 1,3W = 1.04LW and from there i got that the L is 12. now i don't know what to do on as every equation or whatever i do with the information
I keep ending up with things like W = W or i get the answer of L again aka I'm stuck. Please help.

Thank you.
• September 7th 2012, 09:59 PM
Prove It
Re: rectangle: L decr 2.4cm W incr 30%, result: area 4% bigger. find original L and W
\displaystyle \begin{align*} 1.3 W \left( L - 2.4 \right) &= 1.04 LW \\ 1.3LW - 3.12W &= 1.04LW \\ 0.26LW - 3.12W &= 0 \\ W \left( 0.26 L - 3.12 \right) &= 0 \end{align*}

So we have \displaystyle \begin{align*} W = 0 \end{align*} (impossible) or \displaystyle \begin{align*} 0.26L - 3.12 = 0 \end{align*}. Solve for L.

Without being given any extra information, like what the original or resulting areas or perimeters actually were, you can't solve for W.
• September 7th 2012, 10:26 PM
Re: rectangle: L decr 2.4cm W incr 30%, result: area 4% bigger. find original L and W
you did a mistake there on the action in line 2-3. you deducted 0.04 from 1.3, not 1.04, so it should result in 0.26L - 3,12 = 0, but that doesnt matter. the problem is taken from 2009 test from my uni which if done >75% grants you "done" or w/e in "basic math". i'm gonna have to do it too, but i hope it's their mistake then. it should be done with pre-uni knowledge.
• September 7th 2012, 10:33 PM


text:
# What jobs can you get after Studying Physics at University? [closed]

What types of jobs can you get after a degree in Physics? My sister is choosing her course and thinking of doing physics but isn't sure what she can do after it.

-

## closed as too broad by Brandon Enright, tpg2114, jinawee, Chris White, Kyle KanosFeb 20 '14 at 1:40

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

Even someone with a physics major can wash windows, mop floors, and take out the trash, so there is always something. For a better job, you have to be able to do something actually useful. – Olin Lathrop Feb 19 '14 at 20:48

During my stint as a scientist working in industry I interviewed many job applicants. My experience is that it is extraordinarily hard to get good people, and that when I did meet good people it was obvious within the first few minutes of the interview that they were something special. With those people I didn't really care what they'd done at college (assuming it wasn't something silly like "swimwear design").

Still, not everyone can be an Ed Witten (including me - Ed the Horse is more my level :-) and I guess your first priority would be to get your CV through the first cut. I didn't do the initial CV filtering so I can't talk from personal experience. However Physics is a hard subject and if you get a good degree this immediately tells me you're a cut above the run of the mill applicants, and your CV would immediately go on the "to be interviewed" pile. The skills you need to be good at Physics apply to virtually any technical job, and even if you're applying for something completely different, e.g. law or finance, you'll still stand out as a clever person.

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Yes, lol, but first you have to finish the studies... – Pygmalion Apr 16 '12 at 12:17
@John Rennie When interviewing people, what made the special ones stand out? Exam results, innovations in the field, etc..? – PPG Feb 19 '14 at 20:56
@PPG: no, it wasn't achievements. It was their ability to think and express themselves clearly. – John Rennie Feb 20 '14 at 6:46

Researcher and lecturer at university are top jobs for those with degree in physics.

However, you can also teach at high schools, many work as programmers or making financial simulations for banks...

Physics is a universal knowledge, and you can apply it to many different disciplines (much easier than vice-versa). Of course in these cases you do not use your physical knowledge in full. But you most probably get better paycheck there ;-).

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"top" jobs in what way? – Martin Beckett Apr 16 '12 at 20:59
Definitely not in money earned ;) – Pygmalion Apr 16 '12 at 21:00
Nor in number of physics graduates employed. – user1504 Nov 19 '13 at 21:37

it is very hard for you to get a physics-related-job without good qualifications :( i myself i am unemployed although i have a degree on physics

i could suggest

• high school teacher

• Ph D researcher, although you will need very very good qualifications in general

• work as a consultor management

• Work as an 'Engineer material'

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To some extent this will depend on the country - since your profile says you are based in Cardiff, I assume this means she is expecting to study and work in the UK?

Obviously there would be a huge range of possible career options. There are some careers for which a physics degree is likely to be a requirement or at least highly desirable (academic physicist, physics teacher, perhaps a medical physicist or working as a physicist in the nuclear or other industries). There are other careers which would be open to physics graduates as well as those from related subjects (finance for highly mathematical subjects, and academic publishing or patent law for science and engineering subjects, for example). There are many careers for which a degree in just about any subject is a prerequisite (many large businesses have graduate schemes).

Perhaps it would be a good idea for her to talk to a careers advisor at her school/college, or to attend some university open days and discuss it with them. I'm sure there is also lots of useful information on the web (the IOP's Careers page looks promising, for a start).

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I agree with Jose that it is not easy to get a job with just a degree in physics. In this economy especially, just about everyone coming out of school is struggling to find work.

The path I chose, and it worked fairly well for me, was to go on to more school. In my case, I ended up in a nuclear engineering program and got my Master's degree from a good school. The job I currently have resulted from an internship position I held between the two years of grad school.

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$E = mc^2$? "Job" = "Money"? In my country graduates in physics can work at the research institute for about USD 50/month salary equivalent. Head of the nuclear spectroscopy lab at the local institute of nuclear physics (a good friend of mine) is paid about USD 150/month. What I admire in these people is that they are doing physics just because they like it.

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Assuming you mean a BS degree in Physics, she can get any job which only requires a BS degree. Since only the 4th year (of a typical 4 year study) is used to specialize (EE, CH, Phy, Math, etc.), these degrees are 75% the "same." Most hiring managers know this, that is why the specialty is not very important in the hiring process. What is most important is the fact that you obtained a degree. This shows that you are an individual capable of completing complex and difficult tasks/projects. Obviously, the higher the GPA, will result in more job offers. Also, to increase her "hire-ability," she can obtain additional degrees in one or more specialty (multiple disciplines).

If she continues and obtains a Masters or a PH degree, the potential for additional jobs is increased.

If she is interested in working in industry, she can work in manufacturing, or research. If she is interested in education, she can be a teacher. The opportunities are endless, and they are not related to the specialty chosen, but to the abilities of the person.

My advice for her would be to chose physics only if she wants to know what makes the Universe "tick" or is intrigued by the mysteries of the Universe and she would like to try to resolve at least one of them.

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While it may still be true that the subject is not important for jobs in industry, many colleges specialize far earlier than senior year. And outside the US this is even more universally true. – Chris White Feb 19 '14 at 19:12

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Open access peer-reviewed chapter

# Renewable Energy Microgrid Design for Shared Loads

By Ibrahim Aldaouab and Malcolm Daniels

Submitted: November 12th 2017Reviewed: February 26th 2018Published: November 5th 2018

DOI: 10.5772/intechopen.75980

## Abstract

Renewable energy resource (RER) energy systems are becoming more cost-effective and this work investigates the effect of shared load on the optimal sizing of a renewable energy resource (RER) microgrid. The RER system consists of solar panels, wind turbines, battery storage, and a backup diesel generator, and it is isolated from conventional grid power. The building contains a restaurant and 12 residential apartments. Historical meter readings and restaurant modeling represent the apartments and restaurant, respectively. Weather data determines hourly RER power, and a dispatching algorithm predicts power flows between system elements. A genetic algorithm approach minimizes total annual cost over the number of PV and turbines, battery capacity, and generator size, with a constraint on the renewable penetration. Results indicate that load-mixing serves to reduce cost, and the reduction is largest if the diesel backup is removed from the system. This cost is optimized with a combination of particle swarm optimization with genetic-algorithm approach minimizes total annual cost over the number of solar panels and micro-turbines, battery capacity, and diesel generator size, with a constraint on the renewable penetration. Results indicate that load-mixing serves to reduce cost, and the reduction is largest if the diesel backup is removed from the system.

### Keywords

• renewable energy system
• PV
• wind turbine
• battery

## 1. Introduction

### 1.1. Distributed electricity generation

Electrical power is historically generated at a few large power stations and transmitted over long distances to end users. However, in recent years, there is an increase in decentralized or distributed electric power generation, where the power is produced and used at the same location [1]. Often, this decentralized power is produced with renewable energy technologies, such as wind and solar, due to the decreasing costs of these technologies [1].

There are many advantages to a power distribution system that relies on many small generation facilities rather than a few large power plants. Transmitting power over long distances is inefficient and requires expensive infrastructure. Smaller facilities that are close to where the power is used can provide higher quality power, with fewer blackouts and a more steady voltage. Since many of the small generating stations are natural gas powered or powered with renewable sources, there is less pollution than large plants that often run on coal. Finally, a distributed energy generation model is more secure than a centralized model.

The challenge to expanding distributed electricity generation is partly economic and partly technical. The capital costs for building a small-scale facility is large relative to the power it can produce. Many renewable technologies produce irregular power that varies with the weather, and this power is difficult to incorporate into the grid [2].

### 1.2. Motivation for expanding distributed generation

Increasing attention is given to designs for integrating renewable energy with traditional power to satisfy electrical loads from individual or multiple buildings. Research in this area is driven by several factors. First, costs for photovoltaic (PV) panels and wind micro-turbines (MT) are steadily dropping, along with battery energy storage systems (BESS). For example, the reported cost of installed solar PV systems fell by an average of 6–12% per year from 1998 to 2014, depending on the scale of the system [3]. Similarly, the price of wind power is dropping significantly, as more turbines are brought online, and currently about 5% of the energy requirements for the United States is supplied through wind power. In the previous decade, more than two-thirds of all wind installations in the United States have been small- or mid-sized wind turbines [4]. Battery storage has also dropped to a level of about $100 per kWh of capacity [5]. Another motivating factor is the various grid-pricing structures available, which often create an incentive for using less power at certain times of the day or making the overall electrical demand more uniform. Microgrids can achieve this, reducing the cost of grid electrical power. In addition to economic motivation, there is increased recognition of the damage to the environment due to CO2 emission from traditional power generation. The benefit of using RER microgrids for buildings is the potential reduction in carbon and other pollutants because buildings consume over 40% of end-use energy worldwide [6]. In order to address the problem of efficient building energy use and to reduce pollution in buildings, the United States has set a zero net energy target on 50% of commercial buildings by 2040 and on all commercial buildings by 2050 [6]. A zero net energy building is one that produces some renewable energy on-site, such that the building sometimes uses grid power and at other times produces extra renewable power. The average amount of renewable power produced annually is the same as the building’s average annual consumption. Finally, another advantage for microgrids is the improved reliability that they offer. There are multiple sources of power in a microgrid, such that there is less chance of a complete power outage. ### 1.3. Problem statement Several studies consider the optimal design and sizing of the RER system for residential or commercial building individually [7, 8]. The RER microgrid design in this work is applied to a mixed commercial and residential building. Residential loads peak in the evening and early morning times, whereas commercial loads peak in the daytime. A shared residential and commercial load therefore has the potential to be more uniform than one or the other alone. A uniform load is easier to efficiently match to the RER supply, and it may also help lower the grid power costs. This chapter studies the effect of combining loads for an RER microgrid on the full cost of the microgrid. The size of the RER system partly depends on the shape of the load profile, such that irregular profiles require larger RER systems than smooth profiles. For example, an irregular load profile requires more energy storage to satisfy peak loads. The concept explored in this work is the benefit of combining different realistic load profiles in order to develop a total profile that is smoother and less costly to satisfy with an RER microgrid. ## 2. Building energy demand model The residential load profile used for this work is generated from measured aggregate hourly consumption data for 12 apartments in a residential building in Columbus, Ohio [8]. The apartments are on the third floor of a three-story building, which means that they will have higher heating loads in the winter and cooling loads in the summer. This choice represents a worst-case scenario in terms of the peaks in the residential load. One year of hourly metered power use for these apartments is available, starting at 12 am on Sunday, June 9, 2013. These apartments use electricity for hot water, heating, and cooling. The hourly commercial load profile is synthesized from typical load profiles for commercial kitchens [9]. The average demand from the commercial load is selected to be nearly the same as the average demand from the 12 residential apartments. ### 2.1. Historical residential demand Figure 1 illustrates the weekly average for the aggregate residential load data. Each day of the week, there is a peak in the morning at about 8 am, representing the electricity consumption as residents prepare for the workday. At the end of the day, at about 8 pm, there is a larger peak as residents return home for dinner and other electricity-consuming activities. To determine the temperature-dependent component of the residential data, a piecewise linear regression is used. Each week, aggregate residential consumption data are averaged to create 52 single values. The same is done for the temperature. Figure 2 shows a plot of the average weekly power versus average weekly outdoor temperature. The fit shown in Figure 2 has five parameters as follows: a heating slope (HS), cooling slope (CS), heating temperature (HT), cooling temperature (CT), and baseline (B) [8]. The baseline component defines the hourly expected weather independent demand. The heating and cooling slopes KW per Fahrenheit degree (kW/°F) enable respective prediction of the hourly heating and cooling demand for a typical weather year. The threshold temperatures and heating/cooling slopes depend on many factors, such as building construction and size. Values for the temperature-dependent five-parameter model were calculated to be: HS=0.16kW/°F.,CS=0.12kW/F, HT=57F, and CT=48F. ### 2.2. Commercial kitchen demand model Figure 3 illustrates the weekly average baseline consumption for the commercial kitchen. The same basic profile shape is used for each day, scaled to represent the different amounts of customer traffic for each day of the week. The peak consumption occurs at 8 pm as the kitchen serves dinner, and there is also a peak at 1 pm for lunch. The kitchen uses power more consistently than the residential load through the afternoon hours. ### 2.3. Controlling load profile characteristics In order to compare the three individual loads, the load factor (LF) is used. LF is defined as the ratio of the average per-month consumption to the peak hourly consumption for that month. For the data used in this study, an average yearly LF is found by averaging the 12 monthly LF values. Figures 4 and 5 show the comparison of the monthly LF values for the three types of loads considered: residential, commercial, and combined. Figure 4 shows the effect of combining the residential load with only the baseline commercial load. In this case, the load factor of the combined is between the residential and commercial load factors each month. Figure 5 shows the effect of combining the residential load with the commercial load, where the weather-dependent load is included with the commercial load. The load factor behavior varies more in this case. The LF for the combined load is still between the LF for the residential and commercial loads, except for January, March, and April. ## 3. Proposed microgrid structure The main components of the renewable energy resource (RER) microgrid examined here are solar photovoltaics (PV), micro-wind turbines (MT), and a battery energy storage system (BESS). These are utilized for electricity generation and energy storage, and they supply energy to a load that is normally satisfied with grid power alone. The microgrid is studied for two types of operating conditions. First is an isolated RER which provides all of the load power. The second system is an RER with a diesel generator backup in which the generator is activated if the RER cannot supply the full load. For each system, the hourly power flow to or from each element of the microgrid is simulated. The capacity of each RER component is chosen to minimize the overall cost. ### 3.1. Isolated RER The first type of microgrid is the isolated RER, which is shown in Figure 6. The RER operates independently from the local utility grid to provide all of the electricity to residential and commercial building. This system requires large battery storage to use during low wind and solar output times. The RER for the isolated mode must be large enough to produce energy to cover 100% of the building’s energy needs. ### 3.2. RER and diesel generator Figure 7 illustrates the second configuration and RER with a diesel generator backup. The diesel generator has the option of charging the battery, which is significantly different from the other scenarios. ### 3.3. RER and battery modeling #### 3.3.1. Photovoltaic model The power generated by a photovoltaic panel depends on two fundamental parameters: the solar irradiation and the ambient temperature. In order to simplify the model, the power produced by a PV panel, the following equation is used [8]. Pst=ηsAsGtE1 whereηs: Energy conversion efficiency (%), As: PV panel area (m2), Gt: Solar irradiation (W/m2), and Pst: Power generated by solar PV (W). The solar irradiation, sampled hourly, is found from typical meteorological year (TMY) data, for Columbus, Ohio [10]. The number of solar PV panels is Ns. #### 3.3.2. Micro-wind turbine model The electrical power generated by a micro-wind turbine depends on the wind speed, air density, area swept by the turbine blades, and an efficiency factor called the Betz coefficient [9]. A constant air density of 1.225 kg/m3is used, and the Betz coefficient is taken to be 59%. Pwt=0.5ρAwCpVw3tE2 where ρ: Air density (kg/m3), Aw: Micro-turbine swept area (m2), Cp: Betz coefficient, Vwt: Wind speed (m/s), and Pwt: Power generated by micro-turbine (W) The wind speed, sampled hourly, is found from typical meteorological year (TMY) data, for Columbus, Ohio [10]. The number of micro-turbines is Nw. #### 3.3.3. Battery energy storage model Whenever the RER output exceeds the load demand, the extra power is stored in a battery. This power is then used whenever the RER is unable to supply the load demand. The charge level of the battery is PBt, which is restricted to be in the range of 20–80% of the battery capacity, Bcap(kWh). 0.2BcapPBt0.8BcapE3 When charging or discharging the battery, the maximum amount of energy that can be removed during a 1-hour interval is BHR. This limit is expressed by the following inequality. PBtPBt1BHRE4 Both the battery capacity and the hourly charge/discharge limit are parameters for the microgrid. For every hour in the simulation of the microgrid, there is a decision made to charge or discharge the battery. This decision is determined in part by the charge level of the battery. If the current charge level is in the range of 20–80% of BCAP, then the battery can be charged or discharged. If the current charge level is less than 20% of BCAP, the battery can only be charged. If the current charge level is more than 80% of BCAP, the battery can only be discharged. ### 3.4. Diesel generator model The diesel generator supports the microgrid by supplying power PDLtdirectly to the load and power PDBtto the battery. The sum of these two must be less than the maximum power output Dmaxfrom the generator, such that. 0PDBt+PDLtDmaxE5 The conditions for activating the generator at each hour are determined by the current load and RER output, the state of the battery’s charge, and whether or not the generator was running the previous hour. If the RER and battery cannot meet the load, the generator is activated. If the generator was running the previous hour and the battery can take more charge, then the generator is allowed to run during the current hour. ## 4. Dynamic microgrid modeling ### 4.1. Isolated RER dispatch algorithm At each hour of the simulation for the grid-isolated system illustrated in Figure 6, the RER power Ptis determined using the TMY data. In addition, the load PLtand the charge level of the battery at the previous hour PBt1are known. From this information, the graph in Figure 8 is used to determine all other quantities. The graph illustrates the process of updating these quantities each hour. The following characteristics are implemented in this graph. • If the RER output is less than the load at a given hour, then all of the available RER output is sent to the load (i.e., no battery charging at this hour), and the battery satisfies the remainder of the load. • If the RER output is greater than the load, then the load is completely satisfied by the RER, and no battery power is used for the load. As much charging power as possible is transmitted to the battery, if its charge level is less than the upper threshold. Any remaining power from the RER is sent back to local grid. • There is a possibility of power outage if the combined RER and battery outputs cannot meet the load. Increasing the size of the system reduces this possibility. ### 4.2. Diesel generator RER dispatch algorithm At each hour of the simulation for the RER with distributed generation (DG) system illustrated in Figure 7, the RER power Ptis determined using the TMY data. In addition, the load PLtand the charge level of the battery at the previous hour PBt1are known. From this information, the graph in Figure 9 is used to determine all other quantities and to determine when the DG is turned on/off. The graph illustrates the process of updating these quantities each hour. The following characteristics are implemented in this graph. • If the RER output is less than the load at a given hour, then all of the available RER output is sent to the load (i.e., no battery charging at this hour), and as much power as possible from the battery is used to meet the load if its charge level is above the minimum threshold. If this is not sufficient to meet the load, the generator is used to make up the difference. • If the RER output is greater than the load, then the load is completely satisfied by the RER and no battery power is used for the load. As much charging power as possible is transmitted to the battery if its charge level is less than the upper threshold. Any remaining power from the RER is sent back to the local grid. • The generator is turned on if the combined RER and battery power cannot meet the load. If it is already running, then it will remain on until the battery is fully charged. • There is a possibility of power outage if the combined RER, battery, and DG outputs cannot meet the load. Increasing the size of the system reduces this possibility. ## 5. Annual cost model The annual cost of the system, ACS, is found with the following equation. ACS=ACC+ARC+AOC+AGCE6 where ACC=capital cost, ARC=replacement costs, AFC=fuel cost, and AOC=operating costs. Each of these categories is described in the following subsections. ### 5.1. Annual capital cost The capital cost is found from the initial costs for the PV array, MT units, BESS, and DG. The PV capital cost is$1.8 per peak watt of PV power. It is assumed that each solar panel in the array is 2 m2with a 20% efficiency, and that the peak radiation intensity is 1000 W/m2. This leads to a per-panel capital cost of $720, with Nspanels in the entire array. Each MT unit has an up-front cost, including installation, of$22,000 (about $2 per peak watt of wind power) [11, 12]. There are NWof these units installed. Each kWh of BESS capacity has a cost of$300 [11, 12], and the DG cost is approximated by a linear function of the capacity Dmax, from $7000 for a 5 kW capacity to$14,000 for a 40 kW capacity, based on advertised prices [12]. The DG cost formula is therefore

CDG=Dmax5kW$700035kW+$7000E7

The total up-front capital cost is then the sum of all four terms.

### 5.3. Annual operating costs

The only component that has significant operating costs is the DG, which requires fuel, oil for lubrication, and periodic maintenance. Each kWh of output from the DG requires 0.13 gallons of diesel fuel at a cost of $2 per gallon, with an additional maintenance cost of$0.05/kWh. The simulation produces the hourly output from the DG as PDt, and summing this quantity gives the yearly energy output. For the entire plant lifetime, the total operating cost TOCcan therefore be given as follows.

TOC=NPDt0.13galKWh2$gal+0.05$KWhE13

This quantity can be amortized using the same amortization factor that was applied to the capital cost. This gives an annual operating cost of.

AOC=TOC·CRFE14

## 6. Optimization problem formulation

In this section, the objective function is the total annual microgrid cost ACSas described in the previous section. This total annual cost is a nonlinear function of the parameters Ns, Nw, and BCAP, and evaluating this function requires execution of the dynamic model.

The nonlinear minimization is achieved with either the particle-swarm optimization (PSO) algorithm or the genetic algorithm (GA). PSO is used in situations in which no nonlinear constraints are needed, while GA is used if there are constraints.

Some of the constraints are simply bounds on the variables. The minimum values for each number cannot be negative, for example, and the upper bounds are chosen to be large enough for the RER to meet a required percentage of the load. The percentage of the annual load that is met by the wind and solar energy is called the renewable energy penetration formed with the following equation:

PEN=P1t+P2tLt×100%E15

where P1is Power from RER sources, P2is Power from battery, and Lis total load.

The system with DG backup has its own dispatch algorithm, which behaves differently than the grid-connected. This is because the DG on/off cycling incurs a maintenance cost. To avoid this, the rules for determining when to turn on and off the DG are designed to minimize the number of DG cycles. To ensure that there are no power losses, the minimum DG size is restricted to be equal to the peak load, such that the DG is capable of supplying the entire load with no RER assistance, if necessary. A nonlinear constraint is used to enforce a minimum RER penetration. The optimization requires the GA, and it is summarized as follows:

minNs,Nw,BCAP,DMAXACSLBSNsUBSLBWNwUBWLBBBCAPUBBLBDGDMAXUBDGPENPENMAXE16

where Nsis the number of solar panels, NWis number of wind turbines, and BCAPis battery capacity and diesel generation capacity.

The grid-isolated system with no backup is similarly optimized, except that the renewable energy penetration is naturally 100% for this system, since there is no diesel generator backup power used. A modified version of the backup code is used to model the grid-isolated system with no backup. To ensure that there are no power losses, the cost function includes a large cost penalty for each hour of power shortage. With this arrangement, there is no nonlinear constraint function, such that the PSO algorithm can be used. The optimization algorithm chooses a sufficiently large RER system in order to avoid the high cost penalty on the power losses.

## 7. Simulation results

The MATLAB simulation results for the optimization are presented in this section, for two versions of the microgrid: grid-isolated with no backup, and grid-isolated with diesel backup. The hourly RER power is generated in the same way for two cases, using hourly TMY profiles for solar radiation and wind speed. Figure 10 shows an example of the RER power, for 10 wind turbines and 100 solar panels, relative to the combined commercial and residential load. The upper plot in this figure shows hourly RER power produced for the entire year, and the lower figure compares the RER power to the load for 1 week. When RER power output is greater than the load, the excess production can be passed to the battery. When the RER power is less than the load, the battery must make up the deficit completely for the grid-isolated with no backup scenario. If the backup is available, then these elements can be used in addition to the battery.

### 7.1. Isolated microgrid with no backup

The dynamic modeling results for the isolated microgrid with no backup are illustrated in Figure 11. This figure shows the hourly flows of power from the RER directly to the load, along with the power from the battery to the load. The RER and battery sizes must be large enough to meet the load each hour, and the figure illustrates that the combined RER and battery flows are equal to the load. This constraint forces the RER and battery sizes to be large enough to meet the highest demand in the year, which means that during most of the year the battery is not fully utilized. This fact is demonstrated by Figure 12, which shows the battery charge level. The charge level remains high for most of the year, dipping down to low values only during a few times of peak demand.

Table 1 summarizes the results that optimize total microgrid annual cost for the residential load alone, the commercial load alone, and the mixture of the two. For comparison, the sum of the results for the residential and commercial loads is included.

(kWh)
cost (×1000 $)%RER output to the load%RER Pen. No backupResidential388662146.2028100 Commercial428748745.1224100 Sum81613110891.3226100 Mixed7901395986.0027100 ### Table 1. Optimization results for the isolated microgrid with no backup power. When added separately, the sum of the residential and commercial load annual costs is$91,320. If these loads are mixed, however, and are satisfied by a single, larger microgrid, then the annual cost decreases to $86,000. This illustrates the cost advantage of combining the two loads into a single mixed load. Separately supplying the two loads would require more solar panels and a significantly larger battery capacity than supplying the mixed load, although the number of MTs remains the same. Because this simulation does not have diesel generator backup, the RER penetration is 100%. For each case, only about 25% of the RER output is transmitted to the load. This low percentage is due to the fact that the RER size must be chosen significantly large to meet peak yearly demand. This means that there will be many hours of energy overproduction that cannot be utilized. ### 7.2. Microgrid with diesel generator backup When the grid-isolated microgrid is augmented with a DG backup system, this allows the relatively few hours of peak demand to be partially met with DG power. When the system is cost optimized, this leads to several interesting features. The size of the RER shrinks significantly, the usage of the battery becomes more regular throughout the year, and the percentage of RER power that makes it to the load nearly doubles. The trade-off for these improvements is a reduced RER penetration. Figure 13 illustrates the power dispatching for a one-week period in January, showing the intermittent operation of the DG. The DG operates during unusually high loads, which can be seen to occur as shown in Figure 14 when the battery energy level dips to low levels. The DG removes the burden of meeting the peak loads from the battery, such that the energy levels in the battery can swing more uniformly over its full range throughout the year. This is illustrated in Figure 14, which shows the battery energy level together with the DG operation for the entire year of the simulation. The pattern in the DG clearly shows a seasonal component, such that it operates more frequently during the heating and cooling seasons. Table 2 summarizes the results that optimize total microgrid annual cost for the residential load alone, the commercial load alone, and the mixture of the two. The sum of the individual results for the residential and commercial loads is included as before, for comparison with the mixed-load results. LoadNsNwBCAP (kWh) DMAX (kW) DG on time (hr)DG Startscost (×1000$)% RER output to the load% RER Pen.
Residential2142228355886721.004982
Commercial2441215247565020.134784
Sum458344359134411741.134883
Mixed4041356506028339.685282

### Table 2.

Optimization results for the isolated microgrid with DG backup power.

The largest effect of adding the DG backup to the microgrid is with the costs, which are about a fourth of the costs for the no-backup case. The optimization was made with the constraint of greater than 80% RER penetration, and it should be mentioned that the costs can be further reduced by lowering this constraint. Another large benefit in having the DG is the increase in the amount of RER power that makes it to the load, which doubles from roughly 25–50%. When comparing the sum of the costs of the residential and commercial loads to the cost of the mixed load, there is less of a difference than with the no-backup case. This indicates that the DG is reducing some of the benefit of mixing the loads.

## 8. Summary and discussion

To illustrate the effect of renewable energy penetration, the cost optimization is constrained to produce a result with a fixed penetration. This applies to the isolated grid with diesel backup. Figure 15 shows a plot of cost per kWh versus penetration, under the assumption that the microgrid’s non-renewable power component has a price of $0.2/kWh. Plots of the average cost per kWh for all three loads (residential, commercial, and mixed) are shown, where the average cost is computed by dividing the annual cost by the total annual load. As the renewable energy penetration approaches 100%, the cost of power from the microgrid becomes rapidly more expensive, approaching the case 1 result. As the renewable energy penetration decreases to zero, the cost of power approaches the price of nonrenewable energy,$0.2/kWh. The minimum cost per kWh occurs in the range of 30–40% of penetration.

Figure 15 also shows the interaction between load mixing and renewable energy penetration. Below 70% penetration, the mixed load cost is between the residential and commercial costs. Above this threshold, however, the mixed load cost is below both the residential and commercial costs.

## 9. Conclusion

The simulation results confirm that using a backup power source to support renewable energy reduces overall microgrid costs. For example, using a DG backup lowers the renewable energy penetration from 100 to 82%, but cuts the overall cost by more than a factor of two. However, the simulation results also indicate a clear benefit to mixing residential and commercial loads, such that the cost for satisfying the mixed load is less than the sum of the costs for the loads individually. Depending on the configuration, the cost for the mixed load is 3–8% less than the sum of the individual costs.

Microgrid designs for building applications involve determining the best mix of building loads for optimizing energy delivery. The result of the modeling work presented here is that combining loads allow for a measure of control over the microgrid costs. This concept is important for moving toward 100% renewable energy penetration.

## More

© 2018 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution 3.0 License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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Ibrahim Aldaouab and Malcolm Daniels (November 5th 2018). Renewable Energy Microgrid Design for Shared Loads, Smart Microgrids, Majid Nayeripour, Eberhard Waffenschmidt and Mostafa Kheshti, IntechOpen, DOI: 10.5772/intechopen.75980. Available from:

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# Understanding the branch cut and discontinuity of the hypergeometric function

DISCLAIMER: This question comes from math.stackexchange (where it has an active bounty). The link is here. UPDATE: the question has been answered on math.stackexchange at the previous link, and the answer is satisfactory.

I am going through this paper, and I am having trouble understanding page 20. I am still learning my way around managing multi valued complex functions, so I'd like your help in understanding what's happening there.

I have the definition of the hypergeometric series $_2F_1$ as $$_2F_1(a,b;c;z)=\sum_{n=0}^\infty \frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}.$$ Here $(a)_n$ is the rising Pochhammer symbol, $(a)_n=\Gamma(a+n)/\Gamma(a)$ (well defined whenever $a$ is not a negative integer or zero). By elementary computations, the radius of convergence of this series is 1. It is then said that $_2F_1$ can be continued analitically in the whole complex plane along any curve not passing through $[1;+\infty)$, and $1$ itself is a branching point and the function has a cut on the previous segment.

Furthermore, equation $2.115$ of the paper computes the discontinuity when crossing the branch cut as (with $x\geq4/3$ to have the real part of the argument bigger than one) $$_2F_1\left(\frac16,\frac56;1;\frac{3x}4+i\epsilon\right)-_2F_1\left(\frac16,\frac56;1;\frac{3x}4-i\epsilon\right)=i\, _2F_1\left(\frac16,\frac56,1;1-\frac{3x}4\right).$$ I am trying to understand those results. It is not clear to me how to understand from the power series form the behaviour on the boundary of the disk of convergence. It is clear to me that the series in $z=1$ diverges, as the coefficients do not go to zero, so there should be no analytical continuation.

A way to study that function would be its integral representation: $$_2F_1(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}dt.$$ Here it is understood that $\arg t=0=\arg(1-t)$. From this representation, it is clear to me that the function can have branches whenever $a$ is not a positive integer, and if $z\in[1;\infty)$ we have $0$ in the integration path, so we can have multiple branches. And that's exactly my case. The problem is that I can't use this form to prove the previous equation about the discontinuity: all I manage to write is (here I write z=x with x real number bigger than one, and I am neglecting some numerical factors that can easily be reinserted) $$\lim_{\epsilon\to0}((1-xt-i\epsilon t)^{-a}-(1-xt+i\epsilon t)^{-a})=-(1-e^{-2i\pi a})(1-xt)^{-a}.$$ This is different from the line that I would like to prove (as an example, I do not see why I should change the argument, but I also see that with that argument changing the function is at least evaluated in a point where well defineteness is guaranteed).

For reference, I'd like to prove 15.2.3 of https://dlmf.nist.gov/15.2 (where the equality is with the regularized hypergeometric function, that is just the hypergeometric function without a factor $\Gamma(c)$).

To summarize, I have three questions:

• How to prove the discontinuity from the integral form?
• More in general: how to treat power series in order to understand if their singularities are poles or branch cuts, and the discontinuity of their analytic continuations? I know that this is a very broad topic, and I'd also like some references to continue my study.
• A last question: how to approach functions in the complex plane that are defined through a parametric integration of functions that may or may not have branch points, depending on the parameter? Even here, I'm more looking for references that a big answer.

Thanks everybody for the time you took reading this.

EDIT: In the comments, it has been asked me to clarify what I'm looking for, and it was pointed that functions with branch cuts are not discontinuous on the cut.

I unfortunately do not know a general framework to discuss functions with branch cuts, so you will have to forgive my lack of precision. I agree that the function is not discontinuous at the cut because it has not to be considered as a function from the complex plane to the complex plane, but as a function from a more general Riemannian surface to the complex numbers. I think my question makes sense even in this framework, but I have to clarify it a bit. As I do not have precise definitions, I will proceed with an example.

The basic example happens by defining the square root of a number $z$ of modulus $r$ and phase $\theta\in[0;2\pi)$ as $$\sqrt{z}=\sqrt{re^{i\theta}}=\sqrt{r}e^{i\frac{\theta}{2}}.$$ I am basically defining a branch of the global function square root, and I can interpret it as a function on the complex plane, cut with a branch cut going from $0$ to $+\infty$, covering the positive real axis.

What I want to do is to compute, for $x$ real positive \begin{align} \lim_{\epsilon\to0^+}(\sqrt{x+i\epsilon}-\sqrt{x-i\epsilon}). \end{align} To compute that, I can write (the $\arctan$ function is here defined having range in $(-\pi/2,\pi/2)$) $$\sqrt{x+i\epsilon}=\sqrt{\sqrt{x^2+\epsilon^2}e^{i\arctan{\frac\epsilon x}}}=\sqrt{x}+o(\epsilon),\\ \sqrt{x-i\epsilon}=\sqrt{\sqrt{x^2+\epsilon^2}e^{-i\arctan{\frac{\epsilon}x}+i2\pi}}=-\sqrt{x}+o(\epsilon).$$ From this computation, I conclude that the "discontinuity" of the square root on the cut is given by $2\sqrt{x}$.

I'm looking for the analogue of this computation in the more difficult case in which the function is given by the integral of a function with branch cuts and a parameter (that is the argument of the function), as in my example. The result should include the second equation of this post.

• I must comment that supposed discontinuities along "branch cuts" are phantoms... at most artifacts of conventions... so cannot possibly be "detected". This is already visible in the case of $\sqrt{z}$ and such, where, despite popular conventions, there is in fact no genuine "discontinuity" of the local square root function at the negative-real axis. That's why we can't "detect" it. That is, the local power series there converge with radius of convergence equal to the distance to $0$, ignoring our convention about the "branch cut". So, can you clarify what you're hoping to have explained? – paul garrett Feb 23 '18 at 0:17
• I think you don't understand it from the power series. You understand it from the differential equation. – Gerald Edgar Feb 23 '18 at 1:09
• paul garrett, thanks for your comment. I do not know how to treat functions with branches in a formal setting, for now I just have to develop some intuition and computation skill. I have expanded my question to try and address your question. Is it clear? – Salvatore Baldino Feb 23 '18 at 14:27
• Gerald Edgar, that would be the first time for me in which I study the analytic properties of the solution of a differential equation from the equation itself. But maybe it will be the best approach, as I almost never will have a function to work on, just the equation. I will look for this approach, but I'd also be interested in an answer based on manipulations of the integral that I wrote. Thanks for your hint. – Salvatore Baldino Feb 23 '18 at 14:29
• I would like to point out that the question has received a satisfactory answer on Math Stackexchange. I will state it in the question for anyone who is interested in the answer. – Salvatore Baldino Mar 13 '18 at 11:01