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# 2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.2

Students can Download Maths Chapter 1 Relations and Functions Ex 1.2 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka 2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.2

### 2nd PUC Maths Relations and Functions NCERT Text Book Questions and Answers Ex 1.2

Question 1.
Show that the function f : R0 → R0 defined by $$f(x)=\frac{1}{x}$$  is one-one and onto, where R0 is the set of all non-zero real numbers. Is the result true, if the domain R0 is replaced by N with co-domain being same as R0?

f : N → R0,  then the function will remains one – one however mot onto.
Here the range of $$\mathrm{f}=\left\{1, \frac{1}{2}, \frac{1}{3}, \ldots \ldots .\right\} \neq \mathrm{R}_{\mathrm{e}}$$

Question 2.
Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f (x) = x2

hence f is injective, for some elements like, 2, 3 etc has no preimage in N such that
f (x) = 2 hence not surjective.

(ii) f : Z → Z given by f(x) = x2
f is not one-one or injective as f (1) = f (-1) ⇒ 1 = -1
f is not on to or surjective as $$f(x)=2 \Rightarrow x=\sqrt{2} \notin Z$$

(iii) f : R → R given by f(x) = x2
f is not injective
as f (x1) = f (x2) ⇒ x1= ± x2
Range off does not contain any negative real. Hence not surjective.

(iv) f : N → N given by f(x) = x2
Let x1, x2 ∈ N
f(x1) = f(x2) ⇒ $$x_{1}^{3}=x_{2}^{3}$$
hence one-one
Range of f= {1,8,27,…… } ≠ N
f is not surjective.

(v) f : Z → Z given by f (x) = x3
Let x1, x2 ∈ Z
\begin{aligned} f\left(x_{1}\right)=f\left(x_{2}\right) & \Rightarrow x_{1}^{3}=x_{2}^{3} \\& \Rightarrow x_{1}=x_{2} \end{aligned}
f is one-one or injective
Range of f = {0, ±1, ±8, ±27,…………….. } ≠Z
f is not surjective

(vi) f : R → R given by f (x) = x3

Question 3.
Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
∀ x ∈ [0, 1)
f (x) = [x] = 0
f is not one-one, or injective
ex: $$0, \frac{1}{2} \in R \Rightarrow f(0)=[0]=0$$
$$f\left(\frac{1}{2}\right)=\left[\frac{1}{2}\right]=0$$
Range of f is the set of integers ≠ R hence f is not onto.

Question 4.
Show that the Modulus Function f : R → R, given by f (x) = | x |, is neither one-one nor onto, where | x | is x, if x is positive or 0 and | x | is – x, if x is negative.
Let x1, x2 ∈ R f (x.) = f (x1)
\begin{aligned}f\left(x_{1}\right)=f\left(x_{2}\right)&\Rightarrow\left|x_{1}\right|=\left|x_{2}\right| \\& \Rightarrow x_{1}=\pm x_{2} \end{aligned}
$$|-2|=|+2|=2$$hence f is not one-one.
Range of functions is only non negative real numbers
Range of f = {0, ∞) ≠ R
∴ f is not onto.

Question 5.
Show that the Signum Function f : R → R, given by
$$f(x)=\left\{\begin{array}{ll}{1} & {\text { if } x>0} \\{0} & {\text { if } x=0} \\{-1} & {\text { if } x<0}\end{array}\right.$$
is neither one-one nor onto.
f (x) = 1 ∀ x ∈ [0, ∞ )
hence f is not one-one Range of f = {-1,0, 1} hence f is not onto.

Question 6.
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1,4), (2,5), (3,6)} be a function from A to B. Show that f is one-one.
x1, x2 ∈ A, f (x1) = f (x2)
⇒ x1 = x2
∴ hence one-one
f (1) = 4, f (2) = 5, f (3) ≠ R, hence f is one-one.

Question 7.
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x
Let x1 , x2 ∈ R
f (x1) = f (x2) ⇒ 3- 4x1, = 3- 4x2
⇒ x1 = x2 hence one-one function

hence f is not onto

(ii) f : R→ R defined by f(x) = 1 + x2
Let x1, x2 ∈ R
f (x1) = f (x2)
⇒ 1 + x12 = 1 + x22
⇒ x1 = ±x2
∴  f is not one-one
Also range of f contains only real numbers which is greater than or equal to 1≠ R
∴ f is not onto hence f is not bijective.

Question 8.
Let A and B be sets. Show that
f : A x B → B x A such that
f (a, b) = (b, a) is bijective function.
Let (p, q), (r, s) g A x B f (p,q) = f (r, s)
⇒ (q, p) = (s, r) ⇒ q = s and p = r
⇒ (p, q) = (r, s), hence one-one
∀ (b, a) ∈ B x A, 3 (a, b) ∈ A x B
such that f (a, b) = f (b, a), hence on to hence f is bijective function

Question 9.
Let f : N → N be defined by
$$\mathbf{f}(\mathbf{n})=\left\{\begin{array}{c} {\frac{\mathbf{n}+\mathbf{1}}{2}, \text { if } \mathbf{n} \text { is odd }} \\{\mathbf{All} \mathbf{n} \in \mathbf{N}} \\{\frac{\mathbf{n}}{2}, \text { if } \mathbf{n} \text { is even }}\end{array}\right.$$
$$f(2 n-1)=\frac{2 n-1+1}{2}=n$$
$$f(2 n)=\frac{2 n}{2}=n$$
f (2n – 1) = f (2n) = n, but 2n -1 ≠ 2n
en, f (3) = 2, f (4) = 2, but 3 ≠ 4
hence f is not one-one
Here range of f = N
as for any N ∈ N, ∋ 2n ∈ N such that
f(2n) = n
hence f is on to. But it is not bijective

Question 10.
Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by $$f(x)=\left(\frac{x-2}{x-3}\right)$$. is f one-one onto? justify your answer.

Question 11.
Let f : R → R be defined as f(x) = x4. Choose the correct answer.
(A) f is one-one onto
(B) f is many-one onto
(c) f is one-one but not onto
(D) f is neither one-one nor onto.
f (x,) = f (x2) ⇒ x14 = x24
= x14 = x24 = 0
= (x1 – x2) (x1 + x2) (x12 = x22)
x1 = x2 or x1 = – x2
∴ f is not one-one
f (x) = x4 ≥ o ∀ x ∈ R
Range of f = [0, ∞) ∈ R
hence not onto


text: 2 added 1 character in body edited Mar 23 '16 at 19:49 AlessioX 27911 gold badge22 silver badges1616 bronze badges Assuming X is the matrix of new data points (size m x n where each row is an example), you can tackle the problem by considering several kernel. if your kernel is linear (i.e. dot product) The predicted labels can be easily evaluated by p = X * model.w + model.b; where model.w is an array containing the free parameters of the hyperplane and model.b is the bias. if your kernel is RBF (i.e. Gaussian) % square and sum new data points by rows X1 = sum(X.^2, 2); % square and sum SVs X2 = sum(model.SV.^2, 2)'; % evaluate exponential for RBF formula K = bsxfun(@plus, X1, bsxfun(@plus, X2, - 2 * X * model.SV')); % evaluate kernel K = model.kernelFunction(1, 0) .^ K; % multiply Gram matrix by SV labels K = bsxfun(@times, model.SVL', K); % multiply again by alphas K = bsxfun(@times, model.alphas', K); % gather final predictions p = sum(K, 2);  where model.alphas is the Lagrange Multiplier vectors, model.kernelFunction() is the kernel function, model.SV are the Support Vectors and model.SVL are the Support Vectors Labels 3. for an arbitrary kernel, apply the standard rule: or an arbitrary kernel, apply the standard rule: for i = 1:m prediction = 0; for j = 1:size(model.SV, 1) prediction = prediction + model.alphas(j) * model.SVL(j) * model.kernelFunction(X(i,:)', model.SV(j,:)'); end p(i) = prediction + model.b; end  After all that you might want to shrink the predicted labels vector p in a friendly binary format: predictedLabels(p >= 0) = 1; predictedLabels(p < 0) = 0;  and predictedLabels is now your final output. Assuming X is the matrix of new data points (size m x n where each row is an example), you can tackle the problem by considering several kernel. if your kernel is linear (i.e. dot product) The predicted labels can be easily evaluated by p = X * model.w + model.b; where model.w is an array containing the free parameters of the hyperplane and model.b is the bias. if your kernel is RBF (i.e. Gaussian) % square and sum new data points by rows X1 = sum(X.^2, 2); % square and sum SVs X2 = sum(model.SV.^2, 2)'; % evaluate exponential for RBF formula K = bsxfun(@plus, X1, bsxfun(@plus, X2, - 2 * X * model.SV')); % evaluate kernel K = model.kernelFunction(1, 0) .^ K; % multiply Gram matrix by SV labels K = bsxfun(@times, model.SVL', K); % multiply again by alphas K = bsxfun(@times, model.alphas', K); % gather final predictions p = sum(K, 2);  where model.alphas is the Lagrange Multiplier vectors, model.kernelFunction() is the kernel function, model.SV are the Support Vectors and model.SVL are the Support Vectors Labels 3. for an arbitrary kernel, apply the standard rule: for i = 1:m prediction = 0; for j = 1:size(model.SV, 1) prediction = prediction + model.alphas(j) * model.SVL(j) * model.kernelFunction(X(i,:)', model.SV(j,:)'); end p(i) = prediction + model.b; end  After all that you might want to shrink the predicted labels vector p in a friendly binary format: predictedLabels(p >= 0) = 1; predictedLabels(p < 0) = 0;  and predictedLabels is now your final output. Assuming X is the matrix of new data points (size m x n where each row is an example), you can tackle the problem by considering several kernel. if your kernel is linear (i.e. dot product) The predicted labels can be easily evaluated by p = X * model.w + model.b; where model.w is an array containing the free parameters of the hyperplane and model.b is the bias. if your kernel is RBF (i.e. Gaussian) % square and sum new data points by rows X1 = sum(X.^2, 2); % square and sum SVs X2 = sum(model.SV.^2, 2)'; % evaluate exponential for RBF formula K = bsxfun(@plus, X1, bsxfun(@plus, X2, - 2 * X * model.SV')); % evaluate kernel K = model.kernelFunction(1, 0) .^ K; % multiply Gram matrix by SV labels K = bsxfun(@times, model.SVL', K); % multiply again by alphas K = bsxfun(@times, model.alphas', K); % gather final predictions p = sum(K, 2);  where model.alphas is the Lagrange Multiplier vectors, model.kernelFunction() is the kernel function, model.SV are the Support Vectors and model.SVL are the Support Vectors Labels or an arbitrary kernel, apply the standard rule: for i = 1:m prediction = 0; for j = 1:size(model.SV, 1) prediction = prediction + model.alphas(j) * model.SVL(j) * model.kernelFunction(X(i,:)', model.SV(j,:)'); end p(i) = prediction + model.b; end  After all that you might want to shrink the predicted labels vector p in a friendly binary format: predictedLabels(p >= 0) = 1; predictedLabels(p < 0) = 0;  and predictedLabels is now your final output. 1 answered Mar 23 '16 at 19:04 AlessioX 27911 gold badge22 silver badges1616 bronze badges Assuming X is the matrix of new data points (size m x n where each row is an example), you can tackle the problem by considering several kernel. if your kernel is linear (i.e. dot product) The predicted labels can be easily evaluated by p = X * model.w + model.b; where model.w is an array containing the free parameters of the hyperplane and model.b is the bias. if your kernel is RBF (i.e. Gaussian) % square and sum new data points by rows X1 = sum(X.^2, 2); % square and sum SVs X2 = sum(model.SV.^2, 2)'; % evaluate exponential for RBF formula K = bsxfun(@plus, X1, bsxfun(@plus, X2, - 2 * X * model.SV')); % evaluate kernel K = model.kernelFunction(1, 0) .^ K; % multiply Gram matrix by SV labels K = bsxfun(@times, model.SVL', K); % multiply again by alphas K = bsxfun(@times, model.alphas', K); % gather final predictions p = sum(K, 2);  where model.alphas is the Lagrange Multiplier vectors, model.kernelFunction() is the kernel function, model.SV are the Support Vectors and model.SVL are the Support Vectors Labels 3. for an arbitrary kernel, apply the standard rule: for i = 1:m prediction = 0; for j = 1:size(model.SV, 1) prediction = prediction + model.alphas(j) * model.SVL(j) * model.kernelFunction(X(i,:)', model.SV(j,:)'); end p(i) = prediction + model.b; end  After all that you might want to shrink the predicted labels vector p in a friendly binary format: predictedLabels(p >= 0) = 1; predictedLabels(p < 0) = 0;  and predictedLabels is now your final output.

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# An object moves in simple harmonic motion with period 5 seconds and amplitude 7 cm. At time displaystyle{t}={0} seconds, its displacement d from rest is -7 cm, and initially it moves in a positive direction. Give the equation modeling the displacement d as a function of time t.

An object moves in simple harmonic motion with period 5 seconds and amplitude 7 cm. At time $t=0$ seconds, its displacement d from rest is -7 cm, and initially it moves in a positive direction.
Give the equation modeling the displacement d as a function of time t.
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crocolylec
Step 1
We need to find the equation that modeling the displacement d as a function of time t for an object moving in simple harmonic motion.
Step 2
At $t=0$ seconds, the displacement d is -7 cm, and moving in a positive direction. Since initially the object is at the lowest point and moving in positive direction, so we take the negative cosine function. So the general equation for displacement as a function of time t as:
$d=-A\mathrm{cos}\left(Bt-C\right)+D$
Step 3
The amplitude is given by A, here since amplitude is 7 cm, so we have $A=7$.
The period of the object is 5 seconds, so we have
Period $=2\frac{\pi }{B}$
$5=2\frac{\pi }{B}$
$B=2\frac{\pi }{5}$
Step 4
Since the minimum displacement is at $t=0$, so there is no phase shift,
so $C=0$.
Also since there is no vertical shift, so $D=0.$
Thus the equation modeling the displacement d as a function of time t is given by:
$d=-A\mathrm{cos}\left(Bt-C\right)+D$
$=-7\mathrm{cos}\left(2\frac{\pi }{5}t-0\right)+0$
$=-7\frac{\mathrm{cos}\left(\left(2\pi t\right)\right)}{5}$
The finalli equation modeling the displacement d as a function of time t is:
$d=-7\mathrm{cos}\left(\frac{2\pi t}{5}\right)$
Jeffrey Jordon


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# Difference between revisions of "Integer"

The integers consist of the positive natural numbers (1, 2, 3, …), their negatives (−1, −2, −3, ...) and the number zero. The set of all integers is usually denoted in mathematics by Z (or Z in blackboard bold, $\displaystyle{ \mathbb{Z} }$), which stands for Zahlen (German for "numbers"). They are also known as the whole numbers, although that term is also used to refer only to the positive integers (with or without zero). Like the natural numbers, the integers form a countably infinite set.

## Algebraic properties

Like the natural numbers, Z is closed under the operations of addition and multiplication, that is, the sum and product of any two integers is an integer. However, with the inclusion of the negative natural numbers, and, importantly, zero, Z (unlike the natural numbers) is also closed under subtraction. Z is not closed under the operation of division, since the quotient of two integers (e.g., 1 divided by 2), need not be an integer.

closure: $\displaystyle{ a + b }$ is an integer $\displaystyle{ a \cdot b }$ is an integer
associativity: $\displaystyle{ a + (b + c) = (a + b) + c }$ $\displaystyle{ a \cdot (b \cdot c) = (a \cdot b) \cdot c }$
commutativity: $\displaystyle{ a + b = b + a }$ $\displaystyle{ a \cdot b = b \cdot a }$
existence of an identity element: $\displaystyle{ a + 0 = a }$ $\displaystyle{ a \cdot 1 = a }$
existence of inverse elements: $\displaystyle{ a + (-a) = 0 }$
distributivity: $\displaystyle{ a \cdot (b + c) = (a \cdot b) + (a \cdot c) }$

## Order-theoretic properties

Z is a totally ordered set without upper or lower bound. The ordering of Z is given by

$\displaystyle{ \ldots \lt -2 \lt -1 \lt 0 \lt 1 \lt 2 \lt \ldots }$

An integer is positive if it is greater than zero and negative if it is less than zero. Zero is defined as neither negative nor positive.

The ordering of integers is compatible with the algebraic operations in the following way:

1. If $\displaystyle{ a \lt b }$ and $\displaystyle{ c \lt d }$, then $\displaystyle{ a + c \lt b + d }$
2. If $\displaystyle{ a \lt b }$ and $\displaystyle{ 0 \lt c }$, then $\displaystyle{ ac \lt bc }$

(From this fact, one can show that if $\displaystyle{ c \lt 0 }$, then $\displaystyle{ ac \gt bc }$.)


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# Math Help - differntiable functions

1. ## differntiable functions

The twice-differentiable function f is defined for all real numbers and satisies the following conditions:

f(0) = 2
f'(0) = -4
f''(0) = 3

The function g is given by g(x) = e^(ax) + f(x) for all real numbers, where a is a constant. Find g'(0) and g''(0) in terms of a.

2. Originally Posted by DINOCALC09
a) The twice-differentiable function f is defined for all real numbers and satisies the following conditions:

f(0) = 2
f'(0) = -4
f''(0) = 3

b) The function g is given by g(x) = e^(ax) + f(x) for all real numbers, where a is a constant. Find g'(0) and g''(0) in terms of a.
Hello,

to a):

Assume that f''(x) = 3 then

$f'(x) = \int(f''(x))dx = 3x+C$. With f'(0) = -4 you'll get C = -4

Thus f'(x) = 3x-4

Then $f(x) = \int(f'(x)dx = \frac32 x^2-4x+C$ . With f(0) = 2 you'll get C = 2 and therefore the function is:

$f(x) = \frac32 x^2 -4x + 2$

to b)
I assume that you have to use the function f from a). Thus

$g(x) = e^{ax}+f(x)~\implies~g'(x)=a \cdot e^{ax}+f'(x)~\implies~g''(x)=a^2 \cdot e^{ax}+f''(x)$ Plug in x = 0 (Remember $e^0 = 1$) and the values for f'(0) and f''(0).

$g'(0) = a \cdot 1 -4 = a-4$

$g''(x) = a^2+3$


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Prove this Vector Operation

1. Apr 28, 2009

BioBabe91

1. The problem statement, all variables and given/known data
if vectors $$\vec{a}$$ and $$\vec{b}$$ have opposite directions, how to show that |$$\vec{a}$$| + |$$\vec{b}$$| = |$$\vec{a}$$ - $$\vec{b}$$|?

2. Relevant equations
quadratic equation, definition of absolute value

3. The attempt at a solution
$$|\vec{a}-\vec{b}| = \sqrt{\vec{a}^{2}-2\vec{a}\bullet\vec{b}+\vec{b}^{2}}$$
and then I got
$$|\vec{a}-\vec{b}| = \sqrt{\vec{a}^{2}-2|\vec{a}||\vec{b}|cos\Phi+\vec{b}^{2}}$$
So then cosine of the angle is equal to -1, and I don't know how to go from there.

Last edited: Apr 28, 2009
2. Apr 28, 2009

bigplanet401

Looks like a notation issue.

\begin{align} \vec{a}^2 &= a^2\notag\\ |\vec{a}|&=a\notag \end{align}

Everything under the radical is a scalar.

3. Apr 29, 2009

Staff: Mentor

This makes no sense. In the first equation you have the square of a vector, which implies multiplication of some kind. The only kinds of multiplication that are available are 1) multiplication by a scalar, 2) the dot product, and 3) (for vectors in R^3) the cross product.

In the second equation, you claim that the magnitude of a vector is equal to the vector itself.

4. Apr 29, 2009

Staff: Mentor

You're given that the vectors have the opposite directions, which means that a = -kb for some positive scalar k. Also, since a.b = |a||b|cos(theta), and theta = pi, you have a.b = -|a||b|.

Put both of these ideas together, and the result you want should fall out pretty readily.

5. Apr 29, 2009

bigplanet401

Again, I think it's a notation issue. You'll often see shorthands like this in the literature. Since you're using boldface:
\begin{align} \mathbf{a}^2 &= \mathbf{a} \cdot \mathbf{a} = a^2 = |\mathbf{a}|^2 \intertext{and} |\mathbf{a}| &= a. \end{align}

I can see where there might be problems, if instead we did something like
$$\mathbf{a}^2 = \mathbf{aa} = \mathbf{a} \otimes \mathbf{a}$$


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# On Markdown

## In the beginning, there was LaTeX

As a former mathematics student, I have an unusual soft spot for LaTeX. Despite it's obscenely verbose and quirky syntax, it was more than a breath of fresh air compared to Word's Equation Editor. (It required a highlighting, 4 clicks, and keyboard input to create an exponent, if I remember correctly.) But not only does LaTeX keep your fingers on the keyboard, it's output is the most beautiful thing I've ever laid eyes on. I mean, just look at this inequality:

Even using a slew of macros and shortcuts, reading LaTeX code is burdensome at best, and this will forever keep it's use out of the main stream. Worse, making formatting changes in LaTeX is onerous. Luckily for math students their are really great one's already created that are perfect for your homework, and journals traditionally provide their own. (My University library did not, however. Spent at least 8 hours and an argument with the librarian working on the Title page with spots for my advisors' signatures'.) Because of this, LaTeX will never be a mainstream word processor.

However, I think there is one core design aspect of LaTeX that is far more ideal than traditional word processors: separation of corpus and design.

## Enter Markdown

Markdown was created in 2004 with the idea to "make writing simple web pages, and especially weblog entries, as easy as writing an email" (quote from the late Aaron Swartz's blog post announcing the release.) It has gained a deal of popularity in web culture on sites such as reddit and GitHub, as well with some excellent hacker tools like iPython Notebook and Octopress***. (This site is powered by Octopress and each post is written in Markdown.)

There is a more thorough example on wikipedia Wikipedia, but for example *italics* renders as italicts, **bold** as bold, and [links](http://skien.cc) as links.

The really clever thing here is that for most documents, there is very limited formatting to add. For web documents, like blog posts and comment sections, you mostly need sections, paragraphs, italics, bold, links, inline images, and inline quotes. Developers like the ability to add in text code and

inline code blocks.

Best of all, you don't have to write in HTML (Who want's to write an article in raw HTML? Ain't nobody got time for that.) and you can achieve a great deal of formatting.


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## The Cost of Introversion

Ask any good psychologist or neuroscientist to explain the contrast between introversion and extroversion, and you’re likely to be told something akin to this: Introverts expend energy in social interaction and recharge by having solitary downtime, while extroverts recharge by being social and lose energy when left alone or with little to do.

But what is this “energy”? Sure, it’s easy enough to intuit. Clearly, extroverts are always so outgoing, very talkative, and need to be with other people. They excel at group situations and meeting new people, and become antsy and frustrated when left alone for any length of time. Introverts, on the other hand, function much better in one-on-one situations or with small groups of people they know, and tend to dislike smalltalk when it’s just to “network” with new people. In addition, introverts love having solitude for periods of time to recharge their internal batteries.

This analogy of charging/expending energy, while a bit oversimplified, is an accurate one. So then, why are large social situations so “costly” to an introvert? And why can they still “recharge” even when in large groups of close friends/family?

Let’s take a step back from this for a moment and look at how we interact in this society. We all wear our proverbial masks every time we step into the real world: hiding the reality of who we are and letting everyone see only the selves we wish to show.

As an introvert myself, I’ve come to theory that this “energy” cost in social situations is merely the necessity to ourselves of maintaining these masks. As we wear the masks and keep our internal filters and censors active, we expend more and more of our mental energy.

At least for me, this is why I “recharge” by spending lots of times with my close friends: Even though I’m being very social and involving myself with other people, it is with friends to whom I’d trust unfalteringly. I don’t have to keep any of my internal filters going or some such. I can just be me. I don’t have to hide who or what I am.

My theory isn’t very good at explaining the other side of this analogy, though. If this is the case, then why do extroverts lose energy by being alone and recharge by being so social and active? This is already beyond my understanding, and something I’d be mildly interested in researching further…But that’s a topic for another day!

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## Some moments will haunt you; the right ones will inspire…

[Originally posted as a Facebook note on April 4; retroactively posted to this blog days later. Note to self: I should really use this instead of FB Notes…]

To those who rejoice every time I change my Facebook relationship status for April Fool’s, your wanting to celebrate my intrapersonal success is very much appreciated. So I mark this as a success in its own way.

I’m not normally sentimental about these things; but this is a one-time specialty. Today, April 4, 2014, marks the one-year anniversary of my first date. Ever.

While I don’t want to write a sappy romance novel here or anything, suffice to say that asking her out in the first place was so ridiculously far out of my comfort zone, and was only thanks to the encouragement of amazing friends. Even though the end result of that date may not have been what I ultimately hoped, it forced me to break free from my comfort zone to really try to find my own happiness, rather than waiting for it to find me as I had done for so much of my life. To this day, I am eternally grateful to my friends for their advice and encouragement (and not just in relationship matters); but moreso I am grateful to my date (nameless here, for her sake) for her acceptance.

We all wear our proverbial masks every time we step into the real world: hiding the reality of who we are and letting everyone see only the selves we wish to show. But with her, it was as though I was reconnecting with an old friend, and that quick rapport of common interests and passion for technology really made that mask unnecessary.

Maybe it started as a simple infatuation; but for so long I had been lost (and often still am) in my own internal dialogue: constantly interpreting, analyzing and over-thinking everything. Ration, logic, and reason seem to work so beautifully for everything else in life; but not for such as these. My choir director so often advises to “listen louder than you sing”; and this seems so apropos to matters of the heart as well. When I asked her out, it was the first time in forever that I had seriously listened to my feelings louder than that dialogue. I broke free of wondering about how the question would be taken, free of the worry of negative result, and just did.

That initial crush is gone; and I am uncertain of the feelings that remain (some emotions defy analysis) …but I do know this: I am, and want to be, a better person by her having been (being?) in my life.

I’ll end this introspection here; I won’t celebrate this anniversary any more than this simple post, because its significance will fade as I grow to understand and embrace these more intricate emotions, bit by bit.

Categories: Life Tags:

## 2013: The Year of One Blog Post

“♫ 525,600 minutes, 525,000 moments so dear…how do you measure a year?”

“Seasons of Love” (Rent)

As I begin writing this, 2013 will end within the hour. I suppose it’s only fitting to post something introspective on my blog, seeing as that’s how I am as a person. It’s hard not to be lost in my own thought sometimes. It’s comfortable in my own mind.

This past year has certainly had its ups and downs. I’ve sadly grown apart from some friends, but have made many new ones. Some of my friends have married or become engaged — and I couldn’t be happier for them! — and from my friends and the friendships we have I’ve learned so much more about others and about myself. I wish I could truly say everything I learned was good; but through good and bad I’ve become a stronger and better person. I’ve had successes which they have celebrated with me, and hardships which they’ve helped me through in more ways than they know. (It is here where I take another swig of this Smirnoff in toast to your awesomeness, friends!)

2013 has certainly had its milestones: first date, first time driving freeways, first metal concert, first time in Europe, first lucid dream, and so on. I could copy/paste so much more from my diary; but suffice to say I’ve made many significant choices about my life and the man I want to be….no, the man I am becoming. In the same vane, I’m sure 2014 will have its own torrent of emotions, insights, and personal growths. I venture into its unknown with an open mind and an eager heart.

So long, 2013 — It’s been a blast!

(Hey, it just ticked midnight! HAPPY NEW YEAR EVERYONE!)

Categories: Life Tags:

## Poem: Insomnia’s Caress

Having trouble sleeping for whatever reason, so I wrote a short freeform poem to help my mind relax a bit. I call it “Insomnia’s Caress.”

Insomnia’s Caress

Rest betrays my weary soul;
Eyes shut, mind wandering.
Finding neither dream nor rest,
I wonder if it occurs at all.

Hours pass, until the dawn
When light breaks through the veil
I awake to the noise of day,
and rise from the embrace of sleep
to face the world, another time anew.

Categories: Life Tags:

## Happy Equestrian New Year (And: Thank You, Lauren Faust & MLP:FiM Team)

Dear Lauren Faust,

Two years ago today aired the series premiere of this show you created: a show that would forever unite millions of fans – both men and women, adults and children alike, myself included – and change their lives for the better. From the inspirational characters to the wonderful artistry, from its excellent musical numbers (thanks Daniel Ingram) to the brilliant voice acting work all around, My Little Pony: Friendship is Magic has touched the hearts of so many, so quickly. And I your fan, cannot thank you enough.

To be honest, if two years ago you’d have told me I’d today be a fan of My Little Pony, I’d probably have laughed and made some snide, sarcastic retort. But since then, these colorful and playful bunch of magic ponies have stolen this geek’s heart.

I must admit feel a bit like Twilight Sparkle in the “Winter Wrap Up” episode: I can’t draw or do anything artistic at all; I am not imaginative enough to make fanfiction; and I won’t even dare to attempt a Pony Music Video or any such feat, since it would probably just be a waste of bandwidth. So how do I be a productive brony? I hope in lieu of anything fancy or creative, these words will suffice to demonstrate some iota of that gratitude.

In closing: Thank you, Lauren Faust. And thank you to the wonderful team of artists, voice actors, musicians, and other staff behind My Little Pony: Friendship is Magic. It’s been an absolutely fantastic first two seasons, and I eagerly await many more. (Oh, lest I forget, and thank you to all the creative fans who make the wait pass by with amazing art, videos, stories, and other media!)

If you would please, a brohoof!

Aleedye’s Brohoof by ~MacchiatoJolt on DeviantArt (CC BY-NC-SA 3.0)

Yours,
Peter Gordon

Categories: Ponies Tags:

## Fedora Planet Feed

Oops. It seems when updating my .planet file earlier, I copy/pasted the feed for my whole blog, and not solely the Fedora category as intended. I’ve adjusted that now, so the next time Fedora People runs it’s feed grabber, you should see this instead of my previous couple of posts (which were both not very Fedora-related). That doesn’t mean I’m any less interesting though — Feel free to read my blog at its full website for more! =)

Categories: Fedora Tags:

## Passion, Not a Ph.D., Makes the Professor

It has been said more than once that a good teacher knows the subject matter, while a great teacher is passionate about it. And nowhere is this statement more true than at a university level, where things like high class sizes, tenure, publications in academic journals, and other research opportunities can often cause professors to become rather lackadaisical about their teaching.

For a lot of professors today, a significant majority of their teaching comes down to their lectures. And while a professor may be very knowledgeable about that subject, if he or she does not know how to lecture well, it is ultimately the students who suffer. I’ve had a physics professor, for example, with a Ph.D in his field and decades of experience under his belt. He’s certainly a well-learned expert; but when he gave lectures, he spoke in a very soft, monotonous voice. And with run-on sentences everywhere, it seemed like he never paused to even take a breath. Not to mention that English wasn’t his native tongue, so he spoke with a slight accent. (This one I can’t fault him for; but that doesn’t mean it detracted from his lectures any less.)

Even when I was sitting in the front and center of the class, it was a mental strain to pay attention. Not only that, but his lectures were almost entirely him simply reading from his Powerpoint presentation, which he would post online after each class meeting. So in addition to the effort required simply to pay attention, there was always the mental trap of just reading those presentations on my own, outside class. Combine all of this, and it’s clear to see to how 9 shots of espresso could not keep me awake in that class. In fact, I had to withdraw from it; and I ended up taking Chemistry instead, the following semesters.

On the other extreme, I’ve had professors who are obviously passionate about their classes — for example, asking questions of the students, providing real-world examples, inviting lots of discussion, and really just having fun talking about the coursework. (And for the more awesome engineering professors, this often involves quite a bit of hilarious puns.) This happens even in more extracurricular things, too.

For instance, as part of my university’s Men’s Chorus, I find myself grinning from ear to ear after every rehearsal. Our conductor is fantastic. He’s so emphatic about the music we sing, and so emotive about everything we do in rehearsals and in our performances, from proper vocal technique to really nailing those more obscure rhythms and lyrics, and sometimes even to life in general. He’s also very encouraging and approachable. He clearly enjoys being at each rehearsal. And he clearly wants us to be better singers, and certainly, better people both on- and off-stage. He sincerely wants our choir to be more than just the sum of its voices — and you can see it in his mannerisms, his energy, the determination of how he teaches. (I suppose that in some ways, it’s a bit of an irony that one of my all-time favorite teachers is in a subject far removed from my majors.)

I hope that my fellow students (and former students) have had professors on that better end of this spectrum. For now, I’ll just have to hope that those end-of-semester evaluations are given their due considerations…

Categories: College Survival Guide Tags:

October 5th, 2012 1 comment

I’ve come to use LaTeX extensively for typesetting my Mathematics homework. It just makes things so beautiful. And like the XHTML/CSS split, it really does encourage strong separation of content from presentation. One of the things I first hated about it though, is that I grew quickly tired of using \left and \right with brackets, parentheses, or anything that needed to be automatically be sized to its content.

And, as any good engineering student might, I sought to encourage my own laziness by finding a clever shortcut. After a few minutes of searching Google, I came across the DeclarePairedDelimiter command in the mathtools package. Its usage is fairly self-explanatory, as I’ll let the following demonstrate.

\usepackage{mathtools}

Then you can create your own paired delimiters with the DeclarePairedDelimiter command:

\DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\norm}{\lVert}{\rVert} \DeclarePairedDelimiter{\innerproduct}{\langle}{\rangle}

Finally, you can use these new commands in math mode to more easily group your expression. (Append an asterisk to make LaTeX automatically insert the necessary \left and \right commands when you need the delimiters to resize according to their content.)

If $$a \lt 0$$ and $$b \gt 0$$, then $$\abs{a} = -a$$, and so $$\abs*{\frac{a}{b}} = -\frac{a}{b}$$.
The norm of a vector $$v$$, denoted $$\norm{v}$$, is defined by $$\norm{v} = \sqrt{\innerproduct{v,v}}$$.

As a nice bonus, it increases readability of the LaTeX source tremendously, since it gives those grouped expressions some semantic name rather than just being a bunch of formatting. For instance with the above delimiters, \left\lvert \frac{a}{b} \right\rvert is more to type, and a lot less obviously an absolute value, than the simpler \abs*{\frac{a}{b}}.

Happy hacking!

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## An Open Letter to Ken Ham (And: Thank You, Bill Nye)

(I don’t normally post rants like this, but every so often I come upon something so frustrating to refrain from doing otherwise. Apologies for the tone.)

This is an open letter to Mr. Ken Ham, and a response to his YouTube video, “Ken Ham Responds to Bill Nye ‘The Humanist Guy’“, since comments are disabled there. Time indices with the quotations are approximate, and mark the end of that quotation in the aforementioned video.

Mr. Ham:

Before I even respond to the video content itself, I want to address one tiny issue: Comments are not enabled on your video. Why would this be? I believe that this is due to the fact that you KNOW in your heart that what you’re saying is neither sensible nor rational; and you simply want to prevent people from flaming you or posting any form of rebuttal to you as a reply. Well, thank goodness people like me have websites where we can post just about anything we want (within legal limits, of course). And here is my rebuttal to you, good sir: YOU ARE AN IMBECILE. Now, let’s examine the content of your video, and I’ll explain to you why I feel this way.

“In fact, Bill Nye doesn’t really understand science.” (0:57, 2:39, 2:42)

Things like this are stated multiple times throughout your video. And I must admit, I am a bit speechless. I don’t even know where to properly begin responding to that. Let me get this straight, then: Someone who doesn’t understand science could earn a Bachelor of Science degree (emphasis mine) from a private, well-reputed, and highly-accredited Ivy-League university? Someone who doesn’t understand science can develop a hydraulic pressure resonance supressor that’s still used in 747s today? Someone who doesn’t understand science can host his own TV show specifically teaching science to a young audience? Someone who doesn’t understand science can make a sundial that also helps with camera color calibration for the Mars probes? Look at Bill Nye’s record through academia and his professional career. OF COURSE that man understands science. He is a scientist. He lives it every day!

“I mean, Bill Nye himself actually is not a scientist. He studied mechanical engineering […]” (1:23)

Well, let’s see here. According to a Google Definition search, the word “scientist” means “A person who is studying or has expert knowledge of one or more of the natural or physical sciences.” So, let’s see if this applies to Bill Nye…is mechanical engineering a natural or physical science? Well, in fact, yes. ME is in essence the science of physical processes and how to apply them for benefits in technology. And does he have expert knowledge of it? Well…he has a Bachelor’s degree from Cornell,  three Honorary Doctorate degrees from Rensselaer Polytechnic Institute, Goucher College, and Johns Hopkins, many years of hands-on experience working for Boeing and as an aeronautics consultant, multiple scientific patents,  and has one of his inventions on the Mars rovers and another in most 747s still in use today (Source). I’d say that qualifies him quite well as an expert. So yeah, he’s an expert in the physical sciences. So, he is a scientist, by your own words. You’re 0-for-2 so far Mr. Ham.

“Bill Nye has an agenda […] to teach them they are the result of evolutionary processes; that they came from slime over millions of years.” (0:53)

“You can divide science into historical science that’s talking about the past, or observational science. That’s the science that builds our technology.” (1:07)

All science is “observational science.” This whole “historical science” talk is nonsense. Bill Nye is a scientist; and science is inherently observational. Science tries to learn and impart knowledge of the truths of the universe; and how do we do this? Through the scientific method, with painstaking attention to detail in reasoning and evidence. We observe what happens in nature and try to understand every tiny piece of the how and the why. And with the multitude of evidence we have for it — fossils, genealogy, tracing disease patterns, microbial growth, etc. — the theory of evolution is quite a sound one. (If you want proper historical science, try a field like paleontology or archaeology.) You’re 0-for-4 so far…

“I mean, the word ‘science’ means ‘knowledge.'” (0:58)

And Mr. Ham scores a point! Oh good. So you are in fact able to produce some iota of reason from that hole in your face.

“He says if you deny evolution to children, they’re going to have problems, because we need engineers. Well wait a minute. Engineering…and evolution? What has evolution got to do with engineering?” (1:19)

I’ll see your irrationality and raise you a “lolwut?”

The problem with denying children evolution is that in so doing, you’re denying them the very concepts of how to think critically. Of how to properly infer knowledge from some data. Of how to support that knowledge with further tests and experiments. Without these critical thinking skills, any attempt at studying engineering will fail before it can even begin. (Heck, I’d wager good money that most of the entire point of engineering is about finding clever solutions to problems in various related fields, like electronics, aerospace, fluid dynamics, architecture, etc.)

“I hope he did not apply any of his evolutionary principles to any of Boeing’s airplanes.” (1:32)

Well, in fact, he did. Not evolutionary processes. But sound technical and scientific principles. And with those he helped develop a hydraulic pressure resonance suppressor for the 747, still in use today.

“I’ll tell you what is real abuse, what is inappropriate for children: When you take generations of kids and you teach them that they’re just animals; [that] there’s no God; you’re a result of millions of years of evolutionary processes.” (2:12)

Call it abuse if you want, but I’ll repeat what I wrote above: Science is about finding and teaching the truths of our universe. Nothing more.

So you then proceed to acknowledge that we can observe and measure radioactivity while at the same time mentioning that bones and fossils aren’t found with photos or timestamps, so we couldn’t know how old they are. But of course, that same radiation-based technology couldn’t, you know, tell us how old those fossils are! (Le’ *gasp*!)

“He doesn’t teach children how to think critically.” (3:06)

Do you actually believe this? Seriously? A scientist who devotes his LIFE to educating the future generations of scientists…doesn’t teach them to think clearly? Are you really this daft?

“If evolution is true, I mean, it’d be so obvious to the kids that it’s true; but it’s not.” (3:29)

Apparently you are that daft! So, let me get this straight…according to you, stuff that’s true is obvious to school children? Hmm…let’s try this. The square of the length of the hypotenuse of a triangle is equal to the sum of the squares of the other two sides after subtracting twice the product of those two side lengths by the cosine of the acute angle between them. (This is the law of cosines, straight from trigonometry.) This is absolutely true; but is it obvious to children? Heck no. In fact, it’s barely clear to the math students when they first see it.

Okay, maybe you’re not so much a fan of mathematics so let’s try another one – this time from the realm of chemistry. The rate of diffusion or effusion of a gas is inversely proportional to the square root of its molecular mass. This is Graham’s Law, straight from a high school or college chemistry course. Again, this has been experimentally shown to be true; but is it obvious to children? Minus a few child prodigies here and there, I’d say this is definitely not an obvious result to kids.

“You have to do what Bill Nye the Humanist Guy wants. You have to protect them from hearing anything about creation. You totally indoctrinate them. You brainwash them. You don’t teach them to think critically at all.” (3:50)

Indoctrination. Brainwashing. Those are some pretty good terms for teaching creationism.

“Isn’t it interesting how Christians are not frightened to teach their children about evolution?” (4:10)

Apparently you’re trying to make them so. And if not, why even have this argument at all?

“[…] and see why you don’t want Bill Nye ‘the Humanist Guy'” teaching your children.” (4:36)

God-willing, I’d love to have Bill Nye teach my kids science. When I was a boy, he was one of a few who really helped inspire in me a deep appreciation for mathematics and sciences, and I would come home from school eager to see another episode of Bill Nye the Science Guy on PBS. (Even nowadays, I try to catch episodes of Bill Nye’s Solving for X when I can. Sure, they’re usually topics I already know quite well, but he covers them in such neat ways, and has very clever and humorous demonstrations to show them.)

So, in closing, I’d like to repeat my opening remark: Mr. Ken Ham, you are an imbecile. It’s people like you that are holding back our society. Holding back our scientific progress. Stop with your lies and stop spreading FUD about things of which you know seemingly so little. Please disconnect your computers entirely from the internet, and make it a better place in so doing.

I would also like to end this open letter by extending a personal debt of gratitude to Bill Nye. If you are reading this, Mr. Nye, thank you for all that you’ve done for science. And thank you for standing up to religious weirdos like Mr. Ham over here. *thumbs up*

Respectfully yours,

Peter Gordon

Categories: Science Tags:

## Cherry-Picking With Git, For Fun And For Profit

(This is more as a reminder to myself in the future, rather than a full-fledged posting.)

As I learn more about Git and how we use it in Fedora packaging, I keep discovering cool new things about it that I love. One particular aspect that I’m quickly growing to adore is the feature to cherry-pick between commits in related branches. Suppose I’m currently working on the master branch of a package which is kept in sync with f16 and f17 branches.

Now, a new version comes up or I make a patch to fix a bug. After committing the changes to master, I can easily copy those commits to the release branches with a simple cherry-pick operation:

$fedpkg switch-branch f17$ git cherry-pick master

This copies the most recent commit from the master branch into the current one. (See the Git documentation for more information on specifying commit IDs.) Compare this to CVS, where we had to manually diff the devel/ branch directory and apply that to each branch by hand…it’s so much nicer with Git.

Then it’s a simple matter to push the changes to the Fedora repository and run builds for them through Koji, by running fedpkg push and fedpkg build in each branch. Sweet!

Also, the fullscreen mode of WordPress’ new post editor is fantastic – I owe someone a beverage for that.

Now to get some sleep, so I can finish this Algorithm Analysis project tomorrow. Ta for now!

Categories: Fedora Tags:

## Bugzilla Spam, Ahoy!

In an effort to organize my Fedora bug squashing, I’ve been going through all the bugs assigned or CC-ed to me on Fedora’s Bugzilla. The first thing I’ve done is to remove myself from the CC and Assignee of many bugs whose packages I no longer maintain or have interest in. Unfortunately, if you’re on these bugs CC-ed or otherwise, you’re about to get an influx of 200+ automated email notifications from bugzilla about these changes. My apologies!

Categories: Fedora Tags:

## CurvyLooks 0.4 RC1: Almost GNOME 3-ready!

After over three years, CurvyLooks has an update! Yes, folks – this time it’s for GNOME 3 support….well, almost. It is based heavily on the Adwaita GNOME theme (in fact, simply copied verbatim in many parts). Unfortunately, I’ve not been able to adequately set the menu item and notebook-tag coloring, so those parts still look a little bit ugly. I’ve been trying to tweak the Adwaita theme on a trial-and-error basis to see how to do this, to no avail. Contributions welcomed!
Also, I need to post more often on here…this I have decided.

Categories: Tags:

## Silence is NOT Golden!

It’s been a long long while (6+ months! Eep.); but to quote the great Monty Python and the Holy Grail, “I’m not dead yet!” – and neither is this blog.

So, what has this geek been up to recently? On the one hand, my search for a decent [part-time] job is…well, it’s still ongoing. On the other, I’ve been very busy with classwork, family obligations, and other real life duties.

In the meantime, I’ve been working on updates to several of my packages in addition to new goodies (such as the successor to the GNOME Music Applet, Panflute). I apologize for the rather lengthy delays in these! But, I should be able to get to all of them sometime within the next several days or so. (Git still puzzles me slightly, so please bare with me as I slowly resume my packaging duties.) And lastly, I would like to shout a huge THANK YOU to all those who have helped triage and maintain my packages during my brief hiatus. Now, to finish this Deluge update…

Categories: Tags:

## Slight E-Mail Hiccup

Having grown more than a little bit tired of how slow Evolution is at filtering emails, I decided to try the new Thunderbird. (I say “new” because I have not used it since it was version 1.5-ish.)

Unfortunately, I accidentally decided to have them both open at the same time – my understanding being that IMAP is okay with multiple simultaneous connections. Apparently this is not the case between Thunderbird and Evolution, as I quickly discovered. Something clashed with something else somewhere along the code, and now much to my dismay my Inbox is empty. (Oops.)

Long story short, if you sent me an email since Saturday morning (which was the last time I thuroughly checked my email) that you need me to read and/or respond to, please resend it.  Thanks.

Categories: Tags:

## ThinkPad T500: Initial Fedora Report – Marvelous!

Well, after receiving my ThinkPad T500, I set about tweaking the pre-installed Windows Vista, and spent most of the weekend attempting to get it dual-booting Vista and Fedora nicely. For better or worse, neither Fedora’s GRUB nor Windows’ BCD bootloader would accept booting the other OS properly. I came to realize that I hadn’t used or needed to use Windows in over 5 years (even the server at work is CentOS), so It’s highly unlikely that I will need it in the foreseeable future. Just in case though, I can always run it in a VM. So, after burning the Product Recovery Discs, I wiped it all and installed Fedora on this wonderful machine.

I configured it with 4 partitions:

1. /boot (about 500 MB)
3. / (FS root, about 35 GB)
4. /home (the remaining ~210 GB)

As far as the hardware goes, everything appears to work out of the (quite literal) box. The processor’s throttling is automatically handled by cpuspeed and ACPI. The LCD was automatically detected at the appropriate resolution (WSXGA+, 1680×1050) and DPI (129), and graphics (integrated Intel GMA X4500HD) work very well – 3D, Compiz, and everything. Virtualization extensions (so-called “VT”) were disabled in the BIOS by default for whatever reason, but enabling them took only a few seconds and KVM works wonderfully (running an Ubuntu 9.04 virtual machine for a Linux class).

The TrackPoint(tm) and TouchPad both work splendidly, although the TrackPoint does take a lot of practice to get comfortable with. I’m tempted to disable the touchpad in the BIOS, since I don’t actually use it for pointing (and I often erroneously swipe it with my palm while using the so-called “nipple mouse” of the TrackPoint); but I do use it for the scrolling, so I’ve not yet found a happy medium. Suggestions appreciated.

The hard disk (Western Digital WDC WD2500BEVS-0) and DVD burner drive (HL-DT-ST DVDRAM GSA-U20N) are surprisingly speedy. I’ve already burned two copies of the Fedora 11 LiveCD and several data discs (backups) with no problems. The Bluetooth works well, and I can easily send and receive files to and from my phone through the included Bluetooth stack. The wifi  (Intel WiFi Link 5300 [AGN]) works beautifully with the included iwlagn driver, enabled by default. Also, NetworkManager makes connectivity trivial. (Thanks, devs!)

The firewire and PC Card slot also seem to work – they are autodetected and drivers loaded. However, as I have no firewire- or PC Card-based devices, I cannot verify their functionality.

I have come to quickly love the GNOME Power Manager stack. According to it, the battery on a full charge should last me about 5 hours. It keeps track of how much charge the battery has, how quickly I’m consuming or charging it depending on if I am plugged into AC power, etc. It even shows me a graph of my recent power history! Wonderful little tool. With the help of Intel’s fantastic PowerTOP utility, I’ve increased that computed estimate to nearly 6 hours. (!)

With regards to multimedia, things worked beautifully without me even trying. The built-in webcam works perfectly with Cheese, and the built-in sound works for both playback (surprisingly loud at maximum volume!) and recording (not fantastic quality, but it’s very good at eliminating surrounding static and other noises). I can’t wait to try video-calling someone with the recent Empathy enhancements!

All of the hotkeys – brightness-switching, volume control, playback/navigation, etc.) work as expected, and ACPI (with GNOME Power Manager) even automatically suspends when I close the lid – wonderful!

I’ve also noticed that the machine stays colder in Fedora than it ever did while running the preinstalled Windows. Maybe this is only subjective though, as I never checked the actual temperatures in Windows. Or perhaps Linux/Fedora is better at staying in deeper C-states for longer intervals. In either case, I can work with it directly on my lap quite comfortably.

One of the main reasons I decided upon the T-series instead of a similar R-series laptop was weight. Even with the battery installed, it is only about 5.5 pounds. (That’s about 2.5 kilograms for the rest of the world who use a less-insane system of measurements.) This is quite comfortable to carry with me around school and work without tiring my shoulder.

The only thing which I’ve been unable to test, aside from the firewire and PC card slots, is the dual-screen capability. According to the included manual, it should be capable of using the DisplayPort interface to attach a secondary screen (such as a projector) and automatically resize appropriately, with [Fn]+[Spacebar]. I suppose that if this does not yet already Just Work(tm), it will be fixed by the time I need to worry about actually using it though.

Overall, I love this laptop, and am extremely happy with my purchase decision. In fact, the only qualm I have with this wonderful computer is that it’s not yet decorated! Oh great lazyweb, doe anyone know where I can get stickers for GNOME, Fedora, Linux/Tux, et al., to adorn it with?

Also: A huge THANK YOU to all of the developers who helped make my first Linux-on-a-laptop experience an absolutely wonderful one. I love it when things work so simply and effectively!

Categories: Tags:

## Fedora 11: Recent X.org/Intel 965 Update Breakage

Those who have compiz (“Desktop Effects”) enabled on 965 GPUs should be very wary of today’s xorg-x11-server-1.6.3-4.fc11 (FEDORA-2009-8766) update. For some reason or another, it makes Compiz prone to crashing when starting up. The solution to this is to revert to 1.4.3-3, or to temporarily disable Compiz. The latter can be done by changing the GConf entry. Switch to a virtual terminal from the GDM login screen by pressing Ctrl + Alt + F2, then login and run the following command:

gconftool-2 --set /desktop/gnome/session/required_components/windowmanager --type string 'metacity'

Again, this does not actually fix the problem in the Xorg/Compiz stack. It instead  disables Compiz entirely (reverting to Metacity – the default GNOME window manager).

This will workaround the bug until it can be solved properly. Note that any attempt to re-enable Compiz once logged in will also recreate this crash. So, it’s best to just leave it off until this is fixed.

Categories: Uncategorized Tags:

## My First Laptop: A ThinkPad

Another summer has passed me by, and I still haven’t learned how to ride a bike nor how to swim properly. Alas, I suppose I’ll get to these eventually. I have kept one of my resolutions though. I finally splurged and bought myself a ThinkPad (specifically, a T500). I decided upon the following specs:

• Intel Core 2 Duo Processor P8400 (2.26GHz 1066MHz 3MBL2) 25W
• Genuine Windows Vista Home Basic (Will be used only to verify hardware functionality. Fedora will replace it thereafter.)
• 15.4″ WSXGA+ TFT, w/ CCFL Backlight (137 DPI…sweet.)
• Intel Graphics Media Accelerator x4500HD with vPro
• 4 GB PC3-8500 DDR3 SDRAM 1067MHz SODIMM Memory (2 DIMM)
• 250 GB Hard Disk Drive, 5400rpm
• DVD Recordable 8x Max Dual Layer, Ultrabay Slim (Serial ATA)
• Integrated Bluetooth PAN
• Intel WiFi Link 5300 (AGN) with My WiFi Technology
• 9 cell Li-Ion Battery

Thankfully, I found an excellent discount through the CPP (Contractor Purchase Program) which cut the cost from nearly $2,200 to just$1,400 (including shipping and taxes, state fees, et al.) – a savings of 40%! The 4-year protection plan was about 30% of the final cost. Well worth it, though.

As it is, this will be a life-safer, as I’ve been having a lot of intermittent hardware issues on my desktop: I get seemingly-random WiFi disconnects; DVD-burning often fails for no discernible reason, and my hard drive is on its last legs, so to speak. (In fact, just today I saw another 12 sectors go bad and need to be reallocated…its count is now at 177. The manufacturer maximum is 188!) – I’m all anxious and impatient now!

So, in the span of one month, I’ve spent almost $4,000 (tuition/registration fees, books/supplies, and ThinkPad). I think that’s quite enough for a while. Categories: Tags: ## Today’s Spanish Lesson, and Inquiry August 12th, 2009 7 comments “Paying attention” is an English idiom which means “to focus one’s attention on” or “to be attentive to.” Like other idioms, it does not translate literally into other languages, especially Spanish. (That is to say, “pagar atención” – from pagar, “to pay,” and (la) atención, “attention,” is incorrect.) However, there are two ways to say it depending on what it is to which is being paid attention. If paying attention to a person, the phrase is “prestar atención,” which in a literal word-for-word translation means “to lend (one’s) attention.” On the other hand, if paying attention to an idea or a thing, one would use “poner atención.” which if taken in a similar literal translation means “to put (or set) attention.” • Correct: Presten atención al maestro cuando les habla. (Pay attention to the teacher when he speaks to you.) • Incorrect: Pongan atención al maestro cuando les habla. Note the different connotations in the following two similar requests made from a a speaker (in our example, a teacher): • Ésto es muy importante. Pongan atención por favor. (“This is very important. Pay attention [to this concept/thing], please.” The teacher is explaining something to his/her students and does not want distractions.) • Ésto es muy importante. Presten atención por favor. (“This is very important. Pay attention [to me], please.” The teacher is demonstrating something, and wants his/her students to watch closely.) I suppose it’s similar, slightly, to how Japanese translate the verb “to have” (or “to be/to exist”) as 有る（ある, “aru”) for objects and concepts but as 居る（いる, “iru”）for living things. Peculiar…but very interesting. Is this difference in Spanish for similar separation of living and non-living things? I wonder…Thanks for the neat Spanish lesson, mom! Categories: Life Tags: ## CSUs to Cut Spring 2010 Admissions July 15th, 2009 2 comments This is absolutely crazy. Someone just posted it to our Facebook group: Lowering Enrollment: Budget Cuts Force California State University to Close 2010 Spring Admissions As part of an overall strategy to address an unprecedented budget reduction of$584 million for 2009-10, California State University campuses will not accept student applications for the 2010 spring term – with very few exceptions.

In addition, quarter campuses that have been accepting admission applications for the 2010 winter term ceased accepting applications as of July 6.

“Only fully-eligible, first-time freshmen, upper-division undergraduate transfers or graduate and post-baccalaureate applicants who have applied for admission prior to July 6 may be offered admission to the 2010 winter term,” said Jeri Echeverria, CSU executive vice chancellor and chief academic officer.

In addition, no admission applications will be accepted for the 2010 spring term at either quarter or semester campuses, for any enrollment category. CSU has typically admitted more than 35,000 freshmen, undergraduate transfer and graduate students during the spring term.

This is just so unbelievably wrong. A strong education is the foundation of a prosperous workforce. I’m all for fixing the economy, but cutting off one of its roots is not the right way to do it! I’m stunned and quite speechless.

Categories: Life Tags:

## Backup Strategies

With my primary hard drive (a three-year old WD Raptor WD740) having been on life support, so to speak, for the last 3 months, I’ve been a lot more diligent about keeping backup copies of my data. Every couple of days, I log out entirely and run a simple rsync script to copy my entire /home directory to a specialized partition on my secondary disk, which I keep at /mnt/backup for simplicity sake.

While its parameter handling can be a bit quirky, I find that it is extremely useful for two reasons: The first more or less negates its quirky parameter handling: Clear and thorough documentation, with lots of example program calls  The second is that it saves me a lot of time in copying the files. Similar to the DeltaRPM feature I raved about with Fedora 11, it copies over only the changed content instead of the entire directory tree. With my home directory at nearly 20 GB, incrementally updating my backup like this prevents a good 90+% of the data from needing to be copied again.

In this way, I know that I have at least two copies of my data at any given time. A major plus to copying the directory tree as-is is that, once the drive does die and I replace it, I merely need to copy it over, without changing anything or unpacking huge tarballs and applying diffs, et al.

The disadvantage to this is that I only have one consistent backup copy of my data at a given time, and that backup is on a hard drive in the same computer. So, should there be a massive system failure of some sort (knock on wood!), then I would lose my data for certain. I also intend to purchase CD-RWs for this purpose – that is, as an additional backup medium – in the near future. But for right now, the second on-disk copy suffices. I also want to setup a RAID system in my next computer build…but that’ll have to wait.

So this simple rsync method, as with any storage decision, has its benefits and downfalls:
Pros:

• Easy to configure;
• Can be automatically run (e.g., in a cron job);
• Updates occur via content deltas, not full copies;
• Backup data is “as-is”, and can be used immediately after copying.

Cons:

• Only one backup copy;
• Physical proximity to original data;
• Requires space for an entire duplicate of the directory tree.

For me, though, this method works out well. Do others have a similar system? Would you suggest any improvements/simplifications? I’d like to hear your thoughts on the matter! Thanks.

Categories: Tags:


text:
# Time-reversal symmetry and fermion parity pumps¶

## Introduction¶

Charles Kane from the University of Pennsylvania will introduce today’s lecture on two dimensional topological insulators with time-reversal symmetry.

## Adding symmetry to a topological insulator¶

In general, there are different approaches to discover new types of topological systems.

We have already used a very powerful method to make a Kitaev chain and the Chern insulator model. We started from guessing what kind of model to use for the edge, such that it is impossible to obtain without the bulk. Then we combined many such edges (dots for the Kitaev chain and wires for the Chern insulator) and tailored the coupling between them to leave exactly the type of model that we want on one edge.

A very skilled researcher in topology (or more specifically K-theory) may also just calculate the expected topological classification of a system starting only from its dimensionality and symmetries. This is also a powerful method, but often it’s too hard and requires a very high skills in math.

Another approach that we can undertake is to start with one topological Hamiltonian and see what happens if we force the Hamiltonian to have some extra symmetry. This is the approach we will use in this chapter.

Let’s start from a simple example involving something which we already know, a quantum dot with the Hamiltonian $$H_0$$. We know that there is a topological invariant, the number of filled energy levels.

Now we can ask what happens if we force the dot to have a particle-hole symmetry. The Hamiltonian becomes

$\begin{split} H_\textrm{BdG} = \begin{pmatrix} H_0 & 0\\ 0 & -H_0^* \end{pmatrix}. \end{split}$

This model is clearly topologically trivial from the point of view of the old invariant, since the number of filled states is constant. However, there are still level crossings that appear. We may ask if these crossings stay protected if we also include a finite superconducting pairing $$\Delta$$ in the Hamiltonian, which couples the two blocks $$H_0$$ and $$-H_0^*$$.

Of course we know the answer: the crossings stay protected due to the change in the Pfaffian invariant. So what we did was to construct a topologically non-trivial superconducting dot by adding particle-hole symmetry to a topological Hamiltonian with a lower symmetry.

Let’s now apply the same logic to a new system. Specifically, let’s add time-reversal symmetry to a Chern insulator. The Chern insulator has chiral edge states whose direction of propagation is flipped by time-reversal symmetry $$\mathcal{T}$$. So let’s consider a Hamiltonian of the form

$\begin{split} H = \begin{pmatrix} H_0 & 0\\ 0 & \mathcal{T}H_0\mathcal{T}^{-1} \end{pmatrix}\,. \end{split}$

If $$H_0$$ is the Hamiltonian of a Chern insulator with $$N$$ edge states, then $$H$$ will have $$N$$ pairs of counterpropagating edge states that transform into each other by time-reversal symmetry. Moreover, the full $$H$$ obeys time-reversal symmetry, which merely exchanges the two blocks.

The following sketch describes the situation in the case $$N=1$$:

The next task which we now face is to understand if such edges stay topologically protected once we add coupling between the two blocks.

## A perfectly transmitted channel and Kramers degeneracy¶

We could try to see if all the edge states can be removed by adding some terms to the Hamiltonian, but instead we will use a closely related fact.

Let’s study transport through such edge states as a function of their total number and let’s only use the fact that time-reversal symmetry is present. Imagine, there is a total of $$N$$ states going in each direction along the edge, and that the edge is composed of a disordered region sandwiched between two clean regions. Again, let’s represent the situation for the case $$N=1$$.

### Scattering states¶

We label incoming states on the left and right with $$\left|n,\textrm{L}\right\rangle$$ and $$\left|n,\textrm{R}\right\rangle$$. The index $$n$$ goes from $$1$$ to $$N$$. The outgoing states are the time-reversed partners of the incoming states, so they are given by $$\mathcal{T}\left|n,\textrm{L}\right\rangle$$ and $$\mathcal{T}\left|n,\textrm{R}\right\rangle$$. Scattering states in the left and right regions are superpositions of incoming and outgoing states,

$\left|\Psi,\textrm{L}\right\rangle = \sum_{n=1}^N \alpha_{n,\textrm{L}}\,\left|n,\textrm{L}\right\rangle + \beta_{n,\textrm{L}}\,\mathcal{T}\left|n,\textrm{L}\right\rangle\,,$
$\left|\Psi,\textrm{R}\right\rangle = \sum_{n=1}^N \alpha_{n,\textrm{R}}\,\left|n,\textrm{R}\right\rangle + \beta_{n,\textrm{R}}\,\mathcal{T}\left|n,\textrm{R}\right\rangle\,.$

We can form vectors out of all the coefficients in the superposition, for instance $$\alpha_\textrm{L} = (\alpha_{1,\textrm{L}},\dots,\alpha_{N,\textrm{L}})^T$$ for the incoming states on the left side. Incoming and outgoing modes are then related by the scattering matrix $$S$$ of the disordered region,

$\begin{split} \begin{pmatrix} \beta_\textrm{L} \\ \beta_\textrm{R} \end{pmatrix} = S \begin{pmatrix} \alpha_\textrm{L} \\ \alpha_\textrm{R} \end{pmatrix}. \end{split}$

There are a total of $$2N$$ incoming and $$2N$$ outgoing modes, so $$S$$ is a $$2N\times 2N$$ matrix. Since we are including all possible initial and final states, $$S$$ is also unitary, $$S=S^\dagger$$. It can be split into reflection and transmission blocks of dimension $$N\times N$$,

$\begin{split} S = \begin{pmatrix} r & t\\ t' & r' \end{pmatrix}\,. \end{split}$

If we can gap out the edges by adding some extra terms to the Hamiltonian, or backscatter them by adding disorder, then we should be able to achieve the situation where there is no transmission at all, $$t = t' = 0$$. In this case, all modes must be reflected back, so the reflection blocks of the scattering matrix become unitary, $$r^\dagger r = r'^\dagger r' = 1$$.

To see whether this is possible at all, we first have to understand the constraints that time-reversal symmetry imposes on $$S$$.

### Scattering matrices with time-reversal symmetry¶

Let’s recall some basic facts about time-reversal symmetry, which we already studied in the first week. Time-reversal symmetry has an antiunitary operator $$\mathcal{T}$$ which commutes with the Hamiltonian. Being antiunitary, $$\mathcal{T}$$ may come in two flavors - either $$\mathcal{T}^2=1$$ or $$\mathcal{T}^2=-1$$. The first case applies to systems with no or integer spin, such that $$\mathcal{T}=\mathcal{K}$$ in the simplest case, where $$\mathcal{K}$$ is the complex conjugation operator. The second case applies to systems with half-integer spin, and in the simplest case we have $$\mathcal{T}=i\sigma_y\mathcal{K}$$.

Let’s apply the time-reversal operator to our scattering states. We get

$\mathcal{T}\left|\Psi,\textrm{L}\right\rangle = \sum_{n=1}^N \alpha^*_{n,\textrm{L}}\,\mathcal{T}\left|n,\textrm{L}\right\rangle + \beta^*_{n,\textrm{L}}\,\mathcal{T}^2\left|n,\textrm{L}\right\rangle\,,$
$\mathcal{T}\left|\Psi,\textrm{R}\right\rangle = \sum_{n=1}^N \alpha^*_{n,\textrm{R}}\,\mathcal{T}\left|n,\textrm{R}\right\rangle + \beta^*_{n,\textrm{R}}\,\mathcal{T}^2\left|n,\textrm{R}\right\rangle\,.$

Now, since time-reversal symmetry does not change the energy of a state, $$\mathcal{T}\left|\Psi,\textrm{R}\right\rangle$$ and $$\mathcal{T}\left|\Psi,\textrm{L}\right\rangle$$ are scattering states with the same energy as $$\left|\Psi,\textrm{R}\right\rangle$$ and $$\left|\Psi,\textrm{L}\right\rangle$$. Hence, the coefficients of incoming and outgoing modes are still related by the same scattering matrix $$S$$ as before. Note, however, that applying $$\mathcal{T}$$ exchanged the role of the $$\alpha$$’s and $$\beta$$’s, such that the $$\alpha$$’s now correspond to outgoing states and the $$\beta$$’s to incoming states. Hence, we have

$\begin{split} S\mathcal{T}^2 \begin{pmatrix}\beta^*_\textrm{L} \\ \beta^*_\textrm{R} \end{pmatrix} = \begin{pmatrix} \alpha^*_\textrm{L} \\ \alpha^*_\textrm{R} \end{pmatrix}\,. \end{split}$

Multiplying both sides by $$\mathcal{T}^2S^\dagger$$ and taking the complex conjugate gives

$\begin{split} \begin{pmatrix} \beta_\textrm{L} \\ \beta_\textrm{R} \end{pmatrix} = \mathcal{T}^2\,S^T \begin{pmatrix} \alpha_\textrm{L} \\ \alpha_\textrm{R} \end{pmatrix}. \end{split}$

By comparing this equation with the one a few lines above, we finally obtain

$S = \mathcal{T}^2 S^T.$

So if $$\mathcal{T}^2=1$$, the scattering matrix is symmetric ($$S=S^T$$), while if $$\mathcal{T}^2=-1$$, it is antisymmetric ($$S=-S^T$$).

What does this imply if we try to set $$t=t'=0$$?

If $$S=S^T$$, it turns out there is really nothing special we can tell. However, if $$S=-S^T$$ and $$t=t'=0$$, the $$N\times N$$ reflection matrix must be both unitary, $$r^\dagger r=1$$, and antisymmetric, $$r=-r^T$$.

If $$N$$ is odd, this isn’t possible at all, since any odd-dimensional antisymmetric matrix must have a single zero eigenvalue, while unitary matrices only have eigenvalues with unit norm!

We are forced to conclude that it is impossible to have $$r$$ unitary, and therefore it is impossible to have $$t=0$$ in this case. Furthermore, this zero eigenvalue of $$r$$ means that there is always a single mode that is transmitted with unit probability.

This is the discovery that Charles Kane described in the introductory video. We can quickly check it by randomly selecting an antisymmetric scattering matrix with odd $$N$$, like the following one with $$N=3$$,

$\begin{split}\begin{pmatrix}0 & 0.38+-0.21i & 0.17+0.39i & 0.24+-0.09i & 0.66+-0.22i & -0.09+0.26i\\-0.38+0.21i & 0 & -0.63+0.03i & -0.05+0.16i & 0.19+-0.34i & 0.46+-0.14i\\-0.17+-0.39i & 0.63+-0.03i & 0 & 0.32+0.21i & -0.48+-0.06i & 0.16+0.06i\\-0.24+0.09i & 0.05+-0.16i & -0.32+-0.21i & 0 & -0.12+-0.31i & -0.8+0.12i\\-0.66+0.22i & -0.19+0.34i & 0.48+0.06i & 0.12+0.31i & 0 & -0.05+0.1i\\0.09+-0.26i & -0.46+0.14i & -0.16+-0.06i & 0.8+-0.12i & 0.05+-0.1i & 0\end{pmatrix}\end{split}$

and looking at the eigenvalues of $$r^\dagger r$$ and $$t^\dagger t$$:

Reflection eigenvalues

$\begin{pmatrix}-0. & 0.77 & 0.77\end{pmatrix}$
Transmission eigenvalues

$\begin{pmatrix}1. & 0.23 & 0.23\end{pmatrix}$

We conclude that if $$\mathcal{T}^2=-1$$ and the number of edge states going in one direction is odd, they cannot be gapped out, and the system is topological. On the other hand, if there is an even number of such edge states, they can be gapped out. Since these are the only two options, the integer invariant of a Chern insulator is reduced to a $$\pm 1$$ invariant in the presence of time reversal symmetry. These topologically protected, counterpropagating edge states are often referred to as helical edge states.

## Helical edge states are Kramers pairs¶

You might ask yourself what makes $$\mathcal{T}^2=-1$$ special, leading to the topological protection of the helical edge states.

As was mentioned in the first week, if $$\mathcal{T}^2=-1$$ then Kramers’ theorem applies. Kramers’ theorem tells us that given an eigenstate $$|\Psi\rangle$$ of the Hamiltonian with energy $$E$$, its time-reversed partner $$|\Psi_\mathcal{T}\rangle\equiv\mathcal{T}|\Psi\rangle$$ has the same energy, and the two states are orthogonal, $$\langle \Psi | \Psi_\mathcal{T}\rangle=0$$. These two states form a so-called Kramers pair. As we already know, this leads to the fact that Hamiltonians with spinful time-reversal symmetry have two-fold degenerate energy levels - Kramers degeneracy.

Now, the two counterpropagating helical modes are time-reversed partners of each other, so they form precisely such a Kramers pair. The condition $$\langle \Psi | \Psi_\mathcal{T}\rangle=0$$ implies that it is impossible to introduce any backscattering between the two states, unless we break time-reversal symmetry. This is the origin of the unit transmission and of the topological protection of helical edge states.

To gain a more intuitive understanding of this fact at a more microscopic level, we can assume that the projection of the electrons’ spin along a given axis is conserved, say the axis $$z$$ perpendicular to the plane. Then at the edge you have, say, a right-moving mode with spin up and a left-moving mode with spin down, and no other modes if $$N=1$$. Let’s draw again the picture of a helical edge state entering the disordered region:

Thus, an electron moving to the right must have spin up by assumption. In order to be reflected, its spin must also be flipped. However, this spin-flip scattering process is forbidden, and again we conclude that the electron is transmitted with probability one.

In the case $$\mathcal{T}^2=1$$, there is no Kramers’ theorem. As a consequence, even though you can construct models which have counterpropagating edge states, you will find that they have no topological protection and can be gapped out without breaking the time-reversal symmetry.

## The quantum spin Hall effect¶

There is no really precise name for the 2D topological insulator with time-reversal symmetry. It is often called “$$\mathbb{Z}_2$$ topological insulator.” However, this simply indicates that there are only two values of the topological invariant, and so it isn’t a very specific name.

The most commonly used name for this system is “quantum spin Hall insulator.” To understand why, let’s analyse a Hall bar made of such a non-trivial insulator. We will only need a Hall bar with four terminals, as shown below:

We have a finite voltage applied to terminal 1, so electrons are injected into the system from there. You can see that because of the helical edge states, there are as many modes connecting terminal 1 to terminal 3 as there are to terminal 4. A moment of thought, or otherwise a quick calculation, should convince you that in this case there is no net current flowing orthogonal to the applied voltage. The Hall conductance is zero, which is the expected result if time-reversal symmetry is preserved, as it is in our system.

However, counterpropagating edge states have to have exactly opposite spin due to Kramers degeneracy. This means that there may be a net spin current across the sample, orthogonal to the applied voltage.

In particular, let’s again make the simple assumption that the spin projection along some axis is conserved. Then, in the figure above, all modes colored in red have spin up, and all modes colored in blue have spin down. So terminal 1 distributes electrons coming out of it according to their spin: all electrons with spin up end up in terminal 4, and all those with spin down in terminal 3. The system has a quantized spin current between terminals 3 and 4, hence the name “quantum spin Hall effect”.

However, the quantized spin Hall current is not a general property of a quantum spin Hall insulator. Here, it arises because we have combined time reversal symmetry with a spin conservation law, and as we learned in the first week, conservation laws are boring from a topological point of view.

#### Consider the simple case where spin is conserved. In the quantum spin Hall bar system above, what happens if, instead of applying a voltage between terminals 1 and 2, you manage to apply a *spin-polarized* current between terminals 1 and 2?

The system will develop an opposite spin-polarized current to compensate the effect.
A spin-polarized current will develop between terminals 3 and 4.
A voltage difference will develop between terminals 3 and 4.
It is impossible to apply such a current unless the bulk gap closes.

## A model for the quantum spin Hall insulator¶

There is an important model which can be used to describe quantum spin Hall insulators, known as the Bernevig-Hughes-Zhang model or, in short, BHZ model. In essence, this model is equivalent to two copies of the Chern insulator Hamiltonian on the square lattice that we studied in the fourth week.

The BHZ Hamiltonian takes the form

$\begin{split} H_\textrm{BHZ}(\mathbf{k}) = \begin{pmatrix} h(\mathbf{k}) & 0 \\ 0 & h^*(-\mathbf{k}) \end{pmatrix}\,, \end{split}$

with

$h(\mathbf{k}) = \epsilon(\mathbf{k}) + \mathbf{d}(\mathbf{k})\cdot \pmb{\sigma}\,.$

Here $$\pmb\sigma = (\sigma_x, \sigma_y, \sigma_z)$$ is a vector of Pauli matrices acting on the electron/hole degree of freedom (the original two bands of the Chern insulator), $$\epsilon(\mathbf{k}) = C - D(k_x^2+k_y^2)$$, the vector $$\mathbf{d} = [A k_x, -A k_y, M(\mathbf{k})]$$, and $$M(\mathbf{k}) = M - B(k_x^2+k_y^2)$$.

You can see that it is basically two copies of the massive Dirac Hamiltonian we used to study Chern insulators. In particular, there is a linear coupling in momentum between the holes and the electrons. The gap in the Hamiltonian is given by the term $$M(\mathbf{k})$$, a momentum-dependent effective mass.

By changing the sign of $$M$$ from negative to positive, you get a gap closing at $$\mathbf{k}=\pmb{0}$$:


text:
## Calculus (3rd Edition)

$$y'+y=x.$$ (Other examples are possible.)
One example is $$y'+y=x.$$ This an example of a linear, non-separable differential equation.


text:
# In Calculas I've learned that 'F means the integral of a function.

In Calculas I've learned that 'F means the integral of a function. SO what does ''F mean, th integral or an integral of a function?

berkeman
Mentor
I hadn't seen that notation before. I suppose it could be the analog of the shorthand notation for derivative?

f = function
f' = derivative of f
f'' = 2nd derivative of f
'f = integral of f
''f = double integral of f

What's a 2nd derivative and double integral?

Line said:
What's a 2nd derivative and double integral?

You didn't learn much in calculus did you? A 2nd derivative is pretty much what it sounds like the derivative of a derivative, and sort of similar for a double integral.

f = x³
f' = 3x²
f'' = 6x

f'' is just the derivative of f'.

Also, f, is the ''f (double intergral) of f'' so if f = 6x then ''f = x³. I hope that wasn't confusing.

Hootenanny
Staff Emeritus
Gold Member
Gelsamel Epsilon said:
Also, f, is the ''f (double intergral) of f'' so if f = 6x then ''f = x³. I hope that wasn't confusing.

''f = x3 +C

If you're asked for the general solution x³ is sufficient. :tongue2:

HallsofIvy
Homework Helper
Hootenanny said:
''f = x3 +C
Actually, if f= 6x, then 'f= 3x2+ C so "f= x3+ Cx+ D.
I'm very puzzled by
Gelsamel Epsilon said:
If you're asked for the general solution x³ is sufficient.
It is precisely when you are asked for the "general solution" that you must have the entire form! x3 does satisfy the conditions but is not the "general solution". Neither is x3+ C. The general function "f such that f= 6x is "f(x)= x3+ Cx+ D.

When I was in calc and I took test the first thing I always did was write +C down the most the right side of the page.

Ha, sorry HallsofIvy I confused myself, I meant if you're asked for "an" (rather then "the") antiderivative then that satisfies.

HallsofIvy


text:
# Changing color of a beamer presentation (Szeged-dove)

I am working on beamer presentation, Szeged theme and dove colortheme. Basically, I'd like to change the color of all the dove-colored lines and the text that's between them to a RGB value set with \definecolor without changing anything else.

Thanks very much in advance for any help!

• Can you maybe ask a separate question about your font problem and include a minimal working example (MWE) of what does not work? It is very hard to diagnose why a code is not working when we don't see the code. – user36296 Oct 24 '18 at 21:24
• I'll try and edit this one to make it clearer but if I don't succed I'll surely write a new one! – lucia de finetti Oct 24 '18 at 21:33
• Please ask for the font problem in a separate question. It is better for the site if a question does not mix different problems. Imagine if some future user also wants to change the colour. Then we can direct him to this question. But if your question contains a mixture if things, it will be more difficult for future users to find a solution. – user36296 Oct 24 '18 at 21:36
• I see, I'll do it now! – lucia de finetti Oct 24 '18 at 21:39

The colour of the lines is controlled by separation line colour. You can change this colour to whatever colour you have defined:

\documentclass{beamer}

\definecolor{mycolor}{RGB}{255,0,0}

\usetheme{Szeged}
\usecolortheme{dove}

\setbeamercolor{separation line}{bg=mycolor}

\title{text}
\author{names}
\institute{institute}

\begin{document}

\section{title}
\begin{frame}
\frametitle{title}
abc
\end{frame}

\end{document}

• Thanks very much! Is there a way so that the text that appears between separation lines (chapter names and dots on the header; title of the presentation and author name on the footer) is of the same color "mycolor" instead of black? – lucia de finetti Oct 24 '18 at 21:32
• @luciadefinetti Sure, you can do things like \setbeamercolor{title in head/foot}{fg=mycolor}. Can you make a minimal working example (MWE) that contains all elements for which you want to change the colour? – user36296 Oct 24 '18 at 21:34
• @luciadefinetti Please see my edited answer. I guessed a few more elements which might be between the lines. – user36296 Oct 24 '18 at 21:44
• @marmot :) Crystal ball factory: \documentclass{standalone} \usepackage{tikzlings} \begin{document} \foreach \x in {1,...,100}{\tikz{\tikzling[crystalball];}} \end{document} – user36296 Oct 24 '18 at 22:22
• @marmot it is not yet on ctan, but I will submit it soon. At the moment it is only available from github.com/samcarter/tikzlings – user36296 Oct 24 '18 at 22:25


text:
## Game

Flowfree is a puzzle game that is available as an android/ios app and online. The game is played on a square grid with $n$ pairs of same-colored squares. The objective is to join each of the same-colored pairs by means of an unbroken chain of squares of the same color. A typical starting position together with the solution is in shown in the figure below:

## IP Formulation

Define the sets $M = 1…m$ and $N = 1…n$, where $m$ is the size of the grid and $n$ is the number of pairs of same-colored cells. Also define variables $x_{ijk}=1$ if cell $(i,j)$ requires color $k$, otherwise $0$, $∀i ∈ M, j ∈ M, k ∈ N$ and parameters $S_{ijk}=1$ if cell $(i,j)$ has color $k$, otherwise $0$. The conditions required by the puzzle are enforced as follows.

1. Ensure the solution is consistent with the starting configuration.

$$x_{ijk} ≥ S_{ijk}, ∀ i ∈ M, j ∈ M, k ∈ N$$

1. Each cell contains a single color.

$$\sum_{k \in N} x_{ijk} = 1, ∀ i ∈ M, j ∈ M$$

1. Ensure a continuous chain for each color. This is done by ensuring that the initially colored squares have exactly one adjacent square of the same color and all other squares will have exactly two adjacent squares of the same color. Define

$$f(x_{ijk}) = \sum_{p=i−1 \\ p \neq i \\ p \in M}^{i+1} x_{pjk} + \sum_{q=j−1 \\ q\neq j \\ q \in M}^{j+1} x_{iqk} + S_{ijk} \\ f(x_{ijk}) ≥ 2x_{ijk}−5(1−x_{ijk}), ∀ i∈M, j∈M, k∈N \\ f(x_{ijk}) ≤ 2x_{ijk}+5(1−x_{ijk}), ∀ i∈M, j∈M, k∈N$$

## Python code using OR-Tools

Here is the code for the above formulation and puzzle:

from ortools.sat.python import cp_model
from enum import IntEnum

class Color(IntEnum):
YELLOW = 0,
BROWN = 1,
GREEN = 2,
BLUE = 3

def puzzle():
m, n = 5, 4
s = {(i,j,k):0 for i in range(m) for j in range(m) for k in range(n)}
s[(0, 1, Color.YELLOW)], s[(4, 0, Color.YELLOW)] = 1, 1
s[(0, 4, Color.BROWN)], s[(1, 1, Color.BROWN)] = 1, 1
s[(3, 2, Color.BLUE)], s[(2, 4, Color.BLUE)] = 1, 1
s[(1, 4, Color.GREEN)], s[(4, 4, Color.GREEN)] = 1, 1
return m, n, s

def flowfree(puzzle):
m, n, s = puzzle
model = cp_model.CpModel()
x = {(i,j,k): model.NewIntVar(0, 1, 'x(%i,%i,%i)' % (i,j,k))
for i in range(m) for j in range(m) for k in range(n)}

for i in range(m):
for j in range(m):
for k in range(n):
model.Add(x[(i,j,k)] >= s[(i,j,k)])

for i in range(m):
for j in range(m):
model.Add(sum(x[(i,j,k)] for k in range(n))==1)

M = list(range(m))
for i in range(m):
for j in range(m):
for k in range(n):
f = s[(i,j,k)] + \
sum(x[(p,j,k)] for p in range(i-1, i+2) if p != i and p in M) + \
sum(x[(i,q,k)] for q in range(j-1, j+2) if q != j and q in M)
model.Add(f >= 2*x[(i,j,k)]-5*(1-x[(i,j,k)]))
model.Add(f <= 2*x[(i,j,k)]+5*(1-x[(i,j,k)]))

solver = cp_model.CpSolver()
status = solver.Solve(model)
if status == cp_model.OPTIMAL or status == cp_model.FEASIBLE:
for i in range(m):
print(" ".join(str(Color(k)).ljust(15) for j in range(m) for k in range(n)
if solver.Value(x[(i,j,k)]) != 0))
else:
print("Couldn't solve.")

flowfree(puzzle())

The output of the above program is as follows:

Color.YELLOW    Color.YELLOW    Color.BROWN     Color.BROWN     Color.BROWN
Color.YELLOW    Color.BROWN     Color.BROWN     Color.GREEN     Color.GREEN
Color.YELLOW    Color.GREEN     Color.GREEN     Color.GREEN     Color.BLUE
Color.YELLOW    Color.GREEN     Color.BLUE      Color.BLUE      Color.BLUE
Color.YELLOW    Color.GREEN     Color.GREEN     Color.GREEN     Color.GREEN


text:
# help with some parts on a circuit board

#### corvairbob

##### New Member
i have a traffic light controller that is a small circuit board maybe 4x4" anyway i was messing around trying to get the controller to work a cross walk light as well anyway i managed to mess one of the light circuits up. so now the yellow light stays on while the green and red keep cycling.

so i can't get parts because the builder scratched the numbers off the parts, my guess was to keep other from building there own. so here are a few photos so someone may be able help me identify them. the output is a t0-220 case device and it has along the bottom on either side of a dot that i think indicates the gate or the collector, it has c3 44 that is all i can get from these and that device has 120vac on both outer pins and the center pin goes to the light and also connects to a resistor and then to a ic module that i think may be a triac anyway when the center pin gets 120vac on it the bulb lights up.

that same center pin also goes to that is module that has an f in the upper right corner it is 6 pins and has a resistor between that module and the output device.

first is the back i put the wire on it to repair the burnt traces the left one is the yellow and that is on all the time if you look at the2nd photo the left and right pins are 120vac all the time and the center one gets 120vac from the white device just below it. based on the f on my is i think it is the last photo a triac. this one is used to power lights and such, but i do not know if the to-220 is a power transistor or a triac. so maybe someone can help me id these 2 devices based on the back side traces.
thanks

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• moc3063-zero-cross-triac-optoisolator-ic-800x800.jpg
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#### Nigel Goodwin

##### Super Moderator
Most Helpful Member
well i'm glad you got your supply but for me i need only a few and i found them on ebay and they say shipped with a 2 week delivery. and as you say they may get here on time and then maybe not. but one seller sent me a note saying not until sept and they show next week in there delivery time frame. so if it fails in that window i will give them neg feedback based on the false delivery date. they should have changed that date then as most other have done already.
thanks.

Yes, keep an eye on it.

In the meantime you could confirm if you even need them? - if a lamp won't light, then putting a short across pins 4 and 6 of the opto, should turn the TRIAC on. If the light is permanently lit, then removing the opto should make it go off.

Anything other than those two results means the triac is faulty - which as I said is a far more likely scenario than the opto.

#### corvairbob

##### New Member
the light stays on i measured voltage at all the pins on the triac and i have 120vac at all 3 triac pins so either the moc3063 is bad or the triac is bad and is applying 120 vac to the gate on the triac. that is why i got both parts. i will just replace both of them and see it it starts to cycle. if not then i will toss it out and get a new one. i could gat a new one now but i want to see if i can fix the old one just because.
thanks.

#### Nigel Goodwin

##### Super Moderator
Most Helpful Member
the light stays on i measured voltage at all the pins on the triac and i have 120vac at all 3 triac pins so either the moc3063 is bad or the triac is bad and is applying 120 vac to the gate on the triac. that is why i got both parts. i will just replace both of them and see it it starts to cycle. if not then i will toss it out and get a new one. i could gat a new one now but i want to see if i can fix the old one just because.
thanks.
Like I said, take the opto out, and see if the light still stays on - if it does the triac is faulty - I wouldn't even think of considering the opto-coupler, they are incredibly reliable. I've repaired hundreds of light units etc. over the decades, and it's pretty well always the triacs - as I said previously, even a bulb failure commonly takes the triac out.

For that matter, as you have two working channels, swap the triacs between one of the good channels and the bad one - but a simple ohms test round the faulty triac should show it's short, compare it to the other two.

For a simple 'comparison', you have a flat tyre on your car - is it the tyre?, or is it the brake disc? - tough decision.

#### corvairbob

##### New Member
ok i went and cut the resistor that goes to the gate on the traic and yes the light stays on so it must not be the opto isolator that failed but the triac as one said may be the defective device. so that is why i have 120vac on all 3 pins so i have both parts in the mail and will install the traic first as i said and it looks like that will be the faulty part and the board will then work again. i will put the opto isolators in the electronics parts bin for some future repairs if needed.

what i was going for is the walk light would go on with the green and then when the yellow and red lights turned on the don't walk would come on. but i have to block the red light on the signal when the yellow light is on and making the don't walk light turn on. so i tried a bridge rectifier and that failed and caused this issue.

red on O = don't walk O
yellow on O = don't walk O
green on O = === walk O

what i tried to do was put a diode in to keep the red light off during the yellow light so the don't walk would turn on and then when the re light cam on i also got the yellow light so i was experimenting. i guess i either get the board that is built for this and be done with it. so if this board fails again i will get the correct board and be done with-it, but for now i have parts in the mail and will repair this board and keep using it as is.

thanks

Last edited:

#### Nigel Goodwin

##### Super Moderator
Most Helpful Member
Cutting the resistor would be one way, but why not just unsolder one end?.

But first measure the triacs on ohms, across the two anodes on each triac. No need to disconnect anything.

Simple way - hold one meter probe on the metal tab, and apply the other probe to the centre pin - if this reads short?, then it's a non-isolated triac, so you can continue. So meter probe on tab, then touch the other probe to the outside two pins in turn - repeat this across all three triacs - they should all read pretty well the same (high, or O/C). If the triac is faulty, it will most likely read low on this test.

#### corvairbob

##### New Member
Cutting the resistor would be one way, but why not just unsolder one end?.

But first measure the triacs on ohms, across the two anodes on each triac. No need to disconnect anything.

Simple way - hold one meter probe on the metal tab, and apply the other probe to the centre pin - if this reads short?, then it's a non-isolated triac, so you can continue. So meter probe on tab, then touch the other probe to the outside two pins in turn - repeat this across all three triacs - they should all read pretty well the same (high, or O/C). If the triac is faulty, it will most likely read low on this test.
cutting the resistor at this time is the fastest and easiest way to test that out. i did it and it worked. i tested the triac and all 3 pins had 120vac on them so i did not know at that time if it was the triac that shorted or maybe the opto that shorted and was turning on the triac. but by just cutting the resistor out of the circuit as a test proved it was the triac. i can re-solder it when i get the new parts. it is fast and easy way to test and saved time taking the board out of the signal and then putting it back in.
thanks.

#### augustinetez

##### Active Member
If the light is permanently lit, then removing the opto should make it go off.
Not necessarily so if it is a sensitive gate Triac, normal static can turn them on - even had a case where the flux wasn't cleaned off properly and it became conductive enough to keep a Triac on.

#### Nigel Goodwin

##### Super Moderator
Most Helpful Member
Not necessarily so if it is a sensitive gate Triac, normal static can turn them on - even had a case where the flux wasn't cleaned off properly and it became conductive enough to keep a Triac on.

It's fed from an opto-triac, so it won't be a sensitive gate version.

#### corvairbob

##### New Member
ok i got the parts installed the parts and connected test bulbs up to the board that bulb came on dim and the resistor started to glow. so i took it apart and tossed it in the trash. i can get a new board for 30$and i have 15$ into parts now. but it was worth a try. thanks guys for the help it was a crap shoot at best without having any real number to go with.

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text:
A classification of Taylor towers of functors of spaces and spectra

Research paper by Gregory Arone, Michael Ching

Indexed on: 07 Nov '14Published on: 07 Nov '14Published in: Mathematics - Algebraic Topology

Abstract

We describe new structure on the Goodwillie derivatives of a functor, and we show how the full Taylor tower of the functor can be recovered from this structure. This new structure takes the form of a coalgebra over a certain comonad which we construct, and whose precise nature depends on the source and target categories of the functor in question. The Taylor tower can be recovered from standard cosimplicial cobar constructions on the coalgebra formed by the derivatives. We get from this an equivalence between the homotopy category of polynomial functors and that of bounded coalgebras over this comonad. For functors with values in the category of spectra, we give a rather explicit description of the associated comonads and their coalgebras. In particular, for functors from based spaces to spectra we interpret this new structure as that of a divided power right module over the operad formed by the derivatives of the identity on based spaces.