text:
# Why is the change in cost wrt bias in neural network equal to error in the neuron?

While reading the book on neural networks by Michael Nielson, I had a problem understanding equation (BP3), which is

$$\frac{\partial C}{\partial b_{j}^{l}}=\delta_{j}^{l} \tag{BP3}\label{BP3},$$ which can be translated to plain English as follows

The change in cost, $$C$$, with respect to the bias of a neuron, $$b_{j}^{l}$$, is equal to the error in that neuron, $$\delta_{j}^{l}$$.

Why would this be the case?

This is just an application of the chain rule. The same chapter has "Proof of the four fundamental equations" section, which proves BP1-2, while PB3-4 are left as exercise to the reader. I agree that it's a good exercise indeed, that's why I encourage you to stop here and try to prove it yourself using a chain rule.

Now, if you decided to read further, here's the sketch of a proof.

Recall equations (25) and (29), both definitions:

• $$z^l$$ is a linear transformation of $$a^{l-1}$$: $$z^l = w^l a^{l-1} + b^l$$

• $$\delta^l$$ is a partial derivative of $$C$$ with respect to $$z^l$$: $$\delta^l = \dfrac{\partial C}{\partial z^l}$$

The chain rule itself:

• the partial derivative of $$C$$ with respect to $$b^l$$:

$$\frac{\partial C}{\partial b^l_j} = \sum_k \frac{\partial C}{\partial z^l_k} \frac{\partial z^l_k}{\partial b^l_j}$$

Almost all elements in this sum will be zero, except for one when $$k=j$$. The first term in it is by definition $$\delta^l_j$$, the second term $$\dfrac{\partial z^l_j}{\partial b^l_j}$$ is $$1$$, because $$z$$ is linear with respect to $$b$$.

The derivative with respect to weights is taken the same way, and differs only in the last term: $$z$$ is linear with respect to $$w$$ as well, but with a coefficient $$a^{l-1}$$.


text:
# On the duality between trees and disks

## David Oury

A combinatorial category Disk was introduced by André Joyal to play a role in his definition of weak $\omega$-category. He defined the category $\Theta$ to be dual to Disk. In the ensuing literature, a more concrete description of $\Theta$ was provided. In this paper we provide another proof of the dual equivalence and introduce various categories equivalent to Disk or $\Theta$, each providing a helpful viewpoint.

Keywords: delta, disk, duality, globular set, omega-category, theta-category, tree

2000 MSC: 18D05, 18D20, 18D35

Theory and Applications of Categories, Vol. 24, 2010, No. 16, pp 418-450.

http://www.tac.mta.ca/tac/volumes/24/16/24-16.dvi
http://www.tac.mta.ca/tac/volumes/24/16/24-16.ps
http://www.tac.mta.ca/tac/volumes/24/16/24-16.pdf
ftp://ftp.tac.mta.ca/pub/tac/html/volumes/24/16/24-16.dvi
ftp://ftp.tac.mta.ca/pub/tac/html/volumes/24/16/24-16.ps

TAC Home

format: plain


text:
# Limitation on irandom() of game maker?

I have three objects. All of them in the create event contain an action each (execute a piece of code):

obj_r01 => r01=0;

obj_r02 => r02=0;

obj_r03 => r03=0;

Image (to clarify what I have tried to explain):

So far so good, but I have one more object that modifies the variables of these other three: obj_control

The obj_control contains two events: create and draw.

crete event (obj_control):

obj_r01.r01=irandom(9);
obj_r02.r02=irandom(9);
obj_r03.r03=irandom(9);

draw event (obj_control):

This event contains three actions, precisely three "execute a piece of code":

piece one:

draw_text(x,y,obj_r01.r01); => applies to obj_r01

piece two:

draw_text(x,y,obj_r02.r02); => applies to obj_r02

piece three:

draw_text(x,y,obj_r03.r03); => applies to obj_r03

The objects are arranged as follows in the room:

In execution this is what happens:

The value of r01 is randomized, but that of r02 and r03 is not. I also tried to modify the obj_control creation event code to:

randomize();
obj_r01.r01=irandom(9);
randomize();
obj_r02.r02=irandom(9);
randomize();
obj_r03.r03=irandom(9);

However the result does not change. With this I could only suspect that the irandom () function can only create a storable random value, would it solve such a problem?

EDIT 1:

I tried putting randomize() only at the beginning of the code as suggested to me by @Philipp and @DH.:

randomize();
obj_r01.r01=irandom(9);
obj_r02.r02=irandom(9);
obj_r03.r03=irandom(9);

But there was no change in the result.

I also did a test to be sure if the value was not even being randomized or if the values of r02 and r03 were being randomized to 0 and for that I modified their values in the creation of each object:

In obj_r02:

r02=10

In obj_r03:

r03=100

The result:

The first value (not surprisingly) was randomized, but the second and third values were not. Thus it is clear that the randomization is not being made or that the draw event is occurring before the radomization of r02 and r03.

• Generally reseeding your randomizer using randomize() before each call will increase correlation between outputs, rather than decrease it. That's because you've tethered them all to be only one shuffle step away from whatever data source is used to drive randomize (generally the system clock). PRNGs are typically optimized to give good decorrelation in depth (sequential results from the same seed) not in breadth (initial results from adjacent seeds - hash functions are better optimized for this use). What happens if you randomize only before the first roll? Mar 30 '17 at 22:31

The random number generator used by GameMaker is not a true random number generator. It is a pseudorandom number generator (or short: PRNG). A PRNG will return a sequence of numbers which look randomly, but are actually a deterministic sequence. The sequence of numbers a PRNG generates depends on its initial state, the so-called seed value. When you initialize a PRNG with the same seed, it will generate the same sequence.

That means that when you want each run of a game to be an unique experience, you need to use a different seed value for the PRNG for each run. As a developer you usually don't care where that value comes from, just that it isn't the same you already used. Game Maker's randomize function has the purpose to re-seed the random number generator with a new seed which is unlikely to be the same as the last time the game was run. I don't know what that function does exactly "under the hood" to achieve that, but it likely calculates the seed value from various entropy sources, like the current time, user input, etc. When you call randomize multiple times, those entropy sources might not have changed, so you re-initialize it again with the same seed. That means the PRNG will return the same sequence.

That means you generally should call randomize() exactly once: when the game starts. Then just call irandom(n) when you need a random number and it should return a different random number each time (or the same number... with a chance of one in n). So your game initialization function would look like this:

randomize();
obj_r01.r01=irandom(9);
obj_r02.r02=irandom(9);
obj_r03.r03=irandom(9);

• So, what should I do to generate three random values so that the three objects display three random values? Mar 31 '17 at 23:57
• Call randomize in the initialization function of your game, and just call obj_r01.r01=irandom(9); obj_r02.r02=irandom(9); obj_r03.r03=irandom(9); where you need them Apr 1 '17 at 4:32
• @BonecoSinforoso answer updated. Apr 1 '17 at 10:32
• I did not get the expected result Apr 1 '17 at 18:25
• @BonecoSinforoso What results did you expect and what results did you get instead? Apr 1 '17 at 18:31

A complement to @DMGregory comment is the real use of randomize().

It's supposed to set the seed, and there is no need to do it more than once as already noted, but also, as said here

Please note, that when using the random number functions in GameMaker: Studio the initial seed is always the same, as this makes tracing errors and debugging far easier. Should you wish to test with true random, you should call this function at the start of your game.

So, not calling will wield the same results over the runs. The best case is to just call once, at the start of the game, and never call it again to make it "truly" random.

• So, what should I do to generate three random values so that the three objects display three random values? I have already excluded two of the three randomize() Apr 1 '17 at 0:02
• @BonecoSinforoso At the create event of an object, run randomize(). Then, on the objects that need to call irandom(), just use it. Nothing more needed to setup.
– DH.
Apr 1 '17 at 3:36
• I did not get the expected result Apr 1 '17 at 18:25


text:
# If $L_1\in R$ and $L_2$ is non-trivial language then $L_1\leq L_2$

Language $$L$$ is trivial if $$L=\varnothing$$ or $$L=\Sigma^*$$. I'm trying to prove the following theorem:

If $$L_1\in R$$ and $$L_2$$ is non-trivial language then $$L_1\leq L_2$$.

If $$L_2$$ is non-trivial the $$L_2\neq\varnothing$$ and $$L_2\neq\Sigma^*$$. If I want to prove that $$L_1\leq L_2$$, then I need to show that there exists $$f:\Sigma^*\to\Sigma^*$$ which is total, can be calculated by a Turing machine and follows: $$\forall x\in\Sigma^*$$, $$x\in L_1$$ iff $$f(x)\in L_2$$. I also know that $$L_1\in R$$ then there is a turing machine $$M_1$$ that decides language $$L_1$$. But how do I continue from here? How $$f$$ should look like?

We know there are two $$x_1,x_2\in \Sigma^*$$ such that $$x_1\in L_2$$ and $$x_2\notin L_2$$, since $$L_2$$ is not trivial. Now, we will build the following reduction:
$$f(x)=\cases{x_1 & x\in L(M_1)=L_1 \\ x_2 & otherwise}$$
Notice that $$f$$ is computable: calculate $$M_1(x)$$, and output either $$x_1$$ or $$x_2$$ according to the answer you got.


text:
# Infinite power series sum $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}$

Using theorems about differentiation or integration of power series calculate infinite sum of

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}$$

The answer should equal to $$\frac{\pi}{2\sqrt3}$$.

I tried using $$f(x) = \arctan(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)}x^{2n+1}$$ with $$x=\frac{1}{3}$$ but that fails, since we have $$3^n$$ and not $$3^{2n+1}$$ in the exponent.

• math.stackexchange.com/questions/1238780/… May 5 '19 at 9:40
• Try a different $x$. May 5 '19 at 9:42
• $3^n=\frac{1}{\sqrt{3}}\cdot \sqrt{3}^{2n+1}$
– Sil
May 5 '19 at 9:46

Factor an $$x$$ out entirely out of the sum, so you’re left with $$x^{2n}$$, then take $$x=\frac1{\sqrt{3}}$$.

Since$$\arctan(x)=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1},$$you have$$\frac\pi6=\arctan\left(\frac1{\sqrt3}\right)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)3^n\sqrt3}$$and therefore$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)3^n}=\frac{\pi\sqrt3}6=\frac\pi{2\sqrt3}.$$


text:
# Multimodality of mixtures of more than two Normal distributions

Let

$$\phi(x;\mu,\sigma) = \frac{1}{\sigma \sqrt{2\pi}} \exp \left(- \frac{(x-\mu)^2}{2\sigma^2}\right)$$

denote the Gaussian density function ($\sigma > 0$). Let

$$f(x) = \sum_{i=1}^N p_i \phi(x;\mu_i,\sigma_i)$$

denote an $N$-component mixture of normal densities, where $p_i\ge0$ and $\sum_i p_i = 1$. Depending on the parameters $p_i,\mu_i,\sigma_i$, the mixture density $f(x)$ can be unimodal, or have more than one peak.

Ref. [1] gives necessary and sufficient conditions for a mixture of two $(N=2)$ normal densities to be unimodal. Have there been any generalizations applicable to mixtures of more than 2 densities $(N > 2)$?

If there are no rigid conditions, at least an approximate condition would satisfy me. I am mainly interested in the parameter regime $0 \le \mu \le 100$, $\sigma \approx \mu / 2$, and $0 \le p_i \le 1$ arbitrary. $N$ can be anything between 1 and 100.

References

[1] Robertson, C. A., and J. G. Fryer. “Some Descriptive Properties of Normal Mixtures.” Scandinavian Actuarial Journal 1969, no. 3–4 (January 1, 1969): 137–46. https://doi.org/10.1080/03461238.1969.10404590.

• You're not going to get any useful closed-form answers that are generally valid. For an approximation, it behooves you to specify the ranges of values of $p_i,$ $\mu_i,$ and $\sigma_i$ for which you need an approximation, because even that is going to get very messy. – whuber Apr 11 '18 at 17:43
• @whuber Say $0 \le \mu \le 100$, $\sigma \approx \mu / 2$, and $0 \le p_i \le 1$ arbitrary. – becko Apr 11 '18 at 18:59


text:
Date and time:
Boyd Room 322

Title: Products in the cohomology of local Hopf algebras

Abstract: This is joint work with Lucho Avramov and Srikanth Iyengar. Suppose that A is a local commutative Hopf algebra over a field of prime characteristic. The question that we investigate is: How does the a change in the coalgebra structure alter the action of the cohomology ring on the cohomology of modules? This question is particularly interesting in the case that A is the group algebra in characteristic p of an abelian p-group. The algebra A can also be viewed as the restricted enveloping algebra of a commutative Lie algebra. But, the coalgebra structures that we would adopt in the tow settings are different. This leads to different actions on the cohomology of modules. We show that the action of a subring of the cohomology ring, that has the same Krull dimension, is independent of the choice of the coalgebra structure.


text:
What does the numerically verified part of the Riemann Hypothesis tell about prime numbers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:41:40Z http://mathoverflow.net/feeds/question/110944 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110944/what-does-the-numerically-verified-part-of-the-riemann-hypothesis-tell-about-prim What does the numerically verified part of the Riemann Hypothesis tell about prime numbers? 36min 2012-10-28T21:57:34Z 2013-02-19T12:41:19Z
I'm curious about the following question:

As of 2005(?) the Riemann hypothesis is verified for the first 10 trillion zeroes, they are all on the critical line. Does this verification gives us any information about prime number?

In particular, are there any results saying if all the non-trivial zeroes whose imaginary part is < N and > 0 are on the critical line, then we understand something about prime number < M, where M is a number depend on N?
 http://mathoverflow.net/questions/110944/what-does-the-numerically-verified-part-of-the-riemann-hypothesis-tell-about-prim/110959#110959 Answer by Stopple for What does the numerically verified part of the Riemann Hypothesis tell about prime numbers? Stopple 2012-10-29T03:42:45Z 2012-10-29T15:16:32Z
The disproof of Mertens' conjecture (cited above) was certainly a computations tour de force using explicit values of the zeros of $\zeta(s)$. Another good example is the paper of Rosser and Schoenfeld "Sharper Bounds for the Chebyshev Functions $\theta(x)$ and $\psi(x)$" Math. Comp., v. 29 1975, pp. 243-269.

We know by the Prime Number Theorem that $\Psi(x)\sim x$. Rosser and Schoenfeld use values of zeros of $\zeta(s)$ to show, for example, that for $\log(x)>105$, we have $|\Psi(x)-x|<x\epsilon(x)$, where, for $X=(\log(x)/9.6459 08801)^{1/2}$ $$\epsilon(x)= 0.257634 \left(1 + \frac{0.96642}{X} \right) X^{3/4}\exp(-X).$$ The paper contains a number of results of this flavor, about the Chebyshev function $\theta(x)$, and about asymptotics of the $n$th prime $p_n$.

The reason it is difficult to convert results about low lying zeros to results about small primes is that the Explicit Formula, (mentioned in comments above) has the primes and zeros lying on opposite sides of a Fourier Transform. The Heisenberg Uncertainty Principle applies

http://en.wikipedia.org/wiki/Fourier_transform#Uncertainty_principle
 http://mathoverflow.net/questions/110944/what-does-the-numerically-verified-part-of-the-riemann-hypothesis-tell-about-prim/111006#111006 Answer by Stopple for What does the numerically verified part of the Riemann Hypothesis tell about prime numbers? Stopple 2012-10-29T15:35:21Z 2012-10-29T15:35:21Z
Another application of the computation of large numbers of Riemann zeros (beyond verification of the Riemann Hypothesis) is towards bounding the deBruijn-Newman constant $\Lambda$:

deBruijn introduced a deformation parameter $t$ in the Riemann $\Xi$ function so that $\Xi_0(x)=\Xi(x)$ and the Riemann zeros $x(t)$ flow according to the "backward heat equation." Together their work shows the existence of a constant $\Lambda$ such that, for $\Lambda\le t$ the function $\Xi_t(x)$ has only real zeros, while for $t<\Lambda$ there exist complex zeros. The Riemann Hypothesis is the conjecture that $\Lambda\le 0$. Newman made the complementary conjecture that $\Lambda\ge 0$, writing "This new conjecture is a quantitative version of the dictum that the Riemann hypothesis, if true, is only barely so." Csordas, Smith, and Varga were able to analyze the ODEs governing the motion of the zeros, and use the fact that an very close pair of zeros, so-called Lehmer pairs, would give a lower bound on $\Lambda.$

The current best bound via this approach, due to Saouter, Gourdon, and Demichel, is that $$-1.14\times 10^{-11}<\Lambda$$ based on a Lehmer pair at height about $7.95\times 10^{12}$
 http://mathoverflow.net/questions/110944/what-does-the-numerically-verified-part-of-the-riemann-hypothesis-tell-about-prim/111012#111012 Answer by Dimitris Koukoulopoulos for What does the numerically verified part of the Riemann Hypothesis tell about prime numbers? Dimitris Koukoulopoulos 2012-10-29T16:37:01Z 2012-10-29T16:37:01Z
If you look at the explicit formula, then you can get a bound for the error term in the PNT: If $$\psi(x) = \sum_{p^k\le x}\log p,$$ then formula (9) in page 109 of Davenport's book (multiplicative number theory) implies that $$\psi(x) = x + \sum_{-T\le \gamma\le T}\frac{x^\rho}{\rho} + O\left(1+\frac{ x\log^2(xT) }{T}\right),$$ for every $T\ge1$, where the implied constant is completely effective. Now, if we know that all the zeroes of $\zeta$ up to height $T$ lie on the critical line, then this automatically implies that $$|\psi(x) - x | \le x^{1/2}\sum_{-T\le \gamma\le T} \frac{1}{\sqrt{1/4+\gamma^2}} + O\left(1+ \frac{ x\log^2(xT)}{T} \right) \ll x^{1/2}\log^2T + \frac{ x\log^2(xT)}{T},$$ for some effective implied constants. So, in certain ranges of $x$, depending on $T$, you can get very good bounds on the size of $\psi(x)$ and therefore on how many primes there are up to $x$.


text:
# Active image of UV/Image Editor

After creating an image using:

bpy.data.images.new(name="NewImage2", width=512, height=512)

How would I set that image to the active image within the UV/Image Editor in Python?

You loop through area till you find the UV/Image editor and set the active image to you image

for area in bpy.context.screen.areas :
if area.type == 'IMAGE_EDITOR' :
area.spaces.active.image = my_img

• Is there a way to do this if that space isn't open? Jul 6 '16 at 7:10
• @user53145 try changing spaces.active to something like spaces["myspace"] Jul 6 '16 at 18:08

Here's the correct way to do it. This will work even if the UV Editor pane isn't open:

bpy.ops.image.new(name='Some Name')
image = bpy.data.images['Some Name']

bpy.data.screens['UV Editing'].areas[1].spaces[0].image = image

• This would only work if you don't change your Blender screen layout at all. @Chebhou's answer is more robust. Jul 22 '16 at 13:48


text:
# Unable to compile with GNU Emacs

I'm trying to use GNU Emacs to write small programs. I use the compile command make -k but i always get the message

make -k
'make' is not recognized as an internal or external command,
operable program or batch file.

I've tried installing MinGW and modifying the path of my environment variables and a lot of other things but it does not work.

-
Since you talk about MinGW I assume you are on MS windows. Its not easy to compile emacs on windows(atleast for a beginner). Its better to use precomipiled binaries. you can get it here. go to the end of the page to get latest version –  kindahero Apr 13 '12 at 5:33
After you change the environment variable PATH to include the 'c:\path\to\mingw\bin' which includes the 'make' program, restart emacs and give it another go. Also. You could evaluate (setenv "PATH" (concat "c:\path\to\mingw\bin;" (getenv "PATH")))' from the scratch to remove the need to restart emacs. If this comment solves your problem I will turn it into an answer. –  kjfletch Apr 13 '12 at 6:45
First verify that you a DOS window can find the make command. This will verify if it is in your PATH environment variable and also installed.
Then, once you know that it is working but still not working from inside of emacs check the values of PATH that emacs is seeing by evaluating (getenv "PATH") in the scratch buffer or via M-:. Also check the value of the exec-path` variable.


text:
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14 Apr 2016, 08:03
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Difficulty:

65% (hard)

Question Stats:

65% (02:32) correct 35% (02:44) wrong based on 985 sessions

### HideShow timer Statistics

Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. The readout indicated which of the following when she had completed 40% of her exercise session?

A. 10 min 48 sec
B. 14 min 52 sec
C. 14 min 58 sec
D. 16 min 6 sec
E. 16 min 12 sec
Veritas Prep GMAT Instructor
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Posts: 8795
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### Show Tags

05 May 2016, 20:22
19
9
CoffeeCreamSugar wrote:
Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. The readout indicated which of the following when she had completed 40% of her exercise session?

A. 10 min 48 sec
B. 14 min 52 sec
C. 14 min 58 sec
D. 16 min 6 sec
E. 16 min 12 sec

Do a more efficient single calculation:

24 min 18 sec is 90% of her exercise time.
When she completes 40% of her exercise session, the readout will indicate 60% of the time.

So we need $$\frac{6}{9} * 24(\frac{18}{60}) = \frac{81}{5} = 16 ( \frac{1}{5}) = 16 (\frac{12}{60}) =$$16 min 12 sec

*Edited a calculation error.
_________________

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14 Apr 2016, 08:13
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13
90% of the exercise should be completed in 24min 18 sec (1458 sec)

When she completes 40% of her exercise timer should show the time remaining for 60% of her exercise.

If 90% requires 1458 sec then 60% requires --> (60 * 1458)/90 = (2/3)*(1458) = 2 * 486 = 972 secs

Required answer = 972/60 = 16 min 12 sec

##### General Discussion
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05 May 2016, 10:01
CoffeeCreamSugar wrote:
Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. The readout indicated which of the following when she had completed 40% of her exercise session?

A. 10 min 48 sec
B. 14 min 52 sec
C. 14 min 58 sec
D. 16 min 6 sec
E. 16 min 12 sec

Let's call 24min and 18sec Y.
24min and 18 sec are: 243/10.
Dara completed 10% of her session, so:
243/10 = 9x/10
x=27 min.
27-Y= 2min+42sec
If Dara completed 40% of her session, the she worked for 4*(2min+42sec) so that the rest of her session will be:
27min-10min&48sec = 16 min 12 sec
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05 May 2016, 19:23
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24.3 minutes/.9=27 minutes=total time of session
(27)(.6)=16.2 minutes=time remaining when session is 40% completed
16.2 minutes=16 min 12 sec
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06 May 2016, 08:03
2
VeritasPrepKarishma wrote:
CoffeeCreamSugar wrote:
Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. The readout indicated which of the following when she had completed 40% of her exercise session?

A. 10 min 48 sec
B. 14 min 52 sec
C. 14 min 58 sec
D. 16 min 6 sec
E. 16 min 12 sec

Do a more efficient single calculation:

24 min 18 sec is 90% of her exercise time.
When she completes 40% of her exercise session, the readout will indicate 60% of the time.

So we need $$\frac{6}{9} * 24(\frac{18}{60}) = \frac{161}{10} = 16 ( \frac{1}{10}) = 16 (\frac{6}{60}) =$$16 min 6 sec

I guess you have a little flaw somewhere. OA is indeed E (16min 12sec), just saying
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08 May 2016, 19:46
broilerc wrote:
VeritasPrepKarishma wrote:
CoffeeCreamSugar wrote:
Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. The readout indicated which of the following when she had completed 40% of her exercise session?

A. 10 min 48 sec
B. 14 min 52 sec
C. 14 min 58 sec
D. 16 min 6 sec
E. 16 min 12 sec

Do a more efficient single calculation:

24 min 18 sec is 90% of her exercise time.
When she completes 40% of her exercise session, the readout will indicate 60% of the time.

So we need $$\frac{6}{9} * 24(\frac{18}{60}) = \frac{161}{10} = 16 ( \frac{1}{10}) = 16 (\frac{6}{60}) =$$16 min 6 sec

I guess you have a little flaw somewhere. OA is indeed E (16min 12sec), just saying

Yes, you are right. Thanks for pointing it out. Edited.
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20 May 2016, 12:03
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1
24 min 18 sec = 1458 sec --> this is 90% of the total workout (w)

.9w = 1458 sec
w = 1620 sec

The readout when she had done 40% of the workout will be 60% of the total

1620*.6 = 972 --> 972/60 = 16 min 12 sec

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06 Jun 2016, 10:13
I did it like this:

If 24 min and 18 seconds is left and 10% is completed then 90% is left.

Therefore, 24 18/60 = 90% x
x = 27 min total time

Then 10% completed = 27 min - 24 min 18 sec = 2 min 42 seconds

Therefore 60% completed = 10% * 6 = 2 min 42 seconds * 6
= (2 + 42/60) * 6 = 162 / 10 = 16 + .2 min = 16 min 20 seconds

= E
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26 Jun 2016, 10:24
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CoffeeCreamSugar wrote:
Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. The readout indicated which of the following when she had completed 40% of her exercise session?

A. 10 min 48 sec
B. 14 min 52 sec
C. 14 min 58 sec
D. 16 min 6 sec
E. 16 min 12 sec

Time remaining when 10% of session completed = 24 mins and 18 seconds.
Let us covert everything in seconds to calculate the total time.

24 mins 18 seconds = 24*60 + 18 = 1440 + 18 = 1458
90% of time = 1458 seconds.

100% of time = 1458*10/9 = 1620 seconds
40% completed = 60% remaining.
Time remaining = 0.6*1620 = 972seconds = 960 + 12 seconds = 16 minutes 12 seconds.

Correct Option: E
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21 Jul 2016, 05:48
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1
I disregarded the minutes and only focused on the seconds since all answer options have a different seconds count.
18 seconds is 90% --> 20 seconds is 100% --> 60% is 12 seconds --> answer E
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22 Jul 2016, 03:21
she completes 90% work in 1458 seconds (24mins and 12 sec)
therefore to complete 60% work (100% Original work - 40% completed)she requires

60*1458= 972 seconds or 16 mins 12 seconds
90
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07 Aug 2016, 03:31
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1
This problem can be done mentally without any calculations

Readout indicated 24 min 18 sec when she had completed 10% of her exercise.

This means remaining 90% of the work will take 24 min 18 sec => 30% will take 8 min 6 sec

so 40% completed i.e. 60% remaining of the work will take (24min 18 sec- 8 min 6 sec)= 16min 12 sec.
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08 Sep 2016, 23:40
X = total time

.9x = 24.3 minutes

x = 24.3/.9
x = 24.3 * 10/9 = 243/9 TOTAL TIME

243/9 * 6/10 = 243/15 = 16 1/5 minutes

16 1/5 minutes = 16 minutes + 12 seconds

Easy math takes 30 seconds.

A very simple conversion to memorize that makes these type of problems super intuitive: .1 hours = 6 minutes. Remembering this equation means you don't need to waste time converting the entire thing to seconds as some have done above.
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19 Sep 2016, 09:15
Used the following method:

24 min 18 sec = 10% Completed = 90% remaining
''x'' = 40% Completed = 60% remaining.

Converting the time to seconds: 1458 = 0.9, x = 0.6. Cross multiplying, 1458 x 6/10 = 9/10(x). Solving for 'x', I get 972 secs which is equal to 16 mins 12 secs.
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06 Dec 2016, 19:57
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convert 24 min 18 sec to min --> 0.18 sec = 0.3 min --> 24.3 min

24.3 min is 90% of the total time --> 24.3 = 0.90x --> x = 27min

Question is asking us what the time will be when the clock she has 60% left to run --> 0.60x27min= 16.2min

0.2x0.60 = 12 sec

E.
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17 Jan 2017, 10:57
I would like an answer from Bunuel right here.
I answered this Q correctly but I dont think that my way of answer is correct.

Let's assume that the 100% of time is an Integer.
Then at 10% of time , x minutes & 42 seconds (as there are 18 seconds remaining) have passed.

So when she will have consumed 40% of her time, there will have passed 4*x minutes + 4*42 seconds.
This solution gives an integer number for minutes for which we do not care about and 4*42 seconds gives us 168 seconds, which is 2 minutes and 48 seconds PASSED in addition to the 4*x minutes.

The only solution that gives us a remainder of X minutes (for which we continue to not care about) and 12 seconds (60-48), is E.

Am I right ? Or I shouldnt have assumed that the total time is an Integer?
Bunuel
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23 Jan 2017, 06:05
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CoffeeCreamSugar wrote:
Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. The readout indicated which of the following when she had completed 40% of her exercise session?

A. 10 min 48 sec
B. 14 min 52 sec
C. 14 min 58 sec
D. 16 min 6 sec
E. 16 min 12 sec

We are given that Dara completed 10% of her exercise session when the readout indicated 24 min and 18 seconds as her time remaining. In other words she had 90% left of her workout at that time. We need to determine the readout after she has completed 40% of her exercise, or when she has 60% left.

We can create a proportion to determine the time when Dara has 60% left. First, let’s convert 24 min 18 sec to minutes.

24 min 18 sec = (24 + 18/60) min = (24 + 3/10) min = 24.3 min

We can also express 90 percent as 90/100 = 0.9 and 60 percent as 60/100 = 0.6. We define x as the number of minutes remaining in her workout when she has 60%, or 0.6, of her workout left to do.

Our proportion is as follows:

0.9/24.3 = 0.6/x

9/243 = 0.6/x

9x = 145.8

x = 145.8/9 = 16.2 min

Thus, x = (16 + 2/10) min = (16 + 12/60) min = 16 min 12 sec

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25 May 2017, 16:40
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Although I usually tend to rely on the algebraic approach described for this question by other posters, I want to provide an additional approach.

We are told that 90% of the remaining time equals 24 min and 18 seconds.

We can divide this one 90% portion into three 30% portions:

24 min divided by 3 = 3 portions of 8 min

18 sec divided by 3 = 3 portions of 6 sec

Since 30% + 30% = 60%, we then can make the sum of two of the portions to estimate the remaining time when Dara has completed 40% of the exercise session (or the remaining 60%):

(2 portions of 8 min = 16 min) + (2 portions of 6 sec = 12 sec)

Total = 16 min and 12 sec. Option E.

Hope it helps.
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20 Sep 2017, 08:51
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poppe50 wrote:
I disregarded the minutes and only focused on the seconds since all answer options have a different seconds count.
18 seconds is 90% --> 20 seconds is 100% --> 60% is 12 seconds --> answer E

This approach is incorrect, it gives the correct answer out of luck only because the 60% of the total minutes is an integer.

If you want you can separate them, but should be like this:

(1) 24 mins in 90%. Means that in 60%: 60*24/90 = 16 mins (simplify to calculate that)
(2) 18 secs in 90%. Means that in 60%: 60*18/90 = 12 secs (also simplifying)

if (1) turned out to be a fraction, then the answer would not just rely on the 12 secs you calculated.

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# Principles on the Effective Prevention and Investigation of Extra-legal, Arbitrary and Summary Executions

Principles on the Effective Prevention and Investigation of Extra-legal, Arbitrary and Summary Executions was recommended by the Economic and Social Council resolution 1989/65 of May 24, 1989.  As per the resolution, governments of the member states shall prohibit by law all extra-legal, arbitrary and summary executions and shall ensure that any such executions are recognized as offenses under their criminal laws, and are punishable by appropriate penalties which take into account the seriousness of such offenses.

Full Text of Principles on the Effective Prevention and Investigation of Extra-legal, Arbitrary and Summary Executions


text:
## Elementary Algebra

Published by Cengage Learning

# Chapter 6 - Factoring, Solving Equations, and Problem Solving - 6.5 - Factoring, Solving Equations, and Problem Solving - Problem Set 6.5 - Page 269: 3

#### Answer

$x^2-10x+25=(x-5)^2$

#### Work Step by Step

RECALL: Perfect square trinomials can be factored using the following formulas: (i) $a^2+2ab+b^2=(a+b)^2$ (ii) $a^2-2ab+b^2=(a-b)^2$ The given perfect square trinomial can be written as: $x^2-2(x)(5)+5^2$ Factor this perfect square trinomial using formula (ii) above where $a=x$ and $b=5$ to obtain: $x^2-2(x)(5)+5^2=(x-5)^2$ Therefore, $x^2-10x+25=(x-5)^2$

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


text:
# Cosmology with Large Scale Structure Probes

Start year
2012
Organizational Unit
Organizing institutions

### Grants related:

General
Description

The Cosmic Microwave Background (CMB) contains the statistical information about the early seeds of the structure formation in our Universe. Its natural counterpart in the local universe is the distribution of galaxies that arises as a result of gravitational growth of those primordial and small density fluctuations. The characterization of the distribution of inhomogeneities at large-scale in the local Universe provides a powerful tool, complementary to the CMB, to determine the origin and the energy content of the Universe, the expansion rate of the Universe during the cosmic history, and the detailed process of formation of the large-scale structures (LSS). The study of the LSS in the coming years will attempt to address the following open questions in cosmology:

What is the dark matter, and which is its detailed contribution to the energy content of the Universe?

What is the dark energy, and how it affects the dynamics of the Universe?

What is the connection between large scale structure and galaxy formation?

Do fundamental constants vary along the history of the Universe?

Is there evidence for primordial non-Gaussianities giving information on the details of the inflationary expansion epoch of the Universe?

In order to contribute to the possible answer to those questions, in this project we will use several large scale structure probes:

The distribution and large-scale clustering of the galaxies, and its evolution with time. The matter power spectrum (P(k)) and the two-point correlation function (ξ(r)) contain certain geometric features associated to some characteristic length-scales in the Universe, as the horizon at matter-radiation equality, or the acoustic horizon at last scattering. In particular, the latter determines the Baryon Acoustic Oscillation (BAO) scale.

The higher order statistics: the three-point statistics characterizes the deviation from Gaussinity and therefore the structure formation through gravitational instability, the galaxy bias, and the primordial non-Gaussianities.

The distribution of the cosmic voids in the Universe. Both the statistics of big voids, as well as the characterization of the void expansion, provides a complementary tool to determine the matter density and the equation of state of the dark energy. Cosmic voids contain information of the higher order statistics of galaxies and can be used to further constrain the BAO scale.

The cosmic web can be used to characterize the formation of structures and relate the large scale structure with galaxy formation processes.

The distribution and abundance of galaxy clusters, as well as the evolution with time. Among other parameters, the cluster mass function depends both on the matter density as well as in the amplitude of the power spectrum. The time evolution of the mass function n(M,z) is also govern by the growth of structures in the Universe, thus being also sensitive to the equation of state of the dark energy.

Principal investigator
Project staff
Collaborators
Dr.
A. Hempel
Dr.
F. Atrio-Barandela
Dr.
S. Patiri
Dr.
A.J. Cuesta Vázquez
Dr.
A.D. Montero Dorta
Dr.
Dr.
M.A. Sánchez Conde
Dr.
S. Gottlaber
Dr.
A. Klypin
Dr.
Sylos-Labini
Dr.
C. Hernández-Monteagudo
Dr.
A. Pérez Garrido
Dr.
A. Díaz Sánchez
Dr.
I. Villo
Dr.
R. Toledo
Dr.
C. Zhao
1. eBOSS: cosmological analysis from the quasar sample. Marcos Pellejero Ibañez and F. S. Kitaura participated in the construction of the likelihood and the cosmological parameter estimation (including as coauthors Kitaura & Pellejero Ibañez: 2018MNRAS.473.4773A).
2. EUCLID: comparison project of mock galaxy catalogue generating codes showing the accuracy and speed of the PATCHY code (including as coauthors Balaguera-Antolínez, Kitaura & Pellejero Ibañez:https://arxiv.org/abs/1806.09497, https://arxiv.org/abs/1806.09477, https://arxiv.org/abs/1806.09499)
3. Development of an accurate Bias mapping method for large scale structure analysis (Balaguera-Antolínez, Kitaura, Pellejero Ibañez et al 2018:https://arxiv.org/abs/1806.05870)
4. Presentation of the UNITSIM project to provide simulations for the theoretical model comparison for DESI and EUCLID (including as coauthors Kitaura & Pellejero Ibañez:http://www.unitsims.org/ https://arxiv.org/abs/1811.02111)
5. Presentation of BARCODE (Bos, Kitaura & Weygaert 2018: https://arxiv.org/abs/1810.05189, http://adsabs.harvard.edu/abs/2018ascl.soft10002B)

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Ferragamo, A. et al.

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• The miniJPAS survey: Photometric redshift catalogue

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• Euclid preparation. XI. Mean redshift determination from galaxy redshift probabilities for cosmic shear tomography

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text:
# Primary variable constraint Dirichlet-type boundary condition

Project file on GitLab

## Problem description

We start with the following parabolic PDE: $$\left( c \rho_R + \phi \frac{\partial \rho_R}{\partial p}\right) \frac{\partial p}{\partial t} - \nabla \cdot \left[ \rho_R \frac{\kappa}{\mu} \left( \nabla p + \rho_R g \right) \right]$$ $$Q_p = 0.$$ where

• $c$ … a constant that characterizes the storage as a consequence that the solid phase is changing
• $\rho_R$ … the density
• $p$ … pressure
• $t$ … time
• $\kappa$ … the intrinsic permeability tensor of the porous medium
• $\mu$ … is the pressure dependent dynamic viscosity
• $Q_p$ … source/sink terms

In order to obtain a unique solution it is necessary to specify conditions on the boundary $\Gamma$ of the domain $\Omega$.

The benchmark at hand should demonstrate the primary variable constraint Dirichlet-type boundary condition. Here, the size of the sub-domain, the Dirichlet-type boundary condition is defined on, is variable and changes according to a condition depending on the value of the primary variable.

$$\Gamma^\ast_D = { x \in \mathbb{R}^d, x \in \Gamma_D, \text{Condition}(p(x)) }$$

## Examples

On the left (x=0) and right side (x=1) of the domain $\Omega = [0,1]^3$ the usual Dirichlet-type boundary conditions are set, i.e., $$p = 1, \quad x=0 \qquad \text{and} \qquad p = 1\quad x=1$$ The initial condition $p_0$ is set to zero. Additionally, a primary variable constraint Dirichlet-type boundary condition (PVCDBC) is specified: $$p = -0.1, \quad \text{for}\quad z = 1 \quad \text{and}\quad p(x,y,1) > 0.$$ At the beginning of the simulation the PVCDBC is inactive. Because of the ‘normal’ Dirichlet-type boundary conditions the pressure is greater than zero after the first time step and the PVCDBC is activated in the second time step. The effect is depicted in the figure:

This article was written by Thomas Fischer. If you are missing something or you find an error please let us know. Generated with Hugo 0.79.0. Last revision: January 27, 2021
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text:
# Collection of useful topics¶

• Exponents
• Logs
• Summations
• Implications
• Proofs
• Set Theory
• Induction

## Rules of exponents¶

\begin{align*} x^m(x^n) &= x^{m+n} & \dfrac{x^m}{x^n} &= x^{m-n} \\ \\ (x^m)^n &= x^{mn} & (xy)^m &= x^m y^m \\ \\ (\dfrac{x}{y})^m &= \dfrac{x^m}{y^m} & \dfrac{1}{x^m} &= x^{-m} \\ \\ \sqrt{x} &= x^{1/2} & \sqrt[m]{x} &= x^{1/m} \\ \\ \sqrt[n]{x^m} &= x^{m/n} & x^{-(m/n)} &= \dfrac{1}{x^{m/n}} \end{align*}

### Exercise¶

Simplify the following, using the rules of exponents.

🐑 $x^4 x^5$

🐑 $x^7 x^{-3}$

🐑 $x^2 x^{\frac{1}{2}}$

🐑 $\dfrac{x^4}{x^7}$

🐑 $\dfrac{x^3}{\sqrt{x}}$

🐑 $(x^4)^{-2}$

## Exponents and Logs¶

Assuming $a,b>0$, $a,b \neq 1$, and $a$ and $y$ are any real numbers:

\begin{align*} a^x(a^y) &= a^{x+y} & \dfrac{a^x}{a^y} &= a^{x-y} & \dfrac{1}{a^x} &= a^{-x} \\ \\ (a^x)^y &= a^{xy} & (ab)^x &= a^x b^x & (\dfrac{a}{b})^x &= \dfrac{a^x}{b^x} \end{align*}

For $a$, $x$, and $y$ positive real numbers, $n$ a real number, and $a \neq 1$:

\begin{align*} \log_a xy &= \log_a x + \log_a y & \log a x^n &= n \log_a x \\ \\ \log_a \dfrac{x}{y} &= \log_a x - \log_a y & \log_a \sqrt[n]{x} &= \dfrac{1}{n} \log_a x \end{align*}

And what base should we use for the log?

Generally use the base $e$.

$e = \lim_{n\to\infty} \bigg( 1 + \dfrac{1}{n} \bigg) ^n \approx 2.71828$

In [19]:
graph('np.exp(x)',-2,10)

In [18]:
graph('np.log(x)',0.01,10)

Why do economists like logs?

• Transform nonlinear functions into linear functions.
• Logs can be intepreted as a percentage change.

### Exercise¶

🐑 Find possible exponential functions for the graphs in Fig. 4 (SHS, page 119).

🐑 With $f(t) = Aa^t$, if $f(t +t^∗) = 2f(t)$, prove that $a^{t^∗} = 2$. (This shows that the doubling time $t^∗$ of the general exponential function is independent of the initial time t.)

🐑 Recall that the doubling time $t^∗$ of an exponential function $f(t) = Aa^t$ is given by the formula $a^{t^∗} =2$. Solve this equation for $t^∗$.

🐑 Prove that $\log_a(x/y) = \log_a x - \log_a y$.

## Summations¶

$$\sum _{i{\mathop {=}}m}^{n}a_{i}=a_{m}+a_{m+1}+a_{m+2}+\cdots +a_{n-1}+a_{n}$$

Identities:

$\sum _{n=s}^{t}C\cdot f(n)=C\cdot \sum _{n=s}^{t}f(n)$, where C is a constant

$\sum _{n=s}^{t}f(n)+\sum _{n=s}^{t}g(n)=\sum _{n=s}^{t}\left[f(n)+g(n)\right]$

$\sum _{n=s}^{t}f(n)-\sum _{n=s}^{t}g(n)=\sum _{n=s}^{t}\left[f(n)-g(n)\right]$

And lots more.

### Exercises¶

Compute the following:

🐑 $\sum_{i=1}^5 i^2$

🐑 $\sum_{k=3}^6 (5k-3)$

🐑 $\sum_{j=0}^2 \dfrac{(-1)^j}{(j+1)(j+3)}$

Write the following sums using summation notation:

🐑 $1 + 3 + 3^2 + 3^3 + ··· + 3^{81}$

🐑 $a^6_i + a^5_i b_j + a^4_i b^2_j + a^3_i b^3_j + a^2_i b^4_j + a_ib^5_j + b^6_j$

🐑 Evaluate the sum $\sum^n_{m=2} \dfrac{1}{(m-1)m}$ by using the identity $\dfrac{1}{(m-1)m} = \dfrac{1}{m-1} - \dfrac{1}{m}$.

## A bit of logic¶

Find a solution for the equation $x + 2 = \sqrt{4 − x}$.

Why did we go wrong?

\begin{align} x+2 &= \sqrt{4-x} &\implies \\ (x+2)^2 &= 4-x &\implies \\ xˆ2 + 4x + 4 &= 4 - x &\implies \\ x^2 + 5x &= 0 &\implies \\ x(x+5) &= 0 &\implies \\ x = 0 \text{ or } x = -5 & & \end{align}

### Some notation to remember¶

$P \implies Q$

• P implies Q
• if P, then Q
• Q is a consequence of P
• P is a sufficient condition for Q

$P \impliedby Q$

• P only if Q
• P is a necessary condition for Q

$P \iff Q$

• P if and only if Q
• P is a necessary and sufficient condition for Q.

### Exercises¶

🐑 SHS Page 65: Exercise 1-8

## Mathematical Proofs¶

Direct vs indirect proof:

• $P \implies Q$ is equivalent to $\text{not } Q \implies \text{not } P$.

If it is raining, the grass is getting wet.

If the grass is not getting wet, then it is not raining.

Use the two methods of proof to prove that $−x2+5x−4>0 \implies x >0$.

## Set Theory¶

Specify all the elements:

$F = \{\text{fish, pasta, omelette, chicken}\}$

Specify property:

$B = \{(x,y): px+qy \leq m, x \geq 0, y \geq 0 \}$

### Set membership¶

$x \in A$ the element $x$ is in $A$.

$A \subseteq B$ subset, all elements of $A$ are in $B$

$A \subset B$ proper subset, all elements of $A$ are in $B$, and $B$ contains at least one element not in $A$

$A \supseteq B$ superset , $A$ contains all elements in $B$

$A \supset B$ proper superset, $A$ contains $B$ as well as at least one element not in $B$.

### Set operations¶

$A \cup B$ The elements that belong to at least one of the sets A and B

$A \cap B$ The elements that belong to both A and B

$A \setminus B$ The elements that belong to A, but not to B

### Exercise¶

Let $A={1,2,3,4,5}$ and $B={3,6}$. Find $A \cup B$, $A \cap B$, $A \setminus B$, and $B \setminus A$.

### Venn Diagrams¶

Graphical representation of three or more sets.

See diagrams on SHS Page 72.

## Homework¶

🐑 SHS Page 126: Exercise 21, 22

🐑 SHS Page 77-78: Exercise 1-10


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# Differential signaling

(Redirected from Differential signalling)
Elimination of noise by using differential signaling.

Differential signaling is a method for electrically transmitting information using two complementary signals. The technique sends the same electrical signal as a differential pair of signals, each in its own conductor. The pair of conductors can be wires (typically twisted together) or traces on a circuit board. The receiving circuit responds to the electrical difference between the two signals, rather than the difference between a single wire and ground. The opposite technique is called single-ended signalling. Differential pairs are usually found on printed circuit boards, in twisted-pair and ribbon cables, and in connectors.

Provided that the source and receiver impedances in a circuit are equal (it is balanced), external electromagnetic interference tends to affect both conductors identically. Since the receiving circuit only detects the difference between the wires, the technique resists electromagnetic noise compared to one conductor with an un-balanced reference (low-Ω connection to ground).

Contrary to popular belief, differential signalling does not affect noise cancellation. Balanced lines with differential receivers will reject noise regardless of whether the signal is differential or single-ended, but since balanced line noise rejection requires a differential receiver anyway, differential signalling is often used on balanced lines. This improves SNR, reduces EMI, and makes the signal more immune to ground currents or differences.[1]

The technique works for both analog signaling, as in balanced audio—and digital signaling, as in RS-422, RS-485, Ethernet over twisted pair, PCI Express, DisplayPort, HDMI, and USB.

### Ground offset tolerance

In a system with a differential receiver, desired signals add and noise is subtracted away.

### Suitability for use with low-voltage electronics

The electronics industry, particularly in portable and mobile devices, continually strives to lower supply voltage to save power and reduce emitted electromagnetic radiation.[citation needed] A low supply voltage, however, reduces noise immunity. Differential signaling helps to reduce these problems because, for a given supply voltage, it provides twice the noise immunity of a single-ended system.

To see why, consider a single-ended digital system with supply voltage ${\displaystyle V_{S}}$. The high logic level is ${\displaystyle V_{S}\,}$ and the low logic level is 0 V. The difference between the two levels is therefore ${\displaystyle V_{S}-0\,\mathrm {V} =V_{S}}$. Now consider a differential system with the same supply voltage. The voltage difference in the high state, where one wire is at ${\displaystyle V_{S}\,}$ and the other at 0 V, is ${\displaystyle V_{S}-0\,\mathrm {V} =V_{S}}$. The voltage difference in the low state, where the voltages on the wires are exchanged, is ${\displaystyle 0\,\mathrm {V} -V_{S}=-V_{S}}$. The difference between high and low logic levels is therefore ${\displaystyle V_{S}-(-V_{S})=2V_{S}\,}$. This is twice the difference of the single-ended system. If the voltage noise on one wire is uncorrelated to the noise on the other one, it takes twice as much noise to cause an error with the differential system as with the single-ended system. In other words, differential signalling doubles the noise immunity.[citation needed]

### Resistance to electromagnetic interference

This advantage is not directly due to differential signaling itself, but to the common practice of transmitting differential signals on balanced lines.[2][3] Single-ended signals are still resistant to interference if the lines are balanced and terminated by a differential amplifier.

## Comparison with single-ended signaling

In single-ended signaling, the transmitter generates a single voltage that the receiver compares with a fixed reference voltage, both relative to a common ground connection shared by both ends. In many instances single-ended designs are not feasible. Another difficulty is the electromagnetic interference that can be generated by a single-ended signaling system that attempts to operate at high speed.[citation needed]

## Uses

The technique minimizes electronic crosstalk and electromagnetic interference, both noise emission and noise acceptance, and can achieve a constant or known characteristic impedance, allowing impedance matching techniques important in a high-speed signal transmission line or high quality balanced line and balanced circuit audio signal path.

Differential pairs include:

Differential pairs generally carry differential or semi-differential signals, such as high-speed digital serial interfaces including LVDS differential ECL, PECL, LVPECL, Hypertransport, Ethernet over twisted pair, Serial Digital Interface, RS-422, RS-485, USB, Serial ATA, TMDS, FireWire, and HDMI etc. or else high quality and/or high frequency analog signals (e.g., video signals, balanced audio signals, etc.).

## Transmission lines

The type of transmission line that connects two devices (chips, modules) dictates the type of signaling. Single-ended signaling is used with coaxial cables, in which one conductor totally screens the other from the environment. All screens (or shields) are combined into a single piece of material to form a common ground. Differential signaling is used with a balanced pair of conductors. For short cables and low frequencies, the two methods are equivalent, so cheap single-ended circuits with a common ground can be used with cheap cables. As signaling speeds become faster, wires begin to behave as transmission lines.

## Use in computers

Differential signaling is often used in computers to reduce electromagnetic interference, because complete screening is not possible with microstrips and chips in computers, due to geometric constraints and the fact that screening does not work at DC. If a DC power supply line and a low-voltage signal line share the same ground, the power current returning through the ground can induce a significant voltage in it. A low-resistance ground reduces this problem to some extent. A balanced pair of microstrip lines is a convenient solution, because it does not need an additional PCB layer, as a stripline does. Because each line causes a matching image current in the ground plane, which is required anyway for supplying power, the pair looks like four lines and therefore has a shorter crosstalk distance than a simple isolated pair. In fact, it behaves as well as a twisted pair. Low crosstalk is important when many lines are packed into a small space, as on a typical PCB.

## High-voltage differential signaling

High-voltage differential (HVD) signaling uses high-voltage signals. In computer electronics, "high voltage" normally means 5 volts or more.

SCSI-1 variations included a high voltage differential (HVD) implementation whose maximum cable length was many times that of the single-ended version. SCSI equipment for example allows a maximum total cable length of 25 meters using HVD, while single-ended SCSI allows a maximum cable length of 1.5 to 6 meters, depending on bus speed. LVD versions of SCSI allow less than 25 m cable length not because of the lower voltage, but because these SCSI standards allow much higher speeds than the older HVD SCSI.

The generic term High-voltage differential signaling describes a variety of systems. Low-voltage differential signaling or LVDS, on the other hand, is a specific system defined by a TIA/EIA standard.


text:
Chapter Chosen

Motion in A Plane

Physics Part I

Book Store

Currently only available for.
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12
Two vectors acting in opposite direction have resultant 7 units and if they act at angle 90° to each other, the resultant is 13 units. Calculate the magnitude of the two vectors.

Let A and B be the magnitude of two vectors  and let A>B.

When  act in opposite direction, the magnitude of resultant is A - B

Now,

A - B  = 7 [Given]              ...(1)

When  act perpendicular to each other, then their resultant is,

...(2)

Now,

∴

and

or                                     ...(3)

On solving equation (1) and (3), we get

A = 12 and B = 5

555 Views

What are the basic characteristics that a quantity must possess so that it may be a vector quantity?

A quantity must possess the direction and must follow the vector axioms. Any quantity that follows the vector axioms are classified as vectors.

814 Views

What is a scalar quantity?

A physical quantity that requires only magnitude for its complete specification is called a scalar quantity.
1212 Views

Give three examples of vector quantities.

Force, impulse and momentum.
865 Views

Give three examples of scalar quantities.

Mass, temperature and energy
769 Views

What is a vector quantity?

A physical quantity that requires direction along with magnitude, for its complete specification is called a vector quantity.
835 Views


text:
## Open book structures on $(n-1)$-connected $(2n+1)$-manifolds

J. Math. Sci. Univ. Tokyo
Vol. 13 (2006), No. 4, Page 439--523.

Neto, Oziride Manzoli; Massago, Sadao ;Saeki, Osamu
Open book structures on $(n-1)$-connected $(2n+1)$-manifolds
We completely classify simple open book structures on $(n-1)$-connected closed $(2n+1)$-dimensional manifolds for $n\ge 4$, $n\ne 7$, and on $(n-1)$-connected rational homology $(2n+1)$-spheres for $n=3,7$, using their algebraic topological invariants. This generalizes some known results about the classification of fibered knots in spheres and the existence of open book structures on manifolds. We also give applications and examples so as to show the effectiveness of our classification.


text:
# The general solutions of ODE for matrix

If $p(x)$ is a polynomial with real coefficients, consider the following ODE:

$$A'(t)=p(A)$$ where $A$ is an $n \times n$ matrix.

My question is, can we have a general solutions for this kind of ODE? (For example, if $n=1$ we can definitely use the separation of variables)

-

The solutions we get in the case $n=1$ tend to be implicit (not solved for $A$), and usually do not exist for all times. If this does not bother you, you can get something similar for general $n$. Indeed, let $F(z)=\int \frac{1}{p(z)}\,dz$ (this makes sense locally; in a larger scale one has to worry about branches of logarithms). Since $A'$ is of the form $p(A)$, it commutes with $A$. This justifies carrying out the usual chain rule: $\frac{d}{dt}\,F(A) = F'(A)\,A' = (p(A))^{-1}\,p(A)=I$. Therefore, $F(A)=tI+F(A_0)$ gives the solution of the IVP $A'=p(A)$, $A(0)=A_0$.
Illustrating on a simple example: $A'=A^2$. Here $F(A)=-A^{-1}$, hence $A(t)=(A_0^{-1}-tI)^{-1}=A_0(I-tA_0)^{-1}$. (The latter version works even if $A_0$ is noninvertible).


text:
# 4. MITgcm Tutorial Example Experiments¶

The full MITgcm distribution comes with a set of pre-configured numerical experiments. Some of these example experiments are tests of individual parts of the model code, but many are fully fledged numerical simulations. Full tutorials exist for a few of the examples, and are documented in sections Section 4.1 - Section 4.14. The other examples follow the same general structure as the tutorial examples, see below. All example experiments are located in subdirectories under the directory verification. A list of additional experiments (i.e, not documented as full tutorials), with brief description, is provided in Section 4.15 and Section 4.16.

Each example experiment directory has the following subdirectories:

• code: contains code specific to the example. At a minimum, this directory includes the following files:

• code/packages.conf: declares the list of packages or package groups to be used. If not included, the default set of packages is located in pkg/pkg_groups. Package groups are simply convenient collections of commonly used packages which are defined in pkg/pkg_groups (see Using MITgcm Packages). Some packages may require other packages or may require their absence (that is, they are incompatible) and these package dependencies are listed in pkg/pkg_depend.

• code/SIZE.h: declares the size of underlying computational grid. This file is compiled instead of the MITgcm repository version model/inc/SIZE.h.

• The code/ directory may include other files and subroutines specific to the experiment, i.e., containing changes from the standard repository version. For example, some experiments contains CPP header options files to enable or disable some parts of the code at compile time; the most common ones would be model/inc/CPP_OPTIONS.h for core model options and «PKG»_OPTIONS.h for individual packages.

• input: contains the input data files required to run the example. At a minimum, the input directory contains the following files:

• input/data: this file, written as a namelist, specifies the main parameters for the experiment.

• input/data.pkg: contains parameters relative to the packages used in the experiment.

• input/eedata: this file contains “execution environment” data. This consists of a specification of the number of threads to use in $$x$$ and $$y$$. For multi-threaded execution,these will be set to numbers greater than 1.

• Forcing and topography file(s), as well as files describing the initial state of the experiment and any other supporting data. Required support files vary from experiment to experiment, depending on the setup.

• results: this directory contains the output file output.txt produced by the simulation example. This file is useful for comparison with your own output when you run the experiment.

• build: this directory is initially empty and should be used to compile the model and generate the executable.

• run: this directory is initially empty and should be used to run the executable. From the (empty) run directory, link files from input using the command ln -s ../input/* ., then execute the file ../input/prepare_run if it exists. If you are running one of the experiment variations, i.e., using input.«OTHER», first link files from input.«OTHER» (running ../input.«OTHER»/prepare_run if it exists) and next link files from input (and run ../input/prepare_run). Following this procedure, file links from input.«OTHER» will NOT be overwritten by identically named files in input.

## The tutorial experiments are as follows:

In directory tutorial_barotropic_gyre: Single layer ocean gyre (barotropic with free-surface), using a Cartesian grid. If you are new to MITgcm, start here, as no prior experience with MITgcm is assumed. Introduces building and running the model, with description of model output files and simple methods to load and plot model output.

In directory tutorial_baroclinic_gyre: Ocean double-gyre using spherical coordinates (i.e., latitude-longitude coordinates) with 15 vertical layers. Second introductory tutorial, assumes you have read through Barotropic Ocean Gyre. Introduces using NetCDF for model output, and how to use the diagnostics package to customize output and its writing frequency.

In directory tutorial_reentrant channel: Reentrant channel in the Southern Ocean with idealized topography, 49 vertical layers. Tutorial compares solution using coarse horizontal resolution, with and without GM parameterization (pkg/gmredi), versus a high-resolution, eddy-permitting configuration. Third and final introductory tutorial, assumes reader is familiar with MITgcm basics described in tutorials Barotropic Ocean Gyre and Baroclinic Ocean Gyre. Also introduces pkg/layers and pkg/rbcs.

In directory tutorial_advection_in_gyre: Short tutorial comparing the results using different advection schemes in a single-layer, ocean double-gyre. Demonstrates the importance of carefully selecting an advection scheme for a specific setup.

In directory tutorial_global_oce_latlon: Coarse resolution (4ox4o) global ocean simulation, using a spherical ocean grid with 15 vertical layers. Monthly climatological forcing of of wind stress, heat and freshwater fluxes is employed, with surface restoring of temperature and salinity. Simulates the large-scale ocean circulation.

In directory tutorial_global_oce_in_p: Global ocean simulation with a similar configuration as Global Ocean Simulation except pressure is used as the vertical coordinate instead of the traditional height coordinate (exploiting MITgcm’s height–pressure coordinate coding isomorphism). In this configuration the model does NOT make the Boussinesq approximation.

In directory tutorial_held_suarez_cs: Simulates (dry) 3-D atmosphere dynamics using Held and Suarez forcing on a (global) cubed sphere grid. The vertical coordinate is a rescaled pressure coordinate ($$p^*$$) with 20 levels; orography is flat. Radiation effects are represented by Newtonian cooling.

In directory tutorial_deep_convection: Non-uniformly surface-forced ocean deep convection in a doubly periodic box. This tutorial showcases MITgcm’s non-hydrostatic capability in a spatially small domain (3 km x 3 km x 1 km deep), exploring the temporal and spatial characteristics of convection plumes as they might exist during a period of oceanic deep convection.

In directory tutorial_plume_on_slope: Non-hydrostatic simulation of a non-rotating gravity plume descending down a continental slope, forced by surface cooling. Model domain is 2-D with open boundaries conditions used in a the deep-water end of the domain.

In directory tutorial_global_oce_biogeo: Global ocean simulation (similar to tutorial Global Ocean Simulation except using 2.8ox2.8o resolution) which includes a dissolved inorganic carbon biogeochemistry model. The biogeochemical model considers the coupled cycles of carbon, oxygen, phosphorus and alkalinity, which are included as passive tracers. A simplified parameterization of biological production is also included.

In directory tutorial_global_oce_optim: This tutorial illustrates the optimization capacity of the MITgcm, running the adjoint of a global ocean simulation (model setup similar to Global Ocean Simulation). This adjoint run optimizes a time-independent surface heat flux (i.e., the control variable) which brings the model climatology closest to observed climatology, using a cost function based on gridpoint error in temperature. TAF and OpenAD adjoint setups.

In directory tutorial_tracer_adjsens: A second experiment demonstrating MITgcm’s adjoint capabilities, here examining the sensitivity of surface outgassing of a passive tracer as a function of tracer injection site location within the ocean interior. The global (4ox4o) ocean setup from tutorial Global Ocean Simulation is again used for this tutorial. TAF and OpenAD adjoint setups.

In directory tutorial_cfc_offline: This tutorial contains an experiment which uses MITgcm in offline mode (i.e., with prescribed ocean dynamics terms, from a prior forward integration of MITgcm), simulating the penetration of CFCs into the ocean interior during the last century. The model domain is global with 2.8ox2.8o resolution.

In directory tutorial_rotating_tank: Laboratory rotating tank simulation, using a cylindrical coordinate system at laboratory scale of 46 cm diameter and 14.5 cm deep. This is a typical laboratory setup for illustrating principles of geophysical fluid mechanics. An annulus of fluid is heated differentially on the interior and exterior walls of the tank.

## 4.15. Additional Example Experiments: Forward Model Setups¶

For many experiments, additional information is provided in a README file located in the respective experiment’s subdirectory.

1. 1D_ocean_ice_column - Oceanic column with seaice on top.

2. adjustment.128x64x1 - Barotropic adjustment problem on latitude-longitude grid with 128x64 grid points (2.8o resolution).

3. adjustment.cs-32x32x1 - Barotropic adjustment problem on cube sphere grid with 32x32 points per face (roughly 2.8o resolution). Also contains a non-linear free-surface adjustment version (input.nlfs).

4. advect_cs - 2-D passive advection test on cube sphere grid (32x32 grid points per face, roughly 2.8o resolution).

6. advect_xz - 2-D (vertical plane) passive advection test on Cartesian grid. Also contains an additional setup using non-linear free-surface with divergent barotropic flow and implicit vertical advection (input.nlfs), and a setup using piecewise quartic (“mono” and “weno” limiter) advection schemes (input.pqm).

7. aim.5l_Equatorial_Channel - 5-level intermediate atmospheric physics, 3-D equatorial channel configuration.

8. aim.5l_LatLon - 5-level intermediate atmospheric physics, global configuration, on latitude-longitude grid with 128x64x5 grid points (2.8o resolution).

9. aim.5l_cs - 5-level intermediate atmospheric physics, global configuration on cube sphere grid (32x32 grid points per face, roughly 2.8o resolution). Also contains an additional setup with a slab-ocean and thermodynamic sea ice (input.thSI).

10. cfc_example - Global ocean with online computation and advection of CFC11 and CFC12.

11. cheapAML_box - Example using cheap atmospheric mixed layer (cheapaml) package.

12. cpl_aim+ocn - Coupled ocean-atmosphere realistic configuration on cubed-sphere cs32 horizontal grid, using intermediate atmospheric physics (pkg/aim_v23) thermodynamic seaice (pkg/thsice) and land packages.

13. deep_anelastic - Convection simulation on a giant planet: relaxes both the Boussinesq approximation (anelastic) and the thin atmosphere approximation (deep atmosphere).

14. dome - Idealized 3-D test of a density-driven bottom current (Denmark Overflow Mixing and Entrainment experiment).

15. exp2 - Old version of the global ocean experiment (no GM, no partial-cells). Also contains an additional setup with rigid lid (input.rigidLid).

16. exp4 - Flow over a Gaussian bump in open-water or channel with open boundaries. Also contains an additional setup using non-linear free-surface (input.nlfs), and a setup using Stevens (1990) [Ste90] boundary conditions (input.stevens).

17. fizhi-cs-32x32x40 - Global atmospheric simulation with realistic topography, 40 vertical levels, a cubed sphere grid and the full atmospheric physics package.

18. fizhi-cs-aqualev20 - Global atmospheric simulation on an aqua planet with full atmospheric physics. Run is perpetual March with an analytical SST distribution. This is the configuration used for the APE (Aqua Planet Experiment) participation experiment.

19. fizhi-gridalt-hs - Global atmospheric simulation Held-Suarez (1994) [HS94] forcing, with the physical forcing and the dynamical forcing running on different vertical grids.

20. flt_example - Example using float package.

21. front_relax - Relaxation of an 2-D ($$y-z$$) ocean thermal front (test of Gent and McWilliams scheme). Also contains additional setups:

• using the boundary-value problem method (Ferrari et al. 2010 [FGNV10]) (input.bvp).

• with mixed-layer eddy parameterization (Ferrari and McWilliams 2008 [FMCD08]) (input.mxl).

22. global_ocean.90x40x15 - Global ocean simulation at 4ox4o resolution. Similar to tutorial_global_oce_latlon, but using $$z^*$$ coordinates with quasi-non-hydrostatic and non-hydrostatic metric terms. This experiment illustrates the use of sbo package. Also contains additional setups:

• using package ggl90 scheme (Gaspar et al. 1990 [GGL90]) with parameterized tidal and wind energy input into vertical mixing (input.idemix).

23. global_ocean.cs32x15 - Global ocean experiment on the cubed sphere grid. Also contains additional setups:

24. global_ocean_ebm - Global ocean experiment on a lat-lon grid coupled to a zonally averaged atmospheric energy balance model. Similar to global_ocean.90x40x15 experiment.

25. global_oce_biogeo_bling - Global ocean biogeochemistry simulation, based on Biogeochemistry Simulation but using package bling instead of the DIC package.

26. global_with_exf - Global ocean experiment (at 4ox4o) on a lat-lon grid using the exf package with exf interpolation. Similar to tutorial_global_oce_latlon experiment. Also contains a secondary setup with yearly exf fields (input.yearly).

27. halfpipe_streamice - Example using package streamice.

28. hs94.128x64x5 - 3-D atmosphere dynamics on lat-lon grid, using Held and Suarez (1994) [HS94] forcing.

29. hs94.1x64x5 - Zonal averaged atmosphere dynamics using Held and Suarez (1994) [HS94] forcing.

30. hs94.cs-32x32x5 - 3-D atmosphere dynamics using Held and Suarez (1994) [HS94] forcing on the cubed sphere, similar to tutorial_held_suarez_cs experiment but using linear free-surface and only 5 levels. Also contains an additional setup with implicit internal gravity waves treatment and Adams-Bashforth 3 (input.impIGW).

31. ideal_2D_oce - Idealized 2-D global ocean simulation on an aqua planet.

32. internal_wave - Ocean internal wave forced by open boundary conditions. Also contains an additional setup using pkg/kl10 (see Section 8.4.5, Klymak and Legg 2010 [KL10]) (input.kl10).

34. isomip - ISOMIP-like setup (Ice Shelf Ocean Model Intercomparison Project experiment 0) including ice-shelf cavities (pkg/shelfice). Also contains additional setups:

• with “htd” (Hellmer’s thermodynamics, Hellmer 1989 [HO89]) (input.htd).

• using package icefront (input.icefront)

• using package OBCS enabled to balance surface mass (freshwater and ice shelf mass flux) input through open boundaries (input.obcs).

35. lab_sea - Regional (2ox2o) Labrador Sea simulation on a lat-lon grid using pkg/seaice. Also contains additional setups:

36. matrix_example - Test of experimental method to accelerate convergence towards equilibrium.

37. MLAdjust - Simple tests of different viscosity formulations in a zonally reentrant, flat-bottom channel. Contains additional setups; see verification/MLAdjust/README for a listing of different viscosity settings in these experiments:

38. natl_box - Eastern subtropical North Atlantic with KPP scheme. Contains additional setup with added tracers (pkg/ptracers) using the package longstep to speed up integration time (input.longstep).

39. offline_exf_seaice - Sea ice on top of oceanic surface layer in an idealized channel. Forcing is computed by bulk-formulae (pkg/exf) with temperature relaxation to prescribed SST (i.e., no momentum timestepping in ocean, so ocean is “offline”, not to be confused with pkg/offline). Also contains additional setups:

40. seaice_itd - Seaice example using ice thickness distribution (ITD); otherwise very similar to offline_exf_seaice. Also contains additional setups; see verification/seaice_itd/README for details of these setups:

41. seaice_obcs - Similar to lab_sea (input.salt_plume) experiment with only a fraction of the domain and open boundary conditions derived from lab_sea experiment. Also contains additional setups:

42. shelfice_2d_remesh - Simple experiment to test (pkg/shelfice) vertical remeshing code in 2-D idealized-geometry setup.

43. short_surf_wave - Short surface wave adjustment (non-hydrostatic) in homogeneous 2-D vertical section ($$x-z$$).

44. so_box_biogeo - Open boundary Southern Ocean box around Drake Passage, using same model parameters and forcing as experiment tutorial_global_oce_biogeo from which initial conditions and open boundary conditions have been extracted. Also contains additional setup using the SolveSAPHE algorithm (Munhoven 2013 [Mun13]) to determine oceanic pH (input.saphe).

45. solid-body.cs-32x32x1 - Solid body rotation test for cube sphere grid.

46. tutorial_deep_convection - Experiment as described in Section 4.8, also contains an additional setup using the Smagorinisky (1963) [Sma63] viscosity scheme (input.smag3d).

47. vermix - Simple test in a small domain (3 columns) for ocean vertical mixing schemes. The standard setup (input) uses the KPP scheme Large et al. (1994) [LMD94]. Also contains additional setups:

Unless stated otherwise, the physical setup of the adjoint run is identical to the forward run, see description above. TAF adjoint setups require building with directory code_ad with input directory input_ad, whereas OpenAD requires directories coad_oad and input_oad respectively.

1. 1D_ocean_ice_column - Based on standard forward experiment, TAF adjoint setup, uses package ecco.

2. bottom_ctrl_5x5 - TAF adjoint test using the bottom topography as the control parameter, uses package ecco.

4. global_ocean.cs32x15 - Based on standard forward experiment, TAF adjoint setup. Also contains additional TAF adjoint setups:

5. global_ocean_ebm - Based on standard forward experiment, TAF adjoint setup.

6. global_oce_biogeo_bling - Based on standard forward experiment, TAF adjoint setup, uses package ecco.

7. global_with_exf - Based on standard forward experiment, TAF adjoint setup.

8. halfpipe_streamice - Based on standard forward experiment, TAF and OpenAD adjoint setups.

9. hs94.1x64x5 - Based on standard forward experiment, TAF and OpenAD adjoint setups.

11. lab_sea - Based on standard forward experiment, TAF adjoint setup, uses package ecco and divided adjoint (DIVA). Also contains additional TAF adjoint setups:

12. obcs_ctrl - Adjoint test using open boundary conditions as control parameters, uses package ecco.

13. offline_exf_seaice - Based on standard forward experiment, TAF adjoint setup. Also contains additional TAF adjoint setup with sea ice thermodynamics-only using pkg/thsice (input_ad.thsice).


text:
When not at work with students, I spend my time in my room either reading, calculating something using pen and paper, or using a computer. I read almost anything: from the pornographic to the profound, although my main interests are mathematics and physics. "When I get a little money I buy books; and if any is left I buy food and clothes." -Erasmus

## Sunday, March 4, 2012

One thing I noticed while talking with some freshmen physics majors is how few of them actually know how to use a physics book. I think it's due to how different high school physics textbooks are from university physics texts-- the high school texts are usually readable on a per chapter basis, as opposed to a text like Young and Freedman's or Resnick's.

It's also due to how low high school physics expectations are here. I recall surviving high school physics without actually learning anything, and doing a cursory reading of my high school physics book when there was a looming exam. So it comes as a surprise to many students how different university physics textbooks are from what they've encountered in high school.

The common reading strategy seems to be memorizing equations or the end of the chapter summary, and so whenever students encounter textbooks without end-of-chapter summaries, they complain how hard it is to read the book. That's one of the most common complaints I encounter whenever I assign  reading from Spacetime Physics.

A similar strategy is looking for boxed equations and then memorizing all of them. When such a student encounters a physics problem, the problem-solving strategy becomes "Try all the equations with the same variables!". What students forget is the boxed equation is actually something that should be understood in context. The surrounding text is there to explain what assumptions underlie a derived result, and once the underlying assumptions are understood, one can also understand the limitations.

One of my favorite ways of teaching students to be sensitive to underlying assumptions is the De Broglie frequency relation $E=hf$. The problem I assign is to find the De Broglie frequency of a free electron in terms of the wavelength. (The mass of the electron is not given within the problem statement but listed on a table of physical constants.) One common mistake is to write, for free particles with nonzero mass, $E=\frac{hc}{\lambda}$; the wrong assumption here is the wavelength frequency relation $\lambda f=c$ which only works for particles with zero mass. To ensure that my students make an effort to understand underlying assumptions, I make multiple choice items with distracters that reproduce their most common mistakes.

So a student who memorizes equations and is not sensitive to underlying assumptions will find the correct answer (which is derivable from invariance of mass) and the wrong one. One student, who merely memorized formulae and then used a trial and error approach, was unpleasantly surprised to find that his trial and error approach generated all the choices in many test items.

Having talked about the wrong way of reading an introductory physics text, what is the correct way? The right way is to read introductory physics texts on a per-section basis. After reading the section, go to the end of the chapter and then try solving all the odd-numbered problems for that section only.

Probem-solving is a sanity check. The only way to find out if you've actually understood what you've read is  to try to use it, and that is what the problems are for. If you can't solve the problems, reread the section (assuming you actually understood the prerequisite knowledge!) and then try to solve the problems again. If it still doesn't help, try talking it over with classmates, a tutor, or with your professor. Bring your problem solving attempts, and then try to find out why you get the wrong answers-- finding out why you're making mistakes is also part of the learning process.

Obviously, this is not something you can cram, and it means doing the reading and problem-solving on a regular basis. But if you do it right, your understanding will last a lot longer than the formula-memorizers. As a bonus, when faced with a new technical situation, you can build on what you know, and eventually become the expert in your field.

#### 1 comment:

Steve Solun said...

Dear Students,

Let me share with you the post about physics:


text:
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In the following question there are sentences that form a paragraph. Identify the sentence(s) or part(s) of the sentence(s) that is/are correct in terms of grammar and usage (including spelling, punctuation and logical consistency). Then choose the most appropriate option.

1. In $1989,$ a poor Bavarian immigrant named Levi Strauss
2. landed in San Fransisco, California,
3. at the invitation of his brother-in-law David Stern
4. owner of dry goods business.
5. This dry goods business would later became known as Levi Strauss & Company.
1. B only
2. B and C
3. A and B
4. A only
5. A, B and D