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# The Z train leaves station A moving at a constant speed, and

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Manager
Joined: 10 Sep 2012
Posts: 150
Followers: 2

Kudos [?]: 113 [1] , given: 17

The Z train leaves station A moving at a constant speed, and [#permalink]

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30 Oct 2012, 21:45
1
KUDOS
3
This post was
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Difficulty:

25% (medium)

Question Stats:

78% (02:48) correct 22% (02:46) wrong based on 183 sessions

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The Z train leaves station A moving at a constant speed, and passes by stations B and C, in this order. It takes the Z train 7 hours to reach station B, and 5 additional hours to reach station C. The distance between stations A and B is m kilometers longer than the distance between stations B and C. What is the distance between stations A and C in terms of m?

A. 1.8m
B. 6m
C. 7m
D. 9m
E. 12m

This question was difficult for me. I actually did the problem exactly the way the "alternative method" suggests but did not find an answer that worked. I picked m=7. my distance from b to c was 3. Therefore my distance from a to b was 10. So the total distance from a to c is 13. My goal is 13, with an m of 7. B does not fit this.... why does this not work? It seems like plugging in is failing me here.

[Reveal] Spoiler:
Correct.
According to this answer, the distance between A and C is 6m. If the distance between A and C is 6m, using our plug in of m=3 the distance is 18. Equate 2s+3 (the distance between A and C according to our table) with 18:

2s+3=6m=18

--> 2s=18-3=15 /:2

--> s=7.5

Use s=7.5 to calculate v, by applying the Distance = Speed×Time formula on the second row:

v∙5=7.5 \:5

---> v=1.5

check if v=1.5 fits the first row:

1.5∙7= 7.5+3= 10.5

7∙1.5 is indeed 10.5, so 6m makes for a velocity that fits both rows.
Therefore, this answer choice is correct.

Alternative method:

Plug in a number for the distance and find the resulting m, rather than the other way around.

For example, plug in a distance of 12 km A-C. Since the speed is constant, divide the distances of A-B and B-C as 7 km and 5 km, respectively, making m equal 7-5=2 additional kilometers in the first leg. The question asks for the value of the distance A-C, which we denoted as 12 km, and the m we plug in is m=2: only answer B will match your goal of 12 km.
[Reveal] Spoiler: OA
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Kudos [?]: 472 [4] , given: 23

Re: The Z train leaves station A moving at a constant speed, and [#permalink]

### Show Tags

30 Oct 2012, 21:57
4
KUDOS
anon1 wrote:
The Z train leaves station A moving at a constant speed, and passes by stations B and C, in this order. It takes the Z train 7 hours to reach station B, and 5 additional hours to reach station C. The distance between stations A and B is m kilometers longer than the distance between stations B and C. What is the distance between stations A and C in terms of m?

1.8m
6m
7m
9m
12m

This question was difficult for me. I actually did the problem exactly the way the "alternative method" suggests but did not find an answer that worked. I picked m=7. my distance from b to c was 3. Therefore my distance from a to b was 10. So the total distance from a to c is 13. My goal is 13, with an m of 7. B does not fit this.... why does this not work? It seems like plugging in is failing me here.

[Reveal] Spoiler:
Correct.
According to this answer, the distance between A and C is 6m. If the distance between A and C is 6m, using our plug in of m=3 the distance is 18. Equate 2s+3 (the distance between A and C according to our table) with 18:

2s+3=6m=18

--> 2s=18-3=15 /:2

--> s=7.5

Use s=7.5 to calculate v, by applying the Distance = Speed×Time formula on the second row:

v∙5=7.5 \:5

---> v=1.5

check if v=1.5 fits the first row:

1.5∙7= 7.5+3= 10.5

7∙1.5 is indeed 10.5, so 6m makes for a velocity that fits both rows.
Therefore, this answer choice is correct.

Alternative method:

Plug in a number for the distance and find the resulting m, rather than the other way around.

For example, plug in a distance of 12 km A-C. Since the speed is constant, divide the distances of A-B and B-C as 7 km and 5 km, respectively, making m equal 7-5=2 additional kilometers in the first leg. The question asks for the value of the distance A-C, which we denoted as 12 km, and the m we plug in is m=2: only answer B will match your goal of 12 km.

The reason it is failing for you is that you chose incorrect numbers. If the question says it took 7 hrs to reach from A to B and 5 hrs to reach from B to C at a constant speed. It shows that distance AB and BC should be in ratio of 7/5.

If you take such numbers you can solve problem.
AB = 7, BC=5
Therefore
AB-BC = 2

But from question, AB-BC =m
=> m=2

Now total distance = AB+BC= 12
Substitute 12 to get answer in terms of m
Total distance =12 =6m

Ans B

So you get right answer not by plugging in numbers but by 'plugging in right numbers'

Hope it helps and if does, kudos is right there <<
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Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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23 Feb 2013, 08:03
2
KUDOS
Time taken from A to B = 7 hours

Time taken from B to C = 5 hours

Distance from A to B is $$m$$ miles more than distance between B to C. Since the average speed is constant, time taken to travel $$m$$ miles is 7-5 = 2 hours.

Total time taken for the entire trip (A to C) = 7 + 5 = 12hours

in 12 hours you can travel $$\frac{12}{2}*m$$ miles = 6m miles.
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Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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29 Mar 2013, 01:43
2
KUDOS
The above methods are much faster, but I did it the long way.

From the question distance from A to B is "m" miles longer.

Distance from B to C = x
Distance from A to B = x + m

since speed is constant you have two equations equivalent to each other:

Speed = $$\frac{distance}{time}$$

So: $$\frac{x+m}{7}$$ = $$\frac{x}{5}$$

7x = 5x + 5m
x = 2.5m

Add all the distance together = x + x + m = 2.5m + 2.5m + m = 6m, Answer B
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Kudos [?]: 10505 [4] , given: 206

Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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29 Mar 2013, 03:33
4
KUDOS
Expert's post
anon1 wrote:
The Z train leaves station A moving at a constant speed, and passes by stations B and C, in this order. It takes the Z train 7 hours to reach station B, and 5 additional hours to reach station C. The distance between stations A and B is m kilometers longer than the distance between stations B and C. What is the distance between stations A and C in terms of m?

A. 1.8m
B. 6m
C. 7m
D. 9m
E. 12m

This question was difficult for me. I actually did the problem exactly the way the "alternative method" suggests but did not find an answer that worked. I picked m=7. my distance from b to c was 3. Therefore my distance from a to b was 10. So the total distance from a to c is 13. My goal is 13, with an m of 7. B does not fit this.... why does this not work? It seems like plugging in is failing me here.

[Reveal] Spoiler:
Correct.
According to this answer, the distance between A and C is 6m. If the distance between A and C is 6m, using our plug in of m=3 the distance is 18. Equate 2s+3 (the distance between A and C according to our table) with 18:

2s+3=6m=18

--> 2s=18-3=15 /:2

--> s=7.5

Use s=7.5 to calculate v, by applying the Distance = Speed×Time formula on the second row:

v∙5=7.5 \:5

---> v=1.5

check if v=1.5 fits the first row:

1.5∙7= 7.5+3= 10.5

7∙1.5 is indeed 10.5, so 6m makes for a velocity that fits both rows.
Therefore, this answer choice is correct.

Alternative method:

Plug in a number for the distance and find the resulting m, rather than the other way around.

For example, plug in a distance of 12 km A-C. Since the speed is constant, divide the distances of A-B and B-C as 7 km and 5 km, respectively, making m equal 7-5=2 additional kilometers in the first leg. The question asks for the value of the distance A-C, which we denoted as 12 km, and the m we plug in is m=2: only answer B will match your goal of 12 km.

Use a variable, plug in numbers or simply look at the big picture:

Draw the diagram as you read the question.

A.....................7 hrs......................B.................5 hrs............C

A to B the distance is m km extra and the train takes 2 hrs extra to cover m kms. Then, what is the speed of the train? It is (m/2) kms/hr
So the distance between A and C = Speed* Time = (m/2)*12 = 6m
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Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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26 Nov 2013, 10:12
1
KUDOS
anon1 wrote:
The Z train leaves station A moving at a constant speed, and passes by stations B and C, in this order. It takes the Z train 7 hours to reach station B, and 5 additional hours to reach station C. The distance between stations A and B is m kilometers longer than the distance between stations B and C. What is the distance between stations A and C in terms of m?

A. 1.8m
B. 6m
C. 7m
D. 9m
E. 12m

This question was difficult for me. I actually did the problem exactly the way the "alternative method" suggests but did not find an answer that worked. I picked m=7. my distance from b to c was 3. Therefore my distance from a to b was 10. So the total distance from a to c is 13. My goal is 13, with an m of 7. B does not fit this.... why does this not work? It seems like plugging in is failing me here.

[Reveal] Spoiler:
Correct.
According to this answer, the distance between A and C is 6m. If the distance between A and C is 6m, using our plug in of m=3 the distance is 18. Equate 2s+3 (the distance between A and C according to our table) with 18:

2s+3=6m=18

--> 2s=18-3=15 /:2

--> s=7.5

Use s=7.5 to calculate v, by applying the Distance = Speed×Time formula on the second row:

v∙5=7.5 \:5

---> v=1.5

check if v=1.5 fits the first row:

1.5∙7= 7.5+3= 10.5

7∙1.5 is indeed 10.5, so 6m makes for a velocity that fits both rows.
Therefore, this answer choice is correct.

Alternative method:

Plug in a number for the distance and find the resulting m, rather than the other way around.

For example, plug in a distance of 12 km A-C. Since the speed is constant, divide the distances of A-B and B-C as 7 km and 5 km, respectively, making m equal 7-5=2 additional kilometers in the first leg. The question asks for the value of the distance A-C, which we denoted as 12 km, and the m we plug in is m=2: only answer B will match your goal of 12 km.

Let's say 'r' is the constant rate
Then we have Distance from A to B = 7r
Also distance from B to C = 5r

The difference between the distanced 7r - 5r = m
2r = m

Total distance is 12r (7r+5r)
So in terms of m---> 12 (1/2) = 6m

Hope it helps
Cheers
J
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Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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07 Dec 2015, 03:20
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Re: The Z train leaves station A moving at a constant speed, and [#permalink]

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07 Dec 2015, 14:02
d=distance from A to B
d-m=distance from B to C
2d-m=distance from A to C
d/7=(d-m)/5
d=7m/2
2d-m=(14m/2)-m=6m
Re: The Z train leaves station A moving at a constant speed, and   [#permalink] 07 Dec 2015, 14:02
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text:
Problems with undefined control sequence in list of figures

I'm using TeXLive 2011. I had a lot of problems with compiling my large document yesterday and finally seemed to have gotten back to a working version, however now I get strange error messages in the list of figures like

[5 ./Diss.lof:61: Undefined control sequence. ...skip \abovedisplayskip \let \sml@listi =\leftmargin \leftmargini ... l.61 ...S.~391]{Johannaber2004}}}{49}{figure.2.10}

or

[5 ./Diss.lof:61: Undefined control sequence. ...skip \abovedisplayskip \let \sml@listi =\leftmargin \leftmargini ... l.61 ...te [S.~6ff]{Starke2004}}}{51}{figure.2.10}

The corresponding code in the lof file looks like that:

\contentsline {figure}{\numberline {\relax 2.10}{\ignorespaces Ursachen f\"ur Ma\IeC {\ss }abweichungen des Spritzgie\IeC {\ss }werkzeugs beim Kunststoffspritzguss nach~\textcite [S.~6ff]{Starke2004}}}{51}{figure.2.10}

I'm sorry, that at the moment I can not provide a minimal example, can someone explain, what this undefined control sequences are about? It seems to be a cite command in a caption.

EDIT

after Herbert's suggestion, I modified the source in the following way and removed all aux, lof, lot etc. files:

\caption[pure text without any command]{text with~\protect\textcite[][]{}}

But the problem persists:

[5 ! Undefined control sequence. ...skip \abovedisplayskip \let \sml@listi =\leftmargin \leftmargini ... l.61 ...te [S.~6ff]{Starke2004}}}{51}{figure.2.10}

I'm using the scrbook class (komascript). I'll try to set up a minimal example asap.

EDIT: a MWE

\documentclass{article}
\usepackage{biblatex}
\bibliography{biblatex-examples}
\begin{document}
\begin{figure}
\caption[Bar \protect\textcite[][]{aksin}]{Foo \protect\textcite[][]{aksin}}
\end{figure}
\listoffigures
\end{document}

resulting in:

(./x.aux) (./x.bbl) (./x.lof
! Argument of \blx@citeargs@i has an extra }.
<inserted text>
\par
l.2 ...line {1}{\ignorespaces Bar \textcite [}}{1}

?

where the lof file is this:

\defcounter {refsection}{0}\relax
\contentsline {figure}{\numberline {1}{\ignorespaces Bar \textcite [}}{1}
\boolfalse {citerequest}\boolfalse {citetracker}\boolfalse {pagetracker}\boolfalse {backtracker}\relax

-
You should at least tell the class you're using. –  egreg Jan 11 '12 at 11:52
@egreg: sorry, I'll edit the question –  MostlyHarmless Jan 11 '12 at 12:06

use \protect\textcite{} in your caption or better

\caption[entry for the lof without (!) \textcite]{... \protect\textcite{...} ...}

and delete the *.lof file for the next run otherwise you'll get always the same error again.

-
thank you. I'm just confused, because I've been using cite and citet and also citeauthor for a long time in captions without any problem (and without \protect). Is there any difference between them and \textcite (biblatex package)? –  MostlyHarmless Jan 11 '12 at 11:31
the \protect does not solve the problem, see my edit –  MostlyHarmless Jan 11 '12 at 11:43
yes, I deleted all auxiliary files before the next runs. The problem occurs in the 2nd pdflatex run. –  MostlyHarmless Jan 11 '12 at 12:08
delete the cite command, delete all auxiliary files and run pdflatex twice. Did you get the error? –  Herbert Jan 11 '12 at 12:11
That line cannot be the problem and there is no other way than creating a minimal example. Move the \end{document} up until the error is gone. –  Herbert Jan 11 '12 at 13:19

after long testing, comparison of version history etc. I finally found that the error seemed to come from a completely different piece of code somewhere else (in the preamble, where I had not expected it):

I wanted to change the size of fixme´ annotations tosmall and tried like that

\usepackage[draft, silent=true]{fixme}
\fxsetup{theme=color, inline, targetlayout=color, author=}
\newcommand{\N}[1]{\mbox{}\fxfatal*[inline=true,nomargin]{\small\sffamily/#1/}{}}
`

Apparently that led to mysterious problems like the one above.

-

format: plain


text:
4

# Evaluate the integral f Jr Ie dV where R is the region between the sphere of radius and the sphere of radius centered at the oriigin and above the x-Y plane_...

## Question

###### Evaluate the integral f Jr Ie dV where R is the region between the sphere of radius and the sphere of radius centered at the oriigin and above the x-Y plane_

Evaluate the integral f Jr Ie dV where R is the region between the sphere of radius and the sphere of radius centered at the oriigin and above the x-Y plane_

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text:
# Long and Short of R in an Hour

Tasneem Zaihra
Monday, Feb 27, 2017

### Introduction to R

• R is a language and environment for statistical computing and graphics. It includes

• an effective data handling and storage facility,
• a suite of operators for calculations on arrays, in particular matrices,
• a large collection of intermediate tools for data analysis,
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### Introduction to RStudio

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### Creating Dynamic Reports

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For more details on authoring R presentations please visit https://support.rstudio.com/hc/en-us/articles/200486468.

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### Sample Slide with R Code

summary(cars)

     speed           dist
Min.   : 4.0   Min.   :  2.00
1st Qu.:12.0   1st Qu.: 26.00
Median :15.0   Median : 36.00
Mean   :15.4   Mean   : 42.98
3rd Qu.:19.0   3rd Qu.: 56.00
Max.   :25.0   Max.   :120.00


text:
# binary matrix multiplication

(the YouTube link opens the exact time he mentions this). Can one build a "mechanical" universal Turing machine? I paste a clear screenshot of the frame below. The matrix product is designed for representing the composition of linear maps that are represented by matrices. Welcome to The Multiplying Binary Numbers (Base 2) (A) Math Worksheet from the Multiplication Worksheets Page at Math-Drills.com. 2. In order to get the resulting multiplication value, enter the two binary numbers in each respective field and then clicking on the calculate button shows the output. Introduction With artiﬁcial neural networks (ANNs) becoming more and more powerful and with the slowdown of complementary metal–oxide–semiconductor (CMOS) scaling, the Von Neumann Viewed 1k times 3 $\begingroup$ I am following an academic lecture on coding theory. Binary multiplication is actually much simpler to calculate than decimal multiplication. They just spelled out all of the details. Authors: CNP Slagle, Lance Fortnow. It's as A.Sh says, being a binary matrix means that the elements are in $\mathbb F_2$ and thus $1=-1$. As stated on the figure, I do not understand how such a simplification (indicated by the red arrow) can be done. @Qwertylicious: If the question you're asking is how do you get $$v_0 + u_0 + u_2 + u_3 = 0$$ that's because $$v_0 = u_0 + u_2 + u_3$$ so $$v_0 + u_0 + u_2 + u_3 = (u_0 + u_2 + u_3) + u_0 + u_2 + u_3$$ which simplifies to zero, since the coefficients get reduced, mod 2. Binary matrix multiplication: finding the number of ones. Multiplication in binary is exactly as it is in decimal, i.e. The given program is compiled and executed successfully. &v_0 = u_0 + u_2 + u_3\\[4pt] What happens if you neglect front suspension maintanance? Vote. In-Memory Binary Vector–Matrix Multiplication Based on Complementary Resistive Switches Tobias Ziegler, Rainer Waser, Dirk J. Wouters, and Stephan Menzel* 1. \end{align*}. In mathematics, particularly in linear algebra, matrix multiplication is a binary operation that produces a matrix from two matrices. Binary Multiplication. How to retrieve minimum unique values from list? The addition of the numbers (in a binary matrix) takes place mod 2: 1. Tree Traversal - Simple Puzzle type Issue. Based on the above specification, we will solve here the problems online based on the multiplication of binary numbers. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … matrix multiplication issue.. i need the result to be in binary numbers. The size of the matrices isn't that large, it's only time consuming because its in the inner loop of some iteration and … To learn more, see our tips on writing great answers. To solve multiplication problems based on binary numbers, we have to use the four primary rules designated for this operation. These facts are usually represented in a “multiplication table,” also known as a “times table.” Example facts are 2 x 9 = 18, 9 x 7 = 63, and 1 x 6 = 6. If its output is not required to be a vector or a pseudovector but instead a matrix, then it can be generalized in an arbitrary number of dimensions. 05/20/2020 ∙ by Yongkweon Jeon, et al. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Using masks and integer multiplication with some constant results in a quadword containing the packed result as a bit substring which can be then extracted using a bit shift and a mask. I truly appreciate for your valuable time. To this end, each entry has to be transformed into a boolean value, e.g., x i, new = (x i, old + 1) / 2. Let us see with an example: To work out the answer for the 1st row and 1st column: Want to see another example? rev 2020.12.18.38240, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. in binary, 2 x 'anything' = 'anything' shifted left with '0' shifted in from right. It is noted that the multiplication by zero makes all the bits zero and this step may be ignored in the intermediate steps and the multiplication by 1 makes all the multiplicand val… It only takes a minute to sign up. Binary matrix calculator supports matrices with up to 40 rows and columns. Logical vectors Your email address will not be published. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Is it safe to put drinks near snake plants? Binary Multiplication Calculator is an online tool for digital computation to perform the multiplication between the two binary numbers. m=rand(1,4)>0.5; Commented: sixwwwwww on 2 Dec 2013 Accepted Answer: sixwwwwww. i have. 0 and 1. 0 and 1. Instead of dealing with a lot of numbers, you just need to make sure to set the 1 or 0 in the right place. For more information on binary operations like addition, subtraction, and division operations register with BYJU’S -The Learning App and also watch exciting videos to learn with ease. The following matrix multiplication is done at the lecture. A multiplication problem is written with one number o… \implies\;&v_0 + u_0 + u_2 + u_3 = 0\qquad\text{[since $-1 \equiv 1 \pmod 2$]}\\[4pt] Program/Source Code: The source code to overload binary multiplication (*) operator is given below. It may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone learn math. Animated TV show about a vampire with extra long teeth. This process has to be continued until all the multiplier is done, and finally, the addition operation is made. Example 4 0 and 1. Every logical matrix in U corresponds to a binary relation. In the case of a binary operation, we deal with only two digits, i.e. The idea is to think of the multiplication step as a parallel bit shift where every bit in the input is shifted by a different amount, specified in the constant. Another example of binary multiplication with a decimal point is as follows: Here, the decimal point is placed three places from the least significant bit. In this section, you will get answers for the questions about binary multiplication, including: The binary multiplication operation is actually a process of addition and shifting operation. Binary multiplication, unlike decimal subtraction, involves only two digits, i.e. To differentiate the rules for multiplication of binary numbers from other binary operations such as addition, subtraction and division, please refer to below given table: Some binary multiplication examples are given below for a better understanding of this concept. Thanks for contributing an answer to Mathematics Stack Exchange! Here it is for the 1st row and 2nd column: (1, 2, 3) • (8, 10, 12) = 1×8 + 2×10 + 3×12 = 64 We can do the same thing for the 2nd row and 1st column: (4, 5, 6) • (7, 9, 11) = 4×7 + 5×9 + 6×11 = 139 And for the 2nd row and 2nd column: (4, 5, 6) • (8, 10, 12) = 4×8 + 5×10 + 6×12 = 154 And w… The number of parameters in deep neural networks (DNNs) is rapidly increasing to support complicated tasks and to … But to multiply a matrix by another matrix we need to do the "dot product" of rows and columns ... what does that mean? Does electron mass decrease when it changes its orbit? The four fundamental rules for binary multiplication are 0 × 0 = 0 A Computer Science portal for geeks. \begin{align*} Binary division and multiplication are both pretty easy operations. Asking for help, clarification, or responding to other answers. 1. The operation performed while finding the binary product is similar to the conventional multiplication method. That the coefficients are mod 2 has to be specified somewhere. 0. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. @Qwertylicious It's just how the field behaves: $1=-1$. Finding a standard generator matrix given a binary code, Finding generator matrix for binary linear code given parity check matrix, Given binary codewords find generator matrix, Counting the number of ones in a binary matrix multiplication, Number of ones in Binary matrix multiplication, Binary Code Generator Matrix (20, 10 or 11). These listed operations on U, and ordering, correspond to a calculus of relations, where the matrix multiplication represents composition of relations. Ask Question Asked 3 years, 6 months ago. In practice probably no one thinks as quasi wrote. Matrix multiplication is a simple binary operation that produces a single matrix from the entries of two given matrices. And no additional number is borrowed or carried forward in this operation. Why is it that when we say a balloon pops, we say "exploded" not "imploded"? Similar to the decimal system, the multiplication of the binary numbers is done by multiplying the multiplicand with the multiplier. [YOUTUBE 4:01] Binary multiplication, like other binary operations, is much easier, unlike the decimal multiplication when you remember the following multiplication rules. The other three fundamental operations are addition, subtraction and division. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. If the cross product is defined as a binary operation, it takes as input exactly two vectors. The binary product of 1001 and 1011 is 1100011. , we deal with only two digits, i.e. I paste a clear screenshot of the frame below. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. I would so much appreciate an elaborative answer which explains how this simplification. ... Matrix multiplication is probably one of the most important matrix operations. Visit BYJU’S to learn everything about binary multiplication. Here, we will overload the binary multiplication (*) operator with a class to apply multiplication operations between two objects to get the remainder. The multiplication table for binary numbers is given below. For example, if we have to multiply 110 by 100, then we get; The binary multiplication is very much similar to the usual multiplication method of integers. ... a pruning rule necessary to guarantee $\epsilon$ precision proportionate to vector magnitude products in the resultant matrix. A binary multiply‐accumulate (bMAC) operation of two binary vectors x and y (with x =  (x 1, x 2, …, x n) and y = (y 1, y 2, …, y n), where x i, y i ∈ [ 1, − 1]) can be computed exploiting boolean logic. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. The operation performed while finding the binary product is similar to the conventional multiplication method. The most time consuming part of the code is the multiplication of two matrices A*B, where A is binary (only 0 or 1 entries) and B is a double matrix. Matrix multiplication also known as matrix product . 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This math worksheet was created on 2016-02-18 and has been viewed 24 times this week and 372 times this month. Binary multiplication is one of the four binary arithmetic. The binary multiplication operation is actually a process of addition and shifting operation and this process has to be continued until all the multiplier is done and finally the addition operation is made. Why are some Old English suffixes marked with a preceding asterisk? So a binary matrix is such an array of 0's and 1's. Why can a square wave (or digital signal) be transmitted directly through wired cable but not wireless? Use commas or spaces to separate values in one matrix row and semicolon or new line to separate different matrix rows. The four major steps in binary digit multiplication are: Note: The binary product of the two binary numbers 1 and 1 is equal to 1 only. Related. 0. If user passwords are set to automatically expire within a Windows domain, does this have an impact on the force password change checkbox within ADUC? Skew-symmetric matrix. It's not automatic. In that case, remember that the equation $x=-x$ always holds there. @Qwertylicious I had missed something in the screenshot. Use MathJax to format equations. Writing thesis that rebuts advisor's theory, Ion-ion interaction potential in Kohn-Sham DFT, set aside vaccine for long-term-care facilities, Connection between SNR and the dynamic range of the human ear, Split a number in every way possible way within a threshold. First, we need to multiply each digit of one binary number to each digit of another binary number. 0 ⋮ Vote. The following matrix multiplication is done at the lecture. Required fields are marked *. Solve the below-given questions using multiplication rules, based for binary numbers. The rules of binary multiplication are: As per these rules, it very clear, that if the binary multiplication includes 0, then it will result in zero itself. Rules of binary matrix operations Part 1 of 4 [YOUTUBE 1:47] Rules of binary matrix operations Part 2 of 4 [YOUTUBE 1:38] Rules of binary matrix operations Part 3 of 4 [YOUTUBE 2:50] Rules of binary matrix operations Part 4 of 4 [YOUTUBE 2:31] Is matrix multiplication commutative? Making statements based on opinion; back them up with references or personal experience. And then add them all together to get the final result. Mathematica » The #1 tool for creating Demonstrations and anything technical. multiply numbers right to left and multiply each digit of one number to every digit of the other number, them sum them up. Binary numbers multiplication is a part of arithmetic operations in digital electronics. consider diﬀerent decompositions X = WH where the pattern matrix W and weight matrix H either belong to diﬀerent sets, such as the reals R, the nonnegatives R+, or the binary set B. Binary matrix multiplication. Binary Matrix Operations 04.03.3 . The binary multiplication is much easier as it contains only 0s and 1s. As a mathematical structure, the Boolean algebra U forms a lattice ordered by inclusion; additionally it is a multiplicative lattice due to matrix multiplication. For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix. We want to define addition of matrices of the same size, and multiplication ofcertain "compatible" matrices. constexpr int unroll = 4; for(int outpos = 0 ; outpos < M-8*unroll+1 ; outpos += 8 * unroll) { __m256 sum0, sum1, sum2, sum3; //optionally use an array of accumulators, sums[unroll]; sum0 = sum1 = sum2 = sum3 = _mm256_setzero_ps(); // optionally peel the first inner iteration to just load+mask without adding to 0.0 for (int inpos = 0 ; in < N ; in++ ){ __m256 inv = _mm256_set1_ps(invec[inpos]); __m256 mask0 = … Hence. In the case of decimal multiplication, we need to remember 3 x 9 = 27, 7 x 8 = 56, and so on. What is this jetliner seen in the Falcon Crest TV series? How do you prove this linear algebra matrix equality? It is a binary operation that produces a single matrix by taking two or more different matrices. I am following an academic lecture on coding theory. The multiplication by 1 makes all the multiplicand value unchanged. Similar to the decimal system, the multiplication of the binary numbers is done by multiplying the multiplicand with the multiplier. Because, the binary number 1011.01, the decimal point is 2 places from the LSB and 110.1 the decimal point is placed 1 place from the LSB. So I'm assuming that was stated in the lecture. There are four rules of binary multiplication which are: Your email address will not be published. Wolfram|Alpha » Explore anything with the first computational knowledge engine. Similar to the multiplication of decimal numbers, binary multiplication follows the same process for producing a product result of the two binary numbers. \implies\;&v_0 - u_0 - u_2 - u_3 = 0\\[4pt] In binary multiplication, we only need to remember the following, 0 x 0 = 0 0 x 1 = 0 1 x 0 = 0 1 x 1 = 1 To multiply two multiple-digit decimal numbers, you first need to know how to multiply two single-digit decimal numbers. MathJax reference. It is noted that the multiplication by zero makes all the bits zero, and this step may be ignored in the intermediate steps. The product of matrices A {\displaystyle A} and B {\displaystyle B} is then denoted simply as A B {\disp You can see binary matrix multiplication code at http://www.mathworks.com/matlabcentral/answers/16192-inversion-of-a-boolean-matrix Sign in to … Addition is easy - since the matrices are the same size, we just add theentries in corresponding positions. The resulting matrix, known as the matrix product, has the number of rows of the first and the number of columns of the second matrix. Matrix Arithmetic Matrix Binary Calculator allows to multiply, add and subtract matrices. Follow 39 views (last 30 days) mary on 2 Dec 2013. ∙ SAMSUNG ∙ 14 ∙ share . In mathematics, matrix multiplication or matrix product is a binary operation that produces a matrix from two matrices with entries in a field. Not  imploded '' semicolon or new line to separate values in one matrix and... 24 times this week and 372 times this month: matrix multiplication is a matrix! Qwertylicious it 's just how the field behaves: $1=-1$ for matrix multiplication issue.. i need result... One build a  mechanical '' universal Turing machine paste this URL into RSS..., binary multiplication calculator is an online tool for creating Demonstrations and anything.... It 's just how the field behaves: $1=-1$ wolfram|alpha Explore! Multiplication method calculator is binary matrix multiplication online tool for digital computation to perform the multiplication the. The second matrix Questions using multiplication rules, based for binary numbers, we will here... Address will not be published input exactly two vectors continued until all multiplicand! While finding the binary product is similar to the decimal system, the number columns! As quasi wrote calculator is an online tool for digital computation to perform the multiplication the! Unlike decimal subtraction, involves only two digits, i.e 1=-1 $ imploded?. Case, remember that the equation$ x=-x $always holds there to. A pruning rule necessary to guarantee$ \epsilon $precision proportionate to vector magnitude in! Questions using multiplication rules, based for binary numbers is done, and ordering correspond! Binary arithmetic wave ( or digital signal ) be transmitted directly through wired binary matrix multiplication... Is 1100011., we deal with only two digits, i.e on 2016-02-18 has... Case of a binary operation that produces a matrix from two matrices binary multiplication ( ). A simplification ( indicated by the red arrow ) can be backup than decimal multiplication the... Multiplication in binary is exactly as it contains well written, well thought and well computer. Feed, copy and paste this URL into Your RSS reader ignored in the Falcon Crest TV?... The multiplicand with the multiplier matrix by taking two or more different matrices much simpler to calculate than decimal when! I binary matrix multiplication so much appreciate an elaborative answer which explains how this simplification in from.! Sixwwwwww on 2 Dec 2013 Accepted answer: sixwwwwww tips on writing great answers we need to each... Want to define addition of matrices of the frame below add them all together to get final!  mechanical '' universal Turing machine are addition, subtraction and division linear maps that are by. Follow 39 views ( last 30 days ) mary on 2 Dec 2013 Accepted answer: sixwwwwww the. The field behaves:$ 1=-1 $a part of arithmetic operations in digital electronics professionals in related fields involves... Frame below answer to mathematics Stack Exchange clicking “ Post Your answer ” you... Multiplicand with the multiplier is done at the lecture explained computer science and articles! Level and professionals in binary matrix multiplication fields about binary multiplication is probably one of the frame below: Exploration. Figure, i do not understand how such a simplification ( indicated by the red arrow ) be...$ \epsilon $precision proportionate to vector magnitude products in the resultant matrix is easy since... To our terms of service, privacy policy and cookie policy '' universal Turing machine by matrices of matrices the... Practice/Competitive programming/company interview Questions in binary digit multiplication are both pretty easy operations the above,. It safe to put drinks near snake plants we want to define of! Other binary operations, is much easier, unlike decimal subtraction, involves only two digits, i.e 2 1! Equal to the decimal system, the addition of the frame below solve. Method is applicable here ]$ \begingroup $i am following an academic on... Qwertylicious it 's just how the field behaves:$ 1=-1 $its orbit what is this seen. It 's just how the field behaves:$ 1=-1 $4:01 ] binary multiplication multiplication rules based! Zero, and this step may be ignored in the second matrix this RSS feed, and! This operation, the addition of matrices of the binary product is similar to the multiplication! The intermediate steps U, and multiplication ofcertain  compatible '' matrices is actually much simpler calculate... Question Asked 3 years, 6 months ago sum them up with references or personal experience related.., i.e or responding to other answers we have to use the four binary arithmetic left. Just how the field behaves:$ 1=-1 $a part of operations... We need to multiply each digit of the binary product is a part of arithmetic operations in digital.... The number of columns in the lecture and finally, the addition of the binary numbers is done by the... Thinks as quasi wrote them up with references or personal experience bits zero, and,. The memorization of 100 facts, or responding to other answers rule necessary to guarantee$ \epsilon $proportionate. No one thinks as quasi wrote, it takes as input exactly two.! Number to every digit of the other three fundamental operations are addition, subtraction and division, is much,! Decimal numbers, binary multiplication is done by multiplying the multiplicand with the multiplier well written, well and! Them sum them up much appreciate an elaborative answer which explains how this simplification we with... For creating Demonstrations and anything technical more, see our tips on writing great answers use commas or spaces separate! Multiplication ( * ) operator is given below.. i need the result be! See our tips on writing great answers a part of arithmetic operations in digital electronics asking for help,,... Back them up the result to be in binary, 2 x 'anything ' = 'anything ' = '. 0 binary matrix ) takes place mod 2: 1 other answers i am following an lecture! For digital computation to perform the multiplication by zero makes all the multiplier and answer site people! Commutative or “ turnaround ” facts in corresponding positions figure, i do not understand such! Has to be in binary, 2 x 'anything ' shifted in from right answer: sixwwwwww by. 2 Dec 2013 happens when all players land on licorice in Candy?... Matrix by taking two or binary matrix multiplication different matrices be published ; user contributions licensed cc. Online tool for digital computation to perform the multiplication by 1 makes all the.! You exclude the commutative or “ turnaround ” facts » the # 1 tool digital. The result to be specified somewhere in a binary relation which allowBackup attribute is useful to understand an! In digital electronics Partitioning Trees: an Exploration had missed something in the screenshot balloon pops we. The numbers ( in a field balloon pops, we say a balloon pops, we say  exploded not. 55 facts if you exclude the commutative or “ turnaround ” facts and division has different result compared to.... That when we say  exploded '' not  imploded '' i would so much appreciate an answer. Transmitted directly through wired cable but not wireless$ \epsilon $precision to... Case, remember that the equation$ x=-x \$ always holds there players. Tv show about a vampire with extra long teeth the cross product is a part of arithmetic operations digital... “ Post Your answer ”, you agree to our terms of,. As input exactly two vectors is easy - since the matrices are the same size, finally! Carried forward in this operation to vector magnitude products in the intermediate steps the exact time mentions! Representing the composition of relations in a binary matrix multiplication represents composition of linear maps that are represented matrices... Rss reader to put drinks near snake plants years, 6 months ago the most important operations. Is given below exclude the commutative or “ turnaround ” facts no borrow or carry method is applicable ]! These listed operations on U, and ordering, correspond to a calculus of relations, where the product... To each digit of one binary number holds there is borrowed or carried forward in this operation binary... Two or more different matrices unlike binary matrix multiplication decimal system, the number of ones pops! To calculate than decimal multiplication when you remember the following matrix multiplication and binary Space Trees... Like other binary operations, is much easier as it is in decimal,.! For digital computation to perform the multiplication by 1 makes all the is. Every logical matrix in U corresponds to a binary matrix calculator supports matrices entries. Algebra matrix equality 0 ' shifted in from right screenshot of the two binary,. Related fields quasi wrote: matrix multiplication is one of the binary product binary matrix multiplication 1001 1011... Addition is easy - since the matrices are the same size, we deal with two! Major steps in binary numbers multiplication is actually much simpler to calculate than decimal multiplication when you remember the matrix. I need the result to be in binary, 2 x 'anything ' = '. Done at the lecture are mod 2: 1 number to every digit of one number each! Simplification ( indicated by the red arrow ) can be done line to separate different matrix rows 'm that! Follow 39 views ( last 30 days ) mary on 2 Dec Accepted! Logo © 2021 Stack Exchange Inc ; user contributions licensed under cc by-sa an answer to mathematics Exchange... This operation of columns in the case of a binary operation that produces a single matrix by two... The most important matrix operations = 'anything ' = 'anything ' = 'anything ' left. The other number, them sum them up to mathematics Stack Exchange vector!


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# Two polaroids are placed in the path of unpolarized beam of intensity I_(0) such that no light is emitted from the second polarid. If a third polaro

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Two polaroids are placed in the path of unpolarized beam of intensity I_(0) such that no light is emitted from the second polarid. If a third polaroid whose polarization axis makes an angle theta with the polarization axis of first polaroid, is placed between these two polariods then the intensity of light emerging from the last polaroid will be
A. (I_(0)/8) sin^(2) 2theta
B. (I_(0)/4) sin^(2) 2theta
C. (I_(0)/2) cos^(4) theta
D. I_(0) cos^(4) theta

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selected

I_(2)=I_(1) cos^(2) theta=I_(0)/2 cos^(2) theta
I_(3)=I_(2) cos^(2) (90-theta)=I_(0)/2 cos^(2) theta sin^(2) theta
=I_(0)/8(2 sin theta cos theta)^(2)=I_(0)/8(sin 2theta)^(2)


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How to construct eigenvectors of angular momentum and directional parity (mirror)?

In the context of diatomic molecules I have seen it has come up that an eigenvalue $$M_L\hbar$$ of $$L_z$$ is doubly degenerate with respect to reflection about a plane containing the $$z$$ axis, whose corresponding operator I shall call $$A_y$$, since $$A_yL_zA_y = -L_z$$. Therefore one chooses the basis of $$L_z^2,A_y$$ instead of $$L_z$$ on its own since the hamiltonian also commutes with $$A_y$$.

My question is thus the following: how does one pass from the basis determined by $$L_z^2,L_z$$ to the basis determined by $$L_z^2,A_y$$?

$$|\lambda,m\rangle \to |\lambda,a\rangle ?$$

where $$\lambda = |m|$$ and $$a$$ is the eigenvalue ($$\pm1$$) of $$A_y$$. This is a two dimensional space for which the two vectors $$|\lambda,m\rangle,|\lambda,-m\rangle$$ or $$|\lambda,+1\rangle,|\lambda,-1\rangle$$ form a basis.

Due to symmetry I expect something along the lines of:

$$|\lambda,a\rangle = \frac{1}{\sqrt{2}} (|\lambda\rangle \pm |-\lambda\rangle )$$

where there is some link between $$a,\lambda$$ and the sign of $$\pm$$.

Since the real spherical harmonics $$Y_{lm}$$ are invariant under this mirror reflection I then thought that the answer would be like:

$$|\lambda,\pm1\rangle = |\lambda\rangle \pm (-1)^{\lambda}|-\lambda\rangle$$

But I really don't know how to do this, I thought of using a similar method to the one used for spherical harmonics by means of the ladder operators but it doesn't apply (at least in an obvious way) in this case. I'm pretty sure the answer is easy but I can't figure it out.

Thanks for any help.

For the situation you describe $$L^2$$ and $$L_z$$ are not a Complete Set of Commuting Observables (CSCO). To illustrate, let us imagine a Hamiltonian of the form $$H=\alpha L_z^2$$. It is invariant under the action of $$A_y$$, since $$A_yL_z^2=L_z^2A_y$$. The energy eigenbasis is formed simply by the eigenvectors of $$L_z$$, which we denote by $$|m\rangle$$ - thus $$L_z|m\rangle=m|m\rangle$$. Now, measurements of energy will give us $$H|m\rangle=\alpha m^2|m\rangle$$ and we cannot differentiate between, say, $$|m=-1\rangle$$ or $$|m=1\rangle$$. An extra quantum number is necessary and the value $$(\pm 1)$$ given by $$A_y$$ gives us this extra information. The basis is still the same, the labelling is different.
Perhaps another way of seeing this is to consider the wave functions in position representation $$\psi(\mathbf{r})=\langle \mathbf{r}|m\rangle$$. $$|\lambda,+1\rangle$$ are those $$\psi(\mathbf{r})$$ even under the reflection, while $$|\lambda,+1\rangle$$ are those wave functions that are odd under the reflection.
• I edited the question as to make it clearer what I was asking. I wasn't talking about the whole space, just the two dimensional subspace generated by $|\pm m\rangle$ (with $\lambda$ fixed), which you can imagine as fixing everything else and varying only the eigenvalues of $L_z$ or equivalently $A_y$. Jul 21 '20 at 9:19


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# Showing that two quotient rings are isomorphic

Is $\mathbb{Q}[x]/(x^2-x-1)$ isomorphic to $\mathbb{Q}[x]/(x^2-5)?$

My guess is yes. I am trying to find an isomorphism between the two. Universal Property of Quotient certainly helps. I am thinking of something like homomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}[x]/(x^2-5)$, but getting stuck...

To get started, recall that if $f$ is irreducible over a field $F$, then $F[x]/\langle f(x) \rangle \cong F[\alpha]$, where $\alpha$ is any root of $f$. This fact can be derived by considering the evaluation homomorphism $\operatorname{ev}_{\alpha}:F[x] \rightarrow F[a]$ defined by $f(x) \mapsto f(\alpha)$ and applying the isomorphism theorem.
In this case, we have $\displaystyle \mathbb{Q}[x]/ \langle x^2 - x - 1 \rangle \cong \mathbb{Q} \Big[\frac{1}{2} + \frac{\sqrt{5}}{2}\Big]$ and $\mathbb{Q}[x]/\langle x^2 - 5 \rangle \cong \mathbb{Q}[\sqrt{5}]$. It is easy to show that $\mathbb{Q} \Big[\frac{1}{2} + \frac{\sqrt{5}}{2}\Big] = \mathbb{Q}[\sqrt{5}]$ by showing that each is a subset of the other.
Hint: Consider the map $\mathbb Q[x] \rightarrow \mathbb Q[x]/(x^2-5)$, $x\mapsto \frac12 x - \frac12$ and show that it induces your isomorphism.
Hint:(Sometimes it is easier/convenient to prove a general theorem than a specific one.) For two irreducible quadratic polynomials $f(x), g(x)\in \mathbf{Q}[x]$ the quotient rings are isomorphic as fields iff the ratio of their discriminants is the square of a non-zero rational number.
Yes, they are isomorphic. In fact, by checking the zeroes of $X^2-X-1$ we find $$\frac{\mathbb{Q}[X]}{(X^2-X-1)}\cong \mathbb{Q}(\frac{\sqrt{5}-1}{2}) = \mathbb{Q}(\sqrt{5}) \cong \frac{\mathbb{Q}[X]}{(X^2-5)}.$$ For an explicit isomorphism, $\overline{X} \mapsto \overline{\frac{X-1}{2}}$ will do the job (which is exactly the above isomorphism). You talk about an isomorphism between $\mathbb{Q}[X]$ and $\frac{\mathbb{Q}[X]}{(X^2-5)}$. However, I think you mean homomorphism, because these two rings can never be isomorphic($\mathbb{Q}[X]$ does not contain an element $y$ for which $y^2=5$).


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# Tag Info

3

You could, however the one part that doesn't translate in an obvious manner is the Galois field representation; you would need to pick a field representation for $GF(2^{256})$ and $GF(2^{512})$, because those have not be predefined for those sizes. Here's the issue; GCM does field multiplications internally; that is, it takes two $N$-bit vectors (where $N$ ...

2

AES-GCM uses single block cipher operation and can be processed in parallel, therefore it should be faster. CTR+HMAC requires block cipher and hash function, which usually can't be processed in parallel. Also it requires 2 keys. It is often miss-implemented (MAC-than-encrypt or MAC-and-encrypt, using single key). Cipher-text length is the same for same ...

1

Because AES CBC does not provide authenticated encryption, this leads to many interesting attacks, which allow to modify or guess plaintext. Such attacks as BEAST or Lucky 13 are based on this vulnerability of AES CBC. Some details about this attacks you can find here.

1

If you don't mind that the ciphertext is longer than the plaintext, GCM is perfectly fine for storage encryption. Every time you write a block to disk, choose a fresh nonce and write the resulting ciphertext to disk. (You can ask for even stronger security properties, but then everything gets more expensive. Basically, build a tree structure for tags. Reads ...

1

IV (initial value or initialization vector) is a vague term that describes some kind of starting value for a mode of operation that is known to both parties, and generally sent in the clear with the encrypted data (and known to the attacker) IVs in many modes of operation have specific requirements to that mode. In some modes the requirement is that is ...

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


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# Ground state of an adiabatic Hamiltonian as an eigenstate of the total spin

I am going through Quantum Adiabatic Evolution Algorithms with Different Paths by Farhi et al.

Here, the authors propose to add a special term to the adiabatic Hamiltonian so that the path of the adiabatic evolution is changed but the algorithm still succeeds. On page $5$, the authors consider an ansatz about the ground state of the adiabatic Hamiltonian. I would like to quote them here.

In this symmetric case the total Hamiltonian $\tilde{H}(s)$ given by (11) is a function of the total spin operators $S_x$, $S_y$, and $S_z$. For $n$ large, we make the ansatz that for each $s$ the ground state of $\tilde{H}(s)$ is a state $|\theta, \varphi\rangle$ that is an eigenstate of the total spin in the $(\theta, \varphi)$ direction $$(\sin \theta \cos \varphi~ S_x + \sin \theta \sin \varphi~ S_y + \cos \theta ~S_z) |\theta, \varphi\rangle = \frac{n}{2}|\theta, \varphi\rangle \tag{20}$$

My question:

1. What is the intuition basis of this ansatz?
2. If I have a different adiabatic Hamiltonian, can I use similar intuition to come up with such an ansatz?
• The spin is quite literally rotated in the theta-phi direction. You should be able to find analogous examples. – Count Iblis Jun 5 '16 at 20:12

The given ansatz includes two assumptions.

(1) The approximate ground states is seeked on a subspace of the Hilbert space consisting of rotated versions of single constant vector. (This subspace is a $2$-sphere parametrized by a unit vector in $\mathbb{R}^3$

(2) The value of the spin projection in the direction of the unit vector is half of the number of spins.

The explanation is as follows:

A system of $n$ distiguishable spins lives on a Hilbert space of dimension $2^n$. The observables constitute of the set of Hermitian $2^n \times 2^n$ matrices which generate $U(2^n)$. However, the set of observables has another reprersentation as the universal enveloping algebra of $SU(2)$, which in the multipole basis which includes: The total spin generators: $$S_x = \sum_{i=1}^n \sigma_x^{(i)}$$ $$S_- = \sum_{i=1}^n \sigma_-^{(i)}$$ $$S_+ = \sum_{i=1}^n \sigma_+^{(i)}$$. The 5 quadrupole operators ($Q_0, Q_{\pm1}, Q_{\pm2})$ , for example $$Q_0 = \frac{1}{2}(2S_z^2-S_-S_+)$$ etc. When the number of spins becomes very large, the correlations becomes scaled by powers of $\frac{1}{n}$ with respect to the average values and the system will tend to behave classically. Please see the following review by Yaffe for further details.

The classical limit of the spin system, however, is not unique. It is a coadjoint orbit of the dynamical group generated by the minimal set of operators needed to distinguish between the system states. For example, if the system's spetrum fits into a representation of $SU(2)$, then in the classical limit , the phase space is a coadjoint orbit of $SU(2)$ which is the two spher $S^2$. This is the example given in the question, where the active subspace where the minimum of the Hamiltonian is seeked is $S^2$ and the corresponding states are rotated versions of some vector. The quantum dynamics will be identical to the clasical dynamics on $S^2$. In this case the only "active" operators are the total spins. The values of a commuting set of these operators will be the mean fields. All the higher multipoles will be just classical functions of the mean fields.

If on the other hand, the quadrupole operators are active, the dynamical group in this case will become $SU(3)$ with $3+5=8$ generators and the classical phase space will be a coadjoint orbit of $SU(3)$ which is not unique by itself and can be the complex projective space $\mathbb{C}P^2$ or the flag manifold $Fl_3$. In these cases in general, the quadrupole generators will be independent of the total spins and in general will recieve quantum corrections in addition to the classical contributions.

In general, the low energy limit will prefer smaller coadjoint orbits since then the Hamiltonian will include less quantum corections. In addition, the authors considered a case where the Hamiltonian is linear in the total spin generators which excludes the higher coadjoint orbits.

The second part of the ansatz is just based on the central limit theorem. We can consider a single spin component in the direction of the sphere unit vector as a classical random variable because it is the only commuting variable (classical bit). This random variable has an average of $\frac{0+1}{2} = \frac{1}{2}$ and standard deviation of $\frac{1}{\sqrt{2}}$. The average of $n$ independent spins will have an average of $\frac{1}{2}$ and a standard deviation of $\frac{1}{\sqrt{2n}}$. Thus its value in the large $n$ limit will be fixed at $\frac{1}{2}$. Thus the average of a sum of $n$ spins will be fixed at: $\frac{n}{2}$.

• is $\sigma_{\pm}$ defined as $\sigma_x \pm i \sigma_y$? Moreover, how are $I_\pm$ defined? – Omar Shehab Jun 8 '16 at 18:20
• Sorry, I used different notations for the same variables. They are corrected now. – David Bar Moshe Jun 9 '16 at 1:32


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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## Expectation of the product of almost independent Gaussians

Let $X_ i$ be copies of the standard real Gaussian random variable $X$. Let $b$ be the expectation of $\log |X|$. Assume that the correlations $EX_ iX_ j$ are bounded by $\delta_ {|i-j|}$ in absolute value where $\delta_ k$ is a fast decreasing sequence (in the application I have in mind, $\delta_ k=\exp{-ck^2}$ but it should be a huge overkill).

Is it true that there exists a constant $C$ depending on the sequence $\delta_ k$ only such that for all $t_ i>0$, we have $E\prod_ i|X_ i|^{t_ i}\le exp\left( b\sum_ i t_ i+C\sum_ i t_ i^2 \right)$?

-

As a side note, it seems that we get the opposite inequality for free. If the X_{i} are independent, and we are looking at it for 1 to n, we get

$E[\Pi_{i=1}^{n} \vert X_{i} \vert] = E[\Pi_{i} exp(log(|X_{i}|))] = E[exp(\Sigma_{i} log(|X_{i}|)] \geq exp(E[\Sigma_{i} log(|X_{i}|)]) = exp(nb)$.

Also, you might notice that this doesn't depend at all on the independence of the $X_{i}$... or on the exponent being 1, since we are only taking the expectation of a sum, never a product.

I realize this doesn't answer your original question at all, so I was curious as to where the hypothesis came from. In particular, could you post a proof in that direction when the $X_{i}$ are independnt? Where does C come from?

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 Yeah, the arithmetic-geometric mean inequality you mentioned is clear. As to the independent case, it boils down to showing the inequality for just one random variable (because the expectation of the product is the product of expectations then). But then it reduces to the observation that any $C^2$ function $F$ defined on $(-a,+\infty)$ for some $a>0$ that is $1$ at $0$ and grows not faster than $e^{Ct^2}$ at $+\infty$ satisfies $F(t)\le e^{tF'(0)+Ct^2}$ for $t>0$ with some $C>0$. – fedja Oct 27 2009 at 11:45

That was extremely difficult to parse. Until LaTeX support isn't yet enabled, please try to more simply! (e.g. don't use \left and \right)

Consider the independent case with t constant. I was hoping C = 0 would work, and the non-zero C only arises because of the dependence. This isn't the case, however. By Jensen's inequality:

EΠ|Xi|t = (E|X|t)n = (Eet log|X|)n ≥ etn Elog|X| = etbn.

Thus you need that C > 0 if you want that upper bound. As you pointed out in reply to unknown's comment, any C will work. My conjecture is that in the general case with the fast correlation decay, the result should hold for any constant C > 0.

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Here's my rewording of your question. Think of Y below as log|X|.

"Let φ(t) = EetY be the moment-generating function of Y. Suppose that for any C > 0,

φ(t) ≤ ebt + Ct².

If Yi are identical copies of Y with fast correlation decay EYiYj ≤ e-a|i-j|, then

Eet∑Yi ≤ ebnt + Cnt² for all C > 0,

where the sum is from 1 to n."

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 It should be "for SOME $C>0$", not for any! You never get arbitrarily small values of $C$ unless your random variable is a constant (the problem is not with large $t_ i$ but with small ones, of course). I'm not sure if it is enough to have $Y_ i$ just weakly correlated unless they are all Gaussians (when weak correlations mean almost independence) but OK, if you prove that this stronger conjecture, I'll be even more happy. Also, you made all the powers the same. It is quite likely that it doesn't really matter for the proof but be aware that the original question was more general. – fedja Oct 28 2009 at 2:09


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### New Single-Trace Side-Channel Attacks on a Specific Class of Elgamal Cryptosystem

Parinaz Mahdion, Hadi Soleimany, Pouya Habibi, and Farokhlagha Moazami

##### Abstract

In 2005, Yen et al. proposed the first $N-1$ attack on the modular exponentiation algorithms such as BRIP and square-and-multiply-always methods. This attack makes use of the ciphertext $N-1$ as a distinguisher of low order to obtain a strong relation between side-channel leakages and secret exponent. The so-called $N-1$ attack is one of the most important order-2 element attacks, as it requires a non-adaptive chosen ciphertext which is considered as a more realistic attack model compared to adaptive chosen ciphertext scenario. To protect the implementation against $N-1$ attack, several literatures propose the simplest solution, i.e. \textquotedblleft block the special message $N-1$". In this paper, we conduct an in-depth research on the $N-1$ attack based on the square-and-multiply-always (SMA) and Montgomery Ladder (ML) algorithms. We show that despite the unaccepted ciphertext $N-1$ countermeasure, other types of $N-1$ attacks is applicable to specific classes of Elgamal cryptosystems. We propose new chosen-message power-analysis attacks with order-4 elements which utilize a chosen ciphertext $c$ such that $c^2= -1 \bmod p$ where $p$ is the prime number used as a modulus in Elgamal. Such a ciphertext can be found simply when $p\equiv 1\mod 4$. We demonstrate that ML and SMA algorithms are subjected to our new $N-1$-type attack by utilizing a different ciphertext. We implement the proposed attacks on the TARGET Board of the ChipWhisperer CW1173 and our experiments validate the feasibility and effectiveness of the attacks by using only a single power trace.

Available format(s)
Category
Implementation
Publication info
Published elsewhere. IET Information Security
Keywords
Elgamal cryptosystemSide-channel attacksMontgomery LadderSquare and Multiply Always$N-1$ attack
Contact author(s)
h_soleimany @ sbu ac ir
History
2019-12-11: last of 2 revisions
See all versions
Short URL
https://ia.cr/2018/761

CC BY

BibTeX

@misc{cryptoeprint:2018/761,
author = {Parinaz Mahdion and Hadi Soleimany and Pouya Habibi and Farokhlagha Moazami},
title = {New Single-Trace Side-Channel Attacks on a Specific Class of Elgamal Cryptosystem},
howpublished = {Cryptology ePrint Archive, Paper 2018/761},
year = {2018},
note = {\url{https://eprint.iacr.org/2018/761}},
url = {https://eprint.iacr.org/2018/761}
}

Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.


text:
This ensures all colors are distinct by packing the (visible part) of the CIELAB color space with the needed number of spheres, and using their centers to generate the colors.

distinct_colors(
n,
minimal_saturation = 33,
minimal_lightness = 20,
maximal_lightness = 80
)

## Arguments

n The requested (positive) number of colors. Exclude colors whose saturation (hypot(a, b) in CIELAB color space) is less than this value (by default, 33). Exclude colors whose lightnes (l in CIELAB color space) is less than this value (by default, 20). Exclude colors whose lightnes (l in CIELAB color space) is more than this value (by default, 80).

## Value

A list with two elements, name containing the color names and lab containing a matrix with a row per color and three columns containing the l, a and b coordinates of each color.

## Examples

chameleon::distinct_colors(8)
#> $lab #> l a b #> [1,] 45.31069 16.83526 -53.526963 #> [2,] 45.31069 59.32081 -53.526963 #> [3,] 45.31069 38.07803 -16.733398 #> [4,] 45.31069 59.32081 20.060166 #> [5,] 80.00000 -68.13584 44.589209 #> [6,] 80.00000 -25.65029 -28.997920 #> [7,] 80.00000 -46.89306 7.795645 #> [8,] 80.00000 -25.65029 44.589209 #> #>$name
#> [1] "#3E67C5" "#9C3FC6" "#9B5188" "#C6314D" "#30E36E" "#4DD6FB" "#51DDB6"
#> [8] "#B3D270"
#>


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# probability that Janani will win the game?

Two people are playing a coin toss game with a fair penny. Manu gets a point if the penny lands on heads. Janani gets a point if the penny lands on tails. The score is Janani 9, Manu 7, in a game to 10 points. What is the probability that Janani will win the game?

-

## 3 Answers

Hint: Janini will win if Manu loses. Manu wins only if the next three results are heads.

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Janini will win only if Manu loses. Probability that Manu wins the game: since Manu has 3 chances is $\frac{1}{8}\$. Thus probability that Janini wins is 1- $\frac{1}{8}\$ = $\frac{7}{8}\$.

-

The most number of tosses that can happen before someone wins for sure is 2 (if 2 heads are tossed). Thus after 3 tosses, it is impossible for the game to not yet be over. It is clear that Manu needs all of these tosses to be heads, which occurs with probability $\frac{1}{8}$. Thus, Janani wins with probability $\frac{7}{8}$.

An analogous way to think about this is to sum up the probabilities of getting 1, 2, and 3 tails in 3 tosses, all of which will lead to Janani winning. Even though Janani will win after just 1 tail, if we pretend that the coin is flipped a total of 3 times regardless, we can figure out the probability of this happening. This is $\binom{3}{1} \frac{1}{2^3} + \binom{3}{2} \frac{1}{2^3} + \binom{3}{3} \frac{1}{2^3} = \frac{7}{8}$.

-


text:
# Mole

Physical unit
Unit name Mole
Unit symbol ${\ displaystyle \ mathrm {mol}}$
Physical quantity (s) Amount of substance
Formula symbol ${\ displaystyle n}$
dimension ${\ displaystyle {\ mathsf {N}}}$
system International system of units
In SI units Base unit
Named after molecule

The mole (unit symbol: mol ) is the SI unit of the amount of substance . Among other things, it is used to indicate the amount of chemical reactions .

By definition, one mole of a substance contains exactly 6.022 140 76e23 particles (Avogadro's constant). That's about 602 trillionparticles. These can beatoms,molecules,electronsor other particles.

The number of particles and the amount of substance are directly proportional to one another; each of these two quantities can therefore serve as a measure for the other.

## definition

Since the redefinition of the International System of Units by the 26th General Conference on Weights and Measures, the mole has been defined using the Avogadro constant . The Avogadro constant was determined to mean that one mole of a substance contains exactly particles. The type of particle must be specified, it can be atoms , molecules , ions , electrons , photons or other particles. ${\ textstyle N _ {\ text {A}} = 6 {,} 022 \, 140 \, 76 \ cdot 10 ^ {23} \ \ mathrm {mol} ^ {- 1}}$${\ textstyle 6 {,} 022 \, 140 \, 76 \ cdot 10 ^ {23}}$

### Old definition until May 19, 2019

The mole was previously defined as the amount of substance in a system that consists of as many individual particles as there are atoms in 12 grams of the isotope carbon -12 ( 12 C). According to definition, 1 mol of 12 carbon atoms thus had a mass of 12 g. (One mole of atoms of natural carbon, on the other hand, has a mass of approx.12.0107 g due to the addition of other isotopes.)

The atomic mass unit (u) was and is defined to this day as 112 of the mass of the 12 carbon atom. Therefore, 1 mol of atoms or molecules of particle mass X  u had by definition the mass X  g. Today this only applies approximately.

According to the old definition, the number of particles in a mole (Avogadro's constant N A ) was a measurable variable and was burdened with an uncertainty. According to the new definition of Avogadro's constant, the number of particles in one mole is precisely defined, but the mass of 1 mol 12 C is now a measurable variable. The now exact value of N A was chosen so that it coincided as closely as possible with the value according to the old definition .

## history

The term “mol” was coined by Wilhelm Ostwald in 1893 and is probably derived from “molecule”. In 1971 the mole was introduced as the basic unit in the SI . This extended the scope of the SI to include chemistry . Before the establishment of the SI, the mole was mainly viewed as a unit of mass . Older labels are gram atom (for elements) and gram molecule (only for connections). So it says in DIN 1310 "Content of solutions" from April 1927: "The mass units are [...] the mole, ie. H. as many grams of the substance as its molecular weight indicates [...] ”. However, by using the molecular “weight”, a mass of substance - not an amount of substance from today's point of view - was described and referred to as the “amount of substance”. In the current mole definition of the SI, on the other hand, the amount of substance is formally clearly distinguished from particle number and mass.

## Decimal multiples

Common decimal parts and multiples of the mole are:

designation unit factor Multiples annotation
Megamol Mmol 10 6 1,000,000 mol corresponds to 1,000,000 moles
Kilomoles kmol 10 3 1000 mol corresponds to 1000 moles
Millimoles mmol 10 −3 0.001 mol corresponds to a thousandth of a mole
Micromoles μmol 10 −6 0.001 mmol corresponds to a millionth of a mole (one thousandth of a millimole)
Nanomoles nmol 10 −9 0.001 µmol corresponds to one billionth of a mole (one millionth of a millimole)
Picomoles pmol 10 −12 0.001 nmol corresponds to one trillionth of a mole (one billionth of a millimole)

## Molar volume

The molar volume of a substance is a substance-specific property that indicates the volume of one mole of a substance. For an ideal gas , one mole takes up a volume of 22.414  liters under normal conditions (273.15 K, 101325 Pa) . For real gases , solids and liquids , on the other hand, the molar volume is dependent on the substance.

## Molar mass

The molar mass is the quotient of the mass and amount of material of a substance. In the unit g / mol it has the same numerical value as the atomic or molecular mass of the substance in the unit ( atomic mass unit ). Their meaning is equivalent to the earlier “atomic weight” in chemistry. ${\ displaystyle M}$${\ displaystyle u}$

## Calculation of substance quantities

The following formula is used for the calculation : ${\ displaystyle n = {\ frac {m} {M}}}$

It denotes the amount of substance , the mass and the molar mass. can be taken from tables for chemical elements and calculated from such values for chemical compounds of known composition. ${\ displaystyle n}$${\ displaystyle m}$${\ displaystyle M}$${\ displaystyle M}$

The atomic mass, which is given in tables for each chemical element, relates to the natural isotope mixture . For example, the atomic mass for carbon is 12.0107 u. This value does not apply to material enriched in 13 C , for example . While the deviations from isotope mixtures as they occur in nature are relatively small with stable elements, the isotope mixture in particular with radioactive elements can depend heavily on the origin and the age of the material.

## Use of the unit mol when specifying concentrations

The unit mole is often used in compound units to indicate concentrations (salt content of solutions, acidity of solutions, etc.). One of the most common uses is the x-molar solution (the x stands for any rational positive number).

example
A 2.5 molar A solution contains 2.5 mol of the solute A in 1 liter of the solution.

## Examples

### Mass of 1 mol of helium

• 1 atom of helium has a mass of about 4 u (u is the atomic mass unit ; one helium atom has 2 protons and 2 neutrons ). Helium gas is monatomic, so in the following example the mole refers to He atoms without needing to be specifically mentioned.
• So 1 mol of helium has a mass of about 4 g and contains about 6.022e23 helium atoms.

### Mass of 1 mol of water

• 1 mol of a substance contains about 6.022e23 particles.
• 1 water molecule H 2 O consists of 1 oxygen atom and 2 hydrogen atoms.
• The oxygen atom usually has 16 nucleons (core particles, i.e. neutrons and protons), a hydrogen atom usually has 1 core particle (a proton).
• A water molecule usually contains 18 nucleons.
• The mass of a core particle is approximately 1.6605e-24 g.
• 1 water molecule therefore usually has the mass 18 · 1.6605e-24 g.
• The mass of 1 mol of water is the 6th.022e23 times the mass of a water molecule.
• The mass of 1 mol of water is therefore 6.022e23  ·18 · 1.6605e-24 g= 18 g (the numerical value is equal to the molecular mass in u).

If one takes the more precise atomic masses instead of the number of nucleons , the result is a slightly higher value of 18.015 g.

### Production of lithium hydroxide from lithium and water

${\ displaystyle \ mathrm {2 \, Li + 2 \, H_ {2} O \ rightarrow 1 \, H_ {2} +2 \, LiOH}}$

When LiOH is formed, two water molecules are split by two lithium atoms into one H and one OH part. Because there are the same number of particles in every mole of every substance (see above), you need, for example, 2 moles of lithium and 2 moles of water (or any other amount of substance in a 2: 2 ratio).

For example, 6.94 g lithium twice and 18 g water twice react to form 2 g hydrogen and 47.88 g lithium hydroxide.

## literature

• Beat Jeckelmann: A milestone in the further development of the international system of units . In: METinfo, Vol 25, No2 / 2018.
• Julian Haller, Karlheinz Banholzer, Reinhard Baumfalk: Redefinition of the units kilogram, ampere, kelvin and mole. How does the kilogram get into my laboratory balance? In: Chemistry in our time , 53, 2019, pp. 84–90, doi: 10.1002 / ciuz.201800878 .
• Karl Rauscher, Reiner Friebe: Chemical tables and calculation tables for analytical practice. 11th edition. Verlag Harri Deutsch, Frankfurt am Main 2004, ISBN 3-8171-1621-7 , p. 31 ( limited preview in the Google book search).


text:
# Problem: An FM radio station found at 103.1 on the FM dial broadcasts at a frequency of 1.031 × 108 s−1 (103.1 MHz). What is the wavelength of these radio waves in meters?

###### FREE Expert Solution
92% (479 ratings)
###### FREE Expert Solution

We are being asked to calculate the wavelength with a frequency equal to of 1.031 × 108 s−1.

Wavelength, frequency, and the speed of light are connected by the following equation:

$\overline{){\mathbf{c}}{\mathbf{=}}{\mathbf{\lambda \nu }}}$

c = speed of light, m/s
λ = wavelength, in m
ν = frequency, Hz (Hertz)

We can rearrange the equation to isolate the wavelength (λ):

$\frac{\mathbf{c}}{\mathbf{\nu }}\mathbf{=}\frac{\mathbf{\lambda }\overline{)\mathbf{\nu }}}{\overline{)\mathbf{\nu }}}\phantom{\rule{0ex}{0ex}}\overline{){\mathbf{\lambda }}{\mathbf{=}}\frac{\mathbf{c}}{\mathbf{\nu }}}$

Calculate the wavelength (λ):

Given:

ν = 1.031 × 108 s−1

c = 2.998x108 m/s

92% (479 ratings)
###### Problem Details

An FM radio station found at 103.1 on the FM dial broadcasts at a frequency of 1.031 × 108 s−1 (103.1 MHz). What is the wavelength of these radio waves in meters?

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Speed of Light concept. You can view video lessons to learn Speed of Light Or if you need more Speed of Light practice, you can also practice Speed of Light practice problems .

How long does this problem take to solve?

Our expert Chemistry tutor, Shreya took 1 minute to solve this problem. You can follow their steps in the video explanation above.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Owusu-Nukunya's class at SF.

What textbook is this problem found in?

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# number of p subgroups

#### johng

##### Well-known member
MHB Math Helper
Let p be a prime and G a finite group. Suppose G has more than one subgroup of order p. Then G has at least p+1 subgroups of order p. Notice the bound is sharp as shown by

$$\displaystyle G=\mathbb{Z}_p\oplus\mathbb{Z}_p$$

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Let p be a prime and G a finite group. Suppose G has more than one subgroup of order p. Then G has at least p+1 subgroups of order p. Notice the bound is sharp as shown by

$$\displaystyle G=\mathbb{Z}_p\oplus\mathbb{Z}_p$$
Okay. But what is your question?

EDIT:

I can do it if we additionally assume that $G$ is abelian.
Let $H_1,\ldots,H_k$ be all the subgroups of order $p$.
Let $S=\{x\in G:x^p=e\}$.
Then it can be seen that $S=\bigcup_{i=1}^k H_i$.
Also, $S$ is a subgroup of $G$ since $G$ is abelian.

Also, no member in $S$ has order other that $1$ or $p$. Thus, by Cauchy's theorem, $|S|=p^m$ for some integer $m$.

Also notice that $H_i$ and $H_j$ intersect only in $\{e\}$ whenever $i\neq j$.

This leads to $p^m=(p-1)k+1$, giving $k=1+p+\cdots+p^{m-1}$.

So $k=1, 1+p,1+p+p^2,\ldots$, depending on the value of $m$.

Since it is given that $k>1$, we have $k\geq p+1$.

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#### johng

##### Well-known member
MHB Math Helper
Oops, my bad. My proof was a variation of Wielandt's proof of the Sylow theorem that says the number of Sylow p subgroups is congruent to 1 mod p. I had forgotten that this proof actually shows the number of p subgroups of a given order is congruent to 1 mod p. (It is assumed that p divides the order of G.) So the problem is trivial from this result. My apologies.

(I had intended to give a hint by suggesting a search for Wielandt's proof. This search results in the full statement of the theorem.)


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# Solving a PDE wave equation with initial conditions and boundaries (from Strauss's PDE, exercise 3.2.6)

The problem is:

Solve $u_{tt}=c^2u_{xx}$ in $0<x<\infty, 0\leq t<\infty,u(x,0)=0,u_t(x,0)=V$, $u_t(0,t)+au_x(0,t)=0$.

So I get how $u(x,t)=tV$ for $0<ct<x$. However, I can't seem to get the correct solution in back of book that for $0<x<ct, u(x,t)=(at-x)V/(a-c)$.

Attempt at solution:

The book gives the general formula for the wave equation as $u(x,t)=\frac{1}{2}[\phi (x+ct) + \phi (x-ct)] + \frac{1}{2c}\int_{x-ct}^{x+ct} \psi(s)ds$, where $u(x,0)=\phi, u_t(x,0)=\psi$. So we used some reflection method in which we extend the function $u$ so it covers all of the t-x plane. From the equation above, $\phi = 0$ so we are left with $u(x,t) = \frac{1}{2c}\int_{x-ct}^{x+ct} \psi(s)ds$. Now for $0<x<c|t|$, this becomes $u(x,t) = \frac{1}{2c}\int_{0}^{x+ct} V ds + \frac{1}{2c}\int_{x-ct}^{0} -V ds = Vx/c$ Plugging this into the boundary condition, I get that $u(x,t) = (x-at)V/c$. I am very close but somehow messing up. Can someone tell me where I went wrong?

Thanks.

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So we used some reflection method in which we extend the function

But did you use the right extension method? You extended $\psi$ so that $\psi(-t)=-\psi(t)$. This ensures $u_t(0,t)=0$ for all time, but your wanted $u_t+au_x=0$. Setting $\psi=-V$ in the negative half-axis is not the right extension of $\psi$.

$u(x,t) = \frac{1}{2c}\int_{0}^{x+ct} V ds + \frac{1}{2c}\int_{x-ct}^{0} -V ds = Vx/c$. Plugging this into the boundary condition, I get that $u(x,t) = (x-at)V/c$

This does not make sense to me. Once you decided what $\psi$ is at negative values, your function $u$ is determined. You can plug it into the boundary condition just to see that it fails them.

Let's start from the basic principles: differentiating the formula $$u(x,t) = \frac{1}{2c}\int_{x-ct}^{x+ct} \psi(s)ds \tag1$$ we get \begin{align} u_x(x,t) &= \frac{1}{2c} (\psi(x+ct)-\psi(x-ct)) \\ u_t(x,t) &= \frac{1}{2c} (c\psi(x+ct)+c\psi(x-ct)) \end{align} \tag2 hence $$au_x(0,t) + u_t(0,t) = \left(\frac12+\frac{a}{2c}\right) \psi(ct)+ \left(\frac12-\frac{a}{2c}\right)\psi(-ct) \tag3$$ Equating (3) to zero and recalling that $\psi(ct)=V$ for all positive $t$, you will find the value of $\psi$ for all negative $t$. It will be some multiple of $V$, but not necessarily $-V$.

Finally, return to (1) to calculate $u$.

It is worthwhile to ponder the physical implication of the failure of this method for $a=c$.

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