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## How to improve Verbal score from 26-28 to 36-38 ?

This topic has 4 expert replies and 3 member replies
pagani88 Newbie | Next Rank: 10 Posts
Joined
15 Dec 2017
Posted:
6 messages

#### How to improve Verbal score from 26-28 to 36-38 ?

Sat Feb 24, 2018 1:27 pm
Hello everyone,

I am curious about improving my Verbal score? I can't go up higher than 28, while managed to boost my Quant to 40+. Any advice?

Another issue is scoring. My last official GMAT PrepTest was Q44 V25 resulting in 570, but all the score tables say it is supposed to be 580-590. Where does the discrepancy come from?

Artur

### GMAT/MBA Expert

ceilidh.erickson GMAT Instructor
Joined
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Thu Mar 01, 2018 4:01 pm
pagani88 wrote:
Another issue is scoring. My last official GMAT PrepTest was Q44 V25 resulting in 570, but all the score tables say it is supposed to be 580-590. Where does the discrepancy come from?
Here's the simple answer: your 44 on quant isn't really a "44." It's rounded from a raw score: think about the range from 43.5 to 44.49999.... So there is a such thing as a "high 44" and a "low 44," although you'd never know which one you had.

Your 200-800 score is actually based on the RAW scores, not the rounded SCALED scores. So, a Q "high 44" + V "high 25" might track to something 20 pts higher than a "low 44" + "low 25."

Make sense?

_________________

Ceilidh Erickson
Manhattan Prep GMAT & GRE instructor
EdM in Mind, Brain, and Education

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Akrita@Jamboree Senior | Next Rank: 100 Posts
Joined
22 Mar 2018
Posted:
46 messages
Fri Apr 13, 2018 9:30 am
ceilidh.erickson wrote:
Akrita@Jamboree wrote:
In other words, as Ceilidh has already mentioned, the same combination of Scaled scores can fetch different Overall Scores (based on your %ile ranking).
This is not correct, and it does not reflect what I said. Your percentile is irrelevant to your score. If one student with a Q50 ends up with a higher overall score than another student with a Q50, it is because that student had a higher raw score. The overall score is based on RAW scores, and has nothing whatsoever to do with percentiles.

GMAC publishes percentiles each year on overall scores based on the proportion of people who got that score. That percentile necessarily changes year to year. The combination of raw scores that yields a particular overall score does not change year to year. Those percentiles are not factored in to re-calibrate scores the following year. If that were the case, we wouldn't be able to compare a Q50 in 2014 to a Q50 in 2018. Scores (both raw and scaled) have to mean the same thing year over year.

The percentile is based on the score, not the other way around.

And here: https://www.manhattanprep.com/gmat/blog/2017/10/03/manhattan-preps-gmat-score-calculator-what-quant-verbal-scores-result-700-score/
Hi Ceilidh,

Firstly, thank you for your detailed response.

However, what I meant by my statement - and it is easy to see as to how it could have been misconstrued - was that the same "scaled score" can fetch a different overall score. The "levels" of a scaled score would be different, sure, and based on a number of factors, such as the difficulty level of the incorrect questions, whether the test taker left any questions, whether he/she made consecutive mistakes, whether he/she made errors on the experimental/marked questions, etc. rather than just the "raw" score. However, the "final scaled score" that a candidate would see, assuming our above example of candidate A getting first 3 incorrect and candidate B getting last 3 incorrect, would remain the same. The only visible difference - from a test taker's perspective [on the score card] - would be in the "%ile" ranking - this was the point I was trying to make. Of course, as you have pointed out, the "level" of the scaled score would determine the "%ile" ranking, and not vice-versa.

For the sake of clarity, I have since gone back and rephrased my earlier post so as to avoid any ambiguities. To summarize and answer Artur's question, although the score you "see" on screen might be a Q44, there are various levels within the same score, and these "levels" would be reflected in the "%ile" ranking rather than the "scaled score"

### GMAT/MBA Expert

ceilidh.erickson GMAT Instructor
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Thu Mar 01, 2018 3:57 pm
pagani88 wrote:
I am curious about improving my Verbal score? I can't go up higher than 28, while managed to boost my Quant to 40+. Any advice?
Whenever students tell me "I've been studying for months but I haven't seen any improvement!" usually the missing factor turns out to be that they haven't been REVIEWING deeply.

To do well on verbal, you need to do the following:
- read strategy guides from some major prep company. You need to know your SC grammar rules backwards & forwards, and get expert insight into CR & RC process.
- practice OG problems, accurately timed.
- keep an Error Log to review your mistakes: https://www.manhattanprep.com/gmat/blog/2013/01/18/the-worst-mistake-you-can-make-in-gmat-studying/

Here's how to thoughtfully review each question type:

For SC, you want to ask yourself:
- was I thinking about the MEANING of the sentence?
- did I identify the overall structure of SUBJECT + VERB?
- did I identify markers of particular grammar issues: parallelism, comparisons, modifiers, etc?
- did I get distracted by meaningless differences in the answer choices, or did I focus on ones that spoke to clear rule violations?

For CR & RC, ask yourself:
- did I identify the question type?
- did I identify the LOGICAL GAP in the argument (for CR assumption, strengthen, weaken, and evaluate questions) before looking at answer choices?
- did I thoughtfully predict an answer before looking at the answer choices?
- can I identify why each wrong answer is wrong?

Start recording & tracking the kinds of mistakes you make, and you'll start to see improvement.

_________________

Ceilidh Erickson
Manhattan Prep GMAT & GRE instructor
EdM in Mind, Brain, and Education

Manhattan Prep instructors all have 99th+ percentile scores and expert teaching experience.
Sign up for a FREE TRIAL, and learn why we have the highest ratings in the GMAT industry!

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Akrita@Jamboree Senior | Next Rank: 100 Posts
Joined
22 Mar 2018
Posted:
46 messages
Fri Apr 13, 2018 4:34 am
pagani88 wrote:
Hello everyone,

I am curious about improving my Verbal score? I can't go up higher than 28, while managed to boost my Quant to 40+. Any advice?

Another issue is scoring. My last official GMAT PrepTest was Q44 V25 resulting in 570, but all the score tables say it is supposed to be 580-590. Where does the discrepancy come from?

Artur
Hi Artur,

Generally, the GMAT algorithm is based on a bunch of convoluted variables such as, but not limited to, the difficulty level of the incorrect questions, whether you made consecutive mistakes, whether you got the marked/experimental questions correct/incorrect, whether you made a lot of errors, say more than 4-5, in the first 8-10 questions - as a result of which the algorithm will give you a lower score than if you got the same number of errors in the last section of the test and had everything correct till that point.

Essentially, the idea is that if two people - say A and B - get 34 questions correct in the Quant section (as per the pre-April 16 version of the test), and if A gets the first 3 questions incorrect (and everything correct after that), and B gets the last 3 incorrect (and everything correct before that point), then there will, at the very least, be a difference in the %ile ranking of the two Quant scores. Therefore, even though the Scaled score for both candidates might be Q50 - usually this is the score people receive when they make between 2 and 4 mistakes on the Quant section - there would be a difference in the "level" of Q50 attained, and the %ile ranking for candidate B would be higher - the rationale being that he got the "more difficult" questions incorrect - than that of candidate A.

In other words, the same combination of Scaled scores can fetch different Overall Scores (out of 800) (based on "the level" of your scaled score[s]). In most cases, however, the difference is fairly minimal and the range of the scores is between 10-30 points max. For example, for the same combination of Q44 and V25, you might expect someone to get a 560 too based on the above-mentioned parameters.

Hope this helps - please let me know if you have any further queries.

Last edited by Akrita@Jamboree on Fri Apr 13, 2018 5:57 am; edited 1 time in total

### GMAT/MBA Expert

ceilidh.erickson GMAT Instructor
Joined
04 Dec 2012
Posted:
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Fri Apr 13, 2018 5:33 am
Akrita@Jamboree wrote:
In other words, as Ceilidh has already mentioned, the same combination of Scaled scores can fetch different Overall Scores (based on your %ile ranking).
This is not correct, and it does not reflect what I said. Your percentile is irrelevant to your score. If one student with a Q50 ends up with a higher overall score than another student with a Q50, it is because that student had a higher raw score. The overall score is based on RAW scores, and has nothing whatsoever to do with percentiles.

GMAC publishes percentiles each year on overall scores based on the proportion of people who got that score. That percentile necessarily changes year to year. The combination of raw scores that yields a particular overall score does not change year to year. Those percentiles are not factored in to re-calibrate scores the following year. If that were the case, we wouldn't be able to compare a Q50 in 2014 to a Q50 in 2018. Scores (both raw and scaled) have to mean the same thing year over year.

The percentile is based on the score, not the other way around.

And here: https://www.manhattanprep.com/gmat/blog/2017/10/03/manhattan-preps-gmat-score-calculator-what-quant-verbal-scores-result-700-score/

_________________

Ceilidh Erickson
Manhattan Prep GMAT & GRE instructor
EdM in Mind, Brain, and Education

Manhattan Prep instructors all have 99th+ percentile scores and expert teaching experience.
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Akrita@Jamboree Senior | Next Rank: 100 Posts
Joined
22 Mar 2018
Posted:
46 messages
Fri Apr 13, 2018 5:37 am
pagani88 wrote:
Hello everyone,

I am curious about improving my Verbal score? I can't go up higher than 28, while managed to boost my Quant to 40+. Any advice?

Another issue is scoring. My last official GMAT PrepTest was Q44 V25 resulting in 570, but all the score tables say it is supposed to be 580-590. Where does the discrepancy come from?

Artur
Hi Artur,

I am happy to provide my two cents.

As with anything, the only way to improve and/or excel at something is mastering the fundamentals and applying those concepts to the more advanced situations. The GMAT, like any other standardized test, is no different. It requires a thorough understanding of the fundamentals - in both Verbal and Maths - and then being able to strategize those fundamentals so as to be able to solve a question in less than 2 minutes.

A score of V28 indicates that while you have a cursory understanding of the basic concepts and can breeze through the easy questions (this is a good start!), you struggle at the intermediate-advanced level questions, which essentially test those same very concepts, but at a more in-depth level and obfuscated within a bunch of non-essential information. To get to the next level, which should be a 650+ score in your case, you need to be able to filter out the extraneous information and recognize the basic concepts tested - say Subject-Verb Agreement/Parallelism/Tenses/Incorrect subject after the modifier in the case of SC. You also need to start eliminating by looking at the options themselves - this is especially true for the Verbal section, since on the Verbal section, it's not necessarily about selecting the "100% correct" answer choice but rather selecting the "best" answer choice. For example, options containing "Some/few/could/may/might/possibly" are very rarely correct in Strengthen/Weaken CR questions, and hence should be actively eliminated. Similarly, in the case of Reading Comprehension, you need to be able to figure out the basic opinions given in the passages without wasting time trying to understand the details - for example, if an experiment A and its results are mentioned in a passage, you should not be trying to understand "how" the experiment was performed (that can always be looked at when/if there is a question on the same!), but should be able to answer questions such as "What were the results of this experiment?", "Does the author agree with the results?", "Is there a flaw in the experiment?", etc.

As with everything however, improvement does require diligent preparation, and improving by around 90-100 points on the GMAT usually takes at least 1.5-2 months of preparation. A good way to improve would be to start keeping an ERROR LOG of your mistakes, including the time taken to solve the question and the concepts tested/question type. In terms of content, I would suggest the following plan for the next 1.5-2 months.

First 30-45 days (at least 5-6 hours of preparation per day):

- Go through the SC, CR, and RC strategy guides of a leading GMAT test-prep company - this will brush up and refine your fundamentals of the various concepts that are tested on the GMAT

- Re-solve the latest editions of the Official Guide(2018) and Quant+Verbal Review (2018) using the strategies learnt - remember, you need to be ACTIVELY looking to apply the various strategies learnt to a particular question. Maintain a log of the concepts/strategies that you were able to identify, and also those that you couldn't recognize when you solved a question. Revisit those strategies/concepts again - practice makes perfect!

Final 15 days:

Take full GMATPrep mocks (1-6), including the AWA and IR sections - Please ensure that you take the test under exam conditions, religiously adhering to the timings for a particular section. Take the tests in the time-slot in which you have your test booked - for example, if you have your test scheduled in the evening, then in order to get an accurate benchmark of your performance, you should take the test in the evening, and not in the afternoon/morning. Again, as during practice, analyse your errors from these tests. Also, ensure that no one disturbs you during these 3.5-4 hours during which you are taking the exam. This will stimulate the actual GMAT test-taking environment as closely as possible, and also get rid of any nerves on the actual test day, weeding out "careless" mistakes.

Note that GMATPrep 1&2 are freely available from the official website . ExamPack 1 (Tests 3&4) and ExamPack 2 (Tests 5&6) can be purchased from the official website for \$49.99 each.

Best of luck and happy studying

### GMAT/MBA Expert

ceilidh.erickson GMAT Instructor
Joined
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Posted:
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Thu Apr 19, 2018 6:40 am
Akrita@Jamboree wrote:
Hi Ceilidh,

Firstly, thank you for your detailed response.

However, what I meant by my statement - and it is easy to see as to how it could have been misconstrued - was that the same "scaled score" can fetch a different overall score. The "levels" of a scaled score would be different, sure, and based on a number of factors, such as the difficulty level of the incorrect questions, whether the test taker left any questions, whether he/she made consecutive mistakes, whether he/she made errors on the experimental/marked questions, etc. rather than just the "raw" score. However, the "final scaled score" that a candidate would see, assuming our above example of candidate A getting first 3 incorrect and candidate B getting last 3 incorrect, would remain the same. The only visible difference - from a test taker's perspective [on the score card] - would be in the "%ile" ranking - this was the point I was trying to make. Of course, as you have pointed out, the "level" of the scaled score would determine the "%ile" ranking, and not vice-versa.

For the sake of clarity, I have since gone back and rephrased my earlier post so as to avoid any ambiguities. To summarize and answer Artur's question, although the score you "see" on screen might be a Q44, there are various levels within the same score, and these "levels" would be reflected in the "%ile" ranking rather than the "scaled score"
Thank you for clarifying. Since the percentile issue is a very common misconception, I wanted to be sure that no students had the wrong idea.

_________________

Ceilidh Erickson
Manhattan Prep GMAT & GRE instructor
EdM in Mind, Brain, and Education

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format: plain


text:
ACT Math: Looking for the “Clever” Solution

The Math portion of the ACT tests how well you’ve mastered the math you’ve been taught in middle school and high school. But on some questions, it does more than that. There are some questions that test how deeply you know the material, how good you are at finding an efficient solution to a difficult problem. In a sense, these problems test how “clever” you are.

Consider the following problem:

If xy = 28 and x + y = 7, what is the sum of the reciprocals of x and y?

A.  1/28
B.  1/7
C.  1/4
D.  4/7
E.  3/4

The “Math Class” Solution
Let’s start by doing this problem the way we would if we were in math class at school. It’s really just a systems of equations problem. We can solve the second equation for x or y, do a little substitution and voila! … out come the values of x and y. Doesn’t seem that bad, right?

If xy = 7, then it follows that y = 7 – x. Let’s substitute that into the first equation and solve for x.

xy = 28
x(7 – x) = 28
7x – x² = 28
0 = x² – 7x + 28

We’d try to factor that …. but, darn! It doesn’t factor. That means we’ll need to use Quadratic Formula. And right about now, you’re starting to realize that the values of x and y are really ugly and their reciprocals are probably worse.

The “Clever” Solution
So maybe there’s another way to do this, a way that doesn’t even require us to figure out what x and y are. Let’s instead play around with the thing that we’re trying to find

$\frac{1}{x}+\frac{1}{y}$.

To add those, we’ll need a common denominator. In this case it’s xy.

$\frac{y}{xy}+\frac{x}{xy}=\frac{x+y}{xy}$

Aha! Those two expressions in the numerator and the denominator are the two things we were given in the problem! We’re almost done.

$\frac{x+y}{xy}=\frac{7}{28}=\frac{1}{4}$

The correct answer is Choice C.

A lot of you are probably thinking “great … but I’m not sure I’d ever think of that.” And, yeah, I agree that coming up with that approach is pretty tricky. I do a lot of math, so that approach popped into my head. That’s how you’ll get good at recognizing when to do something “clever” as well. You need to practice a lot of problems, and you need to be flexible in your approach. If the traditional approach looks like it’s going to take you a whole bunch of steps or lead you to really ugly answers, consider another approach. Sometimes being clever can get you to the right answer a whole lot faster.

If you have questions about this problem or anything to do with the ACT, send us an email at info@cardinalec.com


text:
Browse Questions

# Given two regression lines $4x+3y+7=0$ and $3x+4y+8=0$,determine :

(i) The regression line of $y$ on $x$. (ii) The regression line of $x$ on $y$. (iii) The coefficient of correlation. This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com


text:
# Find the Reference Angle -pi/40

Find the Reference Angle -pi/40
Find an angle that is positive, less than , and coterminal with .


text:
## The Vigenere Cipher

The Vigenere Cipher is a clever variation on the Caeser shift cipher that is both easy to implement and resistant to very simple frequency analysis attacks. As such, it is more secure than any of the letter-for-letter substitution ciphers.

To encrypt a message like "here is how it works" using the Vigenere cipher, we first need a key. Typically this key is an easily remembered word or phrase. Suppose, for the sake of this example, our key is the word "vector". Translating the letters of our keyword into integers $\pmod{26}$ in the normal way, we find "vector" corresponds to $(21,4,2,19,14,17)$.

We then take our message and shift it's first letter by 21, it's second letter by 4, it's third letter by 2, it's fourth letter by 19, and so on... Once we get to the end of the key, we start back at the beginning, so the 7th letter is shifted by 21, the 8th is shifted by 4, and so on... to obtain the encrypted message "citxwjcsybhnjvml".

\begin{array}{lcccccccccccccccc}
&h &e &r &e &i &s &h &o &w &i &t &w &o &r &k &s\\
&21 &4 &2 &19 &14 &17 &21 &4 &2 &19 &14 &17 &21 &4 &2 &19\\
&c &i &t &x &w &j &c &s &y &b &h &n &j &v &m &l
\end{array}

Decrypting is easy if the key is known. One just shifts back by the same amounts.

However, note the monkey-wrench this throws in the method of trying to decrypt the message without the key by analyzing the frequency with each letter occurs in the message. The letter "e" might be the most frequent letter in our plaintext, but in the example above, it gets encoded sometimes as an "i", sometimes as an "x", and if our message was longer, we would see still other letters an "e" gets turned into...

This is not enough to stop us, however.

For the purposes of having a good example to talk about, consider the following plaintext message:

 "The method used for the preparation and reading of code messages is simple in the extreme and at the same time impossible of translation unless the key is known. The ease with which the key may be changed is another point in favor of the adoption of this code by those desiring to transmit important messages without the slightest danger of their messages being read by political or business rivals. etc"

As an interesting tidbit, the passage above is used by Wade Trappe and Lawrence C. Washington in their book "Introduction to Cryptography with Coding Theory", and is taken from a short article in a 1917 issue of Scientific American that explains the Vigenere cipher and expresses an opinion as to its security.

Suppose someone has encrypted this message using a Vigenere cipher with a keyword of "codes" to obtain:

 VVHQWVVRHMUSGJGTHKIHTSSEJCHLSFCBGVWCRLRYQTFSVGAHWKCUHWAUGLQHNSLRLJSHBLTSPISP RDXLJSVEEGHLQWKASSKUWEPWQTWVSPGOELKCQYFNSVWLJSNIQKGNRGYBWLWGOVIOKHKAZKQKXZ GYHCECMEIUJOQKWFWVEFQHKIJRCLRLKBIENQFRJLJSDHGRHLSFQTWLAUQRHWDMWLGUSGIKKFLRYV CWVSPGPMLKASSJVOQXEGGVEYGGZMLJCXXLJSVPAIVWIKVRDRYGFRJLJSLVEGGVEYGGEIAPUUISFPBT GNWWMUCZRVTWGLRWUGUMNCZVILE

Now, suppose we are trying to decrypt the message above.

If we had guessed or otherwise determined the keyword was 5 letters long, then we could grab every 5th letter starting with the first V above to get

 VVUTTCCCQGCUNJTPJGKUQPKNJKYGKKGCJFQRKQJRQUDUKVPKVGGJJIVGJGGPFNCWUCE

Remember, all of these letters in this sampling should have been shifted by the same amount. Our first job is to figure out by how much these letters have been shifted -- that is to say, to figure out the first letter of our keyword is "C" (Recall the "C" in "Codes" represents a shift right of 2 letters)

Unfortunately, we can't just try every shift possible (there are only 26, remember). The reason for this is one of recognition -- how do you recognize which shift produced something in English? Remember, we only have every fifth letter. If you take any sentence -- like this one -- and yank out every fifth letter and look at it, it looks a lot like gibberish! (e.g., "ITNTLHEATYHELTLAIBS")

However, we can take advantage of the fact that one of these possibilities should "mostly agree" with English in the frequencies of the letters seen.

We know that the frequencies of the letters in English are given by:
$$\begin{array}{cccccccc} a &b &c &d &e &... &y &z\\ 0.082 &0.015 &0.028 &0.043 &0.127 &... &0.020 &0.001 \end{array}$$
Suppose we denote the frequencies of letters expected in an English message that has been Caeser-shifted $i$ letters to the right by a 26-dimensional vector $v_i$ as shown below. So for example,
$$\begin{array}{llllllll} v_0 &= (&.082, &.015, &.028, &..., &.020, &.001)\\ v_1 &= (&.001, &.082, &.015, &.028, &..., &.020)\\ v_2 &= (&.020, &.001, &.082, &.015, &.028, &...,)\\ \vdots \end{array}$$
One of these vectors should agree fairly closely with the frequencies of letters we see in our sampling of every 5th letter. Which vector that is tells us the shift amount for our sampling, and the first letter of our keyword.

So we compute the frequencies of letters seen in our sampling to find
$$u = (0,0,0.104,0.0149,0.0149,0.0298,...)$$
To find the vector in the previous list above that most closely matches the vector $u$, we recall that the dot product of two vectors is connected to the angle $\theta$ between those two vectors in the following way:
$$u \cdot v = |u||v| \cos \theta$$
If we want to find the two vectors $u$ and $v_i$ that most closely match, we want to find the two vectors with the smallest angle between them.

Noting that smaller angles produce larger cosine values and also noting that the magnitude of the denominator is the same for every $v_i$ as the same 26 numbers are involved each time (just in different orders), we can simply seek the two vectors $u$ and $v_i$ whose dot product is largest.

Computing these dot products we find:
$$\begin{array}{ll} u \cdot v_0 &= 0.025\\ u \cdot v_1 &= 0.039\\ u \cdot v_2 &= 0.071 \leftarrow \textrm{Note how much bigger this one is!}\\ u \cdot v_3 &= 0.039\\ u \cdot v_4 &= 0.027\\ u \cdot v_5 &= 0.038\\ u \cdot v_6 &= 0.051\\ u \cdot v_7 &= 0.030\\ u \cdot v_8 &= 0.032\\ u \cdot v_9 &= 0.043\\ u \cdot v_{10} &= 0.034\\ u \cdot v_{11} &= 0.030\\ u \cdot v_{12} &= 0.034\\ u \cdot v_{13} &= 0.045\\ u \cdot v_{14} &= 0.036\\ u \cdot v_{15} &= 0.040\\ u \cdot v_{16} &= 0.043\\ u \cdot v_{17} &= 0.050\\ u \cdot v_{18} &= 0.039\\ u \cdot v_{19} &= 0.030\\ u \cdot v_{20} &= 0.033\\ u \cdot v_{21} &= 0.039\\ u \cdot v_{22} &= 0.037\\ u \cdot v_{23} &= 0.032\\ u \cdot v_{24} &= 0.049\\ u \cdot v_{25} &= 0.035\\ \end{array}$$
Seeing the biggest dot product above was between $u$ and $v_2$, we conclude that all of the letters in our sampling were shifted 2 letters to the right, and our keyword must hence start with a C. [Recall C = 2 when written in the usual way $\pmod{26}$.]

We find the rest of the letters of the keyword in a similar manner, sampling every 5th letter starting with second, third, fourth, and fifth letters of our overall encrypted message, respectively. Then, by shifting back the letters by these same amounts, we arrive at the decrypted message.

There is a lingering question, however. How did we know at the onset that the keyword had five letters?

$\require{amsmath}$To find the key length, we can take advantage of a very clever trick. Suppose we take the entire encrypted message and line it up with an offset of itself. So appealing to the same example used above, and using an offset of $k=4$ letters, we have:
$$\begin{array}{ccccccccccccccc} V&V&H&Q&W&\fbox{V}&V&R&H&M&U&S&G&\cdots\\ &&&&V&\fbox{V}&H&Q&W&V&V&R&H&\cdots\\ \end{array}$$
Notice how the two boxed letters above agree. What is the probability of this happening? Specifically, what is the probability that $i^{th}$ letter of our encrypted message agrees with the $(i+k)^{th}$ letter of our encrypted message?

There are, of course, lots of ways in which these letters might agree. They could both be $A$'s, or they could both be $B$'s, or they could both be $C$'s, and so on... We'll have to find the probabilities of each one of these disjoint possibilities and add them up.

Suppose the $i^{th}$ letter was encoded with a Caeser-shift of $m$ letters to the right, while the $(i+k)^{th}$ letter was encoded with a Caeser-shift of $n$ letters to the right. Recalling the vectors $v_0,v_1,v_2,...$ that we used earlier, note that the letters that occur in the $i^{th}$ position then do so with frequencies found in $v_m$, and letters that occur in the $(i+k)^{th}$ position do so with frequencies found in $v_n$.

Suppose $v_m = (a_m, b_m, c_m, ..., z_m)$ and $v_n = (a_n, b_n, c_n, ..., z_n)$.

The probability that the $i^{th}$ and $(i+k)^{th}$ letters are both $A$'s is then the product $a_m a_n$. Similarly, the probability that the $i^{th}$ and $(i+k)^{th}$ letters are both $B$'s is then the product $b_m b_n$. As you might expect, this pattern continues -- resulting in the sum of all of these probabilities (and consequently the probability that these letters match) being
$$P(\textrm{match}) = a_m a_n + b_m b_n + \cdots + z_m z_n$$
With some degree of amazement, notice that we can write this another way...
$$P(\textrm{match}) = v_m \cdot v_n$$

Now, suppose we ask a follow-up question: "When is the probability of a match the highest?" The trivial answer is when the dot product $v_m \cdot v_n$ is maximized. But as we have seen before,
$$v_m \cdot v_n = |v_m||v_n| \cos \theta$$
is maximized when $\theta$, the angle between $v_m$ and $v_n$ is minimized.

Noting that when $v_m = v_n$, the angle between them must be zero, we can expect the highest probability of a match when $m=n$. Now observe that this happens whenever $k$ (the offset) is a multiple of the keyword length!

Since a higher probability of a match suggests it occurs more frequently, we simply need to count the number of matches seen in our encrypted message for several possible offsets $k$. The offset $k$ that produces the highest number of matches should be either our keyword length, or a multiple of that length. Ta-Da! Isn't that an amazingly clever attack?


text:
# Functors ¶

Go back

You can give modules as parameters to other modules. These modules are called functors. For instance, you could make a module implementing methods for a set, while not knowing if you are elements are int, strings, etc. This module ("set") will take in argument another module ("element type") describing the type of an element, which you will use when coding your module.

(* declaring a module having a type t.*)
module type SetElementType = sig
type t
end

(* declaring a module describing a set *)
module type AbstractSet = sig
(* types *)
type elt
type set
(* functions *)
val add : set -> elt -> set
val empty : set
val first : set -> elt
end

(*
As always, the set "GenericSet" will have to implements "AbstractSet" types and methods.

We are adding an argument that will have to be provided when creating the generic set: S.
And we are saying that the type of elt (=type of an element) will be equals to the type
stored in the type t of S.
*)
module GenericSet (S: SetElementType) : AbstractSet with type elt = S.t

(* copy-paste of the public parts aside from the functions and the modules that are implemented *)
module type SetElementType = sig
type t
end

module type AbstractSet = sig
(* types *)
type elt
type set
(* functions *)
val add : set -> elt -> set
val empty : set
val first : set -> elt
end

(*
You do not know the type of the elt, aside from the fact
that the function s will give you the type.

You could use this complete definition, but I'm opting for the shorter one
module GenericSet (S: SetElementType) : AbstractSet with type elt = S.t = struct
*)
module GenericSet (S: SetElementType) = struct
(* the type elt is equals to the type inside t *)
type elt = S.t

(* implement it hovewer you want *)
type set = elt list
let add set e = e::set
let empty = []
let first = List.hd
end

open Example

module Int_type = struct
type t = int
end

module Int_Set = GenericSet(Int_type)
(* you could do that too, no need for a ; if you have multiple fields *)
module String_Set = GenericSet(struct type t = string end)

(* list with five *)
let set = Int_Set.add Int_Set.empty 5
let _ = Format.printf "%d@." (Int_Set.first set)

(* list with "five" *)
let set = String_Set.add String_Set.empty "five"
let _ = Format.printf "%s@." (String_Set.first set)

You can compile with these commands

$ocamlc -c example.mli$ ocamlc -c example.ml
$ocamlc -c example_test.ml$ ocamlc example.cmo example_test.cmo -o example_test
\$ ./example_test
5
five


text:
# Software Transactional Memory "for real"

## Introduction

• This talk is about programming multi- or many-core machines

• Armin Rigo
• "Language implementation guy"

|pause|

• PyPy project
• Python in Python
• includes a Just-in-Time Compiler "Generator" for Python and any other dynamic language

## Motivation

• A single-core program is getting exponentially slower than a multi-core one

|pause|

• Using several processes exchanging data
• works fine in some cases
• but becomes a large mess in others

|pause|

• this talk!

## Synchronization with locks

|pause|

• How do you know if you missed a place?
• hard to catch by writing tests
• instead you get obscure rare run-time crashes

|pause|

• Issues when scaling to a large program
• order of acquisition

## Synchronization with TM

• TM = Transactional Memory

|pause|

----------------   --------------------
Locks              Transactional Memory
----------------   --------------------

mylock.acquire();    atomic {
x = list1.pop();       x = list1.pop();
list2.append(x);       list2.append(x);
mylock.release();    }

• Locks
• TM

## Locks versus TM

• Locks
• TM in case of conflict

## Synchronization with TM

• "Optimistic" approach:
• no lock to protect shared data in memory
• instead, track all memory accesses
• detect actual conflicts
• if conflict, restart the whole "transaction"

|pause|

• Easier to use
• no need to name locks
• "composability"

## HTM versus STM

• HTM = Hardware Transactional Memory
• Intel Haswell CPU, 2013
• and others

|pause|

• STM = Software Transactional Memory
• various approaches
• large overhead (2x-10x), but getting faster
• experimental in PyPy: read/write barriers, as with GC

## The catch

|pause|

• You Still Have To Use Threads

|pause|

• Threads are hard to debug, non-reproductible

|pause|

• TM does not solve this problem:
• How do you know if you missed a place to put atomic around?
• hard to catch by writing tests
• instead you get obscure rare run-time crashes

|pause|

• What if we put atomic everywhere?

## Analogy with Garbage Collection

• Explicit Memory Management:
• messy, hard to debug rare leaks or corruptions

|pause|

• Automatic GC solves it
• common languages either have a GC or not
• if they have a GC, it controls almost all objects
• not just a small part of them

## Proposed solution

• Put atomic everywhere...
• in other words, Run Everything with TM

## Proposed solution

• Really needs TM. With locks, you'd get this:
• With TM you can get this:

## In a few words

• Longer transactions
• Corresponding to larger parts of the program
• The underlying multi-threaded model becomes implicit

## Typical example

• You want to run f1() and f2() and f3()

|pause|

• Assume they are "mostly independent"
• i.e. we expect that we can run them in parallel
• but we cannot prove it, we just hope that in the common case we can

|pause|

• In case of conflicts, we don't want random behavior
• i.e. we don't want thread-like non-determinism and crashes

## Pooling and atomic statements

• Solution: use a library that creates a pool of threads
• Each thread picks a function from the list and runs it with atomic

## Results

• The behavior is "as if" we had run f1(), f2() and f3() sequentially
• The programmer chooses if he wants this fixed order, or if any order is fine
• Threads are hidden from the programmer

## More generally

• This was an example only
• TM gives various new ways to hide threads under a nice interface

## Not the Ultimate Solution

• Much easier for the programmer to get reproducible results
• But maybe too many conflicts

|pause|

• "The right side" of the problem


text:
Aside

It is tempting to assume that with the appropriate choice of weights for the edges connecting the second and third layers of the NN discussed in this post, it would be possible to create classifiers that output $1$ over any composite region defined by unions and intersections of the 7 regions shown below.


text:
# Question: What do V, E, and F represent here?

• What does V, E, and F represent here? I'm a bit confused because I'm pretty sure F is the middle pentagon, but I don't know about the rest.

• Lol my strategy doing this question:

pentagon in middle = 5 faces
pyramid with 3 faces per edge of pentagon = 15 faces
side triangles with 2 faces each = 10 faces
assuming there is a whole back layer to the shape = x2
(5+15+10) x 2 = 30 x 2 = 60

• @The-Blade-Dancer Thank you for this post; it's probably a good idea for us to annotate the video with the definitions of these variables, $$V, E, \text{ and } F,$$ and this is something we will put down to do in the future!

I think you figured this out already! At any rate, I'll just give a quick answer.

For any 3-D shape made up of flat regular polygons (called a polyhedron), it's convenient to count its number of faces, vertices, and edges.

$$V$$ is the number of vertices in the polyhedron.
$$E$$ is the number of edges in the polyhedron.
$$F$$ is the number of faces in the polyhedron.

For a cube, $$V = 8, E = 12,$$ and $$F = 6.$$

An edge is usually shared by two faces; a vertex can be shared by multiple faces.

If you count the number of faces and multiply by the sides per face, you will be counting each edge of the polyhedron twice.

If you count the number of faces and multiply by the number of corners per face, you will be counting each vertex of the polyhedron multiple times.

• @The-Blade-Dancer That's a really good way! I like it!

Another slight variation on this method is to say that the 3-D figure is composed of the same repeated building-block, the pyramid with five triangular slant faces.

On our side of the shape, there is a big 5-pointed-star looking at us, and it's composed of $$6$$ of these building-block pyramids, so the 5-pointed-star contains $$30$$ faces. The side facing opposite from us contains the same 5-pointed-star, simply rotated, so it provides another $$30$$ faces.

The total is $$30 + 30 = \boxed{60 \text{ faces}}.$$


text:
# Thread: Help me please with setsettings().

1. ## Help me please with setsettings().

I have it.

PHP Code:
 local ITEMS = {"halberd")if x<cap thensetsettings("Settings\\Looting\\List"ITEMS[0]"\\Destination","ground")end
What to do to set settings item from table loot destination using name from table "ITEMS"?

I want change current loot destination to "ground" using setsettings() function?

Thx for help

2. if u dont have much items just make different lines with item names

3. PHP Code:
 if cap < X thensetsettings("Settings\\Looting\\List\\halberd\\Destination","ground")setsettings("Settings\\Looting\\List\\item2\\Destination","ground")setsettings("Settings\\Looting\\List\\item3\\Destination","ground")...elsesetsettings("Settings\\Looting\\List\\halberd\\Destination","2")setsettings("Settings\\Looting\\List\\item2\\Destination","2")setsettings("Settings\\Looting\\List\\item3\\Destination","2")...end
or just same
PHP Code:
 if cap < X thensetsettings("Settings\\Looting\\List\\halberd\\Destination","ground")setsettings("Settings\\Looting\\List\\item2\\Destination","ground")setsettings("Settings\\Looting\\List\\item3\\Destination","ground")...end

4. i have much item. This is only for example.

5. PHP Code:
 local ITEMS = {"item1","item2"} if cap < 100 then     for ITEM = 1, #ITEMS do         setsettings("Settings\\Looting\\List\\"..ITEMS[ITEM].."\\Destination","ground")     end else     for ITEM = 1, #ITEMS do         setsettings("Settings\\Looting\\List\\"..ITEMS[ITEM].."\\Destination","2")     end end

6. Ty man very much. I solve this problem.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•


text:
# If $$f(x,y) = \frac{xy}{x^2 + y^2}$$, does f(x, y) exist, as (x, y) → (0, 0)?

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Solution

Here

$$f(x,y) = \frac{xy}{x^2 + y^2}$$

As (x, y) → (0, 0), we get f(x, y) → $$\frac{0}{0}$$ form

So, set y = mx for x ≠ 0 where m is some constant value then

$$f(x,y) = \frac{mx^2}{x^2 + (mx)^2}$$

$$= \frac{mx^2}{x^2(1 + m^2)}$$

$$= \frac{m}{1 + m^2}$$

Since y is variable, so m has no fixed value. That means f(x, y) may not have a fixed value as (x, y) → (0, 0).

This means, f(x, y) is not necessarily zero at (x, y) ≠ (0, 0). That is f(x, y) is not continuous at (x, y) = (0, 0).

But $$f(x,y) = \frac{xy}{x^2 + y^2}$$ for f(x, y) ≠ (0, 0) that is defined single form function except at (0, 0).

Since xy and $$x^2 + y^2$$ both are polynomial functions which are continuous, so f(x, y) is continuous except at $$x^2 + y^2 = 0$$. This means f(x, y) is not continuous at (0, 0).


text:
# NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

Written by Team Trustudies
Updated at 2021-05-07

## NCERT solutions for class 7 Maths Chapter 13 Exponents And Powers Exercise 13.1

Q.1 Find the value of
(i) ${2}^{6}$
(ii)${9}^{3}$
(iii) ${11}^{2}$
(iv) ${5}^{4}$

NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

(i) ${2}^{6}=2×2×2×2×2×2=64$
(ii) ${9}^{3}=9×9×9=729$
(iii) ${11}^{2}=11×11=121$
(iv)${5}^{4}=5×5×5×5=625$

Q.2 Exress the following in exponential form:
(i) 6 × 6 × 6 × 6
(ii) t × t
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) a × a × a × c × c × c× c × d

NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

(i) $6×6×6×6={6}^{3}$
(ii)$t×t={t}^{2}$
(iii) $b×b×b×b={b}^{4}$
(iv) $5×5×7×7×7={5}^{2}×{7}^{3}={5}^{2}·{7}^{3}$
(v) $2×2×a×a={2}^{2}×{a}^{2}={2}^{2}·{a}^{2}$
(vi)$a×a×a×c×c×c×c×d={a}^{3}×{c}^{4}×d={a}^{3}·{c}^{4}·d$

Q.3 Express each of the following numbers using exponential notation:
(i) 512
(ii) 343
(iii) 729
(iv) 3125

NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

(i) 512= 2×2×2×2×2×2×2×2×2 = ${2}^{9}$
(ii) 343 = 7×7×7 = ${7}^{3}$
(iii) 729 = 3×3×3×3×3×3 = ${3}^{6}$
(iv) 3125 = 5×5×5×5×5= ${5}^{5}$

Q.4 Identify the greater number, wherever possible, in each of the following?
(i) ${4}^{3}or{3}^{4}$
(ii)${5}^{3}or{3}^{5}$
(iii)${2}^{8}or{8}^{2}$
(iv)${100}^{2}or{2}^{100}$
(v) ${2}^{10}or{10}^{2}$

NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

(i) ${4}^{3}or{3}^{4}$
${4}^{3}=4×4×4=64,$
${3}^{4}=3×3×3×3=81$
Since 81 > 64
$?{3}^{4}>{4}^{3}.$
(ii)${5}^{3}or{3}^{5}$
${5}^{3}=5×5×5=125$
${3}^{5}=3×3×3×3×3=243$
Since 243 > 125
${3}^{5}>{5}^{3}.$
(iii)${2}^{8}or{8}^{2}$
${2}^{8}=2×2×2×2×2×2×2×2=256$
${8}^{2}=8×8=64$
Since 256 > 64
${2}^{8}>{8}^{2}.$
(iv) ${100}^{2}or{2}^{100}$ ${100}^{2}=100×100=10000$
${2}^{100}=2×2×2×\dots 100times=214=16384$
Since 16384 > 10,000
${2}^{100}>{100}^{2}.$
(v)${2}^{10}or{10}^{2}$
${2}^{10}=2×2×2×2×2×2×2×2×2×2=1024$
${10}^{2}=10×10=100$
$?{2}^{10}>{10}^{2}.$

Q.5 Express each of the following as the product of powers of their prime
(i) 648
(ii) 405
(iii) 540
(iv) 3600

NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

(i) 648= ${2}^{3}×{3}^{4}$
(ii) 405= ${3}^{4}×{5}^{1}$
(iii) $540={2}^{2}×{3}^{3}×{5}^{1}$
(iv) $3600={2}^{4}×{3}^{2}×{5}^{2}$

Q.6 Simplify:
(i) $2×{10}^{3}$
(ii) ${7}^{2}×{2}^{2}$
(iii) ${2}^{3}×5$
(iv) $3×{4}^{4}$
(v) $0×{10}^{2}$
(vi)${5}^{2}×{3}^{3}$
(vii) ${2}^{4}×{3}^{2}$
(viii)${3}^{2}×{10}^{4}$

NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

(i) $2×{10}^{3}=2×10×10×10==2000$
(ii)${7}^{2}×{2}^{2}==7×7×2×2=196$
(iii)${2}^{3}×5=2×2×2×5=40$
(iv) $3×{4}^{4}=3×4×4×4×4=768$
(v)$0×{10}^{2}=0×10×10==0$
(vi) ${5}^{2}×{3}^{3}=5×5×3×3×3=675$
(vii) ${2}^{4}×{3}^{2}=2×2×2×2×3×3=144$
(viii) ${3}^{2}×{10}^{4}=3×3×10×10×10×10=90000$

Q.7 Simplify:
(i)$\left(?4{\right)}^{3}$
(ii)$\left(?3\right)×\left(?2{\right)}^{3}$
(iii)$\left(?3{\right)}^{2}×\left(?5{\right)}^{2}$
(iv)$\left(?2{\right)}^{3}×\left(?10{\right)}^{3}$

NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

(i)$\left(?4{\right)}^{3}=\left(?4\right)×\left(?4\right)×\left(?4\right)=?64$ [$?$(-a)odd number = (-a)odd number]
(ii) $\left(?3\right)×\left(?2{\right)}^{3}=\left(?3\right)×\left(?2\right)×\left(?2\right)×\left(?2\right)$
$=\left(?3\right)×\left(?8\right)$= 24 [$?$(-a)odd number = (-a)odd number]
(iii)$\left(?3{\right)}^{2}×\left(?5{\right)}^{2}=\left[\left(?3\right)×\left(?5\right){\right]}^{2}$
$={15}^{2}=225\left[?{a}^{m}×{b}^{m}=\left(ab{\right)}^{m}\right]$
(iv)$\left(?2{\right)}^{3}×\left(?10{\right)}^{3}=\left[\left(?2\right)×\left(?10\right){\right]}^{3}$
$={20}^{2}=8000\left[?{a}^{m}×{b}^{m}=\left(ab{\right)}^{m}\right]$

Q.8 Compare the following:
(i)$2.7×{10}^{12};1.5×{10}^{8}$
(ii) $4×{10}^{14};3×{10}^{1}7$

NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

(i) $2.7×{10}^{12}=2.7×{10}^{4}×{10}^{8}=27000×{10}^{8}$
$27000×{10}^{8}>1.5×{10}^{8}$
$2.7×{10}^{12}>1.5×{10}^{8}$
(ii) $3×{10}^{17}=3×{10}^{3}×{10}^{14}=3000×{10}^{14}$
$3000×{10}^{14}>4×{10}^{14}$
$4×{10}^{14}<3×{10}^{17}$

## NCERT solutions for class 7 Maths Chapter 13 Exponents And Powers Exercise 13.2

Q.1 Using laws of e×ponents, simplify and write the answer in e×ponential form:
(i) ${3}^{2}×{3}^{4}×{3}^{8}$
(ii)${6}^{15}÷{6}^{10}$
(iii)${a}^{3}×{a}^{2}$
(iv)${7}^{x}×{7}^{2}$
(v)$\left({5}^{2}{\right)}^{3}÷{5}^{3}$
(vi) ${2}^{5}×{5}^{5}$
(vii) ${a}^{4}×{b}^{4}$
(viii)$\left({3}^{4}{\right)}^{3}$
(ix)$\left({2}^{20}÷{2}^{15}\right)×{2}^{3}$
(x)${8}^{t}÷{8}^{2}$

NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

(i) ${3}^{2}×{3}^{4}×{3}^{8}={3}^{2+4+8}={3}^{14}\left[{a}^{m}÷{a}^{n}={a}^{m+n}\right]$
(ii) ${6}^{15}÷{6}^{10}={6}^{15?10}={6}^{5}\left[{a}^{m}÷{a}^{n}={a}^{m?n}\right]$
(iii) ${a}^{3}×{a}^{2}={a}^{3+2}={a}^{5}\left[{a}^{m}×{a}^{n}={a}^{m+n}\right]$
(iv)${7}^{x}×{7}^{2}={7}^{x+2}\left[{a}^{m}×{a}^{n}={a}^{m+n}\right]$
(v)$\left({5}^{2}{\right)}^{3}÷{5}^{3}={5}^{2×3}÷{5}^{3}={5}^{6}÷{5}^{3}={5}^{6?3}$
$={5}^{3}\left[\left({a}^{m}{\right)}^{n}={a}^{mn},{a}^{m}÷{a}^{n}={a}^{m?n}\right]$
(vi) ${2}^{5}×{5}^{5}=\left(2×5{\right)}^{5}={10}^{5}\left[{a}^{m}×{b}^{m}=\left(ab{\right)}^{m}\right]$
(vii) ${a}^{4}×{b}^{4}=\left(ab{\right)}^{4}\left[{a}^{m}×{b}^{m}=\left(ab{\right)}^{4}\right]$
(ix)$\left({2}^{20}÷{2}^{15}\right)×{2}^{3}={2}^{20?15}×{2}^{3}={2}^{5}×{2}^{3}={2}^{5+3}={2}^{8}$
(x) ${8}^{t}÷{8}^{2}={8}^{t?2}$

Q.2 Simplify and express each of the following in exponential form:
(i) $\frac{\left({2}^{3}×{3}^{4}×4\right)}{\left(3×32\right)}$
(ii) $\left[\left({5}^{2}{\right)}^{3}×{5}^{4}\right]÷{5}^{7}$
(iii) ${25}^{4}÷{5}^{3}$
(iv) $\frac{3×{7}^{2}×{11}^{8}}{21×{11}^{3}}$
(v) $\frac{{3}^{7}}{{3}^{4}×{3}^{3}}$
(vi) ${2}^{0}+{3}^{0}+{4}^{0}$
(vii) ${2}^{0}×{3}^{0}×{4}^{0}$
(viii) $\left({3}^{0}+{2}^{0}\right)×{5}^{0}$
(ix)$\frac{{2}^{8}×{a}^{5}}{{4}^{3}×{a}^{3}}$
(x) $\left(\frac{{a}^{5}}{{a}^{3}}\right)×{a}^{8}$
(xi) $\frac{{4}^{5}×{a}^{8}{b}^{3}}{{4}^{5}×{a}^{5}{b}^{2}}$
(xii) $\left({2}^{3}×2{\right)}^{2}$

NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

(i) $\frac{{2}^{3}×{3}^{4}×4}{3×32}$ = $\frac{{2}^{3}×{3}^{4}×{2}^{2}}{3×{2}^{5}}$
= $\frac{{2}^{3+2}×{3}^{4}}{3×{2}^{5}}$=$\frac{{2}^{5}×{3}^{4}}{3×{2}^{5}}$
= ${2}^{5?5}×{3}^{4?1}$ = ${2}^{0}×{3}^{3}$= ${3}^{3}$
(ii)$\left[\left({5}^{2}{\right)}^{3}×{5}^{4}\right]÷{5}^{7}$ = $\left({5}^{6}×{5}^{4}\right)÷{5}^{7}$
= $\left({5}^{6+4}÷{5}^{7}$ = ${5}^{10?7}$ = ${5}^{3}$
(iii) ${25}^{4}÷{5}^{3}$ = $\left({5}^{2}{\right)}^{4}÷{5}^{3}$= ${5}^{8}÷{5}^{3}$ = ${5}^{8?3}$= ${5}^{5}$
(iv) $\frac{3×{7}^{2}×{11}^{8}}{21×{11}^{3}}$ = $\frac{3×{7}^{2}×{11}^{8}}{7×3×{11}^{3}}$
= ${7}^{2?1}×{11}^{8?3}$ = $7×{11}^{5}$ \) (v) $\frac{{3}^{7}}{{3}^{4}×{3}^{3}}$ = $\frac{{3}^{7}}{{3}^{4+3}}$ = $\frac{{3}^{7}}{{3}^{7}}$ = 1
(vi) ${2}^{0}+{3}^{0}+{4}^{0}$ = 1+1+1 = 0$\left[?{a}^{0}=1\right]$
(vii) ${2}^{0}×{3}^{0}×{4}^{0}$ = 1×1×1 = 1
(viii) $\left({3}^{0}+{2}^{0}\right)×{5}^{0}$ = (1+1)×1=2×1=2
(ix) $\frac{{2}^{8}×{a}^{5}}{{4}^{3}×{a}^{3}}$ = $\frac{{2}^{8}×{a}^{5}}{\left({2}^{2}{\right)}^{3}×{a}^{3}}$
= $\frac{{2}^{8}×{a}^{5}}{{2}^{6}×{a}^{3}}$ = $$2^{8-6}{×a^{5-3}$$ = ${2}^{2}×{a}^{2}$ = $\left(2a{\right)}^{2}$
(x) $\left(\frac{{a}^{5}}{{a}^{3}}\right)×{a}^{8}$ = $\frac{{a}^{5+8}}{{a}^{3}}$ = $\frac{{a}^{13}}{{a}^{3}}$
= ${a}^{13?3}$ = ${a}^{10}$
(xi) $\frac{{4}^{5}×{a}^{8}{b}^{3}}{{4}^{5}×{a}^{5}{b}^{2}}$ = $\frac{{a}^{8}{b}^{3}}{{a}^{5}{b}^{2}}$
= ${a}^{8?5}{b}^{3?2}$ = ${a}^{3}b$
(xii) $\left({2}^{3}×2{\right)}^{2}$ = ${2}^{6}×{2}^{2}$ = ${2}^{6+2}$ = ${2}^{8}$

Q.3 Say true or false and justify your answer: (i)$10×{10}^{1}1={100}^{1}1$
(ii) ${2}^{3}>{5}^{2}$
(iii)${2}^{3}×{3}^{2}={6}^{5}$
(iv) ${3}^{2}0=\left(1000{\right)}^{0}$

NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

(i)LHS = $10×{10}^{1}1={10}^{1+11}={10}^{12}$
RHS$={100}^{1}1=\left({10}^{2}{\right)}^{1}1={10}^{2}2$
${10}^{1}2?{10}^{2}2$
$?$ Statement is false.
(ii)${2}^{3}>{5}^{2}$
LHS = ${2}^{3}=8$
RHS =${5}^{2}=25$
8 < 25
$?{2}^{3}<{5}^{2}$
Thus, the statement is false.
(iii) ${2}^{3}×{3}^{2}={6}^{5}$
LHS $={2}^{3}×{3}^{2}=8×9=72$
RHS $={6}^{5}=6×6×6×6×6=7776$
$72?7776$
$?$The statement is false.
(iv)${3}^{0}=\left(1000{\right)}^{0}$
LHS = ${3}^{0}=1$
RHS = $\left(1000{\right)}^{0}=1$
LHS = RHS

Q.4 Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
(ii) 270
(iii) 729 × 64
(iv) 768

NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

(i) 108 × 192 = 2 × 2 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
=${2}^{8}×{3}^{4}$
(ii) $270=2×3×3×3×5=2×{3}^{3}×5$
(iii) 729 × 64 = 3 × 3 × 3 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2
$={3}^{6}×{2}^{6}$
(iv) 768 = 2×2×2×2×2×2×2×2×3 = ${2}^{8}×3$

Q.5 Simplify:
(i) $\frac{\left({2}^{5}{\right)}^{2}×{7}^{3}}{{8}^{3}×7}$
(ii) $\frac{25×{5}^{2}×{t}^{8}}{{10}^{3}×{t}^{4}}$
(iii) $\frac{{3}^{5}×{10}^{5}×25}{{5}^{7}×{6}^{5}}$

NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers

(i) $\frac{\left({2}^{5}{\right)}^{2}×{7}^{3}}{{8}^{3}×7}$ = $\frac{\left({2}^{10}×{7}^{3}}{\left({2}^{3}{\right)}^{3}×7}$
= $$\frac{(2^{10}× 7^3}{{2^9 × 7}$$ = ${2}^{10?9}×{7}^{3?1}$ = ${2}^{1}×{7}^{2}$ =2× 49 = 98
(ii) $\frac{25×{5}^{2}×{t}^{8}}{{10}^{3}×{t}^{4}}$ = $\frac{{5}^{2}×{5}^{2}×{t}^{8}}{{2}^{3}×{5}^{3}×{t}^{4}}$
= $\frac{{5}^{4}×{t}^{8}}{{2}^{3}×{5}^{3}×{t}^{4}}$ = $\frac{{5}^{4?3}×{t}^{8?4}}{{2}^{3}}$ = $\frac{5{t}^{4}}{{2}^{3}}$
(iii) $\frac{{3}^{5}×{10}^{5}×25}{{5}^{7}×{6}^{5}}$= $\frac{{3}^{5}×{2}^{5}×{5}^{5}×{5}^{2}}{{5}^{7}×{2}^{5}×{3}^{5}}$


text:
# The geometry of $$\mathcal{N}=3$$ AdS4 in massive IIA

Journal of High Energy Physics, Aug 2018

Abstract The geometry of the $$\mathcal{N}=3$$, SO(4)-invariant, AdS4 solution of massive type IIA supergravity that uplifts from the $$\mathcal{N}=3$$ vacuum of D = 4 $$\mathcal{N}=8$$ dyonic ISO(7) supergravity is investigated. Firstly, a D = 4, SO(4)-invariant restricted duality hierarchy is constructed and used to uplift the entire, dynamical SO(4)-invariant sector to massive type IIA. The resulting consistent uplift formulae are used to obtain a new local expression for the $$\mathcal{N}=3$$ AdS4 solution in massive IIA and analyse its geometry. Locally, the internal S6 geometry corresponds to a warped fibration of S2 and a hemisphere of S4. This can be regarded as a warped generalisation of the usual twistor fibration geometry. Finally, the triplet of Killing spinors corresponding to the $$\mathcal{N}=3$$ solution are constructed and shown to obey the massive type IIA Killing spinor equations.

This is a preview of a remote PDF: https://link.springer.com/content/pdf/10.1007%2FJHEP08%282018%29133.pdf

G. Bruno De Luca, Gabriele Lo Monaco, Niall T. Macpherson, Alessandro Tomasiello, Oscar Varela. The geometry of $$\mathcal{N}=3$$ AdS4 in massive IIA, Journal of High Energy Physics, 2018, 133, DOI: 10.1007/JHEP08(2018)133


text:
# Carathéodory metric

In mathematics, the Carathéodory metric is a metric defined on the open unit ball of a complex Banach space that has many similar properties to the Poincaré metric of hyperbolic geometry. It is named after the Greek mathematician Constantin Carathéodory.

## Definition

Let (X, || ||) be a complex Banach space and let B be the open unit ball in X. Let Δ denote the open unit disc in the complex plane C, thought of as the Poincaré disc model for 2-dimensional real/1-dimensional complex hyperbolic geometry. Let the Poincaré metric ρ on Δ be given by

${\displaystyle \rho (a,b)=\tanh ^{-1}{\frac {|a-b|}{|1-{\bar {a}}b|}}}$

(thus fixing the curvature to be −4). Then the Carathéodory metric d on B is defined by

${\displaystyle d(x,y)=\sup\{\rho (f(x),f(y))|f:B\to \Delta {\mbox{ is holomorphic}}\}.}$

What it means for a function on a Banach space to be holomorphic is defined in the article on Infinite dimensional holomorphy.

## Properties

• For any point x in B,
${\displaystyle d(0,x)=\rho (0,\|x\|).}$
• d can also be given by the following formula, which Carathéodory attributed to Erhard Schmidt:
${\displaystyle d(x,y)=\sup \left\{\left.2\tanh ^{-1}\left\|{\frac {f(x)-f(y)}{2}}\right\|\right|f:B\to \Delta {\mbox{ is holomorphic}}\right\}}$
• For all a and b in B,
${\displaystyle \|a-b\|\leq 2\tanh {\frac {d(a,b)}{2}},\qquad \qquad (1)}$
with equality if and only if either a = b or there exists a bounded linear functional ℓ ∈ X such that ||ℓ|| = 1, ℓ(a + b) = 0 and
${\displaystyle \rho (\ell (a),\ell (b))=d(a,b).}$
Moreover, any ℓ satisfying these three conditions has |ℓ(a − b)| = ||a − b||.
• Also, there is equality in (1) if ||a|| = ||b|| and ||a − b|| = ||a|| + ||b||. One way to do this is to take b = −a.
• If there exists a unit vector u in X that is not an extreme point of the closed unit ball in X, then there exist points a and b in B such that there is equality in (1) but b ≠ ±a.

## Carathéodory length of a tangent vector

There is an associated notion of Carathéodory length for tangent vectors to the ball B. Let x be a point of B and let v be a tangent vector to B at x; since B is the open unit ball in the vector space X, the tangent space TxB can be identified with X in a natural way, and v can be thought of as an element of X. Then the Carathéodory length of v at x, denoted α(xv), is defined by

${\displaystyle \alpha (x,v)=\sup {\big \{}|\mathrm {D} f(x)v|{\big |}f:B\to \Delta {\mbox{ is holomorphic}}{\big \}}.}$

One can show that α(xv) ≥ ||v||, with equality when x = 0.


text:
# Try spelling the method you need

Calculus Level 2

The value of the integral $\int_0^{\pi/2}\dfrac{\text dx}{2+\cos x}$ can be expressed in the form $$\dfrac{\pi^A\sqrt{B}}{C}$$ where $$A$$, $$B$$, $$C$$ are positive integers and $$B$$ is not divisible by the square of any prime. Find the value of $$A+B+C.$$

×


text:
# The Binomial Theorem

1. May 26, 2005

### Cincinnatus

It seems like this shouldn't be too difficult and yet i'm stumped.
I am trying to prove the binomial theorem.

(x+y)^n = the sum from k=0 to n of (x^k)*(y^n-k)*(The binomial coefficient n,k)

Anyway, I figure the best way to go about proving this is by induction.
It is easy to show that its true for n=1.
Then I assume that there exists an n_0 such that it is true for all n < n_0.
Now I want to show that the existence of this n_0 implies that the proposition is also true for n=n_0+1.

This is where I get stuck...
My question is, is this even the best way to go about proving this? If so, how can I finish the proof?

Maybe it would be better to give me a hint so I can figure it out on my own...

2. May 26, 2005

### Dr.Brain

http://planetmath.org/encyclopedia/InductiveProofOfBinomialTheorem.html [Broken]


text:
# What is the key word that indictes that the Addition Rule for Probabilities will be used?

If two events, $A$ and $B$, are mutually exclusive, the probability of either one or another to happen equal to a sum their probabilities (notice different symbols used to describe logical condition $\mathmr{and}$:
$P \left(A \mathmr{and} B\right) = P \left(A + B\right) = P \left(A \cup B\right) = P \left(A\right) + P \left(B\right)$


text:
1. Dec 16, 2016

Are photons(when described as a particle) considered objects? You can't see them on an individual level, but they hit you and interact with gravity. Might be a silly question, but was just debating it with a friend.

2. Dec 16, 2016

### Drakkith

Staff Emeritus
That pretty much depends on what you define as an object. I doubt there's a hard answer to your question.

3. Dec 16, 2016

### phinds

Photons are real and I think it's fair to consider them as objects, but they are not the little billiard balls that you sometimes see them represented as and light does not travel as photons. It travels as a wave and it is the wave that "hits" you. A photon is the result of the wave interacting with a particle (of your skin or your retina or a phosphorous screen, etc.)

4. Dec 16, 2016

### Delta²

A photon doesn't have dimensions like an object big or small has(of course then again it depends how you define an object). A photon is a point particle. It does NOT have mass either (though it interacts with gravity but this is due to the curvature of space time due to gravitational field). So a photon doesn't have two of the most (at least according to me) important properties of an object, that is dimensions and mass, so I doubt we can consider a photon to be an object.

5. Dec 16, 2016

### DennisN

It's not a silly question.
That's true. But there are other ways to detect them, see e.g. these links
• Nondestructive Detection of an Optical Photon

6. Dec 23, 2016

What about Gluons?(the other massless particle) do they have dimensions?

7. Dec 24, 2016

### Delta²

No all the elementary particles of the standard model regardless if they are fermions like electron and quark, or bosons like photon and gluon are all considered to be point like particles.

Even if we are to make a modification of the standard model where the elementary particles have dimensions they would be very small for the current technology to see it (dimensions of the order of $10^{-100}$ or even smaller). However the main argument against such modification is that if an elementary particle has dimensions, then it should consist from other smaller particles which have even smaller dimensions and these smaller particles if they have smaller dimensions they should consist from other particles with even smaller dimensions and so on, so we are like forced to accept that if a particle is elementary then it has no dimensions, it is point like.

8. Dec 25, 2016

### electro-physics

Does that mean photon is part of matter to which light hits and not part of light wave itself?

And if it's true that photon is part of matter then does it shows that this photon should always have some tiny mass?

If photon has no mass then can we say it's the discrete property of a shock wave generated when light bounces with matter?

9. Dec 25, 2016

### Drakkith

Staff Emeritus
It does not. It just means that the interaction of the EM wave with matter is quantized, with each interaction transferring a discrete amount of energy instead of a continuous amount.

10. Dec 25, 2016

Photons are particle as well as wave
This phenomenon is known as wave-particle duality
Every thing (including you) behaves as a wave . Very object has a wavelength but bigger objecys like us have very very very small wavelength but as the things goes on decreasing in size there wavelength starts increasing and at a very small level like at photon level they starts behaving like a wave
This phenomenon is known as wave larticle duality

11. Dec 25, 2016

### weirdoguy

No they are not. Wave-particle duality is an outdated concept.

12. Dec 25, 2016

### SleepDeprived

But seriously, if wave-particle duality is an outdated concept, what has replaced it? Could you provide a link?
There is a online video at World Science Festival where Brian Green, Sean Carrol ... and others talk about how they are still struggling trying to explain the wave-particle duality I guess there are still a lot of "we just don't know yet".

[Mentor's note: This post has been edited to remove some off-topic comments]

Last edited by a moderator: Dec 26, 2016
13. Dec 25, 2016

### phinds

Well, "replaced" isn't quite the right concept, it was just dropped since it was not correct. Quantum Mechanics says that EM radiation is not a particle or a wave, it is a quantum object and will exhibit wave characteristics if you measure for wave characteristics and it will exhibit particle characteristics if you measure for particle characteristics but neither of those types of measurements means that the quantum object is a particle or a wave.

14. Dec 26, 2016

So because all of these point like elementary particles don't have dimensions, does this also mean that they have no volume as well?

15. Dec 26, 2016

### Staff: Mentor

The longer answer: They are called "particles" for historical reasons, but they are not anything like what the ordinary English-language word means. That word suggests something like a tiny grain of sand with some tiny but definite volume that's located somewhere in space (although we may not know exactly where); and quantum particles are nothing like that.

16. Dec 26, 2016

### Staff: Mentor

Wave-particle duality is part of the "old quantum theory" which was replaced by the modern theory of quantum mechanics; the key papers of the modern theory were published starting in 1925, and the new theory was largely complete by 1933 or thereabouts. Even then, however, the theory did not allow for the effects of special relativity, so could only be used when the speeds involved were small compared with the speed of light and the energies involved were small compared with the $E=mc^2$ energies of the particles involved. Of course, neither condition applies to photons ($m=0$, and their speed is the speed of light); it took another decade or so of work to hammer out quantum electrodynamics and quantum field theories to arrive at the modern understanding of photons.

That video is not exactly wrong (it's more of a lie to children) but it's also not a serious explanation of what the science really says. Sources like this are not acceptable references under the Physics Forums rules - sometimes there's no substitute for the real thing.

17. Dec 27, 2016

### lychette

There are no outdated concepts.....please qualify by quoting the current concept. All current physics txt books refer to wave-particle duality in their explanation of modern physics. Which ones should we ignore ??

18. Dec 27, 2016

### Staff: Mentor

You will find the modern form of the theory (which isn't that new - about 85 years old new) in many texts - https://www.amazon.com/Quantum-Mechanics-Development-Leslie-Ballentine/dp/9810241054 is a good one. Or, you can just look at the Schrodinger equation - there's no wave-particle duality there, just quantum states with more or less variance in the possible outcomes of a position measurement.

There are certainly modern textbooks that refer to wave-particle duality, but most of them do so only as part of the historical summary that they use to motivate the development of QM.

Last edited by a moderator: May 8, 2017
19. Dec 27, 2016

### lychette

Are you suggesting that wave - particle duality has no place in modern teaching of physics. Should it be transferred to teaching of history?

Last edited by a moderator: May 8, 2017
20. Dec 27, 2016

Exactly.


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# eig¶

## Purpose¶

Computes the eigenvalues of a general matrix.

## Format¶

va = eig(x)
Parameters: x (NxN matrix or KxNxN array) – data used to compute the eigenvalues. va (Nx1 vector or KxNx1 array) – the eigenvalues of x.

## Examples¶

x = {  0.5  1.2  0.3,
0.6  0.9  0.2,
0.8  1.5  0.0 };

va = eig(x);

After the above code, va will equal:

1.8626
-0.1871
-0.2754

To calculate eigenvalues and eigenvectors see eigv(). To calculate generalized eigenvalues and eigenvectors, see lapgeig(), or lapgeigv().

## Remarks¶

If x is an array, va will be an array containing the eigenvalues of each 2-dimensional array described by the two trailing dimensions of x. For example, if x is a 10x4x4 array, va will be a 10x4x1 array containing the eigenvalues of each of the 10 4x4 arrays contained in x.

Errors

If the eigenvalues cannot all be determined, va[1] is set to an error code. Passing va[1] to the scalerr() function will return the index of the eigenvalue that failed. The eigenvalues for indices $$scalerr(va[1])+1 \to N$$ should be correct.

Error handling is controlled with the low bit of the trap flag.

 trap 0 set va[1] and terminate with message trap 1 set va[1] and continue execution

Invalid inputs, such as an $$\infty$$, missing value or NaN will cause an error. If the trap is set to 1, va will be set to a scalar error code and program execution will continue. Passing this scalar error code to the scalerr() function will return -1.

Eigenvalue ordering

The eigenvalues are unordered except that complex conjugate pairs of eigenvalues will appear consecutively with the eigenvalue having the positive imaginary part first.


text:
# Why is the supply of RTG's running short?

I always hear that NASA has only a limited amount of radioisotope thermoelectric generators (RTGs) remaining. I did some superficial research on how they work and what is needed to produce them and couldn't really understand what is the problem with just manufacturing new RTGs.

So why can't NASA's stock of RTGs simply be resupplied?

The Plutonium isotope 238 used in RTGs is highly specialized. It's not produced in large quantities routinely. Not very many radioisotope applications need that much of a highly radioactive isotope, and it's only produced in certain reactors. In fact, there was only one reactor in the USA that produced it. Nuclear stuff is expensive in general and, now that the 1950s period of nuclear optimism is behind us, it's very burdened by regulation.

The Plutonium isotope 238 may be used for RTGs, but not for nuclear weapons. Nuclear weapons mainly use the isotope 239 which can not be used for RTGs.

See Wikipedia for 238Pu and 239Pu.

Most of the USA's supply until recently was a byproduct of the nuclear weapons program in the Cold War. We are (mercifully) no longer building large quantities of nuclear weapons, so RTG plutonium now must be produced as a special job.

For typically political reasons, no strong effort to start making plutonium for RTGs was established until we were nearly out. However, it's now in progress. Ramping up has been slow. The extreme radioactivity and the small concentration of useful product make things harder.

• I agree this is correct, but some references would make it superb. – Organic Marble Jul 16 '20 at 12:06

Any isotope used as the basis for a radioisotope thermoelectric generator (RTG) has to have a short but not too short half life. A half life of several decades is ideal. Such isotopes effectively do not exist in nature. (Tritium, with a half life of 12.32 years, does exist in nature in trace amounts due to generation by cosmic rays, but the half life is a bit too short.)

Plutonium-238 ($$^{238}\text{Pu}$$), with a half life of 87.7 years, is ideal. While extremely trace amounts of $$^{238}\text{Pu}$$ do exist naturally due to extremely rare decays of primordial elements, these are atypical decays. The amount of naturally occurring $$^{238}\text{Pu}$$ is so small that it is essentially non-existent. The only practical source of $$^{238}\text{Pu}$$ is a breeder reactor.

Breeder reactors are typically used to produce products that can be used in nuclear bombs or in nuclear reactors. $$^{238}\text{Pu}$$ is not good for either of those purposes. The only practical use of $$^{238}\text{Pu}$$ is in RTGs and their equivalent. With an apparent surplus back in the 1980s, the US stopped production of $$^{238}\text{Pu}$$ in 1988. This surplus is now gone, and so is Russia's. Production has started again as of late, but the production rate is very low.

• Now that you mention the half-life, I guess it also doesn't help that since 1988 about 20% of the Pu-238 produced then will already be gone due to natural decay. – mlk Jul 17 '20 at 7:11
• @mlk and because of that it probably has to be re-refined to be usable today. – Mark Ransom Jul 17 '20 at 17:57
• As well as the half-life, you want an element which is an alpha-emitter (and whose decay producets are either fairly stable or also alpha emitters) so that the radiation doesn't impact the rest of your probe, and is easily converted into heat in the RTG. – Steve Linton Jul 17 '20 at 18:01
• Pu-238 actually is used in triggers for nukes. – ikrase Jul 20 '20 at 4:53

It's not just the half-life of the material (short enough to make heat, long enough to have a slow changing energy curve) It's also the type of radiation.

For instance Cs137 is plentiful and easy to separate, and it's a beta emitter, which is shieldable. But its decay products are gamma emitters, which is not shieldable. So nope.

Pu238 is an alpha emitter, which is the easiest stuff to shield... and its decay products are (only) alpha emitters, or stable. It's the right stuff.

The problem is, you have to make it by exposing Neptunium-237 to neutron bombardment inside a reactor. Same thing you do to U238 to make Pu239 for weapons. And you get great gobs of Neptunium-237 when you run almost any reactor, and it's easy to separate if you're already running a PUREX line to extract plutonium-239 for nuclear weapons.

But there'd be no earthly reason to do that, unless you are breeding plutonium on purpose for weapons, and that requires the output of a special reactor capable of 90-day fuel changes. That is a short list consisting of Hanford style purpose-built reactors, CANDU, or RBMK.

Aha... suddenly you understand the importance of the rather oddball, silly and dangerous RBMK in Soviet nuclear strategy.

By the way, loading neptunium into a CANDU reactor for short term neutron exposure is exactly what Canada is doing for us, to help us manufacture some RTG fuel. The Russians, our friends in space exploration, could do the same with their RBMK reactors... and again I suspect that’s the reason Russia still keeps them. Not for Pu238 but Pu239.

I suppose you could also drop a Neptunium rod in a pure civilian BWR or PWR type, and stop it every 30-90 days to change fuel. But it's a huge production to change fuel: you must cool the reactor and remove the top of the pressure vessel, control rod drives and all. The question, is if you leave the Neptunium rod in there for the normal refueling interval of several years, will that overexpose it, have the lovely Pu238 capture another neutron or three, and yield the useless-to-an-RTG Pu239 or the makes-an-RTG-dangerous Pu240 or 241?

I don't know the answer to that question, but I can tell you it doesn't work for weapons Pu239 breeding. Pu240/241-contaminated Pu239 is useless for bomb making, because its spontaneous fission will create neutrons at inopportune times shall we say. That is why we can let countries like Iran have BWR/PWR type reactors; they would have to change fuel every 30-90 days to prevent accumulation of Pu240/241, and everyone with an IR satellite watches their reactors' cooling systems to make sure they are not doing that.

I’ve been asked for a sidebar on what’s bad about Pu240 and Pu241? 240 after all has a decay chain that is all alpha, so what’s the prob? The spontaneous fission releases gamma, that you must then shield, and also creates 2 daughter isotopes out of at least 8 possibilities, and those have their own decay chains, often releasing gamma or beta. It’s out of control at that point. Other than that, Mrs. Lincoln, if Pu240 chooses to alpha-decay instead of split, the subsequent decays are all alpha, which would be alright... but Pu241 can’t say the same. If you’re leaving the Neptunium-237 in the reactor long enough to get Pu240, you’re also getting Pu241.

Aside from the other way Pu240/1 affect bombs, the gamma emitters are problematic for both bombs and RTGs. Because humans have to handle those bombs, sailors have to sleep right next to them, and spacecraft need RTG shielding to be liftable. In fact the Navy uses special Pu239 in their bombs, for the sake of crews. Their U238 “ore” spent far less time in the reactor, so has far less Pu240/1 at the expense of far less Pu239 also, meaning lower yield “ore”.

....

And then, you have to set up a PUREX line to separate out both the neptunium and plutonium, and IAEA is just gonna love that.

By the way, the reason it's so important to make the correct plutonium isotope in the reactor is that separating plutonium chemically from other stuff is a straightforward chemical task; but separating plutonium isotopes from each other is effectively impossible. It's only possible with natural uranium because it's 3 units apart instead of 1 (235 vs 238), and even then it's ridiculously hard.

• +1 for the excellent explanation of the importance of chemical separation instead of isotope separation. An explanation (or a link to Wikipedia) of the abbreviations like CANDU or RBMK would be nice. – Uwe Jul 18 '20 at 14:15
• why are Pu240/241 poison? – Noone AtAll Jul 18 '20 at 18:05
• @NooneAtAll Second this. Pu240 is poison to a bomb because it's a neutron emitter that's prone to causing a predetonation (not that it explodes sitting there, but that it explodes after the detonator is fired but before full assembly is reached--this robs the bomb of most of it's power.) Other than being a neutron emitter I don't see how it's a problem for an RTG, though. – Loren Pechtel Jul 19 '20 at 2:15
• @LorenPechtel My first paragraph. Pu238 decays alpha into 2 other things with long half-lives that also decay alpha. Pu240, granted, does the same thing. However Pu241 is a mess; it decays beta into something which decays gamma and sometimes spontaneously fissions, releasing more gamma. If you left your Neptunium in the reactor long enough to make Pu240, you’re also making Pu241. – Harper - Reinstate Monica Jul 19 '20 at 4:13
• @Harper-ReinstateMonica That makes the RTG hazardous to deal with, it doesn't poison it. It will still produce power. – Loren Pechtel Jul 19 '20 at 4:15


text:
# Bayesian interpretation of ridge regression

Assume that we are in the standard supervised learning setting, where we have a response vector $y \in \mathbb{R}^n$ and a design matrix $X \in \mathbb{R}^{n \times p}$. Ordinary least squares seeks the coefficient vector $\beta \in \mathbb{R}^p$ which minimizes the residual sum of squares (RSS), i.e.

$\hat{\beta} = \underset{\beta}{\text{argmin}} \; (y- X\beta)^T (y - X\beta).$

Ridge regression is a commonly used regularization method which looks for $\beta$ that minimizes the sum of the RSS and a penalty term:

$\hat{\beta} = \underset{\beta}{\text{argmin}} \; (y- X\beta)^T (y - X\beta) + \lambda \| \beta\|_2^2,$

where $\|\beta\|_2^2 = \beta_1^2 + \dots + \beta_p^2$, and $\lambda \geq 0$ is a hyperparameter.

The ridge regression estimate has a Bayesian interpretation. Assume that the design matrix $X$ is fixed. The ordinary least squares model posits that the conditional distribution of the response $y$ is

$y \mid X, \beta \sim \mathcal{N}(X\beta, \sigma^2 I),$

where $\sigma > 0$ is some constant. In frequentism we think of $\beta$ as being some fixed unknown vector that we want to estimate. In Bayesian statistics, we can impose a prior distribution on $\beta$ and perform any estimation we want using the posterior distribution of $\beta$.

Let’s say our prior distribution of $\beta$ is that the $\beta_j$‘s are independent normals with the same variance, i.e. $\beta \sim \mathcal{N}(0, \tau^2 I)$ for some constant $\tau$. This allows us to compute the posterior distribution of $\beta$:

\begin{aligned} p(\beta \mid y, X) &\propto p(\beta) \cdot p(y \mid X, \beta) \\ &\propto \exp \left[ - \frac{1}{2} (\beta - 0)^T \frac{1}{\tau^2} I (\beta - 0) \right] \cdot \exp \left[ -\frac{1}{2}(y - X\beta)^T \frac{1}{\sigma^2} (y - X\beta) \right] \\ &= \exp \left[ -\frac{1}{2\sigma^2}(y-X\beta)^T (y - X \beta) - \frac{1}{2\tau^2} \|\beta\|_2^2 \right]. \end{aligned}

From this expression, we can compute the mode of the posterior distribution, which is also known as the maximum a posteriori (MAP) estimate. It is

\begin{aligned} \hat{\beta} &= \underset{\beta}{\text{argmax}} \quad \exp \left[ -\frac{1}{2\sigma^2}(y-X\beta)^T (y - X \beta) - \frac{1}{2\tau^2} \|\beta\|_2^2 \right] \\ &= \underset{\beta}{\text{argmin}} \quad \frac{1}{\sigma^2}(y-X\beta)^T (y - X \beta) + \frac{1}{\tau^2} \|\beta\|_2^2 \\ &= \underset{\beta}{\text{argmin}} \quad (y-X\beta)^T (y - X \beta) + \frac{\sigma^2}{\tau^2} \|\beta\|_2^2, \end{aligned}

which is the ridge regression estimate when $\lambda = \dfrac{\sigma^2}{\tau^2}$.

## 8 thoughts on “Bayesian interpretation of ridge regression”

1. Hi, it seems that you omit 1/2 in the exponent of your formula, although it doesn’t affect the results.

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• The 1/2 is hidden in the \propto sign.

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• sorry, I mean the coefficient in the exponent, which I think it is not true to omit it.

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• Ah yes you are right. We can’t drop the 1/2 in the exponent when computing the posterior (but we can drop it when computing the MAP estimate). I’ve fixed the post to reflect this.

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2. Hi! I wonder what is ‘I’ in normal distribution notation N(XB, sigma^2 I)?

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• It refers to the identity matrix (of the correct size)!

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• Thank you!

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# Why is second-order Gaussian called Laplacian Gaussian?

We usually use Laplacian of Gaussian as the filter for edge detection or blob detection. But the filter itself is essentially a second order Gaussian. So why do we call it Laplacian of Gaussian? Is there some other meanings behind it?

• HINT: A second order of what operation on a Gaussian? – A_A Feb 28 '18 at 11:41

## 2 Answers

Applying a Gaussian filter is a linear operation, also taking any type of derivative (finite difference) like Laplacian are also linear operations. Now considering the order of applying these operations doesn't matter (a property of convolution), you could combine them in many different ways, e.g. take Laplacian at first then take Gaussian or Gaussian first and Laplacian next, also you could apply Gausian on Laplacian or Laplacian on Gaussian and find a LoG Kernel, then apply the obtained Kernel on image once, which is more efficient in number of computations.

• More efficient in number of operations, but also much more precise. With the LoG you compute the exact Laplacian of the blurred image, rather than a finite difference approximation of the Laplacian. – Cris Luengo May 29 '18 at 18:15

I think it's calles LoG because as you wrote you take the second derivation of the gaussian filter.

A laplace operation is a derivation in 2 (or more) dimensions . In 2 dimensions it's what you do when creating the kernel. You derive the Gausskernel with the laplacian operation.

For more details on the laplacian operator see laplace operator wolfram alpha.


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# 8.7: Appendix - Felsenstein's Pruning Algorithm

Felsenstein’s pruning algorithm (1973) is an example of dynamic programming, a type of algorithm that has many applications in comparative biology. In dynamic programming, we break down a complex problem into a series of simpler steps that have a nested structure. This allows us to reuse computations in an efficient way and speeds up the time required to make calculations.

The best way to illustrate Felsenstein’s algorithm is through an example, which is presented in the panels below. We are trying to calculate the likelihood for a three-state character on a phylogenetic tree that includes six species.

Figure 8.4A. Each tip and internal node in the tree has three boxes, which will contain the probabilities for the three character states at that point in the tree. The first box represents a state of 0, the second state 1, and the third state 2. Image by the author, can be reused under a CC-BY-4.0 license.

1. The first step in the algorithm is to fill in the probabilities for the tips. In this case, we know the states at the tips of the tree. Mathematically, we state that we know precisely the character states at the tips; the probability that that species has the state that we observe is 1, and all other states have probability zero:
1. Next, we identify a node where all of its immediate descendants are tips. There will always be at least one such node; often, there will be more than one, in which case we will arbitrarily choose one. For this example, we will choose the node that is the most recent common ancestor of species A and B, labeled as node 1 in Figure 8.2B.

2. We then use equation 7.6 to calculate the conditional likelihood for each character state for the subtree that includes the node we chose in step 2 and its tip descendants. For each character state, the conditional likelihood is the probability, given the data and the model, of obtaining the tip character states if you start with that character state at the root. In other words, we keep track of the likelihood for the tipward parts of the tree, including our data, if the node we are considering had each of the possible character states. This calculation is:

$$L_P(i) = (\sum\limits_{x \in k}Pr(x|i,t_L)L_L(x)) \cdot (\sum\limits_{x \in k}Pr(x|i,t_R)L_R(x)) \label{8.7}$$

where i and x are both indices for the k character states, with sums taken across all possible states at the branch tips (x), and terms calculated for each possible state at the node (i). The two pieces of the equation are the left and right descendant of the node of interest. Branches can be assigned as left or right arbitrarily without affecting the final outcome, and the approach also works for polytomies (but the equation is slightly different). Furthermore, each of these two pieces itself has two parts: the probability of starting and ending with each state along the two branches being considered, and the current conditional likelihoods that enter the equation at the tips of the subtree (LL(x) and LR(x)). Branch lengths are denoted as tL and tR for the left and right, respectively.

One can think of the likelihood "flowing" down the branches of the tree, and conditional likelihoods for the left and right branches get combined via multiplication at each node, generating the conditional likelihood for the parent node for each character state (LP(i)).

Consider the subtree leading to species A and B in the example given. The two tip character states are 0 (for species A) and 1 (for species B). We can calculate the conditional likelihood for character state 0 at node 1 as:

$$L_P(0) = \left(\sum\limits_{x \in k}Pr(x|0,t_L=1.0)L_L(x)\right) \cdot \left(\sum\limits_{x \in k}Pr(x|0,t_R=1.0)L_R(x)\right) \label{8.8}$$

Next, we can calculate the probability terms from the probability matrix P. In this case tL = tR = 1.0, so for both the left and right branch:

$$\mathbf{Q}t = \begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \\ \end{bmatrix} \cdot 1.0 = \begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \\ \end{bmatrix} \label{8.9}$$

So that:

$$\mathbf{P} = e^{Qt} = \begin{bmatrix} 0.37 & 0.32 & 0.32 \\ 0.32 & 0.37 & 0.32 \\ 0.32 & 0.32 & 0.37 \\ \end{bmatrix} \label{8.10}$$

Now notice that, since the left character state is known to be zero, LL(0)=1 and LL(1)=LL(2)=0. Similarly, the right state is one, so LR(1)=1 and LR(0)=LR(2)=0.

We can now fill in the two parts of equation 8.2:

$$\sum\limits_{x \in k}Pr(x|0,t_L=1.0)L_L(x) = 0.37 \cdot 1 + 0.32 \cdot 0 + 0.32 \cdot 0 = 0.37 \label{8.11}$$

and:

$$\sum\limits_{x \in k}Pr(x|0,t_R=1.0)L_R(x) = 0.37 \cdot 0 + 0.32 \cdot 1 + 0.32 \cdot 0 = 0.32 \label{8.11B}$$

So:

LP(0)=0.37 ⋅ 0.32 = 0.12.

This means that under the model, if the state at node 1 were 0, we would have a likelihood of 0.12 for this small section of the tree. We can use a similar approach to find that:

(eq. 8.13)

LP(1)=0.32 ⋅ 0.37 = 0.12.

LP(2)=0.32 ⋅ 0.32 = 0.10.

Now we have the likelihood for all three possible ancestral states. These numbers can be entered into the appropriate boxes:

Figure 8.4C. Conditional likelihoods entered for node 1. Image by the author, can be reused under a CC-BY-4.0 license.

1. We then repeat the above calculation for every node in the tree. For nodes 3-5, not all of the LL(x) and LR(x) terms are zero; their values can be read out of the boxes on the tree. The result of all of these calculations:
1. We can now calculate the likelihood across the whole tree using the conditional likelihoods for the three states at the root of the tree.

$$L = \sum\limits_{x \in k} \pi_x L_{root} (x) \label{8.14}$$

Where πx is the prior probability of that character state at the root of the tree. For this example, we will take these prior probabilities to be uniform, equal for each state (πx = 1/k = 1/3). The likelihood for our example, then, is:

$L = 1/3 ⋅ 0.00150 + 1/3 ⋅ 0.00151 + 1/3 ⋅ 0.00150 = 0.00150 \label{8.15}$

Note that if you try this example in another software package, like GEIGER or PAUP*, the software will calculate a ln-likelihood of -6.5, which is exactly the natural log of the value calculated here.

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