RSA is an asymettric cryptographic algorithm by Ronal Rivest, Adi Shami and Leonard Adleman. RSA is an implementation based on Diffie & Hellman public key cipher ideas. The implementation allows each user to have their own keyset and allows simpler distribution of those keys between members. Each member has a public and private key. The public key is what is shared to others, and is used to encrypt the messages using RSA. The private key is kept secret by the key-pair owner and is used to decrypt the message.
The idea behind RSA is the use of prime numbers. Prime numbers are numbers that can only be divided by themselves or 1. Generating large prime numbers is incredibly time consuming and is hard to find. This fact alone is what RSA relys on for its security. RSA in this case is "computationaly secure", as the information being encrypted will be obsolete by the time the user can crack it. One of the largest prime numbers: 2^(657,885,161)-1 for example is 17,425,170 digits long, fills a 17MB file and takes 4-6 days to prove it is prime.
Another factor behind RSA is the use of coprime numbers. Coprime numbers are two numbers that's greatest common divisor is 1 ( GCD(A,B)==1 ). Note that A and B do not have to be prime numbers to be coprime numbers.
The RSA algorithm is quite simple. First you must generate a public and private key. This is done by answering a number of equations as follows:
- Choose 2 prime numbers P and Q
- Calculate N by multiplying P and Q ( N = P*Q )
- Calculate the totient T by multiplying 1 less of P with 1 less then Q ( T = (P-1)*(Q-1) )
- Choose a 3rd prime number E which is greater then prime P or prime Q but less then T. E when compared with T must also have a GCD of 1 ( E where E > (P || Q ) && E < T && GCD(E,T)==1 )
- Calculate D by stabilizing the following congruency: E * D = 1(mod T). This can be calculated through manual trial and error or using the Extended Euclidean Algorithm
- After this, test that E and D are coprime
After completing this you will have all the components needed for creating a public and private key. In the most simplistic form, E and N make up the public key. D and N make up the private key. This information alone is all that is needed to carry out the RSA algorithm, however for transferring this information a standard ASN.1 format is used most commonly when circulating private and public keys.
Encryption is simply done by running each segment of data through an equation using the public key components created above, and retrieving the ciphertext segments.
The general equation is as follows:
C = M^E(mod N)
C is the ciphertext. M is the plaintext. E and N come from the public key generated in the above section. Note that size of M matters when encrypting with RSA. The value of M as an integer cannot be larger then the value of N as it will otherwise be lost (a consequence of modular arithmetic). Thus when encryption plaintext, some earlier calculations must be done to ensure the M value is small enough.
Given ASCII characters map to 8-bit this means that N must always be larger then 2^8 = 256. This is not always possible if you are using smaller prime numbers such as P = 5 and Q = 11, N = 55. In this case the ASCII letter need to be cut up into smaller segments. The segment size can be calculated with: Math.floor(Log(N)/Log2). This will give the number of bits that the message must be broken into and the maximum number of plaintext bits that can be used in M.
Using our example of P = 5 and Q = 11, we know N = 55. Using the above mentioned equation we know that Math.floor(Log(55)/Log(2)) = 5. That means whatever plaintext value M is, cannot be larger then the maximum value that can be represented by 5 bits or 2^5 = 32. We can also prove this limitation: if we used 6 bits the maxiumum value would be 2^6 = 64 which is larger then N = 55, thus 5 is the maximum we can use.
After the encryption, you will need to do an extra calculation to determine how large the encrypted values should be. This is important as some of the encrypted values may need padding, so that they can be properly parsed at decrytpion. To determine how many bits should be in the encrypted value use the equation Math.ceil(Log(N)/Log(2)). Using our example Math.ceil(Log(55)/Log(2)) = 6. So with our encrypted values in binary, ensure that they are left padded to 6-bits before doing any appending actions of the ciphertext bits.
Decryption is done in the exact same way as Encryption except with the inverse equation using the private key components.
M = C^D(mod N)
C is the ciphertext. M is the plaintext. D and N come from the private key generated from earlier. Like encryption the size of N matters when decrypting with RSA. It differs slightly from encryption though as the encrypted values can represent inclusively up to the value N. Thus our equation for determining how to parse apart the ciphertext segments differs slightly. To determine the the number of bits to represent the ciphertext we use the equation Math.ceil(Log(N)/Log(2)). Using N = 55, we know that the ciphertext must be broken up into Math.ceil(Log(55)/Log(2)) = 6 bit segments.
Like encryption, after decryption you can use Math.floor(Log(N)/Log(2)) to calculate how many bits should be in the plaintext value. Left-pad the value until it is the length calculated, and then append the values together. If the message received is ASCII characters, you can now process the binary in 8-bit segments. Note that to work with N during encryption there may be extra padding on the end. Read in the ASCII data from left to right, parsing it into 8-bit segments. If you reach remaining data, which should all be zeros, simply drop it, before converting the binary to characters.
Lets take the simple example of the message "YO" and encrypt it. First we need to generate all of our key components. Lets use P = 5 and Q = 11. Therefor:
N = P * Q = 5 * 11 = 55 T = (P-1)*(Q-1) = 4 * 10 = 40 E = 13 (note E could be other numbers that meet the requirements but we chose this one to keep it simple)
Next we need to calculate D. This is done using the Extended Euclidean Algorithm. See the bottom of this page for additional resources on the Extended Euclidean Algorithm. It looks like this:
A = 13 B = 40 1) 40 = 3*13 + 1 2) 13 = 13*1 + 0 ---------------- 1 = 40 - 3*13 = (1)40 + (-3)13 (-3) + 40 = 37 D = 37
We can test this is correct by checking the congruency of the equation:
E * D = 1(mod T) 13 * 37 = 1(mod 40) (13*37) - 1 = 480 which is a factor of 40
So D = 37
We now have all of the components. Are public key is comprised of E and N (13, 55) and our private key is D and N (37, 55)
Now to encrypt our message:
To encrypt we use the encryption algorithm above, but first we must convert this message to binary. In binary the message "YO" looks as follows:
010110001 01010001 Y O
Normal ASCII uses 8-bits which is much larger then our N value. Thus we must calculate the maximum that can be used for our N value. By calculating Math.floor(Log(55)/Log(2)) we get 5. So our M value must be 5 bits long. Recutting our message into 5-bit segments looks like this
01011 00101 01000 10000
Note that we have appended 4 zero's onto the end of the last segment. This is to make it even out into 5 bits. for RSA, POST appending is the most effective way to even out the segments. When decryption occurs we will be able to determine what to remove as the message will be regrouped into 8-bit segments as before. When this happens we will have leftovers and know to drop it.
Now we take the integer values those binary numbers represent and run them through the equation. After encrypting, we left pad the binary until it contains Math.ceil(Log(55)/Log(2)) = 6 bits. It looks like this:
01011 00101 01000 10000 11 5 8 16 11^13(mod 55) = 11 = 001011 5^13(mod 55) = 15 = 001111 8^13(mod 55) = 28 = 011100 16^13(mode 55) = 26 = 011010
We now have our encrypted message:
To decrypt the above encrypted message we use the D and the N value in another equation. Again using N we can calculated how to split up the binary. Using Match.ceil(Log(55)/Log(2)) = 6 bits. We cutup the binary into 6-bit segments. This looks as follows:
001011 001111 011100 011010
Using then the decryption equation we can convert these segments back to their plaintext values. Afterwards we calculate Math.floor(Log(55)/Log(2)) = 5 bits to determine how much to pad the plaintext values.
001011 = 11 = 11^37(mod 55) = 11 = 01011 001111 = 15 = 15^37(mod 55) = 5 = 00101 011111 = 28 = 28^37(mod 55) = 8 = 01000 011010 = 26 = 26^37(mod 55) = 16 = 1000
Now to convert back we append all of these together and parse them into 8-bit ASCII
0101100101010001000 01011001 01010001 000 Y O [extra]
Note the extra we had talked about that would be appended. We know we can get rid of this because it does not equate to 8-bits of a valid ASCII character and it is also all zeros.
Python Code Example
Below is a python walkthrough example of an encryption and then decryption using python. Note that this implementation is not the most efficient implementation as it converts the binary representation of the numbers into string, thus allowing each binary value to be accessed at an index. Conversion between binary and binary strings in python is quite trivial, and it does make the below example a bit more legible if you are unfamiliar with bit manipulation actions, but note this is at the expense of performance.
The code sample below includes print statements, copy the the code snippet below and execute with Python3 to get output through each stage of the algorithm
1 import math 2 import sys 3 4 unencryptedMessage = "Hello World!" 5 6 p1 = 5 # prime number 1 7 p2 = 11 # prime number 2 8 9 t = 40 # totient. AKA (p1-1)*(p2-1) 10 n = 55 # n. AKA p*q - part of public key 11 e = 13 # e. prime number - part of public key 12 d = 37 # d. private key 13 14 15 plaintext_message_seg_length = int(math.floor(math.log(float(n), 2))) 16 encrypted_message_seg_length = int(math.ceil(math.log(float(n), 2))) 17 18 print("Plaintext: " + str(plaintext_message_seg_length)) 19 print("Encrypted: " + str(encrypted_message_seg_length)) 20 print("Original Message: " + unencryptedMessage) 21 22 # convert the message to all binary bits - padd out to make sure they all are 8 bits long for the character 23 binaryUnencryptedMessage = ''.join(format(ord(x), '08b') for x in unencryptedMessage) 24 print(binaryUnencryptedMessage) 25 26 # post pad the string to get an even number 27 while len(binaryUnencryptedMessage) % plaintext_message_seg_length != 0: 28 binaryUnencryptedMessage += '0' 29 30 print(binaryUnencryptedMessage) 31 32 # split it up into segments of plaintext_message_seg_length 33 unencryptedMessageSegments = list() 34 for i in range(0, len(binaryUnencryptedMessage), plaintext_message_seg_length): 35 unencryptedMessageSegments.append(binaryUnencryptedMessage[i: i + plaintext_message_seg_length]) 36 37 print(unencryptedMessageSegments) 38 39 #encrypt each segment using RSA 40 encryptedMessageSegments = list() 41 for i in unencryptedMessageSegments: 42 print("------------------") 43 print(i) 44 segmentInt = int(i, 2) # converts string to int, interpreting it as in base 2 45 print(str(segmentInt) + " - " + bin(segmentInt)) 46 encryptedSegmentInt = (segmentInt ** e) % n 47 print(str(encryptedSegmentInt) + " - " + bin(encryptedSegmentInt)) 48 encryptedSegmentBinary = format(encryptedSegmentInt, '0' + str(encrypted_message_seg_length) + 'b') 49 print(encryptedSegmentBinary) 50 encryptedMessageSegments.append(encryptedSegmentBinary) 51 52 53 print("***********************") 54 print(encryptedMessageSegments) 55 encryptedMessageBinaryString = ''.join(encryptedMessageSegments) 56 print(encryptedMessageBinaryString) 57 58 encryptedMessageInt = int(encryptedMessageBinaryString, 2) 59 print(encryptedMessageInt) 60 print(bin(encryptedMessageInt)) 61 62 63 encryptedMessage = encryptedMessageInt.to_bytes(byteorder=sys.byteorder, 64 length=math.ceil(len(encryptedMessageBinaryString) / 8 )) 65 print(encryptedMessage) 66 67 # -- AT THIS POINT THE MESSAGE IS ENCRYPTED AS A BYTE ARRAY-- 68 69 number = int.from_bytes(encryptedMessage, byteorder=sys.byteorder, signed=False) 70 print(number) 71 72 print (" ** BEGINNING DECRYPTION **") 73 74 binaryEncryptedMessage = str(bin(number))[2:] 75 print(binaryEncryptedMessage) 76 77 # pre pad encrypted until is appropriate length to be cut up 78 while len(binaryEncryptedMessage) % encrypted_message_seg_length != 0: 79 binaryEncryptedMessage = '0' + binaryEncryptedMessage 80 81 # cut into decryptable segments 82 encryptedMessageSegments = list() 83 for i in range(0, len(binaryEncryptedMessage), encrypted_message_seg_length): 84 encryptedMessageSegments.append(binaryEncryptedMessage[i: i + encrypted_message_seg_length]) 85 86 print(encryptedMessageSegments) 87 88 unencryptedSegments = list() 89 for i in encryptedMessageSegments: 90 print("------------") 91 segmentInt = int(i, 2) # converts string to int, interpreting it as in base 2 92 print(i) 93 print(str(segmentInt) + " - " + bin(segmentInt)) 94 unencryptedSegmentInt = int((segmentInt ** d) % n) 95 print(unencryptedSegmentInt) 96 97 # left pad with 0 to return segment to decrypted segment length 98 unencryptedSegmentBinary = format(unencryptedSegmentInt, '0' + str(plaintext_message_seg_length) + 'b') 99 print(unencryptedSegmentBinary) 100 unencryptedSegments.append(unencryptedSegmentBinary) 101 102 print(unencryptedSegments) 103 joinedSegments = ''.join(unencryptedSegments) 104 print(joinedSegments) 105 106 107 letters = list() 108 for i in range(0, len(joinedSegments), 8): 109 letters.append(joinedSegments[i: i + 8]) 110 111 print(letters) 112 113 plainMessage = "" 114 for letter in letters: 115 letterInt = int(letter, 2) 116 character = chr(letterInt) 117 plainMessage += character 118 119 print(plainMessage)
Extended Euclidean Algorithm


As mentioned in a previous post, there are many statistics that can be derived from the network visualizations that I have been generating from the course data I have been collecting. At the moment, these are the particular numbers that I have been paying attention to:
- Mean Degree of Nodes – The mean amount of connections per node on the graph.
- Mean Weighted Degree of Nodes – The mean weight of connections per node on the graph.
- Graph Density – A ratio of the number of edges per node to the number of possible edges.
- Modularity – a measure of the strength of division of a network into modules. Networks with high modularity have dense connections between the nodes within modules but sparse connections between nodes in different modules.
- Mean Clustering Coefficient – the degree to which nodes in the graph tend to cluster together.
So, in terms of applying these to the networks generated with awards data:
- Mean Degree of Nodes – The mean amount of connections for each award. i.e. the mean amount of awards that each award is connected to.
- Mean Weighted Degree of Nodes – The mean weight of connections for each award. i.e. the mean amount of modules shared by that award with other awards.
- Graph Density – The amount of connections per award when compared to the total amount of awards in the network. (more affected by an increase in awards offered than others)
- Modularity – a higher modularity suggests that awards are very highly connected with specific other awards, but have very few ‘odd’ connections to other awards in the network. A very high modularity would suggest that a group of awards shared a lot of modules between themselves.
- Mean Clustering Coefficient – a low coefficient would suggest that awards did not group together, and therefore did not share modules between them. A high coefficient would suggest that most of the awards in the network formed clusters with other awards.
The numbers generated for the weighted connections between awards for the academic year 2006/07 through to 2012/13 are as follows:
|Academic Year||Mean Degree||Mean Weighted Degree||Graph Density||Modularity||Mean Clustering Coefficient|
|2006 – 2007||0.804||1.821||0.069||0.657||0.357|
|2007 – 2008||0.763||1.711||0.041||0.726||0.408|
|2008 – 2009||0.500||1.324||0.030||0.588||0.224|
|2009 – 2010||0.405||1.432||0.023||0.574||0.124|
|2010 – 2011||0.720||1.880||0.029||0.777||0.212|
|2011 – 2012||0.716||2.486||0.020||0.810||0.259|
|2012 – 2013||0.651||4.349||0.021||0.847||0.267|
So what do these numbers show and are they actually useful? Well….
Mean degree shows the amount of awards that each award is connected to, on average. If we look at mean weighted degree instead, we then take into consideration the weight of a connection between a pair of nodes, i.e. the amount of joins between them, rather than just the fact that a join exists. Plotting this graphically helps to show the pattern that emerges.
From the graph above it becomes clear that there is a definite drop on MWD (mean weighted degree) from the academic year 07/08 to the year 08/09 (around 22%), showing that the average amount of links between awards dropped fairly considerably. Through looking back at the university’s history, this can be explained as this was the point in time that the amount of points per module of study was altered, meaning that, essentially, multiple version of the same award were running in tandem: some with the old weighting of awards, some the new. This also explains the steady increase in MWD up to 11-12 which is the first year that the old weighted degrees would not have been active at all. From the highest point of the old weighting, to this point in the new weighting, there is an increase of over 36% in the amount of joins between awards offered at the university. This shows that (assuming an increased modularity is good in terms of curriculum design) that the provision has been improved through the alteration of module weightings. Taking into account the overall increase in the amount of awards offered, this also shows that the restructuring of the modules had a significant impact on the sharing of teaching and assessment across different awards.
The number given for the ‘modularity’ of the graphs shows a couple of interesting things.
As noted above, the modularity shows how well the nodes on the graph (i.e. the awards) form into self contained clusters. A value of 1 would suggest that the awards form perfectly into self-contained clusters, having lots of connections between themselves but no connections with other clusters, a value of 0 would suggest the opposite. As you can see from the graph above, in 06/07, the modularity was reasonably high, quite possibly due to the smaller amount of awards offered at the university. This figure rises over the next year, and then drops for two consecutive years as the weighting of modules at the university goes through a period of change. As the change is fully implemented, the modularity rises significantly and continues to rise, almost at a constant rate from 2010-11 through to 2012-13. This would suggest (though is not necessarily the case) that, either by design or good fortune, the awards offered at the university are starting to form into self-contained groups or areas of specialism. This is interesting to note, as the university has recently gone through an organizational restructuring whereby three colleges were formed – could these clusters be contained within the colleges?
Though this has only looked at two series of numbers generated for each of these visualizations, it does show that visualizing course data produces extra data that cannot be collected when the data is in its raw form. Further to this, it also shows that this data accurately reflects historical changes in provision within the university. If these principles can be applied retrospectively to show changes, in which ways can they be applied to decision making processes, to help assess the impact of potential changes?


By Karp R.
Read Online or Download Combinatorics, Complexity and Randomness (Turing Award lecture) PDF
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Chapter 2 One Step at a Time. 1 Weak Induction Let us assume it is almost midnight, and it has not rained all day today. If, from the fact that it does not rain on a given day, it followed that it will not rain the following day, it would then also follow that it would never rain again. Indeed, from the fact that it does not rain today, it would follow that it will not rain tomorrow, from which it would follow that it will not rain the day after tomorrow, and so on. This simple logic leads to another very powerful tool in mathematics: the method of mathematical induction.
If we temporarily disregard A, we have n players left, so by the induction hypothesis there will be one of them, say B, who will list the names of the other n— 1 players. Now if B defeated A, or if anyone defeated by B defeated A, then B lists the name of A, too, and we are done. If not, then A has defeated B, and all the players defeated by B, so A won more games than B, a contradiction. (3) Induction on n. For n = 1, the statement is trivially true. Now assume the statement is true for n and prove it for n + 1.
In other words, a bijection matches the elements of X with the elements of Y, so that each element will have exactly one match. 9. Let / : X —> Y be a function. 8, then we say that / is one-to-one or injective, or is an injection. 8, then we say that / is onto or surjective, or is a surjection. 10. Let X and Y be two finite sets. If there exists a bijection f from X onto Y, then X and Y have the same number of elements. Proof. The bijection / matches elements of X to elements of Y, in other words it creates pairs with one element from X and one from Y in each pair.


An Introduction to Complex Analysis and Geometry (Pure and Applied Undergraduate Texts); 2010; John P. D’Angelo (Author); ISBN-10: 0-8218-5274-4; ISBN-13: 978-0-8218-5274-3; [Hardcover]
Author: John P. D’Angelo, University of Illinois, Urbana, IL
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At DAA, we focus on the social and emotional development of our students in addition to their academic growth. To do so, we teach, practice, and reflect upon six Habits of Character (which includes compassion, responsibility, and perseverance). We talk about these Habits of Character daily at our morning crew meeting, closing crew meeting, and throughout the day. My goal was to give students more specific strategies for one target within one Habit of Character: I can persevere on challenging math work.
I found that the math content in our unit was very rigorous and challenging for students. I know that they absolutely could tackle the difficult problems using various strategies, but I wanted THEM to know that they could tackle these problems and have the confidence, mindset, and skills to do so.
We started by talking about how we could show perseverance every day in math. We developed a chart that we hung up in our classroom that included such strategies as "going slowly", "doing one step at a time", and "learning from my mistakes". I then had students set specific goals. At the beginning of each lesson, and again right before the exit ticket (a check to see what students have mastered at the end of the lesson), students wrote down a goal for the day by completing the sentence frame: "Today I will show perseverance by..." We would share as a class before jumping into the difficult work so that the perseverance strategies were at the forefront of student's thinking. At the end of each lesson we would reflect; how did we show perseverance today? What worked? What didn't work? What can we continue to work on for tomorrow?
Finally, I wanted to build a system to celebrate students for showing perseverance in math. When students showed perseverance on their exit ticket and were able to show mastery of the math material, they were able to hang their exit ticket up on our "Wall of Fame." This gave them a lot of pride and allowed us to celebrate each other's success and growth. The "Wall of Fame" is in the hallway which also allows parents, teachers, and visitors to DAA to recognize and celebrate our students for their perseverance and achievement as well.


Statistics are everywhere–and they’re increasingly used to drive decisions around the globe. Baseball statistics guru Bill James is coming to Longwood for the annual Simkins Lecture on April 23, so the math department put together some of their favorite uses of statistics in the real and sports worlds.
Because we can prove that Babe Ruth is the greatest of all time.
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Because our favorite NBA stars know where to stand to hit shots.
Because it can get us into March Madness.
Because we can see how the world is changing.
Because Jonah Hill can outshine Brad Pitt.


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And no, Mac用のPCゲームをダウンロード answer is not 100. CreditCredit By Steven Strogatz Aug. Others heard Yanny, not Laurel, and insisted the right answer トールゲーム無料ローミングpc 1. Another posted a photo showing that even two different electronic calculators disagreed. The normally reassuring world of math, where right and wrong exist, and logic must prevail, started to seem troublingly, perhaps tantalizingly, fluid. Sign up for the Science Times newsletter. When confronted with 8 ÷ より多くの数学ゲーム無料 2+2everyone on Twitter agreed that the 2+2 in parentheses should be evaluated first. So the question boils down to 8÷2×4. Which way is correct? The standard convention holds that multiplication and division have equal priority. To break the tie, we work from left to right. So the division goes first, followed by より多くの数学ゲーム無料 multiplication. Thus, the right answer is 16. More generally, the conventional order of operations is to evaluate expressions in parentheses first. Then you deal with any exponents. Next come multiplication and division, which, as I said, are considered to have equal priority, with ambiguities dispelled by working from left to right. Finally come addition and subtraction, which are also of equal priority, with ambiguities broken again by working from left to right.


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The Importance of Spiral Review in Math
My name is Miranda Briere, and I teach 4th grade on the Lynn campus. One of the subjects I teach is Math, covering everything from adding and fractions to measurement and geometry! One challenge in teaching math is balancing teaching new lessons that align with curriculum and standards and reviewing previously learned concepts so students don’t lose any understanding. One key strategy I use to maintain this balance is called “spiral review”.
Spiral review is a review of previously learned concepts, either during an entire lesson or in a short portion of a lesson. After learning foundational concepts, students continue to practice throughout the year to maintain understanding. This helps reinforce previously learned concepts and promotes the retention of skills in later grades.
I use spiral review in my classroom at the beginning of every lesson, during a five-minute period I call “Quick Math”. I use worksheets with multiple types of problems that cover math concepts we have already covered previously in the year. These worksheets are aligned with state standards and cover a variety of topics. These worksheets allow an opportunity for students to refresh their memory as well as an opportunity for me, as the teacher, to address misconceptions that have carried over from previous units.
Everyday Math writes in their article Why Spiral Review Distributes Learning, “Spiraling is effective with all learners, including struggling learners. Learning difficulties can be identified when skills and concepts are encountered in the early phases of the spiral, and interventions can be implemented when those skills and concepts are encountered again later in the spiral.”
At North Shore Christian School, we support all learners by implementing a Spiral Review Strategy when making progress through our Math curriculum.


Sometimes, in fitness exercises, the exercising individual would have to raise a part of their body (body submass) as an exercise. Eventually the body submass raised can be associated with an extra free weight to add intensity to the exercise. A typical problem is to log this type of exercises properly.
Exercises raising purely a free external weight can be logged as such and do not need any precaution. On the contrary, for exercises involving a part of the mass of the body (the so called body submass) one should always log the TOTAL WEIGHT for proper and reusable data records.
The knowledge of total pulled weight (not body weight, only the part of it that acts as a weight) is typically crucial for a controlled decrease or increase of intensity based on percentage. For example, it is now a well known fact that the most efficient progression is obtained for 5-10% increases in exercise weight.
A common mistake is to apply say a 5-10% change in only the extra free weight. It necessarily does not produce the targeted effect because it would induce a different percentage of change at the scale of the total raised weight.
The solution for this problem is elementary mathematically speaking but I am surprised I found in nowhere online at this point of time.
The small bits of maths to get this problem solved
The modeling of the problem is based on the WEIGHT (w) vs. MAX_REPS (r) graph. If you never heard about this : Two-set 1-RM calculator. Basically, experimental studies taken over by Matt. Brizcky agrees that, under certain hypothesis, a good approximation of the connection between WEIGHT pulled and the MAX amount of repetitions one can perform with that weight, is a straight line :
This equation presents 3 unknowns, the total raised weight w_tot, and the parameters a and b which the literature presents as being unique for each person and each exercise.
We therefore need 3 equations :
In this set of equations, w_tot refers to the total weight moved during the exercise. It consists of a submass of the body weight (w) and an eventual free weight (w_fw), and a and b described earlier.
Solving this set, you get w which is the submass of your body that is always involved. Add up whatever extra free weight you like to use and the total (w_tot) is the weight you can log for this exercise. Every change of intensity should be done on w_tot and never on the free extra weight alone.
In practice, the protocol is the following:
- Take 3 extra free weights. (zero is an acceptable value). These will be w_fw1, w_fw2, and w_fw3.
- For each weight, perform repetition until fatigue, and write down the respective maximum amount of reps performed. These values will respectively be r_1, r_2 and r_3.
- Use a solver to solve the equation (online calculator in project)
- Always use w_tot (= w + w_fw) to log your progress.
- Accessorily, you can use a and b to plot your w vs. r straight line to link weight and reps to failure.
Examples of use
Typical critical examples of use are presented below:
Example 1 :
You perform fix bar pull-ups (solicitates the back) with no extra weight and you manage to perform 12 reps (fatigue). Your program is an alternate program telling to switch from 70% to 95% of your 1RM. What is your 1RM ? (What would you input in a 1RM calculator ? zero ? because you pulled zero extra weight ? how about your own weight ?)
Example 2 :
You perform fix bar pull-ups (sollicitates the back) with no extra weight and you manage to perform 12 reps (fatigue). You therefore want to add weight to add intensity and get back to a maximum potential of 8 reps (this usually corresponds to an increase of about 10% in total raised weight). What weight should you add ?
Recommendations and limits on the use of this method
The values of a and b change with your progress. Similarly, if you progress it makes sense that your body mass changes, therefore potentially inducing a change the weight you raise. Although this method does not need to be used every time you log data, remember to renew the data every time obvious progress is observed.
This simplified method also is better for exercises involving a translation. Exercises involving a rotation induce momentum. Weight involved in a momentum cannot be added just like that: the contribution of its distance to the center of rotation must be considered.


An ice cream company collected data on their ice cream cones sales over a month in July in a Chicago suburb, along with daily temperature and the weather. The company is interested to develop a correlation between ice cream sales to the hot weather. Market research showed that more people come out in certain neighborhoods, to either enjoy the nice weather, or venture out if they do not have air conditioning in their apartments. The Chicago Police also tracked crime statistics during the same period. Crime statistics included murder, assault, robbery, battery, burglary, theft and motor vehicle theft. The data are shown below:
|July||Day Temp (F)||Weather||Ice cream sales (units)||Crime stats reported|
[ Select ] [“-3892.2 + 40.1(x); r^2 = .78”, “-3462.4 + 49.4(x); r^2 = .78”, “-2362.5 + 39.2(x); r^2 = 0”, “-3432.6 + 41.3(x); r^2 = .61”] Develop a linear regression model for ice cream sales over daily temperature. Show the linear equation in the form of y = ax + b, and the correlation of determination (r^2).
[ Select ] [“1201”, “1101”, “1001”, “1181”] What would be the projected forecast of ice cream sales in units, for daily temperature of 94 F?
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[ Select ] [“-2808.1 + 41.9(x); r^2 = .86”, “-1362.5 + 33.2(x); r^2 = .67”, “-3932.6 + 51.3(x); r^2 = .93”, “-2892.2 + 60.1(x); r^2 = .55”] On July 15 & 16 there were heavy down pour of rain, which might have prevented some to venture out to purchase ice cream during the day. If you were to override those 2 data points, what would be the linear regression model be (by deleting July 15 & 16 data).
[ Select ] [“second correlation is a better forecast”, “need more data”, “no different”, “first correlation is a better forecast”] Compare the two correlation coefficients, which would be considered a better forecast for ice cream sales
[ Select ] [“92.2 + 10.1(x); r^2 = .61”, “50.1 + 11.8(x); r^2 = .96”, “32.6 + .39(x); r^2 = .71”, “46.8 + .30(x); r^2 = .82”] Develop a linear regression on ice cream sales to crime statistics. Show the linear equation in the form of y = ax + b, and the correlation of determination (r^2).
[ Select ] [“no, correlation does not imply causality”, “yes, strong correlation does imply causality”] Does this correlation demonstrate causation, that high ice cream sales cause crime statistics to go up?


Türkçe kelime anlamı çarpmak olan Multiply kelimesinin kullanıldığı toplam 12 adet cümle bulundu. Multiply ile ilgili cümleleri ve bu örnek cümlelerin türkçe anlamlarını altında bulabilirsiniz.
To calculate the volume, multiply the length by the width by the depth.
Hacimi hesaplamak için uzunluğu genişlikle ve derinlikle çarp.
To calculate the capacity, multiply the length by the width by the depth.
Kapasiteyi hesaplamak için, uzunluğu genişlikle derinlikle çarpın.
Be fruitful and multiply.
But that your fear of me was greater than his life, now I bless you and your offspring, just so that they multiply like the stars in the heavens and just as grains of sand on the sea.
For a rough world estimate, multiply by 40.
If you're asked what is, say, thirty-three by eleven, you still don't know what they meant: multiply or divide?
Multiply three by five to get fifteen.
Phosphorus provides nourishment for cyanobacteria (blue algae) which multiply and release toxins.
The only way on Earth to multiply happiness is to divide it.
This new investment will multiply our profit.
"And besides," Dima made sure to add, taking out his calculator and dividing 0.99 by 3,000,000, before multiplying by 100. "You do realize that you would only lose 0.0033%, right?"
Cheap apartment houses are multiplying rapidly.


Measures of multiple spike train synchrony are essential in order to study issues such as spike timing reliability, network synchronization, and neuronal coding. These measures can broadly be divided in multivariate measures and averages over bivariate measures. One of the most recent bivariate approaches, the ISI-distance, employs the ratio of instantaneous interspike intervals (ISIs). In this study we propose two extensions of the ISI-distance, the straightforward averaged bivariate ISI-distance and the multivariate ISI-diversity based on the coefficient of variation. Like the original measure these extensions combine many properties desirable in applications to real data. In particular, they are parameter-free, time scale independent, and easy to visualize in a time-resolved manner, as we illustrate with in vitro recordings from a cortical neuron. Using a simulated network of Hindemarsh–Rose neurons as a controlled configuration we compare the performance of our methods in distinguishing different levels of multi-neuron spike train synchrony to the performance of six other previously published measures. We show and explain why the averaged bivariate measures perform better than the multivariate ones and why the multivariate ISI-diversity is the best performer among the multivariate methods. Finally, in a comparison against standard methods that rely on moving window estimates, we use single-unit monkey data to demonstrate the advantages of the instantaneous nature of our methods.


Roman numerals have a rich history and are still used today in various contexts, such as clock faces, book chapters, and movie credits. Understanding how to write and count Roman numerals can be helpful, whether decoding ancient inscriptions or adding a touch of classic elegance to your designs. This article will guide you through writing and counting Roman numbers and exploring their symbols and rules.
Roman numerals originated in ancient Rome and were used extensively in the Roman Empire. Unlike the decimal system we use today, Roman numerals are based on a combination of letters from the Latin alphabet. Each letter represents a specific value, and by combining these letters, you can represent different numbers.
To write and count Roman numerals, it’s essential to familiarize yourself with the symbols and their corresponding values. Here are the basic symbols and their values:
– I: 1
– V: 5
– X: 10
– L: 50
– C: 100
– D: 500
– M: 1000
Roman numerals are written by combining symbols to represent a specific value. Here are some rules to keep in mind when writing Roman numerals:
1. Addition Rule: You add the values together when a smaller numeral is placed after a larger numeral. For example:
– II represents 1 + 1 = 2.
– VI represents 5 + 1 = 6.
2. Subtraction Rule: When a smaller numeral is placed before a larger one, you subtract the smaller value from the larger one. For example:
– IV represents 5 – 1 = 4.
– IX represents 10 – 1 = 9.
3. Repetition Rule: A numeral can be repeated thrice in a row. Beyond that, a subtraction rule must be used. For example:
– III represents 1 + 1 + 1 = 3.
– XXX represents 10 + 10 + 10 = 30.
4. Larger Numeral Before Smaller Numeral: Larger numerals should always be placed before smaller numerals when using the subtraction rule. For example:
– XL represents 50 – 10 = 40.
– XC represents 100 – 10 = 90.
If you find manually converting numbers into Roman numerals challenging, you can use online tools like a Roman numeral converter. These converters allow you to input Arabic (decimal) numbers and instantly convert them into Roman numerals. They provide a convenient way to obtain a given number’s Roman numeral representation quickly and can be a valuable resource for accurate conversions.
Counting Roman numerals involves adding up the values of the symbols. Start from the left and work to the right, adding or subtracting the values as needed. Remember the rules mentioned earlier, especially the subtraction rule, which requires subtracting a smaller value from a larger value.
To reinforce your understanding of writing and counting Roman numerals, let’s look at a few examples:
1. Write the number 39 in Roman numerals:
– We can represent 30 with XXX and subtract 1 with IX. Therefore, 39 is written as XXXIX.
2. Write the number 76 in Roman numerals:
– We can represent 50 with L and add 10 twice with XX. Therefore, 76 is written as LXXVI.
3. Write the number 105 in Roman numerals:
– We can represent 100 with C and add 5 with V. Therefore, 105 is written as CV.
4. Count the Roman numeral LXVII:
– Starting from the left, we have 50 (L) + 10 (X) + 5 (V) + 1 (I). Therefore, LXVII represents the number 67.
Mastering the art of writing and counting Roman numbers opens up a world of historical exploration and aesthetic possibilities. Practice with various numbers and challenge yourself with more complex examples to sharpen your skills. Whether deciphering ancient texts or adding a touch of elegance to your designs, Roman numerals add a timeless charm and evoke a sense of classical beauty.


The title of my unit is "It's Golden!". Students will interpret and use ratios in different contexts to show the relative sizes of two quantities, using appropriate notations. They will use proportions to solve problems. They will also use cross-multiplication as a method for solving proportions, understanding it as the multiplication of both sides of an equation by a multiplicative inverse
What is change?
How can an understanding of ratios and rates help you solve proportions?
Why do you need to convert between fractions, decimals, and percents?
How can you show that two ratios are equivalent?
How can you change a "rate" to a "unit rate"?
How do you solve a "proportion"?
How are "scale factors" used to create scale drawings?
Tuesday, September 15, 2009
This module has made me think about my role as an instructional designer by making me look from the outside at how I teach. Am I providing a "real life" learning experience for my students? Am I encouraging them to use higher order thinking skills to solve problems independently? I must provide probing questions to encourage independent problem-solving and thinking.


The International Mathematics Olympiad, or IMO, is a major mathematics Olympiad held by the Science Olympiad Basis for students to lay a mathematical foundation for future classes. By posing high-order thinking problems to students, IMO helps them improve their problem-solving ability and analytical thinking. They will be better prepared for future competitive tests as a result of this.
Because the IMO test is held on a national level stage, students who are taking it should practice more on the subject. IMO Maths Olympiad Class 6 Question Papers can help students build a solid basis for higher-level competitive exams. Subject experts have created a sample paper for IMO Math Olympiad Class 6 that takes into account the most recent exam format. The answers to various problems are included in the IMO Math Olympiad Model Papers for Class 6 so that students can double-check their answers and improve in the areas where they are weak. IMO Math Olympiad Sample Papers are available for free download in PDF format. Everyone has access to Class 6, and students can download it immediately from the internet. IMO Math Olympiad Second Level Sample Papers Class 6 can be used to help students succeed in any competitive test by improving their learning abilities and academic achievement.
Solving IMO sample problems for Class 6 might help you prepare for the IMO test faster. Students can gain a deep knowledge of the Olympiad topics, the pattern of exam ,and marking scheme by practicing Class 6 Olympiad sample papers. The IMO syllabus is matched with the school curriculum, so practising these will help you remember what you’ve already learnt. This is also necessary in order to prepare for the annual school math exam. The Math Olympiad exam for Class 6 is divided into many stages in order to prepare pupils for competitive challenges. The goal of this competition is to motivate pupils to excel in arithmetic.
The Olympiad exams follow a curriculum that follows the school’s guidelines. It serves all boards, including ICSE, CBSE, and state boards. As a result, kids from throughout the country can participate in the Math Olympiad, regardless of which board their school belongs to. In mathematics, repetition does not help. Students must master their principles to flourish in math. Math necessitates analytical and logical thinking abilities. The logical reasoning portion of the Indian Talent Olympiad is included in all Olympiads. It ensures that all students are exposed to critical thinking abilities, regardless of whatever topic they are enrolled in.
Syllabus and pattern
Chapters in Math in class 6: –
Chapter 1 – Concept of Geometry
Chapter 2 – Number System
Chapter 3 – Factors & Multiples
Chapter 4 – Integers
Chapter 5 – Decimal Fractions
Chapter 6 – Ratio & Proportion
Chapter 7 – Arithmetic
Chapter 8 – Basic Algebra
Chapter 9 – Data Handling
Chapter 10 – Mensuration
Chapter 11 – Symmetry
Chapter 12 – Logical Reasoning
Tips on IMO preparation.
- Before tackling these sample papers, students should go over the entire IMO curriculum for class 6 because this will help them build a firm knowledge of the major themes and concepts.
- Students should review class 6 IMO sample papers to ensure that they comprehend all of the crucial features, such as the exam pattern and marking method. It’s also useful to know the mark distribution by section and the number of questions in each part.
- After comprehending the IMO syllabus and test structure, students should appropriately arrange their preparation timetable. Students should practice IMO sample papers in a timed setting to improve their problem-solving skills as well as their speed and accuracy.
- Class 6 IMO sample papers are a good way to figure out what kinds of challenges each area has. Students are encouraged to practice each issue in the sample papers to fully comprehend the structure of math Olympiad questions.
- Students can evaluate their performance and progress by consulting the answer keys provided with the class 6 IMO practice papers.
- Time Management -After students have learned topics and solved various problems, the next logical step is to take a mock test to help them improve their time management abilities. To gain a handle on tackling complex issues in a set amount of time, they should practice with timed mock tests, IMO sample papers, and previous Olympiad question papers. Students’ confidence in tackling the IMO exam will be boosted by time-based preparation, which will increase speed and accuracy. Furthermore, these practice sessions will assist students in self-evaluation and evaluation of their preparation. Students can prepare the entire course while also assessing their strengths and weaknesses. Students will be able to strengthen their capabilities for improved performance by identifying their strengths and limitations.
The Advantages of Solving IMO Sample Papers for Students in Class 6
- Solving IMO sample problems for Class 6 is essential for gaining a thorough understanding of the IMO exam’s format and pattern.
- Students’ basic knowledge required for mathematical problem-solving will be developed by tackling Class 6 Math Olympiad sample papers.
- IMO Example Papers for Class 6 are based on the ICSE, CBSE, and several state boards’ syllabuses, therefore students will be well prepared to attain good academic results by completing these sample papers.
- Answer keys with complete solutions are included in the Class 6 math Olympiad sample papers so that students can gain a thorough understanding of all topics and concepts.
- With the help of IMO, kids in Class 6 would be able to plan their preparation wisely, saving a lot of time and effort.
Last-minute tip for a candidate
As the exam date approaches, dread and stress may begin to surface. As a result, you must relax and remain cool to properly implement all of your preparations on exam day. Of fact, stress can cause you to perform below your true potential.


In computer graphics, too much is not enough. Jon Peddie does some simple math on the ridiculous amount of computer power available for rendering.
By Jon Peddie
President, Jon Peddie Research
The number of processors, both x86 and GPU, available for rendering has been increasing exponentially. Rendering is one of the applications that can soak up all the cycles that are available to it, which is an example of Peddie’s First Law: In computer graphics, too much is not enough.
We looked at the installed base of x86 and GPU processors, applied a factor for the average number of cores and developed the following chart.
Cores alone don’t tell the whole story; the real measure is how many millions of operations per second the processor can execute. A general figure of merit is to multiple the processor’s clock by its word size. A GPU running at one-fourth the speed of a CPU, and with just a 32-bit processor compared to the 64-bit CPU, still deliver the most MIPS because of the overwhelming number of cores.
The chart indicates there is at least 450 billion computer cycles available every second. Carrying that to its ridiculous extreme, and assuming the processors run 24 hours a day, there are 38 quadrillion processor cycles available a day for rendering.
These computer cycles are almost free and they are going to be available just by the sheer momentum of the industry. They represent new opportunities and developments. Think about how fast and cheap rendering is going to be. Think about all the background functions that will done automatically and invisibly. It’s Peddie’s Third Law: the technology works when it’s invisible.


Our drywall calculator, also known as a sheetrock calculator, simplifies the process by helping you calculate the exact number of drywall sheets required. Whether you're covering walls or ceilings, this tool ensures that you purchase the right amount of drywall sheets, eliminating waste and saving you time and money.
Calculating the number of drywall sheets you need for a room, including the ceiling and accounting for waste, involves a straightforward formula.
Here are the steps to calculate it:
1. Measure the Room:
- Measure the height of the walls from the floor to the ceiling.
- Measure the length of each wall.
- Measure the width of the room.
2. Calculate the Wall Area:
- For each wall, multiply its height by its length to calculate the square footage of that wall.
- Sum up the square footage of all the walls in the room.
3. Calculate the Ceiling Area:
- Multiply the length by the width of the room to calculate the square footage of the ceiling.
4. Add Wall and Ceiling Areas:
- Add the total wall area to the ceiling area to get the total square footage of the surfaces you need to cover with drywall.
5. Account for Waste:
- It's common to add a waste buffer to account for cutting, mistakes, and unforeseen issues. A 10% buffer is a good starting point, but you can adjust it based on your experience and the complexity of the project. To add a 10% buffer, multiply the total square footage by 1.10.
6. Calculate the Total Drywall Area:
- Multiply the total square footage (including the waste buffer) by 1.10 to get the total drywall area needed.
7. Determine Drywall Sheet Size:
- Check the size of the drywall sheets you plan to use. Common sizes are 4'x8', 4'x10', and 4'x12'.
8. Calculate the Number of Sheets:
- Divide the total drywall area (including the waste buffer) by the square footage of one drywall sheet. This will give you the number of sheets needed.
Number of Drywall Sheets = (Total Drywall Area with Buffer) / (Drywall Sheet Size)
Let's go through an example.
Suppose you have a room with:
- Wall height: 8 feet
- Four walls length: 12 feet, 12 feet, 10 feet, and 10 feet
- Room width: 10 feet
- Ceiling length: 12 feet
- Ceiling width: 10 feet
- Wall area = [(8 feet x 12 feet) + (8 feet x 12 feet) + (8 feet x 10 feet) + (8 feet x 10 feet)] = 288 sq. ft.
- Ceiling area = 12 feet x 10 feet = 120 sq. ft.
- Total wall and ceiling area = 288 sq. ft. + 120 sq. ft. = 408 sq. ft.
- Adding a 10% waste buffer: 408 sq. ft. x 1.10 = 448.8 sq. ft.
- If using 4'x8' sheets: 448.8 sq. ft. / (4 ft x 8 ft) = 14.025 sheets
Round up to the nearest whole number, so you would need 15 sheets of drywall to cover the walls and ceiling of this room taking into account a 10% waste buffer.


Everyday maths & English
Everyday maths and English will help you master these subjects and allow you to apply what you have learned to everyday life. They are great way to re enter the learning environment whilst learning these subjects at a basic level.
This course is an entry level 3 accredited course.
Course Date: Please call to register your interest.
Work related benefits fee: FREE
Co-funded fee: £75
Full fee: £150
The learning curve
Unit 3 Challymead Business Park


At Thinkster our coaches are all certified, experienced math teachers. Your dedicated coach will work diligently to provide one-on-one coaching, daily grading and feedback tailored to your child. We strive to help students perform at their best and achieve their academic goals.
Thinkster’s math’s world-class curriculum is aligned with Common Core and State Standards and designed to build analytical and critical reasoning skills.
We then customize our curriculum to meet the learning patterns and needs of each student, either for extra help.
Tutors shape a student’s learning plan to mirror classroom work, provide extra practice or more of a challenge.
Tutors grade worksheets daily providing personalized comments for Students. Tutors are ready to support students through videos and notes.
With weekly or bi-weekly contact via phone, FaceTime or Skype, students and parents connect with their tutor to discuss a learning plan.
“I have always known I wanted to become a teacher from a very young age. I love being in the classroom and helping my students understand difficult topics. I would be honored to help your child improve his/her math skills.”
Master’s Degree in Curriculum Development Instructional Technology, Certified Tutor and a mother of 4
“With over 30 years of teaching, for me, the best part of Thinkster is conferencing with students. It gives the tutors, as well as the students, someone with whom to connect. I also appreciate the ability to customize the worksheet selections for each student.”
Bachelors Degree in Elementary Education, Masters Degree in Literacy, National Board Certification in Literacy
“My aim is to build my student’s maths confidence. I use my experience to build a study plan to suit each of my student’s mathematical personalities and then work directly with them and their families to guide them through the Thinkster program.”
Cambridge trained Maths Teacher, Former Kumon instructor, Maths blogger and a mother of 4


For problems that require more than one step, a thorough stepbystep explanation is provided. The exercises are designed to aid your study of mathematics by reinforcing important mathematical skills needed to succeed in the everyday world. Printable grade 5 math mathematics worksheets based on the. One book was read by all three children, but every other book was different. Mixed word problems with key phrases worksheets these word problems worksheets will produce addition, multiplication, subtraction and division problems using clear key phrases to give the student a clue as to which type of. Below are eight grade 5 math worksheets with mixed word problems including the 4 basic operations addition, subtraction, multiplication and division, fractions, decimals, lcm gcf and variables. Below are three additional grade 5 math worksheet with word problems involving fractions and the 4 basic operations. Printable grade 5 math mathematics worksheets based on the singapore math curriculum. These word problems are solved with the help of block diagrams or bar models, addition, subtraction, multiplication and division, singapore math, examples and step by step solutions, beforeandafter division problems, fraction problems with decimals, model drawing for 5th grade.
Word problems are one of the first ways we see applied math, and also one of the most anxiety producing math challenges many grade school kids face. We recommend that your student completes one page of practice problems each day. This 5th grade word problems quiz requires problem solving skills. It provides examples and templates of math word problems for 1st to 8th grade classes. Fifth grade common core word problems operations and.
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Please can you help us find shapes at home as well?
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Gain in dB = (40 * log10(cc7/127)) + (40 * log10(cc11/127))
Reading the formula, it appears to me that even if I have expression set at 0 (no bellows movement), the controller will still play the notes, because volume is >0. Is my understanding correct?
Be aware that the output of the formula is gain in decibels, with a range of −∞ (silence) to 0 (maximum). log₁₀(0) is normally undefined, but is treated as −∞ in this formula. If one or both sides of the addition is −∞ then the output of the entire formula is also −∞.
Here are two places that the volume-and-expression formula appears:
General MIDI 2 Specification
, PDF page 12 (printed page 8):
Gain in dB = (40 * log₁₀(cc7/127)) + (40 * log₁₀(cc11/127))
General MIDI Level 1 Developer Guidelines
PDF page 13 (printed page 9):
L(dB) = 40 log (V/127²) where V = (volume × expression)
Ignoring the differences in variable names, these versions of the formula are mathematically equivalent because of the property log(a) + log(b) = log(ab). More specifically, I'll use the following variable names:
G is the gain in decibels from −∞ (silence) to 0 (maximum).
V is the MIDI volume value from 0 to 127.
E is the MIDI expression value from 0 to 127.
G = (40 × log₁₀(V/127)) + (40 × log₁₀(E/127))
then factoring out the 40 from both sides of the addition
G = 40 × ( log₁₀(V/127) + log₁₀(E/127) )
then using the property log(a) + log(b) = log(ab) gets
G = 40 × log₁₀( (V/127) × (E/127) )
I find this version of the formula easier to understand that if one or both of the velocity V or expression E is 0, then the result will be silence.
As the velocity is constant, shall I just use default 127 for note On, and 0 for note Off?
My understanding is that Note On velocity value mainly controls volume, but can also add other effects. On some MIDI devices, when the Note On velocity is 127 or otherwise very large, the resulting instrument sound might sound harsher or have some other exaggerated emphasis. If you use a constant Note On velocity of 127, you might always hear the harsher version of the instrument sound.
So if you want to be able to use various MIDI devices to generate the instrument sound, you may want to avoid a constant Note On velocity of 127, in case it causes an unwanted change in the instrument sound. When I make MIDI files, I usually use a Note On velocity of 100 as a default value.
On the other hand, if you know you are only going to use a specific MIDI device to make the instrument sound, you can test if Note On velocity 127 sounds okay to you and decide to use that if you want.
By the way, my understanding is the Note On velocity will combine the same way into the formula...
Y is the Note On velocity from 1 to 127
G = (40 × log₁₀(V/127)) + (40 × log₁₀(E/127)) + (40 × log₁₀(Y/127))
G = 40 × log₁₀( (V/127) × (E/127) × (Y/127) )
My understanding is Note Off velocity is rarely used by MIDI devices for anything. When I make MIDI files, I usually end notes with a Note On with velocity 0, or sometimes with a Note Off with velocity 0.
I am aware that there is a MIDI Master Volume Universal SysEx message that sets volume for all channels, but I cannot find how this variable is used in conjunction with the others (cc7 and cc11).
I believe the MIDI Master Volume value would combine in the same way...
M is the MIDI Master Volume value from 0 to 16383
G = (40 × log₁₀(M/16383)) + (40 × log₁₀(V/127)) + (40 × log₁₀(E/127)) + (40 × log₁₀(Y/127))
G = 40 × log₁₀( (M/16383) × (V/127) × (E/127) × (Y/127) )
...but I don't use the MIDI Master Volume System Exclusive message when I make MIDI files, so I have no experience with how consistently it behaves on various MIDI devices.