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November 24, 2022

This document aims to serve as an entry point for the various items on the Ethereum roadmap, with a quick summary along with links for those who want to dive deeper.

It is meant as a living document, feel free to contact me if any of the information presented here is unclear, inaccurate, outdated or missing better links.

Note: As indicated by arrows on the roadmap, the various stages listed are not consecutive, various efforts are happening in parallel.

## The Merge

Goal: Have an ideal, simple, robust and decentralized proof-of-stake consensus

### What’s next

• Withdrawals – Enabling validators to withdraw all or part of their stake
• Distributed validators – “multisig but for staking”, where n people share the same validator and m-of-n have to agree on how it behaves
• Enhances staking by protecting against accidental slashing, and making it more accessible (e.g. by trustlessly splitting the 32 ETH required among multiple participants)
• This is not an in-protocol thing, teams such as SSV and Obol are working on it
• View merge – Tweaks the fork-choice rule (the way validators vote) to mitigate a class of attacks
• Essentially enables honest validators to “impose” their view of what the correct head of the chain is, to reduce the chance that malicious validators can split the vote and later reorganize blocks in their favor
• ethresear.ch post with a lot of (very technical) background on the research
• Improved aggregation — Ethereum strives to support as many validators as possible, but having every validator vote on every block (and verify every other validator’s vote) is too bandwidth-intensive. The next best thing is aggregating signatures, but that too has its limits and can be better
• Single slot finality (SSF) — Finalize the chain every slot (12 seconds) instead of every other epoch (12.8 minutes)
• Paths toward SSF
• Along with improved signature aggregation, we still have to figure out two more things:
• SSF consensus algorithm - Existing algorithms compatible with SSF aren’t sufficient, we want one that keeps the chain live even if over $\frac{1}{3}$ of validators are offline
• SSF validator economics - If we end up having to limit the number of validators, how do we limit participation, and what sacrifices do we make?
• Today, the validator selected to propose a block (the leader of the slot) is known a bit ahead of time, enabling a potential DoS attack specifically targetting leaders of upcoming blocks
• ethresear.ch post about a Single SLE protocol based on random shuffling: No one knows who will be the slot’s leader except the leader themselves, until they reveal their block along with a proof of their leadership
• non-single secret leader election might be an option too
• Support even more validators - Ongoing long-term effort: safely supporting more validators is always desirable
• Quantum-safe aggregation-friendly signatures - Part of the long-term effort to make Ethereum safe from quantum computers before it becomes a plausible concern
• The cryptography underlying the BLS signature scheme used is known to be broken by quantum computers, but the alternative signatures schemes known to be quantum-safe aren’t as efficiently aggregated as BLS (Hence the need for a scheme that is both quantum-safe and aggregation-friendly)
• The two leading quantum-safe approaches are STARK-based and Lattice-based

## The Scourge

Goal: ensure reliable and credibly neutral transaction inclusion and avoid centralization and other protocol risks from MEV

### What’s done

• Extra-protocol MEV markets – The MEV-Boost middleware allows the average validator to profit from MEV without having to run sophisticated MEV strategies themselves.

### Distributed builder track

With block proposal staying decentralized, we now have a separate problem where block building becomes centralized. Even with all the other items on the roadmap aiming at minimizing the worst possible downsides of centralized block building, it would still be a major benefit to be able to distribute block building across many nodes.

• Blob construction - Finding ways to alleviate the high bandwidth and processing requirements of data sharding across many nodes that average consumer hardware can run
• Pre-confirmation services - Giving users strong assurances that their transaction will be included in the next block
• Frontrunning protection - Minimizing toxic MEV like sandwiching to keep distributed building credibly neutral

It is still an active area of research with very open design considerations, so it is unclear if the previous two items should get enshrined in the protocol (hence the question marks on the roadmap diagram)

## The Verge

Goal: verifying blocks should be super easy - download N bytes of data, perform a few basic computations, verify a SNARK and you’re done

This section is essentially about filling “the client gap” by making light clients finally viable: Not everyone wants to or can run a full node. The Verge aims to introduce trustless or trust-minimized alternatives that are easy to run and don’t require a lot of storage and bandwidth. The ultimate endgame of The Verge is having these light clients provide security guarantees that are equal to today’s full nodes.

Everything relies on zero-knowledge technology such as SNARKs and STARKs, which themselves rely on polynomial commitment schemes. Here are some links about that:

### What’s done

• Most serious EVM DoS issues solved – Mainly gas pricing issues, fixed in the Berlin upgrade
• Basic light client support (sync committees) – Thanks to sync committees, it is easy to build light clients that follow the consensus layer
• See how Helios client is leveraging sync committees (with a good write-up on how these committees work)

### What’s next

• EIP-4844 implementation – roll out EIP-4844 to mainnet
• Basic rollup scaling - relies on the following:
• EIP-4844 - The scaling is still deemed basic/limited, due to the nature of “every node downloads all the data” restricting the viable capacity of blobspace
• Limited training wheels for rollups (see the proposed milestones)
• Full rollup scaling - relies on:
• P2P design for Data Availability Sampling: Involves all the effort and research into the networking required for data sharding
• DA sampling clients: Development of light-weight clients that can quickly tell if data is available or not through random sampling a few kilobytes
• Efficient DA self-healing: Being able to efficiently reconstitute all the data in the harshest network conditions (e.g. malicious validators attacking, or prolonged downtime of a big chunk of nodes)
• No training wheels for rollups: fully decentralized sequencers, trustless fraud provers, immutable contracts, etc.
• Quantum-safe and trusted-setup-free commitments — Part of the long-term effort to make Ethereum safe from quantum computers before it becomes a plausible concern
• While efficient and powerful, the polynomial commitment used everywhere (KZG) is not quantum-safe and requires a trusted setup. Research into a more ideal commitment suitable for the long term is ongoing, with the eventual goal to “hot swap” KZG under the hood
• SNARK / STARK ASICs – Hardware built specially to create proofs
• Verkle trees - Replace the data structure used for the global state by a more efficient one
• The key benefit is having very short proofs that are easily verified by light clients to validate things like account balances using only the block header – they can already leverage sync committees to validate that a given block header is actually part of the main chain
• Relies on figure out the proper specification, how to safely transition, and how it will affect the EVM gas costs of updating/editing the state (also relies on banning SELF-DESTRUCT from The Purge)
• SNARK-based light clients – SNARKify the sync committee transition to quickly prove which validators form the current sync committee
• Fully SNARKed Ethereum – The following 3 items put together constitute a major milestone toward Ethereum’s Endgame of having extremely efficient and trustless block verification:
• SNARK for Verkle proofs – By merging Verkle proofs into a single SNARK, blocks will contain a short standalone proof about the parts of the state they modify, so it won’t be necessary to verify the whole state of block N-1 to verify that block N modified it correctly.
• SNARK for consensus state transition – Move away from trust-minimized sync committees onto fully trustless verification of everything happening on the consensus layer
• SNARK for L1 EVM – Leveraging the efforts done by rollup teams on zk-EVM by integrating it in L1 directly
• Increase L1 gas limits – By removing today’s burden of “every node needs to store everything” to trustlessly verify blocks, it will be easier to have bigger blocks for more L1 scalability (which will automatically compound all the L2 scaling)
• Move to quantum-safe SNARKs (e.g. STARKs) – Part of the long-term effort to make Ethereum safe from quantum computers before it becomes a plausible concern
• SNARKs are efficient be rely on cryptography known to be broken by quantum computers, while STARKs aren’t

## The Purge

Goal: simplify the protocol, eliminate technical debt and limit costs of participating in the network by clearing old history

### What’s done

• Eliminate most gas refunds - All the gas repricings done in the Berlin upgrade
• Beacon chain fast sync – All the development effort towards syncing from a recent finalized epoch rather than from genesis (known as “checkpoint sync” in most consensus clients)
• EIP-4444 specification – See the EIP specification

### What’s next

• History Expiry — Reduces storage requirements, sync time and code complexity by letting old history expire
• State expiry – Fix the whole “pay once, have your data stored forever” problem regarding the state
• The idea is to automatically expire unused portions of the state and only keeping a verkle tree root that users can use to revive expired state should they need it
• Vitalik’s AMA on State Expiry
• Relies on:
• Base state expiry spec — How we actually do it, see a potential roadmap (and other options)
• Address space extensionIncrease the size of addresses from 20 bytes to 32 bytes to protect against collisions and add data about the state’s period
• Application analysis — Figure out how it might break current applications/contracts and how they can adapt
• LOG reform — Simplify the way event logs work to allow more efficiently searching of historical events
• Serialization harmonization — The execution layer uses RLP for data serialization, while the consensus layer uses SSZ this would get rid of RLP in favor of using SSZ everywhere
• Remove old transaction types — Stop supporting old transaction types (see EIP-2718) to remove code complexity from clients (at the cost of some backwards compatibility)
• EVM simplification track
• Ban SELFDESTRUCT — This opcode is the root of many problems
• Simplify gas mechanics — Involves removing a lot of gas-related EVM features mentioned here
• Precompiles -> EVM implementations — Get rid of precompiled contracts in favor of direct EVM implementations (namely big modular arithmetic, see The Splurge)

## The Splurge

Goal: Fix everything else

All the nice-to-have things that aren’t required for the higher priority stuff belong in The Splurge. The biggest item is account abstraction, but also small tweaks to existing things.

### What’s done

• EIP-1559 — This famous EIP came with many benefits beyond just burning ETH
• ERC-4337 specification — This ERC aims at introducing Account Abstraction without modifying the core protocol

### What’s next

• Endgame EIP-1559 – Enhance EIP-1559 by making it multidimensional, more like an AMM-curve and time-aware
• EVM improvement track along with the simplification track from The Purge leading to the EVM endgame
• EVM Object Format (EOF) — A set of multiples EIPs allowing validating and versioning EVM bytecode when it is deployed. Various explainers: [1], [2], [3]
• Big modular arithemetic – A lot of the roadmap’s cryptography relies on modular arithmetic over very large numbers, which could be done more efficiently in the EVM directly
• Further EVM improvements — Anything else that’s worth adding to improve the EVM – or removing to get rid of complexity
• Account abstraction track leading to the Endgame account abstraction. See Vitalik’s descriptions on the following items for more detail:
• ERC-4337 – Developing compliant smart wallets that actually gain adoption
• Voluntary EOA conversion — With an EIP, allow a normal account to irreversibly add code to convert into it into a contract, namely to become a 4337-compliant smart wallet.
• In-protocol enshrining — Make the above conversion mandatory for all existing accounts
• Verifiable Delay Functions (VDFs) — Essentially “non-parallelizable proof of work” which would enhance the randomness used in PoS and other things
• Explore solution for dust accounts – Rescuing “dust funds” that costs more gas fees to move than they are worth. See a bunch of ideas here


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# Stronger negation of AC given by rejecting “infinite hat” puzzles

Some of the strangest implications of AC are the "infinite hat" puzzles, which are on Wikipedia, and have been talked about on MO several times, including some variants.

There are different ways to formalize these puzzles, but most of them follow the same basic principle:

1. There is a countably-indexed sequence of things (reals, naturals, bits, etc).
2. You only have information about some partial subsequence: usually with one value missing, or an initial segment missing, etc.
3. You get to use the axiom of choice to "mysteriously" infer some value from the original sequence that isn't in your partial subsequence.
4. Informally, this inference can be made to agree with the original "arbitrarily often", such as failing in only a finite set of cases, despite having no knowledge of what the original sequence was.

The basic gist is usually that you use AC to get a "chosen" representative from each equivalence class of sequences with the same infinite "tail." Then, for your partial sequence, you interpolate some missing value based on what the representative says. This strategy is correct for all but a finite set of cases in some initial segment, leading to bizarre consequences.

My question:

It would be interesting to see what happens if you take, as an axiom, the deliberate negation that "infinite hat" puzzles admit any such strategy, and see how strong it is compared to just the negation of $$AC$$.

The simplest way to formalize this, that I can think of, is as follows:

• Let $$a$$ be some sequence $$\Bbb N \to S$$, for some arbitrary set $$S$$.
• For some $$k \in \Bbb N$$, let the knockout sequence $$a \setminus k$$ be the restriction of $$a$$ to the domain $$\Bbb N \setminus k$$. (i.e., one (index, value) pair is missing)
• Let an interpolation function be some $$f$$ that takes as input some $$a \setminus k$$, and returns some value $$s \in S$$, which we interpret as its "suggestion" for the missing value at $$k$$.
• For any such interpolation function $$f$$ and sequence $$a$$, we can define an agreement sequence $$A_{f,a}[k]$$, for which $$A_{f,a}[k] = 1$$ if $$f(a \setminus k) = a[k]$$ and $$A_{f,a} = 0$$ otherwise.
• Likewise, we can define the agreement set $$A_{f,a} = \{k: A_{f,a}[k] = 1\}$$, the set of indices for which the agreement sequence is 1.

AC, then, says that for each choice of $$S$$, there exists an interpolation function $$f$$ such that for all $$S$$-sequences $$a$$, $$A_{f,a}[k]$$ is eventually constant at 1. Just let $$f(a \setminus k) = r(a)[k]$$, where $$r(a)$$ is the chosen representative from the equivalence class for $$a$$.

However, our negation can be much stronger: we can declare that no interpolation function $$f$$ can be guaranteed, for all sequences $$a$$, to have an agreement set with natural density greater than $$1/|S|$$ if $$S$$ is finite, or $$0$$ if it is infinite. That is, any such $$f$$ may do better for some particular sequence $$a$$, but cannot be guaranteed to do better for all $$a$$.

For uncountable $$S$$ we can get even stronger: not only should the natural density of the agreement set be $$0$$, but the agreement set can never be guaranteed to even be any larger than the empty set for all $$a$$. That is, given a sequence of reals, there can be no interpolator $$f$$ that is guaranteed, for all such sequences, to interpolate even one missing value correctly.

There is probably some better way to formalize the above, but at least it isn't that complicated, and formalizes the basic reservation that people often have with infinite hat puzzles. So I'm simply interested in what set theory looks like if we take that perspective and formalize it.

• Have axioms similar to these been studied?
• Are they known to be equivalent to, to imply, or to be implied by other well-known axioms that are incompatible with AC?
• Is there some simpler way to formalize the above?
• You know, it's really great that you ask these things. But many of these questions were discussed previously either here or on math.SE. Specifically, the infinite hat "paradox" implies there are irregular sets. Namely, sets without the Baire property, sets which are not Lebesgue measurable, etc. – Asaf Karagila Jan 31 '19 at 21:36
• I would appreciate links on the implications of the infinite hat "paradox" if you have them, but this is a totally different question than what I'm asking. I'm not asking about the implications of the paradox, I'm asking about the implications of its negation. For example, Banach-Tarski implies non-measurable sets of reals, but "not-Banach-Tarski" is not strong enough to imply the nonexistence of non-measurable sets of reals. – Mike Battaglia Jan 31 '19 at 21:51
• Some of these infinite hat results can be derived from "there exists a nonprincipal ultrafilter on $\mathbb N$." I don't know how this fits into the bigger picture of consistency results. – Jalex Stark Jan 31 '19 at 22:12

Naturally, the more generalizations of the infinite hats puzzle we consider, the stronger it is to assert that none of them have a paradoxical solution. One of the variants you linked in your question is the box variation, where 100 mathematicians take turns entering a room with a countable infinity of boxes, each containing a real number, and after opening arbitrarily many boxes, must guess the contents of some unopened box. With AC, it can be shown there is a strategy where at most one mathematician fails.

One way to generalize the box puzzle is to have there be uncountably many boxes. For example, let's consider the box puzzle where there is one box for every set of reals, each box contains a real, and there are countably infinitely many mathematicians.

It doesn't seem like the increased number of boxes should make things easier on the mathematicians (they still have to guess the contents of some box without any apparently useful knowledge). But it turns out there are explicit winning strategies to this variant, where by "winning" I mean that all but (at most) one mathematician makes a correct guess. This is a theorem of fourth-order arithmetic. I sketch the details for a similar variant here https://mathoverflow.net/a/300556/109573.

If you accept that this is a paradox of the type you are interested in, then a consequence of "there are no paradoxical strategies for any guessing game" is $$\mathcal{P}(\mathbb{R})$$ does not exist. We would have to weaken our foundations to third-order arithmetic, with some antichoice principles like nonexistence of nonprincipal ultrafilters on $$\mathbb{N}.$$

• Hi Elliot - if you look at the way I use interpolators, it's strong enough to rule out both the usual infinite hat puzzle and the 100 mathematician variant, since both depend on an interpolator $f$ that is guaranteed to make perfect guesses for some missing value eventually as the index goes to infinity. My negation was stronger than just saying no "eventually perfect" interpolator exists, it was saying no interpolator should be guaranteed for all sequences to give correct guesses with natural density greater than 1/|S| for finite S, or produce any correct guesses at all for uncountable S. – Mike Battaglia Feb 1 '19 at 6:29
• Is it strong enough to rule out the variant I describe in the second paragraph? My answer isn't about the standard 100 mathematician puzzle. – Elliot Glazer Feb 1 '19 at 6:31
• Sorry, writing comments at the same time - I thought about using $\Bbb R$ as an index set instead of $\Bbb N$, or to let there be arbitrary index sets, but wasn't certain how that would make anything stronger. But your version seems to use well-ordered subsets of $\Bbb R$, "modded out by isomorphism" - what do you mean by that last part? – Mike Battaglia Feb 1 '19 at 6:38
• The variant in the linked post is slightly different. Rather than there being one box for every set of reals (i.e. the index set is $\mathcal{P}(\mathbb{R})$), I have the index set $X$ be the Hartogs number of $\mathbb{R}.$ I was just describing how to construct this set. – Elliot Glazer Feb 1 '19 at 6:42
• Ok, I got tripped up at the part where you say $X \cong 2 \times \omega \times X$. Does $\times$ refer to ordinal multiplication? I thought you were splitting $X$ into two interleaved subsequences, but I may be confused here. – Mike Battaglia Feb 1 '19 at 7:04

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# Mastering in Latex (4)

During the writing of my PhD mid-term exam text, I dealt with some common and usual issues with latex language. Because of this, I decide to share the most relevant and those which are difficult to find an alternative in the internet.

1) How can I put a citation at the caption in a figure?

Probably, you have already dealt with this situation once during your writing process. This is really easy to do if you are used Word but at the Latex, you will need some additional command. Below, there is an example of how I made in my text:

\begin{figure}[h]
  \centering
  \includegraphics[width=1.00\textwidth]{./figs/2013.png}
  \caption[Map of 2013.]{Map 2013.}
  \captionsetup{aboveskip=0pt,font=it}
  \caption*{Source: IBGE \cite{IBGE}.}
  \label{fig:map201}
\end{figure}

Some important comments. The command \caption has two parameters: a) text that will appear in the table of contents and b) text of the caption.

2) How can I insert two figures in parallel?

I know that there is a lot of alternatives to do this but most of them didn’t work for me. The only which works was through the command below using \subfigure:

\begin{figure*}[h]
 	\centering
 	\subfigure[Front View]{
 		\includegraphics[width=.55\linewidth]{./figs/basket.jpg}
	}
	\hspace*{0.5cm}
 	\subfigure[Restroom]{
 		\includegraphics[width=.35\linewidth]{./figs/basket1.jpg}
	}
 	\caption{Basket Stadium.}
 	\label{fig:stadium}
\end{figure*}

If you decide to use it, don’t forget to insert its package. You can do this using this command:

\usepackage{subfigure}

3) How can I insert a footnote in a table?

This is another common situation during the writing, e.g. if you want to describe an acronym that is present in the table. In my case, I do this using the threeparttable command, you can find the package and command below:

\usepackage[flushleft]{threeparttable}
\begin{table}[h!]
	\begin{threeparttable}
	\caption{Reference Caption}
	\centering
	\begin{tabular}{C{0.5cm} C{1.0cm} }
		\hline
		No	& Hour	\\ \hline
		1   	& 13:00 \\
		2   	& 13:04 \\
		3   	& 13:14  \\
		4   	& 13:20\footnotemark[1] \\ \hline
	\end{tabular}
	\begin{tablenotes}
    	  \small
      	  \item \footnotemark[1]{text to reference}.
    	\end{tablenotes}
 	\end{threeparttable}
 	\label{tbl:table1}
\end{table}

In all of the cases, I don’t know if I chose the best or most beautiful alternative to solve the problem but they worked for me and could work for you!


text: I’m not sure what this means, but it … This is demonstrated by encrypting ATTACKATDAWN with IOZQGH, to produce the same ciphertext as in the original example. that any two randomly chosen source language letters are the same (around 0.067 for monocase English) and the probability of a coincidence for a uniform random selection from the alphabet 0 A {\displaystyle m} The known section and its location is verified. [17] Earlier attacks relied on knowledge of the plaintext or the use of a recognizable word as a key. {\displaystyle \Sigma =(A,B,C,\ldots ,X,Y,Z)} However, by using the VigenÃ¨re cipher, E can be enciphered as different ciphertext letters at different points in the message, which defeats simple frequency analysis. denotes the offset of the i-th character of the plaintext For successive letters of the message, successive letters of the key string will be taken and each message letter enciphered by using its corresponding key row. It uses a series of Caesar ciphers to encrypt the text. Checkout wiki page for more information. Cipher Challenge. Later, Johannes Trithemius, in his work Polygraphiae (which was completed in manuscript form in 1508 but first published in 1518),[5] invented the tabula recta, a critical component of the VigenÃ¨re cipher. It's very similar to dealing with a keyword and substitution. , Method of encrypting alphabetic text by using a series of interwoven Caesar ciphers based on the letters of a keyword. In a Caesar cipher, each letter in the passage is moved a certain number of letters over, to be replaced by the corresponding letter. , the calculation would result in Keys were typically single words or short phrases, known to both parties in advance, or transmitted "out of band" along with the message. The rest of the plaintext is enciphered in a similar fashion: Decryption is performed by going to the row in the table corresponding to the key, finding the position of the ciphertext letter in that row and then using the column's label as the plaintext. K To improve the security, the greater the size of the code word, Therefore, to decrypt The Vigenère cipher can also be described and then decrypted algebraically, by assigning each letter from A to Z a value from 0 to 25, with addition being performed modulo 26.  analysis, Answer; There is an easier way to use the Vigenere cipher, using modular arithmetic. ℓ Each team is expected to write, rehearse, and perform an approximately 5 – 10 minute newscast. C Using methods similar to those used to break the Caesar cipher, the letters in the ciphertext can be discovered. is the length of key, VigenÃ¨re encryption and decryption can be written: M {\displaystyle i\in [1,n-m]} M Congratulations to Euchre Mutt for solving Challenge 5 on September 22 at 8:06 UTC. The Gronsfeld cipher is strengthened because its key is not a word, but it is weakened because it has just 10 cipher alphabets. , The Vigenère cipher is a method of encrypting alphabetic text by using a series of interwoven Caesar ciphers, based on the letters of a keyword. How to Break It . [7] The cipher now known as the VigenÃ¨re cipher, however, is that originally described by Giovan Battista Bellaso in his 1553 book La cifra del Sig. In 1854, Charles Babbage was goaded into breaking the VigenÃ¨re cipher when John Hall Brock Thwaites submitted a "new" cipher to the Journal of the Society of the Arts. Ciphertext    NVPPBNFAEEKPSY. ^ {\displaystyle C=C_{1}\dots C_{n}} Also, it is not possible to 4 In fact, the two ciphers were often confused, and both were sometimes called le chiffre indÃ©chiffrable. κ The person sending the message chooses a keyword and repeats it until it matches the length of the plaintext, for example, the keyword "LEMON": Each row starts with a key letter. Agent Madness has drafted a guide to breaking the Vigenere cipher which you may find very helpful with Mission Briefing 6A. (All factors of the distance are possible key lengths; a key of length one is just a simple Caesar cipher, and its cryptanalysis is much easier.) Cipher Challenge - Prepared for a company near Dallas, TX - by D. Dante Lorenso. If a cryptanalyst correctly guesses the key's length, the cipher text can be treated as interwoven Caesar ciphers, which can easily be broken individually. Original Challenge Instructions. Keyword        GREENGREENGREE , K κ Vigenère cipher is the sequence of Caesar ciphers with different transformations (ROTX, see Caesar cipher). Plaintext     hellohowareyou cipher-challenge. A [8] He built upon the tabula recta of Trithemius but added a repeating "countersign" (a key) to switch cipher alphabets every letter. The Kasiski examination, also called the Kasiski test, takes advantage of the fact that repeated words are, by chance, sometimes encrypted using the same key letters, leading to repeated groups in the ciphertext. it involves more than one alphabet, such as a Vigenère cipher (see earlier challenges this year), then the IoC score will drop below c. 0.065, but will still be higher than the completely random text score 0.0385. times in which = [13] The Confederate States of America, for example, used a brass cipher disk to implement the VigenÃ¨re cipher during the American Civil War. L In 1863, Friedrich Kasiski was the first to publish a successful general attack on the VigenÃ¨re cipher. with key letter The Beaufort cipher is a reciprocal cipher. = {\displaystyle E} Despite the VigenÃ¨re cipher's apparent strength, it never became widely used throughout Europe. Was the alphabet keyed? Whereas Alberti and Trithemius used a fixed pattern of substitutions, Bellaso's scheme meant the pattern of substitutions could be easily changed, simply by selecting a new key. is the keyword length. Therefore, if the key length is known (or guessed), subtracting the cipher text from itself, offset by the key length, will produce the plain text subtracted from itself, also offset by the key length. MD5 hash Variant Beaufort cipher Babbage actually broke the much-stronger autokey cipher, but Kasiski is generally credited with the first published solution to the fixed-key polyalphabetic ciphers. To encrypt, a table of alphabets can be used, termed a tabula recta, VigenÃ¨re square or VigenÃ¨re table. Z The VigenÃ¨re cipher, with normal alphabets, essentially uses modulo arithmetic, which is commutative. . That is, however, only an approximation; its accuracy increases with the size of the text. , In the 19th century the scheme was misattributed to Blaise de VigenÃ¨re (1523â1596), and so acquired its present name. I actually spent less time on this challenge than I did on part 1! Anyone can learn computer science. John Hall Brock Thwaites, Bristol â Improvements in apparatus to facilitate communication by cypher.". That method is sometimes referred to as "Variant Beaufort". The distance between the repetitions of QUCE is 30 characters. Challenge Quizzes Cryptography: Level 1 Challenges Cryptography: Level 3 Challenges Vigenère Cipher A certain message is encoded with a 2 letter key. They will be assigned to teams of 3-4. It actually becomes interesting now! D How long is the key? Vigenère cipher: Encrypt and decrypt online. For example, first letter of text is transformed using ROT5, second - using ROT17, et cetera. is the message, The Kasiski examination and Friedman test can help to determine the key length (see below: Â§ Kasiski examination and Â§ Friedman test). Once every letter in the key is known, all the cryptanalyst has to do is to decrypt the ciphertext and reveal the plaintext. … In [38], Ma et al. Test your Wolfram Language coding skills with programming puzzles spanning computation, math and language. Encryption. i For a keyword of KING, determine the following VigenÃ¨re codes: If you are struggling, here the mapping for KING: For example By taking the intersection of those sets, one could safely conclude that the most likely key length is 6 since 3, 2, and 1 are unrealistically short. The primary weakness of the VigenÃ¨re cipher is the repeating nature of its key. For example, if the ^ The idea behind the VigenÃ¨re cipher, like all other polyalphabetic ciphers, is to disguise the plaintext letter frequency to interfere with a straightforward application of frequency analysis. This can be understood as the point where both keys line up. i = Instead it … The letter at the intersection of [key-row, msg-col] is the enciphered letter. {\displaystyle 13\,{\widehat {=}}\,N} m For example, in row L (from LEMON), the ciphertext L appears in column A, which is the first plaintext letter. , For example, the first letter of the plaintext, A, is paired with L, the first letter of the key. Phrase LEMON, for example, defines the sequence of ROT11-ROT4-ROT12-ROT14-ROT13, which is repeated until all block of text is encrypted. Studies of Babbage's notes reveal that he had used the method later published by Kasiski and suggest that he had been using the method as early as 1846.[21]. 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Its key is known 30 characters  Thwaites, Bristol â Improvements in apparatus to facilitate communication by cypher !


text:
Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Continuous uniform distribution ▷ Mean

Theorem: Let $X$ be a random variable following a continuous uniform distribution:

$\label{eq:cuni} X \sim \mathcal{U}(a, b) \; .$

Then, the mean or expected value of $X$ is

$\label{eq:cuni-mean} \mathrm{E}(X) = \frac{1}{2} (a+b) \; .$

Proof: The expected value is the probability-weighted average over all possible values:

$\label{eq:mean} \mathrm{E}(X) = \int_{\mathcal{X}} x \cdot f_X(x) \, \mathrm{d}x \; .$

With the probability density function of the continuous uniform distribution, this becomes:

$\label{eq:cuni-mean-qed} \begin{split} \mathrm{E}(X) &= \int_a^b x \cdot \frac{1}{b-a} \, \mathrm{d}x \\ &= \left[ \frac{1}{2} \, \frac{x^2}{b-a} \right]_a^b \\ &= \frac{1}{2} \, \frac{b^2 - a^2}{b-a} \\ &= \frac{1}{2} \, \frac{(b+a)(b-a)}{b-a} \\ &= \frac{1}{2} (a+b) \; . \end{split}$
Sources:

Metadata: ID: P82 | shortcut: cuni-mean | author: JoramSoch | date: 2020-03-16, 16:12.


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# A-level Mathematics/MEI/S1/Discrete Random Variables

< A-level Mathematics | MEI | S1

## Discrete Random Variables

Discrete random variables are the models for the probabilities of particular results. They are represented by the notation, ${\displaystyle P(X=R)}$ (Where P is the probability of the Discrete Random Variable, X , taking on the result, R.). A discrete random variable can only take on the values r1,r2,r3...rn with the probabilities p1, p2, p3...pn respectively. Discrete random variables allow us to produce a Discrete Frequency Distribution. Discrete meaning that the value of the results can only take on positive whole integers. In S1, Discrete distributions are limited to a finite number of outcomes.

## Exemplar Question

Questions often involve working with algebra in order to find the probabilities, in OCR(MEI), you will often have to work out the value of k in order to work out the probabilities.

The probability distribution of a discrete random variable, X, is given by

${\displaystyle P(X=R)={\frac {kr}{8}}R=2,4,6,8.}$

Find the value of k and tabulate the probability distribution.

To find the value of k, we need to tabulate our probabilities and substitute the values of R into the discrete random variable, X.

    R || 2 | 4 | 6 | 8 ||
p(x=R)|| ${\displaystyle {\frac {k2}{8}}}$


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# Real number calculus vs complex calculus

I have started studying complex integration recently and i just can't seem to get the things in my head .
the biggest problem i am facing is that :
when solving real number integrals the area under the curve of the function is what integration means ...
but i can't seem to find an analogy between this and complex integration ......

micromass
Staff Emeritus
Homework Helper
I have started studying complex integration recently and i just can't seem to get the things in my head .
the biggest problem i am facing is that :
when solving real number integrals the area under the curve of the function is what integration means ...
but i can't seem to find an analogy between this and complex integration ......

Consult the book "visual complex analysis" by Needham.

Also, it is a variant of path integration, so you should first be comfortable with that. Path integrals can be easily interpreted as an area under a curve. http://en.wikipedia.org/wiki/Line_integral#mediaviewer/File:Line_integral_of_scalar_field.gif

Also, it is a variant of path integration, so you should first be comfortable with that. Path integrals can be easily interpreted as an area under a curve. http://en.wikipedia.org/wiki/Line_integral#mediaviewer/File:Line_integral_of_scalar_field.gif
i thought complex plane is a 2 dimensional plane just like our cartesian plane ... so why can't i just take the area under the curve in complex plane here too ?
and while approaching non complex integration why don't we integrate for any two points on the plane with an arbitrary path ? ( sorry if i am wrong about everything i said , i am still a noob at complex integration )

Svein
i thought complex plane is a 2 dimensional plane just like our cartesian plane ... so why can't i just take the area under the curve in complex plane here too ?
and while approaching non complex integration why don't we integrate for any two points on the plane with an arbitrary path ? ( sorry if i am wrong about everything i said , i am still a noob at complex integration )

Yes and no. Yes it is a 2-dimensional cartesian plane when viewed a certain way. No because it is a generalization of the real line when viewed another way.

Example: In the 2-dimensional cartesian plane you can define addition and subtraction of points easily, but there is no straightforward way of defining a way to multiply two points and get a new point. In the complex plane, multiplication is defined from the beginning.

As I said, the complex plane is a generalization of the real line. Complex functions, though, have stricter requirements than real functions. For a real function to be differentiable, the right- and left-hand derivative must both exist and be equal. For a complex function, the derivatives must be equal no matter how we approach the point in question. Such functions are called analytic, and they have several interesting properties, one of them being that the integral from one point to another is not dependent on the path used. This again means that the integral of an analytic function along a closed curve is zero.

Integration is one of the most important tools in complex analysis and the basis for several important theorems.

lavinia
Gold Member
I have started studying complex integration recently and i just can't seem to get the things in my head .
the biggest problem i am facing is that :
when solving real number integrals the area under the curve of the function is what integration means ...
but i can't seem to find an analogy between this and complex integration ......

If a function in the plane were real valued then one could draw its graph in three dimensions. It would look like a surface and it definitely makes sense to talk about the volume underneath it (not area). But the only real valued continuous complex differentiable functions are constants. So for them computing these volumes is uninteresting.

You should try to convince yourself that complex differentiable functions that are real valued are constants. It is not hard.

lavinia
Gold Member
i thought complex plane is a 2 dimensional plane just like our cartesian plane ... so why can't i just take the area under the curve in complex plane here too ?
and while approaching non complex integration why don't we integrate for any two points on the plane with an arbitrary path ? ( sorry if i am wrong about everything i said , i am still a noob at complex integration )

There is nothing to stop you from taking the area under a curve if the function is real valued. If it is complex valued the idea makes no sense.

Last edited:
There is nothing to stop you from taking the area under a curve if the function is real valued. If it is complex valued the idea makes no sense.
but why doesn't it make sense when the function is complex valued ?

lavinia
Gold Member
but why doesn't it make sense when the function is complex valued ?

A complex number is not a height.

A complex line integral can be thought of as two regular integrals added together (after multiplying one of them by i) to get a complex number

mathwonk


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# [Solution] Sending a Sequence Over the Network solution codeforces

## Sending a Sequence Over the Network solution codeforces

Sending a Sequence Over the Network solution codeforces – The sequence 𝑎a is sent over the network as follows:

1. sequence 𝑎a is split into segments (each element of the sequence belongs to exactly one segment, each segment is a group of consecutive elements of sequence);
2. for each segment, its length is written next to it, either to the left of it or to the right of it;
3. the resulting sequence 𝑏b is sent over the network.

For example, we needed to send the sequence 𝑎=[1,2,3,1,2,3]a=[1,2,3,1,2,3]. Suppose it was split into segments as follows: [1]+[2,3,1]+[2,3][1]+[2,3,1]+[2,3]. Then we could have the following sequences:

• 𝑏=[1,1,3,2,3,1,2,3,2]b=[1,1,3,2,3,1,2,3,2],
• 𝑏=[1,1,3,2,3,1,2,2,3]b=[1,1,3,2,3,1,2,2,3],
• 𝑏=[1,1,2,3,1,3,2,2,3]b=[1,1,2,3,1,3,2,2,3],
• 𝑏=[1,1,2,3,1,3,2,3,2]b=[1,1,2,3,1,3,2,3,2].

If a different segmentation had been used, the sent sequence might have been different.

## [Solution] Sending a Sequence Over the Network solution codeforces

The sequence 𝑏b is given. Could the sequence 𝑏b be sent over the network? In other words, is there such a sequence 𝑎a that converting 𝑎a to send it over the network could result in a sequence 𝑏b?

Input

The first line of input data contains a single integer 𝑡t (1𝑡1041≤t≤104) — the number of test cases.

Each test case consists of two lines.

The first line of the test case contains an integer 𝑛n (1𝑛21051≤n≤2⋅105) — the size of the sequence 𝑏b.

The second line of test case contains 𝑛n integers 𝑏1,𝑏2,,𝑏𝑛b1,b2,…,bn (1𝑏𝑖1091≤bi≤109) — the sequence 𝑏b itself.

It is guaranteed that the sum of 𝑛n over all test cases does not exceed 21052⋅105.

Output

For each test case print on a separate line:

• YES if sequence 𝑏b could be sent over the network, that is, if sequence 𝑏b could be obtained from some sequence 𝑎a to send 𝑎a over the network.
• NO otherwise.

You can output YES and NO in any case (for example, strings yEsyesYes and YES will be recognized as positive response).

## [Solution] Sending a Sequence Over the Network solution codeforces

input

Copy
7
9
1 1 2 3 1 3 2 2 3
5
12 1 2 7 5
6
5 7 8 9 10 3
4
4 8 6 2
2
3 1
10
4 6 2 1 9 4 9 3 4 2
1
1

output

Copy
YES
YES
YES
NO
YES
YES
NO

## [Solution] Sending a Sequence Over the Network solution codeforces

In the first case, the sequence 𝑏b could be obtained from the sequence 𝑎=[1,2,3,1,2,3]a=[1,2,3,1,2,3] with the following partition: [1]+[2,3,1]+[2,3][1]+[2,3,1]+[2,3]. The sequence 𝑏b[1,1,2,3,1,3,2,2,3][1,1,2,3,1,3,2,2,3].

In the second case, the sequence 𝑏b could be obtained from the sequence 𝑎=[12,7,5]a=[12,7,5] with the following partition: [12]+[7,5][12]+[7,5]. The sequence 𝑏b[12,1,2,7,5][12,1,2,7,5].

In the third case, the sequence 𝑏b could be obtained from the sequence 𝑎=[7,8,9,10,3]a=[7,8,9,10,3] with the following partition: [7,8,9,10,3][7,8,9,10,3]. The sequence 𝑏b[5,7,8,9,10,3][5,7,8,9,10,3].

In the fourth case, there is no sequence 𝑎a such that changing 𝑎a for transmission over the network could produce a sequence 𝑏b.


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# galpy.df.impulse_deltav_hernquist_curvedstream¶

galpy.df.impulse_deltav_hernquist_curvedstream(v, x, b, w, x0, v0, GM, rs)[source]

NAME:

impulse_deltav_plummer_hernquist

PURPOSE:

calculate the delta velocity to due an encounter with a Hernquist sphere in the impulse approximation; allows for arbitrary velocity vectors, and arbitrary position along the stream

INPUT:

v - velocity of the stream (nstar,3)

x - position along the stream (nstar,3)

b - impact parameter

w - velocity of the Hernquist sphere (3)

x0 - point of closest approach

v0 - velocity of point of closest approach

GM - mass of the Hernquist sphere (in natural units)

rs - size of the Hernquist sphere

OUTPUT:

deltav (nstar,3)

HISTORY:

2015-08-13 - SANDERS, using Wyn Evans calculation


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# Rearrange the formula for the sum of a geometric series to find the value of its common ratio?

What I'm attempting to do is to rearrange the formula for the sum of a geometric series so as to find the value of its common ratio $r$. I've tried several different methods, all of which have failed; though I can't understand why.

This was my last attempt using logarithms:

$$S_n=\frac{a(1-r^n)}{1-r}$$ $$S_n(1-r) = a(1-r^n)$$ $$S_n-S_nr = a - ar^n$$ $$\frac{S_n}{a}-1 = \frac{S_n}{a}r-r^n$$ $$\log{\frac{S_n}{a}} = \log{\frac{S_n}{a}}+\log{r}-n\log{r}$$ $$0 = (\log{r})(n-1)$$ $$r = 10^{\frac{0}{n-1}}$$ $$r=1$$

I can't understand where I'm going wrong. Any advice would be great. Also, if you know a formula to find $r$ using $S_n$ and $a$, then that would be equally fantastic.

Thanks.

-
There is a mistake in the application of log to $(\dfrac{S_n}{a}-1).$ –  Américo Tavares Oct 31 '10 at 23:04
Do you want the sum of the (infinite) geometric series, or do you want the $n$th partial sum of the series? Your question implies the former, but the formula you give for $S_n$ implies the latter. –  Mike Spivey Nov 1 '10 at 2:17

HINT $\ \$ Consider $\ S_{n+1} - r\ S_n$
$$S_{n+1} - rS_n = a$$ (by def. of geometric series)


text:
Coloring

Grunbaum's Conjecture ★★★

Author(s): Grunbaum

\begin{conjecture} If $G$ is a simple loopless \Def[triangulation]{triangulation (topology)} of an \Def{orientable surface}, then the dual of $G$ is 3-edge-colorable. \end{conjecture}

Keywords: coloring; surface

5-local-tensions ★★

Author(s): DeVos

\begin{conjecture} There exists a fixed constant $c$ (probably $c=4$ suffices) so that every embedded (loopless) graph with edge-width $\ge c$ has a 5-local-tension. \end{conjecture}

Keywords: coloring; surface; tension

Degenerate colorings of planar graphs ★★★

Author(s): Borodin

A graph $G$ is $k$-\emph{degenerate} if every subgraph of $G$ has a vertex of degree $\le k$.

\begin{conjecture} Every simple planar graph has a 5-coloring so that for $1 \le k \le 4$, the union of any $k$ color classes induces a $(k-1)$-degenerate graph. \end{conjecture}

Keywords: coloring; degenerate; planar

3-Colourability of Arrangements of Great Circles ★★

Consider a set $S$ of great circles on a sphere with no three circles meeting at a point. The arrangement graph of $S$ has a vertex for each intersection point, and an edge for each arc directly connecting two intersection points. So this arrangement graph is 4-regular and planar.
\begin{conjecture} Every arrangement graph of a set of great circles is $3$-colourable. \end{conjecture}


text:
## Section2.3Vector Fields

Examples:

See the list of example Vector Fields on the Examples submenu of the CalcPlot3D main menu. See more about the Examples menu in Section 4.6.

To add a vector field to the plot, select the option Vector Field on the Add to graph drop-down menu. You'll see an object dialog appear like the following:

Now let's plot the default vector field.

Clear the plot by clicking the clear plot button.

Then check the checkbox in the upper-left corner of this object to plot the default vector field.

This vector field happens to be a 2D vector field, but it is plotted in a 3D plot by default. It should look like Figure 2.3.2. To quickly change the view to be above the $xy$-plane, select the Restrict view to 2D option. Now your view should look like Figure 2.3.3. Note that in 2D, the Use Constant Primary Color option is selected by default. When this option is not selected, the vectors are colored based on their three components similarly to the way faces of a surface are colored based on the three components of their normal vectors.

This vector field is still fairly easy to read, but sometimes the vectors may be very difficult to read because of how they overlap.

There are a couple options for dealing with this. One option is to force all vectors in the vector field to have the same fixed length. Select the checkbox labeled Use fixed length for all vectors to see how this looks. It should appear like Figure 2.3.4.

Another way to vary the vector lengths in a vector field so that they do not overlap is to adjust the scalefactor, labeled Scale vectors by dividing by. By default, vectors in a vector field are divided by 4 to scale them down. You can adjust this value to see how it affects the plot. To see all vectors in the vector field at the natural length, change this value to 1.

The Number of vectors along each axis can be adjusted to change the appearance and refinement of the vector field. When in 2D mode, we just need to adjust the number of vectors along the $x$- and $y$-axes, and keep just one vector along the $z$-axis. But when in 3D mode, you'll want to change the number of levels along the $z$-axis to be more than one.

Note that if you are considering a system of differential equations, you also have the option of changing the notation of the components of the vector field to use differential notation. Select the option to Show system of DEs notation and the components of the vector field should look like those here:

$\Large\textbf{3D Vector Fields:}$

To obtain a 3D vector field, deselect the option to Restrict view to 2D and change the number of vectors along the $z$-axis to be more than 1. For example, if you change this value to 4 and also deselect the option to Use Constant Primary Color, the view will look like Figure 2.3.5.

Changing the third component to be non-zero, say $P = z - x\text{,}$ and rotating the view a little, we obtain a plot like Figure 2.3.6. The definition of the vector field is now:

Note that 3D Glasses mode would be quite helpful for 3D vector fields. Give it a try!

$\Large\textbf{Vector Field Modes:}$

Every vector field we've looked at so far has been in Rectangular Array mode. Now let's consider the other mode options provided here. Although these modes can be used in 2D mode, I think they are most useful when viewed in 3D.

Spherical Array:

Instead of showing vectors at a rectangular grid of points as is done in the Rectangular Array mode, Spherical Array mode places vectors at points on spheres (or circles in 2D).

Instead of specifying the number of vectors in each axis direction, Spherical Array mode uses the number of vectors on each circle (parallel to the $xy$-plane), the number of concentric spheres (or concentric circles in 2D), and the number of circular steps in the vertical $z$-direction.

Cylindrical Array:

In this mode places vectors at points on cylinders (or circles in 2D).

Cylindrical Array mode uses the number of vectors on each circle (parallel to the $xy$-plane), the number of concentric circles, and the number of levels in the vertical $z$-direction.

Note that the Settings button ⚙ in Vector Fields does not yet do anything.


text:
# Definition:Uncountable/Set

(Redirected from Definition:Uncountable Set)

## Definition

A set which is infinite but not countable is described as uncountable.

## Also known as

An uncountable collection can also be seen as:

non-denumerable or nondenumerable
non-countable.

## Also see

• Results about uncountable sets can be found here.


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# A computer operator can type 5,555 words in an hour. How many words will he type in 8 hours?

Last updated date: 23rd Mar 2023
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Hint: We can use the unitary method to solve this type of problem. The unitary method can be explained as - first finding the single unit’s value and then multiplying it with the desired number to get the final value.

Suppose that the cost of y bananas is Rs. x. If we are asked to find how many bananas can be bought for Rs. z, then we must find out the number of bananas that can be bought for Rs. 1 first. So, we get the number of bananas that can be bought for Rs.1 as $\dfrac{y}{x}$. Now, to find the number of bananas that can be bought for Rs. z, we just have to multiply the number of bananas bought for Rs. 1. Therefore, the number of bananas that can be bought for Rs. z is $\dfrac{y}{x}\times z$.
Number of words typed by the computer operator in 8 hours $=5,555\times 8$.
After multiplying $5,555\times 8$, we get 44,440.


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# $$\textit{Ab initio}$$ and group theoretical study of properties of the $$\text{C}_\text{2}\text{C}_\text{N}$$ carbon trimer defect in h-BN

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### Abstract

Hexagonal boron nitride (h-BN) is a promising platform for quantum information processing due to its potential to host optically active defects with attractive optical and spin properties. Recent studies suggest that carbon trimers might be the defect responsible for single-photon emission in the visible spectral range in h-BN. In this theoretical study, we combine group theory together with density functional theory (DFT) calculations to predict the properties of the neutral $$\text{C}_\text{2}\text{C}_\text{N}$$ carbon trimer defect. We find the multi-electron states of this defect along with possible radiative and non-radiative transitions assisted by the spin-orbit and the spin-spin interactions. We also investigate the Hamiltonian for external magnetic field and ground-state hyperfine interactions. Lastly, we use the results of our investigation in a Lindblad master equation model to predict an optically detected magnetic resonance (ODMR) signal and the $$g^2(\tau)$$ correlation function. Our findings can have important outcomes in quantum information applications such as quantum repeaters used in quantum networks and quantum sensing.

### Author and article information

###### Journal
18 October 2021
###### Article
2110.09628
21b7396f-4506-422a-be10-574a81b9cc3f


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# Exact Semidefinite Formulations for a Class of (Random and Non-Random) Nonconvex Quadratic Programs

We study a class of quadratically constrained quadratic programs (QCQPs), called {\em diagonal QCQPs\/}, which contain no off-diagonal terms $x_j x_k$ for $j \ne k$, and we provide a sufficient condition on the problem data guaranteeing that the basic Shor semidefinite relaxation is exact. Our condition complements and refines those already present in the literature and can be checked in polynomial time. We then extend our analysis from diagonal QCQPs to general QCQPs, i.e., ones with no particular structure. By reformulating a general QCQP into diagonal form, we establish new sufficient conditions for the semidefinite relaxations of general QCQPs to be exact. Finally, these ideas are extended to show that a class of random general QCQPs has exact semidefinite relaxations with high probability as long as the number of variables is significantly larger than the number of constraints. To the best of our knowledge, this is the first result establishing the exactness of the semidefinite relaxation for random general QCQPs.

## Citation

Manuscript, Department of Management Sciences, University of Iowa, Feb 2018.


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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

In the last post took a very long time series, and we summarize it. Now we will do something that seems related when we look at the regime bar: regime change detection.

For the next question, we will still be using the datasets available at https://github.com/matrix-profile-foundation/mpf-datasets so you can try this at home.

The original code (MATLAB) and data are here.

Now let’s start:

1. When does the regime change in this time series?

This task exploits another interesting characteristic of Matrix Profile. We know that each position in the Profile Index points to its nearest neighbor. Let’s imagine all these connections as “arcs,” and we will see something interesting: in some regions, it looks like several arcs are pointing to that place. Why’s that?

Imagine, if there is a change in the regime in that time series, we can infer that the nearest neighbor of the previous pattern will not be in the new pattern. It is logical to think that these arcs will tend to be grouped inside each regime region.

Then, how to deal with this feature? Arc counts. Take any point in the time series, and count how many arcs are passing thru that point. The fewer arcs, the greater is the probability that we are changing regimes.

Let’s look to the plot below, where we plot the arcs (not all of them), and in the middle, we see the ‘arc count.’ There we see that there are two valleys, and they match almost precisely where the data changes pattern.

Well, I think I’ve already spoiled all the fun. But, let’s do all the steps anyway:

First, let’s load the tsmp library and import our dataset:

# install.packages('tsmp')

library(tsmp)

baseurl <- "https://raw.githubusercontent.com/matrix-profile-foundation/mpf-datasets/4d99d1bee9b36ca552aa75631748825eda0c064c/"
header = FALSE)\$V1

And plot it as always:

plot(dataset, lwd = 1, type = "l", main = "Pig Internal Bleeding", ylab = "ABP (mmHg)", xlab = "time")

This is not the same dataset from the first figure. This dataset changes only once because it has registered the arterial blood pressure of a pig before and after some bleeding. Well, we can imagine that there will be a regime change. So let’s find it quickly.

Here is the code:

# I'll do this in steps, first compute the matrix profile
mp_obj <- tsmp(dataset, window_size = 100, n_workers = 4, verbose = 0)

# The matrix profile doesn't need to be recomputed, we can apply any mining function on top
# of it, and is fast!
tic <- Sys.time()
mp_obj %>% fluss(num_segments = 1) %T>% plot

## Matrix Profile
## --------------
## Profile size = 14874
## Window size = 100
## Exclusion zone = 50
## Contains 1 set of data with 14973 observations and 1 dimension
##
## Arc Count
## ---------
## Profile size = 14874
## Minimum normalized count = 0.25 at index 7460
##
## Fluss
## -----
## Segments = 1
## Segmentation indexes = 7460
toc <- Sys.time() - tic
cat("\nFinished in", toc, "seconds\n")
##
## Finished in 0.1740022 seconds

We can barely see in the dataset where is the regime change, but I tell you that it is at timestamp 7500->7501.

We used a window size of 100 because it is approximately the length of one period of arterial pressure (or the period of whatever repeated patterns you have in your data); however, up to half or twice that value would work just as well. Then the fluss() function managed to extract where was located the minimum arc count, at 7460 (that is incredible near the truth; it is in the middle of the rolling window).

To finalize, see that we set num_segments to 1. This is the only parameter we should care about here. That informs the algorithm of how many changes it needs to look for.

So we did it again!

This ends our sixth question of one hundred (let’s hope so!).

Until next time.

previous post in this series


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## Keywords (1)

Complete monotonicity, Laguerre and Tur\'an type inequalities are established for the so-called Kr\"atzel function $Z_{\rho}^{\nu},$ defined by $$Z_{\rho}^{\nu}(u)=\int_0^{\infty}t^{\nu-1}e^{-t^{\rho}-\frac{u}{t}}\dt,$$ where $u>0$ and $\rho,\nu\in\mathbb{R}.$ Moreover, we prove the complete monotonicity of a determinant function of which entries involve the Kr\"atzel function.


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# Determine the open intervals on which the graph of the function y = 3x-\tan 2x, (-\frac{\pi}{4},...

## Question:

Determine the open intervals on which the graph of the function {eq}y = 3x-\tan 2x, (-\frac{\pi}{4}, \frac{\pi}{4}) {/eq} is concave upward or concave downward.

A. concave upward {eq}\left (-\frac{\pi}{2}, 1 \right ) {/eq}; concave downward {eq}\left ( 1, \frac{\pi}{2} \right ) {/eq}

B. concave upward {eq}\left ( -\frac{\pi}{2},-\frac{\pi}{6} \right ), \left ( \frac{\pi}{6}, \frac{\pi}{2} \right ) {/eq}; concave downward {eq}\left ( -\frac{\pi}{6}, \frac{\pi}{6} \right ) {/eq}

C. concave upward {eq}\left ( -\frac{\pi}{4}, 0 \right ) {/eq}; concave downward {eq}\left (0, \frac{\pi}{4} \right ) {/eq}

D. concave upward {eq}\left ( 0, \frac{\pi}{4} \right ) {/eq}; concave downward {eq}\left (-\frac{\pi}{4}, 0\right ) {/eq}

E. concave upward {eq}\left (-\frac{\pi}{6}, \frac{\pi}{6}\right ) {/eq}; concave downward {eq}\left (-\frac{\pi}{2},-\frac{\pi}{6}\right ),\left (\frac{\pi}{6}, \frac{\pi}{2}\right ) {/eq}

A ball bearing is placed on an inclined plane and begins to roll. The angle of elevation of the plane is {eq}\theta = \frac{3\pi}{11} {/eq} radians. The distance (in meters) the ball bearing rolls in f seconds is {eq}s(t) = 4.9(sin \theta)t^2 {/eq}. Determine the value of {eq}s'(t) {/eq} after one second. Round numerical values in your answer to one decimal place.

A. {eq}s'(t) = 8.5 {/eq}

B. {eq}s'(t) = 6.4 {/eq}

C. {eq}s'(t) = 9.7 {/eq}

D. {eq}s'(t) = 5.7 {/eq}

E. {eq}s'(t) = 7.4 {/eq}

## Concavity

The concavity of the curve tells us if a certain interval is concave up or concave down. This is calculated by first getting the values that makes the second derivative zero or undefined. These values will become the endpoints of the intervals that we will look into.

## Answer and Explanation:

1.

Let's start by getting the second derivative of the given function.

{eq}\displaystyle y(x) = 3x - \tan(2x) \\ \displaystyle y'(x) = (1)3x^{1-1} - \sec^2(2x) \cdot 2 \\ \displaystyle y'(x) = 3x^{0} - 2\sec^2(2x) \\ \displaystyle y'(x) = 3(1) - 2\sec^2(2x) \\ \displaystyle y'(x) = 3 - 2\sec^2(2x) \\ \displaystyle y''(x) = 0 - (2)2\sec (2x) \cdot \sec(2x)\tan(2x) \cdot 2 \\ \displaystyle y''(x) = -8\sec^2 (2x) \tan(2x) {/eq}

Equate this to zero and solve for {eq}x {/eq}.

{eq}\displaystyle 0 = -8\sec^2 (2x) \tan(2x) \\ \displaystyle \frac{0}{-8} = \sec^2 (2x) \tan(2x) \\ \displaystyle 0 = \sec^2 (2x) \tan(2x) \\ \displaystyle 0 = \sec^2 (2x),\;\; 0 = \tan(2x) \\ \displaystyle \sqrt{0 }= \sqrt{\sec^2 (2x)},\;\; 0 = \tan(2x) \\ \displaystyle 0 = \sec(2x),\;\; 0 = \tan(2x) \\ \displaystyle \textrm{arcsec}(0)= 2x,\;\; \arctan(0) = 2x \\ \displaystyle \text{No solution},\;\; \arctan(0) = 2x \\ \displaystyle \arctan(0) = 2x \\ \displaystyle 0 + \pi n = 2x \\ \displaystyle \pi n = 2x \\ \displaystyle \frac{\pi n}{2} = x {/eq}

The only solution found in the interval {eq}\displaystyle \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) {/eq} is at {eq}n = 0 {/eq}, hence the critical point of the function is

{eq}\displaystyle x = \frac{\pi (0)}{2} \\ \displaystyle x = 0 {/eq}

This will become the endpoint of the intervals that we will look into. The intervals are {eq}\displaystyle \left(-\frac{\pi}{4}, 0\right) \text{ and } \left(0, \frac{\pi}{4}\right) {/eq}. We will choose a value within these intervals and substitute them to the second derivative. If it results to a positive value, then the interval is concave up. If it results to a negative value, then the interval is concave down.

For the first interval {eq}\displaystyle \left(-\frac{\pi}{4}, 0\right) {/eq}, let's choose {eq}\displaystyle x = - \frac{\pi}{6} {/eq}:

{eq}\displaystyle y''(x) = -8\sec^2 (2x) \tan(2x) \\ \displaystyle y''\left ( -\frac{\pi}{6} \right ) = -8\sec^2 (2\left ( -\frac{\pi}{6} \right )) \tan(2\left ( -\frac{\pi}{6} \right )) \\ \displaystyle y''\left ( -\frac{\pi}{6} \right ) = -8(\sec \left ( -\frac{\pi}{3} \right ))^2 \tan\left ( -\frac{\pi}{3} \right ) \\ \displaystyle y''\left ( -\frac{\pi}{6} \right ) = -8(2)^2 (-\sqrt{3}) \\ \displaystyle y''\left ( -\frac{\pi}{6} \right ) = -8(4) (-\sqrt{3}) \\ \displaystyle y''\left ( -\frac{\pi}{6} \right ) = 32\sqrt{3} \\ \displaystyle y''\left ( -\frac{\pi}{6} \right ) = 32\sqrt{3} > 0 \therefore \left(-\frac{\pi}{4},0 \right )\text{ is concave up} \\ {/eq}

For the last interval {eq}\displaystyle \left(0, \frac{\pi}{4}\right) {/eq}, let's choose {eq}\displaystyle x = \frac{\pi}{6} {/eq}:

{eq}\displaystyle y''(x) = -8\sec^2 (2x) \tan(2x) \\ \displaystyle y''\left ( \frac{\pi}{6} \right ) = -8\sec^2 (2\left ( \frac{\pi}{6} \right )) \tan(2\left (\frac{\pi}{6} \right )) \\ \displaystyle y''\left (\frac{\pi}{6} \right ) = -8(\sec \left (\frac{\pi}{3} \right ))^2 \tan\left (\frac{\pi}{3} \right ) \\ \displaystyle y''\left (\frac{\pi}{6} \right ) = -8(2)^2 (\sqrt{3}) \\ \displaystyle y''\left ( \frac{\pi}{6} \right ) = -8(4) (\sqrt{3}) \\ \displaystyle y''\left ( \frac{\pi}{6} \right ) = -32\sqrt{3} \\ \displaystyle y''\left ( \frac{\pi}{6} \right ) = -32\sqrt{3} < 0 \therefore \left(0, \frac{\pi}{4}\right )\text{ is concave down} {/eq}

The answer is C.

2.

Let's start by getting the first derivative of the function for distance.

{eq}\displaystyle s(t) = 4.9 (\sin \theta ) t^2 \\ \displaystyle s'(t) = (2)4.9 (\sin \theta ) t^{2-1} \\ \displaystyle s'(t) = 9.8 (\sin \theta ) t^{1} \\ \displaystyle s'(t) = 9.8 (\sin \theta ) t {/eq}

Then, substitute {eq}\displaystyle \theta = \frac{3\pi}{11} {/eq} and {eq}t = 1 {/eq}.

{eq}\displaystyle s'(1) = 9.8 (\sin \left ( \frac{3\pi}{11} \right ) ) (1) \\ \displaystyle s'(1) = 9.8(0.75574) \\ \displaystyle s'(1) = 7.40 {/eq}

The answer is E.


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# Homework Help: Question if expectation value is considered a measurement?

1. Sep 13, 2012

### JordanGo

Hello,

I was just curious about expectation values. One of the postulates of quantum mechanics state:
The only possible results of a measurement is an eigenvalue of the operator.

Now, is the expectation value considered a measurement, thus considered an eigenvalue?

Thanks!

2. Sep 13, 2012

### Chopin

An expectation value is a weighted average of all of the eigenvalues. So it may or may not actually be a valid eigenvalue itself.

As an example, the X component of the spin operator has two eigenvalues: -1/2 and 1/2. However, if you take the X expectation value of a Z state (which is a linear combination of both of those eigenstates), the expectation value will be 0, since both -1/2 and 1/2 states appear in equal combination. 0 is not actually a valid eigenvalue of the X operator, but it's the average of the two actual eigenvalues.


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Measuring the growth of galaxy clusters - Archive ouverte HAL Access content directly
Preprints, Working Papers, ... Year :

## Measuring the growth of galaxy clusters

Antonaldo Diaferio
• Function : Author

#### Abstract

We suggest how we can use the mass profile of galaxy clusters beyond their virial radius to measure their mass accretion rate, a key prediction of structure formation models. The mass profile can be estimated by applying the caustic technique to dense redshift surveys of clusters and their outskirts, where dynamical equilibrium does not necessarily hold. An additional probe of the mass growth of clusters is their mass fraction in substructures. We show that the caustic technique, that identifies cluster substructures as a by-product, returns catalogs of substructures with mass larger than a few $10^{13}h^{-1}M_\odot$ that are between 60% and 80% complete, depending on the density of the redshift survey.

#### Domains

Physics [physics] Astrophysics [astro-ph]

### Dates and versions

hal-02000560 , version 1 (30-01-2019)

### Identifiers

• HAL Id : hal-02000560 , version 1
• ARXIV :

### Cite

Antonaldo Diaferio. Measuring the growth of galaxy clusters. 2019. ⟨hal-02000560⟩

### Export

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AAS 197, January 2001
Session 3. QSO Absorbers and the IGM
Display, Monday, January 8, 2001, 9:30am-7:00pm, Exhibit Hall

## [3.03] A Cluster of MgII Absorbers in a Large Quasar Group at 1.20 < z < 1.38

G.M. Williger (NASA/GSFC), L.E. Campusano (U. de Chile), R.G. Clowes (U. Central Lancashire, UK)

We have conducted a survey for MgII absorber systems toward the largest structure in the sky at z>0.2, a large quasar group consisting of at least 15 quasars at 1.20\circ \times 2.5\circ on the sky (~200 \times 100 h-2 Mpc2) and depth 156 (210) h-1 comoving Mpc for \Omega0=1 (\Omega0=0.2), H0\equiv 100h km s-1 Mpc-1, \Lambda=0. It is one of a handful of large quasar groups (Graham, Clowes, & Campusano 1995), which are believed to indicate a physical enhancement in the underlying density of the universe when the universe was about 1/3 of its present age. We can test for this mass excess by searching for spatially correlated metal absorption line systems, seen in the spectra of quasars situated in the background of the group. We have obtained spectra of 23 quasars with z>1.2 distributed over a ~2\times 2\circ area embedded in the quasar group, to survey for MgII at 0.70


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# How do you show that an $L^p$ whole (holomorphic on the facility aircraft) function is $0$?

Simply to make clear, I intend to show that:

If $f$ is whole and also $\int_{\mathbb{C}} |f|^p dxdy <\infty$, after that $f=0$.

I assume I can show that this holds true for $p=2$, yet I'm not exactly sure concerning various other values of $p$ ...

0
2019-05-18 22:56:19
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If $f: \RR \to \RR$ is an analytic boundless Lebesgue integrable function, after that it has to specifically please this building (due to the fact that it is continual):. \ [ \ forall \ e, M > 0 \ ; \ exists x_0: |x | > |x_0 | \ Rightarrow \ lbg (\ > M \ ) < \ e,. \ ] where $\lbg$ is the common Lebesgue action.

After that you simply make use of that $f$ is analytic iff $\ forall \ message portable K \ part \ RR \ ; \ exists C > 0: x \ in K, n \ in \ NN \ Rightarrow \ left \ vert \ frac \ partial ^ n f \ partial x ^ n (x) \ appropriate \ vert \ leq C ^ n+1 n!$

But it is very easy to see that this is difficult given that $|f'|$ will certainly need to be randomly huge - or to put it simply that you can go much sufficient bent on the exactly on the x - axis that $|f|$ needs to go from being bigger than some large $M$ to smaller sized than some tiny $m$ on a really tiny period.

This is missing out on plenty of information, yet it needs to be rather very easy to fill in.

0
2019-05-21 11:41:28
Source

Right! Thanks, I really did not think of making use of Hölder's.

Basically,

$$|f(0)| = \left|\int_{|z|=R} \frac{f(z)dz}{z} \right| < \left(\int_{|z|=R} |f(z)|^p |dz|\right)^{1/p} \left(\int_{|z|=R} |z|^{-q} |dz|\right)^{1/q}$$

And $\left(\int_{|z|=R} |z|^{-q} |dz|\right)^{1/q} = R^{-1+1/q} (2\pi)^{1/q} = R^{-1/p} (2\pi)^{1/q}$

$$\int_0^\infty \left((2\pi)^{1/q} |f(0)| R^{1/p}\right)^p dR < || f ||_p^p <\infty$$

And the only means for this to fly is for $f(0)=0$, for $p\le 1$.

0
2019-05-21 10:19:05
Source

Use Hölder is inequality and also Cauchy is indispensable formula to show the function and also its by-products all disappears at absolutely no.

0
2019-05-21 09:40:16
Source


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chapter  9
49 Pages

## Fixturing Accuracy Analysis and Verification

Manufacturing accuracy depends on relative positions of the machining tool and the workpiece (Rong and Wu, 1988). Fixtures are used to locate and hold a workpiece in proper position during machining processes. The de velopment of computer-aided fixture-design (CAFD) systems is becoming increasingly important within flexible manufacturing systems (FMS) and computer-integrated manufacturing systems (CIMS) (Thompson and Gandhi, 1986). Basically, two major approaches exist in CAFD. The first one is the rule-based (or knowledge-based) automated fixture design where geometric reasoning, kinematics analysis, or screw theory may be applied (Pham and de Sam Lazaro, 1990; Trappey and Liu, 1990a; Chou et al., 1989). The second is group technology (GT)-based search and retrieval of existing fix ture designs (Grippo et al., 1987; Rong and Zhu, 1992). The former is ideal for total automation but usually can only be applied to simple workpiece geometry because of the difficulties in geometric modeling and rule extrac tion. The latter is practical for industrial applications due to the use of ex isting knowledge in fixture designs.


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In the 2-satisfiability problem, or 2SAT, the input is a system of constraints between pairs of Boolean variables, and the problem is to find a truth assignment satisfying all the constraints. Although the usual formulation writes the constraints as disjunctions (Boolean or), it’s convenient and equivalent to use implications instead, where each side of an implication can be either one of the variables or its negation. If you encode the variables as numbers, with $1$ for true and $0$ for false, then the negation of a variable $x$ is $1-x$ and the implication $x\rightarrow y$ can be encoded as the linear inequality $x\le y$. If we forget the restriction that the values are $0$ or $1$, this system of inequalities defines a convex polytope, which can be thought of as a relaxation of the given 2SAT problem. Each solution to the 2SAT problem corresponds to a vertex of this polytope but there might be other vertices as well, with fractional coordinates. We don’t want those; it’s more useful to have a polytope all of whose coordinates are integers, coming from solutions to the underlying 2SAT problem. When does this happen (without adding extra inequalities)?

I’m pretty sure the answer has been known, at least implicitly, for a long time, and that there are no new ideas in the rest of this post. See e.g. the answer to this very closely related question on relaxations of the vertex cover problem and note that vertex cover can be formulated as 2SAT, with a variable per vertex and two constraints per edge of the input graph. But when I looked for it, I found it difficult to find references that give the answer explicitly. So here it is. The following conditions are equivalent:

• The given 2SAT instance has an integral relaxation.
• The given 2SAT instance has two solutions that are the complements of each other (each variable is true in one of the solutions and false in the other).
• The compatibility graph of the given 2SAT instance is bipartite.

Here, the compatibility graph is an undirected graph defined by Feder in “Network Flow and 2-Satisfiability” (Algorithmica 1994). It has a vertex for each literal (a variable or its negation). The two vertices for a variable and its negation are connected by an edge, called a “trivial edge”. Additionally, for each implication $x\rightarrow y$, the two literals $x$ and $\lnot y$ cannot both be true, and we add an edge between them.

If the compatibility graph is bipartite, two-color it, make the vertices of one color be true, and make the vertices of the other color be false. Then this gives a solution to the 2SAT instance whose complement is also a solution. So the third condition implies the second condition.

The relaxation of any 2SAT instance contains the point whose coordinates all equal $\tfrac12$, and if the given instance has an integral relaxation then this point can be represented as a weighted average of integer solutions. An odd cycle in the compatibility graph would prevent that from being possible. For any point in the relaxation, and any literal, we consider the value of the literal to be either its coordinate value (if it represents a variable) or one minus that value (if it represents a negated variable). If some cycle has odd length $\ell$, then each integer point in the weighted average could only have total value at most $(\ell-1)/2$ on the literals of the cycle, so their weighted average would also be at most $(\ell-1)/2$, less than the cycle’s total value of $\ell/2$ in the all-$\tfrac12$ solution. So the first condition implies the third condition.

Finally, suppose that there are two complementary solutions, and flip the variables if necessary so that these are the all-false and all-true assignments to the variables. (This flipping simplifies the problem without changing any of the three conditions.) Consider any fractional solution $S$, and construct from it a probability distribution on truth assignments as follows: choose a uniformly random number $x\in [0,1]$, set to true the variables whose coordinate value is at least $x$, and set to false the remaining variables. All truth assignments generated in this way solve the 2SAT problem, because any violation of an implication constraint in the 2SAT problem would translate to a violation of the corresponding coordinate inequality constraint in the relaxation. The given fractional solution $S$ is a weighted average of the resulting collection of 2SAT solutions, where the weight of a solution is the probability of generating it. Therefore, the second condition implies the first condition.


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# Laws of Exponents

The laws of exponents are explained here along with their examples.

### 1. Multiplying powers with same base

For example: x² × x³, 2³ × 2⁵, (-3)² × (-3)⁴

In multiplication of exponents if the bases are same then we need to add the exponents.

Consider the following:

1. 2³ × 2²= (2 × 2 × 2) × (2 × 2) = 2$$^{3 + 2}$$ = 2⁵

2. 3⁴ × 3² = (3 × 3 × 3 × 3) × (3 × 3) = 3$$^{4 + 2}$$ = 3⁶

3. (-3)³ × (-3)⁴ = [(-3) × (-3) × (-3)] × [(-3) × (-3) × (-3) × (-3)] = (-3)$$^{3 + 4}$$ = (-3)⁷

4. m⁵ × m³ = (m × m × m × m × m) × (m × m × m) = m$$^{5 + 3}$$ = m⁸

From the above examples, we can generalize that during multiplication when the bases are same then the exponents are added.

aᵐ × aⁿ = a$$^{m + n}$$

In other words, if ‘a’ is a non-zero integer or a non-zero rational number and m and n are positive integers, then

aᵐ × aⁿ = a$$^{m + n}$$

Similarly, (a/b)ᵐ × (a/b)ⁿ = (a/b)$$^{m + n}$$

Note:

(i) Exponents can be added only when the bases are same.

(ii) Exponents cannot be added if the bases are not same like

m⁵ × n⁷, 2³ × 3⁴

For example:

1. 5³ ×5⁶

= (5 × 5 × 5) × (5 × 5 × 5 × 5 × 5 × 5)

= 5$$^{3 + 6}$$, [here the exponents are added]

= 5⁹

2. (-7)$$^{10}$$ × (-7)¹²

= [(-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7)] × [( -7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7)].

= (-7)$$^{10 + 12}$$, [exponents are added]

= (-7)²²

3. (1/2)⁴ × ( 1/2)³

=[(1/2) × ( 1/2) × ( 1/2) × ( 1/2)] × [ ( 1/2) × ( 1/2) × ( 1/2)]

=(1/2)$$^{4 + 3}$$

=(1/2)⁷

4. 3² × 3⁵

= 3$$^{2 + 5}$$

= 3⁷

5. (-2)⁷ × (-2)³

= (-2)$$^{7 + 3}$$

= (-2)$$^{10}$$

6. (4/9)³ × (4/9)²

= (4/9)$$^{3 + 2}$$

= (4/9)⁵

We observe that the two numbers with the same base are

multiplied; the product is obtained by adding the exponent.

### 2. Dividing powers with the same base

For example:

3⁵ ÷ 3¹, 2² ÷ 2¹, 5(²) ÷ 5³

In division if the bases are same then we need to subtract the exponents.

Consider the following:

2⁷ ÷ 2⁴ = 2⁷/2⁴ = (2 × 2 × 2 × 2 × 2 × 2 × 2)/(2 × 2 × 2 × 2) = 2$$^{7 - 4}$$ = 2³

5⁶ ÷ 5² = 5⁶/5² = (5 × 5 × 5 × 5 × 5 × 5)/(5 × 5) = 5$$^{6 - 2}$$ = 5⁴

10⁵ ÷ 10³ = 10⁵/10³ = (10 × 10 × 10 × 10 × 10)/(10 × 10 × 10) = 10$$^{5 - 3}$$ = 10²

7⁴ ÷ 7⁵ = 7⁴/7⁵ = (7 × 7 × 7 × 7)/(7 × 7 × 7 × 7 × 7) = 7$$^{4 - 5}$$ = 7$$^{-1}$$

Let a be a non zero number, then

a⁵ ÷ a³ = a⁵/a³ = (a × a × a × a × a)/(a × a × a) = a$$^{5 - 3}$$ = a²

again, a³ ÷ a⁵ = a³/a⁵ = (a × a × a)/(a × a × a × a × a) = a$$^{-(5 - 3)}$$

= a$$^{-2}$$

Thus, in general, for any non-zero integer a,

aᵐ ÷ aⁿ = aᵐ/aⁿ = a$$^{m - n}$$

Note 1:

Where m and n are whole numbers and m > n;

aᵐ ÷ aⁿ = aᵐ/aⁿ = a$$^{-(n - m)}$$

Note 2:

Where m and n are whole numbers and m < n;

We can generalize that if ‘a’ is a non-zero integer or a non-zero rational number and m and n are positive integers, such that m > n, then

aᵐ ÷ aⁿ = a$$^{m - n}$$ if m < n, then aᵐ ÷ aⁿ = $$\frac{1}{a^{n - m}}$$

Similarly, (a/b)ᵐ ÷ (a/b)ⁿ = (a/b)$$^{m - n}$$

For example:

1. 7$$^{10}$$ ÷ 7⁸ = $$\frac{7^{10}}{7^{8}}$$

= (7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7)/(7 × 7 × 7 × 7 × 7 × 7 × 7 × 7)

= 7$$^{10 - 8}$$, [here exponents are subtracted]

= 7²

2. p⁶ ÷ p¹=p⁶/p¹

= (p × p × p × p × p × p)/p

= p$$^{6 - 1}$$, [here exponents are subtracted]

= p⁵

3. 4⁴ ÷ 4² = 4⁴/4²

= (4 × 4 × 4 × 4)/(4 × 4)

= 4$$^{4 - 2}$$, [here exponents are subtracted]

= 4²

4. 10² ÷ 10⁴ = 10²/10⁴

= (10 × 10)/(10 × 10 × 10 × 10)

= 10$$^{-(4 - 2)}$$, [See note (2)]

= 10$$^{-2}$$

5. 5³ ÷ 5¹

= 5$$^{3 - 1}$$

= 5²

6. 3⁵/3²

= 3$$^{5 - 2}$$

= 3³

7. (-5)⁹/(-5)⁶

= (-5)$$^{9 - 6}$$

= (-5)³

8. (7/2)⁸ ÷ (7/2)⁵

= (7/2)$$^{8 - 5}$$

= (7/2)³

### 3. Power of a power

For example: (2³)², (5²)⁶, (3² )$$^{-3}$$

In power of a power you need multiply the powers.

Consider the following

(i) (2³)⁴

Now, (2³)⁴ means 2³ is multiplied four times

i.e. (2³)⁴ = 2³ × 2³ × 2³ × 2³

=2$$^{3 + 3 + 3 + 3}$$

=2¹²

Note: by law (l), since aᵐ × aⁿ = a$$^{m + n}$$.

(ii) (2³)²

Similarly, now (2³)² means 2³ is multiplied two times

i.e. (2³)² = 2³ × 2³

= 2$$^{3 + 3}$$, [since aᵐ × aⁿ = a$$^{m + n}$$]

= 2⁶

Note: Here, we see that 6 is the product of 3 and 2 i.e,(2³)² = 2$$^{3 × 2}$$= 2⁶

(iii) (4$$^{- 2}$$)³

Similarly, now (4$$^{-2}$$)³ means 4$$^{-2}$$

is multiplied three times

i.e. (4$$^{-2}$$)³ =4$$^{-2}$$ × 4$$^{-2}$$ × 4$$^{-2}$$

= 4$$^{-2 + (-2) + (-2)}$$

= 4$$^{-2 - 2 - 2}$$

= 4$$^{-6}$$

Note: Here, we see that -6 is the product of -2 and 3 i.e, (4$$^{-2}$$)³ = 4$$^{-2 × 3}$$ =4$$^{-6}$$

For example:

1.(3²)⁴ = 3$$^{2 × 4}$$ = 3⁸

2. (5³)⁶ = 5$$^{3 × 6}$$ = 5¹⁸

3. (4³)⁸ = 4$$^{3 × 8}$$ = 4²⁴

4. (aᵐ)⁴ = a$$^{m × 4}$$ = a⁴ᵐ

5. (2³)⁶ = 2$$^{3 × 6}$$ = 2¹⁸

6. (xᵐ)$$^{-n}$$ = x$$^{m × -(n)}$$ = x$$^{-mn}$$

7. (5²)⁷ = 5$$^{2 × 7}$$ = 5¹⁴

8. [(-3)⁴]² = (-3)$$^{4 × 2}$$ = (-3)⁸

In general, for any non-integer a, (aᵐ)ⁿ=a$$^{m × n}$$ = aᵐⁿ. Thus where m and n are whole numbers.

If ‘a’ is a non-zero rational number and m and n are positive integers, then {(a/b)ᵐ}ⁿ = (a/b) ᵐⁿ

For example:

[(-2/5)³]²

= (-2/5)$$^{3 × 2}$$

= (-2/5)⁶

### 4. Multiplying powers with the same exponents

For example: 3² × 2², 5³ × 7³

We consider the product of 4² and 3², which have different bases, but the same exponents.

(i) 4² × 3² [here the powers are same and the bases are different]

= (4 × 4)×(3 × 3)

= (4 × 3)×(4 × 3)

= 12 × 12

= 12²

Here, we observe that in 12², the base is the product of bases 4 and 3.

We consider,

(ii) 4³ × 2³

=(4 × 4 × 4)×(2 × 2 × 2)

=(4 × 2)× ( 4 × 2)× (4 × 2)

=8 × 8 × 8

=8³

(iii) We also have, 2³ × a³

= (2× 2 × 2)×(a × a × a)

= (2 × a)×(2 × a)×(2 × a)

= (2 × a)³

= (2a)³ [here 2 × a = 2a]

(iv) Similarly, we have, a³ × b³

= (a × a × a)×(b × b × b)

= (a × b)× (a × b)× (a × b)

= (a × b)³

= (ab)³ [here a × b = ab]

Note: In general, for any non-zero integer a, b.

aᵐ × bᵐ

= (a × b)ᵐ

= (ab)ᵐ [here a × b = ab]

Note: Where m is any whole number.

(-a)³ × (-b)³

= [(-a) × (-a) × (-a)]×[(-b) × (-b) × (-b)]

= [(-a) × (-b)]× [(-a) × (-b)]× [(-a) × (-b)]

= [(-a)×(-b)]³

= (ab)³ [here a × b = ab and two negative become positive,

(-) × (-) = +]

### 5. Negative Exponents

If the exponent is negative we need to change it into positive exponent by writing the same in the denominator and 1 in the numerator.

If ‘a’ is a non-zero integer or a non-zero rational number and m is a positive integers, then a$$^{-m}$$ is the reciprocal of aᵐ, i.e.,

a$$^{-m}$$ = $$\frac{1}{a^{m}}$$, if we take ‘a’ as p/q then (p/q)$$^{-m}$$ = 1/(p/q)ᵐ = (q/p)ᵐ

again, $$\frac{1}{a^{-m}}$$ = aᵐ

Similarly, (a/b)$$^{-n}$$ = (b/a)ⁿ, where n is a positive integer

Consider the following

2$$^{-1}$$ = 1/2

2$$^{-2}$$ = 1/2² = 1/2 × 1/2 = 1/4

2$$^{-3}$$ = 1/2³ = 1/2 × 1/2 × 1/2 = 1/8

2$$^{-4}$$ = 1/2⁴ = 1/2 × 1/2 × 1/2 × 1/2 = 1/16

2$$^{-5}$$ = 1/2⁵ = 1/2 × 1/2 × 1/2 × 1/2 × 1/2 = 1/32

[So in negative exponent we need to write 1 in the numerator and in the denominator 2 multiplied to itself five times as 2$$^{-5}$$. In other words negative exponent is the reciprocal of positive exponent]

For example:

1. 10$$^{-3}$$

= $$\frac{1}{10^{-3}}$$, [here we can see that 1 is in the numerator and in the denominator 10³ as we know that negative exponent is the reciprocal]

= 1/10 × 1/10 × 1/10 [here 10 is multiplied to itself 3 times]

= 1/1000

2. (-2)$$^{-4}$$

= 1/(-2)⁴ [here we can see that 1 is in the numerator and in the denominator (-2)⁴]

=(- 1/2) × (- 1/2) × (- 1/2) × (- 1/2)

= 1/16

3. 2$$^{-5}$$

= 1/2⁵

= 1/2 × 1/2

= 1/4

4. $$\frac{1}{3^{-4}}$$

= 3⁴

= 3 × 3 × 3 × 3

= 81

5. (-7)$$^{-3}$$

= 1/(-7)³

6. (3/5)$$^{-3}$$

= (5/3)³

7. (-7/2)$$^{-2}$$

= (-2/7)²

### 6. Power with exponent zero

If the exponent is 0 then you get the result 1 whatever the base is.

For example: 8$$^{0}$$, ( a/b)$$^{0}$$, m$$^{0}$$…....

If ‘a’ is a non-zero integer or a non-zero rational number then,

a$$^{0}$$ = 1

Similarly, (a/b)$$^{0}$$ = 1

Consider the following

a$$^{0}$$ = 1 [anything to the power 0 is 1]

(a/b)$$^{0}$$ = 1

(-2/3)$$^{0}$$ = 1

(-3)$$^{0}$$ = 1

For example:

1. (2/3)³ × (2/3)$$^{-3}$$

= (2/3)$$^{3 + (-3)}$$, [here we know that aᵐ × aⁿ = a$$^{m + n}$$]

= (2/3)$$^{3 - 3}$$

= (2/3)$$^{0}$$

= 1

2. 2⁵ ÷ 2⁵

= 2⁵/2⁵

= (2 × 2 × 2 × 2 × 2)/(2 × 2 × 2 × 2 × 2)

= 2$$^{5 - 5}$$, [here by the law aᵐ ÷ aⁿ =a$$^{m - n}$$]

= 2

= 1

3. 4$$^{0}$$ × 3$$^{0}$$

= 1 × 1, [here as we know anything to the power 0 is 1]

= 1

4. aᵐ × a$$^{-m}$$

= a$$^{m - m}$$

= a$$^{0}$$

= 1

5. 5$$^{0}$$ = 1

6. (-4/9)$$^{0}$$ = 1

7. (-41)$$^{0}$$ = 1

8. (3/7)$$^{0}$$ = 1

### 7. Fractional Exponent

In fractional exponent we observe that the exponent is in fraction form.

Consider the following:

2$$^{\frac{1}{1}}$$ = 2 (it will remain 2).

2$$^{\frac{1}{2}}$$ =√2 (square root of 2).

2$$^{\frac{1}{3}}$$ =∛2 (cube root of 2).

2$$^{\frac{1}{4}}$$ =∜2 (fourth root of 2).

2$$^{\frac{1}{5}}$$ =⁵√2 (fifth root of 2).

For example:

1. a$$^{\frac{1}{n}}$$, [Here a is called the base and 1/n is called the exponent or power]

= ⁿ√a [nth root of a]

2. 3$$^{\frac{1}{2}}$$ = √3 [square root of 3]

3. 5$$^{\frac{1}{3}}$$ = ∛5 [cube root of 5]

`

Exponents

Exponents

Laws of Exponents

Rational Exponent

Integral Exponents of a Rational Numbers

Solved Examples on Exponents

Practice Test on Exponents

Exponents - Worksheets

Worksheet on Exponents


text:
# How to find a marginal distribution from a joint distribution?

So I have this joint dist

$$f(x,y) = \frac{1}{2\pi}\exp(-\frac{x^2}{2} - \frac{y^2}{2} + x^2y - \frac{x^4}{2})$$

I'd like to find $f_x(x)$. So I know that means I need to integrate function wrt y. So my strategy is to pull out whatever terms I can so some of it will look like the normal dist then it'll integrate to one and I'll be left with the marginal. So I have $$f_x(x) = \frac{1}{\sqrt{2\pi}}\int\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2} - \frac{y^2}{2} + x^2y - \frac{x^4}{2})dy$$

This issue then is I'm not sure how I can modify $\exp(-\frac{x^2}{2} - \frac{y^2}{2} + x^2y - \frac{x^4}{2})$ so it will take the form $\exp(-\frac{(y-\mu)^2}{2\sigma^2})$.

Any help would be great, thanks. Sorry for the uberitalicized tex, kinda hacked it together from what I know.

Also I know this is a lot to ask, but this is homework so if you could provide hints or tips as opposed to a complete solution that would be better.

Edited for clarity.

-

First re-write the joint density as

$$f(x,y) = \frac{1}{2\pi} {\rm exp} ({-x^{2}/2}) \cdot {\rm exp} \Big( -\frac{1}{2} \left( y^2 -2x^{2}y + x^4 \right) \Big)$$

Note that $y^2 -2x^{2}y + x^4 = (y-x^{2})^2$. Now the marginal density of $x$ is

$$f(x) = \int_{-\infty}^{\infty} f(x,y) dy = \frac{1}{\sqrt{2 \pi}} {\rm exp} ({-x^{2}/2}) \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} {\rm exp} \Big( -\frac{(y-x^{2})^2}{2} \Big) dy$$

The integrand is a $N(x^{2},1)$ density and therefore it integrates to 1. Therefore,

$$f(x) = \frac{1}{\sqrt{2 \pi}} {\rm exp} ({-x^{2}/2})$$

In other words, the marginal distribution of $X$ is standard normal.

-
+1 Great answer. But did you read the penultimate paragraph about homework and the desire to avoid a complete solution? –  whuber Sep 9 '11 at 19:49
Yeah, I see that now. I thought I read it carefully at the outset, maybe that was an edit. If not, sorry about that, OP. –  Macro Sep 9 '11 at 19:58