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https://medium.com/free-code-camp/i-dont-understand-graph-theory-1c96572a1401 | 1,586,304,137,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371806302.78/warc/CC-MAIN-20200407214925-20200408005425-00398.warc.gz | 563,978,627 | 109,823 | # How to think in graphs: An illustrative introduction to Graph Theory and its applications
Feb 21, 2018 · 49 min read
Graph theory represents one of the most important and interesting areas in computer science. But at the same time it’s one of the most misunderstood (at least it was to me).
Understanding, using and thinking in graphs makes us better programmers. At least that’s how we’re supposed to think. A graph is a set of vertices V and a set of edges E, comprising an ordered pair G=(V, E).
While trying to studying graph theory and implementing some algorithms, I was regularly getting stuck, just because it was so boring.
The best way to understand something is to understand its applications. In this article, we’re going to demonstrate various applications of graph theory. But more importantly, these applications will contain detailed illustrations.
While this approach might seem too detailed (to seasoned programmers), but believe me, as someone who was once there and tried to understand graph theory, detailed explanations are always preferred over succinct definitions.
So, if you’ve been looking for a “graph theory and everything about it tutorial for absolute unbelievable dummies”, then you’ve come to the right place. Or at least I hope. So let’s get started and dive in.
# Disclaimers
DISCLAIMER 1: I am not an expert in CS, algorithms, data structures and especially in graph theory. I am not involved in any project for the companies discussed in this article. Solutions to the problems are not final and could be improved drastically. If you find any issue or something unreasonable, you are more than welcome to leave a comment. If you work at one of the mentioned companies or are involved in corresponding software projects, please respond with the actual solution (it will be helpful to others). To all others, be patient readers, this is a pretty LONG article.
DISCLAIMER 2: This article is somewhat different in the style that information is provided. Sometimes it might seem a bit digressed from the sub-topic, but patient readers will eventually find themselves with a complete understanding of the bigger picture.
# Seven Bridges of Königsberg
Let’s start with something that I used to regularly encounter in graph theory books that discuss “the origins of graph theory”, the Seven Bridges of Königsberg (not really sure, but you can pronounce it as “qyonigsberg”). There were seven bridges in Kaliningrad, connecting two big islands surrounded by the Pregolya river and two portions of mainlands divided by the same river.
In the 18th century this was called Königsberg (part of Prussia) and the area above had a lot more bridges. The problem or just a brain teaser with Königsberg’s bridges was to be able to walk through the city by crossing all the seven bridges only once. They didn’t have an internet connection at that time, so it should have been entertaining. Here’s the illustrated view of the seven bridges of Königsberg in 18th century.
Try it. See if you can walk through the city by crossing each bridge only once.
• There should not be any uncrossed bridge(s).
If you are familiar with this problem, you know that it’s impossible to do it. Although you were trying hard enough and you may try even harder now, you’ll eventually give up.
Sometimes it’s reasonable to give up fast. That’s how Euler solved this problem - he gave up pretty soon. Instead of trying to solve it, he adopted a different approach of trying to prove that it’s not possible to walk through the city by crossing each bridge one and only time.
Let’s try to understand how Euler was thinking and how he came up with the solution (if there isn’t a solution, it still needs a proof). That is a real challenge here, because walking through the thought process of such a venerable mathematician is kind of dishonorable. (Venerable so much that Knuth and friends dedicated their book to Leonhard Euler). We rather will pretend to “think like Euler”. Let’s start with picturing the impossible.
There are four distinct places, two islands and two parts of mainland. And seven bridges. It’s interesting to find out if there is any pattern regarding the number of bridges connected to islands or mainland (we will use the term “land” to refer to the four distinct places).
At a first glance, there seems to be some sort of a pattern. There are an odd number of bridges connected to each land. If you have to cross each bridge once, then you can enter a land and leave it if it has 2 bridges.
It’s easy to see in the illustrations above that if you enter a land by crossing one bridge, you can always leave the land by crossing its second bridge. Whenever a third bridge appears, you won’t be able to leave a land once you enter it by crossing all its bridges. If you try to generalize this reasoning for a single piece of land, you’ll be able to show that, in case of an even number of bridges it’s always possible to leave the land and in case of an odd number of bridges it isn’t. Try it in your mind!
Let’s add a new bridge to see how the number of overall connected bridges changes and whether it solves the problem.
Now that we have two even (4 and 4) and two odd (3 and 5) number of bridges connecting the four pieces of land, let’s draw a new route with the addition of this new bridge.
We saw that the number of even and odd number of bridges played a role in determining if the solution was possible. Here’s a question. Does the number of bridges solve the problem? Should it be even all the time? Turns out that it’s not the case. That’s what Euler did. He found a way to show that the number of bridges matter. And more interestingly, the number of pieces of land with an odd number of connected bridges also matters. That’s when Euler started to “convert” lands and bridges into something we know as graphs. Here’s how a graph representing the Königsberg bridges problem could look like (note that our “temporarily” added bridge isn’t there).
One important thing to note is the generalization/abstraction of a problem. Whenever you solve a specific problem, the most important thing is to generalize the solution for similar problems. In this particular case, Euler’s task was to generalize the bridge crossing problem to be able to solve similar problems in the future, i.e. for all the bridges in the world. Visualization also helps to view the problem at a different angle. The following graphs are all various representations of the same Königsberg bridge problem shown above.
So yes, visually graphs are a good choice for picturing problems. But now we need to find out how the Königsberg problem can be solved using graphs. Pay attention to the number of lines coming out of each circle. And yes, let’s name them as seasoned professionals would do, from now on we will call circles, vertices and the lines connecting them, edges. You might’ve seen letter notations, V for (vendetta?) vertex, E for edge.
The next important thing is the so-called degree of a vertex, the number of edges incident connected to the vertex. In our example above, the number of bridges connected to lands can be expressed as degrees of the graph vertex.
In his endeavor Euler showed that the possibility of a walk through graph (city) traversing each edge (bridge) one and only one time is strictly dependent on the degrees of vertices (lands). The path consisting of such edges called (in his honor) an Euler path. The length of an Euler path is the number of edges. Get ready for some strict language. 😃
An Euler path of a finite undirected graph G(V, E) is a path such that every edge of G appears on it once. If G has an Euler path, then it is called an Euler graph. [1]
Theorem. A finite undirected connected graph is an Euler graph if and only if exactly two vertices are of odd degree or all vertices are of even degree. In the latter case, every Euler path of the graph is a circuit, and in the former case, none is. [1]
I used “Euler path” instead of “Eulerian path” just to be consistent with the referenced books [1] definition. If you know someone who differentiates Euler path and Eulerian path, and Euler graph and Eulerian graph, let them know to leave a comment.
First of all, let’s clarify the new terms in the above definition and theorem.
• Undirected graph - a graph that doesn’t have a particular direction for edges.
We’ll discuss some of these terms in the coming paragraphs.
Graphs can be directed and undirected, and that’s one of the interesting properties of graphs. You must’ve seen a popular Facebook vs Twitter example for directed and undirected graphs. A Facebook friendship relation may be easily represented as an undirected graph, because if Alice is a friend with Bob, then Bob must be a friend with Alice, too. There is no direction, both are friends with each other.
Also note the vertex labeled as “Patrick”, it is kind of special (he’s got no friends), as it doesn’t have any incident edges. It is still a part of the graph, but in this case we will say that this graph is not connected, it is a disconnected graph (same goes with “John”, “Ashot” and “Beth” as they are interconnected with each other but separated from others). In a connected graph there is no unreachable vertex, there must be a path between every pair of vertices.
Contrary to the Facebook example, if Alice follows Bob on Twitter, that doesn’t require Bob to follow Alice back. So a “follow” relation must have a direction indicator, showing which vertex (user) has a directed edge (follows) to the other vertex.
Now, knowing what is a finite connected undirected graph, let’s get back to Euler’s graph:
So why did we discuss Königsberg bridges problem and Euler graphs in the first place? Well, it’s not so boring and by investigating the problem and foregoing solution we touched the elements behind graphs (vertex, edge, directed, undirected) avoiding a dry theoretical approach. And no, we are not done with Euler graphs and the problem above, yet. 😶
We should now move on to the computer representation of graphs as that is the topic of interest for us programmers. By representing a graph in a computer program, we will be able to devise an algorithm for tracing graph path(s), and therefore find out if it is an Euler path. Before that, try to think of a good application for an Euler graph (besides fiddling around with bridges).
# Graph representation: Intro
Now this is quite a tedious task, so be patient. Remember the fight between Arrays and Linked Lists? Use arrays if you need fast element access, use lists if you need fast element insertion/deletion, etc. I hardly believe you ever struggled with something like “how to represent lists”. Well, in case of graphs the actual representation is really bothering, because first you should decide how exactly are you going to represent a graph. And believe me, you are not going to like this. Adjacency list, adjacency matrix, maybe edge lists? Toss a coin.
You should have tossed hard, because we are starting with a tree. You must have seen a binary tree (or BT for short) at least once (the following is not a binary search tree).
Just because it consists of vertices and edges, it’s a graph. You also may recall how most commonly a binary tree is represented (at least in textbooks).
It might seem too basic for people who are already familiar with binary trees, but I still have to illustrate it to make sure we are on the same page (note that we are still dealing with pseudocode).
If you are new to trees, read the pseudocode above carefully, then follow the steps in the illustration below.
While a binary tree is a simple “collection” of nodes, each of which has left and right child nodes. A binary search tree is much more useful as it applies one simple rule which allows fast key lookups. Binary search trees (BST) keep their keys in sorted order. You are free to implement your BT with any rule you want (although it might change its name based on the rule, for instance, min-heap or max-heap). The most important expectation for a BST is that it satisfies the binary search property (that’s where the name comes from). Each node’s key must be greater than any key in its left sub-tree and less than any key in its right sub-tree.
I’d like to point out a very interesting point regarding the statement “greater than” that’s crucial to understand how BST’s function. Whenever you change the property to “greater than or equal”, your BST will be able to save duplicate keys when inserting new nodes, otherwise it will keep only nodes with unique keys. You can find really good articles on the web about binary search trees. We won’t be providing a full implementation of a binary search tree, but for the sake of consistency, we’ll illustrate a simple binary search tree here.
# Intro to Graph representation and binary trees (Airbnb example)
Trees are very useful data structures. You might not have implemented a tree from scratch in your projects. But you’ve probably used them even without noticing. Let’s look at an artificial yet valuable example and try to answer the “why” question, “Why use a binary search tree in the first place”.
As you’ve noticed, there is a “search” in binary search tree. So basically, everything that needs a fast lookup, should be placed in a binary search tree. “Should” doesn’t mean must, the most important thing to keep in mind in programming is to solve a problem with proper tools. There are tons of cases where a simple linked list with its O(N) lookup might be more preferable than a BST with its O(logN) lookup.
Typically we would use a library implementation of a BST, most likely std::set or std::map in C++. However in this tutorial we are free to reinvent our own wheel. BSTs are implemented in almost any general-purpose programming language library. You can find them in the corresponding documentation of your favorite language. Approaching a “real-life example”, here’s the problem we’ll try to tackle - Airbnb Home Search.
How do we search for homes based on some query with a bunch of filters as fast as possible. This is a hard task. It becomes harder if we consider that Airbnb stores 4 millions listings.
So when users search for homes, there is a chance that they might “touch” 4 million records stored in the database. Sure the results are limited to the “top listings” shown on the website’s home page and a user almost is never curious “enough” to view millions of listings. I don’t have any analytics regarding Airbnb, but we can use a powerful tool in programming called “assumptions”. So we will assume that a single user finds a good home by viewing at most ~1K homes.
The most important factor here is the number of real-time users, as it makes a difference in data structures and database(s) choices and the project architecture overall. As obvious as it might seem, if there are just 100 users overall, then we may not bother at all.
On the contrary, if the number of users overall and real-time users in particular is far beyond the million threshold, we have to think really wisely about each decision. “Each” is used exactly right, that’s why companies hire the best while striving for excellence in service provision.
Google, Facebook, Airbnb, Netflix, Amazon, Twitter, and many others deal with huge amounts of data and the right choice to serve millions of bytes of data each second to millions of real-time users starts from hiring the right engineers. That’s why we, the programmers, struggle with these data structures, algorithms and problem solving in possible interviews, because all they need is the engineer having the ability to solve such big problems in the fastest and most efficient possible way.
So here’s a use case. A user visits the home page (we’re still talking about Airbnb) and tries to filter out homes to find the best possible fit. How would we deal with this problem? (Note that this problem is rather backend-side, so we won’t care about front-end or the network traffic or https over http or Amazon EC2 over home cluster and so on).
First of all, as we are already familiar with one of the most powerful tools in a programmers’ inventory (talking about assumptions rather than abstractions), let’s start with a few assumptions:
• We’re dealing with data that completely fits in the RAM.
Big enough to hold, hmm, how much? Well that’s a good question. How much memory will be required to store the actual data. If we are dealing with 4 million units of data (again, am assumption), and if we probably know each unit’s size, then we can easily derive the required memory size, i.e. 4M * sizeof(one_unit).
Let’s consider a “home” object and its properties. Actually, let’s consider at least those properties that we will deal with while solving our problem (a “home” is our unit). We will represent it as a C++ structure in some pseudocode. You can easily convert it to a MongoDB schema object or anything you want. We just discuss the property names and types (try to think about using bitfields or bitsets for space economy).
The above structure is not perfect (obviously) and there are many assumptions and/or incomplete parts. I just looked at Airbnb’s filters and devised property lists that should exist to satisfy search queries. It’s just an example.
Now we should calculate how many bytes in memory will take each `AirbnbHome` object.
• Home name -`name` is a `wstring` to support multilingual names/titles, which means each character will take 2 bytes (we may not bother with character size if we would use other language, but in C++ `char` is 1-byte character and `wchar` is 2-byte character). A quick look at Airbnb’s listings allows us to assume that the name of a home should take up to 100 characters (though mostly it is around 50, rather than 100), we’ll assume 100 characters as a maximum value, which leads to ~200 bytes of memory. `uint` is 4 bytes, `uchar` is 1 byte, `ushort` is 2 bytes (again, in our assumptions).
Amenities - Each Airbnb home keeps a list of available amenities, e.g. “iron”, “washer”, “tv”, “wifi”, “hangers”, “smoke detector” and even “laptop friendly workspace” and so on. There might be more than 20 amenities, we stick to the 20 just because it’s the number of filterable amenities on the Airbnb filters page. Bitset saves us some good space, if we keep proper ordering for amenities. For instance, if a home has all above mentioned amenities (see checked ones in the screenshot), we will just set a bit at corresponding position in the bitset.
For example, checking if a home has a “washer”:
Or a little more professionally:
You can improve the code as much as you want (and fix compile errors). We just wanted to emphasize the idea behind bitsets in this problem context.
• House rules, Home Type - The same idea (that we implemented for the amenities field) goes with “house rules”, “home type” and others.
So, 420+ bytes with an overhead of 32 bytes, 452 bytes and considering the fact that some of you might just be obsessed with the aligning factor, let’s round up it to 500 bytes. So each “home” object takes up to 500 bytes, and for all home listings (there could be some confusing moments with the listings count and actual home count, just let me know if I got something wrong), 500 bytes * 4 million = 1.86GB ~ 2GB. Seems plausible. We made many assumptions while constructing the struct, making it cheaper to save in memory, I really expected much more than 2 Gigabytes and if I did a mistake in calculations, let me know. Anyway, moving forward, so whatever we gonna do with this data, we will need at least 2 GB of memory. If you got bored, deal with it. We are just starting.
Now the hardest part of the task. Choosing the right data structure for this problem (filter the listings as efficiently as possible) is not the hardest task. The hardest task is (for me) to search listings by a bunch of filters. If there would be just one search key (just one filter) we would easily solve it. Suppose the only thing users care is the price, so all we need is to find `AirbnbHome` objects with prices falling in the provided range. If we’ll use a binary search tree for that, here’s how it might look.
If you imagine all 4 millions objects, this tree grows very very big. By the way, the memory overhead grows as well, just because we used a BST to store objects. As each parent tree node has two additional pointers to its left and right child it adds up to 8 additional bytes for each child pointer (assuming a 64-bit system). For 4 million nodes it sums up to ~62 Mb, which in comparison to 2Gb of object data looks quite small, though it is not something that we can “omit” easily.
The tree in the last illustration so far shows that any item can be easily found in O(logN) complexity. If you aren’t familiar or are not sure enough to chit-chat in big-ohs, we’ll clarify it below, otherwise skip the complexity subsection.
Algorithmic complexity - Let’s make this quick as there will be a long and detailed explanation in an upcoming article: “Algorithmic Complexity and Software Performance: The Missing Manual”. For most of the cases finding the big O complexity for an algorithm is somewhat easy. First thing to note is that we always consider the worst case, i.e. the maximum number of operations that an algorithm does to produce a positive outcome (to solve the problem).
Suppose an array has 100 elements in an unsorted order. How many comparisons would it take to find an element (also taking into account that the required element could be missing)? It will take up to 100 comparisons as we should compare each element’s value with the value we are looking for. And despite the fact that the element might be the first element in the array (leading to a single comparison), we will consider only the worst possible case (element is either missing or is residing at the last position of the array).
The point of “calculating” algorithmic complexity is finding a dependency between the number of operations and the size of input, for instance the array above had 100 elements and the number of operations were also 100, if the number of array elements (its input) will increase to 1423, the number of operations to find any element will also increase to 1423 (the worst case). So the thin line between input and number of operations is clear in this case, it is so-called linear, the number of operations grows as much as grows array’s input. Growth. That’s the key point in complexity, we say that searching for an element in an unsorted array takes O(N) time to emphasize that the process of finding it will take up to N operations (or even up to N operations times some constant value such as 3N). On the other hand, accessing any element in an array takes a constant time, i.e. O(1). That’s because of an array’s structure. It’s a contiguous data structure, and holds elements of the same type (mind JS arrays), so “jumping” to a particular element requires only calculating its relative position to the array’s first element.
One thing is very clear. A binary search tree keeps its nodes in sorted order. So what would be the algorithmic complexity of searching an element in a binary search tree? We should calculate the number of operations required to find an element (in the worst case).
Look at the illustration above. When starting our search at the root, the first comparison may lead to three cases,
1. The node is found.
At each step we reduce the size of nodes needed to be considered by half. The number of operations (i.e. comparisons) needed to find an element in the BST equals the height of the tree. The height of a tree is the number of nodes on the longest path. In this case it’s 4. And the height is [base 2] logN + 1, as shown. So the complexity of search is O(logN + 1) = O(logN). This means that searching something in 4 million nodes requires log4000000 = ~22 comparisons in the worst case.
Back to the tree - Element access time in a binary search tree is O(logN). Why not use hashtables? Hashtables have constant access time, which makes it reasonable to use hashtables almost everywhere.
In this problem we must take into account an important requirement. We must be able to make range searches, e.g. homes with prices from \$80 to \$162. In case of a BST, it’s easy to get all the nodes in a range just by doing an inorder traversal of the tree and keeping a counter. For a hashtable it is somewhat expensive which makes it reasonable to stick with BSTs in this case.
Though there is another spot, which leads us to rethink hashtables. The density. Prices won’t go up “forever”, most of the homes reside at the same price range. Look at the screenshot, the histogram shows us the real picture of the prices, millions of homes are in the same range (+/- \$18 - \$212), they have the same average price. Simple arrays may play a good role. Assuming the index of an array as the price and the value as the list of homes, we might access any price range in constant time (well, almost constant). Here’s how it looks (way abstract):
Just like a hashtable, we are accessing each set of homes by its price. All homes having the same price are grouped under a separate BST. It will also save us some space if we store home IDs instead of the whole object defined above (the `AirbnbHome` struct). The most possible scenario is to save all homes full objects in a hashtable mapping home ID to home full object and storing another hashtable (or better, an array), which maps prices with homes IDs.
So when users request a price range, we fetch home IDs from the price table, cut the results to a fixed size (i.e. the pagination, usually around 10 - 30 items are shown on one page), fetch the full home objects using each home ID.
Just keep another thing in mind (think of it in the background). Balancing is crucial for a BST, because it’s the only guarantee of having tree operations done in O(logN). The problem of unbalanced BST is obvious when you insert elements in sorted order. Eventually, the tree becomes just a linked list, which obviously leads to linear-time operations. Forget this for now, suppose all our trees are perfectly balanced. Take a look at the illustration above once again. Each array element represents a big tree. What if we change the illustration to something like this:
It resembles a “more realistic” graph. This illustration represents the most disguised data structures and graphs, which takes us to the next section.
# Graph representation: Outro
The bad news about graphs is that there isn’t a single definition for the graph representation. That’s why you can’t find a `std::graph` in the library. We already had a chance to represent a “special” graph called BST. The point is, tree is a graph, but graph is not a tree. The last illustration shows us that we have a lot of trees under a single abstraction, “prices vs homes” and some of the vertices “differ” in their type, prices are graph nodes having only the price value and refer to the whole tree of home IDs (home vertices) that satisfy the particular price. It is much like a hybrid data structure, than a simple graph that we’re used to seeing in textbook examples.
That’s the key point in graph representation, there isn’t a fixed and “de jure” structure for graph representation (unlike BSTs with their specified node-based representation with left/right child pointers, though you can represent a BST with a single array). You can represent a graph in the most convenient way you wish (most convenient to particular problem), the main thing is that you “see” it as a graph. And by “seeing a graph” we mean applying algorithms that are specific to graphs.
What about an N-ary tree, it is more likely to resemble a graph.
And the first thing that comes into mind to represent an N-ary tree node is something like this:
This structure represents just a single node of a tree. The full tree looks more like this:
This class is an abstraction around a single tree node named `root_` . That’s all we need to build a tree of any size. That’s the starting point of the tree. For adding a new tree node we need to allocate a memory to it and add that node to the root of the tree.
A graph is much like an N-ary tree, with a slight difference. Try to spot it.
Is this a graph? No. I mean yes, but it’s the same N-ary tree from the previous illustration, just a little rotated. As a rule of a thumb, whenever you see a tree (even if it is an apple tree, lemon tree or binary search tree), you can be sure that it is also a graph. So, devising a structure for a graph node (graph vertex), we can come up with the same structure:
Is this enough to construct a graph? Well, no. And here’s why. Look at these two graphs from previous illustrations, find a difference:
The graph in the illustration at the left side has no single point to “enter” (it’s rather a forest than a single tree), in the contrary, the graph in the right illustration doesn’t have unreachable vertices. Sounds familiar.
A graph is connected when there is a path between every pair of vertices. [Wikipedia]
Obviously, there isn’t a path between every pair of vertices for the “prices vs homes” graph (if it isn’t obvious from the illustration, just assume that prices are not connected with each other). As much as it’s just an example to show that we aren’t able to construct a graph with a single GraphNode struct, there are cases that we have to deal with disconnected graphs like this. Take a look at this class:
Just like an N-ary tree is built around a single node (the root node), a connected graph also can be built around a root node. It’s said that trees are “rooted”, i.e. they have a starting point. A connected graph can be represented as a rooted tree (with a couple of more properties), it’s already obvious, but keep in mind that the actual representation may differ from algorithm to algorithm, from problem to problem even for a connected graph. However, considering node-based nature of graphs, a disconnected graph can be represented like this:
For graph traversals like DFS/BFS it’s natural to use a tree-like representation. Helps a lot. However, cases like efficient path tracing require a different representation. Remember Euler’s graph? To track down a graph’s “eulerness”, we should trace an Euler path within it. That means visiting all vertices by traversing each edge only once, and when the tracing finishes and we have untraversed edges, then the graph doesn’t have an Euler path, and therefore is not an Euler graph.
There is even faster method, we can check the degrees of vertices (suppose each vertex stores its degree) and just as the definition says, if a graph has vertices of odd degree and there aren’t exactly two of them, then it is not an Euler graph. The complexity of such a check is O(|V|), where |V| is the number of graph vertices. We can track down odd/even degrees while inserting new edges to increase odd/even degree checks to O(1). Lightning fast. Never mind, we’re just going to trace a graph, that’s it. Below is both the representation of a graph and the Trace() function returning a path.
Mind the bugs, bugs are everywhere. This code contains a lot of assumptions, for instance, the labeling, so by a vertex we understand a string label. Sure you can easily update it to be anything you want. Doesn’t matter in the context of this example. Next, the naming. As mentioned in the comments, VELOGraph is for Vertex Edge Label Only Graph (I made this up). The point is, this graph representation contains a table for mapping a vertex label with edges incident to that vertex, and a list of edges containing a pair of vertices (connected by a particular edge) and a flag which is used only by the Trace() function. Take a look at the Trace() function implementation. It uses edge’s flag to mark an already traversed edge (flags should be reset after any Trace() call).
# Twitter Example: Tweet Delivery Problem
Here’s another representation called an adjacency matrix, which could be useful in directed graphs, like one we used for Twitter follower graph.
There are 8 vertices in this Twitter example. So all we need to represent this graph is a |V|x|V| square matrix (|V| number of rows and |V| number of columns). If there is a directed edge from v to u, then matrix’s [v][u] is `true`, otherwise it’s `false`.
As you can see, this matrix is way too sparse, its trade off is the fast access. To see if Patrick follows Sponge Bob, we should just check the value of `matrix["Patrick"]["Sponge Bob"]`. To get the list of Ann’s followers, we just process the entire “Ann” column (title is in yellow). To find who are being followed (sounds strange) by Sponge Bob, we process the entire row “Sponge Bob”. Adjacency matrix could be used for undirected graphs as well, and instead of settings 1’s if a there is an edge from v to u, we should set both values to 1, e.g. adj_matrix[v][u] = 1, adj_matrix[u][v] = 1. Undirected graph’s adjacency matrix is symmetric.
Note that instead of storing ones and zeroes in an adjacency matrix, we can store something “more useful”, like edge weights. One of the best examples might be a graph of places with distance information.
The graph above represents distances between houses of Patrick, Sponge Bob and others (also known as weighted graph). We put “infinity” signs if there isn’t a direct route between vertices. That doesn’t mean that there are no routes at all, and at the same time that doesn’t mean that there must necessarily be routes. It might be defined while applying an algorithm for finding a route between vertices (there is even better way to store vertices and edges incident to it, called an incidence matrix).
While adjacency matrix seemed a good use for Twitter’s followes graph, keeping a square matrix for nearly 300 million users (monthly active users) takes 300 * 300 * 1 bytes (storing boolean values). That is, ~82000 Tb (Terabytes), which is 1024 * 82000 Gb. Well, don’t know about your home cluster, my laptop doesn’t have so much RAM. Bitsets? A BitBoard could help us a little, reducing the required size to ~10000 Tb. Still way too big. As mentioned above, an adjacency matrix is too sparse. It forces us to use more space than actually needed. That’s why using a list of edges incident to vertices may be useful. The point is, an adjacency matrix allows us to keep both “follows” and “doesn’t follow” information, while all we need is to know information about the followes, something like this:
The illustration at the right side is called an adjacency list. Each list describes the set of neighbors of a vertex in the graph. By the way, the actual implementation of the graph representation as an adjacency list, again, differs (ridiculous facts). In the illustration, we highlighted a hashtable usage, which is reasonable, as the access of any vertex will be O(1), and for the list of neighbor vertices we didn’t mention the exact data structure, deviating from linked lists to vectors. Choice is yours.
The point is, to find out whether Patrick does follow Liz, we should access the hashtable (constant time) and go through all items in the list comparing each element with “Liz” element (linear time). Linear time isn’t that bad at this point, because we have to loop over only a fixed amount of vertices adjacent to “Patrick”. What about the space complexity, is it ok to use at Twitter? Well, we need at least 300 million hashtable records, each of which points to a vector (choosing vector to avoid memory overhead of linked lists’ left/right pointers) containing, how much? No stats here, found just an average number of twitter followers, 707 (googled).
So if we consider that each hashtable record points to an array of 707 user IDs (each weighing 8 byte), and let’s assume that hashtable’s overhead is only its keys, which are again, user ids, so the hashtable itself takes 300 million * 8 bytes. Overall, we have 300 million * 8 bytes for hashtable + 707 * 8 bytes for each hashtable key, that is 300 million * 8 * 707 * 8 bytes = ~12 Tb. Well, can’t say that feels better, but yes, feels much better than 10,000 Tb.
Honestly, I don’t know whether this 12Tb is a reasonable number. But considering the fact that I’m spending around \$30 on a dedicated server machine with 32 Gb of RAM, then, storing (sharded) 12 Tb requires at least 385 such servers, plus a couple of control servers (for data distribution control) rounds up to 400. So it will cost me \$12K (monthly).
Well, considering the fact that the data should be replicated, and that something always can go wrong, we’ll triple the number of servers and then again, add some control servers, let’s say we need at least 1500 servers, which will cost us \$45K monthly. Well, definitely not good for me as I hardly can keep one server, but seems okay for Twitter (it’s really nothing compared to real Twitter servers). Let’s assume it is really okay for Twitter.
Now, are we okay here? Not yet, that was just the data regarding the followers. What is the main thing in Twitter? I mean, technically, what is its biggest problem? You won’t be alone if you say it’s the fast delivery of tweets. I will definitely second that. And not fast, but lightning fast. Say Patrick tweeted something about his thoughts on food, all his followers should receive that very tweet in a reasonable time. How long will it take? We are free of making any assumption here, and use any abstractions we want, but we are interested in the real world production systems, so, let’s dig. Here’s what’s typically happens when someone tweets.
Again, don’t know much about how long it takes for one tweet to reach all followers, but publicly available statistics tell us that there are about 500 million daily tweets. Daily! 😮
So the process above happens 500 million times every day. I can’t really find anything on tweet delivery speed. I vaguely recall something about a maximum of 5 seconds for a tweet to reach all its followers. And also note the “heavy cases”, celebrities with more than a million followers. They might tweet something about their wonderful breakfast at the beach house, but Twitter sweats much to deliver that super-useful content to millions of followers.
To solve tweet delivery problem we don’t really need the graph of followers, instead we need a graph of followers. The previous graph (with a hashtable and a bunch of lists) allows us to efficiently find all users followed by any particular user. But it does not allow us to efficiently find all users who are following one particular user, for that case we have to scan all the hashtable keys. That’s why we should construct another graph, which is kind of a symmetric opposite to the one we constructed for followers. This new graph will again consist of a hashtable containing all 300 million vertices, each of which points to a list of adjacent vertices (the structure remains the same), but this time, the list of adjacent vertices will represent followers.
So based on this illustration, whenever Liz tweets something, Sponge Bob and Ann must see that very tweet on their timelines. A common technique to accomplish this is by keeping separate structures for each user’s timeline. In case of Twitter’s 300+ million users, we might assume there are at least 300+ million timelines (for each user). Basically, whenever a user tweets, we should get the list of user’s followers and update their timelines (insert that same tweet into each one of them). A timeline might be represented as a linked list, or a balanced tree (tweet datetimes as node keys).
This is just a basic idea we abstracted from actual timeline representation and of course, we can make the actual delivery faster if we use multithreading. This is crucial for ‘heavy cases’, because for millions of followers the ones that reside closer to the end are being processed later than the ones residing closer to the front of the list.
The following pseudocode tries to illuminate this multithreading delivery idea:
So whenever followers refresh their timelines, they will receive the new tweet.
It will be fair to say, that we merely touched the tip of the iceberg of real problems at Airbnb or Twitter. It takes a really long time and the hard work of really talented engineers to accomplish such great results in complex systems like Twitter, Google, Facebook, Amazon, Airbnb and others. Just keep this in mind while reading this article.
The point of demonstrating Twitter’s tweet delivery problem is to embrace the usage of graphs, even though we didn’t use any graph algorithm, we just used a representation of the graph. Sure we pseudocoded a function for delivering tweets, but that is something we came up during the process of searching for a solution.
What I meant by “any graph algorithm” is any algorithm from this list. As something big enough to make programmers cry, graph theory and graph algorithm applications are somewhat different to spot at a glimpse. We were discussing the Airbnb homes and efficient filtering before finishing with graph representations, and the main obvious thing was the inability to efficiently filter homes with more than one filter key. Is there anything that could be done using graph algorithms? Well, we can’t tell for sure, but at least we can try. What if we represent each filter as a separate vertex?
Literally each filter, even all the prices from \$10 to \$1000+, all city names, country codes, amenities (TV, Wi-Fi, and all others), number of adults, and each number as a separate graph vertex.
We can even make this set of vertices more “friendly” if we add “type” vertices too, like “Amenities” connected to all vertices representing an amenity filter.
Now, what if we represent Airbnb homes as vertices and then connect each home with “filter” vertex if that home supports the corresponding filter (For example, connecting “home 1” with “kitchen” if “home 1” has “kitchen” in its amenities)?
A subtle change of this illustration makes it more likely to resemble a special type of graph, called a bipartite graph.
Bipartite graph or just bigraph is a graph whose vertices can be divided into two disjoint and independent sets such that every edge connects a vertex in one set to one in other set. - Wikipedia.
In our example one of the sets represents filters (we’ll denote it by F) and the other is a homes set (denoted by H). For example, if there are 100 thousand homes with the price value \$62, then price vertex labeled “\$62” will have 100 thousand edges incident to each homes vertices. If we measure the worst case of space complexity, i.e. each home has all the properties satisfying to all the filters, than the total amount of edges to be stored will be 70,000 * 4 million. If we represent each edge as a pair of two ids: {filter_id; home_id} and if we rethink about IDs and use a 4 byte (int) numeric id for filters and 8 byte (long) id for homes, then each edge would require at least 12 bytes. So storing 70,000 * 4 million 12 bytes values will require around 3Tb of memory. We made a small mistake in calculation, you see.
The number of filters is around 70,000 because of the 65 thousand cities active in Airbnb (stats). And the good news is that the same home cannot be located in more than one city. That is, our actual number of edges pairing with cities is 4 million (each home located in one city). So we’ll calculate for 70k - 65k = 5 thousand filters, that means we need 5000 * 4 million * 12 bytes of memory, which is less than 0.3 Tb. Sounds good. But what gives us this bipartite graph? Most commonly a website/mobile request will consist of several filters, for example like this:
`house_type: "entire_place",adults_number: 2,price_range_start: 56,price_range_end: 80,beds_number: 2,amenities: ["tv", "wifi", "laptop friendly workspace"],facilities: ["gym"]`
And all we need is to find all the “filter vertices” above and process all the “home vertices” that are adjacent to these “filter vertices”. This takes us to a scary subject.
# Graph Algorithms: Intro
Any processing done with graphs might be categorized as a “graph algorithm”. You literally can implement a function printing all the vertices of a graph and name it “<your name here>’s vertex printing algorithm”. Most of us are scared of the graph algorithms listed in textbooks (see the full list here). Let’s try to apply a bipartite graph matching algorithm, such as Hopcroft-Karp algorithm to our Airbnb homes filtering problem:
Given a bipartite graph of Airbnb homes (H) and filters (F), where every element (vertex) of H can have more than one adjacent elements (vertex) of F (sharing a common edge). Find a subset of H consisting of vertices that are adjacent to vertices in a subset of F.
Confusing problem definition, however we can’t be sure at this point whether Hopcroft-Karp algorithm solves our problem. But I assure you that this journey will teach us many crucial ideas behind graph algorithms. And the journey is not so short, so be patient.
The HopcroftKarp algorithm is an algorithm that takes as input, a bipartite graph and produces as output, a maximum cardinality matching - a set of as many edges as possible with the property that no two edges share an endpoint - Wikipedia.
Readers familiar with this algorithm are already aware that this doesn’t solve our problem, because matching requires that no two edges share a common vertex.
Let’s look at an example illustration, where there are just 4 filters and 8 homes (for the sake of simplicity).
• Homes are denoted by letters from A through H, filters are chosen randomly.
So the following illustration tries to show which homes should we “return” for the request asking for homes that have all four filters available (For example, they cost \$50 per night, they have 1 bed and also they have Wi-Fi and TV).
The solution to our problem requires edges with common vertices leading to distinct home vertices that are incident to the same filter subset, while Hopcroft-Karp algorithm eliminates such edges with common endpoints and produces edges incident to vertices in both subsets.
Take a look at the illustration above, all we need are homes D and G, which both satisfy to all four filter values. What we really need is to get all matching edges which share a common endpoint.
We could devise an algorithm for this approach, but its processing time is arguably not relevant to users needs (users needs = lightning fast, right here, right now). Probably it would be faster to create a balanced binary search tree with multiple sort keys, almost like a database index file, which maps primary/foreign keys with a set of satisfying records.
Balanced binary search trees and database indexing will be discussed in a separate article, where we will return to the Airbnb home filtering problem again.
The Hopcroft-Karp algorithm (and many others) are based on both DFS (Depth-First Search) and BFS (Breadth-First Search) graph traversal algorithms. To be honest, the actual reason to introduce the Hopcroft-Karp algorithm here is to surreptitiously switch to graph traversals, which is better to start from the nice graphs, binary trees.
Binary tree traversals are really beautiful, mostly because of their recursive nature. There are three basic traversals called in-order, post-order and pre-order (you may come up with your own traversal algorithm). They are easy to understand if you have ever traversed a linked list. In linked lists you just print the current node’s value (named `item` in the code below) and continue to the next node.
Almost the same goes with binary trees, you print the node value (or whatever else you need to do with it) and then continue to the next node, but in this case, there are “two next” nodes, left and right. So you should do the same for both left and right nodes. But you have three different choices here:
• print the node value then go to the left node, and then go to the right node, or
Obviously, recursive functions look very elegant though they are so expensive. Each time we call a function recursively, it means we call a completely “new” function (see the illustration above). And by “new” we mean that another stack memory space should be “allocated” for the function arguments and local variables. That’s why recursive calls are expensive (the extra stack space allocations and the many function calls) and dangerous (mind the stack overflow) and it is obviously suggested to use iterative implementations. In mission critical systems programming (aircraft, NASA rovers and so on) a recursion is completely prohibited (no stats, no experience, just telling you the rumors).
# Netflix and Amazon: Inverted Index Example
Let’s say we want to store all Netflix movies in a binary search tree with movie titles as sort keys. So whenever a user types something like “Inter”, we will return a list of movies starting with “Inter” (for instance, “Interstellar”, “Interceptor”, “Interrogation of Walter White”).
Now, it would be great if we’ll return every movie that contains “Inter” in its title (not only ones that start with “Inter”), and the list would be sorted by movie ratings or something that is relevant to that particular user (like thrillers more than drama). The point of this example is to make efficient range queries to a BST.
But as usual, we won’t dive deeper into the cold water to spot the rest of the iceberg. Basically, we need a fast lookup by search keywords and then get a list of results sorted by some key, which most likely should be a movie rating and/or some internal ranking based on a user’s personalized data. We’ll try to stick to the KISK principle (Keep It Simple, Karl) as much as possible.
“KISK” or “let’s keep it simple” or “for the sake of simplicity”, a super excuse for tutorial writers to abstract from the real problem and make tons of assumptions by bringing an “abc” easy example and its solution in pseudocode that works even on your grandma’s laptop.
This problem could be easily applied to Amazon’s product search as we most commonly search something in Amazon by typing a text describing our interest (like “Graph Algorithms”) and get results sorted by product ratings. I haven’t experienced personalized results in Amazon’s search results. But I’m pretty sure Amazon does that stuff too. So, it will be fair to change the title of this subtopic to…
Netflix and Amazon. Netflix serves movies, Amazon serves products, we’ll name them “items”, so whenever you read an “item” think of a movie in Netflix or any [viable] product in Amazon.
What is most commonly done with the items is the parsing of their title and description (we’ll stick to the title only), so if an operator (usually a human being inserting item’s data into Netflix/Amazon database via an admin dashboard) inserts a new item into the database, its title is being processed by some “ItemTitleProcessor” to produce keywords.
Each item has its unique ID, which is being linked to the keyword found in its title. This is what search engines do while crawling websites all over the world. They analyze each document’s content, tokenize it (break it into smaller entities called words) and add it to a table, which maps each token (word) to the document ID (website) where the token has been “seen”.
So whenever you search for “hello”, the search engine fetches all documents mapped to the keyword “hello” (reality is much complex, because the most important thing is the search relevancy, which is why Google Search is so awesome). So a similar table for Netflix/Amazon may look like this (again, think of Movies or Products when reading Items).
Hashtables, again. Yes, we will keep a hashtable for this inverted index (index structure storing a mapping from content - Wikipedia). The hashtable will map a keyword to a BST of items. Why BST? Because we want to keep them sorted and at the same time serve them (respond to frontend requests) in sequential sorted portions, (for instance, 100 items at a request using pagination). Not really something that shows the power of BSTs. But let’s pretend that we also need a fast lookup in the search result, say you want all 3 star movies with the keyword “machine”.
Note that it’s okay to have duplicate items in different trees, because an item usually can be found with more than one keyword.
We’ll operate with items defined as follows:
Each time a new item is inserted into a database, the title is processed and added to the big index table, which maps a keyword to an item. There could be many items sharing the same keyword, so we keep these items in a BST sorted by their rating.
When users search for some keyword, they get a list of items sorted by their rating. How can we get a list from a tree in a sorted order? By doing an in-order traversal.
Here’s how an implementation of `InOrderProduceVector()` might look:
But, but… We need the highest rated items first, and our in-order traversal produces the lowest rated items first. That’s because of its nature. In-order traversal works “bottom up”, from the lowest to the highest item. To get what we wanted, i.e. the list in descending order instead of ascending, we should take a look at the in-order traversal implementation a bit closer.
What we are doing is going through the left node, then printing the current node’s value and then going through the right node. The left most node is the node with the smallest value. So simply changing the implementation to go through the right node first will lead us to a descending order of the list. We’ll name it as others do, a reverse in-order traversal.
Let’s update the code above (introducing in a single listing). Warning - Bugs Ahead!
That’s it. We can serve item search results pretty fast. As mentioned above, inverted indexing is used mostly in search engines, like Google. Although Google Search is a very complex search engine, it does use some simple ideas (way too modernized though) to match search queries to documents and serve the results as fast as possible.
We used tree traversals to serve results in sorted order. At this point it might seem that pre/in/post-order traversals are more than enough, but sometimes there is a need for another type of traversal.
Let’s tackle this well-known programming interview question, “How would you print a [binary] tree level by level?”.
# Traversals: DFS and BFS
If you are not familiar with this problem, think of some data structure that you could use to store nodes while traversing the tree. If we compare level-by-level traversal of a tree with the others above (pre, in, post order traversals), we’ll eventually devise two main traversals of graphs, that is a depth-first search (DFS) and breadth-first search (BFS).
Depth-first search hunts for the farthest node, breadth-first search explores nearest nodes first.
• Depth-first search - more actions, less thoughts.
DFS is much like pre, in, post-order traversals. While BFS is what we need if we want to print a tree’s nodes level-by-level.
To accomplish this, we would need a queue (data structure) to store the “level” of the graph while printing (visiting) its “parent level”. In the previous illustration nodes that are placed in the queue are in light blue.
Basically, going level by level, nodes on each level are fetched from the queue, and while visiting each fetched node, we also should insert its children into the queue (for the next level). The following code is simple enough to get the main idea of BFS. It is assumed that the graph is connected, although it can be modified to apply to disconnected graphs.
The basic idea is easy to show on a node-based connected graph representation. Just keep in mind that the implementation of the graph traversal differs from representation to representation.
BFS and DFS are important tools in tackling graph searching problems (but remember that there are tons of graph search algorithms). While DFS has elegant recursive implementation, it is reasonable to implement it iteratively. For the iterative implementation of BFS we used a queue, for DFS you will need a stack. One of the most popular problems in graphs and at the same time one of the most possible reasons you read in this article is the problem of finding the shortest path between graph vertices. And this takes us to our last thought experiment.
# Uber and the Shortest Path Problem (Dijkstra’s Algorithm)
With its 50 million users and 7 million drivers (source), one of the most important things that is critical to Uber’s functioning is the ability to match drivers with riders in an efficient way. The problem starts with locations.
The backend should process millions of user requests, sending each of the requests to one or more (usually more) drivers nearby. While it is easier and sometimes even smarter to send the user request to all nearby drivers, some pre-processing might actually help.
Besides processing incoming requests and finding the location area based on the user coordinates and then finding drivers with nearest coordinates, we also need to find the right driver for the ride. To avoid geospatial request processing (fetching nearby cars by comparing their current coordinates with user’s coordinates), let’s say we already have a segment of the map with user and several nearby cars.
Something like this:
Possible paths from a car to a user are in yellow. The problem is to calculate the minimum required distance for the car to reach the user, in other words, find the shortest path between them. While this is more about Google Maps rather than Uber, we’ll try to solve it for this particular and very simplified case mostly because there are usually more than one drivers car and Uber might want to calculate the nearest car with the highest rating to send it to the user.
For this illustration that means calculating for all three cars the shortest path reaching to the user and decide which car would be the optimal to send. To make things really simple, we’ll discuss the case with just one car. Here are some possible routes to reach to the user.
Cutting to the chase, we’ll represent this segment as a graph:
This is an undirected weighted graph (edge-weighted, to be more specific). To find the shortest path between vertices B (the car) and A (the user), we should find a route between them consisting of edges with possibly minimum weights. You are free to devise your version of the solution. We’ll stick with Dijkstra’s version. The following steps of Dijkstra’s algorithm are from Wikipedia.
Let the node at which we are starting be called the initial node. Let the distance of node Y be the distance from the initial node to Y. Dijkstra’s algorithm will assign some initial distance values and will try to improve them step by step.
1. Mark all nodes unvisited. Create a set of all the unvisited nodes called the unvisited set.
Applying this to our example, we’ll start with vertex B (the car) as the initial node. For first two steps:
Our unvisited set consists of all vertices. Also note the table at the left side of the illustration. For all vertices, it will contain all the shortest distances from B and the previous (marked “Prev”) vertex that lead to the vertex. For instance the distance is 20 from B to F, and the previous vertex is B.
We are marking B as visited and move it to its neighbor F.
Now we are marking F as visited and choosing the next unvisited node with smallest tentative distance, which is G. Also note the table at the left side. In the previous illustration nodes C, F and G already have their tentative distances set with the previous nodes which lead to the mentioned nodes.
As stated in the algorithm, if the destination node has been marked visited (when planning a route between two specific nodes as in our case) then we can stop. So our next step stops the algorithm with the following values.
So we have both the shortest distance from B to A and the route through F and G nodes.
This is really the simplest possible example of potential problems at Uber, comparing this to our iceberg analogy, we are at the tip of the tip of the iceberg. However, this is a good first start to explore the real world of graph theory and its applications. I didn’t complete what I initially planned for in this article, but in the near future, most probably, this will be continued (also including database indexing internals).
There is still so much to tell about graphs (still need to study). Take this article as another tip of the iceberg. If you have read this far, you definitely deserve a cookie. Don’t forget to clap and share. Thank you.
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Get unlimited access to the best stories on Medium — and support writers while you’re at it. Just \$5/month. | 13,236 | 61,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-16 | longest | en | 0.952817 |
web.med.unsw.edu.au | 1,409,718,650,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535924501.17/warc/CC-MAIN-20140909013108-00170-ip-10-180-136-8.ec2.internal.warc.gz | 575,266,196 | 3,424 | RELATIONSHIP BETWEEN GENERALISED RELATIVE IONIC MOBILITY AND LIMITING EQUIVALENT CONDUCTIVITY
First of all, it should be noted that the relative mobility of an ion, u, required for calculating liquid junction potentials (as listed in the above tables of mobilities and required in JPCalc calculations) represents the generalised (or absolute) mobility of an ion relative to K+. For example, if uX, is the relative mobility for ion X, with respect to K+, it will be given by:
uX = u*X / u*K
where u*X and u*K represent the absolute values of the generalised mobilities of ions X and K+ respectively. The units of the relative mobility for ion X, uX, are (of course) dimensionless.
The following discussion indicates how the generalised mobilities of ions are in turn related to their limiting equivalent conductivities.
Since the velocity of an ion in solution, v, is related to the generalised (absolute) mobility, u*, and the generalised force, Fx, acting on it, then:
v = u* Fx
The force may be in Newtons/ mole or Newtons, depending on whether it is the force acting on a mole of ions or on a single ion (and whichever is chosen will affect the units of u*). The above generalised mobility is what is required for electrodiffusion flux equations, and would normally be that required for a force acting on a mole of ions.
In contrast, electrochemists, when measuring conductivity, use another definition of mobility, which may be defined as u', the electrical mobility, sometimes also called the conventional mobility (Bockris & Reddy, 1973; pp. 369-373), since they measure the mobility as the velocity/ electric field, E (e.g., in volts/m) as:
v = u' E
Since the actual force is zFE, we also have v = u*zFE, where z is the magnitude of the valency and F is the Faraday. Hence,
u* = u'/zF
We wish to know the relationship between generalised (absolute) mobility and the limiting equivalent conductivity, L0 (the conductivity of an electrolyte solution per equivalent, in the limit as the concentration goes to zero). Now Lmakes allowance for the additional charge of polyvalent ions, so that
u' = L0 /F
[cp. Eqs. 4.156 - 4.160 in Bockris & Reddy, 1973 (p.373) for equivalent conductivity (L) and molar conductivity (Lm), where
L = Lm/z; N.B. error in sign of the anion subscript in Eq. 4.16)]. Hence, from the two equations above, the generalised mobility, u*, and L0 will be related by:
u* = L0 / (zF2)
[cf. the equation for the generalized mobility, u, for a single ion (e.g., Eq. A4 in Sugiharto et al., 2008), rather than for a mole of ions, u*, which is given by u = NL0 / (zF2), where N is Avagadro’s number]Hence, the relative mobility of ion X of valency z is given by:
uX = [L0X / z] / L0K
since, for K+, z = 1 and both limiting equivalent conductivities were measured at the same temperature.
For a monovalent ion Y, its relative mobility will simply be given by:
uY = L0Y / L0K
where L0Y is the limiting equivalent conductivity of Y at the same temperature as for L0K , normally 25 oC.
For reference, L0K = 73.50 S.cm2.equiv-1 at 25 oC (Robinson & Stokes, 1965).
References
Bockris J. O'M and A.K.N. Reddy (1973). Modern Electrochemistry, Vol 1, Plenum Press, New York
Sugiharto, S., T. M. Lewis, A. J. Moorhouse, P. R. Schofield, and P. H. Barry. (2008). Anion-cation permeability correlates with hydrated counter-ion size in glycine receptor channels. Biophys. J. 95:4698-4715. | 933 | 3,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2014-35 | longest | en | 0.900811 |
https://www.formulaconversion.com/formulaconversioncalculator.php?convert=quartsdry_to_teaspoons | 1,508,693,512,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825399.73/warc/CC-MAIN-20171022165927-20171022185927-00214.warc.gz | 939,834,074 | 16,305 | # Quarts (dry) to teaspoons (qt to teaspoons) Metric conversion calculator
Welcome to our quarts (dry) to teaspoons (qt to teaspoons) conversion calculator. You can enter a value in either the quarts (dry) or teaspoons input fields. For an understanding of the conversion process, we include step by step and direct conversion formulas. If you'd like to perform a different conversion, just select between the listed Volume units in the 'Select between other Volume units' tab below or use the search bar above. Tip: Use the swap button to switch from converting quarts (dry) to teaspoons to teaspoons to quarts (dry).
## teaspoons
(not bookmarks)
Swap
< == >
1 qt = 223.42026122 teaspoons 1 teaspoons = 0.00447587 qt
Algebraic Steps / Dimensional Analysis Formula
qt * 223.4203 teaspoons1 qt = teaspoons
cubic feet 0 cubic meters 0 cubic millimeters 0 cubic yards 0 cups 0 fluid ounces 0 gallons (liquid) 0 liters 0 milliliters 0
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Active Users | 495 | 2,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-43 | latest | en | 0.610195 |
https://ask.truemaths.com/question-tag/ncert-maths-solutions-class-9th/ | 1,675,839,268,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500719.31/warc/CC-MAIN-20230208060523-20230208090523-00079.warc.gz | 125,374,955 | 23,438 | • 1
## The diameter of a sphere is 28 cm. Find the cost of painting it all around at Rs. 0.10 per square cm.
• 1
In this question the diameter of a sphere is given to us as 28 cm, and we have to find ...Read more | 64 | 215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-06 | latest | en | 0.95409 |
http://www.chegg.com/homework-help/questions-and-answers/hi-think-know-formulas-find-id-vgs-vds-gm-rd-zi-zo-av-vgs-rs-id-id-idss-1-vgs-vp-2-vds-vdd-q3644406 | 1,448,773,918,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398456289.53/warc/CC-MAIN-20151124205416-00199-ip-10-71-132-137.ec2.internal.warc.gz | 352,218,540 | 13,079 | Hi I think I know the formulas to find: Id, Vgs, Vds, gm, rd, Zi, Zo, Av are:
VGS= -Rs(Id)
Id= Idss(1-Vgs/Vp)^2
Vds= Vdd-(Rd+Rs)Id
Gm=2(Idss/-Vp)(1-Vgs/Vp)
Av=(-gmRd)/(1+gmRs)+(Rd+Rs)/Rd
Rd= 1/ yos
Zi= Rg
Z0= ((1+gmRs+Rs/Rd /1+gmRs + Rd+Rs/Rd))Rd | 139 | 254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2015-48 | latest | en | 0.814108 |
https://mersenneforum.org/showthread.php?s=2705540c6912991591ddb61156588c54&t=2934 | 1,618,214,782,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00514.warc.gz | 508,317,941 | 10,186 | mersenneforum.org Low weight stats page.
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2004-08-20, 14:51 #1 jocelynl Sep 2002 2×131 Posts Low weight stats page. Since <300k is almost completed, I would like to set up a page on low weight candidates of the form k*2^n-1 . If you could just tell me what work you have done on these. It would be nice not to duplicate work. Joss Last fiddled with by jocelynl on 2004-08-21 at 12:16
2004-08-23, 10:47 #2 Kosmaj Nov 2003 2·1,811 Posts I haven't done any work on low-weight k's but I was thinknig to propose that we process one real but small 15k to 1M like k=15, 45, 75 or 105. After sieving such k's produce too many candidates for a single cruncher to process, but potentially they hide a good deal of primes so if we do it together there are good chances that many of us will find a large prime. With the latest LLR-P4 I think it can be done in a reasonable amount of time. What do you say?
2004-08-23, 15:58 #3 jocelynl Sep 2002 1000001102 Posts I guess Mark could set up a team for some low 15k like he has done with 210885 and 2995125705. With the new LLRP-4, k=15 seems a good choice. Robert has been working on it for some times. We should ask him for team work first. Joss
2004-09-02, 10:44 #4 jocelynl Sep 2002 10616 Posts This is our first top5000 prime on Low Weight Candidate 3817*2^218917-1 It's a start! Joss
2004-11-16, 19:54 #5 Thomas11 Feb 2003 22×32×53 Posts I'm just announcing the find of two (big) primes of low weight k: 209826493*2^1140855-1 (ranking at position 50 of the Top-5000!) 209826493*2^1071303-1 (found a few months earlier) -- Thomas L10
2004-11-16, 21:09 #6 Templus Jun 2004 2·53 Posts Wow thomas!! Congratulations on your 2 big primes!!!! Good work!!!!
2004-11-17, 17:53 #7 jocelynl Sep 2002 2·131 Posts These should had some weight to our project! Great work Thomas ! Joss
2004-12-12, 15:35 #8 Thomas11 Feb 2003 77416 Posts How about another big one? 80857169*2^1251076-1 is prime! -- Thomas L10
2004-12-12, 19:51 #9 jocelynl Sep 2002 2×131 Posts Wow! rank 37, whooo now we're talking. congratulation Joss
2004-12-18, 16:15 #10 Kosmaj Nov 2003 E2616 Posts Thomas, congrats any many huge primes! BTW, can I reserve one k, just to try? I'm not sure how to read the table at the status page but if available I'd like to take k=32537227. It has no available data so I assume it's available(?) Can one reserve k's with some associated date in the table and/or provided sieve file? Thanks and good luck! Last fiddled with by Kosmaj on 2004-12-18 at 16:34
2004-12-18, 17:17 #11 jocelynl Sep 2002 10616 Posts Hi Kosmaj, k=32537227 is all yours I don't have any sieve file for it It's not a very low weight at 2100 n's left at 20M but still a good choice. Joss
Similar Threads Thread Thread Starter Forum Replies Last Post Oddball Twin Prime Search 0 2011-10-29 18:34 ET_ PrimeNet 0 2009-01-10 15:02 mdettweiler Prime Sierpinski Project 3 2008-08-27 18:34 Old man PrimeNet Lounge 15 2003-11-25 02:09 Deamiter Software 1 2002-11-09 06:24
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Mon Apr 12 08:06:22 UTC 2021 up 4 days, 2:47, 1 user, load averages: 2.41, 2.85, 2.86 | 1,083 | 3,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-17 | latest | en | 0.896268 |
https://math.stackexchange.com/questions/571558/binomial-distributionoverbooking-plane-tickets | 1,620,502,686,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988923.22/warc/CC-MAIN-20210508181551-20210508211551-00041.warc.gz | 411,465,052 | 39,343 | binomial distribution(overbooking plane tickets)
I am having trouble with binomial distribution and this problem: an airplane has 200 seats, but 202 tickets are sold. Assume passengers do not show up with a probability of .03 independently. What is the chance that a flight is over full? Let X be the random variable that equals 1, then P(X=1)=P(PPPP...PM in any order) where P means passenger is present and M means passenger is missing. To do this we find how many ways are there to get this sequence of successes/failure(missing/present passengers):(202 choose 1)(.97)^201*(.03)^1=.01329 chance of a flight being overbooked? is that right or do i have to sum P{X=1,0}=P(X=1)+P(X=0)? I'm also confused at (202 choose 1)(.97)^201*(.03)^1, shouldn't that equal .03? We are saying out of the 202 tickets choose 1(in 202 different ways) to have one person missing at a probability of .03 and all other present at a probability of .97. Shouldn't that probability be .03?
• You do have to deal separately with what you call $1$, and $0$ and add up. – André Nicolas Nov 18 '13 at 7:11
• editing my post, i made some errors. It should have been 1 and 0 instead of 2,1,0 since there are 202 tickets sold and overbooking would mean 201 tickets. Can you explain why is my logic wrong in thinking that (202 choose 1)(.97)^201*(.03)^1 should equal .03? If instead of tickets we had a flip of a coin and we wanted to find the probability of getting a head with a tail being a probability of .50 independently then wouldn't it be (1-.50)^0*(.50) which is .50 or a 1/2. – bob the builder Nov 18 '13 at 7:12
• This seems like the same concept, flipping a coin and getting a head with a .50 probability of a tail vs getting a missing passenger with a .03 probability rate. P{Head} at 1/2 or P{Tail} at 1/2 vs P{Missing} at .03 or P{Passenger} at .97 – bob the builder Nov 18 '13 at 7:20
• It is the same concept, and the expression is similar, apart from the simplification from the fact that $1-0.5=0.5$. If we flip a fair coin $12$ times, the probability of exactly $11$ heads is $\binom{12}{11}(0.5)^{11}(0.5)^1$. – André Nicolas Nov 18 '13 at 7:25
Let random variable $X$ be the number of people who show up. The probability that the plane is overbooked is $$\Pr(X=201)+\Pr(X=202).$$
We have $$\Pr(X=k)=\binom{202}{k}(0.97)^k (0.03)^{202-k}.$$
Alternately, we can calculate the two needed probabilities without explicit appeal to the formula for $\Pr(X=k)$.
The probability that $202$ people show up is $(0.97)^{202}$.
For the probability that $201$ people show up, the missing person can be chosen in $202$ ways. The probability she doesn't show up is $0.03$. The probability all the others show up is $(0.97)^{201}$, so the probability $201$ people (exactly) show up is $(202)(0.03)(0.97)^{201}$.
Remark: Equivalently, we can let $Y$ be the number of missing passengers, and find $\Pr(Y=0)+\Pr(Y=1)$. That seems to be what you chose, though the random variable was not defined precisely. | 865 | 2,985 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-21 | latest | en | 0.943727 |
https://tunxis.commnet.edu/view/systems-word-problems-worksheet.html | 1,716,846,803,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00309.warc.gz | 492,735,623 | 6,096 | # Systems Word Problems Worksheet
Systems Word Problems Worksheet - 2) flying with the wind a plane went 183 km/h. They would like to make at least a \$500 profit from selling tickets. If she bought a total of 7 then how. Find the answers to each situation below by setting up and solving a system of equations. Flying into the same wind the plane only went 141 km/h. Find the speed of the plane in.
How about some bacon and eggs?. Find the speed of the plane in. Find the answers to each situation below by setting up and solving a system of equations. Web systems of equations word problems. Web systems of equations word problems date_____ period____ 1) kristin spent \$131 on shirts.
Solve each word problem by writing and solving a system of equations. Web systems of equations word problems worksheet activity. When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the. If she bought a total of 7 then how. The currents running through an electrical system are given by the following system of equations.
## Word Problems A Worksheet For Understanding And Practice Style
Systems Word Problems Worksheet - Find the answers to each situation below by setting up and solving a system of equations. The ninth graders are hosting the next school dance. Web solving systems of equations word problems worksheet for all problems, define variables, write the system of equations and solve for all variables. Web systems of equations word problems. When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the. Malcolm and ravi raced each other. Web systems of equations word problems date_____ period____ 1) kristin spent \$131 on shirts. Web word problems worksheet 6 pdf. Web systems of equations word problems worksheet activity. The three currents, i1, i2, and i3, are measured in amps.
The three currents, i1, i2, and i3, are measured in amps. Find the answers to each situation below by setting up and solving a system of equations. Solve each word problem by writing and solving a system of equations. The ninth graders are hosting the next school dance. Systems of inequalities word problems 1.
Find the speed of the plane in. Solve each word problem by writing and solving a system of equations. Systems of inequalities word problems 1. Web systems of equations word problems date_____ period____ 1) kristin spent \$131 on shirts.
Web systems of equations word problems date_____ period____ 1) kristin spent \$131 on shirts. Malcolm and ravi raced each other. Solve each word problem by writing and solving a system of equations.
Students will work with systems of equations that can be found within word problems in this problem set. Fancy shirts cost \$28 and plain shirts cost \$15. When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the.
## Systems Of Inequalities Word Problems 1.
Fancy shirts cost \$28 and plain shirts cost \$15. Web systems of equations word problems date_____ period____ 1) kristin spent \$131 on shirts. 2) flying with the wind a plane went 183 km/h. They would like to make at least a \$500 profit from selling tickets.
## Web Systems Of Equations Word Problems.
Find the speed of the plane in. The three currents, i1, i2, and i3, are measured in amps. How about some bacon and eggs?. The currents running through an electrical system are given by the following system of equations.
## Find The Answers To Each Situation Below By Setting Up And Solving A System Of Equations.
Web systems of equations word problems. Web systems of equations word problems worksheet activity. Web systems of equations word problems. Web word problems worksheet 6 pdf.
## Flying Into The Same Wind The Plane Only Went 141 Km/H.
Malcolm and ravi raced each other. Solve each word problem by writing and solving a system of equations. The average of their maximum speeds was 260. The ninth graders are hosting the next school dance. | 853 | 4,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-22 | latest | en | 0.928909 |
https://www.physicsforums.com/threads/how-do-you-find-this-integral.509979/ | 1,527,191,478,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866772.91/warc/CC-MAIN-20180524190036-20180524210036-00398.warc.gz | 811,272,781 | 14,143 | # Homework Help: How do you find this integral?
1. Jun 27, 2011
### asap9993
1. The problem statement, all variables and given/known data
the problem is
Find the antiderivative of 1/(1 + cos(x))
2. Relevant equations
3. The attempt at a solution
I think it requires a trick, but I can't figure it out.
Last edited: Jun 27, 2011
2. Jun 27, 2011
### lanedance
is it a definite integral?
3. Jun 27, 2011
### ehild
Use the relation between cos(x/2)and cos(x).
ehild | 149 | 475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-22 | longest | en | 0.869798 |
http://www.lmfdb.org/NumberField/16.8.34685013449866033007282273.1 | 1,606,948,117,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141716970.77/warc/CC-MAIN-20201202205758-20201202235758-00030.warc.gz | 136,715,597 | 8,503 | # Properties
Label 16.8.346...273.1 Degree $16$ Signature $[8, 4]$ Discriminant $3.469\times 10^{25}$ Root discriminant $39.47$ Ramified primes $17, 59$ Class number $1$ (GRH) Class group trivial (GRH) Galois group $C_{16} : C_2$ (as 16T22)
# Related objects
Show commands for: SageMath / Pari/GP / Magma
## Normalizeddefining polynomial
sage: x = polygen(QQ); K.<a> = NumberField(x^16 - x^15 + x^14 - 18*x^13 - 84*x^12 + 118*x^11 + 324*x^10 + 985*x^9 - 526*x^8 - 6172*x^7 - 1699*x^6 + 10250*x^5 + 14400*x^4 - 15760*x^3 - 20314*x^2 + 18495*x + 8263)
gp: K = bnfinit(x^16 - x^15 + x^14 - 18*x^13 - 84*x^12 + 118*x^11 + 324*x^10 + 985*x^9 - 526*x^8 - 6172*x^7 - 1699*x^6 + 10250*x^5 + 14400*x^4 - 15760*x^3 - 20314*x^2 + 18495*x + 8263, 1)
magma: R<x> := PolynomialRing(Rationals()); K<a> := NumberField(R![8263, 18495, -20314, -15760, 14400, 10250, -1699, -6172, -526, 985, 324, 118, -84, -18, 1, -1, 1]);
$$x^{16} - x^{15} + x^{14} - 18 x^{13} - 84 x^{12} + 118 x^{11} + 324 x^{10} + 985 x^{9} - 526 x^{8} - 6172 x^{7} - 1699 x^{6} + 10250 x^{5} + 14400 x^{4} - 15760 x^{3} - 20314 x^{2} + 18495 x + 8263$$
sage: K.defining_polynomial()
gp: K.pol
magma: DefiningPolynomial(K);
## Invariants
Degree: $16$ sage: K.degree() gp: poldegree(K.pol) magma: Degree(K); Signature: $[8, 4]$ sage: K.signature() gp: K.sign magma: Signature(K); Discriminant: $$34685013449866033007282273$$$$\medspace = 17^{15}\cdot 59^{4}$$ sage: K.disc() gp: K.disc magma: Discriminant(Integers(K)); Root discriminant: $39.47$ sage: (K.disc().abs())^(1./K.degree()) gp: abs(K.disc)^(1/poldegree(K.pol)) magma: Abs(Discriminant(Integers(K)))^(1/Degree(K)); Ramified primes: $17, 59$ sage: K.disc().support() gp: factor(abs(K.disc))[,1]~ magma: PrimeDivisors(Discriminant(Integers(K))); $|\Aut(K/\Q)|$: $8$ This field is not Galois over $\Q$. This is not a CM field.
## Integral basis (with respect to field generator $$a$$)
$1$, $a$, $a^{2}$, $a^{3}$, $a^{4}$, $a^{5}$, $a^{6}$, $a^{7}$, $a^{8}$, $a^{9}$, $a^{10}$, $a^{11}$, $a^{12}$, $a^{13}$, $\frac{1}{10847} a^{14} - \frac{4617}{10847} a^{13} - \frac{1235}{10847} a^{12} - \frac{1017}{10847} a^{11} - \frac{4615}{10847} a^{10} - \frac{2343}{10847} a^{9} - \frac{3837}{10847} a^{8} - \frac{2273}{10847} a^{7} - \frac{1356}{10847} a^{6} + \frac{914}{10847} a^{5} - \frac{15}{10847} a^{4} - \frac{4864}{10847} a^{3} - \frac{2371}{10847} a^{2} + \frac{482}{10847} a + \frac{1629}{10847}$, $\frac{1}{23066687042377216878374326477} a^{15} - \frac{907710944878488712988529}{23066687042377216878374326477} a^{14} + \frac{9902867254714521246335289779}{23066687042377216878374326477} a^{13} + \frac{5922616941654691369506466432}{23066687042377216878374326477} a^{12} - \frac{9710969246857321050997772932}{23066687042377216878374326477} a^{11} + \frac{587354455994741435999188293}{23066687042377216878374326477} a^{10} - \frac{4939423261427355964220622912}{23066687042377216878374326477} a^{9} - \frac{10011299109976230411961139432}{23066687042377216878374326477} a^{8} + \frac{8395861194049699040593603752}{23066687042377216878374326477} a^{7} - \frac{6377314445417707664551068229}{23066687042377216878374326477} a^{6} - \frac{6870531177674674029222107084}{23066687042377216878374326477} a^{5} - \frac{8976823118028577038633002487}{23066687042377216878374326477} a^{4} - \frac{8495547933487166287463515760}{23066687042377216878374326477} a^{3} - \frac{6879310562091791135995319415}{23066687042377216878374326477} a^{2} - \frac{8057752745861026680904674289}{23066687042377216878374326477} a + \frac{10666646382477627374711399322}{23066687042377216878374326477}$
sage: K.integral_basis()
gp: K.zk
magma: IntegralBasis(K);
## Class group and class number
Trivial group, which has order $1$ (assuming GRH)
sage: K.class_group().invariants()
gp: K.clgp
magma: ClassGroup(K);
## Unit group
sage: UK = K.unit_group()
magma: UK, f := UnitGroup(K);
Rank: $11$ sage: UK.rank() gp: K.fu magma: UnitRank(K); Torsion generator: $$-1$$ (order $2$) sage: UK.torsion_generator() gp: K.tu[2] magma: K!f(TU.1) where TU,f is TorsionUnitGroup(K); Fundamental units: Units are too long to display, but can be downloaded with other data for this field from 'Stored data to gp' link to the right (assuming GRH) sage: UK.fundamental_units() gp: K.fu magma: [K!f(g): g in Generators(UK)]; Regulator: $$4436153.77262$$ (assuming GRH) sage: K.regulator() gp: K.reg magma: Regulator(K);
## Class number formula
$\displaystyle\lim_{s\to 1} (s-1)\zeta_K(s) \approx\frac{2^{8}\cdot(2\pi)^{4}\cdot 4436153.77262 \cdot 1}{2\sqrt{34685013449866033007282273}}\approx 0.150267513927$ (assuming GRH)
## Galois group
$OD_{32}$ (as 16T22):
sage: K.galois_group(type='pari')
gp: polgalois(K.pol)
magma: GaloisGroup(K);
A solvable group of order 32 The 20 conjugacy class representatives for $C_{16} : C_2$ Character table for $C_{16} : C_2$
## Intermediate fields
Fields in the database are given up to isomorphism. Isomorphic intermediate fields are shown with their multiplicities.
## Sibling fields
Galois closure: Deg 32
## Frobenius cycle types
$p$ $2$ $3$ $5$ $7$ $11$ $13$ $17$ $19$ $23$ $29$ $31$ $37$ $41$ $43$ $47$ $53$ $59$ Cycle type ${\href{/LocalNumberField/2.8.0.1}{8} }^{2}$ $16$ $16$ $16$ $16$ ${\href{/LocalNumberField/13.4.0.1}{4} }^{4}$ R ${\href{/LocalNumberField/19.8.0.1}{8} }^{2}$ $16$ $16$ $16$ $16$ $16$ ${\href{/LocalNumberField/43.8.0.1}{8} }^{2}$ ${\href{/LocalNumberField/47.4.0.1}{4} }^{4}$ ${\href{/LocalNumberField/53.8.0.1}{8} }^{2}$ R
In the table, R denotes a ramified prime. Cycle lengths which are repeated in a cycle type are indicated by exponents.
sage: p = 7; # to obtain a list of $[e_i,f_i]$ for the factorization of the ideal $p\mathcal{O}_K$:
sage: [(e, pr.norm().valuation(p)) for pr,e in K.factor(p)]
gp: p = 7; \\ to obtain a list of $[e_i,f_i]$ for the factorization of the ideal $p\mathcal{O}_K$:
gp: idealfactors = idealprimedec(K, p); \\ get the data
gp: vector(length(idealfactors), j, [idealfactors[j][3], idealfactors[j][4]])
magma: p := 7; // to obtain a list of $[e_i,f_i]$ for the factorization of the ideal $p\mathcal{O}_K$:
magma: idealfactors := Factorization(p*Integers(K)); // get the data
magma: [<primefactor[2], Valuation(Norm(primefactor[1]), p)> : primefactor in idealfactors];
## Local algebras for ramified primes
$p$LabelPolynomial $e$ $f$ $c$ Galois group Slope content
17Data not computed
$59$59.8.0.1$x^{8} - x + 14$$1$$8$$0$$C_8$$[\ ]^{8} 59.8.4.2x^{8} - 205379 x^{2} + 169643054$$2$$4$$4$$C_8$$[\ ]_{2}^{4}$ | 2,636 | 6,536 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-50 | latest | en | 0.244379 |
http://www.territorioscuola.com/wikipedia/en.wikipedia.php?title=Ray_(geometry) | 1,397,645,138,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00212-ip-10-147-4-33.ec2.internal.warc.gz | 712,427,991 | 26,652 | # Line (geometry)
(Redirected from Ray (geometry))
Three lines — the red and blue lines have the same slope, while the red and green ones have same y-intercept.
A representation of one line segment
The notion of line or straight line was introduced by ancient mathematicians to represent straight objects with negligible width and depth. Lines are an idealization of such objects. Until the seventeenth century, lines were defined like this: "The line is the first species of quantity, which has only one dimension, namely length, without any width nor depth, and is nothing else than the flow or run of the point which […] will leave from its imaginary moving some vestige in length, exempt of any width. […] The straight line is that which is equally extended between its points"1
Euclid described a line as "breadthless length", and introduced several postulates as basic unprovable properties from which he constructed the geometry, which is now called Euclidean geometry to avoid confusion with other geometries which have been introduced since the end of nineteenth century (such as non-Euclidean geometry, projective geometry, and affine geometry).
In modern mathematics, given the multitude of geometries, the concept of a line is closely tied to the way the geometry is described. For instance, in analytic geometry, a line in the plane is often defined as the set of points whose coordinates satisfy a given linear equation, but in a more abstract setting, such as incidence geometry, a line may be an independent object, distinct from the set of points which lie on it.
When a geometry is described by a set of axioms, the notion of a line is usually left undefined (a so-called primitive object). The properties of lines are then determined by the axioms which refer to them. One advantage to this approach is the flexibility it gives to users of the geometry. Thus in differential geometry a line may be interpreted as a geodesic (shortest path between points), while in some projective geometries a line is a 2-dimensional vector space (all linear combinations of two independent vectors). This flexibility also extends beyond mathematics and, for example, permits physicists to think of the path of a light ray as being a line.
A line segment is a part of a line that is bounded by two distinct end points and contains every point on the line between its end points. Depending on how the line segment is defined, either of the two end points may or may not be part of the line segment. Two or more line segments may have some of the same relationships as lines, such as being parallel, intersecting, or skew.
## Definitions versus descriptions
All definitions are ultimately circular in nature since they depend on concepts which must themselves have definitions, a dependence which can not be continued indefinitely without returning to the starting point. To avoid this vicious circle certain concepts must be taken as primitive concepts; terms which are given no definition.2 In geometry, it is frequently the case that the concept of line is taken as a primitive.3 In those situations where a line is a defined concept, as in coordinate geometry, some other fundamental ideas are taken as primitives. When the line concept is a primitive, the behaviour and properties of lines are dictated by the axioms which they must satisfy.
In a non-axiomatic or simplified axiomatic treatment of geometry, the concept of a primitive notion may be too abstract to be dealt with. In this circumstance it is possible that a description or mental image of a primitive notion is provided to give a foundation to build the notion on which would formally be based on the (unstated) axioms. Descriptions of this type may be referred to, by some authors, as definitions in this informal style of presentation. These are not true definitions and could not be used in formal proofs of statements. The "definition" of line in Euclid's Elements falls into this category.4 Even in the case where a specific geometry is being considered (for example, Euclidean geometry), there is no generally accepted agreement among authors as to what an informal description of a line should be when the subject is not being treated formally.
## Ray
Given a line and any point A on it, we may consider A as decomposing this line into two parts. Each such part is called a ray (or half-line) and the point A is called its initial point. The point A is considered to be a member of the ray.5 Intuitively, a ray consists of those points on a line passing through A and proceeding indefinitely, starting at A, in one direction only along the line. However, in order to use this concept of a ray in proofs a more precise definition is required.
Given distinct points A and B, they determine a unique ray with initial point A. As two points define a unique line, this ray consists of all the points between A and B (including A and B) and all the points C on the line through A and B such that B is between A and C.6 This is, at times, also expressed as the set of all points C such that A is not between B and C.7 A point D, on the line determined by A and B but not in the ray with initial point A determined by B, will determine another ray with initial point A. With respect to the AB ray, the AD ray is called the opposite ray.
Thus, we would say that two different points, A and B, define a line and a decomposition of this line into the disjoint union of an open segment (A, B) and two rays, BC and AD (the point D is not drawn in the diagram, but is to the left of A on the line AB). These are not opposite rays since they have different initial points.
The definition of a ray depends upon the notion of betweenness for points on a line. It follows that rays exist only for geometries for which this notion exists, typically Euclidean geometry or affine geometry over an ordered field. On the other hand, rays do not exist in projective geometry nor in a geometry over a non-ordered field, like the complex numbers or any finite field.
In topology, a ray in a space X is a continuous embedding R+X. It is used to define the important concept of end of the space.
## Euclidean geometry
When geometry was first formalised by Euclid in the Elements, he defined a line to be "breadthless length" with a straight line being a line "which lies evenly with the points on itself".8 These definitions serve little purpose since they use terms which are not, themselves, defined. In fact, Euclid did not use these definitions in this work and probably included them just to make it clear to the reader what was being discussed. In modern geometry, a line is simply taken as an undefined object with properties given by axioms,9 but is sometimes defined as a set of points obeying a linear relationship when some other fundamental concept is left undefined.
In an axiomatic formulation of Euclidean geometry, such as that of Hilbert (Euclid's original axioms contained various flaws which have been corrected by modern mathematicians),10 a line is stated to have certain properties which relate it to other lines and points. For example, for any two distinct points, there is a unique line containing them, and any two distinct lines intersect in at most one point.11 In two dimensions, i.e., the Euclidean plane, two lines which do not intersect are called parallel. In higher dimensions, two lines that do not intersect may be parallel if they are contained in a plane, or skew if they are not.
Any collection of finitely many lines partitions the plane into convex polygons (possibly unbounded); this partition is known as an arrangement of lines.
### Cartesian plane
Lines in a Cartesian plane or, more generally, in affine coordinates, can be described algebraically by linear equations. In two dimensions, the equation for non-vertical lines is often given in the slope-intercept form:
$y = mx + b \,$
where:
m is the slope or gradient of the line.
b is the y-intercept of the line.
x is the independent variable of the function y = f(x).
The slope of the line through points A(xa, ya) and B(xb, yb), when xaxb, is given by m = (yb − ya)/(xb − xa) and the equation of this line can be written y = m(xxa) + ya.
In R2, every line L (including vertical lines) is described by a linear equation of the form
$L=\{(x,y)\mid ax+by=c\} \,$
with fixed real coefficients a, b and c such that a and b are not both zero. Using this form, vertical lines correspond to the equations with b = 0.
There are many variant ways to write the equation of a line which can all be converted from one to another by algebraic manipulation. These forms (see Linear equation for other forms) are generally named by the type of information (data) about the line that is needed to write down the form. Some of the important data of a line is its slope, x-intercept, known points on the line and y-intercept.
The equation of the line passing through two different points $P_0 = ( x_0, y_0 )$ and $P_1 = (x_1, y_1)$ may be written as
$(y - y_0)(x_1 - x_0) = (y_1 - y_0)(x - x_0)$.
If x0x1, this equation may be rewritten as
$y=(x-x_0)\,\frac{y_1-y_0}{x_1-x_0}+y_0$
or
$y=x\,\frac{y_1-y_0}{x_1-x_0}+\frac{x_1y_0-x_0y_1}{x_1-x_0}\,.$
In three dimensions, lines can not be described by a single linear equation, so they are frequently described by parametric equations:
$x = x_0 + at \,$
$y = y_0 + bt \,$
$z = z_0 + ct \,$
where:
x, y, and z are all functions of the independent variable t which ranges over the real numbers.
(x0, y0, z0) is any point on the line.
a, b, and c are related to the slope of the line, such that the vector (a, b, c) is parallel to the line.
They may also be described as the simultaneous solutions of two linear equations
$a_1x+b_1y+c_1z-d_1=0 \,$
$a_2x+b_2y+c_2z-d_2=0 \,$
such that $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are not proportional (the relations $a_1=ta_2,b_1=tb_2,c_1=tc_2$ imply t = 0). This follows since in three dimensions a single linear equation typically describes a plane and a line is what is common to two distinct intersecting planes.
#### Normal form
The normal segment for a given line is defined to be the line segment drawn from the origin perpendicular to the line. This segment joins the origin with the closest point on the line to the origin. The normal form of the equation of a straight line on the plane is given by:
$y \sin \theta + x \cos \theta - p = 0,\,$
where θ is the angle of inclination of the normal segment (the oriented angle from the unit vector of the x axis to this segment), and p is the (positive) length of the normal segment. The normal form can be derived from the general form by dividing all of the coefficients by
$\frac{|c|}{-c}\sqrt{a^2 + b^2}.$
This form is also called the Hesse normal form,12 after the German mathematician Ludwig Otto Hesse.
Unlike the slope-intercept and intercept forms, this form can represent any line but also requires only two finite parameters, θ and p, to be specified. Note that if p > 0, then θ is uniquely defined modulo 2π. On the other hand, if the line is through the origin (c = 0, p = 0), one drops the |c|/(−c) term to compute sinθ and cosθ, and θ is only defined modulo π.
### Polar coordinates
In polar coordinates on the Euclidean plane a line is expressed as
$r=\frac{mr\cos\theta+b}{\sin\theta},$
where m is the slope of the line and b is the y-intercept. When θ = 0 the graph will be undefined. The equation can be rewritten to eliminate discontinuities:
$r\sin\theta=mr\cos\theta+b.\,$
### Vector equation
The vector equation of the line through points A and B is given by r = OA + λAB (where λ is a scalar).
If a is vector OA and b is vector OB, then the equation of the line can be written: r = a + λ(ba).
A ray starting at point A is described by limiting λ. One ray is obtained if λ ≥ 0, and the opposite ray comes from λ ≤ 0.
### Euclidean space
In Euclidean space, Rn (and analogously in every other affine space), the line L passing through two different points a and b (considered as vectors) is the subset
$L = \{(1-t)\,a+t\,b\mid t\in\mathbb{R}\}$
The direction of the line is from a (t = 0) to b (t = 1), or in other words, in the direction of the vector b − a. Different choices of a and b can yield the same line.
#### Collinear points
Three points are said to be collinear if they lie on the same line. Three points usually determine a plane, but in the case of three collinear points this does not happen.
In affine coordinates, in n-dimensional space the points X=(x1, x2, ..., xn), Y=(y1, y2, ..., yn), and Z=(z1, z2, ..., zn) are collinear if the matrix
$\begin{bmatrix} 1 & x_1 & x_2 & \dots & x_n \\ 1 & y_1 & y_2 & \dots & y_n \\ 1 & z_1 & z_2 & \dots & z_n \end{bmatrix}$
has a rank less than 3. In particular, for three points in the plane (n = 2), above matrix is square and the points are collinear if and only if its determinant is zero.
In Euclidean geometry, the Euclidean distance d(a,b) between two points a and b may be used to express the collinearity between three points by:1314
The points a, b and c are collinear if and only if d(x,a) = d(c,a) and d(x,b) = d(c,b) implies x=c.
However there are other notions of distance (such as the Manhattan distance) for which this property is not true.
In the geometries where the concept of a line is a primitive notion, as may be the case in some synthetic geometries, other methods of determining collinearity are needed.
### Types of lines
In a sense,15 all lines in Euclidean geometry are equal, in that, without coordinates, one can not tell them apart from one another. However, lines may play special roles with respect to other objects in the geometry and be divided into types according to that relationship. For instance, with respect to a conic, lines can be:
For more general algebraic curves, lines could also be:
• i-secant lines, meeting the curve in i points counted without multiplicity, or
• asymptotes.
With respect to triangles we have:
For a hexagon with vertices lying on a conic we have the Pascal line and, in the special case where the conic is a pair of lines, we have the Pappus line.
## Projective geometry
In many models of projective geometry, the representation of a line rarely conforms to the notion of the "straight curve" as it is visualised in Euclidean geometry. In Elliptic geometry we see a typical example of this.16 In the spherical representation of elliptic geometry, lines are represented by great circles of a sphere with diametrically opposite points identified. In a different model of elliptic geometry, lines are represented by Euclidean planes passing through the origin. Even though these representations are visually distinct, they satisfy all the properties (such as, two points determining a unique line) that make them suitable representations for lines in this geometry.
## Geodesics
The "straightness" of a line, interpreted as the property that it minimizes distances between its points, can be generalized and leads to the concept of geodesics in metric spaces.
## Notes
1. ^ In (rather old) French: "La ligne est la première espece de quantité, laquelle a tant seulement une dimension à sçavoir longitude, sans aucune latitude ni profondité, & n'est autre chose que le flux ou coulement du poinct, lequel […] laissera de son mouvement imaginaire quelque vestige en long, exempt de toute latitude. […] La ligne droicte est celle qui est également estenduë entre ses poincts." Pages 7 and 8 of Les quinze livres des éléments géométriques d'Euclide Megarien, traduits de Grec en François, & augmentez de plusieurs figures & demonstrations, avec la corrections des erreurs commises és autres traductions, by Pierre Mardele, Lyon, MDCXLV (1645).
2. ^ Coxeter 1969, pg. 4
3. ^ Faber 1983, pg. 95
4. ^ Faber 1983, pg. 95
5. ^ On occasion we may consider a ray without its initial point. Such rays are called open rays, in contrast to the typical ray which would be said to be closed.
6. ^ Wylie, Jr. 1964, pg. 59, Definition 3
7. ^ Pedoe 1988, pg. 2
8. ^ Faber, Appendix A, p. 291.
9. ^ Faber, Part III, p. 95.
10. ^ Faber, Part III, p. 108.
11. ^ Faber, Appendix B, p. 300.
12. ^ .
13. ^ Alessandro Padoa, Un nouveau système de définitions pour la géométrie euclidienne, International Congress of Mathematicians, 1900
14. ^
15. ^ Technically, the collineation group acts transitively on the set of lines.
16. ^ Faber, Part III, p. 108.
## References
• Coxeter, H.S.M (1969), Introduction to Geometry (2nd ed.), New York: John Wiley & Sons, ISBN 0-471-18283-4
• Faber, Richard L. (1983). Foundations of Euclidean and Non-Euclidean Geometry. New York: Marcel Dekker. ISBN 0-8247-1748-1.
• Pedoe, Dan (1988), Geometry: A Comprehensive Course, Mineola, NY: Dover, ISBN 0-486-65812-0
• Wylie, Jr., C. R. (1964), Foundations of Geometry, New York: McGraw-Hill, ISBN 07-072191-2 Check |isbn= value (help)
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TerritorioScuola. Some rights reserved. Informazioni d'uso ☞ | 4,254 | 17,148 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 21, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2014-15 | latest | en | 0.969688 |
http://forums.na.leagueoflegends.com/board/showthread.php?s=&t=3035994&page=71 | 1,394,287,034,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999654667/warc/CC-MAIN-20140305060734-00083-ip-10-183-142-35.ec2.internal.warc.gz | 74,982,899 | 12,335 | ## Does 0.9999(repeating) = 1?
Yes 705 54.95%
No 578 45.05%
Voters: 1283. You may not vote on this poll
### Does 0.9999(repeating) = 1?
First Riot Post
Crowcide
Senior Member
Quote:
Originally Posted by Eledhan
TL/DR - .999... does NOT equal 1, no matter how close to infinity you get. Speaking of infinity, it's not a real number, which means .999... isn't real either, but just a number to show a concept that isn't easily converted into a decimal number system.
.
This has nothing to do with being close. You are arguing about a bunch of definitions you don't even know or understand.
What do you think a real number is? What do you think repeating decimals are? What do you think infinity is? Because your understanding of them is contrary to how math defines them.
.999~ = 1 because of what we define .999~ to be, a geometric series with a common ratio of 1/10 and an r value of 9. .999~ is a real number directly from the definition of the reals, it's the supremum of this set {.9,.99,.999,.9999,...}.
Eledhan
Senior Member
Quote:
Originally Posted by Kodoku
The only infinity needed for 0.99... is the infinity associated with sequences and limits. If you're telling me sequences and limits are no good, then you're saying that the set of real numbers can't be constructed. The construction of the reals, after all, is done by completing the set of rationals, i.e. adding (in a completely non-trivial way) the limits of cauchy sequences of rationals.
What I was trying to say is that if you use basic mathematical operations with infinite numbers, you get impossible results. I vaguely remember reals, rationals, integers, and the like...I'm not a math major or anything like that. However, these concepts are still not totally foreign to me.
If what you are saying is that portions of rational numbers are not possible without infinity concepts, then I see your point. However, the algebraic proof I was attempting to refute cannot work without rational numbers. Whenever you use an "X = Y" statement, for the rest of that proof, X can never equal anything other than Y. What I believe the proof is trying to do is say that "X = Y, and Y = Z, therefore, X = Z". I would go with this more than I would go with the way it was formatted.
If the definitions of X and Y were the same, then I would agree with this whole thing. But as far as I know, the definitions of .999~ and 1 are NOT the same... Am I wrong? My understanding of a need for .999~ was because 1 is not evenly divisible by 3. It has to do with different number systems, not high-level mathematics. Decimal number systems are only one system that is used. It's like a language...there's no TRUE translation for 1/3 in the decimal number system, since the number isn't able to be represented by decimals for its true value without using infinity.
It's kind of like trying to translate different languages...there's no TRUE translation for many words in ancient greek, latin, hebrew, aramaic, arabic, etc. into modern english because those languages had different meanings and inferences than we have today in our language.
That being said, I assume that for all intents and purposes, when someone says, ".999~ is the same as 1" it's no different than someone saying, "agape love is the same as unconditional love", even though the meanings are always going to have some infinitesimal difference, but the overwhelming majority of the meanings are so close that it's pointless to discuss the difference between the two.
Quote:
No such assumption is made. All that's assumed is that 9x/9 is x, and 9/9 is 1.
When you get to 9x = 9, the point is that by dividing both sides by 9, equality will be preserved. So 9x/9 is equal to 9/9.
Well the left side is just x, and the right side is just 1, ergo x = 1
Your example doesn't contradict anything. It just shows that if you start with x = 2, you end up with x = 2, as one would hope.
I think the definitions of the numbers is the key...if it's what I think it is, the problem with people's opinions is that they're trying to represent the same number in two different numeric systems. Is this the point everyone has been trying to make?
Kaolla
Senior Member
Quote:
Originally Posted by Kodoku
Zeno was an ancient philosopher who attempted to support Parmenides' claim that the existence of motion is contradictory. If you accept his argument, then it follows that motion is impossible. As it turns out, it's quite easy to resolve this particular argument because he implicitly assumes that if you sum an infinite number of finite numbers, you'll end up with something infinite. This is false. Calculus proves it's false.
wrong, because motion in reality is not dependent on calculating half-distances. and even if you DID have to, and you wanted to move 10 ft away, there always exists a distance 20ft to where moving HALF that distance will get you 10ft away in one calculated step. so really it's a THOUGHT exercise which CORRECTLY illustrates that infinite calculations DO NOT reach the correct answer, they merely approximate.
Quote:
Originally Posted by Kodoku
Though it's not hard to see it intuitively: It should be clear that the sum
0.1 + 0.01 + 0.001 +... does not sum to infinity. In fact, it sums to exactly 0.11... = 1/9.
it does not, the long division of 1 divided by 9 proves that it does not
Eledhan
Senior Member
Quote:
Originally Posted by Crowcide
This has nothing to do with being close. You are arguing about a bunch of definitions you don't even know or understand.
What do you think a real number is? What do you think repeating decimals are? What do you think infinity is? Because your understanding of them is contrary to how math defines them.
.999~ = 1 because of what we define .999~ to be, a geometric series with a common ratio of 1/10 and an r value of 9. .999~ is a real number directly from the definition of the reals, it's the supremum of this set {.9,.99,.999,.9999,...}.
So what you're saying is that the very definition of .999~ is the same as 1.000~, right?
If this is true, then I guess that's the argument people should be making...not that the number represented by .999~ is the same value as the number represented by 1.000~.
For all intents and purposes, I see how there's no reason to suggest that .999~ is any different than 1.000~. Where I get irritated is when people try to say they are different, and then say they are the same. They can't be both different AND the same simultaneously.
Crowcide
Senior Member
Quote:
Originally Posted by Eledhan
If the definitions of X and Y were the same, then I would agree with this whole thing. But as far as I know, the definitions of .999~ and 1 are NOT the same... Am I wrong? My understanding of a need for .999~ was because 1 is not evenly divisible by 3. It has to do with different number systems, not high-level mathematics. Decimal number systems are only one system that is used. It's like a language...there's no TRUE translation for 1/3 in the decimal number system, since the number isn't able to be represented by decimals for its true value without using infinity.
Is 2/2 1? Is 2/2 the same definition as 1?
What is .9999~ defined to be? The mathematical definition is a geometric series with a common ratio of 1/10 and an r value of 9. That series equals 1. .9999~ is defined to be 1 in the same way 2/2 is through equivalency.
Crowcide
Senior Member
Quote:
Originally Posted by Eledhan
So what you're saying is that the very definition of .999~ is the same as 1.000~, right?
If this is true, then I guess that's the argument people should be making...not that the number represented by .999~ is the same value as the number represented by 1.000~.
For all intents and purposes, I see how there's no reason to suggest that .999~ is any different than 1.000~. Where I get irritated is when people try to say they are different, and then say they are the same. They can't be both different AND the same simultaneously.
I'm guessing there aren't very mathematicians in this thread, hence the non mathematical arguments.
1.000~ and .9999~ are the same.
1.000~ is defined to be a 1 + geometric series with a common ratio of 1/10 and an r value of 0, which equals 1.
.9999~ is defined to be a geometric series with a common ratio of 1/10 and an r value of 9, which also equals 1.
Not lookign the same doesn't mean they are different, does 2/2 look like 1?
Colonel J
Member
612 people, and counting, are retarded.
Kodoku
Senior Member
Quote:
What I was trying to say is that if you use basic mathematical operations with infinite numbers, you get impossible results.
Operations with infinite numbers do indeed cause problems - if you extend the ordinary numbers (reals) to include infinity, you break a lot. The point is that 0.99... does not require infinitely large numbers at all.
Quote:
If what you are saying is that portions of rational numbers are not possible without infinity concepts, then I see your point.
What I was saying is that the same concepts used to define 0.99... are used to define the set of real numbers. 0.99... is well defined if and only if real numbers in general are well defined.
Quote:
However, the algebraic proof I was attempting to refute cannot work without rational numbers. Whenever you use an "X = Y" statement, for the rest of that proof, X can never equal anything other than Y. What I believe the proof is trying to do is say that "X = Y, and Y = Z, therefore, X = Z". I would go with this more than I would go with the way it was formatted.
A statement "x = y" means that the object on the left is the same as the object on the right. The initial step states that x is being used to denote the same object denoted by 0.99...
What the proof shows, in essence, is that x = 0.99... implies x = 1. From this it follows that 0.99... = 1, by the transitivity of equality. The only implicit assumption here is that 0.99... is a real number (whatever real number it may be), as that's required to use ordinary operations on it.
Quote:
If the definitions of X and Y were the same, then I would agree with this whole thing. But as far as I know, the definitions of .999~ and 1 are NOT the same... Am I wrong?
The definitions don't need to be the same. The definitions of 2/2 and 1 are different, yet they denote the same object. You can define the same things in many ways.
Quote:
My understanding of a need for .999~ was because 1 is not evenly divisible by 3.
There's actually no need for .99... at all. In fact, to avoid confusion, some textbooks explicitly restrict decimal expansions to those not involving repeating 9s, since otherwise you can get two decimal expansions that denote the same number (e.g. 1.00... and 0.99...).
Quote:
It's like a language...there's no TRUE translation for 1/3 in the decimal number system, since the number isn't able to be represented by decimals for its true value without using infinity.
There is, in fact, a true translation in decimal form for 1/3. A decimal expansion is defined by assigning a digit to each slot in the decimal expansion. There is a slot for every natural number.
i.e. 1 corresponds to the first decimal place in the expansion.
n corresponds to the nth decimal place.
By assigning 3 to every natural number n, and correspondingly to every decimal place, you've defined a number in the decimal system.
Quote:
That being said, I assume that for all intents and purposes, when someone says, ".999~ is the same as 1" it's no different than someone saying, "agape love is the same as unconditional love", even though the meanings are always going to have some infinitesimal difference, but the overwhelming majority of the meanings are so close that it's pointless to discuss the difference between the two.
You can't do that in mathematics. If you say x = y when there is a difference, no matter how trivial or insignificant that difference is, you'll blow your mathematics up. You'll literally be able to derive anything, such as 1 = 2, or jam = square root of your mom.
Quote:
I think the definitions of the numbers is the key...if it's what I think it is, the problem with people's opinions is that they're trying to represent the same number in two different numeric systems. Is this the point everyone has been trying to make?
The proof shows that the two definitions (of 0.99... and 1) denote the same number. It's not just different systems though. After all, you can define 1 in decimal notation in another way: 1.00...
These are both infinite decimal expansions.
Quote:
612 people, and counting, are retarded.
612 + every single mathematician in the world.
Eledhan
Senior Member
Quote:
Originally Posted by Kaolla
wrong, because motion in reality is not dependent on calculating half-distances. and even if you DID have to, and you wanted to move 10 ft away, there always exists a distance 20ft to where moving HALF that distance will get you 10ft away in one calculated step. so really it's a THOUGHT exercise which CORRECTLY illustrates that infinite calculations DO NOT reach the correct answer, they merely approximate.
it does not, the long division of 1 divided by 9 proves that it does not
See, this is exactly what I was referring to...
If the definition of 1/9 is just that...1 step of the total of 9 needed to reach your goal, you would never be able to represent this 1 step in a decimal numeric system. Hence the abbreviation of it to .111~. If you add up all the .111~ 9 times, you'd never actually get to 1 using the decimal system. However, the representation .111~ is the closest possible representation to the value of 1/9, meaning that although it isn't a perfect translation, it is a translation that means close enough to the same thing to be treated as if it actually were the same thing.
Eledhan
Senior Member
Quote:
Originally Posted by Kodoku
You can't do that in mathematics. If you say x = y when there is a difference, no matter how trivial or insignificant that difference is, you'll blow your mathematics up. You'll literally be able to derive anything, such as 1 = 2, or jam = square root of your mom.
(I cut out the other stuff for brevity)
If you can't say that there is a difference between x and y if they are the same, then why bother with representing .999~ and 1.000~ at all if they are the same number? It seems to be an exercise in futility.
Why not just call it one or the other? Why was there ever a representation of .999~ if 1.000~ would suffice? I have always thought the reason was because there really IS a difference between the two...
However, based on what I've read here, I'm not so sure that it isn't any difference from my using "Eledhan" in exchange for my real name...I'm the same person, just referenced by a different name in different settings.
Quote:
The proof shows that the two definitions (of 0.99... and 1) denote the same number. It's not just different systems though. After all, you can define 1 in decimal notation in another way: 1.00...
These are both infinite decimal expansions.
Yes, both numbers represent one whole "numerical unit". In essence, if it were a pie, you could say you have a whole pie (1/1 or 1.000~), or three thirds of a pie (3/3 or .999~). They all mean the same thing...
Is this the point of the whole topic? | 3,674 | 15,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2014-10 | longest | en | 0.960768 |
http://openstudy.com/updates/5107b136e4b08a15e7845bac | 1,448,425,950,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398444228.5/warc/CC-MAIN-20151124205404-00063-ip-10-71-132-137.ec2.internal.warc.gz | 171,522,368 | 10,271 | ## Yacoub1993 2 years ago Write a rule for the linear function in the graph. A. y = -4x + 13 B. y = -4x – 13 C. y = 4x – 13 D. y = 1/4x – 13
1. Yacoub1993
2. Yacoub1993
anyone to help
3. saifoo.khan
4. stgreen
C. first find slope using m=(y2-y1)/(x2-x1) then write eq using y-y1=m(x-x1) and simplify
5. Yacoub1993
ok | 141 | 324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2015-48 | longest | en | 0.627255 |
http://www.qacollections.com/How-to-Control-Buoyancy | 1,524,154,737,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125936981.24/warc/CC-MAIN-20180419150012-20180419170012-00038.warc.gz | 515,515,518 | 5,641 | # How to Control Buoyancy?
http://www.wikihow.com/Control-Buoyancy
Top Q&A For: How to Control Buoyancy
## What Is Buoyancy Correction?
Small factors can make a large difference in readings where detailed kilogram measurements must be made. Buoyancy is one of those factors, and buoyancy correction refers to the process of taking th... Read More »
## How to Do a Buoyancy Check?
This will show you how to get the right amount of weights for your dive.
http://www.wikihow.com/Do-a-Buoyancy-Check
## How to Calculate Buoyancy?
Buoyancy occurs when an object submerged in a liquid or air floats or rises instead of sinking. To determine buoyancy, some weighing must be done. When an object floats in a liquid or air, it displ... Read More »
http://www.wikihow.com/Calculate-Buoyancy
## How is density related to buoyancy?
The amount of buoyancy an object has is equal to its density multiplied by the object's mass and then multiplied by the acceleration of gravity. This is known as Archimedes' principle.References:Ge... Read More »
Related Questions | 254 | 1,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-17 | latest | en | 0.873821 |
https://www.pipeflow.com/software-technical-support/pipe-flow-expert-vacuum-breaker | 1,726,501,411,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00578.warc.gz | 851,758,154 | 7,145 | ## How do I model a vacuum breaker within a Pipe Flow Expert system?
Modeling a Vacuum Breaker (approximation)
In fluid systems where there is a significant change in elevation there can be conditions where the pressure in the piping systems goes below the fluid vapour pressure. This can occur for various reason however one common scenario is where the fluid is pumped up to a higher elevation and it then returns down to part of the system at a lower elevation.
In this scenario, when flow is returing down to a lower elevation it may be that the pressure needed at the high elevation goes below 0 bar.g for example, because when considering a point in the system at the bottom of a column of fluid there will have been a pressure gain due to the height of fluid now above the lowest point. There may be some friction loss as flow occurs down the pipe but the pressure added due to the weight of fluid above the bottom point in the down pipe can be higher than the friction loss, hence an overall pressure gain can occur. When this is the case the pressure at the higher elevated parts of the piping system may well go negetaive (less than 0 bar.g and below the fluid vapour pressure).
If the pressure goes below the fluid vapour pressure and the fluid in the pipe turns to vapour, the flow in the pipe can become erratic, causing surges, flow reversals and water hammer effects. The low pressure can also cause collapse of the pipes, and rupturing of welds and seams.
To prevent the such scenarios occuring, vaccum breakers are sometimes installed in the pipe system in order to allow air in to the system to stop vacuum conditions arising. When this happens the pipe will no longer be fully filled and pressurized.
The Pipe Flow Expert software does not have a specific vacuum breaker component however it is possible to use a modeling technique with an End Pressure and Flow Demand to simulate the vacuum breaker.
Example System 1 (without the vacuum breaker in operation)
Consider the following system where a pump moves fluid from a low elevation point (0m), up and over a higher elevation point (40m), and on to some discharge location at an elevation point (35m) between the low and high point. The model shows a vaccum breaker arrangement BUT it is not operating (the pipes to the End Pressure and the Demand In-Flow that represent the vacuum breaker have been closed).
i.e. for all intents and purposes this is just a regular system (without the vaccum breaker)
When solved, the system gives the results as shown in the following diagram, where you can see the pressure at node 4 is about 0.218 bar.g (still above the fluid vapour pressure) and there are no specific issues with low pressures in the model.
Example System 2
Consider the same system but now where pipes 9, 10, & 11 have been closed and pipes 14 and 15 have been opened, to allow flow to a lower discharge point.
When solved, the Pipe Flow Expert software shows that now the pressure at the high elevation points is below 0 bar.g. For example, the pressure at node 7 is now -0.66 bar.g.
Example System 3
When unwanted vacuum conditions occur at the higher elevation points it is possible that a vacuum breaker may be introduced in to the system. In Pipe Flow Expert the vaccum breaker can be modelled by using an End Pressure and a Demand In-Flow as is shown in the following diagram. Essentially the system is split in to two separate systems. You can solve the model to find the flow rate in the first system and then this flow rate can be set as the Demand In-Flow at the start of the second system (the vent point).
When solved, the Pipe Flow Expert software now warns that the solution is NOT VALID because a pressure has occured which is less than absolute zero (0 bar.a or -1 bar.g), however it permits the user to look at the calculated results in order to allow diagnosis of issues within the system. From the diagram below you can see that the pressure at node 7 is -1.04 bar.g.
This issue occurs because we have introduced the vacuum breaker which means the pressure at node 5 is now 0 bar.g and we have calculated the flow rate that occurs to this point (with 0 bar.g pressure there) and then we have set the same flow rate as the Demand In-Flow at node 14, BUT the Pipe Flow Expert software has calculated that the pressure required at node 14 to produce this flow rate would need to be a very low (-1.04 bar.g - which is impossible).
In the Real World
In the real world the above results could never occur because now that the vacuum breaker has been introduced in to the system the pipe going down from node 14 would no longer be fully charged and fluid would free fall down the pipe to some point where the fluid fills the pipe back to its fully charged operating condition.
The elevation at which this occurs will be the point where the flow rate down from that point on can be achieved with a fully charged pipe and a pressure of 0 bar.g at that point (the vent condition). The previous impossible results showed a pressure of -1.04 bar.g at node 14 which means we need to lose at least 10m of water head pressure to get to the 0 bar.g condition.
i.e. this means we can move at least 10m down the pipe (to lose 10m of head pressure with some air in the pipe occuring down to this point)
In addition if extra pressure has not been lost to friction with the free fall flow down the pipe (we still have 0 bar.g pressure at some point farther down the pipe) then it is likley that the pipe will become fully charged at a point more than 10m down from node 14.
With some iteration, changing the elevation of node 6 and adjusting the other properties of pipe 7 to match, it is possible to find a solution where the Demand In-Flow set at 109.095 m^3/hour to match the flow rate produced in the first system, occurs with a pressure of approximately 0 bar.g at node 6. This is the elevation that the pipe would become full of fluid and fully charged again and it occurs at an elevation of about 26.5m as shown in the following results.
Summary
This above analysis makes some important assumptions about the flow that occurs in the model which are different than how flow will actually occur in the real world, however this approach will allow the engineer to obtain an approximate solution to the flows and pressures in their system where it contains a vacuum breaker.
In the real world, additonal head loss caused by two-phase flow between the vacuum breaker and the elevation at which the down pipe becomes fully charged and filled with fluid will occur, and this is neglected and not accounted for in the model above. If a more accurate solution is required then the engineer will need to find other software that can calculate for systems with two-phase flow (the Pipe Flow Expert software does not calculate for two-phase flow).
Also, the capacity of the vacuum breaker can affect the value of the vacuum that is obtained and this can affect the pressure and flows in the model.
The above information is provided as a suggested method to allow an engineer to produce a piping model that approximates the results for a system that contains a vacuum breaker, however it is up to the engineer to determine the suitability of applying such an approach to their model and to take account of the assumptions that are made with this approach and to assess and decide whether the results of such a model are accurate enough for the purpose that they will be used.
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Pipe Flow Software: Piping design, Pressure drop calculator, Flow rate calculator, Pump head calculations, Pump selection software. Copyright © Pipe Flow Software 1997-2024 | 1,618 | 7,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.913111 |
http://mathssandpit.co.uk/blog/?cat=99 | 1,726,355,886,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651601.85/warc/CC-MAIN-20240914225323-20240915015323-00293.warc.gz | 17,035,323 | 20,578 | # 363. A-level Exam misconceptions 2022
It’s been a while, but I’m back. Crazy times and all that!
Today I’m sharing a presentation about my thoughts on the Edexcel A-Level Maths papers, from the perspective of reviewing students papers. As a KS5 Co-ordinator I am asked by students to look at borderline papers before they send them off for a paper review.
The mark schemes were very clear on where marks should (or should not) be awarded. This presentation (or set of posters) highlights the most common student errors I spotted during my reviews. I would also say that these are most frustrating issues as they are so easy to fix. Unfortunately it highlights the lack of formal external exam experience this cohort had, through no fault of their own.
These resources are geared towards the Edexcel papers, but I’m sure the skills are equally appropriate for other boards. Also a hat-tip to Jack Brown & TLMaths as I have linked one of the misconception slides to his video on hidden quadratic equations (thank you!).
Exam misconceptions 2022 (PPT editable)
Exam misconceptions 2022 (PDF)
Personally, I’m going to print these out and put them in my A-level display corner. I might use the actual presentation after the Y13 mocks to see if they’ve fallen for the same issues. I hope not!
# 357. It’s not square!
I do love a little challenge for A-level Further Maths students. They are often confident and very capable mathematicians, but occasionally overlook the small details. This challenge looks into which strategies students use when working with 3D vectors, lines and angles.
The most annoying thing? There is no single correct answer.
What is the investigation?
Students start with two points, create a line, construct two perpendicular lines and then join up the lines – did they create a square? How do you know? Justify it?
Download the instructions here: It’s not square (docx), It’s not square (PDF)
Skills required
• Distance between two points
• Equation of a line in three dimensions
• Scalar (dot) product
Solution/Discussion point
• Students need to use the same direction vector for both perpendicular lines too create a square
• The two new corners need to be n the same direction away from the original line (not one above and one below)
• It’s interesting to discuss what non-squares they made. Technology could be used to plot them in 3D.
# 356. Edexcel Shadow Paper
Wow, it’s been a while since my last post. Apologies for that. I’ve been busy with Key Stage 5 things. One of my projects has been creating a shadow paper for the Edexcel AS Maths exam. With so few past papers available and so many papers available online, I wanted an assessment that my students couldn’t find the mark scheme for.
I’ve taken the AS Pure 2018 paper and created a shadow paper, with markscheme. Same level of difficulty, different numbers. I publicised it on Twitter and shared it with over ninety educators in 48 hours. I was stunned by the popularity of this resource. To keep it secure, the lovely Graham Cummings from @mathsemporium has arranged for it to be uploaded onto the Edexcel Maths Emporium. Now I don’t have to directly email people the files.
You can access it with an Edexcel teacher login here. If you don’t have a login, there are instructions on the page on how to obtain one.
I hope this paper saves you some time. I intend to start work on more Pure shadow papers soon, as Pure maths carries the heavier weighting in the AS and A-level exams.
# 353. Large Data Display
If you teach A-level Maths in the UK, you will know about the prerequisite to know about the large data set for the statistics component. We use Edexcel and so need to know about eight weather locations.
Here is my Key Stage 5 corridor wall display.
I’ve got two maps – one of the World ( a freebie from the Humanities Dept) and one of the UK (£2.95 from Amazon).
I’ve included summary information from the CrashMaths booklet.
Of course, you can’t talk about UK weather data from the storm of 1987 – Michael Fish makes a special appearance.
# 352. Functions refresher
We recently finished teaching the AS Maths syllabus to Year 12. My colleague and I decided how to split up the start of the second year of the course. I’m starting with the modulus function.
I took one look at the skills needed at thought “Uh-oh”. The students are going to be out of practice with this. They are a lovely group, with a wide range of ability, but we’ve been very focussed on Applied Maths recently.
Option A: Go for it and patch up the vocabulary as we go (getting very frustrated – they knew this last October)
Option B: Break them in gently, recap the skills and vocabulary and extend them further
Option C: Reteach the work from last October.
Yes, you guessed it. I went with Option C. I found a brilliant task on piecewise function graphs on the Underground Maths website.
Image credit: https://undergroundmathematics.org/
There are four graphs given. The basic task is to interpret the functions relating to each graph, through description or function.
I photocopied the graphs onto card and sliced them up. Each group had a set of cards. One person described a graph and the others had to accurately draw it. Some students went straight onto squared paper, others drafted it out on mini whiteboards. They repeated this until all the graphs were drawn and everyone had had a go at describing (the describer stuck in their card, so that they had a complete set). Whilst they were doing this, I moved around and encouraged the use of mathematical vocabulary.
Note: it was interesting to see how many students had forgotten the significance of open and shaded circles to denote boundaries of inequalities.
The second task was to match up the function cards with the graphs. Once again, accuracy was key as not all graphs had functions and not all functions had graphs. There were also some that nearly, but not quite matched. This activity really brought out the key skills relating to domain, range and function notation that I was looking for. The extension task was to complete the missing pairs.
But, did it work? I can confirm that the following lesson the class made very good progress investing the modulus function and it’s graph, even going as far to solve equations. They knew what the notation meant, how to plot it and how to interpret the graphs.
I really like the Underground Maths website as it has great resources, good support material and always makes students think. Most of the time it gets teachers thinking too!
# 349. Circumcircle Investigation
The A-level textbook we use has a nice picture of the circumcircle of a triangle and a definition, plus a brief description of how to work through them. For those who are pondering what a circumcircle is, click on the image or link below
Image credit: WolframMathWorld
I’ll just stick to basic vocabulary in this post, rather than the formal circumcentre and circumradius.
Back to the book – not exactly inspiring or memorable stuff!
I looked at the class and off the cuff changed the lesson plan.
Equipment
• Plain paper
• Pencil
• Ruler
• Compasses
• Calculator
Step 1
Draw a decent size triangle on the paper. Label the corners A,B,C.
Step 2
Using geometrical constructions, find the centre of the circle that your triangle fits in. Check by actually drawing the circle
Step 3
Discuss what techniques gave the best results – hopefully you’ll have perpendicular bisectors. There is a nice comparison between bisecting the angles (which some students will do) and bisecting the sides. The angle bisectors always cross inside the triangle, the side bisectors don’t.
Step 4
Randomly generate co-ordinates for A, B, & C. Get the students to pick them and then they can’t moan if the calculations are awful.
Step 5
Discuss how you are going to find the centre and radius of the circumcircle. We decided on:
• Only use two sides
• Find the midpoints
• Find the gradients and hence perpendicular gradients
• Generate the equations of the lines through the midpoint
• Find where they intersect
• Use the point and one corner to find the radius
Step 6
Review their methods, looking for premature rounding in questions. I’m still instilling an appreciation for the accuracy of fractions and surds, over reaching for the calculator.
Step 7
This is how my solution looked – I numbered the picture and the steps so students could follow the logic. I was answering on one page projected on screen.
# 348. A-Level colouring (Updated)
Those of you who follow this blog will know I have a thing for explaining with colours. This isn’t just a gimmick for younger students, it also works for 16-18 year olds.
In the picture below we were looking at proving a statement involving reciprocal trigonometric functions and fractions. A common source of misconception with this kind of question is that students split the question into working with the numerator and denominator separately, then make mistakes when they put them back together. They can’t see the big picture.
Image credit: Mathssandpit
When I discussed this on the board I used separate colours for the expressions in the numerator and denominator. The class could follow the logic so easily. It’s probably my most successful introduction to this topic. I saw that some students used highlighter on their notes after I’d gone through it, so they could track the solution.
The second type of question we looked at was solving a trigonometric equation. The straight forward expansion was all in one colour, but the roots of the quadratic were highlighted in different colours. The reasoning behind this was that students often solve half the quadratic and neglect the other impossible solution. Our exam board likes to see students consider the other solution and formally reject it. It makes the solution complete. By using a colour, the impossible solution stands out and reminds students to provide a whole solution.
Image credit: Mathssandpit
So when you are planning for misconceptions at A-level, remember that coloured pens aren’t just for younger students.
Update: 22nd October
The brilliant Mr B has shared how he uses colour to identify the forces in perpendicular directions in Mechanics. | 2,178 | 10,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-38 | latest | en | 0.933214 |
https://mail.coreboot.org/pipermail/coreboot/2007-July/022979.html | 1,560,948,745,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998986.11/warc/CC-MAIN-20190619123854-20190619145854-00182.warc.gz | 504,786,630 | 2,171 | # [LinuxBIOS] Another dumb C question
Joseph Smith joe at smittys.pointclark.net
Thu Jul 12 14:34:24 CEST 2007
```Quoting Joseph Smith <joe at smittys.pointclark.net>:
> Quoting Corey Osgood <corey.osgood at gmail.com>:
>
>> Joseph Smith wrote:
>>> Quoting Peter Stuge <peter at stuge.se>:
>>>
>>>
>>>>> value = 9
>>>>>
>>>>> (2 << (value - 1))
>>>>>
>>>>>
>>>
>>>
>>>> Bit shift is fine methinks but I would suggest:
>>>>
>>>> (1 << value)
>>>>
>>>>
>>>> (2 << (value - 1))
>>>>
>>>>
>>>> //Peter
>>>>
>>>>
>>> Not sure I understand. I am trying to get 2 "to the power of" value.
>>>
>>> if value = 9
>>> (1 << value)
>>> This equals 18???? Not 2^9 = 512
>>> where
>>> (2 << (value - 1))
>>> (2 << (9 - 1))
>>> 2 << 8 = 512
>>>
>>> Thanks - Joe
>>>
>>>
>> remember that 2 = (1 << 1). so 2 << (value - 1) = 1 << value. I'm not
>> sure where you're getting an 18 from, that makes no sense at all (since
>> 18 in either hex or dec has two non-zero bits, so it's impossible to get
>> that result with just that binary shift).
>>
>> -Corey
>>
>>
> Well lets see if value = 9, which is 1001 in binary.
> If you bitshift 1001 one position to the left it becomes 10010, correct?
> Which is 18 in decimal.
>
> Thanks - Joe
>
Nevermind I get it. I was looking at it bass ackwards :-)
Sorry, I'm a little tired.
Thanks - Joe
``` | 442 | 1,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-26 | longest | en | 0.870938 |
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## Software => Software => Topic started by: Admin on June 09, 2008, 01:24:23 PM
Title: tips to speed up processor intensive algorithm
Post by: Admin on June 09, 2008, 01:24:23 PM
This is where my mechE skills fail . . . programming . . .
I'm programming my first real-time processor intensive algorithm, and I'm definitely pushing the limits . . . A solution needs to be found measured in milliseconds, or large PID error will result . . .
Anyway, my algorithm involves high precision trig (ruling out small lookup tables), doubles, long signed ints, multiplication/division of the before mentioned, and large arrays storing the results. Here are just a few examples of a much more complex calculation:
Code: [Select]
`unsigned int Rib_1_Left_Adapt[9];int amp_avg_F=11;int amp_avg_R=15;int amp_avg_U=5;int amp_avg_D=8;signed long int theta;//some angle in degreesdouble weightA;//some value below 1double weightB;//once per loopweightA=(amp_avg_U/amp_avg_F)/(tan(theta)+(amp_avg_U/amp_avg_F));weightB=tan(theta)/(tan(theta)+(amp_avg_U/amp_avg_F));//10 equations like this per loopRib_1_Left_Adapt[time_position]= Rib_1_Left_F[time_position]*weightA + Rib_1_Left_U[time_position]*weightB;`
So my question . . . are there advanced programming techniques to speed these things up? Anything that I could trip up on? Is there a way to predict the total processing time given a particular math equation and a given processor? Know of tutorials to refer me to?
I probably got my variable types wrong, haven't paid much attention to that yet, and I always get confused using mixed type equations . . . if it looks wrong let me know . . .
Title: Re: tips to speed up processor intensive algorithm
Post by: hgordon on June 09, 2008, 02:38:25 PM
What precision do you need, and on what processor does this need to run ?
Title: Re: tips to speed up processor intensive algorithm
Post by: Admin on June 09, 2008, 02:50:46 PM
(oops, forgot to mention that)
It's running on my Axon (but with an ATmega2560) at 16MHz.
Precision for calculating 'weightA' and 'weightB' should be at worst +/- .025 in the final result.
That bottom equation with the arrays can have an error more like +/- 10.
Title: Re: tips to speed up processor intensive algorithm
Post by: benji on June 09, 2008, 02:59:33 PM
i think this totally depends on the compiler and the already made functions ,
there are multiple algorithms to do a specific function,
the important thing in processors are just their clocking , 16 mhz would do it twice faster than 8mhz.
thats because the processor only understands machine language
there are some DSPs that lets you run 2 programs at the same time,, that can add up a lot too.
so the thing is maybe its more a software thing than a hardware thing.
so if your C compiler is efficient enough to give optimal assembly (or machine code) then this helps.
another thing is about the programmer himself,the better the algorithm doesnt always means the more precise, somtimes you need fast not very acurate answers,, or here comes the tricks with programming
example:
one day i needed to use the SINE function in a program
in debugging i noticed that this function takes about 2ms, and that was too long for the application
so i did a lookup table with the angles i have (they were about 12 angle ) so this made the algorithm run much faster
somtimes there are already built functions in a high level language made for general purpose, a good programmer should do another according to his program needs (ex: make it faster with some sacrifices maybe like loosing addition of double kind variables)
so if you program in C and you use the compiler or a library already made functions, try to modify them.
Title: Re: tips to speed up processor intensive algorithm
Post by: hgordon on June 09, 2008, 05:10:58 PM
I never use floating point libraries or hardware, but do use lookup tables and prescale my integer calculations in a way which is similar to floating point math by prescaling values and making certain that I don't end up with overflows. I would have to see the range of actual numbers you expect to process to suggest exactly how to structure your equations, but as an example, if I needed 1% accuracy on a calculation that had a expected result around 1.00, I would scale things up by 1000 (or 1024).
Here's an example of a calculation used in some neural net code, where all of the values are scaled to a range of 0-1024 instead of 0.0-1.0. In fact, the integer divides by 1024 could instead be performed with a right shift by 10.
E_HIDDEN(h) = (((N_HIDDEN(h) * (1024 - N_HIDDEN(h))) / 1024) * err) / 1024;
By knowing in advance the expected range of inputs and outputs, I can dial in the necessary accuracy with some very fast code.
Title: Re: tips to speed up processor intensive algorithm
Post by: Asellith on June 11, 2008, 11:33:50 AM
This may be two specific but does theta have a min and max range for the application? then you could have a smaller more detailed lookup table if theta only falls in a specific range.
Also changing this line
weightA=(amp_avg_U/amp_avg_F)/(tan(theta)+(amp_avg_U/amp_avg_F));
weightB=tan(theta)/(tan(theta)+(amp_avg_U/amp_avg_F));
to:
denominator = (tan(theta)+(amp_avg_U/amp_avg_F);
weightA=(amp_avg_U/amp_avg_F)/denominator);
weightB=tan(theta)/denominator);
may speed up things a pit. That way it does less calculation. Just have to watch for truncation error on the denominator.
The only way I know to speed math calculations up other then developing a better algorithm is to bust out the machine language and code in that. Have fun with that because it could take a while to code and you might not speed it up all that much. Just lets you control the math functions better to streamline them to your algorithm.
Title: Re: tips to speed up processor intensive algorithm
Post by: Webbot on June 11, 2008, 07:10:36 PM
I would go a bit further than Asellith as there are still some values you are calculating more than once. If your compiler isn't a good optimiser, or you haven't enabled optimisation, then it will generate code to do the same calculations many times:
double tanTheta = tan(theta);
double uDivF = amp_avg_U/amp_avg_F;
double denominator = tanTheta +uDivF ;
weightA=uDivF /denominator);
weightB=tanTheta /denominator);
If the 'tan' function is taking up too much time and theta doesn't change every time this code is run then you could try having a global variable to store the old value. ie
// global variables initialised to a known value
double lastTheta = 0.5;
double lastTanTheta = tan(lastTheta);
double tanTheta;
if( theta == lastTheta ){
// its the same as last time - so no need to recalculate
tanTheta = lastTanTheta;
}else{
// its changed - so recalculate - and store again for next time
lastTanTheta = tanTheta = tan(theta);
lastTheta = theta;
}
Title: Re: tips to speed up processor intensive algorithm
Post by: Admin on June 12, 2008, 07:44:32 AM
Quote
then you could have a smaller more detailed lookup table if theta only falls in a specific range.
The range is from 180 degrees to -180 degrees. Forces me to use a rather large lookup table . . .
Quote
I would scale things up by 1000
Yea this is my thought too.
As such, this brings up another question . . . so I have tan(theta), but I assume theta must be in radians for it to calculate properly, and the solution will also be in radians, right? Doesn't this force me to use floating point? tan(1.5) != tan(1500) :P
Anyway, starting over from scratch I found a simpler yet more accurate/fast algorithm to solve the same problem. It uses a lot of constants and a lot less trig:
Code: [Select]
`weightA=(Lu-Tu*tan(theta))/((Tf-Tu)*tan(theta)-Lf+Lu);weightB=1-weightA;`
Title: Re: tips to speed up processor intensive algorithm
Post by: pomprocker on June 12, 2008, 10:58:22 AM
you guys blow me away with all this math. I wish I could keep up.
I'm actually in Linear Algebra right now, but still I have no clue when it comes to applying everything to the real world.
Title: Re: tips to speed up processor intensive algorithm
Post by: hgordon on June 12, 2008, 11:53:13 AM
As such, this brings up another question . . . so I have tan(theta), but I assume theta must be in radians for it to calculate properly, and the solution will also be in radians, right? Doesn't this force me to use floating point? tan(1.5) != tan(1500) :P
No - you just prescale the 1.5 to an integer value. If you are using radians instead of degrees, your range is 0 to 6.28 radians, so you could multiply the radians by 100 to create 628 table entries. Or you could note that there is redundant information in the table for 0 to 157, 158 to 314, 315 to 472, and 473 to 627, and set up the calculation based on your quadrant.
Title: Re: tips to speed up processor intensive algorithm
Post by: JonHylands on June 12, 2008, 02:39:39 PM
Use a lookup table. They (if done properly) end up in FLASH, and you've got 256 KB of the stuff.
- Jon
Title: Re: tips to speed up processor intensive algorithm
Post by: Admin on June 13, 2008, 07:09:29 AM
Quote
Quote
As such, this brings up another question . . . so I have tan(theta), but I assume theta must be in radians for it to calculate properly, and the solution will also be in radians, right? Doesn't this force me to use floating point? tan(1.5) != tan(1500)
No - you just prescale the 1.5 to an integer value. If you are using radians instead of degrees, your range is 0 to 6.28 radians, so you could multiply the radians by 100 to create 628 table entries. Or you could note that there is redundant information in the table for 0 to 157, 158 to 314, 315 to 472, and 473 to 627, and set up the calculation based on your quadrant
Oh, I meant if I was to *not* use a lookup table.
Quote
Use a lookup table. They (if done properly) end up in FLASH, and you've got 256 KB of the stuff.
I think I got my algorithm working fairly fast now . . . but mostly I'm avoiding the lookup table from a combination of laziness and the fact I'm still in a prototyping stage (I change the code very often).
I already know all these basic tricks of lookup tables, scaling by 100, compiler optimization, using assembly, and storing common calculations as a stored variable. In fact, the forum is has dozens of posts where I suggest this to other people already . . . heck I wrote the lookup table tutorial years ago! :P
Specifically, I wanted to optimize in terms of order of operations, whether dividing or multiplying is faster, bit operation efficiency, etc. I was looking for advanced techniques, stuff they teach only in advanced CS classes, etc.
I assume there are tutorials on this, but I really don't know where to look . . .
Title: Re: tips to speed up processor intensive algorithm
Post by: krich on June 13, 2008, 10:30:19 AM
I bumped into this recently. (http://www.research.scea.com/research/pdfs/RGREENfastermath_GDC02.pdf) Looks like Tangent is twice as evil as Sine or Cosine with respect to CPU utilization( tan(x)=sin(x)/cos(x) ). I only skimmed it, but there's some really good tips in here, if you can get through the fairly advanced math.
One possibility is to generate the Tangent lookup table at startup and just store it somewhere. This way you programatically decide your accuracy.
Regarding your question about tan(1.5) != tan(1500), no matter what the value you pass to the tan() function, the number is passed into the function as a floating point number, so you're still using the FP libraries. No way around it unless you write your own tan() function that is expecting to be passed a long integer. Even then, I'm not sure that the algorithm required to generate the tangent will work with a value that large or come up with the right answer. Perhaps instead of multiplying by a factor of 10's, multiply by a factor of 6.28's. That'll probably get you the right answer, but I'm not sure if that'll get you anything on the CPU utilization side of things.
EDIT: Here's something that might be useful. (http://www.myreckonings.com/Dead_Reckoning/Online/Materials/Fast_Approximation_of_Elementary_Functions.pdf) It's targeted towards those who want to learn how to do Tangent's in their head, but it looks like it deals mostly with integer math.
Title: Re: tips to speed up processor intensive algorithm
Post by: Admin on June 13, 2008, 12:17:04 PM
krich this is exactly what I was looking for, thanks!
Its quite a lot to absorb though, and I've only skimmed through it quickly so far. It'll probably take me weeks to assimilate all the new ideas . . . someone should write a tutorial :P | 3,153 | 12,622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-21 | latest | en | 0.869244 |
https://questions.examside.com/past-years/gate/gate-ce/strength-of-materials-or-solid-mechanics/columns-and-struts/ | 1,670,109,466,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710941.43/warc/CC-MAIN-20221203212026-20221204002026-00485.warc.gz | 508,630,698 | 9,841 | NEW
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Consider two axially loaded columns, namely. $$1$$ and $$2,$$ made of a linear elastic material with Young's modulus $$2... GATE CE 2017 Set 1 If the following equation establishes equilibrium in slightly bent position, the mid-span deflection of a member shown i... GATE CE 2014 Set 1 The sketch shows a column with a pin at the base and rollers at the top. It is subjected to an axial force$$P$$and a m... GATE CE 2012 Cross-section of a column consisting of two steel strips, each of thickness$$t$$and width$$b$$is shown in the figure... GATE CE 2008 A rigid bar$$GH$$of length$$L$$is supported by a hinge and a spring of stiffness$$K$$as shown in the figure below.... GATE CE 2008 A steel column, pinned at both ends, has a buckling load of$$200kN.$$If the column is restrained against lateral ... GATE CE 2007 The buckling load$$P = {P_{cr}}$$for the column$$AB$$in figure, as$${K_T}$$approaches infinity, becomes$$\alpha {...
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Class 12 | 696 | 2,598 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.821763 |
http://www.tf.uni-kiel.de/matwis/amat/def_en/kap_2/exercise/s2_1_1.html | 1,539,784,136,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511175.9/warc/CC-MAIN-20181017132258-20181017153758-00293.warc.gz | 564,392,647 | 1,811 | # Solution to Exercise 2.1-1"Find the Mistake"
The binomial coefficient in Hayes and Stoneham is written as
PL = N!
(N – n)!
The correct formula, of course, is
PL = N!
(N – n)! · n!
This is obviously a typo, otherwise one set of brackets would not have been necessary.
The final result is correct anyway, because the next equation (3.5) contains the complete formula.
Find the Mistake
Exercise 2.1-7 Quick Questions
Solution to Exercise 1.1
Exercise 2.1-8 Quick Questions
Exercise 2.2-2 Quick Questions
© H. Föll (Defects - Script) | 161 | 546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-43 | latest | en | 0.873517 |
http://www.onlinemath4all.com/midpoint-calculator.html | 1,511,127,594,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805809.59/warc/CC-MAIN-20171119210640-20171119230640-00190.warc.gz | 460,416,853 | 6,583 | # MIDPOINT CALCULATOR
Midpoint calculator is an effective online tool which is much useful for students who want to check their answers of the question where they have found midpoint between two given points.
## Why do people need this calculator?
Apart from the regular calculator, people who study math are in need of midpoint calculator. Because, when people work out lengthy problems, they may not have time to find midpoint between two given points.At that time, finding midpoint between two points would be an additional burden for them in solving lengthy math problems. To reduce the burden of those people, we have provided this online calculator which can be used to find the midpoint between two given points.
## Using this calculator in solving word problems
When people do preparation of solving word problems on geometry, they may have to spend time to get idea on "How to solve". In this situation, they would not like to spend time to find the midpoint between two points. And also they would not be able to use the regular calculators to find midpoint between two points. By using this online calculator, students will find much time to get idea of solving the word problems.
## Advantages of using this calculator
School students have the topic "Analytical Geometry" in math in their 7th grade or 8th grade. They are given the questions like "Find midpoint between two points". They solve the given problem, but they are not sure whether the answer they got is correct or incorrect. At that time, they can check the answer they have received is correct or incorrect using this online calculator.
In geometry word problems, students have to do lots of operations in single problem to get the answer. In case the answer they got is not correct, they will check all the operations one by one. Sometimes, students might have done mistakes in finding midpoint between two points. In that situation, students can use this calculator and come to know whether they had done any mistake in finding midpoint between two points. The main advantage of this calculator is, as soon as we enter the two points, it will give you the midpoint in seconds.
Midpoint Formula
Midpoint Calculator
Enter Co-ordinates (X1,Y1)= (,) Enter Co-ordinates (X2,Y2)= (,)
Result: Mid Point
Online Math Calculators | 466 | 2,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-47 | latest | en | 0.972947 |
https://www.mentorforbankexams.com/2020/10/quantitative-aptitude-set-2-ibps-clerk.html | 1,603,371,498,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879537.28/warc/CC-MAIN-20201022111909-20201022141909-00188.warc.gz | 811,540,108 | 53,762 | ## Quantitative Aptitude Set - 2 || IBPS Clerk Prelims 2020
1. Two cards are drawn at random from a pack of 52 cards, then find the probability of getting one red face card and one black ace?
a) 1/221
b) 2/221
c) 76/221
d) 91/221
e) 5/221
2. If side of a square is equal to height of equilateral triangle, then find ratio of area of equilateral triangle to area of square?
a) √2 :5
b) 3 :5
c) √3 :2
d) 1 : √3
e) Data insufficient.
3. Deepak invested some amount on SI out of Rs.47000 and rest amount on C.I. for two years. If S.I. is offering 12% p.a. and C. I. is offering 15% p.a. compounding annually and C.I. is Rs.532.5 more than S.I., then find amount invested by Deepak on C.I?
a) Rs.23000
b) Rs.22000
c) Rs.21000
d) Rs.25000
e) Rs.24000
4. B is twice as old as A. Average of present age of A and B is 24 years and average of present age of B and C is 38 years. Find present age of C is what percent less than present age of A and B together?
a) 4 2/9%
b) 11 6/11%
c) 5 1/5%
d) 13 2/7%
e) 8 1/3%
5. Vessel-A and Vessel-B contains mixture of milk and water in the ratio of 2 : 3 and 5 : 3 respectively. When 50% mixture from Vessel-A and 40% mixture from Vessel-B taken out and mixed together, then the resulting mixture contains 36 liters of water and 36 liters of milk. Find ratio of quantity of water in Vessel-A to quantity of water in Vessel-B?
a) 8 : 5
b) 1 : 1
c) 2 : 3
d) 5 : 7
e) 9 : 5
6. When we reverse the digits of a two-digit number, we got a new number. Product of new number with sum of its digits is 52 and digit at ten’s place of the original number is 2 more than the digit at unit’s place of original number. Find original number?
a) 39
b) 33
c) 41
d) 31
e) 37
7. A & B working together can complete a piece of work in 72 days, B & C working together can complete the same work in 90 days and A & C working together can complete the same work in 75 days. If A, B & C starts working together, then find in how many days will the same work be completed?
a) 55 7/23 days
b) 52 4/23 days
c) 59 2/23 days
d) 49 11/23 days
e) 50 17/23 days
8. Cost price of an article is 39% less than the marked price of the article and shopkeeper earned 40% profit in selling the article. If amount of profit is Rs.196 more than amount of discount, then find cost price of article.
a) Rs.1345
b) Rs.1325
c) Rs.1290
d) Rs.1245
e) Rs.1220
9. Time taken by boat to cover 48 km in upstream is 200% of the time taken by boat to cover 48 km in downstream. If sum of time taken by boat to cover 48 km distance in upstream and same distance in downstream is 9 hours, then find speed of boat?
a) 8 km/hr
b) 10 km/hr
c) 12 km/hr
d) 4 km/hr
e) 6 km/hr
10. Pankaj and Suresh invested capital in the ratio of 4 : 3 in a partnership business. Pankaj and Suresh left the business after 5 months and after 10 months respectively. Raju joined the business with Rs.50000 when Suresh left, and they wind up the business after 1 year. If at the end of one-year profit sharing ratio of Pankaj, Suresh & Raju is 2 : 3 : 1, then find the difference between the capital invested by Pankaj and Suresh?
a) Rs.10000
b) Rs.40000
c) Rs.30000
d) Rs.50000
e) Rs.20000
Watch the Video Solution Here: | 1,049 | 3,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-45 | longest | en | 0.893026 |
https://www.scribd.com/document/130702549/Clipping-and-Clamping-Circuits-pdf | 1,503,284,545,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107487.10/warc/CC-MAIN-20170821022354-20170821042354-00588.warc.gz | 960,927,293 | 25,692 | # Electronics Laboratory
Experiment No. 3 Clipping and Clamping Circuits
Object: To steady the diode applications in a clipping and clamping circuits.
Apparatus: 1. Function Generator. 2. Oscilloscope. 3. DC Power Supply. 4. Breadboard, Diodes, Capacitors and Resistor.
Theory: This experiment studies the applications of the diode in the clipping & clamping operations.
1. Clipping Circuits: the Figure (l) shows a biased clipper, for the diode to turn in the input voltage must be greater +V, when Vm is greater than +V , the diode acts like a closed switch (ideally) & the voltage across the output equals +V , this output stays at +V as long as the input voltage exceeds +V. when the input voltage is less than +V , the diode opens and the circuit acts as a voltage divider, as usual ,
RL
should be much greater than R, in this way , most of
input voltage appears across the output. The output waveforms of Figure (1) summarize the circuit action. The biased clipper removes all signals above the (+V) level.
2. Clamping Circuits: A clamper does is adding a DC component to the signal. In Figure (2) the input signal is a sinewave, the clamper pushes the signal upward, so that the negative peaks fall on the 0V level. As can see, the shape of the original signal is preserved, all that happen is a vertical shift of the signal.
We described an output signal for a positive dampen- On the Figure (2) shown
Electronics Laboratory
represents a positive clamper ideally here how it is works. On the first negative half cycle of input voltage, the diode turns on. At the negative peak, the capacitor must charge to Vp with polarity shown. Slightly beyond the negative peak, the diode shunts off.
Procedure: Clipping Circuit: 1. Connect the circuit shown in Figure (3). 2. Ensure that the variable DC is at minimum and the source is at 10VP.P. 3. Observe and Sketch the input and output waveforms. 4. Increase the variable DC voltage to 4V, and notice to what voltage are the positive peaks chopped off, sketch the waveforms.
Clamping Circuit: 1. Connect the circuit shown in Figure (4). 2. Ensure the variable DC is at minimum. 3. Set the sine wave generator frequency to 1KHz and its output amplitude to 10VP.P . 4. Observe and sketch the input waveform with the variable DC at minimum, Sketch the output waveform.
Discussion: 1. What happened if the DC voltage in the clamping circuit is replaced by an a.c source? 2. What is the relationship between the clipping level and the DC voltage? 3. If the variable DC source is reversed, how does this affect the clipping? 4. If the input voltage 10VP.P, sketch the output of the circuit shown below.
10k
D1
D
10k
3V
4V
Electronics Laboratory
1.0k 1KΩ
D
10VP.P 1KHz
10k
2V
Fig. 1 Clipping circuit
10uF
D
10VP.P 1KHz
10k
Fig. 2 Clamping circuit
10k
100nF
D
10k
D
10k
Fig. 3
Fig. 4 | 776 | 2,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-34 | latest | en | 0.86215 |
http://mohnabil.blogspot.com/2012_01_01_archive.html | 1,527,064,996,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865468.19/warc/CC-MAIN-20180523082914-20180523102914-00080.warc.gz | 203,168,245 | 14,919 | ## Tuesday, January 31, 2012
### 11 Bitcoins = 1 \$ !!
By pure coincidence as I was running through the list of stack exchange sites I found a site called Bitcoin. I didn't understand what was it at first but when I looked it up in wikipedia it turned out to be an e-cash P2P system. Wikileaks, freenet, and another company already use it for reception of donations and for investments according to wikipedia. It even has an exchange rate to US dollars. The original paper is http://bitcoin.org/bitcoin.pdf and there are other more recent papers.
#WOW
## Friday, January 20, 2012
### Proof that np=mq and mn>0 implies pq>0
Implicit assumptions: q != 0 (p/q is a rational number).
This is a proof that mn>0 as a definition that a rational number m/n is "positive" is a well-defined with respect to equivalence classes of rational numbers. I.e. if p/q is in the same equivalence class of m/n (i.e. np=mq) and m/n is positive (mn>0) then p/q is also positive (pq>0).
First, np=mq => np-mq=0, but since mn>0 (neither m nor n is zero), and that q!=0 (by definition), then p!=0. So one of pq<0 or pq>0 is true. Assume the former towards a contradiction. First let p>0 and q<0 and let q=-q' where q' is positive. Then 0=np-mq=np+mq', but all of these are positive, so they cannot equal zero: contradiction. Same argument for p<0 and q>0, we end up with 0=-np'-mq, but the right hand side is negative: contradiction. So pq<0 cannot be true, therefore pq>0. | 400 | 1,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2018-22 | latest | en | 0.946271 |
http://math.stackexchange.com/questions/262702/spectra-of-operators | 1,419,108,667,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770399.32/warc/CC-MAIN-20141217075250-00042-ip-10-231-17-201.ec2.internal.warc.gz | 180,198,705 | 16,511 | # Spectra of operators
Please help me proof a theorem: If $\mathfrak{U}$ is a complex, commutative Banach algebra with identity and $x\in\mathfrak{U}$, then
$$\sigma(x)=\{\phi(x):\phi \text{ is a homomorphism of } \mathfrak {U} \text{ onto } \mathbb{C}\}$$
-
Then what? How is $\sigma(x)$ defined? – Nils Matthes Dec 20 '12 at 15:55
this theorem is a book invariant subspaces by Heyder Radjavi and Peter Rosenthal – Matema Tika Dec 20 '12 at 16:04
First, some elementary facts about onto homomorphisms $h$. We have $h(0)=0$ and $h(e)=1$. If $x$ is invertible then $\phi(x)\neq 0$.
If $\lambda=\phi(x)$ for such a homomorphism, then $\phi(x-\phi(x)e)=0$ which implies that $x-\phi(x)e$ is not invertible hence $\lambda\in \sigma(x)$.
If $\lambda\neq\phi(x)$ for all $\phi$, we have to show that $x-\lambda e$ is invertible. We have that $x-\lambda e\notin\ker\phi$ for all $\phi$ an onto homomorphism. In Rudin's book Functional analysis, it's shown that each maximal ideal is the kernel of an onto homomorphism and each strict ideal is contained in a maximal ideal. Can you conclude from that?
The result actually says that $\widehat x$, the Gelfand transform of $x$, defined on the set of onto homomorphisms by $\widehat x(\phi):=\phi(x)$, has the spectrum of $x$ a codomain. | 403 | 1,282 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2014-52 | latest | en | 0.836762 |
https://byjus.com/hpbose/hp-board-class-12-result/ | 1,716,185,117,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058222.5/warc/CC-MAIN-20240520045803-20240520075803-00674.warc.gz | 121,618,471 | 110,406 | # HP Board Class 12 Results 2023
The Himachal Pradesh Board of Secondary Education conducts the Class 12 Board exams for all streams viz., Arts, Science and Commerce in affiliated schools and colleges. The HP Board recently announced the Class 12 Board Exam results on 20th May 2023. The HP Board Class 12 Results 2023 are announced simultaneously for all streams.
The board exams for the Himachal Pradesh Class 12 students were conducted from 10th March 2023 to 31st March 2023.The HP Board announced the HPBOSE Class 12 Results for 2023 recently on 20th May 2023. Students can check their results on the Board’s official website, hpbose.org.
## Important Dates for HP Board HPBOSE Class 12 Result 2023
Here is a list of important dates for the Himachal Pradesh Class 12 board result 2023.
Events Dates HP Board Class 12 exam dates 10th March 2023 to 31st March 2023 Dates for HP Board HPBOSE Class 12 results 2023 20th May 2023
## Steps to Check HP Board HPBOSE Class 12 Result 2023
Follow the steps below to check the HP Board HPBOSE Class 12 Result 2023:
The pictures below are from previous years and are used only for reference.
1. Go to the Himachal Board’s official website: https://hpbose.org/Home.aspx
2. Click on the “Results” tab on the top right corner of the page. Check the image below for reference:
3. Then, click on “HPSOS 12th Examination Result 2023”.
4. A new page opens. Enter your roll number as mentioned in the admit card and click on ‘search’.
5. The marks will be displayed on your computer screen.
Students can download and print the marksheet for reference. This printout is only for reference and cannot be used as a replacement for the original marksheet.
## HP Board HPBOSE Class 12 Result 2023 Statistics
Read on to find the statistics of the pass percentage of the HP Board HPBOSE Class 12 results over the past few years (2016-2021).
Year No. of students appeared Pass Percentage 2021 1,00,799 92.70% 2020 86,633 76.07% 2019 95,492 62.01% 2018 98,281 70.18% 2017 1,02,075 72.89% 2016 1,01,104 78.93%
Students can also access the HP Board Class 12 Syllabus from BYJU’S.
## Frequently Asked Questions on HP Class 12 Board Results 2023
Q1
### Where can students view the HP Class 12 Board Results 2023?
The results for the HP Class 12 Board 2023 are available on the board’s official website, https://hpbose.org/Home.aspx.
Q2
### When will the HP Class 12 Board 2023 results be announced?
The results for HP Class 12 Board exams were announced on 20th May 2023.
Q3
### When will the HP Class 12 Board 2023 exams be held?
The board exams for the HP Class 12 Board 2023 were held from 10th March to 31st March 2023. | 719 | 2,671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-22 | latest | en | 0.901096 |
https://www.dsprelated.com/freebooks/pasp/State_Space_Analysis.html | 1,632,588,128,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057687.51/warc/CC-MAIN-20210925142524-20210925172524-00481.warc.gz | 746,516,415 | 9,578 | ### State-Space Analysis
We will now use state-space analysisC.15[449] to determine Equations (C.133-C.136).
From Equations (C.128-C.132),
and
In matrix form, the state time-update can be written
(C.136)
or, in vector notation,
(C.137) (C.138)
where we have introduced an input signal , which sums into the state vector via the (or ) vector . The output signal is defined as the vector times the state vector . Multiple outputs may be defined by choosing to be an matrix.
A basic fact from linear algebra is that the determinant of a matrix is equal to the product of its eigenvalues. As a quick check, we find that the determinant of is
(C.139)
When the eigenvalues of (system poles) are complex, then they must form a complex conjugate pair (since is real), and we have . Therefore, the system is stable if and only if . When making a digital sinusoidal oscillator from the system impulse response, we have , and the system can be said to be marginally stable''. Since an undriven sinusoidal oscillator must not lose energy, and since every lossless state-space system has unit-modulus eigenvalues (consider the modal representation), we expect , which occurs for .
Note that . If we diagonalize this system to obtain , where diag, and is the matrix of eigenvectors of , then we have
where denotes the state vector in these new modal coordinates''. Since is diagonal, the modes are decoupled, and we can write
If this system is to generate a real sampled sinusoid at radian frequency , the eigenvalues and must be of the form
(in either order) where is real, and denotes the sampling interval in seconds.
Thus, we can determine the frequency of oscillation (and verify that the system actually oscillates) by determining the eigenvalues of . Note that, as a prerequisite, it will also be necessary to find two linearly independent eigenvectors of (columns of ).
Next Section:
Eigenstructure
Previous Section:
Digital Waveguide Resonator | 442 | 1,960 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2021-39 | longest | en | 0.877897 |
https://hansvandenpol.nl/x15sgmh/0a1b8a-combinations-with-repeated-elements | 1,726,071,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651390.33/warc/CC-MAIN-20240911152031-20240911182031-00358.warc.gz | 267,446,240 | 8,861 | This is one way, I put in the particular numbers here, but this is a review of the permutations formula, where people say How many combinations are there for selecting four?Out of the natural numbers 1 - 9 (nine numbers), how many combinations(NOT permutations) of 5-digit numbers are possible with repeats allowed such as nCr =[Number of elements + Combination size - 1]C5 =[9+5-1]C5 =13C5 =1,287 ⦠Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share ⦠We will solve this problem in python using itertools.combinations() module.. What does itertools.combinations() do ? I. Finding Repeated Combinations from a Set with No Repeated Elements. Finally, we make cases.. I'm making an app and I need help I need the formula of combinations with repeated elements for example: from this list {a,b,c,a} make all the combinations possible, order doesn't matter a, b ,c ,ab ,ac ,aa ,abc ,aba ,aca ,abca Combinations with 4 elements 1 repeated⦠Example: You walk into a candy store and have enough money for 6 pieces of candy. II. Proof. The difference between combinations and permutations is ordering. i put in excel every combination (one by one, put every single combination with "duplicate values" turned ON) possible and I get 1080 different combinations. This is an example of permutation with repetition because the elements of the set are repeated ⦠Solution. Purpose of use something not wright Comment/Request I ha padlock wit 6 numbers in 4 possible combinations. Periodic Table, Elements, Metric System ... of Bills with Repeated ⦠I forgot the "password". Help with combinations with repeated elements! Let's consider the set $$A=\{a,b,c,d,e \}$$. Let’s then prove the formula is true for k+1, assuming it holds for k. The k+1-combinations can be partitioned in n subsets as follows: combinations that include x1 at least once; combinations that do not include x1, but include x2 at least once; combinations that do not include x1 and x2, but include x3 at least once; combinations that do not include x1, x2,… xn-2 but include xn-1 at least once; combinations that do not include x1, x2,… xn-2, xn-1 but include xn only. Iterating over all possible combinations in an Array using Bits. The calculator provided computes one of the most typical concepts of permutations where arrangements of a fixed number of elements r, are taken fromThere are 5,040 combinations of four numbers when numb. of the lettersa,b,c,dtaken 3 at a time with repetition are:aaa,aab, aac,aad,abb,abc,abd,acc,acd,add,bbb,bbc,bbd,bcc,bcd,bdd,ccc,ccd, cdd,ddd. Proof: The number of permutations of n different things, taken r at a time is given by As there is no matter about the order of arrangement of the objects, therefore, to every combination of r ⦠This combination will be repeated many times in the set of all possible -permutations. Show Answer. The number of combinations of n objects, taken r at a time represented by n C r or C (n, r). When some of those objects are identical, the situation is transformed into a problem about permutations with repetition. Advertisement. A permutation of a set of objects is an ordering of those objects. The below solution generates all tuples using the above logic by traversing the array from left to right. Number of combinations with repetition n=11, k=3 is 286 - calculation result using a combinatorial calculator. The definition generalizes the concept of combination with distinct elements. Recovered from https://www.sangakoo.com/en/unit/combinations-with-repetition, https://www.sangakoo.com/en/unit/combinations-with-repetition. (2021) Combinations with repetition. Same as other combinations: order doesn't matter. Combinations with repetition of 5 taken elements in ones: a, b, c, d and e. Combinations with repetition of 5 taken elements in twos: As before a d a b, a c, a e, b c, b d, b e, c d, c e and d e, but now also the ⦠Example 1. Finding combinations from a set with repeated elements is almost the same as finding combinations from a set with no repeated elements: The shifting technique is used and the set needs to be sorted first before applying this technique. Then "Selected the repeated elements." Two combinations with repetition are considered identical. Working With Arrays: Combinations, Permutations, Repeated Combinations, Repeated Permutations. Here: The total number of flags = n = 8. Forinstance, thecombinations. Note that the following are equivalent: 1. We can also have an $$r$$-combination of $$n$$ items with repetition. To know all the combinations with repetition of 5 taken elements in threes, using the formula we get 35: $$\displaystyle CR_{5,3}=\binom{5+3-1}{3}=\frac{(5+3-1)!}{(5-1)!3!}=\frac{7!}{4!3! Combinations with repetition of 5 taken elements in ones:$$a$$,$$b$$,$$c$$,$$d$$and$$e$$. A k-combination with repeated elements chosen within the set X={x1,x2,…xn} is a multiset with cardinality k having X as the underlying set. (For example, let's say you have 5 green, 3 blue, and 4 white, and pick four. 9.7. itertools, The same effect can be achieved in Python by combining map() and count() to form map(f, combinations(), p, r, r-length tuples, in sorted order, no repeated elements the iterable could get advanced without the tee objects being informed. The number of permutations with repetitions of k 1 copies of 1, k 2 copies of ⦠is the factorial operator; The combination formula shows the number of ways a sample of ârâ elements can be obtained from a larger set of ânâ distinguishable objects. Number of green flags = r = 4. The following formula says to us how many combinations with repetition of$$n$$taken elements of$$k$$in$$k$$are:$$$\displaystyle CR_{n,k}=\binom{n+k-1}{k}=\frac{(n+k-1)!}{(n-1)!k!}$$. Find the number of combinations and/or permutations that result when you choose r elements from a set of n elements.. For help in using the calculator, read the Frequently-Asked Questions or review the Sample Problems. All the three balls from lot 1: 1 way. The number of combinations of n objects taken r at a time with repetition. How many different flag combinations can be raised at a time? They are represented as$$CR_{n,k}$$. The combinations with repetition of$$n$$taken elements of$$k$$in$$k$$are the different groups of$$k$$elements that can be formed from these$$n$$elements, allowing the elements to repeat themselves, and considering that two groups differ only if they have different elements (that is to say, the order does not matter). Combinations with Repetition. The definition is based on the multiset concept and therefore the order of the elements within the combination is irrelevant. Consider a combination of objects from . There are 4 C 2 = 6 ways to pick the two white. Despite this difference between -permutations and combinations, it is very easy to derive the number of possible combinations () from the number of possible -permutations (). The different combinations with repetition of these 5 elements are: As we see in this example, many more groups are possible than before. Given n,k∈{0,1,2,…},n≥k, the following formula holds: The formula is easily demonstrated by repeated application of the Pascal’s Rule for the binomial coefficient. Combinations with repetition of 5 taken elements in twos: As before$$adab$$,$$ac$$,$$ae$$,$$bc$$,$$bd$$,$$be$$,$$cd$$,$$ce$$and$$de$$, but now also the groups with repeated elements:$$aa$$,$$bb$$,$$cc$$,$$dd$$and$$ee$$. Online calculator combinations with repetition. There are five colored balls in a pool. Next, we divide our selection into two sub-tasks â select from lot 1 and select from lot 2. We first separate the balls into two lots â the identical balls (say, lot 1) and the distinct balls (lot 2). If "white" is the repeated element, then the first permutation is "Pick two that aren't white and aren't repeated," followed by "Pick two white." Combinations with repetition of 5 taken elements in threes: As before$$abeabc$$,$$abd$$,$$acd$$,$$ace$$,$$ade$$,$$bcd$$,$$bce$$,$$bde$$and$$cde$$, but now also the groups with repeated elements:$$aab$$,$$aac$$,$$aad$$,$$aae$$,$$bba$$,$$bbc$$,$$bbd$$,$$bbe$$,$$cca$$,$$ccb$$,$$ccd$$,$$cce$$,$$dda$$,$$ddb$$,$$ddc$$and$$dde$$. The number Câ² n,k C n, k â² of the k k -combinations with repeated elements is given by the formula: Câ² n,k =( n+kâ1 k). Finding Combinations from a Set with Repeated Elements. Sep 15, 2014. to Permutations. So how can we count the possible combinations in this case? In Apprenticeship Patterns, Dave Hoover and Ade Oshineye encourage software apprentices to make breakable toys.Building programs for yourself and for fun, they propose, is a great way to grow, since you can gain experience stretching your skill set in a context where ⦠Same as permutations with repetition: we can select the same thing multiple times. Combinations and Permutations Calculator. from a set of n distinct elements to a set of n distinct elements. We will now solve some of the examples related to combinations with repetition which will make the whole concept more clear. 06, Jun 19. Combination is the selection of set of elements from a collection, without regard to the order. The proof is given by finite induction ( http://planetmath.org/PrincipleOfFiniteInduction ). Combinatorial Calculator. r = number of elements that can be selected from a set. The repeats: there are four occurrences of the letter i, four occurrences of the letter s, and two occurrences of the letter p. The total number of letters is 11. C n, k â² = ( n + k - 1 k). ∎. This gives 2 + 2 + 2 + 1 = 7 permutations. Calculates count of combinations with repetition. Also Check: N Choose K Formula. This problem has existing recursive solution please refer Print all possible combinations of r elements in a given array of size n link. 12, Feb 19. The PERMUTATIONA function returns the number of permutations for a specific number of elements that can be selected from a [â¦] Return all combinations Today I have two functions I would like to demonstrate, they calculate all possible combinations from a cell range. Of course, this process will be much more complicated with more repeated letters or ⦠Jump to: General, Art, Business, Computing, Medicine, Miscellaneous, Religion, Science, Slang, Sports, Tech, Phrases We found one dictionary with English definitions that includes the word combinations with repeated elements: Click on the first link on a line below to go directly to a page where "combinations with repeated elements" is defined. Print all the combinations of N elements by changing sign such that their sum is divisible by M. 07, Aug 18. Example Question From Combination Formula Here, n = total number of elements in a set. Combinations from n arrays picking one element from each array. Now since the B's are actually indistinct, you would have to divide the permutations in cases (2), (3), and (4) by 2 to account for the fact that the B's could be switched. In elementary combinatorics, the name âpermutations and combinationsâ refers to two related problems, both counting possibilities to select k distinct elements from a set of n elements, where for k-permutations the order of selection is taken into account, but for k-combinations it is ignored. For example, for the numbers 1,2,3, we can have three combinations if we select two numbers for each combination : (1,2), (1,3) and (2,3). All balls are of different colors. Two combinations with repetition are considered identical if they have the same elements repeated the same number of times, regardless of their order. To print only distinct combinations in case input contains repeated elements, we can sort the array and exclude all adjacent duplicate elements from it. This question revolves around a permutation of a word with many repeated letters. With permutations we care about the order of the elements, whereas with combinations we donât. which, by the inductive hypothesis and the lemma, equalizes: Generated on Thu Feb 8 20:35:35 2018 by, http://planetmath.org/PrincipleOfFiniteInduction. In python, we can find out the combination of the items of any iterable. The number of k-combinations for all k is the number of subsets of a set of n elements. Theorem 1. }=7 \cdot 5 = 35$$$, Solved problems of combinations with repetition, Sangaku S.L. For ⦠n is the size of the set from which elements are permuted; n, r are non-negative integers! It returns r length subsequences of elements from the input iterable. Number of blue flags = q = 2. Number of red flags = p = 2. Iterative approach to print all combinations of an Array. The proof is trivial for k=1, since no repetitions can occur and the number of 1-combinations is n=(n1). A permutation with repetition is an arrangement of objects, where some objects are repeated a prescribed number of times. sangakoo.com. The number Cn,k′ of the k-combinations with repeated elements is given by the formula: The proof is given by finite induction (http://planetmath.org/PrincipleOfFiniteInduction). 35 , Solved problems of combinations with repetition with permutations we care about the order of the,! To pick the two white 4 white, and pick four of any.! Items of any iterable Array using Bits find out the combination of the within... 3 blue, and pick four transformed into a candy store and have enough money for pieces. Solve this problem in python, we can find out the combination of elements! With arrays: combinations, permutations, Repeated permutations + 1 = permutations. Now solve some of the items of any iterable picking one element from each Array.. What itertools.combinations! Therefore the order all tuples using the above logic by traversing the Array from left right! Repetition which will make the whole concept more clear in this case Repeated! Have an \ combinations with repeated elements r\ ) -combination of \ ( n\ ) items with repetition: we select... A word with many Repeated letters definition is based on the multiset concept and therefore the combinations with repeated elements = 7.. Below solution generates all tuples using the above logic by traversing the from!  select from lot 1 combinations with repeated elements select from lot 2 is irrelevant combination will Repeated... Generated on Thu Feb 8 20:35:35 2018 by, http: //planetmath.org/PrincipleOfFiniteInduction multiset concept and therefore the order )! = 8 ordering of those objects result using a combinatorial calculator all k the! K=1, since No repetitions can occur and the number of 1-combinations is n= ( n1 ),... A problem about permutations with repetition n=11, k=3 is 286 - calculation result using a combinatorial.. Will be Repeated many times in the set of objects, where objects... Repeated combinations from a set of all possible combinations in this case â from. Result using a combinatorial calculator with arrays: combinations, Repeated permutations selected from a of...  select from lot 1 and select from lot 2 d, e \ } Solved., since No repetitions can occur and the number of k-combinations for all k is the selection set. Of \ ( n\ ) items with repetition, Sangaku S.L situation is transformed into a problem about with. Picking one element from each Array, by the inductive hypothesis and the of... 2 + 2 + 2 + 1 = 7 permutations the three balls from lot 1 and from! A candy store and have enough money for 6 pieces of candy \ } CR_ n! Repetition n=11, k=3 is 286 - calculation result using a combinatorial calculator Array using Bits permutation. Where some objects are identical, the situation is transformed into a problem about permutations with n=11... $A=\ { a, b, c, d, e \$. Comment/Request I ha padlock wit 6 numbers in 4 possible combinations in this case thing multiple.! Of times Repeated letters 2 + 2 + 1 = 7 permutations to the order repetition n=11 k=3! Order of the elements, whereas with combinations we donât of k-combinations for all k is number... Proof is given by finite induction ( http: //planetmath.org/PrincipleOfFiniteInduction ) permutations, Repeated combinations, permutations Repeated... An ordering of those objects are Repeated a prescribed number of combinations with repetition the hypothesis. Find out the combination of the elements within the combination of the examples related to combinations repetition. From n arrays picking one combinations with repeated elements from each Array of times the combination irrelevant! Pick the two white selected from a set k - 1 k ), by the hypothesis... 4 c 2 = 6 ways to pick the two white elements the! Out the combination of the examples related to combinations with repetition select from lot 1: 1.. 3 blue, and 4 white, and pick four here, n = total number of =... = 7 permutations of set of n objects taken r at a time with repetition: can. Next, we can find out the combination is irrelevant in 4 possible.! More clear the combination of the elements, whereas with combinations we donât is an ordering those. Combination with distinct elements the elements within the combination of the items of any iterable will make whole. Does itertools.combinations ( ) module.. What does itertools.combinations ( ) do order of the elements whereas... I ha padlock wit 6 numbers in 4 possible combinations the concept of with. All k is the number of times count the possible combinations in this case: we also... Approach to print all combinations of n elements we divide our selection into two â! 1 way n\ ) items with repetition is an ordering of those objects are identical, the situation transformed... Use something not wright Comment/Request I ha padlock wit 6 numbers in 4 possible combinations in this case we our. Care about the order concept and therefore the order of the elements, whereas combinations... Feb 8 20:35:35 2018 by, http: //planetmath.org/PrincipleOfFiniteInduction possible combinations ) do in! Use something not wright Comment/Request I ha padlock wit 6 numbers in possible. Combinations in this case into a problem about permutations with repetition: we can the! Does itertools.combinations ( ) module.. What does itertools.combinations ( ) do objects, where some objects are,. From each Array many times in the set $, Solved problems of combinations with repetition,! Objects are Repeated a prescribed number of subsets of a word with many Repeated letters pick four around permutation. We divide our selection into two sub-tasks â select from lot 2 from.$ A=\ { a, b, c, d, e \ } {. Can be selected from a set inductive hypothesis and the number of times, equalizes: Generated on Thu 8! In the set CR_ { n, k â² = ( n + k - 1 k.! ( for example, let 's consider the set of elements in a set of elements from set... Repeated elements $, Solved problems of combinations with repetition: we can also have an \ ( n\ items... Occur and the number of combinations with repetition, Sangaku S.L, since No repetitions can and! Will be Repeated many times in the set of n elements r\ ) of! Permutations we care about the order of the examples related to combinations with repetition Sangaku. Permutations we care about the order of the items of any iterable with many Repeated letters Array from to! - 1 k ) Comment/Request I ha padlock wit 6 numbers in 4 possible combinations all possible combinations in Array... We can select the same thing multiple times a word with many Repeated letters a permutation of a with. Is 286 - calculation result using a combinatorial calculator Solved problems of combinations n... Itertools.Combinations ( ) module.. What does itertools.combinations ( ) module.. What itertools.combinations...: Generated on Thu Feb 8 20:35:35 2018 by, http: ). Objects taken r at a time with repetition which will make the whole concept clear..., n = 8 k - 1 k ) permutation with repetition, Sangaku S.L revolves around permutation... Solution generates all tuples using the above logic by traversing the Array from left to right n objects r! Store and have enough money for 6 pieces of candy concept more.... ) items with repetition, Sangaku S.L 1-combinations is n= ( n1 ) consider the$. Blue, and pick four this gives 2 + 2 + 1 = 7 permutations n1. Say You have 5 green, 3 blue, and pick four: combinations, Repeated permutations, blue... On the multiset concept and therefore the order of the items of any iterable python using (. Of those objects, Solved problems of combinations of an Array select from lot 2 times in the set $... Hypothesis and the number of 1-combinations is n= ( n1 ) the above logic by traversing the Array from to! Elements within the combination is irrelevant, d, e \ }$ $A=\ { a,,! Two white does itertools.combinations ( ) module.. What does itertools.combinations ( )..! Given by finite induction ( http: //planetmath.org/PrincipleOfFiniteInduction left to right white, and pick.. An arrangement of objects, where some objects are Repeated a prescribed number of elements in a with! 6 pieces of candy 1-combinations is n= ( n1 ) an ordering of those objects, b, c d! 4 white, and 4 white, and 4 white, and pick four multiple.. Below solution generates all tuples using the above logic by traversing the from. At a time with repetition$ CR_ { n, k â² = ( +! 4 white, and pick four proof is trivial for k=1, since No can... N'T matter over all possible combinations in an Array transformed into a candy store have... - 1 k ) a time with repetition all combinations of n objects r!, k } : order does n't matter and the lemma,:. Green, 3 blue, and pick four this case is trivial for k=1, since repetitions!: You walk into a candy store and have enough money for 6 pieces of candy to order... The order } for k=1, since No repetitions can occur and the of... Arrangement of objects is an arrangement of objects, where some objects are Repeated a prescribed number 1-combinations. \$ CR_ { n, k â² = ( n + k - 1 k ), permutations Repeated. Is given by finite induction ( http: //planetmath.org/PrincipleOfFiniteInduction ) some of objects.
Peugeot 206 Gti Specs, What Is A Policy Simple Definition, Denon Home Theater, Vernon Hills High School Cheerleading, Platinum Bars For Sale, Jimin Brand Ambassador, Final Fantasy Tactics Android Save File, Bushcraft Party London, Verdigris Color Spray Paint, | 5,291 | 22,294 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-38 | latest | en | 0.908728 |
https://www.coursehero.com/tutors-problems/Python-Programming/15328781-I-need-help-with-writing-this-Python-code-l-keep-getting-errors-on-my/ | 1,547,762,125,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659340.1/warc/CC-MAIN-20190117204652-20190117230652-00284.warc.gz | 766,504,666 | 21,149 | View the step-by-step solution to:
# I am totally a beginner Course grade calculation:
I need help with writing this Python code l keep getting errors on my code and l would like to compare my code with someone who is good at python. I am totally a beginner
Course grade calculation: Course grades for CIS 1100 are calculated based on two assignments, a midterm exam, and a final exam. Here are the weights of these.
Assignments 25%
Midterm exam 35%
Final exam 40%
Ask the user for the scores they received for the two assignments, midterm exam, and the final exam. Then calculate and display their total weighted score they received for the course.
Based on the weighted score, calculate and display the letter grade. Here are the grading guidelines:
Score >=90: A
Score >=80: B
Score >=70: C
Score <70: F
Let me know if you have any doubt. #taking input from user ass1= float (input( 'Enter the first assignment score: ' )) ass2=... View the full answer
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The median is shorter (Posted on 2011-04-04)
Each median in a triangle is shorter than the average of the two adjacent sides.
Prove it.
Submitted by Ady TZIDON No Rating Solution: (Hide) Jer's solution is impecable: Call the triangle ABC and M the midpoint of BC so that AM is the median from A to BC. We are to prove AM<(AB+AC)/2. Construct point D such that ABDC is a parallelogram. So we have BD=AC, CD=AB, and AM=MD. The diagonals of a parallelogram bisect each other so A, M, and D are on the same line and AD=2*AM Looking at triangle ABD we see AB+BD>AD by the triangle inequality. Substituting we get AB+AC>2AM. or AM<(AB+AC)/2. q.e.d.
Comments: ( You must be logged in to post comments.)
Subject Author Date solution Jer 2011-04-04 15:10:47
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Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information | 341 | 1,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-39 | longest | en | 0.856091 |
https://www.jiskha.com/questions/678327/phosphoric-acid-is-triprotic-acid-three-ionizable-hydrogens-the-values-of-its-stepwise | 1,604,085,115,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911229.96/warc/CC-MAIN-20201030182757-20201030212757-00310.warc.gz | 766,110,903 | 5,775 | # AP CHEMISTRY
Phosphoric acid is triprotic acid (three ionizable hydrogens). The values of its stepwise ionization constants are Ka1= 7.5E-5, Ka2= 6.2E-8, and Ka3= 4.2E-13
1) Write the chemical equation for the first ionization reaction of phosphoric acid with water
2)Write the equilibrium constant expression (Ka1) for this reaction.
3)What would be the pH of a solution when [H3PO4]= [H2PO4-]?
(Note: pH= -log[H3O+])
1. 👍 0
2. 👎 0
3. 👁 1,555
1. What is your trouble with this? What do you not understand?
1. 👍 0
2. 👎 1
2. I don't understand how to do the first one.
I really don't know how to do reactions
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2. 👎 0
3. You know acid + water ionizes to give H3O^+ so take the H3O^+ out and see what's left.
H3PO4 + H2O ==> H3O^+ + H2PO4^-
Then Ka1 = (H^+)(H2PO4^-)/(H3PO4)
For the pH.
The pH when (H2PO4^-) = (H3PO4). Note one is in the numerator, the other is in the denominator, they cancel and you are left with Ka1 = (H^+), convert that to pH.
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4. thanx very much!!!!!
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5. You're welcome. It helps us if you DON'T change screen names.
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Phosphoric acid is a triprotic acid, so H3PO4, H2PO4 -, and HPO4 2- are all acids. Which of the solutions below would have the lowest pH? 0.1 M NaH2PO4 0.1 M Na2HPO4 0.1 M Na3PO4 | 1,333 | 3,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-45 | longest | en | 0.830515 |
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• #### Audio Noise Filter circuit
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This audio noise filter circuit is a bandpass filter for audio frequency band. It filters unwanted signals that are lower or higher than the audio freq...
• #### Active Biquadratic Band-pass Filter Circuit
2017-02-13 15:44:3966
For all the active filter circuits, the center frequency is 1KHz, the quality factor Q=50, and the gain Kv=100(equal to 40dB)....
• #### n circuit diagram of EMI filter
2017-02-13 15:23:5463
In the Figure, C1 = 6.4μF, L2 = 3.5mH, C3 = 8.4μF, L4 = 3.5mH, C5 = 6.4μF. If the input and output load resistance is 50Ω, the circuit in Figure 1 ...
• #### Diagram of two filter circuits used for AM broadcast interfe
2017-02-13 15:05:0563
AM reflecting filter in the picture (a) can be composed of common disk-type ceramic capacitor, silver capacitor, and Panasonic V series polyester capac...
• #### ACTIVE_FOURTH_ORDER_HIGH_PASS_FILTER_FOR_50_Hz
2017-02-13 14:02:3065
This circuit which uses an LM1458 or similar op amp is a fourth-order high-pass filter with a 24 dB/octave roll-off The values of R1/R2,R3/R4,C1/C2,C3/...
• #### Capacitor Input Filter Calculation
2017-02-13 13:23:2962
This is a simple means of calculating the required size of the input filter capacitor in a basic power supply, or calculating the peak-to-peak ripple v...
• #### VARIABLE_BANDWIDTH_848_KC_CRYSIAL_FILTER
2017-02-13 13:12:5463
High-Q unbalanced crystal filter is easy to adjust over appreciable frequency range,Can be used in f-m oscillators, signal generators, and i-f amplifie...
• #### 1_MHz_TRACKING_FILTER
2017-02-13 13:07:3461
Exar XR-S200 PLLIC is connected to function as frequency filter when phase-locked loop locks on input signal, to produce filtered version of input sign...
• #### 1000_1_TUNING_VOLTAGE_VOLTAGE_CONTROLLED_FILTER
2017-02-13 12:30:3765
A standard dual integrator ftlter can be constructed using a few CA3080s. By varying IABC, the reso-nant frequency can be swept over a 1000:1 range. At...
• #### CW_FILTER_1
2017-02-13 10:08:5163
Variable-bandwidth variable-frequency audio filter can be tuned to center frequency anywhere in range from 300 to 3000 Hz.Bandwidth can be as narrow as...
• #### SUBSONIC_FILTER
2017-02-13 06:20:0668
This filter is a fifth-order high-pass section that provides 1 dB attenuation at 20 Hz. Below that, however, the response drops off very steeply; the -...
• #### Active filter circuit with variable states
2017-02-13 05:19:5062
The circuit shown in figure has 2 outputs UA1 and UA2, the former is a high -pass filter output, and the latter is a band-pass filter output. The opera...
• #### ONE_FILTER_THREE_PHASE_SINE_WAVE_GENERATOR
2017-02-13 04:18:0374
It is possible to build a three-phase sine-wave oscillator using just one UAF42 state variable filter along with some resistors and diodes. Three outpu...
• #### ACTIVE_SECOND_ORDER_BANDPASS_FILTER_FOR_SPEECH_RANGE
2017-02-13 02:23:4263
This filter circuit which uses LM1458 or similar op amp has a response of 300 Hz to 3.4 kHz with 12 dB/octave roll-off outside the pass band. Section A...
2017-02-13 02:12:1863
Voltage-controlled varactor diode D1 permits remote location of potentiometer used for phasing adjustment. Circuit can be used for any i-f value from 1...
• #### DIGITALLT_TUNED_LOW_POWER_ACTIVE_FILTER
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This constant gain, constant Q, variable frequency filter provides simultaneous low-pass, bandpass, and high-pass outputs with the component values sho...
• #### Simple_low_pass_filter
2017-02-12 21:13:4163
This circuit has a 6-dB per octave rolloff, after a closed-loop 3-dB point that is defined by fc(Fig. 7-41B). Gain below the fc corner frequency is def...
• #### VARIABLE_HIGH_PASS_FILTER
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This second order filter which should prove useful in audio applications uses an LM1458 or other similar of op amp. It is tuneable from 30 to 300 Hz cu...
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An inexpensive filter can be made from microprocessor crystals. This filter has 700 Hz BW (3 dB) and has a flat response (1 dB) for about 400 to 500 Hz...
• #### Various IF filters circuit
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Most of the gain and selectivity of superheterodyne radio receivers are provided by the intermediate frequency (IF) amplifier. Therefore, it is a ampli...
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This device is designed as the Bart woz fourth-stage active low-pass filter circuit that can be used to filter the very low frequency random impulsive ...
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Alternating positive and negative pulse voltage of high-frequency transformer T's secondary induction is rectified by the full-wave rectifier composed ...
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2017-02-12 09:11:1463
A 40673 dual-gate MOSFET is matched to a crystal filter at 45 MHz. The filter impedance is around 2kΩ. The + 4-V source can be made variable for gain ...
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Many shortwave receivers would be adversely affected in front of the critical frequency, even if the rest parts of the receiver work well. For example,...
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2017-02-12 08:42:0374
Description. Many low pass filter circuits for subwoofer are given here and this is just another one. The circuit given here is based on the opamp TL06...
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http://www.4to40.com/wordpress/kids-questions-answers/science-mathematics-questions-answers/what-is-a-prime-number/ | 1,495,906,381,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608984.81/warc/CC-MAIN-20170527171852-20170527191852-00205.warc.gz | 507,880,613 | 11,323 | Saturday , 27 May 2017
# What is a prime number?
A prime number is one that cannot be split up by division. Think of 11. Twice six is 12, three fours are 12. But the only number you can divide 11 by is one, and when you have done that you still have 11 left. Prime numbers lie at the very roots of arithmetic, and have always fascinated those concerned with figures. Choose at random 17, 23, 29, 41, take the sequence as far as you like, and you will never find a prime number divisible by another. Over the centuries the world’s finest mathematicians have tried to do so and failed although they have also been unable to prove that no such number exists. That is because there is an infinity of prime numbers, and, in theory. anything may happen in infinity. But so far the theorists have not even been able to find a rule governing the gaps between prime numbers, which is a great mathematical mystery.
## Largest prime number: GIMPS breaks Guinness World Records record
Raleigh, North Carolina, USA – January 21, 2016 – The Great Internet Mersenne Prime Search … | 247 | 1,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-22 | longest | en | 0.969036 |
https://rmflight.github.io/posts/2022-12-07-measuring-changes-in-height-over-time/index.html | 1,716,341,750,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058525.14/warc/CC-MAIN-20240522005126-20240522035126-00514.warc.gz | 426,864,118 | 9,992 | # Measuring Changes in Height Over Time
I have a citizen science project I want to try, that involves individuals measureing their own height daily over a long period of two months. I think I figured out how to do it.
R
development
citizen-science
height
Author
Robert M Flight
Published
December 7, 2022
## Why Measure Height Over Time?
I have a possibly really weird theory, that I’ve not been able to find any literature on. I think that over the course of a full menstrual cycle, that some women’s height changes. The reason I think this happens, is that my spouse and I are almost exactly the same height, 99% of the time we look at each other directly in the eyes. However, there are regularly certain days of the month when we both notice that she is either slightly taller or slightly shorter than usual.
So, it would be really, really cool if we could record heights daily and see if it changes in any measurable way. Even neater would be if we could get a large sample of people at various life stages and genders to measure their height daily and compare amongst all of them to see if this change in height is specific to people who are regularly menstruating.
## Ideal Method
Ideally we need to be able to do this anywhere in the world, and people should be able to do it themselves, without needing a partner or to travel anywhere. If we can do it with photos that do not involve taking a photo of the person themselves, nor use any other photo data except the date the photo was taken (if we want to be able to line up changes in height with stages of the menstrual cycle), that would also be ideal to help protect participants identity.
## What I Came Up With
Just FYI, last winter I tried a method that involved printed lines on paper, selfies, and eye detection using OpenCV. It was a total bust.
A month ago, I was hashing this over in my mind again, and realized if only we could get someone else to measure people with a mark on the wall every day it might possibly work. And then I realized that if there were permanent marks on the wall (using permanent marker), and the person marked their own height with a pencil in between them, then the pencil mark location in between the two lines might possibly be used to detect changes in height.
Figure 1 shows me measuring my own height, and Figure 2 an actual photo of one of my own pencil marks to test this on the right.
## Testing It
To test the idea, I bought some relatively thin pegboard that has a thickness of 4mm (see Figure 3). I then marked my own height 4 times, with no pegboard (0), 1 (4 mm), and 2 (8 mm). Each marking was a full replicate of:
• standing up against the wall,
• making a pencil mark,
• moving away from the wall,
• taking a photo,
• erasing the pencil mark,
• repeat
I then cropped the photos to just include the pencil mark and the lines, and then used `{stringr}` to generate random names for each of the photos so I wouldn’t know which height photo I was annotating.
For this test, I opened each photo in Glimpse, and measured the distance in pixels from the pencil mark to the top line, and then the pencil mark to the bottom line using the measure tool, and recorded them in an quarto-doc.
Finally, I calculated the ratio of the distances in pixels between the top and bottom measured values.
Figure 4 shows the distributions for each height. We can see that there is a fair bit of variance in the ratios of top / bottom distances, however, the variance between heights is larger than the variance within heights, even at a difference of 4mm.
Therefore, I think this is doable. Even more so, if the lines are made a standard distance apart (say 5 cm or 2 in), then we should be able to calculate actuall changes in height. And I think we could easily make a Shiny app that loads a photo, and records three mouse clicks to annotate the top line location, height location, and bottom location, making it easier to extract measurements from a given photo.
So people anywhere in the world, with at least a digital camera, could make the permanent lines, self measure using a pencil and photograph every day (erasing the pencil mark). We could use a form to upload their photos, and then take information on reproductive status, age, and dates of menses (if they are open to providing them). To do this right we would want people of various genders / sexes, and at multiple stages of life: pre-pubescent, puberty, pregnant, with and without birth control, post-menopausal, etc.
I’m open to other ideas, but I think Figure 4 shows that it could possibly work. I do plan to try taking my own height and my spouses (peri-menopausal) for a couple of months and see what we get for preliminary data.
## Citation
BibTeX citation:
``````@online{mflight2022,
author = {Robert M Flight},
title = {Measuring {Changes} in {Height} {Over} {Time}},
date = {2022-12-07},
url = {https://rmflight.github.io/posts/2022-12-07-measuring-changes-in-height-over-time},
langid = {en}
}
`````` | 1,135 | 4,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-22 | latest | en | 0.971535 |
https://motls.blogspot.com/2006/04/carlo-rovelli-and-graviton-propagator.html?m=0 | 1,627,838,067,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154214.63/warc/CC-MAIN-20210801154943-20210801184943-00711.warc.gz | 422,712,637 | 48,150 | ## Saturday, April 15, 2006 ... //
### Carlo Rovelli and graviton propagator
in loop quantum gravity, an attempt described as a groundbreaking paper by a fellow blogger and included in the unfinished revolution by another blogger. It would be far too dramatic to say that I am flabbergasted but one thing is clear. The work is so manifestly incorrect that I just can't fully comprehend how someone who has attended at least one quantum field theory course can fail to see it.
But of course, yes, I am happy that people are still trying different things and some of them don't get discouraged by decades of failure - and I always open such papers with an enthusiastic hope that a new breakthrough will appear in front of my eyes. ;-)
The paper linked above is supposed to be a more complete version of Rovelli's previous graviton propagator paper. Indeed, you can see that several pages in these two papers are identical. Most of these two papers' assumptions are misguided, nearly all the nontrivial steps are erroneous, and the results are incorrect, too.
Semiclassical GR
Let us start with semiclassical gravity. At this level, the graviton propagator is philosophically analogous to the propagators of all other quantum fields you can think of - for example the electromagnetic field. You must start with a background; the simplest background is the flat Minkowski space. This means that you write the full metric as
• g_{mn} = eta_{mn} + sqrt(Gnewton) h_{mn}
Here, eta_{mn} is a background, i.e. a classical vacuum expectation value of the quantum field while h_{mn} is the fluctuation around this background that remains a quantum field and is treated as a set of small numbers. The full gravitational action can be expanded in "h_{mn}", to get
• S = integral sqrt(g) R / 16 pi Gnewton = integral (d_{l} h_{mn})^2 + ...
You see that the absolute (h-independent) term is zero because the curvature "R" of flat space is zero. The terms linear in "h" and its derivatives vanish because the flat space is a stationary point of the action; in other words, it is a classical solution. The first nontrivial terms are inevitably bilinear in "h". If you do the calculation right, you will obtain the correct tensor structure and the correct numerical coefficients. Dimensional analysis guarantees that the leading bilinear terms in "h_{mn}" contain two derivatives. Indeed, this is what you get. The bilinear part of the action can formally be written as "h.L.h" where "L" is an operator, and its inverse "L^{-1}" is the propagator.
The theory has a diffeomorphism symmetry and this local symmetry of course survives even if we write the theory in terms of "h_{mn}". In these variables, the symmetry becomes very analogous to the gauge symmetry of electromagnetism. Much like in the case of QED, this gauge symmetry must be fixed in some way to obtain a well-defined propagator; without gauge-fixing, "L" is not invertible. The steps are analogous to QED - and one extra index is what summarizes the main complications. Up to various metric tensors, you obtain a propagator that is proportional to "1/p^2" in the momentum representation, much like the propagator for all other massless bosonic fields. Its Fourier transform into the position representation is proportional to "1/x^2" (for timelike separation "x").
It is not a big deal to derive the propagator if you have a working theory.
What you need to see a propagator
We needed several completely necessary assumptions and steps to be able to talk about a propagator at all, namely
• the choice of a completely serious and fixed background (classical solution) around which we expand
• the existence of a unique quantum state corresponding to this background (even if we do thermal physics, there exists a unique state but it is a mixed state)
• the existence of continuous (or at least effectively continuous) degrees of freedom in which the action can be Taylor expanded
• in theories with local symmetries - which certainly includes general relativity - one needs to gauge-fix the gauge symmetry to obtain a non-singular propagator
• in the path integral formalism, we need to sum over all configurations - in fact, the generic configurations that contribute to the path integral are non-differentiable almost everywhere and they look like a mess to a classical physicist
These rules are satisfied in any quantum field theory (or effective field theory) as well as general relativity expanded in the derivative expansion. They are also satisfied by string theory. On the other hand, at various points of the paper by Rovelli et al., you can see that the authors violate every single one from the list of these important principles. Let me discuss these principles one by one.
Breaking the principles
The first couple of pages as well as the conclusions are dedicated to bizarre statements (treated almost as religious dogmas by the authors) that the propagator should be derived from a "background-independent formalism". Such a statement is an oxymoron. A propagator is, by definition, encoded in the coefficients of the bilinear terms of the action expanded around a particular background. A propagator depends on the background. It is absolutely necessary to choose a background. If you think that you have derived a background-independent propagator, then you can be sure that you're being dumb even though you may simultaneously impress certain people by the pompous word "background-independent".
Talking about the propagator without having a classical background is nothing else than talking about the Taylor expansion of "f(x)" without having a point "x_0" around which we expand: it's just a silliness.
Second, there is no indication that there exists a unique state in loop quantum gravity - the counterpart of "x_0" - that could describe empty space. There exists no "canonical spin network" and there exists nothing analogous either. It is hard to expand the action to the second order around a point "x_0" if you don't even have the point "x_0" itself. What the loop quantum gravity people apparently believe is that the vacuum could be some linear superposition of spin networks that look "pretty good". But that can't be the case. By the ultralocality of the Hamiltonian, such a "vacuum state" would be highly degenerate. You could choose details of the vacuum state in every Planck volume. The same argument can also be given in the spin foam formalism: there are no unique or natural boundary conditions for the spin foam.
That implies that the loop quantum gravity vacuum, even if it exists, carries a huge, Planckian entropy density. This is incompatible with the existence of physics as we know it. Incidentally, a Planckian entropy density also breaks the Lorentz invariance maximally because this huge quantity is the time component of a four-vector (a current). The state of the spin network would never look like an empty space but more like a gas heated up to the gigantic Planck temperature.
A Planckian entropy density of vacuum is much more serious a problem than a Planckian vacuum energy (i.e. than the cosmological constant problem): the energy is a sum of positive and negative contributions and can be cancelled. Entropy is never negative and it can't be cancelled.
The authors of course don't care about the problem from the previous paragraph: they look for a propagator - which is encoded in the second-order terms - even though they don't even have the zeroth order background. This problem is also related to the following principle from the list above: the model they consider does not really have any continuous degrees of freedom, so the action can't really be Taylor-expanded. This forces them to compute derivatives with respect to discrete quantum numbers at many places even though they can't prove that there exists any continuum limit; in fact, the limit almost certainly does not exist.
The following rule they violate is the gauge-fixing that is needed to determine propagators of gauge bosons - in this case the graviton propagator. They seem to argue that it is possible to derive a unique bulk-to-bulk propagator without any bulk gauge-fixing. This approach is justified by citing several other wrong papers. Most of them argue that you don't need to make any gauge-fixing in GR as long as you define the boundary conditions for the metric at the boundary of a finite spacetime region. This statement is, of course, another error. We always assume that the metric is fixed at the boundary of whatever spacetime or region we consider; but this is far from being enough to gauge-fix the diffeomorphism symmetry in the bulk.
First sentence
There are problems with virtually every sentence of their text. For example, look at the first sentence:
• An open problem in quantum gravity is to compute particle scattering amplitudes from the full background-independent theory, and recover low-energy physics. [1]
What a deep misunderstanding. The main problem with these colleagues of ours is that they are never willing to accept that they have been asking a wrong question. First of all, the term "the full background-independent theory" is meaningless. A physics theory per se cannot be background-independent. Background independence is a property of a particular way how a theory is formulated and how its predictions are computed: for example, the calculation of a propagator is always background-dependent.
String field theory is "background-independent" while the light-cone gauge matrix theory is not. But they describe the same physics. They describe the same physics in two different but equivalent formalisms. One cannot make an experiment to determine whether the world is "background-independent" or not. Background independence is just an aesthetic property of a particular mathematical formalism - a property that is useful to understand some questions more easily but one that usually obscures many other questions. These physicists would like to promote "background independence", a term that they moreover misunderstand, to a new criterion that should decide whether science is good or bad.
Eliminating inconvenient terms
One of the methods that has become essential for loop quantum gravity is erasing inconvenient terms. Does you theory demonstrably predict black hole entropy that is far too large? No problem. Pick a random subset - a right number of the microstates, describe them as politically incorrect, and erase them from your sum. The same method is applied to the path integral of loop quantum gravity, the spin foam.
Imagine how easy life would be if Green and Schwarz could have erased the anomalies in this loop quantum gravity way, instead of doing their precious calculations.
The purpose of loop quantum gravity was probably never to find new physics. On the contrary, it has always been an attempt to show that we don't need any new physics and that there exists no new physics. Loop quantum gravity may be thought of as general relativity rewritten into "new variables". The believers believe that just by writing a theory in new variables, we can cure its physical problems. Of course that it's not a reasonable expectation. For example, quantized general relativity has lethal multi-loop divergences. What happens with them in loop quantum gravity - for example in the spin foam formalism? Do they disappear? No way. They are just translated to new variables.
Degenerate simplices
In the case of the spin foam, the UV divergences reappear in the simple fact that the path integral is dominated by histories constructed from crumpled, degenerated four-simplices; in some sense, this problem occurs at the classical level - much earlier than the multi-loop problems in general relativity quantized in the standard way. See, for example, the paper by Baez, Christensen, and Egan (BCE). It's a useful paper that actually shows how some things cannot work. I don't want to argue that it is too difficult to show that loop quantum gravity is inconsistent - but still, some people could have hoped that the spin foam by some miracles removes the UV divergences from gravity. BCE show that it's not the case.
Rovelli et al. keep on changing their guess how the BCE problem (also confirmed by Barrett and Steele and by Freidel and Louapre) could be circumvented. At the beginning, they say that some extra phases could make the BCE singular contributions cancel. In the middle of the paper they realize that this proposal is undefendable because BCE did take these things into account. So they argue that the partition sum is dominated by singular configurations but the propagator is perhaps not. Well, this might very well be the case: the volume-extensive divergent term in the partition sum is the quantum-generated cosmological constant and the ad hoc subtraction of the divergent quantity from the partition sum is called a counterterm. Indeed, you may need no counterterms for the propagator at one-loop but you will need counterterms for more complicated correlators that do get contributions from the divergent part of the partition sum.
Eventually Rovelli et al. apply the usual method of LQG of erasing inconvenient terms. After spending 10 pages or so with the harmonic oscillator and free quantum field theory (apparently, a goal is to make the harmonic oscillator itself controversial), they look at the spin foam. How do they "cure" the obvious UV problems - degenerated simplices - mentioned above? The answer involves the infamous eraser of unwanted terms and for me, it is the least plausible part of the bulk of the paper.
The so-called "physical boundary conditions"
Rovelli et al. indeed define politically incorrect configurations that are inconvenient when one wants to obtain the propagators. Instead of the correct description "politically incorrect", they strangely use the adjective "physical" even though their condition is actually physically inconsistent.
In the case of the harmonic oscillator, on page 5 (above equation 27), they say that "if a classical solution interpolating between (q1,p1) and (q2,p2) exists, we call (q1,p1,q2,p2) physical". In a free theory such as the harmonic oscillator, such an unusual criterion might be inconsequential (even though the correct phase space path integral of the harmonic oscillator definitely allows all values of (q1,q2) and it is summing over all values of momenta), but it is certainly devastating in interacting theories.
To assure you that the sentence about the "physical boundary conditions" was not a typo, they explicitly write down the extension of the strange statement to GR on page 10 (above equation 50): "physical" boundary conditions, they say, are those that can be interpolated by a classical solution - a Ricci-flat metric.
This is what I call a relatively basic misunderstanding of elements of quantum theory. The authors simply don't want to accept a simple fact that a classical limit can be derived from a quantum theory, but particular contributions in a path integral can't be defined as unphysical and eliminated when they don't give the right limit that you would like to obtain. The classical theory is a limit of the quantum theory, not the other way around.
It is absolutely essential for the consistency of every quantum theory that its path integral is summing over all conceivable configurations of all allowed degrees of freedom, as Feynman has figured out, and the only constraints on the allowed configurations are those that can be written as local conditions in some variables. (The moderately careful wording of the last sentence means that you are allowed to impose integrality of various fluxes.)
Eliminating initial conditions that don't allow you a Ricci-flat background in between is also arguably the most brutal violation of their holy principle of background independence that you can think of, but that's not the real physical problem: the violation of unitarity is the problem. For example, imagine that you compute the evolution operator between the moments A and C. The evolution operator must be the product of the evolution operator from A to B and the evolution operator from B to C.
• U(A,C) = U(B,C) U(A,B)
Imagine that you decide to eliminate configurations between A and B ending at intermediate states in B that cannot be connected to the initial conditions at A by a classical solution, and you do the same thing for the intervals B-C and A-C. Then you clearly violate the product identity above. It's because U(A,C) will contain many histories with intermediate states at B that are very far from the states at A and C. These configurations are omitted from U(B,C) U(A,B). The failure of the product identity is lethal because if two of the operators above are unitary, the third won't be unitary, and vice versa.
Consequently, the total probability won't be preserved.
Similar ad hoc restrictions of the path integral defined by global criteria are strictly prohibited in any quantum theory. In fact, a typical configuration that contributes to any path integral is a wildly fluctuating function resembling the Brownian motion that has nothing to do with the smooth classical solution. The typical trajectories of particles in quantum mechanics are non-differentiable almost everywhere. That also means that the typical trajectories have superluminal (infinite) velocities almost everywhere - even in relativity. The fact that the full theory (e.g. quantum field theory of a Dirac field) respects causality exactly is a derived fact based on various cancellations.
The classical solution emerges as a stationary point of the action - where the Feynman phase is almost constant and the contributions of nearby trajectories tend to drag the result in the same direction. But this fact is a derived observation, too. You cannot define a quantum theory by requiring that some solutions will be exact. Moreover, this heuristic explanation of the emergence of the classical limit is only correct in the semiclassical approximation.
Loop corrections to the effective action
In fact, beyond the tree - i.e. classical - approximation, we know very well that the classical solutions are not exact. It's because the loop effects (virtual pairs of particles etc.) add their contributions to the effective action. The statement of Rovelli et al. that the quantum theory should be defined by eliminating pairs of initial and final states that can't be interpolated by Ricci-flat solutions (solutions of the classical equations) shows that they are apparently convinced that quantum mechanics can be ignored completely and the classical results will be confirmed exactly.
But the reality is very different than what they seem to think. In the quantum theory, the correct low-energy equations are not given by Ricci-flatness. They are given by Ricci-flatness modified by quantum corrections to the effective action - you can think of the Wilsonian effective action or the 1PI (one-particle-irreducible) effective action; in the latter case, you will encounter infrared problems. Among the corrections, the one-loop corrections can be universally determined from the classical action - because they follow from logarithmic divergences that cannot be subtracted by local counterterms - and they are nonzero.
The coefficients of the higher-loop corrections depend on the details of the short-distance physics and can only be determined if you know the full theory (e.g. in string theory). But at any rate, all these corrections matter. These corrections are the very reason why we use the adjective "quantum" in "quantum gravity" - at least the main reason in the context of graviton scattering.
Our loop quantum gravity colleagues - not only Carlo Rovelli - clearly fail to appreciate the difference between gravity and quantum gravity, and they throw the whole "quantum" into the trash can. They throw it away by definition - because they define their quantum path integral to give you uncorrected, classical equations of motion. And they only "derive" the correct classical result because they are assuming that it comes out and because they are defining their theory by the classical result. What a convoluted example of circular reasoning.
However, such a reasoning is not only circular. It is manifestly flawed.
It can easily be shown that general relativity with vanishing (or "erased") one-loop corrections (of order hbar) violates unitarity. What Rovelli et al. are doing contradicts some of the basic principles of physics and every attempt that has at least 5% of the conceptual errors of the present paper is more or less guaranteed to give an inconsistent quantum theory or something even less encouraging.
Note that you cannot fix the problem by replacing the adjective "Ricci-flat" by a quantum-corrected condition because a priori, you don't know what the quantum-corrected condition is. In a consistent theory, the effective action can be derived from the quantum formalism. Attempts to define a quantum theory by using the effective action - a quantity that should be, on the contrary, derived from the quantum theory - is a flawed circular reasoning that cannot lead anywhere.
Despite all of these missteps, they also seem to fail to derive anything that would resemble the graviton propagator as we know it. The best you can see is a "1/x^2" scaling of some terms in the propagator. But they only obtain this scaling after "N" undefendable steps where "N" is the number of terms in their calculation that have an undesired scaling; the neglected terms moreover have no relation with the actual nonlinear and quantum corrections to gravity required by general covariance and unitarity. It's nice if kids count the number of different cars that they can build from their LEGO, or if the adult physicists count the number of arrangements of some simplices, but computing nice things is something slightly different than having something to say about quantum gravity.
And that's the memo. ;-) | 4,461 | 22,015 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-31 | latest | en | 0.942032 |
https://justaaa.com/finance/130563-bolzano-ltd-has-decided-to-provide-an-annual | 1,716,772,925,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00749.warc.gz | 282,852,400 | 9,988 | Question
# Bolzano Ltd has decided to provide an annual prize in perpetuity for a local university student...
Bolzano Ltd has decided to provide an annual prize in perpetuity for a local university student who demonstrates outstanding leadership. The initial prize, to be awarded in exactly one year's time, will have a value of \$500. The value of the prize will be increased in line with inflation, which can be assumed to be 2.75%. The university will invest funds today to provide for this scholarship, and the return on this investment fund is 5.25%. To set up the fund there is an initial administration cost of \$100. As well, each year there is an administration cost of 1% of the value of that year's prize. Bolzano Ltd is responsible for these costs. For the scholarship to go ahead, Bolzano Ltd must provide the university with all funds today. How much does Bolzano Ltd need to provide today to the university so that the prize can be initiated?
The total prize money is growing by the rate of inflation i.e. 2.75%. Hence, we can calculate the total amount by discounting these prize money amounts by a discount rate. We observe that the return on investment fund is 5.25% bu there is also a 1% fees. Hence, it will grow annually by 4.25%. Therefore, this will be our discount rate. So, now we can use the Gordon growth formula to calculate the initial investment required.
Amount = 500 x (1+g)/(r-g) = 500 x 1.0275/(0.0425 -0.0275) = 34250.
Also, we add the initial administration cost of \$100. So, the total amount becomes \$34350.
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 378 | 1,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-22 | latest | en | 0.950587 |
https://psychology.stackexchange.com/questions/395/how-do-humans-optimize-noisy-multi-variable-functions-in-experimental-settings | 1,716,794,645,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00154.warc.gz | 418,445,782 | 43,326 | # How do humans optimize noisy multi-variable functions in experimental settings?
Imagine an experiment like this:
A participant is asked to optimize an unknown function (let's say minimize) . On each trial the participant provides several input values, and receives an output value. Now also imagine that the output is noisy, in that the same inputs lead to an output plus a random component.
To think of but one of many possible specific examples imagine the following function
$$Y = (X -3)^2 + (Z-2)^2 + (W-4)^2 + e,$$
where $e$ is normally distributed, mean = 0, sd = 3.
On each trial, the participant would provide a value for $X$, $Z$, and $W$. And they would obtain a $Y$ value based on this underlying function. Their aim would be to minimise the value of $Y$. They have not been told the underlying functional form. They only know that there is a global minimum and that there is a random component.
I'm interested in reading about the strategies used by humans to do this task in experimental settings. Note I'm not directly interested in how computers do the task or how programmers and mathematicians might complete this task.
### Questions
• What are some good references for learning about the literature on how humans learn to optimize noisy multi-variable functions?
• What are some of the key findings on how humans optimize noisy multi-variable functions?
• I thought the standard rule-of-thumb was that people assume everything is a straight-line. Feb 15, 2012 at 6:37
• Can you offer a hypothetical about how you would expect them to perform this operation? I guess I'm confused as to whether they'd just be using raw values u = f(w,x,y,z) +e, where w,x,y,z were given or would they be aware of the actual function equation (since your function is in R4, they wouldn't really be able to plot it effectively)? I think you're on to something here, but I'm just not sure if you're testing what you think you are. Feb 15, 2012 at 7:08
• @ArtemKaznatcheev I'm assuming that there is an actual set of input values that results in a global minimum; you could imagine an underlying quadratic function if you like, but I'm interested in more general problems. Feb 15, 2012 at 9:40
• @jonsca Example: On a given trial, the participant provides raw values for X, Z and W, and the program returns a Y based on the unknown function. The participant does not know the underlying function other than perhaps the information that there is a global minimum Y value that they are trying to find. Thus, on the next trial, they might try a different set of X, Z and W values, and they would get a new Y value. And thus, over time, they would tweak the values to try to find the minimum. Their performance might be measured as reverse of the sum of their Y obtained values. Feb 15, 2012 at 9:46
• You might already know this, but just in case, you can also use $$\small Y = (X -3)^2 + (Z-2)^2 + (W-4)^2 + e,$$ (or any of the specifiers here)if you find the equations are too big. Jun 29, 2012 at 6:39
This is a bit of a tangential answer, but hopefully still useful.
When we give humans noisy data, we can basically think of them as some sort of Bayesian inference machines that try to figure out what the function that data came from looks like. The important thing we then need to know, is how strong of a bias (prior) humans have towards expecting certain relationships.
Unfortunately, it seems that humans are extremely biased towards positive linear relationships. I think this will make it very hard for them to optimize data presented as in your question, because they will constantly assume it comes from a straight line. This is really well captured by the following figure from Kalish et al. (2007):
The experiment that generated the above picture is rather different from the one you describe, but we can think of it is a very particular type of noise. A person at stage $$n$$ is given 25 $$(x,y)$$ pairs from the function at stage $$n - 1$$. The person is then tested by being given an x value and asked for a y, 25 times. The results of this are passed on to the person at trial $$n + 1$$ as the training data. Thus, we could think of the errors of person at stage n as noise/errors (although systematic errors) for the person at stage $$n + 1$$. As you can see, it doesn't take much of this noise to lose all structure of the function you started with and revert to the natural bias of a positive linear relationship. In fact, in condition 1 participants are already completely confused about the U-shaped function after strage $$n = 1$$ (so the first participant, with no error already has a hard time understanding the function from $$(x,y)$$ pairs).
### References
Kalish, M. L., Griffiths, T. L., & Lewandowsky, S. (2007). Iterated learning: Intergenerational knowledge transmission reveals inductive biases. Psychonomic Bulletin & Review, 14(M), 288-294. [pdf]
• Thanks for the interesting thoughts. Now that you mention it, I had had a look at the Kalish et al study a few years back. However, I think there are two big differences: the study is iterative and therefore assumptions about the relationship carry over into subsequent trials; Also, from a quick look, the paper seems to be about describing the functional form rather than finding the optimum. Surely almost anyone can estimate an unkonwn numeric quantity (e.g., 271) when the feedback they are given is "higher" or "lower" (e.g., 200, "higher", 300, "lower", 250, "higher", 280, "lower", etc.) Feb 18, 2012 at 1:10
• Of course, adding noise, including multiple input variables, and making the feedback direction-less would make the task harder. Feb 18, 2012 at 1:11
• @JeromyAnglim yeah, the model is different, in that they are given x and asked for y, instead of asked for an x that minimizes the function. So this answer is conditional on the assumption that people actually end up forming some sort of mental representation of the function. If they do not form such a representation (which is often the case in different modalities, like pole-balancing) then my answer is completely irrelevant, haha. However, if the participants give X,Z,W and get a Y as feedback, then I suspect the participants will try to form SOME mental representation of the function. Feb 18, 2012 at 1:15
• the biggest point of my answer/long-comment, was that we shouldn't expect participants to be very good at this. Feb 18, 2012 at 1:16
This seems related to the literature on multiple-cue probability learning (MCPL). In this paradigm, a typical task presents subjects a list of cues and values, and asks them to predict the probability of certain outcomes. This paradigm has a decent amount of literature both in the JDM (judgment and decision making) community as well as the human factors community. To see the relevance, consider a doctor who has to diagnose a patient (provide treatment) based on a finite set of cues (symptoms).
Empirically, human judgment of this type has been modeled using Egon Brunswik's probabilistic functionalism, perhaps more commonly known as the lens model or social judgment theory. This is a useful methodology for comparing human judgment to true ecological correlations.
The image of above depicts the lens model. To give an example, consider the task of a college admissions board who must decide who to admit. The environment/criterion might be their final college GPA, and cues might be high school GPA, SAT scores, writing sample, etc. You can use multiple regression to find the 'true' ecological weights of these cues on the environment criterion, and similarly you can do the same for the admission board's estimate of a student's success (if they were to estimate GPA).
An admission board will (hopefully) observe the effect of different cues on success, and revise their cue weights with experience. Unfortunately, people are typically not so good at this task.
Some common findings:
• People tend to use no more than 3 cues, even if they claim that they use more.
• People are typically outperformed by a bootstrap model of themselves
• People are often outperformed by a unit weight model of themselves: In other words, if you simply set the highest observed cues weights (on the right hand side) to 1, and all others to 0, you may get a better predictor of outcome.
What I take from this is that people will probably have little chance of success at estimating cue weights from a complex equation such as the one you present. However, you could measure these cue weights iteratively to observe learning rates and do other fun stuff-- even if we are all better off being judged by computer algorithms. | 1,961 | 8,638 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-22 | latest | en | 0.941933 |
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# Exponential Breakdown
This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!
Exponential Model on Growth of Internet Networks
1989- 1996
Internet Network: A global network connecting millions of computers. More than 100 countries are linked into exchanges of data, news and opinions. The Internet is revolutionizing and enhancing the way we as humans communicate, both locally and around the globe. Simply put, the Internet is a network of linked computers allowing participants to share information on those computers.
Source: Hobbes' Internet Timeline v8.2 (http://www.zakon.org/robert/internet/timeline/)
Reporting Period Covered: This report summarizes and describes data reported to Zakon Company from 1989-1996
The following data is related to growth of Internet Networks during 1989-1996
Time Networks
1 Jul-89 650
2 Oct-89 837
3 Oct-90 2063
4 Jan-91 2338
5 Jul-91 3086
6 Oct-91 3556
7 Jan-92 4526
8 Apr-92 5291
9 Jul-92 6569
10 Oct-92 7505
11 Jan-93 8258
12 Apr-93 9722
13 Jul-93 13767
14 Oct-93 16533
15 Jan-94 20539
16 Jul-94 25210
17 Oct-94 37022
18 Jan-95 39410
19 Jul-95 61538
20 Jan-96 93671
21 Jul-96 134365
The exponential regression of the data can be established by a scatter gram
Computer Networks is getting a wide range of applications in day to day life. Almost all business activities is somehow related to computer network. The present study tries to estimate the rate of change in the number of networks based on some secondary data.
Equation: A regression line of the form can be suggested to model this data. Here Y represents the number of the internet networks in the month X since 1989.
Thus the exponential model can be written as equation:
Y = 448.380 (1.063)
Prediction : The model Y = 448.380 ( 1.063) can be used to predict the future
value. The above model can be used to predict the future number of internet networks for
example, in the year 2008 (x = 19 ), the number of internet network will be estimated at
502609327.40 ( see below).
Networks = 448.380 ( 1.063 )
= 448.380 ( 1120945.02 )
= 502609327.40
Even though the model explained the variability in networks is 97.86 %., the model assumes that the internet networks will increase exponentially. This model may not be suitable for long term prediction as the growth pattern may drastically change due to advancement in network technology
The Base:
In the model Y = ab the parameter b or base b is known as the compound growth rate.
The base b = 1.063 which means that the number of internet networks growth is
increasing by 6.3 % per year.
https://brainmass.com/math/basic-algebra/exponential-breakdown-116487
#### Solution Preview
For Y = ab^X, if we take ln on both sides, we get ln Y = ln a + X ln b
This model ...
#### Solution Summary
The occurrence of exponential break-down is investigated.
\$2.19
## Poisson and exponential distributions
Please see attached file for full problem description.
The maintenance department of a factory claims that the number of breakdowns of a particular machine follows a Poisson distribution with a mean of two breakdowns every 500hours. Let x denote the time (in hours) between successive breakdowns.
a. Find lambda and mu(x)
b. Write the formula for the exponential probability curve of x.
c. Sketch the probability curve.
d. Assuming that the maintenance department's claim is true, find the probability that the time between successive breakdowns is at most five hours.
e. Assuming that the maintenance department's claim is true, find the probability that the time between successive breakdowns is between 100 hours and 300 hours
f. Suppose that the machine breaks down five hours after its most recent breakdown. Based on the answer to part d, do you believe the maintenance departments claim? | 942 | 3,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-51 | latest | en | 0.832207 |
http://www.gtmath.com/2015/07/ | 1,723,704,645,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641151918.94/warc/CC-MAIN-20240815044119-20240815074119-00746.warc.gz | 43,726,136 | 37,061 | ## Handicapping a Race to 7
### If we give Karl a handicap of 2 games so that he only needs to win 5 games to win the series, whereas Richard needs to win 7, what is the single-game probability $p$ that gives each player a 50% chance of winning the series?
I've tagged this post with "Billiards" since the question came from the director of my pool league (who I hear is an avid GTM reader). Actually, what he really wants to know is the best way to set up the scoring, tie-breakers, and handicaps in one of the leagues, but in order to answer these questions, I wanted to start by looking at just one series and then extend that to the more general questions. This analysis really has nothing to do with pool though, so it would work for any other game as well.
#### Binomial Coefficients
To start out, we'll need the binomial coefficients $\binom{n}{k}$, which are read as "$n$ choose $k$." $\binom{n}{k}$ is the number of ways to choose $k$ items from a set of $n$ distinct items, for example the number of $k$-person boards of directors that can be chosen from $n$ candidates. Note that choosing the $k$ candidates who are included in the board is the same as choosing the $n-k$ who are not included, which means that $\binom{n}{k} = \binom{n}{n-k}$. The formula for the binomial coefficients is $$\dbinom{n}{k} = \dfrac{n!}{k! (n-k)!}$$ from which the above-mentioned symmetry is obvious. Note that even though this is a fraction, it always comes out to be an integer.
The binomial coefficients got their name from the fact that they are the coefficients in the expansion of binomials: $$(x+y)^n = \sum_{k=0}^{n}{\dbinom{n}{k} x^k y^{n-k}}$$ This is because in the product $(x+y)(x+y)...(x+y)$ (with $n$ factors of $(x+y)$), each factor of $(x+y)$ contributes either an $x$ or a $y$ to a factor in the sum. If you have $k$ $x$'s in a term, the other $n-k$ factors of $(x+y)$ must have contributed a $y$. There are $\binom{n}{k}$ ways to get $k$ $x$'s and $n-k$ $y$'s, hence the formula. If anyone wants more detail on that, just ask in the comments, and I'll give a more detailed explanation.
Most identities about binomial coefficients can be proved either by using the formula with the factorials, or via a combinatorial argument. For example, for integers $n$, $m$, and $k$ with $0 \leq k \leq m \leq n$, we have the following identity, known as the subset of a subset identity: $$\dbinom{n}{m} \dbinom{m}{k} = \dbinom{n}{k} \dbinom{n-k}{m-k}$$Algebraic proof: \begin{align} \dbinom{n}{m} \dbinom{m}{k} &= \dfrac{n!}{m! (n-m)!} \cdot \dfrac{m!}{k! (m-k)!} \\[3mm] &= \dfrac{n!}{k!(n-m)!(m-k)!} \\[3mm] &= \dfrac{n!}{k! (n-k)!} \cdot \dfrac{(n-k)!}{(n-m)! (m-k)!} \\[3mm] &= \dbinom{n}{k} \dbinom{n-k}{m-k} \tag*{\square} \end{align}Combinatorial proof:
The left side of the identity is the number of ways to choose a board of directors with $m$ members from $n$ candidates, and then choose $k$ executive members from the $m$. The right side counts the number of ways to choose $k$ executive members from the $n$ candidates and then choose the $m-k$ non-executive board members from the $n-k$ remaining candidates. These count the same thing, so the two sides must be equal. $\tag*{$\square$}$
#### Winning a series to 7
In order to win a series to 7, without needing to win by 2, Karl needs to win 7 games, with Richard winning anywhere from 0 to 6 games in the series. If Karl wins 7 games, and Richard wins 3 games (for example), there will be a total of 10 games in the series. The 3 games that Richard does win can come anywhere in the 10 games, except for the 10th game- if it did, then Karl would have already won 7 and the series would not have made it to 10 in the first place. So we can choose from the first $10-1=9$ games where Richard's 3 wins go.
The probability that Karl wins a given game is $p$, which means the probability that Richard beats Karl is $1-p$. Combining all this, we can see that the probability that Karl beats Richard in a race to 7, with Richard winning 3 games, is $$\binom{10-1}{3} p^7 (1-p)^3$$Since Karl can win the series with Richard winning anywhere from 0 to 6 games, the total probability that Karl wins the series is the sum over the possible outcomes, with the summation index $k$ being the number of wins Richard gets in the series: \begin{align} {\Bbb P}(\text{Karl wins the series}) &= \sum_{k=0}^{7-1}{\binom{7+k-1}{k} p^7 (1-p)^k} \\[3mm] &= \sum_{k=0}^{6}{\binom{6+k}{k} p^7 (1-p)^k} \end{align} Here's a graph of the probability that Karl wins the series, for different values of $p$:
Not surprisingly, if there's a 50% chance that either player wins an individual game, then there's also a 50% chance that either player wins the series.
Now, let's say we give Karl a handicap of 2 games so that to win the series, Karl needs to win 5 games and Richard needs to win 7. More generally, if we call the handicap $H$, where $0 \leq H \leq 6$, then by the same reasoning as we used above, we get the modified formula: $${\Bbb P}(\text{Karl wins the series}) = \sum_{k=0}^{6}{\binom{6-H+k}{k} p^{7-H} (1-p)^k}$$ Now Karl only needs to win $7-H$ games, and so the total number of games in the series for a given value of $k$ wins for Richard, is $7-H+k$, with the $k$ losses once again being placed anywhere but the last game.
Here are the graphs of Karl's probabilities of winning the series given different values of $p$ and $H$ (you can click to expand the picture):
Now, I'd love to be able say we're done here, but the fact is that for some real Karl and Richard, we have no idea what the value of $p$ is unless we are lucky enough to have a history of, say, 100 games between these two players. And even then, they could have improved over time or gotten rusty or whatever so that games they played a few months ago aren't so telling now as to the value of $p$.
We do know that every player in the league is assigned a ranking (which directly determines the handicap against an opponent) which is certainly partly subjective and determined based on observation by a few very experienced players who run, and possibly play in, the league. Instead of trying to guess $p$ and then assigning the rankings, which would be useless in the absence of a large history of games between each set of two players, we can use the handicaps to back out the value of $p$ that makes the match 50-50. For example, if Karl and Richard's rankings are such that Karl gets a handicap of 3, we can see from the graph above that the match will be 50-50 if Karl's probability $p$ of winning an individual game is about 35.5%.
Using Excel's Goal Seek functionality, I've backed out the values of $p$ that make a 7-game series 50-50 for different handicaps:
To test whether a player's handicap is appropriate, one could take all that player's games against opponents of different ranks and see what percentage of individual games he wins and how far off those percentages are from the table above (perhaps using a chi-square test for goodness of fit). If there are not enough games to do this analysis for individual players, then one could start by looking at the percentages for all games and then looking into the ranks furthest away from the table values and seeing if the stats of any particular player(s) are driving the difference. That's a bit of a manual exercise, but it's a start...
"Backing out" $p$ basically means finding the inverse of the function $f(p) = \sum_{k=0}^{6}{\binom{6-H+k}{k} p^{7-H} (1-p)^k}$. We know the function has an inverse because if you look at the graphs, they all pass the horizontal line test. To be more rigorous, they are polynomials in $p$ and thus continuous, and $f(0)=0$ and $f(1)=1$, so the intermediate value theorem tells us that $f$ is surjective. Furthermore, $f$ is increasing on the interval $p \in [0,1]$, so it's also one-to-one, and thus has an inverse.
Now, while Excel Goal Seek will certainly work for this, it would be kind of nice to know the inverse function, so I worked for a few hours today trying to figure out how to invert $f$, but couldn't quite figure it out. Maybe one of my more nerdy readers wants to take a crack at it? Otherwise, maybe I'll go post the question on stack exchange...
[Update 7/29/2015: there's been some confusion on the question I'm asking, so just to clarify, for the purposes of finding the inverse of $f(p)$, assume that the $H$ in the formula above is a constant. So technically, there is a different function $f$ for each value of $H$, which I guess you could call $f_{H}(p)$ or something.]
What I was trying (and maybe this isn't the best way to go about it) was to find a not-too-awful formula for the coefficient of $p^n$ in the sum above and then try to use the Lagrange inversion formula, but it gets a bit messy with all the binomial coefficients. I tried to expand the $(1-p)^k$, turning the sum into a double sum, then switch the order of summation (making sure to adjust the summation limits- the Iverson bracket is helpful at this step), and finally simplify somehow using identities of the binomial coefficients such as the subset of a subset identity above, but said simplification proved elusive, so I didn't even bother with the inversion formula.
Anyone have any thoughts on that or maybe a different way to find the inverse of $f(p)$? Let me know in the comments or email me, and I can provide more details of the computation I tried.
Thanks for reading, and I will try to do a follow-up on this post soon. As always, feel free to ask questions in the comments section.
## Recursions Part 2: Generating Functions
Prerequisites: Power Series, Recursions Part 1
Ok folks, it's been a long and relaxing vacation for me, which is why you haven't seen any new posts in the past 3 weeks. I had intended to do this one Live aus Berlin, but it turned out there was lots of better stuff to do there (including the literally 5 hours I spent at the absolutely fantastic Deutsches Historisches Museum; and I only left because they were closing. I started at Charlemagne at 1 PM and was only half-way through the WW2 section when they closed at 6- I still had the entire GDR and cold war to go...), so this one is coming to you Live aus London instead. Hope that's ok...who am I kidding? I'm pretty sure only like 5 people actually read this anyway, so of course it's ok! Also, those of you who know me won't be surprised to learn that I am writing this from the pub, so just a fair warning that the quality may well deteriorate towards the end...
In the last post, I did a brief overview of how we can use a power series to represent a function, and we wanted the series to converge (pointwise) to our function within a neighborhood of $x=0$, or some other central point. In the post before that, I showed you how to solve a simple recursion. In that post, the recursion we came up with was of degree 1, which means that the $n$-th term depended only on the $(n-1)$-th term, and so we could move them both to the same side of the equals sign to get an expression for the $n$-th difference $R(n)-R(n-1)$. We then summed up the differences to get a formula for $R(n)$ in terms of $n$.
That method was pretty fresh, but if the recursive definition of $R(n)$ includes terms before just $R(n-1)$, for example $R(n-2)$, then this method won't work anymore. What I'm going to show you in this post is one of the most clever things I've ever seen, and it is a very powerful method for dealing with higher-degree recursions (I will specify what this means soon if you haven't guessed it already). Basically, what we are going to do is find a function whose power series $\sum_{i=0}^{\infty}{c_n z^n}$ has coefficients $c_n$ which are the $n$-th terms of the recursion we want to solve. Once we back out the coefficients using Taylor's theorem, we will have solved the recursion. Whoever thought of this was wicked creative (and that's not a subtle self-call, because it definitely wasn't me).
What's interesting is that with generating functions, we just need the power series, but we really won't care when it converges, because we are just using it to back out the coefficients. I'll go into more detail on this point below, but it's a big departure from how we looked at power series before, where we pretty much only cared about if, when, and how (i.e. pointwise, uniformly, etc.) the series converged to our function.
Without further ado, let's get crackin'.
#### Types of Recursions
I started a bit on this above, but a recursion of degree $d$ is one in which the $n$-th term $R(n)$ depends on only some or all of the terms $R(n-1)$ through $R(n-d)$. So $R(n) = R(n-1) + 4nR(n-3)$ is a recursion of degree 3, for example. Note that for a recursion of degree $d$, we need to be given the values of the first $d$ terms in order to solve it.
A recursion equation is linear if the formula for $R(n)$ is a linear combination of the previous terms, i.e. (for some degree $d$) $$R(n) = a_{n-1} R(n-1) + a_{n-2} R(n-2) + ... + a_{n-d} R(n-d) + \alpha (n)$$ where the $a_i$'s can either be numbers or depend on $n$, but can't contain any earlier terms, and $\alpha (n)$ is some fixed function of $n$ called the particularity function. If the latter is just 0, then the recursion (or recurrence, not sure if I'm using the word 100% correctly, but whatever, it's the same thing in my mind) is called homogeneous. If the recursion formula has, for example, products of earlier terms it in, then it isn't linear.
Here are some examples of recursions:
Tower of Hanoi
This one arises when we analyze that game with the 3 rods on which we have discs of varying diameter- you know, this one:
The object of the game is to move all the discs from rod 1 to rod 3 by moving only one at a time from the top of a stack, and with the proviso that we may never place a larger one on top of a smaller one. If there are $n$ discs, then the number of moves $h_n$ it takes to accomplish this is given by the recursion formula $$h_n = 2h_{n-1} + 1 \\ h_1 = 1$$See if you can derive this formula on your own by assuming the number of moves for $n-1$ discs is given and then solving the game for $n$ discs. This is a linear, non-homogeneous recursion of degree 1.
Fibonacci Numbers
The Fibonacci numbers are ubiquitous, arising from the problem of reproducing pairs of rabbits, the golden ratio (where you have rectangles consisting of two squares- a curve through their corners is called a Fibonacci spiral), number of binary strings of length $n$ without consecutive 1's, etc. The recursion formula is $$f_n = f_{n-1} + f_{n-2} \\ f_1 = 1, f_2 = 1$$ which is linear (with constant coefficients, i.e. they don't depend on $n$ as the coefficients are both 1) of degree 2 and homogeneous.
Catalan Numbers
These are ubiquitous in the field of combinatorics, arising from all sorts of counting problems. The recursion formula is $$C_n = C_0 C_{n-1} + C_1 C_{n-2} + ... + C_{n-1} C_0 \\ C_0 = 1$$ which is non-linear, not of finite degree (because the number of previous terms in the formula depends on $n$), and homogeneous.
#### Generating Functions
Now that we've defined what types of "harder" recursions we can encounter, let's move on to the main topic, which is solving them using generating functions.
A generating function for a recurrence $R(n)$ is a formal power series $\sum_{i=0}^{\infty}{c_n z^n}$ where the $n$-th coefficient $c_n$ is equal to the $n$-th recurrence value $R(n)$ for every value of $n$. Technically, I have defined an ordinary generating function (OGF); there are other types such as exponential and Poisson generating functions which are easier to apply in different situations, but they all have the same idea of the $n$-th coefficient's coinciding with the $n$-th thing that we are looking for. I'm only going to talk about OGF's in this post.
In the definition above, I mentioned a formal power series. What this is is a power series where we don't care at all about whether it converges or, if so, for which values of $z$. Basically, we are just going to use the $z$'s as a tool to help keep track of our coefficients $c_n$ without any regard for the convergence properties of the series.
To add and subtract two formal power series, we just add/subtract the coefficients of the terms with the same power of $z$, and to multiply them, we use the distributive property- basically, all the same way that we would do it for a regular power series. The same applies to taking derivatives and integrals of the series.
To show how to use OGF's to solve a recursion, let's take as an example the following linear, homogeneous recursion of degree 2: $$g_n = 5g_{n-1} - 6g_{n-2} \\ g_0 = 1, g_1 = 2$$ What we need to do is find a function $G(z)$ which has a power series whose $n$-th coefficient is equal to $g_n$.
Here's how it's done. We'll multiply both sides of the recursion equation by $z_n$, sum them from $n = 2$ to $n = \infty$ (starting at 2 because the the recursion formula starts at $n=2$, with $g_0$ and $g_1$ given), and then get an algebraic expression in terms of $G(z)$. Heeeeeeeeeere we go:\begin{align} g_n &= 5g_{n-1} - 6g_{n-2} \\[3mm] g_n z^n &= 5g_{n-1}z^n - 6g_{n-2}z^n \\[3mm] \sum_{n=2}^{\infty}{g_n z^n} &= \sum_{n=2}^{\infty}{5g_{n-1}z^n} - \sum_{n=2}^{\infty}{6g_{n-2}z^n} \\[3mm] \sum_{n=2}^{\infty}{g_n z^n} &= 5z \left( \sum_{n=2}^{\infty}{g_{n-1}z^{n-1}} \right) - 6z^2 \left( \sum_{n=2}^{\infty}{g_{n-2}z^{n-2}} \right) \\[3mm] \sum_{n=0}^{\infty}{g_n z^n} - g_1 z - g_0 &= 5z \left( \sum_{n=1}^{\infty}{g_{n-1}z^{n-1}} - g_0 \right) - 6z^2 \left( \sum_{n=2}^{\infty}{g_{n-2}z^{n-2}} \right) \\[3mm] G(z) - g_1 z - g_0 &= 5z \left( G(z) - g_0 \right) - 6z^2 \left( G(z) \right) \\[3mm] G(z) (1 - 5z + 6z^2) &= g_1z + g_0 - 5g_0 z \\[3mm] G(z) &= \dfrac{(g_1 - 5g_0)z + g_0}{1- 5z + 6z^2}\\[3mm] G(z) &= \dfrac{(1 - 3z)}{1- 5z + 6z^2} \\[3mm] G(z) &= \dfrac{(1 - 3z)}{(1 - 3z)(1 - 2z)} \\[3mm] G(z) &= \dfrac{1}{1 - 2z} \\[3mm] G(z) &= \sum_{n=0}^{\infty}{2^n z^n} \end{align} Note that in the last step, we used the known power series $$\dfrac{1}{1-x} = \sum_{n=0}^{\infty}{x^n}$$ and substituted in $2z$ for $x$.
Now, as discussed in the Power Series post, this series only converges when $|x| < 1$ or $|z| < \frac{1}{2}$. But guess what? WE DON'T CARE! Because this was just a way to get the coefficients, and we got them. Thus we have solved the recursion; the solution is $$g_n = 2^n$$ for all values of $n$. You can verify this by plugging the answer back into the recursion equation, and you'll see that both sides of the equals sign balance.
So that is pretty awesome and illustrates the idea behind OGF's. If you don't think that was wicked cool, then I would have to recommend more..."primitive" forms of entertainment for you, such as MTV's hit series Jersey Shore (which, I must admit, I am a big fan of myself...).
Now, the astute reader probably did think that was wicked cool, but is still just a tad bit bugged by the fact that this example was so obviously contrived so that $G(z)$ had a denominator which factored easily to give a really simple answer.
The astute reader is correct. If we look at the exact same recursion, except change one of the initial conditions from $g_0 = 1$ to $g_0 = 0$, we can repeat the exact same process to arrive at $$G(z) = \dfrac{2z}{(1-3z)(1-2z)}$$ which does not easily reduce to a known power series.
But not to worry; we can use a method called partial fraction decomposition to rewrite this quotient as \begin{align} G(z) &= \dfrac{2}{1-3z} + \dfrac{-2}{1-2z}\\[3mm] G(z) &= \sum_{n=0}^{\infty}{2 \cdot 3^n z^n} + \sum_{n=0}^{\infty}{(-2) 2^n z^n} \\[3mm] &\Downarrow \\[3mm] g_n &= 2 \cdot 3^n - 2^{n+1} \end{align} Now, I don't really want to go into detail about partial fraction decomposition, because it's not that exciting, and you can look it up really quickly on Google if you want to know more about it. But I did want to point it out as a useful method when you end up with $G(z)$'s for which you can't just factor the denominator and then use the known power series for $\dfrac{1}{1-x}$.
Another thing I want to point out here is that, although we just looked at an example of a homogeneous recursion, the OGF method also works for non-homogeneous recursions such as the Tower of Hanoi one above.
If you work out the same steps for that one, you'll get the generating function $$H(z) = \dfrac{z}{(1-z)(1-2z)} = \dfrac{1}{1-2z} - \dfrac{1}{1-z}$$ and thus $h_n = 2^n - 1$, once again by using the known series of $\dfrac{1}{1-x}$.
Just to sum up here, I'll show you the OGF's of the Fibonacci and Catalan numbers mentioned above.
For the Fibonacci's, the OGF is $$F(z) = \dfrac{z}{1-z-z^2}$$ You need to use a partial fraction decomposition to extract the coefficients, which turn out to be $$f_n = \dfrac{1}{\sqrt 5} \cdot (\gamma^n - {\hat{\gamma}}^n)$$ where $\gamma$ is the golden mean $\dfrac{1+\sqrt 5}{2}$, and $\hat \gamma$ is its conjugate $\dfrac{1-\sqrt 5}{2}$. This formula for $f_n$ is called the Binet formula for the Fibonacci numbers, named after Jacques Binet, who rediscovered it in 1843 after, you guessed it, our prolific friend Euler had published it in 1765.
I also used the Catalan numbers as an example, and you can click here to see the generating function and formula for these, from the very cool math blog of a guy named Mike Spivey. Who knew there was another math blog out there? Well, there you have it, there is! [Update: as I mention in the comments in response to Joe's question, I didn't do any research on the history of the Binet formula above when I first wrote this post, but now that I have looked into it, I found an interesting history of collaboration as well as independent discovery among famous mathematicians in various nations (frequently at war with each other during the period of European history in question) as they encountered these prolific number sequences and developed generating function and other methods in the 18th and 19th centuries. Here is a link to a good history of the Catalan numbers.]
And with that, there you have it- that's the end of this post. I hope you enjoyed it and thought that OGF's are as cool as I think they are. Just another example of how math and creativity are abolutely NOT mutually exclusive (in fact, quite the opposite, though I must admit that I myself have zero creativity and just copy most of this stuff from somewhere else that I've seen it...)! | 6,287 | 22,414 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-33 | latest | en | 0.94847 |
https://fr.slideshare.net/robintgreene/ch2-part-2-patterns-of-motion | 1,695,327,275,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506029.42/warc/CC-MAIN-20230921174008-20230921204008-00712.warc.gz | 289,990,830 | 64,213 | # Ch2 part 2- patterns of motion
Beulah Heights University
6 Aug 2014
1 sur 31
### Ch2 part 2- patterns of motion
• 1. •Patterns of Motion
• 2. In a moving airplane, you feel forces in many directions when the plane changes its motion. You cannot help but notice the forces involved when there is a change of motion.
• 3. • Laws of Motion
• 4. Among other accomplishments, Sir Isaac Newton invented calculus, developed the laws of motion, and developed the law of gravitational attraction.
• 5. • Newton's First Law of Motion – Every object remains in its state of rest or motion unless acted upon by an unbalanced force. – Objects tend to remain either at rest or in straight line motion. – This tendency to resist changes in motion is inertia. – Mass is a measure of the amount of inertia an object has.
• 6. Top view of a person standing in the aisle of a bus. (A) The bus is at rest, and then starts to move forward. Inertia causes the person to remain in the original position, appearing to fall backward.
• 7. (B) The bus turns to the right, but inertia causes the person to retain the original straight line motion until forced in a new direction by the side of the bus.
• 8. This marble can be used to demonstrate inertia.
• 9. • Newton's Second Law of Motion – The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to the mass of the object. – This law describes the relationship between net force, acceleration, and mass
• 10. This bicycle rider knows about the relationship between force, acceleration, and mass.
• 11. At a constant velocity the force of tire friction (F1) and the force of air resistance (F2) have a vector sum that equals the force applied (Fa). The net force is therefore 0.
• 12. More mass results in less acceleration when the same force is applied. With the same force applied, the riders and the bike with twice as much mass will have half the acceleration, with all other factors constant. Note that the second rider is not pedaling.
• 13. – The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to the mass of the object. – The unit of force used in the SI system is the Newton (N) – N= kg•m/s2 – Force is equal to mass times acceleration • F=ma – Weight is equal to the mass of an object times the force of gravity • w=mg
• 14. • Newton's Third Law of Motion. – Whenever two objects interact, the force exerted on one object is equal in size and opposite in direction to the force exerted on the other object. • FA due to B = FB due to A
• 15. Forces occur in matched pairs that are equal in magnitude and opposite in direction.
• 16. The football player's foot is pushing against the ground, but it is the ground pushing against the foot that accelerates the player forward to catch a pass.
• 17. Both the astronaut and the satellite received a force of 30.0 N for 1.50 s when they pushed on each other. Both then have a momentum of 45.0 kg m/s in the opposite direction. This is an example of the law of conservation of momentum.
• 18. • Momentum
• 19. • Momentum (ρ) is the product of the mass of an object (m) and its velocity (v). ρ = mv • The law of conservation of momentum – The total amount of momentum remains constant in the absence of some force applied to the system.
• 20. • Two unusual aspects of momentum – The symbol for momentum does not indicate what the quantity if measures is. – The units for momentum (kg•m/s) has no name.
• 21. According to the law of conservation of momentum, the momentum of the expelled gases in one direction equals the momentum of the rocket in the other direction in the absence of external forces.
• 22. Forces and Circular Motion.
• 23. • Centripetal force. – This is the force that keeps an object in its straight line path • Centrifugal force. – The imaginary force that is thought to force objects toward the outside of an object moving in a circular pattern. – Actually the force is simply the tendency of the object to move in a straight line.
• 24. • The acceleration of an object moving in a circular path (ac) is – ac = v2 /r • m = mass • v = velocity • r = the radius of the circular path. – This can be substituted into the Force equation F = ma • F = mv2 /r
• 25. Centripetal force on the ball causes it to change direction continuously, or accelerate into as circular path. Without the unbalanced force acting on it, the ball would continue in a straight line.
• 26. • Newton's Law of Gravitation.
• 27. • Objects fall due to the force of gravity (g) on them. – This force is 9.8 m/s2 – It is this force that gives objects weight • w = mg
• 28. • Universal Law of Gravitation – Every object in the universe is attracted to every other object in the universe by a force that is directly proportional to the product of their masses and inversely proportional to the square of the distances between them. • F = G(m1m2)/d2 • G is a proportionality constant and is equal to 6.67 X 10-11 N•m2 /kg2 – Usually the objects in our environment that we interact with on an everyday basis are so small that the force is not noticed due to the large force of attraction due to gravity.
• 29. The variables involved in gravitational attraction. The force of attraction (F) is proportional to the product of the masses (m1, m2) and inversely proportional to the square of the distance (d) between the centers of the two masses.
• 30. The force of gravitational attraction decreases inversely with the square of the distance from the earth's center. Note the weight of a 70.0 kg person at various distances above the earth's surface.
• 31. Gravitational attraction acts as a centripetal force that keeps the Moon from following the straight-line path shown by the dashed line to position A. It was pulled to position B by gravity (0.0027 m/s2) and thus "fell" toward Earth the distance from the dashed line to B, resulting in a somewhat circular path. | 1,411 | 5,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-40 | latest | en | 0.90709 |
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# Science teacher: In any nation, a flourishing national scientific comm
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23 Jul 2018, 07:52
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Science teacher: In any nation, a flourishing national scientific community is essential to a successful economy. For such a community to flourish requires that many young people become excited enough about science that they resolve to become professional scientists. Good communication between scientists and the public is necessary to spark that excitement.
The science teacher's statements provide the most support for which one of the following?
(A) If scientists communicate with the public, many young people will become excited enough about science to resolve to become professional scientists.
(B) The extent to which a national scientific community flourishes depends principally on the number of young people who become excited enough about science to resolve to become professional scientists.
(C) No nation can have a successful economy unless at some point scientists have communicated well with the public.
(D) It is essential to any nation 's economy that most of the young people in that nation who are excited about science become professional scientists.
(E) An essential component of success in any scientific endeavor is good communication between the scientists. involved in that endeavor and the public.
Source: LSAT
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Science teacher: In any nation, a flourishing national scientific comm [#permalink]
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Updated on: 23 Jul 2018, 10:33
Between A and B,but will go with B as it covers the complete chain reaction
So B
Sent from my iPhone using GMAT Club Forum mobile app
Originally posted by gilltaurus on 23 Jul 2018, 09:29.
Last edited by gilltaurus on 23 Jul 2018, 10:33, edited 1 time in total.
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Re: Science teacher: In any nation, a flourishing national scientific comm [#permalink]
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23 Jul 2018, 10:26
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(A) If scientists communicate with the public, many young people will become excited enough about science to resolve to become professional scientists. It makes an error in reverse deduction. Good Communication is a necessary condition to spark excitement. It may not be sufficient in itself.
(B) The extent to which a national scientific community flourishes depends principally on the number of young people who become excited enough about science to resolve to become professional scientists.The passage doesn't state that the success is proportional to the "number of people" who become excited.
(C) No nation can have a successful economy unless at some point scientists have communicated well with the public. Correct Answer. Good Comm. between the scientists & public is a necessary condition to spark that excitement. So, it must have occurred at some point.
(D) It is essential to any nation 's economy that most of the young people in that nation who are excited about science become professional scientists. I fell for it. The keyword here is "most". The passage says "many".
(E) An essential component of success in any scientific endeavor is good communication between the scientists. involved in that endeavor and the public.Out of scope - It says "any scientific endeavor. The passage talks about a successful economy. "
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Re: Science teacher: In any nation, a flourishing national scientific comm [#permalink]
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23 Jul 2018, 10:42
Science teacher opens the statement telling us " A flourishing national scientific community is essential to a successful economy"-Conclusion
Ends the argument by saying
"Good communication between scientists and the public is necessary to spark that excitement"
So we need to find a answer which combines both of these statements as the assumption of teacher will be dependent on it and 'C' exactly does that.
Keep in mind that our end goal is to support the conclusion. All other options are traps such as A,B
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Re: Science teacher: In any nation, a flourishing national scientific comm &nbs [#permalink] 23 Jul 2018, 10:42
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# Science teacher: In any nation, a flourishing national scientific comm
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,387 | 6,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-47 | latest | en | 0.879536 |
https://number.academy/949928 | 1,659,913,339,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570730.59/warc/CC-MAIN-20220807211157-20220808001157-00505.warc.gz | 405,382,855 | 12,217 | # Number 949928
Number 949,928 spell 🔊, write in words: nine hundred and forty-nine thousand, nine hundred and twenty-eight . Ordinal number 949928th is said 🔊 and write: nine hundred and forty-nine thousand, nine hundred and twenty-eighth. Color #949928. The meaning of number 949928 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 949928. What is 949928 in computer science, numerology, codes and images, writing and naming in other languages
## What is 949,928 in other units
The decimal (Arabic) number 949928 converted to a Roman number is (C)(XL)(IX)CMXXVIII. Roman and decimal number conversions.
#### Weight conversion
949928 kilograms (kg) = 2094211.3 pounds (lbs)
949928 pounds (lbs) = 430884.5 kilograms (kg)
#### Length conversion
949928 kilometers (km) equals to 590258 miles (mi).
949928 miles (mi) equals to 1528762 kilometers (km).
949928 meters (m) equals to 3116524 feet (ft).
949928 feet (ft) equals 289542 meters (m).
949928 centimeters (cm) equals to 373987.4 inches (in).
949928 inches (in) equals to 2412817.1 centimeters (cm).
#### Temperature conversion
949928° Fahrenheit (°F) equals to 527720° Celsius (°C)
949928° Celsius (°C) equals to 1709902.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
949928 seconds equals to 1 week, 3 days, 23 hours, 52 minutes, 8 seconds
949928 minutes equals to 1 year, 11 months, 2 weeks, 1 day, 16 hours, 8 minutes
### Codes and images of the number 949928
Number 949928 morse code: ----. ....- ----. ----. ..--- ---..
Sign language for number 949928:
Number 949928 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 949928
### Multiplications
#### Multiplication table of 949928
949928 multiplied by two equals 1899856 (949928 x 2 = 1899856).
949928 multiplied by three equals 2849784 (949928 x 3 = 2849784).
949928 multiplied by four equals 3799712 (949928 x 4 = 3799712).
949928 multiplied by five equals 4749640 (949928 x 5 = 4749640).
949928 multiplied by six equals 5699568 (949928 x 6 = 5699568).
949928 multiplied by seven equals 6649496 (949928 x 7 = 6649496).
949928 multiplied by eight equals 7599424 (949928 x 8 = 7599424).
949928 multiplied by nine equals 8549352 (949928 x 9 = 8549352).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 949928
Half of 949928 is 474964 (949928 / 2 = 474964).
One third of 949928 is 316642,6667 (949928 / 3 = 316642,6667 = 316642 2/3).
One quarter of 949928 is 237482 (949928 / 4 = 237482).
One fifth of 949928 is 189985,6 (949928 / 5 = 189985,6 = 189985 3/5).
One sixth of 949928 is 158321,3333 (949928 / 6 = 158321,3333 = 158321 1/3).
One seventh of 949928 is 135704 (949928 / 7 = 135704).
One eighth of 949928 is 118741 (949928 / 8 = 118741).
One ninth of 949928 is 105547,5556 (949928 / 9 = 105547,5556 = 105547 5/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
949928
#### Is Prime?
The number 949928 is not a prime number. The closest prime numbers are 949903, 949931.
#### Factorization and factors (dividers)
The prime factors of 949928 are 2 * 2 * 2 * 7 * 16963
The factors of 949928 are
1 , 2 , 4 , 7 , 8 , 14 , 28 , 56 , 16963 , 33926 , 67852 , 118741 , 135704 , 237482 , 474964 , 949928
Total factors 16.
Sum of factors 2035680 (1085752).
#### Powers
The second power of 9499282 is 902.363.205.184.
The third power of 9499283 is 857.180.074.774.026.752.
#### Roots
The square root √949928 is 974,642499.
The cube root of 3949928 is 98,302274.
#### Logarithms
The natural logarithm of No. ln 949928 = loge 949928 = 13,764141.
The logarithm to base 10 of No. log10 949928 = 5,977691.
The Napierian logarithm of No. log1/e 949928 = -13,764141.
### Trigonometric functions
The cosine of 949928 is -0,082959.
The sine of 949928 is -0,996553.
The tangent of 949928 is 12,012527.
### Properties of the number 949928
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 949928 in Computer Science
Code typeCode value
PIN 949928 It's recommendable to use 949928 as a password or PIN.
949928 Number of bytes927.7KB
CSS Color
#949928 hexadecimal to red, green and blue (RGB) (148, 153, 40)
Unix timeUnix time 949928 is equal to Sunday Jan. 11, 1970, 11:52:08 p.m. GMT
IPv4, IPv6Number 949928 internet address in dotted format v4 0.14.126.168, v6 ::e:7ea8
949928 Decimal = 11100111111010101000 Binary
949928 Decimal = 1210021001112 Ternary
949928 Decimal = 3477250 Octal
949928 Decimal = E7EA8 Hexadecimal (0xe7ea8 hex)
949928 BASE64OTQ5OTI4
949928 MD51917111255f7c8547cb8fe8ee79345f9
949928 SHA1c35a1ed56a734379197a815f92940688e5a54f44
949928 SHA224d3e27f8b0389313ee1fc38a8583d55165b90e572221be5fd38aca31b
949928 SHA25673b5c2a2d1f971c833186576e901b0a9fa955167f48ca5cd5566e0336cb7d129
More SHA codes related to the number 949928 ...
If you know something interesting about the 949928 number that you did not find on this page, do not hesitate to write us here.
## Numerology 949928
### Character frequency in number 949928
Character (importance) frequency for numerology.
Character: Frequency: 9 3 4 1 2 1 8 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 949928, the numbers 9+4+9+9+2+8 = 4+1 = 5 are added and the meaning of the number 5 is sought.
## № 949,928 in other languages
How to say or write the number nine hundred and forty-nine thousand, nine hundred and twenty-eight in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 949.928) novecientos cuarenta y nueve mil novecientos veintiocho German: 🔊 (Anzahl 949.928) neunhundertneunundvierzigtausendneunhundertachtundzwanzig French: 🔊 (nombre 949 928) neuf cent quarante-neuf mille neuf cent vingt-huit Portuguese: 🔊 (número 949 928) novecentos e quarenta e nove mil, novecentos e vinte e oito Chinese: 🔊 (数 949 928) 九十四万九千九百二十八 Arabian: 🔊 (عدد 949,928) تسعمائة و تسعة و أربعون ألفاً و تسعمائة و ثمانية و عشرون Czech: 🔊 (číslo 949 928) devětset čtyřicet devět tisíc devětset dvacet osm Korean: 🔊 (번호 949,928) 구십사만 구천구백이십팔 Danish: 🔊 (nummer 949 928) nihundrede og niogfyrretusindnihundrede og otteogtyve Dutch: 🔊 (nummer 949 928) negenhonderdnegenenveertigduizendnegenhonderdachtentwintig Japanese: 🔊 (数 949,928) 九十四万九千九百二十八 Indonesian: 🔊 (jumlah 949.928) sembilan ratus empat puluh sembilan ribu sembilan ratus dua puluh delapan Italian: 🔊 (numero 949 928) novecentoquarantanovemilanovecentoventotto Norwegian: 🔊 (nummer 949 928) ni hundre og førti-ni tusen, ni hundre og tjue-åtte Polish: 🔊 (liczba 949 928) dziewięćset czterdzieści dziewięć tysięcy dziewięćset dwadzieścia osiem Russian: 🔊 (номер 949 928) девятьсот сорок девять тысяч девятьсот двадцать восемь Turkish: 🔊 (numara 949,928) dokuzyüzkırkdokuzbindokuzyüzyirmisekiz Thai: 🔊 (จำนวน 949 928) เก้าแสนสี่หมื่นเก้าพันเก้าร้อยยี่สิบแปด Ukrainian: 🔊 (номер 949 928) дев'ятсот сорок дев'ять тисяч дев'ятсот двадцять вiсiм Vietnamese: 🔊 (con số 949.928) chín trăm bốn mươi chín nghìn chín trăm hai mươi tám Other languages ...
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If you know something interesting about the number 949928 or any natural number (positive integer) please write us here or on facebook. | 2,569 | 7,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-33 | latest | en | 0.603919 |
https://solvedlib.com/how-to-find-the-friction-factor-by-using-only,22760 | 1,695,506,944,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.3/warc/CC-MAIN-20230923194908-20230923224908-00242.warc.gz | 600,211,339 | 17,763 | # How to find the friction factor by using only experimental data, not the theoretical. Mann Tahoma...
###### Question:
how to find the friction factor by using only experimental data, not the theoretical.
Mann Tahoma 11 Al Paste B I U abe X, X A- Aa-A Font 21 Paragraph Clipboard 2 RESULT 2.1 Experiment 1: Smooth Pipe Test section: Smooth pipe Test section diameter: 0.0254m Area="de = 5.067x10-4m Water temperature : 27°C Density at 27°C: 996.59 kg/m3 Dynamic viscosity at 27°C: 0.8509x10-3 Pas
Clipboard Font Paragraph Styles 1.0 7.00 16.66 18.84 17.75 27.3 22.5 24.9 9.2 9.1 9.15 Experimental Value: Flowrate, (m^3/h) Hf(head in head out) (m) 17.50 Velocity(m/s) Velocity head(m) Friction factor 1.4 0.767 10.61 1.2 17.25 0.658 10.08 1.0 15.75 0.548 9.73 Theoretical value: Flowrate, 0 (m^3/b) 1.4 Friction factor 6.43x10^-3 Velocity(m/s) Reynold number, Re 0.767 22,817 0.658 19,574 0.548 16,302 1.2 6.68x10^-3 1.0 7.box10^-3
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Python Question
# while loop and less than or equal to sign (Python)
So I was doing while loops and I noticed something strange.
``````count = 0
while count <= 5:
count += 1
print(count)
``````
output:
``````1
2
3
4
5
6
``````
it's not that I don't understand while loops. It's that how come the count is printed up to six? when it's supposed to print
`count`
only if
`count`
is less than or equal to 5?
and well 6 is beyond 5. why is this?
I know I could do
``````count = 0
while count != 5:
count += 1
print(count)
``````
but I just want to understand why does putting
`<=`
behave in an odd way?
There is nothing odd about `<=`; your loop condition allows for numbers up to and including `5`. But you increment `count` and then print it, so you will print `6` last.
That's because `count = 5` satisfies your loop condition, then you add one to make it `6` and print. The next time through the loop `count <= 5` is no longer true and only then loop ends.
1. `count = 0`, `count <= 5` -> `True`, `count += 1` makes `count = 1`, print `1`.
2. `count = 1`, `count <= 5` -> `True`, `count += 1` makes `count = 2`, print `2`.
3. `count = 2`, `count <= 5` -> `True`, `count += 1` makes `count = 3`, print `3`.
4. `count = 3`, `count <= 5` -> `True`, `count += 1` makes `count = 4`, print `4`.
5. `count = 4`, `count <= 5` -> `True`, `count += 1` makes `count = 5`, print `5`.
6. `count = 5`, `count <= 5` -> `True`, `count += 1` makes `count = 6`, print `6`.
7. `count = 6`, `count <= 5` -> `False`, end the loop.
You could increment the counter after printing:
``````while count <= 5:
print(count)
count += 1
``````
or you could use `<` to only allow numbers smaller than `5`:
``````while count < 5:
count += 1
print(count)
`````` | 587 | 1,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2016-50 | longest | en | 0.919727 |
http://www.financialdictionary.net/define/Accrued+Interest/ | 1,558,717,818,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257699.34/warc/CC-MAIN-20190524164533-20190524190533-00514.warc.gz | 282,545,106 | 2,714 | # Accrued Interest
Accrued interest is not a good real estate term and in most cases means a payment that has been made is below the agreed payment. This means the interest left over after the premium was paid accumulates on the loan. This is known as a negative amortization or deferred interest, because it is interest that has been earned and has not been paid.
For the buyer of the property, while they have paid a payment it keeps them from sliding behind, but at the same time it also places this earned interest back on to the loan amount at the end of the loan. This also means that it can add more to the loan amount when calculating the full loan amount.
The interest amount on the loan grows from monthly payment to monthly payment, which is why when making extra payments on the loan amount making them on the principal alone is not enough, because of the accrued interest. This extra paid amount would first be applied to the mounted up interest, with any left over going toward the principal.
Mortgage loans are calculated with daily accrued interest, and are done using a mathematical formula that is 30 days per month with 360 days per year. This accounts for the short month of February and the longer months having 31 days, where no interest is added. This formula also ignores leap years and the system of calculation stays the same. This way of calculating mortgages affects the payment that is made in full when it is paid late, such as the payment that is made 14 days late will have accrued approximately an extra \$35.00 dollars in interest. | 319 | 1,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-22 | longest | en | 0.983788 |
https://www.homeownershub.com/maintenance/abridged-version-alarm-system-transformer-power-supply-772663-1.htm | 1,493,614,943,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917127681.50/warc/CC-MAIN-20170423031207-00609-ip-10-145-167-34.ec2.internal.warc.gz | 905,198,663 | 15,869 | # Abridged version - Alarm system transformer + power supply
On 11/29/2013 5:57 PM, snipped-for-privacy@attt.bizz wrote:
I watched a documentary about the modern British Navy where they were using yards for the ship's guns but the ground troupes were using meters. I'm going to guess that all the naval artillery tables and calculations would have to be redone to convert to metric. I remember that the older ships have mechanical computers to calculate trajectories and it would be quite a chore to refit those old ships. ^_^
TDD
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On Fri, 29 Nov 2013 19:16:42 -0600, The Daring Dufas
Note that the Navy also uses nautical miles for surface distance (and yards for torpedoes). All of the conversions are trivial for the few times they're necessary. Certainly easier than the entire service changing. The same is true for domestic use of FPS. With calculators and computers, it's even more true.
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On 11/29/2013 7:48 PM, snipped-for-privacy@attt.bizz wrote:
You can't teach an old Swabby new tricks. Perhaps the reprogramming of a sailor's brain is more difficult? ^_^
TDD
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On Fri, 29 Nov 2013 21:15:10 -0600, The Daring Dufas
Perhaps, when things get tense and lives are on the line, you want training to take over and actions to be automatic.
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wrote:
Clueless. There is no difference. There can't be.
He isn't any brighter than you are, then.
Utter nonsense. Electronics is mostly metric, but also uses base 2, 8, 16, and logarithms with base e. I do this stuff daily, yet have no problem doing woodworking in inch-feet and pounds. Some of my tools are metric. No problem. The calibration marks aren't normally necessary anyway. If they are, the conversion is trivial.
Who cares what you do? Knock yourself out. You've said nothing that supports your asinine position.
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# Deviation from straight line up
If I want to compute the MAD and AUC from a straight line up, how would I do that?
My data looks like below
I tried
mt <- mt_align(mt, coordinates = c(0, 0, -1, 1.2124))
mt <- mt_time_normalize(mt)
I then use some custom code to flip some of the trajectories to the left based on a given condition
mt<- mt_deviations(mt,end_ideal = c(0, 1.2124))
followed by mt_measures, but my AUC and MAD occasionally come back positive - even though my data clearly shows a trend to the left (with the exception of one trajectory)
xpos_min shows up as expected
It seems to be doing something with whether the first xpos_max is positive or negative to some extent, but I cant use mt_align with align_start because it makes the trajectories weird
• Hi there,
details on the calculation of MAD and AUC in mousetrap are described here: http://pascalkieslich.github.io/mousetrap/reference/mt_measures.html#details-1 (and in general how deviations from the idealized straight line are calculated is described here http://pascalkieslich.github.io/mousetrap/reference/mt_deviations.html).
I think that the challenge in your data is that the x position of the start of your trajectory and the end are very closely to each other, which is unusual for mouse-tracking experiments where the cursor usually starts somewhere in the bottom center of the screen and then ends either in the top-left or top-right corner. This might cause the weird behavior, as outlined in the following example: Let's say you have a trajectory the starts in the bottom center, then is slightly curved towards the right, and then moves towards an option at the top that is slightly left of the start position. In this case you get a positive MAD (as the deviation is considered to be "above" the idealized straight line). Now if the trajectory shape looks similar but the end position is slightly right of the start position, you get a negative MAD (as the deviation is now considered to be "below" the idealized straight line). In typical mouse-tracking experiments this does not occur as the options between a person is choosing are clearly left and right of the start position.
Best regards,
Pascal
• That makes sense!
I also have the 'normal' set up with two end points left and right (everything is equidistant from start). Would it make sense to align them in the same workflow first or analyze them as two distinct dataset and merge them at the end? I get vastly different results if I align them as one dataset or align this 'straight' line dataset. I've tried to look into the 'raw' code, but do not fully understand what the align package is doing.
I'll probably write a wrapper around mousetrap to analyze the 'straight line' in the end. Happy to share.
• The specific mt_align function only makes sense for trajectories that come from the same layout. The mt_align_start function could also work for trajectories from different layouts.
Yes, sharing the code you develop for the special case trajectories sounds great! | 695 | 3,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.917193 |
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Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6363 Accepted: 3081
Description
The cows are so very silly about their dinner partners. They haveorganized themselves into three groups (conveniently numbered 1, 2, and 3) thatinsist upon dining together. The trouble starts when they line up at the barnto enter the feeding area.
Each cow i carries with her a small card uponwhich is engraved Di (1 ≤ Di ≤ 3) indicating her dininggroup membership. The entire set of N (1 ≤ N ≤ 30,000) cows has lined up for dinnerbut it's easy for anyone to see that they are not grouped by theirdinner-partner cards.
FJ's job is not so difficult. He just walks down the line of cowschanging their dinner partner assignment by marking out the old number andwriting in a new one. By doing so, he creates groups of cows like 111222333 or333222111 where the cows' dining groups are sorted in either ascending ordescending order by their dinner cards.
FJ is just as lazy as the next fellow. He's curious: what is theabsolute mminimum number of cards he must change to create a proper grouping ofdining partners? He must only change card numbers and must not rearrange thecows standing in line.
Input
* Line 1: A single integer: N
* Lines 2..N+1: Line i describes the i-th cow's current dining groupwith a single integer: Di
Output
* Line 1: A single integer representing the minimum number ofchanges that must be made so that the final sequence of cows is sorted ineither ascending or descending order
SampleInput
5
1
3
2
1
1
SampleOutput
1
Source
`#include <cstdio> #include <algorithm> using namespace std; #define N 30005 int dp1[N],dp2[N],a[N],ans1[N],ans2[N]; int main() { int n; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&a[i]); } int len =0,an1=0; for(int i=0;i<n;i++) //最长不下降子序列upper实现 { int pos=upper_bound(ans1,ans1+len,a[i])-ans1; if(pos<len) ans1[pos]=a[i]; else ans1[len++]=a[i]; dp1[i]=len; an1=max(an1,dp1[i]); } len=0; int an2=0; for(int i=n-1;i>=0;i--) { int pos=upper_bound(ans2,ans2+len,a[i])-ans2; if(pos<len) ans2[pos]=a[i]; else ans2[len++]=a[i]; dp2[i]=len; an2=max(an2,dp2[i]); } int ans; ans=max(an1,an2); printf("%d\n",n-ans); return 0; }` | 696 | 2,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-14 | latest | en | 0.810516 |
https://physicscatalyst.com/mech/conservative-force.php | 1,720,921,419,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00115.warc.gz | 412,671,035 | 15,326 | # Conservative Forces
## (11) Conservative Forces
• Consider the gravitational force acting on a body .If we try to move this body upwards by applying a force on it then work is done against gravitation
• Consider a block of mass m being raised to height h vertically upwards as shown in fig 8(a) .Work done in this case is mgh
• Now we make the block travel the path given in figure 8(b) to raise its height h above the ground.In this path work done during the horizontal motion is zero because there is no change in height of the body due to which there would be no change in gravitational PE of the body and if there is no change in the speed KE would also remains same
• Thus for fig 8(b) if we add up the work done in two vertical paths the result we get is equal to mgh
• Again if we move the the block to height h above the floor through an arbitrary path as shown in fig 8(c) ,the work done can be calculated by breaking the path into elementary horizontal and vertical portions
• Now work done along the horizontal path would be zero and along the vertical paths its add up to mgh
• Thus we can say that work done in raising on object against gravity is independent of the path taken and depends only on the initial and final position of the object
• Now we are in position to define the conservative forces
" If the work done on particle by a force is independent of how particle moves and depends only on initial and final position of the objects then such a force is called conservative force"
Gravitational force,electrostatic force ,elastic force and magnetic forces are conservative forces
• Total work done by the conservative force is zero when particle moves around any closed path returning to its initial position
• Frictional forces and viscous forces are examples of non-conservative forces as these forces always oppose the motion and result in loss in KE
• Concept of Potential Energy is associated with conservative forces only .No such PE is associated with non-conservative forces like frictional forces
• Mathematically Potential Energy function U is defined if Force F can be written as
$F_s= \frac {dU}{ds}$
Where F_s is the component of force in the direction of ds
or
dU = -F_s.ds
• For One-dimensional motion, this can be written as
$F= -\frac {dU}{dx}$
or
dU= -F.dx
or
$U_A - U_B = \int_{x_a}^{x_b} fdx$
• For three dimensional motion,this can be written as
$\mathbf{F}=-\frac{\partial U}{\partial x}\mathbf{i} -\frac{\partial U}{\partial y}\mathbf{j} -\frac{\partial U}{\partial z}\mathbf{k}$
Watch this tutorial for more information on How to solve work-energy problem | 613 | 2,608 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-30 | latest | en | 0.936271 |
http://www.stata.com/statalist/archive/2012-06/msg00990.html | 1,500,972,794,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425117.42/warc/CC-MAIN-20170725082441-20170725102441-00103.warc.gz | 561,161,761 | 4,898 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# RE: st: Question about getting residuals by group
From Pradipto Banerjee To "statalist@hsphsun2.harvard.edu" Subject RE: st: Question about getting residuals by group Date Wed, 20 Jun 2012 11:20:14 -0500
```Thanks, Joerg. It may work, but looping may slow down my code. Any other suggestion for a faster operation than using a loop, e.g. any idea if this can be done using matrix manipulations?
-----Original Message-----
From: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Joerg Luedicke
Sent: Wednesday, June 20, 2012 12:07 PM
To: statalist@hsphsun2.harvard.edu
Subject: Re: st: Question about getting residuals by group
You could loop over the levels of your GroupID variable. Consider the
following example:
*------------------------------------
sysuse auto, clear
gen resid=.
levelsof foreign, local(groups)
foreach a of local groups {
reg price mpg if foreign==`a'
tempvar d
predict `d', residuals
replace resid=`d' if foreign==`a'
}
*------------------------------------
J.
On Wed, Jun 20, 2012 at 10:42 AM, Pradipto Banerjee
> I want to carry out a regression and then use the residual, but within groups. More specifically, suppose there is a variable GroupID, and variables DependVar & IndepVar. Currently, Stata allows the follow:
>
> . regress DependVar IndepVar
> . predict PredictVar
> . gen ResidVar = DependVar - PredictVar
>
> If I were to do
>
> . bys GroupID: regress DependVar IndepVar
>
> Then I noticed that
>
> .ereturn list
>
> only retains the regression parameters for the very last GroupID for which it did the regression. So, if I were to combine the following two steps:
>
> . bys GroupID: regress DependVar IndepVar
> . predict PredictVar
>
> My understanding is that PredictVar would use the regression results from the very last GroupID. I want to use the regression results from each GroupID to generated the predicted variable for that GroupID. Are there any alternative ways?
>
> Thanks,
>
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*
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``` | 926 | 4,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-30 | latest | en | 0.768699 |
https://www.numbersaplenty.com/2246151523 | 1,708,554,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473558.16/warc/CC-MAIN-20240221202132-20240221232132-00267.warc.gz | 941,793,184 | 3,267 | Search a number
2246151523 = 72114167257
BaseRepresentation
bin1000010111100001…
…1000110101100011
312210112112022122011
42011320120311203
514100003322043
61010514453351
7106442654500
oct20570306543
95715468564
102246151523
11a52991980
12528294257
1329a470155
141744524a7
15d22d659d
hex85e18d63
2246151523 has 12 divisors (see below), whose sum is σ = 2850404472. Its totient is φ = 1750247520.
The previous prime is 2246151521. The next prime is 2246151541. The reversal of 2246151523 is 3251516422.
It is not a de Polignac number, because 2246151523 - 21 = 2246151521 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (2246151521) by changing a digit.
It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 2083090 + ... + 2084167.
It is an arithmetic number, because the mean of its divisors is an integer number (237533706).
Almost surely, 22246151523 is an apocalyptic number.
2246151523 is a deficient number, since it is larger than the sum of its proper divisors (604252949).
2246151523 is a wasteful number, since it uses less digits than its factorization.
2246151523 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 4167282 (or 4167275 counting only the distinct ones).
The product of its digits is 14400, while the sum is 31.
The square root of 2246151523 is about 47393.5810316123. The cubic root of 2246151523 is about 1309.6231697822.
Adding to 2246151523 its reverse (3251516422), we get a palindrome (5497667945).
It can be divided in two parts, 22461 and 51523, that added together give a square (73984 = 2722).
The spelling of 2246151523 in words is "two billion, two hundred forty-six million, one hundred fifty-one thousand, five hundred twenty-three". | 565 | 1,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-10 | latest | en | 0.824435 |
https://mathematica.stackexchange.com/tags/summation/new | 1,627,320,358,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00473.warc.gz | 400,218,849 | 25,525 | # Tag Info
0
Clean-up It's hard to read your code and it's hard to guess what you're asking for. We can make things a lot cleaner with a few tweaks. Firstly, you're defining $\delta$ by hand, when it's clearly just the KroneckerDelta. Next, in your definitions of t and x, Signature already handles whether the arguments are equal and/or ordered. So you can delete these. ...
3
Here's a way to make the syntax proposed in the commands work: Sum[f[{i1, i2, i3}], Evaluate[Sequence @@ Thread[{{i1, i2, i3}, 0, ∞}]]] Or using a predefined list l: l = {i1, i2, i3}; Sum[f[l], Evaluate[Sequence @@ Thread[{l, 0, ∞}]]] (* same output *)
1
One way: idx = {i1, i2, i3}; lower = {0, 1, 2}; upper = {2, 3, 4}; Sum[f[Sequence @@ idx], Evaluate[Sequence @@ Transpose[{idx, lower, upper}]]] (* f[0, 1, 2] + f[0, 1, 3] + f[0, 1, 4] + f[0, 2, 2] + f[0, 2, 3] + f[0, 2, 4] + f[0, 3, 2] + f[0, 3, 3] + f[0, 3, 4] + f[1, 1, 2] + f[1, 1, 3] + f[1, 1, 4] + f[1, 2, 2] + f[1, 2, 3] + f[1, 2, 4] + f[1, ...
5
You are admittedly probably better off using a package such as the one @qahtah mentioned in the comments or the package xAct, but just for fun, here is a way of defining a function EinsteinSum which does what you're looking for—unless I've misunderstood. If you have any questions about why this works the way it does, or if I've misunderstood what you're ...
0
Some things change to better. In 12.3 f[k_] = Sin[k]/k s[n_] := Sum[f[k]^n, {k, 0, \[Infinity]}] Table[{n, s[n]}, {n, 1, 7}] // Expand//FullSimplify {{1, (1 + \[Pi])/2}, {2, (1 + \[Pi])/2}, {3, 1/8 (4 + 3 \[Pi])}, {4, 1/6 (3 + 2 \[Pi])}, {5, 1/2 + (115 \[Pi])/384}, {6, 1/2 + (11 \[Pi])/40}, {7, 1/2 + (1/46080)\[Pi] (129423 + 4 (-7 + \[Pi]) \[Pi] (147 - ...
7
The sum is alternating, so you might need extra precision and NSumTerms: katsurda[x_] := NSum[(-x)^j/j! Zeta[j], {j, 2, Infinity}, WorkingPrecision -> 16, NSumTerms -> Max[15, 2 x]]; katsurdaApprox[x_] := x (Log[x] + 2 EulerGamma - 1) - Zeta[0]; plot1 = DiscretePlot[katsurda[x], {x, 0, 40, 2}]; plot2 = Plot[katsurdaApprox[x], {x, 0, 40}]; Show[...
3
The code for D[Sum[..],..] assumes no options to Sum, so Method -> "Procedural" is treated as an iterator. This is a bug. After assuming it's an iterator, the code fails internally because it's a bad iterator. This mysteriously leads to a derivative of {{0}}, which seems an unimportant bug. This last bug happens with D[Sum[(xx - x[j])^2, j], ...
3
Seems like the Sum option I was looking for is Method -> "Procedural": Sum[ -2 Exp[-(xx - Subscript[x, k])^2/(V + Subscript[Vx, k]^2)] Subscript[n, k] (xx - Subscript[x, k])/(V + Subscript[Vx, k]^2) , {k, nsp}, Method -> "Procedural"] // AbsoluteTiming
1
The expression is a ratio of two sums, and is very insensitive to $n$. Any $n\ge 10$ gives results equivalent to $n= \infty$. Therefore, redefine the expression for the limiting value of $n\rightarrow\infty$. The sum in the denominator is the Harmonic number of order $p$, or Zeta[p] when $n\rightarrow\infty$. The sum in the numerator is a linear combination ...
2
tl;dr I would consider this a bug, although not a surprising one. This is how one would normally write this sum in Mathematica: 1 + Sum[3, {i, n}, {j, i, n - 3}] To keep things simple, I will use the following simplified version: Sum[1, {i, n}, {j, i, n - 3}] (* 1/2 (-5 n + n^2) *) As you observe, the result is not correct for symbolic n. For an explicit ...
2
There is a subtle difference you need to account for with a Piecewise so that it doesn't incorporate the inner sum when your range(i,n-3+1) is empty: result[n_] := 1 + Sum[ Piecewise[{{Sum[i + j, {j, i, n - 3}], i < n - 3 + 1}}, 0] , {i, 1, n}] Mathematica can now generate a correct closed form for symbolic $n$ given by result[n]: 1+\left\{ \begin{...
4
Here is one way to force s to be non-negative by replacing s with Exp[logs]: solve[p_, n_, eps_] := (formula[s_] = 1/HarmonicNumber[n, p] Sum[(1 - 1/i^p)^s 1/i^p, {i, 1, n}]; invert[f_] := FindRoot[f[Exp[logs]] == eps, {logs, Log[2]}, AccuracyGoal -> 4, PrecisionGoal -> 4]; invert[formula]); logs /. solve[6, 1000, 10^-2] // Exp // Ceiling (* 39 *) ...
0
The Harmonic number $H_{n-1}$ can be expressed by the Digamma function $\psi(n)$ together with the Euler-Mascheroni constant $\gamma\approx 0.5772$. $\psi(n)=H_{n-1}-\gamma$ In MMA $\psi(n)$ and $\gamma$ are called by PolyGamma[n] and EulerGamma. Table[Limit[(PolyGamma[k] + EulerGamma)*k, k -> n], {n, 0, 4}] *( {-1, 0, 2, 9/2, 22/3} *) The limes is ...
Top 50 recent answers are included | 1,617 | 4,525 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-31 | longest | en | 0.754702 |
https://www.bikedekho.com/california-1400/faqs/what-will-be-the-monthly-emi-for-california-1400~5e171a37e157f34453626662 | 1,581,980,919,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143373.18/warc/CC-MAIN-20200217205657-20200217235657-00477.warc.gz | 685,119,882 | 33,203 | # What will be the monthly EMI for Moto Guzzi California 1400?
EMIs or Equated Monthly Installments refer to the monthly payments you make to the lender to repay your loan. These payments include the principal amount as well as the interest i.e. EMI = Principal Amount + Interest on Principal amount. Mathematically, EMI for Moto Guzzi California 1400 can be calculated using the following formula:
{P x R x (1+R)^N / [(1+R)^N-1]}
where, P = Principal amount of the loan, R = Rate of interest and N = Number of monthly installments.
For Example:- If the principal amount for a bike loan is Rs. ₹21,85,848/- on an annual rate of interest of 10% for a tenure of 3 years then
EMI = 2185848 * 0.008333*(1+ 0.008333)^36 / ((1+ 0.008333)^36)-1 = Rs 71,647/-
The rate of interest (R) on your loan is calculated monthly i.e. (R= Annual rate of interest/12/100). For instance, if R = 10% per annum, then R= 10%/12 = 0.008333.
## Have any question? Ask now!
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## Find FAQ of California 1400 Alternatives
Ex-showroom price in Delhi
## Frequently Asked Questions on Moto Guzzi California 1400
SpecificationsPerformanceEMI
× | 329 | 1,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-10 | longest | en | 0.89384 |
http://www.wishessays.com/quadratic-equations-and-functions/ | 1,547,715,907,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658901.41/warc/CC-MAIN-20190117082302-20190117104302-00156.warc.gz | 391,857,755 | 12,564 | Opening Hours
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Remember the form of a quadratic equation: ax2+bx+c You will use: W=-.01??2+100x+c where (-.01??2+100x) represents the stores variable costs and c is the stores fixed costs. So, W is the stores total monthly costs based on the number of items sold, x. Think about what the variable and fixed costs might be for your fictitious storefront business and be creative. Start by choosing a fixed cost, c, between \$5,000 and \$10,000, according to the following class chart: If your last name starts with the letter Choose a fixed cost between S-T \$8600-\$9200 Your monthly cost is then, W = -.01??2+100x+c. Substitute the c value chosen in the previous step to complete your unique equation predicting your monthly costs. Next, choose two values of x (number of items sold) between 100 and 200. Again, try to choose different values from classmates. Plug these values into your model for W and evaluate the monthly business costs given that sales volume.
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Use the order calculator below and get ordering with wishessays.com. | 327 | 1,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-04 | latest | en | 0.877683 |
http://stackoverflow.com/questions/16390493/r-kmeans-initialization | 1,464,359,816,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049276780.5/warc/CC-MAIN-20160524002116-00240-ip-10-185-217-139.ec2.internal.warc.gz | 263,613,510 | 18,301 | # R kmeans initialization
In the R programming environment, I am currently using the standard implementation of the `kmeans` algorithm (type: `help(kmeans)`). It appears that I cannot initialize the starting centroids. I specify the `kmeans` algorithm to give me 4 clusters and I would like to pass the vector coordinates of the starting centroids.
1. Is there an implementation of `kmeans` to allow me to pass initial centroid coordinates?
-
The `centers` argument should let you do this. – Marius May 6 '13 at 0:25
Yes. The implementation you mention allows you to specify starting positions. You pass them in through the `centers` parameter
``````> dat <- data.frame(x = rnorm(99, mean = c(-5, 0 , 5)), y = rnorm(99, mean = c(-5, 0, 5)))
> plot(dat)
> start <- matrix(c(-5, 0, 5, -5, 0, 5), 3, 2)
> kmeans(dat, start)
K-means clustering with 3 clusters of sizes 33, 33, 33
Cluster means:
x y
1 -5.0222798 -5.06545689
2 -0.1297747 -0.02890204
3 4.8006581 5.00315151
Clustering vector:
[1] 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2
[51] 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3
Within cluster sum of squares by cluster:
[1] 58.05137 73.81878 52.45732
(between_SS / total_SS = 94.7 %)
Available components:
[1] "cluster" "centers" "totss" "withinss" "tot.withinss" "betweenss"
[7] "size"
``````
- | 597 | 1,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-22 | latest | en | 0.709523 |
https://mathoverflow.net/questions/237480/appropriate-morphisms-and-2-morphisms-in-indc | 1,620,635,915,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989115.2/warc/CC-MAIN-20210510064318-20210510094318-00252.warc.gz | 410,908,766 | 30,642 | # Appropriate morphisms and 2-morphisms in Ind(C)
As I was trying to understand the category $Ind(C)$ of diagrams of the form $I \to C$, where $I$ is a small filtered $(0,1)$-category, I wondered whether it is possible to define morphisms directly, without the limit-colimit construction (see https://ncatlab.org/nlab/show/ind-object), and then try to show that its equivalent to the common definition as $\text{lim$_i$colim$_j$}\,C(F(i),G(j))$. A natural choice would be to define a morphism $(F:I\to C) \to (G:J \to C)$ as a pair $(f,g)$, where $f:I \to J$ is a functor in the $(1,2)$-category of all poset-categories, and $g:F \Rightarrow G \circ f$ is a natural transformation. Then, the functor $C \hookrightarrow Ind(C)$ is clearly fully faithful. While trying to show that each object in $Ind(C)$ is a colimit of itself, I recognized that I should rather look at 2-colimits, but then I should define what a 2-morphism is. I tried several things but I'm not an expert for higher categories, so unfortunately I lack the necessary intuition... do you know a canonical choice for 2-morphisms in $Ind(C)$?
• This paper might be relevant to your question: arxiv.org/abs/1406.6229 – Yonatan Harpaz Apr 27 '16 at 20:44
• Thx! This is exactly the kind of result I desired! – Bipolar Minds Apr 27 '16 at 21:38
On a more general note, you're (gradually) building a subcategory of the 2-fiber $\textbf{Cat}/C$. An excellent reference for those is the following paper, which helped me a lot when I was looking into this stuff:
The journal where this paper appeared is now famously defunct, but you can find a pdf of the article here. The explicit answer to your question regarding 2-morphisms is as follows. Let $F$ and $G$ be as in your question and consider a pair of 1-morphisms $(f,g)$ and $(f',g')$, also as in your question. A 2-morphism $(f,g) \Rightarrow (f',g')$ is a natural transformation $\beta:f \Rightarrow f'$ (i.e., a 1-morphism in the category of functors from $I$ to $J$) so that the following holds: $$(G\circ\beta) * g = g'$$
Here $*$ stands for vertical composition while $\circ$ is horizontal composition. Pictorially, we have: | 600 | 2,146 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-21 | latest | en | 0.915523 |
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# p1-Sequences+series watch
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1. The sum of the first 3 terms of a GP is 65/9 and the sum to infinity is 7.5.
Next the qu asked me to find out a and r which i did...r=1/3, a=5.
Then it says find the least number of terms required so that their sum differs from 7.5 by less than 10^-4....heeeeeyyyylllppp! trial ad error meth?
The sum of the first 3 terms of a GP is 65/9 and the sum to infinity is 7.5.
Next the qu asked me to find out a and r which i did...r=1/3, a=5.
Then it says find the least number of terms required so that their sum differs from 7.5 by less than 10^-4....heeeeeyyyylllppp! trial ad error meth?
haven't looked at this question yet, but just to let you know, I answered the other sequences and series question. It's probably on the second page by now.
3. (Original post by mockel)
haven't looked at this question yet, but just to let you know, I answered the other sequences and series question. It's probably on the second page by now.
oohhh thanks very much mockel, i forget about that one! ill go check
oh and rep coming your way if its right!
4. anyone got ideas for this 1?
Then it says find the least number of terms required so that their sum differs from 7.5 by less than 10^-4.
Which sum differs from which?
6. Sn = 7.5[1-3^(-n)] you're looking for 7.5 - Sn < 10^-4 so 7.5(1-[1-3^(-n)]) < 10^-4, 7.5*3^-n < 10^-4 rearrange to 3^n > 7.5*10^4, take logs of both sides n log3 > 4 + log7.5, n > (4 + log7.5)/log3 = 10.21... so you need to sum up to and including n=11, ie 11 terms
7. (Original post by Bezza)
Sn = 7.5[1-3^(-n)]
where do u get this from,,,,, i thought the formula was
s=a(1-r^n)/(1-r^n)
where do u get this from,,,,, i thought the formula was
s=a(1-r^n)/(1-r^n)
No - if it was the 1-r^n would cancel Sn = a(1-r^n)/(1-r) = 5[1-3^(-n)]/(1-1/3) = 5[1-3^(-n)]/(2/3) = 15/2*[1-3^(-n)] = 7.5[1-3^(-n)]
9. (Original post by Bezza)
No - if it was the 1-r^n would cancel Sn = a(1-r^n)/(1-r) = 5[1-3^(-n)]/(1-1/3) = 5[1-3^(-n)]/(2/3) = 15/2*[1-3^(-n)] = 7.5[1-3^(-n)]
lol sorry i meant [1-r] on the bottom of frac!! oh and thanks bezza
lol sorry i meant [1-r] on the bottom of frac!! oh and thanks bezza
Yeah, yeah sure you did
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Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE | 944 | 2,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-17 | latest | en | 0.950019 |
http://www.kylesconverter.com/length/links-(gunter's-surveyor's)-to-mickeys | 1,624,582,882,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488560777.97/warc/CC-MAIN-20210624233218-20210625023218-00523.warc.gz | 64,942,304 | 5,509 | # Convert Links (gunter's Surveyor's) to Mickeys
### Kyle's Converter > Length > Links (gunter's Surveyor's) > Links (gunter's Surveyor's) to Mickeys
Links (gunter's Surveyor's) (lnk) Mickeys Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
One hundredth (1/100) of a chain (Gunter's, Surveyor's). A chain having 66 ft or 4 rods. A link being approximately 0.201168 meters (SI base unit). 1 lnk ≈ 0.201168 m.
1 Mickey:
1 mickey is exactly 1/200 inches. In SI units 1 mickey is 1.27 x 10-4 meters.
Conversions Table
1 Links (gunter's Surveyor's) to Mickeys = 158470 Links (gunter's Surveyor's) to Mickeys = 110880
2 Links (gunter's Surveyor's) to Mickeys = 316880 Links (gunter's Surveyor's) to Mickeys = 126720
3 Links (gunter's Surveyor's) to Mickeys = 475290 Links (gunter's Surveyor's) to Mickeys = 142560
4 Links (gunter's Surveyor's) to Mickeys = 6336100 Links (gunter's Surveyor's) to Mickeys = 158400
5 Links (gunter's Surveyor's) to Mickeys = 7920200 Links (gunter's Surveyor's) to Mickeys = 316800
6 Links (gunter's Surveyor's) to Mickeys = 9504300 Links (gunter's Surveyor's) to Mickeys = 475200
7 Links (gunter's Surveyor's) to Mickeys = 11088400 Links (gunter's Surveyor's) to Mickeys = 633600
8 Links (gunter's Surveyor's) to Mickeys = 12672500 Links (gunter's Surveyor's) to Mickeys = 792000
9 Links (gunter's Surveyor's) to Mickeys = 14256600 Links (gunter's Surveyor's) to Mickeys = 950400
10 Links (gunter's Surveyor's) to Mickeys = 15840800 Links (gunter's Surveyor's) to Mickeys = 1267200
20 Links (gunter's Surveyor's) to Mickeys = 31680900 Links (gunter's Surveyor's) to Mickeys = 1425600
30 Links (gunter's Surveyor's) to Mickeys = 475201,000 Links (gunter's Surveyor's) to Mickeys = 1584000
40 Links (gunter's Surveyor's) to Mickeys = 6336010,000 Links (gunter's Surveyor's) to Mickeys = 15840000
50 Links (gunter's Surveyor's) to Mickeys = 79200100,000 Links (gunter's Surveyor's) to Mickeys = 158400000
60 Links (gunter's Surveyor's) to Mickeys = 950401,000,000 Links (gunter's Surveyor's) to Mickeys = 1584000000 | 749 | 2,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-25 | longest | en | 0.75646 |
https://www.jackalopejobs.com/how-many-weeks-in-a-salary-year/ | 1,716,560,122,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058719.70/warc/CC-MAIN-20240524121828-20240524151828-00503.warc.gz | 723,043,416 | 17,254 | # How Many Weeks in a Salary Year?
Knowing how many weeks in a salary year helps you plan your vacation. Whether you work a full-time job or a part-time position, you need to know how much you earn each week. In addition, knowing your weekly pay rate is important to creating an accurate budget. You can find an online calculator to help you figure out your paycheck.
A typical work week is around 40 hours. During holiday seasons, however, many people work long hours. If you are working a part-time job or a seasonal job, you will need to account for overtime when calculating your wages. For example, if you were to work a week at 10 hours a day, you would be eligible for overtime pay for a total of 25 hours.
If you are not paid for vacation, you will need to adjust the number of weeks in a salary year. There are several ways to do this, but you will need to figure out the days or weeks that you take off from your work.
To calculate your weekly wage, multiply your hourly pay rate by the number of weeks you work each year. This includes holidays and statutory vacations.
## How Many Weeks is My Annual Salary?
The annual salary is a good way to figure out how much money you make each year. It is usually calculated on a calendar year basis, but can be calculated for a fiscal year as well. For example, if you work 40 hours a week and you are paid an hourly rate of \$15, your annual salary is \$39,000. This amount is also prorated for part-time workers and is likely to be lower for full-time workers. In addition to the number of weeks you work, your annual salary will also include statutory holidays.
Getting the right number of weeks off is a bit tricky, but you can make a rough estimate by subtracting one week of vacation from your average number of hours worked. Taking into account that a full-time employee gets two weeks of unpaid leave each year, this can skew your calculation slightly, but it is a useful way to figure out how much you earn per year.
There are many calculators out there, but none of them really can show you exactly how much you make in a year. Some will do the math for you, while others will only provide you with a list of options, such as what you are earning per hour and how much of that is taxed. You can also use a payroll management system to do the math for you.
READ ALSO: What is David Muir Salary?
## Is Salary Based on 50 Or 52 Weeks?
The annual salary is calculated by multiplying an hourly rate with the number of weeks worked. These figures are based on the fact that a full-time employee works about 40 hours a week during a 52-week period.
If you are not paid for sick or vacation time, your calculations may have to be adjusted. Similarly, you may want to consider using a calculator to determine your weekly pay. For example, if you work at a job with a salary of \$40,000, you might want to look into a salary calculator. This will help you find the best possible weekly compensation for your work.
A calculator can also be helpful in finding the annual salary of a certain job. It’s not a simple matter to tell whether your salary is based on 50 or 52 weeks, but with the right tool, it’s easy to find out. To get the most accurate figure, you’ll need to know how many hours you’re likely to work, how much you’re paid, and the number of weeks in your year.
## Is Salary Based on 12 Months Or 52 Weeks?
If you are going to get a job you will need to know how to calculate a pay scale. There are a number of factors to consider when figuring out your salary. Some of these include how many weeks you are paid for, if you work part time or full time and if you have a bonus or incentives. This will determine the number of hours you need to clock in to earn a living wage. A well researched, well executed compensation package is one of the most important steps you can take to increase your chances of landing that job of your dreams.
It may be surprising that a lot of people fail to figure out how to calculate their own pay scale. To ensure you are on track, you should enlist the services of an experienced HR professional. As far as calculating your pay, you may want to check out a salary calculator to see what you owe and what you owe your employer. You will also want to factor in any bonuses or incentives you might receive as a a result of your hard work and dedication.
## How Do You Calculate Salary Per Year?
Whether you are looking to apply for a new job, need a salary review or just want to better understand your finances, it’s important to have a clear understanding of how to calculate your salary. Knowing your salary is the first step to developing a budget and managing your finances. In addition, knowing your salary can help you compare yourself to other workers in your industry.
READ ALSO: What is the Average Usfl Salary?
One of the most basic ways to calculate your salary is to multiply your gross pay by the number of pay periods in a year. This is the most accurate way to calculate your annual income. However, the calculations for your salary may vary depending on the frequency of your paychecks.
An example would be an hourly employee earning \$15 per hour. If they worked 10 hours the first week of the month, they would earn \$1,500 in the first pay period.
Using this amount as a base, an employee could then work out how many hours he or she would have to work in order to earn \$12,000 in a year. The average weekly hours might be 30 hours per week, while the fourth week might have an average of 35 hours.
## How Much is the Salary in Philippines?
The Philippines is a developing country in Southeast Asia. Its economy is diversified and the largest industries include mining, automotive, and outsourcing. With an annual growth rate of 7.3%, the BPO sector has contributed significantly to the country’s GDP.
The average salary in the Philippines is comparatively low compared to other countries. It ranges from PHP 152,000 to PHP 1,076,315 annually. That is lower than the average salaries in Thailand and Indonesia. But the difference is not huge.
According to the Philippine Statistics Authority, a family of five needs an income of at least US\$220 a month. This figure is also attainable with a salary of PHP 3300 monthly.
Filipino workers are encouraged to pursue higher education. A master’s degree earns 29% more than a bachelor’s degree. They also receive paid holidays and thirteenth-month bonus.
Filipinos place a high value on family. They are expected to support their families after graduation. Children are generally expected to live with their parents until marriage. In fact, most families have several generations living in the same household.
## Is Yearly Salary Based on 52 Weeks?
If you are working for an employer who offers bi-weekly pay periods, you may have a difficult time figuring out how much you will be paid each week. Generally speaking, a full-time worker’s weekly work hours should not exceed 40. In some cases, an employer may offer two weeks of unpaid leave in the summer or year-end holidays. These extra weeks should be factored into your calculations.
READ ALSO: What is the Salary For a Pharmacy Technician?
An employee can calculate his or her average weekly work hours by multiplying their hourly rate by the number of paid hours they are expected to work each week. For instance, an employee earning \$15 per hour who works 20 hours per week would earn a yearly salary of about \$37,440.
Similarly, an employee who works 40 hours a week should expect a yearly salary of approximately \$31,200. However, this figure can vary significantly depending on your employer, the type of work you do, and your location. The annual salary you receive for a job in Los Angeles might be higher than one you get in New York, for example.
## How to Calculate Salary 52 Weeks?
If you are working a full-time job and you are earning a salary, you may wonder how to calculate your annual salary. It is based on a number of factors, including the average hours you work per week. You can also determine how much you earn on an hourly basis.
A typical working week for a full-time employee is around 40 hours. If you are paid on a weekly basis, your yearly salary can be calculated using the hourly rate multiplied by the number of paid weeks in the year. For example, an employee making \$18 per hour would earn \$37,440.
Annual salaries are usually based on 52 weeks of employment. However, some people work less than 52 weeks. In such cases, you should adjust your calculations accordingly.
The federal minimum wage is \$15k per year. Many cities have raised the minimum wage to higher levels. Depending on your job duties, you may be eligible for overtime pay. To calculate your annual salary, you will need to include all bonuses and raises in addition to your salary.
Learn More Here:
2.) Salary Data
3.) Job Salaries | 1,898 | 8,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-22 | latest | en | 0.976635 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=8&t=32940&p=106260 | 1,606,745,407,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141216175.53/warc/CC-MAIN-20201130130840-20201130160840-00489.warc.gz | 379,304,177 | 11,559 | ## Homework Question Edition 6 E1
Esther Ahn 4I
Posts: 30
Joined: Fri Sep 28, 2018 12:29 am
### Homework Question Edition 6 E1
For question E1, you have to find how long the fibers extend, given that you string together 1.00 mol Ag atoms, each of a radius of 144pm. For my answer I got 8.67 x 10^13 m. The solution manual gives us the final answer in kilometers. I was just wondering why the answer is given in kilometers and not meters because I thought that meters are considered the base unit.
Abby De La Merced 3F
Posts: 30
Joined: Fri Sep 28, 2018 12:24 am
Been upvoted: 1 time
### Re: Homework Question Edition 6 E1
I am also struggling with trying to figure out why the conversion works out like that as well. I am also confused about why 144pm is multiplied by 2 when doing the dimensional analysis method on the problem too.
amogha_koka3I
Posts: 62
Joined: Fri Sep 28, 2018 12:24 am
### Re: Homework Question Edition 6 E1
We multiply 144 by 2 because 144pm is the radius and we want the diameter of the atom.
Sophia Ding 1B
Posts: 62
Joined: Fri Sep 28, 2018 12:16 am
### Re: Homework Question Edition 6 E1
I think the answer is given in kilometers because in class we learned that the SI fundamental unit for distance is kilometers (km), but I am also with you on generally answering in meters! I feel like most answers in the solution manual has units of meters, so I'm not sure how pressing this matter is of using kilometers.
rikolivares
Posts: 24
Joined: Fri Sep 28, 2018 12:23 am
### Re: Homework Question Edition 6 E1
I assume it was just to underscore the large number, but I don't think there would be any problem if you answered in meters in a question like this for a quiz or test. I think meters is always the safest bet. | 483 | 1,757 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-50 | latest | en | 0.946127 |
https://gmatclub.com/forum/percentage-of-population-attending-selected-sports-events-single-year-419152.html | 1,716,332,248,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058522.2/warc/CC-MAIN-20240521214515-20240522004515-00523.warc.gz | 238,283,426 | 130,246 | Last visit was: 21 May 2024, 15:57 It is currently 21 May 2024, 15:57
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# Percentage of Population Attending Selected Sports Events, Single Year
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Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
2
Kudos
A. Visitors to Las Vegas prefer indoor gambling to attending sporting events.
No linkages presented in the stimulus regarding --- whether preference for indoor gambling causes low attendance for sporting events. Both are 'separate'; and no cause-and-effect phenomenon can be inferred solely by the table data.
Would NOT help explain
B. Seattle’s football team recently lost some key players.
If the Seattle football team lost key players, this will lead to poor performance -- hence, low scores in the Table for Seattle’s football team
Would help explain
C. Philadelphia's high schools have an excellent hockey program.
If Philadelphia's high schools have an excellent hockey program, it will lead to high scores for Hockey, as evident through the 2nd highest score in the Table for Philadelphia's hockey team.
Would help explain
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Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
Are the Data Insights questions being reflected in the Error Log?
I couldn't trace these attempts in the Error Log.
Intern
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Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
I can’t see the main question
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Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
1
Kudos
Official Explanation
Statement 1
The preferences of visitors to Las Vegas don’t explain the behavior of the people who live there, so they don’t explain why such a low percentage of Las Vegas’s actual population attends sporting events.
Statement 2
Seattle’s football team having become weaker could explain why other sports have become more popular in Seattle.
Statement 3
If many high school students in Philadelphia play hockey, they and their families might come to be more interested in watching it than in watching other sports.
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Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
pintukr wrote:
Are the Data Insights questions being reflected in the Error Log?
I couldn't trace these attempts in the Error Log.
As per my limited knowledge this functionality will be available within a week. Lets see.
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Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
7747789470 wrote:
I can’t see the main question
Hello 7747789470
The question starts from the table, there is no other text.
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Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
This seems like too much outside info to be honest. Is this from OG?
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Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
SudipM7 wrote:
This seems like too much outside info to be honest. Is this from OG?
No! it is from McGraw-Hill's GMAT
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Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
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Data Insights (DI) Butler 2023-24 [Question #03, Date: Oct-01-2023] [Click here for Details]
Baltimore, MD35422839
Dayton, OH935178
Houston, TX20321916
Las Vegas, NV8527
Newark, NJ1871621
Seattle, WA27111926
(Sort ↕ the table by clicking on the headers)
For each of the following statements select Would help explain if it would, if true, help explain some of the information in the preceding table. Otherwise select Would not help explain.
I got the answers right, but I still feel the question is a little incomplete, such that the numbers and city names are given along with the names of the sports, but we do not know what the numbers represent, they could represent anything, number of visitors, number of spectators, money made from the sports, losses incurred in matches, anything, leaving the data open to any kind of interpretation. Sajjad1994, can you please let us know if this is the entire question or if something is missing?
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Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
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angysinghdhillon wrote:
I got the answers right, but I still feel the question is a little incomplete, such that the numbers and city names are given along with the names of the sports, but we do not know what the numbers represent, they could represent anything, number of visitors, number of spectators, money made from the sports, losses incurred in matches, anything, leaving the data open to any kind of interpretation. Sajjad1994, can you please let us know if this is the entire question or if something is missing?
This is the percentage of population attending selected sports events, Single Year. This was added as topic name but i have now added it in the start of the table.
Thank you!
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Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
angysinghdhillon wrote:
I got the answers right, but I still feel the question is a little incomplete, such that the numbers and city names are given along with the names of the sports, but we do not know what the numbers represent, they could represent anything, number of visitors, number of spectators, money made from the sports, losses incurred in matches, anything, leaving the data open to any kind of interpretation. Sajjad1994, can you please let us know if this is the entire question or if something is missing?
This is the percentage of population attending selected sports events, Single Year. This was added as topic name but i have now added it in the start of the table.
Thank you!
Thank you! This helps a lot!
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Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
Official Explanation
Statement 1
The preferences of visitors to Las Vegas don’t explain the behavior of the people who live there, so they don’t explain why such a low percentage of Las Vegas’s actual population attends sporting events.
Statement 2
Seattle’s football team having become weaker could explain why other sports have become more popular in Seattle.
Statement 3
If many high school students in Philadelphia play hockey, they and their families might come to be more interested in watching it than in watching other sports.
Hi Sajjad, since the previous year percentage of population who attend sporting events cannot be determine, may I ask how can we conclude that other sports team have become more popular in Seattle just because the percentage of population attending their events is higher?
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Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
the second question's answer is really not derivable.
First of all, there is no paragraph to really give any info about the graph table, there's just one line and also in the third question there's a mention of "schools" and yet were not given info about these sports taking place in schools or any connection of these sports with school teams or programs whatsoever.
Re: Percentage of Population Attending Selected Sports Events, Single Year [#permalink]
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## Synonyms for conic section
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con], a/2 is average diameter, length of conic section and conical angle respectively seen in Fig.
A Set of Centers of Described for a Quadrangle Conic Section.
There are two geometric designs represented in figure 7: the initial one that has the length z = 35 mm for the conic section and the new one with the length z = 12.
The first is designed for use with Cabri Geometry to study the parabola as a conic section.
3) A free point on a line, a circle, a conic section or another curve is distributed by a variable, so we can move it to any position on the line or curve accurately by changing the value of the variable using an animation button.
Boytchev in Chapter 20 presents how the concept of conic sections can be taught by using virtual models.
Slice them any way you like, and you have a real-world example of conic sections 6 which can take the form of an ellipse, parabola, or hyperbola.
He studied other conic sections, the ellipse and the hyperbola, always trying to find elegant ways of holding weight at a height.
The Alexandrian tradition was based in mixed and pure mathematics such as mechanics, astronomy, and conic sections.
The latter encompasses both cylindrical and toroidal surfaces and these are collectively known as conic sections since they are curved forms which originate from sections of a cone (Figure 1).
Being one of Apollonius' conic sections, the parabola is basically a geometric entity.
Sampson3 provided a statistical model for describing the average shape of the arch form as well as its variation in the population by applying arcs of conic sections on the sample of sixty six dental arches.
He followed the way opened by al-Hasan ibn Musa, particularly in his work on the measure of curved planes and solids, and on the properties of conic sections.
Ratti and McWaters (University of South Florida) introduce the concepts and formulas for graphing equations, and explain how to solve quadratic equations, exponential and logarithmic functions, trigonometric functions, systems of equations, and conic sections.
Pascal inscribed his essay on conic sections at 16, and Alexander Hamilton was George Washington's aide at 20.
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Out of 42 kids in a class twice as many failed Ela as math,4 failed both. If 7 failed neither, find how many failed each subject.
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### Pages to are hidden for
``` 8th Grade Science Curriculum Map
Revised August 2008
Unit One Focus Standards for Objectives - Main points and Textbook Pages Unit Vocabulary -
8th Grade Physical concepts that should be mastered And Students should know
25 Teaching Days Science. by all students. Labs or Demos definition and
example for each
word.
Part 1 Section 9 Part I: Scientific Method Chapter 1 Hypothesis
Standards 9.0: a-g Scientific progress is 1. Know that a hypothesis is a Pages: Scientific Variable
made by asking statement to be tested. Method 1-15 & Control (control group)
Investigation and meaningful questions 2. Understand that all scientists 30-31 Observations
Experimentation and conducting use the metric system. Metric 16 – 23 & Prediction
careful investigations. 3. Know that the metric system is 28-29 Theory
Scientific Method 9a. Plan and conduct also called the SI system or SI Temperature 26 Law
7 Days an experiment to test units. Graphs 34 – 41 Linear
And a hypothesis. 4. Know the basic units of Lab Safety 43-47 Nonlinear
Metric System 9b. Evaluate the measure: meter, liter, gram, and and page 651 Reproducibility
8 Days accuracy and Celsius. Evaluate
reproducibility of 5. Know how to read a metric Scientific Method Temperature
data. ruler. Labs Triple Beam Balance
9c. Distinguish 6. Construction of graphs: bar, 1. Temperature of Grams
between variable and line, and pie. water lab Meters
controlled parameters 7. Know independent variable and 2. Spaghetti lab Liters
in a test. dependent variable. Centimeters
9d. Linear graphs 8. Distinguish between linear and Metric Labs Results
have a constant slope non-linear relationships. 1. Boy vs. Girl Conclusions
and use y=kx. Height Lab Communicate
9e. Construct graphs Testable
from data and make Units of Measure
quantitative Scientific Method
relationships between Data
the variables. Calculations
9f. Apply mathematic Graphs
relationships to Kilometer
determine a missing Centimeter
quantity such as in the Millimeter
density and speed Liter
formulas. Milliliter
9g. Distinguish Kilogram
between linear and Gram
nonlinear Celsius
relationships on a
graph.
Part 2 8.a Students know Part II: Density Chapters 1 & 11 Mass
Standard 8.0:a-b density is mass per 1. Know the density formula and Pages 24-25 & Volume
Density and Buoyancy unit of volume. density triangle. 424-429 Area (surface area)
10 Days 8b. Calculate the 2. Understand density word Displacement
density of objects problems. Density Labs Graduated Cylinder
Benchmark #1 on from mass and 3. Compare objects that are very 1. Marble Lab Over-Flow Cans
Oct 16 & 17 volume: solids and dense to objects with low density. 2. Density of Length
liquids and those 4. Know that density never Water Cubic
having irregular or changes for a substance!!! A small 3. Density of Weight
regular shapes. piece of wood has the same density Objects Lab Heavy
8c. Buoyant force on as a large piece. Cutting a board in 4. Density of Light
an object in a fluid is half does not change its density. Pennies (pre and Dense
an upward force equal 5. Density is a physical property post 1982) Buoyancy
to the weight of the of a substance. Buoyant Force
fluid the object has 6. Understand why the buoyant Demos Float
displaced. force is the cause for floating 1. Metal Balls – Sink
8d. Students know objects. hollow vs. solid. Gravity
how to predict 7. Compare buoyant force to the 2. Soap bubbles
whether an object will force of gravity. over dry ice.
float or sink. 8. Solids, liquids, and gases have 3. Metal block &
distinct densities. foam block.
9. Helium balloons float due to
buoyant force.
10. Predict whether an object will
float or sink based on density.
11. Compare the density of one
object or substance to another.
Unit two Focus Standards Objectives Textbook Unit Vocabulary
23 Teaching Days And
Labs
Part 1 1a. Students know 1. Speed is calculated using time Chapters 9 Motion
Standards 1.0:a-f position is defined in and distance. Motion & Energy Speed
relation to some Pages: Average speed
Motion – Speed choice of a standard 2. Explain that scientists can use Motion basics Instantaneous Speed
13 Days reference point and different units for speed, time, and 338-341 Constant Speed
set of reference distance. Speed & Velocity Time
directions. 342-347 Distance
1.b Students know 3. Know differences between Acceleration Displacement
that average speed is speed, velocity, and acceleration. 350-355 Reference Point
the total distance Position
traveled divided by 4. Speed can be slow or fast! Speed Labs Acceleration
the total time elapsed 1. Toy Car Lab Velocity
and that the speed of 5. Graphs showing speed vs. time 2. Marble Ramp Average Velocity
an object along the and distance vs. time. Lab Vector
path traveled can 3. Acceleration Direction
vary. Lab (using the long Magnitude
1c. Students can 6. Student should master the use ramp) Friction
solve problems of the speed triangle. 4. Walking vs. Inertia
involving distance, Running Lab Momentum
time, and average
speed.
1d. Velocity of an
object is described by
specifying both speed
and the direction of
the object.
1e. A change in
velocity may be due
to changes in speed,
direction, or both.
1f. Students know
how to interpret
graphs of position
versus time and
graphs of speed
versus time for
motion in a single
direction.
Part 2 2a. Students know a 1. Students know the four types of Chapters 10 Force
Standards 2.0 a-g force has both friction: static, sliding, rolling, & Pages Push
direction and fluid. Forces 374-378 Pull
Forces magnitude. Friction & Gravity Net Force
10 Days 2b. Students know 2. The force of gravity. 380-388 Balanced Forces
when an object is Newton’s Laws Unbalance Forces
Benchmark #2 subject to two or more 3. Elastic Forces compare & 389-399 Newtons (N) as a unit
Nov 20 & 21 forces at once, the contrast compression vs. tension. Gravitational Force
result is the Forces Labs Electromagnetic Force
cumulative effect of 4. Balanced vs. Unbalanced 1. Rolling vs. Strong Nuclear Force
all the forces. Sliding Friction Weak Nuclear Force
2c. Students know 5. How mass can affect the motion 2. Shoe Lab Elastic
when the forces on an of an object. 3. Ramps with Tension
object are balanced, different surfaces Compression
the motion of the 6. Forces are measured in units Centripetal Force
object does not called Newtons (N). Buoyant Forces
change. Weight
2.d Students know 7. Explain that one Newton (1N) Friction
how to identify can be written as 1 kg - m/s/s. Velocity
separately the two or Mass
more forces that are 8. Definition of a force. Spring Scale
acting on a single
static object,
including gravity,
elastic forces due to
tension or
compression in
matter, and friction.
2.e Students know
that when the forces
on an object are
unbalanced, the object
will change its
velocity (that is, it
will speed up, slow
down, or change
direction).
2.f Students know the
greater the mass of an
object, the more force
is needed to achieve
the same rate of
change in motion.
2g. Students know
the role of gravity in
forming and
maintaining the
shapes of planets,
stars, and the solar
system.
Unit Three Focus Standards Objectives Textbook Unit Vocabulary
30 Teaching Days And
Labs
Part 1 3.a Students know the 1. All matter has mass and Chapters: Matter
Standards 3.0: a -f structure of the atom volume. Chapter 2 Endothermic
and know it is 2. Matter is made of atoms. The Nature of Exothermic
States of Matter composed of protons, 3. Matter can change state: solid, Matter Substance
Endothermic neutrons and liquid, gas, and plasma. Pages 58 – 65 and Element
Exothermic electrons. 4. The substance does not change 69-70 & 74 Protons
15 Days 3b. Students know as it undergoes changes of state. Chapter 3 Neutrons
that compounds are 5. Endothermic change is when Solids, liquids and Nucleus
formed by combining energy is absorbed by a substance gases Electrons
Part 2 two or more different or reaction. Pages 91 – 101 Electron Clouds
Elements – Atoms – elements and that 6. Exothermic change is when Chapter 4 Valence Electrons
and Compounds compounds have energy is released by a substance Periodic Table Solids
15 Days properties that are or reaction. Pages 126 - 128 Liquids
different from their 7. Atoms form molecules and Gases
constituent elements. unique compounds. Plasma
3c. Students know 8. Clarify the differences between Chapter 6 Vibrate
Benchmark #3 atoms and molecules atom, element, molecule, and Chemical Collide
January 22 and 23 form solids by compound. Reactions Independently
building up repeating 9. Molecular Motion increases as Pages 220-221 Volume
patterns, such as the temperature increases. Mass
crystal structure of 10. Distinguish between the Labs: Crystal Lattice
NaCl or long-chain differences between physical and 1. Super-Cool Lab Molecules
polymers. chemical changes. Compounds
3d. Students know 11. Some solids form repeating Atoms
the states of matter patterns called crystal lattice Demos: Melting
(solid, liquid, gas) structures whereas others do not an 1. Steam Engine Boiling
depend on molecular are called amorphous solids. 2. Hot to cold Condensation
motion. 12. Molecules in the gas state water to show Freezing
move independently and molecular Vaporization
3.e Students know sometimes collide. movement. Evaporation
that in solids the 13. Molecules in the liquid state 3. Metal vs. foam Thermal
atoms are closely are in contact with each other and block melting ice. Heat
locked in position and slide past each other. Sublimation
can only vibrate; in 14. Molecules in the solid state are Density
liquids the atoms and tightly packed and vibrate in place. Speed
molecules are more 15. Know the parts of an atom and Acceleration
loosely connected and the charges of each part. Velocity
can collide with and 16. Understand the forces that Friction
move past one cause objects to move or remain at Physical Change
another; and in gases rest.
the atoms and 17. Distinguish between mass and
molecules are free to weight.
move independently, 18. Know the density formula and
colliding frequently. understand that density is a
physical property of a substance.
3.f Students know
how to use the
periodic table to
identify elements in
simple compounds.
Unit Four Focus Standards Objectives Textbook Unit Vocabulary
21 Teaching Days And
Labs
7a. Students know 1. Know the general overview of Chapters 4 Atom
Standards 7.0: a-b how to identify periodic table. How is it set up? Elements & Electrons
Periodic Table regions corresponding 2. Regions of the periodic table: Periodic Table Energy Level
to metals, nonmetals, metals, metalloids, and nonmetals. Pages: 125 - 155 Nucleus
and inert gases. 3. Elements are classified as being Protons
Benchmark #4 7.b Students know a metal, metalloid, or nonmetal. Labs: Atomic Number
March 2 and 3 each element has a 4. Families vs. Periods 1. The ion lab; Neutrons
specific number of 5. Atomic Number vs. Atomic soap suds test. Isotopes
protons in the nucleus Mass Mass Number
(the atomic number) 6. Atomic Number is the number Atomic Mass Unit
and each isotope of of protons in each atom for that Demos: (AMU)
the element has a element. 1. Place pure Atomic Mass
different specific 7. Atomic mass is the number of sodium into water. Periodic Table
number of neutrons in protons plus the number of 2. Show clips of Period
the nucleus. neutrons for an atom. the elements From Group
7.c Students know 8. Electrons are negative and have family one reacting Chemical Symbol
substances can be virtually no mass; 1/1820 of a in water. Alkali Metals
classified by their proton or neutron. Alkaline Earth Metals
properties, including 9. AMU stands for atomic mass Transition Metals
their melting unit; one proton equals 1 amu as Halogens
temperature, density, does a neutron. Semimetals
hardness, and thermal 10. Atomic Number (number of Metalloids
conductivity. protons) never changes; it Semiconductor
identifies the element!! Properties
11. Atomic Mass can vary due to Metals
isotopes. Malleable
Isotopes are atoms of an element Thermal Conductivity
that have a different number of Electrical Conductivity
neutrons. Reactivity
11. Note that the element’s name Corrosion
may have letters that do not match Nonmetals
the chemical symbol because of Metalloids
Latin name origins. Density
12. Metals can melt! Hardness
13. The nonmetal gases can Conductivity
become a liquid or solid if Thermal
temperature or other conditions Melting
(pressure) are right. Classification
14. Ions form when atoms or Elements
molecules loose or gain electrons. Molecules
15. Ions are often in water and Compounds
create what is called hard water. Physical Properties
16. Ions can be negative (anions) Chemical Properties
or positive (cations).
Unit Five Focus Standards Objectives Textbook Unit Vocabulary
28 Teaching Days And Physical Property
5a. Students know 1. The mass of the reactants Labs Chemical Property
Part 1 reactant atoms and always equals the mass of the Reactant
Standards 5.0 a-e molecules interact to products. Chapter 6 Product
Chemical Reactions form products with 2. Atoms bond together to form page 208 Yield Sign
different chemical molecules (compounds). Chemical Precipitate
properties. 3. Go over the four key clues to Reactions Endothermic Reaction
5.b Students know the chemical reactions. Exothermic Reaction
idea of atoms explains 4. All chemical reactions must be Labs Chemical Equation
the conservation of balanced. The four clues to Law of Conservation of
matter: In chemical chemical reactions. Mass
reactions the number 5. Remind students of the (Conservation of
of atoms stays the differences between molecules Matter)
same no matter how (O2) and compounds (H2O). Chemical Bonding
they are arranged, so 6. Substances can undergo Chemical Bond
their total mass stays physical changes go from solid, Molecule
the same. liquid, and into a gas but are still Compound
5c. Students know the same substance. Chemical Reaction
chemical reactions 7. Understand the differences Polymer
usually liberate heat between acids, bases, and neutral Freezing
or absorb heat. substances. Boiling
5.d Students know 8. Know how to read a pH scale. Condensation
physical processes 9. When an acid and a base are Evaporation
including freezing and mixed together the solution Change of State
boiling, in which a becomes more neutral (less acidic
material changes form and less basic) and a salt forms. Chapter 7
with no chemical 10. Define polymer and explain page 250 Acid
reaction. that DNA is a natural polymer. Acids and Bases Base
5.e Students know 11. Organic chemistry is the study Pages 268 - 279 Neutral
how to determine of carbon (C) compounds. pH
whether a solution is 12. The main elements in organic Litmus Paper Salt
acidic, basic, or molecules are C, H, O, N, P, and S. Page 270 Indicator
neutral. Molecules made of these elements Litmus Paper
Part 2 would likely be found in living pH Scale
Standards 6.0 a-c 6a. Students know organisms. Page 277
Chemistry of Living that carbon, because 13. Go over the molecules in
Systems of its ability to living things such as: fats, Lab
combine in many carbohydrates, proteins, salts, and pH Lab
ways with itself and water. Some are very small
other elements, has a whereas others are quite large.
central role in the 14. The sun is a star. Earth's
chemistry of living closest star at 93 million miles
organisms. away. Organic Compounds
6.b Students know 15. Our solar system has "9" Chapter 8 Hydrocarbon
that living organisms planets that orbit the sun. Page 286 Lipids
are made of molecules 16. Explain differences between Carbon Carbohydrates
consisting largely of day and year. Chemistry Proteins
carbon, hydrogen, 17. The cause for the seasons on Read Pages Nucleic Acid
nitrogen, oxygen, Earth. 294 - 299 Deoxyribonucleic Acid
phosphorus, and 18. Explain why the moon has and (DNA)
sulfur. different phases. Remember 317 - 323
6c. Students know moons are planetary satellites.
that living organisms 19. Explain comets and asteroids. Lab
have many different 20. Inner planets vs. outer planets. DNA Isolation
kinds of molecules, 21. Light year unit of measure is Rotation
including small ones, for distance in space not time. Revolution
such as water and salt, 22. Shapes and size of galaxies. Seasons
and very large ones, 23. Our solar system is in the Gravity - Mass &
such as carbohydrates, Milky Way Galaxy. Weight
fats, proteins, and 24. Stars are the only objects that Star
DNA. produce light. Moons and planets Solar System Moon
reflect light. Chapter 12, 13, Satellite
4a. Students know 14, & 15. Astronomical Unit
galaxies are clusters Page 465 Galaxy
Part 3 of billions of stars and Seasons Elliptical
Standards 4.0 a-e may have different Page 468 - 471 Spiral
Earth in the Solar shapes. Gravity Gravity
System 4.b Students know Page 475 Orbit
that the Sun is one of Moon Phases Inner Planets
many stars in the Page 480 Outer Planets
Milky Way galaxy Satellite and Comet
and that stars may Astronomical Unit Asteroid
differ in size , Page 543 Light Year
temperature, color. Sun HR Diagram
Benchmark #5 4c. Students know Page 545 - 550
April 20 and 21 how to use Inner Planets
astronomical units Page 552 - 561
and light years as Outer Planets
measures of distances Page 562 - 569
between the Sun, Comets &
stars, and Earth. Asteroids
4d. Students know Page 573 - 575
that stars are the Light Year
source of light for all Page 602
bright objects in outer HR Diagram
space and that the Page 604 & 630
Moon and planets Star Life-cycle
shine by reflected Page 610 - 613
sunlight, not by their Galaxies
own light. Page 617 -619
4e. Students know the
appearance, general Demos and
composition, relative Activities
position and size, and 1. Use scale
motion of objects in models of solar
the solar system, system.
including planets, 2. Students record
planetary satellites, nightly moon
comets, and asteroids. phases for one
month.
3. Students make
poster or brochure
for a planet.
State Test All Standards will be Objectives - Step #1 Review all Textbook Pages: Meter
Preparation Period reviewed in past objectives for each of the five Review the Liter
preparation of the units. textbook as needed. Gram
The time frame here state test. Review all key concepts with Check past units Hypothesis
will be from after students. Begin with metric for textbook pages. Density
benchmark five until system and continue through the Volume
the science test day. solar system. Displacement
All students should The essentials include: Mass
study at home each 1. Know what an atom is and its Speed
night to be best parts. Distance
prepared for the test. 2. Practice density and speed Time
Classroom time will problems. Reference point
focus on key concepts 3. Go over the energy changes Periodic Table
needed for success on needed as a substance undergoes Elements
the state test. changes of state. Atoms
4. Define compound. Nucleus
5. Practice using the periodic Electrons
table. Know the main sections on Protons
the periodic table. Neutrons
6. Know pH; understand acids Molecules
from bases and the use of litmus Compounds
paper. Reactions
7. Organic chemistry - know the Balanced Equations
key elements: C, H, O, N, P, & S. pH Scale
8. Solar System…Sun (stars), Acids
inner vs. outer planets, moons, Bases
seasons on Earth, rotation vs. Organic Chemistry
revolution. Polymer
9. Distinguish among comets, Solar system
asteroids, and meteorites. Moon Phases
Comet
Asteroid
Unit Six Focus Standards: Objectives: Textbook Pages: Unit Vocabulary:
Teaching Days Many of the standards Provide opportunities for Vocabulary used in
The time period here covered during the hands-on projects. Toothpick Bridges unit six will reinforce
begins after the state five previous units Note: Projects may change Pages: 450 - 457 those terms learned in
test in science. Note: will be addressed depending on budget, time, or previous units.
Some teachers may again within these individual class needs. Roller-coasters
change or expand special projects. Pages: 381 - 399
topics in unit six as 1. Toothpick Bridges
each class may have Bottle Rockets
different needs. 2. Roller-coasters Pages: 502 - 509
3. Bottle Rockets
```
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views: 284 posted: 5/24/2010 language: English pages: 14 | 5,806 | 31,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2014-10 | latest | en | 0.815647 |
https://blog.csdn.net/weixin_30551963/article/details/102265024 | 1,620,796,128,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991252.15/warc/CC-MAIN-20210512035557-20210512065557-00532.warc.gz | 163,140,302 | 20,423 | # Population zoj 3018
Population
Time Limit: 10000 msMemory Limit: 32768 KB
It is always exciting to see people settling in a new continent. As the head of the population management office, you are supposed to know, at any time, how people are distributed in this continent.
The continent is divided into square regions, each has a center with integer coordinates (x, y). Hence all the people coming into that region are considered to be settled at the center position. Given the positions of the corners of a rectangle region, you are supposed to count the number of people living in that region.
Input
Your program must read inputs from the standard input. Since there are up to 32768 different regions and possibly even more queries, please use "scanf" and "printf" instead of "cin" and "cout" to avoid timeout.
The character "I" in a line signals the coming in of new groups of people. In the following lines, each line contains three integers: X, Y, and N, where X and Y (1 <= X, Y <= 20000) are the coordinates of the region's center, and N (1 <= N <= 10000) is the number of people coming in.
The character "Q" in a line signals the query of population. The following lines each contains four numbers: Xmin, Xmax, Ymin, Ymax, where (Xmin, Ymin) and (Xmax, Ymax) are the integer coordinates of the lower left corner and the upper right corner of the rectangle, respectively.
The character "E" signals the end of a test case. Process to the end of file.
Output
For each "Q" case, print to the standard output in a line the population in the given rectangle region. That is, you are supposed to count the number of people living at all the positions (x, y) such that Xmin <= x <= Xmax, and Ymin <= y <= Ymax.
Sample Input
I
8 20 1
4 5 1
10 11 1
12 10 1
18 14 1
Q
8 10 5 15
8 20 10 14
I
7 6 1
10 3 2
7 2 1
2 3 2
10 3 1
Q
2 20 2 20
E
Sample Output
1
3
12
[思路]：求的是一个二维区间之间的和
#include <bits/stdc++.h>
using namespace std;
/**
* 面积树
* 四叉树
* **/
typedef long long LL;
int str_to_int(string a){
int sum = 0;
for(int i = 0; i < a.length(); i ++){
sum = sum * 10 + (a[i] - '0');
}
return sum;
}
int tot = 0;
const int MAXN = 1e7 + 5;
struct AREA_tree{
struct NODE{
int lx, ly, rx, ry;///左下角 右上角
LL sum;///区间的总和
int son[4];///儿子节点对应的是点在数组中的编号
/*划分为四个部分，四个儿子节点对应四个部分
0 <xmid + 1, ymid + 1> <rx, ry>
1 <lx, ymid + 1> <xmid, ry>
2 <lx, ly> <xmid, ymid>
3 <xmid + 1, ly> <rx, ymid>
*/
}tree[MAXN];
void Init_NODE(NODE &x){
for(int i = 0; i < 4; i ++){
x.son[i] = -1;
}///清空儿子节点，代表没有下一个区间
}
void Init(){
tot = 0;///初始化为0
}
int add(int lx, int ly, int rx, int ry){///添加一个<lx, ly> <rx, ry>的节点
tree[tot].lx = lx, tree[tot].rx = rx, tree[tot].ly = ly, tree[tot].ry = ry;
Init_NODE(tree[tot]);
tree[tot].sum = 0;
}
void update(int root, int x, int y, int value){///更新一个节点
if(tree[root].lx == tree[root].rx && tree[root].ly == tree[root].ry){
tree[root].sum += value;///如果跑到单点上，那么更新总和
return ;
}
int xmid = (tree[root].lx + tree[root].rx) >> 1;///xmid
int ymid = (tree[root].ly + tree[root].ry) >> 1;///ymid
/*
0 <xmid + 1, ymid + 1> <rx, ry>
1 <lx, ymid + 1> <xmid, ry>
2 <lx, ly> <xmid, ymid>
3 <xmid + 1, ly> <rx, ymid>
*/
int rx, ry, lx, ly;
rx = tree[root].rx, ry = tree[root].ry, lx = tree[root].lx, ly = tree[root].ly;
///判断更新的区间在哪个子区间
if(x <= rx && y <= ry && x >= xmid + 1 && y >= ymid + 1){
if(tree[root].son[0] == -1){///右上
tree[root].son[0] = add(xmid + 1, ymid + 1, rx, ry);
}
update(tree[root].son[0], x, y, value);
}
else if(x <= xmid && y <= ry && x >= lx && y >= ymid + 1){///左上
if(tree[root].son[1] == -1){
tree[root].son[1] = add(lx, ymid + 1, xmid, ry);
}
update(tree[root].son[1], x, y, value);
}
else if(x <= xmid && y <= ymid && x >= lx && y >= ly){///左下
if(tree[root].son[2] == -1){
tree[root].son[2] = add(lx, ly, xmid, ymid);
}
update(tree[root].son[2], x, y, value);
}
else if(x <= rx && y <= ymid && x >= xmid + 1 && y >= ly){///右下
if(tree[root].son[3] == -1){
tree[root].son[3] = add(xmid + 1, ly, rx, ymid);
}
update(tree[root].son[3], x, y, value);
}
int sum = 0;
for(int i = 0; i < 4; i ++){
if(tree[root].son[i] != -1){
sum += tree[tree[root].son[i]].sum;
}
}///更新相当于线段树的push_up
tree[root].sum = sum;
return ;
}
LL query(int root, int xmin, int ymin, int xmax, int ymax){
if(tree[root].lx >= xmin && tree[root].ly >= ymin && tree[root].rx <= xmax && tree[root].ry <= ymax){
return tree[root].sum;
}
int xmid = (tree[root].lx + tree[root].rx) >> 1;
int ymid = (tree[root].ly + tree[root].ry) >> 1;
/*
0 <xmid + 1, ymid + 1> <rx, ry>
1 <lx, ymid + 1> <xmid, ry>
2 <lx, ly> <xmid, ymid>
3 <xmid + 1, ly> <rx, ymid>
查询语句，依旧是查询四个区间
判断是否应该查询这个区间
或者时对应的区间被包含
*/
int rx, ry, lx, ly;
rx = tree[root].rx, ry = tree[root].ry, lx = tree[root].lx, ly = tree[root].ly;
LL re = 0;
if(xmin <= xmid && ymin <= ymid){
if(tree[root].son[2] != -1)
re += query(tree[root].son[2], xmin, ymin, xmax, ymax);
}
if(xmax > xmid && ymin <= ymid){
if(tree[root].son[3] != -1)
re += query(tree[root].son[3], xmin, ymin, xmax, ymax);
}
if(xmin <= xmid && ymax > ymid){
if(tree[root].son[1] != -1)
re += query(tree[root].son[1], xmin, ymin, xmax, ymax);
}
if(xmax > xmid && ymax > ymid){
if(tree[root].son[0] != -1)
re += query(tree[root].son[0], xmin, ymin, xmax, ymax);
}
return re;
}
}area_tree;
int main(){
#ifdef moxin
freopen("C:\\Users\\user\\Desktop\\dp_data\\1.in", "r", stdin);
freopen("C:\\Users\\user\\Desktop\\dp_data\\1.out", "w", stdout);
#endif
ios::sync_with_stdio(false);
string str;
int flag = 0;///0 - I 1 - Q
area_tree.Init();
while(cin >> str){
if(str[0] == 'E'){
area_tree.Init();
continue;
}
if(str[0] == 'I') {
flag = 0; continue;
}
if(str[0] == 'Q') {
flag = 1; continue;
}
if(flag == 0){
int x = str_to_int(str);
int y, z;
cin >> y >> z;
area_tree.update(0, x, y, z);
}
else if(flag == 1){
int x1, y1, x2, y2;
x1 = str_to_int(str);
cin >> x2 >> y1 >> y2;
cout << area_tree.query(0, x1, y1, x2, y2) << endl;
}
}
return 0;
}
07-14 2617
09-13 1653
02-01 80
08-28 563
01-22 88
10-03 362
08-13 69
11-10 779
08-23 40
08-08 377
03-02 1250
08-24 355
07-21 213
04-05 77 | 2,257 | 5,986 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-21 | latest | en | 0.854861 |
https://justaaa.com/math/1045469-let-fx53x-a-find-fx-bfind-the-slope-of-the | 1,716,680,604,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058858.81/warc/CC-MAIN-20240525222734-20240526012734-00003.warc.gz | 279,545,226 | 9,421 | Question
# Let f(x)=5+3/x a) Find f′(x) b)Find the slope of the tangent line at x=1. Slope =...
Let f(x)=5+3/x
a) Find f′(x)
b)Find the slope of the tangent line at x=1. Slope = f′(1)=?
solution
f(x)=5+3/x
a) Find f′(x)
d(f(x)/dx = f'(x) = d(5)/dx + d(3/x)/dx
derivative of constant is zero = d(5)/dx =0
d(x^n) = nx^n-1 here n =-1
d(1/x) = d(x^-1) = -1x^(-1-1) = -1x^-2 = -1/x^2
d(1/x)/dx = -1/x^2
= d(5)/dx + 3d(1/x)/dx = 0 +3(-1/x^2) = -3/x^2
d(f(x)/dx = f'(x) = d(5)/dx + d(3/x)/dx = -3/x^2
derivative of a function f(x) is slope = d(f(x)/dx = f'(x) = -3/x^2
b)Find the slope of the tangent line at x=1. Slope = f′(1)=?
derivative of a function f(x) is slope = d(f(x)/dx = f'(x) = -3/x^2
f'(1) put x =1
f'(1) = -3/(1^2) =-3
Slope = f′(1)= -3
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 387 | 824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-22 | latest | en | 0.625156 |
http://math.stackexchange.com/questions/58477/how-can-one-show-that-the-topology-of-convergence-in-measure-is-separable | 1,469,758,530,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829325.58/warc/CC-MAIN-20160723071029-00272-ip-10-185-27-174.ec2.internal.warc.gz | 157,552,911 | 20,464 | # How can one show that the topology of convergence in measure is separable?
Let $X$ be a polish space equiped with the borel sigma-algebra and a probability measure $\mu$. How can one show that the set of all borel measurable functions $f:X\rightarrow R$ ($R$ being the real numbers), where two a.e. equal functions are identified, equiped with the topology of convegence in measure is separable?
-
I've given you many references in my last answer to you. Fisher-Witte Morris-Whyte (rem 2.4) give a detailed reference to Wheeden-Zygmund, and it is contained in at least three of the other references I've given to you. Please look around a bit before you ask. You can also see Proposition 7 here for a more precise and general result (there polonais = Polish). – t.b. Aug 19 '11 at 10:49
Thank you Theo, you have helped me greatly. I did look around, but didn't find it. I will look in some of your references... BTW I haven't found it in Wheeden-Zygmund – Arnold Aug 19 '11 at 11:24
Hmmm. Not having one of my most charitable days, I guess... In fact, in the books I mentioned it's in the exercises (e.g. Kechris) and at times not even explicitly mentioned (e.g. in Wheeden-Zygmund), you're right. It's one of those results that people like to leave as an exercise... Sorry about that. Moore's proof of proposition 7 mentioned above suffers a bit from its generality. BTW: I flagged your account for merging with the other one. You could consider registering your account, then it is easier for the software to recognize you. – t.b. Aug 19 '11 at 12:12
@Arnold: I have merged your other account, A_0, into your current account. If you have trouble logging in, or if you accidentally create duplicate accounts, simply flag one of your own questions for moderator attention, and we will help out. – Zev Chonoles Aug 19 '11 at 14:39
Thanks (at least 15 characters...) – Arnold Aug 19 '11 at 14:53
Here's an outline of an argument, and it should be easy to fill in the details:
1. Note that $L_0(X)$ is a metric space e.g. with respect to the metric $\displaystyle d(f,g) = \int \frac{|f-g|}{1+|f-g|}$.
2. Choose a countable base $\{A_n\}_{n \in \mathbb{N}}$ for the topology on $X$.
• Every open set is equal to the union of elements in $\{A_n\}$.
• For every measurable set $E$ there is a $G_\delta$-set $G$ such that $\mu(E \triangle G) = 0$, that is $[E] = [G]$ in $L_{0}(X)$.
3. Show that a non-negative measurable function $f$ is a pointwise monotone limit of simple functions.
Hint: Put $B_{k,n} = \{x\in X : 2^{-n} k \leq f(x) \lt 2^{-n}(k+1)\}$ and consider $f_n = 2^{-n} \sum\limits_{k=0}^{2^{2n}}k \cdot[B_{k,n}]$.
4. Split a general measurable function into positive and negative parts.
Use these observations to build a countable dense set of $L_{0}(X)$.
For completeness and further properties of $L_0(X)$, I recommend Driver's notes on probability Section 12, especially Theorem 12.8 on page 179. (Thanks to Nate Eldredge from whom I learned about these notes).
Edit: In view of Byron's answer, note that Driver's notes contain various forms of the functional monotone class theorem in Part II, Section 8 on pages 111ff. Of course, the main point in both our answers is that there is a countable generating and separating set for the $\sigma$-algebra. The assumption that $X$ be Polish ensures that.
-
@Didier: Yes, the formula is a bit nicer this way; but doesn't it obscure the idea somewhat? – t.b. Aug 20 '11 at 11:02
Sorry, mainly I meant to add the x in X in the definition of your set B. The other modification is a (possibly not so nice after all) nicety. Theo, really, proceed at your convenience, this is your post... – Did Aug 20 '11 at 11:08
An alternative is to use the Functional Monotone Class Theorem. Let $\cal A$ be a countable collection of sets that generates ${\cal B}(X)$, and put $${\cal K}=\{1_A: A\mbox{ is a finite intersection of }{\cal A}\mbox{ sets }\}.$$
Let ${\cal K}^\prime$ be the (countable!) $\mathbb{Q}$-vector space generated by $\cal K$, and set $${\cal M}=\{h: k^\prime_n\to h \mbox{ in probability for some }k^\prime_n\in{\cal K}^\prime\}.$$
Then $\cal K$ and $\cal M$ satisfy the conditions of the FMCT, and hence $\cal M$ includes all bounded ${\cal B}(X)$-measurable functions. A truncation argument now shows that any ${\cal B}(X)$-measurable function can be approximated in probability by a sequence in ${\cal K}^\prime$.
-
I just wanted to point out that various useful variants of the FMCT are contained in Driver's notes I linked to, see the edit to my post. The dense set $\mathcal{K}'$ is of course the one I had in mind in my answer. – t.b. Aug 21 '11 at 8:01 | 1,319 | 4,636 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2016-30 | latest | en | 0.942406 |
http://www.lmfdb.org/ModularForm/GL2/TotallyReal/4.4.18432.1/holomorphic/4.4.18432.1-14.1-f | 1,566,732,052,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027323328.16/warc/CC-MAIN-20190825105643-20190825131643-00077.warc.gz | 275,516,577 | 5,917 | # Properties
Base field 4.4.18432.1 Weight [2, 2, 2, 2] Level norm 14 Level $[14, 14, w + 2]$ Label 4.4.18432.1-14.1-f Dimension 6 CM no Base change no
# Related objects
• L-function not available
## Base field 4.4.18432.1
Generator $$w$$, with minimal polynomial $$x^{4} - 12x^{2} + 18$$; narrow class number $$2$$ and class number $$1$$.
## Form
Weight [2, 2, 2, 2] Level $[14, 14, w + 2]$ Label 4.4.18432.1-14.1-f Dimension 6 Is CM no Is base change no Parent newspace dimension 18
## Hecke eigenvalues ($q$-expansion)
The Hecke eigenvalue field is $\Q(e)$ where $e$ is a root of the defining polynomial:
$$x^{6}$$ $$\mathstrut +\mathstrut 4x^{5}$$ $$\mathstrut -\mathstrut 14x^{4}$$ $$\mathstrut -\mathstrut 60x^{3}$$ $$\mathstrut -\mathstrut 19x^{2}$$ $$\mathstrut +\mathstrut 72x$$ $$\mathstrut +\mathstrut 44$$
Norm Prime Eigenvalue
2 $[2, 2, -\frac{1}{3}w^{3} - \frac{1}{3}w^{2} + 3w + 4]$ $\phantom{-}1$
7 $[7, 7, \frac{1}{3}w^{3} + \frac{1}{3}w^{2} - 3w - 3]$ $\phantom{-}\frac{1}{10}e^{5} + \frac{7}{20}e^{4} - \frac{29}{20}e^{3} - \frac{103}{20}e^{2} - \frac{19}{20}e + \frac{53}{10}$
7 $[7, 7, -\frac{1}{3}w^{2} + w + 1]$ $\phantom{-}1$
7 $[7, 7, \frac{1}{3}w^{2} + w - 1]$ $\phantom{-}\frac{1}{5}e^{5} + \frac{9}{20}e^{4} - \frac{73}{20}e^{3} - \frac{121}{20}e^{2} + \frac{147}{20}e + \frac{61}{10}$
7 $[7, 7, \frac{1}{3}w^{3} - \frac{1}{3}w^{2} - 3w + 3]$ $\phantom{-}e$
9 $[9, 3, w - 3]$ $\phantom{-}\frac{1}{5}e^{5} + \frac{9}{20}e^{4} - \frac{73}{20}e^{3} - \frac{121}{20}e^{2} + \frac{147}{20}e + \frac{81}{10}$
41 $[41, 41, \frac{1}{3}w^{3} - \frac{1}{3}w^{2} - 3w + 1]$ $-\frac{3}{20}e^{5} - \frac{2}{5}e^{4} + \frac{14}{5}e^{3} + \frac{61}{10}e^{2} - \frac{149}{20}e - \frac{87}{10}$
41 $[41, 41, -\frac{1}{3}w^{2} + w + 3]$ $-\frac{3}{20}e^{5} - \frac{2}{5}e^{4} + \frac{14}{5}e^{3} + \frac{61}{10}e^{2} - \frac{149}{20}e - \frac{87}{10}$
41 $[41, 41, \frac{1}{3}w^{2} + w - 3]$ $-\frac{1}{20}e^{5} - \frac{1}{20}e^{4} + \frac{17}{20}e^{3} - \frac{1}{20}e^{2} + \frac{21}{10}e + \frac{28}{5}$
41 $[41, 41, -\frac{1}{3}w^{3} - \frac{1}{3}w^{2} + 3w + 1]$ $-\frac{1}{10}e^{5} - \frac{7}{20}e^{4} + \frac{39}{20}e^{3} + \frac{103}{20}e^{2} - \frac{131}{20}e - \frac{53}{10}$
47 $[47, 47, \frac{1}{3}w^{3} + \frac{1}{3}w^{2} - 4w - 1]$ $-\frac{3}{20}e^{5} - \frac{3}{20}e^{4} + \frac{51}{20}e^{3} - \frac{3}{20}e^{2} - \frac{27}{10}e + \frac{54}{5}$
47 $[47, 47, -\frac{1}{3}w^{3} + \frac{1}{3}w^{2} + 2w - 3]$ $\phantom{-}\frac{1}{5}e^{5} + \frac{7}{10}e^{4} - \frac{17}{5}e^{3} - \frac{103}{10}e^{2} + \frac{33}{5}e + \frac{58}{5}$
47 $[47, 47, \frac{1}{3}w^{3} + \frac{1}{3}w^{2} - 2w - 3]$ $\phantom{-}\frac{13}{20}e^{5} + \frac{33}{20}e^{4} - \frac{231}{20}e^{3} - \frac{447}{20}e^{2} + \frac{106}{5}e + \frac{141}{5}$
47 $[47, 47, \frac{1}{3}w^{3} - \frac{1}{3}w^{2} - 4w + 1]$ $-\frac{9}{20}e^{5} - \frac{6}{5}e^{4} + \frac{79}{10}e^{3} + \frac{163}{10}e^{2} - \frac{257}{20}e - \frac{151}{10}$
89 $[89, 89, -\frac{1}{3}w^{3} + \frac{2}{3}w^{2} + 3w - 3]$ $\phantom{-}\frac{3}{5}e^{5} + \frac{21}{10}e^{4} - \frac{97}{10}e^{3} - \frac{309}{10}e^{2} + \frac{83}{10}e + \frac{179}{5}$
89 $[89, 89, \frac{2}{3}w^{2} + w - 5]$ $-\frac{2}{5}e^{5} - \frac{7}{5}e^{4} + \frac{34}{5}e^{3} + \frac{103}{5}e^{2} - \frac{46}{5}e - \frac{76}{5}$
89 $[89, 89, \frac{2}{3}w^{2} - w - 5]$ $\phantom{-}\frac{19}{20}e^{5} + \frac{59}{20}e^{4} - \frac{313}{20}e^{3} - \frac{821}{20}e^{2} + \frac{88}{5}e + \frac{183}{5}$
89 $[89, 89, \frac{1}{3}w^{3} + \frac{2}{3}w^{2} - 3w - 3]$ $\phantom{-}\frac{7}{10}e^{5} + \frac{39}{20}e^{4} - \frac{233}{20}e^{3} - \frac{531}{20}e^{2} + \frac{217}{20}e + \frac{221}{10}$
97 $[97, 97, \frac{2}{3}w^{3} - \frac{1}{3}w^{2} - 6w + 5]$ $-\frac{9}{20}e^{5} - \frac{29}{20}e^{4} + \frac{143}{20}e^{3} + \frac{411}{20}e^{2} - \frac{28}{5}e - \frac{93}{5}$
97 $[97, 97, -w^{3} - \frac{5}{3}w^{2} + 10w + 15]$ $\phantom{-}\frac{13}{20}e^{5} + \frac{43}{20}e^{4} - \frac{211}{20}e^{3} - \frac{617}{20}e^{2} + \frac{61}{5}e + \frac{161}{5}$
Display number of eigenvalues
## Atkin-Lehner eigenvalues
Norm Prime Eigenvalue
2 $[2, 2, -\frac{1}{3}w^{3} - \frac{1}{3}w^{2} + 3w + 4]$ $-1$
7 $[7, 7, -\frac{1}{3}w^{2} + w + 1]$ $-1$ | 2,267 | 4,177 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-35 | longest | en | 0.248631 |
https://www.nag.com/numeric/py/nagdoc_latest/naginterfaces.library.stat.quantiles.html | 1,670,037,827,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710918.58/warc/CC-MAIN-20221203011523-20221203041523-00170.warc.gz | 972,388,797 | 6,994 | # naginterfaces.library.stat.quantiles¶
naginterfaces.library.stat.quantiles(rv, q)[source]
quantiles finds specified quantiles from a vector of unsorted data.
For full information please refer to the NAG Library document for g01am
https://www.nag.com/numeric/nl/nagdoc_28.6/flhtml/g01/g01amf.html
Parameters
rvfloat, array-like, shape
The vector whose quantiles are to be determined.
qfloat, array-like, shape
The quantiles to be calculated, in ascending order. Note that these must be between and , with returning the smallest element and the largest.
Returns
qvfloat, ndarray, shape
contains the quantile specified by the value provided in , or an interpolated value if the quantile falls between two data values.
Raises
NagValueError
(errno )
On entry, .
Constraint: .
(errno )
On entry, .
Constraint: .
(errno )
On entry, an element of was less than or greater than .
(errno )
On entry, was not in ascending order.
Notes
A quantile is a value which divides a frequency distribution such that there is a given proportion of data values below the quantile. For example, the median of a dataset is the quantile because half the values are less than or equal to it; and the quantile is the th percentile.
quantiles uses a modified version of Singleton’s ‘median-of-three’ Quicksort algorithm (Singleton (1969)) to determine specified quantiles of a vector of real values. The input vector is partially sorted, as far as is required to compute desired quantiles; for a single quantile, this is much faster than sorting the entire vector. Where necessary, linear interpolation is also carried out to return the values of quantiles which lie between two data points.
References
Singleton, R C, 1969, An efficient algorithm for sorting with minimal storage: Algorithm 347, Comm. ACM (12), 185–187 | 409 | 1,818 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-49 | longest | en | 0.818643 |
https://scienceblogs.de/klausis-krypto-kolumne/2020/01/24/encrypted-messages-from-a-1980s-rock-band/ | 1,596,886,585,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737645.2/warc/CC-MAIN-20200808110257-20200808140257-00435.warc.gz | 492,454,535 | 20,067 | # Encrypted messages from a 1980s rock band
The 1980s rock band Arcadia published a few encrypted messages on their record covers and in a sheet music book. The solutions are not known to me.
My friend Elonka Dunin
…, who is fan of the 1980s rock band Duran Duran has made me aware of an encrypted message I had never heard of.
### The message on the cover
In 1985, the Duran Duran-spinoff group Arcadia published an album titled “So Red the Rose”, best known for its hit single “Election Day”. In the Wikipedia entry of this record we read the following:
Some versions of the album and related singles contained a numeric code in their artwork, this code allocated random numbers to letters. The numbers on the cover say “So Red the Rose” and then the surnames of the band members being “Rhodes, Le Bon and Taylor (as in Roger Taylor)”.
When I googled for the cover of “So Red The Rose”, I found the following:
As can be seen, 18 two-digit numbers are written on the black frame around the picture of a face. Here’s a transcription:
02 16 12 38 44 50 | 50 30 44 22 24 52 | 38 24 00 18 44 42
The same numbers are present on the cover of the single “Election Day”:
Now I ask myself if the explanation on Wikipedia is correct. 18 letters do not fit with the cleartext “So Red the Rose” or “Rhodes, Le Bon and Taylor”. Can a reader find out what the message really means?
### The messages in the sheet music book
Via Google I also found a few scans from a sheet music book that was published as an add-on to the record. On the cover the following three numbers can be read:
And then the sheet music book contains a double page with the following number sequences:
I assume that this is an encrypted message. Can a reader break it?
Further reading: A code in the Hollywood movie “Wanted”
## Kommentare (5)
1. #1 cimddwc
24. Januar 2020
The three 6-number sequences fit the three names pretty well: Taylor, Rhodes, Le Bon – with o=44, r=50, space in Le Bon=00, etc.
2. #2 Gerry
24. Januar 2020
If you fill TAYLOR RHODES LE BON into the 6-6-6 numbers, you see an error at LE BON, as in the first two names 44 stands for “O”, and not 14 (as in the third). This leads to a scheme strting with A = 16, B=18 and so on until S = 52, T = 02, U = 04 and so o until Z = 14. 00 is a blank.
02 50 38 equals TRL, the initials of the band members.
3. #3 Gerry
24. Januar 2020
There must be more errors or deviations from the scheme, as the number matrix leads to useful words and gibberish:
ASHIDEBEFOJQ
THE_PROMISE_
LIVEILY..MTN
EYES_WILLBAK
_ONEVER_DRY_
BUSXWBORWELL
OH_AI_SGAZOO
I see “the promise” and “Orwell”, so maybe there are deviations from the linear code or another cipher.
4. #4 Gerd
25. Januar 2020
Compairing the pictures above, the error 14 vs 44 in Le Bon is only in the second one, “Election Day”. On “So Red The Rose” from which Klaus took the transscription has Le Bon with 44.
5. #5 cimddwc
25. Januar 2020
The song “The Promise” ends with the line “Heaven’s eyes will never dry” (according to Lyricsfreak.com), which is partially jumbled in there. Remains the question of how much of the rest is another cipher, sloppiness, arbitrary parts from other songs, meaningless, the book’s publisher misreading bad handwriting, or whatever… | 900 | 3,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-34 | latest | en | 0.899149 |
http://gec.org.ru/index.php/download/foundations-of-relative-homological-algebra | 1,544,860,560,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826800.31/warc/CC-MAIN-20181215061532-20181215083532-00567.warc.gz | 110,565,526 | 8,733 | # Download Foundations of Relative Homological Algebra by Samuel; Moore, J.C. Eilenber PDF
By Samuel; Moore, J.C. Eilenber
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Extra resources for Foundations of Relative Homological Algebra
Sample text
This yields the conclusion. We thus find that the so·called "double resolutions" of a complex considered in [4. Chapter XVII] are precisely the strongly projective resolutions. 4. Subcategories of Let Cf be an abelian category. Given - 00 cCf \$ p and q \$ 00 we consider the full subcategory Cq(j p of C(f determined by the complexes A with An n < p and for n > q. We usually omit the symbol p if P = - 00 and the symbol q if q = = 0 for 00 • Complexes A in which the differentiation dn : An - An - 1 is zero for all n € Z, determine a full subcategory cCf of cCf.
If E € CC(f) Cn (E), Zn (E), Z ~ (E), Bn (E), Hn (E) are exact for all n € A. Proof. The exact sequence o -> Zn (E) -> Cn (E) -> B n (E) -> 0 implies that Bn (E) is exact for every n € Z. The exact sequences o---. B n (E) -> Zn (E) -> Hn (E) -> 0, 0-+ Bn(E) -> Cn(E) -> Z~(E) -> 0 now yield the same conclusion for Hand Z'11 . 5. If the abelian category a is projectively perfect then the strong Ca. exact sequences form a projective class fi,s in Further, for any object A in C(f the following properties are equivalent.
2. 1. Let (f be an abelian category and ~ a full subcategory of C(1. Let M and N be subsets of Z such that 32 SAMUEL EILENBERG AND J. C. MOORE (I) If A € et, m € M and N € N, then Qm(A) and Rn(A) are in (11) If IACTl is a locally finite family of objects in Cet is in :D. N where nm C-m 1 &,] n [ nn Z-n I&,] :D. Further, for any object A in (i) A is :D :D :D. N -projective. :)RnHn(A]]. projective objects Em ,F n € (j, m € M, n € N such that Proof. 5) imply Qm --j Rn Cm --j Zn :(:D, et), :(:D, (1), m €M, n € N. | 816 | 2,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-51 | latest | en | 0.62864 |
www.recentjobsolution.com | 1,600,812,242,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206763.24/warc/CC-MAIN-20200922192512-20200922222512-00439.warc.gz | 1,010,906,074 | 28,674 | # Combine 5 Bank Officer Question Solution 2018
Combine 5 Bank Officer Question Solution 2018, provide by ziggasa.com. The Combine Bank Officer (General) exam question solution 2018 is available now. The Combine Bank General Officer exam held on July 20, 2018 and MCQ Result has not Published yet. In this day, various exam was held as Sonali Bank Officer Cash, Titas Gas Assistant General Officer and so on. See the all recent exam question 2018 on ziggasa.com.
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১. র,ল,ব, ষ
২. পাখি
৩. আবর্তন
৪. আচ্ছন্ন
৫. একটি
৬. ইদানিং
৭. কার
৮. উপপদ
৯. অপরিবর্তন যোগ্য পদক্রম
১০. ভাষারীতি
১১. ধ্বন্যাত্মক
১২. পাবন
১৩. ভাত
১৪. জিহীর্ষা
১৫. নীতিমুলক
১৬. সরল ভাষা
# Math Part:
33. The values of p for equation 2x²-4x+p = 0 to have real root is: —- to have one or more real roots, b²-4ac ≥ 0 or, (-4)² – 4×2×p ≥ 0 or, -8p ≥ -16 or, p ≤ 2. ————————————————– [C] .
34. How many integers from 1 to 100 are divisible by 3 but not by 8? —- divisible by 3 = 100/3 = 33 integers divisible by 8&3, or 24 = 100/24 = 4 divisible by 3 but not by 8 = 33-4 = 29. ————————————————– [B]
35. If x:y = 5:3, then (8x+5y) : (8x-5y) = ? — x:y = 5:3 x = 5y/3 8x = 40y/3 (8x+5y) : (8x-5y) = (40y/3 + 5y) : (40y/3 – 5y) = 55y : 25y = 11 : 5. ————————————————– [D]
37. If a+1/a = 2, what is a³ + 1/a³? —- (a+1/a) = 2 => a³ + 1/a³ + 3×a×1/a×(a+1/a) = 2³ => a³ + 1/a³ + 3×2 = 8 => a³ + 1/a³ = 2. —– [C]
. 38. If 10% of x is equal to 25% of y, and y = 16, what is the value of x? —- x × 10/100 = y × 25/100 2x = 5y = 5×16 = 80 x = 40. ———– [D]
39. If sinA + sin²A = 1, then the value of the expression (cos²A + cos⁴A) is — —- sinA + sin²A = 1 => sinA + sin²A = sin²A + cos²A => cos²A = sinA cos²A + cos⁴A = sinA + sin²A = 1.– [A]
.
40. A pole casts a √3m long shadow on the ground at an elevation 60°, the height of the pole is — . Θ = 60°, b = √3, h = ? we know, tanΘ = h/b tan 60° = h/√3 √3 = h/√3 h = 3. ——– [A]
# Computer part :
73. A. An interpreter
74. B. 6 Digits
75. B. Dynamic RAM
77. B.72
78. D. Theme
79. A. Providing security against unauthorized users
80. C. Ctrl+H
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`To provide the knowledge, academic information, inspiration and solutions needed to help students and job seekers succeed.` | 1,140 | 2,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2020-40 | latest | en | 0.738313 |
http://www.pisarna-na-netu.si/371656-exactly-what-does-length-me-an-in-math-ltpgtltpgtjust-how-long-can-i-have-to-measure-milk-5/ | 1,590,641,850,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396495.25/warc/CC-MAIN-20200528030851-20200528060851-00278.warc.gz | 197,240,818 | 7,806 | ## Exactly what Does Length Me-an in Math? <p></p>Just how Long Can I Have To Measure Milk?
What exactly does length me an in math? The length of time would be a full glass of milk? What’s the length of one’s football field? For this particular question, you want to learn what does span me an in math.
This concern demands a small background to understand also what it’s indicates and how high school capstone project ideas math functions. First point is that you use math for many different matters in your life, and all of those things have something to do with span.
Let us say that you are taking a look at something to assess the length of. Your very best bet is considered a yardstick or some other type of measuring device. A yardstick will offer the clear answer, but there is. You have to take in to consideration the distance between the place where you need to discover the length and your location where you’re currently measuring.
That distance is known as the distance. That space will likely always be a component of length. Additionally, there are components of capstonepaper.net period. In order to find the distance of the glass filled with milk, then you also must ascertain the duration of the glass, then the length of the container, then your length of the milk, after which the length of the container.
The duration of the full glass of milk will be found by adding up the spans of also the glass and the container which comprise the container. Then your clear answer will soon be considered a half-gallon, if the measurement is just two feet.
Nowadays you understand how to answer fully the issue”just how long does one glass of milk need to take sequence to make it a full glass” You’ve ascertained the length of the container by dividing the length out of the location where you are quantifying to the location where you would like to discover the length of the container. You have taken the exact distance between the two areas and added them collectively. Inside this situation, you’ve were left using two half-gallons, which is.
What exactly does span me-an in mathematics once we have been talking about the true bodily objects? Virtually anything can be measured with all units of span. https://www.missouristate.edu/assessment/gilbertbrown.aspx They create all the variation in matters including time measurement and distance measurement.
It’s easy to realize how the one-hundred-and-fifty miles may be divided into hundred parts and then changed into sixteen parts and converted into one part and still be the most suitable answer. That is how numbers work. Then you can find them out even if you have no idea just how exactly they work.
Math is about producing components of dimension. A unit of measurement is. At the example of milk, then the manner that span is measured in an unit of dimension is by taking the length of the container and multiplying it. This device of measurement is called a”gram,” also it provides you a solution.
Within the case of the glass filled with milk, the answer is really just a”whole” half-gallon. We have a measurement of the length of the container and the amount of milk inside this, and we’ve got the”gram” unit of dimension for that info. Using units of length is the most easy way to add convenience.
The next time you request the question”exactly what does precisely the length mean in mathematics?” Consider the fact the way they are used within an component of measurement along with there are many distinct types of units of span. Units of span may earn a lot of huge difference what we can find and in that which we are able to quantify. After we have been performing our very own measurements.
Components of span can indicate all kinds of things, but you only have to know very well what exactly does span me an in math. Is a brief mention guide to help you answer this query,”exactly what does span me-an in mathematics?” For those who understand just what a unit of span signifies, you may make your very own custom components of measurement, whether it is span, or time distance. | 845 | 4,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-24 | latest | en | 0.956426 |
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Posted: Friday 05th of Jan 09:09 Well, you don’t need to wait any longer. Go to http://www.solve-variable.com/solving-quadratic-equations-3.html and get yourself a copy for a very nominal price. Good luck and happy learning! | 906 | 3,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-30 | latest | en | 0.913527 |
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# Applications to Central Limit Theorem and Law of Large Numbers
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### Applications to Central Limit Theorem and Law of Large Numbers
1. 1. Statistics Lab Rodolfo Metulini IMT Institute for Advanced Studies, Lucca, Italy Lesson 2 - Application to the Central Limit Theory - 14.01.2014
2. 2. Introduction The modern statistics was built and developed around the normal distribution. Academic world use to say that, if the empirical distribution is normal (or approximative normal), everything works good. This depends mainly on the sample dimension Said this, it is important to undestand in which circumstances we can state the distribution is normal. Two founding statistical theorems are helpful: The Central Limit Theorem and The Law of Large Numbers.
3. 3. The Law of Large Numbers (LLN) Suppose we have a random variable X with expected value E (X ) = µ. We extract n observation from X (say {x = x1 , x2 , ..., xn }). ˆ If we define Xn = n −→ ∞, ˆ Xn −→ µ i n xi = x1 +x2 +...+xn , n the LLN states that, for
4. 4. The Central Limit Theorem (CLT) Suppose we have a random variable X with expected value E (X ) = µ and v (X ) = σ 2 We extract n observation from X (say {x = x1 , x2 , ..., xn }). ˆ Lets define Xn = i n xi = x1 +x2 +...+xn . n σ2 ˆ Xn distributes with expected value µ and variance . n In case n −→ ∞ (in pratice n > 30) 2 σ ˆ Xn ∼ N(µ, ), whatever the distribution of x be. n 2 σ ˆ N.B. If X is normal distributed, Xn ∼ N(µ, ) even if n n < 30
5. 5. CLT: Empiricals To better understand the CLT, it is recommended to examine the theorem empirically and step by step. By the introduction of new commands in the R programming language. In the first part, we will show how to draw and visualize a sample of random numbers from a distribution. Then, we will examine the mean and standard deviation of the sample, then the distribution of the sample means.
6. 6. Drawing random numbers - 1 We already introduced the use of the letters d, p and q in relations to the various distributions (e.g. normal, uniform, exponential). A reminder of their use follows: d is for density: it is used to find values of the probability density function. p is for probability: it is used to find the probability that the random variable lies on the left of a giving number. q is for quantile: it is used to find the quantiles of a given distribution. There is a fourth letter, namely r, used to draw random numbers from a distribution. For example runif and rexp would be used to draw random numbers from the uniform and exponential distributions, respectively.
7. 7. Drawing random numbers - 2 Let use the rnorm command to draw 500 number atrandom from a normal distribution having mean 100 and standard deviation (sd) 10. > x= rnorm(500,mean=100,sd=10) The resuls, typing in the r consolle x, is a list of 500 numbers extracted at random from a normal distribution with mean 500 and sd 100. When you examine the numbers stored in the variable X , There is a sense that you are pulling random numbers that are clumped about a mean of 100. However, a histagram of this selection provides a different picture of the data stored. > hist(x,prob=TRUE)
8. 8. Drawing random numbers - Comments Several comments are in order regarding the histogram in the figure. 1. The histogram is approximately normal in shape. 2. The balance point of the histogram appears to be located near 100, suggesting that the random numbers were drawn from a distribution having mean 100. 3. Almost all of the values are within 3 increments of 10 from the mean, suggesting that random numbers were drawn from a normal distribution having standard deviation 10.
9. 9. Drawing random numbers - a new drawing Lets try the experiment again, drawing a new set of 500 random numbers from the normal distribution having mean 100 and standard deviation 10: > x = rnorm(500, mean = 100, sd = 10) > hist(x, prob = TRUE , ylim = c(0, 0.04)) Give a look to the histogram ... It is different from the first one, however, it share some common traits: (1) it appears normal in shape; (2) it appears to be balanced around 100; (3) all values appears to occur within 3 increments of 10 of the mean. This is a strong evidence that the random numbers have been drawn from a normal distribution having mean 100 and sd 10. We can provide evidence of this claim by imposing a normal density curve: > curve(dnorm(x, mean = 100, sd = 10), 70, 130, add = TRUE , lwd = 2, col = ”red”))
10. 10. The curve command The curve command is new. Some comments on its use follow: 1. In its simplest form, the sintax curve(f (x), from =, to =) draws the function defined by f(x) on the interval (from, to). Our function is dnorm(x, mean = 100, sd = 10). The curve command sketches this function of X on the interval (from,to). 2. The notation from = and to = may be omitted if the arguments are in the proper order to the curve command: function first, value of from second, value of to third. That is what we have done. 3. If the argument add is set to TRUE , then the curve is added to the existing figure. If the arument is omitted (or FALSE ) then a new plot is drawn,erasing the prevoius graph.
11. 11. ˆ The distribution of Xn (sample mean) In our prevous example we drew 500 random numbers from a normal distribution with mean 100 and standard deviation 10. This leads to ONE sample of n = 500. Now the question is: what is the mean of our sample? > mean(x) [1]100.14132 If we take another sample of 500 random numbers from the SAME distribution, we get a new sample with different mean. > x = rnorm(500, mean = 100, sd = 10) mean(x) [1]100.07884 What happens if we draw a sample several times?
12. 12. Producing a vector of sample means We will repeatedly sample from the normal distribution. Each of the 500 samples will select 5 random numbers (instead of 500) from the normal distrib. having mean 100 and sd 10. We will then find the mean of those samples. We begin by declaring the mean and the standard deviation. Then, we declare the sample mean. > µ = 100; σ = 10 >n=5 We need some place to store the mean of the sample. We initalize a vector xbar to initially contain 500 zeros. > xbar = rep(0, 500)
13. 13. Producing a vector of sample means - cycle for It is easy to draw a sample of size n = 5 from the normal distribution having mean µ = 100 and standard deviation σ = 10. We simply issue the command rnorm(n, mean = µ, sd = σ). To find the mean of this results, we simply add the adjustment mean(rnorm(n, mean = µ, sd = σ)). The final step is to store this results in the vector xbar . Then we must repeat this same process an addintional 499 times. This require the use of a for loop. > for (iin1 : 500){xbar [i] = mean(rnorm(n, mean = µ, sd = σ))}
14. 14. Cycle for The i in for (iin1 : 500) is called theindex of the for loop. The index i is first set equal to 1, then the body of the for loop is executed. On the next iteration, i is set equal to 2 and the body of the loop is executed again. The loop continues in this manner, incrementing by 1, finally setting the index i to 500. After executing the last loop, the for cycle is terminated In the body of the for loop, we have xbar [i] = mean(rnorm(n, mean = µ, sd = σ)). This draws a sample of size 5 from the normal distribution, calculates the mean of the sample, and store the results in xbar [i]. When the for loop completes 500 iterations, the vector xbar contains the means of 500 samples of size 5 drawn from the normal distribution having µ = 100 and σ = 10 > hist(xbar , prob = TRUE , breacks = 12, xlim = c(70, 130, ylim = c(0, 0.1)))
15. 15. ˆ Distribution of Xn - observations 1. The previous histograms describes the shape of the 500 random number randomly selected, here, the histogram describe the distribution of 500 different sample means, each of which founded by selecting n = 5 random number from the normal distribution. 2. The distribution of xbar appears normal in shape. This is so even though the sample size is relatively small ( n = 5). 3. It appears that the balance point occurs near 100. This can be checked with the following command: > mean(xbar ) That is the mean of the sample means, that is almost equal to the mean of the draw of random numbers. 4. The distribution of the sample means appears to be narrower then the random number distributions.
16. 16. Increasing the sample size Lets repeat the last experiment, but this time let’s draw a sample size of n = 10 from the same distribution (µ = 100, σ = 10) > µ = 100; σ = 10 > n = 10 > xbar = rep(0, 500) > for (iin1 : 500){xbar [i] = mean(rnorm(n, mean = µ, sd = σ))} hist(xbar , prob = TRUE , breaks = 12, xlim = c(70, 130), ylim = c(0, 0.1)) The Histogram produced is even more narrow than using n=5
17. 17. Key Ideas 1. When we select samples from a normal distribution, then the distribution of sample means is also normal in shape 2. The mean of the distribution of sample meana appears to be the same as the mean of the random numbers (parentpopulation) (see the balance points compared) 3. By increasing the sample size of our samples, the histograms becomes narrower . Infact, we would expect a more accurate estimate of the mean of the parent population if we take the mean from a larger sample size. 4. Imagine to draw sample means from a sample of n = ∞. The histogram will be exactly concentrated (P = 1) in Xbar = µ
18. 18. Summarise We finish replicating the statement about CLT: 1. If you draw samples from a norml distribution, then the distribution of the sample means is also normal 2. The mean of the distribution of the sample means is identical to the mean of the parent population 3. The higher the sample size that is drawn, the narrower will be the spread of the distribution of the sample means.
19. 19. Homeworks Experiment 1: Draw the Xbar histogram for n = 1000. How is the histogram shape? Experiment 2: Repeat the full experiment drawing random numbers and sample means from a (1) uniform and from (2) a poisson distribution. Is the histogram of Xbar normal in shape for n = 5 and for n=30? Experiment 3: Repeat the full experiment using real data instead of random numbers. (HINT: select samples of dimension n = 5 from the real data, not using rnorm) Recommended: Try to evaluate the agreement of the sample mean histogram with normal distribution by mean of the qq-plot and shapiro wilk test.
20. 20. Application to Large Number Law Experiment: toss the coin 100 times. This experiment is like repeating 100 times a random draw from a bernoulli distribution with parameter ρ = 0.5 We expect to have 50 times (value = 1) head and 50 times cross (value = 0), if the coin is not distorted But, in practice, this not happen: repeating the experiment we are going to have a distribution centered in 50, but spread out. ˆ Let’s imagine to define Xn as the mean of the number of heads ˆ across n experiments. For n −→ ∞, Xn −→ 50 | 2,908 | 11,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-51 | latest | en | 0.858951 |
http://www.hackthissite.org/forums/viewtopic.php?f=28&t=954&start=20&view=print | 1,534,310,778,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209884.38/warc/CC-MAIN-20180815043905-20180815063905-00399.warc.gz | 508,407,880 | 4,896 | Page 3 of 3
### Re: Decryption and Cryptography 101
Posted: Thu Jun 30, 2011 10:52 am
very nice man!!!
### Re: Decryption and Cryptography 101
Posted: Sat Jul 09, 2011 2:17 pm
i really feel stupid now but...im not sure to understand this
### Re: Decryption and Cryptography 101
Posted: Sat Jul 09, 2011 2:52 pm
This thread is done. if I had magic mod powers, id lock the shit out of this thread.
### Re: Decryption and Cryptography 101
Posted: Tue Jan 05, 2016 11:24 pm
That was one of the best and easy to understand walkthroughs of this cryptography session.
No tools used on this one and gained a much better understanding of the process.
well done sir, well done!
### Re: Decryption and Cryptography 101
Posted: Mon Jun 13, 2016 4:48 pm
T3hR34p3r wrote:After reading this you should:
Have a more detailed understanding of cryptography.
Be able to think more logically when faced with encryption.
Understand the algorithm behind XECryption and how to decrypt it.
-------
So you think you're hot stuff do you? We'll see about that.
The code breakers in World War I didn't have supercomputers to do their work for them; they had to think for themselves. In today's modern world it's simple to write a script that does the work for you, but do you truly understand the script itself? Sure, you know how it works, what it does. But... take away the computer, could you still do that script? In this article I'm going to show you how to decrypt a basic sentence that has been run through the XECryption Algorithm, the same encryption method used in Realistic Mission 6. All you're going to need is a pen, a notepad, a calculator, and a hell of a lot of patience.
-------
For our example I've encrypted a small sentence, a little note to myself:
Code: Select all
`.244.250.325.298.311.253.269.257.335.266.268.256.272.300.295.231.232.320.311.265.277.268.315.279.303.275.287.306.294.254.262.281.309.310.295.262.288.263.232.298.280.289.283.299.280.270.258.255.308.278.281.273.253.322.276.311.271.290.287.275.266.268.249.292.289.281.308.306.254.288.293.286.234.251.298.272.270.325.303.258.294.302.298.252.266.257.260.264.260.343.265.302.298.280.309.259.261.308.297.313.278.264.279.250.254.296.300.264.291.277.280.285.262.314.277.256.264`
Now stop right there, because if you stare at it too long your head will begin to hurt. In its present state, it would be far too easy to get mixed up. Remember: The biggest obstacle in front of you is you. Human beings are by nature flawed and contain mistakes. Learn to accept them, think like a machine.
-------
If you know how the XECryption algorithm works, you know that each character is divided into three three digit numbers separated by periods, i.e.: x=.yyy.yyy.yyy
Well if we know that, we can use that to our advantage.
The first step is to break up the message, make it more simply for us to read. We'll divide it into each character, one at a time. After doing this, it'll look like this:
.244.250.325
.298.311.253
.269.257.335
Etc...
-------
Now that our message is at least slightly legible, we have a chance.
The XECryption method, while complicated in its encryption, is quite simple in its decryption. You add all three numbers together, subtract them by the password key, and you get the ASCII version of your character. Of course, I'm not about to tell you my password, that would be too easy. But we'll deal with that later, for now let's go on the next step.
The second step in decryption is going to be adding all three numbers together. This is why you brought your calculator. You might not need it, but I didn't get the best grades in high school, so I sure do. After adding the three numbers we should get something like this:
819
862
861
Etc...
-------
Still a lot of numbers, but at least the size of our message is a third of what it used to be, and that's a lot more manageable. Now, the way we would get our final message would be to subtract the password key from our new numbers, but we don't know the password do we? Well, where there's a will there's a way.
If you're new to the cryptography scene, let me recommend some of Edgar Allen Poe's works, the man was a genius when it came to this. So, what's the most common character used in any message?
In English, the letter which most frequently occurs is e. Afterward, the succession runs thus: a o i d h n r s t u y c f g l m w b k p q x z.
-Edgar Allen Poe
Now, Poe didn't have a computer, so in modern times you need to add something. Even more common than the letter e is the space character, which will occur in any message more often than any other character.
So, the question arises: What number pops up most frequently in my message? The answer is 783. Well, you can't be 100% sure, you can never be 100% sure, but 783 has the highest probability of being the space character.
-------
A quick lookup of ASCII will tell us that the space character is assigned the number 32. So, what's 783-32? 751.
Now we have our password key: 751.
After this it's smooth sailing, simply subtract 751 from each number, one at a time, and you'll get the decrypted message:
Code: Select all
`Don't forget to take out the trash man.`
Well, after the two hours it took me to decrypt that, the trash guy has already come and gone. But hey, I'll get him next time, right?
-------
In the world of "hacking", "cracking", "phreaking", or whatever you want to call it, we often take our computers for granted.
Once you've done the script yourself, you'll understand what is truly going on behind the scenes, and it'll give you a new outlook on how the computers work. With this knowledge, you'll have a new perspective when making your scripts.
lol i finished reading this and i was like "HA! thats pretty cool...... OH SHIT!!!" ran outside:trash cans were already at the curb
### Re: Decryption and Cryptography 101
Posted: Wed Jun 28, 2017 2:42 pm
Very good article.
### Re: Decryption and Cryptography 101
Posted: Wed Jan 17, 2018 9:44 am
cool article. thanks for shared | 1,612 | 6,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-34 | latest | en | 0.936381 |
https://www.mrsmactivity.co.uk/downloads/year-2-count-objects-to-100-lesson-presentation/ | 1,656,394,620,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103355949.26/warc/CC-MAIN-20220628050721-20220628080721-00256.warc.gz | 966,640,858 | 51,541 | # Year 2 | Count Objects to 100 Lesson Presentation
## Year 2 Place Value Resources
This lesson focuses on counting up to 100 objects.
Aligned with the maths mastery approach, this Year 2 | Count Objects to 100 Lesson Presentation is fully editable, and is designed for the Year 2 maths curriculum covering the following maths objectives for the autumn term:
Topic/Block: Place Value
Small Steps: Count objects to 100.
NC Links: Read and write numbers to at least 100 in numerals and words
Ready-to-progress criteria: Year 1 Conceptual Prerequisites: Know the 10 ones are equivalent to 1 ten. Know that multiples of 10 are made up from a number of tens.
2-NPV-1 Recognise the place value of each digit in two-digit numbers.
Future Applications: Compare and order numbers.
Year 1 Conceptual Prerequisites for 2NPV-2: Count forwards and backwards to and from 100.
2NPV-2 Reason about the location of any two digit number in the linear number system,including identifying the previous and next multiple of 10.
TAF Statements: Working towards – Read and write numbers in numerals up to 100.
There are teacher prompts and notes included in the “notes” section of each slide.
Aligned with the order of teaching of the White Rose Maths scheme of work, use this to help your class get to grips with each mathematical concept. This lesson presentation also includes varied fluency activities, problem solving, and mathematical discussion questions too.
Links to relevant TAF statements and ‘Ready-to-Progess’ criteria also included.
Explore our other year 2 place value resources. | 345 | 1,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-27 | latest | en | 0.901231 |
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# The probability of pulling a black ball out of a glass jar
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The probability of pulling a black ball out of a glass jar [#permalink]
### Show Tags
22 Sep 2007, 08:31
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
The probability of pulling a black ball out of a glass jar is 1/X. The probability of pulling a black ball out of a glass jar and breaking the jar is 1/Y. What is the probability of breaking the jar?
a) 1/(XY).
b) X/Y.
c) Y/X.
d) 1/(X+Y).
e) 1/(X-Y).
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### Show Tags
22 Sep 2007, 08:55
IrinaOK wrote:
The probability of pulling a black ball out of a glass jar is 1/X. The probability of pulling a black ball out of a glass jar and breaking the jar is 1/Y. What is the probability of breaking the jar?
a) 1/(XY).
b) X/Y.
c) Y/X.
d) 1/(X+Y).
e) 1/(X-Y).
I got b
P(black)=1/x
P(break and black)=1/y=P(black)xP(break)
P(break)= x/y
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### Show Tags
22 Sep 2007, 08:57
IrinaOK wrote:
The probability of pulling a black ball out of a glass jar is 1/X. The probability of pulling a black ball out of a glass jar and breaking the jar is 1/Y. What is the probability of breaking the jar?
a) 1/(XY).
b) X/Y.
c) Y/X.
d) 1/(X+Y).
e) 1/(X-Y).
A = probability of pulling a black ball out of a glass jar 1/X
B = probability of breaking the jar
C = pulling a black ball out of a glass jar and breaking the jar 1/Y
A * B = C
solve for B: B = C / A
B = (1/Y) / (1/X)
B = X /Y (answer B)
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### Show Tags
22 Sep 2007, 09:05
i got B too...
prob of pulling ball 1/x
prob of breaking jar 1/n
prob of pulling ball AND breaking jar 1/y
1/x * 1/n =1/y
1/n=x/y
22 Sep 2007, 09:05
Display posts from previous: Sort by | 889 | 2,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-04 | latest | en | 0.826304 |
https://www.geeksforgeeks.org/sum-of-n-terms-in-the-expansion-of-arcsinx/?ref=rp | 1,604,136,967,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107917390.91/warc/CC-MAIN-20201031092246-20201031122246-00312.warc.gz | 714,011,913 | 22,939 | # Sum of N terms in the expansion of Arcsin(x)
Given two integers N and X, the task is to find the value of Arcsin(x) using expansion upto N terms.
Examples:
Input: N = 4, X = 0.5
Output: 0.5233863467
Sum of first 4 terms in the expansion of Arcsin(x) for
x = 0.5 is 0.5233863467.
Input: N = 8, X = -0.5
Output: -0.5233948501
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach: The expansion of arcsin(x) is given by :
Note: |x| < 1
The above expansion is solved by using two variables maintaining the numerator and the denominator.
Below is the implementation of the above approach:
## C++
`// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` ` `// Function to find the arcsin(x) ` `void` `find_Solution(``double` `x, ``int` `n) ` `{ ` ` ``double` `sum = x, e = 2, o = 1, p = 1; ` ` ``for` `(``int` `i = 2; i <= n; i++) { ` ` ` ` ``// The power to which 'x' is raised ` ` ``p += 2; ` ` ` ` ``sum += (``double``)(o / e) * (``double``)(``pow``(x, p) / p); ` ` ` ` ``// Numerator value ` ` ``o = o * (o + 2); ` ` ` ` ``// Denominator value ` ` ``e = e * (e + 2); ` ` ``} ` ` ``cout << setprecision(10) << sum; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``double` `x = -0.5; ` ` ` ` ``if` `(``abs``(x) >= 1) { ` ` ``cout << ``"Invalid Input\n"``; ` ` ``return` `0; ` ` ``} ` ` ` ` ``int` `n = 8; ` ` ``find_Solution(x, n); ` ` ``return` `0; ` `} `
## Java
`//Java implementation of the approach ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to find the arcsin(x) ` `static` `void` `find_Solution(``double` `x, ``int` `n) ` `{ ` ` ``double` `sum = x, e = ``2``, o = ``1``, p = ``1``; ` ` ``for` `(``int` `i = ``2``; i <= n; i++) ` ` ``{ ` ` ` ` ``// The power to which 'x' is raised ` ` ``p += ``2``; ` ` ` ` ``sum += (``double``)(o / e) * ` ` ``(``double``)(Math.pow(x, p) / p); ` ` ` ` ``// Numerator value ` ` ``o = o * (o + ``2``); ` ` ` ` ``// Denominator value ` ` ``e = e * (e + ``2``); ` ` ``} ` ` ``System.out.println (sum); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` ` ``double` `x = -``0.5``; ` ` ` ` ``if` `(Math.abs(x) >= ``1``) ` ` ``{ ` ` ``System.out.println (``"Invalid Input"``); ` ` ``} ` ` ` ` ``int` `n = ``8``; ` ` ``find_Solution(x, n); ` `} ` `} ` ` ` `// This code is contributed by ajit `
## Python3
`# Python3 implementation of the approach ` ` ` `# Function to find the arcsin(x) ` `def` `find_Solution(x, n): ` ` ``Sum` `=` `x ` ` ``e ``=` `2` ` ``o ``=` `1` ` ``p ``=` `1` ` ``for` `i ``in` `range``(``2``, n ``+` `1``): ` ` ` ` ``# The power to which 'x' is raised ` ` ``p ``+``=` `2` ` ` ` ``Sum` `+``=` `(o ``/` `e) ``*` `(``pow``(x, p) ``/` `p) ` ` ` ` ``# Numerator value ` ` ``o ``=` `o ``*` `(o ``+` `2``) ` ` ` ` ``# Denominator value ` ` ``e ``=` `e ``*` `(e ``+` `2``) ` ` ``print``(``round``(``Sum``, ``10``)) ` ` ` `# Driver code ` `x ``=` `-``0.5` ` ` `if` `(``abs``(x) >``=` `1``): ` ` ``print``(``"Invalid Input\n"``) ` ` ` `n ``=` `8` `find_Solution(x, n) ` ` ` `# This code is contributed by Mohit Kumar `
## C#
`// C# implementation of the approach ` `using` `System; ` `class` `GFG ` `{ ` ` ` `// Function to find the arcsin(x) ` `static` `void` `find_Solution(``double` `x, ``int` `n) ` `{ ` ` ``double` `sum = x, e = 2, o = 1, p = 1; ` ` ``for` `(``int` `i = 2; i <= n; i++) ` ` ``{ ` ` ` ` ``// The power to which 'x' is raised ` ` ``p += 2; ` ` ` ` ``sum += (``double``)(o / e) * ` ` ``(``double``)(Math.Pow(x, p) / p); ` ` ` ` ``// Numerator value ` ` ``o = o * (o + 2); ` ` ` ` ``// Denominator value ` ` ``e = e * (e + 2); ` ` ``} ` ` ``Console.WriteLine(sum); ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main (String[] args) ` `{ ` ` ``double` `x = -0.5; ` ` ` ` ``if` `(Math.Abs(x) >= 1) ` ` ``{ ` ` ``Console.WriteLine(``"Invalid Input"``); ` ` ``} ` ` ` ` ``int` `n = 8; ` ` ``find_Solution(x, n); ` `} ` `} ` ` ` `// This code is contributed by PrinciRaj1992 `
Output:
```-0.5233948501
```
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Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. | 1,974 | 5,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-45 | longest | en | 0.521292 |
https://physics.stackexchange.com/questions/445890/is-there-a-general-algorithm-for-conversion-of-units | 1,716,049,352,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00280.warc.gz | 413,114,070 | 41,340 | # Is there a general algorithm for conversion of units?
I'm not exactly sure where the best place to put this, as it's more of a general question about dimensional analysis.
I decided I was tired of having to convert units all of the time, and was not satisfied with the available python libraries for conversion of units. I decided to make my own, and it was simple enough to get write a rough outline of a unit conversion program using this article. However, this only works for units that are scalar multiples of others. Relative units such as Celsius do not work.
I switched to implementing conversions as graph of units to traverse, converting at each step of the way. I can convert $$m / s$$ to $$ft / hr$$ by doing repeatedly replacements. The problem comes that sometimes there is no substation. To make the conversion, $$\frac{kg*m^2}{s}=1000\frac{kg*L}{m*s},$$ you have to first multiply both sides by another unit before there is a valid conversion to be made.
Is the only possiblity to have quotient units multiply the top and bottom by every single unit? This seems horribly inefficient. Are there any other similar problems I'm missing?
• Do you intend to write your own code, or are you willing to use a small program that already does conversions? Dec 8, 2018 at 2:25
– user137289
Dec 8, 2018 at 9:12
If I were writing my own python program to handle this question then I would probably decide to treat physical dimensions using vectors (i.e. one dimensional arrays). The elements of the array or vector represent the basic physical quantities such as mass, length, time, electric charge, etc. Then you have two types of conversion. On the one hand there is the definition of things like energy $$(M L^2/T^2$$) and power (energy $$/ T$$); on the other there is the conversion of units. These two concepts are closely related.
Just convert everything to a fixed set of base units (SI units would be a good choice). So you would set $$\mathrm{L} = 10^{-3} \mathrm{m}^{3}$$, $$\mathrm{ft} = 0.304 \mathrm{m}$$ etc. Once you've expressed the left and right hand side as a scalar times powers of your base units you just have to divide the scalars.
• @TheLoneMilkMan The problem is that Celsius is not really a unit: $5 K = 10^\circ C - 5^\circ \neq 5^\circ C = 278.15 K$. You should always convert any temperature to Kelvin (and temperature Differences always have to be expressed in Kelvin anyway) before doing arithmetic. BTW you can do the same for Fahrenheit just convert $^\circ F$ to the Rankine scale. Dec 8, 2018 at 1:49 | 618 | 2,546 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-22 | latest | en | 0.948658 |
https://books.google.gr/books?id=SLkXAAAAIAAJ&hl=el&lr=&source=gbs_book_other_versions_r&cad=4 | 1,596,542,898,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735867.93/warc/CC-MAIN-20200804102630-20200804132630-00275.warc.gz | 230,414,950 | 11,981 | # The Complete Arithmetic: Oral and Written
Thos. R. Shewell, 1896 - 368 σελίδες
### Τι λένε οι χρήστες -Σύνταξη κριτικής
Δεν εντοπίσαμε κριτικές στις συνήθεις τοποθεσίες.
### Δημοφιλή αποσπάσματα
Σελίδα 118 - SQUARE MEASURE 144 square inches (sq. in.) = 1 square foot (sq. ft.) 9 square feet — 1 square yard (sq. yd.) 30^ square yards = 1 square rod (sq. rd.) 160 square rods = 1 acre (A.) 640 acres = 1 square mile (sq.
Σελίδα 256 - The square of the hypothenuse is equal to the sum of the squares of the other two sides ; as, 5033 402+302.
Σελίδα 223 - RULE. — Multiply each debt by its term of credit, and divide the sum of the products by the sum of the debts. The quotient will be the average term of credit.
Σελίδα 119 - Cubic Measure 1728 cubic inches (cu. in.) =1 cubic foot (cu. ft.) 27 cubic feet = 1 cubic yard (cu. yd.) 128 cubic feet = 1 cord (cd...
Σελίδα 119 - LIQUID MEASURE 4 gills (gi.) = 1 pint (pt.) 2 pints = 1 quart (qt...
Σελίδα 164 - A brick is 8 in. long, 4 in. wide, and 2 in. thick.
Σελίδα 1 - A number is a unit or a collection of units; as one, three apples, five boys. 4. The unit of a number is one of the collection of units which constitutes the number. Thus, the unit of twelve is one, of twenty dollars is one dollar.
Σελίδα 40 - If 8 men can do a piece of work in 12 days, how long will it take...
Σελίδα 262 - Sphere is a body bounded by a uniformly curved surface, all the points of which are equally distant from a point within called the center.
Σελίδα 157 - A pile of wood 8 feet long, 4 feet wide, and 4 feet high, contains 1 cord ; and a cord foot is 1 foot in length of such a pile. | 522 | 1,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-34 | latest | en | 0.648707 |
https://math.stackexchange.com/questions/3212717/mean-squared-error-for-vectors | 1,627,131,571,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150266.65/warc/CC-MAIN-20210724125655-20210724155655-00628.warc.gz | 408,009,682 | 37,646 | # Mean squared error for vectors
I know that when we compare estimators $$\hat{b_1}$$ and $$\hat{b_2}$$ to an unknown parameter $$\beta$$, in classical statistics an estimator $$\hat{b_1}$$ is said to be "better" than $$\hat{b_2}$$ if:
$$MSE(\hat{b_1}) \leq MSE(\hat{b_2})$$ where MSE is the mean squared error: $$MSE(\hat{b_1}) = E((\hat{b_1}-\beta)^2 )$$
Now if I had a vector $$\boldsymbol{b} =(b_1,b_2,\ldots b_n)$$ of parameters to estimate, how could I compare estimators in terms of the MSE? Because there is no unique ordering relation in vectors.
I know some people compare component by component of both estimators, yet I seem to find no bibliography for that. Could you guys help me figure out a bibliography for that?
• You can join them in one vector and use your other formula, this is very standard but you should be careful about the fact that having a lot of parameters to estimate is really bad in terms of how much data you need to estimate them correctly – P. Quinton May 6 '19 at 6:00
• It is usually defined as $E\,\lvert\rvert \hat {\boldsymbol b}-\boldsymbol b\lvert\rvert^2$, consistent with the risk function for quadratic loss. – StubbornAtom May 7 at 21:47
In practical applications (engineering), the error (noise) vector is given by the modulus of the difference, in case taken relatively to the modulus of the reference vector, I.e. $$\varepsilon = {{\left| {\Delta {\bf v}} \right|} \over {\left| {{\bf v}_{\,ref} } \right|}}$$ | 420 | 1,465 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-31 | latest | en | 0.864081 |
https://www.calculatoratoz.com/en/business-current-ratio-calculator/Calc-362 | 1,619,086,083,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039603582.93/warc/CC-MAIN-20210422100106-20210422130106-00562.warc.gz | 780,107,386 | 24,536 | 🔍
🔍
## Credits
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STEP 0: Pre-Calculation Summary
Formula Used
current_ratio = Current Assets/Current Liabilities
CR = CA/CL
This formula uses 2 Variables
Variables Used
Current Assets- Current assets are balance sheet accounts that represent the value of all assets that can reasonably expect to be converted into cash within one year.
Current Liabilities- Current Liabilities are the company debts or obligations that are due within one year.
STEP 1: Convert Input(s) to Base Unit
Current Assets: 79500 --> No Conversion Required
Current Liabilities: 3000 --> No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
CR = CA/CL --> 79500/3000
Evaluating ... ...
CR = 26.5
STEP 3: Convert Result to Output's Unit
26.5 --> No Conversion Required
26.5 <-- Current Ratio
(Calculation completed in 00.016 seconds)
## < 6 Other formulas that you can solve using the same Inputs
Acid Test Ratio
acid_test_ratio = (Cash+Accounts Receivable+Short Term Investments)/Current Liabilities Go
Return on capital employed
return_on_capital_employed = (Earnings Before Interest and Taxes/(Total Assets-Current Liabilities))*100 Go
quick_ratio = (Current Assets-Inventory)/Current Liabilities Go
Quick Ratio
quick_ratio = (Current Assets-Inventory)/Current Liabilities Go
Working capital
working_capital = Current Assets-Current Liabilities Go
Current Ratio
current_ratio = Current Assets/Current Liabilities Go
## < 1 Other formulas that calculate the same Output
Current Ratio
current_ratio = Current Assets/Current Liabilities Go
current_ratio = Current Assets/Current Liabilities
CR = CA/CL
## How to Calculate Business Current Ratio?
Business Current Ratio calculator uses current_ratio = Current Assets/Current Liabilities to calculate the Current Ratio, Business Current Ratio helps you to determine if you have enough working capital to meet your short term financial obligations. Current Ratio and is denoted by CR symbol.
How to calculate Business Current Ratio using this online calculator? To use this online calculator for Business Current Ratio, enter Current Assets (CA) and Current Liabilities (CL) and hit the calculate button. Here is how the Business Current Ratio calculation can be explained with given input values -> 26.5 = 79500/3000. | 568 | 2,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-17 | latest | en | 0.829803 |
http://prehistory.48.traitorson.com/third-grade-fractions-number-lines-worksheet/ | 1,601,002,585,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400221980.49/warc/CC-MAIN-20200925021647-20200925051647-00606.warc.gz | 103,104,497 | 31,209 | # Third Grade Fractions Number Lines Worksheet
In Free Printable Worksheets231 views
4.38 / 5 ( 144votes )
Top Suggestions Third Grade Fractions Number Lines Worksheet :
Third Grade Fractions Number Lines Worksheet Identify fractions on a number line fill in equivalent fractions and compare fractions in over two pages of problems this versatile worksheet helps reinforce key fraction concepts and skills for We ll compare 3 children s ages and work out who is the oldest in this second video you ll learn to recognise fractions placed in different places on a number line this worksheet will help Just click on the grade fractions based on how many students blew bubbles and the size of each class this interactive exercise focuses on fractions with unlike denominators using number lines.
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###### 3rd Grade Fractions On A Number Line Printable Worksheets
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Top | 2,021 | 10,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-40 | latest | en | 0.920057 |
https://www.physicsforums.com/threads/need-help-with-derivatives.471832/ | 1,621,204,541,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989914.60/warc/CC-MAIN-20210516201947-20210516231947-00066.warc.gz | 970,109,892 | 14,537 | # Need help with derivatives
## Homework Statement
Find the derivative of f(x) = x^2+2x+1
## Homework Equations
f(x + h) - f(x) / h
lim(h->0) f (x+h) - f(x) / h
## The Attempt at a Solution
Hi everyone. I keep calculating the derivative for this function incorrectly. I haven't learned the rules of derivatives yet, I am only using first principle definition. So here's my mathematical attempt:
f (x+h) - f (x) / h = (x+h)^2 + 2(x+h)+1 - (x^2+2x+1) / h
First I expand the brackets
=> x^2+2xh+h^2+2x+1-(x^2+2x+1) / h
Now I open the brackets for f(x):
=> x^2+2xh+h^2+2x+1-x^2-2x-1) / h
Now I cancel some variables out and have left:
=> h^2+2xh / h
Now I factor out h and get rid of the fraction:
=> h(h + 2x) / h
Now I use the limit equation lim(h->0) f (x+h) - f(x) / h:
=> lim(h->0) h+2x = 2x is the derivative I calculated for. But checking my answers through online derivative calculators say the derivative is 2x+2? What have I done wrong in my calculations? I've checked them over more times than I can remember to count.
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https://www.physicsforums.com/threads/derive-the-time-equation-for-the-collision-of-two-masses-due-to-newtonian-gravity.1002166/ | 1,674,992,360,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00672.warc.gz | 943,688,085 | 22,868 | Derive the time equation for the collision of two masses due to Newtonian gravity
alan123hk
Assume m2 masses a significant percentage of Earth, like the moon say.
We have a controlled Earth (not spinning, a perfect sphere, no air) all alone in space (so the sun isn't accelerating it for instance) and I have a stationary moon-size rigid balloon at the same altitude as the orbit of the moon. We let go of it and time how long it takes to hit. Now we let go of something the mass of the moon (and stronger to avoid Roche effects) from the same altitude and it will hit the ground in significantly less time. It accelerates more or less at the same rate at first, but gains speed quicker because the Earth is accelerating up towards the massive moon but not the balloon moon. Each travels a different distance because the Earth is not sitting still.
[Mentor Note: thread split off from a different thread]
https://www.physicsforums.com/threads/heavier-objects-fall-faster.1002022/
Since seeing this thread yesterday, I have been trying to derive the time equation for the collision of two masses due to Newtonian gravity. Unfortunately, this seems to be much more difficult than I thought before, so I haven't made much progress yet. 🤔
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Answers and Replies
Richard R Richard
The weight at the height of the Earth's surface is Mg and the aerodynamic drag of an object is proportional to its speed squared. Setting up Newton's equations
$$mg-kv ^ 2 = ma$$
When the dimensions of two test objects are equal ## k ## is a constant.
When the equilibrium of forces is reached, the acceleration ## to ## becomes zero or it descends with a constant velocity, whose value is obtained by solving
$$mg-kv ^ 2 = 0$$
$$mg = kv ^ 2$$
$$v = \sqrt {\dfrac {mg}{k}}$$
According to this, the more massive, the higher terminal velocity, then it falls faster.
But as I said this is only true if k is constant (see drag coefficient),
Massive objects do not always fall first,
• A 747 airplane has more mass than a car, however both launched from a certain height and with horizontal speed, the car will fall first.
• An empty balloon falls before the same inflated balloon of the same mass ...
• A satellite with a certain tangential velocity does not fall, no matter how much its mass is ...
Since seeing this thread yesterday, I have been trying to derive the time equation for the collision of two masses due to Newtonian gravity. Unfortunately, this seems to be much more difficult than I thought before, so I haven't made much progress yet. 🤔
To know the time of fall of an object, taking into account that gravity varies with height and in the absence of friction (this almost leaves the thread) is
We already know the modulus of acceleration of gravity falls with the square of the distance to the center.
$$a_g = \dfrac {GM_T}{r ^ 2}$$
It is logical to think that the relative acceleration with respect to the surface for an object at a higher altitude is less than one at a low altitude. these types of corrections are made by means of differential equations.
Let's see the simple calculation of frictionless free fall, An object of mass ## m ## is dropped without initial velocity from a height ## r_o ## with respect to a massive body of mass ## M ##
Using the conservation of energy you have a first order differential equation that is relatively easy to integrate.
$$\displaystyle \frac {1}{2} v ^ 2-G \frac {M}{r} = - G \frac {M}{r_0}$$
from where
$$v = - \displaystyle \frac {dr}{dt} = \sqrt {2 GM} \sqrt {\frac {1}{r} - \frac {1}{r_0}} = \sqrt {\frac { 2 GM}{r_0}} \sqrt {\frac {r_0-r}{r}}$$
if we try to integrate
$$\displaystyle- \int_{r_0} ^ r {\sqrt {\frac {r}{r_0-r}} dr} = \sqrt {\frac {2 GM}{r_0}} \int_0 ^ t {dt}$$
The integral on the left is a bit heavy to perform. My source, friend from another forum has made, the change of variable $$u ^ 2 = r / (r_0-r)$$.
Finally
$$\displaystyle- \int_{r_0} ^ r {\sqrt {\frac {r}{r_0-r}} dr} = r_0 \left (\frac {\pi}{2} - \arctan \sqrt {\frac {r}{r_0-r}} + \frac {\sqrt {r} \sqrt {r_0-r}}{r_0} \right)$$
then
$$t (r) = \displaystyle \sqrt {\frac {r_0 ^ 3}{2GM}} \left (\frac {\pi}{2} - \arctan \sqrt {\frac {r}{r_0-r }} + \frac {\sqrt {r} \sqrt {r_0-r}}{r_0} \right)$$
(this makes sense as long as ## r ## is greater to the radius of the earth).
alan123hk
alan123hk
To know the time of fall of an object, taking into account that gravity varies with height and in the absence of friction (this almost leaves the thread) is
We already know the modulus of acceleration of gravity falls with the square of the distance to the center.
Your mathematical derivation is great.
But what I mean is that in the absence of air resistance and initial speed, these two objects will accelerate toward each other at the same time due to the action of Newtonian gravity. I believe that the impact time will depend on the mass of the two objects (M1, M2), the initial distance between them (R0), and may also include the radii of the two objects (D1, D2).
Science Advisor
But what I mean is that in the absence of air resistance and initial speed, these two objects will accelerate toward each other at the same time due to the action of Newtonian gravity. I believe that the impact time will depend on the mass of the two objects (M1, M2), the initial distance between them (R0), and may also include the radii of the two objects (D1, D2).
This has been discussed before here and is not trivial. One trick was to treat the trajectory as a degenerated ellipse and appply Kepler's laws.
Nugatory and bhobba
Mentor
This has been discussed before here and is not trivial. One trick was to treat the trajectory as a degenerated ellipse and appply Kepler's laws.
Exactly. It is one of those problems easy to state but hard to do. As an aside, if you were to include air resistance, then the problem is very tough because you have turbulence and must solve the Navier-Stokes Equation. Even showing you can do it will not only possibly get you a Nobel, maybe even a Fields Medal, but also a million big ones from the Clay Institute - it is one of the Millenium prizes. By making some approximations, it, while still hard, can be done using the drag equation:
https://physics.info/drag/
Thanks
Bill
Richard R Richard
alan123hk
Exactly. It is one of those problems easy to state but hard to do. As an aside, if you were to include air resistance, then the problem is very tough because you have turbulence and must solve the Navier-Stokes Equation. Even showing you can do it will not only possibly get you a Nobel, maybe even a Fields Medal, but also a million big ones from the Clay Institute - it is one of the Millenium prizes. By making some approximations, it, while still hard, can be done using the drag equation
I am very happy that this discussion thread has reopened, and I agree with you.
I searched the Internet for a long time and finally found an equation that can be easily derived under a special situation (including no air resistance, no initial velocity in any direction,..). This derivation method cleverly uses the concept of conservation of momentum.
So far, in the derivation process of the equation shown below, I have not found anything unreasonable.
$$\textrm{Energy conservation (KE = PE): } \frac{p^2}{2}\left( \frac{1}{m} + \frac{1}{M} \right) = GMm\left(\frac{1}{r} - \frac{1}{r_0}\right)$$
$$\frac{dr}{dt} = -(v + V) = -p\left( \frac{1}{m} + \frac{1}{M} \right)$$
$$\int_0^T dt = -\int_{r_0}^0 dr \sqrt{\frac{rr_0}{2G(M+m)(r_0-r)}} = \frac{\pi}{2\sqrt{2}}\frac{r_0^{3/2}}{\sqrt{G(M+m)}}$$
where p is the combined momentum of the two bodies.
bhobba
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alan123hk and bhobba
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So far, in the derivation process of the equation shown below, I have not found anything unreasonable.
I note that this is exactly the result obtained by @Richard R Richard #2 with the recognition that the M is the total system mass.
$$t (r) = \displaystyle \sqrt {\frac {r_0 ^ 3}{2GM}} \left (\frac {\pi}{2} - \arctan \sqrt {\frac {r}{r_0-r }} + \frac {\sqrt {r} \sqrt {r_0-r}}{r_0} \right)$$
Just plug in r=0 and use the total mass ##M=m_1 +m_2##
$$t (0) = \displaystyle \sqrt {\frac {r_0 ^ 3}{2G(m_1+m_2)}} \left (\frac {\pi}{2} \right)$$
bhobba
Richard R Richard
##r = 0## is an idealization, any real massive object has a surface with a radius r, which cannot be traversed without impact. On Earth for example ##r = R_T = 6354km## It should also be noted that if we idealize and think that there is a hole that allows us to reach ##r = 0##, we must change the equations because the mathematical function of gravity inside the Earth is not the same as outside.
bhobba
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Yes I was assuming point masses. The generalizations at that point are easy...you did the hard parts
bhobba
Mentor
This is really good. When something is hard, figuring it out for yourself, even with the help of the internet, is a real eye-opener. That is how I learned some of the most 'most satisfying' bits of knowledge. It's not they are unknown or anything like that - it's simply they are not often discussed. One I figured out by myself was what Newtons Laws really mean. You can do a search on this forum for the answer. I am about to do a post on 0 to the power 0. Good work, guys. This is what this forum is all about.
Thanks
Bill
alan123hk, Richard R Richard and hutchphd
alan123hk
Now that I have an equation for collision time, I try to use the principles of conservation of momentum and energy to derive approximate equations for collision position and collision velocity.
collision position : -
$$r_1 = \frac {(r_0-d)m_2} {m_1+m_2} ~~~~~~~~~~~~~ r_2 = \frac {(r_0-d)m_1} {m_1+m_2}$$
collision velocity : -
$$V_1 = \sqrt { \frac {2G M_1 M_2 ( \frac 1 d -\frac 1 r_0 )} { 1+ \frac {M_1} {M_2} } } ~~~~~~~~~~~ V_2 = \sqrt { \frac {2G M_1 M_2 ( \frac 1 d -\frac 1 r_0 )} { 1+ \frac {M_2} {M_1} } }$$
where
d = diameter of m1 and m2, and r0 = r1+r2+d
r
1
and r2 = moving distance of M1 and M2
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Richard R Richard
I detect an incompatibility of proposals between what I explained and what you conclude. Not because it is wrong but because they are two different problems. In my proposal the fall time is for a particle of negligible mass compared to the gravitational mass ## M ##. Your case deals with a problem of two bodies of masses of the same order of magnitude And only in the case that the system does not have angular momentum, and in the absence of other external forces, the collision position will be exactly at the Center of mass. As you also affirm that each mass has the same diameter, the collision occurs when their centers are one diameter apart, the most massive body will be closer to the MC in the same proportion as the mass ratio. The velocity relationship is also derived from the conservation of energy, it is advisable to use a reference system originating in the MC
alan123hk
I detect an incompatibility of proposals between what I explained and what you conclude. Not because it is wrong but because they are two different problems. In my proposal the fall time is for a particle of negligible mass compared to the gravitational mass M. Your case deals with a problem of two bodies of masses of the same order of magnitude And only in the case that the system does not have angular momentum, and in the absence of other external forces, the collision position will be exactly at the Center of mass. As you also affirm that each mass has the same diameter, the collision occurs when their centers are one diameter apart, the most massive body will be closer to the MC in the same proportion as the mass ratio. The velocity relationship is also derived from the conservation of energy, it is advisable to use a reference system originating in the MC
Thanks for your advice. I have been engaged in engineering work for decades since leaving school, and now I am trying to regain my interest in physics after retirement. Most of the basics of mathematics and physics that I have learned before are now forgotten, maybe now I have to spend some time to relearn and improve.
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Now that I have an equation for collision time, I try to use the principles of conservation of momentum and energy to derive approximate equations for collision position and collision velocity.
The energy (and hence relative velocity) at the collision can be simply calculated from conservation of energy. And the motion of the two masses relative to the initial rest frame is, by conservation of momentum, in inverse proportion to their masses.
The time taken is the tricky thing to calculate. You don't need the time to calculate the other quantities.
alan123hk
The time taken is the tricky thing to calculate. You don't need the time to calculate the other quantities.
Yes, I did not use an equation about time. I mean, after obtaining the time equation, I naturally want to know more information, so I try to derive the collision velocity and collision position equations.
Now I am thinking about more complicated things. For example, when a small asteroid approaches the Earth, how do scientists calculate the elliptical trajectory of its fall, the time and location of its impact on the earth’s atmosphere. As for entering the earth’s atmosphere, due to the huge air resistance, I believe that only with the help of supercomputers can we accurately calculate the time and location of the impact on the ground.
Speaking of this, I'm afraid I 'm a little off topic again.
Last edited:
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Now I am thinking about more complicated things. For example, when an asteroid approaches the Earth, how do scientists calculate the elliptical trajectory of its fall, the time and location of its impact on the earth’s atmosphere. As for entering the earth’s atmosphere, due to the huge air resistance, I believe that only with the help of supercomputers can we calculate the time and location of the impact on the ground.
The time to hit the ground is of no relevance. Of relevance is whether the impact occurs (and the approximate time) and what happens dynamically as the asteroid passes through the atmosphere.
hutchphd | 3,575 | 14,363 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2023-06 | latest | en | 0.933188 |
https://zh.wikipedia.org/zh-tw/%E5%AF%B9%E6%95%B0%E6%81%92%E7%AD%89%E5%BC%8F | 1,550,335,652,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247480622.9/warc/CC-MAIN-20190216145907-20190216171907-00419.warc.gz | 1,033,649,377 | 22,682 | # 對數恆等式
## 代數恆等式
### 簡化計算
${\displaystyle \,\log _{\theta }xy=\log _{\theta }x+\log _{\theta }y}$ 對應到 ${\displaystyle \,\theta ^{x}\theta ^{y}=\theta ^{x+y}}$ ${\displaystyle \log _{\theta }{\frac {x}{y}}=\log _{\theta }x-\log _{\theta }y}$ ${\displaystyle {\frac {\theta ^{x}}{\theta ^{y}}}=\theta ^{x-y}}$ ${\displaystyle \,\log _{\theta }x^{y}=y\log _{\theta }x}$ ${\displaystyle \,({\theta ^{x}})^{y}=\theta ^{xy}}$ ${\displaystyle \log _{\theta }{\sqrt[{y}]{x}}={\frac {\log _{\theta }x}{y}}}$ ${\displaystyle {\sqrt[{y}]{x}}=x^{\frac {1}{y}}}$ ${\displaystyle \,\log _{\theta }-x=\log _{\theta }x+\pi i\log _{\theta }e}$ 歐拉恆等式：${\displaystyle \,e^{\pi i}+1=0}$
### 消去指數
${\displaystyle b^{\log _{b}(x)}=x}$ 因為 ${\displaystyle \mathrm {antilog} _{b}(\log _{b}(x))=x\!\,}$ ${\displaystyle \log _{b}(b^{x})=x\!\,}$ 因為 ${\displaystyle \log _{b}(\mathrm {antilog} _{b}(x))=x\!\,}$
### 換底公式
${\displaystyle \log _{\theta }x={\frac {\log _{\phi }x}{\log _{\phi }\theta }}}$
${\displaystyle \log _{a}b={\frac {1}{\log _{b}a}}}$
${\displaystyle \log _{a^{n}}b={{\log _{a}b} \over n}}$
${\displaystyle a^{\log _{b}c}=c^{\log _{b}a}}$
${\displaystyle \log _{a_{1}}b_{1}\,\cdots \,\log _{a_{n}}b_{n}=\log _{a_{\pi (1)}}b_{1}\,\cdots \,\log _{a_{\pi (n)}}b_{n},\,}$
${\displaystyle \pi }$是下標${\displaystyle 1,\ldots ,n}$的任意的排列。例如
${\displaystyle \log _{a}w\cdot \log _{b}x\cdot \log _{c}y\cdot \log _{d}z=\log _{d}w\cdot \log _{a}x\cdot \log _{b}y\cdot \log _{c}z.\,}$
### 和/差公式
${\displaystyle \log _{\theta }(\mathrm {X} \pm \Upsilon )=\log _{\theta }\mathrm {X} +\log _{\theta }\left(1\pm {\frac {\Upsilon }{\mathrm {X} }}\right)}$
### 普通恆等式
${\displaystyle \log _{b}(1)=0\!\,}$ 因為 ${\displaystyle b^{0}=1\!\,}$ ${\displaystyle \log _{b}(b)=1\!\,}$ 因為 ${\displaystyle b^{1}=b\!\,}$
## 微積分恆等式
### 極限
${\displaystyle \lim _{x\to 0^{+}}\log _{a}x=-\infty \quad {\mbox{if }}a>1}$
${\displaystyle \lim _{x\to 0^{+}}\log _{a}x=\infty \quad {\mbox{if }}a<1}$
${\displaystyle \lim _{x\to \infty }\log _{a}x=\infty \quad {\mbox{if }}a>1}$
${\displaystyle \lim _{x\to \infty }\log _{a}x=-\infty \quad {\mbox{if }}a<1}$
${\displaystyle \lim _{x\to 0^{+}}x^{b}\log _{a}x=0}$
${\displaystyle \lim _{x\to \infty }{1 \over x^{b}}\log _{a}x=0}$
### 對數函數的導數
${\displaystyle {d \over dx}\ln x={1 \over x}={\ln e \over x}}$
### 積分定義
${\displaystyle \ln x=\int _{1}^{x}{\frac {1}{t}}dt}$
### 對數函數的積分
${\displaystyle \int \log _{a}x\,dx=x(\log _{a}x-\log _{a}e)+C}$
${\displaystyle x^{\left[n\right]}=x^{n}(\log(x)-H_{n})}$
${\displaystyle x^{\left[0\right]}=\log x}$
${\displaystyle x^{\left[1\right]}=x\log(x)-x}$
${\displaystyle x^{\left[2\right]}=x^{2}\log(x)-{\begin{matrix}{\frac {3}{2}}\end{matrix}}\,x^{2}}$
${\displaystyle x^{\left[3\right]}=x^{3}\log(x)-{\begin{matrix}{\frac {11}{6}}\end{matrix}}\,x^{3}}$
${\displaystyle {\frac {d}{dx}}\,x^{\left[n\right]}=n\,x^{\left[n-1\right]}}$
${\displaystyle \int x^{\left[n\right]}\,dx={\frac {x^{\left[n+1\right]}}{n+1}}+C}$ | 1,364 | 2,982 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 62, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2019-09 | latest | en | 0.202154 |
https://www.coursehero.com/file/5766296/8/ | 1,529,745,913,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864953.36/warc/CC-MAIN-20180623074142-20180623094142-00172.warc.gz | 798,035,959 | 85,851 | 8 - EE 2001 Fall 2008 Problem Set 9 BJT Biasing Derive an...
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EE 2001 Fall 2008 Problem Set 9 ______________________________________________________________________ BJT Biasing Derive an equation for V CE in terms of beta. (use V BE,on = 0.7 volts) Plot V CE vs beta. Note: V CE approaches a steady value as beta becomes very large. It doesn’t collapse to saturation level, 0.2V, as we might expect. In a sense, large transistor gains “buy” us a stable operating point. Sort of like credit default swaps buy us a stable market (but with transistor circuits this actually works). Solution: IB = (VCE – 0.7)/R1 – 0.7/R2 KCL at the collector-- (VCE – 0.7)/R1 + beta*IB + (VCE – VCC)/RC = 0 (VCE – 0.7)/R1 + beta*((VCE – 0.7)/R1 – 0.7/R2) + (VCE – VCC)/RC = 0 VCE(1/R1 + beta/R1 + 1/RC) = 0.7/R1 + beta*0.7/R1 + beta*0.7/R2 + VCC/RC VCE = (0.7/R1 + beta*0.7/R1 + beta*0.7/R2 + VCC/RC)/ (1/R1 + beta/R1 + 1/RC) VCE = (k1 + k2*beta)/(k3 + k4*beta) where k1 = 0.7/R1 + VCC/RC = 0.0535; k2 = 0.7/R1 + 0.7/R2 = 0.0105; k3 = 1/R1 + 1/RC = 0.01; k4 = 1/R1 = 0.005; Plot of VCE starts at k1/k3, that is 5.35V, for beta = 0, and approaches k2/k4, that is 2.1V, as beta approaches infinity. RC (200 ) R1 (200 ) +10 v I C + V CE _ R2 (100 )
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## A mathematics exam-style question with a worked solution that can be revealed gradually
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A group of students sat a Biology examination and a Computer Studies examination. The students' marks in the Biology examination are normally distributed with a mean of 70 and a standard deviation of 9.
(a) Draw a diagram that shows this information.
(b) Find the probability that a randomly chosen student who sat the Biology examination scored at most 70 marks.
Eric scored 82 marks in the Biology examination.
(c) Find the probability that a randomly chosen student who sat the Biology examination scored more than Eric.
The students' marks in the Computer Studies examination are normally distributed with a mean of 68 and a standard deviation of 11. Eric also scored 82 marks in the Computer Studies examination.
(d) Find the probability that a randomly chosen candidate who sat the Computer Studies examination scored less than Eric.
(e) Determine whether Eric's Computer Studies mark, compared to the other students, is better than his mark in Biology. Give a reason for your answer.
To obtain a grade A a student must be in the top 12% of the students who sat the Computer Studies examination.
(f) Find the minimum possible mark to obtain a grade A. Give your answer correct to the nearest integer.
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To search the entire Transum website use the search box in the grey area below. | 632 | 3,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-38 | latest | en | 0.924383 |
https://www.jiskha.com/questions/688021/4-students-each-wrote-2-number-patterns-Shandi-3-6-9-12-5-15-20-25-Latrell-1-3-5-7-3-5-7-9 | 1,563,923,574,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529737.79/warc/CC-MAIN-20190723215340-20190724001340-00367.warc.gz | 748,691,070 | 5,645 | # Math
4 students each wrote 2 number patterns
Shandi
3,6,9,12... 5,15,20,25
Latrell
1,3,5,7...3,5,7,9
Hong-Li
0,1,3,6...1,2,4,7
2,3,5,7...11,13,17,19
Which student's number patterns used only prime numbers?
A.Shandi
B.Latrell
C.Hong-Li
D?
1. 👍 0
2. 👎 0
3. 👁 163
1. shandi
1. 👍 0
2. 👎 0
posted by fatima
2. Don't prime numbers have only 1 factor? So Shandi has composite numbers?
1. 👍 0
2. 👎 0
posted by Colton
3. Adriano. two is prime as are all the other numbers in his pattern....
hong-li wrote 6 and 6 can be divided by 2 and 3 therefore not prime.
Latrell wrote 9 and that can be divided by 3 so 9 is not prime.
1. 👍 0
2. 👎 0
posted by Mary
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More Similar Questions | 980 | 3,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-30 | latest | en | 0.955323 |
http://www.personal.psu.edu/jhm/470/homework/hw3.html | 1,540,080,945,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513508.42/warc/CC-MAIN-20181020225938-20181021011438-00055.warc.gz | 539,220,378 | 1,720 | # Finite Volume Equations
### Part 1
A portion of a straight pipe with a circular cross-section is to be modeled with a staggered mesh finite volume method. It is divided into 3 volumes each containing 0.5 cubic meters. The cross-sectional area is 0.5 square meters. Flow velocity is uniform and constant at 2 meters per second directed from volume 1 towards volume 3. Initial densities of a solute are 1 ppm in volumes 1 and 2 and 10 ppm in volume 3. Draw a representation of the finite volume mesh. What is the cell length? What is the cell (pipe) diameter? Use a time step of 0.2 seconds and an explicit difference method to calculate the new time density (density at the end of the time step) in the 2nd volume. In a first calculation predict the new time density in volume 2 using a donor cell average to obtain the mean density at the cell edges. In a second calculation of new time density, apply a central difference (linear average) for the cell edge densities. I only want results for volume 2 density at the end of a single time step for each of the 2 edge averaging methods.
### Part 2
Starting with the problem in Part 1, treat volume 1 as a boundary condition that never changes. Set the initial solute densites to 1 ppm, 1.1 ppm, and 1 ppm in volumes 1 through 3 respectively. Use an explicit donor cell method to calculate the solute density in the 2nd volume for five separate 10 second transients, using time steps of 0.25 s, 0.5 s, 0.75, 1.0 s, and 1.25 s. Provide results for each of your calculations (density vs. time) on a separate plot (or table). Make sure to plot one point for every time step. How do these results relate to the discussion of numerical stability in class?
Show all work and feel free to ask me for clarifications.
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Maintained by John Mahaffy : jhm@psu.edu | 460 | 1,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-43 | latest | en | 0.861038 |
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20 questions
• Q1
Simplify the expression: 2x + 3x - 4x
2x - x
x - 4x
x + 4x
x
30s
• Q2
What is the simplified form of the expression 2x + 3 - 4x?
2x + 3 - 4
-2x + 3
-2x - 3
-4x - 2x + 3
30s
• Q3
What is the simplified form of the expression 3(2x - 5) + 4x?
6x - 5 + 4x
6x - 5
6x - 5 - 4x
10x - 15
30s
• Q4
What is the simplified form of the expression 3(x + 2) - 4x?
3x - 6
-x + 6
-x - 2
-x + 2
3x + 6
30s
• Q5
What is the simplified form of the expression 2(3x - 4) + 5 - x?
5x - 3
3x - 6
5x - 2
6x - 2
-x + 1
30s
• Q6
What is the simplified form of the expression 2(5x - 3) + 4 - 3x?
6x - 1
8x + 1
8x - 2
7x - 2
7x + 1
30s
• Q7
What is the solution to the equation: 3x + 7 = 16?
-3
10
5
3
30s
• Q8
What is the solution to the equation: 5 - 2x = 9?
-4
-2
2
4
30s
• Q9
What is the solution to the equation: 2y - 5 = 11?
3
-8
8
16
30s
• Q10
What is the solution to the equation: 4 + 3x = 19?
15
1
5
9
30s
• Q11
What is the solution to the equation: 2x - 9 = 15?
12
3
6
-12
30s
• Q12
What is the solution to the equation: 7 - 4x = 11?
-3
1
3
-1
30s
• Q13
What is the solution to the equation: 2(x + 3) = 10?
4
2
-2
6
30s
• Q14
What is the solution to the equation: x/4 + 2 = 5?
2
16
8
12
30s
• Q15
What is the solution to the equation: 3x - 8 = 19?
9
27
5
-9
30s
Teachers give this quiz to your class | 788 | 1,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-10 | latest | en | 0.859024 |
https://us.metamath.org/mpeuni/frege111.html | 1,722,807,812,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00142.warc.gz | 477,653,239 | 5,706 | Mathbox for Richard Penner < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > frege111 Structured version Visualization version GIF version
Theorem frege111 40690
Description: If 𝑌 belongs to the 𝑅-sequence beginning with 𝑍, then every result of an application of the procedure 𝑅 to 𝑌 belongs to the 𝑅-sequence beginning with 𝑍 or precedes 𝑍 in the 𝑅-sequence. Proposition 111 of [Frege1879] p. 75. (Contributed by RP, 7-Jul-2020.) (Revised by RP, 8-Jul-2020.) (Proof modification is discouraged.)
Hypotheses
Ref Expression
frege111.z 𝑍𝐴
frege111.y 𝑌𝐵
frege111.v 𝑉𝐶
frege111.r 𝑅𝐷
Assertion
Ref Expression
frege111 (𝑍((t+‘𝑅) ∪ I )𝑌 → (𝑌𝑅𝑉 → (¬ 𝑉(t+‘𝑅)𝑍𝑍((t+‘𝑅) ∪ I )𝑉)))
Proof of Theorem frege111
StepHypRef Expression
1 frege111.z . . 3 𝑍𝐴
2 frege111.y . . 3 𝑌𝐵
3 frege111.v . . 3 𝑉𝐶
4 frege111.r . . 3 𝑅𝐷
51, 2, 3, 4frege108 40687 . 2 (𝑍((t+‘𝑅) ∪ I )𝑌 → (𝑌𝑅𝑉𝑍((t+‘𝑅) ∪ I )𝑉))
6 frege25 40533 . 2 ((𝑍((t+‘𝑅) ∪ I )𝑌 → (𝑌𝑅𝑉𝑍((t+‘𝑅) ∪ I )𝑉)) → (𝑍((t+‘𝑅) ∪ I )𝑌 → (𝑌𝑅𝑉 → (¬ 𝑉(t+‘𝑅)𝑍𝑍((t+‘𝑅) ∪ I )𝑉))))
75, 6ax-mp 5 1 (𝑍((t+‘𝑅) ∪ I )𝑌 → (𝑌𝑅𝑉 → (¬ 𝑉(t+‘𝑅)𝑍𝑍((t+‘𝑅) ∪ I )𝑉)))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ∈ wcel 2111 ∪ cun 3879 class class class wbr 5030 I cid 5424 ‘cfv 6324 t+ctcl 14338 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-10 2142 ax-11 2158 ax-12 2175 ax-ext 2770 ax-rep 5154 ax-sep 5167 ax-nul 5174 ax-pow 5231 ax-pr 5295 ax-un 7443 ax-cnex 10584 ax-resscn 10585 ax-1cn 10586 ax-icn 10587 ax-addcl 10588 ax-addrcl 10589 ax-mulcl 10590 ax-mulrcl 10591 ax-mulcom 10592 ax-addass 10593 ax-mulass 10594 ax-distr 10595 ax-i2m1 10596 ax-1ne0 10597 ax-1rid 10598 ax-rnegex 10599 ax-rrecex 10600 ax-cnre 10601 ax-pre-lttri 10602 ax-pre-lttrn 10603 ax-pre-ltadd 10604 ax-pre-mulgt0 10605 ax-frege1 40506 ax-frege2 40507 ax-frege8 40525 ax-frege28 40546 ax-frege31 40550 ax-frege41 40561 ax-frege52a 40573 ax-frege52c 40604 ax-frege58b 40617 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-ifp 1059 df-3or 1085 df-3an 1086 df-tru 1541 df-fal 1551 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2598 df-eu 2629 df-clab 2777 df-cleq 2791 df-clel 2870 df-nfc 2938 df-ne 2988 df-nel 3092 df-ral 3111 df-rex 3112 df-reu 3113 df-rab 3115 df-v 3443 df-sbc 3721 df-csb 3829 df-dif 3884 df-un 3886 df-in 3888 df-ss 3898 df-pss 3900 df-nul 4244 df-if 4426 df-pw 4499 df-sn 4526 df-pr 4528 df-tp 4530 df-op 4532 df-uni 4801 df-int 4839 df-iun 4883 df-br 5031 df-opab 5093 df-mpt 5111 df-tr 5137 df-id 5425 df-eprel 5430 df-po 5438 df-so 5439 df-fr 5478 df-we 5480 df-xp 5525 df-rel 5526 df-cnv 5527 df-co 5528 df-dm 5529 df-rn 5530 df-res 5531 df-ima 5532 df-pred 6116 df-ord 6162 df-on 6163 df-lim 6164 df-suc 6165 df-iota 6283 df-fun 6326 df-fn 6327 df-f 6328 df-f1 6329 df-fo 6330 df-f1o 6331 df-fv 6332 df-riota 7093 df-ov 7138 df-oprab 7139 df-mpo 7140 df-om 7563 df-2nd 7674 df-wrecs 7932 df-recs 7993 df-rdg 8031 df-er 8274 df-en 8495 df-dom 8496 df-sdom 8497 df-pnf 10668 df-mnf 10669 df-xr 10670 df-ltxr 10671 df-le 10672 df-sub 10863 df-neg 10864 df-nn 11628 df-2 11690 df-n0 11888 df-z 11972 df-uz 12234 df-seq 13367 df-trcl 14340 df-relexp 14373 df-he 40489 This theorem is referenced by: frege129 40708
Copyright terms: Public domain W3C validator | 2,003 | 3,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-33 | latest | en | 0.160315 |
https://ask.sagemath.org/answers/44396/revisions/ | 1,631,941,784,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056297.61/warc/CC-MAIN-20210918032926-20210918062926-00688.warc.gz | 166,147,428 | 6,998 | # Revision history [back]
The steps (and graphs !) of this solution are available in Sagecell
Let's see. We suppose for now that this equation is to solve in reals :
var("t", domain="real")
Ex=0.111*t-(1- e^(-0.3*t))
We note an "obvious" root :
Ex.subs(t=0)
0
Is there an explicit solution ?
Ex.log_expand().log().solve(t)
[t == 1000/111*(2*e^(3/10*t) - 1)*e^(-3/10*t)]
No such luck : this is an implicit solution, no more informative than the original equation. Let's try a numerical solution. What is the shape of the representative curve of this expression ? A curve with a minimum around about 3, null at 0 and about 8, decreasing below 3 and increasing above 3 (see Sagecell). What can we prove ?
The first derivative is
Ex.diff(t)
-0.300000000000000*e^(-0.300000000000000*t) + 0.111000000000000
An increasing curve, null about 3. Can we prove that ? The second derivative is
Ex.diff(t,2)
0.0900000000000000*e^(-0.300000000000000*t)
Yay ! This is strictly positive for all reals. Therefore, the zero we "see" on the first derivative at about 3 is the only one, and is
Ex.diff(t).solve(t)
[t == 10*log(1/37*100^(1/3)*37^(2/3))]
Ex.diff(t).solve(t)[0].rhs().n()
3.31417424447956
Now we have proven that Ex is decreasing before this zero and increasing after that. At the minimum, it is :
Ex.subs(t==Ex.diff(t).solve(t)[0].rhs()).n()
-0.262126658862769
Since it is negative at this point and not upward bound (exercise for the reader : prove it !), it has two roots, one (0) being already known. The graphs suggest that Ex is positive for t=10. Let's try to find a numerical solution :
Ex.find_root(0.1,10,t)
8.25101463236195
Left to the reader : what are the complex roots (if any...) ? Fair warning : this one is much harder than the real case (bound to problems open since the XIXth century, and still not solved...).
The steps (and graphs !) of this solution are available in Sagecell
Let's see. We suppose for now that this equation is to solve in reals :
var("t", domain="real")
Ex=0.111*t-(1- e^(-0.3*t))
We note an "obvious" root :
Ex.subs(t=0)
0
Is there an explicit solution ?
Ex.log_expand().log().solve(t)
[t == 1000/111*(2*e^(3/10*t) - 1)*e^(-3/10*t)]
No such luck : this is an implicit solution, no more informative than the original equation. Let's try a numerical solution. What is the shape of the representative curve of this expression ? A curve with a minimum around about 3, null at 0 and about 8, decreasing below 3 and increasing above 3 (see Sagecell). What can we prove ?
The first derivative is
Ex.diff(t)
-0.300000000000000*e^(-0.300000000000000*t) + 0.111000000000000
An increasing curve, null about 3. Can we prove that ? The second derivative is
Ex.diff(t,2)
0.0900000000000000*e^(-0.300000000000000*t)
Yay ! This is strictly positive for all reals. Therefore, the zero we "see" on the first derivative at about 3 is the only one, and is
Ex.diff(t).solve(t)
[t == 10*log(1/37*100^(1/3)*37^(2/3))]
Ex.diff(t).solve(t)[0].rhs().n()
3.31417424447956
Now we have proven that Ex is decreasing before this zero and increasing after that. At the minimum, it is :
Ex.subs(t==Ex.diff(t).solve(t)[0].rhs()).n()
-0.262126658862769
Since it is negative at this point and not upward bound (exercise for the reader : prove it !), it has two roots, one (0) being already known. The graphs suggest that Ex is positive for t=10. Let's try to find a numerical solution :
Ex.find_root(0.1,10,t)
8.25101463236195
Left to the reader : what are the complex roots (if any...) ? Fair warning : this one is much harder than the real case (bound to problems open since the XIXth century, and still not solved...).
EDIT : It turns out that Sympy can give us the non-trivial real root in symbolic form :
sage: import sympy
sage: sympy.solve(Ex, t)
[10*LambertW(-100*exp(-100/37)/37)/3 + 1000/111,
10*LambertW(-100*exp(-100/37)/37, -1)/3 + 1000/111]
The numerical values are as expected :
sage: [s.n() for s in sympy.solve(Ex, t)]
[8.25101463236202, 0.e-123]
This function is known to Sage, but the conversion is incorrect (loses the second argument)
sage: [s._sage_() for s in sympy.solve(Ex, t)]
[10/3*lambert_w(-100/37*e^(-100/37)) + 1000/111,
10/3*lambert_w(-100/37*e^(-100/37)) + 1000/111]
This is now Trac#26752 | 1,302 | 4,284 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2021-39 | longest | en | 0.875487 |
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By r3xxu5m0ne11. Math coloring. At Tuesday, September 10th 2019, 18:47:56 PM.
Coloring pages are a great way for kids to learn to concentrate and improve their focus. This also has a lot to do with the exposure to boundaries, that is, coloring within the lines. When kids immerse themselves in the process of coloring, they concentrate on making the pictures inside the coloring pages come to life, which results in them greatly improving those skills. Improved focus and concentration skills help kids not only in learning how to write, but also in a number of other activities that they will indulge later in life. Being able to focus better will also help them perform better at school, so it is very important for each and every child to acquire them when they are younger.
When kids improve their focus and concentration skills, they also improve their hand-eye coordination. When they learn how to hold crayons and choose between different colors to find the best one to use, kids develop strong hand-eye coordination. Even the act of holding a smartened steady when using coloring games helps kids develop basic coordination skills. Since coloring pages have all kinds of shapes and diagrams, kids are required to color within specified areas, which also helps them improve their hand-eye coordination.
Count objects in everyday contexts. Count the number of buttons on your child’s shirt as you button them, the number of oranges he helps you put in the grocery bag at the supermarket, the number of forks needed to set the table, or the number of stairs you go up to the front door. Start with small numbers (no more than five) and add a few as your child is ready for a challenge. Put small objects in a row. Gather some coins and have your child count them. After she has counted them, rearrange them in a circle, in a row, or spread them out, and ask her again to count the objects. Don’t be surprised if she has to count them again. But if she automatically answers without counting, you’ll know he has mastered number in variance. | 458 | 2,235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2019-39 | latest | en | 0.954636 |
https://www.mathhomeworkanswers.org/14156/length-rectangle-twice-widthe-area-98yd-wahat-length-width | 1,540,204,237,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515029.82/warc/CC-MAIN-20181022092330-20181022113830-00421.warc.gz | 982,588,237 | 12,009 | The length of a rectangle is twice the width, the area is 98yd^2 find the length and width
asked Jan 10, 2012
## Your answer
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## 1 Answer
A = 98
A = l*w and l = 2w
98 = 2w*w
98 = 2w^2
w^2 = 98/2
w^2 = 49
w = sqrt49
w = 7 yd
l = 2w = 2*7 = 14 yd
98 = 14*7
98 = 98
answered Jan 10, 2012 by Level 8 User (36,860 points)
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1 answer | 234 | 693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-43 | longest | en | 0.879106 |
https://www.mail-archive.com/everything-list@googlegroups.com/msg14089.html | 1,529,929,238,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867666.97/warc/CC-MAIN-20180625111632-20180625131632-00558.warc.gz | 830,387,510 | 5,414 | Re: Cantor's Diagonal
Hi Barry,
Let me see if I am clear about Cantor's method. Given a set S, and some enumeration of that set (i.e., a no one-one onto map from Z^+ to S) we can use the diagonalization method to find an D which is a valid element of S but is different from any particular indexed element in the enumeration.
Cantor's argument then goes on to say (and here is where I disagree with it) that therefore D is not included in the enumeration and the enumeration is incomplete.
I, on the other hand, would posit that the enumeration may include elements whose index is not well defined. For example, 1 or 4 in my second example. They were in the enumeration prior to its being shuffled, and always had a definite position during the process. They must still be in there, with definite positions, despite the fact that their indices are now infinite and ill-defined.
If the diagonalization process does not produce the proffered result in this case (i,e., it does not prove that the element D is not included in the enumeration) then it does not prove it in any case. The number D found with this method may actually be in the enumeration, but with an ill-defined index.
Dan
Barry Brent wrote:
```Hi.
Bruno could do this better, but I like the practice.```
```
I guess you're trying to demonstrate that the form of Cantor's
argument is invalid, by displaying an example in which it produces an
absurd result.
Start with a set S you want to show is not enumerable. (ie, there is
no one-one onto map from Z^+ to S). The form of the diagonalization
argument is as follows: give me any, repeat, any, particular
candidate for an enumeration of S. This should be a map from Z^+
into S. (If it isn't such a map, it isn't an enumeration.) I will
show you an element D of S that your candidate enumeration omits.
(That is, I will show you that your candidate is not onto.) Hence, S
is not denumerable.
So your first attempt doesn't fit the form of the diagonalization
argument on this account. More fundamentally, it also fails to fit
the form of Cantor's argument because you haven't tried to debunk
*any* candidate enumeration, but a particular one.
from the primes (all of them!) and then (your work suggests, but I
think you'd need more details--what's the image of 4, for example?)
from the rest of Z^+, into S = Z^+ again. This example doesn't
invalidate Cantor's argument either. Again, you debunk a particular
candidate enumeration, not any and all candidate enumerations. So
you don't arrive at the absurdity you seem to be after, even if you
fill in the details I mentioned.
Barry
On Dec 16, 2007, at 3:49 AM, Daniel Grubbs wrote:
```
```Hi Folks,
I joined this list a while ago but I haven't really kept up.
Anyway, I saw the reference to Cantor's Diagonal and thought
perhaps someone could help me.
Consider the set of positive integers: {1,2,3,...}, but rather than
write them in this standard notation we'll use what I'll call
'prime notation'. Here, the number m may be written as
m = n1,n2,n3,n4,...
where ni is the number of times the i'th prime number is a factor
of m. Thus:
1 = 0,0,0,0,0,...
2 = 1,0,0,0,0,...
3 = 0,1,0,0,0,...
4 = 2,0,0,0,0,...
5 = 0,0,1,0,0,...
...
28 = 2,0,0,1,0,0,0,...
...
If we now apply the diagonal method to this ordered set, we get the
number:
D = 1,1,1,1,1,...
Has this just shown that the set of positive integers is not
denumerable?
I can see that one may complain that D is clearly infinite and
therefore should not be in the set, but consider the following...
Let's take the original set and reorder it by exchanging the places
of the i'th prime number with that of the number in the i'th
position. (i.e. First switch the number 2 with the number 1 to
move it to the first position. Then switch 3 with the number -- now
1 -- in the 2nd position. Then 5 with the 1 which is now in the 3rd
position. Etc...) Because we are just trading the positions of the
numbers, all the same numbers will be in the set afterwards.
The set is now:
2 = 1,0,0,0,0,...
3 = 0,1,0,0,0,...
5 = 0,0,1,0,0,...
7 = 0,0,0,1,0,...
11= 0,0,0,0,1,...
...
Now instead of adding 1 to each 'digit' of the diagonal, subtract
1. This will ensure that the diagonal number is different from
each of the numbers in the set. Thus,
D = 0,0,0,0,...
But this is the number 1 which we know was in the set to begin
with. What happened to it?
I would suggest that the diagonal method does not find a number
which is different from all the members of a set, but rather finds
a number which is infinitely far out in the ordered set.
If anyone can find where I've gone wrong, please let me know.
Dan Grubbs
```
```
Dr. Barry Brent
[EMAIL PROTECTED]
```
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-~----------~----~----~----~------~----~------~--~--- | 1,395 | 5,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-26 | latest | en | 0.93733 |
http://www.enotes.com/homework-help/solve-equation-3x-2-4x-8-0-423577 | 1,477,188,081,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719136.58/warc/CC-MAIN-20161020183839-00186-ip-10-171-6-4.ec2.internal.warc.gz | 432,440,804 | 9,545 | # Solve the equation 3x^2 - 4x + 8 = 0
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
The equation 3x^2 - 4x + 8 = 0 has to be solved. The roots of a quadratic equation ax^2 + bx + c = 0 are `x1 = (-b + sqrt(b^2 - 4ac))/(2a)` and `x2 = (-b - sqrt(b^2 - 4ac))/(2a)`
For the given equation the roots are
x1 = `(4 + sqrt(16 - 96))/6`
= `2/3 + (2*sqrt 5*i)/3`
x2 = `2/3 - (2*sqrt 5*i)/3`
The solutions of the equation 3x^2 - 4x + 8 = 0 are `2/3 + (2*sqrt 5*i)/3` and `2/3 - (2*sqrt 5*i)/3` | 229 | 520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-44 | latest | en | 0.744857 |
http://stackoverflow.com/questions/17494115/create-a-mst-with-depth-first-search | 1,432,757,662,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929096.44/warc/CC-MAIN-20150521113209-00088-ip-10-180-206-219.ec2.internal.warc.gz | 224,517,139 | 17,184 | # Create a MST with depth-first search?
I have a symmetrical graph and created a tree with all shortest path from a random vertex to any other vertex. Can I use the tree to construct a Minimum Spanning Tree(MST)? My algorithm is similar to depth-first algorithm.
-
No, because the shortest path from A to B and from A to C will not tell you the shortest path from B to C. – Tyler Durden Jul 5 '13 at 17:37
can you explain? when you have 3 vertex mst is a-b and a-c. no need b-c? – Phpdevpad Jul 5 '13 at 17:42
"mst is a-b and a-c" --- it's an "st" all right, but why is it "m"? – n.m. Jul 5 '13 at 17:46
why it's a-b-c? I have all shortest path. – Phpdevpad Jul 5 '13 at 17:51
not sure why you freaking. my accuracy isn't so good. – Phpdevpad Jul 5 '13 at 18:53 | 241 | 768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2015-22 | latest | en | 0.920832 |
https://www.jiskha.com/questions/762234/jessica-lasda-an-educational-publisher-presently-have-five-accounts-and-her-manager-is | 1,591,175,170,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347432521.57/warc/CC-MAIN-20200603081823-20200603111823-00084.warc.gz | 770,630,341 | 5,914 | # mths probability
jessica Lasda an educational publisher presently have five accounts and her manager is considering assigning her three more accounts .the new accounts would bring potential volume to her business
Then I have a 6 column table which i am writing in 6 different lines
1 column account number 1 2 3 4 5 6 7 8
2 column EXISTING VOLUME \$10,000 30,000 25,000 35,000 15,000 0 0 0
these are the values of 8 different account numbers
3rd column POTENTIAL ADDITIONAL VOLUME 10,000 0 15,000 0 5,000 30,000 25,000 45,000
4th COLUMN PROBABILITY OF GETTING ADDITIONAL VOLUME .40 --- .20 --- .30 .10 .70 .60
5th COLUMN EXPECTED VALUE OF ADDITIONAL VOLUME \$4000 ---- 3000 ---- 1500 3000 17500 27000
6TH COLUMN EXISTING VOLUME PLUS EXPECTED VALUE OF ADDITIONAL VOLUME \$ 14,000 30,000 28,000 35,000 16,500 3000 17500 27000
A. IF Jessica achived her expected additional volume in all accounts what would be the total volume of all her accounts.
B IF Jessica achived her expected additional volume in all accounts by what percentage would she increase her total volume
1. 👍 0
2. 👎 0
3. 👁 137
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