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https://techstalking.com/programming/python/hash-value-for-directed-acyclic-graph/	[ "How do I transform a directed acyclic graph into a hash value such that any two isomorphic graphs hash to the same value? It is acceptable, but undesirable for two isomorphic graphs to hash to different values, which is what I have done in the code below. We can assume that the number of vertices in the graph is at most 11.\n\nI am particularly interested in Python code.\n\nHere is what I did. If `self.lt` is a mapping from node to descendants (not children!), then I relabel the nodes according to a modified topological sort (that prefers to order elements with more descendants first if it can). Then, I hash the sorted dictionary. Some isomorphic graphs will hash to different values, especially as the number of nodes grows.\n\nI have included all the code to motivate my use case. I am calculating the number of comparisons required to find the median of 7 numbers. The more that isomorphic graphs hash to the same value the less work that has to be redone. I considered putting larger connected components first, but didn’t see how to do that quickly.\n\n``````from tools.decorator import memoized # A standard memoization decorator\n\nclass Graph:\ndef __init__(self, n):\nself.lt = {i: set() for i in range(n)}\n\ndef compared(self, i, j):\nreturn j in self.lt[i] or i in self.lt[j]\n\ndef withedge(self, i, j):\nretval = Graph(len(self.lt))\nimplied_lt = self.lt[j] \| set([j])\nfor (s, lt_s), (k, lt_k) in zip(self.lt.items(),\nretval.lt.items()):\nlt_k \|= lt_s\nif i in lt_k or k == i:\nlt_k \|= implied_lt\nreturn retval.toposort()\n\ndef toposort(self):\nmapping = {}\nwhile len(mapping) < len(self.lt):\nfor i, lt_i in self.lt.items():\nif i in mapping:\ncontinue\nif any(i in lt_j or len(lt_i) < len(lt_j)\nfor j, lt_j in self.lt.items()\nif j not in mapping):\ncontinue\nmapping[i] = len(mapping)\nretval = Graph(0)\nfor i, lt_i in self.lt.items():\nretval.lt[mapping[i]] = {mapping[j]\nfor j in lt_i}\nreturn retval\n\ndef median_known(self):\nn = len(self.lt)\nfor i, lt_i in self.lt.items():\nif len(lt_i) != n // 2:\ncontinue\nif sum(1\nfor j, lt_j in self.lt.items()\nif i in lt_j) == n // 2:\nreturn True\nreturn False\n\ndef __repr__(self):\nreturn(\"[{}]\".format(\", \".join(\"{}: {{{}}}\".format(\ni,\n\", \".join(str(x) for x in lt_i))\nfor i, lt_i in self.lt.items())))\n\ndef hashkey(self):\nreturn tuple(sorted({k: tuple(sorted(v))\nfor k, v in self.lt.items()}.items()))\n\ndef __hash__(self):\nreturn hash(self.hashkey())\n\ndef __eq__(self, other):\nreturn self.hashkey() == other.hashkey()\n\n@memoized\ndef mincomps(g):\nprint(\"Calculating:\", g)\nif g.median_known():\nreturn 0\nnodes = g.lt.keys()\nreturn 1 + min(max(mincomps(g.withedge(i, j)),\nmincomps(g.withedge(j, i)))\nfor i in nodes\nfor j in nodes\nif j > i and not g.compared(i, j))\n\ng = Graph(7)\nprint(mincomps(g))\n``````\n\nTo effectively test for graph isomorphism you will want to use nauty. Specifically for Python there is the wrapper pynauty, but I can’t attest its quality (to compile it correctly I had to do some simple patching on its `setup.py`). If this wrapper is doing everything correctly, then it simplifies nauty a lot for the uses you are interested and it is only a matter of hashing `pynauty.certificate(somegraph)` — which will be the same value for isomorphic graphs.\n\nSome quick tests showed that `pynauty` is giving the same certificate for every graph (with same amount of vertices). But that is only because of a minor issue in the wrapper when converting the graph to nauty’s format. After fixing this, it works for me (I also used the graphs at http://funkybee.narod.ru/graphs.htm for comparison). Here is the short patch which also considers the modifications needed in `setup.py`:\n\n``````diff -ur pynauty-0.5-orig/setup.py pynauty-0.5/setup.py\n--- pynauty-0.5-orig/setup.py 2011-06-18 20:53:17.000000000 -0300\n+++ pynauty-0.5/setup.py 2013-01-28 22:09:07.000000000 -0200\n@@ -31,7 +31,9 @@\n\next_pynauty = Extension(\nname = MODULE + '._pynauty',\n- sources = [ pynauty_dir + \"https://stackoverflow.com/\" + 'pynauty.c', ],\n+ sources = [ pynauty_dir + \"https://stackoverflow.com/\" + 'pynauty.c',\n+ os.path.join(nauty_dir, 'schreier.c'),\n+ os.path.join(nauty_dir, 'naurng.c')],\ndepends = [ pynauty_dir + \"https://stackoverflow.com/\" + 'pynauty.h', ],\nextra_compile_args = [ '-O4' ],\nextra_objects = [ nauty_dir + \"https://stackoverflow.com/\" + 'nauty.o',\ndiff -ur pynauty-0.5-orig/src/pynauty.c pynauty-0.5/src/pynauty.c\n--- pynauty-0.5-orig/src/pynauty.c 2011-03-03 23:34:15.000000000 -0300\n+++ pynauty-0.5/src/pynauty.c 2013-01-29 00:38:36.000000000 -0200\n@@ -320,7 +320,7 @@\nPyObject p;\n\n- int i,j;\n+ Py_ssize_t i, j;\nint x, y;\n``````\n\nGraph isomorphism for directed acyclic graphs is still GI-complete. Therefore there is currently no known (worst case sub-exponential) solution to guarantee that two isomorphic directed acyclic graphs will yield the same hash. Only if the mapping between different graphs is known – for example if all vertices have unique labels – one could efficiently guarantee matching hashes.\n\nOkay, let’s brute force this for a small number of vertices. We have to find a representation of the graph that is independent of the ordering of the vertices in the input and therefore guarantees that isomorphic graphs yield the same representation. Further this representation must ensure that no two non-isomorphic graphs yield the same representation.\n\nThe simplest solution is to construct the adjacency matrix for all n! permutations of the vertices and just interpret the adjacency matrix as n2 bit integer. Then we can just pick the smallest or largest of this numbers as canonical representation. This number completely encodes the graph and therefore ensures that no two non-isomorphic graphs yield the same number – one could consider this function a perfect hash function. And because we choose the smallest or largest number encoding the graph under all possible permutations of the vertices we further ensure that isomorphic graphs yield the same representation.\n\nHow good or bad is this in the case of 11 vertices? Well, the representation will have 121 bits. We can reduce this by 11 bits because the diagonal representing loops will be all zeros in an acyclic graph and are left with 110 bits. This number could in theory be decreased further; not all 2110 remaining graphs are acyclic and for each graph there may be up to 11! – roughly 225 – isomorphic representations but in practice this might be quite hard to do. Does anybody know how to compute the number of distinct directed acyclic graphs with n vertices?\n\nHow long will it take to find this representation? Naively 11! or 39,916,800 iterations. This is not nothing and probably already impractical but I did not implement and test it. But we can probably speed this up a bit. If we interpret the adjacency matrix as integer by concatenating the rows from top to bottom left to right we want many ones (zeros) at the left of the first row to obtain a large (small) number. Therefore we pick as first vertex the one (or one of the vertices) with largest (smallest) degree (indegree or outdegree depending on the representation) and than vertices connected (not connected) to this vertex in subsequent positions to bring the ones (zeros) to the left.\n\nThere are likely more possibilities to prune the search space but I am not sure if there are enough to make this a practical solution. Maybe there are or maybe somebody else can at least build something upon this idea.\n\nHow good does the hash have to be? I assume that you do not want a full serialization of the graph. A hash rarely guarantees that there is no second (but different) element (graph) that evaluates to the same hash. If it is very important to you, that isomorphic graphs (in different representations) have the same hash, then only use values that are invariant under a change of representation. E.g.:\n\n• the total number of nodes\n• the total number of (directed) connections\n• the total number of nodes with `(indegree, outdegree) = (i,j)` for any tuple `(i,j)` up to `(max(indegree), max(outdegree))` (or limited for tuples up to some fixed value `(m,n)`)\n\nAll these informations can be gathered in O(#nodes) [assuming that the graph is stored properly]. Concatenate them and you have a hash. If you prefer you can use some well known hash algorithm like `sha` on these concatenated informations. Without additional hashing it is a continuous hash (it allows to find similar graphs), with additional hashing it is uniform and fixed in size if the chosen hash algorithm has these properties.\n\nAs it is, it is already good enough to register any added or removed connection. It might miss connections that were changed though (`a -> c` instead of `a -> b`).\n\nThis approach is modular and can be extended as far as you like. Any additional property that is being included will reduce the number of collisions but increase the effort necessary to get the hash value. Some more ideas:\n\n• same as above but with second order in- and outdegree. Ie. the number of nodes that can be reached by a `node->child->child` chain ( = second order outdegree) or respectively the number of nodes that lead to the given node in two steps.\n• or more general n-th order in- and outdegree (can be computed in O((average-number-of-connections) ^ (n-1) #nodes) )\n• number of nodes with eccentricity = x (again for any x)\n• if the nodes store any information (other than their neighbours) use a `xor` of any kind of hash of all the node-contents. Due to the `xor` the specific order in which the nodes where added to the hash does not matter.\n\nYou requested “a unique hash value” and clearly I cannot offer you one. But I see the terms “hash” and “unique to every graph” as mutually exclusive (not entirely true of course) and decided to answer the “hash” part and not the “unique” part. A “unique hash” (perfect hash) basically needs to be a full serialization of the graph (because the amount of information stored in the hash has to reflect the total amount of information in the graph). If that is really what you want just define some unique order of nodes (eg. sorted by own outdegree, then indegree, then outdegree of children and so on until the order is unambiguous) and serialize the graph in any way (using the position in the formentioned ordering as index to the nodes).\n\nOf course this is much more complex though.\n\nYears ago, I created a simple and flexible algorithm for exactly this problem (finding duplicate structures in a database of chemical structures by hashing them).\n\nI named it “Powerhash”, and to create the algorithm it required two insights. The first is the power iteration graph algorithm, also used in PageRank. The second is the ability to replace power iteration’s inside step function with anything that we want. I replaced it with a function that does the following on each step, and for each node:\n\n• Sort the hashes of the node’s neighbors\n• Hash the concatenated sorted hashes\n\nOn the first step, a node’s hash is affected by its direct neighbors. On the second step, a node’s hash is affected by the neighborhood 2-hops away from it. On the Nth step a node’s hash will be affected by the neighborhood N-hops around it. So you only need to continue running the Powerhash for N = graph_radius steps. In the end, the graph center node’s hash will have been affected by the whole graph.\n\nTo produce the final hash, sort the final step’s node hashes and concatenate them together. After that, you can compare the final hashes to find if two graphs are isomorphic. If you have labels, then add them in the internal hashes that you calculate for each node (and at each step).\n\nFor more on this you can look at my post here:\n\nThe algorithm above was implemented inside the “madIS” functional relational database. You can find the source code of the algorithm here:\n\nImho, If the graph could be topologically sorted, the very straightforward solution exists.\n\n1. For each vertex with index i, you could build an unique hash (for example, using the hashing technique for strings) of his (sorted) direct neighbours (p.e. if vertex 1 has direct neighbours {43, 23, 2,7,12,19,334} the hash functions should hash the array of {2,7,12,19,23,43,334})\n2. For the whole DAG you could create a hash, as a hash of a string of hashes for each node: Hash(DAG) = Hash(vertex_1) U Hash(vertex_2) U ….. Hash(vertex_N);\nI think the complexity of this procedure is around (NN) in the worst case. If the graph could not be topologically sorted, the approach proposed is still applicable, but you need to order vertices in an unique way (and this is the hard part)\n\nI will describe an algorithm to hash an arbitrary directed graph, not taking into account that the graph is acyclic. In fact even counting the acyclic graphs of a given order is a very complicated task and I believe here this will only make the hashing significantly more complicated and thus slower.\n\nA unique representation of the graph can be given by the neighbourhood list. For each vertex create a list with all it’s neighbours. Write all the lists one after the other appending the number of neighbours for each list to the front. Also keep the neighbours sorted in ascending order to make the representation unique for each graph. So for example assume you have the graph:\n\n``````1->2, 1->5\n2->1, 2->4\n3->4\n5->3\n``````\n\nWhat I propose is that you transform this to `({2,2,5}, {2,1,4}, {1,4}, {0}, {1,3})`, here the curly brackets being only to visualize the representation, not part of the python’s syntax. So the list is in fact: `(2,2,5, 2,1,4, 1,4, 0, 1,3)`.\n\nNow to compute the unique hash, you need to order these representations somehow and assign a unique number to them. I suggest you do something like a lexicographical sort to do that. Lets assume you have two sequences `(a1, b1_1, b_1_2,...b_1_a1,a2, b_2_1, b_2_2,...b_2_a2,...an, b_n_1, b_n_2,...b_n_an)` and `(c1, d1_1, d_1_2,...d_1_c1,c2, d_2_1, d_2_2,...d_2_c2,...cn, d_n_1, d_n_2,...d_n_cn)`, Here c and a are the number of neighbours for each vertex and b_i_j and d_k_l are the corresponding neighbours. For the ordering first compare the sequnces `(a1,a2,...an)` and `(c1,c2, ...,cn)` and if they are different use this to compare the sequences. If these sequences are different, compare the lists from left to right first comparing lexicographically `(b_1_1, b_1_2...b_1_a1)` to `(d_1_1, d_1_2...d_1_c1)` and so on until the first missmatch.\n\nIn fact what I propose to use as hash the lexicographical number of a word of size `N` over the alphabet that is formed by all possible selections of subsets of elements of `{1,2,3,...N}`. The neighbourhood list for a given vertex is a letter over this alphabet e.g. `{2,2,5}` is the subset consisting of two elements of the set, namely `2` and `5`.\n\nThe alphabet(set of possible letters) for the set `{1,2,3}` would be(ordered lexicographically):\n\n`{0}, {1,1}, {1,2}, {1,3}, {2, 1, 2}, {2, 1, 3}, {2, 2, 3}, {3, 1, 2, 3}`\n\nFirst number like above is the number of elements in the given subset and the remaining numbers- the subset itself. So form all the 3 letter words from this alphabet and you will get all the possible directed graphs with 3 vertices.\n\nNow the number of subsets of the set `{1,2,3,....N}` is `2^N` and thus the number of letters of this alphabet is `2^N`. Now we code each directed graph of `N` nodes with a word with exactly `N` letters from this alphabet and thus the number of possible hash codes is precisely: `(2^N)^N`. This is to show that the hash code grows really fast with the increase of `N`. Also this is the number of possible different directed graphs with `N` nodes so what I suggest is optimal hashing in the sense it is bijection and no smaller hash can be unique.\n\nThere is a linear algorithm to get a given subset number in the the lexicographical ordering of all subsets of a given set, in this case `{1,2,....N}`. Here is the code I have written for coding/decoding a subset in number and vice versa. It is written in `C++` but quite easy to understand I hope. For the hashing you will need only the code function but as the hash I propose is reversable I add the decode function – you will be able to reconstruct the graph from the hash which is quite cool I think:\n\n``````typedef long long ll;\n\n// Returns the number in the lexicographical order of all combinations of n numbers\n// of the provided combination.\nll code(vector<int> a,int n)\n{\nsort(a.begin(),a.end()); // not needed if the set you pass is already sorted.\nint cur = 0;\nint m = a.size();\n\nll res =0;\nfor(int i=0;i<a.size();i++)\n{\nif(a[i] == cur+1)\n{\nres++;\ncur = a[i];\ncontinue;\n}\nelse\n{\nres++;\nint number_of_greater_nums = n - a[i];\nfor(int j = a[i]-1,increment=1;j>cur;j--,increment++)\nres += 1LL << (number_of_greater_nums+increment);\ncur = a[i];\n}\n}\nreturn res;\n}\n// Takes the lexicographical code of a combination of n numbers and returns the\n// combination\nvector<int> decode(ll kod, int n)\n{\nvector<int> res;\nint cur = 0;\n\nint left = n; // Out of how many numbers are we left to choose.\nwhile(kod)\n{\nll all = 1LL << left;// how many are the total combinations\nfor(int i=n;i>=0;i--)\n{\nif(all - (1LL << (n-i+1)) +1 <= kod)\n{\nres.push_back(i);\nleft = n-i;\nkod -= all - (1LL << (n-i+1)) +1;\nbreak;\n}\n}\n}\nreturn res;\n}\n``````\n\nAlso this code stores the result in `long long` variable, which is only enough for graphs with less than 64 elements. All possible hashes of graphs with 64 nodes will be `(2^64)^64`. This number has about 1280 digits so maybe is a big number. Still the algorithm I describe will work really fast and I believe you should be able to hash and ‘unhash’ graphs with a lot of vertices.\n\nAlso have a look at this question.\n\nI’m not sure that it’s 100% working, but here is an idea:\n\nLet’s code a graph into a string and then take its hash.\n\n1. hash of an empty graph is “”\n2. hash of a vertex with no outgoing edges is “.”\n3. hash of a vertex with outgoing edges is concatenation of every child hash with some delimiter (e.g. “,”)\n\nTo produce the same hash for isomorphic graphs before concatenation in step3 just sort the hashes (e.g. in lexicographical order).\n\nFor hash of a graph just take hash of its root (or sorted concatenation, if there are several roots).\n\nedit While I hoped that the resulting string will describe graph without collisions, hynekcer found that sometimes non-isomorphic graphs will get the same hash. That happens when a vertex has several parents – then it “duplicated” for every parent. For example, the algorithm does not differentiate a “diamond” {A->B->C,A->D->C} from the case {A->B->C,A->D->E}.\n\nI’m not familiar with Python and it’s hard for me to understand how graph stored in the example, but here is some code in C++ which is likely convertible to Python easily:\n\n``````THash GetHash(const TGraph &graph)\n{\nreturn ComputeHash(GetVertexStringCode(graph,FindRoot(graph)));\n}\nstd::string GetVertexStringCode(const TGraph &graph,TVertexIndex vertex)\n{\nstd::vector<std::string> childHashes;\nfor(auto c:graph.GetChildren(vertex))\nchildHashes.push_back(GetVertexStringCode(graph,c));\nstd::sort(childHashes.begin(),childHashes.end());\nstd::string result=\".\";\nfor(auto h:childHashes)\nresult+=*h+\",\";\nreturn result;\n}\n``````\n\nI am assuming there are no common labels on vertices or edges, for then you could put the graph in a canonical form, which itself would be a perfect hash. This proposal is therefore based on isomorphism only.\n\nFor this, combine hashes for as many simple aggregate characteristics of a DAG as you can imagine, picking those that are quick to compute. Here is a starter list:\n\n1. 2d histogram of nodes’ in and out degrees.\n2. 4d histogram of edges a->b where a and b are both characterized by in/out degree.\n\nLet me be more explicit. For 1, we’d compute a set of triples `<I,O;N>` (where no two triples have the same `I`,`O` values), signifying that there are `N` nodes with in-degree `I` and out-degree `O`. You’d hash this set of triples or better yet use the whole set arranged in some canonical order e.g. lexicographically sorted. For 2, we compute a set of quintuples `<aI,aO,bI,bO;N>` signifying that there are `N` edges from nodes with in degree `aI` and out degree `aO`, to nodes with `bI` and `bO` respectively. Again hash these quintuples or else use them in canonical order as-is for another part of the final hash.\n\nStarting with this and then looking at collisions that still occur will probably provide insights on how to get better.\n\nWhen I saw the question, I had essentially the same idea as @example. I wrote a function providing a graph tag such that the tag coincides for two isomorphic graphs.\n\nThis tag consists of the sequence of out-degrees in ascending order. You can hash this tag with the string hash function of your choice to obtain a hash of the graph.\n\nEdit: I expressed my proposal in the context of @NeilG’s original question. The only modification to make to his code is to redefine the `hashkey` function as:\n\n``````def hashkey(self):\nreturn tuple(sorted(map(len,self.lt.values())))\n``````\n\nWith suitable ordering of your descendents (and if you have a single root node, not a given, but with suitable ordering (maybe by including a virtual root node)), the method for hashing a tree ought to work with a slight modification.\n\nExample code in this StackOverflow answer, the modification would be to sort children in some deterministic order (increasing hash?) before hashing the parent.\n\nEven if you have multiple possible roots, you can create a synthetic single root, with all roots as children." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87073827,"math_prob":0.9438986,"size":21466,"snap":"2022-40-2023-06","text_gpt3_token_len":5383,"char_repetition_ratio":0.12682882,"word_repetition_ratio":0.02063896,"special_character_ratio":0.26120377,"punctuation_ratio":0.1449016,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9918125,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-04T01:21:53Z\",\"WARC-Record-ID\":\"<urn:uuid:919c89bc-7ea2-4da8-9db5-9ca84e43a66c>\",\"Content-Length\":\"114906\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fe81d7c0-1f58-4b19-b1a5-ea24eeba2d07>\",\"WARC-Concurrent-To\":\"<urn:uuid:914923b6-07be-448e-b1ad-e02ee1803375>\",\"WARC-IP-Address\":\"104.21.42.107\",\"WARC-Target-URI\":\"https://techstalking.com/programming/python/hash-value-for-directed-acyclic-graph/\",\"WARC-Payload-Digest\":\"sha1:TMYPUH4AUYUDWUV3ENR52GJH27O4MDVX\",\"WARC-Block-Digest\":\"sha1:CR4FOH7YVQDA24MTPAMJ3SLNNCUW4RWO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337446.8_warc_CC-MAIN-20221003231906-20221004021906-00580.warc.gz\"}"}
https://ctan.org/pkg/pi	[ "# pi – Calculate pi\n\nGenerates pi, using the formula: ``` Pi=16arctan(1/5)-4arctan(1/239) ``` and leaves the result in an array \\xr, printing what is calculated as it goes along.\n\nThe number of digits you can compute depends on your implementation of . The last digit may be wrong, and if it's 0 or 9, the penultimate digit may also be wrong.\n\n Sources `/macros/plain/contrib/misc/pi.tex` Version 0.993 Maintainer Denis B. Roegel Topics Calculation" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.72225934,"math_prob":0.97581285,"size":761,"snap":"2021-43-2021-49","text_gpt3_token_len":197,"char_repetition_ratio":0.09643329,"word_repetition_ratio":0.0,"special_character_ratio":0.23127463,"punctuation_ratio":0.1294964,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96961117,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-20T13:51:41Z\",\"WARC-Record-ID\":\"<urn:uuid:2fd99763-b7f8-4317-86fb-a0776aa9082f>\",\"Content-Length\":\"14467\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:20809366-3626-48ec-bb96-4bfeb4653fe7>\",\"WARC-Concurrent-To\":\"<urn:uuid:092d32a6-7044-4ff4-a946-b4c04e288a6c>\",\"WARC-IP-Address\":\"5.35.249.60\",\"WARC-Target-URI\":\"https://ctan.org/pkg/pi\",\"WARC-Payload-Digest\":\"sha1:5VC3ZITYBSVLJ7KXKGW7BFGQBBKAOCTM\",\"WARC-Block-Digest\":\"sha1:WRFNW7YAYK5APASOCUWMN7JITE73X42F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585321.65_warc_CC-MAIN-20211020121220-20211020151220-00679.warc.gz\"}"}
https://www.junhaow.com/lc/problems/linked-list/cycle-detection/141_linked-list-cycle.html	[ "Reference: LeetCode\nDifficulty: EasyTwo Pointers\n\n## Problem\n\nGiven a linked list, determine if it has a cycle in it.\n\nNote:\n\n• To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.\n• There could be duplicates.\n\nExample:", null, "", null, "Follow up: Can you solve it using $O(1)$ (i.e. constant) memory?\n\n• Yes. Use two pointers or modify the next.\n\n## Analysis\n\nNote: In terms of duplicate nodes, they are different node objects. So it is okay to use hashCode() and ==.\n\nTest Case:\n\n### Hash Set\n\nUse Set<ListNode> instead of Set<Integer>.\n\nTime: $O(N)$.\nSpace: $O(N)$.\n\n### Two Pointers\n\nThe space complexity can be reduced to $O(1)$ by considering two pointers at different speed. A slow pointer and a fast pointer. The slow pointer moves one step at a time while the fast pointer moves two steps at a time. If there is no cycle in the list, the fast pointer will eventually reach the end and we can return false in this case.", null, "• No cycle\nThe fast pointer reaches the end first and the run time depends on the list’s length, which is $O(N)$.\n• Cycle exists\n• Non-cyclic part:\n• The slow pointer takes non-cyclic length steps to enter the cycle. The length is $L1$.\n• Cyclic part:\n• When two pointers enter into the cycle, it will take:\n• (distance between two pointers (at most $C$) / difference of speed) moves for the fast pointer to catch up with the slow runner.\n\nTherefore, the worst case runtime is $O(L1 + C)$, which is $O(N)$.\n\nWhy will the fast pointer eventually meet the slow pointer?\n\nConsider the one-step case:\n\nIf the fast pointer is two steps behind the slower runner, this case will change into the one-step case above (each time the distance decreases by $1$):\n\nNote: So many corner cases.\n\nMy originally bad code (using flag):\n\nImprovement:\n\nTime: $O(N)$\nSpace: $O(1)$\n\n### Mark Node\n\nBy Argondey: I used a slightly faster destructive approach. I created a new node called “mark” and iterated through the list, setting the next value of each node to the mark node. When reaching the new node the first thing I did was check if the node is the mark node. If it ever is the mark node, we have looped. Brilliant!\n\nTime: $O(N)$\nSpace: $O(1)$\n\nComment", null, "Junhao Wang\na software engineering cat" ]	[ null, "https://bloggg-1254259681.cos.na-siliconvalley.myqcloud.com/jtr5l.jpg", null, "https://bloggg-1254259681.cos.na-siliconvalley.myqcloud.com/vlrjw.jpg", null, "https://bloggg-1254259681.cos.na-siliconvalley.myqcloud.com/c7fpv.png", null, "https://www.junhaow.com/resources/avatar/avatar_2020.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.823454,"math_prob":0.9769121,"size":2657,"snap":"2023-40-2023-50","text_gpt3_token_len":732,"char_repetition_ratio":0.120618165,"word_repetition_ratio":0.05636743,"special_character_ratio":0.2754987,"punctuation_ratio":0.1390845,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9977006,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-11T06:38:23Z\",\"WARC-Record-ID\":\"<urn:uuid:a7fe833f-2eae-47c4-a677-f95fa8c42f8f>\",\"Content-Length\":\"38697\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:336cb3fc-79c1-4c36-a29e-b0d684d5e8ca>\",\"WARC-Concurrent-To\":\"<urn:uuid:f60f9f8f-4931-4317-ac3f-89ea28d414e2>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"https://www.junhaow.com/lc/problems/linked-list/cycle-detection/141_linked-list-cycle.html\",\"WARC-Payload-Digest\":\"sha1:7LQ3Z6C5XAQOQ36WLDBXJL2CAT4MFDZP\",\"WARC-Block-Digest\":\"sha1:SSZLKCSGFCWK4DNZADQAFFIQEDZVOH7E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679103558.93_warc_CC-MAIN-20231211045204-20231211075204-00059.warc.gz\"}"}
https://www.mathcity.org/msc/notes/partial-differential-equations-m-usman-hamid	[ "# Partial Differential Equations by M Usman Hamid\n\nThe course provides a foundation to solve PDE’s with special emphasis on wave, heat and Laplace equations, formulation and some theory of these equations are also intended. We are really very thankful to Prof. Muhammad Usman Hamid for providing these notes and appreciates his effort to publish these notes on MathCity.org", null, "Name Partial Differential Equations (PDE’s) Muhammad Usman Hamid 113 pages PDF (see Software section for PDF Reader) 2.22 MB\n• INTRODUCTION\n• Basic Concepts and Definitions, Superposition Principle, Exercises\n• FIRST-ORDER QUASI-LINEAR EQUATIONS AND METHOD OF CHARACTERISTICS\n• Classification of first-order equations, Construction of a First-Order Equation, Method of Characteristics and General Solutions, Canonical Forms of First-Order Linear Equations, Method of Separation of Variables, Exercises\n• MATHEMATICAL MODELS\n• Heat Equations and consequences, Wave Equations and consequences.\n• CLASSIFICATION OF SECOND-ORDER LINEAR EQUATIONS\n• Second-Order Equations in Two Independent Variables, Canonical Forms, Equations with Constant Coefficients, General Solutions, Summary and Further Simplification, Exercises\n• FOURIER SERIES, FOURIER TRANSFORMATION AND INTEGRALS WITH APPLICATIONS\n• Introduction, Fourier Transform, Properties of Fourier Transform , Convolution theorem, Fourier Sine and Fourier Cosine, Exercises, Fourier Series and its complex form\n• LAPLACE TRANSFORMS\n• Properties of Laplace Transforms, Convolution Theorem of the Laplace Transform, Laplace Transforms of the Heaviside and Dirac Delta Functions, Hankel Transforms, Properties of Hankel Transforms and Applications, Few results about finite fourier transforms, Exercises\n\nPlease click on View Online to see inside the PDF.\n\n• msc/notes/partial-differential-equations-m-usman-hamid" ]	[ null, "https://www.mathcity.org/_media/msc/notes/partial-differential-equations-m-usman-hamid.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8406814,"math_prob":0.90174013,"size":3572,"snap":"2023-14-2023-23","text_gpt3_token_len":857,"char_repetition_ratio":0.12724215,"word_repetition_ratio":0.046728972,"special_character_ratio":0.19428891,"punctuation_ratio":0.08007449,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99573386,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-06T08:54:37Z\",\"WARC-Record-ID\":\"<urn:uuid:bbce194c-2c8b-4687-b0e2-5b6573a5e630>\",\"Content-Length\":\"36623\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef45f7a7-5b21-4338-8594-e860313406b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:98700cf4-4bc5-4cb4-ac57-d5f1b759f03a>\",\"WARC-IP-Address\":\"172.67.200.195\",\"WARC-Target-URI\":\"https://www.mathcity.org/msc/notes/partial-differential-equations-m-usman-hamid\",\"WARC-Payload-Digest\":\"sha1:OU3Z2BYSRJW5EDU4KNHBLPXQRJWLPOSV\",\"WARC-Block-Digest\":\"sha1:7PQ4TJCV5SFGSBL3WG2K5AMVYSESS6KL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652494.25_warc_CC-MAIN-20230606082037-20230606112037-00401.warc.gz\"}"}
https://testbook.com/objective-questions/mcq-on-dynamic-programming--5eea6a0c39140f30f369e0db	[ "# Dynamic Programming MCQ Quiz - Objective Question with Answer for Dynamic Programming - Download Free PDF\n\nLast updated on Sep 9, 2023\n\nMultiple Choice Questions (MCQs) on dynamic programming are valuable for assessing knowledge and understanding of this algorithmic problem-solving technique. Dynamic Programming MCQ help evaluate familiarity with dynamic programming principles, concepts, and applications. By attempting these Dynamic Programming MCQ, individuals can enhance their comprehension of topics such as overlapping subproblems, optimal substructure, and memoization. These questions cover various aspects, including solving optimization problems, finding optimal paths, and efficient resource allocation. Dynamic Programming MCQs enable learners to consolidate their understanding of this powerful algorithmic approach and its practical applications in computer science, mathematics, and other fields.\n\n## Latest Dynamic Programming MCQ Objective Questions\n\n#### Dynamic Programming Question 1:\n\nLet X = {a/25, b/20, c/10, d/20, e/50} be the alphabet and its frequency distribution. What is the optimum prefix code for this distribution?\n\n1. 275\n2. 220\n3. 230\n4. 245\n\nOption 1 : 275\n\n#### Dynamic Programming Question 1 Detailed Solution\n\nHuffman Coding: It is a lossless data compression algorithm. The input characters are assigned with variable length codes and the lengths of the assigned codes are based on the frequencies of corresponding characters. The most frequent character gets the smallest code and the least frequent character gets the largest code.\n\nExplanation:\n\n• Let us have a min-heap with frequencies as key.\nX =", null, "• According to the Huffman algorithm, we will merge the two minimum frequencies, that is, c and d. Now, n1 will be inserted back into the heap.\nX =", null, "", null, "• Now, again we will merge the two minimum frequencies, that is, a and b. Now, n2 will be inserted back into the heap.\nX =", null, "", null, "• Now, we will merge the two minimum frequencies among the remaining ones, that is, n2 and n1. Now, n3 will be inserted back into the heap.\nX =", null, "", null, "• Finally, we will merge the two last frequencies remaining, that is, n3 and e, and root will be inserted in the tree.", null, "• Now the Huffman code for a is 111, b is 110, c is 100, d is 101 and e is 0.\nHence, the minimum cost or optimum prefix code = $$B(C) = \\sum_{p=1}^{n} f(p_i)L(c(p_i))$$ where\n• $$f(p_i) \\rightarrow$$ frequency distribution of alphabet $$p = \\{p_1, p_2, p_3, ...... p_n\\}$$\n• $$L(c(p_i)) \\rightarrow$$ length of code word $$c(p_i)$$\n• Hence, minimum cost or optimum prefix code =\n$$\\rightarrow 10 * 3 + 20 * 3 + 20 * 3 + 25 * 3 + 50 * 1\\\\ \\rightarrow 30 + 60 + 60 + 75 + 50\\\\ \\rightarrow 275$$\n\n#### Dynamic Programming Question 2:\n\nConsider 4 matrices Q, R, S and T with dimensions 13 × 12, 12 × 30, 30 × 15 and 15 × 18 respectively. What is the least number of scalar multiplications needed to find the product QRST using the basic matrix multiplication method?\n\n1. 11150\n2. 11250\n3. 11350\n4. 11450\n\nOption 2 : 11250\n\n#### Dynamic Programming Question 2 Detailed Solution\n\nGiven matrices Q, R, S, and T with dimensions 13 × 12, 12 × 30, 30 × 15, and 15 × 18 respectively. From the given matrices sequence of dimensions can be seen as:\n\n0, P1, P2, P3, P4> = <13, 12, 30, 15, 18>\n\nLet c[i, j] be the minimum number of scalar multiplications needed to compute the product QRST.\n\nUsing Dynamic Programming recursive definition of c[i, j] is given as-\n\n$$c[i,j]=\\begin{array}{l} \\left\\{ {\\begin{array}{{20}{c}} 0\\\\ {\\min }\\\\{i \\le k < j\\left\\{ {c\\left[ {i,k} \\right] + c\\left[ {k + 1,j} \\right] + {P_i}_{ - 1}{P_k}{P_j}} \\right\\}if\\,i = j} \\end{array}if\\,i = j} \\right.\\\\ \\end{array}$$\n\nIn the given scenario i and j can be 1, 2 or 3, and j can be 2, 3 or 4.\n\n$$\\begin{array}{l} c\\left[ {1,2} \\right] = \\begin{array}{{20}{c}} {\\min }\\\\ {1 \\le K < 2} \\end{array}\\left\\{ {k = 1:\\begin{array}{{20}{c}} {c\\left[ {1,1} \\right] + c\\left[ {2,2} \\right] + {P_0}{P_1}{P_2}}\\\\ {0 + 0 + 4680} \\end{array}} \\right\\} \\end{array}$$\n\nc[1, 2] = min {4680} = 4680\n\n$$c\\left[ {2,3} \\right] = \\begin{array}{{20}{c}} {\\min }\\\\ {2 \\le k < 3} \\end{array}\\left\\{ {k = 2:\\begin{array}{{20}{c}} {c\\left[ {2,2} \\right] + c\\left[ {3,3} \\right] + {P_1}{P_2}{P_3}}\\\\ {0 + 0 + 5400} \\end{array}} \\right\\}$$\n\nc[2, 3]= min {5400} = 5400\n\n$$c\\left[ {3,4} \\right] = \\begin{array}{{20}{c}} {\\min }\\\\ {3 \\le k < 4} \\end{array}\\left\\{ {k = 3:\\begin{array}{{20}{c}} {c\\left[ {3,3} \\right] + c\\left[ {4,4} \\right] + {P_2}{P_3}{P_4}}\\\\ {0 + 0 + 8100} \\end{array}} \\right\\}$$\n\nc[3, 4]= min {8100} = 8100\n\n$$c\\left[ {1,3} \\right] = \\begin{array}{{20}{c}} {\\min }\\\\ {1 \\le k < 3} \\end{array}\\left\\{ {\\begin{array}{{20}{c}} {k = 1:}\\\\ {K = 2:} \\end{array}\\begin{array}{{20}{c}} {c\\left[ {2,2} \\right] + c\\left[ {2,3} \\right]{P_0}{P_1}{P_3}}\\\\ {0 + 5400 + 2340}\\\\ {c\\left[ {1,2} \\right] + c\\left[ {3,3} \\right]{P_0}{P_2}{P_3}}\\\\ {4680 + 0 + 5850} \\end{array}} \\right\\}$$\n\n$$c[1,3] = \\min \\left\\{ {\\frac{{7740}}{{10530}}} \\right\\} = 7740$$\n\n$$c\\left[ {2,4} \\right] = \\begin{array}{{20}{c}} {\\min }\\\\ {2 \\le k < 4} \\end{array}\\left\\{ {\\begin{array}{{20}{c}} {k = 2:}\\\\ {K = 3:} \\end{array}\\begin{array}{{20}{c}} {c\\left[ {2,2} \\right] + c\\left[ {3,4} \\right]{P_1}{P_2}{P_4}}\\\\ {0 + 8100 + 6480}\\\\ {c\\left[ {2,3} \\right] + c\\left[ {4,4} \\right]{P_1}{P_3}{P_4}}\\\\ {5400 + 0 + 3240} \\end{array}} \\right\\}$$\n\n$$c[2,4] = \\min \\left\\{ {\\frac{{14580}}{{8640}}} \\right\\} = 8640$$\n\n$$c\\left[ {1,4} \\right] = \\begin{array}{{20}{c}} {\\min }\\\\ {1 \\le k < 4} \\end{array}\\left\\{ {\\begin{array}{{20}{c}} {k = 1:}\\\\ {K = 2:}\\\\ {K = 3:} \\end{array}\\begin{array}{{20}{c}} {c\\left[ {1,1} \\right] + c\\left[ {2,4} \\right] + {P_0}{P_1}{P_4}}\\\\ {0 + 8640 + 2808}\\\\ {c\\left[ {1,2} \\right] + c\\left[ {3,4} \\right] + {P_0}{P_2}{P_4}}\\\\ {4680 + 8100 + 7020}\\\\ {c\\left[ {1,3} \\right] + c\\left[ {4,4} \\right] + {P_0}{P_3}{P_4}}\\\\ {7740 + 0 + 3510} \\end{array}} \\right\\}$$\n\n$$c[1,4]= \\min \\left\\{ {\\begin{array}{{20}{c}} {11448}\\\\ {19800}\\\\ {11250} \\end{array}} \\right\\} = 11250$$\n\nHence the minimum number of scalar multiplications needed to compute the product QRST is 11250 using (Q(RS))T).\n\n#### Dynamic Programming Question 3:\n\nNote:\n\nWhere n is the number of items and W is the weight of the Knapsack .\n\nWhich of the following statements are true?\n\n1. Time complexity of Knapsack is O(n W) where W is the weight of the Knapsack and there are n items.\n2. Time complexity of Knapsack is min( O(nW) , O(2^n) ) where W is the weight of the Knapsack and there are n items.\n3. Knapsack can be implemented in O(nW) space .\n4. Knapsack can be implemented in O(W) space.\n\nOption :\n\n#### Dynamic Programming Question 3 Detailed Solution\n\nThe correct answer is option 2, option 3, and option 4.\n\nConcept:\n\nOption 1 and option 2:\n\nIf the Weight of the Knapsack is in theta (2^n) , then we should use a brute force approach rather than the dynamic programming approach. So the time complexity of the Knapsack problem is min( O(nW) , O(2^n) ).\n\nNow, the normal dynamic programming approach takes O(nW) space, but we can optimize it using some observation.\n\nThe observation is when we are trying to fill the Knapsack with ith object, we only need the optimum result after filling the Knapsack with (i-1)th object, we don’t need entries for (i-2)th , (i-3) th ……. 1 th object. The pseudo-code for implementing the Knapsack problem in O(W) space is\n\nInput : int wt[n] - weight of items to be filled\nint value[n] - value of the items\nint W - Weight of the Knapsack\n\nAlgorithm:\n\n1. Declare two 1 D A[] and B[] arrays of size W+1 each (indexed 0 to W).\n\n2. Initialise them to 0.", null, "Though the time complexity of the above implementation is O(nW), but the space complexity reduces. We use two arrays of size W each. Thus space complexity is O(2W) =O(W)\n\nHence the correct answer is option 2, option 3, and option 4.\n\n#### Dynamic Programming Question 4:\n\nConsider the following items with their associated weights and values.\n\n Item Weight Value 1 4 44 2 7 21 3 5 38 4 2 11 5 7 40 6 10 37 7 3 13\n\nIf a knapsack of capacity 22 units of weight is available and it is allowed to take either the item completely or leave it, the maximum possible profit using the dynamic approach will be _____.\n\n#### Dynamic Programming Question 4 Detailed Solution\n\nThis is a case of 0-1 knapsack because we have to either take the item or completely leave it, we cannot take partial items here.\nInitially, we have 22 units weight of knapsack.\n\nNow calculate the value by weight ratio and sort it in decreasing order.\n\n Item Weight Value Value/Weight 1 4 44 11 3 5 38 7.6 5 7 40 5.7 4 2 11 5.5 7 3 13 4.3 6 10 37 3.7 2 7 21 3\n\nEach stage of the algorithm is described below:\n\n1. In the value column, 11 is the maximum profit ratio, So, we have to take 4kg weight for that. Capacity left=22-4=18 unit. Total profit=44\n2. Next highest ratio is 7.6, So, by taking 5 kg weight for that. Capacity left=18-5=13 unit left. Total profit=44+38=82\n3. Next highest ratio is 5.7, So, by taking 7 kg weight for that. Capacity left=13-7=6 unit. Total profit=82+40=122\n4. Next highest ratio is 5.5, So, by taking 2 kg weight, for that. Capacity left=6-2=4 unit. Total profit=122+11=133\n5. Next highest ratio is 4.3, So, by taking 3 kg weight, for that. Capacity left=4-3=1 unit. Total profit=133+13=146\n6. Now, we have 1 unit of weight space left, but none of the items from 6 and 2 can fit this empty space, so the optimal solution is with 21 units with 146 profit value.\n\nSo, 146 will be the maximum possible profit.\n\n#### Dynamic Programming Question 5:\n\nThe time complexity of an efficient algorithm to find the longest monotonically increasing subsequence of n numbers is\n\n1. O(n)\n2. O(n Ig n)\n3. O(n2)\n4. O(log n)\n5. None of the above\n\nOption 2 : O(n Ig n)\n\n#### Dynamic Programming Question 5 Detailed Solution\n\nThe correct answer is option 2.\n\nConcept:\n\nThe Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order.\n\nAlgorithm:\n\nAlgorithm LongestIncreasingSubsequenceLength( vector& v)\n\n{\nif (v.size() == 0)\nreturn 0;\n\nvector tail(v.size(), 0);\nint length = 1; // always points empty slot in tail\n\ntail = v;\nfor (size_t i = 1; i < v.size(); i++)\n\n{\n\n// new smallest value\nif (v[i] < tail)\ntail = v[i];\n\n// v[i] extends largest subsequence\nelse if (v[i] > tail[length - 1])\ntail[length++] = v[i];\n\n// v[i] will become end candidate of an existing subsequence or Throw away larger elements in all LIS, to make room for upcoming greater elements than v[i] and also, v[i] would have already appeared in one of LIS, identify the location and replace it.\nelse\ntail[CeilIndex(tail, -1, length - 1, v[i])] = v[i];\n\n//uses the binary search (note boundaries in the caller)\n\n}\n\nreturn length;\n}\n\nAnalysis:\n\nTime Complexity:\n\nThe loop runs for N elements. In the worst case, we may end up querying ceil value using binary search (log i) for many A[i]. Therefore, T(n) < O( log N! ) = O(N log N). Analyze to ensure that the upper and lower bounds are also O( N log N ). The complexity is θ(N log N).\n\nHence the correct answer is O(n Ig n).\n\n## Top Dynamic Programming MCQ Objective Questions\n\n#### Dynamic Programming Question 6\n\nMatch the following with respect to algorithm paradigms:\n\n List - I List - II (a) The 8-Queen’s problem (i) Dynamic programming (b) Single-Source shortest paths (ii) Divide and conquer (c) STRASSEN’s Matrix multiplication (iii) Greedy approach (d) Optimal binary search trees (iv) Backtracking\n\n1. (a) - (iv), (b) - (i), (c) - (iii), (d) - (ii)\n2. (a) - (iv), (b) - (iii), (c) - (i), (d) - (ii)\n3. (a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)\n4. (a) - (iv), (b) - (iii), (c) - (ii), (d) - (i)\n\nOption 4 : (a) - (iv), (b) - (iii), (c) - (ii), (d) - (i)\n\n#### Dynamic Programming Question 6 Detailed Solution\n\n8 - queen’s problem:\n\n• In this, 8 queens are placed on a 8 × 8 chessboard such that none of them can take same place.\n• It is the backtracking approach. In backtracking approach, all the possible configurations are checked and check whether result is obtained or not.\n• A queen can only be attacked if it is in the same row or column or diagonal of another queen.\n\nSingle source shortest path:\n\n• It is used to find the shortest path starting from a source to the all other vertices.\n• It is based on greedy approach.\n• Example of single source shortest path algorithm is Dijkstra’s algorithm which initially consider all the distance to infinity and distance to source is 0.\n\nStrassen’s matrix multiplication:\n\n• It is a method used for matrix multiplication. It uses divide and conquer approach to multiply the matrices.\n• Time consumption is improved by using Strassen’s method as compared to standard multiplication method.\n• It is faster than standard method. Strassen’s multiplication can be performed only on square matrices.\n\nOptimal binary search tree:\n\n• It is also known as weight balanced binary tree. In optimal binary search tree, dummy key is added in the tree by first constructing a binary search tree.\n• It is based on dynamic programming approach. Total cost must be as small as possible.\n\n#### Dynamic Programming Question 7\n\nDefine Rn to be the maximum amount earned by cutting a rod of length n meters into one or more pieces of integer length and selling them. For i > 0, let p[i] denotes the selling price of a rod whose length is i meters. Consider the array of prices:\n\np = 1, p = 5, p = 8, p = 9, p = 10, p = 17, p = 18\n\nWhich of the following statements is/are correct about R7?\n\n1. R7 cannot be achieved by a solution consisting of three pieces.\n2. R7 = 19\n3. R7 = 18\n4. R7 is achieved by three different solutions.\n\nOption :\n\n#### Dynamic Programming Question 7 Detailed Solution\n\nAnswer: Option 3 and Option 4\n\nData:\n\np = 1, p = 5, p = 8,\n\np = 9, p = 10, p = 17, p = 18\n\nCalculation\n\nR7= max amount and length can be integer\n\n Pieces Amount 1, 1, 1, 1, 1, 1, 1 7 1, 1, 1, 1, 1, 2 10 1, 1, 1, 1, 3 12 1, 1, 1, 4 12 1, 1, 5 12 1, 6 18 1, 1, 2, 3 15 1, 3, 3 17 1,2,4 15 2, 2, 3 18 7 18\n\nmaximum cost possible is 18 and there are 3 solutions for R7.\n\n#### Dynamic Programming Question 8\n\nLet A1, A2, A3, and A4 be four matrices of dimensions 10 × 5, 5 × 20, 20 × 10, and 10 × 5, respectively. The minimum number of scalar multiplications required to find the product A1A2A3A4 using the basic matrix multiplication method is ______.\n\n#### Dynamic Programming Question 8 Detailed Solution\n\nConcept:\n\nIf we multiply two matrices of order l × m and m × n,\n\nthen number of scalar multiplications required = l × m × n\n\nData:\n\nA1 = 10 × 5, A2 = 5× 20, A3 = 20 × 10, A4 = 10 × 5\n\nCalculation:\n\nThere are 5 ways in which we can multiply these 4 matrices.\n\n(A1A2)(A3A4) , A1A2(A3A4), A1 ((A2A3) A4) , (A1(A2A3))A4, ((A1A2)A3)A4\n\nMinimum number of scalar multiplication can be find out using A1 ((A2A3)A4)\n\nFor A2A3 (order will become 5 ×10) = 5 × 20 × 10 = 1000\n\nFor (A2A3)A4 (order will become 5 × 5) = 5 × 10 × 5 = 250\n\nFor A1 ((A2A3)A4) [Order will become 10 × 5] = 10× 5 × 5 = 250\n\nMinimum number of scalar multiplication required = 1000 + 250 + 250 = 1500\n\nIn all other cases, scalar multiplication are more than 1500.\n\n#### Dynamic Programming Question 9\n\nAssume that multiplying a matrix G1 of dimension 𝑝 × 𝑞 with another matrix G2 of dimension 𝑞 × 𝑟 requires 𝑝𝑞𝑟 scalar multiplications. Computing the product of n matrices G1G2G3…Gn can be done by parenthesizing in different ways. Define Gi Gi+1 as an explicitly computed pair for a given paranthesization if they are directly multiplied. For example, in the matrix multiplication chain G1G2G3G4G5G6 using parenthesization (G1(G2G3))(G4(G5G6)), G2G3 and G5G6 are the only explicitly computed pairs.Consider a matrix multiplication chain F1F2F3F4F5, where matrices F1, F2, F3, F4 and F5 are of dimensions 2 × 25, 25 × 3, 3 × 16, 16 × 1 and 1 × 1000, respectively. In the parenthesization of F1F2F3F4F5 that minimizes the total number of scalar multiplications, the explicitly computed pairs is/are\n\n1. F1F2 and F3F4 only\n2. F2F3 only\n3. F3F4 only\n4. F1F2 and F4F5 only\n\nOption 3 : F3F4 only\n\n#### Dynamic Programming Question 9 Detailed Solution\n\nAs F5 is 11000 matrix so 1000 will play vital role in cost. So it is good to multiply F5 at very last step.So, the sequence giving minimal cost: (((F1(F2(F3F4))(F5)) = 48+75+50+2000 = 2173\n\nExplicitly computed pairs is (F3F4).\n\n#### Dynamic Programming Question 10\n\nWhich of the following is a correct time complexity to solve the 0/1 knapsack problem where n and w represents the number of items and capacity of knapsack respectively?\n\n1. O(n)\n2. O(w)\n3. O(nw)\n4. O(n+w)\n\nOption 3 : O(nw)\n\n#### Dynamic Programming Question 10 Detailed Solution\n\nKnapsack problem has the following two variants-\n\n1. Fractional Knapsack Problem\n2. 0/1 Knapsack Problem\n\nTime Complexity-\n\n• Each entry of the table requires constant time θ(1) for its computation.\n• It takes θ(nw) time to fill (n+1)(w+1) entries. Therefore, O(nw + n + w +1) = O(nw )\n• It takes θ(n) time for tracing the solution since the tracing process traces the n rows.\n• Thus, overall θ(nw) time is taken to solve 0/1 knapsack problem using dynamic programming.\n\n#### Dynamic Programming Question 11\n\nAssembly line scheduling and Longest Common Subsequence problems are an example of __________\n\n1. Dynamic Programming\n2. Greedy Algorithms\n3. Greedy Algorithms and Dynamic Programming respectively\n4. Dynamic Programming and Branch and Bound respectively\n\nOption 1 : Dynamic Programming\n\n#### Dynamic Programming Question 11 Detailed Solution\n\nThe longest common subsequence(LCM)\n\nLCS problem is the problem of finding the longest subsequence common to all sequences in a set of sequences.\n\nLongest Common Subsequence problems is an example of Dynamic Programming.\n\nIn LCS:\n\nIf there is match, A[i, j] = A[i – 1, j – 1] + 1\n\nIf not match: max(A[i – 1, j], A[i, j – 1])\n\nAssembly line scheduling\n\nThe main goal of assembly line scheduling is to give the best route or can say fastest from all assembly line.\n\nAssembly line schedulingproblems is an example of Dynamic Programming.\n\n#### Dynamic Programming Question 12\n\nConsider product of three matrices M1, M2 and M3 having w rows and x columns, x rows and y columns, and y rows and z columns. Under what condition will it take less time to compute the product as (M1M2)M3 than to compute M1 (M2M3)?\n\n1. Always take the same time\n2. (1/x + 1/z) < (1/w + 1/y)\n3. x > y\n4. (w + x) > (y + z)\n\nOption 2 : (1/x + 1/z) < (1/w + 1/y)\n\n#### Dynamic Programming Question 12 Detailed Solution\n\nConcept:\n\nOrder of M1 = w × x\n\nOrder of M2 = x × y\n\nOrder of M3 = y × z\n\nFor cost of (M1M2)M3 = wxy + wyz. Detailed steps below.\n\n• Cost of M1M2 = w × x × y\n• This gives us a new matrix. Let the new matrix be M.\n• Order of M is w × y\n• Cost of MM3 = w × y × z\n• Total cost = M1M2 cost + MM3 cost\n\nSimilarly, cost of M1 (M2M3) = xyz + wxz\n\nFor (M1M2)M3 to take less time than M1(M2M3)\n\n(wxy + wyz) < (xyz + wxz)\n\nDividing both the sides of above equation, we get\n\n(1/x + 1/z) < (1/w + 1/y)\n\n#### Dynamic Programming Question 13\n\nConsider the following two sequences :\n\nX = < B, C, D, C, A, B, C >\n\nand Y = < C, A, D, B, C, B >\n\nThe length of longest common subsequence of X and Y is :\n\n1. 5\n2. 3\n3. 4\n4. 2\n\nOption 3 : 4\n\n#### Dynamic Programming Question 13 Detailed Solution\n\nConcept:\n\nIf there is match, A[i, j] = A[i – 1, j – 1] + 1\n\nIf not match: max(A[i – 1, j], A[i, j – 1])\n\nCALCULATION\n\nLet M = length of X and N = length of Y\n\nint dp[N+1][M+1]\n\nTable of longest common subsequence:\n\n X → B C D C A B C Y ↓ 0 0 0 0 0 0 0 0 C 0 0 1 1 1 1 1 1 A 0 0 1 1 1 2 2 2 D 0 0 1 2 2 2 2 2 B 0 1 1 2 2 2 3 3 C 0 1 2 2 3 3 3 3 B 0 1 2 2 3 3 4 4\n\nThe length of longest common subsequence of X and Y = A[N][M] = 4\n\n#### Dynamic Programming Question 14\n\nThe following paradigm can be used to find the solution of the problem in minimum time:\n\nGiven a set of non-negative integer, and a value K, determine if there is a subset of the given set with sum equal to K:\n\n1. Divide and Conquer\n2. Dynamic Programming\n3. Greedy Algorithm\n4. branch and Bound\n\nOption 2 : Dynamic Programming\n\n#### Dynamic Programming Question 14 Detailed Solution\n\nConcept:\n\n“Subset sum problem”:\n\nGiven a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum(K).\n\nExplanation:\n\nAssume array of non-negative integer of size n. This problem can u solved by -\n\nMethod 1: Using recursion\n\n• Time complexity - O(2n)\n• Space complexity - O(n)\n\nMethod 1: Using Dynamic Programming (Optimal solution of “Subset sum problem”)\n\n• Time complexity - O(n.k)\n• Space complexity - O(n.k)\n\nDynamic Programming paradigm can be used to find the solution of the problem in minimum time.\n\nSubset sum problem is NP Complete Problem, because 0/1 Knapsack problem polynomial reduced to sum of\n\nsubset problem and 0/1 Knapsack problem is NP complete problem.\n\n#### Dynamic Programming Question 15\n\nA Young tableau is a 2D array of integers increasing from left to right and from top to bottom. Any unfilled entries are marked with ∞, and hence there cannot be any entry to the right of, or below a ∞. The following Young tableau consists of unique entries.\n\n 1 2 5 14 3 4 6 23 10 12 18 25 31 ∞ ∞ ∞\n\nWhen an element is removed from a Young tableau, other elements should be moved into its place so that the resulting table is still a Young tableau (unfilled entries may be filled in with a ∞). The minimum number of entries (other than 1) to be shifted, to remove 1 from the given Young tableau is ________.\n\n#### Dynamic Programming Question 15 Detailed Solution\n\nAs it is given here, that there can’t be any entry to the right of or below a ∞\n\nSo, when a 1 is removed, it is replaced by its right-side position value i.e. 2 keeping the 5 and 14 intact.\n\n 2 5 14 3 4 6 23 10 12 18 25 31 ∞ ∞ ∞\n\nNow we replace the vacant position by 4 below it is keeping other numbers at same position.\n\n 2 4 5 14 3 6 23 10 12 18 25 31 ∞ ∞ ∞\n\nNow, fill the vacant position by shifting 6 to its left.\n\n 2 4 5 14 3 6 23 10 12 18 25 31 ∞ ∞ ∞\n\nNow, similarly shift 18 to up to fill vacant position keeping other number at same position.\n\n 2 4 5 14 3 6 18 23 10 12 25 31 ∞ ∞ ∞\n\nNow, shift 25 to it’s left\n\n 2 4 5 14 3 6 18 23 10 12 25 31 ∞ ∞ ∞\n\nNow, fill this vacant position with ∞.\n\n 2 4 5 14 3 6 18 23 10 12 25 ∞. 31 ∞ ∞ ∞\n\nSo, in this way minimum entries that needs to be shifted after removal of 1 from table are 5." ]	[ null, "https://storage.googleapis.com/tb-img/production/22/09/F1_Vinanti_Engineering_22.09.22_D7.png", null, "https://storage.googleapis.com/tb-img/production/22/09/F1_Vinanti_Engineering_22.09.22_D8.png", null, "https://storage.googleapis.com/tb-img/production/22/09/F1_Vinanti_Engineering_22.09.22_D9.png", null, "https://storage.googleapis.com/tb-img/production/22/09/F1_Vinanti_Engineering_22.09.22_D10.png", null, "https://storage.googleapis.com/tb-img/production/22/09/F1_Vinanti_Engineering_22.09.22_D11.png", null, "https://storage.googleapis.com/tb-img/production/22/09/F1_Vinanti_Engineering_22.09.22_D12.png", null, "https://storage.googleapis.com/tb-img/production/22/09/F1_Vinanti_Engineering_22.09.22_D13.png", null, "https://storage.googleapis.com/tb-img/production/22/09/F1_Vinanti_Engineering_22.09.22_D14.png", null, "https://storage.googleapis.com/tb-img/production/23/04/F1_Madhuri_Engineering_17.04.2023_D12.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.78524476,"math_prob":0.99647975,"size":20758,"snap":"2023-40-2023-50","text_gpt3_token_len":6871,"char_repetition_ratio":0.13607016,"word_repetition_ratio":0.1611411,"special_character_ratio":0.3652086,"punctuation_ratio":0.110371076,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994413,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T08:50:51Z\",\"WARC-Record-ID\":\"<urn:uuid:b492c6a3-2b68-4d0a-97c5-d94559b52874>\",\"Content-Length\":\"588438\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:00f2206d-67f8-4ec9-8e39-925393b4cade>\",\"WARC-Concurrent-To\":\"<urn:uuid:8b38f7e9-d8c5-483e-8ad5-020ee9d92394>\",\"WARC-IP-Address\":\"104.22.44.238\",\"WARC-Target-URI\":\"https://testbook.com/objective-questions/mcq-on-dynamic-programming--5eea6a0c39140f30f369e0db\",\"WARC-Payload-Digest\":\"sha1:LENPXAEXSFPT76K2YDGMJYX5SIXNJF54\",\"WARC-Block-Digest\":\"sha1:PDX7A2XHBYY4XQBH3LBM2ZIBHGWEB6OW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100873.6_warc_CC-MAIN-20231209071722-20231209101722-00693.warc.gz\"}"}
https://codeclimate.com/github/poliastro/poliastro/src/poliastro/constants/general.py/source	[ "# poliastro/poliastro\n\nsrc/poliastro/constants/general.py\n\n### Summary\n\nB\n7 hrs\n###### Test Coverage\n``````\"\"\"Astronomical and physics constants.\n\nThis module complements constants defined in `astropy.constants`,\n\nNote that `GM_jupiter` and `GM_neptune` are both referred to the whole planetary system gravitational parameter.\n\nUnless otherwise specified, gravitational and mass parameters were obtained from:\n\n* Luzum, Brian et al. “The IAU 2009 System of Astronomical Constants: The Report of the IAU Working Group on Numerical\nStandards for Fundamental Astronomy.” Celestial Mechanics and Dynamical Astronomy 110.4 (2011): 293–304.\nCrossref. Web. `DOI: 10.1007/s10569-011-9352-4`_\n\n* Archinal, B. A. et al. “Report of the IAU Working Group on Cartographic Coordinates and Rotational Elements: 2009.”\nCelestial Mechanics and Dynamical Astronomy 109.2 (2010): 101–135. Crossref. Web. `DOI: 10.1007/s10569-010-9320-4`_\n\n.. _`DOI: 10.1007/s10569-011-9352-4`: http://dx.doi.org/10.1007/s10569-011-9352-4\n.. _`DOI: 10.1007/s10569-010-9320-4`: http://dx.doi.org/10.1007/s10569-010-9320-4\n\nJ2 for the Sun was obtained from:\n\n* https://hal.archives-ouvertes.fr/hal-00433235/document (New values of gravitational moments J2 and J4 deduced\nfrom helioseismology, Redouane Mecheri et al)\n\n\"\"\"\n\nfrom astropy import time\nfrom astropy.constants import Constant\nfrom astropy.constants.iau2015 import (\nM_earth as _M_earth,\nM_jup as _M_jupiter,\nM_sun as _M_sun,\n)\n\n__all__ = [\n\"J2000\",\n\"J2000_TDB\",\n\"J2000_TT\",\n\"GM_sun\",\n\"GM_earth\",\n\"GM_mercury\",\n\"GM_venus\",\n\"GM_mars\",\n\"GM_jupiter\",\n\"GM_saturn\",\n\"GM_uranus\",\n\"GM_neptune\",\n\"GM_pluto\",\n\"GM_moon\",\n\"M_earth\",\n\"M_jupiter\",\n\"M_sun\",\n\"R_mean_earth\",\n\"R_mean_mercury\",\n\"R_mean_venus\",\n\"R_mean_mars\",\n\"R_mean_jupiter\",\n\"R_mean_saturn\",\n\"R_mean_uranus\",\n\"R_mean_neptune\",\n\"R_mean_pluto\",\n\"R_mean_moon\",\n\"R_earth\",\n\"R_mercury\",\n\"R_venus\",\n\"R_mars\",\n\"R_jupiter\",\n\"R_saturn\",\n\"R_sun\",\n\"R_uranus\",\n\"R_neptune\",\n\"R_pluto\",\n\"R_moon\",\n\"R_polar_earth\",\n\"R_polar_mercury\",\n\"R_polar_venus\",\n\"R_polar_mars\",\n\"R_polar_jupiter\",\n\"R_polar_saturn\",\n\"R_polar_uranus\",\n\"R_polar_neptune\",\n\"R_polar_pluto\",\n\"R_polar_moon\",\n\"J2_sun\",\n\"J2_earth\",\n\"J3_earth\",\n\"J2_mars\",\n\"J3_mars\",\n\"J2_venus\",\n\"J3_venus\",\n\"H0_earth\",\n\"rho0_earth\",\n\"Wdivc_sun\",\n]\n\n# HACK: sphinx-autoapi variable definition\nM_earth = _M_earth\nM_jupiter = _M_jupiter\nM_sun = _M_sun\n\n# See for example USNO Circular 179\nJ2000_TT = time.Time(\"J2000\", scale=\"tt\")\nJ2000_TDB = time.Time(\"J2000\", scale=\"tdb\")\nJ2000 = J2000_TT\n\nGM_sun = Constant(\n\"GM_sun\",\n\"Heliocentric gravitational constant\",\n1.32712442099e20,\n\"m3 / (s2)\",\n0.0000000001e20,\n\"IAU 2009 system of astronomical constants\",\nsystem=\"si\",\n)\n\nGM_earth = Constant(\n\"GM_earth\",\n\"Geocentric gravitational constant\",\n3.986004418e14,\n\"m3 / (s2)\",\n0.000000008e14,\n\"IAU 2009 system of astronomical constants\",\nsystem=\"si\",\n)\n\n# Anderson, John D. et al. “The Mass, Gravity Field, and Ephemeris of Mercury.” Icarus 71.3 (1987): 337–349.\n# Crossref. Web. DOI: 10.1016/0019-1035(87)90033-9\nGM_mercury = Constant(\n\"GM_mercury\",\n\"Mercury gravitational constant\",\n2.203209e13,\n\"m3 / (s2)\",\n0.91,\n\"IAU 2009 system of astronomical constants\",\nsystem=\"si\",\n)\n\n# Konopliv, A.S., W.B. Banerdt, and W.L. Sjogren. “Venus Gravity: 180th Degree and Order Model.”\n# Icarus 139.1 (1999): 3–18. Crossref. Web. DOI: 10.1006/icar.1999.6086\nGM_venus = Constant(\n\"GM_venus\",\n\"Venus gravitational constant\",\n3.24858592e14,\n\"m3 / (s2)\",\n0.006,\n\"IAU 2009 system of astronomical constants\",\nsystem=\"si\",\n)\n\n# Konopliv, Alex S. et al. “A Global Solution for the Mars Static and Seasonal Gravity, Mars Orientation, Phobos and\n# Deimos Masses, and Mars Ephemeris.” Icarus 182.1 (2006): 23–50.\n# Crossref. Web. DOI: 10.1016/j.icarus.2005.12.025\nGM_mars = Constant(\n\"GM_mars\",\n\"Mars gravitational constant\",\n4.282837440e13,\n\"m3 / (s2)\",\n0.00028,\n\"IAU 2009 system of astronomical constants\",\nsystem=\"si\",\n)\n\n# Jacobson, R. A. et al. “A comprehensive orbit reconstruction for the galileo prime mission in the JS200 system.”\n# The Journal of the Astronautical Sciences 48.4 (2000): 495–516.\n# Crossref. Web.\nGM_jupiter = Constant(\n\"GM_jupiter\",\n\"Jovian system gravitational constant\",\n1.2671276253e17,\n\"m3 / (s2)\",\n2.00,\n\"IAU 2009 system of astronomical constants\",\nsystem=\"si\",\n)\n\n# Jacobson, R. A. et al. “The Gravity Field of the Saturnian System from Satellite Observations and Spacecraft\n# Tracking Data.” The Astronomical Journal 132.6 (2006): 2520–2526.\n# Crossref. Web. DOI: 10.1086/508812\nGM_saturn = Constant(\n\"GM_saturn\",\n\"Saturn gravitational constant\",\n3.79312077e16,\n\"m3 / (s2)\",\n1.1,\n\"IAU 2009 system of astronomical constants\",\nsystem=\"si\",\n)\n\n# Jacobson, R. A. et al. “The Masses of Uranus and Its Major Satellites from Voyager Tracking Data and Earth-Based\n# Uranian Satellite Data.” The Astronomical Journal 103 (1992): 2068.\n# Crossref. Web. DOI: 10.1086/116211\nGM_uranus = Constant(\n\"GM_uranus\",\n\"Uranus gravitational constant\",\n5.7939393e15,\n\"m3 / (s2)\",\n13.0,\n\"IAU 2009 system of astronomical constants\",\nsystem=\"si\",\n)\n\n# Jacobson, R. A. “THE ORBITS OF THE NEPTUNIAN SATELLITES AND THE ORIENTATION OF THE POLE OF NEPTUNE.”\n# The Astronomical Journal 137.5 (2009): 4322–4329. Crossref. Web. DOI:\n# 10.1088/0004-6256/137/5/4322\nGM_neptune = Constant(\n\"GM_neptune\",\n\"Neptunian system gravitational constant\",\n6.836527100580397e15,\n\"m3 / (s2)\",\n10.0,\n\"IAU 2009 system of astronomical constants\",\nsystem=\"si\",\n)\n\n# Tholen, David J. et al. “MASSES OF NIX AND HYDRA.” The Astronomical Journal 135.3 (2008): 777–784. Crossref. Web.\n# DOI: 10.1088/0004-6256/135/3/777\nGM_pluto = Constant(\n\"GM_pluto\",\n\"Pluto gravitational constant\",\n8.703e11,\n\"m3 / (s2)\",\n3.7,\n\"IAU 2009 system of astronomical constants\",\nsystem=\"si\",\n)\n\n# Lemoine, Frank G. et al. “High-Degree Gravity Models from GRAIL Primary Mission Data.”\n# Journal of Geophysical Research: Planets 118.8 (2013): 1676–1698.\n# Crossref. Web. DOI: 10.1002/jgre.20118\nGM_moon = Constant(\n\"GM_moon\",\n\"Moon gravitational constant\",\n4.90279981e12,\n\"m3 / (s2)\",\n0.00000774,\n\"Journal of Geophysical Research: Planets 118.8 (2013)\",\nsystem=\"si\",\n)\n\n# Archinal, B. A., Acton, C. H., A’Hearn, M. F., Conrad, A., Consolmagno,\n# G. J., Duxbury, T., … Williams, I. P. (2018). Report of the IAU Working\n# Group on Cartographic Coordinates and Rotational Elements: 2015. Celestial\n# Mechanics and Dynamical Astronomy, 130(3). doi:10.1007/s10569-017-9805-5\n\nR_mean_earth = Constant(\n\"R_mean_earth\",\n6.3710084e6,\n\"m\",\n0.1,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_mean_mercury = Constant(\n\"R_mean_mercury\",\n2.4394e6,\n\"m\",\n100,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_mean_venus = Constant(\n\"R_mean_venus\",\n6.0518e6,\n\"m\",\n1000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_mean_mars = Constant(\n\"R_mean_mars\",\n3.38950e6,\n\"m\",\n2000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_mean_jupiter = Constant(\n\"R_mean_jupiter\",\n6.9911e7,\n\"m\",\n6000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2009\",\nsystem=\"si\",\n)\n\nR_mean_saturn = Constant(\n\"R_mean_saturn\",\n5.8232e7,\n\"m\",\n6000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_mean_uranus = Constant(\n\"R_mean_uranus\",\n2.5362e7,\n\"m\",\n7000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_mean_neptune = Constant(\n\"R_mean_neptune\",\n2.4622e7,\n\"m\",\n19000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_mean_pluto = Constant(\n\"R_mean_pluto\",\n1.188e6,\n\"m\",\n1600,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_mean_moon = Constant(\n\"R_mean_moon\",\n1.7374e6,\n\"m\",\n0,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_sun = Constant(\n\"R_sun\",\n6.95700e8,\n\"m\",\n0,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_earth = Constant(\n\"R_earth\",\n6.3781366e6,\n\"m\",\n0.1,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_mercury = Constant(\n\"R_mercury\",\n2.44053e6,\n\"m\",\n40,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_venus = Constant(\n\"R_venus\",\n6.0518e6,\n\"m\",\n1000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_mars = Constant(\n\"R_mars\",\n3.39619e6,\n\"m\",\n100,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_jupiter = Constant(\n\"R_jupiter\",\n7.1492e7,\n\"m\",\n4000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2009\",\nsystem=\"si\",\n)\n\nR_saturn = Constant(\n\"R_saturn\",\n6.0268e7,\n\"m\",\n4000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_uranus = Constant(\n\"R_uranus\",\n2.5559e7,\n\"m\",\n4000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_neptune = Constant(\n\"R_neptune\",\n2.4764e7,\n\"m\",\n15000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_pluto = Constant(\n\"R_pluto\",\n1.1883e6,\n\"m\",\n1600,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_moon = Constant(\n\"R_moon\",\n1.7374e6,\n\"m\",\n0,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_polar_earth = Constant(\n\"R_polar_earth\",\n6.3567519e6,\n\"m\",\n0.1,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_polar_mercury = Constant(\n\"R_polar_mercury\",\n2.43826e6,\n\"m\",\n40,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_polar_venus = Constant(\n\"R_polar_venus\",\n6.0518e6,\n\"m\",\n1000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_polar_mars = Constant(\n\"R_polar_mars\",\n3.376220e6,\n\"m\",\n100,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_polar_jupiter = Constant(\n\"R_polar_jupiter\",\n6.6854e7,\n\"m\",\n10000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2009\",\nsystem=\"si\",\n)\n\nR_polar_saturn = Constant(\n\"R_polar_saturn\",\n5.4364e7,\n\"m\",\n10000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_polar_uranus = Constant(\n\"R_polar_uranus\",\n2.4973e7,\n\"m\",\n20000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_polar_neptune = Constant(\n\"R_polar_neptune\",\n2.4341e7,\n\"m\",\n30000,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_polar_pluto = Constant(\n\"R_polar_pluto\",\n1.1883e6,\n\"m\",\n1600,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nR_polar_moon = Constant(\n\"R_polar_moon\",\n1.7374e6,\n\"m\",\n0,\n\"IAU Working Group on Cartographic Coordinates and Rotational Elements: 2015\",\nsystem=\"si\",\n)\n\nJ2_sun = Constant(\n\"J2_sun\",\n\"Sun J2 oblateness coefficient\",\n2.20e-7,\n\"\",\n0.01e-7,\n\"HAL archives\",\nsystem=\"si\",\n)\n\nJ2_earth = Constant(\n\"J2_earth\",\n\"Earth J2 oblateness coefficient\",\n0.00108263,\n\"\",\n1,\n\"HAL archives\",\nsystem=\"si\",\n)\n\nJ3_earth = Constant(\n\"J3_earth\",\n\"Earth J3 asymmetry between the northern and southern hemispheres\",\n-2.5326613168e-6,\n\"\",\n1,\n\"HAL archives\",\nsystem=\"si\",\n)\n\nJ2_mars = Constant(\n\"J2_mars\",\n\"Mars J2 oblateness coefficient\",\n0.0019555,\n\"\",\n1,\n\"HAL archives\",\nsystem=\"si\",\n)\n\nJ3_mars = Constant(\n\"J3_mars\",\n\"Mars J3 asymmetry between the northern and southern hemispheres\",\n3.1450e-5,\n\"\",\n1,\n\"HAL archives\",\nsystem=\"si\",\n)\n\nJ2_venus = Constant(\n\"J2_venus\",\n\"Venus J2 oblateness coefficient\",\n4.4044e-6,\n\"\",\n1,\n\"HAL archives\",\nsystem=\"si\",\n)\n\nJ3_venus = Constant(\n\"J3_venus\",\n\"Venus J3 asymmetry between the northern and southern hemispheres\",\n-2.1082e-6,\n\"\",\n1,\n\"HAL archives\",\nsystem=\"si\",\n)\n\nH0_earth = Constant(\n\"H0_earth\",\n\"Earth H0 atmospheric scale height\",\n8_500,\n\"m\",\n1,\n\"de Pater and Lissauer 2010\",\nsystem=\"si\",\n)\n\nrho0_earth = Constant(\n\"rho0_earth\",\n\"Earth rho0 atmospheric density prefactor\",\n1.3,\n\"kg / (m3)\",\n1,\n\"de Pater and Lissauer 2010\",\nsystem=\"si\",\n)\n\nWdivc_sun = Constant(\n\"Wdivc_sun\",\n\"total radiation power of Sun divided by the speed of light\",\n1.0203759306204136e14,\n\"kg km / (s2)\",\n1,\n\"Howard Curtis\",\nsystem=\"si\",\n)``````" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.64152265,"math_prob":0.88697076,"size":13253,"snap":"2020-34-2020-40","text_gpt3_token_len":4418,"char_repetition_ratio":0.2113367,"word_repetition_ratio":0.26223564,"special_character_ratio":0.38210216,"punctuation_ratio":0.2979066,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96666104,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-05T07:45:48Z\",\"WARC-Record-ID\":\"<urn:uuid:af9fa8b2-824e-4ce7-8a92-98258f7c18ed>\",\"Content-Length\":\"30445\",\"Content-Type\":\"application/http; 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http://azfoo.net/gdt/babs/numbers/n/number51315.html	[ "### About the Number 51,315 (fifty-one thousand three hundred fifteen)\n\nThis nBAB for the number 51315 was created on the palindromic date of 5/13/15.\n\n```MathBabbler Number Analyst (MBNA) output:\n=========================================\n51315 is a natural, whole, integer\n51315 is odd\n51315 proper divisors are: 1,3,5,11,15,33,55,165,311,933,1555,\n3421,4665,10263,17105,\n51315 has 15 proper divisors\n51315 is deficient (sum of divisors is 38541; ratio: 0.751067)\n51315 is unhappy\n51315 is a Squarefree Number\n51315 is Harshad (Niven) number\n51315 is composite (not prime)\n51315 has the prime factors: 3511*311 (sum=330)\n51315 is a joke (Smith) number\n51315 is a hoax number\n51315 is palindromic\n51315 is Arithmetic (A003601)\n51315 is not in A038853\n51315 in octal is 0144163\n51315 in hexadecimal is 0xc873\n51315 in binary is 1100100001110011 (is evil)\n51315 nearest square numbers: -239...214 (51076...51529 )\nsqrt(51315) = 226.528\nln(51315) = 10.8457\nlog(51315) = 4.71024\n51315 reciprocal is .00001948747929455324953717236675\n51315 is an apocalyptic power\n51315! is inf\n51315 is 16334.1 Pi years\n51315 is 2565 score and 15 years\n51315 is 25658^2 - 25657^2 and 25658 + 25657 = 51315\n51315 is a multiple of 1 & it contains a 1 (A011531)\n51315 is a multiple of 3 & it contains a 3 (A121023)\n51315 is a multiple of 5 & it contains a 5 (A121025)\n```\n\nCreator: Gerald Thurman [[email protected]]\nCreated: 13 May 2015", null, "" ]	[ null, "http://i.creativecommons.org/l/by/3.0/us/88x31.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7858999,"math_prob":0.9915835,"size":1340,"snap":"2019-35-2019-39","text_gpt3_token_len":482,"char_repetition_ratio":0.22080839,"word_repetition_ratio":0.014563107,"special_character_ratio":0.56044775,"punctuation_ratio":0.14652015,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9837414,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-19T19:27:20Z\",\"WARC-Record-ID\":\"<urn:uuid:130fd47e-9991-48fb-bfef-ed652bfa83ee>\",\"Content-Length\":\"3620\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ad9283c4-52ef-4079-8094-dc33920aa0b7>\",\"WARC-Concurrent-To\":\"<urn:uuid:bf59699c-0780-478b-8c6e-6bf284a39b50>\",\"WARC-IP-Address\":\"70.38.112.232\",\"WARC-Target-URI\":\"http://azfoo.net/gdt/babs/numbers/n/number51315.html\",\"WARC-Payload-Digest\":\"sha1:RQOHZSRWQJVRT5XSDFX5EWAVUNHO4AB4\",\"WARC-Block-Digest\":\"sha1:EM5DL7IF3SBYRC5WQU3U2HQU7C6UL77G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573570.6_warc_CC-MAIN-20190919183843-20190919205843-00464.warc.gz\"}"}
https://admin.clutchprep.com/chemistry/practice-problems/86511/the-temperature-inside-a-pressure-cooker-is-115-c-calculate-the-vapor-pressure-o	[ "Chemistry Practice Problems Clausius-Clapeyron Equation Practice Problems Solution: The temperature inside a pressure cooker is 115 ˚C...\n\n⚠️Our tutors found the solution shown to be helpful for the problem you're searching for. We don't have the exact solution yet.\n\n# Solution: The temperature inside a pressure cooker is 115 ˚C. Calculate the vapor pressure of water inside the pressure cooker. What would be the temperature inside the pressure cooker if the vapor pressure of water was 3.50 atm?\n\n###### Problem\n\nThe temperature inside a pressure cooker is 115 ˚C. Calculate the vapor pressure of water inside the pressure cooker. What would be the temperature inside the pressure cooker if the vapor pressure of water was 3.50 atm?\n\nClausius-Clapeyron Equation\n\nClausius-Clapeyron Equation\n\n#### Q. The enthalpy of vaporization for acetone is 32.0 kJ/mol. The normal boiling point for acetone is 56.5 ˚C. What is the vapor pressure of acetone at 23....\n\nSolved • Thu Oct 18 2018 16:58:27 GMT-0400 (EDT)\n\nClausius-Clapeyron Equation\n\n#### Q. Carbon tetrachloride, CCl 4, has a vapor pressure of 213 torr at 40. ˚C and 836 torr at 80. ˚C. What is the normal boiling point of CCl4?\n\nSolved • Thu Oct 18 2018 15:40:17 GMT-0400 (EDT)\n\nClausius-Clapeyron Equation\n\n#### Q. In Breckenridge, Colorado, the typical atmospheric pressure is 520. torr. What is the boiling point of water (ΔHvap = 40.7 kJ/mol) in Breckenridge?\n\nSolved • Thu Oct 18 2018 15:38:06 GMT-0400 (EDT)\n\nClausius-Clapeyron Equation\n\n#### Q. From the following data for liquid nitric acid, determine its heat of vaporization and normal boiling point.\n\nSolved • Thu Oct 18 2018 15:37:19 GMT-0400 (EDT)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.742682,"math_prob":0.91506916,"size":487,"snap":"2020-10-2020-16","text_gpt3_token_len":126,"char_repetition_ratio":0.17184265,"word_repetition_ratio":0.12307692,"special_character_ratio":0.18069816,"punctuation_ratio":0.074074075,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98453814,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-29T10:09:53Z\",\"WARC-Record-ID\":\"<urn:uuid:58447667-c5a2-4db4-966a-9535e76e2eaa>\",\"Content-Length\":\"80086\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:72a6ddd5-0652-4c37-813b-34941242ef5c>\",\"WARC-Concurrent-To\":\"<urn:uuid:0edebe72-8d69-4b6e-a2c1-75b0aa5fbf16>\",\"WARC-IP-Address\":\"34.228.134.34\",\"WARC-Target-URI\":\"https://admin.clutchprep.com/chemistry/practice-problems/86511/the-temperature-inside-a-pressure-cooker-is-115-c-calculate-the-vapor-pressure-o\",\"WARC-Payload-Digest\":\"sha1:TCESSRU633A2BVGUZFM6P7KY2RTW2HIV\",\"WARC-Block-Digest\":\"sha1:6NFR2RMWDW5QICCIVLWWX7LERGBWG6PG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370494064.21_warc_CC-MAIN-20200329074745-20200329104745-00203.warc.gz\"}"}
https://forums.ni.com/t5/LabVIEW/Wavelet-Transformation-Advanced-signal-processing/td-p/4186926	[ "# LabVIEW\n\ncancel\nShowing results for\nDid you mean:\n\n## Wavelet Transformation- Advanced signal processing\n\nGoal: To find the defect frequencies of the bearing (rolling-element) from the \"simulated\" acceleration signal.\n\nI am trying to use \"Wavelet transform\" functions of Advanced signal processing toolkit for the defect identification.\n\nBelow shown is the acceleration signal in green (cosine of 10 Hz frequency), which has noise and defect signal (in RED plot, having 5 Hz frequency). I tried to get back the red plot with the help of Wavelet analysis, but was not possible as it is hard for me to comprehend these wavelet functions.", null, "Below is the graph showing what I have got in (a) time domain and in (b) frequency domain", null, "the above graph should get closer to the defect (Red plot shown in the first graph) signal.", null, "I have used 1. BP Filter (see below for config.) -> white Plot", null, "2. Wavelet packet analysis tool (see below for config.) -> red Plot. used db14", null, "3. Band pass filter (IIR filter with LPF- 3 Hz and HPF-7 Hz (since i know my defect frequency!))-> Green plot (PERFECT!! I need this but from WPT analysis with that fact that defect frequency is unkown, is that possible ?)\n\n4. Wavelet denoise (see below for config.) -> Blue Plot. Just used Bior3_X, since it was more closer to the defect waveform shape than daubechies.", null, "I know that to find the defect signal, I need to use a mother wavelet which is very similar to the defect waveform. But I dont know to choose that. below is my defect signal with 500Hz frequency, which is repeating at 5 Hz.", null, "I have attached the labview file (saved as html, if any trouble opening it, let me know) that I developed along with the acceleration signal dataset (with 3 Y-axes columns- Signal + noise, defect, 1st and 2nd columns superimposed) .\n\nI try to organise the question as clear as possible, but I can do more if you cannot understand my points.\n\nMessage 1 of 2\n(224 Views)\n\n## Re: Wavelet Transformation- Advanced signal processing\n\nHello Everyone,\n\nas I did not get any reply for my post, I ask doubts regard to more concrete implementation of Wavelet tools.\n\nWavelet packet analysis: How do I get wavelet denoised signal for each of the node levels. For eg., I used Wavelet: db04 with Level: 3 then there are 8 nodes possible 3.0, 3.1, 3.2, ... 3.7, then I do I get the signala for each of these 8 nodes. When I display the node coefficients (refer to my block diagram in my precious post), i get a (only one) co-effciient of -0,661829 for node \"0\". What does this number signify?\n\nAlso, is it possible to define a \"morely\" mother wavelet in labview when I know the center frequency(fc) and bandwidth (sigma)? IF so, let me know. And which tool to use to filter my signal with this wavelet ? any eg. implementation." ]	[ null, "https://forums.ni.com/t5/image/serverpage/image-id/293770i28E5199A3770869D/image-size/medium", null, "https://forums.ni.com/t5/image/serverpage/image-id/293771iA42A53C7AC9770E1/image-size/medium", null, "https://forums.ni.com/t5/image/serverpage/image-id/293773i7B4667D0C9650BF7/image-size/medium", null, "https://forums.ni.com/t5/image/serverpage/image-id/293774i48D9238E4CCDD79B/image-size/medium", null, "https://forums.ni.com/t5/image/serverpage/image-id/293775i38DE6E3364A137F4/image-size/medium", null, "https://forums.ni.com/t5/image/serverpage/image-id/293777i0C52DC683B66F38F/image-size/medium", null, "https://forums.ni.com/t5/image/serverpage/image-id/293780i85F103EEBF8BD72C/image-size/medium", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9290297,"math_prob":0.68358725,"size":2779,"snap":"2022-05-2022-21","text_gpt3_token_len":697,"char_repetition_ratio":0.123603605,"word_repetition_ratio":0.006060606,"special_character_ratio":0.2529687,"punctuation_ratio":0.1303602,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97442365,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-23T08:19:38Z\",\"WARC-Record-ID\":\"<urn:uuid:10e87402-e2ca-4a8a-bb07-af6f4560d70c>\",\"Content-Length\":\"272409\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:01bfda15-f417-48de-ac09-c4ef2206c2f2>\",\"WARC-Concurrent-To\":\"<urn:uuid:cdf054d9-5633-4fe6-bdb1-9d7752803e84>\",\"WARC-IP-Address\":\"104.16.35.15\",\"WARC-Target-URI\":\"https://forums.ni.com/t5/LabVIEW/Wavelet-Transformation-Advanced-signal-processing/td-p/4186926\",\"WARC-Payload-Digest\":\"sha1:J66DILABYP3UBPXZJS567CQWCNFU5MO2\",\"WARC-Block-Digest\":\"sha1:SDVAWQB4GXBEQE7K7CYLRGOAFLQG6AJ4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662556725.76_warc_CC-MAIN-20220523071517-20220523101517-00773.warc.gz\"}"}
https://hackeradam.com/post/bubble-sort/	[ "Bubble sort is the easiest possible sorting algorithm you could imagine. It works by simply iterating over an array of data and comparing adjacent items. If those items are in the wrong order, then they are swapped. In doing this, the largest items “bubble up” to the correct position in the array (assuming ascending order, anyway).\n\nWhile this is great from a simplicity standpoint, it’s a pretty awful solution in terms of efficiency. The worst-case performance of bubble sort is O(n2). The best-case turns out to be O(n).\n\nAs you can see, this algorithm is not suitable for any sufficiently large datasets.\n\n## Implementation\n\nThis algorithm is so simple that I think we can just jump right into the implementation without worrying about too much more explanation.\n\nHere’s a bubble sort implementation in Java:\n\n 1package com.hackeradam.sorting;\n2public class Main {\n3 // Performs a bubble sort on a int array\n4 public static void bubblesort(int arr[]) {\n5 int size = arr.length;\n6 int tmp;\n7 // This boolean flag is a small optimization\n8 // that how long the loop needs to run in some\n9 // situations\n10 boolean swapped = false;\n11 for (int i = 0; i < size - 1; i++) {\n12 swapped = false;\n13 for (int j = 0; j < size - i - 1; j++) {\n14 if (arr[j] > arr[j+1]) {\n15 // Swap the items\n16 tmp = arr[j];\n17 arr[j] = arr[j + 1];\n18 arr[j + 1] = tmp;\n19 swapped = true;\n20 }\n21 }\n22 // If we didn't need to do a swap in that inner loop\n23 // then we must be sorted, so break\n24 if (!swapped)\n25 break;\n26 }\n27 }\n28 public static void printArray(int arr[]) {\n29 for (int i=0; i < arr.length; i++) {\n30 System.out.print(arr[i] + \" \");\n31 }\n32 System.out.println();\n33 }\n34 public static void main(String... args) {\n35 int items[] = {5, 2, 88, 100, 3, 0, -1, 20, 30, 7, 99};\n36 printArray(items);\n37 System.out.println(\"After bubble sort: \");\n38 bubblesort(items);\n39 printArray(items);\n40 }\n41}" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6882553,"math_prob":0.96234924,"size":1904,"snap":"2022-40-2023-06","text_gpt3_token_len":506,"char_repetition_ratio":0.1,"word_repetition_ratio":0.0,"special_character_ratio":0.34033614,"punctuation_ratio":0.15776698,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96511894,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-01T23:16:38Z\",\"WARC-Record-ID\":\"<urn:uuid:f19c981f-2975-48b3-9fd4-2c8318fc782f>\",\"Content-Length\":\"17193\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:712fce01-2a05-491c-bd62-a980898fb29d>\",\"WARC-Concurrent-To\":\"<urn:uuid:e47f5416-8c30-4b2e-9e38-918487fc971f>\",\"WARC-IP-Address\":\"104.21.63.22\",\"WARC-Target-URI\":\"https://hackeradam.com/post/bubble-sort/\",\"WARC-Payload-Digest\":\"sha1:UMMR3J4APVKH2WG5OWF644LNZEXGMUCV\",\"WARC-Block-Digest\":\"sha1:OM66S4HIKDR6W5O23ZX7F3A6SXZPQ6ZK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030336978.73_warc_CC-MAIN-20221001230322-20221002020322-00496.warc.gz\"}"}
https://kodey.co.uk/2020/07/30/normal-distribution-and-the-empirical-rule/	[ "# Normal Distribution And The Empirical Rule\n\nA normal distribution looks a little bit like the below. It’s where the mean, median and mode are on top of one another.\n\nRemember in a previous article, we discussed Chebyshev’s theorem, which gave us a guideline for what percentage of datapoints fell between two intervals? Well, the empirical rule is the same, but better. It gives us a much more accurate approximation of the percentage of data that falls between certain intervals – but, unlike Chebyshev’s theorem, only works for normally distributed data.\n\nHere’s how they compare:\n\nIf we look at the empirical rule, it states the approximate amount of data that will fall between our sigma intervals. The below chart shows this graphically.\n\nSo, we simply take the mean and +/- sigma to define our upper and lower boundaries.\n\n##### 1 comment\n\nPrevious Article", null, "## Standard Normal Distributions Understood\n\nNext Article", null, "## Measures Of Variance\n\n##### Related Posts", null, "## Standard Normal Distributions Understood", null, "## A Guide To Basic Linear Algebra Notation For Machine Learning", null, "## Z Scores & Probability", null, "" ]	[ null, "https://kodey.co.uk/wp-content/uploads/2020/09/pexels-elevate-1267338.jpg", null, "https://kodey.co.uk/wp-content/uploads/2020/09/pexels-daniel-reche-1556707.jpg", null, "https://kodey.co.uk/wp-content/uploads/2020/09/pexels-elevate-1267338.jpg", null, "https://kodey.co.uk/wp-content/uploads/2020/12/pexels-rfstudio-3825462-260x195.jpg", null, "https://kodey.co.uk/wp-content/uploads/2020/09/pexels-matthias-groeneveld-4200740.jpg", null, "https://kodey.co.uk/wp-content/uploads/2020/09/pexels-karolina-grabowska-4021883.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9270274,"math_prob":0.77556133,"size":810,"snap":"2022-27-2022-33","text_gpt3_token_len":168,"char_repetition_ratio":0.094292805,"word_repetition_ratio":0.0,"special_character_ratio":0.20123456,"punctuation_ratio":0.12101911,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9719068,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,5,null,5,null,5,null,7,null,7,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-26T04:12:37Z\",\"WARC-Record-ID\":\"<urn:uuid:6efef3eb-5d73-4004-abcb-f31416de8461>\",\"Content-Length\":\"149206\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d9e641a4-8efa-452d-82e6-ee3fb300c2b5>\",\"WARC-Concurrent-To\":\"<urn:uuid:68792e44-fda4-45c0-bbc8-85d2d65a1530>\",\"WARC-IP-Address\":\"35.214.29.137\",\"WARC-Target-URI\":\"https://kodey.co.uk/2020/07/30/normal-distribution-and-the-empirical-rule/\",\"WARC-Payload-Digest\":\"sha1:CKIZPCWPW2I7NQ6IAZ5U7IBRKEDK2NUP\",\"WARC-Block-Digest\":\"sha1:VJGZ4QHZLR7S5N22EPTXDKTIPXTKLDJK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103037089.4_warc_CC-MAIN-20220626040948-20220626070948-00587.warc.gz\"}"}
https://wiki.docking.org/index.php?title=Google_sheets_hit_picking&diff=prev&oldid=11516	[ "# Difference between revisions of \"Google sheets hit picking\"\n\nTo use Google sheets for hit-picking\n\n1. Put zinc_ids in sorted order in a column, e.g. B2-B100\n\n2. To depict the molecules copy and paste the following in each cells in new column. E.g. put this in cell C2 to depict the molecule in B2\n\n``` =IMAGE(CONCATENATE(\"http://zinc15.docking.org/substances/\", B2, \"-small.png\"))\n```\n\n3. Use IMPORTDATA to populate additional columns of interest in the header of the next columns. It looks like it can only load 3 columns at a time, so for example put the first cell D1, the next . Look up additional columns of interest here `http://zinc15.docking.org/substances/help/`\n\n``` Put in cell D1: =IMPORTDATA(CONCATENATE(\"http://zinc15.docking.org/substances.csv?zinc_id-in=\", JOIN(\"+\", B2:B100),\"&output_fields=smiles mol_formula logp\"))\nPut in cell G1: =IMPORTDATA(CONCATENATE(\"http://zinc15.docking.org/substances.csv?zinc_id-in=\", JOIN(\"+\", B2:B100),\"&output_fields=mwt num_chiral_centers reactivity\"))\nPut in cell K1: =IMPORTDATA(CONCATENATE(\"http://zinc15.docking.org/substances.csv?zinc_id-in=\", JOIN(\"+\", B2:B100),\"&output_fields=predicted_gene_names purchasability supplier_codes\"))\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6307402,"math_prob":0.5151164,"size":2821,"snap":"2022-40-2023-06","text_gpt3_token_len":820,"char_repetition_ratio":0.12034079,"word_repetition_ratio":0.5148515,"special_character_ratio":0.29493088,"punctuation_ratio":0.2359155,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9590532,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-25T05:33:28Z\",\"WARC-Record-ID\":\"<urn:uuid:baed6199-f0bb-4719-98ff-ebcf70cec053>\",\"Content-Length\":\"22570\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c5a97333-9467-481b-aa22-768d683ad9b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:8b638bf9-4b2c-4d27-b449-06498a7df43f>\",\"WARC-IP-Address\":\"169.230.26.169\",\"WARC-Target-URI\":\"https://wiki.docking.org/index.php?title=Google_sheets_hit_picking&diff=prev&oldid=11516\",\"WARC-Payload-Digest\":\"sha1:BDWVWFJGYPRO6ARKTYXUSBKAAUDXR3FG\",\"WARC-Block-Digest\":\"sha1:FSPZLALN2AJZFTAHNYYFJF4XGXGVV7GS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334514.38_warc_CC-MAIN-20220925035541-20220925065541-00430.warc.gz\"}"}
https://www.planetanalog.com/random-number-generator-using-leap-forward-techniques/	[ "# Random Number Generator Using Leap-Forward Techniques\n\nA wide variety of applications, including data encryption and circuit testing, require random numbers. As the cost of the hardware become cheaper, it is feasible and frequently necessary to implement the random number generator directly in hardware itself. Ideally, the generated random number should be uncorrelated. A generator can be either “truly random” or “pseudo random.” The former exhibits true randomness and the value of next number is unpredictable. The later only appears to be random. The sequence is based on specific mathematical algorithms, and thus the pattern is repetitive and predictable. However, if the cycle period is very large, the sequence appears to be non-repetitive and random. The focus of this paper is on pseudo-random number generators.\n\nThis paper provides an overview on the methods that are suitable for hardware implementation. The paper also describes single-bit and shows a leap-forward LFSR implementation.\n\nSingle Bit Random Number Generator Using LFSR\n\nA single bit random number generator produces a value of 0 or 1. The most efficient implementation is to use a Linear Feedback Shift Register (LFSR)", null, "", null, "", null, ". It is based on the recurrence equation:", null, "(1)\n\nHere xi is the ith number generated, ai is a pre-determined constant that can be either 0 or 1 and * and", null, "are And and Xor operator respectively. This equation implies that a new number (xn) can be obtained by utilizing m previous values (xn-1 , xn-2 ,…xn-m ) through a sequence of AND-XOR operations.\n\nIn pseudo-random number generator, the generated pattern will repeat after a certain number of cycles. It is know as the period of the generator. In an LFSR, the maximum achievable period, determined by m , is 2m-1 . We use a special set of ai s in order to achieve the maximum period. Despite their simplicity, the recurrence equations are different for different values of m . The table in Figure 1 lists the recurrence equation for m with values from 2 to 8.", null, "Figure 1: Sample Recurrence Equations\n\nFor example, when m is 4, the equation becomes:", null, "(2)\n\nAssume the initial seed (i.e., s0, s1, s2, s3) is 1000. We can obtain the random number sequence by the equation: 100110101111000. This pattern repeats itself after 15 numbers. The new value, sn , depends on the m previous values, sn-1 , sn-2 ,?sn-m . Therefore, m 1-bit registers are required to store these values. After a new value is generated, the oldest stored value is no longer needed for future generation and can be done by an m -slot shift register, which shifts out the oldest value and shifts in a new value in every clock cycle. In addition to registers, a few XOR gates are also required to perform exclusive operations. Let's use the previous example for m =4 again. We need four 1-bit registers to store the required values. Let q3 , q2 , q1 , and q0 be the outputs of the registers and q3_next , q2_next , q1_next , and q0_next be their next values. The Boolean equation can be written as:", null, "Figure 2\n\nAn LFSR random number generator is very efficiently implemented. It only needs an m-bit shift register and 1 to 3 XOR gates, and thus the resulting circuit is very small and its operation is extremely simple and fast. Furthermore, since the period grows exponentially with the size of the register, we can easily generate a large non-repetitive sequence. For example, with a 64-bit generator running at 1GHz, the period is more than 500 years.\n\nMultiple-Bit Leap Forward LFSR\n\nSome applications require more accuracy and need more that a single bit random number. Since the numbers produced by a single bit LFSR random number generator are correlated, one way to obtain a multi bit random number is to accumulate several single bit numbers.\n\nThe leap-forward LFSR method utilizes only one LFSR and shifts out several bits. This method is based on the observation that LFSR is a linear system and the register state can be written in vector format:", null, "(3)\n\nIn this equation, q(i + 1) and q(i) are the content of the shift register at (i+1)th and it h steps and A is the transition matrix. After the LFSR advances k steps, the equation becomes:", null, "(4)\n\nWe can calculate Ak and determine the Xor structure accordingly. The new circuit leaps k steps in one clock cycle. It still consists of the identical shifter register, although the feedback combinational circuitry becomes more complex. The implementation with 4 bit LFSR can be written as:", null, "(5)\n\nAssume that we need a four-bit random number generator and therefore have to advance four steps at a time. The new transition matrix becomes:", null, "(6)\n\nFor the purpose of circuit implementation, the equation can be written as:", null, "(7)\n\nAfter performing the operations, we can derive the feedback equation for each signal as mentioned below:", null, "(8)\n\nFigure 3 shows the corresponding block diagram.", null, "Figure 3: Block diagram\n\nFigure 4 shows the four-bit output random sequence of the single LFSR and leap-forward. The sequence generated is based on the seed value of 1000.", null, "Figure 4: Four-bit output random sequence of a single LFSR and leap-forward\nResults\n\nThe simulation results are shown in Figure 5 . The first (top) figure shows the simulation results of Leap Forward LFSR techniques while the second (bottom) image is the LFSR result. It looks as though there is a high correlation between the two consecutive outputs of LFSR; while this is not the case, this is at the cost of extra hardware. The synthesis details are shown in Table 1 .\n\n Synthesis Details Tool Xilinx ISE Device family Xilinx5x Spartan2 Target device 2s30pq208\n\nTable 1: Synthesis details for simulation results\n\nConclusion\n\nSeveral design techniques have been examined for hardware random number generators and this feasibility for FPGA devices. LFSR is the most effective method for single bit random number generator. When multiple bits are required, LFSR can be extended by utilizing extra circuitry. For a small number of bits, leap-forward LFSR method is ideal because it balances the combinational circuitry and register, and balances the FPGA resource. We can extend this technique to a greater number of bits where one can get even better uncorrelated sequences for random number generation." ]	[ null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-old-features-numbers-two-blue.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-old-features-numbers-comma.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-old-features-numbers-three-blue.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-feature-bola-equation1.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-feature-bola-circle.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-feature-bola-figure1.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-feature-bola-equation2.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-feature-bola-figure2.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-feature-bola-equation3.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-feature-bola-equation4.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-feature-bola-equation5.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-feature-bola-equation6.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-feature-bola-equation7.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-feature-bola-equation8.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-feature-bola-figure3.gif", null, "https://www.planetanalog.com/wp-content/uploads/images-common-techonline-images-community-content-feature-bola-figure4.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9030761,"math_prob":0.99159384,"size":5951,"snap":"2020-45-2020-50","text_gpt3_token_len":1292,"char_repetition_ratio":0.14831007,"word_repetition_ratio":0.002002002,"special_character_ratio":0.21391363,"punctuation_ratio":0.106984966,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9935509,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32],"im_url_duplicate_count":[null,6,null,null,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-20T15:08:54Z\",\"WARC-Record-ID\":\"<urn:uuid:0117e229-8cfa-448c-898e-247f3ea73d13>\",\"Content-Length\":\"154943\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:61ad4637-bb46-44c8-a621-0d2ee417bd20>\",\"WARC-Concurrent-To\":\"<urn:uuid:fc1d626e-4b33-464b-8566-d9773d90b636>\",\"WARC-IP-Address\":\"192.0.66.120\",\"WARC-Target-URI\":\"https://www.planetanalog.com/random-number-generator-using-leap-forward-techniques/\",\"WARC-Payload-Digest\":\"sha1:VQYYR4YDSFYHL2OWG2P72JHN2ODGU7UT\",\"WARC-Block-Digest\":\"sha1:TTROFLVAIOQVBLRODQMYXWJFQQYOTPRI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107872746.20_warc_CC-MAIN-20201020134010-20201020164010-00137.warc.gz\"}"}
https://www.cornerstonecurriculum.com/product-page/making-math-meaningful-level-4	[ "top of page", null, "`In LEVEL 4 your child is working mostly on their own. Level 4 uses word problems to expand the understanding of addition and subtraction (0 - 9,999), and teaches multiplication of 2 and 3 digit numbers and division. Fractions are expanded from Level 3 to include whether two fractons are equal-not equal, and less than-greater than. Algebra is further developed. The Student Book offers opportunities to apply and practice the concepts and skills covered in the lessons from Level 4. Making Math Meaningful is an exceptional program that elevates a child’s conceptual reasoning! This curriculum teaches the child to understand not only how to add, subtract, multiply, and divide, but more importantly why and when. The student will be equipped to solve every kind of math problem or word problem, will have the understanding to apply their knowledge to everyday life, and will be prepared for algebra and other higher math. Making Math Meaningful trains children in advanced reasoning, to understand math concepts, and to think mathematically. `\n\n# Making Math Meaningful: Level 4\n\n\\$60.00Price\nbottom of page" ]	[ null, "https://static.wixstatic.com/media/0c502d_058190403fda4b94ae2769c9cd8ffe8f~mv2.jpg/v1/fill/w_980,h_1280,al_c,q_85,usm_0.66_1.00_0.01,enc_auto/0c502d_058190403fda4b94ae2769c9cd8ffe8f~mv2.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9344321,"math_prob":0.82728565,"size":1082,"snap":"2023-40-2023-50","text_gpt3_token_len":220,"char_repetition_ratio":0.10204082,"word_repetition_ratio":0.0,"special_character_ratio":0.19963032,"punctuation_ratio":0.10552764,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9521055,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-30T10:49:46Z\",\"WARC-Record-ID\":\"<urn:uuid:2bd072aa-f878-43e3-a02e-2974b1f82ce9>\",\"Content-Length\":\"1050548\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8883b149-f561-4f80-b56d-541b566faf95>\",\"WARC-Concurrent-To\":\"<urn:uuid:48bb8e05-b726-4f57-abef-e2107c96fc86>\",\"WARC-IP-Address\":\"146.75.33.84\",\"WARC-Target-URI\":\"https://www.cornerstonecurriculum.com/product-page/making-math-meaningful-level-4\",\"WARC-Payload-Digest\":\"sha1:DETV6OLBDHTBB7PLRF42FQAGMMH45DP5\",\"WARC-Block-Digest\":\"sha1:XHF4BMI74YMBKSE7HO3MXLWXNLOY22AS\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100184.3_warc_CC-MAIN-20231130094531-20231130124531-00311.warc.gz\"}"}
https://cotejer.github.io/2017/06/12/translation	[ "# Translation\n\nIf there’s a bane to everyone’s existence in mathematics, the word problem seems to be it. These are routinely the most tricky problems to solve, and so many students find themselves understanding the material, only to stumble on this kind of problem. If you ask a student, chances are that they will say these problems are difficult because they aren’t as straightforward as a question that asks to solve for $x$ or to find the area of a figure. I don’t want to spend today debating whether or not it’s a good idea to have word problems, but instead I want to address the fundamental difficulty that I’ve found many students to have in this area.\n\nIn one word, here’s the difficulty: translation.\n\nLet’s face it. Mathematics is a language, just like any other. This means that mathematics has its own grammatical structure, as well as ways to construct sentences. Additionally, there are ways to make your mathematical sentences clear, and there are many other ways to make them incomprehensible. Unfortunately, many students don’t get the experience of seeing what a “good” mathematical sentence looks like, which means they have difficult knowing what expresses total nonsense and what conveys meaning.\n\nTo combat this, one of the skills that can help is an ability to translate between the mathematics and your preferred language. My mother tongue is English, so I’ll be referring to it here, but there’s nothing special about English. The important part is to be able to move between the mathematics and your language. If you can express a mathematical equation in plain English, and (arguably, more importantly) vice versa, it’s so helpful in decoding problems that you will come across. This ability seems so scarce in schools now that it’s almost like a superpower.\n\nI think there’s no better way to show this than through an example:\n\nAn airplane flies against the wind from A to B in 8 hours. The same airplane returns from B to A, in the same direction as the wind, in 7 hours. Find the ratio of the speed of the airplane (in still air) to the speed of the wind.\n\nIf you want to have any hope of solving problems like this in a systematic way, you need to know how to translate between the words and the mathematics you’ll use to answer the question. If you want, I encourage you to try this question out, and make a detailed solution. This doesn’t mean you need to write ten pages of explanation, but it means that everything you do should be clear.\n\nGot it? Okay, let’s go through the way I would solve this.\n\nFirst, you should identify the quantities you’re looking for in this problem. From what I can see, the quantities we are looking for are the speed of the plane when there’s no wind, and the speed of the wind itself. Do we know the numerical values of these quantities? We do not, so let’s give them symbols. Let’s call the speed of the plane without any wind $v$ and the speed of the wind itself $w$. Note that the units of both are something akin to $m/s$.\n\nNow, we don’t know how far the distance is from point A to B, so we’ll just call this distance $d$, with units of metres.\n\nWith all of the variables of the question in hand, it’s time to translate from the words in the problem to mathematical equations. This is a crucial part of solving these problems, and it’s good to take it sentence by sentence. But first, let’s state clearly what we are trying to determine. We want the ratio of the plane’s speed in still air to the wind speed. In our variables, this corresponds to the expression $\\frac{v}{w}$.\n\nHere’s the first sentence again: An airplane flies against the wind from A to B in 8 hours.\n\nFor this sentence, we need to relate the time traveled (8 hours) and the distance from A to B ($d$). The most natural way to do this is through the speed of the trip. Note that, since the airplane is flying against the wind, it’s “net” speed is $v-w$. This is due to the fact that the wind is slowing the plane down. We then know that the speed of anything is given by a distance over a time (here, we are talking about average speed), so we have the speed being $\\frac{d}{8}$. Therefore, we get the following equation:\n\n$v-w=\\frac{d}{8}$\n\nLet’s parse the second sentence: The same airplane returns from B to A, in the same direction as the wind, in 7 hours.\n\nThis is similar to the above, except now we have a time of 7 hours, and a “net” speed of $v+w$. You can think of it as the wind “helping” the airplane as it moves to its destination. The distance is once again the same, so we have the following relation:\n\n$v+w=\\frac{d}{7}$\n\nThis is by far the most difficult part of a word problem. For the most part, students can do the actual computations, but the difficult part is the translation. In this problem, we see that the tricky aspects include knowing that speed is given by a distance divided by a time, and being able to relate the sentences into that form. This isn’t a skill that’s developed within a week. It’s something that you need to focus on for many problems in order to understand how these kinds of situations go.\n\nHere are the big parts of translating from English to mathematics:\n\n• Knowing what equality means. But I know what equality means, you say. Of course, we know implicitly what it means, but the reality is that many students will claim to know what equalities mean, yet write things that aren’t equalities. This is easily seen in terms of units, where many students will make equations whose units don’t agree. Indeed, being able to look at the units of your quantities is a great way to help you come up with equations. This is called dimensional analysis.\n• Knowing what the words “more”, “less”, “at most”, “in total”, and so on, mean. Each of those words has a precise mathematical meaning, and it’s critical that you know the difference between them.\n• Relational words like “double”, and “half”. This closely follows from the above, but this also gets a lot of students tripped up. If I said to you that I (let’s call me $x$) had double the amount of something than you (which we’ll label $y$), is the inequality $x\\ge 2y$ or is it $y \\ge 2x$? It’s very important to know how to differentiate between these two. (Hint: it’s the former.)\n• Knowing how to parse through negation. When you see the word “not”, can you infer the correct meaning? It’s an unfortunate reality that many questions throw in a bunch of extra (and confusing) negation in order to trick students for no good reason, and so being able to deal with this is a must.\n\nIn the end, the important thing is that you can comfortably go between English and mathematics. The arrows should go both ways, even if you’re only tested on one way. The reason is that it’s helpful when learning a new concept to express it in words, since it can make the concept more clear than in abstract notation. Of course, that notation is important when you’re trying to precisely answer a question, but it’s good to be able to talk more informally about an idea.\n\nMy advice is therefore this: when you encounter a word problem, go through it line by line, asking yourself, “How can I translate these words into mathematics?” Similarly, do it the other way when you encounter equations, so you get a sense of the other side of the coin. By doing this, it becomes easier to move between the two languages, without having to carry an English-to-mathematics dictionary." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.95369655,"math_prob":0.92990565,"size":7375,"snap":"2021-04-2021-17","text_gpt3_token_len":1648,"char_repetition_ratio":0.12345679,"word_repetition_ratio":0.04236006,"special_character_ratio":0.22440678,"punctuation_ratio":0.104139715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98547024,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-26T09:06:14Z\",\"WARC-Record-ID\":\"<urn:uuid:f5220388-d3f6-4f6c-83ed-3e0f9ea0bd8f>\",\"Content-Length\":\"10456\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dc1503e3-0d2c-4e33-83e6-b91e2ef3ee65>\",\"WARC-Concurrent-To\":\"<urn:uuid:b1bb4302-78cf-4021-9d83-3a9249032a2c>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"https://cotejer.github.io/2017/06/12/translation\",\"WARC-Payload-Digest\":\"sha1:I6WR4HQ6HVPRPBVR3I4PRNO2BGJKNGDO\",\"WARC-Block-Digest\":\"sha1:2X5ZY5IKKJ5QDZYLKMEVXZ7N6FGOJGLF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704799711.94_warc_CC-MAIN-20210126073722-20210126103722-00291.warc.gz\"}"}
https://www.excelhow.net/how-to-calculate-average-ignore-non-numeric-values-and-errors.html	[ "# How to Calculate Average Ignore Non-Numeric Values and Errors\n\nIn daily work we often need to calculate the average of some numbers in a range for further analysis. Thus, we need to know some basic functions of calculate average in Excel. From now on, we will introduce you some basic functions like AVERAGE, AVERAGEIF, and show you some common formulas to calculate average value.\n\nIn this article, we will let you know how to calculate average by AVERAGE function easily and the way to calculate average based on a given criteria by AVERAGEIF function. We will introduce you the syntax, arguments, and basic usage about the two functions, and let you know the working process of our created formulas.\n\n## 1. EXAMPLE\n\nRefer to “Numbers” column, some numbers are listed in range A2:A11. C2 is used for entering a formula which can calculate the average of numbers in range A2:A11. In fact, Excel has amount of built-in functions and they can execute most simple calculations properly, and AVERAGE is one of the most common used functions.\n\nIn C2, enter “=AVERAGE(A2:A11)”.\n\nThen press Enter, AVERAGE function returns 63.8.\n\nYou can adjust decimal places by “Increase Decimal” or “Decrease Decimal” in “Number” section.\n\nBut if some errors or blank or non-numeric values exist in the list, how can we only calculate average for numbers ignoring the invalid values? See example below.\n\nWe get an error when calculating average for range A2:A11, the difference is there are some invalid values in the list. To calculate average ignoring the errors, that means we need to calculate average of numbers in a range with one or more criteria, thus, we can apply AVERGAEIF function instead of basic AVERAGE function.\n\n## 2. CREATE A FORMULA with AVERAGEIF FUNCTION\n\nStep1: In C2, enter the formula =AVERAGEIF(A2:A11,”>0″).\n\nYou can also name range “Numbers” for A2:A11, then you can enter =AVERAGEIF(Numbers,”>0″).\n\nStep2: Press Enter after typing the formula.\n\nIgnore the improper cells, and only calculate the average of numbers 54, 40, 88, 76, 100, 90 and 44, total 7 numbers. (54+40+88+76+100+90+44)/7=70.3. The formula works correctly.\n\n### a. FUNCTION INTRODUCTION\n\nAVERAGEIF function is AVERAGE+IF. It returns the average of some numbers in a range based on given criteria.\n\nSyntax:\n\n``=AVERAGEIF(range, criteria, [average_range])``\n\nIt supports wildcards like asterisk ‘*’ and question mark ‘?’, also supports logical operators like ‘>’,’<’. If wildcards or logical operators are required, they should be enclosed into double quotes (““), in this case we entered “>0” to show the criteria.\n\nAVERAGEIF – RANGE\n\nThis example is very simple, we have only one list. A2:A11 is criteria range and average range.\n\nIn the formula bar, select “A2:A11”, press F9, values in this range are expanded in an array.\n\nIf average range=criteria range, average range can be omitted.\n\nAVERAGEIF – CRITERIA\n\nObviously, the criteria in our case is “>0”, this condition can filter numbers from criteria range.\n\n### b. HOW THE FORMULA WORKS\n\nAfter expanding values, the formula is displayed as:\n\n``=AVERAGEIF({54;40;#N/A;88;\"ABC\";76;0;100;90;44},\">0\") ``\nNote: blank cell is recorded as 0 in the array.\n\nBecause of criteria “>0”, so invalid cells are ignored. Replace all invalid values with 0. Keep all numbers.\n\n``{54;40;#N/A;88;\"ABC\";76;0;100;90;44} -> {54;40;0;88;0;76;0;100;90;44}``\n\nNow this new array only contains numbers. We can calculate the average now.\n\n## 3. Calculate Average Ignore Non-Numeric Values and Errors with VBA Code\n\nYou can also create a User Defined Function to calculate the average of a range of cells, ignoring any non-numeric values or error values, just do the following steps:\n\nStep1: Click on the “Visual Basic” button in the Developer tab to open the Visual Basic Editor.\n\nStep2: In the Visual Basic Editor, click on “Insert” in the menu and select “Module” to create a new module.\n\nStep3: Paste the VBA code provided in the new module.\n\n``````Function AverageIgnoreNonNumeric(rng As Range) As Variant\nDim cell As Range\nDim sum As Double\nDim count As Long\n\nsum = 0\ncount = 0\n\nFor Each cell In rng\nIf IsNumeric(cell.Value) And Not IsError(cell.Value) And Not IsEmpty(cell.Value) Then\nsum = sum + cell.Value\ncount = count + 1\nEnd If\nNext cell\n\nIf count > 0 Then\nAverageIgnoreNonNumeric = sum / count\nElse\nAverageIgnoreNonNumeric = CVErr(xlErrNA)\nEnd If\nEnd Function``````\n\nStep4: Save the workbook as a macro-enabled workbook by selecting “Excel Macro-Enabled Workbook” from the “Save as type” drop-down menu.\n\nStep5: Close the Visual Basic Editor and return to the Excel worksheet.\n\nStep6: Enter the below formula in a blank cell where you want to calculate the average\n\n``=AverageIgnoreNonNumeric(A2:A11)``\n\nStep7: Press “Enter” to calculate the average ignoring non-numeric values.\n\n## 4. Related Functions\n\n• Excel AVERAGE function\nThe Excel AVERAGE function returns the average of the numbers that you provided.The syntax of the AVERAGE function is as below:=AVERAGE (number1,[number2],…)….\n• Excel AVERAGEIF function\nThe Excel AVERAGEAIF function returns the average of all numbers in a range of cells that meet a given criteria.The syntax of the AVERAGEIF function is as below:= AVERAGEIF (range, criteria, [average_range])…." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7827958,"math_prob":0.97629184,"size":5101,"snap":"2023-40-2023-50","text_gpt3_token_len":1274,"char_repetition_ratio":0.16853051,"word_repetition_ratio":0.017412934,"special_character_ratio":0.24877475,"punctuation_ratio":0.157129,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989886,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T04:07:10Z\",\"WARC-Record-ID\":\"<urn:uuid:bdeb4119-35d6-40ae-b164-67c1f78efb7f>\",\"Content-Length\":\"95016\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7528657e-1325-4b0d-aefa-8ab6a90c7be2>\",\"WARC-Concurrent-To\":\"<urn:uuid:914d5108-1724-48cb-b035-f262a3251cce>\",\"WARC-IP-Address\":\"18.213.98.197\",\"WARC-Target-URI\":\"https://www.excelhow.net/how-to-calculate-average-ignore-non-numeric-values-and-errors.html\",\"WARC-Payload-Digest\":\"sha1:YHGRBXQ5OJCP7WWIN2FYU6XRRBR6J5MJ\",\"WARC-Block-Digest\":\"sha1:ZCOFZKVNXVRCADJCSY6KRMBOJCSYTMSL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506479.32_warc_CC-MAIN-20230923030601-20230923060601-00752.warc.gz\"}"}
https://math.stackexchange.com/questions/1684223/formula-for-a-geometric-series-weighted-by-binomial-coefficients-sum-over-the-u	[ "# Formula for a geometric series weighted by binomial coefficients (sum over the upper index):$\\sum_{i=0}^L {n+i\\choose n}\\ x^i =\\ ?$\n\nThe binomial sum is $$\\sum\\limits_{i=0}^n {n\\choose i}\\ x^i = (1+x)^n,$$ where $\\displaystyle{n\\choose i}=\\frac{n!}{(n-i)!i!}.$\n\nIs there a corresponding formula when you sum over the upper index of the binomial coefficients, not the lower index: $$\\sum_{i=0}^L {n+i\\choose n}\\ x^i =\\ ?$$ Equivalently, I am looking for the generating function of the sequence $$a_i={r+i\\choose k}\\qquad i\\ge0,$$ where $r\\ge k\\ge0$ are fixed parameters.\n\n• When you change upper index you continously change powers i think there isnt any closed form – Archis Welankar Mar 5 '16 at 14:46\n• There is a closed form which contains the incomplete beta function. – Claude Leibovici Mar 5 '16 at 14:49\n\n$$\\sum_{i=0}^L {n+i\\choose n}\\ x^i =\\frac{1-(L+1) \\binom{L+n+1}{n} B_x(L+1,n+1) }{(1-x)^{n+1}}$$ where appears the incomplete beta function." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.55636156,"math_prob":0.99972576,"size":435,"snap":"2021-21-2021-25","text_gpt3_token_len":150,"char_repetition_ratio":0.118329465,"word_repetition_ratio":0.0,"special_character_ratio":0.33103448,"punctuation_ratio":0.11578947,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999944,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-08T03:51:38Z\",\"WARC-Record-ID\":\"<urn:uuid:8191fb44-8555-46cd-955d-5cc0c2f366c9>\",\"Content-Length\":\"166531\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:78abcc4b-a315-47e3-be6a-bd95ee82edf0>\",\"WARC-Concurrent-To\":\"<urn:uuid:bb22bd58-eb53-4182-813d-3b697c1c2b1f>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1684223/formula-for-a-geometric-series-weighted-by-binomial-coefficients-sum-over-the-u\",\"WARC-Payload-Digest\":\"sha1:ARAKTPHO3VA645W3J2CIEDWQOWJX7P2D\",\"WARC-Block-Digest\":\"sha1:UN2HXPPJ27UHOOJUGKBQEGQ2GKZGICKM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988837.67_warc_CC-MAIN-20210508031423-20210508061423-00096.warc.gz\"}"}
https://pwg.gsfc.nasa.gov/stargaze/Slog5.htm	[ "# Disclaimer: The following material is being kept online for archival purposes.\n\n## Although accurate at the time of publication, it is no longer being updated. The page may contain broken links or outdated information, and parts may not function in current web browsers.\n\n Site Map Math Index Glossary Timeline Questions & Answers Lesson Plans\n\n# (M-17) The Number \"e\"\n\nHere begins an optional extension of the sections on logarithms. Its starting point seems completely unrelated--namely, the interest paid on money deposited in a bank. It starts with a very elementary discussion of interest, and users familiar with this subject may quickly skim this part. However, be assured --a connection with logarithms soon appears!\n\n### Interest and Banking--a Very Elementary Introduction\n\nMost people are familiar with interest, the extra amount paid by borrowers to temporarily use other people's money.\n\nSuppose you want to buy an expensive item--car, house, business, farm machinery--but cannot pay the full price right away. You then borrow what you need from a bank, and gradually pay it back, adding a little extra as interest, in return for the privilege of borrowing.\n\nInterest is traditionally measured in percent, hundredths of the amount borrowed, also denoted by the symbol %. If the interest rate is, say, 10 dollars per year for every 100 dollars borrowed, your interest is \"10 percent\" or 10% (\"per cent\" means, \"for each hundred\").\n\nBanks usually lend money for only part of the cost of the purchased item--because then, if the borrower is unable to pay (\"defaults\"), they can legally \"repossess\" the item, and having paid only part of its cost, they suffer no loss.\n\nBecause a \"down payment\" is usually needed before borrowing, one needs to save money to cover it. After putting the money aside for such savings, people often lend it to a bank. The bank will pay a lower interest--say, only 6%--while using the same money to lend it out at some higher rate, such as 10%. That way the bank makes a profit and covers its own costs; but it is still better to lend out money at a lower rate and make a profit of \\$6 for every \\$100--better than not making any profit at all.\n\n### Simple Interest and Compound Interest\n\n(Below, after the * mark, you are expected here to use a calculator with a yx button)\n\nSuppose you lend the bank \\$ 1000 at 6% interest. The amount you lend is known as \"the principal.\"\n\nAfter a year you have earned \\$60 and now have \\$1060\nAfter 2 years, you have earned another \\$60 and now have \\$1120\nAfter 3 years, you have earned another \\$60 and now have \\$1180\nAfter 4 years, you have earned another \\$60 and now have \\$1240\n\n(To save for an expensive purchase, most people would of course save \\$1000 every year and accumulate cash much faster. In this calculation, however, we concentrate on just the first \\$1000)\n\nYou can actually do better! The \\$60 earned in the first year can be added to the \"principal\" amount, so that in the second year it, too, earns interest. In fact, in any year you can add the profits to the principal, and earn more.\n\nThis changes the rules. Until now, what you earned in any year were \\$60, six percent of the amount originally invested. That is called \"simple interest.\" The amount you have at the end of the year is just \\$60 more.\n\nNow the amount you have at the end of the year (assuming you took nothing out) is 1.06 times what you had at the beginning of the year. This is called compound interest, compounded at the end of each year, Let us calculate how much you have at the end of each year. Suppose you started with x dollars. With simple interest\n\nAfter 1 year 1.06 x\nAfter 2 years 1.12 x\nAfter 3 years 1.18 x\nAfter 4 years 1.24 x\n\nWith compound interest as described above, to an accuracy of 4 decimals\n\n After one year you have 1.06 x After another year 1.06 [1.06 x] = (1.06)2 x = 1.1236 x After 3 years 1.06 [(1.06)2 x] = (1.06)3 x = 1.191 x After 4 years 1.06 [(1.06)3 x] = (1.06)4 x = 1.2624 x\n\nSo yes, you are increasing your profit, though the increase is rather moderate.\n\nYou get still more if you add the earned interest to the principal not at the end of each year, but at the end of each half year. The interest earned each half year is only 3%, but the number of \"compounding periods\" is doubled--two per half year. Let's see what you get from an original amount of x dollars (calculating to an accuracy of 5 decimals)\n\n After half a year 1.03 x After 1 year 1.03 [1.03 x] = (1.03)2 x = 1.0609 x After 1.5 years 1.03 [(1.03)2x] = (1.03)3 x = 1.10927 x After 2 years 1.03 [(1.03)3 x] = (1.03)4 x = 1.12551 x After 3 years (1.03)6x = 1.19405 x After 4 years (1.03)8x = 1.26677 x\n\nAnd so on\n\nBut why wait half a year? One may just as well add the money monthly, assuming here all months are equal and each earns 0.5% interest. (* you better have a yx button for the next steps, though with a lot of patience the same results can also be derived without one. They are given to an accuracy of 6 decimals.)\n\nAfter 1 year --12 periods, 0.5% each--you have (1.005)12 x = 1.061678 x\nAfter 2 years--24 periods, 0.5% each--you have (1.005)24 x = 1.127159 x\nAfter 3 years--36 periods, 0.5% each--you have (1.005)36 x = 1.196680 x\nAfter 4 years--48 periods, 0.5% each--you have (1.005)48 x = 1.279489 x\n\nIt is instructive to compare the total amount after, say, 4 years:\n\n Simple interest 1.24 x Compounded yearly 1.2624 x Compounded twice a year 1.26677 x Compounded 12 times a year 1.278489 x\n\nSo yes, you keep getting more each time. You might get even more if you compounded every day (as some banks have brazenly promised) but the number seems to be creeping towards a limit, which you can never pass. You will never get rich this way!\n\nIn the next section we'll try derive that limit. Stand by for some algebra!\n\n### The Limit\n\nSuppose you lend out x dollars at a percentage of p percent (it was 6 in the above example). After one year we have\n(1 + p/100) x\n\nLet us divide the year into N equal parts, each of which earns interest p/N . After each such period, the money which was invested grows by a factor\n\n[1 + p/(100 N)] (1)\n\nThere exist N such periods in the year, so after one year, our investment has grown by a factor\n\nF = [1 + p/(100 N)]N (2)\n\nDividing the power N by some number and also multiplying by it does not change a thing--it's like multiplying by (Q/Q)=1 (whatever Q might be). So if Q=p/100\n\nF = [1 + p/(100 N)][(100N/p) (p/100)] (3)\n\nIf you cancel the fractions, you are back where we started. However, instead of doing so, let us introduce a new variable quantity y: let\n\n100N/p = y (4)\nThen (3) becomes\nF = (1 + 1/y)(y.p/100) (5)\n\nand remembering that one power raised to another is like raising to a power that is the product of both exponents\nF = [(1 + 1/y)y](p/100) (6)\nThe reason for introducing the new variable y is to put at the core of our expression the rather interesting quantity:\n[(1 + 1/y)y] (7)\n\n(raised to power p/100). Once again, by (4) y is defined as\ny = N (p/100)\n\nIf interest is compounded every second of the year, N is a bit over 31 million. Suppose N, the number of compounding periods, grows without limit, and so, therefore, does y. Then (7) becomes a rather strange expression!\n\nOn one hand, the expression we are raising to the yth power is (1 + 1/y), very close to 1, and we know that any power of 1 is still 1--no matter how many times you multiply 1 by itself, the result stays the same. On the other hand, a high power y of any number larger than 1 (even larger by just a tiny bit) will grow without limit.\n\nWhich case is the one here? Let's try the calculator, and assume it has buttons for both (1/x) and x2. We can then easily derive the expression (7) for values of y which are powers of 2:\n\n(1 + 1/2) 2 = 2.25\n(enter 0.5, add 1, hit squaring button)\n(1 + 1/4) 4 = 2.441406...\n(enter 0.5, hit squaring button, add 1, hit x2 button 2 times more)\n(1 + 1/16) 16 = 2.6379284...\n(enter 0.5, hit squaring button 2 times, add 1, hit x2 button 4 times)\n(1 + 1/256) 256 = 2.7129916...\n(enter 0.5, hit squaring button 3 times, add 1, hit x2 button 8 times, since 256 = 28)\n(1+ 1/65536) 65536 = 2.71826039...\n(enter 0.5, hit squaring button 4 times, add 1, hit x2 button 16 times, as 65536 = 216)\n\nYou can see that the result approaches a limit which is neither zero nor infinity, but a number between 2 and 3. Mathematicians denote it by the letter e. And the limiting factor, beyond which compounded interest with percentage p cannot rise, no matter how frequently the compounding is performed, is by (6)\ne(p/100) (8)\n\nProcesses which are compounded naturally--for instance, the number of bacteria (or other living creatures) given an unlimited supply of food, or the number of neutrons in an uncontrolled chain reaction--all these \"grow exponentially\" following a law like the one above.\n\nMany properties of e involve calculus, or an expanded definition of numbers including the square root of (–1), also denoted i. For instance, if the symbol N! (\"factorial of N\") defines the product of whole numbers up to N\n\n1! = 1 2! = 1.2 = 2 3! = 1.2.3 = 6 4! = 24 5! = 120 etc.\n\nand so on, it may be shown that\n\ne = 1 + 1/(1!) + 1/(2!) + 1/(3!) + 1/(4!) + 1/(5!) + ...\n\nand because N! grows extremely fast with increasing values of N, one soon gets pretty accurate values of e. Furthermore, the number e is also \"the base of natural logarithms.\" That, however, is the story of the next section.\n\n### Exploring further\n\nYou saw how\n(1 + 1/y)y\n\ngoes to a limit e = 2.71828... as y gets larger and larger. How about\n\n(1 – 1/y)y\n\n--does it go to a limit too? Indeed, it does. With your calculator you can derive approximations to it, as was done with e. If your calculator has a sign-reversing button \"+/–\" you can use the same steps as before, except that before adding \"1\", reverse the sign of the power of 0.5 by means of that button.\n\nNext, the obvious question--how is that limit related to \"e\"? Try to discover on your own!\n\nAuthor and Curator: Dr. David P. Stern\nMail to Dr.Stern: stargaze(\"at\" symbol)phy6.org .\n\nLast updated 10 October 2007" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94417524,"math_prob":0.980382,"size":8621,"snap":"2022-27-2022-33","text_gpt3_token_len":2244,"char_repetition_ratio":0.11163978,"word_repetition_ratio":0.028427036,"special_character_ratio":0.27862197,"punctuation_ratio":0.12994653,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9848186,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-09T02:21:01Z\",\"WARC-Record-ID\":\"<urn:uuid:590e11ca-eca1-439e-be97-13dfedc70aa0>\",\"Content-Length\":\"19458\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2dbaaad5-1d43-4217-9bfb-fc39675246c1>\",\"WARC-Concurrent-To\":\"<urn:uuid:f866cdf1-3bb6-4532-9f34-e10b8816db6a>\",\"WARC-IP-Address\":\"169.154.154.72\",\"WARC-Target-URI\":\"https://pwg.gsfc.nasa.gov/stargaze/Slog5.htm\",\"WARC-Payload-Digest\":\"sha1:FLARX3OYIAHCP2KOOOJ4WZFWW7PRTWBA\",\"WARC-Block-Digest\":\"sha1:2VET5VK6WKO7ZY5BTET5RVLTDZRK5AOD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570879.37_warc_CC-MAIN-20220809003642-20220809033642-00262.warc.gz\"}"}
https://www.emathhelp.net/en/calculators/pre-algebra/prime-factorization-calculator/?i=4607	[ "# Prime factorization of $4607$\n\nThe calculator will find the prime factorization of $4607$, with steps shown.\n\nIf the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.\n\nFind the prime factorization of $4607$.\n\n### Solution\n\nStart with the number $2$.\n\nDetermine whether $4607$ is divisible by $2$.\n\nSince it is not divisible, move to the next prime number.\n\nThe next prime number is $3$.\n\nDetermine whether $4607$ is divisible by $3$.\n\nSince it is not divisible, move to the next prime number.\n\nThe next prime number is $5$.\n\nDetermine whether $4607$ is divisible by $5$.\n\nSince it is not divisible, move to the next prime number.\n\nThe next prime number is $7$.\n\nDetermine whether $4607$ is divisible by $7$.\n\nSince it is not divisible, move to the next prime number.\n\nThe next prime number is $11$.\n\nDetermine whether $4607$ is divisible by $11$.\n\nSince it is not divisible, move to the next prime number.\n\nThe next prime number is $13$.\n\nDetermine whether $4607$ is divisible by $13$.\n\nSince it is not divisible, move to the next prime number.\n\nThe next prime number is $17$.\n\nDetermine whether $4607$ is divisible by $17$.\n\nIt is divisible, thus, divide $4607$ by ${\\color{green}17}$: $\\frac{4607}{17} = {\\color{red}271}$.\n\nThe prime number ${\\color{green}271}$ has no other factors then $1$ and ${\\color{green}271}$: $\\frac{271}{271} = {\\color{red}1}$.\n\nSince we have obtained $1$, we are done.\n\nNow, just count the number of occurences of the divisors (green numbers), and write down the prime factorization: $4607 = 17 \\cdot 271$.\n\nThe prime factorization is $4607 = 17 \\cdot 271$A." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.866643,"math_prob":0.999951,"size":1569,"snap":"2023-14-2023-23","text_gpt3_token_len":426,"char_repetition_ratio":0.20127796,"word_repetition_ratio":0.37068966,"special_character_ratio":0.39515615,"punctuation_ratio":0.13937283,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999703,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-09T14:49:45Z\",\"WARC-Record-ID\":\"<urn:uuid:7712e0fc-664f-43bf-ab54-a344b9e59ab9>\",\"Content-Length\":\"26060\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ba896591-9e1a-4303-acd6-1554b249c270>\",\"WARC-Concurrent-To\":\"<urn:uuid:52e4347e-dab9-4af4-a5e3-5307151a85f0>\",\"WARC-IP-Address\":\"69.55.60.125\",\"WARC-Target-URI\":\"https://www.emathhelp.net/en/calculators/pre-algebra/prime-factorization-calculator/?i=4607\",\"WARC-Payload-Digest\":\"sha1:WJFAD3K4Y5BNJPJLTXQFG6ET7UZIONHZ\",\"WARC-Block-Digest\":\"sha1:QKPSYOLGCQND7M5SG2OOBW25PBDNAIH6\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224656737.96_warc_CC-MAIN-20230609132648-20230609162648-00497.warc.gz\"}"}
https://www.asknumbers.com/inch-to-mm/199-inches-to-mm.aspx	[ "# How Many Millimeters in 199 Inches?\n\n199 Inches to mm converter. How many millimeters in 199 inches?\n\n199 Inches equal to 5054.6 mm or there are 5054.6 millimeters in 199 inches.\n\n←→\nstep\nRound:\nEnter Inch\nEnter Millimeter\n\n## How to convert 199 inches to mm?\n\nThe conversion factor from inches to mm is 25.4. To convert any value of inches to mm, multiply the inch value by the conversion factor.\n\nTo convert 199 inches to mm, multiply 199 by 25.4, that makes 199 inches equal to 5054.6 mm.\n\n199 inches to mm formula\n\nmm = inch value * 25.4\n\nmm = 199 * 25.4\n\nmm = 5054.6\n\nCommon conversions from 199.x inches to mm:\n(rounded to 3 decimals)\n\n• 199 inches = 5054.6 mm\n• 199.1 inches = 5057.14 mm\n• 199.2 inches = 5059.68 mm\n• 199.3 inches = 5062.22 mm\n• 199.4 inches = 5064.76 mm\n• 199.5 inches = 5067.3 mm\n• 199.6 inches = 5069.84 mm\n• 199.7 inches = 5072.38 mm\n• 199.8 inches = 5074.92 mm\n• 199.9 inches = 5077.46 mm\n\nWhat is a Millimeter?\n\nMillimeter (millimetre) is a metric system unit of length. The symbol is \"mm\".\n\nWhat is a Inch?\n\nInch is an imperial and United States Customary systems unit of length, equal to 1/12 of a foot. 1 inch = 25.4 mm. The symbol is \"in\".\n\nCreate Conversion Table\nClick \"Create Table\". Enter a \"Start\" value (5, 100 etc). Select an \"Increment\" value (0.01, 5 etc) and select \"Accuracy\" to round the result." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.76966774,"math_prob":0.99279356,"size":934,"snap":"2023-40-2023-50","text_gpt3_token_len":333,"char_repetition_ratio":0.23978494,"word_repetition_ratio":0.0,"special_character_ratio":0.44218415,"punctuation_ratio":0.18454936,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98610955,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-06T01:53:08Z\",\"WARC-Record-ID\":\"<urn:uuid:833b5061-f1df-487e-9007-acb6a9a3c18b>\",\"Content-Length\":\"44607\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e90f3d29-b848-4389-b1e0-8c0560c24aba>\",\"WARC-Concurrent-To\":\"<urn:uuid:c1623a4f-aa89-41d1-939b-5605b730e1d2>\",\"WARC-IP-Address\":\"104.21.33.54\",\"WARC-Target-URI\":\"https://www.asknumbers.com/inch-to-mm/199-inches-to-mm.aspx\",\"WARC-Payload-Digest\":\"sha1:52M4IUA7VS33QWVNLHJQBAX53EBAMBDY\",\"WARC-Block-Digest\":\"sha1:OMJO5EOZTWNLWI3XGTW7YXU73CPE2BIM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100575.30_warc_CC-MAIN-20231206000253-20231206030253-00595.warc.gz\"}"}
https://www.thelearningpoint.net/computer-science/learning-python-programming-and-data-structures/learning-python-programming-and-data-structures--tutorial-19--self-referential-data-structures-linked-lists-single-double-and-circular-binary-search-trees	[ "## 1. Stacks in Python\n\nFor a detailed explanation about the general ideas and principles behind a stack data structure, you might find this tutorial useful. Default Python lists can easily be used as stacks in Python. Appending to a list is the equivalent on pushing an element onto a stack. Popping from a list is the equivalent of de-queing an element from the stack.\n\n ```# You can use Lists as Stacks in Python stack = [10,9,8,7,6,5] # Original contents of the stack print \"Original Contents of the Stack\" print stack # Appending to a list is the same as pushing to a stack stack.append(1) stack.append(2) # In the two steps above we push 1 and 2 onto the stack print \"After pushing 1 and 2 onto the stack it looks like:\" print stack # Now we explore the pop operation poppedValue = stack.pop() # Display the popped value print \"Popped Value:\" print poppedValue # Now display what the stack looks like: print \"After the pop operation the stack looks like:\" print stack # Now we explore the pop operation again poppedValue = stack.pop() # Display the popped value print \"Again, we pop a value from the top of the stack. Popped Value:\" print poppedValue # Now display what the stack looks like: print \"After the second pop operation the stack looks like:\" print stack ```\n\n### Output of the Program above:\n\n`~/work/pythontutorials\\$ python DataStructuresStacks.py `\n`Original Contents of the Stack`\n`[10, 9, 8, 7, 6, 5]`\n`After pushing 1 and 2 onto the stack it looks like:`\n`[10, 9, 8, 7, 6, 5, 1, 2]`\n`Popped Value:`\n`2`\n`After the pop operation the stack looks like:`\n`[10, 9, 8, 7, 6, 5, 1]`\n\n## 2. Queues in Python\n\nFor a detailed explanation about the functioning of the Queue data structure (which is a FIFO Data Structure) you might find this tutorial very useful.\nFor creating Queues in Python we can use the collections.deque data structure because it was designed to have efficient pops and pushes from both ends unlike lists which are designed to work efficiently from the right end.\n ```# Deques from collections are convenient to use as Queues # It is not efficient to use lists because they are efficient for reading/appending/popping from the end # But they are not so efficient for dequeing from the beginning from collections import deque queue = deque([\"London\",\"Paris\",\"New York\",\"Delhi\"]) print \"The Original Queue:\" print queue # Now we also queue a few more cities queue.append(\"Mumbai\") queue.append(\"Kolkata\") # Now display the queue after en-queueing these print queue # You will observe that Mumbai and Kolkata have been en-queued at the end of the queue # Now let us start to De-que element from this queue dequedElement1 = queue.popleft() dequedElement2 = queue.popleft() # Let us display the Dequeued Elements. # Given that a Queue is a First In First Out (FIFO) data structure the de-queued elements will be London and Paris print \"Two cities were de-queued\" print \"First Dequeued city:\" print dequedElement1 print \"First Dequeued city:\" print dequedElement2 # You will notice how the first two cities have been removed from the Queue, in FIFO order print \"Current state of the queue after dequeing two cities:\" print queue ```\n\n### Output of the Program above:\n\n`~/work/pythontutorials\\$ python DataStructuresQueues.py `\n`The Original Queue:`\n`deque(['London', 'Paris', 'New York', 'Delhi'])`\n`deque(['London', 'Paris', 'New York', 'Delhi', 'Mumbai', 'Kolkata'])`\n`Two cities were de-queued`\n`First Dequeued city:`\n`London`\n`First Dequeued city:`\n`Paris`\n`Current state of the queue after dequeing two cities:`\n`deque(['New York', 'Delhi', 'Mumbai', 'Kolkata'])`\n\n## 3. Linked Lists in Python\n\nFor a detailed explanation about Linked Lists and the common operations on them, check out the tutorial here. Linked lists are linear, self referential data structures. Common operations on them are insertion, deletion, traveral/display, and finding elements.\n```class Node:\n\ndef __init__(self,data=None,next=None):\nself.data = data\nself.next = next\n\ndef __str__(self):\nreturn \"Node[Data=\" + `self.data` + \"]\"\n\ndef __init__(self):\n\n# Inserting new data at the end of the list\n# Iterate through the list till we encounter the last node.\n# A new node is created for this data element\n# And the last pointer points to this\ndef insert(self,data):\nelse:\nwhile (current.next != None) and (current.data == data): current = current.next\t\t\tcurrent.next = Node(data)\n\n# Deleting a given data value from the linked list\n# If the head contains this data value\n# Set head = node which comes next after the current head\n# Otherwise go to the node such that the node after it (next to it) contains the value we're looking for\n# set node.next = node.next.next\n# so, the node which dontains the specified value/data; is skipped\ndef delete(self,data):\nif current.data == data:\nif current == None:\nreturn False\nelse:\nwhile (current != None) and (current.next != None) and (current.next.data != data):\ncurrent = current.next\nif (current != None) and (current.next != None) :\ncurrent.next = current.next.next\n\n# Find a given data value in the linked list\n# Traverse the linked list till you either find the data value or you come to the end of the list\n\ndef find(self,data):\nfound = False\nwhile ((current != None) and (current.data != data) and ( current.next != None)):\ncurrent = current.next\nif current != None:\nfound = True\nreturn found\n\n# Traverse the linked list till you reach its end\n# Display each node which you traverse\ndef display(self):\nstring_representation = \" \"\nwhile current != None:\nstring_representation += str(current) + \"--->\"\ncurrent = current.next\nprint string_representation\n\n# Initialize a new linked list\n\n# Insert values in the linked list\nprint \"Inserting values 1,2,3,9 in the Linked List\"\nll.insert(1)\nll.insert(2)\nll.insert(3)\nll.insert(9)\n\nll.display()\n\n# Delete an element from the linked list. Demonstrate the Delete function\nprint \"Delete an element (data = 3) from the linked list\"\nll.delete(3)\n\nprint \"Display the linked list again. The value 3 is deleted. \"\nll.display()\n\n# Try to find the value 2 in the linked list (Demonstrating the Find function)\nprint \"Try to find the value 2 in the linked list\"\nfound = ll.find(2)\nif found == True:\nprint \"The value 2 is present in the Linked List\"\nelse:\nprint \"The value 2 is not present in the linked list\"\n\n# Try to find the value 20 in the linked list\nprint \"Try to find the value 20 in the linked list\"\nfound = ll.find(20)\nif found == True:\nprint \"The value 20 is present in the Linked List\"\nelse:\nprint \"The value 20 is not present in the linked list\"\n```\n\n### Output of the Program above:\n\n`~/work/pythontutorials\\$ python DataStructuresLinkedLists.py `\n`Initializing linked list`\n`Inserting values 1,2,3,9 in the Linked List`\n`Displaying the linked list`\n` Node[Data=1]--->Node[Data=2]--->Node[Data=3]--->Node[Data=9]--->`\n`Delete an element (data = 3) from the linked list`\n`Display the linked list again. The value 3 is deleted. `\n` Node[Data=1]--->Node[Data=2]--->Node[Data=9]--->`\n`Try to find the value 2 in the linked list`\n`The value 2 is present in the Linked List`\n`Try to find the value 20 in the linked list`\n`The value 20 is not present in the Linked List`\n\n## 4. Double Linked Lists in Python\n\n ```class Node: # Constructor to initialize data # If data is not given by user,its taken as None def __init__(self, data=None, next=None, prev=None): self.data = data self.next = next self.prev = prev # __str__ returns string equivalent of Object def __str__(self): return \"Node[Data = %s]\" % (self.data,) class DoubleLinkedList: def __init__(self): self.head = None self.tail = None def insert(self, data): if (self.head == None): # To imply that if head == None self.head = Node(data) self.tail = self.head else: current = self.head while(current.next != None): current = current.next current.next = Node(data, None, current) self.tail = current.next def delete(self, data): current = self.head # If given item is the first element of the linked list if current.data == data: self.head = current.next self.head.prev = None return True # In case the linked list is empty if current == None: return False # If the element is at the last if self.tail == data: self.tail = self.tail.prev self.tail.next = None return True # If the element is absent or in the middle of the linked list while current != None: if current.data == data : current.prev.next = current.next current.next.prev = current.prev return True current = current.next # The element is absent return False def find(self, data): current = self.head while current != None: if current.data == data : return True current = current.next return False def fwd_print(self): current = self.head if current == None: print(\"No elements\") return False while (current!= None): print (current.data) current = current.next return True def rev_print(self): current = self.tail if (self.tail == None): print(\"No elements\") return False while (current != None): print (current.data) current = current.prev return True # Initializing list l = DoubleLinkedList() # Inserting Values l.insert(1) l.insert(2) l.insert(3) l.insert(4) # Forward Print l.fwd_print() # Reverse Print l.rev_print() # Try to find 3 in the list if (l.find(3)): print(\"Found\") else : print(\"Not found\") # Delete 3 from the list l.delete(3) # Forward Print l.fwd_print() # Reverse Print l.rev_print() # Now if we find 3, we will not get it in the list if (l.find(3)): print(\"Found\") else : print(\"Not found\") ```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7386524,"math_prob":0.7643894,"size":9442,"snap":"2023-14-2023-23","text_gpt3_token_len":2424,"char_repetition_ratio":0.18022886,"word_repetition_ratio":0.19513798,"special_character_ratio":0.2656217,"punctuation_ratio":0.1540107,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9907615,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-07T18:50:35Z\",\"WARC-Record-ID\":\"<urn:uuid:9155a222-2846-45c7-be51-9a6453c35c6b>\",\"Content-Length\":\"152217\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b8a0532f-1aad-4a10-850a-64f2bc5eb6a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:76e431b9-6966-4314-b72e-45103eff4fa5>\",\"WARC-IP-Address\":\"172.67.209.130\",\"WARC-Target-URI\":\"https://www.thelearningpoint.net/computer-science/learning-python-programming-and-data-structures/learning-python-programming-and-data-structures--tutorial-19--self-referential-data-structures-linked-lists-single-double-and-circular-binary-search-trees\",\"WARC-Payload-Digest\":\"sha1:S3O4JLIP6KKAHURYWESXBR4ZAOBHMLML\",\"WARC-Block-Digest\":\"sha1:KSIFVVZATHCZ3W7ZWQ37OCFIAMARRYZU\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224654012.67_warc_CC-MAIN-20230607175304-20230607205304-00597.warc.gz\"}"}
https://mda.tools/docs/plsda-classification-plots.html	[ "## Classification plots\n\nMost of the plots for visualisation of classification results described in SIMCA chapter can be also used for PLS-DA models and results. Let’s start with classification plots. By default it is shown for cross-validation results (we change position of the legend so it does not hide the points). You can clearly spot for example three false positives and one false negatives in the one-class PLS-DA model for virginica.\n\npar(mfrow = c(1, 2))\nplotPredictions(m.all)\nplotPredictions(m.vir)", null, "In case of multiple classes model you can select which class to show the predictions for.\n\npar(mfrow = c(1, 2))\nplotPredictions(m.all, nc = 1)\nplotPredictions(m.all, nc = 3)", null, "" ]	[ null, "https://mda.tools/docs/_main_files/figure-html/unnamed-chunk-152-1.png", null, "https://mda.tools/docs/_main_files/figure-html/unnamed-chunk-153-1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8638602,"math_prob":0.9633679,"size":529,"snap":"2023-40-2023-50","text_gpt3_token_len":105,"char_repetition_ratio":0.13714285,"word_repetition_ratio":0.0,"special_character_ratio":0.18336484,"punctuation_ratio":0.052083332,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9582886,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T04:35:26Z\",\"WARC-Record-ID\":\"<urn:uuid:eb2329f0-7d08-414a-b0f2-f9f8b902bfb7>\",\"Content-Length\":\"28472\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6a061dbc-aa44-438b-ba0d-d40c59200fb8>\",\"WARC-Concurrent-To\":\"<urn:uuid:10be7cf1-e5c2-4a38-8652-1b5111b4a8b9>\",\"WARC-IP-Address\":\"172.66.0.96\",\"WARC-Target-URI\":\"https://mda.tools/docs/plsda-classification-plots.html\",\"WARC-Payload-Digest\":\"sha1:TIXYBOHCHXNGLWKHGEEPF2L6A5D3NWVI\",\"WARC-Block-Digest\":\"sha1:XXY4NJLEBZRSGBLVGTSRL2LCZGJWJHJG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510481.79_warc_CC-MAIN-20230929022639-20230929052639-00417.warc.gz\"}"}
https://chemistry.stackexchange.com/questions/153175/calculating-the-apparent-molecule-length-from-density-of-the-bulk	[ "# Calculating the apparent molecule length from density of the bulk [closed]\n\nI'm interested to estimate the apparent length of the water molecule assuming it could be approximated as a cube. This is an intentional simplification.\n\n## Calculating molecule length from volume in density equation\n\nI used this equation to find the volume for a mole of water molecules with $$\\varrho$$ (water density, assumed $$\\pu{0.997 g cm^{-3}}$$), $$m$$ (mass), and $$V$$ (volume).\n\nFor the determination of the molar volume, I substituted mass $$m$$ by molar mass $$M$$ and rearranged\n\n$$\\varrho = m / V$$\n\ninto the equivalent form\n\n$$V_{\\mbox{molar}} = \\frac{M} {\\varrho} = \\frac {\\pu{18.01 g mol^{-1}}} {\\pu{0.997 g cm^{-3}}} = \\pu{18.06 cm^3 mol^{-1}}$$\n\nBecause $$\\pu{1 m^3} = \\pu{10^6 cm^3}$$, the molar volume equates to $$\\pu{18.06 \\times 10^{-6} m^3 mol^{-1}}$$.\n\nGiven Avogadro's number, I computed the volume of the individual volume with\n\n$$V_{\\mbox{molecule}} = \\frac{ \\pu{18.06 \\times 10^{-6} m^3 mol^{-1}}} { \\pu{6.022 \\times 10^{23} mol^{-1}} } = \\pu{2.999 \\times 10^{-29} m^3}$$\n\nThen I took the cube root to find the length, as I want $$m^3 \\rightarrow m$$:\n\n$$\\ell(\\text{molecule}) = \\sqrt{\\pu{2.999 \\times 10^{-29} m^3}} = \\pu{3.10 \\times 10^{-10} m} = \\pu{0.31 nm}.$$\n\n#### What I need help with\n\nThis isn't taking intermolecular forces into account, but I want to know if it's an ok method for getting a rough idea of the molecule length, and if it's done correctly.\n\n• Density often is denoted by $\\rho$ ($\\rho$) or $\\varrho$ ($\\varrho$ in LaTeX / mhchem syntax available to you here on chemistry.se), which is different to $p$ then typically used about pressure. But this is smallish compared to the initial equation, because density is not mass times volume, but mass divided by volume (e.g., water, 1 g per cubic centimetre, or about a [metric] ton per cubic metre. The length you obtain is about twice as wide as the (intramolecular) H-H distance (about 0.15 nm). – Buttonwood Jun 20 at 20:29\n• Assuming molecules as small cubes is a simplification and it depends a lot on the circumstances if this is acceptable, or not. It may be a start of a model later to be refined (e.g. limiting the number of interactions modelling crystal growth) for water's bent, or benzene's flat shape. For future reference: if you use an equality sign, then both numbers and dimensions must equate, too. This shows e.g., when they cancel out (dimension analysis) even if this demands a bit more to type / write. – Buttonwood Jun 20 at 21:58\n• Over time (and repeated exposure) you will pick up skills new to you. Once you have enough reputation, for example, you may e.g. access to the edits made by other users. This equally is a source of inspiration (and eventually, training) by reference for how to format questions and answers in the community of chemistry.se (maybe you know the expression of «when in Rome, do as the Romans do»). – Buttonwood Jun 21 at 19:47\n• Strange this question was closed for the reason given but from things you have said @buttonwood my question is answered (about if it can be used roughly to estimate length), would gladly have given you the answer if your stuff wasn’t in a comment. – Nickotine Jun 22 at 11:40\n• It actually is fascinating how close you get with this absolutely simplistic approach. I#m not sure why this was closed, I wouldn't have. But I'm also not confident enough to overrule this decision. Also (but not entirely sure how helpful this is): researchgate.net/post/… – Martin - マーチン Jun 24 at 21:51" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7666839,"math_prob":0.99579144,"size":1865,"snap":"2021-31-2021-39","text_gpt3_token_len":591,"char_repetition_ratio":0.12466416,"word_repetition_ratio":0.0,"special_character_ratio":0.35603216,"punctuation_ratio":0.1064935,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992737,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-23T20:15:27Z\",\"WARC-Record-ID\":\"<urn:uuid:f129f8a9-af4d-42b6-8e8f-507e1a7f53ad>\",\"Content-Length\":\"144609\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:53561313-0fee-44d4-a7e1-a05548767349>\",\"WARC-Concurrent-To\":\"<urn:uuid:e9d092e6-33dd-4e46-a45b-b8ec3e2807b0>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://chemistry.stackexchange.com/questions/153175/calculating-the-apparent-molecule-length-from-density-of-the-bulk\",\"WARC-Payload-Digest\":\"sha1:4PDS7QSPL4VUJFSD6CJDV2N2DHFMJRB2\",\"WARC-Block-Digest\":\"sha1:BYE4FLE6A7OSGGDDFTYWZGG4K55KF6AD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046150000.59_warc_CC-MAIN-20210723175111-20210723205111-00056.warc.gz\"}"}
https://bitesizebio.com/10951/how-does-it-move-interpreting-motion-of-an-object-with-the-mean-squared-displacement/	[ "# How Does it Move? Interpreting Motion of an Object with the Mean-Squared Displacement\n\nStuff moves. It is useful to study how stuff moves, because motion analysis can tell us a lot about the object that is moving. For example, we can learn if an object’s motion is aimless, diffusive wandering, or directed towards some goal, free to explore the available environment, or restricted to a confined space.\n\nStudying the motion of a single object requires different tools than studying the average motion of a population of objects. Here, I want to introduce you to one of my favorite tools, that mainstay of biophysics, the mean-squared displacement. In this article, I’ll first explain what mean-squared displacement (MSD) is and show you a few of the easiest things to learn from a MSD plot. In a follow-up article, I’ll show you how to make your own plot from a series of measurements of an object’s location.\n\nWhat kind of object you study doesn’t really matter; I use MSD to analyze motion of proteins and nucleic acids in cells, but the analysis could just as validly be applied to study the motion of individual insects, of amoeboid cells in a petri dish, of a macrophage in a tissue, or really whatever you’re interested in. The only measure required is the object’s location over time.\n\n## So what is MSD?\n\nMSD is a statistic that tells you how much your object displaces, on average, in a given interval of time, squared. So if you saw something like this:", null, "You would interpret it as saying that your object displaces 1000nm2 in a 5 second time interval, on average. For easier interpretation, you could take the square root of that, which gives you the “root mean squared displacement.” Then, you’d be able to say that your object displaces about 100nm in 5 seconds, on average.\n\nIf it is easier to think about specific data points by taking their square root, why don’t we show root MSD plots instead of MSD plots? Well, some people do. But most still use MSD, because it turns out that looking at motion^2 makes it easy to learn some cool things about the kind of motion the object is doing, as you’ll see below.\n\nYou might have noticed I keep saying that MSD describes how much your object displaces, not how much your object moves. This is because usually when you take movies of your object, there is time between the frames. And though you can measure the difference in the object’s position between two frames (i.e. displacement), you don’t really know how much the object moved in that time. For all you know, it may have traveled to the sun and back when your camera wasn’t looking.\n\nIt’s also good to keep in mind that MSD is a mean. Your object might have moved 1 nm in one 5 second interval and 1000 nm in another 5 second interval and have the same MSD as an object that moved 500nm in both 5 second intervals. It’s important to look at the standard deviation of MSDs to see how consistent the displacements are.\n\nSo MSD tells you how much your object displaced in a given interval of time. That’s great, but the really cool thing about MSD is what you can learn when you look at graphs that show many MSD data points over increasing intervals of time. Let’s do that:", null, "The x-axis is what frequently trips people up: the x-axis in in units of a time interval. Often, this is represented as delta-t, a change in time (although sometimes confusingly is just represented as “time”). People often see a “t” or an “s” on the x-axis and then assume the plot is showing a time series. IT IS NOT A TIME SERIES! The x-axis shows time intervals. I cannot stress this enough.\n\nNow that you know what MSD is and the axes, let’s briefly go through the four main things you can detect with a fast glance at an MSD plot. The details behind the following truths are a bit beyond the scope of this article, but can be found in the primary lit.\n\n## 1. Diffusion\n\nIf the MSD plot follows a linear trend, then the object whose motion you are observing moves diffusively; i.e. without direction, as in a random walk. For example, an unbound GFP particle will move diffusively through a cell’s cytoplasm.", null, "## 2. Active motion\n\nIf the MSD plot does not follow a linear trend but instead follows an increasing slope, then the object exhibits directed motion. For example, the motor protein dynein moves along microtubules in a directed fashion, always going from the “plus” end of microtubules to the “minus” end.", null, "## 3. Constrained motion\n\nIf the plot plateaus, then the motion of the object is constrained. Furthermore, the square root of the height of the plateau (minus error) defines the size of the region of constraint. For example, if you labeled a centromere in a budding yeast cell, it could never move outside of the nucleus, and this would be reflected in an MSD plot by a constraint likely with the plateau at the height of the diameter of the nucleus, squared.", null, "## 4. Measurement error\n\nThe intercept of an MSD plot is the measurement error (which can include a number of sources, including microscopy errors and errors in particle tracking in your image analysis software). This makes intuitive sense: the intercept of an MSD plot is where the time interval = 0. Since an object can’t move in zero time, the intercept must reflect measurement error.", null, "With this foundation, you’re equipped to do a basic interpretation of MSD plots. Keep in mind that while we have learned a lot from a simple plot, we can go deeper still, and learn all sorts of cool things like mechanical properties of molecular polymers. Stay tuned for the follow-up, when we learn to make MSD plots from a time series of object locations!\n\n1.", null, "Najib on July 14, 2017 at 8:09 pm\n\n2.", null, "j on January 7, 2016 at 11:31 am\n\nPretty useful article for understanding these graphs! But isn’t the square root of 1000 31.6 instead of 100?\n\nCheers,\n\nJ\n\n•", null, "Joe on November 12, 2019 at 7:56 pm" ]	[ null, "https://bitesizebio.com/wp-content/uploads/2013/08/First-figure-MSD-copy.jpg", null, "https://bitesizebio.com/wp-content/uploads/2013/08/Figure-2-MDS1.jpg", null, "https://bitesizebio.com/wp-content/uploads/2013/09/diffusion-353x265.png", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20width='282'%20height='212'%20viewBox='0%200%20282%20212'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20width='282'%20height='212'%20viewBox='0%200%20282%20212'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20width='282'%20height='212'%20viewBox='0%200%20282%20212'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20width='80'%20height='80'%20viewBox='0%200%2080%2080'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20width='80'%20height='80'%20viewBox='0%200%2080%2080'%3E%3C/svg%3E", null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20width='80'%20height='80'%20viewBox='0%200%2080%2080'%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9328218,"math_prob":0.889003,"size":5912,"snap":"2022-27-2022-33","text_gpt3_token_len":1311,"char_repetition_ratio":0.12626946,"word_repetition_ratio":0.011417697,"special_character_ratio":0.21532476,"punctuation_ratio":0.10288066,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9710382,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,4,null,4,null,4,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-26T23:57:54Z\",\"WARC-Record-ID\":\"<urn:uuid:b94dec8e-ee7b-438b-a090-aef3738f01b3>\",\"Content-Length\":\"98977\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:481a0688-8e2d-45c5-ba6a-30d4c9cb11ee>\",\"WARC-Concurrent-To\":\"<urn:uuid:4fa6aa74-8aba-47a5-a3b6-c6e1c5befbf5>\",\"WARC-IP-Address\":\"162.159.135.42\",\"WARC-Target-URI\":\"https://bitesizebio.com/10951/how-does-it-move-interpreting-motion-of-an-object-with-the-mean-squared-displacement/\",\"WARC-Payload-Digest\":\"sha1:BUT5ZSRGGS7L5NWTMV6AVH4WUDJJ6DGS\",\"WARC-Block-Digest\":\"sha1:NTW5HRB7N6CFV2ZZLW3ICEVCEHZO5QOS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103322581.16_warc_CC-MAIN-20220626222503-20220627012503-00195.warc.gz\"}"}
https://everything2.com/title/Green%2527s+Theorem	[ "Theorem 1: Green's Theorem: Let D be a simple closed region in two dimensions, and let its boundary be C. For a vector field F(x,y) = (P(x,y), Q(x,y)), the line integral of F on the positive orientation of C is equal to the surface integral over the region D of the partial derivative of Q with repsect to x minus the partial derivative of P with respect to y\n\nTheorem 2: Area of a Region: If C is a simple closed curve that bounds a region to which Theorem 1 applies, then the area of the region D bounded by C is one-half the line integral along C of F(x,y)=(-y,x)\n\nTheorem 3: Vector Form of Green's Theorem: Let D be a subset of R-2 be a region to which Green's theorem applies, let C be its boundary (oriented counter-clockwise), and let F=(P,Q) be continuously differentiable vector field on D. Then, the line integral along C is equal to the surface integral of <curl(F),k> over the region D.\n\ngreen machine = G = greenbar\n\nGreen's Theorem prov.\n\n[TMRC] For any story, in any group of people there will be at least one person who has not heard the story. A refinement of the theorem states that there will be exactly one person (if there were more than one, it wouldn't be as bad to re-tell the story). [The name of this theorem is a play on a fundamental theorem in calculus. --ESR]\n\n--The Jargon File version 4.3.1, ed. ESR, autonoded by rescdsk.\n\nLog in or register to write something here or to contact authors." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9273084,"math_prob":0.9924244,"size":896,"snap":"2020-10-2020-16","text_gpt3_token_len":229,"char_repetition_ratio":0.1412556,"word_repetition_ratio":0.073619634,"special_character_ratio":0.25,"punctuation_ratio":0.11165048,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989956,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-30T17:23:20Z\",\"WARC-Record-ID\":\"<urn:uuid:fb692609-d081-4988-8cb7-c195fa08d7d4>\",\"Content-Length\":\"21835\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e7f50127-bd4e-4e73-b153-7b0d0077986c>\",\"WARC-Concurrent-To\":\"<urn:uuid:cf1a5286-57f4-4fcb-bef7-0c77aa2e9d7f>\",\"WARC-IP-Address\":\"35.165.66.52\",\"WARC-Target-URI\":\"https://everything2.com/title/Green%2527s+Theorem\",\"WARC-Payload-Digest\":\"sha1:ATBOYFGVTKJ6GKJSLJECYZMBXT7EZM2P\",\"WARC-Block-Digest\":\"sha1:NGDZP4MP34RAA2NYBNKIVGXYE4ZOOUEI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370497171.9_warc_CC-MAIN-20200330150913-20200330180913-00502.warc.gz\"}"}
https://shichaoxin.com/2021/11/09/%E6%9C%BA%E5%99%A8%E5%AD%A6%E4%B9%A0%E5%9F%BA%E7%A1%80-%E7%AC%AC%E4%BA%8C%E5%8D%81%E4%B9%9D%E8%AF%BE-%E9%9B%86%E6%88%90%E5%AD%A6%E4%B9%A0%E4%B9%8BBagging%E4%B8%8E%E9%9A%8F%E6%9C%BA%E6%A3%AE%E6%9E%97/	[ "# 【机器学习基础】第二十九课：集成学习之Bagging与随机森林\n\n## Bagging，“包外估计”（out-of-bag estimate），随机森林（Random Forest）\n\nPosted by x-jeff on November 9, 2021\n\n【机器学习基础】系列博客为参考周志华老师的《机器学习》一书，自己所做的读书笔记。\n\n# 1.前言\n\n【机器学习基础】第二十七课：集成学习之个体与集成可知，欲得到泛化性能强的集成，集成中的个体学习器应尽可能相互独立；虽然“独立”在现实任务中无法做到，但可以设法使基学习器尽可能具有较大的差异。给定一个训练数据集，一种可能的做法是对训练样本进行采样，产生出若干个不同的子集，再从每个数据子集中训练出一个基学习器。这样，由于训练数据不同，我们获得的基学习器可望具有比较大的差异。然而，为获得好的集成，我们同时还希望个体学习器不能太差。如果采样出的每个子集都完全不同，则每个基学习器只用到了一小部分训练数据，甚至不足以进行有效学习，这显然无法确保产生出比较好的基学习器。为解决这个问题，我们可考虑使用相互有交叠的采样子集。\n\n# 2.Bagging", null, "$\\mathcal{D}_{bs}$是自助采样产生的样本分布。\n\nBagging是一个很高效的集成学习算法。值得一提的是，自助采样过程还给Bagging带来了另一个优点：由于每个基学习器只使用了初始训练集中约63.2%的样本，剩下约36.8%的样本可用作验证集来对泛化性能进行“包外估计”（out-of-bag estimate）。为此需记录每个基学习器所使用的训练样本。不妨令$D_t$表示$h_t$实际使用的训练样本集，令$H^{oob}(\\mathbf x)$表示对样本$\\mathbf x$的包外预测，即仅考虑那些未使用$\\mathbf x$训练的基学习器在$\\mathbf x$上的预测，有：\n\n$H^{oob}(\\mathbf x)=\\arg \\max \\limits_{y \\in \\mathcal{Y}} \\sum^T_{t=1} \\mathbb{I} (h_t(\\mathbf x)=y) \\cdot \\mathbb{I} (\\mathbf x \\notin D_t)$\n\n$\\epsilon^{oob} = \\frac{1}{\\lvert D \\rvert} \\sum_{(\\mathbf x,y)\\in D} \\mathbb{I} (H^{oob}(\\mathbf x) \\neq y)$\n\n# 3.随机森林", null, "" ]	[ null, "https://github.com/x-jeff/BlogImage/raw/master/MachineLearningSeries/Lesson29/29x1.png", null, "https://shichaoxin.com/img/shanghaiwaitan.jpeg", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.94303924,"math_prob":0.9991246,"size":2095,"snap":"2022-27-2022-33","text_gpt3_token_len":1990,"char_repetition_ratio":0.09803922,"word_repetition_ratio":0.0,"special_character_ratio":0.1718377,"punctuation_ratio":0.030303031,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99579203,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-28T00:26:13Z\",\"WARC-Record-ID\":\"<urn:uuid:5777043e-0078-492d-a1b3-a9f81f409275>\",\"Content-Length\":\"30367\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cf92f5cb-e7ad-483a-ba47-ecb4cf7e574e>\",\"WARC-Concurrent-To\":\"<urn:uuid:8c1d928f-8862-4e99-a8cb-a1c579d9b1ba>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"https://shichaoxin.com/2021/11/09/%E6%9C%BA%E5%99%A8%E5%AD%A6%E4%B9%A0%E5%9F%BA%E7%A1%80-%E7%AC%AC%E4%BA%8C%E5%8D%81%E4%B9%9D%E8%AF%BE-%E9%9B%86%E6%88%90%E5%AD%A6%E4%B9%A0%E4%B9%8BBagging%E4%B8%8E%E9%9A%8F%E6%9C%BA%E6%A3%AE%E6%9E%97/\",\"WARC-Payload-Digest\":\"sha1:OAWHQIJTKYXS7EBNVMVWPGRE22YNHU67\",\"WARC-Block-Digest\":\"sha1:J7DDEGABR6MOHHGNPGDKE7ACBMJ7EWTU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103344783.24_warc_CC-MAIN-20220627225823-20220628015823-00186.warc.gz\"}"}
https://www.intechopen.com/books/optimization-algorithms-examples/piecewise-parallel-optimal-algorithm	[ "Open access peer-reviewed chapter\n\n# Piecewise Parallel Optimal Algorithm\n\nBy Zheng Hong Zhu and Gefei Shi\n\nSubmitted: October 14th 2017Reviewed: March 20th 2018Published: September 5th 2018\n\nDOI: 10.5772/intechopen.76625\n\n## Abstract\n\nThis chapter studies a new optimal algorithm that can be implemented in a piecewise parallel manner onboard spacecraft, where the capacity of onboard computers is limited. The proposed algorithm contains two phases. The predicting phase deals with the open-loop state trajectory optimization with simplified system model and evenly discretized time interval of the state trajectory. The tracking phase concerns the closed-loop optimal tracking control for the optimal reference trajectory with full system model subject to real space perturbations. The finite receding horizon control method is used in the tracking program. The optimal control problems in both programs are solved by a direct collocation method based on the discretized Hermite–Simpson method with coincident nodes. By considering the convergence of system error, the current closed-loop control tracking interval and next open-loop control predicting interval are processed simultaneously. Two cases are simulated with the proposed algorithm to validate the effectiveness of proposed algorithm. The numerical results show that the proposed parallel optimal algorithm is very effective in dealing with the optimal control problems for complex nonlinear dynamic systems in aerospace engineering area.\n\n### Keywords\n\n• optimal control\n• parallel onboard optimal algorithm\n• discretizing Hermite–Simpson method\n• nonlinear dynamic system\n• aerospace engineering\n\n## 1. Introduction\n\nSpace tether system is a promising technology over decades. It has wide potential applications in the space debris mitigation & removal, space detection, power delivery, cargo transfer and other newly science & technic missions. Recently, there is continuous interest in the space tether systems, in leading space agencies such as, NASA’s US National Aeronautics and Space Administration, ESA’s European Space Agency, and JAXA’s Japan Aerospace Exploration Agency . Their interest technologies include the electrodynamic tether (EDT) propulsion technology, retrieval of tethered satellite system, multibody tethered system and space elevator system. Compared with existing technologies adopted by large spacecraft such as the rocket or thruster, the space tether technology has the advantages of fuel-efficiency (little or no propellant required), compact size, low mass, and ease-of-use . These advantages make it reasonable to apply the space tethered system for deorbiting the fast-growing low-cost micro/nano-satellites and no-fuel cargo transfer. The difficulty associated with space tether system is to control & suppress its attitudes during a mission process for the technology to be functional and practical. Many works have been devoted to solving this problem, and one effort is to use the optimal control due to its good performances in the complex and unstable nonlinear dynamic systems. In this chapter, a new piecewise onboard parallel optimal control algorithm is proposed to control and suppress the attitudes of the space tether system. To test its validity, two classical space tether systems, the electrodynamic tether system (EDT) and partial space elevator (PSE) system are considered and tested.\n\nAn EDT system with constant tether length is underactuated. The electric current is the only control input if there are no other active forces such as propulsion acting on the ends of an EDT. The commonly adopted control strategy in the literature is the current regulation using energy-based feedback in this underactuated control problem. Furthermore, many efforts have been done to solve this problem with optimal control. Stevens and Baker studied the optimal control problem of the EDT libration control and orbital maneuverer efficiency by separating the fast and slow motions using an averaged libration state dynamics as constraints instead of instantaneous dynamic constraints in the optimal control algorithm. The instantaneous states are propagated from the initial conditions using the optimal control law in a piecewise fashion. Williams treated the slow orbital and fast libration motions separately with two different discretization schemes in the optimal control of an EDT orbit transfer. The differential state equations of the libration motion are enforced at densely allocated nodes, while the orbital motion variables are discretized by a quadrature approach at sparsely allocated nodes. The two discretization schemes are unified by a specially designed node mapping method to reflect the coupling nature of orbital and libration motions. The control reference, however, is assumed known in advance.\n\nA PSE system is consisted with one main satellite and two subsatellites (climber & end body) connected to each other by tether(s). The difficulty associated to such a system is to suppress the libration motion of the climber and the end body. This libration is produced by the moving climber due to the Coriolis force, which will lead the system unstable. While the climber is fast moving along the tether, the Coriolis force will lead to the tumbling of the PSE system. Thus, the stability control for suppressing such a system is critical for a successful climber transfer mission. To limit the fuel consuming, tension control is widely used to stable the libration motion of the space tethered system due to it can be realized by consuming electric energy only . Many efforts have been devoted to suppressing the libration motion of space tethered system such as, Wen et al. stabled the libration of the tethered system by an analytical feedback control law that accounts explicitly for the tension constraint. The study shows good computational effect, and the proposed method requires small data storage ability. Ma et al. used adaptive saturated sliding mode control to suppress the attitude angle in the deployment period of the space tethered system. Optimal control [8, 9] is also proved as a way to overcome the libration issue. The above tension control schemes are helpful for both two-body and three-body tethered system. Up to data, limited devotions have been done on the libration suppression of a PSE system using tension control only. Williams used optimal control to design the climber’s speed function of a climber for a full space elevator . Modeled by simplified dynamic equations, an optimal control problem is solved, and the solution results in zero in-plane libration motion of the ribbon in the ending phase of climber motion. The study shows that to eliminate the in-plane oscillations by reversing the direction of the elevator is possible. Kojima et al. extended the mission function control method to eliminate the libration motion of a three body tethered system. The proposed method is effective when the total tether length is fixed and the maximum speed of the climber no more than 10 m/s.Although these efforts are useful to suppress the libration motion of the PSE system, it still difficult to control the attitudes of such a system in the transfer period.\n\nTo overcome the challenges in aforementioned works, we propose a parallel onboard optimal algorithm contains two phases. Phase 1 concerns the reference state trajectory optimization within a given time interval, where an optimal control model is formulated based on the timescale separation concept [3, 12] to simplify the dynamic calculations of the EDT & PSE system. An open-loop optimal state trajectory is then obtained by minimizing a cost function subject to given constraints. The state trajectory of paired state and control input variables is solved approximately by the direct collocation method that is based on the Hermite-Simpson method . In this phase, the simplified dynamic model is by used. Phase 2 concerns the tracking of the open-loop optimal state trajectory within the same interval. A closed-loop optimal control problem is formulated in a quadrature form to track the optimal state trajectory obtained in phase 1. Unlike phase 1, all the major perturbative forces are included, and more realistic geomagnetic and gravitational field models are considered. While the system is running the process in phase 2 with one CPU, the next phase 1 calculation is running in another CPU with data modification based on the errors obtained in the last calculation program. The simulation results demonstrate the effectiveness of the approach in fast satellite deorbit by EDTs in equatorial orbit. Furthermore, for fast transfer period of the partial space elevator, the propose method also shows good effect on suppression the libration angles of the climber and the end body with tension control only.\n\n## 2. Optimal control algorithm\n\n### 2.1. Control scheme\n\nAssume two CUPs are used to process the calculation. CUP-1 is used to determine the open-loop optimal control trajectory of dynamic states employing the simple dynamic equations. The obtained optimal state trajectory will be tracked by CPU-2 using closed-loop RHC. While the system is tracking the i-th interval, the (i + 1)-th optimal trajectory is being calculated in CPU-1. Once the tracking for the i-th interval is finished (implemented by CPU-2), the real finial state Siwill be stored in the memory and the (i + 1)-th optimal trajectory can be tracked. By repeating the above process, the optimal suppression control problem is solved in a parallel piecewise manner until the transfer period is over.\n\nThe calculation of the optimal trajectory in CPU-1 is a prediction state trajectory whose initial state is estimated as S˜ii+1=S˜ii+12ei1, where S˜iidenotes the estimated finial state of the i-th interval obtained by CUP-1, the superscript and subscript denote the internal number and the states node number. S˜ii+1is the estimated initial state of the (i + 1)-th interval, and ei1=Si1S˜i1i1is the error between the real and the estimated final state of the (i-1)-th interval, the data used to calculate the error is picked up from the memory. For the first interval, i = 1, S0=S˜01and e0=0. The computational diagram of the entire control strategy is given as shown in Figure 1.\n\n### 2.2. Open-loop control trajectory\n\nThe libration angles of the climber and the end body are required to be kept between the desired upper/lower bounds in the climber transfer process. To make the calculation convenient and simple. The accessing process is divided into a series of intervals, such that, the transfer process titi+1is discretized into nintervals, where tiand ti+1are the initial and final time, respectively. ti+1can be obtained in terms of the transfer length of the climber and its speed. To make calculation convenient to be realized in practical condition, the transfer process is divided evenly. The optimal trajectory should be found to satisfy the desired cost functions for each time intervals as\n\nJi=titi+1ΠxudtE1\n\nsubject to the simplified dynamic equations. All the errors between simple model and the entire model are regarded as perturbations. The above cost function minimization problem is solved by a direct solution method, which uses a discretization scheme to transform the continuous problem into a discrete parameter optimization problem of nonlinear programming within the interval, to avoid the difficulty usually encountered when standard approaches are used on derivation of the required conditions for optimality . There are a number of efficient discretization schemes, such as, Hermite-Legendre-Gauss-Lobatto method and Chebyshev pseudospectral method , in the literature for discretizing the continuous problem. In the current work, a direct collocation method, based on the Hermite-Simpson scheme [14, 17], is adopted because of its simplicity and accuracy.\n\nAssume that the time interval titi+1is discretized into nsubintervals with n + 1 nodes at the discretized time τk(k=0,1,.,n).\n\nγk=τk+1τk,k=1nγk=ti+1tiE2\n\nThe state vectors and control inputs are discretized at n + 1 nodes, x0, x1, x2, …, xnand υ0, υ1, υ2, …, υn. Further, denote the state vectors and the control inputs at mid-points between adjacent nodes by x0.5, x1.5, x2.5, …, xn0.5and υ0.5, υ1.5, υ2.5, …, υn0.5and the mid-point state vectors, xk+0.5, can be derived by the Hermite interpolation scheme,\n\nxk+0.5=12xk+xk+1+γ8ΓxkυkτkΓ(xk+1υk+1τk+1)E3\n\nAccordingly, the cost function in Eq. (1) can be discretized by the Simpson integration formula as\n\nJγ6k=0n1Πxkυkτk+4Π(xk+0.5υk+0.5τk+0.5)+Π(xk+1υk+1τk+1)E4\n\nThe nonlinear constraints based on the tether libration dynamics, the first-order states, can also be denoted by discretized equations using the Simpson integration formula, such that\n\nγ6Πxkυktk+4Π(xk+0.5υk+0.5τk+0.5)+Π(xk+1υk+1τk+1)+xkxk+1=0E5\n\nThe left-hand side of Eq. (5) is also named as the Hermite-Simpson Defect vector in the literature. Finally, the discretization process is completed by replacing the constraints for the initial states and the continuous box constraints with the discretized constraints,\n\nx0=xstart,xminxkxmax,υminυkυmax,υminυk+0.5υmaxE6\n\nThe minimization problem of a continuous cost function is now transformed to a nonlinear programming problem. It searches optimal values for the programming variables that minimize the discretized form of cost function shown in Eq. (4) while satisfying the constraints of Eqs. (5) and (6). The subscript index “k” will be refreshed in the next time interval.\n\n### 2.3. Closed-loop optimal control for tracking open-loop optimal state trajectory\n\nThe RHC is implemented by converting the continuous optimal control problem into a discrete parameter optimization problem that can be solved analytically. Like the open-loop trajectory optimization problem, the same direct collocation based on the Hermite-Simpson method is used to discretize the RHC problem.\n\nBy using the similar notations of discretization above, the cost function is discretized using the Simpson integration formula as\n\nG=12δxi+1TSδxi+1+γ12k=0n1δxkTQδxk+4δxk+0.5TQδxk+0.5+δxk+1TQδxk+1+Rδυk2+4δυk+0.52+δυk+12E7\n\nand the constraints are discretized into\n\nδxkδxk+1+γ6Akδxk+Bkδυk+4Ak+0.5δxk+0.5+4Bk+0.5δυk+0.5+Ak+1δxk+1+Bk+1δυk+1=0E8\nδxk=xτkxoptτkE9\nδxk+0.5=12δxk+δxk+1+γ8Akδxk+BkδυkAk+1δxk+1Bk+1δυk+1E10\n\nwhere, Ak=Aτk, Ak+0.5=Aτk+0.5, Bk=Bτk, Bk+0.5=Bτk+0.5.\n\nThe derivation of Eq. (8a) finally leads to a quadratic programming problem to find a programming vector Z=δx0Tδx1TδxnTδυ0δυ1δυnδυ0.5δυ1.5δυn0.5T, which minimizes the cost function:\n\nG=12ZTMZE11\n\nsubject to.\n\nCZ=X X=δx0T000TE12\n\nwhere the matrices Cand Mare given in the Appendix.\n\nIt is easy to find the solution analytically to this standard quadratic programming problem by .\n\nZ=M1CTCM1CT1XE13\n\nand the control correction at the current time can be obtained as\n\nδυti=VZ=VM1CTCM1CT1XKtinthxtixopttiE14\n\nwhere the row vector Vis defined to “choose” the target value from the optimal solution, and the position of “1” in the row vector Vis the same as the position of δυ0in the column vector Z. Finally, the control input of the closed-loop control, υti, is\n\nυti=υoptti+δυti=υoptti+KtinthxtixopttiE15\n\nIt is apparent that the closed-loop control law derived here is a linear proportional feedback control law, and the feedback gain matrix Kis a function of time. Without any explicit integration of differential equations, Kcan either be determined offline or online depending on the computation and restoration ability onboard the satellite.\n\nIt is worth to point out some advantages of this approach. Firstly, the matrices Mand Care both formulated by the influence matrices at certain discretization notes (Ak, Bk, Ak + 0.5, Bk + 0.5). If tiand tiare both set to be coincident with the discretization nodes used in the open-loop control problem, then most of the influence matrices calculated previously can be used directly in the tracking control process to reduce computational efforts. This is the advantage of using the same discretization method in the current two-phased optimal control approach. Secondly, the matrix Mis unchanged if treating the terminal horizon time ti+thas the time-to-go and keep the future horizon interval titi+1unchanged. This means the inverse of Mcould be calculated only once in the same interval. It is attractive for the online implementation of HRC, where the computational effort is critical. As the entire interval can be discretized into small intervals, if these intervals are sufficiently small relative to the computing power of the satellite, then the calculation process can be carried by the onboard computer. Furthermore, for small intervals, the computation for the open-loop optimal trajectory of the next interval can be done by CPU-1 while the tracking is still in process, see Figure 1. This makes of the proposed optimal suppression control a parallel online implementation, which is another advantage of this control scheme.\n\n## 3. Cases study\n\n### 3.1. Parallel optimal algorithm in attitudes control of EDT system\n\nIn order to test the validity of the proposed optimal algorithm, a case study of the attitudes control of EDT system in aerospace engineering is used. The obtained results are compared with some existing control methods.\n\n#### 3.1.1. Problem formulation\n\nThe EDT system’s orbital motion is generally described in an Earth geocentric inertial frame (OXYZ) with the origin Oat the Earth’s centre, see Figure 2(a). The X-axis directs to the point of vernal equinox, the Z-axis aligns with the Earth’s rotational axis, and the Y-axis completes a right-hand coordinate system, respectively. The equation of orbital dynamic motion can be written in the form of Gaussian perturbation , which is a set of ordinary differential equations of six independent orbital elements (a, Ω, i, ex, ey, φ)\n\ndΩdt=σyrsinuna21e2siniE17\ndidt=σyrcosuna21e2E18\ndexdt=1e2naσzsinu+σx1+rpcosu+rpex+dΩdteycosiE19\ndeydt=1e2naσzcosu+σx1+rpsinu+rpeydΩdtexcosiE20\ndt=n1naσz2ra+1e21+1e2ecosνσx1+rp1e21+1e2esinνσyrcosisinuna21e2siniE21", null, "Figure 2.Illustration of coordinate system for the EDT’s orbital (a) and libration (b) motion.\n\nThe components of perturbative accelerations are defined in a local frame. The components σxand σzare in the orbital plane, and σzis the radial component pointing outwards. The out-of-plane component σycompletes a right-hand coordinate system. The components of perturbative accelerations depend on the tether attitude and the EDT’ orbital dynamics is coupled with the tether libration motion.\n\nThe libration motion of a rigid EDT system is described in an orbital coordinate system shown in Figure 2(b). The z-axis of the orbital coordinate system points from the Earth’s center to the CM of the EDT system, the x-axis lies in the orbital plane and points to the direction of the EDT orbital motion, perpendicular to the z-axis. The y-axis completes a right-hand coordinate system. The unit vectors along each axis are expressed as eox, eoy, and eoz, respectively. Then, the instantaneous attitudes of the EDT system are described by an in-plane angle α(pitch angle, rotating about the y-axis) followed by an out-of-plane angle β(roll angle, rotating about the x’-axis, the x-axis after first rotating about the y-axis). Thus, the equations of libration motion of the EDT system can be derived as,\n\nα¨+ν¨2α̇+ν̇β̇tanβ+3μr3sinαcosα=Qαm˜L2cos2βE22\nβ¨+α̇+ν̇2sinβcosβ+3μr3cos2αsinβcosβ=Qβm˜L2E23\n\nwhere m˜=m1m2+m1+m2mt/3+mt2/12mEDT1is the equivalent mass, (Qα, Qβ) are the corresponding perturbative torques by the perturbative forces to be discussed below.\n\nThe perturbative accelerations (σx, σy, σz) and torques (Qα, Qβ) in Eq. (15) are induced by multiple orbital perturbative effects, namely, (i) the electrodynamic force exerting on a current-carrying EDT due to the electromagnetic interaction with the geomagnetic field, (ii) the Earth’s atmospheric drag, (iii) the Earth’s non-homogeneity and oblateness, (iv) the lunisolar gravitational perturbations, and (v) the solar radiation pressure, respectively. The EDT system is assumed thrust-less during the deorbit process, while the atmosphere, geomagnetic and ambient plasma fields are assumed to rotate with the Earth at the same rate. The geodetic altitude, instead of geocentric altitude, should be used in the evaluation of the environmental parameters, such as, atmospheric and plasma densities, to realistically account for the Earth’s ellipsoidal surface, such that,\n\nhg=rrpo1eE2cos2θ1/2E24\n\nwhere the polar radius rpoand the Earth’s eccentricity eEare provided by NASA .\n\nMoreover, the local strength of geomagnetic field is described by the IGRF2000 model [20, 21, 22] in a body-fixed spherical coordinates of the Earth, such that\n\nBϕ=1sinθn=1r0rn+2m=0nmgnmsinhnmcosPnmθcBθ=n=1r0rn+2m=0ngnmcos+hnmsinPnmθcθcBr=n=1r0rn+2n+1m=0ngnmcos+hnmsinPnmθcE25\n\nwhere r0 = 6371.2 × 103 km is the reference radius of the Earth, respectively.\n\nThe average current in the EDT is defined as\n\nIave=1L0LIsdsE26\n\nThe open-loop optimal control problem for EDT deorbit can be stated as finding a state-control pair xtυtover each time interval titi+1to minimize a cost function of the negative work done by the electrodynamic force\n\nJ=titi+1Fe·vdtttti+1ΠxυtdtE27\n\nsubject to the nonlinear state equations of libration motion\n\nx1=x2x2=2η3esinν+2x2+η2x4tanx33η3sinx1cosx1+sinitanx32sinucosx1cosusinx1cosiςλIaveμmμm˜η3E28\nx3=x4x4=x2+η22sinx3cosx33η3cos2x1sinx3cosx3sini2sinusinx1+cosucosx1ςλIaveμmμm˜η3E29\n\nwhere x1x2x3x4=ααββ, η=1+ecosν, ν̇=μ/p30.51+ecosν2, r=p1+ecosν1, λ=m1+0.5mt/mEDTis determined by the mass ratio between the end-bodies, and ςis determined by the distribution of current along the EDT, such that, ς=Iave1L20LsIsds. Accordingly, ς=0.5is used for the assumption of a constant current in the EDT. The initial conditions xti=xstartand the box constraint ααmax,ββmax,IminIaveImax. The environmental perturbations are simplified by considering only the electrodynamic force with a simple non-tilted dipole model of geomagnetic field, such that,\n\nB=μmr3cosusinieox+μmr3cosieoy2μmr3sinusinieozE30\n\nAccordingly, the electrodynamic force Feexerting on the EDT can be obtained as,\n\nFe=0LB×Ilds=IaveLB×lE31\n\n#### 3.1.2. Results and discussion\n\nThe initial and boundary conditions of box constraints of the case are shown in Tables 1 and 2.\n\nParametersValues\nMass of the main satellite5 kg\nMass of subsatellite1.75 kg\nMass of the tether0.25 kg\nDimensions of main satellite0.2 × 0.2 × 0.2 m\nDimensions of subsatellite0.1 × 0.17 × 0.1 m\nTether length500 m\nTether diameter0.0005 m\nTether conductivity (aluminum)3.4014 × 107 Ω−1 m−1\nTether current lower/upper limits0 ~ 0.8 A for the equatorial orbit\nOrbital altitudes700 ~ 800 km\n\n### Table 1.\n\nParameters of an EDT system.\n\nParametersValues\nImax(equatorial)0.4 A\nImax(inclined)0.1sinisinβ¯BsinΩG+α¯BΩ+cosicosβ¯Bcos1iβ¯BA\nImin(equatorial & inclined)0 A\nαmax(equatorial & inclined)45 degrees\nβmax(equatorial & inclined)45 degrees\n\n### Table 2.\n\nBoundary Values of Box Constraints in the Open-Loop Trajectory Optimization.\n\nFirstly, the validity of the proposed optimal control scheme in the equatorial orbit where the EDT system gets the highest efficiency is demonstrated. The solid line in Figure 4 shows the time history of the EDT’s average current control trajectory obtained from the open-loop optimal control problem. It is clearly shown that the average current in the open-loop case reaches the upper limit most of the time, which indicates the electrodynamic force being maximized for the fast deorbit. As expected, the current is not always at the upper limit in order to avoid the tumbling of the EDT system. This is evident that the timing of current reductions coincides with the peaks of pitch angles shown in Figure 5. The effectiveness of the proposed control scheme in terms of keeping libration stability is further demonstrated by the solid lines in Figure 5, where the trajectory of libration angles is no more than 45 degrees. It is also found that the amplitude of the pitch angle nearly reaches 45 degrees, the maximum allowed value, whereas the roll angle is very small in the whole deorbit process. As a comparison, tracking optimal control with the non-tilted dipole model and the IGRF 2000 model of the geomagnetic fields are conducted respectively.\n\nThe dashed lines in Figures 3 and 4 show the tracking control simulations where all perturbations mentioned before are included with the non-tilted dipole geomagnetic field model. As expected, the closed-loop tracking control works well in this case since the primary electrodynamic force perturbation is the same as the one used in the open-loop trajectory optimization. It is shown clearly in Figure 4 that the pitch angle under the proposed closed-loop control tracks the open-loop optimal trajectory very closely with this simple environment model. Figure 4 also shows the roll angle is almost zero even if it is not tracked. At the same time, Figure 3 shows that the current control modification to the optimal current trajectory is relative small, i.e., 12% above the maximum current, for the same reason. Now the same cases are analyzed again using a more accurate geomagnetic field model – the IGRF 2000 model with up to 7th order terms (Figures 5 and 6). The solid line in Figure 5 is the open-loop current control trajectory while the dashed line is the modified current control input obtained by the receding horizon control. Compared with Figure 3, it shows more current control modifications are needed to track the open-loop control trajectory because of larger differences in dynamic models between the open-loop and closed-loop optimal controls, primarily due to the different geomagnetic field models. Because of the same reason, it is noticeable in Figure 6 that the instantaneous states of the EDT system, controlled by the closed-loop optimal control law, are different from the open-loop reference state trajectory at the end of the interval. The instantaneous states are used for the next interval as initial conditions for the open-loop optimal control problem to derive the optimal control trajectory in that interval. This is reflected in Figure 6 that the solid lines are discontinuous at the beginning of each interval. The dashed line in Figure 7 shows that the pitch and roll angle under the closed-loop control has been controlled to the open-loop control trajectory, indicating the effectiveness of the proposed optimal control law. The roll angle is not controlled in this case as mentioned before. Compared Figure 4 with Figure 6, it shows that the roll angle increases significantly since there is an out-of-plane component of the electrodynamic force resulting from the IGRF 2000 geomagnetic model. However, the amplitude of the roll angle is acceptable within the limits and will not lead to a tumbling of the EDT system.", null, "Figure 3.Time history of average current in the equatorial orbit (non-tilted dipole geomagnetic model). Solid line: Open-loop state trajectory. Dashed line: Close-loop tracking trajectory.", null, "Figure 4.Time history of pitch and roll angles in the equatorial orbit (non-tilted dipole geomagnetic model). Solid line: Open-loop state trajectory. Dashed line: Close-loop tracking trajectory.", null, "Figure 5.Time history of average current in the equatorial orbit (IGRF 2000 model). Solid line: Open-loop state trajectory. Dashed line: Close-loop tracking trajectory.", null, "Figure 6.Time history of pitch and roll angles in the equatorial orbit (IGRF 2000 model). Solid line: Open-loop control trajectory. Dashed line: Close-loop tracking trajectory.", null, "Figure 7.Comparison of EDT deorbit rates using different control laws and geomagnetic field models.\n\nFinally, we make a comparison to show the performance of the proposed onboard parallel optimal control law from the aspect of deorbit rate. A simple current on–off control law from a previous work of Zhong and Zhu is used here as baselines for the comparison of EDT deorbit efficiency. The current on–off control becomes active only if the libration angles exceed the maximum allowed values. Furthermore, it will turn on the current only in the condition that the electrodynamic force does negative work in both pitch and roll directions. In this paper, the maximum allowed amplitude for pitch and roll angles was set to 20° and the turned-on current was assumed to be 0.4 A, roughly the average value of the current control input into the closed-loop optimal control. Besides, a minimum interval of 10 minutes for the switching was imposed to avoid equipment failure that might happen due to the frequent switching. Figure 8 shows the comparisons of the deorbit rates in different cases (the present optimal control and the current switching control with the non-tilted dipole or the IGRF 2000 model of geomagnetic field). It is shown that the EDT deorbit under the proposed optimal control scheme is faster than the current on–off control regardless which geomagnetic field model is used. The deorbit time of proposed optimal control based on the IGRF2000 model is about 25 hours, which equals approximately 15 orbits, whereas the deorbit time of simple current on–off control based on the same geomagnetic field model is about 55 hours, which equals approximately 33 orbits. The results also indicate that in the optimal control scheme, the effect is mostly shown in the current control input, instead of the deorbit rate, where Figure 6 shows much more current control effort is required due to the different magnetic field models were used in the open-loop control trajectory optimization and the closed-loop optimal tracking control.\n\n### 3.2. Parallel optimal algorithm in libration suppression of partial space elevator\n\nFor further test of the effect of the onboard parallel algorithm, the proposed control method is used to suppress the libration motions of the partial space elevator system. As studied in , this system is a non-equilibrium nonlinear dynamic system. It is difficult to suppress such a system in the mission period by using the common control design methods. In this case, we mainly concern obtaining the local time optimization.\n\n#### 3.2.1. Problem formulation\n\nConsider an in-plane PSE system in a circular orbit is shown in Figure 8, where the main satellite, climber and the end body are connected by two inelastic tethers L1and L2, respectively. The masses of the tethers are neglected. Assuming the system is subject to a central gravitational field and orbiting in the orbital plane. All other external perturbations are neglected. The main satellite, climber and end body are modeled as three point masses (M, m1and m2) since the tether length is much greater than tethered bodies [5, 23]. Thus, the libration motions can be expressed in an Earth inertial coordinate system OXYwith its origin at the centre of Earth. Denoting the position of the main satellite (M) by a vector rmeasuring from the centre of Earth. The climber m1is connected to the main satellite Mby a tether 1 with the length of L1and a libration angle θ1measured from the vector r. The distance between them is controlled by reeling in/out tether 1 at main satellite. The end body m2is connected to m1by a tether 2 with the length of L2and a libration angle θ2measured from the vector r. The length of tether 2 L2is controlled by reeling in or out tether 2 at end body. The mass of the main satellite is assumed much greater than the masses of the climber and the end body. Therefore, the CM of the PSE system can be assumed residing in the main satellite that moves in a circular orbit. Based on the aforementioned assumptions, the dynamic equations can be written as.\n\nθ¨1=3ω2sin2θ122ω+θ̇1L̇1L1sinθ1θ2T2L1m1E32\nθ¨2=3ω2sin2θ222ω+θ̇2L̇cL̇1L0L1+Lc+sinθ1θ2T1L0L1+Lcm1E33\nL¨1=3ω2L1cos2θ1+2ωL1θ̇1+L1θ̇12T1m1+cosθ1θ2T2m1E34\nL¨c=3ω2L0L1+Lccos2θ2+3ω2L1cos2θ1+2ω+θ̇2L0L1+Lcθ̇2+2ω+θ̇1L1θ̇1+cosθ1θ21T1m1m1m2cosθ1θ2+m2T2m1m2E35\n\nwhere L0is the initial total length of two pieces of the tethers and Lcis the length increment relates to L0.\n\nThe libration angles are required to be kept between the desired upper/lower bounds in the climber transfer process. The accessing process is divided into a series of intervals. In this case, we modified the aforementioned parallel optimal algorithm. The total transfer length L10L1fof the climber is discretized evenly. The optimal trajectory should be found to make the transfer time minimize in each equal tether transferring length. Then the cost function can be rewritten as\n\nJi=tfE36\n\nsubject to the simplified dynamic equations\n\nθ¨1=3ω2θ12ω+θ̇1L̇1L1θ1θ2T2L1m1E37\nθ¨2=3ω2θ22ω+θ̇2L̇cL̇1L0L1+Lc+θ1θ2T1L0L1+Lcm1E38\nL¨1=3ω2L1+2ωL1θ̇1+L1θ̇12+T2T1m1E39\nL¨c=3ω2L0+Lc+2ω+θ̇2L0L1+Lcθ̇2+2ω+θ̇1L1θ̇1T2m2E40\n\nwhere u=T1T2, x=θ1θ̇1θ2θ̇2L1L̇1and idenotes the interval number. In (26) the gravitational perturbations and the trigonometric functions are ignored, then they can be simplified following the assumptions: sinθjθj,cosθj1j=12. All the errors between simple model and the entire model are regarded as perturbations.\n\nTo ensure the availability and the suppression of the libration angles, following constrains are also required to be subjected 0T1T1max,0T2T2max,θ1θ1max,θ2θ2max,LcLcLimitL̇1mL̇1L̇1M,L̇cmL̇cL̇cM, where T1maxand T2maxare the upper bounds of the tension control inputs T1 and T2, respectively. θ1max, θ2maxand LcLimitare the magnitudes of libration angles θ1, θ2 and the maximum available length scale of Lc, respectively. L̇1mand L̇1Mare the lower and upper bounds of climber’s moving speed L̇1, respectively. L̇cmand L̇cMare the lower and upper bounds of end-bodies’ moving speed L̇c, respectively. It should be noting that, to avoid the tether slacking, the control tensions are not allowed smaller than zero. Dividing the time interval titi+1evenly into nsubintervals. The cost function minimization problem for each time interval can be solved by Hermite–Simpson method, due its simplicity and accuracy . Then nonlinear programming problem is to search optimal values for the programming variables that minimize the cost function for each interval shown in (25). The closed-loop optimal tracking control method, is same as that in case 1. Direct transcription methods are routinely implemented with standard nonlinear programming (NLP) software. The sparse sequential quadratic programming software SNOPT is used via a MATLAB-executable file interface.\n\n#### 3.2.2. Results and discussion\n\nThe proposed control scheme is used to suppress the libration angles of the PSE system in the ascending process with following system parameters and initial conditions: r = 7100 km, m1 = 500 kg, m2 = 1000 kg, θ1(0) = θ2(0) = 0, L0 = 20 km, L1(0) = 19,500 m, Lc(0) = 0, θ̇10=θ̇20=0, L̇10=20m/s, and L̇c0=0for the ascending process. The whole transfer trajectory is divided into 50 intervals. The climber’s ascending speed along tether 1 is allowed to be controlled to help suppress the libration angles and keep the states of the system in an acceptable area. The constrains are set as T1max=T2max=200N, θ1max=θ2max=0.3rad, L̇1m=15m/s, L̇1M=25m/s, L̇cm=10m/sand L̇cM=10m/s.\n\nThe simulation results of this case are shown in Figures 911. The climber’s open-loop libration angle approaches its upper bound at 850 s. After 850 sθ1is kept at 0.3 radby the end of the ascending period, see the dashed line in Figure 10. Using the closed-loop control, the tracking trajectory of θ1matches the open-loop trajectory very well overall, see solid line in Figure 9. A short gap appears between 875 s – 880 s, this is caused by the errors of the model and computation. Figure 9 also shows the changes of the trajectories of θ2.The trajectory of θ2obtained by closed-loop control tracks the open-loop trajectory well and reaches 0.1 radby the end of the ascending period. The closed-loop trajectories of L1and Lcare shown in Figure 10. They are the reflections of the control inputs. Both L1and Lcshow smooth fluctuations between 40s and 350 s. Figure 10 shows the time history of trajectories of L̇1and L̇c, respectively. In the first 140 s the trajectory of L̇cincreases continuously until reaches its upper bound. Then, it keeps at 10 m/sby 270 s with some slight fluctuations. From 270 s to 355 s, it reduces continuously to −3 m/s. After that, L̇cfluctuates around −3 m/s by the end of the transfer period. As a reflection of the control input, L̇1also shows fluctuation during the transfer phase with obvious small-scale fluctuations appear in the period of 120 s – 260 s and 360 s – 750 s. This impacts during the whole transfer period, the changeable speed of the climber has the ability to help the suppression of the libration angles and states trajectory tracking. This time history of the control inputs is shown in Figure 11with frequent changes between its lower bound and upper bound.\n\n## 4. Conclusions\n\nThis chapter investigated a piecewise parallel onboard optimal control algorithm to solve the optimal control issues in complex nonlinear dynamic systems in aerospace engineering. To test the validity of the proposed two-phase optimal control scheme, the long-term tether libration stability and fast nano-satellite deorbit under complex environmental perturbations and the libration suppression for PSE system are considered. For EDT system, instead of optimizing the control of fast and stable nano-satellite deorbit over the complete process, the current approach divides the deorbit process into a set of intervals. For the PSE system, each time interval is set depends on the minimize transfer time for equal transfer length interval. Within each interval, the predicting phase simplifies significantly the optimal control problem. The dynamic equations of libration motion are further simplified to reduce computational loads using the simple dynamic models. The trajectory of the stable libration states and current control input is then optimized for the fast deorbit within the interval based on the simplified dynamic equations. The tracking optimizes the trajectory tracking control using the finite receding horizon control theory within the time interval corresponding to the open-loop control state trajectory with the same interval number. By applying the close-loop control modification, the system motions are integrated without any simplification of the dynamics or environmental perturbations and the instantaneous states of the orbital and libration motions. The i-th time interval’s closed-loop tracking is processed in tracking phase while the (i + 1)-th time interval’s optimal state trajectory is predicted in the predicting phase. This prediction is based on the error between the real state and the predicting state in the (i-1)-th time interval. By repeating the process, the optimal control problem can be achieved in a piecewise way with dramatically reduced computation effort. Compared with the current on–off control where the stable libration motion is the only control target, numerical results show that the proposed optimal control scheme works well in keeping the libration angles within an allowed limit.\n\n## Acknowledgments\n\nThis work is funded by the Discovery Grant of the Natural Sciences and Engineering Research Council of Canada, the National Natural Science Foundation of China, Grand No. 11472213 and the Chinese Scholarship Council Scholarship No. 201606290135.\n\n## A. Appendix\n\nThe detailed expressions for the matrixes Cand Mare shown as followed,\n\nC=C1C2C3,M=M11M120M12TM22000M33\nC1=[E0χ01χ02χ11χ12χn11χn120E],C2=[00χ03χ04χ13χ14χn13χn1400],C3=23γ¯[00B0.5Bn0.500]χj1=E+γ¯6Aj+γ¯3Aj+0.5+γ¯212Aj+0.5Aj,χj2=E+γ¯6Aj+1+γ¯3Aj+0.5γ¯212Aj+0.5Aj+1χj3=γ¯212Aj+0.5Bj+γ¯6Bj,χj4=γ¯212Aj+0.5Bj+1+γ¯6Bj+1\nM11=ϖ0011ϖ01110ϖ0111Tϖ1111ϖn1n111ϖn1n110ϖn1n11Tϖnn11M22=ϖ0022ϖ01220ϖ0122Tϖ1122ϖn1n122ϖn1n220ϖn1n22Tϖnn22\nM12=ϖ0012ϖ01120ϖ1012ϖ1112ϖn1n112ϖn1n120ϖnn112ϖnn12M33=23γ¯R00R\nϖjj11=γ¯64Q+γ¯28AjTQAj,ϖjj+111=γ¯6Qγ¯216AjTQAj+1+γ¯4AjTQγ¯4QAj+1ϖ0011=γ¯62Q+γ¯216A0TQA0+γ¯4A0TQ+γ¯4QA0,ϖnn11=γ¯62Q+γ¯216AnTQAnγ¯4AnTQγ¯4QAn+Sfϖjj22=γ¯62R+γ¯28BjTQBj,ϖjj+122=γ¯BjTQBj+1ϖ0022=γ¯6R+γ¯216B0TQB0,ϖnn22=γ¯6R+γ¯216BnTQBnϖjj12=γ¯348AjTQBj,ϖjj+112=γ¯224QBj+1+γ¯4AjTQBj+1ϖj+1j12=γ¯224QBjγ¯4Aj+1TQBj,ϖ0012=γ¯224QB0+γ¯4A0TQB0,ϖnn12=γ¯224QBN¯γ¯4AnTQBn\n\nwhere Eis the unit matrix which has the same dimension as Aj, and j = 0,1,2,…,n-1.\n\nchapter PDF\nCitations in RIS format\nCitations in bibtex format\n\n## More\n\n© 2018 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution 3.0 License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\n## How to cite and reference\n\n### Cite this chapter Copy to clipboard\n\nZheng Hong Zhu and Gefei Shi (September 5th 2018). Piecewise Parallel Optimal Algorithm, Optimization Algorithms - Examples, Jan Valdman, IntechOpen, DOI: 10.5772/intechopen.76625. Available from:\n\n### chapter statistics\n\n1Crossref citations\n\n### Related Content\n\n#### Optimization Algorithms\n\nEdited by Jan Valdman\n\nNext chapter\n\n#### Bilevel Disjunctive Optimization on Affine Manifolds\n\nBy Constantin Udriste, Henri Bonnel, Ionel Tevy and Ali Sapeeh Rasheed\n\n#### Applications from Engineering with MATLAB Concepts\n\nEdited by Jan Valdman\n\nFirst chapter\n\n#### Digital Image Processing with MATLAB\n\nBy Mahmut Sinecen\n\nWe are IntechOpen, the world's leading publisher of Open Access books. Built by scientists, for scientists. Our readership spans scientists, professors, researchers, librarians, and students, as well as business professionals. We share our knowledge and peer-reveiwed research papers with libraries, scientific and engineering societies, and also work with corporate R&D departments and government entities." ]	[ null, "https://www.intechopen.com/media/chapter/61007/media/F2.png", null, "https://www.intechopen.com/media/chapter/61007/media/F3.png", null, "https://www.intechopen.com/media/chapter/61007/media/F4.png", null, "https://www.intechopen.com/media/chapter/61007/media/F5.png", null, "https://www.intechopen.com/media/chapter/61007/media/F6.png", null, "https://www.intechopen.com/media/chapter/61007/media/F7.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8972564,"math_prob":0.94817185,"size":23108,"snap":"2021-21-2021-25","text_gpt3_token_len":4554,"char_repetition_ratio":0.17010042,"word_repetition_ratio":0.02739726,"special_character_ratio":0.18746755,"punctuation_ratio":0.07555221,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98484504,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-17T01:48:56Z\",\"WARC-Record-ID\":\"<urn:uuid:5f848fdc-0fdd-4acb-b3c1-a5afdf52063c>\",\"Content-Length\":\"712229\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5e1d028d-6526-422e-8cd2-32eea9532378>\",\"WARC-Concurrent-To\":\"<urn:uuid:3e296546-ebe8-4aca-bd11-83977afef45d>\",\"WARC-IP-Address\":\"35.171.73.43\",\"WARC-Target-URI\":\"https://www.intechopen.com/books/optimization-algorithms-examples/piecewise-parallel-optimal-algorithm\",\"WARC-Payload-Digest\":\"sha1:ARVLLW7C7FH5C3BDUOYDDA3CBL26P4OC\",\"WARC-Block-Digest\":\"sha1:7WGLRB7DWR3GV52COTBXZQ44DIPH6FRE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991921.61_warc_CC-MAIN-20210516232554-20210517022554-00353.warc.gz\"}"}
https://quant.stackexchange.com/questions/41605/libor-market-model-lmm-under-risk-neutral-measure	[ "# Libor Market Model (LMM) under risk neutral measure\n\nI would like to establish the equations of forward libors under risk neutral measure. Here is how I do it, and what I get :\n\nUnder the $P_{T_j}$ measure, forward Libor $L_j$ is martingale. Thus:\n\n$$dL_j = L_j \\times \\sigma_j(t) \\times dW^{T_j}$$\n\nThe change of numéraire implies that:\n\n$$\\frac{dQ^{T_j}}{dQ^} = \\frac{P(t,T)\\times\\exp{\\left(-\\int_0^T{r(s)ds}\\right)}}{P(0,T)}$$\n\nUnder $Q^$, $P(t,T)$ discounted is martingale, which means that:\n\n$$\\frac{dP(t,T)}{P(t,T)} = r(t) dt + \\eta(t) dW^$$\n\nSolving this gives:\n\n$$P(t,T) = P(0,T) \\times \\exp\\left(\\int_0^T\\left(r(s)-\\frac{1}{2} \\eta(s)^2\\right)ds + \\int_0^T{\\eta(s)}dW^{T_j}\\right)$$\n\nThus:\n\n$$\\frac{dQ^{T_j}}{dQ^} = \\exp\\left(-\\int_0^T{\\frac{1}{2}} \\eta(s)^2ds + \\int_0^T{\\eta(s)}dW^{T_j}\\right)$$\n\nGirsanov shows that:\n\n$$dW^{T_j} = dW^* - \\eta(t) dt$$\n\nUnder $Q^$:\n\n$$\\frac{dL_j}{L_j} = \\alpha(t) dt + \\sigma_j(t) dW^{}$$\n\nWriting $P(t,T)$ as $\\exp(-\\int_t^T{f(t,s)ds}) = \\exp(Y_t)$:\n\n$$\\frac{dP(t,T)}{P(t,T)} = dY_t +\\frac{1}{2}<Y_t>dt$$\n\nwith $dY_t = f(t,t) dt -\\int_t^T{\\alpha(t) dt ds } - \\int_t^T{\\sigma(t)dt dW_s}$. Identifying, I conclude that:\n\n$$\\eta(t) = \\sigma(t)$$\n\nFinally:\n\n$$dL_j(t) = -L_j(t) \\sigma_j(t) \\int_t^{T_j}{\\sigma_j(s)ds} dt + L_j\\sigma_j(t) dW^.$$\n\nIs it correct ?\n\n• It is not clear why $dY_t = f(t,t) dt -\\int_t^T{\\alpha(t) dt ds } - \\int_t^T{\\sigma(t)dt dW_s}$. Note that $f(t, T)$ is the instantaneous forward rate, while $L_j$ is a forward rate over a given time period, for example, $[T_{j-1}, \\, T_j]$. – Gordon Sep 7 '18 at 18:02\n\nWe assume that, under the $T_j$-forward probability measure $P_{T_j}$, \\begin{align} \\frac{dP(t, T_j)}{P(t, T_j)} = \\mu_P(t, T_j) dt + \\sigma_P(t, T_j) dW_t^{T_j}, \\end{align} where $\\mu_P(t, T_j)$ and $\\sigma_P(t, T_j)$ are the respective drift and volatility functions. Let $Q$ be the risk-neutral probability measure. Then \\begin{align} \\frac{dQ}{dP_{T_j}}\\big\|_t &= \\frac{e^{\\int_0^t r_s ds}P(0, T_j)}{P(t, T_j)}\\\\ &=e^{\\int_0^t \\big(r_s -\\mu_P(s, T_j)+\\frac{1}{2} \\sigma_P(s, T_j)^2 \\big) ds - \\int_0^t \\sigma_P(s, T_j) dW_s^{T_j}}. \\end{align} Since $\\frac{dQ}{dP_{T_j}}\\big\|_t$ is a martingale under $P_{T_j}$, \\begin{align} \\int_0^t \\Big(r_s -\\mu_P(s, T_j)+\\frac{1}{2} \\sigma_P(s, T_j)^2 \\Big) ds = -\\frac{1}{2}\\int_0^t \\sigma_P(s, T_j)^2 ds. \\end{align} That is, \\begin{align} \\mu_P(t, T_j) = r_t + \\sigma_P(t, T_j)^2, \\end{align} and \\begin{align} \\frac{dQ}{dP_{T_j}}\\big\|_t &= e^{-\\frac{1}{2} \\int_0^t\\sigma_P(s, T_j)^2 ds - \\int_0^t \\sigma_P(s, T_j) dW_s^{T_j}}. \\end{align} Then, under the risk-neutral probability measure $Q$, $\\{W_t, \\, t \\ge 0\\}$, where, for $t \\ge 0$, \\begin{align} W_t = W_t^{T_j} + \\int_0^t \\sigma_P(s, T_j) ds, \\end{align} is a standard Brownian motion. Moreover, \\begin{align} \\frac{dL_j}{L_j} = -\\sigma_j(t) \\sigma_P(t, T_j) dt + \\sigma_j(t) d W_t. \\end{align*}" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.57163346,"math_prob":1.0000083,"size":1257,"snap":"2019-51-2020-05","text_gpt3_token_len":559,"char_repetition_ratio":0.14445332,"word_repetition_ratio":0.0,"special_character_ratio":0.45266506,"punctuation_ratio":0.12087912,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000099,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-29T20:27:29Z\",\"WARC-Record-ID\":\"<urn:uuid:b15c53d7-c03e-4958-9291-fefecaa8d9c2>\",\"Content-Length\":\"134472\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dd6e66c2-1105-4a7c-b136-481a2629852b>\",\"WARC-Concurrent-To\":\"<urn:uuid:cb80121f-3567-4841-8e06-198986e468b2>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://quant.stackexchange.com/questions/41605/libor-market-model-lmm-under-risk-neutral-measure\",\"WARC-Payload-Digest\":\"sha1:2LHZS5SNYLEBEQNCSLLG33UOJ2GLYN62\",\"WARC-Block-Digest\":\"sha1:WJP6YZQXQOEFIVRWOKZDHAVPHV5W5DRG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251802249.87_warc_CC-MAIN-20200129194333-20200129223333-00305.warc.gz\"}"}
https://www.elibrary.imf.org/view/journals/001/2015/139/article-A001-en.xml	[ "Does Easing Monetary Policy Increase Financial Instability?\n• 1 0000000404811396https://isni.org/isni/0000000404811396International Monetary Fund\n\n## Contributor Notes\n\nThis paper develops a model featuring both a macroeconomic and a financial friction that speaks to the interaction between monetary and macro-prudential policies. There are two main results. First, real interest rate rigidities in a monopolistic banking system have an asymmetric impact on financial stability: they increase the probability of a financial crisis (relative to the case of flexible interest rate) in response to contractionary shocks to the economy, while they act as automatic macro-prudential stabilizers in response to expansionary shocks. Second, when the interest rate is the only available policy instrument, a monetary authority subject to the same constraints as private agents cannot always achieve a (constrained) efficient allocation and faces a trade-off between macroeconomic and financial stability in response to contractionary shocks. An implication of our analysis is that the weak link in the U.S. policy framework in the run up to the Global Recession was not excessively lax monetary policy after 2002, but rather the absence of an effective regulatory framework aimed at preserving financial stability.\n\n## Abstract\n\nThis paper develops a model featuring both a macroeconomic and a financial friction that speaks to the interaction between monetary and macro-prudential policies. There are two main results. First, real interest rate rigidities in a monopolistic banking system have an asymmetric impact on financial stability: they increase the probability of a financial crisis (relative to the case of flexible interest rate) in response to contractionary shocks to the economy, while they act as automatic macro-prudential stabilizers in response to expansionary shocks. Second, when the interest rate is the only available policy instrument, a monetary authority subject to the same constraints as private agents cannot always achieve a (constrained) efficient allocation and faces a trade-off between macroeconomic and financial stability in response to contractionary shocks. An implication of our analysis is that the weak link in the U.S. policy framework in the run up to the Global Recession was not excessively lax monetary policy after 2002, but rather the absence of an effective regulatory framework aimed at preserving financial stability.\n\n## I Introduction\n\nThe global financial crisis and ensuing Great Recession of 2007-09 have ignited a debate on the role of policies for the stability of the financial system or the economy as a whole (i.e., so called macro-prudential policies). In advanced economies, this debate is revolving around the role of monetary and regulatory policies in causing the global crisis and how the conduct of monetary policy and the supervision of financial intermediaries should be altered in the future to avoid the recurrence of such a catastrophic event.\n\nIn this paper we develop a simple model featuring both a macroeconomic and a financial friction—i.e., a real interest rate rigidity that give rise to a traditional macroeconomic stabilization objective and a pecuniary externality that give rise to a more novel financial stability objective—which speaks to the necessity to complement monetary policy with macro-prudential policies in response to contractionary shocks.\n\nThe prime objective of macro-prudential policy is to limit build-up of system-wide financial risk in order to reduce the frequency and mitigate the impact of a financial crash.1 Most commonly used prudential tools, however, interact with other policy instruments. The overlap between different policy areas is a major challenge for policy-makers, who have to consider the unintended impact of their instruments on other policy objectives as well as the unintended impact of other policy-makers’ instruments on their own policy objective (Svensson, 2012).\n\nFor instance, monetary policy can affect financial stability: investors may be induced to substitute low-yielding, safe assets for higher-yielding, riskier assets (Rajan, 2005, Dell’Ariccia, Laeven, and Marquez, 2011); investors may also be encouraged to take greater risks if they perceive that monetary policy is being used asymmetrically to support asset prices during downturns (Issing, 2009); and asset price increases induced by falling interest rates might cause banks to increase their holdings of risky assets through active balance sheet management (Adrian and Shin, 2009, 2010). On the other hand, macro-prudential policy instruments can affect macroeconomic stability. In fact, by affecting variables such as asset prices and credit, macro-prudential policy is likely to affect a key transmission mechanism of monetary policy (see e.g., Ingves, 2011). This overlap in the respective areas of influence entails the possibility of the instruments having offsetting or amplifying effects on their objectives if they are implemented in an uncoordinated manner, possibly leading to worse outcomes than if the instruments had been coordinated (see Bean, Paustian, Penalver, and Taylor, 2010, Angelini, Neri, and Panetta, 2011, between others).\n\nAgainst this background, some observers have assigned to monetary policy a key role in exacerbating the severity of the global financial crisis of 2007-09. Taylor (2007), in particular, noticed that during the period from 2002 to 2006 the U.S. federal funds rate was well below what a good rule of thumb for U.S. monetary policy would have predicted. Figure 1 displays the actual federal funds rate (solid line) and the counterfactual policy rate that would have prevailed if monetary policy had followed a standard Taylor rule (dashed line). Indeed, the interest rate implied by the Taylor rule is well above the actual federal funds rate, starting from the second quarter of 2002. Taylor (2007) argues that such a counterfactual policy rate would have contained the housing market bubble; moreover, Taylor also supports the idea that deviating from this rule-based monetary policy framework has been a major factor in determining the likelihood and the severity of the 2007-09 crisis (Taylor, 2010).\n\nDespite a somewhat widely shared common sentiment that the Federal Reserve is partly to blame for the housing bubble, the issue is highly controversial in academia and the policy community. Besides Taylor (2007, 2010), Borio and White (2003), Gordon (2005), and Borio (2006) support the idea that monetary policy contributed significantly to the boom that preceded the global financial crisis. In contrast, Posen (2009), Bean (2010), and Svensson (2010) argue against this thesis.2\n\nTo address some of these issues, we develop a simple model of consumption-based asset pricing with collateralized borrowing, monopolistic banking, real interest rates rigidities and pecuniary externalities. The presence of real and financial frictions give rise to both a traditional macroeconomic stabilization role for policy and a more novel financial stability objective.\n\nThe macroeconomic stabilization objective arises from the presence of monopolistic competition and real interest rates rigidities in the banking sector. Due to monopolistic power, banks apply a markup on lending rates. Moreover, when banks cannot fully adjust their lending rates in response to macroeconomic shocks, the economy displays distortions typical of models with staggered price setting, generating equilibrium allocations that are not Pareto efficient (Hannan and Berger, 1991, Kwapil and Scharler, 2010, Gerali, Neri, Sessa, and Signoretti, 2010).\n\nThe financial stability objective stems from the fact that the model endogenously generates financial crises when the borrowing constraint occasionally binds. When access to credit is subject to an occasionally binding collateral constraint, a pecuniary externality arises. Private agents do not internalize the effect of their individual decisions on the market price of collateral, thus borrowing and consuming more than socially efficient, and increasing the frequency and the severity of financial crises.\n\nThe main results of the analysis are two. First, the analysis of our model economy shows that real interest rate rigidities have a different impact on financial stability depending on the sign of the shock hitting the economy. In response to positive shocks to interest rates, aggregate lending rates rise, too. However, because of interest rate stickiness, they increase less than in the flexible interest rate case. This affects next period net worth through two effects. On the one hand, lower lending rates prompt consumers to borrow more than in the flexible rate case, and thus lowering next period net worth; on the other hand, interest rate repayments are lower relative to the flexible case, thus increasing next period net worth. As the second effect dominates the first one in equilibrium (for a wide range of parameter values), the probability of a crisis is lower with interest rate stickiness. Thus, interest rate rigidity acts as an automatic macro-prudential stabilizer in response to shocks that require interest rates to increase. In contrast, when interest rates are hit by a negative shock, aggregate lending interest rates decrease relatively less. Because of the same mechanisms working in reverse, real interest rate rigidity leads to a higher probability of a crisis in response to shocks that lower interest rates (relative to the flexible interest rates case).\n\nSecond, the model shows that a policy authority, facing the same constraints faced by private agents and with only one policy instrument (namely, the policy interest rate), may not achieve efficiency when a shock that lowers interest rates hits the economy. Specifically, in response to shocks that lower interest rates, achieving both macroeconomic and financial stability entails a trade-off because the two objectives require interventions of opposite direction on the same policy tool; in our case, the interest rate. However, when two different instruments are at the policymaker’s disposal (as, for example, a tax on debt and the policy interest rate), efficiency can be achieved in response to both positive and negative shocks to the risk-free interest rate.\n\nOur analysis has important implications regarding the role of U.S. monetary policy for the stability of the financial system in the run-up to the Great recession. In particular, we show that Taylor’s argument—i.e., that higher interest rates would have reduced both the probability and the severity of the great recession—is supported by our theoretical model only if we make the auxiliary assumption that the Fed had to address all distortions in the economy with only one instrument, namely the policy interest rate. However, Taylor’s argument cannot be rationalized in the context of our model when the policy authority has two different instruments. In this case, in response to a negative aggregate demand shock, interest rates ought to be lowered as much as needed without concerns for financial stability. As suggested by Bernanke (2010) and Blanchard, Dell’Ariccia, and Mauro (2010), this implies that the same monetary policy stance as the one adopted by the Fed during the 2002-06 period, accompanied by stronger regulation and supervision of the financial system, might have been more effective in reducing the likelihood and the severity of the crisis, relative to a tighter monetary policy stance with the same financial supervision and regulation observed during the 2002-06 period.\n\nThis paper is related to several strands of literature. The first is the branch of the New Keynesian literature that considers financial frictions and Taylor-type interest rate rules (see Angelini, Neri, and Panetta, 2011, Beau, Clerc, and Mojon, 2012, Kannan, Rabanal, and Scott, 2012, for example). These papers consider either interest rules augmented with macro-prudential arguments—such as credit growth, asset prices, loan-to-value limits—or a combination of interest and macro-prudential rules in order to allow monetary policy to “lean against financial winds.” However, in this class of models, macro-prudential regulation is taken for granted, in the sense that it does not target a clearly identified market failure giving rise to a well defined financial stability objective. In our model, there is a well defined pecuniary externality that justify government intervention for financial stability purposes.\n\nThe second is a growing literature on pecuniary externalities that interprets financial crises as episodes of financial amplification in environments where credit constraints are only occasionally binding (see, between others, Korinek, 2010, Bianchi, 2011, Jeanne and Korinek, 2010a, b, Benigno, Chen, Otrok, Rebucci, and Young, 2013). In this class of models the need for macro-prudential policies stems from a well-defined market failure: a pecuniary externality originating from the presence of the price of collateral in the aggregate borrowing constraint faced by private agents. However, in all these models, the financial friction is the only distortion in the economy. The question of how the pursuit of financial stability may affect macroeconomic stability is therefore novel relative to this literature.\n\nThe third and final is a small, but growing literature that considers both macroeconomic and financial frictions at the same time. Benigno, Chen, Otrok, Rebucci, and Young (2011) analyze a fully specified new open economy macroeconomics 3-period model that features the same financial friction analyzed here and Calvo-style nominal rigidities. The solution of the fully non-linear version of that model (i.e., without resorting to approximation techniques) shows that there is a trade-off between macroeconomic and financial stability, but it is quantitatively too small to warrant the use of a second policy instrument in addition to the interest rate. Kashyap and Stein (2012) use a modified version of the pecuniary externality framework of Stein (2012) where the central bank has both a price stability and a financial stability objective. Similar to our findings, a trade-off emerges between the two objectives when the policy interest rate is the only instrument and it disappears when there is a second instrument (a non-zero interest rate on reserves, in their case). However, they do not model the price stability objective explicitly. Woodford (2012), in contrast, sets up a New Keynesian model with credit frictions, where the probability of a financial crisis is endogenous (i.e., it is a regime-switching process that depends on the model variables). Woodford characterizes optimal policy in this environment, showing that—under certain circumstances—the central bank may face a trade-off between macroeconomic and financial stability. However, he does not explicitly model financial stability.\n\nIn contrast, in our paper, both the macroeconomic and the financial stability objective are well defined and each objective originates from a friction that we model explicitly. The interaction between the macroeconomic and the financial friction delivers a stark trade-off between macroeconomic and financial stability, that helps rationalize the role of monetary policy and macro-prudential policy (or the lack of thereof) in the run-up to the Great Recession in the United States.\n\nThe rest of the paper is organized as follows. Section 2 describes the model economy. Sections 3 and 4 characterize the decentralized and the socially planned equilibrium of the economy, respectively. In Section 5 we discuss the implications of our model in terms of the role played by U.S. monetary policy for the stability of the financial system in the run-up to the Great Recession. Section 6 concludes.\n\n## II The Model\n\nWe include monopolistic banking and real interest rate rigidities in the pecuniary externality framework of Jeanne and Korinek (2010a). In Jeanne and Korinek (2010a)’s set up, consumers borrow directly from international capital markets (or foreign banks). In our model, consumers must borrow from a stylized monopolistic banking sector that intermediates foreign saving. Assuming that some of the changes in US interest rates ultimately originate abroad is in line with the view that the global external imbalances (i.e., the behavior of foreign savings) influenced significantly the US economy in the run-up to the Great Recession. Alternatively, borrowers could be interpreted as entrepreneurs/households in a closed economy enjoying a comparative advantage in owning certain assets.\n\nThe financial friction is given by the presence of collateralized borrowing. Strictly speaking, the real frictions are two: the first is the presence of market power in loan markets, exercised by monopolistically competitive banks, and the second is infrequent adjustment of interest rates by banks.\n\nThe economy is populated by two sets of agents: a continuum of monopolistically competitive banks and a continuum of identical atomistic individuals who borrow from banks and consume. Each set of agents has a mass normalized to one. There are only three periods, denoted t = 0, 1, 2.\n\nAt the beginning of period 0 consumers own an asset whose available stock is normalized to 1. In order to consume they can either sell a fraction of the asset (1 – θi, 1) at market prices or borrow from banks (bi, 1). They have a well-defined demand function for loans which is decreasing in the lending interest rate (RL1). Monopolistic banks freely borrow from foreign lenders at the risk free interest rate (Rt = R) and—given loans demand—optimally set their lending rates. The risk-free interest rate can be hit by a temporary shock (R ± ν) at the beginning of period 0. We assume that only a fraction of banks (μ) can reset their lending rates conditional on this shock, while the remaining banks (1 – μ) need to keep their lending rates fixed. The purpose of this assumption is to introduce macroeconomic stabilization considerations in relatively simple manner.3 The credit market clears after the realization of the shock, which is observed by all agents. At the end of period 0, households consume (ci, 0).\n\nIn period 1, consumers are endowed with a stochastic endowment (e), they repay their debt (bi, 1 RL1), borrow an additional amount from banks (bi, 2), realize banks profits (πi, 1), and consume (ci, 1). Note that debt rollover is subject to a collateral constraint: additional borrowing (bi, 2) is limited to a fraction of consumers’ assets (at their market value). The purpose of this assumption is to introduce financial stability considerations. If hit by a shock in period 0, the level of the risk-free interest rate returns to its pre-shock value (R) in period 1.\n\nPeriod 2 represents the long run. Consumers get a deterministic return on the asset that they own (y), repay their debt (bi, 2 RL2), realize banks profits (πi, 2), and consume (ci, 2).\n\nWe now discuss the consumers’ and banks’ problems in turn.\n\n### A Consumers and Loan Demand\n\nThe utility of each consumer, indexed by i ϵ [0,1], is given by:\n\n$\\begin{array}{cc}u\\left({c}_{i,0}\\right)+u\\left({c}_{i,1}\\right)+{c}_{i,2},& \\left(1\\right)\\end{array}$\n\nwhere, for simplicity, we assume a unitary discount factor. The period utility function, u(·), is a standard CES function:\n\n$\\begin{array}{cc}u\\left(c\\right)=\\frac{{c}^{1-\\rho }}{1-\\rho }.& \\left(2\\right)\\end{array}$\n\nThe budget constraint can be written as:\n\n$\\begin{array}{cc}\\left\\{\\begin{array}{l}{c}_{i,0}={b}_{i,1}+\\left(1-{\\theta }_{i,1}\\right)p0,\\\\ {c}_{i,1}+{b}_{i,1}{R}_{L1}=e+{b}_{i,2}+\\left({\\theta }_{i,1}-{\\theta }_{i,2}\\right){p}_{1}+{\\pi }_{i,1},\\\\ {c}_{i,2}+{b}_{i,2}{R}_{L2}={\\theta }_{i,2y}+{\\pi }_{i,2}.\\end{array}& \\left(3\\right)\\end{array}$\n\nInitially, each consumer owns θi, 0 = 1 unit of the asset, where the price of the asset in period t is denoted by pt. Consumers can buy or sell the asset in a perfectly competitive market, but they cannot sell it to the lenders and rent it back. As in Jeanne and Korinek (2010b), we assume that consumers derive some important benefits from owning the asset.4 Note that consumers are identical and, in a symmetric equilibrium, we must have θi, 0 = θi, 1 = θi, 2 = 1.\n\nAs it is evident from the budget constraint, in order to consume in period 0, consumers need to either sell a fraction of their assets (1 – θi, 1) or borrow from banks (bi, 1). Moreover, each consumer, in period 1, faces a collateral constraint of the form:\n\n$\\begin{array}{cc}{b}_{i,2}\\le {\\theta }_{i,1}{p}_{1},& \\left(4\\right)\\end{array}$\n\nwhere θi, 1 is the quantity of domestic collateral held by the consumer at the beginning of period 1.\n\nThe microfoundation of the collateral constraint follows the spirit of Kiyotaki and Moore (1997). However, for tractability, while in Kiyotaki and Moore (1997) borrowing capacity is an increasing function of the future value of the collateral, we assume that borrowing capacity is an increasing function of the current value of the collateral. The same modelling choice has been made by Mendoza (2010), Jeanne and Korinek (2010b) and Mendoza and Smith (2006), and is justified by the work of Cordoba and Ripoll (2004) and Kocherlakota (2000) who show that collateral constraints specified with next-period price of collateral asset do not yield quantitatively significant differences in response to shocks.\n\nConsumers maximize (1) subject to the budget constraint (3) and the collateral constraint (4). The utility maximization problem of the representative consumer (i.e., variables without the subscript i) can be written as:\n\n$\\begin{array}{cc}\\underset{{b}_{1},{b}_{2},{\\theta }_{2}}{\\text{max}}\\left\\{\\begin{array}{c}u\\left({b}_{1}+\\left(1-{\\theta }_{1}\\right){p}_{0}\\right)+{\\mathbb{\\text{E}}}_{0}\\left[u\\left(e+{b}_{2}+\\left({\\theta }_{1}-{\\theta }_{2}\\right){p}_{1}+{\\pi }_{1}-\\\\ -{b}_{1}{R}_{L1}\\right)+{\\theta }_{2y}+{\\pi }_{2}-{b}_{2}{R}_{L2}-\\lambda \\left({b}_{2}-{\\theta }_{1}{p}_{1}\\right)\\end{array}\\right\\}.& \\left(5\\right)\\end{array}$\n\nSolving this problem backwards, the first order conditions are:\n\n$\\begin{array}{cc}\\begin{array}{l}{p}_{1}=\\frac{y}{{u}^{\\prime }\\left({c}_{1}\\right)},\\\\ {u}^{\\prime }\\left({c}_{1}\\right)={R}_{L2}+\\lambda ,\\\\ {u}^{\\prime }\\left({c}_{0}\\right)={R}_{L1}{\\mathbb{\\text{E}}}_{0}\\left[{u}^{\\prime }\\left({c}_{1}\\right)\\right].\\end{array}& \\left(6\\right)\\end{array}$\n\nThe first equation represents the asset pricing condition for the economy. The second and third equations are the Euler equation for consumption in period 1 and 0, respectively.5\n\n#### Consumers’ demand of loans\n\nIn order to allow for market power in the banking sector, we model the market for loans in a Dixit and Stiglitz (1977) framework.6 That is, we assume that loan contracts bought by consumers are a constant elasticity of substitution composite basket of slightly differentiated financial products—each supplied by a bank j—with an elasticity of substitution ζ (which will be the main determinant of the spread between bank rates and the risk-free rate).\n\nIn particular, the consumer i, in order to obtain a loan of a given size bi,t, needs to take out a continuum of loans bi, j, t from all existing banks j, such that:\n\n$\\begin{array}{cc}{b}_{i,t}\\le {\\left({\\int }_{0}^{1}{b}_{i,j,t}^{\\frac{\\zeta -1}{\\zeta }}dj\\right)}^{\\frac{\\zeta -1}{\\zeta }}& \\left(7\\right)\\end{array}$\n\nwhere ζ > 1 is the elasticity of substitution between differentiated loans (or banking services, in general). Demand by consumer i seeking a real amount of loans equal to bi,t can be derived by minimizing the total repayment due to the continuum of banks j over bi,j,t. Aggregating over symmetric households, this minimization problem yields downward-sloping loans demand curves of the kind:\n\n$\\begin{array}{cc}{b}_{j,t}=\\left(\\frac{{R}_{Lj,t}}{{R}_{Lt}}{\\right)}^{-\\zeta }{b}_{t}.& \\left(8\\right)\\end{array}$\n\nwith the aggregate interest rate on loans given by:\n\n$\\begin{array}{cc}{R}_{Lt}=\\left({\\int }_{0}^{1}{R}_{Lj,t}^{1-\\zeta }dj{\\right)}^{\\frac{1}{1-\\zeta }}.& \\left(9\\right)\\end{array}$\n\n### B Banks and Loan Supply\n\nThere is a continuum of monopolistically competitive domestic banks indexed by j ϵ [0,1] owned by households. Microeconomic theory typically considers market power as a distinctive feature of the banking sector (Freixas and Rochet, 2008).7 In particular, we assume that each bank j supplies slightly differentiated financial products, and no other bank produces the same variety: each bank has, therefore, some monopoly power over its product. However, each bank competes with all other banks, since consumers consider each bank’s variety as imperfect substitutes. As banks have market power over the supply of their variety, they set prices to maximize profits, taking into account the elasticity of demand for their variety.\n\nEach bank j collects fully insured deposits dj,t from foreign investors at the risk-free interest rate Rt = R, where R* is exogenous and given. We further assume that foreign lenders have an infinite supply of deposits, so that banks can satisfy any demand for loans. Finally, banks use deposits to supply loans to consumers with the following constant return to scale production function:\n\n$\\begin{array}{cc}{b}_{j,t}={d}_{j,t}.& \\left(10\\right)\\end{array}$\n\nIn each period, bank j maximizes its profits choosing prices and quantities:\n\n$\\underset{{R}_{Lj,t}{b}_{j,t}}{\\text{max}}{b}_{j,t}{R}_{Lj,t}-{d}_{j,t}{R}_{t},$\n\nsubject to the demand schedule in (8) and to the production function in (10). The first order condition for this problem implies that the optimal lending rate applied by banks is a positive constant gross markup ((M)) over the marginal cost:\n\n$\\begin{array}{cc}{R}_{Lt}\\left(j\\right)=\\frac{\\zeta }{\\zeta -1}{R}_{t}=M{R}_{t}.& \\left(11\\right)\\end{array}$\n\nNote that, together with consumers’ optimality conditions, equation (11) determines the equilibrium of the economy. That is, once the lending rate has been set by banks, households make their consumption (and, therefore, borrowing) decisions. The market clearing in the loan market closes the model.\n\nWe also assume that the banking sector displays short-run interest rate stickiness. In particular, we assume that banks cannot immediately adjust their lending rates in response to macroeconomic developments. The presence of interest rate stickiness in the banking sector can be justified by the presence of adjustment costs of and monopolistic power. For example, Hannan and Berger (1991) show that, in the presence of fixed adjustment costs, banks re-set their lending rates only if the costs of changing the interest rate are lower than the costs of maintaining a non-equilibrium rate (see also Neumark and Sharpe, 1992). Empirically, it is a well documented fact that the adjustment of banks lending rates to changes in the risk-free rate is only partial and heterogeneous, in particular in the short run. For example, Kwapil and Scharler (2010) show that interest rate pass-through of consumer loans in the U.S. can be as low as 0.3, implying that interest rates charged on consumer loans are smoothed heavily by banks. For tractability, we implement interest rate stickiness by means of a simple one-period real rigidity—while we assume that in the long-run interest rates are fully flexible.\n\nIn particular we assume that, if the risk-free interest rate is hit by a temporary shock (ν) in period 0, only a fraction μ of the banks can reset their rates, whereas the remaining 1 – μ banks cannot. This entails that, following a shock to the risk free interest rate, the aggregate lending rate will be in general different from the one desired by banks: remembering that consumers are price takers and that their loans demand depends on the average interest rate in the economy, this friction will lead to a distortion in the competitive equilibrium and will create the policy scope for restoring efficiency. Moreover, given that the incomplete pass-through of changes in the risk-free rate on lending rates is a realistic assumption only in the short run, we assume that from period 1 interest rates are again fully flexible.\n\nFinally, note that shocks to the interest rate in period 0 are observed by all agents before they make their decisions. We will consider three different scenarios: no shock to the risk-free interest rate (ν = 0), a temporary increase in the risk-free rate (ν < 0), and a temporary reduction in the risk-free rate (ν < 0). We can interpret these three scenarios as the result of a realized temporary “shock” to the risk-free interest rate at the beginning of period 0. Specifically, shock ν can be interpreted as a demand shock—such as a preference shock or a government spending shock—in a closed economy or as a foreign demand shock in a small open economy (see Harrison and Oomen (2010) and Cook and Devereux (2011), for example).\n\n### C Shocks and Parameter Values\n\nTo be able to solve and simulate the model we need to make assumptions about key parameters: the distribution of the stochastic endowment (e), the return of the asset (y), households’ preferences (ρ), the degree of monopolistic competition in the banking sector (ζ), the risk-free interest rate (R), the degree of interest rates stickiness (μ), and the size of the shocks to the interest rate (ν). Table 1 summarizes the assumptions we make on these processes and parameters.\n\nTable 1.\n\nCalibration of Model’s Parameter", null, "Note. 3M US T-Bill is the the average 3-Month Treasury Bill deflated with US CPI; RL is the 15-Year mortgage fixed rate deflated with US CPI. U.S. monthly data from 1985:M1 to 2007:M3.\n\nWe assume that endowment e has a deterministic component $\\overline{e}$ and a stochastic component $\\stackrel{˜}{\\epsilon }$:\n\n$\\begin{array}{cc}e=\\overline{e}+\\stackrel{˜}{\\epsilon },& \\left(12\\right)\\end{array}$\n\nwhere $\\stackrel{˜}{\\epsilon }$ is uniformly distributed over the [-ε, +ε] interval. This implies that the endowment e is uniformly distributed over the [$\\overline{e}$ε, $\\overline{e}$ + ε] interval. We will analyze the model’s properties for different values of the maximum size of the shock to the endowment (ε). In particular, we will consider values for ε such that the economy may be constrained for a sufficiently large negative realization of the shock, but would not be constrained in the absence of disturbances. As shown in Appendix A, under these assumptions the model can be solved largely in closed form.\n\nWhile it is possible to make reasonable assumptions for the majority of parameters, two degrees of freedom are left for the solution of the model: the return of the asset (y) and the expected value of the endowment ($\\overline{e}$). Following Jeanne and Korinek (2010a), we assume $\\overline{e}$ = 1.3 and y = 0.8. Jeanne and Korinek (2010a) choose these two parameters jointly with the maximum size of the shock to the endowment (ε) to control when the borrowing constraint binds. Note also that, given the stylized nature of the model, we do not use it for quantitative analysis. The parametrization chosen is to study the solution in the case in which the borrowing constraint does not bind today, but can bind tomorrow. The analysis of a model with an occasionally binding constraint adds a layer of complexity that is not present in standard New Keynesian models and banking models, and thus justifies simplifying in other dimensions. In the last section of the paper we shall use the qualitative predictions of the model to interpret the recent U.S. experience in the run-up to the Great Recession.\n\nWe calibrate the remaining parameters using U.S. data from 1985 to 2007, i.e., from the beginning of the Great Moderation to the beginning of the Great Recession. The gross risk-free real interest rate is set to R = 1.015 in order to match the average yield of the 3-Month Treasury Bill (deflated with US CPI) over the period 1985-2007. We set the elasticity of substitution between financial products to ζ = 33.3, which implies a gross markup of (M) ≃ 1.03. This markup yields approximately a spread of 250 basis points over the risk-free interest rate, which is consistent with the average spread of the 15-year mortgage fixed rate over the 3-Month Treasury Bill rate.8 Household preferences are given by a constant elasticity of substitution utility function, with a relative risk aversion coefficient ρ = 2, which is a conventional value.\n\nUnder these assumptions, the model economy is never constrained when εεb = 0.095. That is, the constraint never binds below the threshold εb, and the probability of observing a crisis in period 1 is zero. In this case, the model has a closed-form solution given by optimality conditions (6) together with λ = 0. In contrast, when ε > 0.095 there exists a positive probability that the constraint will bind in period 1: in this case the model does not have a closed-form solution and, therefore, the levels of debt and consumption have to be solved numerically (as shown in Appendix A).\n\nThe calibration of the degree of interest rate stickiness (μ) is more difficult. Even if there is compelling evidence on the imperfect adjustment of retail interest rates rate to movements in the risk free rate, the degree of such rigidity is not consistently quantified. For the U.S., Kwapil and Scharler (2010) estimate a short-run pass through of 0.3 for consumer loans.9 Based on this evidence we assume that, in the short run, only 50 percent of the banks can adjust their lending rates conditional to a movement in the interest rate. In the long-run, in contrast, pass through is assumed to be complete.10\n\nFinally, we assume that the risk-free interest rate is affected by a shock in period 0, such that:\n\n$\\begin{array}{cc}{R}_{1}={R}^{}+\\upsilon ,& \\left(13\\right)\\end{array}$\n\nwhere υ can take three values, namely ν = {0, +0.02, −0.02}. The size of the shock matches the standard deviation of the yield on the U.S. 3-Month Treasury Bill over the 1985-2007 period.\n\n## III Decentralized Equilibrium\n\nWe can now analyze the decentralized equilibrium of the economy. In order to build intuition, we will consider first the effects of the financial friction (which manifests itself conditional on shocks to the endowment) by comparing the allocation in our model economy with an economy in which the collateral constraint is never binding. Second, we will analyze the effect of the macroeconomic friction (which manifests itself conditional on shocks to the risk free interest rate) by comparing the allocation in our model economy with an economy with fully flexible interest rates. Third, and finally, we will analyze the full model, when both frictions are at work simultaneously.\n\n### A Financial Friction\n\nThe financial friction affects the economy only when the collateral constraint is not binding today but can bind tomorrow with a positive probability. In particular, a shock (ε) to the endowment received by households, if large enough to make the collateral constraint binding, will lead to a downward spiral of declining consumption, falling asset prices, and tighter borrowing constraints typical of financial accelerator models, such as Bernanke, Gertler, and Gilchrist (1996) and Kiyotaki and Moore (1997). We label states in which the collateral constraint is binding as “crisis states” and define the probability that the constraint will bind in period 1 (i.e., the crisis probability) as our measure of financial stability.11\n\nWe consider different values of the maximum size of the shock (ε) so that i) the collateral constraint never binds (i.e., the shock $\\stackrel{˜}{\\epsilon }$ is never large enough to push the economy in the constrained region); and ii) the collateral constraint is occasionally binding (i.e., for large enough realizations of the shock $\\stackrel{˜}{\\epsilon }$ the economy can enter the constrained region and experience a financial crisis). As we discussed earlier, the threshold for ε that makes the collateral constraint bind with positive probability is εb ≃ 0.095.\n\nFigure 2 displays the behavior of some endogenous variables in our model for different values of the maximum size of the shock (ε, displayed on the horizontal axis). The upper-left panel of Figure 2 plots the equilibrium level of borrowing in period 0 (b1). Conditional on b1, it is possible to compute net worth (eb1 RL1), consumption (c1), and the probability of observing a crisis (π) in period 1.\n\nWhen εεb the economy is never constrained, households’ decisions are not affected by the size of the shock ε: if hit by a negative endowment shock, households can borrow from banks to smooth consumption. In contrast, when the maximum size of the shock is above its threshold (εb = 0.095), consumers take into account that there is a positive probability that the constraint will bind in period 1. They insure by reducing borrowing in period 0 (so that their net worth next period will be higher) and by reducing their consumption in period 1. The probability of a crisis (π) is positive and increases in a non-linear way with the maximum size of the shock to the endowment.\n\nThe intuition for the comparative statics in Figure 2 is the following. The Lagrangian multiplier (λ) in the Euler equation (6) represents the shadow value of the collateral constraint. When the shock to the endowment is not large enough to push the economy in the constrained region, λ = 0 and the economy achieves its efficient allocation (as we are not explicitly considering here other distortions). In contrast, when the maximum size of the shock (ε) is large enough, λ may be positive and increasing in ε. Therefore, the larger ε, the larger is λ and the level of precautionary savings undertaken by consumers.\n\n### B Macroeconomic Friction\n\nLet us now analyze how the macroeconomic friction affects our model economy. As it is well known from the standard New Keynesian literature, there are two potential distortions in models with monopolistic competition and staggered pricing. First, monopolistic power forces average output below the socially optimal level. Second, staggered pricing implies that both the economy’s average markup and the relative price of different goods will vary over time in response to shocks, violating efficiency conditions.12 As we shall see below, our model displays a similar behavior.\n\nLet us assume for the moment that interest rates can freely adjust and that lending rates at the beginning of period 0 are set at the desired level, as a markup over the marginal cost (RL1 = (M) R). If a positive shock υ > 0 hits the economy, banks face a new, higher marginal cost and update their lending interest rates such that RL1 = (M) (R* + ν). Households update their loans demand accordingly and the loans market clears at a higher lending rate. In response to the higher interest rate, consumption and borrowing in period 0 fall relative to the case in which ν = 0. This allocation (henceforth “flex-rates” allocation) is efficient, conditional on the shock ν.\n\nIn a sticky-rates environment, not all banks can reset their lending rate so as to be consistent with the new marginal cost. The fraction μ of banks that can reset lending rates will set:\n\n${R}_{L1}^{u}=M\\left({R}^{}+\\nu \\right).$\n\nIn contrast, the remaining 1 – μ banks will not be allowed to reset their lending rates, implying that:\n\n${R}_{L1}^{1-\\mu }=M{R}^{}<{R}_{L1}^{\\mu }.$\n\nAs a consequence, the aggregate lending rate in the economy would differ from its flex-rates counterpart. According to equation 9, the aggregate lending rate in the sticky-rates economy becomes:\n\n${R}_{L1}=M\\left({R}^{}+\\mu \\nu \\right),$\n\nwhich is higher than the lending rate prevailing under flex-rates in the case of positive shocks to the interest rate. A similar gap of opposite sign emerges when the ν shock is negative.\n\nThe model properties analyzed in this section can be summarized as follows. In general, interest rates stickiness results in an average interest rate, RL1, which differs from the one required to obtain the flex-rates allocation, therefore affecting the aggregate level of borrowing and consumption. More specifically, when a positive shock hits the interest rate, debt and consumption are higher than in the flex-rates economy, because interest rates increase by less than they would in a fully flexible world. But, when a negative shock hits the economy, debt and consumption are lower than in the flex-rates economy, because interest rates decrease by less than they would in a fully flexible world. As we shall see, this property has crucial implications for the results of our analysis when the macroeconomic frictions interact with the financial friction.\n\n### C The Interaction between Financial Friction and Macroeconomic Friction\n\nIn this section we show that the impact of staggered interest rates setting on the crisis probability (our measure of financial stability, i.e. the crisis probability) depends on the sign of the shock hitting the economy. Specifically, we shall see that in response to positive shocks to the risk-free interest rate, the probability of a crisis in the sticky price economy is lower (increases less) than in the in the flex-price economy. Instead, in response to negative shocks to the risk-free interest rate, the crisis probability is higher (it falls less) than in the flex-rates economy. In this sense, we say that interest rate rigidities have an asymmetric impact on financial stability.\n\nWe first analyze the effect of a positive shock to the risk-free interest rate (Figure 3). The benchmark is the economy with both frictions but no interest rate shocks (solid line, i.e., the same allocation as in Figure 2). The thin line with asterisk markers and the thin line with circle markers display the equilibrium after the interest rate shock has hit, under flexible and sticky interest rates respectively.\n\nAs we showed above, under the assumption of sticky interest rates, the aggregate lending rate in the economy does not increase as much as the risk-free rate following a positive shock. What are the implications for the probability of a crisis, our measure of financial stability? On the one hand, lower lending rates—relative to the flex-rates case—prompt consumers to borrow more (b1) in period 0 and to consume more (c1) in period 1, as shown by the difference between the circles line and the asterisks line. All else equal, this implies higher expected next-period refinancing needs (b2) and, therefore, a higher expected probability that the constraint will be binding in period 1. On the other hand, and despite the higher level of borrowing in period 0, expected net worth (eb1RL1) in period 1 is larger under sticky rates than under flex rates, because of lower interest rate repayments. All else equal, this implies a relaxation of the borrowing constraint in period 1. The net effect is displayed in the bottom right-hand panel of Figure 3: when a positive shock hits the economy, with sticky interest rates the probability that the constraint will bind in period 1 increases by less than in the flex-rates case. This is because, as long as the coefficient of relative risk aversion (ρ) is larger than 1, the effect of the lower interest rates on net worth dominates the effect on borrowing and consumption. Note that this result is robust to assuming different values for all other parameters of the model, including the size of the shock to the interest rate (ν) and the degree of interest rate stickiness (μ). Changing these parameters does not affect the mechanisms driving the result, but only the magnitude of the effects. In other words, for every possible value of ν and μ the allocation under sticky-rates (circles line) is bounded between the allocation under flex-rates (asterisks line) and the allocation where no shock hits the economy (solid line).\n\nConsider now a negative shock. In the case of a negative shock, sticky interest rates exacerbate the effects of the financial friction rather than dampening it. To see that, Figure 4 displays how the model equilibrium and the crisis probability vary in response to a negative shock to the risk-free interest rate.\n\nUnder interest rate stickiness (circles line), the average lending rate now falls by less than the risk-free interest rate. In the sticky rate economy, consumption and borrowing are lower (or increase less) than in the flex-rate economy (asterisks line) in response to the shock, but next period interest payments are higher. As a result next-period net worth in the sticky rates economy is lower than in the flex-rate economy, and the crisis probability is higher (or it falls less).\n\nIn conclusion, the analysis of the interaction between the macroeconomic and financial friction can be summarized as follows: when both the macroeconomic and the financial friction are present, interest rate stickiness, conditional on positive shocks to the interest rate reduces the crisis probability relative to the flex-rate equilibrium; conditional on negative shocks to the interest rate, it increases the crisis probability relative to the flex-rate equilibrium. In this sense, interest rate rigidities have an asymmetric impact on financial stability.\n\n## IV Restoring Efficiency\n\nIn this section we discuss how policy intervention, and in particular monetary policy, can address the market failures of our model economy.\n\nTo build understanding and intuition for the main results, we first analyze the case in which there is only the financial friction or the macroeconomic friction. Then, we consider the case in which the policy authority faces both frictions with either one or two policy instruments.\n\nA key result is that a policy-maker with a macro-prudential instrument (a tax on borrowing) and a monetary policy instrument (the policy interest rate) can address both distortions induced by the financial friction and the macroeconomic friction. In contrast, if the interest rate is the only available instrument, the policy-maker faces a trade-off between macroeconomic and financial stability when the economy is hit by negative shocks.\n\n### A Addressing the Pecuniary Externality\n\nAs it is well known, the occasionally binding constraint that is in our model generates a pecuniary externality. This pecuniary externality drives a wedge between private and socially optimal outcomes because private agents do not internalize the effect of their decisions on the asset price that enters the specification of the borrowing constraint. A social planner, unlike private agents, can internalize that consumption decisions affect the asset price—as shown by the asset price equation in (6)—which, in turn, affects the aggregate collateral constraint in (4).13\n\nFollowing Jeanne and Korinek (2010a), the planner’s problem for this economy can be written as:\n\n$\\underset{{b}_{1},{b}_{2}}{\\text{max}}\\left\\{\\begin{array}{c}u\\left({b}_{1}\\right)+{\\mathbb{\\text{E}}}_{0}\\left[u\\left(e+{b}_{2}+{\\pi }_{1}-{b}_{1}{R}_{L1}\\right)+y-{b}_{2}{R}_{L2}\\\\ -{\\lambda }^{sp}\\left({b}_{2}-{p}_{1}\\left(e+{b}_{2}-{b}_{1}{R}_{L1}\\right)\\right)\\right]\\end{array}\\right\\},$\n\nwhere the maximization is subject to the budget constraint (3), the aggregate borrowing constraint (4), and the pricing rule of the competitive equilibrium allocation:\n\n${p}_{1}\\left({c}_{1}\\right)=\\frac{y}{{u}^{\\prime }\\left({c}_{1}\\right)},$\n\nwhere the asset price, p1 (c1), is now a function of aggregate consumption.\n\nThe corresponding first order conditions are:\n\n$\\begin{array}{cc}{u}^{\\prime }\\left({c}_{0}\\right)={R}_{L1}{\\mathbb{\\text{E}}}_{0}\\left[{u}^{\\prime }\\left({c}_{1}\\right)+{\\lambda }^{sp}{p}^{\\prime }\\left({c}_{1}\\right)\\right],\\hfill & \\left(14\\right)\\\\ {u}^{\\prime }\\left({c}_{1}\\right)={R}_{L2}+{\\lambda }^{sp}\\left(1-{p}^{\\prime }\\left({c}_{1}\\right)\\right).\\hfill & \\end{array}$\n\nBy comparing (6) and (14) and noting that p(c1) > 0, it is clear that there is a wedge between the decentralized and the social planner allocation: the social planner saves more than private agents whenever the borrowing constraint is expected to bind in period 1 with positive probability (i.e., whenever λsp > 0). This reflects the fact that the social planner internalizes the endogeneity of next period’s asset price to this period’s aggregate saving. As a consequence, when the constraint never binds, the allocation of resources in the economy is efficient (ignoring the other frictions in the model). However, when there is a positive probability that the constraint binds in period 1, the allocation is not efficient. Consumption and borrowing in the decentralized equilibrium are excessive relative to the allocation chosen by the social planner (i.e., there is overborrowing in the parlance of the literature). As a result, the crisis probability is also higher in the decentralized equilibrium relative to the social planner equilibrium.\n\n#### A Pigouvian Tax on Borrowing\n\nIn this set-up, Jeanne and Korinek (2010a) show that efficiency can be restored in the decentralized economy by imposing a Pigouvian tax on borrowing in period 0, namely b1 (1—τ), which is rebated with transfers (TR) in a lump-sum fashion. The optimal tax is given by:\n\n$\\begin{array}{cc}\\tau =\\mathbb{\\text{E}}\\left[\\frac{{\\lambda }^{sp}{p}^{\\prime }\\left({c}_{1}\\right)}{{u}^{\\prime }\\left({c}_{1}\\right)}\\right],& \\left(15\\right)\\end{array}$\n\nThis equation states that whenever the borrowing constraint binds in period 1 with positive probability, the policy-maker imposes a positive tax on borrowing in period 0, prompting private agents to issue less debt in period 0 than under decentralized equilibrium. This is because both the shadow value of the collateral constraint (λsp) and the derivative ${p}_{1}^{\\prime }\\left({c}_{1}\\right)$ are positive.\n\n#### An Interest Rate Policy\n\nA Pigouvian tax on borrowing may be difficult to implement. But the constrained efficient allocation can also be decentralized with the interest rate. The policy-maker can equally curtail households’ borrowing by increasing lending interest rates. For instance, the policy-maker (e.g., a central bank in this specific case) can increase the interest rate at the beginning of period 0, affecting banks marginal cost and, therefore, consumers’ borrowing and consumption decisions.\n\nThis increase in interest rates—if rebated with lump sum transfers (TR)—has the same effect of the Pigouvian tax analyzed above. To see this, assume for simplicity that the central bank can affect the interest rate by an additive factor ψ, so that the marginal cost for banks would be given by R + ψ (see Stein, 2012). The consumers’ maximization problem becomes:\n\n$\\underset{{b}_{1},{b}_{2}}{\\text{max}}\\left\\{\\begin{array}{c}u\\left({b}_{1}\\right)+{\\mathbb{\\text{E}}}_{0}\\left[u\\left(e+{b}_{2}+{\\pi }_{1}-{b}_{1}M\\left({R}^{}+\\psi \\right)+TR\\right)+\\\\ +y-{b}_{2}{R}_{L2}-{\\lambda }^{sp}\\left({b}_{2}-{p}_{1}\\right)\\right]\\end{array}\\right\\}.$\n\nBy equalizing the first order condition with respect to b1 of the decentralized equilibrium and the social planner equilibrium, we can derive the level of ψ which closes the wedge:\n\n$\\left\\{\\begin{array}{l}{u}^{\\prime }\\left({c}_{0}\\right)={R}_{L1}{\\mathbb{\\text{E}}}_{0}\\left[{u}^{\\prime }\\left({c}_{1}\\right)+{\\lambda }^{sp}{p}^{\\prime }\\left({c}_{1}\\right)\\right],\\\\ {u}^{\\prime }\\left({c}_{0}\\right)=M\\left({R}^{}+\\psi \\right){U}^{\\prime }\\left({c}_{1}\\right),\\end{array}$\n\nSolving for ψ yields:\n\n$\\begin{array}{cc}\\psi ={\\mathbb{\\text{E}}}_{0}\\left[\\frac{{\\lambda }^{sp}p\\prime \\left({c}_{1}\\right)}{{u}^{\\prime }\\left({c}_{1}\\right)}\\right]{R}^{}.& \\left(16\\right)\\end{array}$\n\nNotice that as long as the shadow value of the collateral constraint (λsp) is different from zero, ψ is positive and can be interpreted as a prudential “markup” factor on the risk-free interest rate. This, in turn, implies that whenever the constraint is binding with positive probability, the central bank would raise interest rates so that households consume less and issue less debt in period 0, reducing the probability of hitting the constraint in case of an adverse shock in period 1.\n\nIn summary, when the borrowing constraint is the only friction in the economy and the policy rate is the only policy instrument, a social planner can achieve constrained efficiency by increasing interest rates in period 0. This allocation is isomorphic to the one obtained with the Pigouvian tax on debt analyzed in the previous section. As we shall see below, however, this is not always the case. When both the financial and the macroeconomic frictions are present it will depend on the sign of the shock hitting the economy.\n\n### B Addressing Monopolistic Competition and Interest Rate Stickiness\n\nAs mentioned in the previous sections, our model embeds two macroeconomic distortions. The first distortion is the presence of market power in loan markets. The second distortion is staggered adjustment of lending rates. In this section we discuss how policy can address them.\n\nMonopolistic competition in the banking sector implies an inefficiently low level of consumption, because lending interest rates are, on average, higher than under perfect competition. As it is standard in the New Keynesian literature, this inefficiency could be eliminated in the decentralized economy through the suitable choice of a subsidy to interest rate repayments such that:\n\n$\\begin{array}{cc}{R}_{Lt}=\\underset{1}{\\underbrace{M\\left(1-{\\eta }_{t}\\right){R}_{t}}}.& \\left(17\\right)\\end{array}$\n\nHence, the optimal allocation can be attained if (M) (1 − ηt) = 1 or, equivalently, by setting η = ζ.\n\nStaggered interest rate setting implies an inefficient level of borrowing, consumption, and net worth because the economy’s aggregate lending rate will generally differ from the one prevailing under flexible rates.\n\nOne way to address the consequences of interest rate stickiness is as follows. Assume that the central bank can affect the interest rate by an additive factor ψ. Thus, the marginal cost of funds for banks—conditional on a shock to the risk free interest rate—would be given by R + ν + ψ. Then, the central bank could set:\n\n$\\psi :{R}_{L1}=M\\left({R}^{}+\\upsilon \\right),$\n\nwhich is the efficient (i.e., without distortions) level of the lending interest rate. Solving this equality yields:\n\n$\\begin{array}{cc}\\psi =\\frac{1-\\mu }{\\mu }\\upsilon .& \\left(18\\right)\\end{array}$\n\nHence, in response to a positive shock to the risk-free rate (ν > 0), the central bank would raise interest rates above the competitive equilibrium level by the factor ψ > 0; in contrast, in response to a negative shock to the risk free rate (ν < 0), the central bank would lower interest rates below the competitive equilibrium level by the factor ψ < 0.\n\nTo see why this would be a constrained-efficient decentralized equilibrium, note the following: in response to such a policy intervention, banks that can adjust their interest rates would do so and make an optimal decision. In contrast, banks that are not allowed to change their interest rates would not be optimizing anyway. But consumers would face the same aggregate interest rate prevailing without sticky rates (that is, as in the undistorted economy) and hence make optimal decisions.\n\n### C Addressing Both Frictions with Two Instruments\n\nWe now analyze how to implement the constrained-efficient allocation in the decentralized economy when both frictions are present. We focus on the case in which the subsidy to interest rate repayments, η in equation (17), is always in place—so as to remove the distortion generated by monopolistic competition—and the policy-maker has two policy instruments at disposal. Specifically, we consider a policy-maker who maximizes the expected utility of consumers (5), subject to their budget constraints (3) and borrowing constraints (4). The two instruments to address the two distortions in the economy are (1) the interest rate wedge (ψ) to address the distortion generated by staggered interest rate setting and (2) a prudential tax on debt (τ) to address the distortion generated by the pecuniary externality.\n\nWhen two instruments are available, the policy-maker can address the macroeconomic and financial stabilization problems separately. This is regardless of whether a single policy authority is in charge of both monetary and financial-stability policy (e.g., a central bank) or whether one authority is in charge of monetary policy and the other is in charge of macroprudential policy. In other words, in our set-up, there are no incentives for a central bank and a financial stability authority to deviate from a coordinated equilibrium.\n\nWith these considerations in mind, we consider first a positive shock to the risk-free interest rate. The dashed line of Figure 5 displays the model equilibrium when the policy-maker restores efficiency. Figure 5 also displays two additional allocations: the competitive equilibrium in which a positive shock hits the economy and interest rates are flexible (triangles line); and the competitive equilibrium in which a positive shock hits the economy and interest rates are sticky (squares line). Note that, unlike in the decentralized equilibria discussed in the previous section, here a subsidy (η) is also removing the distortionary effect of the markup and restores the level of interest rates that would prevail in a perfectly competitive banking sector. For this reason, borrowing under flexible interest rates in Figure 5 (triangles line) is now larger than borrowing in Figure 3 (asterisks line).14\n\nThe policy-maker undertakes two independent policy actions, one to address the distortion generated by staggered interest rate setting (the macroeconomic friction) and another one to address the distortion generated by the occasionally binding borrowing constraint (the financial friction). Consider first the macroeconomic friction and then the financial friction.15 The policy-maker first raises interest rates by a factor ψ > 0 to restore the aggregate lending rate that would prevail under flex rates, moving the economy from the sticky-rates competitive equilibrium (squares line) to the flex-rates competitive equilibrium (triangles line). Then, the policy-maker imposes a distortionary tax on debt (τ) to restore the efficient level of borrowing, moving the economy to the constrained efficient equilibrium (dashed line).\n\nAs shown in equation 15, the optimal level of τ is zero when the constraint never binds and positive when the constraint is expected to bind with positive probability in period 1. Figure 5 shows that when εεb the triangles line and dashed line coincide. However, when ε > εb the tax on borrowing is positive, borrowing in period 0 is lower than in the flex-rates competitive equilibrium (upper-right panel of Figure 5), while consumption in period 1 is larger (lower-right panel of Figure 5). That is, whenever the collateral constraint is expected to bind with a positive probability, the policy-maker forces private agents to borrow less in period 0—therefore increasing their net worth next period—and to consume more in period 1, thereby reducing the probability of a financial crisis.\n\nNote here that, like before, in the flex-rate equilibrium the net worth (and the crisis probability) is lower (higher) than in the sticky-rate one because of the higher debt repayment in period 1. With flexible interest rates, borrowing is lower but it is more costly to service, so net worth in period 1 is lower and the probability of a crisis is higher. Adding the tax on debt curtails borrowing without increasing debt service costs. As the maximum size of the endowment shock increases, the optimal level of debt falls. And above a certain threshold, the crisis probability falls even below the one in the sticky-rate equilibrium.\n\nConsider now a negative shock to the risk-free interest rate (Figure 6). To address the pecuniary externality, the policy-maker can impose a tax on debt whenever there is a positive probability that the constraint will bind in period 1 regardless of the sign of the shock. To address the interest rate rigidity, when a negative shock hits the economy, the policy-maker can lower interest rates by ψ. In this case—and differently from a positive shock to the risk-free rate—achieving the flex-rate equilibrium already reduces the probability of a crisis, as the economy moves from the sticky-rates competitive equilibrium (squares line) to the flex-rates competitive equilibrium (triangles line). This is because the higher borrowing than in the sticky-rate case is more than compensated by the lower interest payment. So, when the policy maker uses also the tax on debt, the probability of a crisis decreases even more relative to the sticky-rate case, and it is always below it.\n\nIn summary, with two instruments, such as a tax on borrowing and the monetary policy interest rate, a policy maker can address both the financial and the macroeconomic friction, thereby achieving constrained efficiency, independently of the sign of the shock hitting the economy.\n\n### D The Trade-off: Addressing Both Frictions with One Instrument\n\nLet us now consider the case in which both frictions are present in the model but the interest rate is the only instrument at the policy-maker’s disposal.\n\nBefore proceeding it is useful to recall that, in our model, the financial friction results in more borrowing than socially desirable in period 0 when the collateral constraint has a positive probability to bind in period 1, regardless of the sign of the shock. In contrast, the macroeconomic friction generates either more or less borrowing than socially desirable depending on whether the economy is hit by a positive or a negative shock. It is thus evident that, if the policy-maker has only one instrument, she/he may face a trade off in the face of negative shocks when the economy requires interventions in opposite direction.\n\nConsider a positive shock to the risk-free interest rate. As we showed before, both the macroeconomic and the financial friction result in higher borrowing in period 0 relative to the socially efficient allocation. To address the macroeconomic friction, the policy-maker can raise interest rates by the factor ψ = (1 − μ) ν/μ > 0, as implied by equation (18); and, to address the financial friction, she/he can further raise interest rates by the factor ψ = E0 [R λsp p(c1))/u(c1)] > 0, as implied by equation (16). Therefore, when a positive shock hits the economy, a single instrument can restore efficiency.\n\nHowever, when a negative shock hits the economy, the macroeconomic friction and the financial friction require opposite actions on the interest rate. The macroeconomic friction requires a decrease in interest rates: given that interest rates fall by less than in the flexible rate case, the social planner intervenes to lower interest rates by the factor ψ = −(1 − μ) ν/μ < 0. In contrast, the financial friction requires an increase in interest rates independently of the sign of the shock. Hence, if the interest rate is the only instrument, the social planner would try to lower interest rates to address the macroeconomic friction and, at the same time, to raise the interest rate to address the financial friction.\n\nSummarizing, when both macroeconomic and financial frictions are present, if the policy interest rate is the only available instrument, a policy maker that aims to achieve both macroeconomic and financial stability faces a policy trade-off. In particular, the trade-off emerges when the economy is hit by negative interest rate shocks, because addressing both frictions requires interventions of opposite sign on the policy instrument.\n\nNote here that this result is consistent with the findings of Kashyap and Stein (2012), who raise the issue of a potential conflicts between price stability and financial stability when the policy rate is the only policy instrument. Formally, they show that the introduction of a second instrument (interest payments on reserves, in their model) can resolve that trade-off. Our analysis above not only corroborates their result in a different setting, but also shows that trade-off emerges depending on the sign of the shock.\n\n## V Implications for US Monetary Policy and Financial Stability\n\nIn this section we look at the U.S. recent experience through the lens of our model in a qualitative way. Specifically, the theoretical results in the previous section have implications for the debate on the role of U.S. monetary policy in the run-up to the Great Recession.\n\nUnder former Chairman Alan Greenspan, the Federal Reserve lowered its benchmark rate from 6.5 percent to about 2 percent in 2000-01 as a response to the burst of the dot-com bubble. It further lowered interest rates to 1 percent in 2002-03 in response to a deflationary scare, and finally started a long sequence of tightening actions that, during the 2004-06 period, brought the Federal fund rate back to 5 percent (see Figure 1).\n\nAgainst this background, Taylor (2007) put forth the idea that the Federal Reserve helped inflate U.S. housing prices by keeping rates too low for too long after 2002. His main argument departed from the observation that the policy rate was well below what implied by a standard Taylor rule, a good approximation to the conduct of monetary policy in the previous several years (Figure 1). As a consequence, “those low interest rates were not only unusually low but they logically were a factor in the housing boom and therefore ultimately the bust.”16 Therefore, according to this view, higher interest rates would have reduced both the probability and the severity of the bust that led to the Great Recession.\n\nIn this section we evaluate this claim against the qualitative predictions of our model. In particular, we will show that Taylor’s argument can be rationalized within the logic of our model only if we make the following auxiliary assumptions: the policy authority is responsible for financial stability—in addition to the traditional objective of price stability—and it has only one instrument at its disposal. However, Taylor’s argument is no longer valid within the logic of our model if the policy authority has two instruments to address the macroeconomic and the financial friction or, as we showed in the previous section, when there are two different policy authorities for macroeconomic and financial stability with one instrument each. In the latter case, which is the institutional set-up prevailing in the United States, in response to a negative aggregate demand shock, the “optimal” response of the central bank is to slash interest rates without concern for financial stability, which is addressed with the second instrument (or by the other authority).\n\nAs we discuss below, the evidence suggests that the U.S. regulators were at best ineffective in curbing the continued expansion of subprime mortgage lending well past the point at which prime lending had started to decline. We conclude from this analysis that Taylor’s claim that U.S. monetary policy is to blame for the Great Recession is not justified within the logic of our model, given the regulatory regime prevailing in the United States and the evidence we report on its inability to curb subprime lending while monetary policy was tightening its stance during the 2004-06 period.\n\nTo assess Taylor’s contention through the lenses of the model, consider a negative shock hitting the economy, such as the one that occurred in March 2000 when the dot-com bubble burst. Set the beginning of period 0 as the year 2000 and assume that the economy comes back to its pre-shock level of activity after four years, namely at the beginning of 2004—consistent with the fact that the policy rate was raised for the first time in July 2004. Therefore, each time period in our model corresponds to about 4 years in the data.\n\nFigure 7 reports the qualitative behavior of the lending interest rate as implied by our model when a negative shock hits the economy. We consider two policy regimes. First, the policy-maker has just one instrument to address both frictions (Panel a). Second, the policy-maker has two separate instruments to address macroeconomic and financial friction (Panel b). Notice that each panel of Figure 7 reports the behavior of two interest rates. The solid line is the lending rate that would prevail when there is no interest rate stickiness and no policy action is undertaken (i.e., in the decentralized economy when interest rates are fully flexible) and will serve as a benchmark. The dashed line is the lending interest rate that would prevail when interest rate stickiness is present under the two policy regimes analyzed.\n\nAs we discussed in the previous section, if the interest rate is the only policy instrument, there is a trade-off between macroeconomic and financial stability conditional on a negative shock to our model economy. To achieve the efficient allocation, the policy-maker ought to move interest rates in opposite directions. On the one hand, she would have to lower interest rates to restore the aggregate lending rate that would prevail in the absence of interest rate stickiness. On the other hand, she/he would have to raise them to contain the excess borrowing generated by the pecuniary externality. As a result, interest rates in this environment would be set higher than the level predicated by focusing only on macroeconomic stability. As illustrated by the left-hand panel of Figure 7, assuming that the weight attached to macroeconomic and financial stability is the same, average lending interest rates would fall by less than in the flex-rates case. Therefore, under this regime, our model is consistent with Taylor’s argument in the sense that it suggests keeping interest rates higher than the flex-rate case to avoid excessive borrowing and large asset price increases, and to reduce the probability of a crisis if the economy is hit by a negative shock in the future.\n\nThe results are different when the policy-maker has two separate instruments to address financial and macroeconomic friction. As noted above, this is equivalent to the case in which there are two separate and independent policy authorities, such as a central bank with the objective of price stability and a financial regulator with the objective of financial stability. As we discussed above, in this case, the policy-maker can achieve efficiency with two independent policy actions, regardless of the sign of the shock. Therefore, once the excess borrowing generated by the financial friction is addressed with a macro-prudential tool, it is optimal for the central bank to lower interest rates in order to address interest rate stickiness and restore the flex-rates allocation. As a matter of fact, the right-hand panel of Figure 7 displays how the average lending rate under this regime (dashed line) is effectively equal to the one prevailing under the flexible interest rates (solid line).\n\nIn the United States, institutional responsibility for financial stability is shared among a multiplicity of agencies. Therefore, for Taylor’s contention to be justified within our model, we would have to observe an effective regulatory clampdown on mortgage lending during the period in which monetary policy was unusually lax by the standard of the Taylor rule. As we shall see below, the regulatory effort to contain mortgage lending during the period 2003-06 was at best ineffective, if not absent altogether. The evidence we report, therefore, provides support for the idea that regulation (or, more exactly, the lack thereof) was a key factor in determining the magnitude of the boom-bust cycle experienced by the U.S. housing market rather than monetary policy per se.\n\nSince the Glass-Steagall Act of 1932, U.S. depository institutions (e.g., banks, thrifts, credit unions, savings and loans, etc.) have been regulated by different federal agencies.17 In contrast, non-depository mortgage originators have enjoyed much more freedom even when they were subsidiaries of bank holding companies (see Engel and McCoy, 2011, Demyanyk and Loutskina, 2012). Moreover, as it is well known, the rise of securitization was accompanied by a shift in the structure of the mortgage industry from an originate-and-hold model to an originate-and-distribute model. Thus, well before the crisis, financial intermediation theory pointed out the risks associated with this shift, with securitization potentially leading to a reduction of financial intermediaries’ incentives to carefully screen borrowers (see Diamond and Rajan, 2001, Petersen and Rajan, 2002).\n\nFigure 8 provides a picture of the evolution of the U.S. mortgage market and monetary policy over the 2000-07 period. Broadly speaking, the picture shows that, after the Federal Reserve started to tighten its monetary policy stance and the prime segment of the mortgage market turned around, the subprime segment of the market continued to boom, with increased perceived risk of loans portfolios and declining lending standards. Despite this evidence, the first restrictive regulatory action was undertaken only in late 2006, after almost two years of steady increases in the federal funds rate.\n\nThe upper-left panel of Figure 8 (Panel a) reports the evolution of the federal funds rate (annual average) together with mortgage originations by category over the period 2001-2007. While prime mortgage originations started to fall in 2003, non-prime mortgage originations continued to increase in 2004 and 2005.18 As a matter of fact, the share of non-prime mortgage over total mortgage originations went from about 20 percent in 2001 to more than 50 percent in 2006, experiencing the largest increase in 2004, while the Federal Reserve was already tightening its monetary policy stance. A similar pattern emerges by looking at the issuance of mortgage backed securities (MBS).19 The upper-right panel of Figure 8 (Panel b) shows how the share of private label MBS sharply increased in the 2003-06 period.\n\nThe lower-left panel of Figure 8 (Panel c) reports the federal funds rate together with the share of mortgage originations with a Loan-to-Value (LTV) ratio greater than 90 percent. Note here that, while the use of countercyclical LTV ratios has been suggested—and in some emerging market economies has already been adopted—as a macro-prudential policy tool, the share of high LTV ratio mortgages in the U.S. spiked in 2005, two years after the beginning of the monetary policy tightening.\n\nFinally, the lower-right panel of Figure 8 (Panel d) reports additional evidence on the fact that, while loan quality was relatively stable or improving from 2000 to 2003, it deteriorated sharply from 2004 to 2007. The Office of the Comptroller of the Currency publishes an annual underwriting survey to identify trends in lending standards and credit risk for the most common types of commercial and retail credit offered by national banks. Using data from the 2009 survey, which covered 52 banks engaged in residential real estate lending, Panel (d) reports the evolution of changes in underwriting standards (dash-dotted line) and the perceived level of credit risk (dashed line) in residential real estate loan portfolios.20 The figure shows that, while the level of perceived risk was sharply increasing starting from 2004, banks started easing their lending standards from 2003 and did even more so in the 2004-05 period.\n\nDespite this evidence, U.S. regulators did not take action while monetary policy was being tightened. On the contrary, for instance, the SEC proposed in 2004 a system of voluntary regulation under the Consolidated Supervised Entities program, allowing investment banks to hold less capital in reserve and increase leverage that might have contributed to fueling the demand for mortgage-backed securities (vertical line in our charts under label SEC).\n\nWhen regulators finally decided to act, it was too late. It was not until September 2006 that regulators agreed on new guidelines (vertical line under label FDIC 1) aimed at tightening “non-traditional” mortgage lending practices. Note however that, even if it served as a signal to the mortgage market of changing direction of regulatory policy, the new underwriting criteria did not apply to subprime loans, whose standards were discussed in a subsequent regulatory action which was introduced in June 2007 (vertical line under label FDIC 2). By that time, more than 30 subprime lenders had gone bankrupt and many more followed suit.\n\nIn summary, the evidence above suggests that Taylor’s contention that excessively lax monetary policy might have contributed to the occurrence and the severity of the great recession does not appear justified within the logic of our model. Indeed, in the context of a framework in which the regulatory and monetary policy functions are assigned to different agencies that can rely on different instruments, the evidence above suggests that monetary policy was appropriately targeting macroeconomic stability. The regulatory function of the system, instead, was at best ineffective in addressing the financial imbalance that continued to grow in the subprime mortgage market while monetary policy was tightened in 2004-05. With the fall in interest rates after the burst of the dot-com bubble and with house prices at bubble-inflated levels, the mortgage industry found creative ways to expand lending and make large profits. Government regulators maintained a hands-off approach for too long: even though the variables plotted are equilibrium outcomes, Figure 8 shows that policy measures aimed at tightening a largely unregulated sector of the U.S. mortgage market kicked in much later than the tightening of monetary policy enacted by the Federal Reserve.\n\n## VI Conclusions\n\nIn this paper, we develop a model featuring both a macroeconomic and a financial friction that speaks to the interaction between monetary and macro-prudential policy and to the role of U.S. monetary policy in the run up to the Great Recession.\n\nThere are two main results. First, we show that real interest rate rigidities have a different impact on financial stability (defined and measured as the probability that a borrowing constraint binds), depending on the sign of the shock hitting the economy. In response to positive shocks to the risk-free interest rate, real interest rate rigidity acts as an automatic macro-prudential stabilizer. This is because higher debt today associated with lower interest rates (relative to the flexible interest rate case) is offset by lower interest repayments, resulting in higher net worth and lower probability of a crisis in the future. In contrast, when the risk-free rate is hit by a negative shock, real interest rate rigidity leads to a relatively higher crisis probability through the same mechanisms working in reverse (borrowing and consumption are relatively lower today, but they are offset by relatively higher debt service tomorrow, resulting in lower future net-worth and higher crisis probability).\n\nSecond, we show that, when the interest rate is the only policy instrument to address both the macroeconomic and the financial friction, and a shock that lowers interest rates hits the economy, a policy trade-off emerges. This is because the two frictions require interventions of opposite direction on the same instrument. Other instruments, however, may be at the policy-maker’s disposal in order to achieve and maintain financial stability. Our model shows that, when two instruments are available, this trade-off disappears and efficiency can be restored.\n\nOur analysis has interesting implications regarding the role of U.S. monetary policy in the runup to the Great Recession. In a series of recent papers Taylor (2007, 2010) suggested that higher interest rates in the 2002-2006 period would have reduced both the likelihood and the severity of the Great Recession. Our findings above support this argument only if we make the auxiliary assumption that the policy authority seeks to address all distortions in the model with a single instrument, namely the policy interest rate. In contrast, when the policy authority has two different instruments, interest rates can be lowered as much as needed in response to a contractionary shock without concerns for financial stability. 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The problem for the representative consumer therefore is:\n\n${\\mathcal{V}}_{1}=\\underset{{b}_{2},{\\theta }_{2}}{\\text{max}}\\left\\{u\\left(e+{b}_{2}+\\left({\\theta }_{1}-{\\theta }_{2}\\right){p}_{1}+{\\pi }_{1}-{b}_{1}{R}_{L1}\\right)+{\\theta }_{2}y+{\\pi }_{2}-{b}_{2}{R}_{L2}-\\lambda \\left({b}_{2}-{\\theta }_{1}{p}_{1}\\right)\\right\\},$\n\nwhere notice that the net worth eb1RL1 is taken as given. The first order conditions read:\n\n$\\left\\{\\begin{array}{l}FOC\\left({b}_{2}\\right):\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}{u}^{\\prime }\\left({c}_{1}\\right)={R}_{L2}+\\lambda ,\\\\ FOC\\left({\\theta }_{2}\\right):\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}{p}_{1}=y/{u}^{\\prime }\\left({c}_{1}\\right).\\end{array}$\n\nIn period 0, consumers solve the following problem:\n\n$\\underset{{b}_{1}}{\\text{max}}\\left\\{u\\left({b}_{1}\\right)+{\\mathbb{\\text{E}}}_{0}\\left[{\\mathcal{V}}_{1}\\right]\\right\\},$\n\nwhere we make use of the fact that, in equilibrium, θt = 1. The maximization yields:\n\n${u}^{\\prime }\\left({c}_{0}\\right)={R}_{L1}{\\mathbb{\\text{E}}}_{0}\\left[{u}^{\\prime }\\left({c}_{1}\\right)\\right].$\n\nPreliminaries. The first order conditions of the competitive equilibrium (CE) therefore are:\n\n$\\left\\{\\begin{array}{l}FOC\\left({b}_{1}\\right):\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}{u}^{\\prime }\\left({c}_{0}\\right)={R}_{L1}{\\mathbb{\\text{E}}}_{0}\\left[{u}^{\\prime }\\left({c}_{1}\\right)\\right],\\\\ FOC\\left({b}_{2}\\right):\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}{u}^{\\prime }\\left({c}_{1}\\right)={R}_{L2}+\\lambda ,\\\\ FOC\\left({\\theta }_{2}\\right):\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}{p}_{1=y/{u}^{\\prime }\\left({c}_{1}\\right)}.\\end{array}$\n\nWhen the economy is not constrained (λ = 0) the model has the following close form solution:\n\n$\\left\\{\\begin{array}{l}{u}^{\\prime }\\left({c}_{1}\\right)={R}_{L2}\\\\ {u}^{\\prime }\\left({c}_{0}\\right)={\\mathbb{\\text{E}}}_{0}\\left[{R}_{L2}{R}_{L1}\\right],\\\\ {p}_{1}=\\frac{y}{{R}_{L2}}.\\end{array}⇒\\left\\{\\begin{array}{c}{c}_{1}^{}=\\left({R}_{L2}{\\right)}^{-\\frac{1}{\\rho }}\\\\ {c}_{0}^{}={b}_{1}^{}=\\left({R}_{L2}{R}_{L1}{\\right)}^{-\\frac{1}{\\rho }},\\\\ {p}_{1}^{}=\\frac{y}{{R}_{L2}}.\\end{array}$\n\nMoreover, by definition, the collateral constraint must hold when the economy is not constrained21:\n\n$\\underset{{c}_{1}^{}+{b}_{1}^{}{R}_{L1}-e}{\\underbrace{{b}_{2}^{}}}\\le \\underset{\\frac{y}{{R}_{L2}}}{\\underbrace{{p}_{1}^{}}},$\n\nwhich we can rewrite as:\n\n$e\\ge {e}^{b}={c}_{1}^{}+{b}_{1}^{}{R}_{L1}-\\frac{y}{{R}_{L2}}$\n\nThat is, whenever the endowment is above a certain threshold (eeb) the economy is not constrained. On the other hand, when the economy is constrained (e < eb) the collateral constraint is binding and consumers would like to borrow b2 > p1. Given that this is not possible, consumers will borrow as much as they can, trying to maximize their consumption in period 1. In this case, the collateral constraint will bind with equality b2 = p1, so that:\n\n${c}_{1}+{b}_{1}{R}_{L1}-e=\\frac{y}{{u}^{\\prime }\\left({c}_{1}\\right)},$\n\nand using the fact that the utility function is in CES form:\n\n$\\begin{array}{cc}{c}_{1}+{b}_{1}{R}_{L1}-e=y{c}_{1}^{\\rho }.& \\left(\\text{A.}1\\right)\\end{array}$\n\nTherefore, depending whether the constraint is binding or not, we can express borrowing in period 0 as:\n\n$\\begin{array}{cc}{b}_{1}=\\left\\{\\begin{array}{cc}{\\begin{array}{c}\\begin{array}{c}\\left({R}_{L2}{R}_{L1}\\right)\\end{array}\\end{array}}^{-\\frac{1}{\\rho }}& e\\ge {e}^{b}\\\\ \\frac{y{c}_{1}^{\\rho }-{c}_{1}+e}{{R}_{L1}}& e<{e}^{b}\\end{array}& \\left(\\text{A.2}\\right)\\end{array}$\n\nFinally, we assume that the endowment is stochastic and follows a uniform distribution $e\\sim U\\left(\\overline{e}-\\epsilon ,\\text{\\hspace{0.17em}}\\overline{e}+\\epsilon \\right)$.\n\nAssumption on parameter values. To be able to solve the model we need to make assumptions on the value of two parameters: y and $\\overline{e}$. In particular, we will consider values such that 1) the economy may be constrained for sufficiently large negative shocks but 2) would not be constrained in the absence of uncertainty.\n\nFirst, we want a condition that is necessary and sufficient for the economy to be constrained with some probability, when $e\\sim U\\left(\\overline{e}-\\epsilon ,\\text{\\hspace{0.17em}}\\overline{e}+\\epsilon \\right)$. Let’s reason the other way round: we already showed that the economy is indeed unconstrained in period 1 if and only if:\n\n$\\overline{e}\\ge {e}^{b}={c}_{1}^{}+{b}_{1}^{}{R}_{L1}-\\frac{y}{{R}_{L2}}.$\n\nWhen e is stochastic, for the economy to be unconstrained, the above inequality must hold for all possible realizations of e (in particular the adverse realizations). In other words it must be the case that:\n\n$\\begin{array}{ccc}\\hfill e-\\epsilon & \\ge & {c}_{1}^{}+{b}_{1}^{}{R}_{L1}-\\frac{y}{{R}_{L2}},\\hfill \\\\ \\hfill \\overline{e}& \\ge & {c}_{1}^{}+{b}_{1}^{}{R}_{L1}-\\frac{y}{{R}_{L2}}+\\epsilon .\\hfill \\end{array}$\n\nTherefore, when $\\overline{e}<{c}_{1}^{}+{b}_{1}^{}{R}_{L1}-\\frac{y}{{R}_{L2}}+\\epsilon$ there exists a non-zero probability that the constraint binds.\n\nSecond, we want a condition that is necessary and sufficient for the economy to be unconstrained when there is no uncertainty around the realizations of e (i.e., ε = 0 and $\\overline{e}=e$). When ε = 0, the constraint is not binding in period 1 if and only if $e=\\overline{e}\\ge {e}^{b}$, that is:\n\n$\\overline{e}\\ge {c}_{1}^{}+{b}_{1}^{}{R}_{L1}-\\frac{y}{{R}_{L2}}.$\n\nTherefore, with no uncertainty, when $\\overline{e}\\ge {c}_{1}^{}+{b}_{1}^{}{R}_{L1}-\\frac{y}{{R}_{L2}}$ the constraint never binds.\n\nSummarizing we choose an $\\overline{e}$ such that would not be constrained in the absence of uncertainty but it economy may be constrained for sufficiently large negative shocks:\n\n$\\left({R}_{L2}{\\right)}^{-\\frac{1}{\\rho }}+\\left({R}_{L2}{R}_{L1}{\\right)}^{-\\frac{1}{\\rho }}{R}_{L1}-\\frac{y}{{R}_{L2}}\\le \\overline{e}<{\\left({R}_{L2}\\right)}^{-\\frac{1}{\\rho }}+\\left({R}_{L2}{R}_{L1}{\\right)}^{-\\frac{1}{\\rho }}{R}_{L1}-\\frac{y}{{R}_{L2}}+\\epsilon .$\n\nThis implies that there will be a threshold for the size of the shock (εb) above which the collateral constraint will start to be binding with positive probability. Specifically, the collateral constraint would be binding for realizations of e in the interval $\\left[\\overline{e}-\\epsilon ,\\text{\\hspace{0.17em}}\\overline{e}-{\\epsilon }^{b}\\right]$. The level of εb can be easily computed as:\n\n${\\epsilon }^{b}=\\overline{e}-{e}^{b}=\\overline{e}-{c}_{1}^{}-{b}_{1}^{}{R}_{L1}+\\frac{y}{{R}_{L2}}.$\n\nCompetitive equilibrium. We will find numerical values for consumption at time 1 (c1) by using is the Euler equation FOC(b1), which gives us an optimal relation between consumption in period 0 and consumption in period 1.22 In order to be able to solve this equation we need 1) to find an expression for borrowing as a function of consumption for both constrained and unconstrained states, as we already did in equation (A.2); and 2) to weight those states for their probability. Combining FOC(b1), the budget constraint, and the expression for b1 derived earlier in equation (A.2) we get the following system of equations:\n\n$\\left\\{\\begin{array}{c}{b}_{1}^{-\\rho }={R}_{L1}{\\mathbb{\\text{E}}}_{0}\\left[{c}_{1}^{-\\rho }\\right],\\\\ {b}_{1}=\\left\\{\\begin{array}{cc}\\left({R}_{L2}{R}_{L1}{\\right)}^{-\\frac{1}{\\rho }}& e\\ge {e}^{b},\\\\ \\frac{\\begin{array}{c}y{c}_{1}^{\\rho }-{c}_{1}+e\\end{array}}{{R}_{L1}}& e<{e}^{b}.\\end{array}\\end{array}$\n\nBy plugging the second equation in the first one we can write:\n\n$\\text{Pr}\\left(e<{e}^{b}\\right)\\cdot {\\left[{b}_{1}^{-\\rho }\\right]}^{\\text{binding}}+\\text{Pr}\\left(e\\ge {e}^{b}\\right)\\cdot {\\left[{b}_{1}^{-\\rho }\\right]}^{\\text{non-}\\text{binding}}={R}_{L1}{\\mathbb{\\text{E}}}_{0}\\left[{c}_{1}^{-\\rho }\\right].$\n\nNow, by effectively substituting for b1, The LHS of the previous equation can be expressed as follows:23\n\n$\\begin{array}{cc}{b}_{1}^{-\\rho }& =\\frac{1}{2\\epsilon }\\underset{\\overline{e}-\\epsilon }{\\overset{\\overline{e}-{\\epsilon }^{b}}{\\int }}\\left(\\frac{y{c}_{1}^{\\rho }-{c}_{1}+e}{{R}_{L1}}{\\right)}^{-\\rho }\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}de+\\frac{1}{2\\epsilon }\\underset{\\overline{e}-{\\epsilon }^{b}}{\\overset{\\overline{e}+\\epsilon }{\\int }}{R}_{L2}{R}_{L1}de=\\hfill \\\\ & =\\frac{1}{2\\epsilon }\\underset{\\overline{e}-\\epsilon }{\\overset{\\overline{e}-{\\epsilon }^{b}}{\\int }}\\left(\\frac{y{c}_{1}^{\\rho }-{c}_{1}}{{R}_{L1}}+\\frac{e}{{R}_{L1}}{\\right)}^{-\\rho }\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}de+\\frac{{R}_{L2}{R}_{L1}}{2\\epsilon }\\left[e{\\right]}_{\\overline{e}-{\\epsilon }^{b}}^{\\overline{e}+\\epsilon }=\\hfill \\\\ & =\\frac{1}{2\\epsilon }{\\left[{R}_{L1}\\frac{\\left(\\frac{y{c}_{1}^{\\rho }-{c}_{1}}{{R}_{L1}}+\\frac{e}{{R}_{L1}}{\\right)}^{-\\rho +1}}{-\\rho +1}\\right]}_{\\overline{e}-\\epsilon }^{\\overline{e}-{\\epsilon }^{b}}+\\frac{{R}_{L2}{R}_{L1}}{2\\epsilon }\\left[\\epsilon +{\\epsilon }^{b}\\right]\\hfill \\\\ & =\\frac{{R}_{L1}^{\\rho }}{2\\epsilon \\left(1-\\rho \\right)}\\left[\\left(y{c}_{1}^{\\rho }-{c}_{1}+e{\\right)}^{-\\rho +1}{\\right]}_{\\overline{e}-\\epsilon }^{\\overline{e}-{\\epsilon }^{b}}+\\frac{{R}_{L2}{R}_{L1}}{2\\epsilon }\\left[\\epsilon +{\\epsilon }^{b}\\right].\\hfill \\end{array}$\n\nBy equalizing LHS to RHS numerically, we obtain the competitive equilibrium level of consumption at time 1, where remember that:\n\n$\\begin{array}{lll}\\text{LHS}& =& \\frac{{R}_{L1}^{\\rho }}{2\\epsilon \\left(1-\\rho \\right)}\\left[\\left(y{c}_{1}^{\\rho }-{c}_{1}+\\overline{e}-{\\epsilon }^{b}{\\right)}^{-\\rho +1}-{\\left(y{c}_{1}^{\\rho }-{c}_{1}+\\overline{e}-\\epsilon \\right)}^{-\\rho +1}\\right]+\\frac{{R}_{L2}{R}_{L1}}{2\\epsilon }\\left[\\epsilon +{\\epsilon }^{b}\\right]\\\\ \\text{RHS}& =& {R}_{L1}{\\mathbb{\\text{E}}}_{0}\\left[{c}_{1}^{-\\rho }\\right].\\end{array}$\n\nFinally, one can also derive the level of optimal debt at time 0, by using again FOC(bT):\n\n${b}_{1}={\\mathbb{\\text{E}}}_{0}\\left[\\left({R}_{L1}{c}_{1}^{-\\rho }{\\right)}^{-\\frac{1}{\\rho }}\\right].$\n\nSocial planner. The social planner problem is solved with the same strategy. The first order conditions are:\n\n$\\left\\{\\begin{array}{cc}FOC\\left({b}_{1}\\right):& {u}^{\\prime }\\left({c}_{0}\\right)={R}_{L1}{\\mathbb{\\text{E}}}_{0}\\left[{u}^{\\prime }\\left({c}_{1}\\right)+\\lambda {p}^{\\prime }\\left({c}_{1}\\right)\\right],\\hfill \\\\ FOC\\left({b}_{2}\\right):& {u}^{\\prime }\\left({c}_{1}\\right)={R}_{L2}+\\lambda \\left(1-{p}^{\\prime }\\left({c}_{1}\\right)\\right),\\hfill \\\\ FOC\\left({\\theta }_{2}\\right):& {p}_{1}=\\frac{y}{{u}^{\\prime }\\left({c}_{1}\\right)}.\\hfill \\end{array}$\n\nFirst we have to find an expression for p(c1). From FOC(θ2) we get:\n\n$p\\left({c}_{1}\\right)=\\frac{y}{{u}^{\\prime }\\left({c}_{1}\\right)}=y{c}_{1}^{\\rho },$\n\nand computing the derivative:\n\n${p}^{\\prime }\\left({c}_{1}\\right)=\\frac{\\partial \\left(y{c}_{1}\\right)}{\\partial {c}_{1}}=\\rho y{c}_{1}^{\\rho -1},$\n\nNotice that the p(c1) is positive and decreasing. Notice also that, by definition, the Lagrange multiplier (λ) is positive only when the constraint is binding. By looking at FOC(b1) of the social planner problem, we can state that the planner limits over-borrowing. In fact, ${u}^{\\prime }{\\left({c}_{1}\\right)}^{sp}>{u}^{\\prime }{\\left({c}_{1}\\right)}^{CE}$ which implies that consumption and, therefore, borrowing at time 1 are lower relative to the competitive equilibrium. On the other hand, the planner increases consumption in period 1: given that p(c1) > 0 from FOC(b2) we see that ${u}^{\\prime }{\\left({c}_{1}\\right)}^{sp}<{u}^{\\prime }{\\left({c}_{1}\\right)}^{CE}$.\n\nWe also need a value of λ. Notice that the Lagrange multiplier of the social planner is numerically different from the one of the competitive equilibrium problem. In fact, from FOC(b2) we get\n\n$\\lambda =\\frac{{c}_{1}^{-\\rho }-{R}_{L2}}{1+y}.$\n\nCombining these two results we can compute:\n\n$\\lambda p\\prime \\left({c}_{1}\\right)=\\left\\{\\begin{array}{cc}0\\hfill & e\\ge {e}^{b},\\\\ \\frac{\\rho y}{1+y}\\left({c}_{1}^{-1}-{R}_{L2}{c}_{1}^{\\rho -1}\\right)\\hfill & e<{e}^{b}.\\end{array}$\n\nWe can now solve for the level of c1. The FOC(b1) can be written:\n\n${b}_{1}^{-\\rho }={R}_{L1}{\\mathbb{\\text{E}}}_{0}\\left[{c}_{1}^{-\\rho }+\\lambda {p}^{\\prime }\\left({c}_{1}\\right)\\right].$\n\nThe LHS has already been computed before. The RHS is:\n\n$\\begin{array}{l}\\frac{{R}_{L1}}{2\\epsilon }\\underset{\\overline{e}-\\epsilon }{\\overset{\\overline{e}-{\\epsilon }^{b}}{\\int }}\\left({c}_{1}^{-\\rho }+\\frac{\\rho y}{1+y}\\left({c}_{1}^{-1}-{R}_{L2}{c}_{1}^{\\rho -1}\\right)\\right)de+\\frac{{R}_{L1}}{2\\epsilon }\\underset{\\overline{e}-{\\epsilon }^{b}}{\\overset{\\overline{e}-{\\epsilon }^{b}}{\\int }}{c}_{1}^{-\\rho }de,\\\\ \\frac{{R}_{L1}}{2\\epsilon }\\left[\\left({c}_{1}^{-\\rho }+\\frac{\\rho y}{1+y}\\left({c}_{1}^{-1}-{R}_{L2}{c}_{1}^{\\rho -1}\\right)\\right)\\left(\\epsilon -{\\epsilon }^{b}\\right)+{c}_{1}^{-\\rho }\\left(\\epsilon +{\\epsilon }^{b}\\right)\\right],\\\\ \\frac{{R}_{L2}}{2\\epsilon }\\left[\\left(\\frac{\\rho y}{1+y}\\left({c}_{1}^{-1}-{R}_{L2}{c}_{1}^{\\rho -1}\\right)\\right)\\left(\\epsilon -{\\epsilon }^{b}\\right)+2{c}_{1}^{-\\rho }\\epsilon \\right]\\end{array}$\n\nAs we have done above, by equalizing LHS to RHS numerically, we obtain the competitive equilibrium level of consumption at time 1, where remember that:\n\n$\\begin{array}{lll}\\text{LHS}& =& \\frac{{R}_{L1}^{\\rho }}{2\\epsilon \\left(1-\\rho \\right)}\\left[\\left(y{c}_{1}^{\\rho }-{c}_{1}+\\overline{e}-{\\epsilon }^{b}{\\right)}^{-\\rho +1}-{\\left(y{c}_{1}^{\\rho }-{c}_{1}+\\overline{e}-\\epsilon \\right)}^{-\\rho +1}\\right]+\\frac{{R}_{L2}{R}_{L1}}{2\\epsilon }\\left[\\epsilon +{\\epsilon }^{b}\\right]\\\\ \\text{RHS}& =& \\frac{{R}_{L1}}{2\\epsilon }\\left[\\left(\\frac{\\rho y}{1+y}\\left({c}_{1}^{-1}-{R}_{L2}{c}_{1}^{\\rho -1}\\right)\\right)\\left(\\epsilon -{\\epsilon }^{b}\\right)+2{c}_{1}^{-\\rho }\\epsilon \\right].\\end{array}$\n\nFinally, one can derive the optimal expression for borrowing at time 1 from the social planner FOC(b1):\n\n${b}_{1}=\\left({R}_{L1}{\\mathbb{\\text{E}}}_{0}\\left[{c}_{1}^{-\\rho }+\\lambda {p}^{\\prime }\\left({c}_{1}\\right)\\right]{\\right)}^{-\\frac{1}{\\rho }}.$\n\nCrisis Probability. The crisis probability is defined as the probability of the constraint to be binding:\n\n$\\begin{array}{cc}{\\text{Pr[b}}_{2}& >{p}_{1}\\right]\\hfill \\\\ & =\\frac{1}{2\\epsilon }\\underset{\\overline{e}-\\epsilon }{\\overset{\\overline{e}-{\\epsilon }^{b}}{\\int }}de=\\frac{1}{2\\epsilon }\\left(\\epsilon -{\\epsilon }^{b}\\right).\\hfill \\end{array}$\n\nwhich, using the optimality conditions and the budget constraint, can be written as\n\n$\\text{Pr}\\left[\\left({c}_{1}-\\left(e-{b}_{1}{R}_{L1}\\right)>\\frac{y}{{u}^{\\prime }\\left({c}_{1}\\right)}\\right].$\n\nNow, knowing that $e=\\overline{e}+\\stackrel{˜}{\\epsilon }$ and that $\\stackrel{˜}{\\epsilon }\\sim \\mathcal{U}\\left(-\\epsilon ,\\epsilon \\right)$, we can write\n\n$\\text{Pr}\\left[\\stackrel{˜}{\\epsilon }<\\underset{x}{\\underbrace{{c}_{1}-\\stackrel{˜}{e}+{b}_{1}{R}_{L1}-\\frac{y}{{u}^{\\prime }\\left({c}_{1}\\right)}}}\\right].$\n\nIn particular, the probability of the constraint to be binding is given by:\n\n$\\text{Pr}\\left[-\\epsilon \\le \\stackrel{˜}{\\epsilon }\n\nBank of England and Johns Hopkins University, respectively. This paper was previously circulated under the title “Coordinating Monetary and Macro-Prudential Policies.” We would like to thank Jihad Dagher and the seminar participants at the 2013 Macro Banking and Finance Workshop, the 2013 MMF Conference, the Cattolica University, the Bank of Italy, the EPFL, the Bank of Portugal, the Banque de France, the Bank of England, the IADB, LACEA 2012 Annual Meetings, and EEA 2012 Annual Meetings for useful discussions and helpful comments. The information and opinions presented in this paper are entirely those of the authors, and not necessarily those of the Bank of England.\n\nBernanke (2010) recently said that “the best response to the housing bubble would have been regulatory, rather than monetary”.\n\nThis assumption may be justified by both theoretical and empirical findings (see, for example, Hannan and Berger, 1991, Neumark and Sharpe, 1992, Kwapil and Scharler, 2010, Gerali, Neri, Sessa, and Signoretti, 2010).\n\nConsumers can be interpreted as households owning durable assets (such as their homes, for example).\n\nDetails on the the derivation of the equilibrium conditions are reported in Appendix A.\n\nThe presence of market power can be justified by the existence of switching costs which lead to long-term relationships between banks and borrowers (see Diamond (1984) for example). Empirically, the presence of market power in the banking sector, as well as its determinants over the business cycle, are well documented. See, for example, Berger, Demirguc-Kunt, Levine, and Haubrich (2004) and Degryse and Ongena (2008).\n\nNotice here that Gerali, Neri, Sessa, and Signoretti (2010) set the elasticity of loan contracts to about 2.5, to match an average spread of 170 basis points of deposit rates on the policy rate. Our number differs from theirs because we assume that the markup is applied to the gross interest rate (i.e., MR) instead of the net interest rate (i.e., 1+Mr).\n\nThese estimates are in line with older studies on interest rate pass-through in the U.S.. For example, Cottarelli and Kourelis (1994) estimate a short run pass through of 0.32 and a long run pass through of 1; Moazzami (1999) and Borio and Fritz (1995) report a short run coefficient of 0.4 and 0.34, respectively.\n\nNote that the calibration of this parameter does not affect the qualitative behavior of our model.\n\nNote that both Woodford (2012) and Benigno, Chen, Otrok, Rebucci, and Young (2013) define financial stability in these terms.\n\nNote here that, if no shock pushes the economy away from its equilibrium, the average markup would be equal to the constant frictionless markup and the price of all goods in the economy would be the same, implying that no efficiency condition would be violated.\n\nSee Korinek (2010), Bianchi (2011), Jeanne and Korinek (2010a,b), Benigno, Chen, Otrok, Rebucci, and Young (2013) for a more detailed discussion.\n\nResults reported in Figure 5 are robust to the case in which the distortions generated by monopolistic competition are not removed.\n\nNote here that changing the order of the policy actions would not alter the results.\n\nJohn Taylor, interviewed by Bloomberg at the American Economic Association’s annual meeting, Atlanta, January 5, 2010, available at: http://www.bloomberg.com/apps/news?pid=newsarchive&sid=a44P5KTDjWWY\n\nFor instance, the Office of the Comptroller of the Currency is in charge of nationally chartered banks and their subsidiaries. The Federal Reserve covers affiliates of nationally chartered banks. The Office of Thrift Supervision oversees savings institutions. The Federal Deposit Insurance Corporation insures deposits of both state-chartered and nationally chartered banks.\n\nBy prime loans we refer to loans that conform to Government Sponsored Enterprises (GSE) guidelines; by nonprime loans we refer to Alt-A, Home Equity, FHA/VA, and subprime mortgages.\n\nMBS which are issued or guaranteed by a government sponsored enterprise (GSE) such as Fannie Mae or Freddie Mac are referred to as “agency MBS.” Some private institutions, such as subsidiaries of investment banks, banks, financial institutions, non-bank mortgage lenders and home builders, also issue mortgage securities, the so-called “private label” MBS.\n\nNet percentage calculated by subtracting the percent of banks tightening from the percent of banks easing. Negative values, therefore, indicate easing.\n\nNote here that we are assuming that profits are realized at the end of the period so that they have no effect on the borrowing constraint.\n\nRember that c0 = b1 from the budget constraint.\n\nIf X is uniformely distributed with U(a, b), then the nth moment of X is given by ${\\mathbb{\\text{E}}}_{0}\\left[{X}^{n}\\right]=\\frac{1}{b-a}\\underset{a}{\\overset{b}{\\int }}{x}^{n}dx$.\n\nDoes Easing Monetary Policy Increase Financial Instability?\nAuthor: Ambrogio Cesa-Bianchi and Mr. Alessandro Rebucci\n•", null, "A Counterfactual Path for the U.S. Policy Rate.This chart replicates the counterfactual federal funds rate reported by Taylor (2007). The counterfactual path for the policy rate from 1996 to 2007 is obtained with a Taylor rule of the type: it=rt+πt+1.5(πt−π)+0.5(yt−yt), where rt, long-run real value of the federal funds rate, is set to 2 percent, πt is CPI inflation, π is target inflation (assumed at 2 percent), yt is real GDP growth, and yt is real potential GDP growth.\n•", null, "Model Equilibrium with Financial Friction.On the horizontal axis is the maximum size of the endowment shock (ε).\n•", null, "Model Equilibrium with Both Frictions: Positive Shock to the Interest Rate.On the horizontal axis is the maximum size of the endowment shock (ε). The thick solid line displays the equilibriums when no shock hits the risk-free interest rate; the thin line with asterisk markers and the thin line with circle markers display the equilibrium after a positive shock hits the risk-free rate under flex-rates and sticky-rates, respectively.\n•", null, "Model equilibrium with both frictions - Negative Shock to The Interest Rate.On the horizontal axis is the maximum size of the endowment shock (ε). The thick solid line displays the equilibriums when no shock hits the risk free interest rate; the thin line with asterisk markers and the thin line with circle markers display the equilibrium after a negative shock hits the risk free rate under flex-rates and sticky-rates, respectively.\n•", null, "Efficient Allocation with Both Frictions: Positive Shock to the Interest Rate.On the horizontal axis is the maximum size of the endowment shock (ε). The thin lines with triangle and square markers display the equilibrium after a positive shock hits the risk-free rate under flex-rates and sticky rates, respectively; the dashed line displays the efficient allocation with two policy instruments. A subsidy is in place to remove the distortionary effect of monopolistic competition.\n•", null, "Efficient Allocation with Both Frictions: Negative Shock to the Interest Rate.On the horizontal axis is the maximum size of the endowment shock (ε). The thin lines with triangle and square markers display the equilibrium after a negative shock hits the risk-free rate under flex-rates and sticky rates, respectively; the dashed line displays the efficient allocation with two policy instruments. A subsidy is in place to remove the distortionary effect of monopolistic competition.\n•", null, "Alternative Path of the Lending Interest Rate under Different Assumptions about the Number of Instruments at the Policy-maker’s Disposal.Competitive Eq. -Flex (ν < 0) displays the lending interest rate in the decentralized economy under fully flexible interest rates; Policy displays the lending interest rate that would prevail with a policy maker addressing both the macroeconomic and the financial frictions with one or two instruments, respectively.\n•", null, "Monetary Policy and the U.S. Housing Sector.The figure provides a picture of the evolution of the U.S. mortgage market and monetary policy over the 2000-07 period." ]	[ null, "https://www.elibrary.imf.org/skin/73d3da6f9c8e812b492dcc7a9cf15c322f6a1944/img/Blank.svg", null, "https://www.elibrary.imf.org/skin/73d3da6f9c8e812b492dcc7a9cf15c322f6a1944/img/Blank.svg", null, "https://www.elibrary.imf.org/skin/73d3da6f9c8e812b492dcc7a9cf15c322f6a1944/img/Blank.svg", null, "https://www.elibrary.imf.org/skin/73d3da6f9c8e812b492dcc7a9cf15c322f6a1944/img/Blank.svg", null, "https://www.elibrary.imf.org/skin/73d3da6f9c8e812b492dcc7a9cf15c322f6a1944/img/Blank.svg", null, "https://www.elibrary.imf.org/skin/73d3da6f9c8e812b492dcc7a9cf15c322f6a1944/img/Blank.svg", null, "https://www.elibrary.imf.org/skin/73d3da6f9c8e812b492dcc7a9cf15c322f6a1944/img/Blank.svg", null, "https://www.elibrary.imf.org/skin/73d3da6f9c8e812b492dcc7a9cf15c322f6a1944/img/Blank.svg", null, "https://www.elibrary.imf.org/skin/73d3da6f9c8e812b492dcc7a9cf15c322f6a1944/img/Blank.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9124814,"math_prob":0.969959,"size":101659,"snap":"2022-40-2023-06","text_gpt3_token_len":22044,"char_repetition_ratio":0.1797944,"word_repetition_ratio":0.17020339,"special_character_ratio":0.21808203,"punctuation_ratio":0.12515937,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98840696,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-26T09:36:43Z\",\"WARC-Record-ID\":\"<urn:uuid:19ab2266-f574-4af5-bee3-73685de7b082>\",\"Content-Length\":\"1019822\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:371e79d1-bd41-49bf-9888-41a99ad5e908>\",\"WARC-Concurrent-To\":\"<urn:uuid:4be37f5a-3108-4fde-8086-b85fc469e92e>\",\"WARC-IP-Address\":\"3.248.172.205\",\"WARC-Target-URI\":\"https://www.elibrary.imf.org/view/journals/001/2015/139/article-A001-en.xml\",\"WARC-Payload-Digest\":\"sha1:Q2YMQQKBHYDVF5BRYNMNEOR7V34LTVYX\",\"WARC-Block-Digest\":\"sha1:6GHX7DNTC5WS3556PXXQJMFXKD5DJO27\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334855.91_warc_CC-MAIN-20220926082131-20220926112131-00098.warc.gz\"}"}
https://mathoverflow.net/questions/326280/6-linear-pde-for-only-3-unknowns	[ "# 6 linear PDE for only 3 unknowns?\n\nLet $$x \\in (0,L)$$, $$t \\in (0,T)$$, and let $$u_0 = u_0(x) \\in \\mathbb{R}^3$$, $$g=g(t) \\in \\mathbb{R}^3$$, $$P = P(x,t) \\in \\mathbb{R}^3$$ and $$Q = Q(x,t) \\in \\mathbb{R}^3$$ be continuously differentiable functions.\n\nDenote by $$\\hat{P}, \\hat{Q}$$ the matrices \\begin{align} \\hat{P} = \\begin{bmatrix} 0 & -P_3 & P_2 \\\\ Q_3 & 0 & -P_1 \\\\ -P_2 & P_1 & 0 \\end{bmatrix}, \\qquad \\hat{Q} = \\begin{bmatrix} 0 & -Q_3 & Q_2 \\\\ Q_3 & 0 & -Q_1 \\\\ -Q_2 & Q_1 & 0 \\end{bmatrix}. \\end{align}\n\nMy question is:\n\nAssuming that the following four conditions hold:\n\nCondition 1: \\begin{align} \\partial_x P - \\partial_t Q = \\hat{P}Q \\qquad \\text{in }(0,L) \\times (0,T), \\end{align} Condition 2: \\begin{align} u_0(0) = g(0), \\end{align} Condition 3: \\begin{align} u_0' = - \\hat{Q}u_0 \\qquad \\text{for } x \\in (0,L), \\end{align} Condition 4: \\begin{align} g' = - \\hat{P}g \\qquad \\text{for } t \\in (0,T), \\end{align} is there a solution $$u = u(x,t) \\in \\mathbb{R}^3$$ to \\begin{align} \\begin{cases} \\partial_t u = -\\hat{P}u &\\text{in }(0,L)\\times(0,T) \\\\ \\partial_x u = - \\hat{Q}u &\\text{in }(0,L)\\times(0,T) \\\\ u(x,0) = u_0(x) & \\text{for } x \\in (0,L)\\\\ u(0,t) = g(t) & \\text{for }t \\in (0,T) \\end{cases} \\end{align} or not ?\n\nIn the above, $$f'$$, $$\\partial_t f$$a and $$\\partial_x f$$ denote the derivative, partial time derivative and partial space derivative respectively; and $$P_k$$, $$Q_k$$, $$u_k$$, $$u_{0k}$$ and $$g_k$$ denote the components of $$P$$, $$Q$$, $$u$$, $$u_0$$ and $$g$$ respectively.\n\nMy question with all terms developed rewrites as follows. Assuming that the following four conditions hold:\n\nCondition 1: \\begin{align} \\begin{cases} \\partial_x P_1 - \\partial_t Q_1 = P_2 Q_3 - P_3 Q_2\\\\ \\partial_x P_2 - \\partial_t Q_2 = P_3 Q_1 - P_1 Q_3\\\\ \\partial_x P_3 - \\partial_t Q_3 = P_1 Q_2 - P_2 Q_1 \\end{cases} \\qquad \\text{in }(0,L) \\times (0,T), \\end{align} Condition 2: \\begin{align} u_0(0) = g(0), \\end{align} Condition 3: \\begin{align} \\begin{cases} u_{01}' = \\ \\ Q_3(\\cdot,0) u_{02} - Q_2(\\cdot,0) u_{03} \\\\ u_{02}' = -Q_3(\\cdot,0) u_{01} + Q_1(\\cdot,0) u_{03} \\\\ u_{03}' = \\ \\ Q_2(\\cdot,0) u_{01} - Q_1(\\cdot,0) u_{02} \\end{cases} \\qquad \\text{for } x \\in (0,L), \\end{align} Condition 4: \\begin{align} \\begin{cases} g_1' &= \\ \\ P_3(0,\\cdot) g_2 - P_2(0,\\cdot) g_3 \\\\ g_2' &= -P_3(0,\\cdot) g_1 + P_1(0,\\cdot) g_3 \\\\ g_3' &= \\ \\ P_2(0,\\cdot) g_1 - P_1(0,\\cdot) g_2 \\end{cases} \\qquad \\text{for } t \\in (0,T), \\end{align} is there a solution $$u = u(x,t) \\in \\mathbb{R}^3$$ to the problem: \\begin{align} \\begin{cases} \\partial_t u_1 = \\ \\ P_3 u_2 - P_2 u_3 &\\text{in }(0,L)\\times(0,T) \\\\ \\partial_t u_2 = -P_3 u_1 + P_1 u_3 &\\text{in }(0,L)\\times(0,T) \\\\ \\partial_t u_3 = \\ \\ P_2 u_1 - P_1 u_2 & \\text{in }(0,L)\\times(0,T) \\\\ \\partial_x u_1 = \\ \\ Q_3 u_2 - Q_2 u_3 & \\text{in }(0,L)\\times(0,T) \\\\ \\partial_x u_2 = -Q_3 u_1 + Q_1 u_3 &\\text{in }(0,L)\\times(0,T) \\\\ \\partial_x u_3 = \\ \\ Q_2 u_1 - Q_1 u_2 & \\text{in }(0,L)\\times(0,T) \\\\ u(x,0) = u_0(x) & \\text{for } x \\in (0,L)\\\\ u(0,t) = g(t) & \\text{for }t \\in (0,T). \\end{cases} \\end{align}\n\nWhat I started. I started reasoning the following way. For $$x \\in (0,L)$$ fixed, a function of the form \\begin{align} u_1(x,t) &= u_{01}(x) + \\int_0^t P_3(x,s)u_2(x,s) - P_2(x,s)u_3(x,s)ds\\\\ u_2(x,t) &= u_{02}(x) + \\int_0^t -P_3(x,s)u_1(x,s) + P_1(x,s)u_3(x,s)ds\\\\ u_3(x,t) &= u_{03}(x) + \\int_0^t P_2(x,s)u_1(x,s) - P_1(x,s)u_2(x,s)ds, \\end{align} satisfies the initial value problem involving time derivatives, with $$u(\\cdot, 0) = u_0$$, while for $$t \\in (0,T)$$ fixed, a solution of the form \\begin{align} \\begin{aligned} u_1(x,t) &= g_1(t) + \\int_0^x Q_3(\\xi,t)u_2(\\xi,t) - Q_2(\\xi, t)u_3(\\xi, t)d\\xi \\\\ u_2(x,t) &= g_2(t) + \\int_0^x -Q_3(\\xi, t)u_1(\\xi, t) + Q_1(\\xi, t) u_3(\\xi, t)d\\xi\\\\ u_3(x,t) &= g_3(t) + \\int_0^x Q_2(\\xi, t)u_1(\\xi, t) - Q_1(\\xi, t)u_2(\\xi, t)d\\xi. \\end{aligned} \\end{align} satisfies the initial value problem involving space derivatives, with $$u(0, \\cdot) = g$$. I wanted to show, using the four conditions, that both expressions for the solution coincide. Using conditions 3 and 4 we can rewrite both expressions as \\begin{align} u_1(x,t) &= u_{01}(0) + \\int_0^x Q_3(\\xi, 0)u_{02}(\\xi) - Q_2(\\xi, 0)u_{03}(\\xi)d\\xi \\\\ &\\quad + \\int_0^t P_3(0,s)g_2(s) - P_2(0,s)g_3(s)ds + \\int_0^t \\int_0^x \\partial_x (P_3 u_2 - P_2 u_3 )(\\xi, s)d\\xi ds\\\\ u_2(x,t) &= u_{02}(0) + \\int_0^x -Q_3(\\xi, 0)u_{01} + Q_1(\\xi, 0)u_{03}(\\xi)d\\xi \\\\ &\\quad + \\int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s)ds + \\int_0^t \\int_0^x \\partial_x(-P_3 u_1 + P_1 u_3)(\\xi, s)ds\\\\ u_3(x,t) &= u_{03}(0) + \\int_0^x Q_2(\\xi, 0)u_{01}(\\xi) - Q_1(\\xi, 0)u_{02}(\\xi)\\\\ &\\quad + \\int_0^t P_2(0,s)g_1(s) - P_1(0,s)g_2(s)ds + \\int_0^t \\int_0^x \\partial_x (P_2 u_1 - P_1 u_2)(\\xi,s)d\\xi ds, \\end{align} and \\begin{align} u_1(x,t) &= g_1(0) + \\int_0^t P_3(0,s)g_2(s) - P_2(0,s)g_3(s) ds \\\\ &\\quad + \\int_0^x Q_3(\\xi,0)u_{02}(\\xi) - Q_2(\\xi, 0)u_{03}(\\xi)d\\xi + \\int_0^x \\int_0^t \\partial_t ( Q_3 u_2 - Q_2 u_3)(\\xi, s)dsd\\xi\\\\ u_2(x,t) &= g_2(0) + \\int_0^t -P_3(0,s)g_1(s) + P_1(0,s)g_3(s) ds\\\\ &\\quad + \\int_0^x -Q_3(\\xi, 0)u_{01}(\\xi) + Q_1(\\xi, 0) u_{03}(\\xi)d\\xi + \\int_0^x \\int_0^t \\partial_t (-Q_3 u_1 + Q_1 u_3)(\\xi,s)dsd\\xi\\\\ u_3(x,t) &= g_3(0) + \\int_0^t P_2(0,s)g_1(s) - P_1(0,s)g_2(s) ds \\\\ &\\quad + \\int_0^x Q_2(\\xi, 0)u_{01}(\\xi) - Q_1(\\xi, 0)u_{02}(\\xi)d\\xi + \\int_0^x \\int_0^t \\partial_t ( Q_2 u_1 - Q_1 u_2)(\\xi, s)ds d\\xi \\end{align} respectively. It seems that both expressions would coincide if the following equalities hold: \\begin{align} \\partial_x (P_3 u_2 - P_2 u_3 ) &= \\partial_t ( Q_3 u_2 - Q_2 u_3)\\\\ \\partial_x(-P_3 u_1 + P_1 u_3) &= \\partial_t (-Q_3 u_1 + Q_1 u_3)\\\\ \\partial_x (P_2 u_1 - P_1 u_2) &= \\partial_t ( Q_2 u_1 - Q_1 u_2). \\end{align} To show that these equalities hold, I would like to use condition 1. However, I do not know how to argue from there.\n\nAny suggestion of how to proceed from here, any alternative way of reasoning, or explanation of why there would be no solution, would be welcome. Thank you.\n\n(This question is also on Mathematics Stack Exchange: https://math.stackexchange.com/questions/3160600/6-linear-pde-for-only-3-unknowns)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5123088,"math_prob":1.0000064,"size":6094,"snap":"2019-35-2019-39","text_gpt3_token_len":2905,"char_repetition_ratio":0.17405583,"word_repetition_ratio":0.15502183,"special_character_ratio":0.50639975,"punctuation_ratio":0.13763441,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999615,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-25T03:57:05Z\",\"WARC-Record-ID\":\"<urn:uuid:77cc0c6a-3cef-458a-a2ea-50d358319c79>\",\"Content-Length\":\"114618\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:21b5901d-a48d-45d6-b9a2-ea1c4a7f7148>\",\"WARC-Concurrent-To\":\"<urn:uuid:4ac4cd53-fbad-4f01-be8a-bb895c21c17e>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/326280/6-linear-pde-for-only-3-unknowns\",\"WARC-Payload-Digest\":\"sha1:ASQFFX6RXLH33MJZQBFWDJW3MEWNCXQS\",\"WARC-Block-Digest\":\"sha1:UXPC4LJJ3TRWIJ3JBIMJQPEHP4NPLIVE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027322170.99_warc_CC-MAIN-20190825021120-20190825043120-00103.warc.gz\"}"}
http://www.fact-index.com/r/re/reflection_coefficient.html	[ "Main Page \| See live article \| Alphabetical index\n\n# Reflection coefficient\n\nIn telecommunication, the term reflection coefficient (RC) has the following meanings:\n\n1. The ratio of the amplitude of the reflected wave and the amplitude of the incident wave.\n\n2. At a discontinuity in a transmission line, the complex ratio of the electric field strength of the reflected wave to that of the incident wave.\n\nNote 1: The reflection coefficient may also be established using other field or circuit quantities.\n\nNote 2: The reflection coefficient is given by the equations below, where Z 1 is the impedance toward the source, Z 2 is the impedance toward the load, the vertical bars designate absolute magnitude, and SWR is the standing wave ratio:\n\nSource: from Federal Standard 1037C and from MIL-STD-188" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8900114,"math_prob":0.92422706,"size":574,"snap":"2022-05-2022-21","text_gpt3_token_len":118,"char_repetition_ratio":0.15789473,"word_repetition_ratio":0.021052632,"special_character_ratio":0.20383275,"punctuation_ratio":0.116071425,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95995635,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-28T17:31:41Z\",\"WARC-Record-ID\":\"<urn:uuid:18805c8a-108e-4163-acaf-d94f69199daa>\",\"Content-Length\":\"4199\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8757b876-44f7-48b8-b991-036caf48576a>\",\"WARC-Concurrent-To\":\"<urn:uuid:ada29835-8676-4a43-8631-35e38c4c5d63>\",\"WARC-IP-Address\":\"23.227.169.68\",\"WARC-Target-URI\":\"http://www.fact-index.com/r/re/reflection_coefficient.html\",\"WARC-Payload-Digest\":\"sha1:6YNU6X6KWGXWPKZEVRPDC43NNNMMRF2W\",\"WARC-Block-Digest\":\"sha1:2DSCMYDSX5XXY7SDODMRWAFMYTL6AB6D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652663016949.77_warc_CC-MAIN-20220528154416-20220528184416-00252.warc.gz\"}"}
https://www.percentagecal.com/answer/560-is-what-percent-of-248000	[ "#### Solution for 560 is what percent of 248000:\n\n560:248000100 =\n\n(560100):248000 =\n\n56000:248000 = 0.23\n\nNow we have: 560 is what percent of 248000 = 0.23\n\nQuestion: 560 is what percent of 248000?\n\nPercentage solution with steps:\n\nStep 1: We make the assumption that 248000 is 100% since it is our output value.\n\nStep 2: We next represent the value we seek with {x}.\n\nStep 3: From step 1, it follows that {100\\%}={248000}.\n\nStep 4: In the same vein, {x\\%}={560}.\n\nStep 5: This gives us a pair of simple equations:\n\n{100\\%}={248000}(1).\n\n{x\\%}={560}(2).\n\nStep 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS\n(left hand side) of both equations have the same unit (%); we have\n\n\\frac{100\\%}{x\\%}=\\frac{248000}{560}\n\nStep 7: Taking the inverse (or reciprocal) of both sides yields\n\n\\frac{x\\%}{100\\%}=\\frac{560}{248000}\n\n\\Rightarrow{x} = {0.23\\%}\n\nTherefore, {560} is {0.23\\%} of {248000}.\n\n#### Solution for 248000 is what percent of 560:\n\n248000:560100 =\n\n(248000100):560 =\n\n24800000:560 = 44285.71\n\nNow we have: 248000 is what percent of 560 = 44285.71\n\nQuestion: 248000 is what percent of 560?\n\nPercentage solution with steps:\n\nStep 1: We make the assumption that 560 is 100% since it is our output value.\n\nStep 2: We next represent the value we seek with {x}.\n\nStep 3: From step 1, it follows that {100\\%}={560}.\n\nStep 4: In the same vein, {x\\%}={248000}.\n\nStep 5: This gives us a pair of simple equations:\n\n{100\\%}={560}(1).\n\n{x\\%}={248000}(2).\n\nStep 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS\n(left hand side) of both equations have the same unit (%); we have\n\n\\frac{100\\%}{x\\%}=\\frac{560}{248000}\n\nStep 7: Taking the inverse (or reciprocal) of both sides yields\n\n\\frac{x\\%}{100\\%}=\\frac{248000}{560}\n\n\\Rightarrow{x} = {44285.71\\%}\n\nTherefore, {248000} is {44285.71\\%} of {560}.\n\nCalculation Samples" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81854653,"math_prob":0.9996181,"size":2288,"snap":"2020-34-2020-40","text_gpt3_token_len":775,"char_repetition_ratio":0.19264448,"word_repetition_ratio":0.41315788,"special_character_ratio":0.4763986,"punctuation_ratio":0.15261044,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999949,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-04T20:37:52Z\",\"WARC-Record-ID\":\"<urn:uuid:d25cf47d-1a9e-412c-b907-0790db8f2460>\",\"Content-Length\":\"10511\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:127ead53-40a6-4e92-90de-b840635dd10f>\",\"WARC-Concurrent-To\":\"<urn:uuid:4c0d3895-03b1-4acf-b5f4-ec175b20832e>\",\"WARC-IP-Address\":\"217.23.5.136\",\"WARC-Target-URI\":\"https://www.percentagecal.com/answer/560-is-what-percent-of-248000\",\"WARC-Payload-Digest\":\"sha1:G42KBJDMDFRQAZ35ITJGTIREODVVKABI\",\"WARC-Block-Digest\":\"sha1:G473BUVBDR5UCZ2FCAZLOCCXH3OPZGBQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735882.86_warc_CC-MAIN-20200804191142-20200804221142-00314.warc.gz\"}"}
https://www.hindawi.com/journals/aa/2020/7421396/	[ "/ / Article\nSpecial Issue\n\n## Recent Trends in Celestial Mechanics\n\nView this Special Issue\n\nResearch Article \| Open Access\n\nVolume 2020 \|Article ID 7421396 \| https://doi.org/10.1155/2020/7421396\n\nA. Mostafa, M. H. El Dewaik, \"Balanced Low Earth Satellite Orbits\", Advances in Astronomy, vol. 2020, Article ID 7421396, 12 pages, 2020. https://doi.org/10.1155/2020/7421396\n\n# Balanced Low Earth Satellite Orbits\n\nGuest Editor: Euaggelos E. Zotos\nAccepted10 Jul 2020\nPublished19 Aug 2020\n\n#### Abstract\n\nThe present work aims at constructing an atlas of the balanced Earth satellite orbits with respect to the secular and long periodic effects of Earth oblateness with the harmonics of the geopotential retained up to the 4th zonal harmonic. The variations of the elements are averaged over the fast and medium angles, thus retaining only the secular and long periodic terms. The models obtained cover the values of the semi-major axis from 1.1 to 2 Earth’s radii, although this is applicable only for 1.1 to 1.3 Earth’s radii due to the radiation belts. The atlas obtained is useful for different purposes, with those having the semi-major axis in this range particularly for remote sensing and meteorology.\n\n#### 1. Introduction\n\nThe problem of the motion of an artificial satellite of the Earth was not given serious attention until 1957. At this time, little was known about the magnitudes of the coefficients of the tesseral and sectorial harmonics in the Earth’s gravitational potential. It was pretty well known at this time (1957–1960) that the contributions of the 3rd, 4th, and 5th zonal harmonics were of order higher than the contribution of the 2nd zonal harmonic, but the values of the coefficients C30, C40, and C50 were not very well established. No reliable information was available for the tesseral or the sectorial coefficients except that the observations of orbiting satellites indicated that these coefficients must be small, certainly no more than the first order with respect to C20.\n\nFor low Earth orbits within an altitude less than 480 km, if the satellite attitude is stabilized, or at least a mean projected area could be estimated, the perturbative effects of atmospheric drag should be included. Unfortunately, the literature is still void of even a mention of this topic of balancing this kind of very low Earth orbits. The reason may be the present increased interest in space communications and broadcasting, which still make use of the geostationary orbits that lie beyond the effects of atmospheric drag, though they still suffer the effects of drift solar radiation pressure.\n\nWith the advance of the space age, it became clear that most space applications require fixing, as strictly as possible, the areas covered by the satellite or the constellation of satellites. In turn, fixing the coverage regions requires fixed nodes and fixed apsidal lines. This in turn leads to the search for orbits satisfying these requirements. The families of orbits satisfying such conditions are called “frozen orbits” . Clearly, the design of such orbits includes the effects of the perturbing influences that affect the motion of the satellite. As the present work is interested in low Earth orbits, only the effect of Earth oblateness is taken into concern. These have been extensively treated in the literature .\n\nThis paper is aiming at constructing an atlas of the balanced low Earth satellite orbits, which fall in the range from 600 km to 2000 km above sea surface, in the sense that the variations of the elements are averaged over the fast angle to keep only long periodic and secular variations that affect the orbit accumulatively with time. In this paper, a model is given for the averaged effects (over the mean anomaly) of Earth oblateness. Then, the Lagrange planetary equations for perturbations of the elements are investigated to get sets of orbital values at which the variations of the elements can be cancelled simultaneously.\n\n#### 2. Earth Potential\n\nThe actual shape of the Earth is that of an eggplant. The center of mass does not lie on the spin axis, and neither the meridian nor the latitudinal contours are circles. The net result of this irregular shape is to produce a variation in the gravitational acceleration to that predicted using a point mass distribution. This variation reaches its maximum value at latitude 45 deg and approaches zero at latitudes 0 and 90 deg.\n\nThe motion of a particle around the Earth can be visualized best by resolving it into individual motions along the meridian and the latitudinal contours. The motion around the meridian can be thought of as consisting of different periodic motions called “zonal harmonics.” Similarly, the motion along a latitudinal contour can be visualized as consisting of different periodic motions called “tesseral harmonics.” The zonal harmonics describe the deviations of a meridian from a great circle, while the tesseral harmonics describe the deviations of a latitudinal contour from a circle.\n\nAt points exterior to the Earth, the mass density is zero, thus, at external points the gravitational potential satisfieswhere V is a scalar function representing the potential. Also, the gravitational potential of the Earth must vanish as we recede to infinitely great distances. With these conditions on the above equation, the potential V at external points can be represented in the following form :\n\nThis expression of the potential is called “Venti potential,” and it was adopted by the IAU (International Astronomical Union) in 1961. The terms arising in the above equation are Cnm and Snm are harmonic coefficients (they are bounded as is always the case in physical problems), R is the equatorial radius of the Earth, is the Earth’s gravitational constant, G is the universal constant of gravity, M is the mass of the Earth, and (r, α, δ) are the geocentric coordinates of the satellite (Figure 1, ) with α measured east of Greenwich, and represents the associated Legendre polynomials.The terms with m = 0 are called “zonal harmonics.”The terms with 0 < m < n are called “tesseral harmonics.”The terms with m = n are called “sectorial harmonics.”\n\nThe case of axial symmetry is expressed by taking m = 0, while if equatorial symmetry is assumed, we consider only even harmonics since P2n+1 (−x) = −P2n+1 (x). Also, the coefficients C21 and S21 are vanishingly small. Further if the origin is taken at the center of mass, the coefficients C10, C11, and S11 will be equal to zero.\n\nConsidering axial symmetry, with origin at the center of mass, we can writewhere Jn = −Cn0.\n\nTaking terms up to j4, we can write V in the following form:where\n\nIt is a purely geometrical transformation to express the potential function V(r, δ), given by the above equations, as a function of the Keplerian orbital elements a, e, i, Ω, ω, and I in their usual meanings (Figure 2, ), where a and e are the semi-major axis and the eccentricity of the orbit, respectively, i is the inclination of the orbit to the Earth’s equatorial plane, Ω and ω describes the position of the orbit in space where Ω is the longitude of the ascending node and ω is the argument of perigee, and finally, l is the mean anomaly to describe the position of the satellite with respect to the orbit.\n\nThen, V (a, e, i, Ω, ω, I) is in a form suitable to use in Lagrange’s planetary equations, and in canonical perturbations methods through the relation. From the spherical trigonometry of the celestial sphere, we havewhere f is the true anomaly, ω is the argument of perigee, and Si= sin i, Ci= cos i.\n\nSubstituting for δ, in Pn (sin δ), we get\n\nThus, we get V2, V3, and V4 as functions of the orbital elements.\n\nWe now proceed to evaluate the effects of Earth oblateness, considering the geopotential up to the zonal harmonic J4.\n\nIn the present solution, we consider only the secular and long periodic terms, averaging over the mean anomaly l.\n\n##### 2.1. The Disturbing Function\n\nThe disturbing function is defined as follows:where V2, V3, and V4 are the 2nd, 3rd, and 4th terms of the geopotential, namelywhere , , and is the true anomaly.\n\nSince terms depending only on the fast variable l will not affect the orbit in an accumulating way with time, we average the perturbing function R over the mean anomaly with its period 2π. The average function is defined by\n\nAs the perturbing function R is a function of the true anomaly f not the mean anomaly l, we use the relationwhere both angles have the same period and the same end points 0 and 2π, and therefore the average function will be given by\n\nApplying the required integrals, we get the averaged disturbing function where\n\n##### 2.2. Lagrange Equations for the Averaged Variations of the Elements\n\nThe Lagrange planetary equations for the variations of the elements for a disturbing potential R are as follows :where n is the mean motion given by .\n\nSubstituting for the averaged disturbing function due to Earth oblateness, the Lagrange equations becomewhere the equation for l is neglected because we concentrate on balancing the orbit position not the satellite motion in the orbit.\n\n##### 2.3. Variations of the Elements due to Earth Oblateness\n\nSubstituting for the required derivatives in equations (16) to (20) yields\n\nDefining\n\nWe getor\n\nDefining\n\nWe get\n\nDefining as in equation (28), we get equation (29):\n\nEquations (22)–(29) give the average effects of Earth oblateness including the zonal harmonics of the geopotential up to J4 on the Keplerian elements of the satellite orbit.\n\n#### 3. Balanced Low Earth Satellite Orbits\n\nIn what follows, we try to find orbits that are balanced in the sense that the averaged (over the fast variable l) variations of the orbit elements are set equal to zero. In equation (23), we put it equal to zero and get a relation between the argument of perigee ω and the inclination i, while treating the eccentricity e and the semi-major axis a as parameters. This will give a range of values for ω and i at different values of e and a, which are all give balanced orbits with respect to both the eccentricity e and the inclination i. The same is done with equations (26) and (29), while putting = 0 and = 0.\n\nThe applicable ranges for this model of the semi-major axis a are , where the range 1.4R–2R is avoided due to the predominance of the radiation belts at these levels, to avoid the damages of the equipment that it may produce, besides its fatal effects on human life (for inhabited spacecrafts). The values for the eccentricity e are taken between 0.01 (almost circular orbit) and 0.5.\n\n##### 3.1. Orbits with Fixed Eccentricity and Inclination\n\nBy equating the variation of e by zero, we get\n\nThis implies\n\nSo either orwhere\n\nEquations (32) and (33) give the family of low orbits that have both the eccentricity and the inclination fixed.\n\nThe condition for the existence of such orbits is clearly thatwhich is guaranteed by\n\nWe put it as a condition on the eccentricity i only when\n\n##### 3.2. Orbits with Fixed Node\n\nThe families of orbits for which dΩ/dt = 0 (i.e with fixed nodes) are obtained from\n\nThis implies\n\nThis can be arranged as a second order equation for sin(ω), giving a family of orbits with fixed argument of perigee for different values of ω as a function of the inclination i, the semi-major axis a, and the eccentricity e.\n\nWhen solving for sin(ω), we get the family of orbits for which the longitude of the node is balanced,where\n\nThe condition for having the orbit if real solution exists is again that , which gives\n\n##### 3.3. Orbits with Fixed Perigee\n\nFor the argument of perigee to balance, we solve .We substitute from equation (26) into equation (29) then expand cos(2ω) and collect terms with respect to sin(ω). We get\n\nThus for , we getwhere\n\nEquations (44)–(47) give the relation between sin(ω) and the inclination i, the semi-major axis a, and the eccentricity e, which gives the family of orbits that balance the argument of perigee ω, subject of course to the restriction that is a real value between −1 and 1.\n\n#### 4. Numerical Results\n\nIn this section, numerical results and graphs are obtained for the case of seasat a = 7100 km by putting ω as a function of e and i from equations (32) and (33). The curves are plotted within the possible range given by condition (36) to give curves of balanced e and i. The curves are against the eccentricity e in the range [0.01, 0.5]. The numerical values involved are J2 = 0.001082645, J3 = −0.000002546, J4 = −0.000001649, R = 6378.165 km, α = 7100 m, and μ = 398600.5 km3sec−2.\n\nThe condition (36) gives the upper and lower bounds of the function F(i) as a function of e, and since the function is increasing with e and has no critical points in the interval [0.01, 0.5], then the minimum value occurs at e= 0.01 and the maximum at e= 0.5, which gives . The graphs are plotted for different values of F(i), which corresponds to specific values of i found by solving the equation resulting from setting F(i) equal to the required values. After that we plot = 0 and = 0 simultaneously for the same values of F(i) at each specific i-value, to find the orbit values at which we have nearest values for = 0 and = 0. In the graphs of and , the relation (32) was kept to ensure that e and i are already balanced.\n\nFive selected values of F(i) are chosen: F(i) = 0.01, 0.5, 0.10, 0.15, and 0.20. Negative values will give the same results with negative sign since F(i) is odd with respect to i. Also for each value, the solution of F(i) = x, with x equals one of the above values we get four real values of i on the form: i1, 180 − i1, i2, and −180 − i2, where i1 and i2 are near the critical inclination one of them is positive and the other is negative. Table 1 gives the values of i corresponding to the selected values of F(i).\n\n F(i) i1 180 − i1 i2 −180 − i2 0.01 63.63 116.37 −63.22 −116.78 0.05 64.27 115.73 −62.11 −117.89 0.10 64.84 115.16 −59.76 −120.24 0.15 65.25 114.75 −55.14 −124.86 0.20 65.56 114.44 −47.16 −132.84\n\nWe note that as F(i) increases, i gets away from the critical inclination, and the curve gets shorter indicating less stability of ω as expected.\n\nThe graphs are plotted for each value of F(i) first for the balanced values of e and i, then for the corresponding four values of i, four graphs are plotted for and to find the nearest values of zero variation for both elements.\n\nFigures 312 show the possibility of balancing ω with e and i, while Ω will have a variation of order 10−6/sec or it must be balanced alone at i = 90 deg (as shown in equation (26)), according to the orbit kind. Figures 3, 5, 7, 9, and 11 show the possible values of ω(e) that balance e and i, while Figures 4, 6, 8, 10, and 12 show the curves of and at the values of i at which .\n\n#### 5. Conclusion\n\nLet balanced orbits be defined as those for which the orbital elements are set equal to zero under the effect of secular and long periodic perturbations. In this work, the effect of Earth oblateness is the considered perturbing force because of dealing with low Earth orbits. The above analysis then shows that such an orbit will be balanced within a reliable tolerance only for few weeks since we are forced to accept the motion of either the node or the perigee by about 10−6 deg/sec. The reason is that under the influence of the Earth oblateness, (exactly) for , while only near the critical inclination deg. Hence, the best procedure is to design a satellite constellation for which the nodal shifts due to the perturbative effects and Earth rotation are modeled to yield continuous coverage. 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View at: Publisher Site \| Google Scholar\n7. S. L. Coffey, A. Deprit, and E. Deprit, “Frozen orbits for satellites close to an earth-like planet,” Celestial Mechanics & Dynamical Astronomy, vol. 59, no. 1, pp. 37–72, 1994. View at: Publisher Site \| Google Scholar\n8. M. Lara, A. Deprit, and A. Elipe, “Numerical continuation of families of frozen orbits in the zonal problem of artificial satellite theory,” Celestial Mechanics & Dynamical Astronomy, vol. 62, no. 2, pp. 167–181, 1995. View at: Publisher Site \| Google Scholar\n9. D. Brouwer, “Solution of the problem of artificial satellite theory without drag,” The Astronomical Journal, vol. 64, p. 378, 1959. View at: Publisher Site \| Google Scholar\n10. Y. Kozai, “The motion of a close earth satellite,” The Astronomical Journal, vol. 64, p. 367, 1959. View at: Publisher Site \| Google Scholar\n11. P. M. Fitzpatrick, Principles of Celestial Mechanics, Academic Press, Cambridge, MA, USA, 1970.\n12. H. D. Curtis, Orbital Mechanics for Engineering Students, Elsevier, Amsterdam, Netherlands, 2005.\n13. D. Brower and G. M. Clemence, Methods of Celestial Mechanics, Academic Press, Cambridge, MA, USA, 1961.\n\n#### More related articles\n\nArticle of the Year Award: Outstanding research contributions of 2020, as selected by our Chief Editors. Read the winning articles." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8887233,"math_prob":0.95736176,"size":18433,"snap":"2021-43-2021-49","text_gpt3_token_len":4493,"char_repetition_ratio":0.14417495,"word_repetition_ratio":0.05844981,"special_character_ratio":0.24222861,"punctuation_ratio":0.13833243,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9905909,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-18T16:08:51Z\",\"WARC-Record-ID\":\"<urn:uuid:f8a78c84-fe19-4ad2-b540-1ed645696c98>\",\"Content-Length\":\"1049303\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bc4a4bad-7858-436f-8f9c-a231ecd36fbd>\",\"WARC-Concurrent-To\":\"<urn:uuid:e4820648-ffb5-461b-8831-ad7d5431cf6b>\",\"WARC-IP-Address\":\"13.32.208.94\",\"WARC-Target-URI\":\"https://www.hindawi.com/journals/aa/2020/7421396/\",\"WARC-Payload-Digest\":\"sha1:2Y6UY5PLIIZ5VZSLU5DXTKTLUMAOVXSG\",\"WARC-Block-Digest\":\"sha1:MHO5JKRTOWICRVZRC46MEMZQSYWNTIBQ\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585204.68_warc_CC-MAIN-20211018155442-20211018185442-00427.warc.gz\"}"}
https://benidormclubdeportivo.org/what-are-the-factor-pairs-of-60/	[ "Factors of 60 room integers that deserve to be divided evenly right into 60. There are 12 factors of 60 of which 60 chin is the best factor and also its prime determinants are 2, 3 and 5 The amount of all components of 60 is 168.\n\nYou are watching: What are the factor pairs of 60\n\nFactors that 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60Negative determinants of 60: -1, -2, -3, -4, -5, -6, -10, -12, -15, -20, -30 and -60Prime factors of 60: 2, 3, 5Prime factorization of 60: 2 × 2 × 3 × 5 = 22 × 3 × 5Sum of components of 60: 168\n 1 What are the determinants of 60? 2 How come Calculate factors of 60? 3 Factors the 60 by element Factorization 4 Factors of 60 in Pairs 5 FAQs on factors of 60\n\n## What are the components of 60?\n\nThe factors of 60 room all the numbers that provide the value 60 as soon as multiplied. Together 60 is an also number, it has much more than one factor. To understand why it is composite, let\"s remind the definition of a composite number. A number is stated to be composite if the has more than two factors.\n\n60 ÷ 2 = 335 and 12 are factors of 60.Factors that 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.\n\n## How to Calculate Factors the 60?\n\nTo calculate the factors of 60, we can use different methods like prime factorization and also the division method. In element factorization, us express 60 as a product the its prime factors and in the department method, we view what numbers divide 60 exactly.\n\nHence, the components of 60 are, 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.\n\nExplore components using illustrations and interactive examples.\n\n## Factors of 60 by prime Factorization\n\nPrime factorization is the process of creating a number as a product the its prime factors. Permit us learn exactly how to uncover the components of a number making use of prime factorization. We begin by trying to find the smallest prime number which can divide 60 leaving the remainder as 0. As 60 is an even number, 2 is among its factors.\n\nThe number 60 is split by the the smallest prime number i beg your pardon divides 60 exactly, i.e. It pipeline a remainder 0. The quotient is then split by the smallest or second smallest element number and also the procedure continues till the quotient i do not care indivisible. Let us divide 60 by the prime number 2⇒ 60 ÷ 2 = 30Now we should divide the quotient 30 by the next least prime number. Since 30 is an even number, it is divisible through 2⇒ 30 ÷ 2 =1515 will be divisible through 3⇒ 15 ÷ 3= 55 is a element number and hence not additional divisible.", null, "Prime factorization of 60 = 2 × 2 × 3 × 5Therefore, the prime factors of 60 are 2, 3, and also 5.\n\n### Prime administer by variable Tree\n\nThe other method of element factorization is acquisition 60 as the root and creating branches by splitting it by the the smallest prime number. This an approach is comparable to the division method above. The distinction lies in presenting the factorization. The figure listed below shows the variable tree the 60.", null, "Collect every the number in circles and also write 60 as the product of this numbers.Hence, 60 = 2 × 2 × 3 × 5\n\nPair factors are the factors of a number given in pairs, which as soon as multiplied together, offer that initial number.6 and also 10 both are determinants of 60. For example, 6 × 10 = 60.The pair components of 60 would be the two numbers which, when multiplied together, an outcome in 60.\n\nThe complying with table represents the pair determinants of 60:\n\n Product Pair factors 1 × 60 (1, 60) 2 × 30 (2, 30) 3 × 20 (3, 20) 4 × 15 (4, 15) 5 × 12 (5, 12) 6 × 10 (6, 10)\n\nWe have the right to have an unfavorable factors likewise for a provided number. The negative pair factors of 60 are (-1, -60), (-2, -30), (-3, -20), (-4, -15), (-5, -12) and also (-6, -10).\n\nImportant Notes:\n\nThe factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.Every also number will have 2 as among its factors.60 is a composite number as it has much more factors various other than 1 and itself.\n\nChallenging Questions:\n\nWhat are the common factors of 60 and also 600?Are 2/3 and 4/5 determinants of 600?\n\nExample 1: Fifteen friends have actually 60 pizzas from a pizza shop and distributed among themselves equally. How many pizzaz would each among them get?\n\nSolution:\n\n15 pizzas are to be divided amongst friends equally.This implies we must divide 60 by 15.60 ÷ 15 = 4Hence, each girlfriend will obtain 4 pizzas.\n\nExample 2: Sia wants to know if over there are any common components of 60 and 80. Help her in recognize the answer to it.\n\nSolution:\n\nFactors that 60 = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and also 60Factors the 80 = 1, 2, 4, 5, 8, 10, 16, 20, 40, 80The typical factors the 60 and 80 = 1, 2, 4, 5, 10, and 20Hence, the typical factors that 60 and 80 are 1, 2, 4, 5, 10, and also 20.\n\nExample 3: uncover the product of every the prime factors of 60.\n\nSolution:\n\nSince, the prime components of 60 are 2, 3, 5. Therefore, the product the prime components = 2 × 3 × 5 = 30.", null, "## FAQs on components of 60\n\n### What space the factors of 60?\n\nThe determinants of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 and also its negative factors space -1, -2, -3, -4, -5, -6, -10, -12, -15, -20, -30, -60.\n\n### What is the Greatest common Factor the 60 and also 21?\n\nThe factors of 60 and 21 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 and 1, 3, 7, 21 respectively.Common determinants of 60 and 21 space <1, 3>.Hence, the GCF the 60 and 21 is 3.\n\n### What Numbers space the Prime components of 60?\n\nThe prime components of 60 are 2, 3, 5.\n\nSee more: If You Have Faith The Size Of A Mustard Seed Kjv, Bible Verses About Mustard Seed\n\n### What space the usual Factors of 60 and 38?\n\nSince, the determinants of 60 room 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 and also the determinants of 38 space 1, 2, 19, 38.Hence, <1, 2> space the common factors of 60 and also 38.\n\n### What is the sum of all the determinants of 60?\n\nSum that all components of 60 = (22 + 1 - 1)/(2 - 1) × (31 + 1 - 1)/(3 - 1) × (51 + 1 - 1)/(5 - 1) = 168" ]	[ null, "https://benidormclubdeportivo.org/what-are-the-factor-pairs-of-60/imager_1_2771_700.jpg", null, "https://benidormclubdeportivo.org/what-are-the-factor-pairs-of-60/imager_2_2771_700.jpg", null, "https://benidormclubdeportivo.org/what-are-the-factor-pairs-of-60/imager_3_2771_700.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92159474,"math_prob":0.99523723,"size":5746,"snap":"2022-27-2022-33","text_gpt3_token_len":1817,"char_repetition_ratio":0.19157088,"word_repetition_ratio":0.094858155,"special_character_ratio":0.35398537,"punctuation_ratio":0.18020679,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99874145,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T11:47:37Z\",\"WARC-Record-ID\":\"<urn:uuid:6fd40328-72f5-4a69-910e-04859607b715>\",\"Content-Length\":\"16547\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d0e266c3-777a-4a6e-be77-fabce0cc5641>\",\"WARC-Concurrent-To\":\"<urn:uuid:db1261e3-7824-484e-b1a7-44ab86918661>\",\"WARC-IP-Address\":\"172.67.177.16\",\"WARC-Target-URI\":\"https://benidormclubdeportivo.org/what-are-the-factor-pairs-of-60/\",\"WARC-Payload-Digest\":\"sha1:OAFNX55P2IOJE6GSEUI447KPIH5RULNU\",\"WARC-Block-Digest\":\"sha1:JOCNXX73TZQD5DUADTMSAXF7LCNQEHCG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104576719.83_warc_CC-MAIN-20220705113756-20220705143756-00086.warc.gz\"}"}
https://wealthofunderstanding.com/portfolio/simple-vs-compound-interest/	[ "", null, "## Simple vs Compound Interest", null, "Recently we discussed that not all loans are created equal because of the way the lender structures your payments. But loans can also differ in the way they charge interest. And this is bigger than loans–this is also a major key in how you earn interest.\n\nThe big difference is whether the loan/investment uses Simple or Compound interest.\n\nSimple Interest Examples:\n– Savings Bonds\n\\$100 invested at 4% for 10 years.\n\\$100 x .04 x 10 = \\$40 interest earned.\n\n– Line of Credit\n\\$100 borrowed at 5% for 5 years.\n\\$100 x .05 x 5 = \\$25 interest owed.\n\nWith Simple Interest, the interest builds in a straight line.\n\nCompound Interest Examples:\n– Stock Market\n\\$100 invested at 4% for 10 years.\n\\$100 x (1 + .04)10 – \\$100 = \\$48 interest earned\n\n-Home Mortgage\n\\$100 borrowed at 5% for 5 years.\n\\$100 x (1 + .05)5 – \\$100 = \\$27 interest owed.\n\nWith Compound Interest, the interest builds in a curved line.\n\nWe used small examples to keep the math easy, but let’s run it again with some larger examples:\n\nSavings Bond: \\$10,000 invested at 4% for 30 years\n\\$10,000 x .04 x 30 = \\$12,000 interest earned.\n\nLine of Credit: \\$100,000 borrowed at 5% for 30 years\n\\$100,000 x .05 x 30 = \\$150,000 interest owed.\n\nStock Market: \\$10,000 invested at 4% for 30 years\n\\$10,000 x (1 + .04)30 – \\$10,000 = \\$22,433 interest earned.\n\nHome Mortgage: \\$100,000 borrowed at 5% for 30 years\n\\$100,000 x (1 + .05)30 – \\$100,000 = \\$332,194 interest owed.\n\nThe act of compounding has the power to double the amount of interest generated by a loan/investment over the course of 30 years." ]	[ null, "https://www.facebook.com/tr", null, "https://149474742.v2.pressablecdn.com/wp-content/uploads/2020/10/SimpleVsCompound_1x1.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9276663,"math_prob":0.9877534,"size":1524,"snap":"2023-40-2023-50","text_gpt3_token_len":434,"char_repetition_ratio":0.17631578,"word_repetition_ratio":0.16546762,"special_character_ratio":0.3753281,"punctuation_ratio":0.14925373,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9529883,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T22:01:13Z\",\"WARC-Record-ID\":\"<urn:uuid:0f1c35ae-4533-4bec-bed4-00866acd2517>\",\"Content-Length\":\"53775\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:90420167-9070-45a3-bae2-4f26c6313293>\",\"WARC-Concurrent-To\":\"<urn:uuid:0cbea702-2b9a-4bca-932e-97567c4c6bbf>\",\"WARC-IP-Address\":\"199.16.172.46\",\"WARC-Target-URI\":\"https://wealthofunderstanding.com/portfolio/simple-vs-compound-interest/\",\"WARC-Payload-Digest\":\"sha1:PW7M6OL3JGRRMLEVGY4V55LNPHFOSCMH\",\"WARC-Block-Digest\":\"sha1:B3WJVYQX4QM4LKHTDFJIU3VSGAENHFYR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100146.5_warc_CC-MAIN-20231129204528-20231129234528-00807.warc.gz\"}"}
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=60&t=38750&p=131017	[ "## Are all pH calculations done to 2 decimal places or do we use sig figs in pH calculations? [ENDORSED]\n\nHenri_de_Guzman_3L\nPosts: 88\nJoined: Fri Sep 28, 2018 12:25 am\n\n### Are all pH calculations done to 2 decimal places or do we use sig figs in pH calculations?\n\nFor example 6B.5c in the 7th edition textbook asks us to calculate the pH for 0.0092 M Ba(OH)2. This is 2 sig figs but the answer in the back gives 12.96. Should the answer instead be 13?\n\nanthony_trieu2L\nPosts: 60\nJoined: Fri Sep 28, 2018 12:29 am\n\n### Re: Are all pH calculations done to 2 decimal places or do we use sig figs in pH calculations? [ENDORSED]\n\nFor pH calculations, the sig figs are only accounted for after the decimal point. In this case, since there are two sig figs in 0.0092 M Ba(OH)2, the final answer should present two digits after the decimal point. Thus, 12.96 is correct.\n\nRandallNeeDis3K\nPosts: 34\nJoined: Fri Sep 28, 2018 12:25 am\n\n### Re: Are all pH calculations done to 2 decimal places or do we use sig figs in pH calculations?\n\npH calculations only follow sig figs after the decimal point! so two sig figs after the decimal point makes sense.\n\nReturn to “Calculating pH or pOH for Strong & Weak Acids & Bases”\n\n### Who is online\n\nUsers browsing this forum: No registered users and 2 guests" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7158797,"math_prob":0.9851308,"size":1257,"snap":"2020-45-2020-50","text_gpt3_token_len":360,"char_repetition_ratio":0.1963288,"word_repetition_ratio":0.33050847,"special_character_ratio":0.291965,"punctuation_ratio":0.13620071,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.979542,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-05T03:56:07Z\",\"WARC-Record-ID\":\"<urn:uuid:fe1a55b3-c63c-4540-ab8a-88dadb623a21>\",\"Content-Length\":\"55892\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:688325a9-39cc-4970-ac7d-85d04bc7e042>\",\"WARC-Concurrent-To\":\"<urn:uuid:fb2d2703-a598-4729-95a0-16a9541a8e67>\",\"WARC-IP-Address\":\"169.232.134.130\",\"WARC-Target-URI\":\"https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=60&t=38750&p=131017\",\"WARC-Payload-Digest\":\"sha1:SVMTN7S6Y5HSQVYFMQNR4GUGSJEHKVEU\",\"WARC-Block-Digest\":\"sha1:T2GALHCACA5OLVB5DDYOVRSESHC5QTUP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141746033.87_warc_CC-MAIN-20201205013617-20201205043617-00642.warc.gz\"}"}
https://dsp.stackexchange.com/questions/41370/interpolation-formula-for-two-dimensional-signal-reconstruction-in-the-frequency	[ "# Interpolation formula for two dimensional signal reconstruction in the frequency domain from polar samples\n\nIn the book, Advanced Topics in Shannon Sampling and Interpolation Theory by Robert J. Marks II, one may find an interpolation formula for reconstructing a two dimensional signal from regular polar samples in the frequency domain:\n\nLet $f(x,y)$ be space-limited to $2A$. Let its Fourier transform in polar coordinates $F(\\rho,\\phi)$ be angularly band-limited to $K$. Then $F(\\rho,\\phi)$ can be reconstructed from its polar samples via\n\n$$F(\\rho,\\phi)=\\sum_{n=-\\infty}^\\infty\\sum_{k=0}^{N-1}\\widetilde{F}\\left(\\frac{n}{2A},\\frac{2\\pi k}{N}\\right)\\operatorname{sinc}\\left[\\frac{2A(\\rho-n)}{2A}\\right]\\frac{\\sin\\left[\\frac{1}{2}(N-1)\\left(\\phi-\\frac{2\\pi k}{N}\\right)\\right]}{N\\sin\\left[\\frac{1}{2}\\left(\\phi-\\frac{2\\pi k}{N}\\right)\\right]},$$ where $N$ is assumed even and $$\\widetilde{F}\\left(\\frac{n}{2A},\\frac{2\\pi k}{N}\\right)=\\begin{cases}F\\left(\\frac{n}{2A},\\frac{2\\pi k}{N}\\right),&n\\ge 0, \\\\ F\\left(-\\frac{n}{2A},\\frac{2\\pi k}{N}+\\pi\\right),&n<0.\\end{cases}$$\n\nI figured that the argument of the cardinal sine function contained a typo, since the $2A$ in the numerator and denominator would appear to cancel. Thus I would guess that the author actually had\n\n$$\\operatorname{sinc}\\left[\\frac{\\rho-2An}{2A}\\right]$$\n\nin mind. Correct me if I'm wrong?\n\nThen the last quotient with the sine functions looks very similar to the Dirichlet kernel:\n\n$$D_N\\left(\\phi-\\frac{2\\pi k}{N}\\right)=\\frac{\\sin\\left[\\left(N+\\frac{1}{2}\\right)\\left(\\phi-\\frac{2\\pi k}{N}\\right)\\right]}{\\sin\\left[\\frac{\\phi-\\frac{2\\pi k}{N}}{2}\\right]}$$\n\nbut is not quite the same.\n\nIncidentally, my attempts to implement the interpolation formula in the theorem via Matlab for specific examples have failed completely with regards to reconstructing a signal. Perhaps someone can point me in the direction of a correct formula for signal reconstruction in the frequency domain from polar samples?", null, "suggests that $$\\operatorname{sinc}\\left[2A(\\rho-\\frac{n}{2A})\\right]$$ might be the correct form of the $\\operatorname{sinc}$ term.\n• Then this is the same as $$\\operatorname{sinc}\\left[\\frac{\\rho-n\\Delta\\rho}{\\Delta\\rho}\\right],$$ with $\\Delta\\rho=\\frac{1}{2A}$. Interestingly I also tried to implement this formula on Matlab too, but with no success. – Jason Born Jun 9 '17 at 18:08" ]	[ null, "https://i.stack.imgur.com/QTkK6.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.72242886,"math_prob":0.9997428,"size":1858,"snap":"2019-51-2020-05","text_gpt3_token_len":577,"char_repetition_ratio":0.16882417,"word_repetition_ratio":0.0,"special_character_ratio":0.2949408,"punctuation_ratio":0.0754717,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999972,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-12T21:47:28Z\",\"WARC-Record-ID\":\"<urn:uuid:f6722b3e-ddc6-4e9a-9cdb-4f60c8cd7e06>\",\"Content-Length\":\"133632\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:62145085-6b09-4628-96b9-29a9f651fc9c>\",\"WARC-Concurrent-To\":\"<urn:uuid:0e31b427-bb81-4917-ba40-6ce6eaa88a03>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://dsp.stackexchange.com/questions/41370/interpolation-formula-for-two-dimensional-signal-reconstruction-in-the-frequency\",\"WARC-Payload-Digest\":\"sha1:L4PBF63MANH3SN2T5QGTPJ2QKHTE3C37\",\"WARC-Block-Digest\":\"sha1:TNUORCPGD2QZATTLTNYYTO43BXK7AQ5Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540547165.98_warc_CC-MAIN-20191212205036-20191212233036-00413.warc.gz\"}"}
https://metanumbers.com/42265	[ "## 42265\n\n42,265 (forty-two thousand two hundred sixty-five) is an odd five-digits composite number following 42264 and preceding 42266. In scientific notation, it is written as 4.2265 × 104. The sum of its digits is 19. It has a total of 3 prime factors and 8 positive divisors. There are 33,072 positive integers (up to 42265) that are relatively prime to 42265.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 5\n• Sum of Digits 19\n• Digital Root 1\n\n## Name\n\nShort name 42 thousand 265 forty-two thousand two hundred sixty-five\n\n## Notation\n\nScientific notation 4.2265 × 104 42.265 × 103\n\n## Prime Factorization of 42265\n\nPrime Factorization 5 × 79 × 107\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 42265 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 42,265 is 5 × 79 × 107. Since it has a total of 3 prime factors, 42,265 is a composite number.\n\n## Divisors of 42265\n\n1, 5, 79, 107, 395, 535, 8453, 42265\n\n8 divisors\n\n Even divisors 0 8 4 4\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 51840 Sum of all the positive divisors of n s(n) 9575 Sum of the proper positive divisors of n A(n) 6480 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 205.585 Returns the nth root of the product of n divisors H(n) 6.52238 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 42,265 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 42,265) is 51,840, the average is 6,480.\n\n## Other Arithmetic Functions (n = 42265)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 33072 Total number of positive integers not greater than n that are coprime to n λ(n) 8268 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 4413 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 33,072 positive integers (less than 42,265) that are coprime with 42,265. And there are approximately 4,413 prime numbers less than or equal to 42,265.\n\n## Divisibility of 42265\n\n m n mod m 2 3 4 5 6 7 8 9 1 1 1 0 1 6 1 1\n\nThe number 42,265 is divisible by 5.\n\n## Classification of 42265\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n• Square Free\n\n### Other numbers\n\n• LucasCarmichael\n• Sphenic\n\n## Base conversion (42265)\n\nBase System Value\n2 Binary 1010010100011001\n3 Ternary 2010222101\n4 Quaternary 22110121\n5 Quinary 2323030\n6 Senary 523401\n8 Octal 122431\n10 Decimal 42265\n12 Duodecimal 20561\n20 Vigesimal 55d5\n36 Base36 wm1\n\n## Basic calculations (n = 42265)\n\n### Multiplication\n\nn×i\n n×2 84530 126795 169060 211325\n\n### Division\n\nni\n n⁄2 21132.5 14088.3 10566.2 8453\n\n### Exponentiation\n\nni\n n2 1786330225 75499246959625 3190975672748550625 134866586808717492165625\n\n### Nth Root\n\ni√n\n 2√n 205.585 34.8332 14.3382 8.41775\n\n## 42265 as geometric shapes\n\n### Circle\n\n Diameter 84530 265559 5.61192e+09\n\n### Sphere\n\n Volume 3.16251e+14 2.24477e+10 265559\n\n### Square\n\nLength = n\n Perimeter 169060 1.78633e+09 59771.7\n\n### Cube\n\nLength = n\n Surface area 1.0718e+10 7.54992e+13 73205.1\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 126795 7.73504e+08 36602.6\n\n### Triangular Pyramid\n\nLength = n\n Surface area 3.09401e+09 8.89767e+12 34509.2\n\n## Cryptographic Hash Functions\n\nmd5 1b9b3786eae526755467c2593d194005 37156b9eddb65ef2207af1e927b7e6c083bfe0e8 b5b7ec8069eae2248ed162d8e0f878ef4e93a3f63c48a6b8323e827356f49358 11a51a0a8e07cc7429775fb63627d21cf968a8a959dd51e3ca327f212d6d4ffac5f7c6153afab0da8efcb7891dd2eac98abaa11ddeec69185b6715e490b438e2 0a800039d9092ad65ed56ce20872df2dfa0bc522" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6073133,"math_prob":0.9844902,"size":4535,"snap":"2020-24-2020-29","text_gpt3_token_len":1609,"char_repetition_ratio":0.11895829,"word_repetition_ratio":0.028106509,"special_character_ratio":0.45181918,"punctuation_ratio":0.076030925,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9958916,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-13T04:35:15Z\",\"WARC-Record-ID\":\"<urn:uuid:5e5b305c-7058-417b-9ce3-48699dff93aa>\",\"Content-Length\":\"48312\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef7959ba-e343-484e-bcd2-185b5935efed>\",\"WARC-Concurrent-To\":\"<urn:uuid:2a2d473a-5166-44b8-ae57-294ee1f4b021>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/42265\",\"WARC-Payload-Digest\":\"sha1:7VRA4JFQNQCCVSRPSFJZWQ7J3IHQ5S6P\",\"WARC-Block-Digest\":\"sha1:HHMELXUASJN6OIN27O6XNPNJ4WFHYZK6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657142589.93_warc_CC-MAIN-20200713033803-20200713063803-00316.warc.gz\"}"}
https://getchemistryhelp.com/chemistry-lesson-significant-digits-rounding/	[ "", null, "", null, "Dr. Kent McCorkle, or \"Dr. Kent' to his student, has helped thousands of students be successful in chemistry.\n\n# Chemistry Lesson: Significant Digits & Rounding\n\n[View the accompanying Lesson on Significant Digits & Measurements here.]\n\n## Significant Digits & Rounding\n\nAll numbers from a measurement are significant. However, we often generate nonsignificant digits when performing calculations. We get rid of nonsignificant digits by rounding off numbers. There are three rules for rounding off numbers.\n\n## Rules for Rounding Numbers\n\n1. If the first nonsignificant digit is less than 5, drop all nonsignificant digits.\n2. If the first nonsignificant digit is greater than or equal to 5, increase the last significant digit by 1 and drop all nonsignificant digits.*\n3. If a calculation has two or more operations, retain at least one nonsignificant digit until the final operation and then round off the answer.\n\n## Examples\n\nA calculator displays 17.846239 and 3 significant digits are justified.\n• The first three significant digits are 17.8. The first nonsignificant digit is 4. According to rule #1, because 4 is less than 5 we round down, discarding all of the nonsignificant digits, leaving us with 17.8.\nA calculator displays 17.856239 and 3 significant digits are justified.\n• The first three significant digits are 17.8. The first nonsignificant digit is 5. According to rule #2, because it is 5 or greater then we add 1 to the last significant digit (the 8 becomes a 9) and drop all of the nonsignificant digits. The answer is thus 17.9.\n\n## Placeholder Zeros\n\nRound the measurement 151 mL to 2 significant digits.\n\n• We keep the first two significant digits, 15. The first nonsignificant digit, 1, tells us to round down leaving us with 15. However, 15 is 10x smaller than the original number so we need to put a 0 in the ones place to maintain the magnitude of the number. So the answer would be 150. Remember, such placeholder zeros (trailing zeros) are not significant so 150 does have two significant digits.\n\nRound the measurement 2788 g to 3 significant digits.\n\n• We keep the first three significant digits, 278. The first nonsignificant digit is the final 8, which indicates we are to round up, adding 1 to the final significant 8 in the tens places making it 279. But again, 279 is much smaller than 2788 so we again need a placeholder zero in the ones place. So the answer would be 2790.\n\n## Examples\n\n1. Round 4.1278 to 3 significant digits.\n• 4.13\n2. Round 4.1278 to 2 significant digits.\n• 4.1\n3. Round 63401 to 3 significant digits.\n• 63400\n4. Round 0.0562 to 2 significant digits.\n• 0.056\n\nAn alternate rule says that if the first nonsignificant digit is a 5 then always round so that last significant digit is even. Why? Since a 0 doesn’t really require ’rounding’, there are only 9 numbers that can be either dropped or rounded up. 5 is exactly in the middle of 1-9 so always rounding up at 5 would lead to more numbers being rounded up (5,6,7,8,9) than are rounded down (1,2,3,4). To avoid this, if a number ends in 5 then some of the time it’s rounded up and some of the time it’s rounded down – whichever way will make the last significant digit even.\n\nFor additional practice problems on significant digits and rounding, visit Significant Digits & Rounding Practice Problems.", null, "" ]	[ null, "https://www.facebook.com/tr", null, "https://getchemistryhelp.com/wp-content/themes/customtheme/images/custom/dr-kent.jpg", null, "https://moderate1-v4.cleantalk.org/pixel/41068a4fab7566008e9d7bf9dfc5bd02.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8871752,"math_prob":0.9501094,"size":2811,"snap":"2023-40-2023-50","text_gpt3_token_len":688,"char_repetition_ratio":0.24474528,"word_repetition_ratio":0.07786885,"special_character_ratio":0.24653149,"punctuation_ratio":0.11785714,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98950243,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T02:00:56Z\",\"WARC-Record-ID\":\"<urn:uuid:049b8762-5e9d-4dc3-8d66-fb77d90f16df>\",\"Content-Length\":\"94465\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:23e604f0-759e-4b9f-b8bc-c00e7a5e9395>\",\"WARC-Concurrent-To\":\"<urn:uuid:d9f14537-dd29-4786-8fbe-94b083bfe7ab>\",\"WARC-IP-Address\":\"104.199.121.102\",\"WARC-Target-URI\":\"https://getchemistryhelp.com/chemistry-lesson-significant-digits-rounding/\",\"WARC-Payload-Digest\":\"sha1:3PU2224UY22NIUF6ZEGHDSTSMAX5RH2V\",\"WARC-Block-Digest\":\"sha1:5CJIVF5LNYFWB4PIBS5P4S3GPKJ3BHK6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510942.97_warc_CC-MAIN-20231002001302-20231002031302-00213.warc.gz\"}"}
https://bv.fapesp.br/en/auxilios/91067/lp-lq-decay-estimates-for-evolution-operators/	[ " Research Grants 15/16038-2 - Equações diferenciais parciais - BV FAPESP\nStart date\nBetweenand\n\n# Lp-Lq decay estimates for Evolution Operators\n\n Grant number: 15/16038-2 Support type: Regular Research Grants Duration: November 01, 2015 - October 31, 2017 Field of knowledge: Physical Sciences and Mathematics - Mathematics - Analysis Principal researcher: Marcelo Rempel Ebert Grantee: Marcelo Rempel Ebert Home Institution: Faculdade de Filosofia, Ciências e Letras de Ribeirão Preto (FFCLRP). Universidade de São Paulo (USP). Ribeirão Preto , SP, Brazil\n\nAbstract\n\nIn this project, we are interested in Lp-Lq decay estimates (not necessarily on the conjugate line) in time for linear hyperbolicequations or more in general, p-evolution equations. The results are derived by developing a suitable WKB analysis.We plan to apply these estimates to study semi-linear problems. In particular, we are interested in proving results about global existence (in time) of the solution, possibly assuming small initial data. Here we plan to understand in which cases the decay rates of solutions to the semi-linear problems coincide with those ones for the corresponding linear problem, and in which other cases a loss of decay appears. Then the question for the exact loss of decay appears.So methods to show optimality should be developed.We plan to study both models with constant coefficients and with time-dependent coefficients as well. In the case of time-dependent coefficients, we will assume suitable regularity and a sufficient control of the oscillations. Also, the interaction of the time-dependent coefficients will be studied to avoid bad influence on the asymptotic profile, or to obtain better decay estimates.In a first moment, we will mainly consider wave-type equations, possibly with damping terms, and with nonlocal terms, like fractional powers of the Laplacian. In this way we cover external and structural damping up to the visco-elastic case. Finally, we plan to study higher order equations and, if possible, first-order systems, \\$p\\$-evolution equations and problems in an abstract setting. (AU)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8103347,"math_prob":0.88973385,"size":386,"snap":"2021-31-2021-39","text_gpt3_token_len":74,"char_repetition_ratio":0.14921466,"word_repetition_ratio":0.0,"special_character_ratio":0.15284973,"punctuation_ratio":0.10169491,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9703283,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-22T12:10:17Z\",\"WARC-Record-ID\":\"<urn:uuid:074f2be8-0a0e-4441-8510-36b8d6d5b0ca>\",\"Content-Length\":\"61292\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a9c778fc-49e8-4b69-9746-2e847c0dd9ee>\",\"WARC-Concurrent-To\":\"<urn:uuid:55df5b68-1ab4-408b-b772-28da3be7b35a>\",\"WARC-IP-Address\":\"143.108.10.177\",\"WARC-Target-URI\":\"https://bv.fapesp.br/en/auxilios/91067/lp-lq-decay-estimates-for-evolution-operators/\",\"WARC-Payload-Digest\":\"sha1:7TOWBMZK6TOV2I3237DSI6NRNLYBXR4O\",\"WARC-Block-Digest\":\"sha1:EVOXVMQIB5G43R5K2VBMZRDEBT5KXD45\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057347.80_warc_CC-MAIN-20210922102402-20210922132402-00475.warc.gz\"}"}