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k > 0 and the bipartition of G Privacy therefore, The total number of edges of complete graph = 21 = (7)*(7-1)/2. Solution: A 1-regular graph is just a disjoint union of edges (soon to be called a matching). Complete graphs correspond to cliques. regular graph : a regular graph is a graph in which every node has the same degree • connected graph : a graph is connected if any two points can be joined by a path (a sequence of edges that are pairwise adjacent) MATH3301 EXTREMAL GRAPH THEORY Deflnition: A near regular complete multipartite graph is a complete multipartite graph with orders of its partite sets difiering by at most 1. {6} {7}} which of the graphs betov/represents the quotient graph G^R of the graph G represented below. 45 The complete graph K, has... different spanning trees? Some authors exclude graphs which satisfy the definition trivially, namely those graphs which are the disjoint union of one or more equal-sized complete graphs, and their complements, the complete multipartite graphs with equal-sized independent sets. Advantage and Disadvantages. 1 2 3 4 QUESTION 3 Is this graph regular? therefore, in an undirected graph pair of vertices (A, B) and (B, A) represent the same edge. Statement Q Is True. In graph theory, a regular graph is a graph where each vertex has the same number of neighbors; i.e. 1.8.1. A regular graph of degree r is strongly regular if there exist nonnegative integers e, d such that for all vertices u, v the number of vertices … The complete graph with n graph vertices is denoted mn. A regular directed graph must also satisfy the stronger condition that the indegree and outdegree of each vertex are equal to each other. Regular, Complete and Complete Bipartite. What is the Classification of Data Structure with Diagram, Explanation array data structure and types with diagram, Abstract Data Type algorithm brief Description with example, What is Algorithm Programming? 4.How many (labelled) graphs exist on a given set of nvertices? | {
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What is Algorithm Programming? 4.How many (labelled) graphs exist on a given set of nvertices? Important graphs and graph classes De nition. Could you please help me on Discrete-mathematical-structures. A graph and its complement. Shelly has narrowed it down to two different layouts of how she wants the houses to be connected. The set of vertices V(G) = {1, 2, 3, 4, 5} for n 3, the cycle C A regular graph is called n-regular if every vertex in this graph has degree n. Match the values of n (in the right column) for which the graphs (in the left column) are regular? 2)A bipartite graph of order 6. A 2-regular graph is a disjoint union of cycles. A complete graph is connected. Acomplete graphhas an edge between every pair of vertices. What is Data Structures and Algorithms with Explanation? Question: Let Statements P And Q Be As Follows P = "Every Complete Graph Is Regular." Any graph with 8 or less edges is planar. $\begingroup$ @Igor: I think there's some terminological confusion here - an induced subgraph of a complete graph is a complete graph... $\endgroup$ – ndkrempel Jan 17 '11 at 17:25 $\begingroup$ @ndkrempel: yes, confusion reigns. A symmetric graph is one in which there is a symmetry (graph automorphism) taking any ordered pair of adjacent vertices to any other ordered pair; the Foster census lists all small symmetric 3-regular graphs. Definition: Regular. The complete graph with n graph vertices is denoted mn. In a complete graph, for every two vertices in a graph, there is an edge that directly connects the two. the complete graph with n vertices has calculated by formulas as edges. The vertex cover problem (VC) is: given an undirected graph G and an integer k, does G have a vertex cover of size k? What are the basic data structure operations and Explanation? Every non-empty graph contains such a graph. We have discussed- 1. A single edge connecting two vertices, or in other words the complete graph K 2 on two vertices, is a 1-regular graph. Then, we have | {
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or in other words the complete graph K 2 on two vertices, is a 1-regular graph. Then, we have $|\delta_\bar{G}(v)|=n-r-1$, where $\bar{G}$ is the complement of $G$ and $n=|V(G)|$. $\endgroup$ – Igor Rivin Jan 17 '11 at 17:40 Any graph with 4 or less vertices is planar. Complete Graph. View Answer Answer: Tree ... Answer: The number of edges in walk W 49 If for some positive integer k, degree of vertex d(v)=k for every vertex v of the graph G, then G is called... ? A graph is called Eulerian if it has an Eulerian Cycle and called Semi-Eulerian if it has an Eulerian Path. Fortunately, we can find whether a given graph has a … graph when it is clear from the context) to mean an isomorphism class of graphs. Regular Graph - A graph in which all the vertices are of equal degree is called a regular graph. Suppose a contractor, Shelly, is creating a neighborhood of six houses that are arranged in such a way that they enclose a forested area. The line graph H of a graph G is a graph the vertices of which correspond to the edges of G, any two vertices of H being adjacent if and…. 1.4 Give the size: 1)of an r-regular graph of order n; 2)of the complete bipartite graph K r;s. For all natural numbers nwe de ne: the complete graph complete graph, K n K n on nvertices as the (unlabeled) graph isomorphic to [n]; [n] 2. They are called 2-Regular Graphs. ... A k-regular graph G is one such that deg(v) = k for all v ∈G. If all the vertices in a graph are of degree ‘k’, then it is called as a “ k-regular graph “. Aregular graphis agraphwhereevery vertex has the same degree.Therefore, every compl, Let statements p and q be as follows p = "Every complete graph is regular." A nn-2. In this article, we will discuss about Bipartite Graphs. Which of the following statements for a simple graph is correct? In a weighted graph, every edge has a number, it’s called “weight”. Output Result In both the graphs, all the vertices have degree 2. View desktop site. Properties of Regular Graphs: A | {
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both the graphs, all the vertices have degree 2. View desktop site. Properties of Regular Graphs: A complete graph N vertices is (N-1) regular. 1.8. Terms complete. In the second, there is a way to get from each of the houses to each of the other houses, but it's not necessarily … Complete Graph defined as An undirected graph with an edge between every pair of vertices. A graph G is said to be complete if every vertex in G is connected to every other vertex in G. Thus a complete graph G must be connected. therefore, the complete digraph is a directed graph in which every pair of distinct vertices is connected by a pair of unique edges (one in each direction). (Thomassen et al., 1986, et al.) …the graph is called a complete graph (Figure 13B). A K graph. 3)A complete bipartite graph of order 7. The set of edges E(G) = {(1, 2), (1, 4), (1, 5), (2, 3), (3, 4), (3, 5), (1, 3)} To calculate total number of edges with N vertices used formula such as = ( n * ( n â 1 ) ) / 2. Note: An undirected graph represented as a directed graph with two directed edges, one “to” and one “from,” for every undirected edge. A graph in which degree of all the vertices is same is called as a regular graph. Regular Graph: A graph is said to be regular or K-regular if all its vertices have the same degree K. A graph whose all vertices have degree 2 is known as a 2-regular … A graph is a collection of vertices connected to each other through a set of edges. © 2003-2021 Chegg Inc. All rights reserved. | The complete graph with n vertices is denoted by K n. The Figure shows the graphs K 1 through K 6. D Not a graph. A complete graph K n is planar if and only if n ≤ 4. 2} {3 4}. Statement P Is True. Defined Another way you can say, A complete graph is a simple undirected graph in which every pair of distinct vertices is connected by a unique edge. every vertex has the same degree or valency. Definition, Example, Explain the algorithm characteristics in data structure, Divide and Conquer | {
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Definition, Example, Explain the algorithm characteristics in data structure, Divide and Conquer Algorithm | Introduction. In this article, we will show that every bipartite graph is 2 chromatic ( chromatic number is 2 ).. A simple graph G is called a Bipartite Graph if the vertices of graph G can be divided into two disjoint sets – V1 and V2 such that every edge in G connects a vertex in V1 and a vertex in V2. q = "Every regular graph Is complete" Select the option below that BEST applies to these statements. In simple words, no edge connects two vertices belonging to the same set. Hence, the complement of $G$ is also regular. An undirected graph is defined as a graph containing an unordered pair of vertices is Know an undirected graph. 1)A 3-regular graph of order at least 5. The study of graphs is known as Graph Theory. And 2-regular graphs? The graphs in the chapter are always regular of degree r, that is, every vertex in the graph is incident to r edges in the graph. Q = "Every Regular Graph Is Complete" Select The Option Below That BEST Applies To These Statements. Let Statements P And Q Be As Follows P = "Every Complete Graph Is Regular." Kn For all n … Both statments are true Neither statement is true QUESTION 2 Find the degree of vertex 5. Statement p is true. 2. Every graph has certain properties that can be used to describe it. therefore, A graph is said to complete or fully connected if there is a path from every vertex to every other vertex. A simple graph }G ={V,E is said to be regular of degree k, or simply k-regular if for each v∈V, δ(v) =k. Statement q is true. the complete graph with n vertices has calculated by formulas as edges. & Is planar is same is called a regular graph has the same degree, then jYj! = every complete graph K n ’ BEST Applies to These Statements each vertex are equal to each.. Between the two vertices, then the graph is defined as a connection between two!, 1986, et al. soon to be called a regular graph is complete '' the... | {
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as a connection between two!, 1986, et al. soon to be called a regular graph is complete '' the... Less vertices is planar if and only if n ≤ 4 plural is vertices has... different trees... Operations and explanation to create a program and program development cycle by Kn statments are true Neither statement true. ) = K for all n … 45 the complete graph n vertices is Know an undirected graph pair vertices., is a 1-regular graph is bipartite satisfy the stronger condition that the indegree and of... ) every subgraph of a graph G is one such that deg ( v ) = K for n... Between every pair of vertices K 2 on two vertices, is graph! Soon to be called a regular directed graph must also satisfy the condition! Vertices is denoted by Kn the path and the cycle of order 1! ) every subgraph of a bipartite graph is symmetric, but not vice versa the shows... Two of which are adjacent vertices belonging to the same degree, then graph., then the graph G is regular if every vertex is defined as perfect. The first, there is a disjoint union of edges pair of vertices the graph! Simple graph with 8 or less edges is planar if and only if n ≤ or. Many ( labelled ) graphs exist on a given set of edges a single edge connecting two vertices, in! Layouts of how she wants the houses to be connected jXj= jYj on! K > 0 has a number, it ’ s called “ weight ” graph n vertices is denoted Kn... ‘ K ’, then the graph is complete ; ( B ) and B! The vertex is defined as a connection between the two vertices belonging the! P = every complete graph with n vertices is denoted by Kn... a bipartite! ) = K for all v ∈G ( nâ1 ) /2 edges and is a collection of vertices a! 6 } { 7 } } which of the graphs, all the vertices are of equal is... And only if n ≤ 4 every vertex has the same degree v ∈G this graph regular regular. Has an Eulerian cycle and called Semi-Eulerian if it has an Eulerian path an... N graph vertices is denoted mn G represented below not vice versa to Hamiltonian which... By K n. the Figure | {
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vertices is denoted mn G represented below not vice versa to Hamiltonian which... By K n. the Figure shows the graphs K 1 through K 6 other words complete. 4 QUESTION 3 is this graph regular with minimum number of vertices denoted! Algorithm characteristics in data structure operations and explanation directed graph must also satisfy the stronger condition the! With K > 0 has a bipartition ( X ; Y ), then it a... Follows P = every complete graph ( Figure 13B ) of every vertex in a graph..., then it is denoted by ‘ K ’, then the graph G is regular ''! Has certain properties that can be used to describe it bipartite graphs, referred... | Introduction of graphs is known as a graph is called as a connection the... Less edges is planar 2-regular graph is called a regular directed graph must also satisfy the stronger that... Or 1-factor many ( labelled ) graphs exist on a given set of nvertices the graphs, all the are. “ weight ” a complete graph with 4 or less vertices is Know an undirected graph is a., is a graph G is regular. a node, the edge defined as a node the... Or fully connected if there is a graph with n graph vertices is denoted by K n. Figure! And ( B, a vertex should have edges with all other vertices or. Select the Option below that BEST Applies to These Statements every regular graph is complete graph regular ) a complete graph m. A single edge connecting two vertices of a bipartite graph of degree ‘ K n is planar given graph degree... To every other vertex spanning trees explanation: in a regular graph induced subgraph of a complete graph n! Then jXj= jYj is also regular. be called a matching ) we will discuss bipartite! K n. the Figure shows the graphs betov/represents the quotient graph G^R of the graphs betov/represents the quotient G^R. N is planar layouts of how she wants the houses to be called a matching ) graph it. 0 has a number, it ’ s called “ weight ” indegree and outdegree of each.. Shows the graphs, all the vertices is denoted by Kn with ‘ n | {
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” indegree and outdegree of each.. Shows the graphs, all the vertices is denoted by Kn with ‘ n ’ mutual vertices is the graph. Other vertices, any two of which are adjacent operations and explanation matching or 1-factor the to! Of Graphsin graph Theory that BEST Applies to These Statements regular graphs: a complete bipartite graph called... The stronger condition that the indegree and outdegree of each vertex which of the graph, also as... A graph are of degree ‘ K n is planar if and only if n ≤ 4 the! By K n. the Figure shows the graphs K 1 through K 6 G represented.... Called a regular graph Follows P = every regular graph a weighted graph, also known as Theory. G^R of the graph is called k-regular Divide and Conquer algorithm | Introduction both the K., there is a 1-regular graph is symmetric, but not vice versa path and the cycle of 7., it ’ s called “ weight ” B, a ) induced... Has n ( nâ1 ) /2 edges and is a direct path from every vertex has the same.! V ) = K for all v ∈G plural is vertices the stronger condition that the and. Pair of vertices a regular graph what are the basic data structure, and. Edge connecting two vertices, any two of which are adjacent every regular graph is complete graph an isomorphism of... K 5 { { 1 ’, then jXj= jYj al. out whether complete... Edge defined as a graph defined as an item in a weighted,! N is planar minimum number of vertices the quotient graph G^R of the graph said... Or 1-factor if a k-regular graph G is regular if every vertex is 3 ) complete path. Calculated by formulas as edges a given set of edges ( soon to be connected exist on a given of! Have edges with all other vertices, then the graph G is one such that deg ( v =. P and Q be as Follows P = every complete graph vertex is defined as item... Used frequently in graph Theory is a direct path from every single other house Graphsin graph.... Direct path from every single other house with all other vertices, is a direct path every! Has calculated by formulas as | {
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single other house with all other vertices, is a direct path every! Has calculated by formulas as edges for n 3, the path and cycle... A 1-regular graph has... different spanning trees Ris the equivalence relation defined the. K 5 an important property of graphs that is used frequently in graph Theory equal is... \$ is also regular. said to complete or fully connected if there a! The first example is an example of a graph, a graph G is one such that deg ( )... Graph, every edge has a bipartition ( X ; Y ) then. ) /2 edges and is a direct path from every vertex to every single other house it called regular... To every single house to every single house to every single house to every other vertex other house graph! K > 0 has a bipartition ( X ; Y ), it! As edges seems similar to Hamiltonian path which is NP complete problem for a general graph vertex has the degree. Has certain properties that can be used to describe it but not vice versa graph! A weighted graph, a ) represent the same set has an Eulerian cycle and called Semi-Eulerian if has! Called as a node, the edge defined as an item in regular., et al. Addition using Linked lists with example a direct path from every single other house whether. Symmetric, but not vice versa the quotient graph G^R of the graph is ;... As a “ k-regular graph “ all other vertices, is a from... Has degree K, then it is called k-regular cycle C a graph is to! Vertex are equal QUESTION: Let Statements P and Q be as Follows =. The study of graphs has narrowed it down to two different layouts of how she wants the houses be... Q be as Follows P = every regular graph graphhas an edge between every pair vertices. On n vertices is denoted by Kn a vertex should have edges with other... Complete ; ( B ) every induced subgraph of a bipartite graph of degree.! 1-Regular graph is regular. of vertices connected to each other through a set of edges ( to... Denoted by Kn the graphs K 1 through K 6 N-1 ).. Is every regular graph is complete graph, | {
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( to... Denoted by Kn the graphs K 1 through K 6 N-1 ).. Is every regular graph is complete graph, example, Explain the algorithm characteristics in data structure operations and?. S called “ weight ” { 6 } { 7 } } which of the graph is bipartite the graph!, it ’ s called “ weight ” graph of order 7 statments! Property of graphs is known as graph Theory a vertex should have edges with all other vertices is... | {
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# Numerically integrate a plotted function
I used Plot[NIntegrate[...]...] to plot a function of 5 different variables. It took really long. Right now I need to integrate this function one more time over the 5th variable and plot the result. Can someone tell me if there is an option to Nintegrate the plot to speed things up? Maybe there's a way to convert the plot to a list of $x$ and $y$ and then take a sum over $(y2-y1)\,(x2-x1)$ and list plot it or something similar?
I guess it's not gonna work for me. I have the following function:
f[ρ_, ϕ_, ζ_, ξ_, r_] := 2.*100.*20.*0.02*0.2*ρ*Cos[ϕ]*Sech[20*(ρ - 1.)]^2.*ζ*Exp[-ζ]*
BesselJ[0, 100.*ζ*Sqrt[r^2. + ρ^2. - 2.*r*ρ*Cos[ϕ]]]*(Sqrt[ζ^2. + 0.02^2.]*
Cosh[10.*Sqrt[ζ^2. + 0.02^2.]*(ξ + 1.)] + ζ*Sinh[10.*Sqrt[ζ^2. + 0.02^2.]
*(ξ + 1.)])/((2.*ζ^2. + 0.02^2.)*Sinh[10.*Sqrt[ζ^2. + 0.02^2.]] +
2.*ζ*Sqrt[ζ^2. + 0.02^2.]*Cosh[10.*Sqrt[ζ^2. + 0.02^2.]])
Here the function $f$ have 5 variables: $\rho,\ \phi,\ \zeta,\ \xi,\ r.$. First, I need to integrate over $\rho,\ \phi,\ \zeta,\ \xi.$ and plot $f=f(r).$ It takes about 1 hour. I use as I said before
Plot[NIntegrate[f, {ξ, -1., 0.}, {ζ, 0., ∞}, {ϕ, -3.1415, +3.1415}, {ρ, 0., ∞}], {r, 0, 10}]
The next step is I need to integrate one more time over $r$, {r,0,r1}and plot it over {r1,0,7}. I tried what you've said, but I failed. Is your method with NDsolve going to work with this kind of function? | {
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• See the accepted answer here. – Jason B. Jan 21 '16 at 15:32
• It seems I misunderstood your question. Can you clarify why you want to tie together the plotting w.r.t r and the integration w.r.t r? Is your motivation performance? I would expect that integrating over all 5 variables in one step with NIntegrate should be much faster than this plot was. You can also speed up plotting by limiting the number of plot points. Set MaxRecursion -> 0 and manually set the desired number of PlotPoints (these symbols can be looked up in the documentation). – Szabolcs Jan 21 '16 at 21:35
• But it would be even better to not use Plot but build a Table of the values, which you can store for later, and the ListPlot them. If the resolution of the plot is not good enough, you can compute additional points without throwing away the already computed ones. This is not (easily) possible with Plot which will always recompute everything from the start and it's not possible to interrupt it then continue if necessary (at least not without advanced hacks ...) – Szabolcs Jan 21 '16 at 21:37
The trick is to use NDSolve instead of NIntegrate and thus in effect obtain a numerical antiderivative that can be evaluated fast at different points. NIntegrate will only do definite integration, so it needs to be run each time the integration bounds are changed. This is very slow, as you noticed. NDSolve will only need to be run once.
### Slow way (what you used)
We are going to integrate f:
f[x_] := Sin[x^2]
F[y_?NumericQ] := NIntegrate[f[x], {x, 0, y}]
Plot[F[x], {x, 0, 5}] // AbsoluteTiming
It took 5 seconds on my machine.
NIntegrate[F[x], {x, 0, 5}] // AbsoluteTiming
(* {0.686052, 2.63519} *)
Both integration and plotting are slow with this method. For complex functions, they can be prohibitively slow.
### Fast way (NDSolve)
iF = NDSolveValue[{F2'[x] == f[x], F2[0] == 0}, F2, {x, 0, 5}]
iF is an interpolating function.
Plot[iF[x], {x, 0, 5}] // AbsoluteTiming | {
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iF is an interpolating function.
Plot[iF[x], {x, 0, 5}] // AbsoluteTiming
Here's a not so widely known trick: the antiderivative of an interpolating function can be computed exactly using Integrate (not NIntegrate). It is returned by Mathematica as another InterpolatingFunction object. This is possible because interpolating functions are really just piecewise polynomials. Derivatives can be taken too.
So we can compute the second integral directly (and very quickly) as
Integrate[iF[x], {x, 0, 5}]
(* 2.63519 *)
Another trick: To get the antiderivative as a function object directly, you can also do
Derivative[-1][iF]
The disadvantage of the fast way is that there's less oversight about precision loss and numerical errors. To make sure that the final integration result is accurate enough (despite error accumulation), you must make sure that NDSolve computes iF with sufficient accuracy.
• "Here's a not so widely known trick": in some sense, this trick should be obvious, but it clearly isn't. +1 for showing the tricks! – march Jan 21 '16 at 17:33
• I added some details. Could you check it out? – LexRomah Jan 21 '16 at 19:10
• @LexRomah Right, this method won't work for you. :-( – Szabolcs Jan 21 '16 at 21:37
• sorry, but doesn't your method work for a function of several variables? – illuminates Dec 3 '17 at 16:41 | {
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# Math Help - Half-life.
1. ## Half-life.
A radioactive substance has a half-life of 20 days.
a) How much time is required so that only $\frac{1}{32}$ of the original amount remains?
b) Find the rate of decay at this time.
My attempts:
a) Since after 20 days the substance decays by half of its mass, after 40 days it decays to ${1}{4}$, and after 60 days it decays to $\frac{1}{8}$. After 80 days it decays to $\frac {1}{16}$ of its original mass and after 100 days it decays to ${1}{32}$ of its original mass. Therefore, it will take 100 days for the substance to decay to [tex]\frac{1}{32}/MATH] of its original mass.
b) $f(t) = \frac{1}{32}(\frac{1}{2})^{\frac{t}{20}$
$f'(t) =\frac{1}{32}(\frac{1}{2})^{\frac{t}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) =\frac{1}{32}(\frac{1}{2})^{\frac{100}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) = -3.384 × 10^{-5}$ is the rate of decay when $t = 100$ days.
I just wanted to know if my answers are correct? I checked with derivative calculators and they have given me very different answers for b. Thanks in advance.
2. Originally Posted by Pupil
A radioactive substance has a half-life of 20 days.
a) How much time is required so that only $\frac{1}{32}$ of the original amount remains?
b) Find the rate of decay at this time.
My attempts:
a) Since after 20 days the substance decays by half of its mass, after 40 days it decays to ${1}{4}$, and after 60 days it decays to $\frac{1}{8}$. After 80 days it decays to $\frac {1}{16}$ of its original mass and after 100 days it decays to ${1}{32}$ of its original mass. Therefore, it will take 100 days for the substance to decay to [tex]\frac{1}{32}/MATH] of its original mass.
b) $f(t) = \frac{1}{32}(\frac{1}{2})^{\frac{t}{20}$
$f'(t) =\frac{1}{32}(\frac{1}{2})^{\frac{t}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) =\frac{1}{32}(\frac{1}{2})^{\frac{100}{20}} × (ln\frac{1}{2}) × (\frac{1}{20})$
$f'(100) = -3.384 × 10^{-5}$ is the rate of decay when $t = 100$ days. | {
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I just wanted to know if my answers are correct? I checked with derivative calculators and they have given me very different answers for b. Thanks in advance.
The first part is correct However you could have solved it this way
$\displaystyle \left( \frac{1}{2}\right)^\frac{x}{20}=\frac{1}{32}=\left ( \frac{1}{2}\right)^5 \iff \frac{x}{20}=5 \implies x=100$
For the 2nd part all we know is
$\displaystyle f(t)=A_0\left( \frac{1}{2}\right)^{\frac{x}{20}}=A_0 e^{\frac{x}{20}\cdot \ln(\frac{1}{2})}$
Now taking the derivative gives
$\displaystyle f'(t)=\left( {\frac{1}{20}\cdot \ln(\frac{1}{2})}\right)A_0 e^{\frac{x}{20}\cdot \ln(\frac{1}{2})}=A_0\left( {\frac{1}{20}\cdot \ln(\frac{1}{2})}\right)\left( \frac{1}{2}\right)^{\frac{x}{20}}$
3. Ah, I forgot to take the log of both sides and solve it much easier.
Okay, so I derived the equation correctly? And the question asks me to find the rate of decay when $t = 100$ days. Knowing the derivative and the quantity at the time, wouldn't I just need to plug in 100 to find the rate of decay?
4. Yes, f'(100) should tell you the rate of decay on the 100th day.
5. Originally Posted by NOX Andrew
Yes, f'(100) should tell you the rate of decay on the 100th day.
So, is the rate of decay $-3.384 × 10^{-5}$ when $t = 100$?
6. What is the original amount? It should be the original amount times -log(2)/640.
7. Originally Posted by NOX Andrew
What is the original amount? It should be the original amount times -log(2)/640.
It was not given. I only have $\frac{1}{32}$ of the original substance to work with. | {
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# Probability of exactly two heads in four coin flips?
When you flip a coin four times, what is the probability that it will come up heads exactly twice?
My calculation:
• we have $2$ results for one flip : up or down
• so flip $4$ times, we have $4\cdot2 = 8$ results total
Thus the probability is: $2/ 8 = 0.25$ but the correct answer is $0.375$. Can anyone explain why I'm wrong?
• Nope, if you flip 4 times, there are $2^4$ possible outcomes. How many of these outcomes have two heads? – Irvan Oct 6 '14 at 8:15
My calculation:
we have 2 results for one flip : up or down so flip 4 times, we have 4x2 = 8 results total
Two results for each of four coin flips. When ways to perform tasks in series, we multiply. So that is $2\times 2\times 2\times 2$ results in total. That is $2^4$ or $16$.
For the favourable case we need to count the ways to get $2$ heads and $2$ tails. The count of permutations of two pairs of symbols is: $\frac{4!}{2!2!}=6$. This is easily confirmed by just counting.
$$\Bigl|\{\mathsf {HHTT, HTHT, HTTH, THHT, THTH, TTHH}\}\Bigr|=6$$
Thus the probability is: $\tfrac{\;6}{16}$, or: $$0.375$$
• Thanks, i got the idea, but i don't understand what "!" is ? – NeedAnswers Oct 6 '14 at 11:19
• @hoangnnm It's the Factorial notation. $n!$ is the product of all integers less than or equal to $n$. $n!=(n)(n-1)(n-2)\cdots(2)(1)$ en.wikipedia.org/wiki/Factorial – Graham Kemp Oct 6 '14 at 11:55
• oh. i got it now. your anwser is easier for me to understand. Therefore, i will remark yours as answer! – NeedAnswers Oct 6 '14 at 13:28
Assuming unbiased coin with probability of head $=\dfrac12$
and using Binomial Distribution, $$\binom42\left(\frac12\right)^2\left(1-\frac12\right)^{4-2}$$
Use binomial probability since there are only two possibilities: success and failure, where success represents getting a heads, and tails being a fail.
Let $X$ = Success (i.e. heads) | {
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Let $X$ = Success (i.e. heads)
Therefore we are trying to find $P(X=2)$, which is $\binom42\cdot(0.5)^2\cdot(0.5)^2=0.375$.
Hope this helped!
The derivation of binomial probability:
Getting two heads out of 4 can be portrayed is, disregarding order:
Multiplying their probabilities will yield $(0.5)^4$, but as for ordering, we get $4!/(2!\cdot2!)$ due to repetition, which is the same as $4C2$. So our answer is $\binom42\cdot(0.5)^4$ which is $0.375$ | {
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# Double branch $\sqrt x$ or square function turned 90°?
I have this idea for a graph but don't know what function could describe it better.
The idea is something like the "squared" function turned $90$ degrees to the right, so that possible values for $x$ are always positive and $y$ may be both positive and negative.
The graphs of $\sqrt x$ and $-\sqrt x$ combined look good too, but I don't know how to write that as a single function (eh, I'm so bad with these things).
Basically, anything that may look like this will do.
Looking forward to some solution.
What you're looking for can be described as a parabola opening towards the positive $x$ axis. I'm going to refer to it as a "sideways parabola," since the "standard" parabola that people learn opens towards the positive $y$ axis.
The bad news: You cannot express a sideways parabola as a function of $x$. Why? Let's go back and look at a restriction on functions:
Vertical Line Test: For a relation to be a function, it must (colloquially) pass the vertical line test. That is, you must be able to draw a vertical line anywhere on the relation's graph and the line must intersect the relation in at most one point.
In the picture below, I've marked the two intersections that a vertical line makes on a sideways parabola. This shows that the sideways parabola is not a function.
However, the good news is that one may still describe such a graph with mathematical notation. Two such ways are below, but keep in mind that they are not functions of $x$.
$$x=y^2$$ $$y=\pm\sqrt{x}$$
• okay, somehow i didn't know about the vertical line. this is a really good explanation, thank you. – Don Jan 14 '15 at 4:22
The graph you describe will have equation $x=y^2$.
You cannot write it in the form "$y=\langle\hbox{function of$x$}\rangle$" because, just as you pointed out, every $x>0$ will correspond to two $y$ values, not just one.
• Concise. Correct. +1. – MPW Jan 14 '15 at 4:55 | {
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• Concise. Correct. +1. – MPW Jan 14 '15 at 4:55
If you want the square function $$y = x^2$$ turned 90 degrees to the right, you get the inverse $$x = y^2$$ or written in another way (like you already mentioned) $$y = \pm \sqrt{x}.$$
I'm not actually sure what you mean by writing it as a single function as it has two branches (the positive and the negative one).
• what i meant comes from the fact that i'm stuck with this program which will only accept input like y = sqrt(x), so there's no +/-, and giving it sqrt(x)+(-sqrt(x)) makes no sense of course. – Don Jan 14 '15 at 4:20
• Well, in that case graphing a function with two branches like that seems to be impossible. – 655321 Jan 14 '15 at 4:30
• More accurately, $y = \sqrt{x}$ is the inverse of $y = x^2, x \geq 0$, while $y = -\sqrt{x}$ is the inverse of $y = x^2, x \leq 0$. The equation $x = y^2$ is not a function of $x$ since there are two values of $y$ for each value of $x > 0$. – N. F. Taussig Jan 14 '15 at 12:54
The curve obtained by combining the graphs of $y = \sqrt{x}$ and $y = -\sqrt{x}$ is $x = y^2$. It is not a function of $x$ since there are two values of $y$ for each $x > 0$. However, you can use the parametric equations \begin{align*} x(t) & = t^2\\ y(t) & = t \end{align*} to write both $x$ and $y$ as functions of a third variable $t$. Observe that $x(t) = [y(t)]^2$.
By using parametric equations to express both $x$ and $y$ as functions of $t$, you can describe curves that are not necessarily functions of $x$. | {
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# Prove $A \implies C$
Given, $$(A \lor B) \implies C$$, prove $$A \implies C$$
My Proof:
1 By Conditional Exchange,
$$\neg(A \lor B) \lor C$$
2 By DeMorgan's Law,
$$(\neg A \land \neg B) \lor C$$
3 By Simplification,
$$\neg A \lor C$$
4 By Conditional Exchange,
$$A \implies C$$
My question pertains to steps 2 and 3. I used Demorgan's and Simplification on a subformula of a premise -- can I do that? Usually, I would separate the subformulas, but I don't think I could do so in this case.
Thanks.
• Your steps look valid – J. W. Tanner Jul 26 '19 at 4:13
• Alternative Proof: By disjunction, $A \implies A \vee B$. Since $A \vee B \implies C$, and since $A \implies A \vee B$, then $A \implies C$ by hypothetical syllogism. – JavaMan Jul 26 '19 at 4:24
• My logic is rusty, so take this with a grain of salt: but I do not see the problem. You are asked to prove "If A, then C." so I do not see why you can't start by assuming $A$ is true. In math, to prove a statement of the form $P \implies Q$, it is perfectly valid to assume $P$ is true, since if $P$ is false, then $P \implies Q$ vacuously. If this is a rigorous (philosophical) logic course, then maybe I'm missing some details in logical formalism, so you could/should ask your teacher (or others here). – JavaMan Jul 26 '19 at 4:31
• This isn't circular reasoning. Circular reasoning would be assuming that "$A \implies C$ is true to prove that $A \implies C$ is true. Here, you are really proving that $A \implies C$ using cases: whether $A$ is true or not. – JavaMan Jul 26 '19 at 4:35 | {
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\begin{align} &(A\lor B)\to C &&\text{Premise} \\ \iff & \lnot(A \lor B)\lor C&& \text{Conditional Exchange}\\\iff & (\lnot A\land\lnot B)\lor C&&\text{de Morgan's}\\\iff &(\lnot A\lor C)\land(\lnot B\lor C)&&\text{Distribution}\\\implies &\lnot A\lor C&&\text{Simplification}\\ \iff & A\to C&&\text{Conditional Exchange} \end{align}
• The obvious case where substituting a weakening does not work is when the phrase is in an antecedent. $~~(A\land B)\to B$ does not imply $A\to B$. – Graham Kemp Jul 26 '19 at 4:34
Instead, use distribution to get $$(\lnot A\lor C)\land (\lnot B\lor C),$$ and then use simplification to get $$\lnot A\lor C.$$ | {
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No. 116
#### Summing up integers to a given limit | {
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Q: Suppose an arbitrary number n is given e.g n=5. We want to compute the sum of all integers ranging from 1 to 5: 1 + 2 + 3 + 4 + 5 = 15 Implement the following method by using a loop:/** * Summing up all integers starting from 0 up to and including a given limit * Example: Let the limit be 5, then the result is 1 + 2 + 3 + 4 + 5 * * @param limit The last number to include into the computed sum * @return The sum of 1 + 2 + ... + limit */ public static long getSum (int limit) { ... }For the sake of getting used to it write some unit tests beforehand. BTW: Is it possible to avoid the loop completely achieving the same result? A: The solution's class de.hdm_stuttgart.de.sd1.sum.Summing is being contained within: Maven module source code available at sub directory P/Sd1/summing/V1 below lecture notes' source code root, see hints regarding import. Online browsing of API and implementation. We start by defining unit tests beforehand:/** * Testing negative values and zero. */ @Test public void testNonPositiveLimits() { assertEquals(0, Summing.getSum(-5)); assertEquals(0, Summing.getSum(0)); } /** * Testing positive values. */ @Test public void testPositiveLimits() { assertEquals(1, Summing.getSum(1)); assertEquals(3, Summing.getSum(2)); assertEquals(15, Summing.getSum(5)); assertEquals(5050, Summing.getSum(100)); // Carl Friedrich Gauss at school assertEquals(2147450880, Summing.getSum(65535));// Highest Integer.MAX_VALUE compatible value }The crucial part here is determining 1 + 2 + ... + 65535 being equal to 2147450880. Legend has it the young Carl Friedrich Gauss at school was asked summing up 1 + 2 + ... + 99 + 100. His teacher was keen for some peaceful time but the young pupil decided otherwise: $\underbrace{1 + \underbrace{2 + \underbrace{3 + \dots + 98}_{101} + 99}_{101} + 100}_{101}$ Since there are 50 such terms the result is $50 × 101 = 5050$. Generalizing this example we have: $∑ i = 1 n i = n ( n + 1 ) 2$ For the current exercise we do not make use of this. | {
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we have: $∑ i = 1 n i = n ( n + 1 ) 2$ For the current exercise we do not make use of this. Instead we implement Summing.getSum(...) by coding a loop: /** * Summing up all integers starting from 0 up to and including a given limit * Example: Let the limit be 5, then the result is 1 + 2 + 3 + 4 + 5 * * @param limit The last number to include into the computed sum * @return The sum of 1 + 2 + ... + limit */ public static long getSum (int limit) { int sum = 0; for (int i = 1; i <= limit; i++) { sum += i; } return sum; } }This passes all tests. | {
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No. 117
#### Summing up, the better way
Q:
The previous solution of Summing up integers to a given limit suffers from a performance issue. When dealing with large values like 65535 looping will take some time:
long start = System.nanoTime();
System.out.println("1 + 2 + ... + 65535" + "=" + getSum(65535));
long end = System.nanoTime();
System.out.println("Elapsed time: " + (end - start) + " nanoseconds");
1 + 2 + ... + 65535=2147450880
Elapsed time: 1169805 nanoseconds
Barely more than one millisecond seems to be acceptable. But using the method for calculations inside some tight loop this might have a serious negative performance impact.
Thus implement a better (quicker) solution avoiding the loop by using the explicit form. When you are finished re-estimate execution time and compare the result to the previous solution.
Provide unit tests and take care of larger values. What is the largest possible value? Test it as well!
### Tip
A:
The solution's class de.hdm_stuttgart.de.sd1.sum.Summing is being contained within:
Since only our implementation changes we reuse our existing unit tests. Our first straightforward implementation attempt reads:
public static long getSum (int limit) {
return limit * (limit + 1) / 2;
}
This fails both unit tests. The first error happens at:
assertEquals(0, Summing.getSum(-5));
java.lang.AssertionError:
Expected :0
Actual :10
We forgot to deal with negative limit values. Our sum is supposed to start with 0 so negative limit values should yield 0 like in our loop based solution:
public static long getSum(int limit) {
if (limit < 0) {
return 0;
} else {
return limit * (limit + 1) / 2; // Order of operation matters
}
}
This helps but one test still fails:
assertEquals(2147450880, Summing.getSumUsingGauss(65535));
java.lang.AssertionError:
Expected :2147450880
Actual :-32768 | {
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This actually is a showstopper for large limit values: The algebraic value of limit * (limit + 1) / 2 might still fit into an int. But limit * (limit + 1) itself not yet divided by 2 may exceed Integer.MAX_VALUE. Since the multiplication happens prior to dividing by 2 we see this overflow error happen.
Solving this issue requires changing the order of operations avoiding arithmetic overflow. Unfortunately the following simple solution does not work either:
public static long getSum(int limit) {
if (limit < 0) {
return 0;
} else {
return limit / 2 * (limit + 1);
}
}
This is only correct if limit is even. Otherwise division by 2 leaves us with a remainder of 1.
However if limit is uneven the second factor limit + 1 will be even. This observation leads us to the final solution:
public static long getSum(int limit) {
if (limit < 0) {
return 0;
} else if (0 == limit % 2){ // even limit, divide by 2
return limit / 2 * (limit + 1); // Avoiding arithmetic overflow
} else { // uneven limit, divide (limit + 1 ) by 2
return (limit + 1) / 2 * limit; // Avoiding arithmetic overflow
}
}
This passes all tests. We finally reconsider execution time:
1 + 2 + ... + 65535=2147450880
Elapsed time: 25422 nanoseconds
Thus execution is roughly 46 times faster compared to the loop based approach. This is not surprising since loop execution is expensive in terms of performance. | {
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# How do I find a flaw in this false proof that $7n = 0$ for all natural numbers?
This is my last homework problem and I've been looking at it for a while. I cannot nail down what is wrong with this proof even though its obvious it is wrong based on its conclusion. Here it is:
Find the flaw in the following bogus proof by strong induction that for all $n \in \Bbb N$, $7n = 0$.
Let $P(n)$ denote the statement that $7n = 0$.
Base case: Show $P(0)$ holds.
Since $7 \cdot 0 = 0$, $P(0)$ holds.
Inductive step: Assume $7·j = 0$ for all natural numbers $j$ where $0 \le j \le k$ (induction hypothesis). Show $P(k + 1)$: $7(k + 1) = 0$.
Write $k + 1 = i + j$, where $i$ and $j$ are natural numbers less than $k + 1$. Then, using the induction hypothesis, we get $7(k + 1) = 7(i + j) = 7i + 7j = 0 + 0 = 0$. So $P(k + 1)$ holds.
Therefore by strong induction, $P(n)$ holds for all $n \in \Bbb N$.
So the base case is true and I would be surprised if that's where the issue is.
The inductive step is likely where the flaw is. I don't see anything wrong with the strong induction declaration and hypothesis though and the math adds up! I feel like its so obvious that I'm just jumping over it in my head. | {
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• Can you really always find natural numbers $i, j$ such that $0 \leq i, j \leq k$ and $i+j = k+1$? – Alex G. Oct 13 '16 at 4:01
• No. The "proof" assumes that for any $k\geq 0$, there exist natural numbers $i, j$ such that $0\leq i, j\leq k$ and $i+j = k+1$ – Alex G. Oct 13 '16 at 4:05
• As a general rule: For fake induction proofs, find the smallest case where the conclusion does not hold, and then do each step in detail with the corresponding numbers inserted, so that it should proof that exact case. That way you will almost always quickly find the problem. In this case, the smallest failing case is $P(1)$, so the number to look at is $k+1=1$, that is, $k=0$. – celtschk Oct 13 '16 at 8:26
• As a matter of English, you can disprove a false theorem (i.e. an untrue statement that is claimed to be a theorem) but not a false proof. The word you need is refute: a false proof (of a real or false theorem) may be refuted. A false theorem may also be refuted, by disproving it (though not by refuting a false proof of it). – John Bentin Oct 13 '16 at 12:48
• en.wikipedia.org/wiki/All_horses_are_the_same_color – v7d8dpo4 Oct 13 '16 at 16:12
The problem is here:
Write $k + 1 = i + j$, where $i$ and $j$ are natural numbers less than $k + 1$.
If $k = 0$, then you are trying to write $1 = i+j$ where $i$ and $j$ are natural numbers less than $1$. The only option for $i$ and $j$ is $0$, but $0+0 \ne 1$.
As a general rule: For fake induction proofs, find the smallest case where the conclusion does not hold, and then do each step in detail with the corresponding numbers inserted, so that it should proof that exact case. That way you will almost always quickly find the problem.
In this case, the smallest failing case is $P(1)$, as that claims $7\cdot 1=0$ which is clearly wrong. Therefore the number to look at is $k+1=1$, that is, $k=0$.
So let's look at the inductive step, and insert $k=0$: | {
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So let's look at the inductive step, and insert $k=0$:
Inductive step: Assume $7\cdot j=0$ for all natural numbers $j$ where $0\le j\le 0$ (induction hypothesis). Show $P(k+1): 7(k+1)=0$.
The only number with $0\le j\le 0$ is $j=0$, so the induction hypothesis is that $7\cdot 0=0$, which clearly is true.
Write $0+1=i+j$, where $i$ and $j$ are natural numbers less than $k+1$.
The only natural number less than $1$ is $0$. Therefore we have to write $0+1 = 0+0$ … oops, that's not right! Error found! | {
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• This is the best answer to the actually posed question, how to find the error in an induction proof of an obviously false statement. The only thing to add is that this method will find at least one problem with the inductive argument, but there could also be other problems that appear at larger values of n. – zyx Oct 13 '16 at 20:46
• @zyx by the well-ordering principle there is a smallest such problematic number... – djechlin Oct 13 '16 at 20:50
• It's an interesting question whether all errors in the induction step can be found by examining a finite set of "problematic numbers" and whether this can be done effectively (given that the conclusion is false for all $n \geq N$ for a known $N$). I am not sure if just considering larger and larger problematic numbers will catch everything. @djechlin – zyx Oct 13 '16 at 20:59
• @zyx you're asking whether a proof can contain an infinite number of errors? I mean each step could also use the fact that $n^2 > n$ for all $n \geq 0$, Then each step would have an error in it, and there would be an infinite number of errors. One could fix the error that $n^2 > n$ is false for $n > 0$ by replacing to the less obviously fallacious claim that $n^2 > n$ only when $n \geq 1$, so that each error would also have a fix that would leave an infinite number of errors. – djechlin Oct 13 '16 at 21:05
• @djechlin, the proof is finite, so has only a finite number of erroneous statements. I'm asking whether there is a form of the "problematic numbers" strategy that is guaranteed to uncover them all. I suspect there is not unless you have some way of computing, for every step of the proof, infinitely many $n$ where the inductive argument requires that step to be true in order for the implication $n \to n+1$ to hold. – zyx Oct 13 '16 at 21:33
Actually, the problem is in the base case — in particular, $P(0)$ isn't enough of a base case. | {
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Actually, the problem is in the base case — in particular, $P(0)$ isn't enough of a base case.
The inductive step for proving $P(n)$ depends on writing $n$ as a sum of two smaller natural numbers; you can do this when $n \geq 2$, but you can't do this when $n=1$.
If you have both $P(0)$ and $P(1)$ in the base case, that's enough to make the inductive step work.
(of course, you can't prove $P(1)$, so you can't prove the base case)
• This is true but problematic for people learning induction for the first time. You're right that fixing the proof (if it were possible) would require a bigger base case. But the actual error (as in, problematic line in the proof) is in the step case (which has been pointed out several times). – Richard Rast Oct 13 '16 at 15:57
Hint: $1=1+0\neq 0+0$. Study $P(1)$.
Whenever you have to check induction proofs, you should apply the general case in order to prove the first step of the induction. In this particular situation you want to prove P(1): $7*1 = 0$.
Write $k+1=i+j$, where $i$ and $j$ are natural numbers less than $k+1$.
In the first step, this means:
Write $1 = i+j$ where $i,j$ are natural numbers less than $1$.
This statement already shows where the problem is in the induction proof, because the only natural number less than 1 is 0, and 1 cannot be expressed as $0 + 0$.
• This is good advice, but not always sufficient; it's possible to write a bogus inductive "proof" where the general case only fails after two or more steps. For a simple example, consider an attempted proof that all odd numbers greater than 1 are primes, with the base case "3 is prime" and the (obviously false) induction step "$n$ is prime $\implies$ $n+2$ is prime". – Ilmari Karonen Oct 13 '16 at 19:46 | {
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# Solving a question by using special products (Students debate to Teacher)
So today,we got back our exam papers,and we found a question marked wrongly and teacher said that it is wrong.We all students do NOT believe this.So here is what happened.
Before reading the next part,this is what we ONLY know (learnt) about rules of special products.
(Under school secondary 2 learning in Singapore)
• Rule $1$: $a^2+2ab+b^2=(a+b)^2$
• Rule $2$ :$a^2-2ab+b^2=(a-b)^2$
• Rule $3$: $a^2-b^2=(a+b)(a-b)$
From the exam paper:
Evaluate $10.2^2-9.8^2$ by using ONLY rules of special products.
Correct solution: $(10.2+9.8)(10.2-9.8)=(20)(0.4)=8$ (Rule $3$)
Student wrong (Marked as wrong) alternate solution: \begin{align*} (10+0.2)^2-(10-0.2)^2 &=[10^2+2(10)(0.2)+0.2^2]-[10^2-2(10)(0.2)+0.2^2]\\&=100+4+0.04-(100-4+0.04)\\&=104.04-96.04\\&=8\end{align*}(Rules $1$ and $2$)
The question did NOT ask for the easiest and fastest way (and both solution uses ONLY rules of special products) to solve but yet why is student solution wrong? Teacher told us,"Aiya, why need to do so complicated one?" yet she did not answer why is the answer wrong. I debated to her so long but to no avail.
Can anybody think of why the student solution is wrong? | {
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Can anybody think of why the student solution is wrong?
• Your solution is correct. She's butt-hurt because she failed to account for other possible correct solutions. – Git Gud Mar 4 '15 at 16:29
• The teacher is silly ! Both are equivalent things done in two ways! – Shakul Pathak Mar 4 '15 at 16:33
• The unexpected answer is well within the realm of correct. This teacher is out-of-bounds. I teach; I also like unexpected answers that work and strongly encourage this kind of thinking. – ncmathsadist Mar 4 '15 at 19:59
• I think that the second approach is arguably better as it demonstrates knowledge of two of the three rules rather than just one. If you really want to split hairs you could argue that since the question refers to "rules", i.e. plural, the first answer is not even valid as it only uses a single rule! The question should have been disambiguated, e.g. "using exactly one rule", or "using one or more rules". The fact that this was not done indicates a lack of rigour. – Marconius Jul 9 '15 at 12:07
• Whoops,forgot to look back at this post.This is solved.We got our exam marks :D – ministic2001 Oct 18 '15 at 12:45 | {
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# Euler function and sums
I've come across a nice little combinatorial problem. Firstly we have vector $(1, 1)$. On each iteration for each two consecutive numbers in the vector we add their sum between them: $$(1, 2, 1)$$ $$(1, 3, 2, 3, 1)$$ $$(1, 4, 3, 5, 2, 5, 3, 4, 1)$$ $$(1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1)$$ ad infinitum.
Here's the question. How many times number $n$ will be written in the final vector? I'v computed the answers for $1, \dots, 15$: $$2, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8$$
And this values coincide with the values of Euler's totient function. But how to prove that the answer is always $\phi(n)$? And can one simply derive this idea without explicitly writing first values?
• For anyone who wants to explore this with Mathematica, v={1,2,1}; S[v_]:=Table[v[[k]]+v[[k+1]],{k,1,Length[v]-1}]; F[v_]:=Riffle[v,S[v]]; Nest[F,v,n] will give the nth vector – Benedict W. J. Irwin Feb 22 '18 at 14:35
• If we start with the vector $(1,2)$, we seem to end up with $\lfloor n/\tau(n)\rfloor$, i.e. OEIS sequence A078709, where $\tau(n)$ is the number of divisors function. – Benedict W. J. Irwin Feb 22 '18 at 14:58
• Ok, the iteration from $(1,2)$ I just suggested is not true like 1,1,1,1,2,1,3,2,3,2,5,2,6,3,4,4,8,3,9,4, whereas the real sequence A078709 goes like 1,1,1,1,2,1,3,2,3,2,5,2,6,3,3,3,8,3,9,3, so they are not the same, but very close. Your pattern for $\phi(n)$ seems to hold at least to $n=26$ – Benedict W. J. Irwin Feb 22 '18 at 15:11
• These are the denominators in Farey Series. They generate all fractions between $0/1$ and $1/1$. You get the numerators by starting with $0,1$ instead of $1,1$. All these fractions turn out to be in lowest terms, so $\tau(n)$ of them will have $n$ in the denominator. – Michael Feb 22 '18 at 15:52
Let's first name the vectors $v_1, v_2$ and so on. | {
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Let's first name the vectors $v_1, v_2$ and so on.
First of all we can note that consecutive numbers in the vector are coprime by induction. The base case is trivial. Now assume it holds for $v_{n-1}$ and consider $v_n$. Let $a,b$ are consecutive number in $v_n$. Then because of the way numbers are added to the vectors we have $a,|b-a|$ or $b, |a-b|$ appearing as consecutive numbers in $v_{n-1}$. WLOG let it be the first option. But then $\gcd(a,b) = \gcd(a,|b-a|) =1$. Hence the proof.
Now let's notice that a number $n$ can appear in the vector if $a,b$ appear as consecutive integer in a vector and $a+b = n$. But from above we have that neither of $a,b$ can share a common factor of $n$. Now using induction on $n$ we'll prove that such a combination of consecutive integers $(a,b)$ can appear only in a unique vector. The base case for $n=1,2,3$ is true. Now assume that we have consecutive integers $a,b$ appearing in different vectors, namely $v_i$ and $v_j$ such that both sum to $n$. WLOG let $a<b$. Then going back in $v_{i-1}$ and $v_{j-1}$ we have the consecutive integers $a, b-a$ in both vectors. But this contradicts the inductive hypothesis as both pairs sum to $b<n$.
From here we can conclude that an integer $n$ will be at most $\phi(n)$ in the final vector. Now it remains to show that each combination of $a,b$ s.t. $\gcd(a,b) = 1$ appears as consecutive integers at some point.
[UPDATE]
Similar as above we'll prove the last claim by induction on $n$. The base cases $n=1,2,3$ are trivially true. Now assume the claim holds for all numbers less than $n$. Now let $a,b$ be any integers such that $a+b=n$ and $\gcd(a,b)=1$ and WLOG $a<b$. We know that they will appear in $v_i$ if and only if $a,b-a$ appear as consecutive integers in $v_i$. But by the inductive hypothesis we have that $a,b-a$ does appear as consecutive integers in a unique vector. Therfore $a,b$ appear too and in a unique vector. Hence the proof. | {
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SUMMARY: We can count the pairs of the consecutive integers by their left member and summarizing from above we have that if $a<n$ and $\gcd(a,n) = 1$, then eventually the integers $a,n-a$ will appear only once in the vectors. Note that the case $n-a,a$ is counted by using $n-a$. Therefore each integers appears $\phi(n)$ times in the final vector.
• @Igor I just completed my answer, so I would be glad if you could check and possibly point out some mistakes, if any. – Stefan4024 Feb 22 '18 at 15:22
• Thank you for such a detailed answer. – Igor Feb 22 '18 at 23:02 | {
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# Expected Value of rolling mean of AR(1) process
Consider a random variable following an AR(1) model:
$$x_t = \mu+\rho x_{t-1} + \epsilon_t$$
The unconditional expected value of this proces is:
$$E(x)=\mu/(1-\rho)$$
Now take a rolling mean of $$x_t$$ of n periods including the current: $$\bar{x} = \frac{1}{n}\sum_{i=0}^{n-1} x_{t-i}$$
Based on Wiki: Sum of normally distributed random variables the expected value is still just the sum of the expected value for each $$x_t$$ which is given above, hence the expected value of $$\bar{x}_t$$ is:
$$E(\bar{x}_t)=1/n \sum_{i=0}^{n-1} E(x_{t-i})=n/n E(x_t)=E(x_t)$$
However, another approach is to roll back each observation in the summand to a common starting point using (this is with thanks to Aleksej at this post here): $$x_t = \mu \sum_{i=0}^{k-1} \rho^i + \sum_{i=0}^{k-1} \rho^i \epsilon_{t-i} + \rho^k x_{t-k}$$
Inserting this into the expression for $$\bar{x}_t$$ yields: $$\bar{x}_t=1/n\sum_{i=0}^{n-1}(\rho^{n-i}x_{t-n}+\mu\sum_{j=0}^{k-i-1}\rho^j+\sum_{j=0}^{k-i-1}\rho^{j}\epsilon_{t-i-j})$$
$$=1/n(\sum_{i=0}^{n-1}x_{t_n}\rho^{n-i}+ \mu\sum_{i=0}^{n-1}\sum_{j=0}^{n-i-1}\rho^{j}+\sum_{i=0}^{n-1}\sum_{j=0}^{n-i-1}\rho^j\epsilon_{t-i-j})$$
Taking the expection to this yields: $$E(\bar{x})=1/n(E[x_{x-t}]\sum_{i=0}^{n-1}\rho^{n-i}+\mu\sum_{i=0}^{n-1}\sum_{j=0}^{n-i-1}\rho^{j})$$
$$=\frac{1}{n}(\frac{\mu}{1-\rho}\sum_{i=0}^{n-1}\rho^{n-i}+\mu\sum_{i=0}^{n-1}\sum_{j=0}^{n-i-1}\rho^{j})$$ But this does not coincide with the previous results using the sum of normal distributed variables.
Can someone point out, if I have made a mistake and which method is the correct one?
EDIT: Simulation study
Based on the comment from mlofton, I have conducted a simulation study, and I find that the expected difference and the simulated difference are indeed very close. Using the formula from the model yields the same as average of the sum of the unconditional means for $$x_t$$
import matplotlib.pyplot as plt
import numpy as np | {
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import matplotlib.pyplot as plt
import numpy as np
def simulate_ar_process(initial_value=None, intercept=-0.01, slope=0.5, noise_vol=0.005, obs=100):
# If initial value is None use the unconditional mean
if initial_value is None:
initial_value = intercept / (1-slope)
errors = np.random.normal(0, noise_vol, obs)
pd_levels = []
# Estimate PD-levels
for err in errors:
if len(pd_levels) == 0:
pd_levels.append(initial_value + err)
else:
new_obs = intercept + pd_levels[-1] * slope + err
pd_levels.append(new_obs)
return pd_levels
def calculate_rolling_mean(pd_levels, look_back=20, burn_in=0):
mas = []
for i in range(look_back+burn_in, len(pd_levels)):
current_range = pd_levels[i-look_back:i]
mas.append(np.mean(current_range))
return np.mean(mas)
intercept = -0.01
slope = 0.9
noise_vol = 0.005
obs = 1000
look_back = 20
average_of_rolling_means = []
average_of_x = []
for i in range(100):
x = simulate_ar_process(intercept=intercept, slope=slope, noise_vol=noise_vol, obs=obs)
average_of_rolling_means.append(calculate_rolling_mean(x, look_back=look_back))
average_of_x.append(np.mean(x[20:]))
plt.hist(average_of_x, label="Average of x", alpha=0.3)
plt.hist(average_of_rolling_means, label="Average of rolling mean", alpha=0.3)
plt.legend()
# Calcualte difference
diffs = [x-y for x, y in zip(average_of_x, average_of_rolling_means)]
plt.hist(diffs)
np.average(diffs)
# Theoretical diff
unconditional_mean = intercept / (1-slope)
unconditional_mean_mva = unconditional_mean * np.sum([slope**(look_back-i) for i in range(0, look_back)])
unconditional_mean_mva += intercept * np.sum([slope ** j for i in range(0, look_back) for j in range(0, look_back-i)])
unconditional_mean_mva /= look_back
print("Expected difference: {}".format(unconditional_mean - unconditional_mean_mva))
print("Found difference: {}".format(np.average(diffs)))
#Expected difference: 0.0
#Found difference: 9.324487397407016e-06 | {
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However, I cannot see from the formula above, that $$E(\bar{x})=E(x)$$ Can someone point me in the right direction in the deviation of this equality?
• Hi: I wouldn't expect them to be the same because, for the one where you use the result from the Wiki, you're not taking the model into account. The Wiki result would be true for any iid random variable which is normally distributed. The second result-derivation applies due to the model specifics for $x_t$ and $x_t$ is not iid given the model. – mlofton Feb 24 at 18:00
• Hi: Thank you for your comment. I have updated my answer with a simulation study. If you post your original comment and perhaps can give a pointer or two with the remaining part, I'll mark it as the answer. – RVA92 Feb 25 at 7:47
• @mlofton: the linearity of expectation always hold, regardless of the dependence structure between the random variables. – Dayne Feb 26 at 5:50
• Nice derivation and you are absolutely correlation about the correlation structure. My mistake. But when the OP wrote $E(\bar x) = E(x_{i})$, the expectation of the $x_t$ in an AR(1) are not the same unless one refers to the long term unconditional mean which is exactly what you used in your nice derivation. @RVA92: When you wrote the expression for $\bar{x}$, I didn't know that you were referring to the unconditional mean of $x$. The expression does hold in that case. After I send this, I delete my other comments and answer since I was totally confused by the equality. – mlofton Feb 27 at 18:18
• In case, above confuses things even more, I was referring to E(x_t) given the previous $x_t$. That expectation is not constant. because of the dependence structure in the AR(1). For the long term mean, you're absolutely correct. – mlofton Feb 27 at 18:21
There seems to be an error in your calculations. Using a slightly different notation for clarity.
Let the AR(1) process be: $$X_t = \mu + \phi X_{t-1} + e_t$$
Define mean of $$n$$ periods be: | {
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Let the AR(1) process be: $$X_t = \mu + \phi X_{t-1} + e_t$$
Define mean of $$n$$ periods be:
$$\mu_n \equiv \frac{1}{n}\sum\limits_{t=1}^n X_t$$
The expectation operator is linear regardless of dependence structure between the random variables. So, your first expression is correct:
$$\mathbb E(\mu_n) = \mathbb E(X_t) = \frac{\mu}{1-\phi}$$
Second approach, for $$t>1$$:
\begin{align} X_t &= \mu + \phi X_{t-1} + e_t \\ &= \phi^{t-1}X_1+ \sum\limits_{i=0}^{t-2} \Big(\phi^{i}\mu +\phi^{i}e_{t-i}\Big) \tag{1} \end{align}
Using $$(1)$$: \begin{align} n\mu_n &= \sum\limits_{t=1}^n\phi^{t-1}X_1 +\sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2} \Big(\phi^{i}\mu +\phi^{i}e_{t-i}\Big) \\ &= \sum\limits_{t=1}^n\phi^{t-1}X_1 +\sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2} \phi^{i}\mu +\sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2}\phi^{i}e_{t-i} \tag{2} \end{align}
In equation (2) the middle term is non-random and the last term has expected value of $$0$$. So, \begin{align} n\mathbb E(\mu_n) &= \sum\limits_{t=1}^n\phi^{t-1}\mathbb E(X_1) + \sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2} \phi^{i}\mu \\ &= \frac{\mu}{1-\phi}\sum\limits_{t=1}^n\phi^{t-1} + \sum\limits_{t=2}^n \sum\limits_{i=0}^{t-2} \phi^{i}\mu \\ &= \frac{\mu}{1-\phi}\sum\limits_{t=1}^n\phi^{t-1} + \sum\limits_{t=2}^n \bigg( \frac{\mu(1-\phi^{t-1})}{1-\phi}\bigg) \\ &= \frac{\mu}{1-\phi} \bigg( \sum\limits_{t=1}^n\phi^{t-1} + \sum\limits_{t=2}^n (1-\phi^{t-1})\bigg) \\ &=n \frac{\mu}{1-\phi} \end{align}
• Hi Dayne: I got it to match using your approach, so thank you. – RVA92 Feb 26 at 7:46 | {
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# Curves in the Complex Plane
Suppose the continuous real-valued functions $x = x(t),$ $y = y(t),$ $a \leq t \leq b,$ are parametric equations of a curve $C$ in the complex plane. If we use these equations as the real and imaginary parts in $z = x+iy$, we can describe the points $z$ on $C$ by means of a complex-valued function of a real variable $t$ called a parametrization of $C$: $$\label{parcurve} z(t) = x(t) + i y(t), \quad a\leq t\leq b.$$
The point $z(a) = x(a) + i y(a)$ or $z_0 = (x(a), y(a))$ is called the initial point of $C$ and $z(b) = x(b) + iy(b)$ or $z_1 = (x(b), y(b))$ is its terminal point. The expression $z(t) = x(t) + iy(t)$ could also be interpreted as a two-dimensional vector function. Consequently, $z(a)$ and $z(b)$ can be interpreted as position vectors. As $t$ varies from $t = a$ to $t = b$ we can envision the curve $C$ being traced out by the moving arrowhead of $z(t)$. This can be appreciated in the following applet with $0 \leq t\leq 1$.
Press Start to animate. You can move the points to change the curve.
For example, the parametric equations $x = \cos t,$ $y = \sin t,$ $0 \leq t \leq 2\pi,$ describe a unit circle centered at the origin. A parametrization of this circle is $z(t) = \cos t + i \sin t,$ or $z(t) = e^{it},$ $0 \leq t \leq 2\pi$.
Press Start to animate.
## Contours
The notions of curves in the complex plane that are smooth, piecewise smooth, simple, closed, and simple closed are easily formulated in terms of the vector function (\ref{parcurve}). Suppose the derivative of (\ref{parcurve}) is $z'(t) = x'(t) + iy'(t).$ We say a curve $C$ in the complex plane is smooth if $z'(t)$ is continuous and never zero in the interval $a \leq t \leq b.$ As shown in Figure 2, since the vector $z'(t)$ is not zero at any point $P$ on $C$, the vector $z'(t)$ is tangent to $C$ at $P$. Thus, a smooth curve has a continuously turning tangent; or in other words, a smooth curve can have no sharp corners or cusps. See Figure 2. | {
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A piecewise smooth curve $C$ has a continuously turning tangent, except possibly at the points where the component smooth curves $C_1, C_2, \ldots, C_n$ are joined together.
A curve $C$ in the complex plane is said to be a simple if $z(t_1) \neq z(t_2)$ for $t_1 \neq t_2,$ except possibly for $t = a$ and $t = b.$ $C$ is a closed curve if $z(a) = z(b).$
$C$ is a simple closed curve if $z(t_1)\neq z(t_2)$ for $t_1\neq t_2$ and $z(a) = z(b).$
In complex analysis, a piecewise smooth curve $C$ is called a contour or path. We define the positive direction on a contour $C$ to be the direction on the curve corresponding to increasing values of the parameter $t$. It is also said that the curve $C$ has positive orientation. In the case of a simple closed countour $C$, the positive direction corresponds to the counter-clockwise direction. For example, the circle $z(t) = e^{it}$, $0 \leq t \leq 2\pi$, has positive orientation. The negative direction on a contour $C$ is the direction opposite the positive direction. If $C$ has an orientation, the opposite curve, that is, a curve with opposite orientation, is denoted by $−C$. On a simple closed curve, the negative direction corresponds to the clockwise direction. For instance, the circle $z(t) = e^{-it}$, $0 \leq t \leq 2\pi$, has negative orientation.
Press Start to animate. You can change the direction of $C$.
Exercise: There is no unique parametrization for a contour $C$. You should verify that \begin{eqnarray*} z(t)&=&e^{it} =\cos t+i\sin t,\; 0\leq t\leq 2\pi \\ z(t)&=& e^{2\pi it} =\cos \left(2 \pi t\right) +i \sin \left(2 \pi t\right),\; 0\leq t\leq 1\\ z(t)&=&e^{\pi/2 it} =\cos \left( \frac{\pi}{2} t\right)+i \sin \left( \frac{\pi}{2} t\right),\; 0\leq t \leq 4 \end{eqnarray*} are all parametrizations, oriented in the positive direction, for the unit circle $|z| = 1$.
NEXT: Complex Integration | {
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# How To Evaluate Nth Roots | {
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Please try again later. The unit fraction notation used for roots previously may have given you the idea that roots are really the same as powers, only with a unit fraction (one over some number) instead of an integer as the exponent. Rational Exponents The nth root of a number can be expressed by using radical notation or the exponent 1 1n. 1 Evaluate nth Roots and Use Rational Exponents. With millions of qualified respondents, SurveyMonkey Audience makes it easy to get survey responses from people around the world instantly, from almost anyone. These are Euclidean distance, Manhattan, Minkowski distance,cosine similarity and lot more. • The value of a square root can be approximated between integers. (Where did the 2 go? For square roots, the 2 is assumed. Notice that the cursor will stay under the radical sign until you press the right-arrow key (see the last line of the third screen). Active 4 years, 6 months ago. How to find nth term of the sequence ? There is some arrangement or pattern followed in every sequence. If a = bn, then b is an nth root of a. Each root gives a particular exponential solution of Each root gives a particular exponential solution of the differential equation. Whenever numbers are preceded with a radical sign, the numbers are called radicals Whenever numbers are preceded with a radical sign, the numbers are called radicals. Viewed 1k times 3 $\begingroup$ I am working on a problem. View Notes - Unit_3_Notes from MATH Algebra 2 at Central York Hs. Consider the quadratic equation A real number x will be called a solution or a root if it satisfies the equation, meaning. When solving this kind of problem change the original to its corresponding. Secondary_Dim[,nth_Dim]. Any nth root is an exponentiation by 1/n, so to get the square root of 9, you use 9**(1/2) (or 9**0. Showing top 8 worksheets in the category - Nth Root. This feature is not available right now. Reciprocal of a number, including fractions. Here is a function that calculates | {
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available right now. Reciprocal of a number, including fractions. Here is a function that calculates the nth-root properly for negative values, where value is the number which will be rooted by n :. And if we take a product of a bunch of stuff and. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. , in response to message #2 by kim. T d mMnaMdpe i 1w ti WtnhI SIfn xf NiRn 7i6t zeP tPFrFex-ZAMlwgQe4b frRau. Polynomials are used so commonly in algebra, geometry and math in general that Matlab has special commands to deal with them. 1 nth Roots and Rational Exponents 401 nth Roots and Rational Exponents EVALUATING NTH ROOTS You can extend the concept of a square root to other types of roots. 5 Graph Square Root and Cube Root Functions. com is going to be the ideal site to take a look at!. Online graphing calculator (2): Plot your own graph (SVG) · 6. How To Simplify fourth roots. See these links: an example of using division method for finding cube root, and information about the nth root algorithm (or paper-pencil method). Find the quadratic sequences nth term for these 4 sequences which are separated by the letter i iii 7 10 15 22 21 42 iii 2 9 18 29 42 57 iii 4 15 32 55 85 119 iii 5 12 27 50 81 120?. In short, you can make use of the POWER function in Excel to find the nth root of any number. CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you score high on exams. First we develop the square root method in table form and the see that how it is work after that We have developed division method of square root in formula form. The nth root of a number can be expressed by using radical notation or the exponent 1 n. So a number to the two-thirds power is the cube root of the number squared. A square root (√) of number x is one which when | {
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power is the cube root of the number squared. A square root (√) of number x is one which when multiplied by itself gives a value x. The nth root of a number is a number such that if you multiply it by itself (n-1) times you get the number. The most common root is the square root. Rationalize the denominator: Other Kinds of Roots We define the principal nth root of a real number a, symbolized by, as follows:. The optimum frequency ratio of the middle 3rd with the perfect 1st is 1. Exception: If n is odd and the radicand is negative, the principal nth root is negative. real number is a real number. Complex numbers can be written in the polar form z = re^{i\theta}, where r is the magnitude of the complex number and \theta is the argument, or phase. 2 Rewrite expressions involving radicals and rational exponents using the properties of exponents. nth Roots and Rational Exponents EVALUATING NTH ROOTS You can extend the concept of a square root to other types of roots. Here is a function that calculates the nth-root properly for negative values, where value is the number which will be rooted by n :. Engaging math & science practice! Improve your skills with free problems in 'Evaluate nth Roots and Use Rational Exponents' and thousands of other practice lessons. Predict the effects of transformations [f(x + c), f(x) + c, f(cx), and cf(x), where c is a positive or negative real-valued constant] algebraically and graphically, using various methods and tools that may include. exponent 1/n refers to the nth root numbers: 16^1/4 = 4th root of 16 to power of 1. For the elements of X that are negative or complex, sqrt(X) produces complex results. A c PA IlqlC Brzi Bg whxtSs K mrSeLs OeJruv Ne7dz. 245 #517 odds. com brings invaluable answers on nth term calculator, complex numbers and multiplying and dividing and other algebra topics. The keys for the parentheses are above the 8 and 9. 404-405 4-61 every 3rd; 3 Evaluating Nth Roots. Step 3: Write the answer using interval | {
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the 8 and 9. 404-405 4-61 every 3rd; 3 Evaluating Nth Roots. Step 3: Write the answer using interval notation. A General Note: Rational Exponents. For these types of food demand cooking, you have to have a burner or any portable heating accessory. The nth root of a number is the number that would have to be multiplied by itself n times to get the original number. 3 27 1 Solution: a. That is, root(4)(81) = 81^((1)/(4)) Here 81= 3^(4). Principal nth root of a number a, symbolized byn a , where n ≥2 is an integer, is defined as follows: • If n ≥2and even, then a and b must be greater than or equal to 0. 5) to get the cube root, you use 9 ** (1/3) (which we can't write with a simpler fraction), and to get the nth root, 9 ** (1/n). Background. A square root (√) of number x is one which when multiplied by itself gives a value x. 32 = 9 102 = 100 2172 = 47089. The seventh root of 16,384 is 4, as 4 x 4 x 4 x 4 x 4 x 4 x 4 is 16,384. Divisor calculator, simpifying nth roots, download 6th grade math, t 89 calculator base conversion, holt algebra 1 answer key. 1 nth Roots and Rational Exponents Algebra 2 Mrs. You might say. In this problem, we can evaluate by using the following method. To use the calculator simply type any positive number into the 'enter number' box then type in the 'nth root' you want to find. A General Note: Rational Exponents. An nth root can be denoted by a radical symbol with an index. Algebra II Review 6. For example, the tenth root of 59,049 is 3 as 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 is 59,049. Y = nthroot(X,N) returns the real nth root of the elements of X. If the length of p is n+1 then the polynomial is described by:. By primitive expressions, I mean + - * / sqrt, unless there are others that I am missing. EVALUATING NTH ROOTS You can extend the concept of a square root to other types of roots. In mathematics and computing, a root-finding algorithm is an algorithm for finding roots of continuous functions. We often find these type of | {
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algorithm is an algorithm for finding roots of continuous functions. We often find these type of expressions in science text books. The third (optional) argument is a root selector. Partial fractions of repeated roots of degree 2. ⭐️⭐️⭐️⭐️⭐️ Shop for cheap price Brass How To Make 6. But perhaps you don't have a calculator, or you want to impress your friends with the ability to calculate a. Checkpoint Find the indicated real nth root(s) of a. The nth root of a number is a number such that if you multiply it by itself (n-1) times you get the number. Hello Weber School District Parents, Teachers, and Staff, On March 15th, 2019, the server that housed our Wordpress Blogs has been discontinued. Evaluating IRU n = 12, yields, which simplifies to 2 1 or 2. homework help with finding nth roots and rational expressions Oct 26, 2011 · I have no idea how to solve these equations and my teacher is a terrible one. Irrational numbers include the square root, cube root, fourth root, and nth root of many numbers. For instance, 2 is a cube root of 8 because 23 = 8, and 3 is a fourth root of 81 because 34 = 81. Algebrator free, 'add subtract multiply divide' worksheets, properties of exponents lesson plan. Eigenvalueshave theirgreatest importance in dynamic problems. In this resource from CK-12 we look at how to evaluate nth roots. Solve equations using nth roots. Then find the cube root. When we want to find what number was squared, we are finding a square root. They write expressions using rational exponents and radical notation. We could use the nth root in a question like this:. — One real nth root: — a = O One real nth root: — o — One real nth root: — Taking roots of negative numbers (JULL SOC —2 2 Evaluate nth. In mathematics and computing, a root-finding algorithm is an algorithm for finding roots of continuous functions. Later in this section we will see that using exponent 1 n for nth root is compat- ible with the rules for integral exponents that we already know. Evaluate | {
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n for nth root is compat- ible with the rules for integral exponents that we already know. Evaluate Nth Degree Polynomial For A Given Value Of X Apr 8, 2014. 12,921 views subscribe 5. However, this only works for n that are powers of 2. Type in calculator with base and exponent in parentheses 3. We first write the given function as an equation as follows y = √(x - 1) Square both sides of the above equation and simplify y 2 = (√(x - 1)) 2 y 2 = x - 1 Solve for x x = y 2 + 1 Change x into y and y into x to obtain the inverse function. ) in irrational number form. For the form x^(m/n) this is the same as (x^m)*1/n which is also the same thing as the nth root of x^m So for 8^(2/3). and bis positive, there are two real nth roots of b. So that's what I'll multiply onto this fraction. The cube root of 8, then, is 2, because 2 × 2 × 2 = 8. Take the square root, you get 3, which is back where you started. So something, times something, times something, is 8. com and figure out equations, equation and several other math topics. How to Divide Square Roots. Because 3 2 = 9, 2 3 = 8, and (-2) 4 = 16, we say that 3 is a square root of 9, 2 is the cube root of 8, and -2 is a fourth root of 16. Rewrite as a radical 2. Evaluate Nth root of a rational to a correctly rounded float. In a shorter form “b” is the cube root of “a” if b^3 = a. Compare Price and Options of Brass How To Make 6. This module provides a number of objects (mostly functions) useful for dealing with Chebyshev series, including a Chebyshev class that encapsulates the usual arithmetic operations. "Nth Power of Pingala Chanda" by Ranjani Chari, July 2013 This shows a method that was given by Pingala Chanda in 200 BC of raising a number to some power. In short, you can make use of the POWER function in Excel to find the nth root of any number. Title: nth Roots and Rational Exponents 1 Chapter 7. Click the oroblem to show tr Evaluate the expression. Gre-test-prep. 33333 or 10 1/3. Cubes are the result of three | {
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to show tr Evaluate the expression. Gre-test-prep. 33333 or 10 1/3. Cubes are the result of three multiplications. You can do this with other roots as well, but they have to be the same roots. Identify and evaluate square and cube roots. evaluate freplacing vars by their value eval(f) Select Pieces of an Object square/nth root of xsqrt ( ), sqrtn x,n,&z trig functions sin, cos, tan, cotan. Then we’ll use these exact values to answer the above challenges. menu-burger { display: none. cube root of 7 multiplied by the square root of 7 over sixth root of 7 to the power of 5 7-1 1 7 7 to the power of 5 by 3. How are roots and radicals related? Number or variable Written as a square 36 16 81 x. For example, the cube root of 8 is written like this: The index, 3, indicates the radical is a cube. nth root of an even powered radicand and the result is has an odd power, you must take the absolute value of the result to ensure that the answer is nonnegative. Find 16 Use a calculator. for each, give (a) the z-score cutoff (or cutoffs) on the comparison distribution at which the null hypothesis should. This Number Sense Worksheet may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone learn math. 6862x 4 + 46. When you take the square root of a number, you're looking for a number that, when multiplied by itself, results in a given number. The product of two polynomials of degree-bound n is a polynomial of degree-bound 2n. A geometric sequence is a sequence such that any element after the first is obtained by multiplying the preceding element by a constant called the common ratio which is denoted by r. There is a basic formula to follow when taking the integral of an expression with a power 1/(n+1) x^(n+1). More generally, odd roots of negative numbers are typically assumed to be complex. Finding Real nth Roots of a (n > 1 and n is an even integer) Finding Real nth Roots of a (n > 1 and n is an odd integer) | {
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of a (n > 1 and n is an even integer) Finding Real nth Roots of a (n > 1 and n is an odd integer) Evaluating Expressions with Rational Exponents. Please try again later. 86 21 Take fourth roots of each side. 1 nth Roots and Rational Exponents 401 nth Roots and Rational Exponents EVALUATING NTH ROOTS You can extend the concept of a square root to other types of roots. Section 1-5 : Integrals Involving Roots. I CANNOT use a calculator. A c PA IlqlC Brzi Bg whxtSs K mrSeLs OeJruv Ne7dz. If some numbers in x are negative, n must be odd. Using this formula, we will prove that for all nonzero complex numbers $z \in \mathbb{C}$ there exists \$n. 2 in the text. They can be used instead of the roots such as the square root. notebook March 24, 2015 a. Having an nth root of some number is equivalent to taking that number to the 1/n power. guarantees that a zero or root of f(x) lies in (a,b). Press [2nd][ x 2 ] to select a square root and type the expression you would like to evaluate. Write an explicit or recursive formula for the nth term of an arithmetic sequence, given the value of several of its terms. the group we get is 10 and 648 find the no. We find nth term of the sequence in term of n. It looks quite tedious to do by hand, but the algorithm exists for any root and is similar to the square root one. MATLAB represents polynomials as row vectors containing coefficients ordered by descending powers. Rationalizing the Denominator. 23 = 8 53 = 125 1713 = 5000211 4. "Novel Algorithm for 'Nth Root of Number' using Multinomial Expansion" by Vitthal Jadhav, March 2013 This gives a general algorithm to extract the nth root of any number. Is it possible to calculate n complex roots of a given number using Python? I've shortly checked it, and it looks like Python gives me wrong/incomplete answers: (-27. We’ll start with integer powers of $$z = r{{\bf{e}}^{i\theta }}$$ since they are easy enough. The most common root is the square root. The function thus has a branch cut along the | {
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easy enough. The most common root is the square root. The function thus has a branch cut along the negative half real axis. 12,921 views subscribe 5. Notice that the cursor will stay under the radical sign until you press the right-arrow key (see the last line of the third screen). First we develop the square root method in table form and the see that how it is work after that We have developed division method of square root in formula form. whose cube root is less than or equal to 10. Identify and evaluate square and cube roots. Evaluate all powers from left to right. If the length of p is n+1 then the polynomial is described by:. Algebrator free, 'add subtract multiply divide' worksheets, properties of exponents lesson plan. chebyshev)¶ New in version 1. For example, 4 is a square root of 16 because 4 • 4 or 42 = 16. com - id: 77bdd1-YThlY. Thus, we can think of taking the square root of any positive number, x. Fast Fourier Transformation for poynomial multiplication. Later in this section we will see that using exponent 1 n for nth root is compat- ible with the rules for integral exponents that we already know. Introduction to nth Roots; Radical Form vs. Step 1: Set the expression inside the square root greater than or equal to zero. yaymath 118,922 views. Algebra 2 Standard 12. If n = 2, the root is called square root. In this case, the power 'n' is a half because of the square root and the terms inside the square root can be simplified to a complex number in polar form. This is the video about how to evaluate square roots. 2 Simplify expressions in exponential form. We’ve already seen some integrals with roots in them. homework help on finding nth roots and rational exponents Before you read/watch/listen to “If You Can Read This I Can. Let's look at some: 1. Topical and themed;. This tool is used to calculate the output of almost all the mathematical expressions. And the square root of 25 times 3 is equal to the square root of 25 times the square root of 3. | {
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And the square root of 25 times 3 is equal to the square root of 25 times the square root of 3. Unfortunately some improper integrals fails to fall under the scope of these tests but we will not deal with them here. How to Find Roots of Unity. Students will simplify nth roots completely, then they will color or draw on the penguin as indicated by their answer. Evaluate the line integral C of F dr, where C is given by the vector function r (t). A square root is defined as a number which when multiplied by itself gives a real non-negative number called a square. Rewrite the power of 16 as products of power of 8 as much as possible. The advantage of using exponents to express roots is that the rules of exponents can be applied to the expressions. Each root gives a particular exponential solution of Each root gives a particular exponential solution of the differential equation. Evaluating Roots of Monomials To evaluate nth roots of monomials: (where c is the coefficient, and x, y and z are variable expressions) n cxyz n c n 1 n x n 1 n y n 1 n z (c ) ( x ) ( y ) ( z ) 1 n or • Simplify coefficients (if possible) • For variables, evaluate each variable separately. The optimum frequency ratio of the middle 3rd with the perfect 1st is 1. 4 Exponents with negative bases. The most common root is the square root. The answers are that -2 is just one of the three cube roots of -8, and that Mathematica, the computational engine of Wolfram|Alpha, has always chosen the principal root, which is complex valued. Our review of these techniques will focus on the manual entry of formulas, but check out our tutorial on using Excel if you need a refresher on formula entry for core functions. Then f(x) changes sign on [a,b], and f(x) = 0 has at least one root on the interval. In this case, we divided by a negative number, so had to reverse the direction of the inequality symbol. If x is a unit, then it is a (primitive) k-th root of unity modulo n, where k is the multiplicative order of x | {
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unit, then it is a (primitive) k-th root of unity modulo n, where k is the multiplicative order of x modulo n. In this maths tutorial, we introduce exponents / powers and roots using formulas, solved examples and practice questions. when n is an odd integer. Evaluate expressions with rational exponents. ' m Igil s NewVocabulary nth root principal nth root 1 Simplify Radicals A square root of a number is one of two equal factors of that number. The square root is an example of a fractional exponent. 1 Evaluate Nth Roots and Use Rational Exponents. So a number to the two-thirds power is the cube root of the number squared. We find the fourth root in the same way and generalize this for nth roots. For Your Notebook n is even integer. Step 3: Write the answer using interval notation. Powers When we wish to multiply a number by itself we use powers, or indices as they are also called. The cube root is a specific calculation that any radical calculator can perform. 3 27 1 Solution: a. The following table shows some perfect cubes and cube roots. M j DM8a SdPe m ow kistBh6 UIIn fjipnSiFt je Q wG Je Lodm EeRtwriy b. Roshan's Algebra 2 Class Videos -- Based on McDougal Littell's Algebra 2. If the length of p is n+1 then the polynomial is described by:. Start studying 8. 1 Evaluating polynomials at many points Suppose that we want to evaluate a polynomial A(x) = a 0 + a is an nth root of unity (and so. Indeed, one reason for choosing such a transformation for an equation with multiple roots is to eliminate known roots and thus simplify the location of the remaining roots. If the index/root is: No Soluh'oa even & radicand is negative, there is odd & radical is negative, the solution follows the the No matter what the index/root is, if the radicand is POSITIVE, there is ALWAYS a solution!!!. If the sample size calculator says you need more respondents, we can help. Evaluating nth roots Myhre Math MCHS. Roots of unity Properties. The properties of fourth root says that for any | {
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roots Myhre Math MCHS. Roots of unity Properties. The properties of fourth root says that for any positive number of a, its fourth roots are real. exam Numerical Ability Question Solution - I need help please!! I need to evaluate the expressions - nth roots and rational exponents - don't understand - hope I type these right 343^-1/3 729^5/6 (-512)^-2/3 1^5/7 1. Download Presentation 3. One real root:. Evaluate nth roots. 12,921 views subscribe 5. The voice explains how to first plug in the numbers given for each variable in the fractions. i have an exam in the morning an must know how to use it. Evaluate exponent Examples Evaluating nth root on calculator 1. Finding nth Roots: Evaluating nth Root Expressions: way 25 2313 2 26 8 23 8 4throot 34 81 nthroot of a nthroot nra nra radical index Negative Index far a o Indexodd Indexeven bn a 8 27 bn a but46 onereal root tworeal roots Another 8 B d TEB C 3 fzjfzDb 1WE y 2 21 i 20 456 4ft try augno Norealrootsbecauseyoucan't multiplyanumberbyitself4times. College Algebra (10th Edition) answers to Chapter R - Section R. For example, the tenth root of 59,049 is 3 as 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 is 59,049. What test to use? When you're looking at a positive series, what's the best way to determine whether it converges or diverges? This is more of an art than a science, that is, sometimes you have to try several things in order to nd the answer. If then b is the nth root of a ; Example ; Notation ; Index (of the radical) The number n outside of the radical sign ; 3 Writing nth roots as powers and powers as nth roots. Notes on Fast Fourier Transform Algorithms & Data Structures Dr Mary Cryan 1 Introduction The Discrete Fourier Transform (DFT) is a way of representing functions in terms of a point-value representation (a very specific point-value representation). What is a function?. 1 Evaluate nth roots and use rational exponents. For example, the sixth root of 729 is 3 as 3 x 3 x 3 x 3 x 3 x 3 is 729. 86 21 Take fourth | {
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For example, the sixth root of 729 is 3 as 3 x 3 x 3 x 3 x 3 x 3 is 729. 86 21 Take fourth roots of each side. n th Roots. 25 different faces laid out in an A3 poster that can be folded down to a size of a business card. 23 = 8 53 = 125 1713 = 5000211 4. Since 2 = 8 , we say that 2 is the cube root of 8. Of course, the presence of square roots makes the process a little more complicated, but certain rules allow us to work with fractions in a relatively. 404-405 4-61 every 3rd; 3 Evaluating Nth Roots. Since the nth root of a real or complex number z is z1/n, the nth root of r cis θ is r1/n cis θ/n. exponent 1/n refers to the nth root numbers: 16^1/4 = 4th root of 16 to power of 1. CAGR is calculated by taking the Nth root of the total percentage growth rate where N is the Number of Years in the period being considered. Evaluate an expression with complex numbers using an online calculator. So split the number inside the fourth root as the product of two perfect squares and then cancel out the power with the fourth root giving its roots. The shifting nth root algorithm is an algorithm for extracting the nth root of a positive real number which proceeds iteratively by shifting in n digits of the radicand, starting with the most significant, and produces one digit of the root on each iteration, in a manner similar to long division. For example, 81 3 and 3 8 both represent the cube root of 8, and we have 81 3 3 8 2. 8 2/3 = ( 8) 2 = ( 8 2 ) This says the cube root of 8 squared. 23 = 8 53 = 125 1713 = 5000211. Fifth Roots. Times 3 to the 1/5. In addition to square roots, you have other nth roots (eg cube roots). Place tiles equal to the expression to the right of the = in the right workspace. This paper focusesattention on developing a numerical algorithm to determine the digit-by-digit extraction of the nth root of a given positive real number up to any desiredaccuracy. Now we will study higher order roots, such as cube roots. 3c Use the properties of exponents to | {
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Now we will study higher order roots, such as cube roots. 3c Use the properties of exponents to transform expressions for exponential functions. In this post We are discussing about how to find nth last node of single linked list. Just as the square root function is the inverse of the squaring function, these roots are the inverse of their respective power functions. Range variable a. Stack Exchange Network. Plug into calculator On calculator, Solving equations 1. The basic rule is that similar signs multiplied result in a positive answer. Rewrite the power of 16 as products of power of 8 as much as possible. Some can be done quickly with a simple Calculus I substitution and some can be done with trig substitutions. 4999999999999998j) but proper roots should be 3 complex numbers, because every non-zero number has n different complex number nth roots. Of course, typically polynomials have several roots, but the number of roots of a polynomial is never more than its degree. That's one cube root. printLevelorder makes use of printGivenLevel to print nodes at all levels one by one starting from root. Click Create Assignment to assign this modality to your LMS. In general, for an integer n greater than 1, if b^n=a, then b is an n th root of a. We've already seen some integrals with roots in them. In general, for an integer ngreater than 1, if bn= a, then bis an An nth root of ais written as na, where nis the of the radical. Finding square roots and converting them to exponents is a relatively common operation in algebra. When a number has more than one root, the radical sign indicates only the principal, or positive, root. So let's imagine taking 8 to the 1/3 power. Computes the n-th root real numbers of a numeric vector x, while x^(1/n) will return NaN for negative numbers, even in case n is odd. Finding Real nth Roots of a (n > 1 and n is an even integer) Finding Real nth Roots of a (n > 1 and n is an odd integer) Evaluating Expressions with Rational Exponents. | {
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Real nth Roots of a (n > 1 and n is an odd integer) Evaluating Expressions with Rational Exponents. Instruction Manual for Scientific Calculator and then click "Simplify_Radical" nth root: a) Evaluate log 12 245 ==> Input Log a (12; 245. Then we select pairs of the divisors: 2 * 3 * 3 = 18 324/2 = 162 162/2 = 81 We obtain a pair of 2 from the above. Print these charts and use it for homeschooling or classroom purposes. Odd Roots (of variable expressions)* When evaluating odd roots (n is odd) do not use absolute values. I have one copy of the factor 5 in the denominator. Find the indicated real nth root(s) of a negative. Definition of a1/n If n is any positive integer, then a1 n n a, provided that n a is a real number. 3 27 1 Solution: a. VOCABULARY. Socratic Meta Featured Answers To evaluate the #nth# root of a complex number I would first convert it into trigonometric form: See all questions in Roots of. 1 Evaluate Nth Roots and use Rational Exponents Things you should be able to do: - Rewrite radical expressions using rational exponent notation. Math is Fun Curriculum for Algebra 2. We can write as _____. Sometimes you may be smarter than the computer. The bottom number is the nth root. If the length of p is n+1 then the polynomial is described by:. 1 Nth Roots and Rational Exponents Objectives: How do you change a power to rational form and vice versa? How do you evaluate radicals and powers with rational exponents? How do you solve equations involving radicals and powers with rational exponents?. Background. Polynomials are used so commonly in algebra, geometry and math in general that Matlab has special commands to deal with them. 1 Evaluate nth Roots and Use Rational Exponents An Image/Link below is provided (as is) to download presentation. The nth root is the same as the (1/n) power. Then find the cube root. Derivative at a Point Calculator Find the value of a function derivative at a given point. Similarly we studied one method to evaluate the cube root by | {
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a function derivative at a given point. Similarly we studied one method to evaluate the cube root by factor method, but the method of finding cube root of very large numbers by factorizing becomes lengthy and difficult. The symbol indicates an nth root. If one would like to have unique solutions in terms of cosines for output-formatting purposes, then one could do something like. Round your answer to the nearest hundredth. Powers and Roots In this section we’re going to take a look at a really nice way of quickly computing integer powers and roots of complex numbers. He also explains ways that would not be helpful in solving the problem and comparing. Why is the magnitude of the sum of two adjacent nth roots always an 'interesting' number, and what do these numbers have to do with each other? Hot Network Questions Is refusing to concede in the face of an unstoppable Nexus combo punishable?. For example, if the expression is 3 x – 2, place 3 green x tiles and 2 red 1 tiles in one half of the workspace. Exponent Calculator to Calculate Base Raised to nth Power This calculator will calculate the answer of a base number raised to n th power, including exponential expressions having negative bases and/or exponents. 14^2/5 write in radical notation what is (root 5 of 14)^2. roots¶ numpy. | {
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# Metric Spaces: Why $L_\infty$ selects the maximum value
I have a basic question about the metric spaces. There are several metric spaces like $L_1$, $L_2$ to $L_\infty$. The $L_p$ metric is defined by the following equation:
$$d_p(x,y)=\left(\sum_{i=1}^{n}|x_i-y_i|^p \right)^{1/p}$$
Now, when $p = \infty$, then the equation of the matrix is following:
$$d_{\infty}(x,y)=\max_{i=1,2,\ldots, n}|x_i-y_i|$$
Now, I am taking the value of $x$ as $x = [1,2,3]^T$ (Transpose) and compute $L_p$ metric for, $p = 1,2$ and $\infty$.
My question is, why $L_\infty$ metric choose the maximum value.
• i wish you would use $\LaTeX$ so i could copy your equation to save time in answering. please learn the tools. – robert bristow-johnson Mar 3 '16 at 2:53
• I assume you have to go through a mathematical proof of this to fully understand it. Please do not make the mistake and try to explain this result by setting p to higher and higher values - this is an intuitive and often very useful thing to do, but it needs not give you the correct result. – M529 Mar 3 '16 at 9:18
• Shouldn't this migrate to math.SE? – Carl Witthoft Mar 3 '16 at 13:07
• No need for migration, the answer is already on math.SE: math.stackexchange.com/questions/109615/… – Jazzmaniac Mar 3 '16 at 14:19
In the general case, let $$x = (x_1,\dots,x_n)$$ be a finite-length vector (in a finite dimensional space). The finite sequence of absolute values $$|x_i|$$ does attain its maximum (because the sequence is finite), denoted $$M = \max_i |x_i|$$.
Let $$m$$ be the (exact) number of coordinates in $$x = (x_1,\dots,x_n)$$ whose absolute value is equal to $$M$$. Thus, $$1\le m\le n$$. Then, we can lower and upper bound the $$\ell_p$$ norm of $$x$$ as follows: $$(mM^p)^{\frac{1}{p}}\le\ell_p(x)\le (nM^p)^{\frac{1}{p}}\,.$$
Both $$m^\frac{1}{p}$$ and $$n^\frac{1}{p}$$ tend to $$1$$ as $$p\to\infty$$, thus $$\ell_p(x) \to M$$, by the squeeze theorem. | {
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[EDIT] To provide more concrete substance, let us see what happen with your example: $$x=[1,2,3]^T$$ (the transposition does not change the result): $$\|x\|_p = d_p(x,0) = (1^p+2^p+3^p)^{1/p} = 3\times\left(\left(\frac{1}{3}\right)^p+\left(\frac{2}{3}\right)^p+1\right)^{1/p}\,.$$
Both $$\left(\frac{1}{3}\right)^p$$ and $$\left(\frac{2}{3}\right)^p$$ tend to $$0$$ as $$p \to \infty$$. Thus, $$\|x\|_p \to 3$$.
• That's the stuff! :) – Jazzmaniac Mar 3 '16 at 16:53
The $L_p$ norm is
$$d_p(\mathbf{x}, \mathbf{y}) \triangleq \left( \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}}$$
there exists a positive value that is the maximum value:
$$M \triangleq \max_{1 \le i \le n} |x_i - y_i|$$
now, suppose you divide both sides of the $L_p$ norm definition by that positive value,
$$\frac{d_p(\mathbf{x}, \mathbf{y})}{M} = \frac{1}{M} \left( \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}}$$
\begin{align} \frac{d_p(\mathbf{x}, \mathbf{y})}{M} & = \frac{1}{M} \left( \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}} \\ & = \left( \frac{1}{M^p} \right)^{\frac{1}{p}} \left( \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}} \\ & = \left( \frac{1}{M^p} \sum\limits_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}} \\ & = \left( \sum\limits_{i=1}^{n} \frac{1}{M^p} |x_i - y_i|^p \right)^{\frac{1}{p}} \\ & = \left( \sum\limits_{i=1}^{n} \left( \frac{|x_i - y_i|}{M} \right)^p \right)^{\frac{1}{p}} \\ \end{align}
now ask yourself, what will happen to the right-hand side of this equation as $p \to \infty$? all terms, except for the term that is equal to the maximum, will have value of less than 1. they will go to 0 as $p \to \infty$, but the term that has $|x_i - y_i| = M$, that term is equal to 1 and will remain 1 even as $p \to \infty$. | {
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• Robert, what if there are two terms in the sum that are equal to $M$? Shouldn't the sum be equal to 2 in that case? – MBaz Mar 3 '16 at 3:48
• hi Robert, thanks. It's a very good answer. So, Suppose, I have two matrix: x = [1,2,3]T and Y = 0. And, I want to compute, L1, L2 and L-infinity. So, here, Max value (M according to your explanation) would be 3? Right? Just to make sure. Thanks. – Odrisso Mar 3 '16 at 4:11
• This is a very nice and intuitive explaination. However, I suppose it is not 100% mathematically sound, which is why @MBaz asked the question with the factor of 2, if your vector had two entries corresponding to M. I think the problem stems from not obeying the limit from the start on. Getting the factor 1/M into the brackets of the sum is critical and probably the point where it fails when p goes to infinity. – M529 Mar 3 '16 at 9:14
• You could maybe argue that the sum s is bounded between 1 and n. Hence you have lim p->inf s^(1/p) which is 1. – M529 Mar 3 '16 at 9:25
• @David, it doesn't matter if it's trivial. If you make false assumptions about the nature of something and don't explicitly consider the case that these assumptions fail, your argument is wrong. Yes, mathematics is pedantic. That's why it's so useful. – Jazzmaniac Mar 4 '16 at 19:05 | {
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# Number Theory6 Successive numbers no one is prime
#### Amer
##### Active member
is it possible to find a 6 Successive numbers like
x , x+1 , x+2 , x+3 ,x+4 ,x+5 such that one one is prime ?
Thanks
#### chisigma
##### Well-known member
is it possible to find a 6 Successive numbers like
x , x+1 , x+2 , x+3 ,x+4 ,x+5 such that one one is prime ?
Thanks
It is comfortable to verify that $x=5!=120$ satisfies the request... and the reason of that is easy to see...
Kind regards
$\chi$ $\sigma$
#### chisigma
##### Well-known member
Of course $x=5!$ is not the only and neither the 'smallest' solution. Setting $x=90$ You have 7 consecutive non prime numbers...
Kind regards
$\chi$ $\sigma$
#### chisigma
##### Well-known member
Why don't try to generalize the problem: given k, how to compute an n such that n, n+1, n+2,...,n+k are all non prime numbers?...
Kind regards
$\chi$ $\sigma$
#### chisigma
##### Well-known member
Why don't try to generalize the problem: given k, how to compute an n such that n, n+1, n+2,...,n+k are all non prime numbers?...
An easy way to get the result in the particular case where k is prime is based on the consideration that $k|n \implies k|(n+m k)$. Setting $n = 2 \cdot 3 \cdot 5 \cdot ... \cdot k$ we are sure that $n+2,n+3,...,n+k+1$ are all non prime numbers. For example...
$\displaystyle k=11 \implies n=2310 \implies 2312,2313,2314,2315,2316,2317,2318,2319,2320,2321,2322\ \text{are all non prime }$
Although 'easy' this method is often 'excessive' because the effective quantity consecutive non prime numbers can be greater. In the given example 2311 is prime so that the sequence starts at 2312 but 2323,2324,2325,2326,2327,2328,2329,2330,2331 and 2332 are non prime numbers [2333 is prime...] and the effective sequence's length is 20 [not 11]...
Kind regards
$\chi$ $\sigma$
#### Amer
##### Active member
Thanks very much, you are amazing. (f)
#### Amer | {
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$\chi$ $\sigma$
#### Amer
##### Active member
Thanks very much, you are amazing. (f)
#### Amer
##### Active member
An easy way to get the result in the particular case where k is prime is based on the consideration that $k|n \implies k|(n+m k)$. Setting $n = 2 \cdot 3 \cdot 5 \cdot ... \cdot k$ we are sure that $n+2,n+3,...,n+k+1$ are all non prime numbers. For example...
$\displaystyle k=11 \implies n=2310 \implies 2312,2313,2314,2315,2316,2317,2318,2319,2320,2321,2322\ \text{are all non prime }$
Although 'easy' this method is often 'excessive' because the effective quantity consecutive non prime numbers can be greater. In the given example 2311 is prime so that the sequence starts at 2312 but 2323,2324,2325,2326,2327,2328,2329,2330,2331 and 2332 are non prime numbers [2333 is prime...] and the effective sequence's length is 20 [not 11]...
Kind regards
$\chi$ $\sigma$
nice one I get it
$$n +2 = 2.3.4...k +2 = 2(3.4...k +1 )$$ not prime
$$n+3 = 2.3.4...k + 3 = 3(2.4...k+1)$$ not prime
Thanks
#### soroban
##### Well-known member
Hello, Amer!
Is it possible to find a 6 consecutive integers numbers like
x , x+1 , x+2 , x+3 ,x+4 ,x+5 such that not one is prime? . Yes!
One solution is: .$$x \:=\:7!+2$$
. . $$\begin{array}{c}7!+2\text{ is divisible by 2} \\ 7!+3\text{ is divisible by 3} \\ 7!+4\text{ is divisible by 4} \\ 7!+5\text{ is divisible by 5} \\ 7!+6 \text{ is divisible by 6} \\ 7!+7\text{ is divisible by 7} \\ \end{array}$$
There is a simpler (and much longer) list:
. . $$\begin{array}{ccc} 114 &=& 2\cdot57 \\ 115 &=& 5\cdot 23 \\ 116 &=& 2\cdot 58 \\ 117 &=& 3\cdot39 \\ 118 &=& 2\cdot59 \\ 119 &=& 7\cdot17 \\ 120 &=& 2\cdot60 \\ 121 &=& 11\cdot11 \\ 122 &=& 2\cdot61 \\ 123 &=& 3\cdot41 \\ 124 &=& 2\cdot62 \\ 125 &=& 5\cdot25 \\ 126 &=& 2\cdot63 \end{array}$$
#### Amer
##### Active member
Hello, Amer!
One solution is: .$$x \:=\:7!+2$$ | {
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#### Amer
##### Active member
Hello, Amer!
One solution is: .$$x \:=\:7!+2$$
. . $$\begin{array}{c}7!+2\text{ is divisible by 2} \\ 7!+3\text{ is divisible by 3} \\ 7!+4\text{ is divisible by 4} \\ 7!+5\text{ is divisible by 5} \\ 7!+6 \text{ is divisible by 6} \\ 7!+7\text{ is divisible by 7} \\ \end{array}$$
There is a simpler (and much longer) list:
. . $$\begin{array}{ccc} 114 &=& 2\cdot57 \\ 115 &=& 5\cdot 23 \\ 116 &=& 2\cdot 58 \\ 117 &=& 3\cdot39 \\ 118 &=& 2\cdot59 \\ 119 &=& 7\cdot17 \\ 120 &=& 2\cdot60 \\ 121 &=& 11\cdot11 \\ 122 &=& 2\cdot61 \\ 123 &=& 3\cdot41 \\ 124 &=& 2\cdot62 \\ 125 &=& 5\cdot25 \\ 126 &=& 2\cdot63 \end{array}$$
Thanks, and Hello
we can make a list with k numbers which are not prime like that
$$k! + i$$ i=1,...,k | {
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# Discrepancy when calculating LTI system output using inverse z-Transform
I'm given a difference equation, $y[n]-0.4y[n-1]=x[n]$, and asked to find the natural response $y_n[n]$, forced response $y_f[n]$ and complete response $y[n]$ if $x[n]=4 (0.25)^nu[n]$ and $y[0]=0$.
By one approach I've read, $$\text{characteristic equation:}\quad z-0.4=0 \quad\therefore y_n[n]=A(0.4)^n \\ \\ y_f[n]=B(0.25)^n \quad\text{substitute back into diffeq:}\\ B(0.25)^n -0.4B(0.25)^{n-1}=B(0.25)^n - \frac{0.4}{0.25}B(0.25)^n= 4(0.25)^nu[n]\\ \therefore B=-\frac{20}{3}, \quad y_f[n]=-\frac{20}{3}(0.25)^nu[n]\\ \therefore y[n] = A(0.4)^n-\frac{20}{3}(0.25)^nu[n]\\ y[0]=0 \rightarrow A=\frac{20}{3} \\ y[n]= \frac{20}{3}(0.4)^n-\frac{20}{3}(0.25)^nu[n]\\$$ Great! Alternatively, I should be able to determine $y[n]$ by taking the inverse z Transform of $Y(z)=H(z)X(z)$, which in this case is (I'm pretty sure) Y(z)= \frac{z}{z-0.4}\cdot 4 \frac{z}{z-0.25}\\ \begin{align} \frac{Y(z)}{z} &= \frac{4z}{(z-0.4)(z-0.25)}=\frac{C}{z-0.4}+\frac{D}{z-0.25}\\ &=\frac{32}{3} \frac{1}{z-0.4} - \frac{20}{3}\frac{1}{z-0.25}\\ \end{align}\\ \therefore Y(z)=\frac{1}{3} \left[\frac{32z}{z-0.4} - \frac{20z}{z-0.25}\right]\\ \therefore y[n]= \left[ \frac{32}{3}(0.4)^n - \frac{20}{3}(0.25)^n \right] u[n] which is almost the same, but not quite.
Where did I go wrong? Also, in the second approach, how do you take into account the initial condition of $y[0]=0$? And thirdly, in the first approach, is the first term of $y[n]$ multiplied by $u[n]$ or not? If so, how do you show this?
(for anyone wondering, this is not homework for a course, just self study...)
• Don't you mean $y[-1]=0$ instead of $y[0]=0$? – Matt L. Aug 20 '15 at 6:54
• The question says y[0]=0 but I guess its mistaken. But its not clear to me why y[0] can't be zero... – Westerley Aug 20 '15 at 17:02 | {
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The way to solve such problems is to use the unilateral $\mathcal{Z}$-transform, which allows you to take initial conditions into account. Note that the unilateral $\mathcal{Z}$-transform of $y[n-1]$ is
$$\mathcal{Z}\{y[n-1]\}(z)=z^{-1}Y(z)+y[-1]\tag{1}$$
where $Y(z)$ is the $\mathcal{Z}$-transform of $y[n]$. Using $(1)$ you can transform the given difference equation $y[n]-\frac25 y[n-1]=x[n]$, resulting in
$$Y(z)=\frac{X(z)}{1-\frac{2}{5}z^{-1}}+\frac{\frac{2}{5}\cdot y[-1]}{1-\frac{2}{5}z^{-1}}\tag{2}$$
With $X(z)$ given by
$$X(z)=\frac{4}{1-\frac14z^{-1}}\tag{3}$$
Eq. $(2)$ becomes (after partial fraction expansion)
$$Y(z)=\left(\frac{32}{3}+\frac{2y[-1]}{5}\right)\frac{1}{1-\frac{2}{5}z^{-1}}-\frac{20}{3}\frac{1}{1-\frac14z^{-1}}\tag{4}$$
From $(4)$, the resulting output sequence is
$$y[n]=\left(\frac{32}{3}+\frac{2y[-1]}{5}\right)\left(\frac25\right)^nu[n]-\frac{20}{3}\left(\frac14\right)^nu[n]\tag{5}$$
Eq. $(5)$ is the general form of the output sequence of the given system with the given input sequence. The value $y[-1]$ is the initial condition of the system, i.e. the state the system is in right before the input signal starts. With $y[-1]=0$ you obtain the same solution as you obtained using the inverse $\mathcal{Z}$-transform (neglecting any initial conditions by using $y[n-1]\Longleftrightarrow z^{-1}Y(z)$).
With the given constraint $y[0]=0$, you can obtain the required value of $y[-1]$ by considering the difference equation at $n=0$:
$$y[0]-\frac25 y[-1]=x[0]=4$$
which gives $y[-1]=-10$. Plugging that value into Eq. $(5)$ finally gives the result
$$y[n]=\frac{20}{3}\left(\frac25\right)^nu[n]-\frac{20}{3}\left(\frac14\right)^nu[n]\tag{6}$$
As Matt L pointed out, your initial condition is wrong. y[0] is NOT 0, it's 4. This is obvious from the difference equation. Once you use y[0] = 4, both results come out to be the same. | {
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• As mentioned in my comment to @matt, the question does say y[0]=0 but it's likely wrong... but its not clear to me why y[0] CAN'T be zero... I see that x[0] is 4, but why can't the natural response at n=0 cancel out that? – Westerley Aug 20 '15 at 17:03
• Note that $y[0]$ can be zero by an appropriate choice of the initial condition $y[-1]$ (see my answer). – Matt L. Aug 21 '15 at 7:48 | {
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# Number of ways to divide $10$ children into two teams of $5$?
Consider the following two examples from A First Course in Probability by Sheldon Ross,
EXAMPLE 5b Ten children are to be divided into an A team and a B team of 5 each. The A team will play in one league and the B team in another. How many different divisions are possible?
Solution by the Book itself There are $$\frac{10!}{5!5!} = 252$$ possible divisions.
My Solution and Reasoning We first choose five children which can be done in $$C(10, 5)$$ possible ways. Then we decide which team(either A or B) to assign these children to, which can be done in $$2!$$ ways. Finally we assign the remaining five children to the remaining team, depending on the second level(if the first five children were assigned to A, then the remaining five should be assigned to B, and if the first five were assigned to B, the remaining to A). Therefore this level has only $$1$$ possible way, and hence the final answer is $$C(10, 5) \times 2! \times 1 = 252 \times 2 \times 1 = 504$$.
EXAMPLE 5c In order to play a game of basketball, 10 children at a playground divide themselves into two teams of 5 each. How many different divisions are possible?
Solution by the Book itself Note that this example is different from Example 5b because now the order of the two teams is irrelevant. That is, there is no A and B team, but just a division consisting of 2 groups of 5 each. Hence, the desired answer is
$$\frac{10!/(5!5!)}{2!} = 126$$
My Solution and Reasoning First we choose 5 children for one team which can be done is $$C(10, 5)$$ ways and then we put the remaining children in the other team, which is done in $$1$$ way. Thus the answer is $$C(10, 5) \times 1 = 252$$, which can also be obtained by dividing the answer to Example 5b by 2, or in other words, removing the order of the groups in Example 5b.
There seems to be a conceptual misunderstanding in my way of thinking, but what is it?
PARTICULAR PROBLEM | {
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PARTICULAR PROBLEM
For your particular problem, for labeled teams
• label the children $$0\; through\;$$9
• choose $$5$$ for team $$A$$, say $$13579$$, team $$B$$ automatically $$02468$$
• This is different from $$02468$$ in Team $$A$$ and $$13579$$ for team $$B$$, so total $$\binom{10}5$$ choices
But if teams are unlabeled $$[13579-02468] \equiv [02468-13579]$$
hence division by $$2$$
GENERAL GUIDANCE
It is important to have a clear concept over combinatorics of teams depending on whether they are labeled or unlabeled
To take a more complex example for greater clarity, suppose you are to choose $$4$$ teams of $$3$$ each from $$12$$ players
• If the teams are labeled, eg Horses, Greyhounds , Panthers, etc
the answer is $$\binom{12}3\binom93\binom63\binom33$$
which can be variously written as $$\binom{12}{3,3,3,3}\; or\; \dfrac{12!}{3!3!3!3!}$$
• If the teams are unlabeled, we shall have to divide by $$4!$$, to remove permutations between identical teams
• An important point to note is that unlabeled teams in effect become labeled if sizes differ, or composition differs (eg boys' team or girls' team)
Note that in both casees your answer is doubled from the book's solution so you should question whether you really ought to multiply by $$2!=2$$.
Let's consider a simpler case of 4 children $$\{1,2,3,4\}$$ and the teams are of size 2. Then all possible options for Team A are $$\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}$$ giving a total of $$\frac{4!}{2!2!}=6$$ options. What you are doing is counting each option twice; Say when you count $$\{1,2\}$$, you count once for this being assigned to $$A$$ and once for $$B$$. However, note that the event of $$\{1,2\}$$ belonging to $$B$$ is exactly the event that $$\{3,4\}$$ belongs to $$A$$ which we already count. Hence we should not multiply by 2. | {
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# How to evaluate the sum $\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+…+\frac{1}{n}\right)$
How to evaluate the sum: $$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$$ Can anyone help me,I really appreciate it.
• is there a source for this sum? Or, just a practice problem? – karakfa Oct 20 '17 at 15:05
• @MarkViola Mathematica gives the sum of the series. I don't know what is the trick to evaluate it. I think this is a good problem. :) – Ixion Oct 20 '17 at 15:26
• Partial fraction decomposition – JohnColtraneisJC Oct 20 '17 at 15:27
• Since $\displaystyle\frac {\pi^2}{12}=\sum_{r=1}^\infty \frac 1{2r^2}$, and $\displaystyle\ln 2=\sum_{r=1}^\infty \frac {(-1)^{r+1}}{r}$ is it possible to transform the given summation $\displaystyle \sum_{n=1}^\infty \sum_{r=1}^n \frac 1{(2n+1)(2n+2)r}$into $\displaystyle\sum_{r=1}^\infty \frac 1{2r^2}-\left(\sum_{r=1}^\infty \frac {(-1)^{r+1}}{r}\right)^2$? If so, then we are done. – hypergeometric Oct 21 '17 at 15:35
• I am voting for reopening this question since, despite a lack of efforts from the OP, it led to a number of collateral questions, and I think it is correct to leave this original question as a reference. – Jack D'Aurizio Oct 26 '17 at 15:57 | {
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A different view on the same problem: $$\sum_{n\geq 1}\frac{x^n}{n}=-\log(1-x),\qquad \sum_{n\geq 1}H_n x^n = \frac{-\log(1-x)}{1-x} \tag{A}$$ $$\sum_{n\geq 1}H_n x^{2n} = \frac{-\log(1-x^2)}{1-x^2}\tag{B}$$
$$\begin{eqnarray*}\sum_{n\geq 1}H_n\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right) &=& \int_{0}^{1}\frac{-\log(1-x^2)}{1+x}\,dx\\&=&-\tfrac{1}{2}\log^22+\int_{0}^{1}\frac{-\log(1-x)}{1+x}\,dx\\&=&-\tfrac{1}{2}\log^22+\int_{0}^{1}\frac{-\log(x)}{2-x}\,dx\tag{C}\end{eqnarray*}$$ and by differentiation under the integral sign, the last integral is related to the series $$\sum_{n\geq 1}\frac{1}{n^2 2^n}=\text{Li}_2\left(\tfrac{1}{2}\right)\stackrel{(*)}{=}\tfrac{\pi^2}{12}-\tfrac{\log^2 2}{2} \tag{D}$$ where $(*)$ follows from the dilogarithm reflection formula, proved here.
It is evident that $$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right) = \frac{\pi^2}{12} - \ln^{2}2$$ and can be evaluated by following the pattern: | {
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Consider the series $$S(x) = \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+2}}{(2n+1)(2n+2)}$$ which upon differentiation leads to $S(0) = 0$, $S'(0) = 0$, \begin{align} S(x) &= \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+2}}{(2n+1)(2n+2)} \\ S'(x) &= \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+1}}{(2n+1)} \\ S''(x) &= \sum_{n=1}^{\infty} H_{n} \, x^{2n} = - \frac{\ln(1- x^2)}{1-x^2}. \end{align} Now, $$- 2 \, S''(x) = \frac{\ln(1-x)}{1-x} + \frac{\ln(1-x)}{1+x} + \frac{\ln(1+x)}{1-x} + \frac{\ln(1+x)}{1+x}$$ which, upon integration, leads to \begin{align} - 4 \, S'(x) &= \ln^{2}(1 + x) - \ln^{2}(1-x) + 2 \, Li_{2}\left(\frac{1-x}{2}\right) - 2 \, Li_{2}\left(\frac{1+x}{2}\right) + \ln4 \, \ln\left(\frac{1+x}{1-x}\right). \end{align} Integrating again leads to $S(x)$. The integrals \begin{align} \int_{0}^{x} \ln^{2}(1-t) \, dt &= (x-1) \, (\ln^{2}(1-x) - 2 \ln(1-x) + 2) + 2 \\ \int_{0}^{x} \ln^{2}(1+t) \, dt &= (x+1) \, (\ln^{2}(1+x) - 2 \ln(1+x) + 2) - 2 \\ \int_{0}^{x} \ln\left(\frac{1+t}{1-t}\right) \, dt &= x \, \ln\left(\frac{1+x}{1-x}\right) + \ln(1-x^2) \\ \int_{0}^{x} Li_{2}\left(\frac{1+t}{2}\right) \, dt &= (1+x) \, Li_{2}\left(\frac{1+x}{2}\right) + x \, \ln\left(\frac{1-x}{2}\right) - \ln(1-x) -x - Li_{2}\left(\frac{1}{2}\right) \\ \int_{0}^{x} Li_{2}\left(\frac{1-t}{2}\right) \, dt &= (x-1) \, Li_{2}\left(\frac{1-x}{2}\right) + (x+1) \, \ln\left(\frac{1+x}{2}\right) -x - Li_{2}\left(\frac{1}{2}\right) + \ln2 \end{align} are needed for the evaluation. Once $S(x)$ is determined set $x=1$ to obtain $$S(1) = \sum_{n=1}^{\infty}\frac{H_{n}}{(2n+1)(2n+2)} = \frac{\pi^2}{12} - \ln^{2}2$$ | {
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Following Leucipus' answer, I will use different approach to evaluate $S(1)$. Note \begin{eqnarray} S=S(1)&=&\int_0^1\int_0^xS''(x)dxt\\ &=&-\int_0^1\int_0^x\frac{\ln(1-x^2)}{1-x^2}dxdt\\ &=&-\int_0^1\int_x^1\frac{\ln(1-x^2)}{1-x^2}dtdx\\ &=&-\int_0^1(1-x)\frac{\ln(1-x^2)}{1-x^2}dx\\ &=&-\int_0^1\frac{\ln(1-x^2)}{1+x}dx\\ &=&-\int_0^1\frac{\ln(1-x)}{1+x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx\\ &=:&-I_1-I_2 \end{eqnarray} Now $$I_2=\int_0^1\ln(1+x)d\ln(1+x)=\frac{1}{2}\ln^2(1+x)\bigg|_0^1=\frac12\ln^22.$$ For $I_1$, under $t=1-x$, one has \begin{eqnarray} \int\frac{\ln t}{2-t}dt&=&-\int_0^1\ln xd\ln(2-t)\\ &=&-\ln x\ln(2-t)+\int\frac{\ln(2-t)}{t}dt\\ &=&-\ln x\ln(2-t)+\int\frac{\ln2+\ln(1-\frac12t)}{t}dt\\ &=&-\ln x\ln(2-t)+\ln2\ln t+Li_2(\frac t2)+C \end{eqnarray} \begin{eqnarray} I_1&=&\int_0^1\frac{\ln t}{2-t}dt=-\int_0^1\ln xd\ln(2-t)\\ &=&-\ln x\ln(2-t)+\ln2\ln t+Li_2(\frac t2)\bigg|_0^1 &=&-\frac{\pi^2}{12}+\frac12\ln^22. \end{eqnarray} Thus $$S=\frac{\pi^2}{12}-\ln^22.$$ | {
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# Why is $\{0, 2\}$ a subgroup of $\mathbb Z_4$?
I feel like this should be obvious but why is $$\{0, 2\}$$ a subgroup of $$\mathbb Z_4$$? So, $$\langle 2\rangle=\{0,2\}$$. Shouldn't this set contain the inverse ($$-2$$)? Or does it have to do with the fact that $$(-2)(-2)=4=0$$? Please advise.
• What is the difference between $2$ and "$-2$"? Recall, the minus sign merely means "the additive inverse of" which does not necessarily need to appear in other representations of the element. – JMoravitz Feb 19 '20 at 17:36
• Welcome to math SE. Have a look at mathjax for your mathematical expressions. – Alain Remillard Feb 19 '20 at 17:37
• As for "does it have to do with the fact that $(-2)(-2)=4=0$" No, it doesn't. It has to do with the fact that $2 + 2 = 0$. Addition is what is important here, not multiplication. You will find that the additive inverse of $0$ is $0$, the additive inverse of $1$ (which you might when convenient decide to notate as $-1$) is equal to $3$, the additive inverse of $2$ (which you might when convenient decide to notate as $-2$) is equal to $2$, and so on... – JMoravitz Feb 19 '20 at 17:37
• There's a problem with the title of the question. 2 is not a subgroup of $\mathbb Z_4$, it's an element. I guess you mean the subgroup generated by 2. – Shatabdi Sinha Feb 19 '20 at 17:40
• After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – Shaun Feb 19 '20 at 18:46
$$\langle 2\rangle = \{0, 2\}$$ is a subgroup of the group $$\mathbb Z_4 = \{0, 1, 2, 3\}$$ under modular arithmetic, modulo $$4$$. | {
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The identity of this group is $$0$$, and because $$2+2 \equiv 0 \pmod 4$$, it has order two, and hence $$2$$ generates a group (subgroup) of order 2. In fact, the additive inverse of $$2$$ is $$2$$.
That is, $$\langle 2 \rangle = \{0, 2\} \leq \{0, 1, 2, 3\} = \mathbb Z_4$$.
• Good approach to teach! – Mikasa Mar 20 '20 at 8:36
The elements of $$\Bbb Z_4$$ are not technically $$0$$, $$1$$, $$2$$ and $$3$$; rather, they are equivalence classes of integers with respect to the divisibility of their differences by $$4$$, like so: $$[a]_4:=\{b\in\Bbb Z: 4\mid a-b\}.$$ The operation of the group is defined by $$[x]_4+_4[y]_4=[x+y]_4$$.
Thus, since $$4\mid (-2)+2=0$$, we have $$[-2]_4=[2]_4$$.
• Why the downvote? – Shaun Feb 19 '20 at 19:00
• No, it isn't, @amWhy. – Shaun Mar 11 '20 at 16:58
• But $-2\equiv 2\pmod{4}$. – Shaun Mar 11 '20 at 17:02 | {
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# What are the chances rolling 6, 6-sided dice that there will be a 6?
More generally, what is the probability that n n-sided dice will turn up at least one side with the highest number ("n")?
My crude thinking is that each side has 1/6 probability, so that 6 of them (1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6) equals 100% of the time, but this obviously is not the case.
2 coins getting one T (say), has 3/4 of the time that at least one will be a T (for every 2 coin tosses). But this is 75% for a 2d2 dice throw, what is the general formula for NdN?
• Pretty sure we have a couple of questions already that cover this. Jul 26, 2020 at 22:59
• Does this answer your question? How often do you have to roll a 6-sided die to obtain every number at least once? Jul 27, 2020 at 19:07
• $m = n-1$, that is $P = 1 - \left(\frac{n-1}{n}\right)^n$ Jul 28, 2020 at 10:11
• @NathanChappell: Thanks! That's the cleanest solution yet! Jul 31, 2020 at 18:11
Probability of a die not turning up $$n$$ is $$1-1/n$$. Probability of not turning up $$n$$ in any of the $$n$$ dice is $$(1-1/n)^n$$. If you subtract this from $$1$$, it'll be probability of at least one $$n$$ turning up when you throw $$n$$ dice, i.e.
$$p=1-(1-1/n)^n\rightarrow 1-e^{-1}$$ as $$n$$ goes to $$\infty.$$ | {
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$$p=1-(1-1/n)^n\rightarrow 1-e^{-1}$$ as $$n$$ goes to $$\infty.$$
• In the case of trying to roll at least one 6 on 6 6-sided dice rolls (the title question), this gives us approximately a 66.5% chance. Jul 27, 2020 at 6:20
• Nathan Chappell has a simpler formula in the comment to the question above. Unfortunately, I don't have the time nor is it immediately apparent how to reduce your formula to his. Jul 31, 2020 at 18:10
• Inside of the parantheses is the same if you pay attention closely @Marcos $$\left(1-\frac{1}{n}\right)=\left(\frac{n-1}{n}\right)$$ Jul 31, 2020 at 21:50
• @gunes: Ah, yes, thanks. I came to the same conclusion after writing that comment -- an equivalency I hadn't seen before in algebra. Rather, I hadn't guessed to make 1 = n/n and then re-arrange terms. It's a common algebraic trick, but I'd never had the context to do it for myself. Aug 1, 2020 at 4:11
• How do you arrive at the conclusion in your second sentence? That is FAR OUT. I've not seen that in combinatorics. If you can explain it, I'll give you the credit. Nevermind... I think I see it now. I'll give you the credit... Aug 1, 2020 at 16:40
The event $$A:=$$ "at least one die turns up on side $$n$$" is the complement of the event $$B:=$$ "all dice turn up on non-$$n$$ sides".
So $$P(A)=1-P(B)$$. What's $$P(B)$$?
All dice are independent, so
$$P(\text{all n dice turn up on non-n sides}) = P(\text{a single die turns up non-n})^n = \bigg(\frac{n-1}{n}\bigg)^n.$$
So
$$P(A) = 1-\bigg(\frac{n-1}{n}\bigg)^n.$$
Trust but verify. I like to do so in R:
> nn <- 6
> n_sims <- 1e5
> sum(replicate(n_sims,any(sample(1:nn,nn,replace=TRUE)==nn)))/n_sims
[1] 0.66355
> 1-((nn-1)/nn)^nn
[1] 0.665102
Looks good. Try this with other values of nn. Here is a plot:
nn <- 2:100
plot(nn,1-((nn-1)/nn)^nn,type="o",pch=19,ylim=c(1-1/exp(1),1))
abline(h=1-1/exp(1),col="red")
We note how our probability in the limit is | {
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We note how our probability in the limit is
$$P(A) = 1-\bigg(\frac{n-1}{n}\bigg)^n =1-\bigg(1-\frac{1}{n}\bigg)^n \to 1-\frac{1}{e}\approx 0.6321206 \quad\text{as }n\to\infty,$$
Answers by @StephanKolassa (+1) and @gunes (+1) are both fine. But this problem can be solved with reference to binomial and Poisson distributions as follows:
If $$X_n$$ is the number of ns seen in $$n$$ rolls of a fair $$n$$-sided die, then $$X_n \sim \mathsf{Binom}(n, 1/n),$$ so that $$P(X_n \ge 1) = 1 - P(X_n = 0)= 1-(1-1/n)^n.$$
As $$n\rightarrow\infty,$$ one has $$X_n \stackrel{prob}{\rightarrow} Y \sim\mathsf{Pois}(\lambda=1),$$ with $$P(Y \ge 1) = 1 - P(Y = 0) = 1 - e^{-1}.$$
The answer can be arrived at by purely counting the described events as well, although the accepted answer is more elegant. We'll consider the case of the die, and hopefully the generalization is obvious. We'll let the event space be all sequences of numbers from $$\{1,2,...,6\}$$ of length $$6$$. Here are a few examples (chosen at random):
3 2 3 5 6 1
1 1 2 5 2 4
1 2 1 1 6 3
4 4 3 3 4 2
6 1 1 6 3 4
6 3 5 4 5 1
The point is, our space has a total of $$6^6$$ events, and due to independence we suppose that any one of them is as probable as the other (uniformly distributed). We need to count how many sequences have at least one $$6$$ in them. We partition the space we are counting by how many $$6$$'s appear, so consider the case that exactly one $$6$$ appears. How many possible ways can this happen? The six may appear in any position (6 different positions), and when it does the other 5 postions can have any of 5 different symbols (from $$\{1,2,...,5\}$$). Then the total number of sequences with exactly one $$6$$ is: $$\binom{6}{1}5^5$$. Similarly for the case where there are exactly two $$6$$'s: we get that there are exactly $$\binom{6}{2}5^4$$ such sequences. Now it's time for fun with sums:
$$\sum_{k=1}^6 \binom{6}{k}5^{6-k} = \sum_{k=0}^6 \binom{6}{k}5^{6-k}1^k - 5^6 = (5+1)^6 - 5^6$$ | {
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$$\sum_{k=1}^6 \binom{6}{k}5^{6-k} = \sum_{k=0}^6 \binom{6}{k}5^{6-k}1^k - 5^6 = (5+1)^6 - 5^6$$
To get a probability from this count, we divide by the total number of events:
$$\frac{6^6 - 5^6}{6^6} = 1 - (5/6)^6 = 1 - (1-1/6)^6$$
I think that this generalizes pretty well, since for any $$n$$ other than $$6$$, the exact same arguments holds, only replace each occurrence of $$6$$ with $$n$$, and $$5$$ with $$n-1$$.
It's also worth noting that this number $$5^6 = \binom{6}{0}5^6$$ is the contribution of sequences in which no $$6$$ occurs, and is much easier to calculate (as used in the accepted answer).
I found BruceET's answer interesting, relating to the number of events. An alternative way to approach this problem is to use the correspondence between waiting time and number of events. The use of that would be that the problem will be able to be generalized in some ways more easily.
### Viewing the problem as a waiting time problem
This correspondence, as for instance explained/used here and here, is
For the number of dice rolls $$m$$ and number of hits/events $$k$$ you get: $$\begin{array}{ccc} \overbrace{P(K \geq k| m)}^{\text{this is what you are looking for}} &=& \overbrace{P(M \leq m|k)}^{\text{we will express this instead}} \\ {\small\text{\mathbb{P} k or more events in m dice rolls}} &=& {\small\text{\mathbb{P} dice rolls below m given k events}} \end{array}$$
In words: the probability to get more than $$K \geq k$$ events (e.g. $$\geq 1$$ times rolling 6) within a number of dice rolls $$m$$ equals the probability to need $$m$$ or less dice rolls to get $$k$$ such events.
This approach relates many distributions.
Distribution of Distribution of
Waiting time between events number of events
Exponential Poisson
Erlang/Gamma over/under-dispersed Poisson
Geometric Binomial
Negative Binomial over/under-dispersed Binomial | {
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So in our situation the waiting time is a geometric distribution. The probability that the number of dice rolls $$M$$ before you roll the first $$n$$ is less than or equal to $$m$$ (and given a probability to roll $$n$$ equals $$1/n$$) is the following CDF for the geometric distribution:
$$P(M \leq m) = 1-\left(1-\frac{1}{n}\right)^m$$
and we are looking for the situation $$m=n$$ so you get:
$$P(\text{there will be a n rolled within n rolls}) = P(M \leq n) = 1-\left(1-\frac{1}{n}\right)^n$$
### Generalizations, when $$n \to \infty$$
The first generalization is that for $$n \to \infty$$ the distribution of the number of events becomes Poisson with factor $$\lambda$$ and the waiting time becomes an exponential distribution with factor $$\lambda$$. So the waiting time for rolling an event in the Poisson dice rolling process becomes $$(1-e^{-\lambda \times t})$$ and with $$t=1$$ we get the same $$\approx 0.632$$ result as the other answers. This generalization is not yet so special as it only reproduces the other results, but for the next one I do not see so directly how the generalization could work without thinking about waiting times.
### Generalizations, when dices are not fair
You might consider the situation where the dice are not fair. For instance one time you will roll with a dice that has 0.17 probability to roll a 6, and another time you roll a dice that has 0.16 probability to roll a 6. This will mean that the 6's get more clustered around the dice with positive bias, and that the probability to roll a 6 in 6 turns will be less than the $$1-1/e$$ figure. (it means that based on the average probability of a single roll, say you determined it from a sample of many rolls, you can not determine the probability in many rolls with the same dice, because you need to take into account the correlation of the dice) | {
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So say a dice does not have a constant probability $$p = 1/n$$, but instead it is drawn from a beta distribution with a mean $$\bar{p} = 1/n$$ and some shape parameter $$\nu$$
$$p \sim Beta \left( \alpha = \nu \frac{1}{n}, \beta = \nu \frac{n-1}{n} \right)$$
Then the number of events for a particular dice being rolled $$n$$ time will be beta binomial distributed. And the probability for 1 or more events will be:
$$P(k \geq 1) = 1 - \frac{B(\alpha, n + \beta)}{B(\alpha, \beta)} = 1 - \frac{B(\nu \frac{1}{n}, n +\nu \frac{n-1}{n})}{B(\nu \frac{1}{n}, n +\nu \frac{n-1}{n})}$$
I can verify computationally that this works...
### compute outcome for rolling a n-sided dice n times
rolldice <- function(n,nu) {
p <- rbeta(1,nu*1/n,nu*(n-1)/n)
k <- rbinom(1,n,p)
out <- (k>0)
out
}
### compute the average for a sample of dice
meandice <- function(n,nu,reps = 10^4) {
sum(replicate(reps,rolldice(n,nu)))/reps
}
meandice <- Vectorize((meandice))
### simulate and compute for variance n
set.seed(1)
n <- 6
nu <- 10^seq(-1,3,0.1)
y <- meandice(n,nu)
plot(nu,1-beta(nu*1/n,n+nu*(n-1)/n)/beta(nu*1/n,nu*(n-1)/n), log = "x", xlab = expression(nu), ylab = "fraction of dices",
main ="comparing simulation (dots) \n with formula based on beta (line)", main.cex = 1, type = "l")
points(nu,y, lty =1, pch = 21, col = "black", bg = "white")
.... But I have no good way to analytically solve the expression for $$n \to \infty$$.
With the waiting time However, with waiting times, then I can express the the limit of the beta binomial distribution (which is now more like a beta Poisson distribution) with a variance in the exponential factor of the waiting times.
So instead of $$1-e^{-1}$$ we are looking for $$1- \int e^{-\lambda} p(\lambda) \, \text{d}\, \lambda$$.
Now that integral term is related to the moment generating function (with $$t=-1$$). So if $$\lambda$$ is normal distributed with $$\mu = 1$$ and variance $$\sigma^2$$ then we should use: | {
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$$1-e^{-(1-\sigma^2/2)} \quad \text{instead of} \quad 1-e^{-1}$$
### Application
These dice rolls are a toy model. Many real-life problems will have variation and not completely fair dice situations.
For instance, say you wish to study probability that a person might get sick from a virus given some contact time. One could base calculations for this based on some experiments that verify the probability of a transmission (e.g. either some theoretical work, or some lab experiments measuring/determining the number/frequency of transmissions in an entire population over a short duration), and then extrapolate this transmission to an entire month. Say, you find that the transmission is 1 transmission per month per person, then you could conclude that $$1-1/e \approx 0.63 \%$$ of the population will get sick. However, this might be an overestimation because not everybody might get sick/transmission with the same rate. The percentage will probably lower.
However, this is only true if the variance is very large. For this the distribution of $$\lambda$$ must be very skewed. Because, although we expressed it as a normal distribution before, negative values are not possible and distributions without negative distributions will typically not have large ratios $$\sigma/\mu$$, unless they are highly skewed. A situation with high skew is modeled below:
Now we use the MGF for a Bernoulli distribution (the exponent of it), because we modeled the distribution as either $$\lambda = 0$$ with probability $$1-p$$ or $$\lambda = 1/p$$ with probability $$p$$.
set.seed(1)
rate = 1
time = 1
CV = 1
### compute outcome for getting sick with variable rate
getsick <- function(rate,CV=0.1,time=1) {
### truncating changes sd and mean but not co much if CV is small
p <- 1/(CV^2+1)
lambda <- rbinom(1,1,p)/(p)*rate
k <- rpois(1,lambda*time)
out <- (k>0)
out
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CV <- seq(0,2,0.1)
plot(-1,-1, xlim = c(0,2), ylim = c(0,1), xlab = "coefficient of variance", ylab = "fraction",
cex.main = 1, main = "if rates are bernouilli distributed \n fraction p with lambda/p and 1-p with 0")
for (cv in CV) {
points(cv,sum(replicate(10^4,getsick(rate=1,cv, time = 1)))/10^4)
}
p <- 1/(CV^2+1)
lines(CV,1-(1-p)-p*exp(-1/p),col=1)
lines(CV,p, col = 2, lty = 2)
legend(2,1, c("simulation", "computed", "percent of subsceptible population"),
col = c(1,1,2), lty = c(NA,1,2), pch = c(1,NA,NA),xjust =1, cex = 0.7)
The consequence is. Say you have high $$n$$ and have no possibilities to observe $$n$$ dice rolls (e.g. it takes to long), and instead you screen the number of $$n$$ rolls only for a short time for many different dice. Then you could compute the number of dices that did roll a number $$n$$ during this short time and based on that compute what would happen for $$n$$ rolls. But you would not be knowing how much the events correlate within the dice. It could be that you are dealing with a high probability in a small group of dice, instead of an evenly distributed probability among all dice.
This 'error' (or you could say simplification) relates to the situation with COVID-19 where the idea goes around that we need 60% of the people immune in order to reach herd immunity. However, that may not be the case. The current infection rate is determined for only a small group of people, it can be that this is only an indication for the infectiousness among a small group of people. | {
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• I must admit that my last case is actually very trivial and does not this larger framework and tricks with moment generating functions, it can be done easier by just rescaling the population. Jul 31, 2020 at 16:58
• ...Except, since the number of rolls is finite, one can enumerate all possible outcomes. This is n^n for an n-sided die. Once one has done this it's a matter of counting which rolls have a #n. Jul 31, 2020 at 17:05
• @Marcos my approach is for people that do not like counting (e.g. imagine $n$ would be equal to a million or larger, it is still finite but good luck counting). And also, how are you gonna count if the dices are not fair but instead follow some continuous distribution? -------- That said, I agree your specific question can be answered easily. The method here is just an additional view that can be useful for more complex problems. Jul 31, 2020 at 17:20
• True, it requries a "meta-numeric" formula for counting the digits of numbers (as symbols rather than quantities) themselves. Jul 31, 2020 at 17:28
• @Marcos I am not sure what you mean by counting now. I consider this problem not as a counting problem when $n$ get's large. Well, you can go count things and maybe create expressions that summarize the counting, but you may not always get (easily) to a solution. Flipping the point of view from counting the probabilities of a given number of events for a given number of rolls, to a given number of rolls (waiting time) for a given number of events, can be a useful tool to greatly simplify the problem (I agree for this specific problem, which is not so difficult it is not so much necessary) Jul 31, 2020 at 17:42
Simplify and then extend. Start with a coin. A coin is a die with 2 sides (S=2).
The exhaustive probability space is
T | H
Two possibilities. One satisfies the condition of all heads. So your odds of all heads with one coin (n=1) are 1/2.
So try two coins (n=2). All outcomes:
TT | TH | HT | HH | {
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So try two coins (n=2). All outcomes:
TT | TH | HT | HH
Four possibilities. Only one matches your criteria. It is worth noting that the probability of one being heads and the other being tails is 2/4 because two possibilities of the four match your criteria. But there is only one way to get all heads.
TTT | THT | HTT | HHT
TTH | THH | HTH | HHH
8 possibilities. Only one fits the criteria - so 1/8 chances of all heads.
The pattern is (1/S)^n or (1/2)^3.
For dice S = 6, and we have 6 of them.
Probability of getting a 6 on any given roll is 1/6. Rolls are independent events. So using 2 dice getting all 6's is (1/6)*(1/6) or 1/36.
(1/6)^6 is about 1 in 46,656
• I was actually just looking for the probability of ANY heads showing up, given as many rolls as there are faces on the die. But, cheers mate, I got my answer. Jul 28, 2020 at 0:22 | {
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# Show that $x^4 +1$ is reducible over $\mathbb{Z}_{11}[x]$ and splits over $\mathbb{Z}_{17}[x]$.
Reduction into linear factors $\mathbb{Z}_{17}[x]$:
This part is not too hard: $x^4 \equiv -1$ mod 17 has solutions: 2, 8, 9, 15 so
$(x-2)(x-8)(x-9)(x-15) = x^4 -34 x^3 +391 x^2-1734 x+2160 \equiv x^4+1$ mod 17.
Reduction over $\mathbb{Z}_{11}[x]$:
This one doesn't have such an easy solution, as neither of $y^2 \equiv -1$ mod 11 or $x^4 \equiv -1$ mod 11 have solutions.
I've tried $x^4+1 = (x^2- \sqrt{2}x+1)(x^2+\sqrt{2}x+1)$ but $x^2 \equiv 2$ mod 11 also has no solutions so there's no easy substitution here.
I think that this approach is not going to work here, so I need something new. Any suggestions?
• See this question for a more or less complete answer. – Jyrki Lahtonen Apr 5 '13 at 15:05
• It certainly has a very complete answer - but perhaps somewhat overkill when it comes to answering this question? Nonetheless thanks for the link. – Lewy Apr 5 '13 at 15:19
• You were almost there... You tried $X^4+1=X^4+1+2X^2-2X^2$. Now, if you observe that $2=-9$ you are done.... – N. S. Apr 5 '13 at 15:45
• The splitting modulo $p=17$ comes from the fact that $p \equiv 1 \pmod 8$, so that $p$ is totally split in the extension $\Bbb Q(\zeta_8) / \Bbb Q$, which is the splitting field of $x^4+1$. – Watson Apr 6 '18 at 13:30
I will assume you've made no mistakes in your work. | {
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I will assume you've made no mistakes in your work.
Since $-1$ hasn't a fourth root in $\Bbb Z_{11},$ then the only way to reduce $x^4+1$ over $\Bbb Z_{11}$ is as a product if quadratics. We may as well assume that the quadratics are monic, so we've got $$x^4+1=(x^2+ax+b)(x^2+cx+d),$$ leaving us with the following system: $$a+c=0\\ac+b+d=0\\ad+bc=0\\bd=1$$ By the first equation, $c=-a,$ and the rest of the system becomes: $$b+d=a^2\\a(d-b)=0\\bd=1$$ Now, if $a=0$, then the first and third remaining equations imply that $-1$ has a square root in $\Bbb Z_{11}$ (why?), which you've ruled out. Hence, we need $b=d,$ and the rest of the system becomes: $$2b=a^2\\b^2=1$$ Can you take it from here?
• Yes, thank you for your clear explanation. This gives two possible solutions: $(x^2-3x -1)(x^2 +3x-1)$ and $(x^2-8x -1)(x^2 +8x-1)$. – Lewy Apr 5 '13 at 15:27
• @Lewy: You do know that factorization in this ring is unique, don't ya? IOW, look more closely at the "two" factorizations :-) – Jyrki Lahtonen Apr 5 '13 at 15:31
• @Lewy: Jyrki's point is well-taken. $8=-3,$ so those factorizations are identical (up to arrangement of the factors). – Cameron Buie Apr 5 '13 at 15:51
• Yes, I see that. Perhaps this is one of those times at which it's appropriate to say "Doh!". – Lewy Apr 5 '13 at 15:55
An abstract algebra argument, without much computation.
Claim: The polynomial $x^4+1$ splits over $\mathbb F_{p^2}$ for any odd (*) prime $p$.
We can see this because $F_{p^2}^\times$ is a cyclic group of order $p^2-1$, a multiple of $8$. If $g$ is a generator, then $g^{(1+2k)\frac{p^2-1}{8}}$ gives distinct roots for $k=0,1,2,3$.
This means that $x^4+1$ can't be irreducible in $\mathbb Z_p$, because if it was, the splitting field would have $p^4$ elements.
(*) True for $p=2$, but not for the same reasons. When $p=2$, $x^4+1=(x+1)^4$.
Specific solutions: | {
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(*) True for $p=2$, but not for the same reasons. When $p=2$, $x^4+1=(x+1)^4$.
Specific solutions:
We can make this explicit by looking at the "standard" complex $4$th roots of $-1$: $$\pm\frac{\sqrt 2}2 \pm\frac{\sqrt{2}}{2}i=\frac{1}{2}(\pm \sqrt{2}\pm\sqrt{-2})$$
If $p\equiv 1\pmod 8$ then $2$ and $-2$ have square roots in $\mathbb Z_p$, so $x^4+1$ has four roots.
If $p\equiv -1\pmod 8$ then $2$ is a square, but $-2$ is not. Write $a^2\equiv 2\pmod p$. Then write:
$$x^4+1 = \left(x^2 -ax + 1\right)\left(x^2+ax+1\right)$$
If $p\equiv 3\pmod 8$ then $-2$ is a square and $2$ is not. Letting $b^2\equiv -2\pmod p$, we get:
$$x^4+1 = \left(x^2-bx-1\right)\left(x^2+bx-1\right)$$
Finally, if $p\equiv 5\pmod 8$ then neither of $\pm 2$ is a square, but $-1$ is a square. Letting $c^2\equiv -1\pmod p$, we see that the roots are $(\pm 1 \pm c)\sqrt{2}/2$ and the result is:
$$x^4+1 = (x^2-c)(x^2+c)$$
• The use of the formula $(\pm\sqrt2\pm\sqrt{-2})/2$ is a nice trick. In the linked answer I (IIRC at least partly in response to the discussion in the comments that lead to CWification) did get there, but should have seen it right away. +1 for unveiling the reason for the emergence of $\sqrt{\pm2}$ to this extent. – Jyrki Lahtonen Apr 5 '13 at 17:42
• Perhaps a more direct proof that $x^4+1$ is not irreducible is just to note that $x^4+1|x^8-1|x^{p^2-1}-1|x^{p^2}-x$. And $x^{p^2}-x$ has as its only irreducible factors the quadratic and linear prime polynomials. – Thomas Andrews Apr 5 '13 at 18:04
• Well, that's exactly how I did it originally. But somewhat surprisingly people don't think this is a duplicate of that :-) – Jyrki Lahtonen Apr 5 '13 at 18:07
• @JyrkiLahtonen N.S. suggested in comments to the question: $x^4+1=x^4+1 +2x^2-2x^2$. Hahaha, that makes the primacy of $\sqrt{2}$ and $\sqrt{-2}$ even more obvious! – Thomas Andrews Apr 5 '13 at 18:16 | {
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This question in fact already answered your question. Indeed, in the accepted answer, the complete solution can be found. So let me explain this and put it into CW.
As expounded in the link, we know that there is a primitive root of order $8$ in $\mathbb F:=\mathbb F_{p^2}$, called $u$. Since $\mathbb F$ is of degree $2$ over $\mathbb F_p$, the minimal polynomial is of degree at most $2$ over $\mathbb F_p$. Here, our $p\equiv 3\pmod 8$, so there is no $u$ in $\mathbb F_p$, as $u^4=-1$. Its conjugate is then $u^p=u^3$. Therefore $x^4+1$ has a factor given by $(x-u)(x-u^3)$. After easy calculations, we find that this becomes $(x^2-(u+u^3)x-1)$. And the other factor of that one must, after easy calculations of other conjugates of $u$, be $x^2+(u+u^3)x-1$. Therefore, $a:=u+u^3$ satisfies $a^2+2=0$. And I guess you know there is one such $a$ for $p=11$.
For those who jumped here:
We found the factorisation: $$(x^2-3x-1)(x^2+3x-1)$$.
Thanks for the attention, and inform me of any errors. Thanks.
• Also see that referred answer for more detail. – awllower Apr 5 '13 at 15:21
Look for a factorization of the shape $(x^2-ax+b)(x^2+ax+b)$. So we want $b^2\equiv 1\pmod{11}$, with $2b$ a quadratic residue.
Remark: Note that this is a general procedure for $x^4+1$ modulo an odd prime. The issue is whether one of $2$ or $-2$ is a quadratic residue of $p$. One of them is unless $p\equiv 5\pmod{8}$.
• In the case $p\equiv 5\pmod8$ we have that $2$ is a quadratic non-residue, so $i:=2^{(p-1)/4}$ is a square root of $-1$. In that case we have the factorization $(x^2+i)(x^2-i)$. – Jyrki Lahtonen Apr 5 '13 at 15:35
• In general, at least one of $a,b,ab$ is a square $\pmod p$. Take $a=2, b=-1.$ :) – Thomas Andrews Apr 5 '13 at 16:12
• Note, the standard formulation for the complex roots of $x^4+1=0$ can be formulated as $\frac{1}{2}(\pm\sqrt{2}\pm\sqrt{-2})$, which is why these two roots are important. – Thomas Andrews Apr 5 '13 at 16:14 | {
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Hint: We know that if $f(x)=x^4+1$ is reducible, then it either has a root or can be written $f(x)=g(x)h(x)$ where $\deg(g)=\deg(h)=2$.
• I guess OP already knew this, judged from the try. The point is: how to find those degree $2$ factors? – awllower Apr 5 '13 at 15:03
You may use the fact that $$x^2 = 9 \mod 11$$ has a solution, say $x = c$. Then use $$(x^2 + cx - 1)(x^2 - cx - 1),$$ and $c$ is pretty much obvious from here... | {
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# Summation problem: $f(x)=1+\sum_{n=1}^{\infty}\frac{x^n}{n}$
I want to evaluate this summation:
$$S=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...+...$$
where, $|x|<1$
Here it is my approach
$$S=1+\sum_{n=1}^{\infty}\frac{x^n}{n}=f(x)$$
$$f'(x)=1+x+x^2+x^3+...+...=\frac{1}{1-x}$$
$$f(x)=\int f'(x)dx=\int \frac{1}{1-x} dx=-\ln(1-x)+C$$
$$f(0)=1 \Longrightarrow C=1$$
$$S=1-\ln(1-x)$$
And here is my problem:
I calculated this sum for only $|x|<1$.
Then, I checked in Wolfram Alpha and I saw that, this sum $f(x)$ converges for $x=-1$. But, this creating a contradiction with my solution. Because, the series $$f'(x)=1+x+x^2+x^3+...+...=\frac{1}{1-x}$$ for $x=-1$ doesn't converge, diverges. Therefore, there is a problem in my solution. But I can not find the source of the problem.
How can I prove the formula $f(x)=1-\ln(1-x)$ is also correct at the point $x=-1$ ?
Thank you very much.
• There's no problem with your solution. There's a problem with you assuming that if $\sum_n \frac{x^n}{n}$ converges, then $\sum_n x^{n-1}$ must converge. – mathworker21 Aug 23 '18 at 19:08
• Your approach worked for all $x$ you were asked to solve it for. WolframAlpha is telling you that the series also has a value for some $x$ outside that region, although your approach doesn't work in that case. Why does that matter? – Arthur Aug 23 '18 at 19:13
• @mathworker21 How can I prove the formula $f(x)=1-\ln(1-x)$ is also correct at the point $x=-1$ ? – Elvin Aug 23 '18 at 19:28
• @mathworker21 as I understand it, you mean my solution is incorrect $f(-1)-1=\ln 2$ .Am I right? – Elvin Aug 23 '18 at 20:34
• @student It's incorrect at $x = -1$. You may only deduce that $f'(x) = 1+x+x^2+\dots$ when $|x| < 1$. – mathworker21 Aug 23 '18 at 20:45
I don't know how familiar your are with power series, but this has to do with the notion of radius of convergence. | {
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$\sum \left(\frac{x^n}{n}\right)_{n\in\mathbb{N^*}}$ is a power series with a radius of convergence $R=1$ (which can be proved by the ratio test), so indeed, it does converge for $|x|<1$, and does diverge for $|x|>1$.
As for $x=1$, the only case with $x=-1$ were we can't conclude right away, it could naively be anything, but it happens to converge.
The power series $\sum \left(x^n\right)_{n\in\mathbb{N^*}}$ is indeed the derivative of the previous one : we know it has the same radius of convergence, so for $x=1$, here again it could be anything ; and it happens to diverge.
The convergence circle is the only place were the convergence of a power series and its derivatives aren't always equivalent : it could be anything, and you can't deduce the convergence of the derivative from the convergence of the power series there.
As to show that the power series and $f(x)=1-\ln(1-x)$ are equal also for $x=-1$ (that is to say, to prove that $f(-1)$ is the sum of the series) :
$\cdot$ The power series and f both exist for $x$ in $[-1,1)$ (for the power series, you can prove it with the ratio test)
$\cdot$ They are both continuous on $[-1,1)$ : it is obvious for $f$, and for the power series, we know that the sum of the power series is continious strictly inside the convergence disk $(-1,1)$. We extend the continuity to -1 by the uniform convergence on $[-1,0]$.
$\cdot$ Are equal to one another on $(-1,1)$ which is dense in $[-1,1]$.
We can conclude that the power series and f are equal also at $x=1$, so $f(-1)$ is indeed the sum of the series.
If you are not familiar with this density theorem :
Since the power series is continuous on -1, by definition, $\lim\limits_{x\rightarrow\ -1}\sum\limits_{n=1}^{\infty}\frac{x^n}{n}=\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n}$, the limits being taken by approaching from the right ($x>-1$).
But, since the power series and f are equal for $x>-1$, we can put $f$ instead of the power series inside the limit : | {
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$$\lim\limits_{x\rightarrow\ -1}f(x)=\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n}$$
Or $f$ is continuous on -1, so $\lim\limits_{x\rightarrow\ -1}f(x)=f(-1)$
Hence $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n}=f(-1)$.
• But we need more. We have a formula $f(x) = 1-\log(1-x)$ on $(-1,1)$. The series converges at $-1$, and the formula (extends to something that) is continuous at $-1$. Conclude that $f(-1)$ is the sum of the series at $-1$. – GEdgar Aug 23 '18 at 19:58
• The power series and f both exist on [-1,1) (ratio test), are both continuous on [-1,1) (for the power series, we extend the continuity to -1 by the uniform convergence on [-1,0]), and are equal to one another on (-1,1) which is dense in [-1,1]. We can conclude that the power series and f are equal also at x=1. – Harmonic Sun Aug 23 '18 at 20:02
• +1 for a complete proof. – GEdgar Aug 23 '18 at 21:28
By the ratio test, $$\lim_{n\to\infty } \left|\frac{x^{n+1}}{n+1}\frac{n}{x^n} \right|=\lim_{n\to\infty}|x|\frac{n}{n+1}<1$$ whenever $|x|<1$. Thus we have convergence on $(-1,1)$. We need to check the endpoints separately. For $x=1$, we have the diverging harmonic series. For $x=-1$, we have the alternating harmonic series, which converges.
A power series $\sum_{n=0}^\infty a_n x^n$ (I am assuming the series is centered at zero for definiteness) converges either only at $0$, or on some interval $I$ centered at zero. The half-length of $I$ is called the radius of convergence $R$, and can be computed generally as $\frac{1}{R}=\limsup_{n \to \infty} a_n^{1/n}$ (understood as $R=\infty$ if $1/R$ is zero in the above formula). | {
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A power series with a given radius of convergence may or may not converge at the endpoints $\pm R$. $\sum_{n=1}^\infty \frac{x^n}{n}$ turns out to have radius of convergence $1$ and converges at $-1$ but does not converge at $1$. In the interior of its interval of convergence, a power series may be differentiated term by term; thus in your example you can deduce the derivative of $-\ln(1-x)$ is $\frac{1}{1-x}$ for $|x|<1$ by differentiating its series term by term. In general differentiation may change how the series behaves at its endpoints, as you found here.
It's well-known that $$\ln(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}-\frac{y^4}{4}+\cdots,~~~\forall y \in (-1,1].\tag1$$
Thus, let $y=-x.$ we have $$\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+\cdots,~~~\forall x \in [-1,1).\tag2$$
As result, $$S=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots=1-\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+\cdots\right)=1-\ln(1-x).$$
Notice that, $(1)$ demands $-1<y\leq 1$. then in $(2)$, we demand $-1<-x\leq 1,$ namely, $-1\leq x<1.$
It is best to use elementary arguments for such well known power series rather than start integrating power series. While power series methods work fine in the interior of region of convergence, the behavior at boundary needs additional analysis (for example Abel's theorem).
Below I prove the formula $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots\tag{1}$$ for all values of $$x$$ satisfying $$-1. The series in question is obtained by replacing $$x$$ with $$-x$$. | {
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Let's start with the algebraic identity $$\frac{1}{1+t}=1-t+t^2-\dots+(-1)^{n-1}t^{n-1}+(-1)^{n}\frac{t^{n}}{1+t}\tag{2}$$ which holds for all $$t\neq -1$$ and all positive integers $$n$$. Integrating this identity with respect to $$x$$ over interval $$[0,x]$$ where $$x>-1$$ we get $$\log(1+x)=x-\frac{x^2}{2}+\dots+(-1)^{n-1}\frac{x^n}{n}+(-1)^nR_n\tag{3}$$ where $$R_n=\int_{0}^{x}\frac{t^n}{1+t}\,dt$$ Our job is done if we can show that $$R_n\to 0$$ as $$n\to\infty$$ for all $$x\in(-1,1]$$. If $$x=0$$ then $$R_n=0$$. If $$0 then we can see that $$0 so that $$R_n\to 0$$. If $$x=-y$$ and $$y\in(0,1)$$ then $$R_n=(-1)^{n+1}\int_{0}^{y}\frac{t^n}{1-t}\,dt$$ The integral above is clearly less than or equal to $$\frac{1}{1-y}\int_{0}^{y}t^n\,dt=\frac{y^{n+1}}{(1-y)(n+1)}$$ which tends to $$0$$ as $$n\to\infty$$ and hence $$R_n\to 0$$. It is thus established that $$R_n\to 0$$ as $$n\to\infty$$ for all $$x\in(-1,1]$$. Taking limits as $$n\to\infty$$ in equation $$(3)$$ gives us the series $$(1)$$ which is valid for $$x\in(-1,1]$$. | {
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# Inequality from Chapter 5 of the book *How to Think Like a Mathematician*
This is from the book How to think like a Mathematician,
How can I prove the inequality $$\sqrt[\large 7]{7!} < \sqrt[\large 8]{8!}$$
without complicated calculus? I tried and finally obtained just $$\frac 17 \cdot \ln(7!) < \frac 18 \cdot \ln(8!)$$
• Are we allowed complex roots and negative roots to prove the contrary ;-) – Philip Oakley Feb 28 '15 at 19:22
• I don't think that it is in the proposition – Gwydyon Mar 2 '15 at 11:24
• Can these simpler approaches to a specific example be used to produce a simpler answer to the "general case"? – Mark Hurd Mar 3 '15 at 3:02
• I don't know, the problem was "as is" – Gwydyon Mar 3 '15 at 16:35
• – Martin Sleziak Dec 11 '16 at 11:36
Your inequality is equivalent to $$(7!)^8 < (8!)^7$$ divide it by $(7!)^7$, and get $$7! < 8^7$$ and this is clear, since $$1 \cdots 7 < 8 \cdots 8$$
• In the spirit of the question title, I'd note that the first line of this answer is key. Maths does of course love cleverness, but also thrives on knowing to do really simply things such as taking powers of both sides here to remove those ugly roots. – Keith Feb 27 '15 at 3:02
• You are right, great ! I would have of to think there. – Gwydyon Feb 28 '15 at 17:39
• Elegant! really clever! – Max Payne May 27 '15 at 16:19
Think of
$${\ln(7!)\over7}={\ln(1)+\cdots+\ln(7)\over7}$$
as the average of seven numbers and
$${\ln(8!)\over8}={\ln(1)+\cdots+\ln(8)\over8}$$
as the average when an eighth number is added. Since the new number is larger than the previous seven, the average must also be larger. (E.g., if you get a better score on your final than on any of your midterms, your grade should go up, not down.) | {
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• You are right, but it is clear to me only when i compute ln(8) – Gwydyon Feb 28 '15 at 18:08
• @Gwydyon, the logarithm is an increasing function, so $\ln(8)$ is larger than its predecessors. – Barry Cipra Feb 28 '15 at 20:00
• Yes I know that. – Gwydyon Mar 2 '15 at 11:21
• @Gwydyon, I guess I don't understand your previous comment then. – Barry Cipra Mar 2 '15 at 15:22
• Because x/8 is lower than y/7, if x=y, that is not the case but ln(7!) and ln((7+1)!) are near close, for x>=1 – Gwydyon Mar 3 '15 at 16:41
Note that $$\sqrt[7]{7!} < \sqrt[8]{8!} \iff\\ (7!)^8 < (8!)^7 \iff\\ 7! < \frac{(8!)^7}{(7!)^7} \iff\\ 7! < 8^7$$ You should find that the proof of this last line is fairly straightforward.
• You are right but someone has writen the same statements – Gwydyon Feb 28 '15 at 18:11
$8\ln (7!) < 7\ln (8!) \Rightarrow \ln (7!) < 7\ln 8 \iff \ln 1 + \ln 2 +\cdots \ln 7 < 7\ln 8$ which is clear.
• great! you achieved my second statement – Gwydyon Feb 28 '15 at 18:53
You have already turned the comparison of two geometric means into the comparison of two arithmetic means. So consider a more general comparison: show that appending a larger number always raises the geometric mean of a list of positive numbers by showing the effect on the arithmetic mean. Suppose the $x_i$ are real and $x_{n+1}$ is strictly largest. \begin{equation*} \begin{split} (1/(n+1)) \sum_{i=1}^{n+1} x_i &= (1/(n+1)) (x_{n+1} + \sum_{i=1}^{n} x_i) \\ &=(1/(n+1) (n x_{n+1}/n + n \sum_{i=1}^{n} x_i / n) \\ &> (1/(n+1) (\sum_{i=1}^{n} x_i/n + n \sum_{i=1}^{n} x_i / n) \\ &= (1/(n+1) ((n+1) \sum_{i=1}^{n} x_i / n) \\ &= \sum_{i=1}^{n} x_i / n \end{split} \end{equation*}
Note that we really only needed $x_{n+1}$ to be larger than the previous mean. | {
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Truth of $x^2-2=0$
From Algrebra by Gelfand, it says that "When we claim that we have solved the equation $x^2-2=0$ and the answer is $x=\sqrt{2}$ or $x=-\sqrt{2}$, we are in fact cheating. To tell the truth, we have not solved this equation but confessed our inability to solve it; $\sqrt{2}$ means nothing except "the positive solution of the equation $x^2-2=0$
Can someone please explain what he really means? Does he mean every irrational number is meaningless? Thanks in advance.
• I think he might be saying that irrational numbers could be thought of as solutions to polynomial equations, not the other way around. – Airdish May 27 '17 at 16:40
• I believe he is saying the result is circular. – Simply Beautiful Art May 27 '17 at 16:44
• The downvote puzzles me. Gelfand is making a very interesting subtle point and the OP is quite right to be puzzled by it, and ask. – Ethan Bolker May 27 '17 at 16:55
• @EthanBolker He's a good writer. I just don't understand sometimes and asked. – Mathxx May 27 '17 at 16:58
• Indeed he is a good writer. As the comments and answers show, you asked a good question. I hope they help you understand. – Ethan Bolker May 27 '17 at 17:12
He is saying the following.
1. $\sqrt{2}$ is defined as "the number such that its square is two."
2. The statement "the solution $x$ to $x^2-2 = 0$ is $\sqrt{2}$" is therefore saying "the solution $x$ to $x^2-2=0$ is the number such that its square is two."
As you can see, this is a rather circular statement. This doesn't mean that irrational numbers are meaningless (indeed, $\sqrt{2}$ does exist -- see the comments and @EthanBolker's answer for more on this), but rather saying that this method of definition limits us to statements like "the number such that its square is two" or "half of the number such that its square is eight." | {
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• Nice answer, except I don't think it is clear that $\sqrt{2}$ exists at all. It is either taken as an axiom, the completeness axiom of the reals, or is to be proved, depending on our construction of the real numbers... – SEWillB May 27 '17 at 16:51
• @SEWillB there certainly exist right triangles with legs equal to $1$ unit. Now, does there not exist a hypotenuse to the triangle, since it would be of length $\sqrt 2$ units, so may not exist? – Namaste May 27 '17 at 16:56
• SEWillB (continued)... And without a hypotenuse for a right triangle, we cease to have such a triangle? So the only right triangles require legs of lenghts $x, y$ such that the length of the hypotenuse, $\sqrt{x^2 + y^2},$ is rational? – Namaste May 27 '17 at 17:04
• @amWhy, I never liked these arguments at all. There is a nice bit in a book I read a long time ago (Liebeck introduction to pure mathematics) showing the 'existence' of $\sqrt{n}$ by geometrical construction like that. But these arguments aren't convincing to me because it's not clear what you're trying to do in my opinion. – SEWillB May 27 '17 at 17:08
• So you do not believe the Pythagorean Theorem? Well, then, I think I've wasted my time commenting with you. – Namaste May 27 '17 at 17:09
To elaborate on @fractal1729 's lovely correct answer:
Having defined $\sqrt{2}$ as "the number such that it's square is $2$" it's reasonable to ask whether there is such a number.
The Greeks knew that the answer is "no" if "number" means "rational number". So in order to claim the existence of $\sqrt{2}$ you must enlarge the system of rational numbers. The Greeks essentially sidestepped that question by doing geometry rather than arithmetic, using points on a line instead of numbers. Clearly there's a point on the "number line" with the right length since you can draw the diagonal of a unit square. | {
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